Solution Collins Cap 8
Solution Collins Cap 8
Solution Collins Cap 8
8-1. A drive shaft for a new rotary compressor is to be supported by two bearings, which are 200 mm apart.
A V-belt system drives the shaft through a V-sheave (see Figure 17.9) mounted at midspan, and the belt is
pretensioned to Po kN, giving an vertically downward force of 2 Po at midspan. The right end of the shaft is
directly coupled to the compressor input shaft through a flexible coupling. The compressor requires a steady
input torque of 5700 N-m. Make a first0cut conceptual sketch of a shaft configuration that would be
appropriate for this application.
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Solution
300
8-2. The drive shaft of a rotary coal grinding miss is to be driven by a gear reducer through a flexible shaft
coupling, as shown in Figure P8.2. The main shaft of the gear reducer is to be supported on two bearings
mounted 10 inches apart at A and C, as shown. A 1:3 spur gear mesh drives the shaft. The 20o spur gear is
mounted on the shaft at midspan between the bearings, and has a pitch diameter of 9 inches. The pitch
diameter of the drive pinion is 3 inches. The grinder is to be operated at 600 rpm and requires 100 horsepower
at the input shaft. The shaft material is to be AISI 1060 cold-drawn carbon steel (see Table 3.3). Shoulders for
gears and bearings are to be a minimum of 1/8 inch (1/4 inch on the diameter). A design safety factor of 1.5 is
desired. Do a first-cut design of the shaft, including a second-cut sketch showing principal dimensions.
-------------------------------------------------------------------------------------------------------------------------
Solution
(3) Assuming infinite life, we estimate the fatigue endurance limit as S 'f = 0.5Su = 45 ksi
T 10,504
FBz = = 2334 = 10,504 lb (tangent, down)
( D p / 2) 4.5
FBx = FBz tan ϕ = 2334 tan 20o ≈ 850 lb (radially, toward gear center)
Now: ∑ Fx = 0 : RAx + FBx + RCx = 0 → 425 + (−850) + (−850) + RCx = 0 ⇒ RCx = 425 lb
∑ Fz = 0 : RAz + FBz + RCz = 0 → 1167 + (−2334) + RCz = 0 ⇒ RCz = 1167 lb
301
Problem 8-2 (continued)
TA − B = 0 , TB −C = 10,504 in-lb
(7) The shaft diameter is determined using (8-8). The fatigue strength will be taken to be approximately 85%
of the endurance limit ( S 'f = 45 ksi ). Later revisions should be review this assumption by using equations (5-
37) and (5-39). A stress concentration factor of Kt = 1.7 will be assumed for the first iteration, and a value of
q = 0.8 will be assumed. We now determine
K fb = q ( Kt − 1) + 1 = 0.8(1.7 − 1) + 1 ≈ 1.6
We note that the bending moment and torque are zero at point A. Therefore, we consider the transverse shear
at this point (see Example 8.1). Therefore, at point A
Next we apply (8-11) to both points B and C . We note that Ta = M m = 0 at each point. Therefore, at point B
16 ⎧ Ma T ⎫ 16 ⎧ 6210 10,504 ⎫
d B3 = ⎨2 K fb (nd ) + 3 m ⎬ = ⎨2(1.7)(1.5) + 3 ⎬ = 5.25
π ⎩ SN Su ⎭ π ⎩ 0.85(45, 000) 90, 000 ⎭
d B = 1.738" ≈ 1.75"
16 ⎧ Ma T ⎫ 16 ⎧ 10,504 ⎫
dC3 = ⎨2 K fb (nd ) + 3 m⎬= ⎨ 3 ⎬ = 1.03 ⇒ dC = 1.0"
π ⎩ SN Su ⎭ π ⎩ 90, 000 ⎭
302
8-3. A belt-drive jack-shaft is sketched schematically in Figure P8.3.
a. Construct load, shear, and bending moment diagrams for the shaft in both the horizontal and the
vertical plane.
b. Develop an expression for the resultant bending moment on the shaft segment between the left pulley
and the right bearing.
c. Find the location and magnitude of the minimum value of bending moment on the shaft segment
between the left pulley and the right bearing.
d. Calculate the torque in the shaft segment between pulleys.
e. If the shaft is to be made of hot-rolled 1020 steel (see Figure 5.31), is to rotate at 1200 rpm, and a
design safety factor of 1.7 is desired, what diameter would be required to provide infinite life?
------------------------------------------------------------------------------------------------------------------------------
Solution
The shear force and bending moment diagrams in the y-x and y-z planes are as shown below. The coordinate q
in the starting at point C in the y-x load diagram is used is part b.
303
Problem 8-3 (continued)
(b) Using the coordinate q in the load diagram above, we can write an expression for the bending moment
between B and C for moments about the x and the z axes. For bending about the x axis we have
( M x ) B −C = -911.3 - 2925q
( M z ) B −C = 2227.5 - 4950q
(c) Differentiating with respect to q , setting the derivative equal to zero, and solving for q
Knowing the torque is TBD = 1710 N-m , using nd = 1.7 , reading S 'f ≈ 33 ksi ≈ 228 MPa from Figure 5.31
and assuming k∞ = 0.85 , which results in S N = 0.85(228) ≈ 194 MPa . In addition, Su ≈ 379 MPa .
Therefore
16 ⎧ Ma T ⎫ 16 ⎧ ⎛ 2407 ⎞ 1710 ⎫
d B3 = ⎨2 K fb (nd ) + 3 m ⎬ = ⎨2(1.7) ⎜ ⎟+ 3 ⎬ ≈ 0.000255
π ⎩ SN Su ⎭ π ⎩ ⎝ 194 × 106 ⎠ 379 × 106 ⎭
d B = 0.0633 m ≈ 63 mm
304
8-4. Repeat problem 8-3, except that the shaft is to be made of AISI 1095 steel, quenched and drawn to
Rockwell C 42 (see Table 3.3)
-------------------------------------------------------------------------------------------------------------------------
Solution
The solution is identical to that of 8-3, except for part (e), where Su ≈ 1379 MPa and S yp ≈ 952 MPa .
Estimating the S-N curve, S 'f = 0.5(1379) ≈ 690 MPa and S N = 0.85(690) ≈ 587 MPa
16 ⎧ ⎛ 2407 ⎞ 1710 ⎫
d B3 = ⎨2(1.7) ⎜ 6 ⎟
+ 3 ⎬ ≈ 0.000082
π ⎩ ⎝ 587 × 10 ⎠ 1379 × 106 ⎭
d B = 0.0434 m ≈ 43 mm
305
8-5. A pinion shaft for a helical gear reducer (see Chapter 15) is sketched in Figure P8.5, where the reaction
forces on the pinion are also shown. The pinion shaft is to be driven at 1140 rpm by a motor developing 14.9
kW.
a. Construct load, shear , and bending moment diagrams for the shaft, in both the horizontal and vertical
plane. Also make similar diagrams for axial load and for torsional moment on the shaft, assuming that the
bearing at the right end (nearest the gear) supports all thrust (axial) loading.
b. If the shaft is to be made of 1020 steel (see Figure 5.31), and a design factor of safety of 1.8 is desired,
what diameter would be required at location B to provide infinite life?
-------------------------------------------------------------------------------------------------------------------------
Solution
Note that the moments applied at point C come from taking moments of the three active force components
about point C.
i j k
∑ M C = 0 : r × FC = 0 0 −0.035 ≈ −72i + 125 j
−3571 −2047 1502
∑ M A = 0 : rAB × R B + rAC × FC + Tm j
i j k i j k
= 0 0.140 0 + 0 0.205 0 + Tm j − 72i + 125 j = 0
RBx 2047 RBz −3571 −2047 1502
306
Problem 8-5 (continued)
In addition, the axial force ( Fy ) and torque ( T = M y ) variations along the shaft are
For hot rolled 1020 steel, reading S 'f ≈ 33 ksi ≈ 228 MPa from Figure 5.31 and assuming k∞ = 0.85 , which
results in S N = 0.85(228) ≈ 194 MPa . In addition, Su ≈ 379 MPa . Therefore
16 ⎧ ⎛ 234 ⎞ 125 ⎫
d B3 = ⎨2(1.8) ⎜ ⎟+ 3 ⎬ ≈ 0.000025
π ⎩ ⎝ 194 × 106 ⎠ 379 × 106 ⎭
d B = 0.0292 m ≈ 29 mm
307
8-6. A power transmission shaft of hollow cylindrical shape is to be made of hot-rolled 1020 steel with
Su = 65, 000 psi , S yp = 43, 000 psi , e = 36 percent elongation in 2 inches, and fatigue properties as shown
for 1020 steel in Figure 5.31. The shaft is to transmit 85 horsepower at a rotational speed of n = 1800 rpm ,
with no fluctuations in torque or speed. At the critical section, midspan between bearings, the rotating shaft is
also subjected to a bure bendign moment of 2000 in-lb, fixed in a vertical plane by virtue of a system of
symmetrical external forces on the shaft. If the shaft outside diameter is 1.25 inches and the incide diameter is
0.75 inch, what operating life would be predicted before fatigue failure occurs?
-------------------------------------------------------------------------------------------------------------------------
Solution
The torque is
From the problem statement, the bending moment is completely reversed (due to shaft rotation) and is
The state of stress at the outside surface is as shown. The stresses are expressed
in terms of the shaft diameters as
Ta 16Tdo 16(2976)(1.25)
τ xy = = = = 8916 psi
J (
π d o − di
4 4
) (
π (1.25) 4 − (0.75) 4 )
Mc 32 Md o 32(2000)(1.25)
σx = = = = 11,983 psi
I (
π d o − di
4 4
) (
π (1.25) 4 − (0.75) 4 )
The equivalent stress for this state of stress is expressed as σ eq = σ x2 + 3τ xy
2
From the loading conditions, the mean and alternating torque and moment are Tm = 2976 , Ta = 0 and
M m = 0 , M a = 2000 . As a result the mean and alternating shear and normal stresses are
τ xy − m = 8916 psi , τ xy − a = 0 and σ x − m = 0 , σ x − a = 11,983 psi . Therefore
σ eq − m = σ x2− m + 3τ xy
2
− m = (0) + 3(8916) = 15, 443 psi
2 2
σ eq − a = σ x2− a + 3τ xy
2
− a = (11,983) + 3(0) = 11,983 psi
2 2
σ max = 11,983 + 15, 443 = 27, 426 psi < S yp = 43, 000 psi . Therefore
σa 11,983
σ eq −CR = = = 15, 717 psi
1 − σ m / Su 1 − 15, 443 / 65, 000
From Figure 5.31, using σ eq −CR = 15, 717 psi , we estimate infinite life ( N = ∞ ).
308
8-7. A solid cylindrical power transmission shaft is to be made of AM 350 stainless steel for operation in an
elevated temperature air environment of 540 o C (see Table 3.5). The shaft is to transmit 150 kW at a
rotational speed of 3600 rpm, with no fluctuation in torque or speed. At the critical section, midspan between
bearings, the rotating shaft is also subjected to a pure bending moment of 280 N-m , fixed in the vertical
plane by a system of symmetrical external forces on the shaft. If the shaft diameter is 32 mm, predict a range
within which the mean operational life would be expected to fall.
--------------------------------------------------------------------------------------------------------------------------------
Solution
Tr 16T Mr 32M
τ xy = = and σ x = =
J π d3 I π d3
This is a relatively simple state of stress and the principal stress can be determined from either the stress cubic
equation or Mohr’s circle. Since it is a state of plane stress, we know that σ eq = σ x2 + 3τ xy
2
, so
σ eq − a = σ x2− a + 3τ xy
2
− a and σ eq − m = σ x − m + 3τ xy − m .
2 2
Noting that Tmax = Tmin = Tm = 398 N-m , Ta = 0 . With M max = +280 N-m , and M min = −280 N-m , we
determine M m = 0 and M a = 280 . Therefore
32(280)
σ x−a = = 87 MPa and σ x − m = 0
π (0.032)3
16(398)
τ xy − a = 0 and τ xy − m = = 61.9 MPa
π (0.032)3
Therefore
σ eq − a = σ x2− a + 3τ xy
2
−a = (87 )2 + 3 ( 0 )2 = 87 MPa
σ eq − m = σ x2− m + 3τ xy
2
−m = ( 0 )2 + 3 ( 61.9 )2 = 107 MPa
309
Problem 8-7 (continued)
From Table 3.5 we approximate the 540 o C material properties as ( Su )540 o C = 821 MPa and
( S yp )540 C = 572 MPa . In addition, eRT ( 50 mm ) = 13% , so the material is considered ductile.
o
The
maximum normal stress is
σa
(σ eq −CR )540 C =
o
σm
=
87
107
= 100 MPa
1− 1−
Su 821
The S-N curve for AM 350 stainless steel at 540 o C is not readily available, so we will approximate the
fatigue failure stress. Assume the guidelines given for nickel based alloys are applicable, giving
For the ultimate strength we are using S 'f = 0.3(821) to 0.5(821) @ 108 cycles , so
(
Comparing this to σ eq −CR )540 C = 100 MPa we conclude that infinite life is expected. A more accurate
o
310
8-8. A shaft of square cross section 2.0 inches by 2.0 inches, is being successfully used to transmit power in a
application where the shaft is subjected to constant steady pure torsion only. If the same material is used and
the same safety factor is desired, and for exactly the same application, what diameter should a solid
cylindrical shaft be made for equivalent performance?
--------------------------------------------------------------------------------------------------------------------------
Solution
Tcirc. T 2T .
(τ max )circ. = = circ. = rect
Qcirc. ⎛ π r 3 ⎞ π r3
⎜⎜ ⎟⎟
⎝ 2 ⎠
2
= 0.6 ⇒ r 3 = 1.061
π r3
r = 1.02"
311
8-9. A shaft with a raised bearing pad, shown in Figure P8.9 must transmit 75 kW on a continuous basis at a
constant rotational speed of 1725 rpm. The shaft material is annealed AISI 1020 steel. A notch-sensitivity
index of q = 0.7 may be assumed for this material. Using the most accurate procedure you know, estimate the
largest vertical midspan bearing force P that can be applies while maintaining a safety factor of 1.3 based on
an infinite life design.
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Solution
P
RL = RR =
2
⎛ L ⎞ P ⎛ L ⎞ PL
M = RL ⎜ ⎟ = ⎜ ⎟ = = 0.125P
⎝2⎠ 2⎝2⎠ 4
Since the shaft is rotating at a constant rate we know that Tm = T and Ta = 0 . Similarly, since the bearing
force is constant the bending stress is completely reversed, resulting in M a = M = PL / 4 and M m = 0 . We
can apply (9-8) to determine the allowable bearing force P.
16 ⎧ Ma T ⎫ 16 ⎧ 0.125P T ⎫
d3 = ⎨2 K fb ( nd ) + 3 m ⎬ = ⎨2 K fb ( nd ) + 3 ⎬
π ⎩ SN Su ⎭ π ⎩ SN Su ⎭
Rearranging this
4S N ⎛ π d 3 T ⎞
P= ⎜ − 3 ⎟
K fb ( nd ) ⎝ 16 Su ⎠
From the given dimensions we establish r / d = 2 / 32 = 0.0625 and D / d = 38 / 32 = 1.1875 . From Chapter 5
we approximate Kt ≈ 2.00 , which results in
9549 ( kw ) 9549 ( 75 )
T = Tm = = = 415 N-m
n 1725
312
Problem 8-9 (continued)
Using the data; d = 0.032 m , L = 0.50 m , K fb = 1.7 , nd = 1.3 , Su = 393 × 106 , S N = 228 × 106 , and
Tm = 415 N-m and (1)
313
8-10. A solid circular cross-section shaft made of annealed AISI 1020 steel (see Figure 5.31) with an ultimate
strength of 57,000 psi and a yield strength of 43,000 psi is shouldered as shown in Figure P8.10. The
shouldered shaft is subjected to a pure bending moment , and rotates at a speed of 2200 rpm. How many
revolutions of the shaft would you predict before failure takes place?
--------------------------------------------------------------------------------------------------------------------------
Solution
Next
Mc 1600(0.5)
σ act = 1.66 = 1.66 = 27, 053 ≈ 27 ksi
I π (1) 4 / 64
314
8-11. A rotating solid cylindrical shaft must be designed to be as light as possible for use in an orbiting space
station. A safety factor of 1.15 has been selected for this design, and the tentative material selection is Ti-
150a titanium alloy. This shaft will be required to rotate a total of 200,000 revolutions during its design life.
At the most critical section of the shaft, it has been determined from force analysis that the rotating shaft will
be subjected to a steady torque of 1024 rpm and a bending moment of 1252 N-m. It is estimated that the
fatigue stress concentration factor for this critical section will be 1.8 for bending and 1.4 for torsion. Calculate
the required minimum shaft diameter at this critical section.
-------------------------------------------------------------------------------------------------------------------------
Solution
8-11. For Ti 150a, for a design life of 2 × 105 cycles, we get S N = 2×105 ≈ 69 ksi ≈ 476 MPa . Approximating
Su ≈ 1000 MPa , the diameter is approximated from
16 ⎧ Ma T ⎫ 16 ⎧ 1252 1024 ⎫
d3 = ⎨2 K fb ( nd ) + 3 m ⎬ = ⎨2(1.8) (1.15 ) + 3 ⎬ = 0.0000645
π ⎩ SN Su ⎭ π ⎩ 476 × 10 6
1000 × 106 ⎭
d = 0.040 m = 40 mm
315
8-12. The sketch in Figure P8.12 shows a shaft configuration determined by using a now-obsolete ASME
shaft code equation to estimate several diameters along the shaft. It is desired to check the critical sections
along the shaft more carefully. Concentrating attention on critical section E-E, for which the proposed
geometry is specified in Figure P8.12, a force analysis has shown that the bending moment at E-E will be
100,000 in-lb, and the torsional moment is steady at 50,000 in-lb. The shaft rotates at 1800 rpm. Tentatively,
the shaft material has been chosen to be AISI 4340 ultra-high strength steel (see Table 3.3). A factor of safety
of 1.5 is desired. Calculate the minimum diameter the shaft should have at location E-E if infinite life is
desired.
---------------------------------------------------------------------------------------------------------------
Solution
From Tables 3.3 and 3.10 Su = 287 ksi , S yp = 270 ksi , and e(2") = 11% . Estimating the S –N curve, since
Su > 200 ksi , we have S 'f = 100 ksi . Since no information is available for calculating k∞ , we assume
k∞ = 1 , which results in S f = S 'f = 100 ksi .
From the problem statement we have a steady torque and completely reversed bending . Using Figure 5.5 (a)
with r / d = 0.25 / 3.5 = 0.07 and D / d = 4 / 3.5 = 1.14 . This gives Kt ≈ 2.1 . With Su = 287 ksi , @ r = 0.25 ,
we use Figure 5.47 and get q ≈ 1.0 . Therefore
K f = q ( K t − 1) + 1 = 1(2.1 − 1) + 1 = 2.1
16 ⎧ Ma T ⎫ 16 ⎧ 100 50 ⎫
d3 = ⎨2 K fb ( nd ) + 3 m ⎬ = ⎨2(2.1)(1.5) + 3 ⎬ = 33.62
π ⎩ SN Su ⎭ π ⎩ 100 287 ⎭
d = 3.22 in
316
8-13. One of two identical drive shafts for propelling a 600 N radio controlled robot is shown in Figure
P8.13. The shaft is supported by bearings at A and C and driven by gear B. The chains attached to sprockets
D and E drive the front and rear wheels (not shown). The tight side chain tensions on sprockets D and E make
an angle θ = 5o with the horizontal z axis. The gear and sprocket forces are as shown. The shaft is to be
made of AISI cold-drawn medium carbon steel with ultimate and yield strengths of 621 MPa and 483 MPa,
respectively. The robot is being designed for a yearly competition, so long term fatigue is not a primary
consideration. However, since the robotic competition generally involves multiple incidents of high impact,
you decide to include fatigue considerations and assume S N = 300 MPa and nd = 1.5 . Neglecting stress
concentration factors, calculate an appropriate shaft diameter.
------------------------------------------------------------------------------------------------------------------------------------
Solution
FBy = 3600 N ↑
FBz = 3200 N ←
TB = 3600(0.020) = 72 N-m
FDz = 1200 cos 5o ≈ 1195 N →
FDy = 1200sin 5o ≈ 105 N ↑
FEz = 1200 cos 5o ≈ 1195 N ←
FEy = 1200sin 5o ≈ 105 N ↓
TD = TE = 1200(0.030) = 36 N-m
317
Problem 8-13 (continued)
Since no axial forces exist, an axial force diagram is not required. The torque diagram and approximate shear
force and bending moment diagrams for the xy and xz planes are shown below
From the moment diagrams it is obvious that the maximum moment occurs at point B. This is also the
location of the maximum torque and represents the critical point along the shaft. Since the shaft is rotating
in order to drive the wheels, this moment is the alternating moment, with a magnitude of
The torque at B is the mean torque and has a magnitude of 72 N-m . The maximum bending moment is.
Knowing that Su = 621 MPa , S N = 300 MPa , nd = 1.5 and by neglecting stress concentrations , K fb = 1 ,
the shaft diameter is approximated using
16 ⎧ Ma T ⎫ 16 ⎧ 65.8 72 ⎫
d3 = ⎨2 K fb ( nd ) + 3 m ⎬ = ⎨2(1)(1.5) + 3 ⎬ = 0.0000044
π ⎩ SN Su ⎭ π ⎩ 300 × 10 6
621× 106 ⎭
d = 0.0164 m = 16.4 mm
318
8-14. At a weekly design review meeting someone suggests that perhaps the shaft in problem 8.13 will
undergo too much deflection at end E. Therefore it is suggested that an addition bearing support be placed 20
mm to the right of the sprocket at E, thus extending the shaft length to 120 mm. Assuming the same material
and design constrains as in Problem 8.13, determine the required diameter for this shaft
----------------------------------------------------------------------------------------------------------------------------
Solution
FBy = 3600 N ↑
FBz = 3200 N ←
TB = 3600(0.020) = 72 N-m
FDz = 1200 cos 5o ≈ 1195 N →
FDy = 1200sin 5o ≈ 105 N ↑
FEz = 1200 cos 5o ≈ 1195 N ←
FEy = 1200sin 5o ≈ 105 N ↓
TD = TE = 1200(0.030) = 36 N-m
∑ Fy = 0 : Ay + C y + Fy + 3600
+ 105 − 105 = 0
Ay + C y + Fy = −3600 (1)
∑ ( M A ) z = 0 : 3600(0.03) + C y (0.06)
+ 105(0.08) − 105(0.10)
+ Fy (0.12) = 0
0.06C y + 0.12 Fy = −105.9 → C y + 2 Fy = −1765 (2)
319
Problem 8-14 (continued)
yC −1 =
Pbx 2
6 EIL
(L − b2 − x2 )
yC −1 =
3600(0.03)(0.06)
6 EI (0.12)
(
(0.12) 2 − (0.03) 2 − (0.06) 2 )
0.0891
yC −1 =
EI
yC − 2 =
Pbx 2
6 EIL
(L − b2 − x 2 )
yC − 2 =
105(0.04)(0.06)
6 EI (0.12)
(
(0.12) 2 − (0.04) 2 − (0.06) 2 )
0.00322
yC − 2 =
EI
yC −3 =
Pbx 2
6 EIL
(
L − b2 − x 2 )
yC −3 =
−105(0.02)(0.06)
6 EI (0.12)
(
(0.12) 2 − (0.02) 2 − (0.06) 2 )
0.00182
yC −3 = −
EI
3
PL3 C y (0.12)
yC − 4 = =
48EI 48 EI
0.000036C y
yC − 4 =
EI
1
yC = yC −1 + yC − 2 + yC −3 + yC − 4 = 0 = ⎡ 0.0891 + 0.0032 − 0.00182 + 0.000036C y ⎤ C y ≈ −2514
EI ⎣ ⎦
For the xz plane we use the model shown and follow the
same procedures as before. The equations of equilibrium
yield
320
Problem 8-14 (continued)
zC −1 =
Pbx 2
6 EIL
(L − b2 − x 2 )
zC −1 =
3200(0.03)(0.06)
6 EI (0.12)
(
(0.12) 2 − (0.03) 2 − (0.06) 2 )
0.0792
zC −1 =
EI
zC − 2 =
Pbx 2
6 EIL
(
L − b2 − x2 )
zC − 2 =
−1195(0.04)(0.06)
6 EI (0.12)
(
(0.12) 2 − (0.04) 2 − (0.06) 2 )
0.0366
zC − 2 = −
EI
zC −3 =
Pbx 2
6 EIL
(L − b2 − x 2 )
zC −3 =
1195(0.02)(0.06)
6 EI (0.12)
((0.12) 2 − (0.02) 2 − (0.06) 2 )
0.0207
zC −3 =
EI
PL3 C (0.12)3
zC − 4 = = z
48EI 48EI
0.000036C z
zC − 4 =
EI
1
zC = zC −1 + zC − 2 + zC −3 + zC − 4 = 0 =
EI
[0.0792 − 0.0366 + 0.0207 + 0.000036Cz ] Cz ≈ −1758
321
Problem 8-14 (continued)
The approximate shear force and bending moment for the xy and xz planes, as well as the torque distribution
are shown below.
From the moment diagrams it is obvious that the maximum moment occurs at point B. This is also the
location of the maximum torque and represents the critical point along the shaft. Since the shaft is rotating
in order to drive the wheels, this moment is the alternating moment, with a magnitude of
The torque at B is the mean torque and has a magnitude of 72 N-m . The maximum bending moment is.
Knowing that Su = 621 MPa , S N = 300 MPa , nd = 1.5 and by neglecting stress concentrations , K fb = 1
, the shaft diameter is approximated using
16 ⎧ Ma T ⎫ 16 ⎧ 59.1 72 ⎫
d3 = ⎨2 K fb ( nd ) + 3 m ⎬ = ⎨2(1)(1.5) + 3 ⎬ = 0.00000403
π ⎩ SN Su ⎭ π ⎩ 300 × 106 621× 106 ⎭
d = 0.01592 m = 15.92 mm
322
8-15. To obtain a quick-and-dirty estimate for the maximum slope and deflection of the steel shaft shown in
Figure P8.15. it is being proposed to approximate the stepped shaft by an “equivalent” shaft of uniform
diameter d = 100 mm . The shaft may be assumed to be simply supported by bearings at locations A and G,
and loaded as shown. Estimate the maximum deflection of the equivalent-uniform-diameter shaft and the
slopes at bearing locations A and G.
----------------------------------------------------------------------------------------------------------------------------
Solution
L =1 m
⎡ π ( 0.1)4 ⎤
EI = 207 × 109 ⎢ ⎥ = 1.016 × 106 N-m 2
⎢ 64 ⎥
⎣ ⎦
P ⎛ b 3
2⎞ −36 × 103 ⎛ (0.725)3 ⎞
θG = ⎜⎜ 2bL + − 3b ⎟⎟ = ⎜⎜ 2(0.725)(1.0) + − 3(0.725) 2 ⎟ = −0.00150 rad
6 EI ⎝ L ⎠ 6 1.016 × 10 (
6
) ⎝ 1.0 ⎟
⎠
Deflections are determined based on information in the table. Using a = 0.725 m and b = 0.275 m ;
Pab ( a + 2b ) 3a ( a + 2b )
ymax =
27 EI
=
( −36 ×10 ) ( 0.725)( 0.275)( 0.725 + 0.55)
3
3(0.725) ( 0.725 + 0.55 )
= −0.000555 m
27 (1.016 × 10 ) 6
At
P ⎛ b3 ⎞ 9 × 103 ⎛ (0.4)3 ⎞
θA = ⎜⎜ bL − ⎟⎟ = ⎜⎜ 0.4(1.0) − ⎟⎟ = 0.000496 rad
6 EI ⎝ (
L ⎠ 6 1.016 × 106 ) ⎝ 1.0 ⎠
P ⎛ b3 2⎞ 9 × 103 ⎛ (0.4)3 ⎞
θG = ⎜⎜ 2bL + − 3b ⎟⎟ = ⎜⎜ 2(0.4)(1.0) + − 3(0.4) 2 ⎟ = 0.000567 rad
6 EI ⎝ L ⎠ 6 1.016 × 10
6
( ) ⎝ 1.0 ⎟
⎠
The deflection is determined using the same properties as for the slopes; P = 9 kN , a = 0.6 m and b = 0.4 m
323
Problem 8-15 (continued)
ymax =
Pab ( a + 2b ) 3a ( a + 2b )
=
(9 ×10 ) ( 0.6 )( 0.4)( 0.6 + 0.8) 3(0.6) ( 0.6 + 0.8) = 0.000175 m
3
27 EI 27 (1.016 × 10 )
6
Since the location of the maximum deflection for both cases is relatively close, the results are superposed and
the location is averaged.
1
at x = ( 0.4449 + 0.5292 ) = 0.4871 m
2
324
8-16. For the stepped steel shaft of problem 8-15, use integration to determine the maximum displacement
and the slope of the shaft at A and G.
----------------------------------------------------------------------------------------------------------------------------
Solution
∑ Fy = 0 : RL + RR + 9 − 36 = 0
RL + RR = 27
Next, we set up a moment expression for the beam. Since singularity functions are simple to set up and use,
that approach will be used here. Using the free body diagram shown and singularity functions, we write
M ( x) − RL x + 36 x − 0.275 − 9 x − 0.6 = 0
Therefore
d2y
EI = M ( x) = RL x − 36 x − 0.275 + 9 x − 0.6
dx 2
dy R 2 36 2 9 2
Integrating twice; EI = EIθ ( x) = L x − x − 0.275 + x − 0.6 + C1
dx 2 2 2
R 3 36 3 9 3
EIy ( x) = L x − x − 0.275 + x − 0.6 + C1 x + C2
6 6 6
Using the boundary condition y (0) = 0 , C2 = 0 . Using the boundary condition y (1) = 0
2 2 2
EIθ ( x) = 11250 x − 18000 x − 0.275 + 4500 x − 0.6 − 1559.5 (1)
3 3 3
EIy ( x) = 3750 x − 6000 x − 0.275 + 1500 x − 0.6 − 1559.5 x (2)
The stepped shaft results in different EI products for various sections of the shaft.
⎛ π (0.075) 4 ⎞
(
EI AB = EI EG = 207 × 109 ⎜
⎜ 64
) ⎟⎟ = 0.3215 × 10
6
⎝ ⎠
325
Problem 8-16 (continued)
⎛ π (0.100) 4 ⎞ ⎛ π (0.125) 4 ⎞
(
EI BD = 207 × 109 ⎜
⎜) 64
⎟⎟ = 1.0161× 10
6
(
EI DE = 207 × 109 ⎜
⎜ )64
⎟⎟ = 2.481× 10
6
⎝ ⎠ ⎝ ⎠
Plotting equations (1) and (2) results in the slope and displacement curves shown. The maximum
displacement is ymax ≈ 1.15 mm .The slopes at A and G are θ A = −0.00499 rad and θG = 0.00304 rad .
0.2 0.004
0.003
0
0 0.2 0.4 0.6 0.8 1 0.002
-0.2
0.001
Displacement (mm)
-0.4
0
Slope (rad)
0 0.2 0.4 0.6 0.8 1
-0.6 -0.001
-0.002
-0.8
-0.003
-1
-0.004
-1.2
-0.005
-1.4 -0.006
x (m) x (m)
326
8-17. A rotating shaft having 5.00-cm outside diameter and a 6.0-mm-thick wall is to be made of AISI 4340
steel. The shaft is supported at its ends by bearings that are very stiff, both radially and in their ability to resist
angular deflections caused by shaft bending moments. The support bearings are spaced 60 cm apart. A solid-
disk flywheel weighing 450 N is mounted at midspan, between the bearings. What limiting maximum shaft
speed would you recommend for this application, based on the need to avoid lateral vibration of the rotating
system?
--------------------------------------------------------------------------------------------------------------------------
Solution
PL3
ymax =
192 EI
In addition, we can determine
Therefore we use P = 450 + 38.22 = 488.2 . In addition, we know that E = 207 GPa , and we calculate
I=
(
π d o4 − di4 ) = π ⎡⎢⎣( 0.05) 4
− ( 0.038 ) ⎤⎥
4
⎦ = 2.04 × 10−7 m 4
64 64
488.2(0.6)3
ymax = = 1.3 × 10−5 m
(
192 207 × 10 9
)( 2.04 ×10 )−7
Noting that g = 9.81 m/s 2 and that ysh = y fw = ymax = 1.3 × 10−5 m we determine
60 ⎡ 488.2(1.3 × 10−5 ) ⎤
ncr = 9.81 ⎢ −5 2 ⎥
= 8295 rpm
2π ⎢⎣ 488.2(1.3 × 10 ) ⎦⎥
Since it is recommended that the operating speed should be no more that 1/3 to 1/2 of ncr , we suggest
327
8-18. Repeat problem 8-17 using a solid shaft of the same outside diameter instead of the hollow shaft.
--------------------------------------------------------------------------------------------------------------
Solution
PL3
ymax =
192 EI
In addition, we can determine
π ( 0.05 )
2
π do2
Wsh = Lw1 = (0.60)(7.68 × 104 ) = 90.48 N
4 4
Therefore we use P = 450 + 90.48 ≈ 540.5 . In addition, we know that E = 207 GPa , and we calculate
π ( 0.05 )
4
π do4
I= = = 3.07 × 10−7 m 4
64 64
540.5(0.6)3
ymax = = 9.57 × 10−6 m
( )(
192 207 × 109 3.07 × 10−7 )
Noting that g = 9.81 m/s 2 and that ysh = y fw = ymax = 9.57 × 10−6 m we determine
60 ⎡ 540.5(9.57 × 10−6 ) ⎤
ncr = 9.81 ⎢ −6 2 ⎥
= 9670 rpm
2π ⎣⎢ 540.5(9.57 × 10 ) ⎦⎥
Since it is recommended that the operating speed should be no more that 1/3 to 1/2 of ncr , we suggest
328
8-19. A 2-inch-diameter solid cylindrical 1020 steel shaft is supported on identical rolling-element bearings
(see Chapter 11) spaced 90 inches apart, as sketched in Figure P8.19. A rigid coupling weighing 80 lb is
incorporated into the shaft at location A, 30 inches form the left bearing, and a small solid-disk flywheel
weighing 120 lb (see Chapter 18) is mounted on the shaft at location B, 70 inches from the left bearing. The
shaft is to rotate at 240 rpm. The bearings are not able to resist any shaft bending moments.
a. Neglecting any radial elastic deflection in the support bearings, and neglecting the mass of the shaft,
estimate the critical speed for lateral vibration of the rotating system shown. If this estimate of critical speed
is correct, is the proposed design acceptable?
b. Reevaluate the critical speed estimate of (a) by including the mass of the shaft in the calculation. If this
new estimate of critical speed is correct, is the proposed design acceptable?
c. Reevaluate the critical speed estimate of (b) if the radial elastic deflections of the bearings (the spring rate
of each bearing has been provided by the bearing manufacturer as 5 × 105 lb-in ) are included in the
calculation. Does this new estimate of ctiotical speed, if correct, support the postulate that the system is
adequately designed from the standpoint of lateral vibration?
------------------------------------------------------------------------------------------------------------------------
Solution
(a) The critical speed for lateral vibration of the shaft may be estimated from (8-18). From the problem
statement, shaft weight and bearing stiffness effects will be neglected for this estimate. We need the
displacement at points A and B. We treat each load independently and add the results. For the shaft we know
E = 30 × 106 psi , I = π (2)4 / 64 = 0.785 in 4 , and L = 90 in. The product EI = 23.55 × 106 lb-in .
PA a 2b 2 80(30) 2 (60) 2
( y A )c = = = 0.04076"
3EIL ( )
3 23.55 × 106 ( 90 )
( y B )c = (
PAbx 2
)
L − b2 − x 2 =
80(30)(20) ⎡ (90) 2 − (30) 2 − (20) 2 ⎤ = 0.00257"
6 EIL ( 6
)
6 23.55 × 10 ( 90 ) ⎣ ⎦
PB a 2b 2 120(70) 2 (20) 2
( yB ) fw = = = 0.03699"
3EIL ( )
3 23.55 × 106 ( 90 )
( y A ) fw = (
PB bx 2
L − b2 − x2 = ) 120(20)(30) ⎡ (90) 2 − (20) 2 − (30) 2 ⎤ = 0.00381"
6 EIL ( 6
)
6 23.55 × 10 ( 90 ) ⎣ ⎦
329
Problem 8-19 (continued)
π (2)3
Wsh = (90)(0.283) ≈ 80 lb
4
The case of a concentrated load at the center is used. For a concentrated load in the center of the
W x
( )
shaft ( y A ) sh = sh 3L2 − 4 x 2 . Using x = 30"
48 EI
80(30)
( y A )sh = ⎡3(90) 2 − 4(30) 2 ⎤ = 0.04395"
(
48 23.55 × 10 6
) ⎣ ⎦
For point B we use x = 30" (working from right to left along the shaft)
80(20)
( yB )sh = ⎡3(90) 2 − 4(20) 2 ⎤ = 0.03213"
(
48 23.55 × 106 ⎣ ) ⎦
We also need to determine the mid-span deflection due to the shaft weight, which is
Wsh L3 80(90)3
( yC )sh = = = 0.05159"
(
48 EI 48 23.55 × 106 )
We also need the deflection at C due to the collar and the flywheel. For the collar a = 70" , b = 20",
and x = 45" .
( yC )c =
Wsh bx 2
(
L − b2 − x 2 = )
80(20)(45) ⎡ (90) 2 − (20) 2 − (45) 2 ⎤ = 0.03213"
6 EIL 6
(
6 23.55 × 10 ( 90 ) ⎣ ) ⎦
( yC ) fw =
Wsh bx 2
(
L − b2 − x2 = )
120(20)(45) ⎡(90)2 − (20) 2 − (45) 2 ⎤ = 0.04820"
6 EIL 6 23.55 × 10 ( 90 ) ⎣
6
( ) ⎦
330
Problem 8-19 (continued)
ncr 618
= = 2.68
nop 240
∑ Fy = 0 : RL − RR − 280 = 0
∑ M L = 0 : 90 RR − 80(30) − 80(45) − 120(70) = 0
RR = 160 lb , RL = 120 lb
Each bearing has a spring stiffness of k = F / y = 5 × 105 . The deflection at each bearing is therefore
y RL = 120 / 5 × 105 = 0.002" and y RR = 160 / 5 × 105 = 0.003" . We approximate the effect of bearing
displacement by averaging the displacement and adding it to the existing displacements. Using
yavg = (0.002 + 0.003) / 2 = 0.0025" results in
ncr 578
= = 2.41
nop 240
331
8-20. For the proposed coupling sketched in Figure P8.20, evaluate the folloeing aspects of the proposed
configuration if a design safety factor of 2.0 is desired.
a. Shear and bearing in the keys.
b. Shear and bearing in the flange attachment bolts.
c. Bearing on the flange at attachment bolt interfaces.
d. Shear in the flange at the hub.
The input shaft has a nominal diameter of 2.25 inches, and supplies a steady input of 50 hp at 150
rpm. The bolt circle diameter is db = 6.0 inches . Cold-drawn AISI 1020 steel is being proposed as the
material for the coupling components, including the bolts, and also the material for the key (see Table 3.3). Is
the coupling design acceptable as proposed?
------------------------------------------------------------------------------------------------------------------------------
Solution
63, 025(50)
(a) The torque is T = ≈ 21, 000 in-lb . For a 2” diameter shaft, a 1/2" square key is recommended
150
(Table 8.1). Since the load does not fluctuate K tτ = 1.0 and
π (2.25)
Leq − str = ≈ 3.5"
2(1.0)
2T 2(21, 000)
τs = = = 10, 666 ≈ 10, 670 psi
DwL 2.25(0.5)(3.5)
Based on distortional energy, τ yp = 0.577 S yp = 0.577(51) = 29, 430 psi . The existing factor of safety is
τ yp 29, 430
nex = = = 2.76 ≈ 2.8 > nd = 2 - acceptable
τs 10, 670
4T 4(21, 000)
σc = = = 21,333 ≈ 21,330 psi
DwL 2.25(0.5)(3.5)
S yp 51, 000
nex = = = 2.39 ≈ 2.4 > nd = 2 - acceptable
σc 21,330
(b) The area of each 0.5” diameters bolt is Ab = π (0.5) 2 / 4 = 0.1963 ≈ 0.196 in 2 . The total shear area is
Asb = 6 Ab = 1.176 in 2 . The torque-induced force at the bolt circle is
2T 2(21, 000)
FB = = = 7000 lb
dB 6
332
Problem 8-20 (continued)
Each bolt in the pattern supports a force of Fb = FB / 6 = 1167 lb . The shear stress in each bolt is
Fb 1167
τb = = = 5954 ≈ 5950 psi
Ab 0.196
τ yp 29, 430
nex = = = 4.95 ≈ 5 > nd = 2 - acceptable
τb 5950
1167
σc = = 3734 ≈ 3730 psi
0.5(0.625)
S yp 51, 000
nex = = = 13.66 >> nd = 2 - acceptable
σc 3730
(c) Since the flange and bolt material are the same, the existing factors of safety for flange and bolt bearing
are acceptable.
2T 2(21, 000)
Fh = = = 9882 ≈ 9880 lb
dh 4.25
The flange shear area at the edge of the hub is Ash = π (4.25)(0.625) = 8.345 in 2 . The shaer stress and
existing factor of safety are
Fh 9880
τ sf = = = 1184 psi
Ash 8.345
τ yp 29, 430
nex = = = 24.85 > nd = 2 - acceptable
τ sh 1184
333
8-21. As a new engineer, you have been assigned the task of recommending an appropriate shaft coupling for
connecting the output shaft of an 8.95 kW gear-motor drive unit, operating at 600 rpm, to the input shaft of a
newly designed seed-corn cleaning machine ordered by a farm-supply depot. Based on the capabilities within
your company’s production facility, it has been estimated that the parallel centerline misalignment between
the motor drive shaft and the input shaft of the seed cleaning machine may be as much as 0.8 mm, and the
angular misalignment between shafts may be as much as 2o . What type of coupling would you recommend?
--------------------------------------------------------------------------------------------------------------
Solution
Referring to Figure 8.4, and reading “flexible couplings” in Section 8.9, the following table is made
Comparing the information in the problem statement; Moderate torque capacity, 0.8 mm parallel alignment
and 2o angular misalignment, We conclude that coupling (d), a spring coupling, is appropriate.
334
8-22. a. A chain drive (see Chapter 17) delivers 110 horsepower to the input shaft of an industrial blower in a
paint manufacturing plant. The drive sprocket rotates at 1700 rpm, and has a bore diameter of 2.50
inches and a hub length of 3.25 inches. Propose an appropriate geometry for a standard square key,
including width and length dimensions, if the key is to be made of 1020 cold-drawn steel having
Su = 61, 000 psi and S yp = 51, 000 psi . The key material may be assumed to be weaker than either the
mating shaft material or hub material. A design safety factor of 3 is desired.
b. Would it be possible to use a standard Woodruff key of the same material in this application?
-----------------------------------------------------------------------------------------------------------------------------
Solution
63, 025(110)
(a) The torque is T = ≈ 4078 in-lb . For a 2.5” diameter shaft, a 5/8" square key is
1700
recommended (Table 8.1). Since the load does not fluctuate K tτ = 1.0 and
π (2.5)
Leq − str = ≈ 3.93"
2(1.0)
The hub length is only 3.25”, so the longest key that can be used is 3.25”. Actually, a 3.0” key would give
end clearance, so we will assume the key length to be 3.0”. The shear stress is
2T 2(4078)
τs = = = 1739.9 ≈ 1740 psi
DwL 2.5(0.625)(3.0)
Based on distortional energy, τ yp = 0.577 S yp = 0.577(51) = 29, 430 psi . The existing factor of safety is
τ yp 29, 430
nex = = = 16.9 > nd = 3 - acceptable
τs 1740
4T 4(4078)
σc = = = 3479.9 ≈ 3480 psi
DwL 2.5(0.625)(3.0)
S yp 51, 000
nex = = = 14.6 > nd = 3 - acceptable
σc 3480
Based on these safety factors a smaller key would work. Rearranging the equation for σ c and replacing σ c
with an allowable stress, (σ c )allow = S yp / nd = 51/ 3 = 17 ksi , the key width resulting is a factor of safety of
3.0 can be determined
4T 4(4078)
w= = = 0.1279 ≈ 0.13"
DL (σ c )allow 2.5(3.0)(17, 000)
335
Problem 8-22 (continued)
(b) To investigate the possible use or a Woodruff key, Figure 8.6 (d) and Table 8.2 provide the information
needed. For a design safety factor of 3.0, τ d = τ yp / nd = 29, 430 / 3 = 9810 psi . Setting τ d = τ s
2T 2(4078)
wL = = = 0.3326 in 2
Dτ d 2.5(9810)
336
8-23. Repeat problem 8-22,except that the drive perocket rotates at 800 rpm.
--------------------------------------------------------------------------------------------------------------
Solution
63, 025(110)
(a) The torque is T = ≈ 8666 in-lb
800
S yp 51 0.577 S yp 0.577(51)
σd = = = 17 ksi τd = = = 9.81 ksi
nd 3 nd 3
The hub length is only 3.25”, so the longest key that can be used is 3.25”. Actually, a 3.0” key would give
end clearance, so we will assume the key length to be 3.0”. Setting τ d = τ s
2T 2(8666)
w= = = 0.2356 ≈ 0.24"
Dτ s L 2.5(9810)(3.0)
Setting σ d = σ c
4T 4(8666)
w= = = 0.2718 ≈ 0.27"
Dσ c L 2.5(17, 000)(3.0)
(b) To investigate the possible use or a Woodruff key, Figure 8.6 (d) and Table 8.2 provide the information
needed. Setting σ d = τ c
⎛D ⎞ 4T 4(8666)
wL = ⎜ − h ⎟ L = = = 0.8156 ≈ 0.82 in 2
⎝2 ⎠ Dσ d 2.5(17, 000)
⎛ 2.5 ⎞
⎜ 2 − 0.641⎟ (1.5) = 0.9135 in > 0.82 in
2 2
⎝ ⎠
Based on this we select #1212 key that could be used for this application.
337
8-24. For the chain drive specifications given in problem 8.22, and for the same sprocket dimensions, select
the minimum size of grooved pin that could be used to attach the sprocket to the shaft, assuming the grooved
pin to be made of 1095 steel quenched and drawn to Rockwell C 42 (see Table 3.3)
-------------------------------------------------------------------------------------------------------------------------------
Solution
For the material specified, Su = 61 ksi , S yp = 51 ksi , e(2") = 22% . The torque is
63, 025(110)
T= ≈ 4078 in-lb
1700
Assuming the shear force is equally distributed between the two shear areas, the shear force Fs is
T 4078
Fs = = = 1631 lb
D 2.5
Since nd = 3.0
Fd = nd Fs = 3(1631) = 4893 lb
From Table 8.6, the smallest “grooved” pin with a capacity of 4893 lb is
338
8-25. The hub of a gear is keyed to an 80-mm diameter shaft using a 30 mm long square key. The shaft is
required to operate at 1800 rpm. The shaft and key are made from the same alloy steel, with S y = 350 MPa
and τ all = 140 MPa .
a. Determine the power that can be transmitted by the key.
b. Determine the power capacity of the shaft assuming Ktτ = 1.8 .
--------------------------------------------------------------------------------------------------------------
Solution
(a) The force transmitted through the interface of the hub and shaft is related to the toque by
T = Fr = Fd / 2 , where F = τ A . For a shear failure of the key
Tshear = τ all A ( d / 2 ) = τ all A ( 2 / 2 ) = τ all A = τ all wl = 140 × 106 ( 0.020 )( 0.030 ) = 84 kN-m
Failure could also result from bearing stress. The torque in this case is defined by
The maximum allowable torque is therefore Tmax = Tshear = 84 kN-m . Therefore the power that can be
transmitted through the key is determined from
(b) The horsepower capacity of the shaft is determined by first defining the allowable torque in the shaft
based on its shear strength, τ all = 140 MPa . The maximum torque supported by the shaft is related to the
maximum shear stress in the shaft by
Since this is significantly smaller than the power that the key will withstand, we conclude that the shaft will
fail before the key.
339
8-26. a. A V-pulley is to be mounted on the steel 1,0-inch-diameter engine drive-shaft of a lawn tractor. The
pulley must transmit 14 horsepower to the drive-shaft at a speed of 1200 rpm. If a cup point setscrew
were used to attach the pulley hub to the shaft, what size setscrew would be required? A design safety
factor of 2 is desired.
b. What seating torque would be recommended to properly tighten the setscrew so that it will not slip
when power is being transmitted?
----------------------------------------------------------------------------------------------------------------------------------
Solution
(a) The rules of thumb in 8.8 suggest that selection is nominally chosen to be almost 1/4 of the shaft diameter,
and set screw length chosen to be about 1/2 the shaft diameter. For a 1” diameter shaft
The torque is
63, 025(14)
T= = 735.29 ≈ 735 in-lb
1200
2T 2(735)
Fs = = = 1470 lb
d 1.0
Fd = nd Fs = 2(1470) = 2940 lb
From Table 8.5, a 1/2" set screw is needed. This size seems too large for the shaft-hub size. It will be
suggested that 2 set smaller screws be used, which support a load of Fd = 2940 / 2 = 1470 lb each. The
recommendations is:
Use two 5/16” set screws that are 1/2" long and spaced 90o apart
340
Chapter 9
9-1. When stresses and strains in a machine element or a structure are investigated, analyses are based on either a
“strength of materials” approach or a “theory of elasticity” model. The theory of elasticity model facilitates
determining the distributions of stresses and strains within the body rather than assuming the distributions are
required by the strength of materials approach. List the basic relationships from elasticity theory needed to
determine the distributions of stress and strains within elastic solids subjected to externally applied forces and
displacements.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
The basic relationships from elasticity theory needed to determine the distributions of stresses and strains within
elastic solids subjected to externally applied forces and displacements include;
341
9-2. Equations for stresses in thin-walled cylinders are less complicated that equations for stress in thick-walled
cylinders because of the validity of two simplifying assumptions made when analyzing thin-walled cylinders. What
are these two assumprtions?
------------------------------------------------------------------------------------------------------------------------------------------
Solution
(1) The wall must be thin enough to satisfy the assumption that the radial stress component ( σ r ) at the wall is
negligibly small compared to the tangential ( σ t ) stress component.
(2) The wall must be thin enough that σ t is uniform across it.
342
9-3. a. A thin-walled cylindrical pressure vessel with closed ends is to be subjected to an external pressure po with
an internal pressure of zero. Starting with the generalized Hooke’s Law equations, develop expressions for
radial, transverse (hoop), and longitudinal (axial) strain in the cylindrical vessel wall as a function of pressure
po , diameter d, wall thickness t, Young’s modulus E, and Poisson’s ratio ν .
b. Assume the vessel is made from AISI 1018 HR steel [ Su = 400 MPa , S yp = 220 MPa , ν = 0.30 ,
E = 207 GPa , and e ( 50 mm ) = 25% ] and if the external pressure is po = 20 MPa . If the vessel has an outer
diameter of 125 mm, wall thickness of 6 mm, and length of 400 mm, determine if the vessel length increases or
decreases and by how much.
c. Determine if the vessel thickness changes (increase or decrease) and by how much.
d. Would you predict yielding of the vessel wall? (Neglect stress concentrations and clearly support your
prediction with appropriate calculations.)
------------------------------------------------------------------------------------------------------------------------------------------
Solution
(a) Given radial ( σ r ) , transverse ( σ t ) , and longitudinal ( σ l ) stress components, the radial, transverse, and
longitudinal strains according to generalized Hooke’ Law are
1 1 1
εr = ⎡σ r −ν (σ t + σ l ) ⎤⎦ , ε t = ⎡⎣σ t −ν (σ r + σ l ) ⎤⎦ , ε l = ⎡⎣σ l −ν (σ t + σ r ) ⎤⎦
E⎣ E E
For a thin-walled pressure vessel σ r = 0 . Since the pressure is external (as opposed to the internal pressure for
which the stress-pressure relationships were developed)
po d p d
σt = − , σl = − o
2t 4t
ν ⎡ po d po d ⎤ 3ν ⎛ po d ⎞
εr = − − − =
E ⎢⎣ 2t 4t ⎥⎦ E ⎝⎜ 4t ⎠⎟
1 ⎡ po d −ν po d ⎤ ⎡ν − 2 ⎤ ⎛ po d ⎞
εt = ⎢− − =
E ⎣ 2t 4t ⎥⎦ ⎢⎣ E ⎥⎦ ⎝⎜ 4t ⎠⎟
1 ⎡ po d −ν po d ⎤ ⎡ 2ν − 1 ⎤ ⎛ po d ⎞
εl = ⎢− − =
E ⎣ 4t 2t ⎥⎦ ⎢⎣ E ⎥⎦ ⎝⎜ 4t ⎠⎟
(b) The change in length of the vessel is determined by ∆L = loε l . Using the date given
(c) The change in wall thickness is determined by ∆t = toε r . Using the date given
343
Problem 9-3 (continued)
(d) The material is ductile and the state of stress is biaxial. The principal stresses are
po d 20 × 106 (0.125)
σ1 = σ t = − =− = −208 MPa
2t 2(0.006)
σ2 = σr = 0
po d 20 × 106 (0.125)
σ3 = σt = − =− = −104 MPa
4t 4(0.006)
( )
Using distortional energy, FIPTOI (σ 1 − σ 2 ) + (σ 2 − σ 3 ) + (σ 3 − σ 1 ) ≥ 2 S yp
2 2 2 2
344
9-4. A thin- walled, closed end pressure vessel has an outer diameter of 200 mm, a wall thickness of 10 mm, and
length of 600 mm. The vessel is subjected to an internal pressure of 30 MPa and an external tensile axial force F.
Assume the vessel is made form a steel alloy with Su = 460 MPa , S yp = 270 MPa , ν = 0.30 , E = 207 GPa , and
e ( 50 mm ) = 25% . Determine the largest force F that can be applied before yielding occurs.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
po d 30 × 106 (0.2)
σt = =− = 300 MPa
2t 2(0.01)
po d 30 × 106 (0.2)
σt = = = 150 MPa
4t 4(0.01)
F F F
≈ = ≈ 0.159 F MPa if F is in kN
A π dt π (0.2)(0.01)
Therefore we have
The principal stresses will be σ1 = 300 , σ 2 = 0 , and σ 3 = 150 + 0.159 F , provided F < 150 / 0.159 = 943 .
Assuming that F < 943 and using the distortional energy failure theory
(σ1 − σ 2 )2 + (σ 2 − σ 3 )2 + (σ 3 − σ1 )2 ≥ 2 ( S yp )
2
345
9-5. Based on the concepts utilized to derive expressions for the stresses in the wall of a thin-walled cylindrical
pressure vessel, derive expressions for the stress in the wall of a thin-walled spherical pressure vessel.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
Considering any thin-walled hemisphere with diameter d and wall thickness t, the
stresses in the wall can be modeled as σ . The force due to pressure acting on the
back wall ( Fpx ) must balance the force due to the stress ( σ ). We can write
∑ Fx = 0 : σ (π dt ) − Fpx = 0
⎛ πd2 ⎞
Since Fpx = pi A = pi ⎜ ⎟⎟
⎜ 4
⎝ ⎠
⎛ πd2 ⎞ pi d
σ (π dt ) − pi ⎜⎜ ⎟⎟ = 0 ⇒ σ=
⎝ 4 ⎠ 4t
346
9-6. A steel hydraulic cylinder, closed at the ends, has an inside diameter of 3.00 inches and an outside diameter of
4.00 inches. The cylinder is internally pressurized by an oil pressure of 2000 psi. Calculate (a) the maximum
tangential stress in the cylinder wall, (b) the maximum radial stress in the cylinder wall, and (c ) the maximum
longitudinal stress in the cylinder wall.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
347
9-7. A cylindrical pressure vessel made of AISI hot-rolled steel plate is closed at the ends. The cylinder wall has an
outside diameter of 12.0 inches and an inside diameter of 8.0 inches. The vessel is internally pressurized to a gage
pressure of 15,000 psi.
a. Determine, as accurately as you can, the magnitudes of maximum radial stress, maximum tangential stress,
and maximum longitudinal stress in the cylindrical pressure vessel.
b. Making the “usual” thin-walled assumptions, and using a mean cylindrical wall diameter of 10.0 inches, for
your calculations, again determine the magnitude of the maximum radial stress, the maximum tangential stress,
and the maximum longitudinal stress in the cylindrical pressure vessel wall.
c. Compare the results of (a) and (b) by calculating the percentage errors as appropriate, and comment on the
results of your comparison.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
(a) The maximum radial stress is at r = a = 4.0" and is σ r − max = − pi = −15, 000 psi . The maximum tangential
stress is at r = a = 4.0" and is
⎛ (6.0) 2 + (4.0) 2 ⎞ ⎛ 52 ⎞
σ t − max = 15, 000 ⎜⎜ 2 ⎟
⎟
= 15, 000 ⎜ ⎟ = 39, 000 psi
⎝ (6.0) − (4.0) ⎠ ⎝ 20 ⎠
2
⎛ (4.0) 2 ⎞ ⎛ 16 ⎞
σ l − max = 15, 000 ⎜⎜ ⎟⎟ = 15, 000 ⎜ ⎟ = 12, 000 psi
⎝ (6.0) − (4.0) ⎠ ⎝ 20 ⎠
2 2
(c) Comparing the thin-wall results with the more accurate thick-wall results, we construct the following table
Comments: There are significant and intolerable errors in the thin-walled estimates of σ r and σ l . The estimates for
σ t not too bad.
348
9-8. A closed end cylindrical pressure vessel made form AISI 1018 HR steel [ Su = 400 MPa , S yp = 220 MPa ,
ν = 0.30 , E = 207 GPa , and e ( 50 mm ) = 25% ] has in inside diameter of 200 mm and a wall thickness of 100 mm.
It is required to operate with a design factor of safety of nd = 2.5 . Determine the largest internal pressure that can
be a applied before yielding occurs.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
Since di = 2a = 200 mm , and t = 100 mm we know a = 100 mm and b = 200 mm , which results in
The largest stresses occur on the inside surface, where r = a = 0.1 m . This gives
pi ⎡ 0.04 ⎤ pi ⎡ 0.04 ⎤ 5 pi
σr = ⎢1 − 2⎥
= − pi σt = ⎢1 + 2⎥
= pi σr =
3 ⎢⎣ (0.1) ⎥⎦ 3 ⎢⎣ (0.1) ⎥⎦ 3 3
5 p
From this we note σ1 = σ t = pi , σ 2 = σ r = i , and σ 3 = σ r = − pi . The design stress is
3 3
σ d = S yp / nd = 220 / 2.5 = 88 MPa . Applying the distortional energy theory
(σ 1 − σ 2 ) 2 + ( σ 2 − σ 3 ) 2 + (σ 3 − σ 1 ) 2 ≥ 2 (σ d ) 2
2 2 2
⎛5 1 ⎞ ⎛1 ⎞ ⎛ 5 ⎞
⎜ 3 pi − 3 pi ⎟ + ⎜ 3 pi − (− pi ) ⎟ + ⎜ − pi − 3 pi ⎟ ≥ 2 ( 88 )
2
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
96 2
pi ≥ 2 ( 88 ) → pi = 38.1 MPa
2
pi = 38.1 MPa
9
349
9-9. Calculate the maximum tangential stress in the steel hub of a press fit assembly when pressed upon the
mounting diameter of a hollow steel shaft. The unassembled hub dimensions are 3.00 inches for the inside diameter
and 4.00 inches for the outside diameter, and the unassembled dimensions of the shaft at the hub mounting site are
3.030 inches outside diameter and 2.00 inches for the inside diameter. Proceed by first calculating the interfacial
pressure at the mating surfaces caused by the press fit, then calculating the hub tangential stress caused by the
pressure.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
Both the hub and the shaft are steel with E = 30 × 106 psi ,ν = 0.30 , and
∆
p=
⎡ a ⎛b +a
2 2 ⎞ d ⎛ d 2 + c2 ⎞⎤
2⎢ ⎜⎜ 2 + ν h ⎟⎟ + ⎜⎜ 2 2 −ν s ⎟⎟ ⎥
⎝b −a ⎠ Es ⎝ d − c
2
⎢⎣ Eh ⎠ ⎥⎦
0.030
p= = 48,928 psi
⎡ 1.5 ⎛ (2.0) + (1.5)
2 2 ⎞ 1.515 ⎛ (1.515) 2 + (1.0) 2 ⎞⎤
2⎢ ⎜⎜ + 0.30 ⎟⎟ + ⎜⎜ − 0.30 ⎟⎟ ⎥
⎣⎢ 30 × 10 ⎝ (2.0) − (1.5) ⎠ 30 × 10 ⎝ (1.515) − (1.0)
6 2 2 6 2 2
⎠ ⎦⎥
⎛ b2 + a 2 ⎞ ⎛ (2.0) 2 + (1.5) 2 ⎞
σ th = p ⎜⎜ 2 2 ⎟⎟ = 48,928 ⎜⎜ 2 ⎟
⎟
= 174, 743 psi
⎝b −a ⎠ ⎝ (2.0) − (1.5) ⎠
2
The tangential stress in the hub is very high, and the design should be carefully reevaluated.
350
9-10. The hub of an aluminum [ Su = 186 MPa , S yp = 76 MPa , ν = 0.33 , E = 71 GPa ] pulley has an inside
diameter of 100 mm and an outside diameter of 150 mm. It is pressed onto a 100.5-mm-diameter hollow steel
[ Su = 420 MPa , S yp = 350 MPa , ν = 0.30 , E = 207 GPa ] shaft with an unknown inner diameter. Determine the
allowable inside diameter of the steel shaft assuming a design factor of safety of nd = 1.25 .
------------------------------------------------------------------------------------------------------------------------------------------
Solution
Since ∆ = 0.1005 − 0.100 = 0.0005 m , the contact pressure between the hub and the shaft is
∆
p=
⎡ a ⎛ b2 + a2 ⎞ d ⎛ d 2 + c2 ⎞⎤
2⎢ ⎜⎜ b 2 − a 2 + ν h ⎟⎟ + E ⎜⎜ d 2 − c 2 −ν s ⎟⎟ ⎥
⎣⎢ Eh ⎝ ⎠ s ⎝ ⎠ ⎦⎥
Knowing that a = 0.05 , b = 0.075 , c = unknown , d = 0.0505 , Eh = 71 GPa , ν h = 0.33 , Es = 207 GPa , and
ν s = 0.3 , the contact pressure is
0.0005
p=
⎡ 0.05 ⎛ (0.075) + (0.05)
2 2 ⎞ 0.05025 ⎛ (0.05025) 2 + c 2 ⎞⎤
2⎢ ⎜⎜ + 0.33 ⎟⎟ + ⎜⎜ + 0.30 ⎟⎟ ⎥
⎣⎢ 71× 10 ⎝ (0.075) − (0.05) ⎠ 207 × 10 ⎝ (0.05025) − c
9 2 2 9 2 2
⎠ ⎦⎥
0.00025
p= (1)
⎡ −12 −12 ⎛ 0.002525 + c
2 ⎞⎤
⎢ 2.063 × 10 + 0.243 × 10 ⎜⎜ + 0.30 ⎟⎟ ⎥
⎝ 0.002525 − c
2
⎢⎣ ⎠ ⎥⎦
⎛ d 2 + c2 ⎞ ⎛ (0.05025) 2 + c 2 ⎞ ⎛ 0.002525 + c 2 ⎞
σ ts = − p ⎜⎜ ⎟
2 ⎟
= − p ⎜⎜ ⎟⎟ = − p ⎜⎜ ⎟⎟ σ rs = − p
⎝ d −c ⎠ ⎝ (0.05025) − c ⎝ 0.002525 − c
2 2 2 2
⎠ ⎠
⎛ 0.002525 + c 2 ⎞
σ1 = 0 , σ 2 = σ rs = − p σ 3 = σ ts = − p ⎜⎜ 2 ⎟
⎟
⎝ 0.002525 − c ⎠
Since the shaft is ductile, we use distortional energy with σ d = S yp / nd = 350 /1.25 = 280 MPa
(σ 1 − σ 2 ) 2 + ( σ 2 − σ 3 ) 2 + (σ 3 − σ 1 ) 2 ≥ 2 (σ d ) 2
2 2
⎛ ⎪⎧ ⎛ 0.002525 + c
2 ⎞ ⎪⎫ ⎞ ⎛ ⎛ 0.002525 + c 2 ⎞ ⎞
( 0 − (− p) ) 2
+ ⎜ − p − ⎨− p ⎜
⎜ ⎜ ⎟⎟ ⎬ ⎟ + ⎜ − p ⎜⎜
⎜
⎟⎟ − 0 ⎟ ≥ 156.8 × 10
⎟
15
⎩⎪ ⎝ 0.002525 − c ⎠ ⎭⎪ ⎟⎠ ⎝ ⎝ 0.002525 − c
2 2
⎝ ⎠ ⎠
351
Problem 9-10 (continued)
2 2
⎛ ⎛ 0.002525 + c 2 ⎞ ⎞ ⎛ ⎛ 0.002525 + c 2 ⎞⎞
p2 + ⎜ − p + p⎜ ⎟⎟ ⎟ + ⎜ − p ⎜⎜ ⎟⎟ ⎟ ≥ 156.8 × 10
15
⎜ ⎜ 0.002525 − c 2 ⎟ ⎜ ⎟
⎠ ⎠ ⎝ ⎝ 0.002525 − c
2
⎝ ⎝ ⎠⎠
⎧ 2⎫
⎪ ⎛ 0.002525 + c ⎞ ⎛ 0.002525 + c 2 ⎞ ⎪
2
2 p 2 ⎨1 + ⎜ ⎟⎟ + ⎜⎜ ⎟⎟ ⎬ ≥ 156.8 × 10
15
(2)
⎜
⎪⎩ ⎝ 0.002525 − c ⎠ ⎝ 0.002525 − c
2 2
⎠ ⎪⎭
Substituting (1) into (2) and iterating to a solution we find that failure is not predicted until c ≈ 0.0325 m . Therefore
we can have a hollow steel shaft with an inside diameter of
di = 32.5 mm
352
9-11. In the design of a jet cargo aircraft, the tail stabilizer, a horizontal flight-control surface, is to be mounted high
up on the tail rudder structure, and it is to be controlled by two actuator units. The aft unit is to provide the major
large-amplitude movement, while the forward unit is to provide the trim action. The forward actuator consists
essentially of a power-screw (see Chapter 12) driven by an electric motor, with , for dual-unit safety purposes, an
alternative drive consisting of a hydraulic motor that can also drive the screw. In addition, a hand drive is provided
in case both the electric drive and the hydraulic drive unit fail.
The screw consists of a hollow tube of high-strength steel with threads turned on the outer surface, and, for
fail-safe dual-load-path purposes, a titanium tube is to be shrink-fitted inside of the hollow steel tube. For
preliminary design purposes, the screw may be considered to be a tube of 4 inches inside diameter and ½-inch wall
thickness. The proposed titanium tube would have a 4-inch nominal outside diameter and 1-inch-thick wall. The
tubes are to be assembled by the hot-and-cold shrink assembly method. The linear coefficient of thermal expansion
for the steel material is 6.5 × 10−6 in/in/ o F , and for linear coefficient of thermal expansion for the titanium is
3.9 × 10−6 in/in/ o F .
a. Determine the actual dimensions at 70 o F that should be specified if the diametral interference must never,
at any temperature within the expected range, be less than 0.002 inch. Expected temperatures range
between the extremes of −60 o F and 140 o F . Also, the steel tube must not exceed a tangential stress level
of 120,000 psi at either temperature extreme.
b. Determine the temperature to which the screw must be heated for assembly purposes if the titanium tube is
cooled by liquid nitrogen to −310 o F , and if the diametral clearance distance between tubes for assembly
purposes should be about 0.005 inch.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
(a) Specified temperatures are Tmin = −60 o F , Tmax = +145 o F, Troom = +70 o F . By problem specification, for al l
temperatures within the stated range ∆ ≥ 0.002" and the stress in the steel tube at all temperatures in the range must
satisfy the relation σ t − steel ≤ 120 ksi. Because α s > α t , the “loss of fit” problem is most serious at Tmax = +145 o F .
Thus
Therefore, the actual dimensions should be ∆Do −ti = 4.0028" and ∆Di − steel = 4.0000" .
Next, the tangential stress level must be checked in the steel outer tube for the most severe case, which occurs at
Tmin = −60 o F . The diametral interference at this temperature is calculated as
353
Problem 9-11. (continued)
0.0042
p=
⎡ 2.0 ⎛ (2.5) + (2.0)
2 2 ⎞ 2.0 ⎛ (2.0) 2 + (1.0) 2 ⎞⎤
2⎢ ⎜⎜ + 0.30 ⎟⎟ + ⎜⎜ − 0.30 ⎟⎟ ⎥
⎢⎣ 30 × 10 ⎝ (2.5) − (2.0) ⎠ 16 × 10 ⎝ (2.0) − (1.0)
6 2 2 6 2 2
⎠ ⎥⎦
0.0042
=
⎡ ⎛ 10.25 ⎞ ⎛5 ⎞⎤
2 ⎢0.333 × 10−6 ⎜ + 0.30 ⎟ + 0.125 × 10−6 ⎜ − 0.30 ⎟ ⎥
⎣ ⎝ 2.25 ⎠ ⎝ 3 ⎠⎦
0.0042
= = 4247 ≈ 4250 psi
2 ⎡0.3237 × 10−6 + 0.1708 × 10−6 ⎤
⎣ ⎦
⎛ (2.5) 2 + (2.0) 2 ⎞
(σ th )−60o = 4250 ⎜
⎜ (2.5) 2 − (2.0) 2 ⎟⎟ = 19,361 psi
⎝ ⎠
(b) The change in outer diameter of the titanium tube from room temperature ( 70 o F ) to −310 o F is
(
∆Do −ti = 4.0028 3.9 × 10−6 ) ( −310 − 70) = −0.0059"
The outer diameter of the titanium tube at −310 o F is therefore
The clearance between the −310 o F titanium diameter and the 70 o F steel inner diameter is
Since the required clearance in 0.005”, the steel diametral increase required is
Therefore
∆s 0.0019
( ∆T )req = = = 73.03o F
Di − steelα s 4.0000(6.5 × 10−6 )
T = 70 + 73 = 143o F
354
9-12. A component in a machine used to assure quality control consists of several disks mounted to a shaft. As parts
pass under the disks, the acceptable parts pass through, while the unacceptable parts do not. The disks themselves
are subject to wear and require frequent replacement. Replacement time has typically been a lengthy process which
affects productivity. In order to decrease replacement time you have been asked to investigate the feasibility of a
“quick change” shaft in which the disks are slid onto a shaft, which is then subjected to internal pressure, causing it
to expand and create a tight fit with the disk. The disk is required to support a friction torque of 100 N-m. The disks
are made of brass [ Su = 510 MPa , S yp = 414 MPa , ν = 0.35 , E = 105 GPa ] and the shaft is made of aluminum
[ Su = 186 MPa , S yp = 76 MPa , ν = 0.33 , E = 71 GPa ]. The hub of the brass disks have an inside diameter of 25
mm and an outside diameter of 50 mm, and a hub length of 25 mm. We initially assume a coefficient of friction
between brass and aluminum to be µ = 0.25 and an outside shaft diameter of 24.5 mm. Perform a “first pass”
assessment of the feasibility of this design idea.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
In order to perform the “quick change” the outside diameter of the shaft must be small enough to allow the disks to
able to freely slide. We begin by assuming the outside diameter of the shaft is 24.5 mm before it is pressurized.
Once pressurized, the expansion must be sufficient to create enough pressure that the friction torque requirement is
met. The problem becomes one of noting that the hub is modeled as a thick-walled cylinder subjected to internal
pressure and the shaft is a thick-walled cylinder subjected to both internal and external pressure. The contact
pressure required to create a friction torque of 100 N-m is
µ pπ d s2 lh
Tf =
2
0.25 pπ (0.0245) 2 (0.025)
100 = → p ≈ 17 MPa
2
∆
p = 17 × 106 =
⎡ a ⎛b +a 2 ⎞ d ⎛ d 2 + c2
2 ⎞⎤
2⎢ ⎜⎜ 2 + ν h⎟
⎟ + ⎜⎜ 2 2 −ν s ⎟⎟ ⎥
⎝b −a ⎠ Es ⎝ d − c
2
⎣⎢ Eh ⎠ ⎦⎥
∆
17 × 106 =
⎡ 0.0125 ⎛ (0.025) 2 + (0.0125) 2 ⎞ 0.01225 ⎛ (0.01225) 2 + c 2 ⎞⎤
2⎢ ⎜⎜ + 0.35 ⎟⎟ + 9 ⎜
⎜ − 0.33 ⎟⎟ ⎥
⎢⎣ 105 × 10 ⎝ (0.025) − (0.0125) ⎠ 71× 10 ⎝ (0.01225) − c
9 2 2 2 2
⎠ ⎥⎦
∆
17 × 106 =
⎡ ⎛ 0.00015 + c 2 ⎞⎤
2 ⎢ 0.240 × 10−12 + 0.173 × 10−12 ⎜⎜ − 0.33 ⎟⎟ ⎥
⎝ 0.00015 − c
2
⎣⎢ ⎠ ⎦⎥
⎛ 0.00015 + c 2 ⎞
∆ = 8.16 × 10−6 + 5.88 × 10−6 ⎜ − 0.33 ⎟ (1)
⎜ 0.00015 − c 2 ⎟
⎝ ⎠
⎛ b +a
2 2 ⎞ ⎛
(0.025) + (0.0125)
2 2 ⎞
σ th = p ⎜⎜ 2 2 ⎟⎟ = p ⎜⎜ 2 ⎟
= 1.667 p = 1.667(17) = 28.3 MPa , σ rh = − p = −17 MPa , σ lh = 0
− ⎟
⎝ (0.025) − (0.0125) ⎠
2
⎝ b a ⎠
355
Problem 9-12. (continued)
This results in
ε th =
1
105 × 109 ⎣
( ⎦ )
⎡ 28.3 × 106 − 0.35 −17 × 106 ⎤ = 326.2 µ m/m (2)
From (9-46) we know ∆ = 2 ( ε th a + ε ts d ) . Begin by assuming the ε th = ε ts = 326.2 µ m/m , which results in
∆ = 2 ( ε th a + ε ts d ) = 2(326.2 × 10−6 ) ( 0.025 + 0.0245 ) = 32.3 × 10−6 m . From (1) we now have
⎛ 0.00015 + c 2 ⎞
32.3 × 10−6 = 8.16 × 10−6 + 5.88 × 10−6 ⎜ − 0.33 ⎟
⎜ 0.00015 − c 2 ⎟
⎝ ⎠
⎛ 0.00015 + c 2 ⎞
4.105 = ⎜ − 0.33 ⎟ → c = 0.00973 m = 9.73 mm
⎜ 0.00015 − c 2 ⎟
⎝ ⎠
For the shaft, we use (10-25) and (10-26) to define the stress components σ rs and σ ts . We also set σ ls = 0
2 2
⎛ 0.01225c ⎞ ⎛ 0.01225c ⎞
pi c 2 − p(0.01225)2 + ⎜ ⎟ ( p − pi ) pi c 2 − p (0.01225) 2 − ⎜ ⎟ ( p − pi )
σ rs = ⎝ r ⎠ σ ts = ⎝ r ⎠
(0.01225) − c
2 2
(0.01225) − c
2 2
Since we are interested in the strains at the interface between the shaft and hub, we set r = 0.01225 , resulting in
This results in
1 1
ε ts = ⎡σ ts −ν s (σ rs + σ ls ) ⎤⎦ =
Es ⎣
[3.418 pi − 4.418 p − 0.33(−1.1 p)]
71× 106
3.418 pi − 4.055 p
ε ts = (3)
71× 106
3.418 pi − 4.055 p
326.2 × 10−6 = → 3.418 pi = 23160 + 4.055(17 × 106 )
71× 106
pi = 20.175 MPa
This pressure produces stresses of σ rs = σ 3 = −22.2 MPa , σ ts = σ 2 = −6.15 MPa , and σ ls = σ 1 ≈ 0 at the shaft/hub
interface. Using distortional energy
356
Problem 9-12. (continued)
(σ1 − σ 2 )2 + (σ 2 − σ 3 )2 + (σ 3 − σ1 )2 ≥ 2 ( S yp )
2
Since the failure condition is not met, we initially conclude that the proposed “quick change” shaft idea is feasible.
Additional refinement of the design is required.
357
9-13. A steel gear is to be shrink-fitted over a mounting diameter on a solid steel shaft, and its hub abutted against a
shoulder to provide axial location. The gear hub has a nominal inside diameter of 1 ½ inches and a nominal outside
diameter of 3 inches. The nominal shaft diameter is 1 ½ inches. To transmit the torque, it has been estimated that a
class FN5 force fit (see Table 6.7) will be required. Stresses in the hub must not exceed the yield strength of the hub
material, and a design safety factor of at least 2, based on yielding, is desired.
Two ductile candidate steel materials have been proposed for the gear: AISI 1095 steel quenched and
drawn to a hardness of Rockwell C 42, and AISI 4620 hot-rolled steel (with case-hardened teeth). Evaluate these
two materials for the proposed application, and decide which material to recommend (see Table 3.3)
------------------------------------------------------------------------------------------------------------------------------------------
Solution
Using Table 3.3, we find the material properties shown Material Su (ksi) S yp (ksi)
for the two candidate materials. From Table 6.7 we find
AISI 1095 @ RC 41 200 138
that for a class FN5 force fit 0.0014 ≤ ∆ ≤ 0.0040 .
AISI 4620 HR 87 63
E ∆ ⎡ a 2 ⎤ 30 × 10 ( 0.0040 ) ⎡ (0.75) 2 ⎤
6
p= ⎢1 − 2 ⎥ = ⎢1 − ⎥ = 30, 000 psi
4a ⎢⎣ b ⎥⎦ 4(0.75) ⎢⎣ (1.5) 2 ⎥⎦
Both σ th and σ rh are maximum at the inner hub, where b = 1.5 and a = 0.75
⎛ (1.5) 2 + (0.75) 2 ⎞
σ th = 30, 000 ⎜⎜ 2 ⎟
⎟
= 50, 000 psi and σ rh = −30, 000 psi
⎝ (1.5) − (0.75) ⎠
2
2
⎛ S yp ⎞
1⎡
( 50 − 0 ) + ( 0 − {−30} ) + ( −30 − 50 ) ⎤ ≥ ⎜ ( )
2 2 2 2
⎟ ⇒ 19, 600 ≥ S yp
2 ⎣⎢ ⎦⎥ ⎝ 2 ⎠
This gives S yp = 140 ksi . Neither candidate material meets this requirement, but AISI 1095 @ RC 4 is quite close.
358
9-14. A steel gear has a hub with a nominal bore diameter of 1.0 inch, outer hub diameter of 2.0 inches, and hub
length of 1.5 inches. It is being proposed to mount the gear on a steel shaft of 1.0-inch diameter using a class FN4
force fit (see Table 6.7).
a. Determine the maximum tangential and radial stresses in the hub and the shaft for the condition of loosest
fit.
b. Determine the maximum tangential and radial stresses in the hub and the shaft for the condition of tightest
fit.
c. Estimate the maximum torque that could be transmitted across the press fit connection before slippage
would occur between the gear and the shaft.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
From Table 6.7 we find that for a class FN4 force fit and a
1” nominal shaft diameter, 0.0010 ≤ ∆ ≤ 0.0023 .
⎛ (1.0) 2 + (0.5) 2 ⎞
σ th = 11, 250 ⎜⎜ 2 ⎟
⎟
= 18, 750 psi σ rh = σ rs = σ ts = − p = −11, 250 psi
⎝ (1.0) − (0.5) ⎠
2
⎛ (1.0) 2 + (0.5) 2 ⎞
σ th = 25,875 ⎜⎜ 2 ⎟
⎟
= 43,125 psi σ rh = σ rs = σ ts = − p = −25,875 psi
⎝ (1.0) − (0.5) ⎠
2
(c) For the maximum dependable torque that can be transmitted across the press fit by friction, without slip, the
tightest fit should be used. From appendix Table A-1, for lubricated mild steel on steel µ = 0.11
359
9-15. The 60-mm-long hub of a steel [ Su = 420 MPa , S yp = 350 MPa , ν = 0.30 , E = 207 GPa ] pulley has a
rectangular strain gage rosette applied. Strain gages A and C are perpendicular and gage B is at 45o to the other two
gages as illustrated in Figure P9.15. The outside diameter of the hub is 50 mm and the inside diameter is 25 mm.
Each strain gage is zeroed prior to the pulley being press fit to the shaft. The pulley is fit onto a solid steel shaft
made from the same material as the pulley with a diametral interference of ∆ = 0.04 mm . Determine the strains
indicated by each strain gage after the shaft and pulley are assembled.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
Since the shaft is solid and is the same material as the pulley, the contact pressure between the pulley and shaft is
approximated by
p=
E ∆ ⎡ ⎛ a ⎞ ⎤ 207 × 10 4 × 10
⎢1 −
2
⎥=
9 −5
( ) ⎢⎡1 − ⎛ 0.0125 ⎞ ⎥⎤ = 124 MPa
2
4a ⎢⎣ ⎜⎝ b ⎟⎠ ⎥⎦ 4(0.0125) ⎢⎣
⎜ 0.025 ⎟
⎝ ⎠ ⎥⎦
The stress components on the outside surface ( r = b = 0.025 m ) of the hub are
a2 p ⎡ ⎛ b ⎞ ⎤
2
σ rh = σ r = ⎢1 − ⎜ ⎟ ⎥=0 σ lh = σ l = 0
b 2 − a 2 ⎢⎣ ⎝ r ⎠ ⎥⎦
a2 p ⎡ ⎛ b ⎞ ⎤
2 ⎡ ( 0.0125)2 ⎤
σ th = σ t = ⎢1 + ⎜r⎟ ⎥ = 2(124) ⎢ ⎥ = 82.7 MPa
b − a ⎢⎣ ⎝ ⎠ ⎥⎦
2 2
⎢ ( 0.025 ) − ( 0.0125 ) ⎥
2 2
⎣ ⎦
Since the strain gages are surface mounted, they can only measure the tangential and longitudinal strain components.
These two strain are determined to be
εl =
1
207 × 109 ⎣ ( ⎦ )
⎡ 0 − 0.3 82.7 × 106 + 0 ⎤ = −118.8 µ m/m − 120 µ m/m
The strains measured by gages A and C are easy to determine based on the strain
gage orientations. These are
ε A = ε l = −120 µ m/m
ε C = ε t = 400 µ m/m
For strain gage B we can use strain transformation equations or Mohr’s circle of
strain to identify the fact that gage B will measure a normal strain that is 90o from
the planes defining ε A and ε C . In addition, we note that ε A and ε C are the
principal strains. From Mohr’s circle we determine
ε B = 140 µ m/m
360
Chapter 10
10-1. Plain bearings are often divided into four categories, according to the prevailing type of lubrication at the
bearing interface. List the four categories, and briefly describe each one.
------------------------------------------------------------------------------------------------------------------------------------
Solution
Hydrodynamic lubrication is characterized by a rotating shaft in an annular journal bearing so configured that a
viscous lubricant may be “pumped” into the wedge-shaped clearance space by the shaft rotation to maintain a stable
thick fluid film through which asperities of the rotating shaft cannot contact surface asperities of the journal.
Boundary lubrication may be characterized by a shaft and journal bearing configuration in which the surface area is
too small or too rough, or if the relative velocity is too low, or if temperatures increase too much (so the velocity is
lowered too much), or if loads become too high, asperity contacts may be induced through the (thin) oil film.
Hydrostatic lubrication may be characterized by a pair of sliding surfaces in which a thick lubricant film is
developed to separate the surfaces by an external source of pressurized lubricant.
Solid film lubrication may is characterized by bearing for which dry lubricants, such as graphite or molybdenum
disulfide, or self-lubricating polymers, such as Teflon or nylon are used.
361
10-2. From a strength-based shaft design calculation, the shaft diameter at on of the bearing sites on a steel shaft has
been found to be 38 mm. The radial load at this bearing site is 675 N, and the shaft rotates at 500 rpm. The operating
temperature of the bearing has been estimated to be about 90o C . It is desired to use a length-to-diameter ratio of 1.5
for this application. Based on environmental factors, the material for the bearing sleeve has been narrowed down to
a choice between nylon and filled Teflon. Which material would you recommend?
------------------------------------------------------------------------------------------------------------------------------------
Solution
From Table 11.1, (Vmax )nylon = 182.9 m/min and (Vmax )Teflon = 304.8 m/min - both meet velocity criteria.
W W 675
P= = = = 0.312 MPa
dL d (1.5d ) 1.5(0.038) 2
From Table 11.1, ( Pmax )nylon = 13.8 MPa and ( Pmax )Teflon = 17.2 MPa - both meet the velocity criteria.
From Table 11.1, ( PVmax )nylon = 6.3 MPa-m/min - does not work, ( PVmax )Teflon = 21.0 MPa-m/min - acceptable
362
10-3. It is being proposed to use a nylon bearing sleeve on a fixed steel shaft to support an oscillating conveyor tray
at equal intervals along the tray, as shown in Figure P10.3. Each bearing bore is top be 12.5 mm, bearing length is to
be 25 mm, and it is estimated that the maximum load to be supported by each bearing is about 2 kN. Each bearing
rotates ±10o per oscillation on its fixed steel journal, at a frequency of 60 oscillations per minute. Would the
proposed nylon bearing sleeve be acceptable for this application?
--------------------------------------------------------------------------------------------------------------------------------------
Solution
⎛ π ⎞
Vosc = ϕ fd = 20 ⎜ ⎟ (60)(0.0125) = 0.262 m/min
⎝ 180 ⎠
W 2000
P= = = 6.4 MPa
dL 0.0125(0.025)
363
10-4. A local neighborhood organization has become interested in replicating a waterwheel-driven grist mill of the
type that had been used in the community during the nineteenth century, but they have not been able to locate any
detailed construction plans. One of their concerns is with the bearings needed to support the rotating waterwheel.
To give an authentic appearance, they would like to use an oak bearing on each side of the waterwheel to support a
cast-iron waterwheel shaft. The waterwheel weight, including the residual weight of the retained water, is estimated
to be about 12,000 lb, and the wheel is to rotate at about 30 rpm. It has been estimated on the basis of strength that
the cast-iron shaft should be no less than 3 inches in diameter. The bearings need to be spaced about 36 inches
apart. Propose a suitable dimensional configuration for each of the two proposed oak bearings so that bearing
replacement will rarely be needed. It is anticipated that 68˚F river water will be used for lubrication.
--------------------------------------------------------------------------------------------------------------------------------------
Solution
π dN π ( 3)( 30 )
Vcont = =
12 12
= 24 fpm
psi-ft
( PV )max = 12, 000 , thus
min
PV 12, 000
P= = = 5, 000 psi
V 24
Checking Table 10.1 we see for wood that Pmax = 2000 psi and thus meets the unit load criterion. Since
R 6, 000
L= = = 4.0 inches
Pd 500 ( 3)
then
L 4
= = 1.333
d 3
1 L
≤ ≤2
2 d
The temperature should not be a problem since 68˚F river water is to be used as the lubricant. Therefore, it should
be satisfactory to use two oak bearings, each nominally 3 inches in bore diameter by 4 inches long.
364
10-5. The shaft shown in Figure P10.5 is part of a transmission for a small robot. The shaft supports two spur gears
loaded as indicated, is supported by bearings at A and D, and rotates at 1200 rpm. A strength based analysis has been
performed and it has been determined that a shaft diameter of 10 mm would be adequate for the steel shaft. You are
considering the use of boundary-lubricated bearings for which L = 1.5d . A porous bronze bearing sleeve has been
proposed. Determine if this bearing selection is adequate.
Figure P10.5
Steel shaft supporting two spur
gears
--------------------------------------------------------------------------------------------------------------------------------------
Solution
∑ Fy = 0 : Ay + Dy = 200
( ∑ M A )z = 0 : 50 Dy + 30(300) − 20(500) = 0
D y = 20 N , Ay = 180 N
∑ Fz = 0 : Az + Dz = −550
( ∑ M A ) y = 0 : 50 Dz + 30(750) − 20(200) = 0
Dz = −370 N , Az = −180 N
( Dy ) + ( Dz )2 = ( 20 )2 + ( −370 )2 ≈ 370 N
2
RD =
365
P = 2.5 < Pmax = 13.8
V = 37.7 < Vmax = 365.8
PV = 94.25 < ( PV )max = 105.1
Therefore the porous bronze bearing sleeve is adequate for this application
366
10-6. From a strength-based analysis, a shaft diameter at one of its support bearing sites must be at least 1.50
inches. The maximum radial load to be supported at this location is estimated to be about 150 lb. The shaft rotates
at 500 rpm. It is desired to use a Nylon bearing sleeve at this location. Following established design guidelines for
boundary-lubricated bearings, and keeping the bearing diameter as near to the 1.50-inch minimum as possible,
propose a suitable dimensional configuration for the bearing.
--------------------------------------------------------------------------------------------------------------------------------------
Solution
π dN 1.5π ( 500 )
Vcont = = = 196 fpm
12 12
We see from Table 10.1 that Nylon (Vmax = 600 fpm) meets the velocity criterion. Try a “square bearing”
configuration with L = d = 1.5 inches. Thus,
W 150
P= = = 67 psi
dL 1.5 (1.5 )
Again checking Table 10.1 we note that Nylon (Pmax = 2000 psi) meets the unit load criterion. For PV we have
psi-ft
PV = ( 67 )(196 ) = 13,130
min
We note that from Table 10.1 that Nylon does not meet the unit load criterion. Try the maximum recommended
bearing length L = 2d = 3.0 inches. Thus, we find
W 150
P= = = 33 psi
dL 1.5 ( 3.0 )
and
psi-ft
PV = ( 33)(196 ) = 6480
min
We again see that this still does not meet the (PV)max = 3000 requirement for Nylon. Therefore, it will be necessary
to increase the bearing diameter. Start with the upper limiting value L = 2d, thus
W W
P= = 2
dL 2d
and
⎛ W ⎞⎛ π dN ⎞ π WN
PV = ⎜ 2 ⎟⎜ ⎟=
⎝ 2d ⎠⎝ 12 ⎠ 24d
π WN 150π ( 500 )
d req ' d = = = 3.27 inches
24 ( PV )max 24 ( 3000 )
If a Nylon bearing is used, it dimensions must be at least 3.3 in. bore by 6.6 in. in length.
367
10-7. A preliminary result obtained as a possible solution for problem 10-6 indicates that the smallest acceptable
bearing diameter for the specifications given is about 3.3 inches. Engineering management would prefer to have a
bearing diameter of about 1.50 inches (the minimum based on shaft strength requirements), and they are asking
whether it would be possible to find another polymeric bearing material that might be satisfactory for this
application. Using Table 10.1 as your resource, can you find a polymeric bearing material other than Nylon that will
meet established design guidelines and function properly with a diameter of 1.50 inches?
--------------------------------------------------------------------------------------------------------------------------------------
Solution
π dN 1.5π ( 500 )
Vcont = = = 196 fpm
12 12
W 150
P= = = 33 psi
dL 1.5 ( 3.0 )
psi-ft
PV = ( 33)(196 ) = 6480
min
From Table 10.1 we see that the potential candidates are based on allowable (PV)max:
Checking allowable unit loads Pmax, all three candidates qualify. Checking the allowable sliding velocity Vmax ,
Phenolics and filled Teflon qualify but Teflon Fabric does not meet the velocity criterion. Thus, either Phenolics or
filled Teflon would be acceptable. Cost would probably govern the choice (Phenolics would probably win).
368
10-8. A plain bearing is to be designed for a boundary-lubricated application in which a 75-mm-diameter steel
shaft rotating at 1750 rpm must support a radial load of 1 kN. Using established design guidelines for boundary-
lubricated bearings and Table 10.1 as your resource, select an acceptable bearing material for this application.
--------------------------------------------------------------------------------------------------------------------------------------
Solution
Using L = d
W 1000
P= = = 0.178 MPa
dL 0.075(0.075)
369
10-9. A plain bearing is to be designed for boundary-lubrication applications in which a 0.5-inch-diameter steel
shaft rotating at 1800 rpm must support a radial load of 75 lb. Using established design guidelines for boundary-
lubricated bearings, and using Table 10.1 as your resource, select an acceptable bearing material for this application
if the operating temperature is estimated to be about 350˚F.
--------------------------------------------------------------------------------------------------------------------------------------
Solution
π dN π ( 0.5 )(1800 )
Vcont = = = 236 fpm
12 12
W 75
P= = = 300 psi
dL ( 0.5 )( 0.5 )
We have that
psi-ft
PV = ( 300 )( 236 ) = 70,800
min
Checking Table 10.1, we see that no material meets the (PV)max criterion. Therefore, make a new assumption on the
L/d ratio using the upper limit, L = 2d. Thus,
W 75
P= = = 150 psi
dL ( 0.5 )(1.0 )
psi-ft
PV = (150 )( 236 ) = 35, 400
min
Checking Table 10.1, materials now meeting all three criteria include:
1. Porous bronze
2. Porous lead-bronze
3. Porous lead-iron
4. Porous aluminum
However, we see from Table 10.1 that Porous aluminum does not meet the specified operating temperature of 300˚F.
Thus, a selection would be made among the first three candidates based on cost.
370
10-10. A proposed flat belt drive system (see Chapter 17) is being considered for an application in which the driven
steel shaft is to rotate at a speed of 1440 rpm, and the power to be transmitted is 800 W. As shown in Figure P10.10,
the power is transmitted to the 10-mm-diameter (driven) shaft by a flat belt running on a shaft-mounted pulley. The
pulley has a nominal pitch diameter of 60 mm, as sketched in Figure P10.10. It is desired to support the driven shaft
using two grease-lubricated plain bearings, one adjacent to each side of the pulley (see Figure P10.10). The two
bearings share the belt load equally. It has been determined that the initial belt tension, T0, should be 150 N ( in
each side of the belt) to achieve optimum performance, and it may reasonably be assumed that the sum of tight side
and slack side belt tension will remain approximately equal to 2T0 for all operating conditions. Select satisfactory
plain bearings for this application, including their diameter, their length, and an acceptable material from which to
make them (see Table 10.1).
--------------------------------------------------------------------------------------------------------------------------------------
Solution
m
Vcont = π dN = π ( 0.010 )(1440 ) = 45
min
W 150
P= = = 1.5 MPa
dL 10 (10 )
MPa-m
PV = (1.5 )( 45 ) = 67.5
min
1. Porous bronze
2. Porous lead-bronze
3. Porous bronze-iron
4. Porous lead-iron
5. Aluminum
The final selection would be based on cost (probably porous bronze). The preliminary recommendation will be:
Use porous bronze bearings, both sides, with bore diameter of 10 mm and length of 10 mm.
371
10.11. It is desired to use a hydrodynamically lubricated 360o babbit-metal plain bearing for use in supporting the
crankshaft (see Chapter 19) of an automotive-type internal combustion engine for an agricultural application. Based
on strength and stiffness calculations, the minimum nominal journal diameter is 50 mm, and a length-to-diameter
ratio of 1.0 has been chosen. The maximum radial load on the bearing is estimated to be 3150 N and the journal
rotates in the bearing sleeve at 1200 rpm. High load-carrying ability is regarded as much more important than low
friction. Tentatively, an SAE 30 oil has been chosen, and the average bearing operating temperature has been
determined to be about 65o C . Estimate the power loss due to bearing friction.
--------------------------------------------------------------------------------------------------------------------------------------
Solution
From Table 11.2 for a 50 mm diameter bearing, c = 38.1× 10−3 mm and c / r = 0.0381/ 25 = 0.00152 .
From Figure 11.14 with L / d = 1.0 , for maximum loading, ε max −load ≈ 0.47 .
F1 ⎛ c ⎞
From Figure 11.9 with L / d = 1.0 and ε = 0.47 , f F1 = ≈ 7.4 . Using this
ηU ⎜⎝ r ⎟⎠
7.4ηU
F1 =
c/r
From Figure 10.3, with 65o C ≈ 150o F and SAE grade 30, we get η ≈ 3.4 × 10−6 rehns . Converting
7.4(0.02344)(3.142)
F1 = = 385.6 Pa-m = 385.6 N/m
0.00152
Ft = F1 L = 385.6(0.05) = 19.28 N
Tn 0.488(1200)
kw = = = 0.258 kw
9549 9549
372
10.12. Text In an automobile crankshaft application, a hydrohydynamic full 360˚ journal bearing must be 2 inches
in nominal diameter based on strength requirements, and the bearing length being considered is 1.0 inch. The
journal is to be made of steel and the bearing sleeve is to be made of a copper-lead alloy (see Table 10.2). The
bearing must support a radial load of 1000 lb, and the journal rotates at 3000 rpm. The lubricant is to be SAE 20 oil,
and the average operating temperature at the bearing interface has been estimated to be about 130˚F. Load-carrying
ability and low friction loss are regarded as about equally important.
--------------------------------------------------------------------------------------------------------------------------------------
Solution
(a)
d 2.0000
r= = = 1.0000 inch
2 2
rev
n = 3000 rpm = 50
sec
From Table 10.2, for an automotive crankshaft application using a copper-lead alloy bearing sleeve and a steel
journal, for a 2-inch diameter bearing
c = 0.0014 inch
c 0.0014
= = 0.0014
r 1.0000
From Figure 10.4, for L/d = 0.5, read ε corresponding to minimum friction drag and maximum load carrying ability
as
ε max −load = 0.57
ε max − friction = 0.89
Since these values are regarded as being of equal importance select a midrange value of ε = 0.7. Then
If the sleeve were reamed, then from Figure 6.11 Rb = 63 µ-inch and the journal roughness should be
R j = 84 − 63 = 21 µ-inch or less
373
(b) Recommendations for acceptable manufacturing are based on the values for Rj and Rb
F1 ⎛ c ⎞
f F1 = = 9.4
ηU ⎜⎝ r ⎟⎠
in
U = 2π rn = 2π (1.0000 )( 50 ) = 314
sec
9.4 ( 3.8 × 10−6 ) ( 314 ) lb
F1 = = 8.0
0.0014 in
Q
fQ = = 5.1
rcnL
in 3
Q = 5.1( rcnL ) = 5.1(1.0 )( 0.0014 )( 50 )(1.0 ) = 0.36
sec
374
10-13. A hydrodynamic journal bearing rotates at 3600 rpm. The bearing sleeve has a 32 mm-diameter and is 32
mm long. The bearing radial clearance is to be 20 µm, and the radial load on the bearing is said to be 3 kN. The
lubricant chosen is SAE oil supplied at an average temperature of 60˚C. Estimate the friction-generated heating rate
for this bearing if the eccentricity ratio has been determined to be 0.65.
--------------------------------------------------------------------------------------------------------------------------------------
Solution
J N-m
H g = F1 LU , or Watts
s s
We have
π dn π ( 0.032 )( 3600 ) m
U= = = 6.0
60 60 s
F1 ⎛ c ⎞
f F1 = = 9.2
ηU ⎜⎝ r ⎟⎠
9.2ηU
F1 =
⎛c⎞
⎜ ⎟
⎝r⎠
375
10-14. It is desired to design a hydrodynamically lubricated 360˚ plain bearing for a special factory application in
which a rotating steel shaft must be at least 3.0 inches nominal diameter and the bushing (sleeve) is to be bronze,
reamed to size. The radial bearing load is to be 1000 lb. The desired ratio of length to diameter is 1.5. The shaft is
to rotate at a speed of 1000 rpm. It has been estimated that an eccentricity ratio of 0.5 should be a good starting
point for designing the bearing, based on an evaluation of the optimal design region of Figure 10.14 for a length-to-
diameter ration of 1.5.
--------------------------------------------------------------------------------------------------------------------------------------
Solution
From Table 10.2, for “general machine practice-continuous rotation motion”, and d ≥ 3.0 in., then
c = 0.004 to 0.007
c
= 0.003 to 0.005
r
Initially select c/r = 0.003. Using Table 6.11, and initially deciding to grind the steel journal, (sleeve is reamed),
Rb = 63 µ-in.
R j = 16 µ-in.
Writing all pertinent expressions as functions of d gives:
⎛L⎞
L = ⎜ ⎟ d = 1.5d
⎝d ⎠
W 1000 667 lb
W1 = = =
L 1.5d d in
π dn π d (1000 ) in
U= = = 16.67π d
60 60 sec
Ah = Ch Ap = 8 (π dL ) = 12π d 2 in 2
c⎛d ⎞ ⎛d ⎞
c = ⎜ ⎟ = 0.003 ⎜ ⎟ = 0.0015d
r⎝2⎠ ⎝2⎠
Use Figures 10.7 and 10.9 to evaluate fW1 and f F1 . Note however that there is no curve presented for the specified
value of L/d = 1.5. Thus, it will be necessary to utilize the interpolation equation to find value of fW1 and f F1 for L/d
= 1.5. Hence,
1 ⎡ 1 1
f = ⎢ − 8 (1 − 1.5 )(1 − 2 × 1.5 )(1 − 4 × 1.5 ) f ∞ + 3 (1 − 2 × 1.5 )(1 − 4 × 1.5 ) f1.0
(1.5 )
3
⎣
From Figure 10.7 with ε = 0.5, values of fW1 at L/d ratios of ∞, 1.0, 0.5, and 0.25 are
f ∞ = 7.0
f1.0 = 1.8
f 0.5 = 0.69
f 0.25 = 0.18
Thus,
376
(f )
W1 L
=1.5
= 0.185 ( 7.0 ) + 0.987 (1.8 ) − 0.185 ( 0.69 ) − 0.012 ( 0.18 ) = 2.94
d
From Figure 10.9 with ε = 0.5, values of f F1 at L/d ratios of ∞, 1.0, 0.5, and 0.25 are
f ∞ = 8.5
f1.0 = 7.8
f 0.5 = 7.5
f 0.25 = 7.3
and
(f )
F1 L
=1.5
= 0.185 ( 8.5 ) + 0.987 ( 7.8 ) − 0.185 ( 7.5 ) − 0.012 ( 7.3) = 7.79
d
Thus,
2
W1 ⎛ c ⎞
= 2.94
ηU ⎜⎝ r ⎟⎠
( 667 d )
( 0.003) = 2.94
2
η (16.67π d )
667 ( 0.003)
2
3.89 × 10−5
d= =
2.94 (16.67π )η η
Combining equations and assuming that the ambient air is Θ a = 75o F we have
⎛ Θ − 75 ⎞
2.31× 10−4 (12π d 2 ) ⎜ o ⎟ 9336
k A ( Θ − Θa ) J Θ ⎝ 2 ⎠
F1 = 1 h s = = 8.62 × 10−3 ( Θ o − 75 )
60UL 60 (16.67π d )(1.5d )
F1 ⎛c⎞
⎜ ⎟ = 7.79
ηU ⎝r⎠
7.79η (16.67π d ) ⎛ lb ⎞
F1 = = 1.36 × 105 η d ⎜ ⎟
0.003 ⎝ in ⎠
Equating yields
377
Solve these by trial and error. As a first try select SAE 10 oil.
Oil Spec. Θo , o F η, rehns from eq. η, rehns from Fig 10.3 Comment
SAE 10 175˚ 1.04x10 -6
1.0x10 -6
Close
SAE 10 176˚ 1.06x10-6 1.05x10-6 Adequate
Thus,
6.39 × 10−8
d= (176 − 75) = 6.14 inch
1.05 × 10−6
d = 6.14 inch
L = 9.2 inch
Oil; SAE 10
Θo = 176˚F
Checking Table 10.2 for this larger shaft we see that it may be desired to increase the clearance. However, we shall
keep c = 0.003 in. for now. Checking the minimum film thickness gives
The existing film thickness is about four times the required film thickness, therefore the recommendations should
hold for a ground journal.
378
10-15. For the design result you found in solving problem 10-14,
a. Find the friction drag torque.
b. Find the power dissipated as a result of friction drag.
--------------------------------------------------------------------------------------------------------------------------------------
Solution
d = 6.14 inch
L = 9.2 inch
Oil; SAE 10
Θo = 176˚F
F1 = 8.62 × 10−3 ( Θo − 75 )
⎛ 6.14 ⎞
T f = Ft r = ( F1 L ) r = 0.87 ( 9.2 ) ⎜ ⎟ = 24.6 in-lb
⎝ 2 ⎠
Tf n 24.6 (1000 )
( hp ) f = = = 0.39 horsepower
63, 025 63, 025
379
10-16. A hydrodynamically lubricated 360˚ plain bearing is to be designed for a machine tool application in which a
rotating steel spindle must be at least 1.00 inch nominal diameter, the bushing is to be bronze, and the steel spindle
is to be lapped into the bronze bushing. The radial bearing load is 40 lb, and the spindle is to rotate at 2500 rpm.
The desired ratio of length to Diameter is 1.0. Conduct a preliminary design study to determine a combination of
dimensions and lubricant parameters for this application.
--------------------------------------------------------------------------------------------------------------------------------------
Solution
The steel spindle is to be lapped into the bronze bushing and the bearing has a 360˚ configuration. The sliding
surface velocity is
π dN π (1.00 )( 2500 ) ft
Vcont = = = 654
12 12 min
W 40
p= = = 40 psi
dL 1.00 (1.0 )
From Table 10.2 we see that for precision spindle practice, with hardened and ground spindle lapped into a bronze
bushing and for diameters under 1 inch; with velocity above 500 ft/min and pressure under 500 psi that the data are
split between the first two lines of the table. As a start let’s pick c = 0.0015 inch. In addition, from Figure 6.11,
lapping produces a finish of Rj = Rb = 8 µ-in. Also, c/r = 0.0015/0.50 = 0.003.
From Figure 10.14, for L/d = 1.0, we read the values of ε corresponding to maximum load carrying ability and
maximum friction drag, respectively as
Since no specification is given for ε, a midrange value will be assumed, i.e., ε = 0.6. From Figure 10.9, for L/d = 1.0
and ε = 0.6
F1 ⎛ c ⎞
f F1 = = 8.5
ηU ⎜⎝ r ⎟⎠
8.5ηU
F1 =
⎛c⎞
⎜ ⎟
⎝r⎠
⎛L⎞
L = ⎜ ⎟ d = d inch
⎝d ⎠
W 40 lb
W1 = =
L d in
π dn π d ( 2500 ) in
U= = = 41.67π d
60 60 sec
Ah = Ch Ap = 8 (π dL ) = 8π d 2 in 2
c⎛d ⎞ ⎛d ⎞
c = ⎜ ⎟ = 0.003 ⎜ ⎟ = 0.0015d
r⎝2⎠ ⎝2⎠
380
⎛ Θ − 70 ⎞
2.31× 10−4 ( 8π d 2 ) ⎜ o ⎟ 9336
k1 Ah ( Θ s − Θ a ) J Θ ⎝ 2 ⎠ lb
F1 = = = 3.45 × 10−3 ( Θo − 70 )
60UL 60 ( 41.67π d )( d ) in
Also, we have
8.5η ( 41.67π d ) lb
F1 = = 3.71× 105η d
0.003 in
2
W1 ⎛ c ⎞
fW1 = = 2.7
ηU ⎜⎝ r ⎟⎠
( 40 d )( 0.003)
2
= 2.7
η ( 41.67π d )
40 ( 0.003)
2
1.02 × 10−6
d= = in.
2.7 ( 41.67π )η η
1.02 × 10−6 9.30 × 10−9
= ( Θo − 70 )
η η
η = 8.48 × 10 −11 ( Θo − 70 ) rehns
2
From Figure 10.3 η = f graph ( Θo , oil ) in rehns . Solve by trial and error. As a first try select SAE 10 oil.
Oil Spec. Θo , o F η, rehns from eq. η, rehns from Fig 10.3 Comment
SAE 10 175˚ 9.35x10 -7
1.1x10-6
Close
SAE 10 180˚ 1.03x10-6 1.0x10-6 Adequate
Thus,
9.30 × 10−9
d= (180 − 70 ) = 1.02 inch
1.0 × 10−6
d = 1.00 inch
L = 1.00 inch
Oil; SAE 10
Θo = 180˚F
381
( ho )existing = c (1 − ε ) = 0.0015 (1 − 0.6 ) = 0.0006 inch
The existing film thickness is about seven times the required film thickness, therefore the recommendations should
hold for a lapped bearing pair.
382
10-17. For your proposed design result found in solving problem 10-16
--------------------------------------------------------------------------------------------------------------------------------------
Solution
The following results for problem 10-16 which are pertinent are;
d = 1.00 inch
L = 1.00 inch
Oil; SAE 10
Θo = 180˚F
lb
F1 = 3.45 × 10−3 ( Θo − 70 ) = 0.38
in
⎛ 1.0 ⎞
T f = Ft r = ( F1 L ) r = 0.38 (1.0 ) ⎜ ⎟ = 0.19 in-lb
⎝ 2 ⎠
Tf n 0.19 ( 2500 )
( hp ) f = = = 0.008 horsepower
63, 025 63, 025
383
10-18. A hydrodynamically lubricated 360˚ plain bearing is to be designed for a conveyor-roller support
application in which the rotating cold-rolled steel shaft must be at least 100 mm nominal diameter and the bushing is
to be made of poured Babbitt, reamed to size. The radial bearing load is to be 18.7 kN. The desired ratio of length
to diameter is 1.0. The shaft is to rotate continuously at a speed of 1000 rpm. Low friction drag is regarded as more
important than high load-carrying capacity. Find a combination of dimensions and lubricant parameters suitable for
this conveyor application.
--------------------------------------------------------------------------------------------------------------------------------------
Solution
The spindle is cold rolled steel and the bushing is poured Babbitt, reamed to size. From table 10.2, for general
machine practice, continuous rotation, cold rolled steel journal in poured Babbitt bushing reamed to size, and for a
100 mm diameter select a clearance as 0.005 inches or 0.127 mm. We have that c/r = 0.127/50 = 0.0025. From
Figure 10.14, for L/d = 1.0, read values of ε corresponding to minimum friction drag and maximum load carrying
capacity as
We note that for this application that low friction drag is regarded as more important than high load carrying
capacity. Thus, select ε = 0.65.
( ho )required ≥ 5.0 ( R j + Rb )
Using Figure 6.11 for a cold rolled journal Rj = 1.6 µm and for a reamed bushing Rb = 1.6 µm. Thus,
It is noted that (ho)existing exceeds (ho)required by a factor of about 3 which is an acceptable margin. From Figure 10.7
and 10.9, for L/d = 1.0 and ε = 0.65 we have
F1 ⎛ c ⎞
f F1 = = 9.3
ηU ⎜⎝ r ⎟⎠
2
W1 ⎛ c ⎞
fW1 = = 3.5
ηU ⎜⎝ r ⎟⎠
W 18.7 × 103 N
W1 = = = 1.87 × 105
L 0.100 m
π dn π ( 0.100 )(1000 ) m
U= = = 5.24
60 60 s
W ⎛ c ⎞ 1.87 × 10 ( 0.0025 )
5 2 2
η= 1 ⎜ ⎟ = = 0.64 Pa ⋅ s
3.5U ⎝ r ⎠ 3.5 ( 5.24 )
0.064
η= = 9.28 × 10−6 rehns
6895
From Figure 10.3, one of several oil selections that would be satisfactory is SAE 50 oil operating at about 152˚F or
67˚C.
384
In summary we have:
d = 100 mm
L = 100 mm
Oil: SAE 50
Θo = 67˚C
385
10-19. For your proposed design result found in solving problem 10-18, find the friction drag torque.
--------------------------------------------------------------------------------------------------------------------------------------
Solution
d = 100 mm
L = 100 mm
Oil: SAE 50
Θo = 67˚C
η = 0.064 Pa-s = 9.28x10-6 rehns
U = 5.24 m/s
c/r = 0.0025
F ⎛c⎞
f F1 = 1 ⎜ ⎟ = 9.3
ηU ⎝ r ⎠
9.3ηU
T f = Ft r = ( F1 L ) r = ( L )( r )
⎛c⎞
⎜ ⎟
⎝r⎠
9.3 ( 0.064 )( 5.24 )( 0.100 )( 0.050 )
= = 6.24 N-m
0.0025
386
Chapter 11
11-1. For each of the following applications, select two possible types of rolling element bearings that might make
a good choice.
(a) High-speed flywheel (see Chapter 18) mounted on a shaft rotating about a horizontal centerline.
(b) High-speed flywheel mounted on a shaft rotating about a vertical centerline.
(c) Low-speed flywheel mounted on a shaft rotating about a vertical centerline.
------------------------------------------------------------------------------------------------------------------------------------
Solution
Utilizing Table 11.1, and deducing primary design requirements from problem statements, the following potential
bearing types may be selected:
(a) Design requirements: moderate to high radial capacity, moderate to low thrust capacity, high limiting
speed, moderate to high radial stiffness, moderate to low axial stiffness. Bearing candidates:
(1) Maximum capacity ball bearing
(2) Spherical roller bearing
(b) Design requirements: moderate radial capacity, moderate –one-direction thrust capacity, high limiting
speed, and moderate radial stiffness, moderate to high axial stiffness. Bearing candidates:
(1) Angular contact ball bearings
(2) Single-row tapered roller bearings
(c) Design requirements: low to no radial capacity, moderate to high thrust capacity- one direction, limiting
speed low, radial stiffness low to none, moderate to high axial stiffness. Bearing candidates:
(1) Roller thrust bearing
(2) Tapered roller thrust bearing
387
11-2. A single-row radial ball bearing has a basic dynamic load rating of 11.4 kN for an L10 life of 1 million
revolutions. Calculate its L10 life if it operates with an applied radial load of 8.2 kN.
------------------------------------------------------------------------------------------------------------------------------------
Solution
3
L ⎛ Cd ⎞
=
106 ⎜⎝ P ⎟⎠
3
⎛ 11.4 ⎞
( L10 )8.2 kN =⎜ ⎟ × 10 = 2.69 × 10 rev
6 6
⎝ 8.2 ⎠
388
11-3. a. Determine the required basic dynamic load rating for a bearing mounted on a shaft rotating at 1725 rpm if
it must carry a radial load of 1250 lb and the desired design life is 10,000 hours.
b. Select a single-row radial ball bearing from table 11.5 that will be satisfactory for this application if the outside
diameter of the bearing must not exceed 4.50 inches.
------------------------------------------------------------------------------------------------------------------------------------
Solution
1
⎛ L ⎞3
[Cd ]req = ⎜ d6 ⎟ Pd
⎝ 10 ⎠
Ld = (104 hr ) (1725 rev min )( 60 min hr ) = 1.04 × 109 rev
1
⎛ 1.04 × 109 ⎞ 3
[Cd ]req =⎜ 6 ⎟ (1250 ) = 12, 700 lb
⎝ 10 ⎠
389
11-4. A single-row radial ball bearing must carry a radial load of 2250 N and no thrust load. If the shaft that the
bearing is mounted to rotates at 1175 rpm, and the desired L10 life of the bearing is 20,000 hr, select the smallest
bearing from Table 11.5 that will satisfy the design requirements.
------------------------------------------------------------------------------------------------------------------------------------
Solution
From (11-1)
1/ 3
[Cd ]req = ⎡⎢
Ld ⎤
6⎥
Pd
⎣ 10 ⎦
and
So
1/ 3
⎡ 1.41× 109 ⎤
[Cd ]req = ⎢ 6 ⎥ (2250) = 25.23 kN
⎣⎢ 10 ⎦⎥
From Table 11.5, the smallest bearing with Cd = 25.23 kN is bearing # 6306, while bearing #6207 is the next
smallest.
390
11-5. In a preliminary design calculation, a proposed deep-groove ball bearing had been tentatively selected to
support one end of a rotating shaft. A mistake has been discovered in the load calculation, and the correct load turns
out to be about 25 percent higher than the earlier incorrect load used to select the ball bearing. To change to a larger
bearing at this point means that a substantial redesign of all the surrounding components will probably be necessary.
If no change is made to the original bearing selection, estimate how much reduction in bearing life would be
expected.
------------------------------------------------------------------------------------------------------------------------------------
Solution
3
L ⎛ Cd ⎞
From (11-1) = . Setting correct load Pc equal to incorrect load Pi, then
106 ⎜⎝ P ⎟⎠
3 3
Lc ⎛ Cd ⎞ Li ⎛ Cd ⎞
=⎜ ⎟ =⎜ ⎟
106 ⎝ Pc ⎠ 106 ⎝ Pi ⎠
Lc Pc3 = Cd3 × 106 Li Pi 3 = Cd3 × 106
Lc Pc3 = Li Pi 3
Lc Pi 3 Pi 3 1
= 3 = = = 0.51
(1.25Pi ) (1.25 )
3 3
Li Pc
391
11-6. A number 6005 single-row radial deep-groove ball bearing is to rotate at a speed of1750 rpm. Calculate the
expected bearing life in hours for radial loads of 400, 450, 500, 550, 600, 650, and 700 lb, and make a plot of life
versus load. Comment on the results.
------------------------------------------------------------------------------------------------------------------------------------
Solution
From Table 11.5, the basic dynamic load rating for a 6005 single-row radial deep groove ball bearing is Cd = 2520
lb. From (11-1)
P, lb P3 (L)hr
400 6.4 x 107 2375
450 9.1 x 107 1670
500 1.3 x 108 1170
550 1.7 x 108 890
600 2.2 x 108 690
650 2.7 x 108 545
700 3.4 x 108 450
Note how rapidly the expected life decreases even for relatively small increases in load.
392
11-7. Repeat problem 11-6, except use a number 205 single-row cylindrical roller bearing instead of the 6005
radial ball bearing.
------------------------------------------------------------------------------------------------------------------------------------
Solution
From Table 11.6, the basic dynamic load rating for a 205 single-row cylindrical roller bearing is Cd = 6430 lb.
P, lb P3 (L)hr
400 6.4 x 107 39,531
450 9.1 x 107 27,800
500 1.3 x 108 19,460
550 1.7 x 108 14,900
600 2.2 x 108 11,500
650 2.7 x 108 9,035
700 3.4 x 108 7,440
Note how rapidly the expected life decreases even for relatively small increases in load.
393
11-8. A number 207 single-row cylindrical roller bearing has tentatively been selected for an application in which
the design life corresponds to 90 percent reliability (L10 life) is 7500 hr. Estimate what the corresponding lives
would be for reliabilities of 50 percent, 95 percent, and 99 percent.
------------------------------------------------------------------------------------------------------------------------------------
Solution
Lp = K R L10
K 50 = 5.0
K 95 = 0.62
K 99 = 0.21
394
11-9. Repeat problem 11-8, except use a number 6007 single-row radial ball bearing instead of the 207 roller
bearing.
------------------------------------------------------------------------------------------------------------------------------------
Solution
L p = K R L10
K 50 = 5.0
K 95 = 0.62
K 99 = 0.21
395
11-10. A solid steel spindle shaft of circular cross section is to be used to support a ball bearing idler pulley as
shown in Figure P11.10. The shaft may be regarded as simply supported at the ends and the shaft does not rotate.
The pulley is to be mounted at the center of the shaft on a single-row radial ball bearing. The pulley must rotate at
1725 rpm and support a load of 800 lb, as shown in the sketch. A design life of 1800 hours is required and a
reliability of 90 percent is desired. The pulley is subjected to moderate shock loading conditions.
(a) Pick the smallest acceptable bearing from Table 11.5 if the shaft at the bearing site must be at least 1.63
inches in diameter.
(b) Again using Table 11.5, select the smallest bearing that would give an infinite operating life, if you can find
one. If you find one, compare its size with the 1800-hour bearing.
------------------------------------------------------------------------------------------------------------------------------------
Solution
⎛ rev ⎞⎛ min ⎞
Ld = ⎜ 1725 ⎟ (1800 hr ) = 1.86 × 10 rev
8
⎟⎜ 60
⎝ min ⎠⎝ hr ⎠
And moderate shock loading exists. A single row radial ball bearing is to be selected. From (11-3)
Pe = X d Fr + Yd Fa
X d1 = 1, Yd1 = 0
X d2 = 0.55, Yd2 = 1.45
Hence,
( Pe )1 = 1(800 ) + 0 ( 0 ) = 800 lb
( Pe )2 = 0.55 (800 ) + 1.45 ( 0 ) = 440 lb
Since (Pe)1 > (Pe)2, Pe = (Pe)1 = 800 lb. Calculating the basic dynamic radial load rating requirement from (11-4),
1
⎡ L ⎤3
⎣⎡Cd ( 90 ) ⎦⎤ req ' d =⎢ ⎥ ( IF ) Pe
d
⎢⎣ K R (106 ) ⎥⎦
From Table 11.3, IF = 1.75 and from Table 11.2 KR = 1.0 for R = 90. Thus, we find
1
⎡1.86 × 108 ⎤ 3
⎡⎣Cd ( 90 ) ⎤⎦ req ' d =⎢ ⎥ (1.75 )( 800 ) = 7990 lb
⎢⎣ 1.0 (106 ) ⎥⎦
The smallest acceptable bearing, with a bore of at least 1.63, from Table 11.5 is bearing No. 6309 (limiting speed
ok). Checking the static load rating Pse, using (11-5) and Table 11.4
Pse = 800 lb
From Table 11.5, for bearing No. 6309, Cs = 7080 lb. Since Pe < Cs the static load rating is also acceptable.
Therefore select bearing No. 6309 and locally increase the shaft diameter at the bearing site to d = 1.7717 inches,
with appropriate tolerances.
396
(b) From Table 11.5, looking for
397
11-11. A helical idler gear (see Chapter 15) is to be supported at the center of a short hollow circular shaft using a
single-row radial ball bearing. The inner race is presses on the fixed non-rotating shaft, and the rotating gear is
attached to the outer race of the bearing. The gear is to rotate at 900 rpm. The forces in the gear produce a resultant
radial force on the bearing of 1800 N and a resultant thrust force on the bearing if 1460 N. The assembly is subjected
to light shock loading conditions. Based on preliminary stress analysis of the shaft, it must have at least a 50-mm
outside diameter. It is desired to use a bearing that will have a life of 3000 hours with 99% reliability. Select the
smallest acceptable bearing (bore) from Table 11.5.
----------------------------------------------------------------------------------------------------------------------------------
Solution
Given ( dbore )min ≥ 50 mm , Fr = 1800 N , Fa = 1460 N , n = 900 rpm , and R = 99% , the design life is
Moderate shock exists and a single-row ball bearing is to be selected. From (11-3)
Pe = X d Fr + Yd Fa
From Table 11.4; X d1 = 1.0 , Yd1 = 0 and X d2 = 0.55 , Yd2 = 1.45 . Therefore
1/ 3
⎡ ⎤
Ld
⎡⎣Cd ( 99 ) ⎤⎦ =⎢ ⎥ ( IF ) Pe
req
⎣⎢ R ( )
⎢ K 106 ⎥
⎦⎥
1/ 3
⎡ 1.62 × 108 ⎤
⎡⎣Cd ( 99 ) ⎤⎦ =⎢ 6⎥
(1.4 )( 3107 ) ≈ 28.5 kN
req
⎢⎣ 0.21× 10 ⎥⎦
From Table 11.5, the smallest acceptable bearing with ( dbore )min ≥ 50 mm a #6210, where the limiting speed is
acceptable (7000 – 8500 rpm). Checking static load
From Table 11.4; X s1 = 1.0 , Ys1 = 0 and X s2 = 0.60 , Ys2 = 0.50 . Therefore
Since ( Pse )2 > ( Pse )1 , Pse = ( Pse )2 = 1810 N . From Table 11.4, Cs = 23.2 kN for a #6210 bearing
398
11-12. An industrial punching machine is being designed to operate 8 hours per day, 5 days per week, at 1750 rpm.
A 10-year design life is desired. Select an appropriate Conrad type single-row ball bearing to support the drive shaft
if bearing loads have been estimated as 1.2 kN radial and 1.5 kN axial, and light impact conditions prevail. Standard
L10 bearing reliability is deemed to be acceptable for this application.
------------------------------------------------------------------------------------------------------------------------------------
Solution
And light impact loading exists. and a single-row ball bearing is to be selected From (11-3)
Pe = X d Fr + Yd Fa
From Table 11.4; X d1 = 1.0 , Yd1 = 0 and X d2 = 0.55 , Yd2 = 1.45 . Therefore
1
⎡ ⎤3
Ld
⎡⎣Cd ( 90 ) ⎤⎦ =⎢ ⎥ ( IF ) P
( )
e
req ⎢ K 106 ⎥
⎣⎢ R ⎦⎥
From Tables 11.2 and 11.3: K R = 1.0 for R = 90% and IF = 1.4 . Therefore
1/3
⎡ ⎤
2.81× 108
⎡⎣Cd ( 90 ) ⎤⎦ =⎢ ⎥ (1.4 )( 2.84 ) = 23.93 kN
req
⎣⎢ ( )
⎢ (1.0 ) 106 ⎥
⎦⎥
From Table 11.5, an appropriate bearing would be bearing 6306, having a bore of 30 mm, outside diameter of 72
mm, and width of 19 mm. Limiting speed of 9000 rpm is ok. Checking static load rating Pse using (11-5) and Table
11.4,
From Table 11.4; X s1 = 1.0 , Ys1 = 0 and X s2 = 0.60 , Ys2 = 0.50 . Therefore
Since ( Pse )2 > ( Pse )1 , Pse = ( Pse )2 = 1.47 kN . From Table 11.5 we find Cs = 16 kN for a No. 6306 bearing.
Assuming that the 30 mm bore is large enough to accommodate the strength-based shaft diameter requirement, the
final selection is bearing No. 6306.
399
11-13. The shaft shown in Figure P11.13 is to be supported by two bearings, one at location A and the other
location B. The shaft is loaded by a commercial-quality driven helical gear (see Chapter 15) mounted as shown.
The gear imposes a radial load of 700 lb and a thrust load of 2500 lb applied at a pitch radius of 3 inches. The thrust
load is to be fully supported by bearing A (bearing B takes no thrust load). It is being proposed to use a single-row
tapered roller bearing at location A, and another one at location B. The devise is to operate at 350 rpm, 8 hours per
day, 5 days per week, for 3 years before bearing replacement is necessary. Standard L10 reliability is deemed
acceptable. A strength-based analysis has shown that the minimum shaft diameter must be 1.375 inches at both
bearing sites. Select suitable bearings for both location A and location B.
------------------------------------------------------------------------------------------------------------------------------------
Solution
Before proceeding with bearing selection, the bearing reactions must be found at both A and B using equilibrium
concepts. Thus,
10 ( RB )r + 6 ( −7000 ) − 3 ( 2500 ) = 0
42, 000 + 7,500
( RB )r = = 4,950 lb (up)
10
10 ( RA )r − 4 ( 7000 ) + 3 ( 2500 ) = 0
28, 000 − 7,500
( RB )r = = 2, 050 lb (up)
10
( RA )a + 2500 = 0
( RA )a = −2500 lb (left)
Bearing A: Bearing A:
Fr = 2050 lb Fr = 4950 lb
Fa = 2500 lb Fa = 0 lb
n = 350 rpm n = 350 rpm
R = 90 percent R = 90 percent
400
For both bearings the design life is to be
From Table 11.3, for commercial gearing IF = 1.2 and from Table 11.2, for R = 90% , K R = 1.0 . For bearing A
then, from (11-3)
Pe = X d Fr + Yd Fa
From Table 11.4, for single row roller bearing (α ≠ 0), which is a good assumption for tapered roller bearings;
X d1 = 1.0 , Yd1 = 0 and X d 2 = 0.4 , Yd 2 = 0.4 cot α . Since α is not known till the bearing is selected, first assume
that Yd2 ≈ 1.5 , and revise later when α becomes known. Hence,
Since ( Pe )2 > ( Pe )1 , Pe = ( Pe )2 = 4570 lb . The basic dynamic radial load rating from (11-4) is
1
⎡ ⎤a
Ld
⎡⎣Cd ( 90 ) ⎤⎦ =⎢ ⎥ ( IF ) P
( )
e
req ⎢ K 106 ⎥
⎣⎢ R ⎦⎥
3/10
⎡ ⎤
1.3 × 108
⎡⎣Cd ( 90 ) ⎤⎦ =⎢ ⎥ (1.2 )( 4570 ) = 23, 620 lb
req
⎣⎢ ( )
⎢ (1.0 ) 106 ⎥
⎦⎥
From Table 11.7, tentatively select bearing No. 32307, which has a value of Cd = 24,100 lb. However, it must be
noted that this bearing has a value of Yd2 = 1.9 which is significantly different from the value assumed before.
Using this value and recalculating Pe = ( Pe )2 as
3/10
⎡ ⎤
1.3 × 108
⎡⎣Cd ( 90 ) ⎤⎦ =⎢ ⎥ (1.2 )( 5570 ) = 28,800 lb
req
⎣⎢ ( )
⎢ (1.0 ) 106 ⎥
⎦⎥
Looking again at Table 11.7, an appropriate selection appears to be bearing No. 32309. Note that for this bearing
that Yd2 = 1.74 . This value is close to the value 1.9 and a little smaller so that this bearing selection is satisfactory.
Further, the bore diameter is 1.7717 inches, greater than the minimum shaft size of 1.375 inches. Thus, the
recommendation for bearing site A is:
401
Width: 1.5059 inches
Since ( Pe )1 > ( Pe )2 , Pe = ( Pe )1 = 4950 lb . Calculating the basic dynamic radial load rating requirement from (11-4),
using a = 10/3,
3/10
⎡ ⎤
1.3 × 108
⎡⎣Cd ( 90 ) ⎤⎦ =⎢ ⎥ (1.2 )( 4950 ) = 25, 600 lb
req
⎣⎢ ( )
⎢ (1.0 ) 106 ⎥
⎦⎥
From Table 11.7, tentatively select bearing No. 32308 which has a value of Cd = 27,700 lb, and Cs = 33,700 lb, both
acceptable values. The recommendation for bearing site B is
402
11-14. From a stress analysis of a rotating shaft, it has been determined that the shaft diameter at one particular
bearing site must be at least 80 mm. Also, from a force analysis and other design specifications, a duty cycle is well
approximated by three segments, each segment having the characteristics defined in Table P11.14.
The total design life for the bearing is to be 40,000 hours and the desired reliability is 95 percent. A single-
row deep groove ball bearing is preferred.
a. Select an appropriate bearing for this application, using the spectrum loading procedure.
b. Compare the result of (a) with bearing selection for this site using the steady load procedure,
assuming that a constant radial load (and corresponding axial load) is applied to the bearing
throughout all segments of its operation.
------------------------------------------------------------------------------------------------------------------------------------
Solution
(a) Following the approach of Example 11.2, the following table may be constructed (for single-row deep-
groove ball bearing)
403
Also, from table 11.2, for R = 95%, KR = 0.62 and therefore the design life is, by problem specification Hd = 40,000
hr. To find Ld, first find the duration of one cycle, as follows:
For segment 1, 100 revolutions at 500 rpm give time t1 for segment 1 as
1000 rev
t1 = = 0.2 min
rev
500
min
Similarly,
500 rev
t2 = = 0.5 min
rev
1000
min
300 rev
t3 = = 0.3 min
rev
1000
min
⎝ hr ⎠ ⎝ min ⎠ ⎝ cycle ⎠
1
⎡ ⎤3
2.16 × 109
⎡⎣Cd ( 95 ) ⎤⎦ =⎢ ⎥ 3 0.11 ⎡1.35 (1.8 ) ⎤ 3 + 0.056 ⎡3.5 ( 3) ⎤ 3 + 0.33 ⎡1.75 ( 5 ) ⎤ 3
req
( )
⎢ ( 0.62 ) 106
⎢⎣
⎥
⎥⎦
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
From Table 11.5, the smallest acceptable bearing is No. 6320. This bearing has
dbore = 100 mm
doutside = 215 mm
width = 47mm
Checking limiting speed for bearing No. 6320, 3000 rpm is acceptable. The basic static load rating of 140 kN > 7
kN is acceptable.
Also, the bore diameter of 100 mm is acceptable because it will govern the strength-based minimum shaft diameter
of 80 mm.
(b) Using the simplified method, choosing segment 2 loading data from the table above, (11-4) gives
1
⎡ ⎤3
2.16 × 109
⎡⎣Cd ( 95 ) ⎤⎦ =⎢ ⎥ ( 3.5 )( 3) = 159.2 kN
req ⎢ ( 0.62 ) 106
⎣⎢ ( ) ⎥
⎦⎥
404
From Table 11.5, the smallest acceptable bearing is No. 6320. In this case the simplified method selects the same
bearing with a lot less work. This result will not always be achieved however, as demonstrated by Example 11.2.
405
11-15. A preliminary stress analysis of the shaft for a rapid-return mechanism has established that the shaft
diameter at a particular bearing site must be at least 0.70 inch. From a force analysis and other design specifications,
one duty cycle for this device last 10 seconds, and is well approximated by two segments, each segment having the
characteristics defined in Table P11.5.
The total design life for the bearing is to be 3000 hours. A single-row tapered roller bearing is preferred,
and a standard L10 reliability is acceptable.
a. Select an appropriate bearing for this application, using the spectrum loading procedure.
b. Compare the result of (a) with a bearing selection for this site using the steady load procedure, assuming
that a constant radial load equal to the largest spectrum load (and corresponding axial load) is applied to
the bearing throughout the full duty cycle.
------------------------------------------------------------------------------------------------------------------------------------
Solution
(a) Following the approach of Example 11.2, the following table may be constructed (for single row tapered
roller bearing).
( Pe )1 = Fr 800 600
( Pe )2 = ( 0.4 Fr + 1.5Fa )
∗ 920 240
Pe, kN 920 600
( Pse )1 = Fr 800 600
( Pse )2 = 0.5Fr + 0.75Fa 700 300
Pse, lb 800 600
ti cycle , sec 2 8
cycle
Nop, rpm 900 1200
ni duty cycle 30 160
α i = ni 900 0.16 0.84
IF 1.35 (light impact) 1.0 (steady)
406
To calculate ni, for segment 1,
⎛ 900 rev ⎞
ni = ( 2 sec ) ⎜ ⎟ = 30 rev
⎝ 60 sec ⎠
For segment 2
⎛ 1200 rev ⎞
ni = ( 8 sec ) ⎜ ⎟ = 160 rev
⎝ 60 sec ⎠
Also, from Table 11.2, for L10 (R = 90), KR = 1.0. The design life is, by problem specification, Hd = 3000 hr, so
3
3
⎡ ⎤ 10 ⎡ 10 10 ⎤ 10
2.05 × 108 ⎥ ⎢( 0.16 ) {(1.35 )( 920 )} 3 + ( 0.84 ) {(1.0 )( 600 )} 3 ⎥
⎡⎣Cd ( 90 ) ⎤⎦ =⎢
req ⎢
( )
⎢⎣ (1.0 ) 10
6 ⎥ ⎢
⎦⎥ ⎣ ⎦⎥
3
⎡⎣Cd ( 90 ) ⎤⎦ = 4.94 ⎡3.30 × 109 + 1.53 × 109 ⎤ 10 = 3971 lb
req ⎣ ⎦
From Table 11.7, the smallest acceptable bearing is No. 30204. Actually a smaller bearing would be acceptable but
this is the smallest bearing in the table. Note that for this bearing Yd2 = 1.74 which is higher than the value assumed
in the tabled value of (Pe)2. Recalculating gives
so bearing 30204 remains acceptable and the bore diameter of 0.7874 will go over the maximum shaft diameter of
0.70 inch, so it is acceptable on that basis too.
The tentative selection then will be bearing No. 30204. However, it would be advisable to search for manufacture’s
catalogs for smaller bearings before making a final choice.
(b) Using the simplified method, choosing segment 1 loading data from the table above, then (11-4) gives
3
⎡ ⎤ 10
2.05 × 108
⎡⎣Cd ( 90 ) ⎤⎦ =⎢ ⎥ (1.35 )(1016 ) = 6770 lb
req
⎢⎣ ( )
⎢ (1.0 ) 106 ⎥
⎦⎥
From Table 11.7, the smallest acceptable bearing is No. 3034. So the simplified method results in a smaller required
bearing.
407
11-16. A preliminary analysis of the metric equivalent of bearing A in Figure P11.13 has indicated that a 30209
tapered roller bearing will provide a satisfactory L10 bearing life of 3 years (operating at 350 rpm for 8 hours per
day, 5 days per week) before bearing replacement is necessary. A lubrication consultant has suggested that if an
ISO/ASTM viscosity-grade-46 petroleum oil is sprayed into the smaller end of the bearing (tapered roller bearings
provide a geometry-based natural pumping action, including oil flow from their smaller ends toward their larger
ends), a minimum elastohydrodynamic film thickness ( hmin ) of 250 nanometers can be maintained. If the bearing
races and the tapered rollers are all lapped into a surface roughness height of 100 nanometers, estimate the bearing
life for the 30209 tapered roller bearing under these elastohydrodynamic conditions.
-------------------------------------------------------------------------------------------------------------------------------------
Solution
From 11-16
hmin 250
Λ= = ≈ 1.77
Ra2 + Rb2 (100 ) + (100 )
2 2
408
11.17. A rotating steel disk, 40 inches in diameter and 4 inches thick, is to be mounted at midspan on a 1020 hot-
rolled solid steel shaft, having Su = 65,000 psi, e = 36 percent elongation in 2 inches, and fatigue properties as shown
in Figure 2.19. A reliability of 90 percent is desired for the shaft and bearings, and a design life of 5 x 108 cycles
has been specified. The shaft length between symmetrical bearing centers [see (b) below for proposed bearings] is
to be 5 inches. The operating speed of the rotating system is 4200 revolutions per minute. When the system
operates at steady-state full load, it has been estimated that about three horsepower of input to the rotating shaft
required.
a. Estimate the required shaft diameter and the critical speed for the rotating system, assuming that the
support bearings and the frame are rigid in the radial direction. The bending fatigue stress concentration
factor has been estimated as Kfb = 1.8, and the composite strength-influencing factor, k5x108, used in (2-28),
has been estimated as 0.55. A design safety factor of 1.9 has been chosen. Is the estimated critical speed
acceptable?
b. Make a second estimate for the critical speed of the rotating system, this time including the bearing
stiffness (elasticity). Based on the procedure outlined in Example 11.1, a separate study has suggested
that a single-row deep-groove ball bearing number 6209 (see Table 11.5), with oil lubrication, may be
used for this application. In addition, an experimental program has indicated that the force-deflection data
shown in Figures 11.8 and 11.9 are approximately correct for the tentatively selected bearing. Is your
second estimate of critical speed acceptable? Comment on your second estimate, and if not acceptable,
suggest some design changes that might make it acceptable.
c. Make a third estimate for critical speed of the rotating system if a medium preload is included by the way
the bearings are mounted. Comment on your third estimate.
-------------------------------------------------------------------------------------------------------------------------------------
Solution
1
⎡16 ⎧ 2nd K fb M a T ⎫⎤ 3
( d s )str =⎢ ⎨ + 3 m ⎬⎥
⎢⎣ π ⎩ SN Su ⎭⎥⎦
⎡ π ( 40 )2 ⎤
WD = 0.283 ⎢ ⎥ ( 4.0 ) = 1, 423 lb
⎢⎣ 4 ⎥⎦
WD L (1423)( 5 )
M max = = = 1779 in-lb
4 4
WD 1423
RR = RL = = = 712 lb
2 2
The torque on the shaft is
63, 025 ( 3)
T= = 45 in-lb
4200
409
Then we have using the fatigue equation
1
⎡16 ⎧⎪1.9 ( 2 )(1.8 )(1779 ) 45 ⎫⎪⎤ 3
( d s )str =⎢ ⎨ + 3 ⎬⎥ = 1.51 inches
⎣⎢ π ⎩⎪ 18,150 65, 000 ⎭⎪⎦⎥
(1423)( 5)
3
WL3
( ym )no − pre = = = 0.00048 inch
48 EI ⎛ π (1.51)4 ⎞
48 ( 30 × 10 ⎜
6
)⎜ ⎟
64 ⎟
⎝ ⎠
The critical shaft frequency, assuming bearing and housing to be infinitely stiff is
1 rev
( ncr )no − pre = 187.7 = 8567
0.00048 min
and
(b) Using Figure 11.9 as the basis, and using the radial bearing reaction of 712 lb, the radial deflection for a
single bearing with no preload may be read as
(y ) brg no − pre
= 0.00048 inch
so the total midspan lateral displacement of the disk center for the unloaded shaft centerline becomes
This is below the recommended guideline of section 8.6, and must be regarded as a risky design, requiring
improvement or experimental verification. To improve, use larger shaft or preload bearings.
(c) Again using Figure 11.9 as a basis, when a medium preload is induced
(y )brg med
= 0.00015 inch
( ym )med = 0.00042 + 0.00015 = 0.00057 inch
410
1 rev
( ncr )light = 187.7 = 7862
0.00057 min
giving
This is slightly below the recommended guidelines of section 8.6, but would probably be acceptable. Note that
preloading has significantly improved the system.
411
Chapter 12
12-1. Figures 12.5, 12.6, and 12.7 depict a power screw assembly in which the rotating screw and
nonrotating nut will raise the load W when the torque TR is applied in the direction shown (CCW rotation of
screw if viewed from bottom end). Based on a force analysis of the power screw system shown in the three
figures cited, the torque required to raise the load is given by (12-7).
a. List the changes that must be made in the free-body diagrams shown in Figures 12.6 and 12.7 if
the load is to be lowered by reversing the sense of the applied torque.
b. Derive the torque equation for lowering the load in this power screw assembly. Compare your
results with (12-8).
---------------------------------------------------------------------------------------------------------------------------------
Solution
∑ M = −T + µ Wr − r F cos θ sin α + r F µ
z L c c p n n p n t cos α = 0
and
W
Fn =
µt sin α + cos θ n cos α
⎡ − cos θ n sin α + µt cos α ⎤
TL = Wrp ⎢ ⎥ + µcWrc
⎣ µt sin α + cos θ n cos α ⎦
412
12-2. The power lift shown in Figure P12.2 utilizes a motor drive Acme power screw to raise the platform,
which weighs a maximum of 3000 lb when loaded. Note that the nut, which is fixed to the platform, does
not rotate. The thrust collar of the power screw presses against the support structure, as shown, and the
motor drive torque is supplied to the drive shaft below the thrust collar, as indicated. The thread is 1 ½ -
inch Acme with 4 threads per inch. The thread coefficient of friction is 0.40. The mean collar radius is 2.0
inches, and the collar coefficient of friction is 0.30. If the rated power output of the motor drive unit is 7.5
hp, what maximum platform lift speed (ft/min) could be specified without exceeding the rated output power
of the motor drive unit? (Note any approximations used in your calculations.)
---------------------------------------------------------------------------------------------------------------------------------
-------
Solution
From (4-39)
Tn
hp =
63, 025
63, 025 ( 7.5 ) 4.73 ×105 rev
nmax = =
TR TR min
⎡ cos θ n sin α + µt cos α ⎤
TR = Wrp ⎢ ⎥ + µcWrc (from 12 − 7)
⎣ cos θ n cos α − µt sin α ⎦
W = 3000 lb
1.50
ro = = 0.75 in.
2
p
rp = ro −
4
1
p = = 0.25 in.
4
0.25
rp = 0.75 − = 0.688 in.
4
Using (12-2)
p 0.25
α = tan −1 = tan −1 = 3.31o
2π rp 2π ( 0.688 )
The lift speed “s” in ft/min is related to the rotational speed nmax as follows:
413
12-3. A power lift similar to the one shown in Figure P12.2 uses a single-start square-thread power screw
to raise a load of 50 kN. The screw has a major diameter of 36 mm and a pitch of 6 mm. The mean radius
of the thrust collar is 40 mm. The static thread coefficient of friction is estimated as 0.15 and the static
collar coefficient of friction as 0.12.
a. Calculate the thread depth.
b. Calculate the lead angle.
c. Calculate the helix angle.
d. Estimate the starting torque required to raise the load.
---------------------------------------------------------------------------------------------------------------------------------
-------
Solution
p 6
= = 3 mm
2 2
(b) Since this is a single-start thread, the lead angle α may be determined from (12-2)
⎛ p ⎞
α = tan −1 ⎜
⎜ 2π r ⎟⎟
⎝ p ⎠
p 36 6
rp = ro − = − = 16.5 mm
4 2 4
⎛ 6 ⎞
α = tan −1 ⎜⎜ ⎟⎟ = tan ( 0.058 ) = 3.31
−1 o
⎝ 2π (16.5 ) ⎠
(c) Since the helix angle ψ is the complement of the lead angle α,
Ψ = 90 – 3.31 = 86.69˚
(d) The starting torque required to raise the load may be obtained from (12-7) as
414
12-4. In a design review of the power lift assembly shown in Figure P12.2, a consultant has suggested that
the buckling of the screw might become a problem if the lift height (screw length) becomes “excessive.”
He also has suggested that for buckling considerations the lower end of the steel screw, where the collar
contacts the support structure, may be regarded as fixed, and at the upper end where the screw enters the
nut, the screw may be regarded as pinned but guided vertically. If a safety factor of 2.2 is desired, what
would be the maximum acceptable lift height Ls?
---------------------------------------------------------------------------------------------------------------------------------
-------
Solution
W = 3000 lb
1.50
ro = = 0.75 in.
2
1
p = = 0.25 in.
4
p 0.25
rroot = ro − = 0.75 − = 0.625 in.
2 2
d r = 2 ( 0.625 ) = 1.25 in.
Using Euler’s equation (2-36), with Le = 0.7Ls (see Figure 2.7 (d)),
π 2 EI
( Pcr ) =
( 0.7 Ls )
2
Since nd = 2.2
( Pcr )req ' d
Pd =
2.2
π (1.25 )
4
πd 4
I≈ r
= = 0.12 in 4 and E = 30 × 106 psi
64 64
( Pcr )req ' d = 2.2 Pd = 2.2 (W ) = 2.2 ( 3000 ) = 6600 lb
π 2 EI
= 6600
( 0.7 Ls )
2
π 2 ( 30 × 106 ) ( 0.12 )
Ls = = 104.8 in. (maximum acceptable lift height)
( 0.7 ) ( 6600 )
2
415
12-5. Replot the family of efficiently curves shown in Figure 12.8, except do the plot for square threads
instead of Acme threads. Use the same array of friction coefficients, and again assume the collar friction to
be negligibly small.
---------------------------------------------------------------------------------------------------------------------------------
-------
Solution
1 − µt tan α
eµc = 0 =
1 + µt cot α
Calculating eµc = 0 as a function of α (0 ≤ α ≤ 90) for each value of µt shown in Figure 12.8, the following
table may be constructed.
Using the format of Figure 12.8, these values may be plotted for square-thread screws, as shown below,
note that negative efficiencies are undefined.
416
417
12-6. A 50-mm single-start power screw with a pitch of 10 mm is driven by a 0.75 kw drive unit at a
speed of 20 rpm. The thrust is taken by a rolling element bearing, so collar friction may be neglected. The
thread coefficient of friction is µt = 0.20 . Determine the maximum load that can be lifted without stalling
the drive, the efficiency of the screw, and determine if the power screw will “overhaul” under maximum
load if the power is disconnected.
---------------------------------------------------------------------------------------------------------------------------------
-------
Solution
rp = ro − p / 4 = 25 − 10 / 4 = 22.5 mm
TR
Wmax =
⎡ cos θ n sin α + µt cos α ⎤
rp ⎢ ⎥
⎣ cos θ n cos α − µt sin α ⎦
p 10
where α = tan −1 = tan −1 = 4.1o . Since α is small, θ n ≈ θ = 14.5o . In addition
2π rp 2π (22.5)
0.75(9549)
TR = = 358 N-m
20
358 358
Wmax = = = 56.4 kN
⎡ cos14.5 sin 4.1 + 0.2 cos 4.1 ⎤
o o o ⎡ 0.26839 ⎤
0.0225 ⎢ 0.0225 ⎢
o o o ⎥ ⎥
⎣ 0.95137 ⎦
⎢⎣ cos14.5 cos 4.1 − 0.2sin 4.1 ⎥⎦
Wmax = 56.4 kN
l cos θ 10 cos14.5o
µt < = = 0.0685
2π rp 2π (22.5)
418
12-7. A standard 1 1/2 –inch rotating power screw with triple square threads is to be used to lift a 4800-lb
load at a lift speed of 10 ft/min. Friction coefficients for both the thread and the collar have been
experimentally determined to be 0.12. The mean thrust collar friction diameter is 2.75 inches.
a. What horsepower would you estimate to be required to drive this power screw assembly?
b. What motor horsepower would you recommend for this installation?
---------------------------------------------------------------------------------------------------------------------------------
-------
Solution
p
(a) ro = 1.50/2 = 0.75 in., rc = 2.75/2 = 1.38 in. For a square thread, rp = ro − . From Table 12.1,
4
for a standard 1- ½ inch square thread should have 3 threads per inch, so p = 1/3 = 0.33 in. and
p 0.33
rp = ro − = 0.75 − = 0.67 in.
4 4
⎛ np ⎞ ⎛ 3 ( 0.33) ⎞
α = tan −1 ⎜ = tan −1 ⎜⎜ ⎟⎟ = 13.2
⎜ 2π rp ⎟⎟
o
⎝ ⎠ ⎝ 2π ( 0.67 ) ⎠
From (12-1) l = np = 3(0.33) = 1.0 in/rev. To find the speed n in rpm to produce a lift of 10 ft/min, then
⎛ rev ⎞ ⎛ 12 in ⎞ ⎛ 10 ft ⎞
n = ⎜ 1.0 ⎟⎜ ⎟⎜ ⎟ = 120 rpm
⎝ in ⎠ ⎝ ft ⎠ ⎝ min ⎠
1968 (120 )
hp = = 3.75 horsepower
63, 025
(b) Installed motor horsepower should incorporate a safety factor on the power required, and should
specify a “standard” available motor probably a 5-horsepower motor in this case. A motor
manufactures catalog should be consulted. In fact, a gear motor would probably be required to
supply 5-horsepower at 120 rpm.
419
12-8. Repeat problem 12-7 if everything remains the same except that the power screw has double square
threads.
---------------------------------------------------------------------------------------------------------------------------------
-------
Solution
p
(a) ro = 1.50/2 = 0.75 in., rc = 2.75/2 = 1.38 in. For a square thread, rp = ro − . From Table 12.1,
4
for a standard 1- ½ inch square thread should have 3 threads per inch, so p = 1/3 = 0.33 in. and
p 0.33
rp = ro − = 0.75 − = 0.67 in.
4 4
⎛ np ⎞ ⎛ 2 ( 0.33) ⎞
α = tan −1 ⎜ = tan −1 ⎜⎜ ⎟⎟ = 8.9
⎜ 2π rp ⎟⎟
o
⎝ ⎠ ⎝ (
2π 0.67 )⎠
From (12-1) l = np = 2(0.33) = 0.66 in/rev. To find the speed n in rpm to produce a lift of 10 ft/min, then
⎛ 1 rev ⎞ ⎛ 12 in ⎞ ⎛ 10 ft ⎞
n=⎜ ⎟⎜ ⎟⎜ ⎟ = 182 rpm
⎝ 0.66 in ⎠ ⎝ ft ⎠ ⎝ min ⎠
1702 (182 )
hp = = 4.91 horsepower
63, 025
(c) Installed motor horsepower should incorporate a safety factor on the power required, and should
specify a “standard” available motor probably a 7.5-horsepower motor in this case. A motor
manufactures catalog should be consulted. In fact, a gear motor would probably be required to
supply 5-horsepower at 182 rpm.
420
12-9. A 40-mm rotating power screw with triple square threads has a pitch of p = 8 mm . The screw is to
be used to lift a 22 kN load at a speed of 4 meters/min. Friction coefficients for both the collar and threads
have been determined to be µt = µc = 0.15 . The mean thrust collar friction diameter is 70 mm. Determine
the power required to drive the assembly.
---------------------------------------------------------------------------------------------------------------------------------
Solution
ro = 40 / 2 = 20 mm , p = 8 mm , µt = µc = 0.15 , rc = 70 / 2 = 35 mm
rp = ro − p / 4 = 20 − 8 / 4 = 18 mm
⎡ sin α + µt cos α ⎤
TR = Wrp ⎢ ⎥ + Wrc µc
⎣ cos α − µt sin α ⎦
np 3(8)
where α = tan −1 = tan −1 ≈ 12o
2π rp 2π (18)
⎡ sin12o + 0.15cos12o ⎤
TR = 22 ( 0.018 ) ⎢ o o ⎥
+ 22 ( 0.035 )( 0.15 ) = 0.1484 + 0.1155 = 0.2639 kN-m
⎣⎢ cos12 − 0.15sin12 ⎦⎥
With l = np = 3(8) = 24 mm/rev , the rotational speed to lift the load at a rate of 4 meters/min is
⎛ 1 rev ⎞ ⎛ m ⎞
n=⎜ ⎟ ⎜ 4 min ⎟ = 167 rpm
⎝ 0.024 m ⎠⎝ ⎠
TR n 263.9(167)
( kw)req = = = 4.62 kw
9549 9549
421
12-10. Find the torque required to drive a 16-mm single-start square thread power screw with a 2 mm
pitch. The load to be lifter is 3.6 kN. The collar has a mean friction diameter of 25 mm, and the coefficients
of collar and thread friction are µc = 0.12 and µt = 0.15 .
---------------------------------------------------------------------------------------------------------------------------------
-------
Solution
⎡ sin α + µt cos α ⎤
TR = Wrp ⎢ ⎥ + Wrc µc
⎣ cos α − µt sin α ⎦
np 2
where α = tan −1 = tan −1 ≈ 1.2o
2π rp 2π (15.5)
⎡ sin1.2o + 0.15cos1.2o ⎤
TR = 3.6 ( 0.0155 ) ⎢ o o ⎥
+ 3.6 ( 0.0125 )( 0.12 ) = 0.00957 + .0054 = .01496 kN-m
⎢⎣ cos1.2 − 0.15sin1.2 ⎥⎦
422
12-11. A mild-steel C-clamp has a standard single-start ½-inch Acme thread and mean collar radius of
5/16 inch. Estimate the force required at the end of a 6-inch handle to develop a 300-lb clamping force.
(Hint: see Appendix Table A.1 for friction coefficients.)
---------------------------------------------------------------------------------------------------------------------------------
-------
Solution
From Table A-1, for mild steel on mild steel, general application, dry sliding, the typical value is given as
µc = µt = 0.35, ro = 0.50/2 = 0.25 in. From Figure 12.2(c), for an Acme thread, rp = ro –p/4. From Table
12.1, a standard ½-inch Acme thread has 10 threads per inch. Thus, p = 1/10 = 0.10 in. and rp = 0.25 –
0.10/4 = 0.225 in. Utilizing (12-2), with n = 1 for a single thread
⎛ np ⎞ ⎛ 0.10 ⎞
α = tan −1 ⎜ = tan −1 ⎜⎜ ⎟⎟ = 4.05
⎜ 2π r ⎟⎟
o
⎝ p ⎠ ⎝ (
2π 0.225 )⎠
For small α θ n ≈ θ and θ = 14.5 o
62.8
F= ≈ 10.5 lb
6
423
12-12. Design specifications for a power screw lifting device require a single-start square thread having a
major diameter of 20 mm and a pitch of 4 mm. The load to be lifted is 18 kN, and it is to be lifter at a rate
of 12 mm/s. The coefficients thread and collar friction are estimated to be µt = µc = 0.15 , and the mean
collar diameter is 25 mm. Calculate the required rotational speed of the screw and the power required to
drive it.
---------------------------------------------------------------------------------------------------------------------------------
---
Solution
ro = 20 / 2 = 10 mm , p = 4 mm , µt = µc = 0.15 , rc = 25 / 2 = 12.5 mm ,
rp = ro − p / 4 = 9 mm
⎡ sin α + µt cos α ⎤
TR = Wrp ⎢ ⎥ + Wrc µc
⎣ cos α − µt sin α ⎦
np 4
where α = tan −1 = tan −1 ≈ 4.1o
2π rp 2π (9)
With l = p = 4 mm/rev , the rotational speed to lift the load at a rate of 12 mm/s is
⎛ 1 rev ⎞ ⎛ mm ⎞ ⎛ sec ⎞
n=⎜ ⎟ ⎜ 12 sec ⎟ ⎜ 60 min ⎟ = 180 rpm n = 180 rpm
⎝ 4 mm ⎠⎝ ⎠⎝ ⎠
TR n 127.4(180)
( kw)req = = = 2.4 kw ( kw)req = 2.4 kw
9549 9549
424
12-13. A 20-mm power screw for a hand-cranked arbor press is to have a single-start square thread with
a pitch of 4mm. The screw is to be subjected to an axial load of 5 kN. The coefficient of friction for both
threads and collar is estimated to be about 0.09. The mean friction diameter for the collar is to be 30 mm.
a. Find the nominal thread width, thread height, mean thread diameter, and the lead.
b. Estimate the torque required to “raise” the load.
c. Estimate the torque required to “lower” the load.
d. Estimate the efficiency of this power screw system.
---------------------------------------------------------------------------------------------------------------------------------
-------
Solution
⎛ np ⎞ ⎛ 4 ⎞
α = tan −1 ⎜ −1
⎟⎟ = tan ⎜⎜ ⎟⎟ = 4.05
o
⎜ 2π r 2π ( 9 )
⎝ p ⎠ ⎝ ⎠
Referring to Figure 12.2(a), Wt = p/2 = 4/2 = 2 mm and ht = p/2 = 4/2 = 2 mm. We have that l =
np = (1)(4) = 4 mm.
⎡ sin α + µt cos α ⎤
TR = Wrp ⎢ ⎥ + Wrc µc
⎣ cos α − µt sin α ⎦
⎡ sin 4.5 + 0.09 cos 4.5 ⎤
= 5000 ( 0.009 ) ⎢ ⎥ + 5000 ( 0.015 )( 0.09 )
⎣ cos 4.5 − 0.09sin 4.5 ⎦
TR = 45 ( 0.17 ) + 6.75 = 14.4 N-m
⎡ − sin α + µt cos α ⎤
TL = Wrp ⎢ ⎥ + Wrc µc
⎣ cos α + µt sin α ⎦
⎡ − sin 4.5 + 0.09 cos 4.5 ⎤
= 5000 ( 0.009 ) ⎢ ⎥ + 5000 ( 0.015 )( 0.09 )
⎣ cos 4.5 + 0.09sin 4.5 ⎦
TL = 45 ( 0.011) + 6.75 = 7.25 N-m
1
e=
⎡ cos 0 + 0.09 cot 4.5 ⎤ ⎛ 15 ⎞
⎢⎣ cos 0 − 0.09 tan 4.5 ⎥⎦ + 0.09 ⎝⎜ 9 ⎠⎟ cot 4.5
1
= = 0.25 (25 percent)
2.16 + 1.91
425
12-14. Based on design specifications and loads, a standard single-start 2 inch Acme power screw
with 4 threads per inch has tentatively been chosen. Collar friction is negligible. The screw is
in tension and the torque require to raise a load of 12,000 lb at the specified lift speed has been
calculated to be 2200 in-lb. Concentrating your attention on critical point B shown in Figure
12.9, calculate the following:
a. Nominal torsional shear stress in the screw.
b. Nominal direct stress in the screw.
c. Maximum transverse shearing stress due to thread bending. Assume that three threads carry the
full load.
d. Principal stresses at critical point B.
---------------------------------------------------------------------------------------------------------------------------------
-------
Solution
We have ro = 2.00/2 = 1.00 in. and p = ¼ = 0.25 in. Referring to critical point B shown in Figure
12.9
2 (TR − 0 ) 2TR
τs = =
π rr3 π rr3
p 0.25
rr = ro − = 1.00 − = 0.875 in.
2 2
2 ( 2200 )
τs = = 2,100 psi ( 2.1 kpsi )
π ( 0.875 )
3
W 12, 000
σ dir = = = 4,990 psi (4.99 kpsi)
π rr π ( 0.875 )2
2
3W 3 (12, 000 )
τ r − max = = = 8, 730 psi (8.73 kpsi)
2π rr pne 2π ( 0.875 )( 0.25 )( 3)
( 4.99 ) + 4 ( 80.62 )
2
4.99 +
σ1 =
2
σ 1 = 2.50 + 9.32 = 11.82 kpsi
σ2 = 0
σ 3 = 2.5 − 9.32 = −6.82 kpsi
(e) From Table 3.3, for 1020 C.D. steel, Syp = 70 ksi. For the specified safety factor nd = 2.3,
426
S yp 70
σd = = = 30.4 ksi
nd 2.3
427
12-15. Based on design specifications and loads, a single-start 48-mm diameter Acme power screw with an
8 mm pitch has been tentatively selected. Collar friction is negligible. The screw is in tension and the
torque required to raise a load of 54 kN at the specified lift speed has bee calculated to be 250 N-m.
Concentrate on point C shown in Figure 13.9and calculate:
---------------------------------------------------------------------------------------------------------------------------------
--
Solution
ro = 48 / 2 = 24 mm , p = 8 mm , µc = 0 , rp = ro − p / 4 = 22 mm , rr = ro − p / 2 = 20 mm
W = 54 kN , TR = 250 N-m
4Tr 4(250)
a. τs = = = 9.95 MPa
π rr3 π (0.020)3
W 54 000
b. σ dir = = = 39.8 MPa
π rr2 π (0.020) 2
c. σb =
(
12W rp − rr ) = 12 ( 54 000 )( 0.022 − 0.020 ) = 107.4 MPa
π ( 0.020 )( 3)( 0.008 )
2 2
π rr ne p
σ 3 − 380σ 2 + (21.89) 2 σ = 0
(
σ σ 2 − 380σ + 479 = 0 )
σ1 ≈ 379 MPa , σ 2 ≈ 1.5 MPa , σ 3 = 0
428
12-16. A special square-thread single-start power screw is to be use to raise a 10-ton load. The screw is
to have a mean thread diameter of 1.0 inch, and four threads per inch. The mean collar radius is to be 0.75
inch. The screw, the nut, and the collar are all to be made of mild steel, and all sliding surfaces are
lubricated. (See Appendix Table A.1 for typical coefficients of friction.) It is estimated that three threads
carry the full load. The screw is in tension.
a. Calculate the outside diameter of this power screw.
b. Estimate the torque required to raise the load.
c. Estimate the torque required to lower the load.
d. If a rolling element bearing were installed at the thrust collar (gives negligible collar friction),
what would be the minimum coefficient of thread friction needed to prevent overhauling of the fully
loaded screw?
e. Calculate, for the conditions of (d), the nominal values of torsional shearing stress in the screw,
direct axial stress in the screw, the thread bearing pressure, maximum transverse shearing stress in the
thread, and thread bending stress.
---------------------------------------------------------------------------------------------------------------------------------
-------
Solution
We note that rp = 1.0/2 = 0.50 in. and from Table A-1, for mild steel on mild steel, lubricated, that µstatic =
0.11 and µrunning = 0.08.
⎡ p⎤ ⎡ 0.25 ⎤
d o = 2ro = 2 ⎢ rp + ⎥ = 2 ⎢ 0.50 + = 1.125 in.
⎣ 4 ⎦ ⎣ 4 ⎥⎦
⎛ np ⎞ ⎛ 0.25 ⎞
α = tan −1 ⎜ −1
⎟⎟ = tan ⎜⎜ ⎟⎟ = 4.55
o
⎜ 2π r 2π ( 0.50 )
⎝ p ⎠ ⎝ ⎠
⎡ sin α + µt cos α ⎤
TR = Wrp ⎢ ⎥ + Wrc µc
⎣ cos α − µt sin α ⎦
⎡ sin 4.55 + 0.08cos 4.55 ⎤
= 20, 000 ( 0.50 ) ⎢ ⎥ + 20, 000 ( 0.75 )( 0.08 )
⎣ cos 4.55 − 0.08sin 4.55 ⎦
TR = 10, 000 ( 0.16 ) + 1200 = 2800 in-lb
⎡ − sin α + µt cos α ⎤
TL = Wrp ⎢ ⎥ + Wrc µc
⎣ cos α + µt sin α ⎦
⎡ − sin 4.55 + 0.08cos 4.55 ⎤
= 20, 000 ( 0.50 ) ⎢ ⎥ + 20, 000 ( 0.75 )( 0.08 )
⎣ cos 4.55 + 0.08sin 4.55 ⎦
TL = 10, 000 ( 0.000672 ) + 1200 = 1207 in-lb
l cos θ p 0.25
µt = = = = 0.08
2π rp 2π rp 2π ( 0.5 )
429
(e) From (12-21), with µc = 0, torsional shearing stress in the screw is
2T
τ s = R3
π rr
(TR )µ =0 = 1600
c
in-lb (TR with µc = 0)
p 0.25
rr = rp − = 0.50 − = 0.4375 in.
4 4
2 (1600 )
τs = = 12,164 psi
π ( 0.4375 )
3
W 20, 000
σ dir = = = 33, 260 psi
π rr π ( 0.4375 )2
2
W 20, 000
σ B = pB = = = 17, 400 psi
π ( ro2 − ri 2 ) ne π ( 0.562 − 0.43752 ) ( 3)
From (12-23), the maximum transverse shearing stress due to thread bending is
3W 3 ( 20, 000 )
τ r − max = = = 29,100 psi
2π rr pne 2π ( 0.4375 )( 0.25 )( 3)
430
12-17. A power screw lift assembly is to be designed to lift and lower a heavy cast-iron lid for a 10-
foot-diameter pressure cooker used to process canned tomatoes in a commercial caning factory. The
proposed lift assembly is sketched in Figure P12.17. The weight of the cast iron lid is estimated to be 4000
lb, to be equally distributed between two support lugs as shown in Figure P12.17. It may be noted that the
screw is in tension, and it has been decided that a standard Acme thread form should be used. Preliminary
calculations indicate that the nominal tensile stress in the screw should not exceed a design stress of 8000
psi, based on yielding. Stress concentration and safety factor have both been included in the specification
of the 8000 psi design stress. Fatigue may be neglected as a potential failure mode because of the
infrequent use of the life assembly. The rotating steel screw is supported on a rolling element bearing
(negligible friction), as shown, and the nonrotating nut is to be made of porous bronze (see Table 10.1).
The coefficient of friction between the screw and the nut has been estimated to be 0.08.
a. Estimate the tentative minimum root diameter for the screw, based on yielding due to direct tensile
load alone as the governing failure mode.
b. From the results of (a), what Acme thread specification would you suggest as a first-iteration
estimate for this application?
c. What would be the maximum driving torque, Td, for Acme thread specified in (b)?
d. What torsional shearing stress would be induced in the root cross-section of the suggested power
screw by driving torque TR.
e. Identify the critical points that should be investigated in the Acme thread power screw.
f. Investigate the contact zone between screw threads and nut threads, and resize the screw if
necessary. Assume that the full load is carried by three threads. If resizing is necessary,
recalculate the driving torque for the revised screw size.
g. What horsepower input would be required to drive the screw, as sized in (f), if it is desired to raise
the lid 18 inches in no more than 15 seconds?
---------------------------------------------------------------------------------------------------------------------------------
-------
Solution
W 4W
σ dir = = = σd
A π d r2
4W 4 ( 4000 )
( d r )req ' d = = = 0.80 in.
πσ d π ( 8000 )
(b) From Figure 12.2(c), for an Acme thread do/2 = ro = rr + p/2. Note that from Table 12.1 that the
standard Acme screws in this size range (see (b)) have around 5 threads per inch,
p ≈ 1 5 = 0.20 in. Thus,
⎛ 0.80 0.20 ⎞
do ≈ 2 ⎜ + ⎟ = 1.00 in.
⎝ 2 2 ⎠
For a first iteration, select a standard 1-inch Acme thread with 5 threads per inch.
(c) From Figure 12.2(c), for a standard 1-inch Acme thread rp = ro – p/4 = 0.5 – 0.20/4 = 0.45 in.
Using (12-2), assuming a single-start thread,
⎛ p ⎞ ⎛ 0.20 ⎞
α = tan −1 ⎜ −1
⎟⎟ = tan ⎜⎜ ⎟⎟ = 4.05
o
⎜ 2π r 2π ( 0.45 )
⎝ p ⎠ ⎝ ⎠
431
⎡ cos θ n sin α + µt cos α ⎤
TR = Wrp ⎢ ⎥
⎣ cos θ n cos α − µt sin α ⎦
⎡ cos14.5sin 4.05 + 0.08cos 4.05 ⎤
= 4000 ( 0.45 ) ⎢ ⎥
⎣ cos14.5cos 4.05 − 0.08sin 4.05 ⎦
= 1800 ( 0.151) = 272 in-lb
(e) The critical points to be investigated are those shown as “A”, “B”, and “C” in Figure 12.9.
(f) The contact zone is represented by critical point “A” of Figure 12.9. The governing wear equation
is given by (12-20) as
W 4, 000
σ B = pB = = = 4, 715 psi
π ( r − ri ) ne
o
2 2
π ( 0.502 − 0.402 ) ( 3)
From Table 10.1, porous bronze has an allowable maximum pressure of pallow = 2000 psi. So the
screw must be resized to bring pB down to 2000 psi or less. Sticking with the standard Acme
screws (Table 12.1) we see that the next larger screw is 1 ½ -inch with 4 threads. For this larger
screw
1.50
ro = = 0.75 in.
2
1
p = = 0.25 in.
4
0.25
rp = 0.75 − = 0.625 in.
2
thus
4, 000
σ B = pB = = 2,563 psi
π ( 0.752 − 0.652 ) ( 3)
This is still too high compared to the 2000 psi allowable, so the next larger standard size is taken .
A 2-inch Acme thread with 4 threads per inch. Thus, we have
432
2.0
ro = = 1.0 in.
2
1
p = = 0.25 in.
4
0.25
rp = 1.0 − = 0.875 in.
2
thus
4, 000
σ B = pB = = 1,810 psi
π (1.02 − 0.8752 ) ( 3)
Thus, the screw to be selected is a 2-inch Acme screw with 4 threads per inch. Using (12-7) to
calculate the torque requires first the following for the 2-inch screw:
p 0.25
rp = ro − = 1.0 − = 0.9375 in.
4 4
⎛ p ⎞ ⎛ 0.25 ⎞
α = tan −1 ⎜ ⎟ = tan −1 ⎜⎜ ⎟⎟ = 2.43
o
⎜ 2π r ⎟
⎝ p ⎠ ⎝ (
2π 0.9375 )⎠
⎡ cos θ n sin α + µt cos α ⎤
TR = Wrp ⎢ ⎥ + Wrc µc
⎣ cos θ n cos α − µt sin α ⎦
⎡ cos14.5sin 2.43 + 0.08cos 2.43 ⎤
= 4000 ( 0.9375 ) ⎢ ⎥
⎣ cos14.5cos 2.43 − 0.08sin 2.43 ⎦
= 3750 ( 0.124 ) = 465 in-lb
inch
l = np = (1.0 )( 0.25 )
rev
⎛ 1 rev ⎞ ⎛ 18 in ⎞⎛ sec ⎞
n=⎜ ⎟⎜ ⎟ ⎜ 60 ⎟ = 288 rpm
⎝ 0.25 in ⎠ ⎝ 15 sec ⎠ ⎝ min ⎠
Tn 465 ( 288 )
( hp )req ' d = = = 2.12 horsepower
63, 025 63, 025
433
Chapter 13
13-1. You have been assigned the task of examining a number of large flood gates installed in 1931 for
irrigation control at a remote site on the Indus River in Pakistan. Several large steel bolts appear to have
developed cracks, and you have decided that they should be replaced to avert a potentially serious failure of
one or more of the flood gates. Your Pakistani assistant has examined flood gate specifications, and has
found that the original bolts may be well characterized as 32-mm medium carbon quenched and tempered
steel bolts, of property class 8.8. you have brought with you only a limited number of replacement bolts in
this size range, some of which are ASTM Class A325, type 3. Which, if either, of these replacement bolts
would you recommend as a substitute for the cracked originals? Justify your recommendation.
------------------------------------------------------------------------------------------------------------------------
Solution
From Table 13.5, the minimum bolt properties for class 8.8 are:
From Table 13.3, for SAE Grade 7 bolts in the size range ¼ - 1 ½ inch diameter, they have the following
minimum properties:
Su = 133 ksi
S yp = 115 ksi
S proof = 105 ksi
From table 13.4, ASTM Class A325, type 3 bolts in the size range 1 1/8 – 1 ½ inch diameter , have the
following minimum properties:
Su = 105 ksi
S yp = 81 ksi
S proof = 74 ksi
Comparing properties, The SAE Grade 7 bolts exceed the original bolt strength specifications; ASTM Class
A325 type 3 bolts fall short. Therefore, recommend SAE Grade 7 bolts.
434
13-2. A high-speed “closing machine” is used in a tomato canning factory to install lids and seal the cans.
It is in the middle of the “pack” season and a special bracket has separated from the main frame of the
closing machine because the 3/8-24 UNF-2A hex-cap screws used to hold the bracket in place have failed.
The head markings on the failed cap screws consist of the letters BC in the center of the head. No cap
screws with this head marking can be found in the storeroom. The 3/8-24 UNF-2A cap screws that can be
found in the “high-strength” bin have five equally spaced radial lines on the heads. Because is so important
to get up-and-running immediately to avoid spoilage, you are being asked, as an engineering consultant,
whether the available cap screws with head markings of five radial lines can be safely substituted for the
broken originals. How do you respond? Justify your recommendation.
------------------------------------------------------------------------------------------------------------------------
Solution
From Figure 13.6, the “BC” head marking identifies ASTM class A354 grade BC bolts, and five equally
spaced radial lines identifies SAE grade 7 bolts.
From Tables 13.3 and 13.4, the minimum strength properties for the two head markings are:
435
13-3. A cylindrical flange joint requires a total clamping force between two mating flanges of 45 kN. It is
desired to use six equally spaced cap screws around the flange. The cap screws pass through clearance
holes in the top flange and thread into tapped holes in the bottom flange.
------------------------------------------------------------------------------------------------------------------------
Solution
45 000
Fb = = 7500 N
6
As a starting point, select a class 4.8 bolt with a proof strength of 310 MPa. Based on proof strength
Fb 7.5 × 103
At = = 6
= 24.19 mm 2
S proof 310 × 10
From Table 13.2 the appropriate screw selection would be a size 8.0
436
13-4. It is desired to use a set of four bolts to attach the bracket shown in Figure P13.4 to a stiff steel
column. For purposes of economy, all bolts are to be the same size. It is desired to use ASTM Class
A307 low-carbon steel material and standard UNC threads. A design safety factor of 2.5 has been
selected, based on yielding as the governing failure mode.
a. What bolt-hole pattern would you suggest and what bolt specification would you recommend?
b. What tightening torque would you recommend if it is desired to produce a preload force in each
bolt equal to 85 percent of the minimum proof strength?
------------------------------------------------------------------------------------------------------------------------
Solution
(a) Based on judgment, it has been decided (somewhat arbitrarily) to place bolt centerlines at 1-inch
in from each edge of the vertical 7” x 3” plate sketched in Figure P13.4. That is,
Using (13-31)
P 6000 1500
τb = 4
= =
4 Ab Ab
∑A
1
i
⎡ ( 6000 )( 5 )( 6 ) ⎤
( Fb )max = Ab ⎢ ⎥ = 2, 432 lb
⎢⎣ 2 Ab (1) + 2 Ab ( 6 ) ⎥⎦
2 2
(F ) 2, 432
σ b = b max =
Ab Ab
Using (x-xx), with yielding as the failure mode and nd = 2.5 (per problem specification)
437
S yp
σ e = σ b2 + 3τ b2 =
nd
2 2
36, 000 ⎛ 2432 ⎞ ⎛ 1500 ⎞
= ⎜ ⎟ + 3⎜ ⎟
2.5 ⎜ A ⎟ ⎝ Ab ⎠
⎝ b ⎠
1
( 2432 ) + 3 (1500 )
2 2
14400 =
Ab
Ab = 0.247 in 2
4 Ab 4 ( 0.247 )
d min = = = 0.561 in.
π π
From Table 13.1, using the UNC series, the above value corresponds to a nominal ¾ inch coarse
thread. The recommended bolt specification, therefore, would be
3
− 10 UNC − 2 A ASTM Class A307
4
(b) From Table 13.4, the proof strength of the bolt specified above is Sproof = 33,000 psi, so the design
strength, specified to be 85 percent of proof strength, is
Using Ar = 0.3020 in2 from Table 13.1, the design load for the bolt is
438
13-5. Estimate the nominal size of the smallest SAE Grade 1 standard UNC bolt that will not yield under a
tightening torque of 1000 in-lb. Neglect stress concentration.
------------------------------------------------------------------------------------------------------------------------
Solution
From Table 13.3, for SAE Grade 1 material Syp = 36,000 psi. From (13-30)
Ti = 0.2 Fb db
Ti 1000 5000
Fb = = =
0.2db 0.2db db
Fb 5000 db 6366
σb = σ x = = = 3 (axial stress in bolt)
Ab π db2 4 db
The torsional shear stress in the bolt due to tightening is from (4-33) and (4-35)
S yp
σ e = σ b2 + 3τ b2 =
nd
2 2
⎛ 6366 ⎞ ⎛ 5093 ⎞
⎜ 3 ⎟ + 3 ⎜ 3 ⎟ = 36, 000
⎝ db ⎠ ⎝ db ⎠
1
( 6366 ) + 3 ( 5093) = 36, 000
2 2
db3
db = 0.671 in. (nominal diameter)
From Table 13.1, the smallest SAE Grade 1 bolt that would not yield is
3
− 10 UNC - 2A SAE Grade1
4
439
13-6. A standard fine-thread metric machine screw made of steel has a major diameter of 8.0 mm and a
head marking of 9.8. Determine the tensile proof force (kN) for this screw. It may be assumed that
the coefficient of friction is about 0.15 for both the threads and the collar.
------------------------------------------------------------------------------------------------------------------------
Solution
From Table 13.5, for “property class” 9.8, Sproof = 650 MPa. From Table 13.2, for a 8.0 mm, fine series
metric screw, the tensile stress area is At = 40 mm2. The proof force for this bolt is
⎛ 40 ⎞
Fproof = S proof Ar = 650 × 106 ⎜ 2 ⎟
= 26, 000 N (26 kN)
⎝ 1000 ⎠
440
13-7. A standard coarse-thread metric cap screw made of steel has a major diameter of 10.0 mm. If a
torque wrench is use to tighten the cap screw to a torque of 35 N-m, estimate the axial preload force
induced in the cap screw. It may be assumed that the coefficient of friction is about 0.15 for both
thread and collar.
------------------------------------------------------------------------------------------------------------------------
Solution
From (13-30)
Ti 35
Fi = = = 1750 N (1.75 kN)
0.2db 0.2 ( 0.010) )
441
13-8. Engineering specifications for a machine tool bracket application call for a nonlubricated M30 x 2
threaded fastener of property class 8.8 to be tightened to 100 percent of proof load. Calculate the
torque required to accomplish this. It may be assumed that the coefficient of friction is about 0.15 for
both the threads and the collar.
------------------------------------------------------------------------------------------------------------------------
Solution
For a fine thread we have from Table 13.2 db = 30 mm, and At = 628 mm2. From table 13.5, for “property
class” 8.8, the proof stress is Sproof = 600 MPa and the proof force for the bolt is
⎛ 628 ⎞
Fproof = S proof At = 600 × 106 ⎜ 2 ⎟ = 376,800 N (376.8 kN)
⎝ 1000 ⎠
Ti = 0.2 ( 376,800 )( 0.030 ) = 2, 260 N-m
442
13-9. A ¾-16 SAE Grade 2 steel bolt is to be used to clamp two 1.00-inch-thick steel flanges together
with a 1/16-inch-thick special lead-alloy gasket between the flanges, as shown in Figure P13.9. The
effective load-carrying area of the steel flanges and of the gasket may be taken as 0.75 sq. in.
Young’s modulus for the gasket is 5.3 x 106 psi. If the bolt is initially tightened to induce an axial
preload force in the bolt of 6000 lb, and if an external force of 8000 lb is then applied as shown,
------------------------------------------------------------------------------------------------------------------------
Solution
The load carrying areas of steel flanges and lead gasket are Astl = Ag = 0.75 in2.. The modulii are
Eg = 5.3 x 106 psi, Estl = Eb = 30 x 106 psi. The spring rate of the bolt flanges and gasket are
π ( 0.75 ) ( 30 × 106 )
2
π db2 Eb lb
kb = = = 6.425 × 106
4 Leff 4 ( 2.0625 ) in
443
The bolt force is then
⎛ kb ⎞
Fb = ⎜ ⎟ P + Fi
⎝ kb + k m ⎠
⎛ 6.425 × 106 ⎞
6 ⎟(
=⎜ 8000 ) + 6000 = 9, 215 lb (tension)
⎝ 6.425 × 10 + 9.56 × 10 ⎠
6
⎛ 9.56 × 106 ⎞
6 ⎟(
Fm = ⎜ 8000 ) − 6000 = −1215 lb (compression)
⎝ 6.425 × 10 + 9.56 × 10 ⎠
6
(c) The gasket is in series with the flange member so Fg = Fm =-1215 lb (compression)
(d) From Table 13.1, for a standard ¾-16 bolt, Ar = 0.3513 in2 so the actual stress at the root is, using
(5-25)
⎛ 9215 ⎞
σ act = K tσ nom = 3.0 ⎜ ⎟ = 78, 690 psi
⎝ 0.3513 ⎠
From Table 13.3, SAE Grade 2 for a ¾ inch size gives Syp = 57,000 psi. Since σact =78,690 > Syp
= 57,000 local yielding at the thread root would be expected.
444
13-10. A special reduced-body bolt is to be used to clamp two ¾-inch-thick steel flanges together with a
1/8-inch-thich copper-asbestos gasket between the flanges in an arrangement similar to the one shown
in Figure P13.9. The effective are for both the steel flanges and the copper-asbestos gasket may be
taken as o.75 square inch. Young’s modulus of elasticity for the copper-asbestos gasket is 13.5 x 106
psi. The special bolt has ¾-16 UNF threads but the body of the bolt is reduced to 0.4375 inch in
diameter and generously filleted, so stress concentration may be neglected. The bolt material is AISI
4620 cold-drawn steel.
a. Sketch the joint, showing the reduced-body bolt, and the loading.
b. If the bolt is tightened to produce a preload in the joint of 5000 lb, what external force Psep could
be applied to the assembly before the joint would start to separate?
c. If the external load P fluctuates from 0 to 555 lb at 3600 cycles per minute, and the desired design
life is 7 years of continuous operation, would you predict failure of the bolt by fatigue?
------------------------------------------------------------------------------------------------------------------------
Solution
(a) The joint configuration may be sketched as shown below. Note the reduced body diameter of the
bolt. Dimensions and loading are also shown.
(b) Utilizing (13-16), the joint will start to separate when Fm = 0, so separation occurs when
⎛ km ⎞
0=⎜ ⎟ Pmax − Fi
⎝ kb + k m ⎠
or when
⎛ k + km ⎞
Psep = ⎜ b ⎟ Fi
⎝ km ⎠
π ( 0.4375 ) ( 30 × 106 )
2
lb
kb = = 2.78 × 106
4 (1.625 ) in
( 0.75) ( 30 ×106 ) lb
k stl = = 15.00 × 106
2 ( 0.75 ) in
( 0.75) (13.5 ×10 )6
6 lb
kg = = 81.00 × 10
0.125 in
445
Combining the spring rates gives for the members
1 lb
km = = 12.7 × 106
1 1 in
+
15.00 × 106 81.00 × 106
⎛ 2.78 × 106 + 12.7 × 106 ⎞
Psep =⎜ ⎟ ( 5000 ) = 6094 lb
⎝ 12.7 × 106 ⎠
(c) Since Pmax 5555 < Psep = 6094 the joint never separates and (13-15) is valid for the whole range of
applied cyclic loading. Hence
⎛ kb ⎞
Fb = ⎜ ⎟ P + Fi
⎝ kb + k m ⎠
⎛ 2.78 × 106 ⎞
=⎜ ⎟ P + 5000
⎝ 2.78 × 10 + 12.7 × 10
6 6
⎠
Fb = 0.18P + 5000
( Fb )max = 0.18 ( 5555 ) + 5000 = 5998 lb
( Fb )min = 0.18 ( 0 ) + 5000 = 5000 lb
The corresponding maximum and minimum stresses in the 0.4375-inch diameter bolt body are
( Fb )max 5998
σ max = = = 39,900 psi
Ab ⎛ π ( 0.4375 )2 ⎞
⎜ ⎟
⎜ 4 ⎟
⎝ ⎠
( Fb )min 5000
σ min = = = 33, 260 psi
Ab⎛ π ( 0.4375 )2 ⎞
⎜ ⎟
⎜ 4 ⎟
⎝ ⎠
39,900 + 33, 260
σm = = 36,580 psi
2
39,900 − 33, 260
σa = = 3,320 psi
2
The bolt material is AISI 4620 cold drawn steel, so from Table 3.3, Su = 101,000 psi, and Syp = 85,000 psi.
Thus, using (5-72)
σa
σ eq −CR = for σ m ≥ 0 and σ max ≤ S yp
σm
1−
Su
we have
3320
σ eq −CR = = 5, 205 psi
36,580
1−
101, 000
446
From “Estimating S-N Curves” in section xx 2.6, Sf = Se ≈ 0.5Su = 50,500 psi. Even with adjustments of
the type shown in (5-55) and (5-56) would suggest that the bolt would be predicted to have infinite life.
Hence failure would not be predicted to occur after 7 years.
447
13-11. A typical bolted joint of the type shown in Figure 13.9 uses a ½-13 UNC bolt, and the length of
the bolt and length of the housing is the same. The threads stop immediately above the nut. The bolt
is steel with Su 101,000 psi, Syp = 85,000 psi, and Sf = 50,000 psi. The thread stress concentration
factor is 3. The effective area of the steel housing is is 0.88 in2. The load fluctuates cyclically from 0
to 2500 lb at 2000 cpm.
a. Find the existing factor of safety for the bolt if no preload is present.
b. Find the minimum required value of preload to prevent loss of compression in the housing.
c. Find the existing factor of safety for the bolt if the preload in the bolt is 3000 lb.
------------------------------------------------------------------------------------------------------------------------
Solution
The load P fluctuates cyclically from Pmin = 0 to Pmax = 2500 lb at n = 2000 cpm.
(a) From Table 13.1 Ar = 0.1257 in2. With no preload, when P is applied the joint separates and the
bolt takes the full loading range. The bolt thread at the inner end of the nut is the critical point,
and has a stress concentration factor of 3, so
⎛ Pmax ⎞ ⎛ 2500 ⎞
σ max = (σ act )max = K tf ⎜ ⎟ = 3⎜ ⎟ = 59, 670 psi
⎝ Ar ⎠ ⎝ 0.1257 ⎠
⎛ Pmin ⎞ ⎛ 0 ⎞
σ min = (σ act )min = K tf ⎜ ⎟ = 3⎜ ⎟=0
⎝ Ar ⎠ ⎝ 0.1257 ⎠
59, 670 + 0
σm = = 29,835 psi
2
59, 670 − 0
σa = = 29,835 psi
2
Thus, we have
σa
σ eq −CR = for σ m ≥ 0 and σ max ≤ S yp
σm
1−
Su
29,835
σ eq −CR = = 42,343 psi
29,835
1−
101, 000
Sf 50, 000
ne = = = 1.18 (existing safety factor)
σ eq −CR 42,343
(b) From (13-16), the minimum preload Fi to prevent loss of compression in the housing (Fm = 0) is
⎛ km ⎞
( Fi )min =⎜ ⎟ Pmax
⎝ kb + k m ⎠
The spring rates are given by
π ( 0.5 ) ( 30 × 106 )
2
π db2 lb
kb = = = 5.89 × 106
4L 4L in
Am Em 0.88 ( 30 × 106 ) lb
km = = = 26.40 × 106
L L in
⎛ 26.4 × 106 ⎞
( Fi )min = ⎜ 6 ⎟
2500 = 2044 lb
⎝ 5.89 × 10 + 26.40 × 10 ⎠
6
448
(c) If Fi = 3000 lb, from (13-15)
⎛ 5.89 × 106 ⎞
( Fb )max =⎜ 6 ⎟
2500 + 3000 = 3456 lb
⎝ 5.89 × 10 + 26.40 × 10 ⎠
6
Note that preloading the joint with a 3000 lb preload would nearly double the safety factor (from 1.18 to
2.18).
449
13-12. A ½-20 UNF-2A SAE Grade 2 steel cap screw is being considered for use in attaching a cylinder
head to an engine block made of 356.0 cast aluminum (see Table 3.3). It is being proposed to engage
the cap screw into an internally threaded hole tapped directly into the aluminum block. Estimate the
required length of thread engagement that will ensure tensile failure of the cap screw before the
threads are stripped in the aluminum block. Assume that all engaged threads participate equally in
carrying the load. Base your estimate on direct shear of the aluminum threads at the major thread
diameter, and the distortion energy theory of failure to estimate the shear yield strength for the
aluminum block.
------------------------------------------------------------------------------------------------------------------------
Solution
Tensile failure load in the ½-20 UNF-2A steel cap screw is Ff = ( Su ) stl Ats . From Table 13.3, for SAE
Grade 2 steel, for ½-inch diameter ( Su ) stl = 74, 000 psi and from Table 13.1, for the ½-20 UNF steel cap
screw Ats = 0.1600 in2. Thus, Ff = (74,000)(0.16) = 11,840 lb. Basing housing thread shear (stripping) on
direct shear of aluminum threads at the major thread diameter,
⎛ 1 ⎞
Athd = π dpne Lh = π ( 0.50 ) ⎜ ⎟ ( 20 ) Lh = 1.57 Lh
⎝ 20 ⎠
From Table 3.3, for 356.0 cast aluminum, (Syp)356.0 = 27,000 psi, and
Ff 11,840 7541
τ thd = τ yz = = =
Athd 1.57 Lh Lh
σ e = 0 + 3τ yz2 = S yp
1
τ thd = S yp = 0.577 S yp = 0.577 ( 27, 000 ) = 15,588 psi
3
7541
= 15,588
Lh
Lh = 0.50 in.
Thus, the minimum length of thread engagement in the aluminum housing should be 0.50 inches. To
insure tensile failure of the cap screw before stripping the aluminum threads, introduce a safety factor of
say 1.5 and recommend
450
13-13. A support arm is to be attached to a rigid column using two bolts located as shown in Figure
P13.13. The bolt at A is to have an MS 20 × 2.5 thread specification and the bolt at B is to have an
MS 10 × 1.5 specification. It is desired to use the same material for both bolts, and the probable governing
failure mode is yielding. No significant preload is induced in the bolts as a result of the tightening process,
and it may be assumed that friction between the arm and the column does not contribute to supporting the
18 kN load. If a design safety factor of 1.8 has been selected, what minimum tensile yield strength is
required for the bolt material?
------------------------------------------------------------------------------------------------------------------------
Solution
For this joint configuration we have a direct shear and a shear due to torsion. The shear stress in each bolt
will be defined by the vector sum of the components, defined as
P (e)(ri ) P
(τ i )T = (τ i )P = 2
↓
Jj
∑ Ai
i =1
Using bolt B as the origin, the centroid of the bolt pattern is located at
100 AA
x =0 y=
AA + AB
π π
where AA = (20) 2 = 314.2 mm 2 AB = (10) 2 = 78.5 mm 2
4 4
100(314.2)
y= = 80 mm
314.2 + 78.5
Therefore the radii from the bolt pattern c.g. to each bolt is rA = 30 mm and rB = 80 mm , which results in
(τ A )T =
(18 ×10 ) (0.375)(0.03) = 257.9 MPa
3
785.2 × 10−9
(τ A )P =
(18 ×10 ) 3
= 45.8 MPa
( 314.2 + 78.5) ×10−6
(τ B )T =
(18 ×10 ) (0.375)(0.08) = 687.7 MPa
3
785.2 × 10−9
(τ B )P =
(18 ×10 ) 3
= 45.8 MPa
( 314.2 + 78.5) ×10−6
The total shear stress at each bolt is
451
Since bolt B sees the largest shear stress we use
τ 1240
( S yp )req = 0.577
yield
=
0.577
≈ 2150 MPa
From Table 3.3 we see that noting meets our needs. Therefore the joint needs to be redesigned.
452
13-14. A steel side plate is to be bolted to a vertical steel column as shown in Figure P13.14, using ¾-10
UNC SAE Grade 8 steel bolts.
a. Determine and clearly indicate the magnitude and direction of the direct shearing stress for the
most critically loaded bolt.
b. Determine and clearly indicate the magnitude and direction of the torsion-like shearing stress for
the most critically loaded bolt.
c. Determine the existing safety factor on yielding for the most critically loaded bolt, assuming that
no significant preload has been induced in the bolt due to tightening.
------------------------------------------------------------------------------------------------------------------------
Solution
For the joint configuration of Figure P13-14, both direct shear and torsion-like shear must be considered.
For direct shear, using (13-31)
P
τb = 3
∑A
i =1
i
π ( 0.75 )
2
Ab = = 0.442 in 2
4
10, 000
τb = = 7541 psi ↓ (vertically down)
3 ( 0.442 )
For torsion like shear, (13-31) and (13-32) must first be used to find the c.g. of the joint. Using A-B as a
reference,
5 ( 0.44 )
y= = 1.67 in.
3 ( 0.44 )
B y symmetry
x = y = 1.67 in.
453
By symmetry
and
so
(
J j = ∑ Ai ri 2 = 0.44 2.362 + 2 ( 3.73)
2
) = 14.69 in 4
We see that (τ t ) A = (τ t )C > (τ t ) B so bolts A and C are equally critical, and both are more critical than B.
From (13-32),
454
13-15. A 1020 hot-rolled steel cantilever support plate is to be bolted to a stiff steel column using four
M16 x 2 bolts of Property Class 4.6, positioned as shown in Figure P13.15. For the 16-kN static
load and the dimensions given, and assuming that none of the load is supported by friction, do the
following:
------------------------------------------------------------------------------------------------------------------------
Solution
(a) For the joint configuration of Figure P13-15, both direct shear and torsion-like shear must be
considered. For direct shear, assuming the shear force F to be equally distributed over the 4 bolts,
Fb = 16/4 = 4 kN ↓ (vertically down). For torsion-like shear, (13-31) and (13-32) must first be
used to find the c.g. of the joint. Since the bolts are all the same, and the bolt pattern is
symmetrical, the c.g. lies at the geometrical center of the bolt pattern as shown here.
By symmetry,
rA = rB = rC = rD = r and
r = 602 + 752 = 96 mm
Peri 16 ( 75 + 50 + 300 ) ri N
Ft = = = 184.5ri
4 ( 96 )
2 2
4r mm
455
The resultant shear force at A and B are given by
(F ) t
x
A
= 184.5 ( 60 ) = 11.1 kN
(F ) t
y
A
= 184.5 ( 75 ) = 13.8 kN
FA = FB = 11.12 + ( 4 + 13.8 ) = 21 kN
2
(F ) t
x
C
= 184.5 ( 60 ) = 11.1 kN
(F ) t
y
D
= 184.5 ( 75 ) = 13.8 kN
FA 21, 000
τ max = = = 104 MPa (at locations A and B)
Ab π ( 0.016 )2
4
FA 21, 000
σ max = = = 87.5 MPa
Abrg ( 0.015 )( 0.016 )
(d) Assuming the maximum bending moment occurs at the plate cross section through holes A and B
12
( I NA )hole = 5.12 × 10−9 + 8.64 × 10−7 = 8.69 × 10−7 m 4
( 0.015 )( 0.200 )
3
456
M max c 5600 ( 0.100 )
σ max = = = 67.8 MPa
I 8.26 × 10−6
(e) To check for yielding, the following data may be extracted from Table 3.3 and 13.5;
(S )
yp 1020 HR = 30, 000 psi (207 MPa)
(S )
yp Class 4.6 = 240 MPa
Comparing the various stresses calculated with these strength values, no yielding would be expected.
457
13-16. For the eccentrically loaded riveted joint shown in Figure P13.16, do the following:
------------------------------------------------------------------------------------------------------------------------
Solution
(a) Using (13-33) and (13-34), and assuming that all the rivets are the same size, and taking a
reference line through rivets 4-5-6 and 3-4 we find
3 Ar ( 2 + 2 )
x= = 2.0 in.
6 Ar
2 Ar ( 6 ) + 2 Ar ( 6 + 3)
y= = 5.0 in.
6 Ar
(b) For the joint configuration of Figure P13-16, both direct shear and torsion-like shear must be
considered. For direct shear, assuming that the shear force F to be equally distributed over the 6
rivets,
10, 000
Fr = = 1667 lb ↓ (vertically down)
6
For the torsion-like shear, the radii from the c.g. for the 6 rivets is given as
(1 + 3)
2
r1 = r6 = + 22 = 4.472 in.
(1)
2
r2 = r5 = + 22 = 2.236 in
( 5)
2
r3 = r4 = + 22 = 5.385 in
The force taken by each rivet depends upon its distance ri from the c.g. Hence,
F1 F2 F
= =L = i
r1 r2 ri
Thus,
Mri
Fti =
r12 + r22 + r32 + r42 + r52 + r62
458
The torsion-like forces can be obtained in the x and y directions by using the appropriate x or y value for ri.
Hence, we find
4 F3 4 ( 3476 )
τs = = = 7868 psi
π Dr (1) π ( 0.75 )2
2
F3 3476
σc = = = 14,830 psi
π Dr N r ⎛ 5 ⎞
⎜ ⎟ ( 0.75 )(1)
⎝ 16 ⎠
459
13-17. For a bracket riveted to a large steel girder, as sketched in Figure P13.17, perform a complete
stress analysis of the riveted joint. The yield stresses are S yp = 276 MPa for the plate and S yp = 345 MPa
for the rivets. Assume the rivet centerline is 1.5 times the rivet diameter away from the edge of the plate
and protruding head rivets are used. The plate is 6 mm thick and the girder is much thicker. Determine the
existing factors of safety on yielding for each of the potential types of failure for the riveted joint, except
edge shear-out and edge tearing.
------------------------------------------------------------------------------------------------------------------------------
Solution
(τ1 )T =
(90 ×10 ) r = ( 90 ×10 ) (0.225)
3
i
3
Jj Jj
(τ 5 )T =
(90 ×10 ) (0.20)
3
Jj
where
(τ1 )T =
(90 ×10 ) (0.225) = 493 MPa ↑
3
(τ 5 )T =
( 90 ×10 ) (0.20) = 438 MPa ↓
3
(τ i )P =
( 90 ×10 ) / 5 = 4 ( 90 ×10 ) = 57.3 MPa ↓
3 3
π (0.02) 2 / 4 5π (0.02) 2
460
τ1 = 493 ↑ +57.3 ↓ = 435.7 MPa ↑ τ 5 = 438 ↓ +57.3 ↓= 495.3 MPa ↓
With the largest shear stress having been defined, we no assess failure modes.
Plate tensile failure: No hole diameter was given, so we arbitrarily assume a diameter of Dh = 22 mm
Fs 90 000
σt = = = 40 MPa
( b − N r Dh ) t ⎡⎣( 0.075 + 0.20 + 0.075 + 0.075 + 2(1.5)(0.02) ) − 5(0.022) ⎤⎦ (0.006)
276
ne = = 6.9
40
Rivet shear stress: The maximum rivet shear stress has been determined to be τ max = τ 5 = 495.3 MPa ↓
0.577(345)
ne = ≈ 0.4
495.3
Bearing failure between rivet and plate: The maximum rivet shear stress has been determined to be
τ max = τ 5 = 495.3 MPa ↓ . Since each rives experiences a different shear stress, the bearing stress at each
will be different. The shear force supported by this rivet is therefore
π (0.02) 2
Fs −5 = Arτ max = (495.3) ≈ 156 kN
4
Fs −5 156 000
σc = = ≈ 1300 MPa
tDr 0.006(0.02)
276
ne = ≈ 0.21
1300
461
13-18. A simple butt-welded strap, similar to the one shown in Figure 13.20, is limited by surrounding
structure to a width of 4 inches. The material of the strap is annealed AISI 1020 steel (see Table
3.3), and an E 6012 welding electrode has been recommended for this application. The applied
load P fluctuates from a minimum of 0 to a maximum of 25,00 lb and back, continuously.
a. If a safety factor of 2.25 has been selected, k∞ is approximately 0.8 [see (5-57)], and infinite life is
desired, what thickness should be specified for the butt-welded strap?
b. If any fatigue failures do occur when these welded straps are placed in service, at what location
would you expect to see the fatigue cracks initiating?
------------------------------------------------------------------------------------------------------------------------
Solution
From Table 3.3, for AISI 1020 steel (annealed); Su = 57,000 psi, Syp = 43,000 psi and from Table 13.13, for
E6012 electrodes; Su = 62,000 psi, Syp = 50,000 psi. From Table 13.9, for HAZ of reinforced butt weld Kf
= 1.2.
⎛ P ⎞
σ = K f σ nom = K f ⎜ ⎟
⎝ tLw ⎠
25, 000 8571
σ = 1.2 =
t ( 4 − 0.5 ) t
(Note that 0.5 inch has been deducted from Lw to account for unsound weld at its ends) and
σ min = 0
8571
+0
4285
σm = t =
2 t
8571
−0
4285
σa = t =
2 t
4285
σa t
σ eq −CR = =
σm ⎛ 4285 ⎞
1− 1 − ⎜⎜ ⎟⎟
⎝ t ( 57, 000 ) ⎠
Su
4285
t= + 0.075
σ eq −CR
From Figure (5.31), for 1020 steel, S’f = 33,000 psi, so from (5-55) Sf = 0.8(33,000) = 26,400 psi and the
design stress is
Sf 26, 400
σd = = = 11, 733 psi
nd 2.25
thus
4285
t= + 0.075 = 0.44 in.
11, 733
462
(b) If fatigue cracks occur, they would be expected to initiate in the HAZ.
463
13-19. A horizontal side plate made of 1020 steel (see Figure 5.31) is to be welded to a stiff steel
column using E 6012 electrode, as specified in Figure P13.19. If the horizontally applied load F
fluctuates cyclically from + 18 kN (tension) to -18 kN (compression) each cycle, k∞ is
approximately 0.75 [see (5-57)], and a design safety factor of 2.5 is desired, what fillet weld size
would you recommend if all fillet welds are to be the same size? Infinite life is desired.
------------------------------------------------------------------------------------------------------------------------
Solution
From (13-48)
F
(τ w )nom =
0.707 sLw
18, 000
(τ w )max = K f (τ w )nom = ( 2.7 )
0.707 s ( 0.060 + 0.060 + 0.050 )
4.055 × 106
(τ w )max = Pa
s
−4.055 × 106
(τ w )min = Pa
s
(τ w )m = 0
4.055 × 106
(τ w )a = Pa
s
The design stress, using the DET is τd = τf/nd = 0.577 Sf/nd . From Figure 5.31, for 1020 steel, S’f = 33,000
psi = 228 MPa. Thus,
So a fillet weld size of 10 mm is recommended. This is compatible with the 10 mm plate thickness.
464
13-20. A steel side plate is to be welded to a vertical steel column according to the specifications in
Figure P13.20. Neglecting stress concentrations effects, calculate the magnitude and clearly
indicate the direction of the resultant shearing stress at the critical point. In selecting the critical
point be sure to consider effects of both torsion-like shear and direct shear.
------------------------------------------------------------------------------------------------------------------------
Solution
Au = 5 + 5 = 10 in.
A = tAu = 0.707 sAu = 0.707 ( 0.375 )(10 ) = 2.65 in 2
To find the weld c.g., using the lines AB and BC as reference lines, yields
x=
∑A ui xi
=
5 ( 2.5 )
= 1.25 in.
Au 10
y=
∑A ui yi
=
5 ( 2.5 )
= 1.25 in.
Au 10
The direct stress is assumed uniform over both welds, and is vertically downward, thus
P 5000
τw = = = 1886 psi
Aw 0.265
For the secondary stress we need to calculate the polar moment of inertia of the weld joint. We can again
treat the weld as unit welds and we see that
J u = I xu + I yu
53
+ 5 (1.25 ) + 5 (1.25 ) = 26.04 in 3
2 2
I xu =
12
I yu = I xu = 26.04 in 3
J u = 2 ( 26.04 ) = 52.08 in 3
J = tJ u = 0.707 ( 0.375 ) 52.08 = 13.8 in 4
465
It is deduced by examining critical points A, B, and C that rB << rA = rC so that B is eliminated as a potential
critical point, and although rA = rC the torsion-like shear opposes the direct shear at A and adds to it at C, so
C is the most critical.
The x and y components of the torsion-like shear stress can be obtained from
Tri
τ tc =
J
⎛ 4265 ⎞
ϕ = tan −1 ⎜ ⎟ = 30.9
o
⎝ 7133 ⎠
466
13-21. A proposed double lap-joint [see Figure 13.1(f)] is to be symmetrically loaded in tension, parallel
to the plane of the straps to be joined. Adhesive bonding is being considered as a means of joining the
straps. The single center strap is titanium and the double outer straps are medium-carbon steel. This
aerospace application involves continuous operation at a temperature of about 350˚ F, moderate impact
loading, and occasional exposure to moisture. What types of structural adhesives would you recommend as
good candidates for bonding this double lap joint?
------------------------------------------------------------------------------------------------------------------------
Solution
Based on Table 13.14, structural adhesives that provide an acceptable response to the attributes required by
problem specifications are:
Based on these data, rule out epoxies for poor impact resistance and choose silicones over Hot Melts and
Anaerobics since Silicones are equal or better for impact resistance, moisture resistance, and applications
involving dissimilar metals.
467
13-22. In an adhesively bonded lap joint (see Figure 13.24) made of two sheets of metal, each having
thickness t, it has been found from experimental testing program that the maximum shearing stress in the
adhesive may be estimated as
2P
(τ max )adh =
bLL
------------------------------------------------------------------------------------------------------------------------
Solution
For adhesive shear failure and metal sheet yielding failure to be equally likely Pf-adh = Pf-metal or using our
given equation
(τ nax )adh ( b )( LL )
= S yp ( b )( t )
2
2S yp t
LL =
(τ nax )adh
468
13-23. It is being proposed to use a lap joint configuration to adhesively bond two 0.9 mm thick strips of
2024-T3 aluminum using an epoxy adhesive. Assuming a stress distribution factor of K s = 2 , what bond
overlap length would you recommend
----------------------------------------------------------------------------------------------------------------------
Solution
K s S yp t
LL =
(τ max )adh
From Table 3.3, S yp = 345 MPa , and form Table 14.16 (τ max )adh = 15 MPa so
2(345)(0.0009)
LL = = 0.0414 m LL = 41.4 mm
15
469
Chapter 14
14-1. You are asked as a consultant, to determine a procedure for finding a “best estimate” for the design stress to
be used in designing the helical-coil springs for a new off-the-road vehicle. The only known information is:
1. The spring material is a ductile high-strength ferrous alloy with known ultimate strength, Su, and known
yield strength, Syp.
2. Spring deflection during field operation is estimated to range from a maximum of ymax to a minimum of ymin
= 0.30ymax.
Based on the known information, write a concise step-by-step procedure for determining a “best estimate” value for
the design stress.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
To estimate the design stress based on problem specification, the following observations and actions would be
pertinent:
1. Since the spring deflection ranges from ymax to 0.30ymax, this application involves fluctuating loads;
consequently the governing failure mode is probably fatigue.
2. Since no fatigue properties are given in the specifications, it will be necessary to estimate fatigue properties
from the static properties given, as discussed in 5.6. This will result in a uniaxial S-N curve for the spring
material, including appropriate strength influencing factors ( see (5-55) and (5-57)) since very long life
(infinite life) is specified.
3. From the S-N curve constructed in step 2, read the fatigue limit S’f.
4. Since stress σ is proportional to deflection δ, and the specified deflection ranges from ymax to 0.30ymax
cyclically, this is a case of non-zero mean cyclic stress. Therefore, the modified Goodman relationship (see
(5-70)) will be required to find a zero-mean fatigue strength, Smax-N, that accounts for the non-zero mean
state of cyclic stress.
5. Since Smax-N is uniaxial, the primary stress τ due to torsion in a helical coil spring is multiaxial (shear stress
is always uniaxial), and the material is ductile, the distortion energy theory will be suggested to relate τf to
Smax-N, that is, τf = 0.577 Smax-N.
6. A design safety factor (see 2.13), nd, appropriate to the application, must be selected.
7. The design stress may be calculated by dividing τf by nd.
470
14-2. An open-coil helical coil compression spring has a spring rate of 80 lb/in. When loaded by an axial
compressive force of 30 lb, its length was measured to be 0.75 inch. Its solid height has been measured as 0.625
inch.
a. Calculate the axial force required to compress the spring from its free length to its solid height.
b. Calculate the free length of the spring.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
∆F Fs − Fi
k= =
∆y Li − Ls
Fs = k ( Li − Ls ) + Fi = 80 ( 0.75 − 0.625 ) + 30 = 40 lb
Fi 30
b. L f = Li + = 0.75 + = 1.125 in.
k 80
471
14-3. An open-coil helical-coil compression spring has a free length of 76.2 mm. When loaded by an axial
compressive force of 100 N, its length is measured as 50 .8 mm.
(a) Calculate the spring rate of this spring.
(b) If this spring, with a free length of 76.2 mm, were loaded by an axial tensile force of 100 N, what would
you predict its corresponding length to be?
------------------------------------------------------------------------------------------------------------------------------------------
Solution
∆F Fi − 0 100 N
k= = = = 3.94
∆y yi − 0 ( 76.2 − 50.8 ) mm
Ft 100
(b) LFt = L f + = 76.2 + = 101.6 mm
k 3.94
472
14-4. A helical-coil compression spring has an outside diameter of 1.100 inches, a wire diameter of 00.085 inch,
and has closed and ground ends. The solid height of this spring has been measured as 0.563 inch.
(a) Calculate the inner coil radius.
(b) Calculate the spring index.
(c) Estimate the Wahl factor.
(d) Calculate the approximate total number of coils, end-of-wire to end-of-wire, in this spring.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
D0 − Di = 2d
Di D0 − 2d 1.100 − 2 ( 0.085 )
Ri = = = = 0.465 in.
2 2 2
⎛ d⎞ ⎛ 1.10 0.085 ⎞
2 ⎜ R0 − ⎟ 2⎜ − ⎟
2R
= ⎝
2⎠ ⎝ 2 2 ⎠
C= = = 11.94
d d 0.085
Kw = 1.12
Ls 0.563
Nt = = = 6.6 coils
d 0.085
473
14-5. An existing helical-coil compression spring has been wound from 3.50-mm peened music wire into a spring
having an outside diameter of 22 mm and 8 active coils. What maximum stress and deflection would you predict if
an axial static load of 27.5 N were applied?
------------------------------------------------------------------------------------------------------------------------------------------
Solution
We have that
2 R = D0 − d
D0 − d 22 − 3.50
R= = = 9.25 mm
2 2
2 R 2 ( 9.25 )
C= = = 5.3
d 3.5
K w = 1.29
From (14-120
⎛ 16 FR ⎞ ⎛ 16 ( 27.5 )( 0.00925 ) ⎞
τ max = K w ⎜ 3 ⎟
= 1.29 ⎜ ⎟ = 39.0 MPa
⎝ πd ⎠ ⎜ π ( 0.0035 )
3
⎟
⎝ ⎠
64 FR 3 N 64 ( 27.5 )( 0.00925 ) ( 8 )
3
474
14-6. An existing helical-coil compression spring has been wound from unpeened music wire of 0.105-inch
diameter into a spring with mean coil radius of 0.40 inch. The applied axial load fluctuates continuously from zero
to 25 lb, and a design life of 107 cycles is desired. Determine the existing safety factor for this spring as used in this
application.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
From (14-1), and data for music wire from Table 14.1
From Table 14.8, for steel alloys under released loading, and for 107 cycles
(τ )
f 107
= 0.38
Sut
(τ )
f 107
= 0.38 ( 266, 250 ) = 101,175 psi
From (14-12)
⎛ 16 FR ⎞
τ max = K w ⎜ 3 ⎟
⎝ πd ⎠
2 R 2 ( 0.4 )
C= = = 7.62
d 0.105
K w ≈ 1.19 (Table 14.5)
⎛ 16 ( 25 )( 0.4 ) ⎞
τ max = 1.19 ⎜ ⎟ = 52,350 psi
⎜ π ( 0.105 )3 ⎟
⎝ ⎠
(τ )
f 107 101,175
nex = = = 1.93
τ max 52,350
475
14-7. A round wire helical-coil compression spring with closed and ground ends must work inside a 60-mm-
diameter hole. During operation the spring is subjected to a cyclic axial load that ranges between a minimum of 650
N and a maximum of 2400 N. The spring rate is to be approximately 26 kN/m. A life of 2 × 105 cycles is required.
Initially, assume k N = 2×105 = 0.85 . A design factor of safety of nd = 1.2 is desired. Design the spring.
----------------------------------------------------------------------------------------------------------------------------------------
Solution
1.10 Do = 1.10(2 R + d ) ≤ Dh = 60
Do = 2 R + d ≈ 54 mm
σd =
S N = 2×105
=
( S 'N =2×10 )( k N =2×10 ) = ( S 'N =2×10 ) ( 0.85) = 0.71S '
5 5 5
N = 2×105
nd nd 1.2
476
Since shear is a multiaxial state of stress, the distortional energy theory gives
54 − 6
R= = 24 mm
2
Refining our approximations form the S-N curve we estimate S 'N = 2×105 ≈ 810 MPa , so
⎛ 16 FR ⎞ ⎛ 16(2400)(0.024) ⎞
τ max = KW ⎜ 3 ⎟
= 1.18 ⎜⎜ 3 ⎟⎟ = 1603 MPa
⎝ πd ⎠ ⎝ π (0.006) ⎠
Noting that Sut = 2153.5(6) −0.1625 ≈ 1610 MPa and form Table 15-12 τ yp = 0.4(1610) = 644 MPa < τ max . Therefore
we need to change dimensions of the spring in order to lower τ max by a factor of about 1603 / 644 ≈ 2.5 . This can be
3
accomplished by increasing the wire size to approximately 2.5(6) = 8.14 mm . Taking a conservative approach ,
we assume d = 10 mm (a standard size), which results in
54 − 10
R= = 22 mm
2
( )
For a 10 mm diameter wire, the yield stress in shear becomes τ yp = 0.4 2153.5(10) −0.1625 ≈ 593 MPa . In addition,
a 10 mm diameter wire results in c = 44 /10 = 4.4 , which is within the range of 4 to 12. Using this value of c, we
find KW ≈ 1.36 , meaning τ max becomes
⎛ 16(2400)(0.022) ⎞
τ max = 1.36 ⎜⎜ ⎟⎟ ≈ 366 MPa < τ yp = 593 MPa
⎝ π (0.01)3 ⎠
⎛ 16(650)(0.022) ⎞
τ min = 1.36 ⎜⎜ 3 ⎟⎟ ≈ 99 MPa
⎝ π (0.01) ⎠
τ max + τ min 366 + 99
τm = = ≈ 233 MPa
2 2
τ −τ 366 − 99
τ a = max min = ≈ 134 MPa
2 2
τa
τ eq −CR =
1−τ m /τu
( )
τ u = 0.577 Su = 0.577 2153.5(10) −0.1625 ≈ 855 MPa
477
134
τ eq −CR = = 184 MPa
1 − 233 / 855
From a refined assessment of the S-N curve, we determine that for a wire diameter of 10 mm,
S 'N = 2×105 ≈ 790 MPa . This results in τ d = 0.41( 790 ) =324 MPa < τ eq −CR = 184 MPa , which is acceptable.
Next we use the spring rate k = 26 kN/m and rewrite (15-22) to get
N=
d 4G
=
(
(0.01) 4 79 × 109 ) = 44.6 active coils
3
64 R k (
64(0.022) 26 × 103
3
)
Selecting closed and ground ends, we add one active coil to each end, meaning Nt = 46.6 . The approximate solid
height of the spring will be
Ls = N t d = 46.6(10) = 466 mm
yclash = 6.73 mm
Checking for buckling, we have L f / 2 R = 565 / 44 = 12.8 . For closed and ground ends (fixed ends) ycr / L f ≈ 0.16 .
Thus
Since yop − max = 92.3 mm and ycr = 90.4 mm , the spring may tend to buckle. It is, however supported within a hole,
so unless other factors arise because of buckling, the final design specs will be
478
14-8. A helical-coil spring with plain ends, ground, is to be used as a return spring on the cam-driven valve
mechanism shown in Figure P14.8. The 1.50-inch-diameter rod must pass freely through the spring. The cam
eccentricity is 0.75 inch (i.e., the total stroke is 1.50 inches). The height of the compressed spring when the cam is
at the head-end-dead-center (HEDC) position is 3.0 inches, as shown. The spring must exert a force of 300 lb when
at the HEDC position shown in the sketch, and must exert a force of 150 lb when the crank-end-dead-center (CEDC)
at the bottom of the stroke. That is, the spring is preloaded into the machine. The spring is to be made of a patented
spring steel wire that has a 200,000-psi ultimate tensile strength, 190,000-psi tension yield strength, and 90,000-psi
fatigue endurance limit. A safety factor of 1.25 is desired, based on infinite life design. Determine the following:
--------------------------------------------------------------------------------------------------------------------------------------
Solution
Following the procedure of Section 14.6, and adopting the 10 % diametral clearance suggest in guideline 12A
1.10 ( Drod ) = Di = ( 2 R − d )
1.10 (1.50 ) = 1.65 = 2 R − d
2 R = 8d
2 R = d + 1.65
d + 1.65 = 8d
1.65
d= = 0.236 in.
7
From Table 14.2 initially select a standard wire size d = 0.250 in., then
1.65 + 0.25
R= = 0.95 in.
2
2 ( 0.95 )
C= = 7.6
0.25
K w = 1.20 (From Table14.5)
⎛ 16 ( 250 )( 0.95 ) ⎞
τ max = 1.20 ⎜ ⎟ = 92,900 psi
⎜ π ( 0.25 )3 ⎟
⎝ ⎠
We have that Syp = 190,000 psi, so based on the distortion energy theory τyp = 0.577(190,000) = 109,630 psi. We
note therefore, that yielding does not occur at maximum load. Thus, for minimum load we have
⎛ 16 ( 200 )( 0.95 ) ⎞
τ min = 1.20 ⎜ ⎟ = 74,316 psi
⎜ π ( 0.25 )3 ⎟
⎝ ⎠
92,900 − 74,316
τa = = 9, 292 psi
2
92,900 + 74,316
τm = = 83, 608 psi
2
The equivalent completely reversed shear stress may be estimated by adapting (5-72)
479
τa 9, 292
τ eq −CR = = = 33, 730 psi
τm 83, 608
1− 1−
τu 0.577 ( 200, 000 )
The fatigue endurance limit is 90,000 psi, using the distortion energy theory τf = 0.577(90,000) = 51,930 psi. The
factor of safety is given as 1.25, hence
τf 51,930
τd = = = 41,544 psi
nd 1.25
Comparing the shear stresses we see that the stress is acceptable for infinite life design.
For plain ends ground, From Figure 14.9, add 0.5 inactive coils to each end, so that the total number of coils is
From guideline 12.E, an appropriate clash allowance would be yclash = 0.10 yop and from Figure 14.5
Fmax − Fmin
yop = yop − max − yi =
k
250 − 200
= = 1.50 inches
33.3
It is noted that yop can also be determined from the specified eccentricity of the cam as yop = 2(0.75) = 1.50 inch. We
find the clash as yclash = 0.10(1.50) 0.15 inch. Thus the free length of the spring should be
480
14-9. A proposed helical-coil compression spring is to be wound from standard unpeened music wire of 0.038-inch
diameter, into a spring with outer coil diameter of 7 16 inch and 12 1 total turns from end of wire to end of wire.
2
The ends are to be closed. Do the following:
--------------------------------------------------------------------------------------------------------------------------------------
Solution
Sut = Bd a
B = 184, 600 psi
a = −0.1625
Sut = 184, 600 ( 0.038 )
−0.1625
≈ 314, 000 psi
τ yp
= 0.4 (Table 14.7)
Sut
τ yp = 0.4 ( 314, 000 ) = 125, 600 psi
⎛ 16 FR ⎞
τ max = K w ⎜ 3 ⎟
⎝ πd ⎠
2R
C=
d
D d 0.4375 0.038
R= − = − = 0.20 in.
2 2 2 2
2 ( 0.20 )
C= = 10.5
0.038
K w = 1.14 (Table 14.5)
481
d 4G
k=
64 R 3 N
N t = 12.5
N i = 2(1.0) = 2.0 (closed ends)
N = N t − N i = 12.5 − 2.0 = 10.5 coils
( 0.038 ) (11.5 × 106 )
4
k= = 4.46 lb/in
64 ( 0.2 ) (10.5 )
3
Fmax 5.94
ymax = = = 1.33 in.
k 4.46
(L )
f max
= Ls + yclash + ymax = 0.475 + 0.10 (1.33) + 1.33 = 1.94 in
(h) For buckling calculate the slenderness ratio and the deflection ratio
Lf 1.94
SR = = = 4.85
2R 2 ( 0.2 )
ymax 1.33
DR = = = 0.685
Lf 1.94
From Figure 14.11, it may be observed that buckling is certainly a potential problem, depending on the details of the
end supports. An experimental investigation would be recommended.
482
14-10. A helical-coil compression spring is to be designed for a special application in which the spring is to be
initially assembled in the mechanism with a preload of 10 kN, and exert a force of 50 N when it is compressed and
additional 140 mm. Tentatively, it has been decided to use music wire, to use closed ends, and to use the smallest
standard wire diameter that will give a satisfactory performance. Also, it is desired to provide a clash allowance of
approximately 10 percent of the maximum operating deflection.
a. Find a standard wire diameter and corresponding mean coil radius that will meet the desired specifications.
b. Find the solid height of the spring.
c. Find the free height of the spring.
--------------------------------------------------------------------------------------------------------------------------------------
Solution
F 50 − 10
kreq ' d = = = 285.7 N / m
y 140 × 10−3
Sut = Bd a
B = 2153.5 ( For music wire, d = mm)
a = −0.1625
Sut = 2153.5d −0.1625
τ yp
= 0.4 (Table 14.7)
Sut
τ yp = 0.4 ( 2153.5 ) d −0.1625 = 861.4d −0.1625 MPa
⎛ 16 Fmax R ⎞
τ max = K w ⎜ ⎟
⎝ πd ⎠
3
⎛d ⎞ 8
R = C ⎜ ⎟ = d = 4d
⎝2⎠ 2
K w = 1.18 for C = 8
τ max ⎜
= 1.18 ⎜
(
⎛ 16 50 ⎡ 4d (10−3 ) ⎤
⎣ ⎦ ) ⎟⎞
⎟
⎜ π ⎡ d (10 ) ⎤
3
−3
⎟
⎝ ⎣ ⎦ ⎠
1201.9
τ max = MPa
d2
1201.9
= 861.4d −0.1625
d2
1201.9
d ( 2 − 0.1625) = = 1.40
861.4
1
d = (1.40 )1.8375 = 1.2 mm
From Table 14.2 we have d = 1.20 mm as a standard wire size, so tentatively select this wire size. Thus, R = 4(1.20)
= 4.80 mm
483
(b) From (14-22)
d 4G
k= = 285.7
64 R 3 N
N=
(1.20 ×10 ) ( 79.3 ×10 ) = 81.3 turns
−3 4 9
64 ( 4.8 × 10 ) ( 285.7 )
−3 3
Based on experience this is probably too many coils. If you have no experience check on buckling potential. (Figure
14.11). One way to reduce the number of coils is to make R larger. To do this, we should cycle all the way back to
the calculation of the maximum shear stress and repeat the calculation with a new value of R, i.e., a new (larger)
spring index C. The largest practical value of spring index is C = 12.
⎛ d ⎞ 12
R = C ⎜ ⎟ = d = 6d
⎝2⎠ 2
K w = 1.12 for C = 12
τ max = 1.12 ⎜ ⎜ (
⎛ 16 50 ⎡6d (10−3 ) ⎤
⎣ ⎦ ) ⎟⎞
⎟
⎜ π ⎡ d (10 ) ⎤
3
−3
⎟
⎝ ⎣ ⎦ ⎠
1711.2
τ max = MPa
d2
1711.2
= 861.4d −0.1625
d2
1711.2
d ( 2 − 0.1625) = = 1.99
861.4
1
d = (1.99 )1.8375 = 1.45 mm
From Table 14.2 we have d = 1.60 mm as the closest (larger) standard wire size, so tentatively select this wire size.
Thus, R = 6(1.60) = 9.60 mm
N=
(1.60 ×10 ) ( 79.3 ×10 ) = 32.1 turns
−3 4 9
64 ( 9.6 × 10 ) ( 285.7 )
−3 3
For closed ends Ni = 2(1.0) = 2.0 coils, thus, Nt = N + Ni = 32.1 + 2.0 = 34.1 coils. Thus,
Fmax 50
ymax = = = 0.175 m = 175 mm
k 285.7
yclash = 0.10(175) = 17.5 mm
L f = yop − max + yclash + Ls = 175 + 17.5 + 54.6 = 247.1 mm
484
14-11. Two steel helical-coil compression springs are to be nested about a common axial centerline. The outer
spring is to have an inside diameter of 38 mm, a standard wire diameter of 2.8 mm and 10 active coils. The inner
spring is to have an outside diameter of 32 mm , a standard wire diameter of 2.2 mm , and 13 active coils. Both
springs are to have the same free length. Do the following:
--------------------------------------------------------------------------------------------------------------------------------------
Solution
d o4 G di4G
a. For the outer spring ko = and for the inner spring ki =
64 Ro3 N o 64 Ri3 Ni
1 1
Ro = ( Di + di ) = ( 38 + 2.8) = 20.4 mm do = 2.8 mm N o = 10
2 2
1 1
Ri = ( Do − d0 ) = ( 32 − 2.2 ) = 14.9 mm di = 2.2 mm Ni = 13
2 2
ko =
(
(0.0028) 4 79 × 109 ) ≈ 894 N/m ki =
(
(0.0022) 4 79 × 109 ) ≈ 672 N/m
3 3
64(0.0204) (10) 64(0.0149) (13)
b. Since the springs are in parallel knest = ko + ki = 894 + 672 = 1566 N/m . Therefore
c. For the inner and out springs, co = Do / do = 40.4 / 2.8 = 14.6 , KW −o = 1.097 , ci = Di / di = 29.8 / 2.2 = 13.6 ,
and KW −i = 1.105 . Therefore
⎛ 16(39.15)(0.0202) ⎞
(τ max )o = 1.097 ⎜⎜ ⎟⎟ ≈ 201 MPa
⎝ π (0.0028)3 ⎠
⎛ 16(39.15)(0.0149) ⎞
(τ max )i = 1.105 ⎜⎜ ⎟⎟ ≈ 311 MPa
⎝ π (0.0022)3 ⎠
485
14-12. A round wire helical-coil tension spring has end loops of the type shown in Figures 14.7 (a) and (b). The
wire diameter of the spring is 0.042 inch, and the mean coil radius is 0.28 inch. Pertinent end-loop dimensions are
(ref. Figure 14.7), riA = 0.25 inch and riB = 0.094 inch. An applied axial static tension force of F = 5.0 lb is to be
applied to the spring.
(a) Estimate the maximum stress in the wire at critical point A.
(b) Estimate the maximum stress in the wire at critical point B.
(c) If the spring wire is ASTM A227 material, and a safety factor of 1.25 is desired, would the stresses at
critical point A and B be acceptable?
--------------------------------------------------------------------------------------------------------------------------------------
Solution
⎛ 32 FR ⎞ 4 F
σ max − A = kiA ⎜ 3 ⎟
+
⎝ πd ⎠ πd
2
rmA d 0.042
kiA = rmA = riA + = 0.25 + = 0.271 in.
riA 2 2
0.271
kiA = = 1.10
0.25
⎛ 32 ( 5.0 )( 0.28 ) ⎞ 4 ( 5.0 )
σ max − A = 1.10 ⎜ ⎟+
⎜ π ( 0.042 ) ⎟ π ( 0.042 )2
3
⎝ ⎠
= 211, 725 + 3609 = 215,334 psi
⎛ 16 FR ⎞
τ max − B = kiB ⎜ 3 ⎟
⎝ πd ⎠
rmB d 0.042
kiB = rmB = riB + = 0.094 + = 0.115 in.
riB 2 2
0.115
kiB = = 1.22
0.094
⎛ 16 ( 5.0 ) 0.28 ⎞
τ max − B = 1.22 ⎜ ⎟ = 117, 411 psi
⎜ π ( 0.042 )3 ⎟
⎝ ⎠
250, 000
nex − A = = 1.16
215, 234
This does not meet the safety factor specification of nd = 1.25, therefore, it is not acceptable at A. Checking point B,
using the distortion energy theory gives
This is close but still does not meet the specification of nd = 1.25, therefore, it is not acceptable.
486
14.13. A battery powered nail gun uses two helical coil compression springs to help propel the nail from the end of
the gun. When the springs are in the fully extended position as shown I Figure 14.13 (a), the 2 oz hammer and nail
are traveling at 90 ft/s. In this position, both springs are in their free length position (for the initial design, these are
assumed to be 2.0” for the upper spring and 2.5” for the lower spring). Due to mechanical advantage, the force
exerted on the hammer by the trigger is 40 lb and the hammer displaces the spring 1.0” when completely
compressed as shown in Figure 14.13 (b). Neither spring can have an outside diameter greater than 3/8”. Assuming
that each spring has a shear modulus G = 12 × 106 psi . Define an initial design the two springs, specifying for each,
the parameters: wire type, d, D, c, and N. Also, note any potential problems and suggest modifications that could be
used.
(a) (b)
Figure 14.13
Schematic of a battery powered nail gun.
--------------------------------------------------------------------------------------------------------------------------------------
Solution
Begin by defining the spring rate for each spring. Knowing the exit
velocity ( 90 ft/s = 1080 ft/s ) and weight, we can write
1 2 1 2 1 ⎛ 2 ⎞⎛ 1 ⎞ 2 1 2
mv = kx : ⎜ ⎟ (1080) = kx
2 2 2 ⎜⎝ 16 ⎟⎠ ⎝ 32.2(12) ⎠ 2
kx 2 = 377.33 lb-in
The springs are in parallel, so the spring rate shown above is a linear
combination of the spring rates for each spring ( k = k1 + k2 ). The total displacement of each spring is identical
( x = 1.0" ), therefore
487
At this point we have the option to have equal or different spring rates for each spring. The basic equation for spring
rate can be written in different forms, as shown below.
d 4G d 4G dG
k= 3
= 3
=
64 R N 8D N 8c3 N
The third form of this equation is perhaps the most useful for our purposes. We can write
d1G d 2G ⎛ d d ⎞ d d −6
k1 = and k2 = , so ⎜ 31 + 3 2
⎜ 8c N 8c N ⎟⎟ G = 377.33 , so 3 1 + 3 2 ≈ 252 × 10
8c13 N1 8c23 N 2 ⎝ 1 1 2 2 ⎠ c N
1 1 c N
2 2
On option is to assume each spring has the same spring rate, so we could write
d1 d2
3
= 3
= 126 × 10−6
c1 N1 c2 N 2
Another option is to assume the springs are identical except for the number of turns. Assuming
d = d1 = d 2 and c = c1 = c2 , which gives
⎛ d ⎞⎛ 1 1 ⎞ −6
⎜ 3 ⎟ ⎜ N + N ⎟ = 252 × 10
⎝ ⎠⎝ 1
c 2 ⎠
The latter expression will be used to define each spring. The spring index range of values is 4 ≤ c ≤ 12 , so we begin
by arbitrarily assuming c = 4 , which yields
⎛ d ⎞⎛ 1 1 ⎞ −6 ⎛ 1 1 ⎞
⎜ 3 ⎟ ⎜ N + N ⎟ = 252 × 10 ⇒ d⎜
N
+
N
⎟ = 0.0161
⎝ 4 ⎠⎝ 1 2 ⎠ ⎝ 1 2 ⎠
The outside diameter can’t be larger than 3/8”. Assuming that buckling may become a problem we select
Do = 0.375 . The mean diameter is D = Do − d = 0.375 − d . Having selected c = D / d = 4 , we determine
0.375 − d
= 4 ⇒ 0.375 = 5d ⇒ d = 0.075"
d
The closest standard wire diameter size to this is d = 0.076" . We tentatively have
d = 0.076" , D = 0.304" , c = 4
⎛ d ⎞⎛ 1 1 ⎞ ⎛ 0.076 ⎞ ⎛ 1 1 ⎞ −6 1 1
⎜ 3 ⎟ ⎜ N + N ⎟ = ⎜ 3 ⎟ ⎜ N + N ⎟ = 252 × 10 ⇒ +
N1 N 2
≈ 0.212
⎝ c ⎠⎝ 1 2 ⎠ ⎝ 4 ⎠⎝ 1 2 ⎠
Since spring 2 (the lower spring) is longer than spring 1, we assume that there are more turns in the lower spring
than in the upper spring. Arbitrarily assume N1 = 8 . This results in
1 1
+ = 0.212 ⇒ N 2 = 11.49 ≈ 12
8 N2
488
With c = 4 , K w = 1.404 . Each spring will experience a force of 20 lb, so the maximum shear stress will be
⎛ 16 FR ⎞ ⎛ 16(20)(0.152) ⎞
τ max = K w ⎜⎜ 3 ⎟
⎟ = 1.404 ⎜⎜ 3 ⎟ ⎟ = 49,519 ≈ 50 ksi
⎝ π (d ) ⎠ ⎝ π (0.076) ⎠
This is sufficiently small so that material selection should not be a problem. We initially assume that music wire will
be used, resulting in
−0.1625
Sut = 184.6 ( 0.076 ) = 280.6 ksi
Approximating τ yp = 0.4 Sut = 0.4 ( 280.6 ) = 112.2 ksi , we get an existing factor of safety of nex = 112.2 / 50 = 2.24 ,
which is adequate.
Buckling
Since the free length of each spring 1is L f = 2.0" , we determine from Figure 14.11 that for possible buckling
analysis we have L f / D ≈ 6.57 and if both ends are fixed ycr / L f ≈ 0.3 ⇒ ycr ≈ 0.6" . Since our spring
displacement is 1”, buckling is probably not a problem. The free length of spring 2 is greater than that of spring 1,
so buckling may be a problem with that spring too. The free lengths of the springs could be shortened so that L f / D
decreases and ycr / L f increases.
489
14-14 A conical compression spring is made from 3-mm-diameter steel wire and has an active coil diameter that
varies from 25 mm at the top to 50 mm at the bottom. The pitch (distance between coils) is p = 8 mm throughout.
There are four active coils. A force is applied to compress the spring and the stress always remains in the elastic
range.
a. Determine which coil (top or bottom, or one in the middle) deflects to zero pitch first as the force is
increased.
b. Determine the force corresponding to the deflection identified in part (a). In other words, determine the
force causing displacement of 8 mm.
---------------------------------------------------------------------------------------------------------------------------------------
F d 4G d 4G
Solution k= = =
δ 64 R 3 N 8 D3 N
a. The spring constant k is proportional to d 4 / D3 and d is constant. At the top of the spring
(d 4
/ D3 ) top
3
(
= 3 ( 3 / 25 ) = 0.005184 and at the bottom d 4 / D3 ) bottom
= 3 ( 3 / 50 ) = 0.000648 . This means that
3
490
14-15. A helical-coil compression is made form music wire has a rectangular cross section with dimensions a × b
as shown in Figure P14.15. Assume the maximum shear stresses due to torsion and transverse shear exist at the
same point on the rectangular cross section. Assume dimensions a and b are related by the relationship b = na ,
where 0.25 ≤ n ≤ 2.5 . Similarly, define a spring index for rectangular cross section springs as c = D / a .
a. Develop the expression for the maximum shear stress as a function of the applied load F, and the parameters
a, n, and c. Reduce the equations to its simplest form.
b. The parameter K in Table 4.4 represents the polar moment of inertia. Beginning with equation (14-19),
develop an expression for the stiffness of this spring in terms of the dimension a and c.
c. Assuming a mid-range spring index ( c = 8 ), the Whal factor for a circular wire spring is K w = 1.184 .
Assuming the cross-sectional area of a circular and rectangular spring are identical, we can show that
π d 2 = 4na 2 . Using this information, plot τ circ / τ rect vs n and kcirc / krect vs n for 0.25 ≤ n ≤ 2.5
Figure P14.15
Helical coil compression spring with rectangular cross
section.
------------------------------------------------------------------------------------------------------------------------------------
Solution
(a) A free body diagram of a spring coil shows that a torque and a transverse shear are
present. Using Table 4.3 for a rectangular section, the direct shear stress due to
transverse shear is
3 F 1.5 F 1.5 F
τ dir = = =
2 A ab na 2
The shear stress due to torsion is τ T = T / Q = FR / Q . Using Table 4.4 for a rectangular section and noting the
differences in dimensions between our problem and those given in Table 4.4, we determine
( )
2
8 ( a / 2) (b / 2)
2 2
0.5 ( ab )
2
0.5 na 2 0.5n 2 a3
Q= = = =
3 ( b / 2 ) + 1.8 ( a / 2 ) 1.5b + 0.9a 1.5na + 0.9a 1.5n + 0.9
F Fa F
τ= 2 3
⎡⎣(ca / 2) ( 3n + 1.8 ) + 1.5na ⎤⎦ = 2 3
⎡⎣(1.5nc + 0.9c ) + 1.5n ⎤⎦ = 2 2 [1.5n(c + 1) + 0.9c]
n a n a n a
491
F
τ= [1.5n(c + 1) + 0.9c ]
( na )2
(b) Given
⎡ 16 a/2⎛ (a / 2)4 ⎞⎤ ⎡
3 16 a ⎛ (a / 2) 4 ⎞⎤
K = J = (b / 2)(a / 2)3 ⎢ − 3.36 ⎜⎜ 1 − ⎟⎟ ⎥ = (na / 2)(a / 2) ⎢ − 3.36 ⎜⎜ 1 − ⎟⎟ ⎥
⎢⎣ 3 b / 2 ⎝ 12(b / 2) 4 ⎠ ⎥⎦ ⎢⎣ 3 an ⎝ 12(na / 2) 4 ⎠ ⎥⎦
na 4 ⎡16 3.36 ⎛ a4 ⎞ ⎤ na 4 ⎛ 1 ⎞ 4 ⎡n 0.0175 ⎤
K= ⎢ − ⎜⎜1 − = − 0.21a 4 ⎜1 − ⎟ = a ⎢ 3 − 0.21 +
16 n ⎝ 12(na ) 4 ⎟⎟ ⎥⎥ n4 n4 ⎦
⎥
⎣⎢ 3 ⎠⎦ 3 ⎝ 12 ⎠ ⎣
From (14-19)
FR 2 L F (ca / 2) ( 2π (ca / 2) N )
2
F (ca)3 π N Fc3π N
y= = = =
GJ ⎡n 0.0175 ⎤ 4 ⎡n 0.0175 ⎤ ⎡n 0.0175 ⎤
Ga 4 ⎢ − 0.21 + 4 ⎥ 4Ga ⎢ 3 − 0.21 + n 4 ⎥ 4Ga ⎢ 3 − 0.21 + n 4 ⎥
⎣ 3 n ⎦ ⎣ ⎦ ⎣ ⎦
k=
(
4Ga n5 − 0.63n 4 + 0.0525 )
3 4
3c n π N
16 F (4d ) 18.944 F F
(c) Since c = 8 = 2 R / d , τ circ = 1.184 = , τ rect = [1.5n(c + 1) + 0.9c ]
π d3 na 2 ( na )2
18.944 F
τ circ na 2 ⎛ 18.944 F ⎞⎛ (na ) 2 ⎞ 18.944n
= =⎜ ⎟ ⎜⎜ F 1.5n(c + 1) + 0.9c ⎟⎟ = 1.5n(c + 1) + 0.9c
τ rect F
1.5n(c + 1) + 0.9c ] ⎝ na
2
⎠⎝ [ ]⎠
2 [
( na )
⎡n 0.0175 ⎤
4Ga ⎢ − 0.21 + ⎡ 5 4 ⎤
n 4 ⎥⎦ 4Ga ⎣ n − 0.63n + 0.0525⎦
4
d G dG ⎣3
kcirc = 3
= krect = =
64 R N 8c 3 N c3π N 3n 4 c3π N
dG
kcirc
= 8 c3 N =
dG 3n 4 c3π N ( =
)
3d π n 4
krect
⎣ ⎦ ⎣⎣ (
4Ga ⎡ n5 − 0.63n 4 + 0.0525⎤ 8c3 N 4Ga ⎡ ⎡ n5 − 0.63n 4 + 0.0525⎤ ⎤
⎦⎦
32a n5 − 0.63n 4 + 0.0525 ) ( )
3n 4 c3π N
Using π d 2 = 4na 2 ⇒ d = 2a n / π
492
1.4
1.2
Tau_circ/Tau_rect
0.8
0.6
0.4
0.2
0
0 0.5 1 1.5 2 2.5
n
1.2
0.8
k_circ/k_rect
0.6
0.4
0.2
0
0 0.5 1 1.5 2 2.5
n
493
14-16. For the helical-coil tension spring of problem 14-12, calculate the maximum stress in the main body of the
spring, away from the ends, and identify where the critical point occurs. If the wire material is ASTM A227, would
failure of the spring wire be expected?
---------------------------------------------------------------------------------------------------------------------------------
Solution
Assuming that the tension spring does not have a built-in preload of more than 5.0 lb, the maximum stress in the
main body of the spring at the inner coil radius at mid-height of the wire, and would be given by (14-12) as
⎛ 16 FR ⎞
τ max = K w ⎜ 3 ⎟
⎝ πd ⎠
2 R 2 ( 0.28 )
C= = = 13.3
d 0.042
K w = 1.107 (Table 14.5)
⎛ 16 ( 5.0 )( 0.28 ) ⎞
τ max = 1.107 ⎜ ⎟ = 106,540 psi
⎜ π ( 0.042 )3 ⎟
⎝ ⎠
From Figure 14.4, for ASTM A227 material, for a wire diameter of 0.042 in, Sut = 245,000 psi. Thus, using the
distortion energy theory
494
14-17. A round wire helical-coil tension spring is to be used as a return spring on a cam-driven lever, as shown in
Figure P14.17. The spring must be pretensioned to exert a 45 N force at the “bottom” of the stroke, and should have
a spring rate of 3500 N/m. The peak-to-peak operating deflection for this spring is 25 mm. The spring is made form
a patented steel-alloy which has Sut = 1380 MPa , S yp = 1311 MPa , and a fatigue endurance limit of
S f = 621 MPa . It is desired to have a spring index of c = 8 , and a design factor of safety of nd = 1.5 . Design a
light weight spring for this application. Specifically determine the wired diameter (d), the mean coil radius (R) and
the number of active coils (N).
(USE EXISTING FIGURE P14.14 – Change numbering to P14.17 and change dimensions as shown)
---------------------------------------------------------------------------------------------------------------------------------------
Solution
Sut = 1380 MPa , S yp = 1311 MPa , S f = 621 MPa , c = 8 , KW = 1.18 (from Table 14.5), nd = 1.5 ,
yop = ymax − ymin = 25 mm , kreq = 3500 N/m , Ppreload = Pmin = 45 N
⎛ 16 Pmax R ⎞
τ max = KW ⎜ 3 ⎟
⎝ πd ⎠
Pmin 45
ymin = = = 0.01286 m = 12.86 mm
k 3500
ymax = ymin + 25 = 37.86 mm
Pmax = kymax = 3500(0.03786) = 132.5 N
⎛ 16(132.5)(4d ) ⎞ 3185
τ max = 1.18 ⎜ ⎟= 2
⎝ π d3 ⎠ d
τ fm 0.577 ( Smax − N )
τd = = = 0.385Smax − N
nd 1.5
From Chapter 5
Sf τ fm
Smax − N = for σ m ≥ 0 and = Smax − N ≤ S yp
1 − mt Rt nd
621
Smax − N = = 983 MPa < S yp = 1311 MPa
1 − 0.55(0.67)
Therefore τ d = 0.385(983) ≈ 379 MPa . This yields
3185
2
= 379 × 106 ⇒ d = 0.00289 m ≈ 3 mm
d
495
This is a nonstandard size. The closest standard size to this (from Table 14.2) is d = 3.5 mm . Therefore
c = 8 = 2R / d ⇒ R = 8(3.5) / 2 = 14 mm
d 4G d 4G (0.0035) 4 ( 79 × 109 )
k= ⇒ N= = = 16.5
64 R3 N 64 R3 k 64(0.014)3 (3500)
496
14--18. The round wire helical-coil tension spring shown in Figure P14.18 is to be used to make a return spring for
a pneumatically actuated lever that operates between fixed stops, as shown. The spring must be pretensioned to 25
lb at the bottom stop (minimum load point), and operates through a total spring deflection of 0.43 inch, where I is
halted by the upper stop, then returns to the lower stop and repeats the cycle. The spring is made of No.12 wire
(0.105- inch diameter., has a mean coil radius of 0.375 inch, and has been wound with 15 full coils plus a turned-up
half loop on each end for attachment. Material properties for the spring wire are Su = 200,000 psi, Syp = 185,000 psi,
e (2 inches) = 9 percent, Sf = 80,000 psi, E = 30 x 106 psi, G = 11.5 x 106 psi, and ν = 0.30. Compute the existing
safety factor for this spring, based on an infinite-life design criterion.
---------------------------------------------------------------------------------------------------------------------------------
Solution
τ fm (τ f ) N =∞
nex = =
τ max τ max
Sf
S max − N for σ m ≥ 0 and S max − N ≤ S yp
1 − mt Rt
where
k=
64 ( 0.375 ) N
3
k= = 27.25 lb/in
64 ( 0.375 ) (15.2 )
3
497
Pmax = 25 + 0.43 ( 27.25 ) = 36.7 lb
36.7 + 25
Pm = = 30.9 lb
2
30.9
Rt = = 0.84
36.7
Sf 80, 000
S max − N =
1 − mt Rt 1 − ( 0.60 )( 0.84 )
= 161,300 psi (< S yp = 185, 000 psi )
(τ )
f N =∞
= 0.577 (161,300 ) = 93,100 psi
2 R 2 ( 0.375 )
C= = = 7.14
d 0.105
K w = 1.21 ( Table 14.5 )
⎛ 16 ( 36.7 ) ( 0.375 ) ⎞
τ max = 1.21⎜ ⎟ = 73, 263 psi
⎜ π ( 0.105 )3 ⎟
⎝ ⎠
93,100
nex = = 1.27
73, 263
498
14--19. A round wire open-coil helical-coil spring is wound using No. 5 patented steel wire (d = 0.207 inch), with a
mean coil radius of 0.65 inch. The spring has 15 active coils, and its free height is 6.0 inches. The material
properties for the spring wire age given in Figure P14.19. The spring is to be used in an application where it is
axially deflected 1.0 inch from its free height into tension, then 1.0 inch from its free height into compression during
each cycle, at a frequency of 400 cycles per min.
---------------------------------------------------------------------------------------------------------------------------------
Solution
(a) Noting that the deflection is completely reversed, the loading is completely reversed, the stressing is
completely reversed, and therefore, the S-N curve of Figure P14-19 is directly applicable by using a failure
theory to relate shearing stress to direct stress. Choosing the distortion energy theory τ max = 0.577σ max .
Using (14-22)
2 R 2 ( 0.65 )
C= = = 6.28
D 0.207
K w = 1.24 ( Table 14.5 )
⎛ 16 ( 75.6 )( 0.65 ) ⎞
τ max = 1.24 ⎜ ⎟ = 34,990 psi
⎜ π ( 0.207 )3 ⎟
⎝ ⎠
τ max 34,990
σ max = = = 60, 640 psi
0.577 0.577
Using this value in Figure P14.9, we see that infinite life is expected.
Lf 6.0
SR = = = 4.6
2R 2 ( 0.65 )
ymax 1.0
= = 0.17
Lf 6.0
If this point is plotted, whether buckling occurs depends on end fixity. If both ends are pivoted, buckling
might occur. Experimental verification should be specified if buckling is an important issue.
f h = 6910 cpm
499
(d) From (14-61)
500
14-20. A helical-coil spring is to be wound using d = 3.5 mm wire made from a proprietary ferrous alloy for
which Sut = 1725 MPa , S yp = 1587 MPa , e ( 50 mm ) = 7% , E = 210 GPa , G = 79 GPa , and ν = 0.35 . It is
desired to use the spring in a cyclic loading situation where the axial load on the spring during each cycle ranges
from 450-N tension to 450-N compression. The spring deflection at maximum load must be 50 mm. A spring with
18 active coils is being proposed for the application.
a. Compute the existing safety factor for this spring based on an infinite-life design criterion. Comment on
the results.
b. If the spring is wound so that when it is unloaded the space between coils is the same as the wire
diameter, would you expect buckling to be a problem? (Support your answer with appropriate
calculations.)
c. Approximately want maximum operating frequency should be specified for this mechanism?
-----------------------------------------------------------------------------------------------------------------------------------
Solution
(a) d = 3.5 mm , Sut = 1725 MPa , S yp = 1587 MPa , E = 210 GPa , G = 79 GPa , N = 18 , Pmax = 450 N ,
Pmin = −450 N , ymax = 50 mm
(τ f ) N =∞
nex =
τ max
where, using the distortional energy theory, τ f ( ) N =∞ = 0.577S f . Since only static properties are known, we
approximate the S-N curve. By rule of thumb, S f = 100 ksi (690 MPa) at 106 cycles if
Sut > 200 ksi (1380 MPa) . Therefore τ f ( ) N =∞ = 0.577(690) = 398 MPa . The required spring rate is
Pmax 450
k= = = 9000 N/m
ymax 0.050
Next
d 4G (0.0035) 4 ( 79 × 109 )
k= 3
⇒ 9000 = 3
⇒ R3 = 1.143 × 10−6
64 R N 64 R (18)
R = 0.0105 m = 10.5 mm
Using c = 2 R / d = 21/ 3.5 = 6 we determine KW = 1.253 (from Table 14.5), which results in
⎛ 16(450)(0.021) ⎞
τ max = 1.253 ⎜⎜ 3 ⎟ ⎟ = 1406 MPa
⎝ π (0.0035) ⎠
398
nex = = 0.283
1406
Since nex = 0.283 we can not expect infinite life and the spring must be redesigned.
501
Lf 2( Nd ) 2 [18(3.5) ] 126
SR = = = = =6
2R 2R 2 (10.5 ) 21
and
ymax 50
= = 0.397
Lf 126
Buckling could be a problem if the ends are hinged. Experimental verification should be considered.
d Gg
fn = 2
2π R N 32 w
fn =
0.0035 ( 79 ×10 ) (9.81) ) = 0.5614
9
502
14--21. For a simply supported flat-beam spring of rectangular cross section, loaded at midspan, answer the
following questions:
---------------------------------------------------------------------------------------------------------------------------------
Solution
503
14-22. Derive a general expression for the maximum stress in an end-loaded multileaf cantilever spring with n
leaves, each having a thickness t and width b1 . Assume a spring with S yp = 1862 MPa , b1 = 100 mm , t = 10 mm ,
and an end load of P = 400 N . Plot the allowable length of the spring as a function of the number of leaves for
1≤ n ≤ 5.
---------------------------------------------------------------------------------------------------------------------------------------
Solution
For flexure σ x = Mc / I x , where for a cantilever beam with an end load M = Px , c = t / 2 , and I x = z x t 3 /12 . In
addition
zx x nb1 x
= ⇒ zx =
nb1 L L
Therefore
nb1t 3
Ix = x
12 L
Resulting in
Px (t / 2) 6 PL
σx = 3
=
nb1t nb1t 3
x
12 L
400
6(400) L
1862 × 106 = 3
⇒ L = 0.0776n
n(0.1)(0.01) 350
Allowable Length (mm)
300
250
200
150
100
50
0
0 2 4 6
Number of Leaves
504
14--23. Derive an equation for the spring rate of an end-loaded multileaf cantilever spring with n leaves, each
having a width b1 and thickness t.
---------------------------------------------------------------------------------------------------------------------------------
Solution
d 2 y Px zxt 3 zx x nb1 x
= , Ix = , = → zx = ( width at x from free end )
dx 2 EI x 123 nb1 L L
nb1t 3
Ix = x
12 L
d 2 y Px (12 L ) 12 PL
= =
dx 2 Enb1t 3 x Enb1t 3
Integrating gives
dy 12 PL
= x + C1
dx Enb1t 3
12 PL ⎛ x 2 ⎞
y= ⎜ ⎟ + C1 x + C2
Enb1t 3 ⎝ 2 ⎠
505
14-24. A horizontal cantilever-beam spring of constant rectangular cross section is loaded vertically across the
free end by a force that fluctuates cyclically from 4.5 kN down to 22.5 kN up. The beam is 125 mm wide and 250
mm long. The material is a ferrous alloy with Sut = 970 MPa , S yp = 760 MPa , and S f = 425 MPa . A design
factor of safety of nd = 1.5 is required, stress concentration factors can be neglected, and an infinite life is required.
Determine the required beam thickness.
-------------------------------------------------------------------------------------------------------------------------------------------
Solution
Fatigue is the probable failure mode and there is non-zero mean loading.
Sf τ fm
Smax − N = for σ m ≥ 0 and = Smax − N ≤ S yp
1 − mt Rt nd
where mt =
Sut − S f
=
970 − 425 σ P
= 0.562 and Rt = m = m =
( 22.5 − 4.5) / 2
= 0.40
Sut 970 σ max Pmax 22.5
425
Smax − N = = 548.2 MPa < S yp = 760 MPa
1 − 0.562(0.4)
Smax − N 548.2
σd = = = 365.5 MPa
nd 1.5
506
14--25. A horizontal simply supported multileaf spring is to be subjected to a cyclic midspan load that fluctuates
from 2500 lb down to 4500 lb down. The spring is to have 8 leaves, each 3.0 inches wide. The distance between
shackles (simply supports) is to be 22.0 inches. Properties of the selected spring material are given in Figure
P14.25.
a. Neglecting stress concentration effects, how thick should the leaves be made to provide infinite life, with a
design safety factor of 1.2?
b. What would be the spring rate of this spring?
---------------------------------------------------------------------------------------------------------------------------------
Solution
(a) Noting that fatigue is the probable failure mode, and non-zero mean loading is imposed, from (5-70)
Sf
S max − N = for σ m ≥ 0 and S max − N ≤ S yp
1 − mt Rt
Su − S f 140, 000 − 70, 000
mt = = = 0.5
Su 140, 000
⎛ 4500 + 2500 ⎞
σ P ⎜ ⎟
2
Rt = m = m = ⎝ ⎠ = 0.78
σ max Pmax 4500
70, 000
S max −∞ = = 114, 750 psi
1 − ( 0.5 )( 0.78 )
Note that 114,750 > Syp = 110,000 psi so the above is not valid and Smax-∞ = Syp = 110,000 psi. Therefore,
the design stress σd is
k= = 3169.2 lb/in
3 ( 22 )
3
507
14-26. A multileaf simply supported truck spring is to be designed for each rear wheel, using AISI 1095 steel
( Sut = 1379 MPa , S yp = 952 MPa , S f = 690 MPa ). The truck weight is 16 kN, with 65% of the weight on the rear
wheels. The static midspan deflection is 100 mm and the maximum midspan deflection during operation is 200 mm.
The loading may be considered to be released cyclic loading. The length of the spring between supports must
between 1.2 meters and 1.6 meters. It has been decided that a design factor of safety of nd = 1.3 should be used.
Design a leaf spring to meet these requirements if infinite life is desired.
------------------------------------------------------------------------------------------------------------------------------------
Solution
Since 65% of the truck weight is on the 2 rear springs, the static load supported by each spring is
0.65(16)
Ps = = 5.2 kN
2
Since the midspan static deflection is 100 mm, the spring rate is
Ps 5.2
k= = = 52 kN/m
ys 0.1
The maximum load will be Pmax = kymax = 52 (0.20) = 10.4 kN . By specification, the loading is considered releases
cyclic, so Pmin = 0 . To assess fatigue we use
Sf τ fm
Smax − N = for σ m ≥ 0 and = Smax − N ≤ S yp
1 − mt Rt nd
690
Smax − N = = 920 MPa < S yp = 952 MPa
1 − 0.50(0.50)
Smax − N 920
σd = = = 708 MPa
nd 1.3
σx =
3PL
=
3(10.4) L
=
(15.6 ×10 ) L3
708 × 10 6
=
(15.6 ×10 ) L 3
⎛ L ⎞
⇒ t 2 = 22 × 10−6 ⎜
2 ⎟
nb1t ⎝ nb1 ⎠
508
The spring rate ( k = 52 kN/m ) is
k = 52 × 103 =
8 Enb1t 3
=
( )
8 207 × 109 nb1t 3
⇒
nb1t 3
= 94.2 × 10−9
3L3 3L3 L3
⎡ ⎛ L ⎞⎤
nb1t ⎢ 22 × 10−6 ⎜ ⎟⎥
⎣⎢ ⎝ nb1 ⎠ ⎦⎥ t
3
= 94.2 × 10−9 ⇒ = 0.00428
L L2
From the problem specifications, 1.2 m ≤ L ≤ 1.6 m . Select the midrange value, so that L = 1.4 m . This give
t = 0.00838 m = 8.38 mm . This is not a standard length, so we select (from Table 14.4) t = 8.5 mm . Using these
values for L and t
⎛ 1.4 ⎞
(0.0085) 2 = 22 × 10−6 ⎜ ⎟ ⇒ nb1 = 0.4263
⎝ nb1 ⎠
0.4263
n= = 6.77 leaves
0.063
Since the number of leaves must be an integer, we try using 7 leaves, with b1 = 63 mm , giving
nb1t 2 7(0.063)(0.0085) 2
L= = = 1.45 m
22 × 10−6 22 × 10−6
This is well within the specified range for length. Therefore we end up with
(σ x )max =
(15.6 ×10 ) (1.45) ≈ 710 MPa
3
7(0.063)(0.0085) 2
The design stress was σ d = 708 MPa , which would result in an existing factor of safety of ne = 0.997 . This is
very close to the desired safety factor, so soliciting a second opinion before proceeding is probably a good idea.
509
14-27. A new revision of an “extreme sport” device that attaches to a persons feet and legs allows them to enhance
the power of their legs for running, jumping, etc. Your company is trying to improve on the design, as conceptually
shown in Figure P14.27 (a). For the initial phase of the design, you are to explore various spring configurations. The
spring (section AB in the drawing) spans an arc of 60o and the arc length AB is initially assumed to be 600 mm long.
Your company intends to make the spring out of a woven composite material with an elastic modulus of
E = 28 GPa and a yield strength of S yp = 1200 MPa . Preliminary investigation has shown that you expect the force
a “normal” user exerts at point A to be 1000 N down (Figure 14.27 (b)), which was established by considering
normal walking and running gates, impact, trick maneuvers, etc. For the most complex maneuver anticipated, you
have approximated the required spring rate to be k = 3 kN/m . Two spring designs are being considered. One is a
rectangular section and the other is cylindrical (Figure 14.27 (c)). Your preliminary estimate for the width of the
rectangular spring is w = 30 mm , which accounts for normal leg widths, possible interference, etc. Spring
attachment point A is offset from attachment point B by an amount δ , which is not considered in this initial phase
of design. Considering only flexure, determine the required thickness of the rectangular spring and the
corresponding diameter for a solid cylindrical spring.
Figure P14.27
Concept design for a new
extreme sport
mechanism.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
M = RP (1 − cos θ )
The spring rate can be defined as k = P / δ A , where is δ A determined from δ A = ∂U AB / ∂P . Expressed in terms of
the moment, this becomes
θ
∂U AB 1
M ( ∂M / ∂P ) Rdθ
EI ∫0
δA = =
∂P
510
where ∂M / ∂P = R (1 − cos θ ) . Using this we write
∂U AB 1 θ PR 3 π /3
∫0 ( RP (1 − cos θ ) ) ( R (1 − cos θ ) ) Rdθ = ∫0 (1 − cosθ ) dθ
2
δA = =
∂P EI EI
π /3
PR 3 ⎧ 3θ 1 ⎫ PR 3
= ⎨ − 2sin θ + sin 2θ ⎬ = 0.055
EI ⎩ 2 4 ⎭0 EI
⎛ 0.03t 3 ⎞
3 × 103 = 2706 × 109 ⎜⎜ ( 9 3
⎟⎟ = 6.765 × 10 t ) ⇒ t 3 = 0.4435 × 10−6
⎝ 12 ⎠
trect = 0.0076 m = 7.6 mm
⎛πd4 ⎞
3 × 103 = 2706 × 109 ⎜⎜ ( 9
⎟⎟ = 1.328 × 10 d
4
) ⇒ d 4 = 22.6 × 10−9
⎝ 64 ⎠
d cyl = 0.01226 m = 12.26 mm
511
14--28. A single-piece torsion bar spring of the type sketched in Figure P14.28 (a) is being considered by a group
of students as a means of supporting the hood of an experimental hybrid vehicle being developed for intercollegiate
competition. The maximum length L that can be accommodated is 48 inches. Figure P14.28 (b) illustrates their
concept. They plan to install the counterbalancing torsion bar spring along the hood-hinge centerline, with one of
the 3-inch integral end-levers in contact with the hood, as shown. The hood will then be raised until it contacts the
45˚ hood stop; the 3-inch support lever on the opposite end will be rotated until the hood is just held in contact with
the hood stop without any other external lifting force on the hood. The support lever will next be given an
additional rotation to lightly preload the hood against the stop, and then clamped to the supporting structure.
a. Determine the diameter do of a solid-steel torsion bar that would counterbalance the hood weight and
provide a 10 ft-lb torque to hold the hood against the stop shown, if the design stress in shear for the
material is τd = 60,000 psi.
b. At what angle, with respect to a horizontal datum, should the clamp for the 3-inch support lever be placed
to provide the desired 10 ft-lb preload torque? Neglect bending of the integral levers.
c. Make a plot showing gravity-induced torque, spring torque, and net torque, all plotted versus hood-opening
angle.
d. Is the operating force, Fop, required to open or close the installed hood, reasonable for this design
configuration? What other potential problems can you foresee with this arrangement?
---------------------------------------------------------------------------------------------------------------------------------
Solution
16Tmax d 0 16Tmax
τ max = =
π d 04 π d 03
For a proper design set τmax = τd, giving
16Tmax 16 (Tmax )
d0 = 3 = 3
πτ d π ( 60, 000 )
Tmax
θ total =
ktor
π ( 0.33) (11.5 × 106 )
4
π d 04 G in-lb
k= = = 279
32 L 32 ( 48 ) rad
438
θ total = = 1.57 rad = 90o
279
And, angle of twist in torsion bar to lift the hood until it just touches the hood stop is,
512
318
θtouch = = 1.14 rad = 65.3o
279
And, angle of twist in torsion bar to provide the 10 ft-lb preload after the hood touches the stop is,
120
θ preload = = 0.43 rad = 24.6o
279
Since the hood stop is at β = 45˚, the clamp angle γ for the 3-inch support lever should be installed at
(c) To plot gravity-induced torque (due to hood weight), spring torque, and net torque, all versus hood opening
angle β, the following relationship may be developed from Figure P14-28, noting that CCW is positive and
β = o when the hood is closed:
(d) The force required to close the hood, Fop, (see Figure P14-28) is
Tnet
Fop =
36
Noting Fop magnitudes shown in the table above, Fop appears to be a reasonable range. One possible
problem might be that when unlatched the hood may “jump” open and injure someone. A damper may be
required to slow the opening event.
513
To r q ue Vs . Op e n in g An g le
800
625
T o rq u e , i n -l b (CCW + )
450
Tsp
275
100 Tnet
-7 5
-2 5 0
Tgr
-4 2 5
-6 0 0
0 10 20 30 40 50
B eta, D egrees
514
14-29. A single-bodied helical-coil torsion spring (see Figure 14.20) has a wire diameter of 1 mm and an outside
coil diameter of 10 mm. The tightly coiled spring has 9.5 coil, with an end extension of a = 12 mm from the coil
center to the point of load application at each end. The spring material is a steel allow with Sut = 2030 MPa .
---------------------------------------------------------------------------------------------------------------------------------------
Solution
(a) ktor =
π d 4E
=
π d 4E
=
π (0.001) 207 ×10 4
( 9
) = 0.034 N-m/rad
64 L 64( N π D) 64(9.5)(0.01π )
T 0.1
(b) θ= = = 2.94 rad = 168.3o
ktor 0.034
(c) Using c = 2 R / d = 10 /1 = 10 , the maximum normal stress, noting that Pa = T = 0.1 N-m , is
⎛ 4c − 1 ⎞ 32 ( Pa ) 32(0.1)
σ max = ⎜ ⎟ = 1.083 = 1103 MPa
⎝ 4c − 4 ⎠ π d 3 π (0.001)3
Sut 2030
(d) ne = = = 1.84
σ max 1103
The desired safety factor is exceeded, so the design is acceptable. One could go through and modify the
specifications to get closer to nd = 1.5 , but it is not essential that this is done.
515
14-30. A matched pair of torsional helical-coil springs, one left-hand and the other right hand, is scheduled for use
to counterbalance the weight of a residential overhead garage door. The arrangement is sketched in Figure P14.30.
The 1-inch-diameter rotating shaft is supported on three bearings near the top of the door, one bearing at each end,
and one at midspan. A small wire rope is wrapped around each of the pulleys to symmetrically support the weight
of the door. The 2.5-inch pulley radii are measured to the wire rope centerlines. Each spring is wound from
standard oil-tempered steel wire having a wire diameter of 0.225 inch, a mean coil radius of 0.89 inch, and 140
turns, closely coiled. The total length of wire-rope excursion from spring-unloaded position to door closed-position
is 85 inches.
a. Calculate the maximum bending stress in the springs when the door is closed, and tell where it occurs.
b. What would be the heaviest garage door that could be counterbalanced using the arrangement of Figure
P14.30 and this matched pair of springs.
---------------------------------------------------------------------------------------------------------------------------------------
Solution
(a) Total shaft revolutions, ns, from door open (springs unloaded) to door closed (springs at max load torque) is
L 85
ns = = = 5.4 rev
2π rpulley 2π ( 2.5 )
θ s = 2π ns = 2π ( 5.4 ) = 34.0 rev
30 × 106 ( 0.225 ) ( 34.0 )
4
EIθ s E π d 4θ s
M = = = = 163.9 in-lb (each spring)
L 64 ( 2 π RN ) 128 ( 0.89 )(140 )
From (14-6)
2 R 2 ( 0.89 )
C= = = 7.91
d 0.225
⎛ 4 ( 7.91) − 1 ⎞ 32 (163.9 )
σ max = ⎜⎜ ⎟⎟ = 162, 473 psi (at inner coil radius)
⎝ 4 ( 7.91) − 4 ⎠ π ( 0.225 )
3
(b) Using the value of M from above and reviewing Figure P14-30
⎛ Wdoor ⎞
⎜ 2 ⎟ rp = M = 163.9
⎝ ⎠
2 (163.9 )
(Wdoor )max = = 131 lb
2.5
(c) Since the springs are tightly wound, the free length may be approximated as
The weight of each spring may be estimated as (for steel springs with w = 0.283 Lb/in3),
⎛πd2 ⎞ ⎛ π ( 0.225 )2 ⎞
Wspr = ( 2π RN ) ⎜ ⎟ w = 2π ( 0.89 )(140 ) ⎜ ⎟ ( 0.283) = 8.8 lb
⎝ 4 ⎠ ⎜ 4 ⎟
⎝ ⎠
516
14-31. As a diversion, a “machine design” professor has built a new rough-sawn cedar screened porch as an
attachment to the back of his house. Instead of using a traditional “screen-door spring” (closed-coil helical-coil
tension spring) to keep his screen door closed, he has decided to use a helical-coil torsion spring of the single-bodied
type shown in Figure 14.20(a). He plans to install the spring so its coil centerline coincides with the door-hinge
centerline. The distance from the hinge centerline to the pull-handle is to be 32 inches, and his goal is to provide a
handle-pull of 1 lb when the door is closed and a pull of 3 lb after the door has been rotated open through an angle of
180˚ about its hinge centerline. Tentatively, a standard No. 6 music wire has been chosen ( d = 0.192 inch) for the
spring. A design stress of σd = 165,000 psi has been calculated, based on yielding as the probable governing failure
mode.
---------------------------------------------------------------------------------------------------------------------------------------
Solution
⎛ 4c − 1 ⎞ 32 P a
σ max = ⎜ ⎟
max
⎝ 4c − 4 ⎠ π d
3
= = = 1.19
4c − 4 32 Pmax a 32 ( 3)( 32 )
c = 4.85
2R
c= = 4.85
d
4.85
R= ( 0.192 ) = 0.47 in.
2
(b) From (14-54),
64 Pclosed aL 64 (1)( 32 ) L
θ pre = = = 0.016 L
πd E π ( 0.192 ) ( 30 × 106 )
4 4
64 ( 3)( 32 ) L
θ open = = 0.048L
π ( 0.192 ) ( 30 × 106 )
4
θ open = θ closed + π
0.048L = 0.016 L + π
π
L= = 98.2 in.
0.048 − 0.016
θ pre = 0.016 ( 98.2 ) = 1.57 rad = 90o
L 98.2
N= = = 33.3 coils
2π R 2π ( 0.47 )
517
14-32. An unlabeled box of steel Belleville spring washers (all washers in the box are the same) has been found in
a company storeroom. As a summer-hire, you have been asked to analytically evaluate and plot the force-deflection
characteristics of the spring washers. The dimensions of the washers in the unlabeled box, with reference to the
sketch of Figure 14.21, are as follows:
Do = 115 mm
Di = 63.9 mm
t = 2.0 mm
h = 3.0 mm
Do the following:
a. Estimate the force required to just “flatten” one of the Belleville washers.
b. Plot a force-deflection curve for one of the washers using force magnitudes ranging from zero up
to the full flattening force. Characterize the curve as linear, nonlinear hardening, or nonlinear
softening.
c. Plot a force-deflection curve for two of the washers stacked together in parallel. Characterize the
curve as linear, nonlinear hardening, or nonlinear softening.
d. Plot a force-deflection curve for two of the washers stacked together in series. Characterize the
curve as linear, nonlinear hardening, or nonlinear softening.
e. Calculate the magnitude of the highest tensile stress that would be expected in the single washer of
(b) above at the time that the applied load just “flattens” the washer.
---------------------------------------------------------------------------------------------------------------------------------------
Solution
4 E ⎛ ht 3 ⎞
Fflat = ⎜ ⎟
1 −ν 2 ⎝ K1 Do2 ⎠
Do 115
= = 1.8
Di 63.9
K1 = 0.65 ( Table 14.9 )
Fflat = ⎜ ⎟ = 2.54 kN
1 − ( 0.3) ⎜ 0.65 (115 × 10−3 ) ⎟
2 2
⎝ ⎠
(b) Using Figure 14.32, the following table may be constructed for h/t = 3/2 = 1.5:
F/Fflat F, kN y/h y, mm
0 0 0 0
0.2 0.51 0.07 0.21
0.4 1.02 0.16 0.48
0.6 1.52 0.25 0.75
0.8 2.03 0.40 1.2
0.97 2.46 0.60 1.8
1.0 2.54 1.0 3.0
The force deflection curve for one washer may be plotted from the table as follows:
518
S ingle W asher
3
F, kN
0
0 1 2 3 4
y, mm
(c) When stacked in parallel as shown here (ref. Fig 4.19), deflections are the same for both washers but forces
add.
y, mm y/h F, kN 2F, kN
0 0 0 0
0.21 0.07 0.51 1.02
0.48 0.16 1.02 2.04
0.75 0.25 1.52 3.04
1.2 0.40 2.03 4.06
1.8 0.60 2.46 4.92
2.5 0.83 2.64 5.28
3.0 1.0 2.54 5.08
519
The force deflection curve may be plotted as shown.
Tw o W asher in P arallel
4
2F, kN
0
0 1 2 3 4
y, mm
It is noted that this curve would be classified as non-linear softening but is closer to linear up to a load around 3 kN.
520
Table P14.32C – Two Washers in Series
F, kN y, mm 2y, mm
0 0 0
0.51 0.21 0.42
1.02 0.48 0.96
1.52 0.75 1.5
2.03 1.2 2.4
2.46 1.8 3.6
2.64 2.5 5.0
2.54 3.0 6.0
Tw o W asher in S eries
2
F, kN
0
0 1 2 3 4 5 6 7
2y, mm
4 EtyDi ⎡ ⎛ h y ⎞⎤
σC = ⎢ K 3 + ( 2 K 3 − K 2 ) ⎜ t − 2t ⎟ ⎥
K1 Do3 (1 −ν 2 ) ⎣ ⎝ ⎠⎦
521
K1 = 0.65
K 2 = 1.17
K 3 = 1.30
which is the highest tensile stress (at c.p. C; outer edge, lower surface) when washer is just flattened.
522
14-33. Considering a solid square bar of steel with side-dimensions s and length-dimension L, would you predict
that more elastic strain energy could be stored in the bar (without yielding) by using it as a direct tension spring
axially loaded in the L-direction, or by using it as a cantilever-bending spring loaded perpendicular to the L-
direction? Make appropriate calculations to support your predictions.
---------------------------------------------------------------------------------------------------------------------------------------
Solution
A direct tension spring, using (14-61) for the case where the force
corresponding to incipient yielding, Pyp, is applied
Pyp ytens
U tens =
2
Pyp L
ytens =
AE
Pyp ⎛ Pyp L ⎞ S yp A ⎛ S yp AL ⎞
U tens = ⎜ ⎟= ⎜ ⎟
2 ⎝ AE ⎠ 2 ⎝ AE ⎠
S yp2 2
S yp
=
2E
( AL ) =
2E
( s L)
2
Pyp ybend
U bend =
2
M yp c ( P L ) ⎛⎜⎝ 2s ⎞⎟⎠
yp
6 Pyp L
S yp = = =
I ⎛ s (s) 3
⎞ s3
⎜ ⎟
⎜ 12 ⎟
⎝ ⎠
S yp ( s )
3
Pyp =
6L
523
Pyp L3 4 Pyp L3
ybend = =
⎛ s4 ⎞ Es 4
3E ⎜ ⎟
⎝ 12 ⎠
⎡ S yp ( s )3 ⎤
4⎢ ⎥ L3
⎢ 6 L ⎥⎦ 2S yp L2
= ⎣ =
Es 4 3Es
1 ⎡ S yp s ⎤ ⎡ 2 S yp L ⎤ 1 ⎡ S yp 2 ⎤
3 2 2
U bend = ⎢ ⎥⎢ ⎥= ⎢
2 ⎢⎣ 6 L ⎥⎦ ⎢⎣ 3Es ⎥⎦ 9 ⎢⎣ 2 E
( s L )⎥
⎥⎦
and
Hence, the direct tension member can store 9 times as much strain energy as the bending member, before yielding
begins.
524
14-34. a. Write the equations from which the form coefficient CF may be found for a simply supported center-
loaded multileaf spring.
b. Find the numerical value of CF for this type of spring and compare it with the value given in Table 14.10.
---------------------------------------------------------------------------------------------------------------------------------------
Solution
⎛σ 2 ⎞
uv = CF ⎜ max ⎟
⎝ 2E ⎠
Fy
U=
2
3FL3
yc =
8 Enb1t 3
3F 2 L3
U=
16 Enb1t 3
⎡1 ⎛ L ⎞ ⎤ nb Lt
v = 2 ⎢ ⎜ ⎟ ( nb1 ) t ⎥ = 1
⎣2 ⎝ 2 ⎠ ⎦ 2
U ⎛ 3F 2 L3 ⎞ ⎛ 2 ⎞ 3F 2 L2
uv = =⎜ ⎟⎜ ⎟=
v ⎝ 16 Enb1t 3 ⎠ ⎝ nb1 Lt ⎠ 8n 2 b12 t 4
3FL
σ max =
2nb1t 2
uv ( 2 E ) 3F 2 L2 ( 2 ) E ( 4 ) n 2 b12 t 4
CF = = = 0.38
σ max
2
8En 2 b12 t 4 ( 9 ) F 2 L2
This value agrees with table 14.10.
525
14-35. Repeat problem 14-34 except for the case of a simply supported beam spring of rectangular cross section.
---------------------------------------------------------------------------------------------------------------------------------------
Solution
From (14-63)
⎛σ 2 ⎞
uv = CF ⎜ max ⎟
⎝ 2E ⎠
3
FL
y=
48 EI
Fy F ⎛ FL3 ⎞ F 2 L3
U= = ⎜ ⎟=
2 2 ⎝ 48EI ⎠ 96 EI
12 F 2 L3 F 2 L3
U= 3
=
96 Ebt 8 Ebt 3
U F 2 L3 F 2 L2
uv = = =
v 8Ebt ( btL ) 8 Eb3t 4
3
Mc FL ( t 2 )(12 ) 3FL
σ max = = =
I 4bt 3 2bt 2
u ( 2 E ) F 2 L2 ( 2 ) E ( 4 ) b 2 t 4
CF = v 2 = = 0.11
σ max 8 Eb 2 t 4 ( 9 ) F 2 L2
526
14-36. Repeat problem 14-34 except for the case of an end loaded cantilever-beam spring of rectangular cross
section.
---------------------------------------------------------------------------------------------------------------------------------------
Solution
From (14-63)
⎛σ 2 ⎞
uv = CF ⎜ max ⎟
⎝ 2E ⎠
3
FL
y= (cantilever beam with end load)
3EI
Fy F ⎛ FL3 ⎞ F 2 L3
U= = ⎜ ⎟=
2 2 ⎝ 3EI ⎠ 6 EI
12 F 2 L3 2 F 2 L3
U= =
6 Ebt 3 Ebt 3
U 2 F 2 L3 2 F 2 L2
uv = = =
v Ebt ( btL ) Eb 2 t 4
3
Mc FL ( t 2 )(12 ) 6 FL
σ max = = = 2
I bt 3 bt
u ( 2E ) 2F L ( 2E ) b t
2 2 2 4
CF = v 2 = = 0.11
σ max Eb 2 t 4 ( 36 ) F 2 L2
527
14-37. Repeat problem 14-34 except for the case of a spiral flat-strip torsion spring.
---------------------------------------------------------------------------------------------------------------------------------------
Solution
From (14-63)
⎛σ 2 ⎞
uv = CF ⎜ max ⎟
⎝ 2E ⎠
Adapting (14-61) to the case of applied torque T acting through angular deflection θ,
Tθ
U=
2
12TL
θ= 3 (from 14 − 57)
bt E
T ⎛ 12TL ⎞ 6T 2 L
U = ⎜ 3 ⎟= 3
2 ⎝ bt E ⎠ bt E
U 6T 2 L 6T 2
uv = = 3 =
v bt E ( btL ) Eb 2 t 4
⎛ 6T ⎞
σ max = ki ⎜ 2 ⎟
ki ≈ 1.1 (assuming the mandrel diameter is at least 5 times t)
⎝ bt ⎠
u ( 2E ) 6T 2 ( 2 E ) b 2 t 4
CF = v 2 = = 0.28
σ max Eb 2 t 4 ( 36 ) ki2T 2
528
14-38. Repeat problem 14-34 except for the case of a round wire helical-coil torsion spring.
---------------------------------------------------------------------------------------------------------------------------------------
Solution
From (14-63)
⎛σ 2 ⎞
uv = CF ⎜ max ⎟
⎝ 2E ⎠
Adapting (14-61) to the case of applied torque T acting through angular deflection θ,
Tθ
U=
2
64TL
θ= (from 14 − 54)
π d 4E
T ⎛ 64TL ⎞ 32T 2 L
U = ⎜ 4 ⎟=
2 ⎝ π d E ⎠ π d 4E
πd2
The volume of the spring wire is v = L , then
4
U 32T 2 L ( 4 ) 128T 2
uv = = =
v π d 4 Eπ d 2 L π 2 d 6 E
⎛ 4c − 1 ⎞ 32T
σ max = ⎜ ⎟ 3
⎝ 4c − 4 ⎠ π d
⎛ 4 ( 8 ) − 1 ⎞ 32T 35.4T
σ max = ⎜⎜ ⎟⎟ 3 =
⎝ 4 (8) − 4 ⎠ π d πd3
u ( 2 E ) 128T 2 ( 2 E ) π 2 d 6
CF = v 2 = ≈ 0.20
σ max Eπ 2 d 6 ( 35.4 ) T 2
2
529
14-39. Repeat problem 14-34 except for the case of a round wire helical-coil compression spring.
---------------------------------------------------------------------------------------------------------------------------------------
Solution
From (14-64)
⎛τ 2 ⎞
uv = CF ⎜ max ⎟
⎝ 2G ⎠
64 FR 3 N
y= (from 14-21)
d 4G
Fy F ⎛ 64 FR 3 N ⎞ 32 F 2 R 3 N
U= = ⎜ ⎟=
2 2 ⎝ d 4G ⎠ d 4G
⎛πd2 ⎞ π 2 RNd 2
The volume of a helical coil spring is v = 2π RN ⎜ ⎟=
⎝ 4 ⎠ 2
U 32 F 2 R 3 N 64 F 2 R 2
uv = = 4 2 =
v d Gπ RNd 2 π 2 d 6G
⎛ 16 FR ⎞
τ max = K w ⎜ 3 ⎟
⎝ πd ⎠
u ( 2G ) 64 F 2 R 2π 2 d 6 ( 2 ) G
CF = v 2 = 2 6
τ max π d GK w2 ( 256 ) F 2 R 2
1
CF =
2 K w2
Selecting a spring index of 8 (typical value), Table 14.5 gives Kw = 1.18, hence
1
CF = = 0.36
2 (1.18 )
2
530
Chapter 15
15-1. For each of the design scenarios presented, suggest one or two types of gears that might make good
candidates for further investigation in terms of satisfying the primary design requirements.
a. In the design of a new-concept agriculture hay-conditioner, it is necessary to transmit power from one
rotating parallel shaft to another. The input shaft is to rotate at a speed of 1200 rpm and the desired output
speed is 350 rpm. Low cost in an important factor. What types (s) of gearing would you recommend?
State your reasons.
b. In the design of a special speed reducer for a laboratory test stand, it is necessary to transmit power from
one rotating shaft to another. The centerlines of the two shafts intersect. The driver shaft speed is 3600
rpm and the desired speed of the output shaft is 1200 rpm. Quiet operation is an important factor. What
types(s) of gearing would you recommend? State your reasons.
c. It is desired to use a 1-hp, 1725-rpm electric motor to drive a conveyer input shaft at a speed of
approximately 30 rpm. To give a compact geometry, the motor drive shaft is to be oriented at 90 degrees to
the conveyor input shaft. The shaft centerline may either intersect or not, depending on designers
judgment. What types(s) of gearing would you recommend? State your reasons.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
Based on discussion of 15.2, straight tooth spur gears and helical gears would be likely candidates. Since low cost is
a factor, and noise is probably not very important in this environment, straight tooth spur gears would be an
excellent choice. The reduction ratio of 3.4 can readily be accommodated.
Based on discussion of 15.2, zerol bevel gears would be likely candidates. The reduction ratio of 3.0 can readily be
accommodated. Spiral bevel gears would probably be quieter, so they should make an excellent choice.
2. Shaft centerlines are to be 90˚ to each other, and may intersect or not.
Based on discussion of 15.2, bevel gears or worm gears would be candidates. However, to achieve a reduction ratio
of 57.5, bevel gears would be unwieldy in size, violating the compact geometry specification. Worm gears would be
an excellent choice.
531
15-2. The compound helical gear train sketched in Figure P15.2 involves three simple helical gears (1,2,5) and one
compound helical gear (3,4). The number of teeth on each gear id indicated in the sketch. If the input gear (1) is
driven clockwise at a speed of n1 = 1725 rpm, calculate the speed and direction of the output gear (5).
------------------------------------------------------------------------------------------------------------------------------------------
Solution
nout ⎡ ∏ ( N driver )i ⎤
= ±⎢ ⎥
nin ⎣⎢ ∏ ( N driven )i ⎦⎥
Noting that the drivers are 1, 2, and 4, and the driven gears are 2,3, and 5, and for the compound gear n3 = n4, thus
nout ⎡⎛ N1 ⎞ ⎛ N 2 ⎞ ⎛ N 4 ⎞ ⎤
= ⎢⎜ − ⎟⎜ − ⎟⎜ − ⎟⎥
nin ⎢⎣⎝ N 2 ⎠ ⎝ N 3 ⎠ ⎝ N 5 ⎠ ⎥⎦
nout ⎡⎛ 20 ⎞ ⎛ 66 ⎞⎛ 88 ⎞ ⎤
= ⎜ − ⎟ ⎜ − ⎟⎜ − ⎟ = −1.60
nin ⎢⎣⎝ 66 ⎠ ⎝ 22 ⎠⎝ 50 ⎠ ⎥⎦
nout = −1.6 ( nin ) = −1.6 (1725 ) = −2760 rpm
n5 = nout = 2760 rpm CCW
532
15-3. The sketch of Figure P15.3 shows a two-stage reverted gear reducer that utilizes two identical pairs of gears
to enable making the input shaft and output shaft collinear. If a 1-kw, 1725-rpm motor operating at full rated power
is used to drive the input shaft in the CW direction, do the following:
c. Assuming a98 percent efficiency of each gear mesh, calculate the torque available for driving the load at
the output shaft.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
nout n2 N 16
= =− 1 =− = −0.333
nin n1 N2 48
n2 = −0.333n1 = −0.333 (1725 ) = −575 rpm
nB = n2 = 575 rpm CCW
nout n4 n4 N 16
= = =− 3 =− = −0.333
nin n3 n2 N4 48
n4 = −0.333n3 = −0.333 ( 575 ) = −191.7 rpm
nout = nc = n4 = 191.7 rpm CW
Since there are two speed reducers meshes in series between the input (Tin) and the output (Tout), each mesh ration
being 3:1 and each having an efficiency of 98 percent.
⎛ 3 ⎞⎛ 3 ⎞
Tout = Tin ⎜ ⎟ ⎜ ⎟ ( 0.98 )( 0.98 ) = 8.64 Tin
⎝ 1 ⎠⎝ 1 ⎠
Tout = 8.64 ( 5.54 ) = 47.9 N-m
533
15-4. A two-planet epicyclic gear train is sketched in Figure P15.4. If the ring gear is fixed, the sun gear is driven
at 1200 rpm in the CCW direction, and the carrier arm is used as output, what would be the speed and direction of
rotation of the carrier arm?
------------------------------------------------------------------------------------------------------------------------------------------
Solution
It may be noted that if the ring gear 3 is fixed (ω3 = 0), the sun gear 1 is used as input (ω1 = ωin), and carrier arm 4 is
used as output (ω4 = ωout), and the first and last gears in the train are taken as sun gear 1 and ring gear 3, (15-4)
becomes
ω3 − ω4 0 − ω4 ⎛ 30 ⎞ ⎛ 20 ⎞
= = −
ω1 − ω4 ω1 − ω4 ⎜⎝ 20 ⎟⎠ ⎜⎝ 70 ⎟⎠
0 − ω4
= −0.43
ω1 − ω4
ω4 ωout n
= = 0.30 = out
ω1 ωin nin
For nin = 1200 rpm (CCW), and noting that nout has the same direction as nin, thus
It is interesting to note that this is the same configuration examined in Example 15.1 except that there are two
identical planet gears in this problem and only one in Example 15.1. It may be properly deduced that the kinematics
are the same whether one planet, two planets, or more are used. The advantage of multiple planets in a balanced
configuration is that gear tooth loads are reduced.
534
15-5. A special reverted planetary gear train is sketched in Figure P15.5. The planet gears (2-3) are connected
together (compound) and are free to rotate together on the carrier shaft. In turn, the carrier shaft is supported by a
symmetrical one-piece pair of carrier arms attached to an output shaft (5) that is collinear with the input shaft (1).
Gear 4 is fixed. If the input shaft (1) is driven at 250 rpm in the CW direction, what would be the output shaft (5)
speed and direction?
------------------------------------------------------------------------------------------------------------------------------------------
Solution
Noting that sun gear 1 is used as input (ω1 = ωin), and carrier arm 5 is used as output (ω5 = ωout), and since 2-3 is a
compound shaft ω2 = ω3, and gear 4 is fixed (ω4 = 0), if first and last gears in the train are taken as sun gear 1 and
ring gear 4, (15-4) becomes
ω4 − ω5 ⎛ 20 ⎞⎛ 16 ⎞
= − − ⎟
ω1 − ω5 ⎜⎝ 30 ⎟⎜ ⎠⎝ 34 ⎠
0 − ω5
= 0.314
ω1 − ω5
ω5 ωout nout
= = = −0.458
ω1 ωin nin
nout = 0.458 ( 250 ) = 114.5 rpm CCW
535
15-6. An annular gear A on shaft S1, has 120 teeth and drives a pinion B having 15 teeth, keyed to shaft S2.
Compounded with B is a 75 tooth gear C which drives a 20 tooth gear D on shaft S3. Compounded with gear D is
gear E having 144 teeth driving gear F on shaft S4. The axis of S4 is collinear with the axis of S1. All shafts are
parallel and in the same plane.
(a) How many teeth must gear F have if all gears have the same diametral pitch?
(c) If the gears have a diametral pitch of 4, what is the distance between shafts S1 and Ss?
------------------------------------------------------------------------------------------------------------------------------------------
Solution
(a) From the geometry of the gear train we have using the diameters
d F d E d D dC d A d B
+ = + + −
2 2 2 2 2 2
d F = d A + dC + d D − d B − d E
Since the gears have the same diametral pitch, then d = N/Pd or
N F = N A + NC + N D − N B − N E
= 120 + 75 + 20 − 15 − 144
N F = 56 teeth
536
N A ⎛ NC ⎞ ⎛ N E ⎞
Train value = ⎜− ⎟⎜ − ⎟
NB ⎝ ND ⎠ ⎝ NF ⎠
120 ⎛ 75 ⎞ ⎛ 144 ⎞
= ⎜ − ⎟⎜ − ⎟ = 77.2
15 ⎝ 20 ⎠ ⎝ 56 ⎠
d A d B dC d D
C1− 3 = − + +
2 2 2 2
N A N B NC N D
= − + +
2 Pd 2 Pd 2 Pd 2 Pd
120 15 75 20
= − + + = 25 inches
2 ( 4) 2 ( 4) 2 ( 4) 2 ( 4)
537
15-7. The gear train shown in Figure P15.7 has an input speed of 1200 rpm (clockwise). Determine the output
speed (rpm) and direction of rotation.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
Note that the first gear and the last gear selected for the train value equation must be gears that have planetary
motion.
ω3 = ω2
ωarm = ω5 = ω7
ω7 N N ⎛ 20 ⎞
=− 2 ω7 = −ω2 2 = −1200 ⎜ ⎟ = −667 rpm
ω2 N5 N5 ⎝ 36 ⎠
ω6 N N ⎛ 32 ⎞
=− 3 ω6 = −ω2 3 = −1200 ⎜ ⎟ = −1600 rpm
ω2 N4 N4 ⎝ 24 ⎠
ωL − ω A
= Train value
ωF − ω A
Let the first gear of the train be ω6 and the last gear of the train ω10 we then have
ωL − ω A
= Train value
ωF − ω A
ω10 + 667 ⎛ N ⎞ ⎛ N ⎞ ⎛ 28 ⎞ ⎛ 23 ⎞
= - ⎜ 6 ⎟ ⎜ 8 ⎟ = ⎜ − ⎟ ⎜ − ⎟ = 0.259
−1600 + 667 ⎝ N 9 ⎠ ⎝ N10 ⎠ ⎝ 54 ⎠ ⎝ 46 ⎠
ω10 = −667 + 933 ( 0.259 ) = −425
ω10 = ωout = 425 (CCW)
538
15-8. The gear train shown in Figure P15.8 has an input speed of 720 rpm clockwise and an input torque of 300 lb-
in. Determine:
(a) The speed and direction of rotation of the output – shaft (rpm).
------------------------------------------------------------------------------------------------------------------------------------------
Solution
Note that the first gear and the last gear selected for the train value equation must be gears that have planetary
motion.
ωarm = ω9 = ω3
ω4 N N6 ⎛ 35 ⎞
=− 6 ω4 = −ω6 = −720 ⎜ ⎟ = −840 rpm (CCW)
ω6 N4 N4 ⎝ 30 ⎠
ω9 N N2 ⎛ 21 ⎞
=− 2 ω9 = −ω2 = −720 ⎜ ⎟ = −240 rpm (CCW)
ω2 N3 N3 ⎝ 63 ⎠
ωL − ω A
= Train value
ωF − ω A
ωF = ω4 = 840 ( ccw ) , ωL = ω10 , ω9 = ω A = 240
ω10 − 240 ⎛ N ⎞⎛ N ⎞ ⎛ 39 ⎞ ⎛ 24 ⎞
= - ⎜ 5 ⎟⎜ 8 ⎟ = - ⎜ ⎟ ⎜ ⎟ = −0.375
840 − 240 ⎝ N 7 ⎠⎝ N10 ⎠ ⎝ 78 ⎠ ⎝ 32 ⎠
ω10 = 240 + 600 ( −0.375 ) = 15 rpm (ccw)
539
15-9. In the gear train shown in Figure P15.9, the planet carrier (2) is turning clockwise at the rate of 500 rpm and
the sun gear (3) is turning counterclockwise at the rate of 900 rpm. All gears have the same diametral pitch.
Determine the speed and direction of the annular gear (7).
------------------------------------------------------------------------------------------------------------------------------------------
Solution
d7 d3 d d
= + d4 + 5 + 6
2 2 2 2
N7 N3 N5 N 6
= + N4 + + (Since they have the same diametral pitch)
2 2 2 2
18 39 21
= + 25 + +
2 2 2
N 7 = 128 teeth
2 3 4 5 6 7
⎛ N 3 ⎞⎛ N 6 ⎞
ω7 = 500 − 1400 ⎜ ⎟⎜ ⎟
⎝ N 5 ⎠⎝ N 7 ⎠
⎛ 18 ⎞ ⎛ 21 ⎞
= 500 − 1400 ⎜ ⎟ ⎜ ⎟
⎝ 39 ⎠ ⎝ 128 ⎠
ω7 = 394 rpm (cw)
540
15-10. What are the kinematic requirements that must be met to satisfy the “fundamental law of gearing”?
------------------------------------------------------------------------------------------------------------------------------------------
Solution
To satisfy the fundamental law of gearing kinematically, the common normal to the curved tooth surfaces at their
point of contact must, for all gear positions, intersect the line of centers at a fixed point P called the pitch point.
541
15-11. Define the following terms, using a proper sketch where appropriate.
a. Line of action
b. Pressure angle
c. Addendum
d. Dedendum
e. Pitch diameter
f. Diametral pitch
g. Circular pitch
h. Pitch point
-----------------------------------------------------------------------------------------------------------------------------------------
Solution
(a) Referring to the sketch below, the line of action, as shown a-b, is a fixed line in space, through pitch point
P, representing the direction of the resultant force transmitted from the driving gear to the driven gear.
542
(g) The circular pitch, pc, is the distance from any selected reference point on one tooth to a corresponding
point on the next adjacent tooth, measured along the pitch circle.
(h) The pitch point P is the point of contact between pitch circles of two meshing gears.
543
15-12. Describe what is meant by a “gear tooth system.”
------------------------------------------------------------------------------------------------------------------------------------------
Solution
A gear tooth system is a standardized set of tooth proportions agreed upon to facilitate interchangeability and
availability of gears. Characteristics specified to define a standardized tooth system depend upon first selecting a
pressure angle, then defining addendum, dedendum, working depth, whole depth, minimum tip clearance, and
circular tooth thickness as functions of diametral pitch (or module).
544
15-13. A straight-tooth spur gearset is being considered for a simple-speed reduction device at an early stage in the
design process. It is being proposed to use standard 20˚ involute full-depth gear teeth with a diametral pitch of 4 and
a 16-tooth pinion. A reduction ratio of 2.50 is needed for the application. Find the following:
b. Circular pitch
c. Center distance
------------------------------------------------------------------------------------------------------------------------------------------
Solution
ω p n p rg N g
= = = = 2.50
ω g ng rp N p
N p = 16, N g = 2.50 (16 ) = 40 teeth
π π
pc = = = 0.785 in.
Pd 4
N p + Ng 16 + 40
C= = = 7.0 in.
2 Pd 2 ( 4)
2π rb
= pc cos ϕ
N
Npc cos ϕ N ( 0.785 ) cos 20
o
rb = =
2π 2π
rb = 0.1174 N
( rb ) p = 0.1174 (16 ) = 1.878 in.
( rb ) g = 0.1174 ( 40 ) = 4.696 in.
(e) From Table 15.3, for a 16 tooth pinion, the maximum number of gear teeth without having interference is
101. Since for this gearset
Ng = 40 < 101
545
15-14. Repeat problem 15-9, except use a diametral pitch of 8 and 14-tooth pinion.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
ω p n p rg N g
= = = = 2.50
ω g ng rp N p
N p = 14, N g = 2.50 (14 ) = 35 teeth
π π
pc = = = 0.393 in.
Pd 8
N p + Ng 14 + 35
C= = = 3.063 in.
2 Pd 2 (8)
2π rb
= pc cos ϕ
N
Npc cos ϕ N ( 0.393) cos 20
o
rb = =
2π 2π
rb = 0.059 N
( rb ) p = 0.059 (14 ) = 0.826 in.
( rb ) g = 0.059 ( 35) = 2.065 in.
(e) From Table 15.3, for a 14 tooth pinion, the maximum number of gear teeth without having interference is
26. Since for this gearset
Ng = 35 > 26
546
15-15. Repeat problem 15-9, except use a reduction ratio of 3.50.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
ω p n p rg N g
= = = = 3.50
ω g ng rp N p
N p = 16, N g = 3.50 (16 ) = 56 teeth
π π
pc = = = 0.785 in.
Pd 4
N p + Ng 16 + 56
C= = = 9.0 in.
2 Pd 2 ( 4)
2π rb
= pc cos ϕ
N
Npc cos ϕ N ( 0.785 ) cos 20
o
rb = =
2π 2π
rb = 0.1174 N
( rb ) p = 0.1174 (16 ) = 1.878 in.
( rb ) g = 0.1174 ( 56 ) = 6.574 in.
(e) From Table 15.3, for a 16 tooth pinion, the maximum number of gear teeth without having interference is
101. Since for this gearset
Ng = 56 < 101
547
15-16. Repeat problem 15-9, except use a diametral pitch of 12 and a reduction ratio of 7.50.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
ω p n p rg N g
= = = = 7.50
ω g ng rp N p
N p = 16, N g = 7.50 (16 ) = 120 teeth
π π
pc = = = 0.262 in.
Pd 12
N p + Ng 16 + 120
C= = = 5.67 in.
2 Pd 2 ( 4)
2π rb
= pc cos ϕ
N
Npc cos ϕ N ( 0.262 ) cos 20
o
rb = =
2π 2π
rb = 0.039 N
( rb ) p = 0.039 (16 ) = 0.624 in.
( rb ) g = 0.039 (120 ) = 4.68 in.
(e) From Table 15.3, for a 16 tooth pinion, the maximum number of gear teeth without having interference is
101. Since for this gearset
548
15-17. Repeat problem 15-9, except use a diametral pitch of 12, a 17-tooth pinion, and a reduction ratio of 7.50.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
ω p n p rg N g
= = = = 7.50
ω g ng rp N p
N p = 17, N g = 7.50 (17 ) = 127.5 teeth
This is unacceptable since an integral number of teeth is mandatory. Ng = 127 teeth will be chosen, resulting in a
small deviation in reduction ratio, i.e.,
127
mR = = 7.47
17
π π
pc = = = 0.262 in.
Pd 12
N p + Ng 16 + 127
C= = = 5.96 in.
2 Pd 2 (12 )
2π rb
= pc cos ϕ
N
Npc cos ϕ N ( 0.262 ) cos 20
o
rb = =
2π 2π
rb = 0.039 N
( rb ) p = 0.039 (17 ) = 0.663 in.
( rb ) g = 0.039 (127 ) = 4.953 in.
(e) From Table 15.3, for a 17 tooth pinion, the maximum number of gear teeth without having interference is
∞ . Therefore interference would not be expected.
549
15-18. A Straight-tooth spur gearset has a 19-tooth pinion that rotates at a speed of 1725 rpm. The driven gear is to
rotate at a speed of approximately 500 rpm. If the gear teeth have a module of 2.5, find the following:
b. Circular pitch
c. Center distance
------------------------------------------------------------------------------------------------------------------------------------------
Solution
np Ng 1725
= = = 3.45
ng Np 500
Since N p = 19
N g = 3.45 (19 ) = 65.55
This is unacceptable since an integral number of teeth is mandatory. Ng = 66 will be chosen, resulting in a small
Np
deviation from the specified speed of the driven gear, i.e., ng =
Ng
( n p ) = 19
66
(1725) = 497 rpm . This will be
assumed to be close enough, so Ng = 66 teeth.
pc = π m = π ( 2.5 ) = 7.85 mm
m 2.5
C=
2
( N p + Ng ) =
2
(19 + 66 ) = 106.25 mm
550
15-19. Repeat problem 15-14, except for a pinion that rotates at 3450 rpm.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
np Ng 3450
= = = 6.9
ng Np 500
Since N p = 19
N g = 6.9 (19 ) = 131.1
This is unacceptable since an integral number of teeth is mandatory. Ng = 131 will be chosen, resulting in a small
Np 19
deviation from the specified speed of the driven gear, i.e., ng =
Ng
( n p ) = 131 ( 3450 ) = 500.4 rpm . This will be
assumed to be close enough, so Ng = 131teeth.
pc = π m = π ( 2.5 ) = 7.85 mm
m 2.5
C=
2
( N p + Ng ) =
2
(19 + 131) = 187.5 mm
551
15-20. Repeat problem 15-14, except for a module of 5.0.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
np Ng 1725
= = = 3.45
ng Np 500
Since N p = 19
N g = 3.45 (19 ) = 65.55
Since an integral number of teeth is mandatory choose Ng = 66 teeth which will result in a driven gear speed of
Np
ng =
Ng
( n p ) = 19
66
(1725 ) = 497 rpm . This will be assumed to be close enough.
pc = π m = π ( 5.0 ) = 15.71 mm
m 5.0
C=
2
( N p + Ng ) =
2
(19 + 66 ) = 212.50 mm
552
15-21. Repeat problem 15-14, except for a driven gear that rotates at approximately 800 rpm.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
np Ng 1725
= = = 2.16
ng Np 800
Since N p = 19
N g = 2.16 (19 ) = 41 teeth
pc = π m = π ( 2.5 ) = 7.85 mm
m 2.5
C=
2
( N p + Ng ) =
2
(19 + 41) = 75 mm
553
15-22. A pair of 8-pitch straight-tooth spur gears is being proposed to provide a 3:1 speed increase. If the gears are
mounted on 6-inch centers, find the following:
c. If power supplied to the driving pinion at full load is 10 hp, and power loss at the gear mesh is negligible,
what is the power available at the output gear shaft?
------------------------------------------------------------------------------------------------------------------------------------------
Solution
(a) From (15-10), C = rp + rg = 6.0 inches. For 3:1 speed increase, using (15-15),
ng rp
= =3
np rg
3rg + rg = 6.0
6.0
rg = = 1.5 in. (d g = 3.0 in.)
4
rp = 3rg = 3 (1.5 ) = 4.5 in. (d p = 9.0 in.)
( Pd ) d = 8d = N
N p = 8d p = 8 ( 9 ) = 72 teeth
N g = 8d g = 8 ( 3) = 24 teeth
554
15-23. Repeat problem 15-18, except for a 3:1 speed decrease.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
(a) From (15-10), C = rp + rg = 6.0 inches. For 3:1 speed decrease, using (15-15),
ng rp 1
= =
np rg 3
3rp + rp = 6.0
6.0
rp = = 1.5 in. (d p = 3.0 in.)
4
rg = 3rp = 3 (1.5 ) = 4.5 in. (d g = 9.0 in.)
( Pd ) d = 8d = N
N p = 8d p = 8 ( 3.0 ) = 24 teeth
N g = 8d g = 8 ( 9.0 ) = 72 teeth
555
15-24. A proposed straight full-depth spur gear mesh is to consist of a 21-tooth pinion driving a 28-tooth gear. The
proposed diametral pitch is to be 3, and the pressure angle is 20˚. Determine the following, and where possible,
show each feature on a simple scale drawing of the gear mesh.
c. Pressure angle
j. Circular pitch
k. Tooth thickness
n. Length of action
o. Base pitch
------------------------------------------------------------------------------------------------------------------------------------------
Solution
Np 21
dp = = = 7.00 inches
Pd 3
Ng 28
dg = = = 9.33 inches
Pg 3
556
Npc cos ϕ N cos ϕ ⎛ π d ⎞ d cos ϕ
rb = = =
2π 2π ⎜⎝ N ⎟⎠ 2
d cos ϕ
rb = = 0.470d
2
( rb ) p = 0.470d p = 0.470 ( 7.0 ) = 3.29 inches
(f) From Table 15.1, for Pd = 3, addendum a is ap = 1.000/Pd = 1.000/3 = 0.333 inch
where
7.0 9.33
C = rp + rg = + = 8.17 inches
2 2
( 3.50 + 0.333) − ( 3.50 cos 20o ) + ( 4.67 + 0.333) − ( 4.67 cos 20o ) − 8.17 sin 20o
2 2 2 2
Z=
Z = 1.56 inches (shown as a-b in sketch)
Z 1.56
mp = = = 1.59
pb 0.984
557
Scale: Half size
558
15-25. A proposed straight full-depth spur gear mesh is to have a reduction ratio of 4:1 and a center distance of
7.50 inches. The proposed diametral pitch is to be 3, and the pressure angle is 20˚. Determine the following, and
where possible, show each feature on a simple scale drawing of the gear mesh.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
ng rp 1
= =
np rg 4
rg = 4rp
rp + 4rp = 7.50 inches
7.50
rp = = 1.5 inches
5
559
(e) ( rb ) g = 0.470d g = 0.470 (12.0 ) = 5.64 inches
(f) From Table 15.1, for Pd = 3, addendum a is ap = 1.000/Pd = 1.000/3 = 0.333 inch
(m) Noting that point a, where tooth contact initiates upon approach, lies outside interference point c,
interference does exist.
560
15-26. Preliminary design calculations have suggested that design objectives may be met by a straight spur gearset
using standard full-depth 2 ½ -pitch involute gear teeth, and a 21-tooth pinion meshing with a 28-tooth gear. A 25˚
pressure angle has been selected for this application, and the gear teeth are to be shaved to AGMA quality number
Qv = 8. Find the following:
a. Addendum
b. Dedendum
c. Clearance
d. Circular pitch
f. Base pitch
g. Length of action
i. Module
------------------------------------------------------------------------------------------------------------------------------------------
Solution
1.000 1.000
(a) a = = = 0.400 inch
Pd 2.5
1.250 1.250
(b) b = = = 0.500 inch
Pd 2.5
0.350 0.350
(c) For shaved teeth c = = = 0.140 inch
Pd 2.5
π π
(d) pc = = = 1.257 inches
Pd 2.5
1.571
(e) t = = 0.628 inch
2.5
dp pc N p 1.257 ( 21)
rp = = = = 4.20 inches
2 2π 2π
dg pc N g 1.257 ( 28 )
rg = = = = 5.60 inches
2 2π 2π
C = rp + rg = 4.20 + 5.60 = 9.80 inches
561
(r + a p ) − ( rp cos ϕ ) + (r + ag ) − ( rg cos ϕ ) − C sin ϕ
2 2 2 2
Z= p g
Z 1.64
(h) m p = = = 1.44
pb 1.139
2.54 2.54 mm
(i) m= = = 1.016
Pd 2.5 teeth
562
15-27. Repeat problem 15-22, except for a 20˚ pressure angle.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
1.000 1.000
(a) a = = = 0.400 inch
Pd 2.5
1.250 1.250
(b) b = = = 0.500 inch
Pd 2.5
0.350 0.350
(c) For shaved teeth c = = = 0.140 inch
Pd 2.5
π π
(d) pc = = = 1.257 inches
Pd 2.5
1.571
(e) t = = 0.628 inch
2.5
dp pc N p 1.257 ( 21)
rp = = = = 4.20 inches
2 2π 2π
dg pc N g 1.257 ( 28 )
rg = = = = 5.60 inches
2 2π 2π
C = rp + rg = 4.20 + 5.60 = 9.80 inches
( 4.20 + 0.40 ) − ( 4.20 cos 20o ) + ( 5.60 + 0.40 ) − ( 5.60 cos 20o ) − 9.80sin 20o
2 2 2 2
Z=
Z = 1.89 inches
Z 1.89
(h) m p = = = 1.66
pb 1.139
2.54 2.54 mm
(i) m= = = 1.016
Pd 2.5 teeth
563
15-28. A straight-tooth one-stage spur gear reducer has been used in a high-production home appliance for many
years. The gear pair constitutes about one-half the $50 production cost of the appliance. Consumer complaints
about gear noise have grown over the years and sales are declining. One young engineer has found data that suggest
the noise level would be significantly reduced if the AGMA quality number could be increased from its current
value of Qv = 8 for hobbed gears to a value of Qv = 11, achieved by shaving the gears.
a. Estimate the increase in production cost of the appliance if gear shaving were used to achieve Qv = 11.
b. Can you suggest any other approach that might accomplish the noise-reduction goal without restoring to
shaving the gears?
------------------------------------------------------------------------------------------------------------------------------------------
Solution
Examining Figure 15.21, it may be noted that the cost scale is logarithmic.
(a) Under current practice, the curve representing hobbing (cutting) intersects the Qv = 8 level at “unit cost”
level (call it point A). In this case, hobbing the gears represents about ½ the cost of producing the appliance
($50) so cost of hobbing the gears is now about Chobb = $25.00. The “shaving” cost curve intersects the Qv
= 11curve at a higher cost level (call it point B). On the log scale from unit cost to 10 times cost, the
shaving cost curve is about 0.3 of the interval corresponding to about 2 times unit cost on the logarithmic
scale (log 2 ≈ 0.3). Thus the cost of shaving would be about Cshave = 2Chobb = 2(25.00) = $50.00 and the
production cost therefore would increase to about Ctotal-shaved = $50.00 + $25.00 = $75.00 and the production
cost, therefore, would increase from about $50 to about $75, a 50 percent increase in cost.
(b) Using helical gears might be a better and less costly approach.
564
15-29. A straight-toothed full-depth involute spur pinion with a pitch diameter of 100 mm is mounted on input
shaft driven by an electric motor at 1725 rpm. The motor supplies a steady torque of 225 N-m.
a. If the involute gear teeth have a pressure angle of 20˚, determine the transmitted force, the radial separating
force, and the normal resultant force on the pinion teeth at the pitch point.
c. Calculate the percent difference in resultant force if the pressure angle were 25˚ instead of 20˚.
d. Calculate the percent difference in resultant force if the pressure angle were 14 ½ ˚ instead of 20˚.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
Tp 225
Ft = = = 4500 N
rp 0.050
Fr = Ft tan ϕ = 4500 tan 20o = 1638 N
Ft 4500
( Fn )20 = = = 4789 N
cos ϕ cos 20o
o
Tn 225 (1725 )
(b) From (4-41), kw = = = 40.65 kilowatts
9549 9549
565
15-30. Referring to the two-stage gear reducer sketched in Figure P15.3, concentrate attention on the first-stage
mesh between the pinion (1) and the gear (2). The pinion is being driven by a 1-kw, 1725-rpm electric motor,
operating steadily at full capacity. The tooth system has a diametrical pitch of 8 and a pressure angle of 25˚. Do the
following for the first-stage gear mesh:
a. Sketch the gearset comprised of pinion 1 (driver) and gear 2 (driven) taken together as a free body, and
assume that shaft support bearings are symmetrically straddle mounted about the gear on each shaft. Show
all external forces and torques on the free body, speeds and directions of the two gears (refer to Figure
P15.3), and the line of action.
b. Sketch the pinion, taken alone as a free body. Show all external forces and torques on the pinion, including
the driving torque, tangential forces, separating force, and bearing reaction forces. Give numerical values.
c. Sketch the driven gear, taken as a free body. Show all external forces and torques on the gear, including
the driving torque, tangential forces, separating force, and bearing reaction forces. Give numerical values.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
(a) the gearset comprised of pinion 1 and gear2, taken together as a free body may be sketched after
determining the following information:
Np 16
dp = = = 2.0 inches = 50.8 mm
Pd 8
Ng 48
dg = = = 6.0 inches = 152.4 mm
Pd 8
dp 2.00
ng = np = (1725 ) = 575 rpm
dg 6.00
kw ( 9549 ) (1)( 9549 )
Tp = = = 5.54 N-m
np 1725
kw ( 9549 ) (1)( 9549 )
Tg = = = 16.61 N-m
ng 575
566
Gearset as a Free Body
(b)
567
Tp 5.54
Ft = = = 218 N
rp 0.0254
Fr = Ft tan 25o = 218 tan 25o = 101.7 N
Ft 218
Fn = = = 240.5 N
cos ϕ cos 25o
R pv = − Fr = −101.7 N (down)
R ph = − Ft = −218 N (down)
Summing moments about O2,
Ft rg = Tg
Tg 16.61
Ft = = = 218 N
rg 0.0762
Fr = Ft tan ϕ = 218 tan 25o = 101.7 N
Ft 218
Fn = = = 240.6 N
cos ϕ cos 25o
Rgv = 101.7 N (up)
Rgv = 218 N (up)
h
Rg = 240.6 N
568
15-31. In the two-stage gear reducer sketched in Figure P15.31, concentrate attention on the compound shaft “B,”
with attached gears 2 and 3, taken together as the free body of interest. The gears have standard 20˚ involute full-
depth teeth, with a diametral pitch of 6. The motor driving the input shaft “A” is a 20-hp, 1725-rpm electric motor
operating steadily at full rated power. For the chosen free body, do the following:
a. Clearly sketch a top view of shaft “B,” and show all horizontal components of the loads and reactions.
b. Sketch a front view (elevation) of shaft “B,” and show all vertical components of the loads and reactions.
c. If shaft “B” is to have a uniform diameter over its whole length, identify potential critical points that should
be investigated when designing the shaft.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
(a) Prior to making the sketch, the following information may be determined. From (15-8), for input pinion
N1 24
d1 = = = 4.0 inches
Pd 6
N 2 36
d2 = = = 6.0 inches
Pd 6
d1 4.0
n2 = ( n1 ) = (1725) = 1150 rpm
d2 6.0
63, 025 ( hp ) 63, 025 ( 20 )
T2 = = = 1096 in − lb
n2 1150
Ft 2 r2 = T2
T2 1096
Ft 2 = = = 365 lb
r2 ( 6.0 2 )
Fr 2 = Ft 2 tan ϕ = 365 tan 20o = 133 lb
N 3 18
d3 = = = 3.0 inches
Pd 6
T3 = T2 = 1096 in-lb
T3 1096
Ft 3 = = = 730 lb
r3 ( 3.0 2 )
Fr 3 = Ft 3 tan ϕ = 730 tan 20o = 266 lb
Assuming both gears to be straddle mounted between closely spaced bearings, the top view of shaft B may be
sketched as follows:
569
(b) For the front view we have
(c) Torsional moment is constant over the whole shaft length, so if the diameter is constant, maximum
torsional shear stress is constant over whole surface of the shaft.
Direct shear (or transverse shear) is generated in the shaft at the edge of each bearing, where it adds to the
torsional shear, stress concentration probably is a concern at this location as well.
The conclusion then is, since bearing reaction forces are larger at gear 3 than at gear 2, and shaft torque is
the same, the governing critical section is adjacent to gear 3, and the governing critical point lies at the
shaft surface.
570
15-32. Referring again to Figure P15.4, note that the ring gear is fixed, the sun gear is driven at 1200 rpm in the
CCW direction with a torque of 20 N-m, and the two-planet carrier arm is used as output. The 20˚ involute gears
have a module of 2.5. Do the following:
b. Determine the pitch diameter of each gear in the train, and verify that they are physically compatible in the
assembly.
c. Find the center distance between planets on the 2-planet carrier arm.
d. Sketch each member of the train as a free body, showing numerical values and directions of all forces and
torques on each free body.
g. Calculate the nominal radial load on each of the bearings in the assembly, neglecting gravitational forces.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
pc N1 2.5π ( 30 )
d1 = = = 75 mm
π π
pN 2.5π ( 20 )
d2 = c 2 = = 50 mm
π π
pN 2.5π ( 70 )
d3 = c 3 = = 175 mm
π π
To be physically compatible, referring to Figure P15.4, d1 + 2d2 = d3 , checking this equation by using the above
values gives 75 + 2(50) = 175 and 175 = 175. Thus, the pitch diameters are physically compatible.
(c) From geometry of Figure P15.4 and using C = d1 + 2r2 = 75 + 2(50/2) = 125 mm
(d) A sketch of each member of the train is given as below. For sun gear 1, taking moments about O1,
T1 = Ft1r1
T1 20
Ft1 = = = 533.3 N
r1 ( 0.075 2 )
Assuming the tangential force is equally divided between the two planet mesh sites,
533.3
( Ft1 )each mesh site = = 266.7 N
2
( Fr1 )each mesh site = Ft1 tan 20o = 266.7 tan 20o = 97 N
By equilibrium resolution the forces and moments on each free body may be found as shown on the sketch
below.
571
(f) Using (15-4), and noting that ring gear 3 is fixed (ω3 = 0), sun gear 1 is used as input (ω1 = ωin), and carrier
arm 4 is used as output (ω4 = ωout), and if first and last gears in the train are taken as sun gear 1 and ring
gear 3 we have
ω3 − ω4 0 − ω4 ⎛ 30 ⎞ ⎛ 20 ⎞
= = − ⎜ ⎟ ⎜ ⎟ = −0.43
ω1 − ω4 ω1 − ω4 ⎝ 20 ⎠ ⎝ 70 ⎠
ω4 ωout n
= = +0.30 = out
ω1 ωin nin
nout = 0.30 (1200 ) = 360 rpm CCW
(g) Examining the sketch above, the only bearings in the system (neglecting gravitational forces) that are
subjected to a nonzero radial load are the two planet bearings. The normal resultant radial load on each
planet bearing is, from the sketch, R2 = 533.3 N.
572
15-33. A 10-pitch 20˚ full-depth involute gearset with a face width of 1.25 inches is being proposed to provide a
2:1 speed reduction for a conveyor drive unit. The 18-tooth pinion is to be driven by a 15-hp, 1725-rpm electric
motor operating steadily at full rated power. A very long life is desired for this gearset, and the reliability of 99
percent is required. Do the following:
a. Using the simplified approach. Estimate the nominal bending stress at the tension-side root fillet of the
driving pinion.
b. Estimate the fatigue stress concentration factor for the tension-side root fillet of the driving pinion.
c. Calculate the actual bending stress at the tension-side root fillet of the driving pinion.
d. Repeat (c) for the tension-side root fillet of the driven gear.
e. Based on the recommendation of an in-house materials specialist, Grade 1 AISI 4620 hot-rolled steel is to
be used for both the pinion and the gear (see Tables 3.3 and 3.13), and the value of k∞ [see (5-57)] has been
estimated for this application to be 0.75, including the 99 percent reliability requirement but not including
stress concentration effects. Estimate the existing safety factor at the tension-side root fillet of whichever
of the gears is more critical, based on tooth bending fatigue.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
(a) By specification
np
=2
ng
np 1725
ng = = = 863 rpm
2 2
Np 18
rp = = = 0.90 inch
2 Pd 2 (10 )
2π rp n p 2π ( 0.9 )(1725 ) ft
V= = = 812.9
12 12 min
33, 000 ( hp ) 33, 000 (15 )
( Ft ) p = = = 609 lb
V 812.9
( Ft ) p Pd 609 (10 )
(σ b )nom = = = 15, 767 psi
root
pinion bYp 1.25 ( 0.309 )
(b) Using Figure 4.23, taking e as working depth (2.000/Pd from Table 15.1), and h as base thickness [circular
thickness (1.1571/Pd from Table 15.1) multiplied by cosφ],
e 2.000 Pd 2.000 10
≈ = = 1.35
h (1.571 Pd ) cos 20o (1.571 10 ) cos 20o
Extrapolating in Figure 4.23 for a standard cutter tooth tip radius, K t ≈ 1.5 . Using an estimate of root fillet radius of
573
0.35 0.35
ρf = = = 0.035 inch
Pd 10
The value of q for AISI 4620 hot rolled steel (Su = 87,000 psi from Table 3.3), from Figure 5.46, is q ≈ 0.72 , so (15-
37) gives Kf = (0.72)(1.5 – 1) + 1 = 1.4
(σ b )actual
root
= K f (σ b )nom
root
= 1.4 (15, 767 ) = 22, 075 psi.
pinion pinion
(d) Following the same reasoning for the gear, from (15-15)
ng Np
=
np Ng
⎛ np ⎞ ⎛ 1725 ⎞
Ng = ⎜ ⎟ N p = ⎜
⎜n ⎟ ⎟ (18 ) = 36 teeth
⎝ g⎠ ⎝ 863 ⎠
36
rg = = 1.80 inches
2 (10 )
( Ft ) g = ( Ft ) p = 609 lb
609 (10 )
(σ b )nom = = 12,889 psi
root
gear 1.25 ( 0.378)
Since the result above will be that same for the gear as for the pinion, using (15-38) for the gear gives
(σ b )actual
root
= K f (σ b )nom
root
= 1.4 (12,889 ) = 18, 045 psi.
gear gear
(e) From the above results it is seen that the pinion is more critical. From Table 3.3, for hot rolled AISI 4620
steel, Su = 87,000 psi, and Syp = 63,000 psi. Since S-N data for AISI 4620 are not available, the method of
section 5.6 for estimating the fatigue endurance limit for the material, S ′f , gives, since Su < 200 ksi,
Since the pinion teeth experience one-way bending, at the pinion root fillet, based on the actual stress at the root
pinion
574
22, 075 − 0
σa = = 11, 038 psi
2
22, 075 + 0
σm = = 11, 038 psi
2
σa 11,308
σ eq −CR = = = 12, 642 psi
σm 11,308
1− 1−
Su 87, 000
Sf 32, 625
nex = = ≈ 2.6
σ eq −CR 12, 642
575
15-34. For the gearset specifications of problem 15-33, do the following:
a. Using the simplified approach, estimate the surface fatigue wear stress for the meshing gear teeth..
b. If the Grade 1 4620 gear teeth are carburized and case hardened (not including the root fillet) to a hardness
of approximately RC 60, maintaining the 99 percent reliability requirement, and recalling that very long life
is desired, determine the surface fatigue strength of the case-hardened teeth.
c. Estimate the existing safety factor based on surface fatigue wear failure.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
⎛ 2 2 ⎞
Ft ⎜ +
⎜ d sin ϕ d sin ϕ ⎟⎟
σ sf = ⎝ p g ⎠
⎛ 1 −ν p2 1 −ν g2 ⎞
π b cos ϕ ⎜⎜ + ⎟
⎝ Ep Eg ⎟⎠
np
=2
ng
1725
ng = = 863 rpm
2
N p 18
dp = = = 1.8 inches
Pd 10
⎛ np ⎞
N g = ⎜ ⎟ N p = 2 (18 ) = 36 teeth
⎜n ⎟
⎝ g⎠
N g 36
dg = = = 3.6 inches
Pd 10
2π rp n p 2π (1.8 2 )(1725 ) ft
Vg = V p = = = 812.9
12 12 min
33, 000 ( hp ) 33, 000 (15 )
Ft = = = 609 lb
V 812.9
From Table 3.9, for steel Ep = Eg = E = 30 x 106 psi, and νp = νg = ν = 0.3, thus with these values we have
⎛ 2 2 ⎞
609 ⎜ o
+ o ⎟
σ sf = ⎝ 1.8sin 20 3.6sin 20 ⎠ = 115,135 psi.
o ⎛ 1 − 0.3 1 − 0.32 ⎞
2
π (1.25 ) cos 20 ⎜ + 6 ⎟
⎝ 30 × 10 30 × 10 ⎠
6
(b) From Figure 15.29, the 90% reliability surface fatigue strength Ssf at a life of N = 1010 cycles, for steel gears
case hardened to RC 60, may be read (by extrapolation) as
(S ) sf N =1010
= 90, 000 psi ( R = 90%)
⎛ 0.81 ⎞
(S ) sf N =1010
=⎜ ⎟ 90, 000 = 81, 000 psi ( R = 99%)
⎝ 0.90 ⎠
576
(S )
sf N =1010 81, 000
(c) nex = = = 0.70 (clearly unacceptable)
σ sf 115,135
577
15-35. Using the simplified approach (do not refine results by using AGMA equations), design a single-reduction
straight spur gear unit to operate from a 5.0-hp electric motor running at 900 rpm to drive a rotating machine
operating at 80 rpm. The motor is to operate steadily at full rated power. Near-infinite life is desired. A reliability
of 90 percent is acceptable for this application. It is proposed to use ASTM A-48 (class 40) gray cast-iron material
for both gears. Using k∞ = 0.70, properties for this material may be based on Chapter 3 data, except for surface
fatigue strength, which may be taken as ( Ssf ) N =108
= 28,000 psi. A safety factor of 1.3 is desired. As part of the
design procedure, select or determine the following so as to safety design specifications:
a. Tooth system
b. Quality level
c. Diametral pitch
d. Pitch diameters
e. Center distance
f. Face width
------------------------------------------------------------------------------------------------------------------------------------------
Solution
Following the design procedure suggested in 15.11, for a single-reduction straight spur gear unit:
A first conceptual sketch may be based on the following criteria and specifications:
hp = 5.0 horsepower
assume no losses
np = 900 rpm
ng = 80 rpm
Lf ≈ ∞
R = 90%
k∞ = 0.70
ν = 0.21-0.27
578
By specification: (S )
sf N =108
= 28, 000 psi , nd = 1.3. By specification np/ng = 900/80 = 11.25
(a) Tentatively, choose a standard full depth AGMA involute profile, 20˚ pressure angle tooth system as
defined in Table 15.1
(b) From Table 15.4, for the application specified, since speeds are low, nominal accuracy seems sufficient.
This corresponds to a AGMA quality level of Qv equal to 6 or 7.
(c) Following the guidance of 15.11, step 6, a tentative selection will be made of Pd = 10.
(d) From Table 15.3, to avoid interference and undercutting, since the reduction ratio is so large, tentatively
select Np = 17 teeth, so from (15-8), for Pd = 10,
Np 17
dp = = = 1.7 inches
Pd 10
⎛ np ⎞
dg = ⎜
⎜n ⎟⎟ d p = 11.25 (1.7 ) = 19.125 inches
⎝ g ⎠
9 14
(f) From (15-16) as a guide line, select b as ≤b≤ , or 0.9 ≤ b ≤ 1.4 . First try b = 1.25 inch.
10 10
579
⎛ dg ⎞ ⎛ 19.125 ⎞
Ng = ⎜ ⎟ N p = ⎜
⎜d ⎟ ⎟ (17 ) ≈ 191 teeth
⎝ p⎠ ⎝ 1.7 ⎠
2π rp n p 2π (1.7 2 )( 900 ) ft
V= = = 401
12 12 min
33, 000 ( hp ) 33, 000 ( 5.0 )
Ft = = = 411.5 lb
V 401
Yp = 0.303 (Table 15.5 for N p = 17)
Ft Pd 411.5 (10 )
(σ b )nom = = = 10,864 psi
pinion
root bYp 1.25 ( 0.303)
2.000 2.000
e= = = 0.2 inch
Pd 10
1.571 1.571
h= cos ϕ = cos 20o = 0.15 inch
Pd 10
e 0.2
= ≈ 1.3
h 0.15
K t ≈ 1.5 (Extrapolating in Figure x.xx (4.23))
q = 1.0
K f = K t = 1.5
(σ b )actual
pinion
= K f (σ b )nom
pinion
= 1.5 (10,864 ) = 16,300 psi
root root
Since from Table 15.5, Yg > Yp, the gear tooth root stress will be smaller (less critical) so the pinion governs. Using
the procedures of section 5.6, the long life fatigue strength for cast iron is, since Sut = 40,000 psi < 88,000 psi,
S ′f = 0.4 ( 40,000 ) = 16, 000 psi and from (5-55), Sf = 0.7(16,000) = 11,200 psi. Since the pinion teeth experience
one-way bending, at the pinion tension-side root fillet
16,300 − 0
σa = = 8150 psi
2
16,300 + 0
σm = = 8150 psi
2
σa 8,150
σ eq −CR = = = 10, 235 psi
σm 8,150
1− 1−
Su 40, 000
Sf 11, 200
nex = = = 1.1
σ eq −CR 10, 235
This does not meets the criterion of nd = 1.3, so the design must be improved, probably by selecting a new value of
Pd, e.g., Pd = 8.
17
dp = = 2.13 inches
8
d g = 11.25 ( 2.13) = 23.96 inches
580
9 14
Again select b as ≤b≤ , or 1.13 ≤ b ≤ 1.75 . This time try b = 1.50 inches.
8 8
2π rp n p 2π ( 2.13 2 )( 900 ) ft
V= = = 502
12 12 min
33, 000 ( hp ) 33, 000 ( 5.0 )
Ft = = = 329 lb
V 502
329 ( 8 )
(σ b )nom = = 5790 psi
pinion
root 1.50 ( 0.303)
5, 790 − 0
σa = = 2895 psi
2
5, 790 + 0
σm = = 2895 psi
2
2895
σ eq − CR = = 3,120 psi
2895
1−
40, 000
Sf 11, 200
nex = = = 3.6
σ eq −CR 3,120
This meets the criteria of nd = 1.3. Before attempting to optimize the design for bending (trying to make
changes that move the value of nex toward 1.3), surface fatigue durability should be checked. Using
midrange values we have E = 18.5 x 106 psi and ν = 0.24. From (15-44) we have
⎛ 2 2 ⎞
Ft ⎜ +
⎜ d sin ϕ d sin ϕ ⎟⎟
σ sf = ⎝ p g ⎠
⎛ 1 −ν p 1 −ν g2 ⎞
2
π b cos ϕ ⎜⎜ + ⎟
⎝ Ep Eg ⎟⎠
⎛ 2 2 ⎞
329 ⎜ o
+ o ⎟
= ⎝ 2.13sin 20 23.96sin 20 ⎠ = 46, 690 psi
⎛ 1 − 0.242 ⎞
π (1.5 ) cos 20o ⎜ 6 ⎟( )
2
⎝ 18.5 × 10 ⎠
(S )
sf N =108 28, 000
nex = = = 0.60
σ sf 46, 690
This is clearly unacceptable. The required nex is 1.3. While an increase face width would lower σsf a little,
it would still be unacceptable. For another iteration, try Pd = 6. Repeating
17
dp = = 2.83 inches
6
d g = 11.25 ( 2.83) = 31.84 inches
9 14
Again select b as ≤b≤ , or 1.5 ≤ b ≤ 2.33 . This time try b = 2.3 inches.
6 6
581
2π rp n p 2π ( 2.83 2 )( 900 ) ft
V= = = 667
12 12 min
33, 000 ( hp ) 33, 000 ( 5.0 )
Ft = = = 247 lb
V 667
⎛ 2 2 ⎞
247 ⎜ o
+ o ⎟
σ sf = ⎝ 2.83sin 20 31.84sin 20 ⎠ = 28,330 psi
⎛ 1 − 0.242 ⎞
π ( 2.3) cos 20o ⎜ 6 ⎟( )
2
⎝ 18.5 × 10 ⎠
(S )
sf N =108 28, 000
nex = = ≈ 1.0
σ sf 28,330
This is a good improvement, but still does not meet the requirement of nd = 1.3. For a third iteration, try Pd
= 4. Repeating the calculations gives
17
dp = = 4.25 inches
4
d g = 11.25 ( 4.25 ) = 47.81 inches
9 14
Again select b as ≤ b ≤ , or 2.25 ≤ b ≤ 3.5 . This time try b = 2.75 inches.
4 4
2π rp n p 2π ( 4.25 2 )( 900 ) ft
V= = = 1001
12 12 min
33, 000 ( hp ) 33, 000 ( 5.0 )
Ft = = = 165 lb
V 1001
⎛ 2 2 ⎞
165 ⎜ o
+ o ⎟
σ sf = ⎝ 4.25sin 20 47.81sin 20 ⎠ = 17, 290 psi
⎛ 1 − 0.242 ⎞
π ( 2.75 ) cos 20o ⎜ 6 ⎟( )
2
⎝ 18.5 × 10 ⎠
28, 000
nex = ≈ 1.6
17, 290
This safety factor is higher than specified. For a forth iteration, try Pd = 5. Thus,
17
dp = = 3.40 inches
5
d g = 11.25 ( 3.40 ) = 38.25 inches
9 14
Again select b as ≤b≤ , or 1.80 ≤ b ≤ 2.80 . This time try b = 2.30 nches.
5 5
2π rp n p 2π ( 3.40 2 )( 900 ) ft
V= = = 801
12 12 min
33, 000 ( hp ) 33, 000 ( 5.0 )
Ft = = = 206 lb
V 801
582
⎛ 2 2 ⎞
206 ⎜ o
+ o ⎟
σ sf = ⎝ 3.40sin 20 38.25sin 20 ⎠ = 23, 615 psi
o ⎛ 1 − 0.24 ⎞
2
π ( 2.30 ) cos 20 ⎜ 6 ⎟( )
2
⎝ 18.5 × 10 ⎠
28, 000
nex = ≈ 1.2
23, 615
Increasing the face width to b = 2.75 inches gives σsf = 21,597 psi. and
28, 000
nex = ≈ 1.3
21,597
This is a good solution; many other good solutions can also be found. The following recommendations should meet
specifications and provide proper operation for a very long time.
Pd = 5
Np = 17 teeth
Ng = 191 teeth
dp = 3.40 inches
dg = 38.25 inches
b = 2.75 inches
3.40 + 38.25
C= = 20.83 inches
2
583
15-36. A 10-pitch 20˚ full-depth involute gearset, with Qv = 10 and a face width of 1.25 inches, is being proposed
to provide a 2:1 speed reduction for a conveyor drive unit. The 18-tooth pinion is to be driven by a 15-hp, 1725-rpm
electric motor operating steadily at full-rated power. A very long life is desired for this gearset, and a reliability of
99 percent is required. Do the following:
a. Using the AGMA refined approach, calculate the tooth bending stress at the tension-side root fillet of the
driving pinion.
d. If the proposed material for both gears is Grade 1 4620 steel, and the teeth are carburized and case
hardened (not including the root fillet) to a hardness of approximately RC 60, maintaining the 99 percent
reliability requirement, determine the AGMA surface fatigue strength (pitting resistance) for the
carburized and case-hardened gear teeth.
b. If the proposed material for both gears is AISI 4620 through-hardened to BHN 207, estimate the existing
safety factor at the tension-side root fillet of whichever gear is more critical, based on tooth bending
fatigue.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
(a) By specification:
R = 99% (required), Life requirement: very long, Tooth system: 20˚ full-depth involute teeth
ng = np/2 = 1725/2 = 863 rpm, Ng = 2(18) = 36 teeth, dp = Np/Pd = 18/10 = 1.8 inches.
From (15-23)
2π rp n p 2π (1.8 2 )(1725 ) ft
V= = = 812.9
12 12 min
33, 000 ( hp ) 33, 000 (15 )
Ft = = = 609 lb
V 812.9
Ft Pd
σb = Ka Kv K m K I
bJ
Substituting yields
609 (10 )
σb = (1.25)(1.17 )(1.6 )(1.0 ) = 47,500 psi
(1.25)( 0.24 )
(b) Same calculations; actually, gear tooth fillet stress is slightly lower in most cases.
584
′ . Using Figure 15.25, assuming Grade 1, with BHN = 207,
(c) From (15-42), Stbf = YN Rg Stbf
′ = 28, 500 psi and from Figure 15.28, for very long life (1010 cycles) YN = 0.8, and from Table 15.13, for
Stbf
99 % reliability, Rg = 1.0. Thus we have Stbf = ( 0.8 )(1.0 )( 28,500 ) = 22,800 psi and from (15-43)
(S )
tbf N =1010 22,800
nex = = ≈ 0.50
(σ b ) 47,500
585
15-37. For the gearset specifications of problem 15-36, do the following:
c. Using the AGMA refined approach, calculate the surface fatigue contact stress for the meshing gear teeth.
d. Repeat (b) for the tension-side root fillet of the driven gear.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
(a) By specification:
R = 99% (required), Life requirement: very long, Tooth system: 20˚ full-depth involute teeth
ng = np/2 = 1725/2 = 863 rpm, Ng = 2(18) = 36 teeth, dp = Np/Pd = 18/10 = 1.8 inches.
From (15-23)
2π rp n p 2π (1.8 2 )(1725 ) ft
V= = = 812.9
12 12 min
33, 000 ( hp ) 33, 000 (15 )
Ft = = = 609 lb
V 812.9
Ft
σ sf = C p K a Kv K m
bd p I
Noting from Table 3.9 that for steel material E = 30 x 106 psi and ν = 0.3, the elastic coefficient Cp may be
evaluated as
1
Cp = = 2, 290
⎛ 1 − 0.32 ⎞
6 ⎟( )
π⎜ 2
⎝ 30 × 10 ⎠
The geometry factor I may be evaluated from the gear ratio mG where mG = N g N p = 36 18 = 2 , giving
Substituting yields
609
σ sf = 2, 290 (1.25)(1.17 )(1.6 ) = 176,185 psi
(1.25 )(1.8 )( 0.107 )
586
(b) From (15-17), S sf = Z n RG S sf′ , using Table 15.15, assuming Grade 1, the surface fatigue strength of
carburized and hardened steel may be read as S sf′ = 180, 000 psi , and from Table 15.13 RG = 1.0 and from
Figure 15.31, for a life of 1010 cycles, Z N =1010 = 0.67 , so
S sf 120, 600
nex = = ≈ 0.7
σ sf 176,185
587
15-38. Using the AGMA refined approach, design a high-precision (Qv = 12) single-reduction straight spur gear
unit to operate from a 50-hp electric motor running at 5100 rpm to drive a rotating machine operating at 1700 rpm.
The motor operates steadily at full rated power. A life of 107 pinion revolutions is desired, and a reliability of 99
percent is required. It is proposed to use Grade 2 AISI 4620 steel carburized and case hardened to RC 60 for both
gears. An important design constraint is to make the unit as compact as practical (i.e., use the minimum possible
number of pinion teeth without undercutting). A safety factor of 1.3 is desired. Select or determine the following so
as to satisfy the design specifications:
a. Tooth system
b. Diametral pitch
c. Pitch diameters
d. Center distance
e. Face width
------------------------------------------------------------------------------------------------------------------------------------------
Solution
Following the design procedure suggested in 15.11, for a single-reduction straight spur gear unit:
A first conceptual sketch may be based on the following criteria and specifications:
hp = 50 horsepower
Qv = 12
np = 5100 rpm
ng = 1700 rpm
(L )
p d = 107 rev
R = 99 %
Material (both gear): AISI 4620, Grade 2, carburized and case hardened to RC 60
nd = 1.3
588
(a) Tentatively, choose a standard full depth AGMA involute profile, 20˚ pressure angle tooth system as
defined in Table 15.1
(b) Following the guidance of 15.11, step 6, a tentative selection will be made of Pd = 10.
(c) From Table 15.9, to avoid interference and undercutting, for a reduction ratio of 3:1, tentatively select Np =
21 teeth, so from (15-8), for Pd = 10,
Np 21
dp = = = 2.1 inches
Pd 10
⎛ np ⎞
dg = ⎜
⎜n ⎟⎟ d p = 3 ( 2.1) = 6.3 inches
⎝ g ⎠
9 14
(e) From (15-16) as a guide line, select b as ≤b≤ , or 0.9 ≤ b ≤ 1.4 . First try b = 1.25 inch.
10 10
⎛ dg ⎞ ⎛ 6.3 ⎞
Ng = ⎜ ⎟ N p = ⎜
⎜d ⎟ ⎟ ( 21) = 63 teeth
⎝ p⎠ ⎝ 2.1 ⎠
2π rp n p 2π ( 2.1 2 )( 5100 ) ft
V= = = 2804
12 12 min
33, 000 ( hp ) 33, 000 ( 50 )
Ft = = = 588 lb
V 2804
FP
σ b = t d K a Kv K m K I
bJ
589
Km = 1.3 (Table 15.7 with precision gears)
Substituting yields
588 (10 )
σb = (1.0 )(1.1)(1.3)(1.0 ) = 19, 780 psi
(1.25)( 0.34 )
′ , from Table 15.10, for carburized and case hardened steel (RC 55-64), Grade 2,
From (15-42), Stbf = YN RG Stbf
′ = 65, 000 psi . From Figure 15.28, for 107 cycles, YN = 1.0, and from Table 15.13, for 99 % reliability, RG =
Stbf
1.0. So Stbf = (1.0 )(1.0 )( 65, 000 ) = 65, 000 psi and from (15-43)
Since this exceeds the desired safety factor of 1.3, it will temporarily be accepted, until surface durability can be
examined. From (15-46)
Ft
σ sf = C p Ka Kv Km
bd p I
Noting from Table 3.9 that for steel material E = 30 x 106 psi and ν = 0.3, the elastic coefficient Cp may be evaluated
as
1
Cp = = 2, 290
⎛ 1 − 0.32 ⎞
6 ⎟( )
π⎜ 2
⎝ 30 × 10 ⎠
The geometry factor I may be evaluated from the gear ratio mG where mG = N g N p = 63 21 = 3 , giving
588
σ sf = 2, 290 (1.0 )(1.1)(1.3) = 118,314 psi
(1.25 )( 2.1)( 0.12 )
From (15-47), S sf = Z n RG S sf′ , using Table 15.15, assuming Grade 2, the surface fatigue strength of carburized and
hardened steel may be read as S sf′ = 225, 000 psi , and from Table 15.13 RG = 1.0 and from Figure 15.31, for a life
of 1010 cycles, Z N =107 = 1.0 , so
590
S sf 225, 000
nex = = = 1.9
σ sf 118,314
Comparing the bending fatigue with the surface fatigue we see that the surface fatigue is more critical (1.9 < 3.3),
and the existing factor of safety is substantially larger than the desired value of 1.3. To reduce nex toward 1.3, try a
second iteration using Pd = 12. Thus,
Np 21
dp = = = 1.75 inches
Pd 12
d g = 3 (1.75 ) = 5.25 inches
9 14
From (15-16) as a guide line, select b as ≤b≤ , or 0.75 ≤ b ≤ 1.17 . Try b = 1.0 inch.
12 12
⎛ dg ⎞ ⎛ 6.3 ⎞
Ng = ⎜ ⎟ N p = ⎜
⎜d ⎟ ⎟ ( 21) = 63 teeth
⎝ p⎠ ⎝ 2.1 ⎠
2π rp n p 2π (1.75 2 )( 5100 ) ft
V= = = 2337
12 12 min
33, 000 ( hp ) 33, 000 ( 50 )
Ft = = = 706 lb
V 2337
706
σ sf = 2, 290 (1.0 )(1.1)(1.3) = 158, 780 psi
(1.0 )(1.75 )( 0.12 )
225, 000
nex = = 1.4
158, 780
706
σ sf = 2, 290 (1.0 )(1.1)(1.3) = 177, 520 psi
( 0.8 )(1.75 )( 0.12 )
225, 000
nex = ≈ 1.3
177,520
706 (12 )
σb = (1.0 )(1.1)(1.3)(1.0 ) = 44,540 psi
( 0.8)( 0.34 )
65, 000
nex = = 1.5
44,540
This too is acceptable. Based on the final iteration, the following responses are offered:
591
(a) Tooth system: AGMA standard full depth 20˚ involute profile.
592
15-39. A right-hand helical gear, found in storage, has been determined to have a transverse circular pitch of
26.594 mm and a 30˚ helix angle. For this gear, calculate the following:
a. Axial pitch
b. Normal pitch
------------------------------------------------------------------------------------------------------------------------------------------
Solution
pt 26.594
(a) From (15-52), px = = = 46.062 mm
tanψ tan 30o
pc pt 26.594 mm
(c) From (15-13), ( m )t = = = = 8.465
π π π tooth
pn 23.031 mm
(d) From (15-13), ( m )n = = = 7.331
π π tooth
593
15-40. The preliminary design proposal for a helical gearset running on parallel shafts proposes a left-hand 18-
tooth pinion meshing with a 32-tooth gear. The normal pressure angle is 20˚, the helix angle is 25˚, and the normal
diametral pitch is 10. Find the following:
c. Axial pitch
------------------------------------------------------------------------------------------------------------------------------------------
Solution
π π
(a) From (15-54), pn = = = 0.314 inch
Pn 10
pn 0.314
(b) From (15-51), pt = = = 0.346 inch
cosψ cos 25o
pn 0.314
(c) From (15-52), px = = = 0.743 inch
sinψ sin 25o
⎡ tan ϕn ⎤ −1 ⎡ tan 20 ⎤
o
(e) From (15-56), ϕt = tan −1 ⎢ ⎥ = tan ⎢ o ⎥
= 21.88o
⎣ cosψ ⎦ ⎣ cos 25 ⎦
N p pt 18 ( 0.346 )
dp = = = 1.98 inch
(f) From (15-50),
π π
32 ( 0.346 )
dg = = 3.52 inch
π
2.250 2.250
(g) From Table 15.17, for Pn < 20, Whole depth = = = 0.225 inch
Pn 10
594
15-41. Repeat problem 15-40, except that the 18-tooth helical pinion is right-hand.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
Solution is the same as for 15-36 above, the “hand” of the pinion plays no role in these calculations.
595
15-42. Repeat problem 15-40, except that the normal diametral pitch is 16.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
π π
(a) From (15-54), pn = = = 0.196 inch
Pn 16
pn 0.196
(b) From (15-51), pt = = = 0.216 inch
cosψ cos 25o
pn 0.196
(c) From (15-52), px = = = 0.464 inch
sinψ sin 25o
N p pt 18 ( 0.216 )
dp = = = 1.24 inch
(f) From (15-50),
π π
32 ( 0.216 )
dg = = 2.20 inch
π
2.250 2.250
(g) From Table 15.17, for Pn < 20, Whole depth = = = 0.141 inch
Pn 16
596
15-43. The sketch of Figure P15.43 shows a one-stage gear reducer that utilizes helical gears with a normal
diametral pitch of 14, normal pressure angle of 20˚, and a helix angle of 30˚. The helix of the 18-tooth drive pinion
1 is left-hand. The input shaft is to be driven in the direction shown (CCW) by a 1/2-hp, 1725 rpm electric motor
operating steadily at full rated power, and the desired output shaft speed is 575 rpm. Determine the following:
f. Center distance
g. Pitch-line velocity
h. Numerical values and directions of tangential, radial, and axial force components on the pinion while
operating at full rated motor horsepower
------------------------------------------------------------------------------------------------------------------------------------------
Solution
⎡ tan ϕ n ⎤ −1 ⎡ tan 20 ⎤
o
(a) From (15-56), ϕt = tan −1 ⎢ ⎥ = tan ⎢ o ⎥
= 22.8o
⎣ cosψ ⎦ ⎣ cos 30 ⎦
Np 18
(c) From (15-57), d p = = = 1.48 inches
Pn cosψ 14 cos 30o
Ng
dg =
Pn cosψ
⎛ np ⎞ ⎛ 1725 ⎞
(d) From (15-57), N g = ⎜ ⎟ N p = ⎜
⎜n ⎟ ⎟ (18 ) = 54 teeth
⎝ g⎠ ⎝ 575 ⎠
54 54
dg = = = 4.45 inches
Pn cosψ 14 cos 30o
(e) Ng = 54 teeth
d p + dg 1.48 + 4.45
(f) From (15-10), C = = = 2.97 inches
2 2
π d p np π (1.48 )(1725 ) ft
(g) From (15-23), V = = = 668
12 12 min
597
Since the driving pinion rotates CCW, the tangential force Ft = 24.7 lb is to the right on the pinion teeth as viewed
by looking down the motor shaft toward the pinion (1) in Figure P15-43, and as sketched below. From (15-66),
using φt = 22.8˚,
Forces and directions are as shown.
598
15-44. Repeat problem 15-43 except that the 18-tooth helical pinion is right-handed.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
The solution is the same as for 15-43 above, except that for a right-hand pinion with a 30˚ helix angle, the direction
of the axial component Fa = 13.3 lb is reversed. That is, in the sketch above in 15-43, Fa changes to the down
direction.
599
15-45. A parallel-shaft helical gearset is driven by an input shaft rotating at 1725 rpm. The 20˚ helical pinion is
250 mm in diameter and has a helix angle of 30˚. The drive motor supplies a steady torque of 340 N-m.
a. Calculate the transmitted force, radial separating force, axial thrust force , and normal resultant force on the
pinion teeth at the pitch point.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
Tp 340
(a) By equilibrium, Ft = = = 2720 N , from (15-66), Fr = Ft tan ϕt = 2720 tan 20o = 990 N and
⎛ dp ⎞ ⎛ 0.250 ⎞
⎜ ⎟ ⎜ ⎟
⎝ 2 ⎠ ⎝ 2 ⎠
from (15-67), Fa = Ft tanψ = 2720 tan 30o = 1570 N . From (15-68),
Ft
Fn =
cos ϕ n cosψ
ϕn = tan −1 ( tan ϕt cosψ ) = tan −1 ( tan 20o cos 30o ) = 17.5o
2720
Fn = = 3293 N
cos17.5o cos 30o
Tn ( 340 )(1725)
kw = = = 61.4 kw
9549 9549
600
15-46. The sketch of Figure P15.46 shows a proposed two-stage reverted gear reducer that utilizes helical gears to
provide quiet operation. The gears being suggested have a module of 4 mm in the normal plane, and a normal
pressure angle of 0.35 rad. The input shaft is driven in the direction shown by a 22-kw, 600-rpm electric motor. Do
the following:
c. Sketch a free-body diagram of the 54-tooth gear (2), showing numerical values and directions of all force
components applied to gear (2) by the 24-tooth pinion (1).
d. Sketch a free-body diagram of the 22-tooth pinion (3), showing numerical values and directions of all force
components applied to the pinion (3) by the 50-tooth gear (4).
------------------------------------------------------------------------------------------------------------------------------------------
Solution
⎛ 24 ⎞
(a) From (15-2), n2 = ⎜ − ⎟ 600 = −267 rpm (CCW)
⎝ 54 ⎠
⎛ 24 ⎞⎛ 22 ⎞
(b) From (15-2), n4 = ⎜ − ⎟⎜ − ⎟ 600 = 117 rpm (CW)
⎝ 54 ⎠⎝ 50 ⎠
9549 ( kw ) 9549 ( 22 )
T1 = = = 350 N-m
n1 600
d1 = mt1 N1 since m = mt
pt p
=π = n
mt mn
⎛ p ⎞
mt = ⎜ t ⎟ mn
⎝ pn ⎠
pn = pt cosψ
⎛ pt ⎞ mn 4 mm
mt = ⎜ ⎟ mn = = = 4.31
⎝ tp cos ψ 1 ⎠ cosψ 1 cos 0.38 tooth
d1 = ( 4.31)( 24 ) = 103.4 mm
T1 350
By equilibrium, Ft1 = = = 6770 N , from (15-66),
⎛ d1 ⎞ ⎛ 0.1034 ⎞
⎜2⎟ ⎜ 2 ⎟
⎝ ⎠ ⎝ ⎠
⎛ tan ϕn ⎞ ⎛ tan 0.35 ⎞
Fr1 = Ft1 tan ϕt = Ft1 ⎜ ⎟ = 6770 ⎜ ⎟ = 2661 N
⎝ cosψ 1 ⎠ ⎝ cos 0.38 ⎠
Sketching the left-hand gear 2, as shown below, the forces Ft1 , Fr1 , and Fa1 have the magnitudes and directions
shown, as applied on the gear.
601
9549 ( 22 )
(d) From (4-41), using n2 = 267 CCW, TB = T2 = T3 = = 787 N-m and
267
mn 4 mm
mt 3 = = = 4.66
cosψ 3 cos 0.54 tooth
d3 = mt 3 N 3 = 4.66 ( 22 ) = 102.5 mm
T3 787
Ft 3 = = = 15,356 N
⎛ d3 ⎞ ⎛ 0.1025 ⎞
⎜ 2⎟ ⎜ 2 ⎟
⎝ ⎠ ⎝ ⎠
⎛ tan 0.35 ⎞
Fr 3 = 15,356 ⎜ ⎟ = 6535 N
⎝ cos 0.54 ⎠
Fa 3 = 15,356 tan 0.54 = 9205 N
Sketching the left-hand pinion 3, as shown below, the forces Ft3 , Fr3 , and Fa3 have the magnitude and direction
shown, as applied on the pinion.
602
15-47. An existing parallel-shaft single reduction spur gear speed reducer is made up of a 21-tooth 8-pitch input
pinion driving a 73-tooth gear mounted on the output shaft. The center distance between pinion and gear is 5.875
inches. The input shaft is driven by a 15-hp, 1725-rpm electric motor operating steadily at full rated capacity. To
reduce vibration and noise, it is desired to substitute a helical gearset that can operate on the the same center distance
and provide approximately the same angular velocity ratio as the existing spur gearset. Study this request and
propose a helical gearset that can perform the function satisfactorily at a reliability level of 99 percent for a very
long lifetime. Assume that the helical gears will be cut by an 8-pitch 20˚ full-depth involute hob. The probable
material is through-hardened Grade 1 steel with a hardness of BHN 350. Determine the following:
a. Using the spur gear data as a starting point, make a preliminary design proposal for a pair of helical gears
with the same center distance and approximately the same angular velocity ratio as the existing spur
gearset. Specifically, determine a combination of transverse diametral pitch, number of teeth on the pinion,
and number of teeth on the mating gear that will satisfy specifications on center distance and angular
velocity ratio.
b. Determine the helix angle. Does it lie within the recommended range of values?
c. Determine the pitch diameter for the pinion and the gear.
f. Calculate the existing safety factor for the tentatively selected helical gear pair, based on tooth bending
fatigue as a potential failure mode.
g. Calculate the existing safety factor for the tentatively selected helical gear pair, based on surface fatigue
pitting wear as a potential failure mode.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
ng Np 21
= = = 0.288
np Ng 73
Pn = 8 (matches hob)
Pt < 8 ( see (15 − 55))
N p + Ng
Pt =
2C
73
Ng = N p = 3.476 N p
21
N p + 3.476 N p 4.476 N p
Pt = = = 0.381N p
2C 2 ( 5.875 )
Iterating to a compatible set of Np, Ng, and Pt to give the same center distance and velocity ratio as the spur gear set:
603
Np Ng Pt Remarks
Pt 7.24
So tentatively select Np = 19, Ng = 66, and Pt = 7.24. From (15-55), cosψ = = = 0.905
Pn 8
(b) ψ = cos −1 0.905 = 25.18o . This helix angle lies in the recommended range of 10˚ to 35˚.(see paragraph
following (15-60)).
Np 19
dp = = = 2.62 inches
Pn cosψ 8cos 25.18o
Ng 66
dg = = = 9.12 inches
Pn cosψ 8cos 25.18o
⎛ 1.000 ⎞
d op = d p + 2a p = 2.62 + 2 ⎜ ⎟ = 2.87 inches
⎝ 8 ⎠
⎛ 1.000 ⎞
d og = d g + 2ag = 9.12 + 2 ⎜ ⎟ = 9.37 inches
⎝ 8 ⎠
2 pn 2π 2π
bmin = = = = 1.85 inches
sinψ Pn sinψ 8sin 25.18o
Ft Pt
(σ b ) p = Ka Kv Km K I
bJ
2π rp n p 2π ( 2.62 2 )(1725 ) ft
V= = = 1183
12 12 min
33, 000 ( hp ) 33, 000 (15 )
Ft = = = 418 lb
V 1183
604
J = 0.33 (Table 15.9; precision gearing; undercutting possible)
Substituting yields
418 ( 7.24 )
(σ b ) p = (1.25)(1.15)(1.6 )(1.0 ) = 11, 400 psi
1.85 ( 0.33)
′ , using Figure 15.25, for Grade 1steel at BHN 350 (as specified), Stbf
From (15-42), Stbf = YN RG Stbf ′ = 40, 000 psi .
From Figure 15.28, for very long life (1010 cycles), YN = 0.8, and from Table 15.13, for 99 % reliability, RG = 1.0.
So Stbf = ( 0.8 )(1.0 )( 40, 000 ) = 32, 000 psi and from (15-43)
(S )
tbf N =1010 32, 000
nex = = = 2.8
(σ b ) p 11, 400
Ft
σ sf = C p Ka Kv Km
bd p I
Noting from Table 3.9 that for steel material E = 30 x 106 psi and ν = 0.3, the elastic coefficient Cp may be evaluated
as
1
Cp = = 2, 290
⎛ 1 − 0.32 ⎞
6 ⎟( )
π⎜ 2
⎝ 30 × 10 ⎠
The geometry factor I may be evaluated from the gear ratio mG where mG = N g N p = 66 19 = 3.47 .
Thus,
418
σ sf = 2, 290 (1.25)(1.15 )(1.6 ) = 89, 450 psi
(1.85 )( 2.62 )( 0.13)
From (15-47), S sf = Z n RG S sf′ , From Figure 15.30 S sf′ = 143, 000 psi , and from Table 15.13 RG = 1.0 and from
Figure 15.31, for a very long life (1010 cycles), Z N =1010 = 0.7 , so
Using (2-89)
605
S sf 100,100
nex = = = 1.1
σ sf 89, 450
(h) Comparing bending fatigue with the surface fatigue we see that the surface fatigue failure mode governs,
with a safety factor of 1.1. This is a marginal safety factor, and should be carefully examined before
proceeding with the design. A face width increase may be in order.
606
15-48. Repeat problem 15-47, except assume that the helical gears will be cut using a 12-pitch 20˚ full-depth
involute hob.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
ng Np 21
= = = 0.288
np Ng 73
Pn = 12 (matches hob)
Pt < 12 ( see (15 − 55))
N p + Ng
Pt =
2C
73
Ng = N p = 3.476 N p
21
N p + 3.476 N p 4.476 N p
Pt = = = 0.381N p
2C 2 ( 5.875 )
Iterating to a compatible set of Np, Ng, and Pt to give the same center distance and velocity ratio as the spur gear set:
Np Ng Pt Remarks
Pt 7.24
So tentatively select Np = 19, Ng = 66, and Pt = 7.24. From (15-55), cosψ = = = 0.603
Pn 12
(j) ψ = cos −1 0.603 = 52.9o . This helix angle does not lies in the recommended range of 10˚ to 35˚.(see
paragraph following (15-60)). Therefore, a new iteration sequence should be tried, using larger Np values.
Implementing this, it may be noted that for ψ = 35˚ (largest acceptable value), then
Hence, using
607
Np Ng Pt Remarks
27 + 94 10.30
So tentatively select Np = 27, Ng = 94, and Pt = = 10.30 and cosψ = = 0.858 , thus
2 ( 5.875 ) 12
ψ = cos −1 0.858 = 30.9o This helix angle lies in the recommended range of 10˚ to 35˚.
Np 27
dp = = = 2.62 inches
Pn cosψ 12 cos 30.9o
Ng 94
dg = = = 9.12 inches
Pn cosψ 12 cos 30.9o
⎛ 1.000 ⎞
d op = d p + 2a p = 2.62 + 2 ⎜ ⎟ = 2.79 inches
⎝ 12 ⎠
⎛ 1.000 ⎞
d og = d g + 2ag = 9.12 + 2 ⎜ ⎟ = 9.30 inches
⎝ 12 ⎠
2 pn 2π 2π
bmin = = = = 1.0 inches
sinψ Pn sinψ 12sin 30.9o
Ft Pt
(σ b ) p = Ka Kv Km K I
bJ
2π rp n p 2π ( 2.62 2 )(1725 ) ft
V= = = 1183
12 12 min
33, 000 ( hp ) 33, 000 (15 )
Ft = = = 418 lb
V 1183
608
Kv = 1.15 (Figure 15.24 with Qv = 10 and V = 1183 ft/min)
Substituting yields
418 (10.30 )
(σ b ) p = (1.25 )(1.15 )(1.6 )(1.0 ) = 26, 060 psi
1.0 ( 0.38 )
′ , using Figure 15.25, for Grade 1steel at BHN 350 (as specified), Stbf
From (15-42), Stbf = YN RG Stbf ′ = 40, 000 psi .
From Figure 15.28, for very long life (1010 cycles), YN = 0.8, and from Table 15.13, for 99 % reliability, RG = 1.0.
So Stbf = ( 0.8 )(1.0 )( 40, 000 ) = 32, 000 psi and from (15-43)
(S )
tbf N =1010 32, 000
nex = = ≈ 1.2
(σ b ) p 26, 060
Ft
σ sf = C p Ka Kv Km
bd p I
Noting from Table 3.9 that for steel material E = 30 x 106 psi and ν = 0.3, the elastic coefficient Cp may be evaluated
as
1
Cp = = 2, 290
⎛ 1 − 0.32 ⎞
6 ⎟( )
π⎜ 2
⎝ 30 × 10 ⎠
The geometry factor I may be evaluated from the gear ratio mG where mG = N g N p = 94 27 = 3.48 .
⎛ tan ϕ n ⎞ −1 ⎛ tan 20
o
⎞
ϕt = tan −1 ⎜ ⎟ = tan ⎜ o ⎟
= 22.98o
⎝ cosψ ⎠ ⎝ cos 30.9 ⎠
Thus,
418
σ sf = 2, 290 (1.25)(1.15)(1.6 ) = 117, 240 psi
(1.0 )( 2.62 )( 0.14 )
609
From (15-47), S sf = Z n RG S sf′ , From Figure 15.30 S sf′ = 143, 000 psi , and from Table 15.13 RG = 1.0 and from
Figure 15.31, for a very long life (1010 cycles), Z N =1010 = 0.7 , so
S sf 100,100
nex = = = 0.85
σ sf 117, 240
(p) This safety factor is clearly unacceptable (since nex < 1). Increasing face width b is one way to increase the
safety factor. For example, using b = 2.0 inches, we have
418
σ sf = 2, 290 (1.25 )(1.15 )(1.6 ) = 82,900 psi
( 2.0 )( 2.62 )( 0.14 )
and
100,100
nex = = 1.2
89, 200
This is probably an acceptable safety factor, but should be carefully reviewed before proceeding.
610
15-49. A newly proposed numerically controlled milling machine is to operate from a helical-gear speed reducer
designed to provide 65 horsepower at an output shaft speed of 1150 rpm. It has been suggested by engineering
management that a 3450-rpm elastic motor be used to drive the speed reducer. An in-house gearing consultant has
suggested that a normal diametral pitch of 12, a normal pressure angle of 20˚, a helix angle of 15˚, an AGMA
quality number of 10, an a safety factor of 1.7 would be appropriate starting point for the design. Design the gears.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
From (15-56),
⎛ tan ϕ n ⎞ −1 ⎛ tan 20
o
⎞
ϕt = tan −1 ⎜ ⎟ = tan ⎜ ⎟ = 20.65
o
⎝ cosψ
o
⎠ ⎝ cos15 ⎠
Pt = Pn cosψ = 12 cos15o = 11.59
Np
dp =
Pn cosψ
Referring to Table 15.9 (higher precision gearing), to avoid under cutting, tentatively select Np = 21 teeth, thus
21
dp = = 1.81 inches
12 cos15o
⎛ np ⎞ ⎛ 3450 ⎞
Ng = ⎜ ⎟ N p = ⎜
⎜n ⎟ ⎟ ( 21) = 63 teeth
⎝ g⎠ ⎝ 1150 ⎠
Ng 63
dg = = = 5.44 inches
Pn cosψ 12 cos15o
d p + dg 1.81 + 5.44
C= = = 3.63 inches
2 2
π d p n p π (1.81)( 3450 ) ft
V= = = 1635
12 12 min
33, 000 ( hp ) 33, 000 ( 65 )
Ft = = = 1312 lb
V 1635
From (15-59),
Ft Pt
(σ b ) p = Ka Kv Km K I
bJ
611
J = 0.34 (Table 15.9; precision gearing)
Substituting yields
1312 (11.59 )
(σ b ) p = (1.25 )(1.15 )(1.6 )(1.0 ) = 51, 430 psi
2.0 ( 0.34 )
Examining Table 15.10, it appears that (σ b ) p should be reduced so that a reasonable material selection may be
made. Trying Pn = 10:
Pt = 10 cos15o = 9.66
21
dp = = 2.17 inches
10 cos15o
Ng 63
dg = = = 6.52 inches
Pn cosψ 10 cos15o
d p + dg
2.17 + 6.52
C= =
= 4.35 inches
2 2
π d p n p π ( 2.17 )( 3450 ) ft
V= = = 1960
12 12 min
33, 000 ( hp ) 33, 000 ( 65 )
Ft = = = 1094 lb
V 1960
Also,
2.0 (π 10 )
bmin = = 2.43 inches
sin15o
1094 ( 9.66 )
(σ b ) p = (1.25 )(1.18 )(1.6 )(1.0 ) = 30,190 psi
2.43 ( 0.34 )
′ , from Figure 15.28, for very long life YN = 0.8 and from Table 15.13, for R = 99%, Rg
From (15-42), Stbf = YN Rg Stbf
= 1.0. Thus,
Stbf 51,325
(S′ ) tbf req ' d
=
YN Rg
=
( 0.8)(1.0 )
= 64,150 psi
From Table 15.10, select Grade 2 carburized and hardened steel (RC 58-64) with minimum core hardness of RC 25,
′ = 65, 000 psi . Next we have from (15-46)
giving Stbf
612
Ft
σ sf = C p Ka Kv Km
bd p I
Noting from Table 3.9 that for steel material E = 30 x 106 psi and ν = 0.3, the elastic coefficient Cp may be evaluated
as
1
Cp = = 2, 290
⎛ 1 − 0.32 ⎞
6 ⎟( )
π⎜ 2
⎝ 30 × 10 ⎠
The geometry factor I may be evaluated from the gear ratio mG where mG = N g N p = 63 21 = 3 .
Thus,
1094
σ sf = 2, 290 (1.25)(1.18)(1.6 ) = 146, 280 psi
( 2.43)( 2.17 )( 0.12 )
From (15-47), S sf = Z n RG S sf′ , From Table 15.15 for Grade 2 steel carburized and hardened to RC 58-64, with core
hardness of RC 25, S sf′ = 225, 000 psi , and from Table 15.13 RG = 1.0 and from Figure 15.31, for a very long life
(1010 cycles), Z N =1010 = 0.75 , so
Using (2-89)
S sf 168,800
nex = = = 1.2
σ sf 146, 280
This does not meet the design safety factor requirement of nd = 1.7. Perhaps the easiest remedy is to increase the
face width. Trying b = 3.0 gives
1094
σ sf = 2, 290 (1.25 )(1.18 )(1.6 ) = 131, 650 psi
( 3.0 )( 2.17 )( 0.12 )
and
168,800
nex = = 1.3
131, 650
613
1094
σ sf = 2, 290 (1.25 )(1.18)(1.6 ) = 114, 000 psi
( 4.0 )( 2.17 )( 0.12 )
and
168,800
nex = = 1.5
114, 000
This will be regarded as close enough, but the proposed design should be reviewed with engineering management.
This is only one of the many possible design configurations. If management insists on nd = 1.7, the next logical step
would be to try Pn = 8, and repeat the whole calculation. Summarizing the proposed design parameters:
Material: Grade 2 carburized and hardened steel (RC 58-64) with minimum core hardness of RC 25.
Reduction ratio: mG = 3
614
15-50. Repeat problem 15-49, except that the suggested helix angle is 30˚.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
From (15-56),
⎛ tan ϕn ⎞ −1 ⎛ tan 20 ⎞
o
ϕt = tan −1 ⎜ ⎟ = tan ⎜ o ⎟
= 22.8o
⎝ cosψ ⎠ ⎝ cos 30 ⎠
Pt = Pn cosψ = 12 cos 30o = 10.39
Np
dp =
Pn cosψ
Referring to Table 15.9 (higher precision gearing), to avoid under cutting, tentatively select Np = 21 teeth, thus
21
dp = = 2.02 inches
12 cos 30o
⎛ np ⎞ ⎛ 3450 ⎞
Ng = ⎜ ⎟ N p = ⎜
⎜n ⎟ ⎟ ( 21) = 63 teeth
⎝ g⎠ ⎝ 1150 ⎠
Ng 63
dg = = = 6.06 inches
Pn cosψ 12 cos 30o
d p + dg 2.02 + 6.06
C= = = 4.04 inches
2 2
π d p np π ( 2.02 )( 3450 ) ft
V= = = 1824
12 12 min
33, 000 ( hp ) 33, 000 ( 65 )
Ft = = = 1176 lb
V 1824
From (15-59),
Ft Pt
(σ b ) p = Ka Kv Km K I
bJ
Substituting yields
615
1176 (10.39 )
(σ b ) p = (1.25)(1.17 )(1.6 )(1.0 ) = 80, 090 psi
1.05 ( 0.34 )
Examining Table 15.10, it appears that (σb)p should be reduced so that a reasonable material selection may be made.
Trying Pn = 10:
Ng 63
dg = = = 7.27 inches
Pn cosψ 10 cos 30o
d p + dg
2.42 + 7.27
C= = = 4.85 inches
2 2
π ( 2.42 )( 3450 )
π d p np ft
V= = = 2186
12 12 min
33, 000 ( hp ) 33, 000 ( 65 )
Ft = = = 981 lb
V 2186
Also,
2.0 (π 10 )
bmin = = 1.26 inches
sin 30o
981( 8.66 )
(σ b ) p = (1.25 )(1.19 )(1.6 )(1.0 ) = 47, 200 psi
1.26 ( 0.34 )
From Table 15.10 it appears that (σb)p, is still too large for a reasonable material selection. Try increasing b to 2.5
inches. Repeating the above calculation
(σ b ) p = ⎛⎜
1.26 ⎞
⎟ 47, 200 = 23, 790 psi
⎝ 2.50 ⎠
Thus, ( Stbf ) = 1.7 ( 23, 790 ) = 40, 440 psi . From (15-42), Stbf = YN Rg Stbf
′ , from Figure 15.28, for very long life
req ' d
From Table 15.10, select Grade 2 carburized and hardened steel (RC 58-64) with minimum core hardness of RC 25,
′ = 65, 000 psi . Next we have from (15-46)
giving Stbf
616
Ft
σ sf = C p Ka Kv Km
bd p I
Noting from Table 3.9 that for steel material E = 30 x 106 psi and ν = 0.3, the elastic coefficient Cp may be evaluated
as
1
Cp = = 2, 290
⎛ 1 − 0.32 ⎞
6 ⎟( )
π⎜ 2
⎝ 30 × 10 ⎠
The geometry factor I may be evaluated from the gear ratio mG where mG = N g N p = 63 21 = 3 .
Thus,
981
σ sf = 2, 290 (1.25)(1.18 )(1.6 ) = 124, 244 psi
( 2.5)( 2.42 )( 0.13)
From (15-47), S sf = Z n RG S sf′ , From Table 15.15 for Grade 2 steel carburized and hardened to RC 58-64, with core
hardness of RC 25, S sf′ = 225, 000 psi , and from Table 15.13 RG = 1.0 and from Figure 15.31, for a very long life
(1010 cycles), Z N =1010 = 0.75 , so
Using (2-89)
S sf 168,800
nex = = = 1.4
σ sf 124, 244
This is too low, perhaps the easiest remedy is to increase the face width. Trying b = 3.0 gives
981
σ sf = 2, 290 (1.25)(1.18 )(1.6 ) = 113, 420 psi
( )( )( 0.13)
3.0 2.42
and
168,800
nex = = 1.5
113, 420
981
σ sf = 2, 290 (1.25 )(1.18)(1.6 ) = 98, 224 psi
( 4.0 )( 2.42 )( 0.13)
617
and
168,800
nex = = 1.7
98, 224
Material: Grade 2 carburized and hardened steel (RC 58-64) with minimum core hardness of RC 25.
Reduction ratio: mG = 3
15-51. A pair f straight bevel gears, similar to those shown in Figure 15.41, has been incorporated into a right-angle
speed reducer (shaft centerlines intersect at 90˚). The straight bevel gears have a diametral pitch of 8 and a 20˚
pressure angle. The gear reduction ratio is 3:1, and the number of teeth on the bevel pinion is 16. Determine the
following:
f. Average pitch cone radius for the pinion, assuming face width is maximum recommended value
g. Average pitch cone radius for the gear, assuming face width is maximum recommended value
h. Pinion addendum
i. Gear addendum
618
j. Pinion dedendum
k. Gear dedendum
------------------------------------------------------------------------------------------------------------------------------------------
Solution
⎛1⎞
(a) From (15-77), γ p = cot −1 mG = cot −1 3 = tan −1 ⎜ ⎟ = 18.4o
⎝3⎠
Np 16
(c) From (15-8), d p = = = 2.0 inches
Pd 8
(d) We have
dg
mG = = 3.0
dp
d g = 3d p = 3 ( 2.0 ) = 6.0 inches
⎛ dp ⎞ ⎛ 2.0 ⎞
bmax −1 = 0.3 ⎜ = 0.3 ⎜ ⎟ = 0.95 inches
⎜ 2sin γ p ⎟⎟ ⎝ 2sin18.4 o
⎠
⎝ ⎠
10 10
bmax − 2 = = = 1.25 inches
Pd 8
b 2.0 ⎛ 0.95 ⎞
(f) From (15-78), ( rave ) p = rp − sin γ p = −⎜ ⎟ sin18.4 = 0.85 inch
o
2 2 ⎝ 2 ⎠
b 6.0 ⎛ 0.95 ⎞
(g) From (15-78), ( rave ) g = rg − sin γ g = −⎜ ⎟ sin 71.6 = 2.55 inch
o
2 2 ⎝ 2 ⎠
⎡ ⎛N ⎞ ⎤
2
⎢ 0.540 + 0.460 ⎜ p ⎟⎟ ⎥
2.000 ⎢ ⎜N
ap = −⎢ ⎝ g ⎠ ⎥
⎥
Pd ⎢ Pd ⎥
⎢ ⎥
⎢⎣ ⎥⎦
⎡ ⎛ 16 ⎞ ⎤
2
⎢ 0.540 + 0.460 ⎜ ⎟ ⎥
2.000 ⎢ ⎝ 48 ⎠ ⎥ = 0.176 inch
= −
8 ⎢ 8 ⎥
⎢ ⎥
⎣ ⎦
619
⎡ ⎛N ⎞ ⎤
2
⎢ 0.540 + 0.460 ⎜ p ⎟⎟ ⎥
⎢ ⎜N
ag = ⎢ ⎝ g ⎠ ⎥
⎥
⎢ P d ⎥
⎢ ⎥
⎢⎣ ⎥⎦
⎡ ⎛ 16 ⎞ ⎤
2
⎢ 0.540 + 0.460 ⎜ ⎟ ⎥
=⎢ ⎝ 48 ⎠ ⎥ = 0.074 inch
⎢ 8 ⎥
⎢ ⎥
⎣ ⎦
⎡ ⎛N ⎞ ⎤
2
⎢ 0.540 + 0.460 ⎜ p ⎟⎟ ⎥
⎤ ⎢ ⎜N
⎡⎛ 0.188 ⎞ ⎝ g ⎠ ⎥
( de ) p = ⎢⎜ ⎟ + 0.002 ⎥ + ⎢ ⎥
⎢⎣⎝ Pd ⎠ ⎥⎦ ⎢ Pd ⎥
⎢ ⎥
⎣⎢ ⎦⎥
0.188
= + 0.002 + [ 0.074] = 0.100 inch
8
⎡ ⎛N ⎞ ⎤
2
⎢ 0.540 + 0.460 ⎜ p ⎟⎟ ⎥
⎤ ⎢ ⎜N
⎡⎛ 2.188 ⎞ ⎝ g ⎠ ⎥
( de )g = ⎢⎜ ⎟ + 0.002 ⎥ − ⎢ ⎥
P
⎣⎢⎝ d ⎠ ⎦⎥ ⎢ P d ⎥
⎢ ⎥
⎣⎢ ⎦⎥
2.188
= + 0.002 − [ 0.074] = 0.202 inch
8
620
15-52. Repeat problem 15-51, except use a diametral pitch of 12 and a reduction ratio of 4:1.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
⎛1⎞
(a) From (15-77), γ p = cot −1 mG = cot −1 4 = tan −1 ⎜ ⎟ = 14o
⎝4⎠
Np 16
(c) From (15-8), d p = = = 1.33 inches
Pd 12
(d) We have
dg
mG = = 4.0
dp
d g = 4d p = 4 (1.33) = 5.32 inches
⎛ dp ⎞ ⎛ 1.33 ⎞
bmax −1 = 0.3 ⎜ = 0.3 ⎜ ⎟ = 0.82 inches
⎜ 2sin γ ⎟⎟ ⎝ 2sin14
o
⎠
⎝ p ⎠
10 10
bmax − 2 = = = 0.83 inches
Pd 12
b 1.33 ⎛ 0.82 ⎞
(f) From (15-78), ( rave ) p = rp − sin γ p = −⎜ ⎟ sin14 = 0.57 inch
o
2 2 ⎝ 2 ⎠
b 5.32 ⎛ 0.82 ⎞
(g) From (15-78), ( rave ) g = rg − sin γ g = −⎜ ⎟ sin 76 = 2.26 inch
o
2 2 ⎝ 2 ⎠
⎡ ⎛N ⎞ ⎤
2
⎢ 0.540 + 0.460 ⎜ p ⎟⎟ ⎥
2.000 ⎢ ⎜N
ap = −⎢ ⎝ g ⎠ ⎥
⎥
Pd ⎢ Pd ⎥
⎢ ⎥
⎣⎢ ⎦⎥
⎡ ⎛ 16 ⎞ ⎤
2
⎢ 0.540 + 0.460 ⎜ ⎟ ⎥
2.000 ⎢ ⎝ 48 ⎠ ⎥ = 0.117 inch
= −
12 ⎢ 12 ⎥
⎢ ⎥
⎣ ⎦
621
⎡ ⎛N ⎞ ⎤
2
⎢ 0.540 + 0.460 ⎜ p ⎟⎟ ⎥
⎢ ⎜N
ag = ⎢ ⎝ g ⎠ ⎥
⎥
⎢ P d ⎥
⎢ ⎥
⎢⎣ ⎥⎦
⎡ ⎛ 16 ⎞ ⎤
2
⎢ 0.540 + 0.460 ⎜ ⎟ ⎥
=⎢ ⎝ 48 ⎠ ⎥ = 0.049 inch
⎢ 12 ⎥
⎢ ⎥
⎣ ⎦
⎡ ⎛N ⎞ ⎤
2
⎢ 0.540 + 0.460 ⎜ p ⎟⎟ ⎥
⎤ ⎢ ⎜
⎡⎛ 0.188 ⎞ ⎝ Ng ⎠ ⎥
( de ) p = ⎢⎜ ⎟ + 0.002 ⎥ + ⎢ ⎥
⎢⎣⎝ Pd ⎠ ⎥⎦ ⎢ Pd ⎥
⎢ ⎥
⎣⎢ ⎦⎥
0.188
= + 0.002 + [ 0.049] = 0.067 inch
12
⎡ ⎛N ⎞ ⎤
2
⎢ 0.540 + 0.460 ⎜ p ⎟⎟ ⎥
⎤ ⎢ ⎜N
⎡⎛ 2.188 ⎞ ⎝ g ⎠ ⎥
( de )g = ⎢⎜ ⎟ + 0.002 ⎥ − ⎢ ⎥
P
⎣⎢⎝ d ⎠ ⎦⎥ ⎢ P d ⎥
⎢ ⎥
⎣⎢ ⎦⎥
2.188
= + 0.002 − [ 0.049] = 0.135 inch
12
622
15-53. It is being proposed to use a Coniflex ® straight bevel gearset to provide a 3:1 speed reduction between a 15-
tooth pinion rotating at 300 rpm and a meshing gear mounted on a shaft whose centerline intersects the pinion shaft
centerline at a 90˚ angle. The pinion shaft is driven steadily by a 3-hp source operating at full rated power. The
bevel gears are to have a diametral pitch of 6, a 20˚ pressure angle, and a face width of 1.15 inches. Do the
following:
g. Determine whether the force magnitudes calculated for the pinion and the gear are consistent with the free-
body equilibrium of the bevel gearset (see Figure 15.41 for geometric arrangement).
------------------------------------------------------------------------------------------------------------------------------------------
Solution
b dp
(r )p avg − sin γ p
=
2 2
N p 15
dp = = = 2.50 inches
Pd 6
⎛1⎞
γ p = cot −1 mG = tan −1 ⎜ ⎟ = 18.4o
3 ⎝ ⎠
2.50 1.15
(r )p avg =
2
−
2
sin18.4o = 1.07 inches
2π ( rp ) np 2π (1.07 )( 300 ) ft
avg
Vavg = = = 168
12 12 min
33, 000 ( hp ) 33, 000 ( 3)
Ft = = = 589 lb
Vavg 168
623
( Fr ) p = Ft tan ϕ cos γ p = ( 589 ) ( tan 20o ) cos18.4o = 203 lb
(g) Referring to Figure 15.41, along with the forces calculated, it may be noted that by equilibrium, forces at
the contact site must be equal and opposite. That is, ( Ft ) p and ( Ft ) g are equal and opposite, ( Fr ) p and
( Fa ) g are equal and opposite, and ( Fr ) g and ( Fa ) p are equal and opposite.
624
15-54. Repeat problem 15-53, except for a diametral pitch of 10.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
b dp
(r ) p avg − sin γ p
=
2 2
N p 15
dp = = = 1.50 inches
Pd 10
⎛1⎞
γ p = cot −1 mG = tan −1 ⎜ ⎟ = 18.4o
3 ⎝ ⎠
1.50 1.15
(r ) p avg =
2
−
2
sin18.4o = 0.57 inches
2π ( rp ) np 2π ( 0.57 )( 300 ) ft
avg
Vavg = = = 90
12 12 min
33, 000 ( hp ) 33, 000 ( 3)
Ft = = = 1100 lb
Vavg 90
(g) Referring to Figure 15.41, along with the forces calculated, it may be noted that by equilibrium, forces at
the contact site must be equal and opposite. That is, ( Ft ) p and ( Ft ) g are equal and opposite, ( Fr ) p and
( Fa ) g are equal and opposite, and ( Fr ) g and ( Fa ) p are equal and opposite.
625
15-55. Repeat problem 15-53, except for a diametral pitch of 16.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
b dp
(r ) p avg− sin γ p
2
=
2
N p 15
dp = = = 0.94 inches
Pd 16
⎛1⎞
γ p = cot −1 mG = tan −1 ⎜ ⎟ = 18.4o
3 ⎝ ⎠
0.94 1.15
(r ) p avg =
2
−
2
sin18.4o = 0.29 inches
2π ( rp ) np 2π ( 0.29 )( 300 ) ft
avg
Vavg = = = 46
12 12 min
33, 000 ( hp ) 33, 000 ( 3)
Ft = = = 2152 lb
Vavg 46
(n) Referring to Figure 15.41, along with the forces calculated, it may be noted that by equilibrium, forces at
the contact site must be equal and opposite. That is, ( Ft ) p and ( Ft ) g are equal and opposite, ( Fr ) p and
( Fa ) g are equal and opposite, and ( Fr ) g and ( Fa ) p are equal and opposite.
626
15-56. For the Coniflex ® bevel gearset described in problem 15-43, the following information has been tabulated
or calculated:
Tp = 630 in-lb
Pd = 6
dp = 2.50 inches
b = 1.15 inches
Np = 15 teeth
Ng = 45 teeth
Further, it is proposed to use Grade 2 AISI 4140 steel nitride and through-hardened to BHN 305 for both the
pinion and the gear. Other known design information includes the following items:
3. The gear is straddle mounted with a closely positioned bearing on each side, but the pinion overhangs its
support bearing.
A reliability of 99 percent is required, and a design safety factor of at least 1.3 is desired. Do the following:
a. Calculate the tooth bending fatigue stress for the more critical of the pinion or the gear.
b. Determine the tooth bending fatigue strength for the proposed AISI 4140 steel material corresponding to a
life of 109 cycles.
c. Calculate the existing safety factor for the proposed design configuration, based on tooth bending fatigue as
the governing failure mode. Compare this with the desired design safety factor, and make any comments
you think appropriate.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
2Tp Pd
σb = K a Kv K m
d p bJ
Ka = 1.25 (Table 15.6; uniform drive, moderate shock assumed to driven machine)
Kv = 1.1 (Figure 15.24 with Qv = 8and Vavg = 168 ft/min data of problem 15-53)
Km = 1.1 (From Figure 15.43; for b = 1.15 inches and one member straddle-mounted)
J = 0.24 (From Figure 15.44, for the pinion teeth, using Np = 15, Ng = 45)
Thus,
627
2 ( 630 )( 6 )
σb = (1.25 )(1.1)(1.1) = 16,570 psi
( 2.5)(1.15 )( 0.24 )
(b) From Figure 15.26, for Grade 2 nitrided and through-hardened AISI 4140 steel, at 99% reliability, at a life
′ = 49, 000 psi . From Figure 15.28, for 109 cycles, YN = 0.9, so that
of 107 cycles, Stbf
′ = 0.9 (1.0 )( 49, 000 ) = 44,100 psi .
Stbf = YN Rg Stfb
Stbf 44,100
nex = = = 2.7
σb 16,570
Based on the desired safety factor nd = 1.3, this is acceptable. Until surface durability is checked, no effort to
optimize the design is warranted.
628
15-57. Based on the specifications and data for the Coniflex ® bevel gearset given in problem 15-56, do the
following:
a. Calculate the surface fatigue durability stress for the Coniflex ® bevel gearset under consideration.
b. Determine the surface fatigue durability strength for the proposed nitride and hardened AISI 4140 steel
material corresponding to a life of 109 cycles.
c. Calculate the existing safety factor for the proposed design configuration based on surface fatigue
durability as the governing failure mode. Compare this with the desired design safety factor, and make any
comments you think appropriate.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
2Tp
σ sf = ( C p )bevel Ka Kv Km
bd p2 I
3 3
(C )p bevel =
⎛ 1 −ν ⎞ 2
=
⎛ 1 − 0.32 ⎞
= 2805
2π ⎜ ⎟ ( 2) 2π ⎜ 6 ⎟( )
2
⎝ E ⎠ ⎝ 30 × 10 ⎠
Ka = 1.25 (Table 15.6; uniform drive, moderate shock assumed to driven machine)
Kv = 1.1 (Figure 15.24 with Qv = 8and Vavg = 168 ft/min data of problem 15-53)
Km = 1.1 (From Figure 15.43; for b = 1.15 inches and one member straddle-mounted)
Thus,
2 ( 630 )
σ sf = 2805 (1.25)(1.1)(1.1) = 166,800 psi
(1.15 )( 2.50 ) ( 0.075)
2
(b) From Table 15.15, for Grade 2 nitrided and through hardened AISI 4140 steel, at 99 % reliability, at a life
of 107 cycles, S sf′ = 163, 000 psi . From Figure 15.31, for 109 cycles, ZN = 0.8. Using (15-47),
S sf = Z N RG S sf′ = ( 0.8 )(1.0 )(163, 000 ) = 130, 400 psi
S sf 130, 400
nex = = ≈ 0.8
σ sf 166,800
This safety factor (nex < 1.0) is clearly not acceptable. Redesign is necessary.
629
15-58. A Coniflex ® , straight-tooth bevel gearset is supported on shafts with centerlines intersecting at a 90˚ angle.
The gear is straddle mounted between closely positioned bearings, and the pinion overhangs its support bearing.
The 15-tooth pinion rotates at 900 rpm, driving the 60-tooth gear, which has a diametral pitch of 6, pressure angle of
20˚, and face width of 1.25 inches. The material for both gears is through-hardened Grade 1 steel with a hardness of
BHN 300 (see Figure 15.25). It is desired to have a reliability of 90 percent, a design life of 108 cycles, and a
governing safety factor of 2.5. Estimate the maximum horsepower that can be transmitted by this gear reducer while
meeting all of the design specifications given.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
Similarly, from (15-47), S sf = Z N Rg S sf′ .From Figure 15.30, for R = 99 % and Nd = 107 cycles, S sf′ = 125, 700 psi .
From Figure 15.31, for 108 cycles, ZN = 0.90. From Table 15.13, for R = 90 %, Rg = 1.18, thus
Based on (2-84),
Stbf 40,360
(σ d )tbf = = = 16,144 psi
nd 2.5
S sf 133,500
(σ d )sf = = = 53, 400 psi
nd 2.5
(σ b )allow d p bJ
(T )
p max − b =
2 Pd K a K v K m
(σ )
2
d p2 bI
( Tp ) =
sf allow
2 (C )
max − sf 2
p bevel Ka Kv Km
Np 15
dp = = = 2.50 inches
Pd 6
1 Np 15
γ p = cot −1 mG = tan −1 = tan −1 = tan −1 = 14o
mG Ng 60
b dp 2.50 1.25
(r )
p avg − sin γ p =
=
2 2 2
−
2
sin14o = 1.10 inches
2π ( rp ) n p 2π (1.10 )( 900 ) ft
avg
V= = = 518
12 12 min
Ka = 1.25 (Table 15.6; assume uniform drive, moderate shock for driven machine)
630
Km = 1.1 (From Figure 15.43; for b = 1.25 inches and one member straddle-mounted)
J = 0.235 (From Figure 15.44, for the pinion teeth, using Np = 15, Ng = 60)
I = 0.077 (From Figure 15.45, for the pinion using Np = 15, Ng = 60)
3 3
(C )
2
p bevel = = = 7.87 × 106
⎛ 1 −ν ⎞ 2
⎛ 1 − 0.32 ⎞
2π ⎜ ⎟ ( 2) 2π ⎜ 6 ⎟( )
2
⎝ E ⎠ ⎝ 30 × 10 ⎠
Setting (σ sf ) allow
= (σ d ) sf gives
( Tp ) max − sf
=
2 ( 7.87 × 106 ) (1.25 )(1.2 )(1.1)
= 66 in-lb
Since (Tp ) is smaller than (Tp )max −b , surface fatigue governs. Hence, from (15-84)
max − sf
Tp n p 66 ( 900 )
( hp )allow
max = = = 0.94 horsepower
63, 025 63, 025
631
15-59. Repeat problem 15-58, except change the material to through-hardened Grade 2 steel with a hardness of
BHN 350 (see Figure 15.25).
------------------------------------------------------------------------------------------------------------------------------------------
Solution
Similarly, from (15-47), S sf = Z N Rg S sf′ .From Figure 15.30, for R = 99 % and Nd = 107 cycles, S sf′ = 156, 450 psi .
From Figure 15.31, for 108 cycles, ZN = 0.90. From Table 15.13, for R = 90 %, Rg = 1.18, thus
Based on (2-84),
(σ b )allow d p bJ
(T )
p max − b =
2 Pd K a K v K m
(σ )
2
d p2 bI
(Tp ) =
sf allow
2 (C )
max − sf 2
p bevel Ka Kv Km
Np 15
dp = = = 2.50 inches
Pd 6
1 Np 15
γ p = cot −1 mG = tan −1 = tan −1 = tan −1 = 14o
mG Ng 60
b dp 2.50 1.25
(r )
p avg =− sin γ p =
2 2 2
−
2
sin14o = 1.10 inches
2π ( rp ) n p 2π (1.10 )( 900 ) ft
avg
V= = = 518
12 12 min
Ka = 1.25 (Table 15.6; assume uniform drive, moderate shock for driven machine)
Km = 1.1 (From Figure 15.43; for b = 1.25 inches and one member straddle-mounted)
J = 0.235 (From Figure 15.44, for the pinion teeth, using Np = 15, Ng = 60)
I = 0.077 (From Figure 15.45, for the pinion using Np = 15, Ng = 60)
632
3 3
(C )
2
p bevel = = = 7.87 × 106
⎛ 1 −ν ⎞ 2
⎛ 1 − 0.32 ⎞
2π ⎜ ⎟ ( 2) 2π ⎜ 6 ⎟( )
2
⎝ E ⎠ ⎝ 30 × 10 ⎠
Setting (σ sf ) allow
= (σ d ) sf gives
(T ) p max − sf =
2 ( 7.87 × 106 ) (1.25 )(1.2 )(1.1)
= 102 in-lb
Since (Tp ) is smaller than (Tp )max −b , surface fatigue governs. Hence, from (15-84)
max − sf
Tp n p 102 ( 900 )
( hp )max = = = 1.46 horsepower
allow 63, 025 63, 025
633
15-60. It is desired to design a long-life right-angle straight bevel gear speed reducer for an application in which an
850-rpm, 5-hp internal combustion engine, operating at full power, drives the pinion. The output gear, which is to
rotate at approximately 350 rpm, drives a heavy-duty industrial field conveyor. Design the gearset, including the
selection of an appropriate material, if a reliability of 95 percent is desired.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
(a) Based on the specifications, a first conceptual layout of the bevel gearset may be sketch as shown below.
(b) Potential primary failure modes appear to be tooth bending fatigue and surface fatigue pitting.
(d) Using specified shaft speed requirements, the gear ratio mG may be calculated from (15-77) as
ω p n p 850
mG = = = = 2.43
ω g ng 350
(f) From Figure (15.47), for Tp = 371 in-lb and mG = 2.43, dp ≈ 1.25 inches.
634
(g) From Figure (15.48), select Np = 18 teeth and for the gear, Ng = mGNp =(2.43)(18) = 43.74 ≈ 44 teeth.
⎛ dp ⎞
bL ≤ 0.3 ⎜ ⎟⎟
⎜ 2sin γ
⎝ p ⎠
⎛ 1 ⎞
γ p = cot −1 mG = tan −1 ⎜ ⎟ = 22.4
o
⎝ 2.43 ⎠
⎛ 1.25 ⎞
bL ≤ 0.3 ⎜ o ⎟
= 0.49 inch
⎝ 2sin 22.4 ⎠
10 10
bPd ≤ = = 0.69 inch
Pd 14.4
dp
b 1.25 0.5
(r )
p avg =
2 2
− sin γ p =
2
−
2
sin 22.4o = 0.53 inch
2π ( rp ) n p 2π ( 0.53)( 850 ) ft
avg
V= = = 236
12 12 min
2Tp Pd
σb = K a Kv K m
d p bJ
Ka = 1.75 (Table 15.6; single cylinder i.e. engine moderate shock in the conveyor)
Km = 1.1 (From Figure 15.43; for b = 0.5 inches and one member straddle-mounted)
J = 0.24 (From Figure 15.44, for the pinion teeth, using Np = 18, Ng = 44)
Thus,
635
2 ( 371)(14.4 )
σb = (1.75)(1.1)(1.1) = 150,833 psi
(1.25)( 0.5)( 0.24 )
Comparing with steel gear strengths given in Table 15.10, σb must be reduced significantly. Try increasing
dp to 2.50 inches, thus repeating the above calculations gives:
⎛ 2.5 ⎞
bL ≤ 0.3 ⎜ o ⎟
= 0.98 inch
⎝ 2sin 22.4 ⎠
10 10
bPd ≤ = = 1.39 inch
Pd 7.2
dp
b 2.5 1.0
(r )
p avg =
2 2
− sin γ p =
2
−
2
sin 22.4o = 1.06 inch
2π ( rp ) n p 2π (1.06 )( 850 ) ft
avg
V= = = 472
12 12 min
2 ( 371)( 7.2 )
σb = (1.75 )(1.1)(1.1) = 18,850 psi
( 2.50 )(1.0 )( 0.24 )
Reviewing Chapter X (5) methodology, a design safety factor will be chosen as nd = 1.4, so based on (2-84)
′ .From Figure 15.28, using Nd = 1010 cycles, YN = 0.85. From Table 15.13, for R = 95
From (15-42), Stbf = YN Rg Stbf
%, Rg ≈ 1.1, thus
26, 400
(S′ )
tbf req ' d
=
( 0.85)(1.1)
= 28, 240 psi
From Figure 15.25, select through-hardened steel, Grade 1, heat treated to BHN 210.
636
2Tp
σ sf = ( C p )bevel Ka Kv Km
bd p2 I
3 3
(C )p bevel =
⎛ 1 −ν ⎞ 2
=
⎛ 1 − 0.32 ⎞
= 2805
2π ⎜ ⎟ ( 2) 2π ⎜ 6 ⎟( )
2
⎝ E ⎠ ⎝ 30 × 10 ⎠
2 ( 371)
σ sf = 2805 (1.75)(1.1)(1.1) = 158, 230 psi
(1.0 )( 2.50 ) ( 0.079 )
2
Comparing this value with Figure 15.30, the Grade 1 through-hardened steel selected above will not be satisfactory,
nor will Grade 2. Instead, try changing to carburized and hardened Grade 2 steel heat treated to surface hardness of
RC 58-64, (minimum core hardness of RC 25), from Table 15.15. This gives, S sf′ = 225, 000 psi . From (15-47),
S sf = Z N Rg S sf′ . From Figure 15.31, for 1010 cycles, ZN = 0.75, and Rg ≈1.1, thus
S sf 185, 625
nex = = ≈ 1.2
σ sf 158, 230
This does not satisfy nd = 1.4. Going back to Table 15.15, select Grade 3 steel carburized and hardened to surface
hardness of RC 58 – 64, (minimum core hardness of RC 30), giving S sf′ = 275, 000 psi and
and
226,875
nex = = 1.4
158, 230
So all design specifications are satisfied. Summarizing the design proposal, the following is suggested:
Material: AISI 4620 (see 15.5) Grade 3, carburized and hardened at surface to RC 58 – 64, with RC 30
minimum core hardness.
Tooth system: Coniflex standard full depth 20˚ straight bevel gear teeth
Np = 18 teeth
Ng = 44 teeth
dp = 2.50 inches
dg = 6.11 inches
Pd = 7.2
637
b = 1.0 inch
638
15-61. Repeat problem 15-60, except use an 850 rpm, 10-hp internal-combustion engine, operating at full power,
to drive the pinion.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
(a) Based on the specifications, a first conceptual layout of the bevel gearset may be sketch as shown below.
(b) Potential primary failure modes appear to be tooth bending fatigue and surface fatigue pitting.
(d) Using specified shaft speed requirements, the gear ratio mG may be calculated from (15-77) as
ω p n p 850
mG = = = = 2.43
ω g ng 350
(f) From Figure (15.47), for Tp = 741 in-lb and mG = 2.43, dp ≈ 2.0 inches. Actually, because of the impact
caused by the I.C. engine drive, a larger pinion pitch diameter will be assumed to start, i.e., dp = 3.0 inches.
(g) From Figure (15.48), select Np = 20 teeth and for the gear, Ng = mGNp =(2.43)(20) = 48.6 ≈ 49 teeth.
639
(i) Using (15-73)
⎛ dp ⎞
bL ≤ 0.3 ⎜ ⎟⎟
⎜ 2sin γ
⎝ p ⎠
⎛ 1 ⎞
γ p = cot −1 mG = tan −1 ⎜ ⎟ = 22.4
o
⎝ 2.43 ⎠
⎛ 3.0 ⎞
bL ≤ 0.3 ⎜ o ⎟
= 1.18 inch
⎝ 2sin 22.4 ⎠
10 10
bPd ≤ = = 1.5 inch
Pd 6.67
2π ( rp ) n p 2π (1.27 )( 850 ) ft
avg
V= = = 565
12 12 min
2Tp Pd
σb = K a Kv K m
d p bJ
Ka = 1.75 (Table 15.6; single cylinder i.e. engine moderate shock in the conveyor)
Km = 1.1 (From Figure 15.43; for b = 1.2 inches and one member straddle-mounted)
J = 0.24 (From Figure 15.44, for the pinion teeth, using Np = 20, Ng = 49)
Thus,
2 ( 741)( 6.67 )
σb = (1.75 )(1.22 )(1.1) = 26,870 psi
( 3.0 )(1.2 )( 0.24 )
(m) Next from (15-86),
640
2Tp
σ sf = ( C p )bevel Ka Kv Km
bd p2 I
3 3
(C )p bevel =
⎛ 1 −ν ⎞ 2
=
⎛ 1 − 0.32 ⎞
= 2805
2π ⎜ ⎟ ( 2) 2π ⎜ 6 ⎟( )
2
⎝ E ⎠ ⎝ 30 × 10 ⎠
2 ( 741)
σ sf = 2805 (1.75 )(1.22 )(1.1) = 174,800 psi
(1.2 )( 3.0 ) ( 0.083)
2
To attempt selection of an acceptable material, first check surface fatigue since it often governs. Reviewing Chapter
2 methodology, a design safety factor will be chosen as nd = 1.4. From (15-47)
S sf Z N Rg S sf′
(σ d )sf = =
nd nd
From Figure 15.31, for 1010 cycles, ZN = 0.75, and from Table 15.13, for R = 95 %, Rg ≈1.1, thus, setting
(σ d )sf = σ sf we have
nd (σ d ) sf 1.4 (174,800 )
(S′ )
sf req ' d
=
Z N Rg
=
0.75 (1.1)
= 296, 630 psi
The only material that comes close is, from Table 15.15, Grade 3 steel carburized and hardened to surface hardness
of RC 58 – 64 ( with RC 30 minimum core hardness), giving S sf′ = 275, 000 psi . Based on (15-47)
and
S sf 226,875
nex = = = 1.3
σ sf 174,800
This does not quite meet the conditions of nd = 1.4, but will be considered acceptable for now. Engineering
management must decide whether nex = 1.3 is acceptable in the final analysis.
Checking on the tooth-bending safety factor, for the Grade 3 material selected from Table 15.10, we have
′ = 75, 000 psi and from (15-42), Stbf = YN Rg Stbf
Stbf ′ .From Figure 15.28, using Nd = 1010 cycles, YN = 0.85. From
Table 15.13, for R = 95 %, Rg ≈ 1.1, thus
and
Stbf 70,125
nex −b = = = 2.6
σb 26,870
641
This is more than adequate. Summarizing the design proposal, the following is suggested:
Material: AISI 4620 (see 15.5) Grade 3, carburized and hardened at surface to RC 58 – 64, with RC 30
minimum core hardness.
Tooth system: Coniflex standard full depth 20˚ straight bevel gear teeth
Np = 20 teeth
Ng = 49 teeth
dp = 3.0 inches
dg = 7.35 inches
Pd = 6.67
b = 1.2 inch
nd = 1.3
642
15-62. A proposed worm gearset is to have a single-start worm with a pitch diameter of 1.250 inches, a diametral
pitch of 10, and a normal pressure angle of 14 ½ ˚. The worm is to mesh with a worm gear having 40 teeth and a
face width of 0.625 inch. Calculate the following:
a. Axial pitch
c. Circular pitch
f. Addendum
g. Dedendum
k. Center distance
l. Velocity ratio
------------------------------------------------------------------------------------------------------------------------------------------
Solution
π dg
px =
Ng
Ng 40
dg = = = 4.00 inches
Pd 10
π ( 4.00 )
px = = 0.314 inch
40
Lw 0.314
(d) From (15-94), λw = tan −1 = tan −1 = 4.57o
π dw π (1.25 )
643
(g) From Table 15.20, de = 1.157/Pd = 1.157/10 = 0.116 inch
2.000 2.000
(h) From Table 15.20, ( d o ) w = d w + = 1.25 + = 1.45 inch
Pd 10
2.314 2.314
(i) From Table 15.20, ( d r )w = d w − = 1.25 − = 1.019 inch
Pd 10
d g + dw 4.00 + 1.25
(k) From (15-91), C = = = 2.63 inches
2 2
ωw N g 40
(l) From (15-89), = = = 40
ωg N w 1
2.314 2.314
(m) From Table 15.20, ( d r ) g = d g − = 4.00 − = 3.769 inch
Pd 10
3.000 3.000
(n) From Table 15.20, ( do )g = dg + = 4.00 + = 4.300 inch
Pd 10
644
15-63. A double-start worm has a lead of 60 mm. The meshing worm gear has 30 teeth, and has been cut using a
hob having a module of 8.5 in the normal plane. Do the following:
c. Calculate the center distance and determine whether it lies in the range of usual practice.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
Lw
dw =
π tan λw
⎛ pn ⎞
λw = cos −1 ⎜ ⎟
⎝ px ⎠
L 60
px = w = = 30 mm
Nw 2
pn = π mn = π ( 8.5 ) = 26.70 mm
⎛ 26.70 ⎞
λw = cos −1 ⎜ ⎟ = 27.13
o
⎝ 30 ⎠
60
dw = = 37.27 mm
π tan 27.13o
N g px 30 ( 30 )
(b) From (15-87), d g = = = 286.48 mm
π π
d g + dw 286.48 + 37.27
C= = = 161.88 mm
2 2
645
C 0.875 C 0.875
≤ dw ≤
3.0 1.6
(161.88) (161.88 )
0.875 0.875
≤ dw ≤
3.0 1.6
85.71 85.71
≤ dw ≤
3.0 1.6
28.57 ≤ d w ≤ 53.57
We see that dw = 37.27 mm, which does lie in the recommended range.
ωw N g 30
(d) From (15-89), = = = 15
ωg N w 2
25.4 25.4
(e) From (15-14), Pd = = = 3.0
mn 8.5
646
15-64. A triple-start worm is to have a pitch diameter of 4.786 inches. The meshing worm gear is to be cut using a
hob having a diametral pitch of 2 in the normal plane. The reduction ratio is to be 12:1. Do the following:
d. Calculate the center distance and determine whether it lies in the range of usual practice.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
⎛ω ⎞
(a) From (15-89), N g = ⎜ w
⎜ω ⎟⎟ N w = (12 )( 3) = 36 teeth
⎝ g ⎠
Lw = N w px
pn
px =
cos λw
π π
pn = = = 1.571
Pn 2
1.571
px =
cos λw
3 (1.571) 4.713
Lw = =
cos λw cos λw
Lw = π d w tan λw = π ( 4.786 ) tan λw = 15.04 tan λw
4.713 ⎛ sin λw ⎞
= 15.04 tan λw = 15.04 ⎜ ⎟
cos λw ⎝ cos λw ⎠
4.713
sin λw = = 0.313
15.04
λw = sin −1 0.313 = 18.24o
pn N g pn N g
dg = =
π cosψ g π cos λw
1.571( 36 )
dg = = 18.95 inches
π cos18.24o
d g + dw 18.95 + 4.786
C= = = 11.87 inches
2 2
647
C 0.875 C 0.875
≤ dw ≤
3.0 1.6
(11.87 ) (11.87 )
0.875 0.875
≤ dw ≤
3.0 1.6
8.71 8.71
≤ dw ≤
3.0 1.6
2.90 ≤ d w ≤ 5.44
We see that dw = 4.876 inches, which does lie in the recommended range.
648
15-65. It is proposed to drive an industrial crushing machine, designed to crush out-of-tolerance scrap ceramic
bearing liners, with an in-stock 2-hp, 1200-rpm electric motor coupled to an appropriate speed reducer. The
crushing machine input shaft is to rotate at 60 rpm. A worm gear speed reducer is being considered to couple the
motor to the crushing machine. A preliminary sketch of the wormset to be used in the speed reducer proposes a
double-start right-hand worm with axial pitch of 0.625 inch, a normal pressure angle of 14 ½ ˚, and a center distance
of 5.00 inches. The proposed material for the worm is steel with a minimum surface hardness of Rockwell C 58.
The proposed gear material id forged bronze.
Calculate or determine the following, assuming the friction coefficient between worm and gear to be 0.09,
and that the motor is operating steadily at full rated power;
------------------------------------------------------------------------------------------------------------------------------------------
Solution
⎛ω ⎞ ⎛n ⎞ ⎛ 1200 ⎞
(a) From (15-89), Ng = ⎜ w ⎟ Nw = ⎜ w ⎟ Nw = ⎜ ⎟ 2 = 40 teeth
⎜ω ⎟ ⎜n ⎟ ⎝ 60 ⎠
⎝ g⎠ ⎝ g⎠
N g px ( 40 )( 0.625 )
dg = = = 7.96 inches
π π
Lw
λw = tan −1
π dw
Lw = N w px = 2 ( 0.625 ) = 1.25 inches
d w = 2C − d g = 2 ( 5 ) − 7.96 = 2.04 inches
1.25
λw = tan −1 = 11.04o
π ( 2.04 )
649
Vw
Vs =
cos λw
2π ( d w 2 ) nw
2π ( 2.04 2 )(1200 ) ft
Vw = = = 641
12 12 min
641 ft
Vs = = 653
cos11.04o min
⎛ sin ϕ n ⎞
Fwr = Fwt ⎜ ⎟
⎝ cos ϕ n sin λw + µ cos λw ⎠
⎛ sin14.5o ⎞
= 103 ⎜ o ⎟
= 94 lb
⎝ cos14.5 o
sin11.04 o
+ 0.09 cos11.04 ⎠
hpout = e ( hpin )
cos ϕ n − µ tan λw cos14.5o − 0.09 tan11.04o
e= = = 0.67
cos ϕ n + µ cot λw cos14.5o + 0.09 cot11.04o
hpout = 0.67 ( 2 ) = 1.34 horsepower
650
15-66. A worm gear speed reducer has a right-hand triple-threaded worm made of hardened steel, a normal
pressure angle of 20˚, an axial pitch of 0.25 inch, and a center distance of 2.375 inches. The gear is made of forged
bronze. The speed reduction from input to output is 15:1. If the worm is driven by a ½-hp, 1200-rpm electric motor
operating steadily at full rated power, determine the following, assuming the coefficient of friction between worm
and gear to be 0.09:
g. An acceptable range for face width that should allow a nominal operating life of 25,000 hours. (Hint: See
footnote 130 relating to equations (15-115).)
------------------------------------------------------------------------------------------------------------------------------------------
Solution
⎛ω ⎞ ⎛n ⎞
(a) From (15-89), N g = ⎜ w ⎟ N w = ⎜ w ⎟ N w = (15 ) 3 = 45 teeth
⎜ω ⎟ ⎜n ⎟
⎝ g⎠ ⎝ g⎠
N g px ( 45)( 0.25)
(b) From (15-87), d g = = = 3.58 inches
π π
Lw
λw = tan −1
π dw
Lw = N w px = 3 ( 0.25 ) = 0.75 inches
d w = 2C − d g = 2 ( 2.375 ) − 3.58 = 1.17 inches
0.75
λw = tan −1 = 11.5o
π (1.17 )
Vw
Vs =
cos λw
2π ( d w 2 ) nw 2π (1.17 2 )(1200 ) ft
Vw = = = 368
12 12 min
368 ft
Vs = o
= 376
cos11.5 min
651
33, 000 ( hp )in 33, 000 ( 0.5 )
(e) From (15-104), Fwt = = = 44 lb
Vw 376
From (15-117), for dg = 3.58 < 8.0 inches, ( K s )d = 1000 , selecting the larger, Ks = 1000
g
652
15-67. It is desired to utilize a worm gearset to reduce the speed of a 1750-rpm motor driving the worm down to
the output gear shaft speed of approximately 55 rpm, and provide 1 ½ horsepower to the load. Design an acceptable
worm gearset, and specify the nominal required horsepower rating of the drive motor.
------------------------------------------------------------------------------------------------------------------------------------------
Solution
(a) mG = 1750/55 = 31.8. Initially select Nw =2 and tentatively choose hardened steel for the worm and forged
bronze for the gear. From (15-89),
⎛n ⎞ ⎛ 1750 ⎞
N g = ⎜ w ⎟ Nw = ⎜
⎜n ⎟ ⎟ ( 2 ) = 64 teeth
⎝ g⎠ ⎝ 55 ⎠
Tentatively select center distance (arbitrary judgment call; may require adjustment) as C = 5.0 inches. From (15-
90),
C 0.875 C 0.875
≤ dw ≤
3.0 1.6
( 5.0 ) ( 5.0 )
0.875 0.875
≤ dw ≤
3.0 1.6
1.36 ≤ d w ≤ 2.56
Select dw = 2.0 inches. From (15-91), dg = 2C – dw = 2(5.0) – 2.0 = 8.0 inches. From (15-88), Pd = Ng/dg = 64/8.0 =
8. From (15-93), Lw = Nwpx, and from (15-87)
π dg π ( 8.0 )
px = = = 0.393 inch
Ng 64
Lw = 2 ( 0.393) = 0.785 inch
Lw 0.785
λw = tan −1 = tan −1 = 7.12o
π dw π ( 2.0 )
Vw
Vs =
cos λw
2π ( d w 2 ) nw 2π ( 2.0 2 )(1750 ) ft
Vw = = = 916
12 12 min
916 ft
Vs = o
= 923
cos 7.12 min
653
hpout = e ( hpin )
cos ϕ n − µ tan λw
e=
cos ϕ n + µ cot λw
From Appendix Table A1, for steel sliding on bronze, µ = 0.09, thus we have
From (15-92), bmax = 0.67 dw = 0.67(2.0) = 1.34 inches. So the acceptable range is 0.36 ≤ b ≤ 1.34 inches. Select b
= 0.75 inch and from (hp)in = 2.6, select a drive motor of 3.0 hp, 1750 rpm motor.
Material:
φn = 20˚, λw = 7.12˚
654
Nw = 2 starts
Ng = 64 teeth
C = 5.0 inches
dw = 2.0 inches
dg = 8.0 inches
Pd = 8
b = 0.75 inch
655
Chapter 16
16-1. A short-shoe block brake is to have the configuration shown in Figure P16.1, with the drum
rotating clockwise at 500 rpm, as shown. The shoe is molded fiberglass, the drum is aluminum-bronze, and
the entire assembly is continuously water-sprayed. Maximum allowable contact pressure is 120 psi and the
coefficient of friction of wet molded fiberglass on aluminum-bronze is 0.15.
a. Using symbols only, derive an expression for actuating force, Fa , expressed as a function of
pmax .
b. If the actuating force must not exceed 30 lb, what minimum lever length d should be used?
c. Using symbols only, write an expression for braking torque T f .
d. Calculate a numerical value for the maximum-allowable braking torque that may be expected
from this design.
---------------------------------------------------------------------------------------------------------------------------------
Solution
∑ M B = 0 : cN + µ Ne − Fa d = 0
( c + µ e ) N max
( Fa )max =
d
⎛ c + µe ⎞
( Fa )max = pmax A ⎜ ⎟
⎝ d ⎠
⎛ c + µe ⎞ ⎛ 4 + 0.15(6) ⎞
(b) d min = pmax A ⎜ ⎟ = 200(2) ⎜ ⎟ = 65.3"
⎜ ( Fa ) ⎟ ⎝ 30 ⎠
⎝ max ⎠
T f = µ Nr
656
16-2. Repeat problem 16-1, except that the drum rotates clockwise at 600 rpm, the shoe lining is woven
cotton, the drum is cast iron, and the environment is dry. In addition, referring to Figure P16.1, e is 3.0
inches, R is 8.0 inches, the contact area is 8.0 in 2 , and Fa = 60 lb is the maximum-allowable value of
applied force, vertically downward.
---------------------------------------------------------------------------------------------------------------------------------
Solution
From specification: n = 600 rpm , pmax = 100 psi (from Table 16.1) , µ = 0.47 (from Table 16.1) ,
A = 8.0 in 2 , ( Fa )max = 60 lb
∑ M B = 0 : 4 N + 3µ N − Fa d = 0
( 4 + 3µ ) N max
( Fa )max =
d
⎛ 4 + µ3 ⎞
( Fa )max = pmax A ⎜ ⎟
⎝ d ⎠
⎛ 4 + 3µ ⎞ ⎛ 4 + 0.47(3) ⎞
(b) d min = pmax A ⎜ ⎟ = 100(8) ⎜ ⎟ = 71.2"
⎜ ( Fa ) ⎟ ⎝ 60 ⎠
⎝ max ⎠
T f = µ Nr
657
16-3. Classify the short-shoe block brake shown in Figure P16.1 as either “self-energizing” or “non-self-
energizing”.
---------------------------------------------------------------------------------------------------------------------------------
Solution
(c + µe) N
Fa = >0
d
658
16-4. Repeat problem 16-3 for the case where the drum rotates counterclockwise at 800 rpm.
---------------------------------------------------------------------------------------------------------------------------------
Solution
∑ M B = 0 : cN − eµ N − Fa d = 0
( c − eµ ) N
Fa =
d
Because of the −eµ term, there is a potential to reduce Fa , the brake is self-energizing and has the
possibility of being self-locking, depending on the magnitude of eµ .
659
16-5. Repeat problem 16.1, except that the shoe lining is cermet and the drum is steel.
---------------------------------------------------------------------------------------------------------------------------------
Solution
From specification: n = 500 rpm , pmax = 150 psi (from Table 16.1) , µ = 0.32 (from Table 16.1) ,
A = 2 in 2
∑ M B = 0 : cN + µ Ne − Fa d = 0
( c + µ e ) N max
( Fa )max =
d
⎛ c + µe ⎞
( Fa )max = pmax A ⎜ ⎟
⎝ d ⎠
⎛ c + µe ⎞ ⎛ 4 + 0.32(6) ⎞
(b) d min = pmax A ⎜ ⎟ = 150(2) ⎜ ⎟ = 59.2"
⎜ ( Fa ) ⎟ ⎝ 30 ⎠
⎝ max ⎠
T f = µ Nr
660
16-6. Repeat problem 16-1, except that the coefficient of friction is 0.2, the maximum-allowable pressure
is 80 psi, the contact area A is 10.0 in 2 1, e is 30.0 inches, R is 9.0 inches, c is 12.0 inches, and the
maximum-allowable value of Fa is 280 lb.
---------------------------------------------------------------------------------------------------------------------------------
Solution
From specification: n = 500 rpm , pmax = 80 psi , µ = 0.2 , A = 10.0 in 2 , e = 30.0 in , c = 12.0 in ,
Fa = 280 lb
∑ M B = 0 : cN + µ Ne − Fa d = 0
( c + µ e ) N max
( Fa )max =
d
⎛ c + µe ⎞
( Fa )max = pmax A ⎜ ⎟
⎝ d ⎠
⎛ c + µe ⎞ ⎛ 12 + 0.2(30) ⎞
(b) d min = pmax A ⎜ ⎟ = 80(10) ⎜ ⎟ = 51.4"
⎜ ( Fa ) ⎟ ⎝ 280 ⎠
⎝ max ⎠
T f = µ Nr
661
16-7. A short-shoe block brake is to have the configuration shown in Figure P16.7, with the drum rotating
clockwise at 600 rpm, as shown. The shoe is molded fiberglass, the drum is stainless steel, and the entire
assembly is submerged in salt water. The maximum-allowable contact pressure is 0.9 MPa and the
coefficient of friction of wet molded fiberglass on stainless steel is µ = 0.18 .
a. Using symbols only, derive an expression for the actuating force Fa as a function of pmax .
b. What is the maximum actuating force that should be used for proper operation and acceptable
design life?
c. Using symbols only, write an expression for braking torque.
d. Calculate a numerical value for maximum braking torque that may be expected from this design.
e. Would you classify the design as “self-energizing” or “non-self-energizing”? Why?
---------------------------------------------------------------------------------------------------------------------------------
Solution
∑ M B = 0 : bN + d µ N − aFa = 0
⎛ b + dµ ⎞
Fa = ⎜ ⎟N
⎝ a ⎠
b + dµ ⎞
(b) ( Fa )max = ⎛⎜
⎝ a ⎠
⎛ 0.05 + 0.075(0.18) ⎞
⎟ pmax A = ⎜
⎝ 0.45 ⎠
(
6 −6
)(
⎟ 0.9 ×10 125 × 10 = 15.88 N )
( Fa )max = 15.9 N
(c) Taking the drum as a free body diagram and summing moments about its center, where radius of the
drum is the dimension c is the problem statement
T = R µ N max = R µ pmax A
( )(
(d) TF = R µ N max = 0.10(0.18) 0.9 × 106 125 × 10−6 = 2.025 N-m)
662
16-8. For the shoe brake shown in Figure P16.8, it is difficult to determine by inspection whether the shor-
shoe assumption will produce a sufficiently accurate estimate of braking torque upon application of the
actuating force Fa .
a. Determine the percent error in calculated braking torque that you would expect in this case if the
short-shoe assumption is used for calculation of the braking torque. Base your determination on the
premise that the long-shoe equations are completely valid.
b. Would the error made by using the short-shoe assumption be on the “conservative” side (braking
torque calculation by short-shoe assumption is less than the true value of braking torque) or on the
“nonconservative” side?
c. Do you consider the error made by using the short-shoe assumption to be significant or negligible
for this particular case?
---------------------------------------------------------------------------------------------------------------------------------
Solution
∑ M B = 0 : 10 N − 4 µ N − 30 Fa = 0
30 Fa 30(650)
N= = = 2119.6 lb
10 − 4 µ 10 − 4(0.2)
From 16-52
663
16-9. Repeat problem 16.8, except that a = 36 inches , Fa = 300 lb , α = 40o , β = 45o , R = 7 inches ,
b = 2 inches , µ = 0.25 , and the drum rotates counterclockwise at a speed of 2500 rpm. An accurate
estimate of 150 psi for the actual value of pmax has already been made.
---------------------------------------------------------------------------------------------------------------------------------
Solution
(a) For the short-shoe brake, we take moments about the pivot point B and find
36 Fa 36(300)
∑ M B = 0 : 10 N − 3µ N − 36 Fa = 0 ⇒ N= = = 1167.6 ≈ 1168 lb
10 − 3µ 10 − 3(0.25)
664
16-10. A short-shoe block brake has the configuration shown in Figure P16.10, with the drum rotating
clockwise at 63 rad/sec, as shown. The shoe is wood and the drum is cast iron. The weight of the drum is
322 lb, and its radius of gyration is 7.5 inches. The maximum-allowable contact pressure is 80 psi, and the
coefficient of friction is µ = 0.2 . Other dimensions are shown in Figure P16.10.
a. Derive an expression for the actuating force Fa , and calculate its maximum-allowable numerical
value.
b. Derive an expression for braking torque, and calculate its numerical value when the maximum-
allowable actuating force is applied.
c. Estimate the time required to bring the rotating drum to a stop when the maximum-allowable
actuating force is applied.
d. Would you expect frictional heating to be a problem in this application?
---------------------------------------------------------------------------------------------------------------------------------
Solution
(12 + 3µ ) N
12 N + 3µ N − 36 Fa = 0 ⇒ Fa =
36
J eωop
2
Mk 2ωop
2
322(7.5) 2 (63) 2
Hf = = = = 9.97 Btu
2 Jθ 2(9336) 2(386)(9336)
For cast iron, C = 0.12 Btu/lb- o F . Assuming that the heat-absorbing mass of the drum is approximately
10% of the drum, W = 0.1(Wd ) = 32.2 lb . Therefore
9.97
∆Θ = = 2.58 o F Î frictional heating does not appear to be a problem
0.12(32.2)
665
16-11. Repeat problem 6-10 for the case where the drum rotates counterclockwise at 63 rad/sec.
---------------------------------------------------------------------------------------------------------------------------------
Solution
(12 − 3µ ) N
12 N − 3µ N − 36 Fa = 0 ⇒ Fa =
36
J eωop
2
Mk 2ωop
2
322(7.5) 2 (63) 2
Hf = = = = 9.97 Btu
2 Jθ 2(9336) 2(386)(9336)
For cast iron, C = 0.12 Btu/lb- o F . Assuming that the heat-absorbing mass of the drum is approximately
10% of the drum, W = 0.1(Wd ) = 32.2 lb . Therefore
9.97
∆Θ = = 2.58 o F Î frictional heating does not appear to be a problem
0.12(32.2)
666
16-12. Figure P16.12 shows a 1000-kg mass being lowered at a uniform velocity of 3 m/s by a flexible
cable wrapper around a drum of 60-cm diameter. The drum weight is 2 kN, and it has a radius of gyration
of 35 cm.
---------------------------------------------------------------------------------------------------------------------------------
Solution
From the specifications: mload = 1000 kg , vload = 3 m/s , d drum = 60 cm , Wdrum = 2 kN , kdrum = 35 cm
2
1 1 1 1 ⎛W ⎞ ⎛ vload ⎞
KE = ( KE )m + ( KE )d = mload vload
2
+ J eωop
2
= mload vload
2
+ ⎜ drum kdrum
2
⎟⎜ ⎟
2 2 2 2⎝ g ⎠ ⎝ rd ⎠
2
1 ⎛ 2000 ⎞
( 0.35 )2 ⎛⎜ 3 ⎞⎟ = 4500 + 1248.7 = 5748.7
1 2
= (1000)(3) 2 + ⎜
2 2 ⎝ 9.81 ⎟⎠ ⎝ 0.6 ⎠
KE ≈ 5749 J
(b) To bring the moving system to a stop, the work required (W) must equal the kinetic energy (KE), or
W = 5749 J . The work is
W = T f θ = 5749
where
⎛ ωop − 0 ⎞ ωop tr
θ = ωavg tr = ⎜ ⎟ tr =
⎝ 2 ⎠ 2
In this
vop
ωop =
rd
So
vop tr
θ=
2rd
Therefore
667
16-13. A short-shoe brake is sketched in Figure P16.13. Four seconds after the 1-kN actuating force is
applied, the CW rotating drum comes to a full stop. During this time, the drum makes 100 revolutions. The
estimated coefficient of friction between drum and shoe is 0.5. Do the following:
---------------------------------------------------------------------------------------------------------------------------------
Solution
(a)
(b) ∑M D = 0:
( 0.3 − 0.15µ ) N
0.3 N − 0.15µ N − 0.7 Fa = 0 ⇒ Fa =
0.7
The brake is self-energizing since the friction moment aids Fa in applying the brake, and if
( 0.3 − 0.15µ ) / 0.7 ≤ 0 , it will be self-locking. To check this
0.7 Fa 0.7(1000)
N= = = 3.111 kN
0.3 − 0.15µ 0.3 − 0.15(0.5)
668
Problem 16-13 (continued)
(e) To bring the system to a stop, the work must equal the kinetic energy, W = T f θ = KE , where
30 30
d≥ = = 60 cm
µ 0.5
669
16-14. A long-shoe brake assembly is sketched in Figure P16.14. The estimated coefficient of friction
between the show and drum is 0.3, and the maximum-allowable pressure for the lining material is 75 psi.
Noting the CCW direction of rotation, determine the following:
a. The maximum actuating force Fa that can be applied without exceeding the maximum-allowable
contact pressure.
b. The friction braking torque capacity corresponding to Fa calculated in (a).
c. The vertical and horizontal components of the reaction force at pin C.
d. Is the brake self-energizing or self-locking?
---------------------------------------------------------------------------------------------------------------------------------
Solution
MN + M f
(a) For a CCW rotation, Fa = , where
a
wc Rr1 pmax
MN = ( 2α − sin 2ϕ2 + sin 2ϕ1 )
4 ( sin ϕ )max
µ wc Rpmax
Mf =
4 ( sin ϕ )max
( r1 ( cos 2ϕ2 − cos 2ϕ1 ) − 4R ( cos ϕ2 − cos ϕ1 ) )
where
wc = 1.6"
R = 5.9"
r1 = (7) 2 + (7.9) 2 = 10.56"
⎛ 7.0 ⎞
β = tan −1 ⎜ ⎟ = 41.54 = 0.725 rad
o
⎝ 7.9 ⎠
α ⎛ 5.9 ⎞
= sin −1 ⎜ ⎟ = 30 = 0.5236 rad
o
2 ⎝ 2(5.9) ⎠
ϑ1 = 60 − β = 18.46o = 0.3222 rad
ϑ2 = α + ϑ1 = 60 + 18.46 = 78.46o = 1.3694 rad
( sin ϕ )max = sin ϕ2 = sin 78.46o = 0.9798
1.6(5.9)(10.56)(75)
MN = ( 2.0944 − 0.3920 + 0.6007 ) = 1907.65(2.3031) = 4393.5
4 ( 0.9798 )
Mf =
0.3(1.6)(5.9)(75)
4 ( 0.9798 )
(10.56 ( −0.91996 − 0.79948) − 4(5.9) ( 0.2001 − 0.9485) )
= 54.195(−18.157 + 17.598) = −30.27
4393.5 − 30.27
Fa = = 221.45 lb Fa ≈ 222 lb
19.7
670
Problem 16-14 (continued)
T f ≈ 957 in-lb
wc Rpmax
Rv = Fa − ⎡( sin β − µ cos β )( cos 2ϕ1 − cos 2ϕ2 )
4 ( sin ϕ )max ⎣
+ ( cos β + µ sin β )( 2α − sin 2ϕ2 + sin 2ϕ1 ) ⎤⎦
1.6(5.9)(75)
= 222 − ⎡( sin 41.54 − 0.3cos 41.54 )( cos 36.92 − cos156.92 )
4(0.9798) ⎣
+ ( cos 41.54 + 0.3sin 41.54 )( 2.0944 − sin156.92 + sin 36.92 ) ⎤⎦
= 222 − 180.65 ⎡⎣( 0.6633 − 0.3(0.7485) )(1.7194 ) + ( 0.7485 + 0.3(0.6631) )( 2.303) ⎤⎦
Rv = −308.4 lb
wc Rpmax
Rh = ⎡( µ sin β − cos β )( cos 2ϕ1 − cos 2ϕ2 )
4 ( sin ϕ )max ⎣
+ ( µ cos β + sin β )( 2α − sin 2ϕ2 + sin 2ϕ1 ) ⎤⎦
1.6(5.9)(75)
= ⎡( 0.3sin 41.54 − cos 41.54 )( cos 36.92 − cos156.92 )
4(0.9798) ⎣
+ ( 0.3cos 41.54 + sin 41.54 )( 2.0944 − sin156.92 + sin 36.92) ⎤⎦
= 180.65 ⎡⎣( 0.3(0.6633) − 0.7485 )(1.7194 ) + ( 0.3(0.7485) + 0.6631)( 2.303) ⎤⎦
Rh = 207.9 lb
(d) Since M f < 0 , it is CW about point C, and the brake is self-energizing, but since Fa ≈ 222 lb > 0 , the
brake is not self-locking.
671
16-15. Repeat problem 16-14 for the case where the drum rotates in the clockwise direction.
---------------------------------------------------------------------------------------------------------------------------------
Solution
MN − M f
(a) For a CW rotation, Fa = , where
a
wc Rr1 pmax
MN = ( 2α − sin 2ϕ2 + sin 2ϕ1 )
4 ( sin ϕ )max
µ wc Rpmax
Mf =
4 ( sin ϕ )max
( r1 ( cos 2ϕ2 − cos 2ϕ1 ) − 4R ( cos ϕ2 − cos ϕ1 ) )
where
wc = 1.6"
R = 5.9"
r1 = (7) 2 + (7.9) 2 = 10.56"
⎛ 7.0 ⎞
β = tan −1 ⎜ ⎟ = 41.54 = 0.725 rad
o
⎝ 7.9 ⎠
α ⎛ 5.9 ⎞
= sin −1 ⎜ ⎟ = 30 = 0.5236 rad
o
2 ⎝ 2(5.9) ⎠
ϑ1 = 60 − β = 18.46o = 0.3222 rad
ϑ2 = α + ϑ1 = 60 + 18.46 = 78.46o = 1.3694 rad
( sin ϕ )max = sin ϕ2 = sin 78.46o = 0.9798
1.6(5.9)(10.56)(75)
MN = ( 2.0944 − 0.3920 + 0.6007 ) = 1907.65(2.3031) = 4393.5
4 ( 0.9798 )
Mf =
0.3(1.6)(5.9)(75)
4 ( 0.9798 )
(10.56 ( −0.91996 − 0.79948) − 4(5.9) ( 0.2001 − 0.9485) )
= 54.195(−18.157 + 17.598) = −30.27
4393.5 − (30.27)
Fa = = 224.6 lb Fa ≈ 225 lb
19.7
T f ≈ 957 in-lb
672
Problem 16-15 (continued)
wc Rpmax
Rv = Fa − ⎡( sin β − µ cos β )( cos 2ϕ1 − cos 2ϕ2 )
4 ( sin ϕ )max ⎣
+ ( cos β + µ sin β )( 2α − sin 2ϕ2 + sin 2ϕ1 ) ⎤⎦
1.6(5.9)(75)
= 225 − ⎡( sin 41.54 − 0.3cos 41.54 )( cos 36.92 − cos156.92 )
4(0.9798) ⎣
+ ( cos 41.54 + 0.3sin 41.54 )( 2.0944 − sin156.92 + sin 36.92 ) ⎤⎦
= 225 − 180.65 ⎡⎣( 0.6633 − 0.3(0.7485) )(1.7194 ) + ( 0.7485 + 0.3(0.6631) )( 2.303) ⎤⎦
Rv = −305.4 lb
wc Rpmax
Rh = ⎡( µ sin β − cos β )( cos 2ϕ1 − cos 2ϕ2 )
4 ( sin ϕ )max ⎣
+ ( µ cos β + sin β )( 2α − sin 2ϕ2 + sin 2ϕ1 ) ⎤⎦
1.6(5.9)(75)
= ⎡( 0.3sin 41.54 − cos 41.54 )( cos 36.92 − cos156.92 )
4(0.9798) ⎣
+ ( 0.3cos 41.54 + sin 41.54 )( 2.0944 − sin156.92 + sin 36.92) ⎤⎦
= 180.65 ⎡⎣( 0.3(0.6633) − 0.7485 )(1.7194 ) + ( 0.3(0.7485) + 0.6631)( 2.303) ⎤⎦
Rh = 207.9 lb
(d) Since M f < 0 ,and the drum rotation is CW (reverse direction from that used to derive 16-45), the
− M f means the moment is CCW and the brake is non self-energizing, but since Fa ≈ 225 lb > 0 , the
brake is not self-locking.
673
16-16. The brake system shown in Figure P16.16 is to be fabricated using a wc = 30 mm wide lining
material at the contact surface. The coefficient of friction between the drum the lining material is µ = 0.2 .
The lining material should not be operated at maximum pressures higher than 0. 8 MPa. Determine the
minimum activation force Fa .
---------------------------------------------------------------------------------------------------------------------------------
Solution
µ wc Rpmax
Mf = ⎡ r1 ( cos 2ϕ2 − cos 2ϕ1 ) − 4 R ( cos ϕ2 − cos ϕ1 ) ⎤⎦
4 ( sin ϕ )max ⎣
0.2(0.03)(0.10)(0.8 ×106 )
= ⎡⎣ 0.25 ( cos196.22 − cos16.26 ) − 4(0.10) ( cos 98.13 − cos8.13) ⎤⎦
4(1.0)
= 120 ⎡⎣ 0.25 ( −0.960 − 0.960 ) − 0.4 ( −0.1414 − 0.9899 ) ⎤⎦ = 120 [ −0.96 + 0.4525] = −0.061 kN-m
MN + M f 0.555 + (−0.061)
Fa = = = 0.988 kN
a 0.5
For the lower arm all relevant parameters are the same with
M N = 0.555 kN-m and M f = −0.061 kN-m . The direction of
the friction force in this model indicates
MN − M f 0.555 − (−0.061)
Fa = = = 1.232 kN
a 0.5
674
16-17. The brake shown in Figure P16-17 is to be fabricated using an impregnated asbestos lining material
at the contact surface. The lining material should not be operated at maximum pressures higher than 100
psi.
a. What is the largest actuating force Fa , that should be used with this braking system as no
designed?
b. If the largest-allowable actuating force is applied, what braking torque is produced on the rotating
drum?
---------------------------------------------------------------------------------------------------------------------------------
Solution
MN + M f
(a) For a CCW rotation, Fa = , where
a
wc Rr1 pmax
MN = ( 2α − sin 2ϕ2 + sin 2ϕ1 )
4 ( sin ϕ )max
µ wc Rpmax
Mf =
4 ( sin ϕ )max
( r1 ( cos 2ϕ2 − cos 2ϕ1 ) − 4R ( cos ϕ2 − cos ϕ1 ) )
2.0(6.0)(14.14)(100)
MN = (π − 0 + 0 ) = 4242π ≈ 13,327 in-lb
4 (1.0 )
Mf =
0.2(2.0)(6.0)(100)
4 (1)
(14.14 ( −1 − 1) − 4(6.0) ( 0 − 1) )
= 60 ( −28.28 + 24 ) = −256.8 ≈ −257 in-lb
13,327 − 257
Fa = = 435.67 ≈ 436 lb Fa ≈ 436 lb
30
T f ≈ 1440 in-lb
675
16-18. Repeat problem 16-17 for the case where the drum rotates in the clockwise direction.
---------------------------------------------------------------------------------------------------------------------------------
Solution
MN − M f
(a) For a CW rotation, Fa = , where
a
wc Rr1 pmax
MN = ( 2α − sin 2ϕ2 + sin 2ϕ1 )
4 ( sin ϕ )max
µ wc Rpmax
Mf =
4 ( sin ϕ )max
( r1 ( cos 2ϕ2 − cos 2ϕ1 ) − 4R ( cos ϕ2 − cos ϕ1 ) )
2.0(6.0)(14.14)(100)
MN = (π − 0 + 0 ) = 4242π ≈ 13,327 in-lb
4 (1.0 )
Mf =
0.2(2.0)(6.0)(100)
4 (1)
(14.14 ( −1 − 1) − 4(6.0) ( 0 − 1) )
= 60 ( −28.28 + 24 ) = −256.8 ≈ −257 in-lb
13,327 − (−257)
Fa = = 452.8 ≈ 453 lb Fa ≈ 436 lb
30
T f ≈ 1440 in-lb
676
16-19. A 16-inch diameter drum has two internally expanding shoes as shown in Figure P16.19 . The
actuating mechanism is a hydraulic cylinder AB, which produces the same actuating force Fa on each shoe
(applied at points A, and B). The width of each shoe is 2 inches, the coefficient of friction is µ = 0.24 , and
the maximum pressure is pmax = 150 psi . Assuming the drum rotates clockwise, determine the minimum
required actuating force, and the friction torque.
---------------------------------------------------------------------------------------------------------------------------------
Solution
The moments due to the normal and friction forces either add or
subtract, based on a clockwise rotation of the drum, as shown in the
figure. The angles ϕ1 and ϕ2 are:
By noting the directions of the normal and friction forces in the figure,
we determine
MN − M f
Shoe AC: ∑ M C = 0 : 12 Fa + M f − M N = 0 Fa =
10
MN + M f
Shoe BD: ∑ M D = 0 : 12 Fa − M f − M N = 0 Fa =
10
( )
The smallest activation force results from Fa = M N − M f /10 . For a clockwise drum rotation, shoe AC
is the primary shoe and experiences the largest pressure, and shoe BD is the secondary shoe, which will not
experience the same pressure as shoe AC. The magnitudes of M N and M f will be independent of the shoe
being considered, therefore we take advantage of symmetry. Since ϕ 2 < 90o for both shoes,
( sin ϕ )max = sin ϕ2 = 0.9836 . The moments due to the normal and friction forces are determined from
equations (16-42) and (16-45), in which α = 0.442π , wc = 2 in. , ϕ1 = 0 , ϕ2 = 79.6o , µ = 0.24 ,
pmax = 150 psi , R = r1 = 8 in. . Therefore
wc Rr1 pmax
MN = [ 2α − sin 2φ2 + sin 2φ1 ]
4 ( sin φ )max
=
(2)(8) 2 (150) ⎡
4(0.9836) ⎣
( )
2 ( 0.442π ) − sin 159.2o + sin ( 0 ) ⎤ = 4880 [ 2.777 − 0.3551] = 11,819 lb-in
⎦
µ wc Rpmax
Mf = ⎡ r1 ( cos 2φ2 − cos 2φ1 ) − 4 R ( cos φ2 − cos φ1 ) ⎤⎦
4 ( sin φ )max ⎣
=
0.24(2)(8)(150) ⎡
4(0.9836) ⎣⎢
( ( ) ) ( ( )
8 cos 159.2o − cos ( 0 ) − 4(8) cos 79.6o − cos ( 0 ) ⎤⎥
⎦ )
= 146.4 [ −7.479 − (−26.223) ] = 2744 lb-in
677
Problem 16-19 (continued)
MN − M f 11,819 − 2744
( Fa )min = ( Fa ) primary = = = 907.5 lb
10 10
T=
µ wc R 2 pmax
sin θ max
[ cos ϕ1 − cos ϕ2 ] =
0.24(2)(8) 2 (150)
0.9836
( (
cos ( 0 ) − cos 76.9o ) ) = 3623 in-lb
T primary = 3623 in-lb
The pressure on the secondary shoe can be determined by noting that pmax / sin θ max = p / sin θ and
sin θ = sin θ max . Therefore the normal and friction moments already computed can be used to determine the
pressure on the secondary shoe. We can express the moments as
11,819 2744
MN = p Mf = p
150 150
MN + M f ⎛ 1 ⎞
Therefore Fa = ⇒ 907.5 = ⎜ ⎟ (11,819 p + 2744 p ) ⇒ p = 93.47 psi
10 ⎝ 10(150) ⎠
T=
µ wc R 2 pmax
sin θ max
[ cos ϕ1 − cos ϕ2 ] =
0.24(2)(8) 2 (93.47)
0.9836
( (
cos ( 0 ) − cos 76.9o ) ) = 2258 in-lb
Tsec in-lb
678
16-20. A simple band brake of the type shown in Figure 16.9 is to be constructed using a lining material
that has a maximum-allowable contact pressure of 600 kPa. The diameter of the rotating drum is to be 350
mm, and the proposed width of the band is 100 mm. The angle of wrap is 270o . Tests of the lining material
indicate that a good estimate for the coefficient of friction is 0.25. Do the following:
---------------------------------------------------------------------------------------------------------------------------------
Solution
P1 10.5
(b) P2 = = = 3.23 kN
e µα e(0.25)(1.5π )
( ) ( )
(c) T f = bR 2 pmax 1 − e − µα = 0.1(0.175) 2 (600 × 103 ) 1 − e− (0.25)(1.5π ) ≈ 1272 N-m
m 45
(d) Fa = P2 = (3.23) = 161.5 N
a 900
679
16-21. Repeat problem 16-20 for the case where the angle of wrap is 180o .
---------------------------------------------------------------------------------------------------------------------------------
Solution
P1 10.5
(b) P2 = µα
= (0.25)(π )
= 4.79 kN
e e
( ) ( )
(c) T f = bR 2 pmax 1 − e − µα = 0.1(0.175) 2 (600 × 103 ) 1 − e−0.25π = 999.7 ≈ 1000 N-m
m 45
(d) Fa = P2 = (4.79) = 239.5 N
a 900
680
16-22. A simple band brake is constructed using a 0.050-inch-thick by 2-iniwide steel band to support the
tensile forces. Carbon-graphite material is bonded to the inside of the steel band to provide the friction
surface for braking, and the rotating drum is to be a solid-steel cylinder, 16 inches in diameter a 2 inches in
axial thickness. The brake band is wrapped around the rotating drum so that it is in contact over 270o of the
drum surface. It is desired to bring the drum to a complete stop in exactly one revolution from its operating
speed of 1200 rpm. What would be the maximum tensile stress induced in the 0.050-inch-thick steel band
during the braking period if the drum were brought to a stop in exactly one revolution? Assume that the
rotating drum is the only significant mass in the system, and that the brake is kept dry.
---------------------------------------------------------------------------------------------------------------------------------
Solution
Specifications: bband = bdrum = 2.0" , tband = 0.050" , R = 16 / 2 = 8" , α = 270o = 1.5π rad ,
θ stop = 2π rad , nop = 1200 rpm , µ = 0.25 (from Table 16.1)
P1 P1 P1
σ max = = = = 10 P1
A bband tband 2.0(0.05)
P1 = P2 e µα = P2 e0.25(1.5π ) ≈ 3.25P2
Tf Tf
P2 = = ≈ 0.0556T f
(
R e µα
−1 ) 8(e 0.25(1.5π )
−1)
Wd ke2ωop
In addition, we know T f = , where
gtr
( )
Wd = wsteel π R 2bdrum = 0.283π (8) 2 (2) = 113.8 lb
From Appendix Table A.2, case 3, ke2 = R / 2 = 4.0 . Therefore, knowing nop = 1200 rpm , we determine
ωop = 2π nop / 60 ≈ 125.7 rad/s , and
113.8(4)(125.7)
Tf = = 148.23 / tr
386tr
T f = 148.23 / tr = 1482.3
681
16-23. A differential band brake is sketched in Figure P16.23. The maximum-allowable pressure for the
band lining material is 60 psi, and the coefficient of friction between the lining and the drum is 0.25. The
band and lining are 4 inches in width. Do the following:
a. If the drum is rotating in the clockwise direction, calculate the tight-side tension and slack-side
tension at maximum-allowable pressure.
b. For clockwise drum rotation, calculate the maximum torque capacity.
c. For clockwise drum rotation, calculate the actuating force corresponding to maximum torque
capacity.
d. If the direction of drum rotation is changed to counterclockwise, calculate the actuating force
corresponding to maximum torque capacity.
---------------------------------------------------------------------------------------------------------------------------------
Solution
Specifications: µ = 0.25 pmax = 60 psi , b = 4.0" , α = 247o = 1.372π rad R = 8" , m1 = 1" , m2 = 4" ,
a = 20"
m2 P2 − m1 P1 Tmax 694.4
Fa = = = = 28.9 lb
20 + 4 24 24
(d) For CCW rotation, P2 becomes the tight side and P1 the slack side. This results in P2 = 1920 lb and
P1 = 653.6 lb for the maximum-allowable pressure of 60 psi. Therefore
Tmax 6985.6
Fa = = ≈ 291 lb
24 24
682
16-24. The differential band brake sketched in Figure P16.24 has a 25-mm-wide band. The coefficient of
friction between the counterclockwise rotating drum and the lining is µ = 0.25 . If the maximum allowable
pressure of 0.4 MPa, determine the activation force, Fa .
---------------------------------------------------------------------------------------------------------------------------------
Solution
⎛ 200 ⎞
β = sin −1 ⎜ ⎟ = 30 so α = 180 + β = 210 = 3.665 rad
o o
⎝ 400 ⎠
( )
P1 = pmax bR = 0.4 × 106 ( 0.025 )( 0.20 ) = 2000 N
P1 2000
P2 = µα
= 0.25(3.665)
= 800 N
e e
Fa = 144.8 N
683
16-25. A 2-m-long arm is attached to a 50-mm-diameter drum, which is free to rotate about an axle at O.
The arm is required to support a 5N force as shown in Figure P16.25. In order to keep the arm from
rotating, a band brake is being suggested. For the initial analysis, we assume the belt is 100 mm wide and is
made from woven cotton. The actuation force is provided by pneumatic cylinder BC, attached to one end of
the belt. The pneumatic cylinder can supply a pressure of 0.3 MPa. Determine the cross-sectional area of
the cylinder in order to supply enough force to keep the required force to keep the arm in the position
shown, and determine the pressure on the pad.
-------------------------------------------------------------------------------------------------------------------------
Solution
For woven cotton, we find (from Table 16.1) µ = 0.47 and pmax = 0.69 MPa . The torque about the axel
resulting from the 5 N load is
8.66
T = 8.66 N-m = R( P1 − P2 ) = 0.025(1 − 0.109) P1 P1 = = 388.8 N
0.02228
388.8
P1 = pA = 0.3 × 106 A : A= = 1.296 × 10−1 m 2 = 1296 mm 2 A = 1296 mm 2
0.3 × 106
684
16-26. In the analysis and design of disk brakes and clutches, it is usual to hypothesize either “uniform
wear” or “uniform pressure” as a basis for making calculations.
a. What important information can be derived on the basis of making a proper choice between these
two hypotheses?
b. How would you decide whether to choose the uniform wear hypothesis of the uniform pressure
hypothesis in any given situation?
---------------------------------------------------------------------------------------------------------------------------------
Solution
(a) A proper choice between the “uniform wear” and “uniform pressure” assumptions provides the basis
for more accurate estimates and calculations for pressure distribution, braking torque, and actuation force
for a disk brake or clutch.
(b) If disks tend to be rigid, the greatest wear will initially occur over the outer circumferential portion of
the disks because tangential velocity is greater there and ultimately the pressure redistribution results in
nearly uniform wear.
If the disks tend to be flexible, the disks tend to be in intimate contact and the result is uniform pressure.
685
16-27. It is desired to replace the long-shoe brake shown in Figure P16.17 with a simple band brake of the
same width b. If the materials used are the same at the friction surface (i.e. µ = 0.2 and pmax = 100 psi are
unchanged) and the drum size must remain unchanged, what wrap angle should be used for the simple band
brake to produce the same braking torque capacity as the long-shoe brake shown in Figure P16.17?
---------------------------------------------------------------------------------------------------------------------------------
Solution
max
µ wc R 2 pmax
sin θ max
(
[cos ϕ1 − cos ϕ2 ] = bR 2 pmax 1 − e− µα )
µ [ cos ϕ1 − cos ϕ 2 ]
1 − e − µα =
sin θ max
0.2(1 − 0)
1 − e −0.2α = = 0.2 ⇒ e−0.2α = 1 − 0.2 = 0.8 or e0.2α = 1/ 0.8 = 1.25
1
686
16-28. A disk clutch is being proposed for an industrial application in which the power-input shaft supplies
12 kw steadily at 650 rpm. The patented friction lining material, which is to be bonded to one or more
annular disk surfaces, is to be brought into contact with stiff steel disks to actuate the clutch. The outside
diameter of the clutch disks must be no larger that 125 mm, and it is desired to configure the annular
friction surfaces so that the inner diameter is about 2/3 of the outer diameter. The coefficient of friction
between the patented friction lining material and steel is µ = 0.32 and the maximum-allowable contact
pressure is pmax = 1.05 MPa . What is the minimum number of friction surfaces required for the clutch to
function properly?
---------------------------------------------------------------------------------------------------------------------------------
Solution
(T f )uw
(T f )uw = n f µπ pmax ri ( ro2 − ri2 ) ⇒ nf =
(
µπ pmax ri ro2 − ri2 )
9549(kw) 9549(12)
Tf = = = 176.3 N-m
n 650
1763
nf = = 2.88
(
0.32π (1.05 × 10 )(0.0417) (0.125) 2 − (0.0417) 2
6
)
Therefore 3 friction surfaces are required.
687
16-29. A disk brake is to be constructed for use on a high-speed rotor balancing machine. It has been
decided that a carbon-graphite friction material be used against a steel-disk material surface to provide the
braking action. The environment is dry. For clearance reasons the inner diameter of the steel brake disk
must be 10.0 inches and its thickness is 0.375 inch. Further, the brake must be able to absorb
2.5 × 106 in-lb of kinetic energy in one-half revolution of the disk brake as it brings the high-speed rotor to
a full stop. Only one braking surface can be used.
---------------------------------------------------------------------------------------------------------------------------------
Solution
(a) From Table 16.2, µ = 0.25 , pmax = 300 psi , Tmax = 1000o F . Since tdisk = 0.375" , they are assumed to
be rigid, so the uniform wear assumption is used.
KE 2.5 × 106
W = T f θ stop ⇒ Tf = = ≈ 0.796 × 106 in-lb
θ stop π
So
( ) (
0.796 × 106 = n f µπ pmax ri ro2 − ri2 = 1(0.25)π (300)(5) ro2 − (5) 2 ) ⇒ ro = 26.47"
KE
(c) The temperature change is ∆T = , where
CWJθ
( )
W = 0.1Vdisk ( w) = 0.1 ⎡π 26.52 − 52 (0.375) ⎤ (0.283) ≈ 22.58 lb
⎣ ⎦
2.5 × 106
∆T = ≈ 99o F
0.12(22.58)(9336)
688
16-30. For use in a specialized underwater hoisting application, it is being proposed to design a disk clutch
with a 20-inch outside diameter. Hard-drawn phosphor bronze is to be used in contact with chrome-plated
hard steel to form the friction interfaces ( µ = 0.03 , pmax = 150 psi ). The clutch is to transmit 150
horsepower continuously at a rotational speed of 1200 revolutions per minute. Following a rule of thumb
that says that for good design practice the inner diameter of a disk clutch should be about 2/3 of the outer
diameter, determine the proper number of friction interfaces to use for this proposed clutch. Since the
device operates under water, neglect temperature limitations.
---------------------------------------------------------------------------------------------------------------------------------
Solution
Specifications: ro = 10" , µ = 0.03 , pmax = 150 psi , ( hp )in = 150 horsepower nop = 1200 rpm ,
ri = 2ro / 3 = 6.67"
Equating these
So use
nf = 2
689
16-31. A pneumatic cylinder with a 25 mm internal diameter operates at a pressure of 0.50 MPa and
supplies the activation force for a clutch that is required to transmit 10 kw at 1600 revolutions per minute.
The friction interfaces of the clutch are a rigid molded nonasbestos with µ = 0.45 and pmax = 1.0 MPa .
The outer diameter of the clutch is initially assumed to be 150 mm and the inner diameter is assumed to be
100 mm. Determine the activation force and the number of friction surfaces assuming both uniform wear
and uniform pressure.
------------------------------------------------------------------------------------------------------------------------------
Solution
The activation force is the same regardless of the uniform wear or uniform pressure assumption.
π (0.025) 2
N a = pA =
4
( 0.50 ×10 ) = 245 N
6
9549(kw) 9549(10) ⎛r +r ⎞
T= = ≈ 60 N-m = n f µ ⎜ o i ⎟ Na
n 1600 ⎝ 2 ⎠
⎛ 0.075 + 0.050 ⎞
60 = n f (0.45) ⎜ ⎟ (245) = 6.9n f
⎝ 2 ⎠
nf = 9
Check to see if the actuation force exceeds the allowable
( )
N a = 2π pmax ri ( ro − ri ) = 2π 1.0 × 106 (0.05)(0.075 − 0.05) = 7854 N > 245 N
The activation force supplied by the cylinder does not exceed the allowable.
T=
(
2n f µ ro3 − ri3 )N
( )
a
3 ro2 − ri2
60 =
(
2n f (0.45) (0.075)3 − (0.050)3 ) (245) = 6.98n
( )
f
3 (0.075) − (0.050) 2 2
nf = 9
( ) ( )(
N a = π pmax ro2 − ri2 = π 1.0 × 106 (0.075) 2 − (0.050)2 = 9817 N > 245 N )
The activation force supplied by the cylinder does not exceed the allowable.
690
16-32. It is desired to replace a single-contact-surface disk brake used on the end of a rotating drum by a
long-shoe block brake, as shown in Figure P16.32, without changing the drum. The materials used at the
friction interface are the same for both cases. It may reasonably be assumed, therefore, that both brakes will
operate at the same limiting pressure pmax during actuation. The original disk brake contact surface had an
outer radius equal to the drum radius, and an inner radius of two-thirds the outer radius. What width b is
required for the new long-shoe brake shown in Figure P16.32 to produce the same braking torque capacity
as the old disk brake?
---------------------------------------------------------------------------------------------------------------------------------
Solution
(T f )uw = n f µπ pmax ri ( ro2 − ri2 ) = (1)µπ pmax (4) ( (6)2 − (4)2 ) = 80µπ pmax
Equating these
b = wc = 6.98"
691
16-33. A disk clutch has a single set of mating annular friction surfaces having an outer diameter of 300
mm and an inner diameter of 225 mm. The estimated coefficient of friction between the two contacting
surfaces is 0.25, and the maximum-allowable pressure for the lining material is 825 kPa. Calculate the
following:
a. Torque capacity under conditions that make the uniform wear assumption more nearly valid.
b. Torque capacity under conditions that make the uniform pressure assumption more nearly valid.
---------------------------------------------------------------------------------------------------------------------------------
Solution
From specifications: ri = 225 / 2 = 112.5 mm , ro = 300 / 2 = 150 mm , n f = 1 , µ = 0.25 , pmax = 825 kPa
(T f )uw = n f µπ pmax ri ( ro2 − ri2 ) = (1)(0.25)π (825 ×103 ) (0.1125) ( (0.15)2 − (0.1125)2 ) = 717.6
(T f )uw ≈ 718 N-m
(b) Uniform pressure
⎛ ⎞ ⎛ ⎞
(T f )up = 2π n f µ pmax ⎜⎜ ro −3 ri ( ) ⎜⎜ (0.15) − (0.1125)3
3 3 3
⎟⎟ = 2π (1)(0.25) 825 × 10 ⎟⎟ = 842.8
3
⎝ ⎠ ⎝ 3 ⎠
692
16-34. A multiple-disk clutch is to be designed to transmit a torque of 750 in-lb while fully submerged in
oil. Space restrictions limit the outside diameter of the disks to 4.0 inches. The tentatively selected
materials for the interposed disks are rigid molded-asbestos against steel. Determine appropriate values for
the following:
---------------------------------------------------------------------------------------------------------------------------------
Solution
From specifications: T f ( )red ' d = 750 in-lb , do = 4.0 in , µ = 0.06 (from Table 16.1), pmax = 300 psi
(a) The usual range of values for ri is 0.6ro ≤ ri ≤ 0.8ro . Selecting a mid-range value, ri = 0.7ro = 1.4 in
di = 2.8 in
(T f )uw = n f µπ pmax ri ( ro2 − ri2 ) = n f (0.06)π ( 300) (1.4) ( (2)2 − (1.4)2 ) = 161.5n f
750 = 161.5n f ⇒ n f = 4.64
(c)
( )uw
2 Tf 2(750)
Na = = = 1470.5
n f µ ( ro + ri ) 5(0.06) ( 2 + 1.4 )
N a ≈ 1471 lb
693
16-35. The wheels of a standard adult bicycle have a rolling radius of approximately 340 mm and a
radius to the center of the hand-actuated caliper disk brake pads (see Figure 16.13) of 310 mm. The
combined weight of the bike plus the rider is 890 kN, equally distributed between the two wheels. If the
coefficient of friction between the tires and the road surface is twice the coefficient of friction between the
caliper brake pads and the metallic wheel rim, calculate the clamping force that must be applied at the
caliper to slide the wheel upon hand brake application.
---------------------------------------------------------------------------------------------------------------------------
Solution
∑ Mo = 0 :
2( µbp N a )(0.310) = µroad (445 × 103 )(0.340)
µbp N a = 2µbp (244 × 103 )
N a = 488 kN
694
16-36. A cone clutch with a cone angle of 12o is disengaged when a spring ( k = 200 lb/in ) is
compressed by means of a lever with a 10 lb load applied as shown in Figure P16.37. The clutch is required
to transmit 4 hp at 1000 rpm. The lining material along an element of the cone is 3.0 inches long. The
coefficient of friction and the maximum pressure for the lining material are µ = 0.38 and pmax = 100 psi ,
respectively. The free length of the spring is L f = 3.0 in. and it is compressed by x inches for operation
(when the clutch is engaged). Determine the amount of spring compression required for the clutch to
engage properly and. the distance a that the 10 lb force has to be away from pivot point A (see Figure
P16.36) in order to compress the spring an additional 0.05 inches to disengage the clutch.
Figure P16.36
Cone clutch
---------------------------------------------------------------------------------------------------------------------------------
Solution
Engaged
F 0.1"
Solution
Disengaged
F + ∆F
695
⎛ 16 Fmax R ⎞ ⎛ 8 Fmax D ⎞ ⎛ 8F ⎞
τ max = KW ⎜ ⎟ = KW ⎜ 3 ⎟
= CKW ⎜ max ⎟
⎝ πd 3
⎠ ⎝ πd ⎠ ⎝ πd2 ⎠
⎛ 8 (195 ) ⎞
τ max = 113, 000 /1.3 = 86,923 = CKW ⎜⎜ ⎟ → CKW = 6.453
2 ⎟
⎝ π (0.192) ⎠
⎛ 4C − 1 0.615 ⎞
where KW = ⎜ + ⎟
⎝ 4C − 4 C ⎠
696
16-37. A cone clutch having a cone angle α of 10o is to transmit 40 horsepower continuously at a
rotational speed of 600 rpm. The contact width of the lining along an element of the cone is 2.0 inches. The
lining material is wound asbestos yarn and wire, operating against steel. Assuming that the uniform wear
assumption holds, do the following:
---------------------------------------------------------------------------------------------------------------------------------
Solution
From specifications: α = 10o , hp = 40 horsepower , n = 600 rpm , Wc = 2.0 inches , µ = 0.38 (from Table
16.1), pmax = 100 psi
697
Chapter 17
17-1. A flat belt drive system is to be designed for an application in which the input shaft speed (driving
pulley) is 1725 rpm, the driven shaft speed is to be approximately 960 rpm, and the power to be transmitted
has been estimated as 3.0 horsepower. The driven machine has been evaluated and found to have
characteristics of moderate shock loading during operation. The desired center distance between driving
and driven pulleys is approximately 18 inches.
a. If 1/8-inch-thick polyamide belt material were chosen for this application, what belt width would
be required?
b. What initial tension would be required for proper operation?
------------------------------------------------------------------------------------------------------------------------------
Solution
33, 000(hp) 2π rs ns
(a) (Tt − Ts ) = , where V =
V 12
From Table 17.1, for a polyamide belt with tb = 0.125" ( ≈ 0.13) , w = 0.042 lb/in 3 , µ = 0.8 ,
Ta = 100 lb/in , and the minimum recommended pulley diameter is d s = 4.3" , which gives
2π (4.3 / 2)(1725)
V= ≈ 1942 ft/min
12
Therefore
33, 000(3.0)
(Tt − Ts ) = = 50.9 ≈ 51 lb ⇒ Ts = Tt − 51
1942
⎛n ⎞ ⎛ 1725 ⎞
From (15-23) d L = d s ⎜ s ⎟ = 4.3 ⎜ ⎟ ≈ 7.73"
⎝ nL ⎠ ⎝ 960 ⎠
⎛ d − ds ⎞ ⎛ 7.73 − 4.3 ⎞
Using Figure 17.1: θ s = π − 2α = π − 2sin −1 ⎜ L ⎟ = π − 2sin −1 ⎜ ⎟ = 2.95 rad
⎝ 2C ⎠ ⎝ 2(18.0) ⎠
w1v 2
From (17-3) Tc = , where w1 = (bb )(0.125)(0.042)12 = 0.063bb , so
g
(0.063bb )(1942 / 60) 2
Tc = = 2.05bb
32.2
From (17-4)
Tt − Tc Tt − 2.05bb Tt − 2.05bb
= e µθ s ⇒ = e0.8(2.95) = 10.6 or = 10.6 (1)
Ts − Tc Ts − 2.05bb (Tt − 51) − 2.05bb
698
Problem 17-1 (continued)
Ta bb 100bb
From (17-7) (Tt )d = , where K a = 1.25 (from Table 17.2), so (Tt )d = = 80bb .
Ka 1.25
80bb − 2.05bb
= 10.6 ⇒ 77.95bb = 10.6 ( 7795bb − 51)
(80bb − 51) − 2.05bb
bb = 0.7224"
To ≈ 32.3 lb
699
17-2. A flat belt drive consists of two cast-iron pulleys, each 4 feet in diameter, spaced 15 feet apart, center-
to-center. The belt is to be made of two-ply oak-tanned leather, each ply being 5/32 inch thick, and the
specific weight of the leather material is 0.040 lb/in 3 . The application involves a water-spray environment
in which the belt is constantly subjected to the water spray (see Appendix Table A.1 for coefficients of
friction). It has been experimentally determined that the tensile stress in the belt should not exceed 300 psi
for safe operation. If 50 horsepower is to be transmitted at a pulley speed of 320 rpm, what belt width
should be specified?
------------------------------------------------------------------------------------------------------------------------------
Solution
Specifications: d1 = d 2 = 48" , t ply = 0.15625" , hp = 50 horsepower , n = 320 rpm , wbelt = 0.040 lb/in 3 ,
C = 15 ft = 180 in , σ allow = 300 psi , µ = 0.20 (from Table A.1), θ = π (from Figure 17.8)
2π rs ns 2π (48 / 2)(320)
V= = = 4021 ft/min
12 12
Tt − Tc Tt − 20.92bb
= e µθ s ⇒ = e0.2(π ) = 1.874
Ts − Tc Ts − 20.92bb
Ta bb
(Tt )d = , where K a = 1.25 (from Table 17.2, assuming moderate shock)
Ka
93.75bb ( bb )
(Tt )d = = 75bb2
1.25
75bb2 − 20.92bb
= 1.874 ⇒ 34.98bb2 − 9.76bb − 410.3 = 0 ⇒ bb = 3.57"
( )
75bb2 − 410.3 − 20.92bb
Typically this would be rounded up to an even inch, so the specified belt width would be
bb = 4.0"
700
17-3. A 1725-rpm 5-horsepower high-torque electric motor is to be used to drive a woodworking table saw
in a storm-windoe manufacturing plant. The saw is to operate 16 hours per day, 5 days per week, at full-
rated motor horsepower. A V-belt drive is to be used nbetween the motor and the saw pulley. Ideally, the
center distance between the motor drive sheave and the driven saw sheave should be about 30 inches, and
the driven sheave should rotate at approximately 1100 rpm. Saw operation will probably produce moderate
shock loading. Propose a V-belt drive arrangement that will provide a mean life of about 1 year between
belt replacement.
------------------------------------------------------------------------------------------------------------------------------
Solution
N p = (1100 passes/min )( 60 min/hr )(16 hr/day )( 5 days/wk )( 52 wk/yr ) = 2.75 × 108 passes
The ratio is R = nmotor / nsaw = 1725 /1100 = 1.57 , giving ( d d ) saw = 1.57 ( d d )mot
V=
( ) (
2π rs ns 2π (d p ) mot / 2 nmot 2π (d p )mot / 2 1725
= =
)
= 451.6(d p ) mot
12 12 12
From figure 17.10, for 6.25 horsepower and nmotor = 1725 rpm , the tentative selection is an A-section belt.
For an A-section belt, the minimum recommended datum diameter (Table 17.4) is 3.0 inches. We begin by
selecting a slightly larger size, say ( d d )mot = 4.5" . From Table 17.4, we use 2hd = 0.25 , giving
( d d )saw = 1.57 ⎡⎣( d d )mot + 0.25⎤⎦ − 0.25 = 1.57 [ 4.5 + 0.25] − 0.25 = 7.21"
456.7 456.7
(Tt − Ts ) = = ≈ 96.2 lb
(d p ) mot 4.5 + 0.25
701
Problem 17-3 (continued)
( d d ) L θ L + ( d d )s θ s
( Ld )nom = 4C 2 − ⎡⎣( d d ) L − ( d d ) s ⎤⎦ +
2
2
7.21(3.23) + 4.5(3.05)
= 4(30)2 − [ 7.21 − 4.5]
2
+ = 59.94 + 18.51 = 78.45"
2
From Table 17.5, the closest standard datum length is Ld = 82.3" for an A-section belt. From Table 17.3 for
d d = 4.5" , we find w1 = 0.065 lb/ft , resulting in
Recalling from above that Ts = Tt − 96.2 , determining from Figure 17.9 that β = 36o , and using (17-9)
Tt − Tc Tt − 2.58
=e s ( )
µθ / sin β / 2
⇒ = e0.3(3.05) / sin18 = 19.32 ⇒ Tt = 104 lb
Ts − Tc [Tt − 96.2] − 2.58
Ts = 104 − 96.2 = 7.8 lb
From Table 17.3, C1 = 5.0 , C2 = 111 , C3 = 0.101× 10−6 , C4 = 0.175 , Ac = 1.73 × 10−3 , K i = 6.13 × 10−8 ,
K o = 19.8 , K m = 26.4 , k = −1 . Therefore, using (17-13), (17-14), and (17-15)
C1 + C2 5.0 + 111
(Tbe )motor = = = 25.78 lb
( d d )motor 4.5
C1 + C2 5.0 + 111
(Tbe )saw = = = 16.09 lb
( d d )saw 7.21
702
Problem 17-3 (continued)
( N f )motor = 6.13 ×10−8 [19,800 − 12,315]2 [ 26, 400 − 13,373]2 (82.3)1.75 (2145)− k = 6.11×108 cycles
( N f )saw = 6.13 ×10−8 [19,800 − 9514]2 [ 26, 400 − 10,572]2 (82.3)1.75 (2145)−k = 1.7 ×109 cycles
From (17-20)
(Np )f =
1
1
1
=
1
1
1
= 4.495 × 108 passes
+ +
( N f )motor ( N f )saw 6.11× 108 1.7 × 109
Comparing this to the number of required passes ( 2.75 × 108 passes ), the selected belt has about 60% more
tha the required life of 1 year, so it is regarded as acceptable and we use an A-series belt with
( d d )motor = 4.5" and ( d d )saw = 7.21"
703
17-4. A D-section V-belt is to be used to drive the main power shaft of an agricultural combine (an
agricultural combine may be regarded as a combination of conveyors, elevators, beaters, and blowers). The
power source is a 6-cylinder, 30-hp internal combustion engine which delivers full-rated power to a 12-
inch-diameter sheave at 1800 rpm. The driven sheave is 26 inches in diameter, and the center distance
between sheaves is 33.0 inches. During the harvest season, combines typically operate continuously 24
hours per day.
a. If a D-section V-belt were specified for use in this application, how often would you predict that
the belt would require replacement?
b. Based on the knowledge that it takes about five hours to change out the main drive belt, would you
consider the replacement interval estimate in (a) to be acceptable?
------------------------------------------------------------------------------------------------------------------------------
Solution
2π rs ns π (12)(1800)
V= = = 5655 ft/min
12 12
( d d )L θ L + ( d d )s θ s
( Ld )nom = 4C 2 − ⎡⎣( d d ) L − ( d d ) s ⎤⎦ +
2
2
25.6(3.57) + 11.6(2.71)
= 4(33) 2 − [ 25.6 − 11.6]
2
+ = 64.598 + 61.414 ≈ 126"
2
From Table 17.5, the closest standard datum length is Ld = 123.3" for an D-section belt. From Table 17.3
for d d = 11.6" , we find w1 = 0.406 lb/ft , resulting in
704
Problem 17-4 (continued)
Recalling that Ts = Tt − 262.5 , determining from Figure 17.9 that β = 36o , and using table A.1 to find
µ = 0.3
Tt − Tc Tt − 112
=e s ( )
µθ / sin β / 2
⇒ = e0.3(2.71) / sin18 = 13.886 ⇒ Tt = 394.8 lb
Ts − Tc [Tt − 262.5] − 112
Ts = 394.8 − 262.5 = 132.3 lb
From Table 17.3, C1 = 26.5 , C2 = 256 , C3 = 0.291× 10−6 , C4 = 0.10 , Ac = 5.15 × 10−3 , K i = 6.76 × 10−8 ,
K o = 10.8 , K m = 14.6 , k = 0 . Therefore, using (17-13), (17-14), and (17-15)
C1 + C2 26.5 + 256
(Tbe )s = = = 24.35 lb
( d d )s 11.6
C1 + C2 26.5 + 256
(Tbe )L = = = 11.04 lb
( dd )L 25.6
Tce = C3V 2 = 0.291× 10−6 ( 5655 ) = 9.31 lb
2
( N f )s = 6.76 ×10−8 [10,800 − 4913]2 [14, 600 − 9289]2 (123.3)1.75 (5655)0 ≈ 3 ×109 cycles
( N f )L = 6.76 ×10−8 [10,800 − 3620]2 [24, 600 − 7997 ]2 (123.3)1.75 (5655)0 ≈ 6.94 ×109 cycles
From (17-20)
(Np )f =
1
1
1
=
1
1
1
≈ 2.1× 109 passes
+ +
( )s ( N f )L
Nf 3 × 109 6.94 × 109
705
Problem 17-4 (continued)
( N p )day = ⎛ LV ft/min
⎞
( 60 min/hr )( 24 hr/day ) =
5655 ft/min
⎛ 126 ⎞
( 60 min/hr )( 24 hr/day )
p
⎜ ⎟ ft/pass ⎜ ⎟ ft/pass
⎝ 12 ⎠ ⎝ 12 ⎠
= 7.75 × 105 passes/day
( N p ) f 2.1×109
BL = = ≈ 2710 days
( N p )day 7.75 ×105
(b) The replacement cycle is more than adequate. The belt is overdesigned and a smaller-section belt
should be investigated.
706
17-5. A portable bucket elevator for conveying sand is to be driven by a single-cylinder internal-
combustion engine operating at a speed of 1400 rpm, using a B-section V-belt. The driving pulley and
driven pulley each have a 5.00-inch diameter. If the bucket elevator is to lift two tons per minute (4000
lb/min) of sand to a height of 15 feet, continuously for 10 hours per working day, and if friction losses in
the elevator are 15 percent of operating power, how many working days until failure would you estimate
for the B-section belt if it has a datum length of 59.8 inches (B 59 belt)?
------------------------------------------------------------------------------------------------------------------------------
Solution
( hp )d = K a ⎡( hp )op + ( hp ) fr ⎤
⎣ ⎦
4000(15)
( hp )op = = 1.82 hp
33, 000
( hp ) fr = 0.15 ( hp )op = 0.15(1.82) = 0.27 hp
( hp )d = 1.75 [1.82 + 0.27 ] = 3.658 horsepower
2π rs ns 2π (5 / 2)(1400)
V= = ≈ 1833 ft/min
12 12
Recalling that Ts = Tt − 65.86 , determining from Figure 17.9 that β = 36o , and using table A.1 to find
µ = 0.3
Tt − Tc Tt − 3.25
=e s ( )
µθ / sin β / 2
⇒ = e0.3π / sin18 = 21.11 ⇒ Tt ≈ 72.4 lb
Ts − Tc [Tt − 65.86] − 3.25
Ts = 72.4 − 65.86 = 6.24 lb
707
Problem 17-5 (continued)
From Table 17.3, C1 = 5.2 , C2 = 123 , C3 = 0.133 × 10−6 , C4 = 0.15 , Ac = 1.73 × 10−3 , K i = 1.78 × 10−7 ,
K o = 17.3 , K m = 26.0 , k = −1 . Therefore, using (17-13), (17-14), and (17-15)
C1 + C2 5.2 + 123
(Tbe )L = (Tbe )s = = = 25.64 lb
( dd )s 5
( N f )L = ( N f )s = 1.78 ×10−7 [17,300 − 10, 405]2 [26, 000 − 11, 224]2 ( 59.8)1.75 (1833)−1 ≈ 1.3 ×109 cycles
From (17-20)
(Np )f =
1
1
1
=
1
1
1
≈ 6.5 × 108 passes
+ +
( )s ( N f )L
Nf 1.3 × 109 1.3 × 109
( N p )day = ⎛ LV ft/min
⎞
( 60 min/hr )(10 hr/day ) =
1833 ft/min
⎛ 59.8 ⎞
( 60 min/hr )(10 hr/day )
p
⎜ ⎟ ft/pass ⎜ ⎟ ft/pass
⎝ 12 ⎠ ⎝ 12 ⎠
= 2.21× 105 passes/day
( N p ) f 6.5 ×108
BL = = ≈ 2941 days
( N p )day 2.21×105
708
17-6. In a set of 5V high-capacity V-belts, each belt has a pitch length (outside length) of 132.0 inches, and
operates on a pair of 12-inch diameter multiply grooved sheaves. The rotational speed of the sheave is 960
rpm. To achieve a mean life expectancy of 20,000 hours, find the number of belts that should be used in
parallel to transmit 200 horsepower.
------------------------------------------------------------------------------------------------------------------------------
Solution
Specifications: d p = 12" (per sheave) , n = 960 rpm , L p = 132" , ( H m )req ' d = 20, 000 hours ,
( hp )op = 200 horsepower
2π rs ns 2π (12 / 2)(960)
V= = ≈ 3016 ft/min
12 12
Since the sheaves are the same, θ s = π , and from Figure 17.9 that β = 36o , and using table A.1 to find
µ = 0.3 , we get
Tt − Tc Tt − 11.1
=e s ( )
µθ / sin β / 2
⇒ = e0.3π / sin18 = 21.11
Ts − Tc ⎡Tt − 10.94 ( hp )belt ⎤ − 11.1
⎣ ⎦
( )
Ts = Tt − 10.94(hp)belt : Ts = 11.48 ( hp )belt + 11.1 − 10.94(hp)belt = 0.54 ( hp )belt + 11.1
From Table 17.3, C1 = 6 , C2 = 200 , C3 = 0.202 × 10−6 , C4 = 0.17 , Ac = 2.88 × 10−3 , Ki = 9.99 × 10−8 ,
K o = 16 , K m = 29.2 , k = −1 . Therefore, using (17-13), (17-14), and (17-15)
C1 + C2 6 + 200
Tbe = = ≈ 17.2 lb
d 12
Tce = C3V 2 = 0.202 × 10−6 ( 3016 ) = 1.84 lb
2
709
17-6 (continued)
σm =
Tte + Tbe + 2Tce + Tse
=
( ) (
1.95 ( hp )belt + 1.89 + 17.2 + 2(1.84) + 0.092 ( hp )belt + 1.89 )
2 Ac (
2 2.88 × 10−3 )
= 354.5 ( hp )belt + 4281
From (17-20)
Nf
( N p )req 'd = 1
1
1
=
2
( )req ' d = 2(3.29 ×108 ) = 6.58 ×108
⇒ N f = 2 Np
+
Nf Nf
( ) ( )
2 2
6.58 × 108 = 9.99 × 10−8 ⎡16, 000 − 354.2 ( hp )belt + 2986 ⎤ ⎡ 29, 200 − 354.5 ( hp )belt + 4281 ⎤ (132 ) (3016) −1
1.75
⎣ ⎦ ⎣ ⎦
3.86 × 1015 = ⎡⎣13,104 − 354.5 ( hp )belt ⎤⎦ ⎡⎣ 24,919 − 354.5 ( hp )belt ⎤⎦
2 2
( hp )belt ≈ 26 horsepower
( hp )op 200
Nbelts = = = 7.69
( hp )belt 26
710
17-7. It is desired to use a compact roller chain drive to transmit power from a dynamometer to a test stand
for evaluation of aircraft auxillary gear boxes. The chain drive must transmit 90 horsepower at a small-
sprocket speed of 1000 rpm.
------------------------------------------------------------------------------------------------------------------------------
Solution
K a ( hp )nom
(a) From (17-24) , ( hp )d = . Assume moderate shock, so from Table 17.2 K a = 1.25 .
K st
Assuming 1 strand, K st = 1
1.25(90)
( hp )d = = 112.5 horsepower
1
For this horsepower and a drive sprocket speed (small sprocket) of 1000 rpm, Figure 17.14 suggests a no.
120 chain, which , from Table 17.6, has a pitch of p = 1.5" and a minimum center distance of Cmin = 33" .
From step 3 of section 17.10, the optimum range of center distances is about 30 – 50 chain pitches, or 45 –
75 inches for p = 1.5" . A nominal center distance of 30 pitches will be adopted to satisfy the “compact”
specification, so
ns 1000
NL = Ns = (12) = 12 teeth
nL 1000
From (17-21)
Compared to the design horsepower of ( hp )d = 112.5 horsepower , this is a bit low. Increasing N s may
increase the horsepower so we will set ( hplim )lp = ( hp )d = 112.5 horsepower and solve for N s
This close enough, so we set N s = 14 teeth . Now, from (17-22), for a no. 120 chain
711
17-7 (continued)
Compared to the design horsepower (112.5 horsepower), the roller bearing limiting fatigue power is too
low, so multiple strand chain may be required. Assuming 4 strands, we revise the design horsepower. From
Table 17.7, K st = 3.3 , so
1.25(90)
( hp )d = = 34.1 horsepower
3.3
With 4 strands of no. 120 chain, both ( hplim )lp = 112.1 horsepower (calculated using N s = 14 teeth ) and
( hplim )rb = 38.95 horsepower are acceptable. From (17-23)
⎛ n pN ⎞ ⎛ n ⎞
( hplim ) g = ⎜ s s ⎟ ( 4.413 − 2.073 p − 0.0274 N L ) − ⎜ ln l ⎟ (1.59 log p + 1.873)
⎝ 110.84 ⎠ ⎝ 1000 ⎠
⎛ 1000(1.5)(14) ⎞ ⎛ 1000 ⎞
=⎜ ⎟ ( 4.413 − 2.073(1.5) − 0.0274(14) ) − ⎜ ln ⎟ (1.59 log(1.5) + 1.873)
⎝ 110.84 ⎠ ⎝ 1000 ⎠
Ns p 14(1.5)
dp = = = 6.68"
π π
π d p ns π (6.68)(1000)
Vchain = = = 1748.8 ft/min
12 12
712
17-8. It is desired to use a roller chain drive for the spindle of a new rotating shaft fatigue testing machine.
The drive motor operates at 1750 rpm and the fatigue machine spindle must operate at 2170 rpm. It is
estimated that the chain must transmit 11.5 hp. Spindle-speed variation of no more than 1 percent can be
tolerated.
------------------------------------------------------------------------------------------------------------------------------
Solution
(a) From Table 17.2, for an electric motor drive (assuming uniform load) and 1 strand, K a = K st = 1
K a ( hp )nom
( hp )d = = 11.5 horsepower
K st
ns 2170
NL = Ns = (21) = 26 teeth
nL 1750
Using (17-21)
( hplim )lp
p(
3.0 − 0.07 p ) 11.5
= = = 0.107 ⇒ p ≈ 3 0.107 = 0.475
K lp N 1.08
s ns
0.9
0.004(21)1.08 (2170)0.9
Using Table 17.6, we select p = 0.5" as the closest standard chain. Tentatively use a no. 40 chain. Next
Comparing this with ( hp )d = 11.5 horsepower , it is not quite close enough, so we try increasing N s to
N s = 25 teeth , which results in
713
Problem 17-8 (continued)
1000 K rb N 1.5
s p
0.8
1000(17)(25)1.5 (0.5)0.8
( hplim )rb = = = 12.1 horsepower
n1.5
s (2170)1.5
ns 2170
NL = Ns = (25) = 31 teeth
nL 1750
Next
⎛ n pN ⎞ ⎛ n ⎞
( hplim ) g= ⎜ s s ⎟ ( 4.413 − 2.073 p − 0.0274 N L ) − ⎜ ln l ⎟ (1.59 log p + 1.873)
⎝ 110.84 ⎠ ⎝ 1000 ⎠
⎛ 2170(0.5)(25) ⎞ ⎛ 1750 ⎞
=⎜ ⎟ ( 4.413 − 2.073(0.5) − 0.0274(31) ) − ⎜ ln ⎟ (1.59 log(0.5) + 1.873)
⎝ 110.84 ⎠ ⎝ 1000 ⎠
Use 1-strand no. 40 precision roller chain with N s = 25 teeth and N L = 31 teeth
Ns p 25(0.5)
dp = = = 3.98"
π π
π d p ns π (3.98)(1000)
Vchain = = = 1042 ft/min
12 12
1000 K rb N 1.5
s p
0.8
1000(3.4)(25)1.5 (0.5)0.8
( hplim )rb = = = 2.4 horsepower
n1.5
s (2170)1.5
714
17-9. A five-strand no. 40 precision roller chain is being proposed to transmit power from a 21-tooth
driving sprocket that rotates at 1200 rpm. The driven sprocket is to rotate at one-quarter the speed of the
driving sprocket. Do the following:
a. Determine the limiting horsepower that can be transmitted by this drive, and state the governing
failure mode.
b. Find the tension in the chain.
c. What chain length should be used if a center distance of approximately 20 inches is desired?
d. Does the 20-inch center distance lie within the recommended range for this application?
e. What type of lubrication should be used for this application?
------------------------------------------------------------------------------------------------------------------------------
Solution
ns 1200
(a) NL = Ns = (21) = 84 teeth
nL 300
⎡( hplim ) ⎤ ⎡⎛ n pN ⎞ ⎛ n ⎞ ⎤
= K st ⎢⎜ s s ⎟ ( 4.413 − 2.073 p − 0.0274 N L ) − ⎜ ln l ⎟ (1.59 log p + 1.873) ⎥
⎣ g ⎦ 5 − strand
⎣⎝ 110.84 ⎠ ⎝ 1000 ⎠ ⎦
⎧⎛ 1200(0.5)(21) ⎞ ⎛ 300 ⎞ ⎫
= 3.9 ⎨⎜ ⎟ ( 4.413 − 2.073(0.5) − 0.0274(84) ) − ⎜ ln ⎟ (1.59 log(0.5) + 1.873) ⎬
⎩⎝ 110.84 ⎠ ⎝ 1000 ⎠ ⎭
The limiting horsepower is therefore ⎡( hplim )lp ⎤ = 31.6 horsepower and he failure mode is link
⎣ ⎦ 5− strand
plate fatigue.
(b) The slack side tension in a chain is equal to zero, so Ts = 0 . For the tight side
715
Problem 17-9 (continued)
2π ( d 2 / 2 ) ns
Where V = and d s = N s p / π = 21(0.5) / π = 3.34" , so
12
2π ( 3.34 / 2 ) (1200)
V= = 1049 ft/min
12
and
33, 000(31.6)
Tt = = 994 lb (5 strands)
1049
994
(Tt )strand = =198.8 ≈ 199 lb/strand
5
⎛ N + Ns ⎞ ( N L − N s ) 84 + 21 ⎛ 20 ⎞ (84 − 21)
2 2
(c) L=⎜ L ⎟ + 2C + = + 2⎜ ⎟+ = 135 pitches
⎝ 2 ⎠ 4π 2C 2 ⎝ 0.5 ⎠ 4π 2 ( 20 / 0.5 )
(d) The optimum center distance is 30 ≤ C ≤ 50 pitches . For this application C = 20 / 0.5 = 40 pitches ,
which is in the recommended range.
(e) V =1049 ft/min , so based on the recommendations of 17.7, we use a Type II lubrication
716
17-10. It is desired to market a small air-driven hoist in which the load is supported on a single line of wire
rope and the rated design load of 3/8 ton (750 lb). The wire rope is to be wrapped on a drum of 7.0 inches
diameter. The hoist should be able to lift and lower the full-rated load 16 times a day, 365 days a year, for
20 years before failure of the rope occurs.
a. If a special flexible 6 × 37 improved plow steel (IPS) rope is to be used, what rope size should be
specified?
b. With the rope size determined in (a), it is desired to estimate the “additional stretch” that would
occur in the rope if a 750-lb load were being lowered at the rate of 2 ft/sec, and when the load reaches
a point 10 feet below the 7.0-inch drum, a brake is suddenly applied. Make such an estimate.
------------------------------------------------------------------------------------------------------------------------------
Solution
(a) Based on judgment, assume nstatic = 5 and n fatigue = 1.25 . For a 6 × 37 wire rope, From Table 17.9:
Ar = 0.427d r2 , Su = 200 ksi
T 750 1765
σt = = 2
= 2
Ar 0.427d r dr
1765 1765
40, 000 = ⇔ dr = = 0.21"
d r2 40, 000
⎛ lifts ⎞⎛ days ⎞
N d = ⎜16 ⎟ ( 20 years ) = 1.168 × 10 cycles
5
⎟⎜ 365
⎝ day ⎠⎝ yaer ⎠
( p )N f
= ( RN ) f ( Su ) = 0.0037(200,000) = 740 psi
( p )N 740
( pd ) fatigue = f
= = 592 psi
n fatigue 1.25
From (17-33)
2T 2(750)
( d r ) fatigue = = = 0.362"
d s ( pd ) fatigue 7(592)
717
Problem 17-10 (continued)
From Table 17.10, for a 6 × 37 rope on a cast carbon-steel sheave (BHN 10), the allowable bearing pressure
based on wear is ( pd ) wear = 1180 psi . Inserting this pressure into (17-33)
2T 2(750)
( d r )wear = = = 0.182"
d s ( pd ) wear 7(1180)
Comparing all of these diameters, ( d r ) fatigue = 0.362" is the governing diameter. Selecting the next larger
rope from Table 17.9
d r = 3 / 8 = 0.375"
Summarizing: (1) select a 3/8-inch diameter 6 × 37 IPS fiber core (FC) rope
(2) Choose a cast carbon-steel sheave material (BHN =-160) with a 7” diameter
( PE )spring = = ≈ 2477δ 2
2 Lo 2(10 × 12)
559.2 = 2477δ 2 ⇒ δ = 559.2 / 2477 = 0.475"
718
17-11. An electric hoist, in which the load is supported on two lines, is fitted with a ¼-inch 6 × 19
improved plow steel (IPS) wire rope that wraps on an 8-inch-diameter drum and carries an 8-inch sheave
with an attached hook for load lifting. The hoist is rated at 1500-lb capacity.
a. If full-rated load were lifted each time, about how many “lifts” would you predict could be made
with the hoist? Use a fatigue safety factor of 1.25. Note that there are 2 “bends” of the rope for each lift
of the load.
b. If the hoist were used in such a way that one-half the time it is lifting full-rated load but the rest of
the time it lifts only one-third of rated load, what hypothesis or theory would you utilize for estimating
the number of lifts that could safely be made under these circumstances?
c. Numerically estimate the number of lifts that could be safely made under the mixed loading
described in (b). Again, use a fatigue safety factor fo 1.25 in your estimating procedure.
------------------------------------------------------------------------------------------------------------------------------
Solution
2T 2(750)
p= = = 750 psi
d r d s 0.25(8)
p 750
( RN ) f = = = 0.00375 and ( RN )d = n f ( RN ) f = 1.25(0.00375) = 0.0046875 ≈ 0.0047
Su 200, 000
92, 000
N bend = = 46, 000 lifts
2
ni N lifts
(c) ∑N = 1 . With 750 lb lifts half the time and 250 lb lifts the rest of the time, n250 = n750 =
2
.
i
Therefore
2
N lifts = , where N 750 = 46, 000
1 1
+
N 250 N 750
2(250)
( RN )F = 250 = = 0.00125 ≈ 0.0013 and ( RN )d − 250 = 1.25(0.0013) = 0.0016 . From Figure
200, 000(0.25)(8)
17.17, N bend − 250 = 600, 000 lifts , and with 2 bends per lift, N 250 = 300, 000
2
N lifts = = 79, 769 N lifts ≈ 79,800
1 1
+
300, 000 46, 000
719
17-12. It is desired to select a wire rope for use in an automotive tow truck application. A single line is to
be used and , considering dynamic loading involved in pulling cars back onto the roadway, the typical load
on the rope is estimated to be 7000 lb. It is estimated that approximately 20 cars per day will be pulled back
onto the highway (i.e., the rope experiences 20 “bends” per day) under full load. If the truck is used 360
days per year, and a design life of 7 years is desired for the rope:
a. What size IPS wire rope would you specify if the rope is to be of 6 × 19 regular lay construction?
b. What minimum sheave diameter would you recommend?
------------------------------------------------------------------------------------------------------------------------------
Solution
Specifications: W = 7000 lb , 6 × 19 IPS rope , Ar = 0.404d r2 , Su = 200 ksi , 20 “bends”/day, 365 days/yr,
for 7 yrs
T 7000 17,327
(a) Assume nstatic = 5 and n fatigue = 1.5 , giving σ t = = =
Ar 0.404d r2 d r2
17,327 17,327
40, 000 = ⇔ d r = ( d r )static = = 0.658 ≈ 0.66"
d r2 40, 000
⎛ bends ⎞⎛ days ⎞
N d = ⎜ 20 ⎟⎜ 365 ⎟ ( 7 years ) = 50,960 bends
⎝ day ⎠⎝ yaer ⎠
From (17-33)
2T 2(7000) 17.5
( d r ) fatigue = = =
d s ( pd ) fatigue 800d s ds
17.5
= 0.5147 , or ( d r ) fatigue = 0.717 ≈ 0.72"
From Table 17.9, for a 6 × 19 rope, d3 = 34d r , so ( d r ) fatigue =
2
34
From Table 17.10, the allowable pressure for a 6 × 19 rope is pmax = 1000 psi . From (17-33)
2T 2T 2(7000)
( d r )wear = = ⇒ ( d r )wear = = 0.412" or ( d r )wear = 0.642"
2
Failure is controlled by ( d r ) fatigue = 0.717 ≈ 0.72" . The next largest size (from Table 17.9) is
d r = 3 / 4 = 0.75"
720
17-13. A deep-mine hoist utilizes a single line of 2-inch 6 × 19 extra improved plow steel (EIPS) wire rope
wrapped on a cast carbon-steel drum that has a diameter of 6 feet. The rope is used to vertically lift loads of
ore weighing about 4 tons from a shaft that is 500 feet deep. The maximum hoisting speed is 1200 ft/min
and the maximum acceleration is 2 ft/s 2 .
a. Estimate the maximum direct stress in the “straight” portion of the 2-inch single-line wire rope.
b. Estimate the maximum bending stress in the “outer” wires of the 2-inch wire rope as it wraps onto
the 6-foot diameter drum.
c. Estimate the maximum unit radial pressure (compressive stress) between the rope and the sheave.
Hint: Model the 2-inch single-line rope wrapped around the 6-foot sheave as a “band brake,” utilizing
equation (16-76) with α = 2π and µ = 0.3 , to find pmax .
d. Estimate the fatigue life of the 2-inch wire rope as used in this application.
------------------------------------------------------------------------------------------------------------------------------
Solution
Wa = ma =
(Wp + Wr ) a =
8000 + 3200
(2) = 695.7 ≈ 696 lb
max
g 32.2
σb =
dw
ds
Er =
(2 /13)
72
( )
10.8 × 106 = 23, 077 psi σ b = 23, 077 psi
(d) RN = p / Su = 1088 / 220, 000 = 0.0049 . From Figure 17.17, for a 6 × 19 rope, we estimat
721
17-14. It is necessary to mount a 7.5 horsepower 3450-rpm electric motor at right angles to a centrifugal
processor as shown in Figure P17.14 (also see arrangement A − B2 of Figure 16.6). It had been planned to
use a 1 : 1 bevel gearset to connect the motor to the processor, but a young engineer suggested that a
flexible shaft connection might be quieter and less expensive. Determine whether a flexible shaft is a viable
alternative, and, if so, specify a flexible shaft that should work.
------------------------------------------------------------------------------------------------------------------------------
Solution
Specifications: hpop = 7.5 horsepower , nop = 3450 rpm , Rb = 12" (from Figure P17.14) , arrangement
A − B2 of Figure 17.6
Since the operation is unidirectional, use Table 17.11 and equation (17-35)
722
17-15. To avoid other equipment on the same frame, the centerline of a 2-horsepower 1725 rpm electric
motor must be offset from the parallel centerline of the industrial mixer that it must drive.
a. For the offset shown in Figure P17.15, select a suitable flexible shaft (also see arrangement
A − B1 of Figure 16.6).
b. To improve mixing efficiency, it is being proposed to replace the standard 2-horsepower motor
with a “reversible” motor having the same specifications. With the “reversible” motor, would it be
necessary to replace the flexible shaft chosen in (a)? If so, specify the size of the replacement shaft.
c. Comparing results of (a) and (b), can you think of any potential operational problems associated
with the flexible shaft when operating in the “reverse” direction?
------------------------------------------------------------------------------------------------------------------------------
Solution
Specifications: hpop = 2 horsepower , nop = 1725 rpm , arrangement A − B1 of Figure 17.6, x = 4" and
y = 12" from Figure P17.15
x 2 + y 2 (4) 2 + (12) 2
(a) From (17-36) Rb = = = 10" .
4x 4(4)
The 73 in-lb torque is close to the limiting torque of 70 in-lb associated with the d min = 0.495" core
diameter. Since (per footnote 1) there is a built in safety factor of 4, we would probably accept the 0.495”
core diameter.
(b) For reversing loads (bidirectional operation), we use Table17.12 and note that for a core diameter of
0.495” and a 10” bend radius, the maximum allowable torque is Tallow = 117 in-lb , so the core diameter is
acceptable.
(c) One reason might be Torsional stiffness; for LOL operation it is 0.081 deg/ft/in-lb vs
0.06 deg/ft/in-lb for (a). This might produce a torsional vibration problem.
723
Chapter 18
18-1. A deep-draw press is estimated to have the load torque versus angular-displacement characteristics
shown in Figure P18.2. The machine is to be driven by a constant-torque electric motor at 3600 rpm. The
total change in angular velocity from its maximum value to its minimum value must be controlled to within
±3 percent of the average angular velocity of the drive.
a. Compute and sketch the motor input torque versus angular displacement curve.
b. Sketch a curve of angular velocity (qualitative) versus angular displacement (qualitative).
c. On the torque versus angular displacement curve, carefully locate angular displacement values
corresponding to maximum and minimum angular velocity.
d. Calculate U max .
e. Calculate the required mass moment of inertia for a flywheel that would properly control the speed
fluctuation to within ±3 percent of the average angular velocity, as specified.
---------------------------------------------------------------------------------------------------------------------------
Solution
(a) nmot = 3600 rpm , ∆ωmax = ±3% of ωavg , ωave = 2π nmot / 60 = 377 rad/sec
= 1400π ft-lb
[Td dθ ]1cycle = 2π Td
2π Td = 1400π
Td = 700 ft-lb
724
Problem 8-1 (continued)
θωmin
U max = ∫
θωmax
(Tl − Td ) dθ
= area of shaded region
ωmax = 1.03ωavg = 1.03(377) = 388 rad/sec and ωmin = 0.97ωavg = 0.97(377) = 366 rad/sec
388 − 366
From (18-11), C f = = 0.058
377
U max 18,850
J req ' d = = = 2.287 ≈ 2.3 in-lb-sec2
C f ωave (0.058) ( 377 )
2 2
725
18-2. A hammermill has the load torque versus angular displacement curve shown in Figure P18.2, and is
to be driven by a constant-torque electric motor at 3450 rpm. A flywheel is to be used to provide proper
control of the speed fluctuation.
a. Compute and plot the motor input torque versus angular displacement.
b. Sketch angular velocity (qualitative) of the shaft-flywheel system as a function of angular
displacement (qualitative). Specifically note locations of maximum and minimum angular velocity on
the torque versus angular displacement curve.
c. Calculate U max .
d. Calculate the mass moment of inertia required for the flywheel to properly control the speed
fluctuation.
---------------------------------------------------------------------------------------------------------------------------
Solution
2π Td = 12.5π
Td = 6.25 kN-m
726
Problem 18-2 (continued)
θωmin
U max = ∫
θωmax
(Tl − Td ) dθ
= area of shaded region
1⎡ π π
U max = ⎢ (18 − 8) ⎤⎥ + (8 − 6.25 ) = 10.6 kN-m
2⎣ 2⎦ 2
U max 10 600
J req ' d = = = 0.406 N-m-sec 2
C f ωave (0.2) ( 361.3)
2 2
727
18-3 A natural gas engine is to be used to drive an irrigation pump that must be operated within ±2 percent
of its nominal operating speed of 1000 rpm. The engine torque angular displacement curve is the sawtooth
Tengine curve shown in Figure P8.3. The pump torque versus angular displacement curve is the stepped
T pump curve shown. It is desired to use a solid-steel flywheel of 10-inch radius to obtain desired speed
control.
a. Sketch angular velocity (qualitative) of the flywheel system as a function of angular displacement
(qualitative), and identify points od maximum and minimum angular velocity on the torque versus
angular displacement curve.
b. Calculate U max .
c. Calculate the mass moment of inertia of the flywheel that would be required to properly control
the speed.
d. Of what thickness should the flywheel be made?
---------------------------------------------------------------------------------------------------------------------------
Solution
(a) nnom = 1000 rpm , ∆ωmax = ±2% of ωavg , ωave = 2π nnom / 60 = 104.7 rad/sec , ro = 10 "
Figure P18.3 is reproduced below along with a sketchs of ω (quantitative) versus θ (quantitative). We
note that θωmax = π / 2 rad and θωmin = 3π / 2 rad .
θωmin
(b) Solving (16-6) graphically, U max = ∫
θωmax
(Tl − Td ) dθ = area of shaded region
⎡1 π⎤
U max = 2 ⎢ (1200 − 600) ⎥ ≈ 300π ft-lb
⎣2 2⎦
728
Problem 8-3 (continued)
ωmax = 1.02ωavg = 1.02(104.7) = 106.8 rad/sec and ωmin = 0.98ωavg = 0.98(104.7) = 102.6 rad/sec
106.8 − 102.6
From (18-11), C f = = 0.04
104.7
Mro2 wπ tro4
J= =
2 2g
729
18-4. A spoke-and-rim flywheel of the type shown in Figure 18.4(a) is made of steel, and each of the six
spokes may be regarded as very stiff members. The mean diameter of the flywheel rim is 970 mm. The rim
cross section is rectangular, 100 mm in the radial dimension and 50 mm in the axial dimension. The
flywheel rotates counterclockwise at a speed of 2800 rpm.
a. Calculate your best estimate of the maximum bending stress generated in the rim.
b. At what critical sections in the rim would this m
---------------------------------------------------------------------------------------------------------------------------
Solution
( d m )rim = 970 mm n = 2800 rpm , rectangular cross section 100 mm radial × 50 mm axial
wAr rm n 2 L2 cr
(σ b )max =
35, 200(12) I r
The 35,200 in the denominator comes from (18-18), in which g = 32.2 ft/s 2 = 386.4 in/s 2 . In our case,
we use g = 9.81 m/s 2 , which results in
wAr rm n 2 L2 cr
(σ b )max =
894.6(12) I r
For steel w = 76.81 kN/m3 . The remaining terms in the equation above are rm = 0.970 / 2 = 0.485 m ,
Ar = (0.10)(0.05) = 5 × 10−3 m 2 , n = 2800 rpm , L = 2π rm / ns = 2π (0.485) / 6 = 0.509 m ,
cr = 0.1/ 2 = 0.05 m , and I r = 0.05(0.10)3 /12 = 4.167 × 10−6 m 4 . Therefore
(σ b )max =
( 76.81×10 )(5 ×10 ) ( 0.485)( 2800) ( 0.509)
3 −3 2 2
( 0.05 )
= 423 MPa
894.6(12) ( 4.167 × 10 ) −6
(b) Based on the approximate rim stresses shown in Figure 18.6, this maximum stress occurs at the outer
fibers of the rim at the “fixed supports”, i.e. adjacent tot the spoke centerline.
730
18-5. A spoke-and-rim flywheel of the type shown in Figure 18.4(a) has a mean rim diameter of 5 ft and
rotates at 300 rpm. The rim cross section is 8 inches by 8 inches. During the duty cycle, the flywheel
delivers energy at a constant torque of 9000 ft-lb over ¼ revolution, during which time the flywheel speed
drops 10 percent. There are six spokes of elliptical cross section, with major axis twice the minor axis. The
major axes of the elliptical spokes are parallel to the circumferential direction. The cast-iron material
weighs 480 lb/ft 3 and has a design-allowable stress of 3000 psi.
---------------------------------------------------------------------------------------------------------------------------
Solution
From specifications;
Flywheel type: spoke-and-rim, material: cast iron,
wCI = 480 lb/ft 3 = 0.28 lb/in 3 , σ allow = 3000 psi , d rim = 5 ft = 60 in , n = 300 rpm ,
rim cross section: square
rim dimensions: radial = 8”, axial = 8”
duty cycle: flywheel delivers energy at a constant torque of 9000 ft-lb over ¼ revolution
speed change: speed drops 10% over the ¼ revolution when energy is supplied by the
flywheel
spoke cross-section: elliptical with major axis parallel to circumference. Major axis is
twice (2a) the minor axis (2b)
ns = 6
π
0.28(60 / 2) 2 (300) 2 (8) 2 sin
As = 6 = 13.74 in 2
17, 600(3000)
731
Problem 18-5 (continued)
wAr rm n 2 L2 Cr
(σ b )max =
35, 200(12)( I r )
2π rm 2π (30) bd 3 8(8)3
L= = = 31.4 in , Cr = 8 / 2 = 4" , I r = = = 341.3 in 4 .
ns 6 12 12
Therefore
(d) Following the rule of thumb of 18.6, paragraph 4. d hub = 2.25d shaft = 2.25d s and
T (d s / 2) T (d s / 2) 16T 16T
τ max = = = ⇒ ds = 3
J π d s4 / 32 π d s3 πτ max
For the material used S yp = 30 ksi . Based on a judgment call, we set nd = 2 . Using distortional energy we
use τ yp = 0.577 S yp = 17.31 ksi and set τ max = τ yp / nd = 17.31/ 2 ≈ 8.66 ksi . Therefore
16(9000 × 12)
ds = 3 = 3.989 ≈ 4"
π (8666)
This results in
732
18-6. A disk flywheel has a 600-mm outside diameter, 75-mm axial thickness, and is mounted on a 60-mm
diameter shaft. The flywheel is made of ultra-high-strength 4340 steel (see Tables 3.3, 3.4, and 3.5). The
flywheel rotates at a speed of 10,000 rpm in a high-temperature chamber operating at a constant
temperature of 425o C . Calculate the existing safety factor for this flywheel, based on yielding as the
governing failure mode.
---------------------------------------------------------------------------------------------------------------------------
Solution
From Table 3.5 we use date for ΘC = 427o C , which specifies Su = 1524 MPa and S yp = 1283 MPa .
Since yielding is the failure mode, ( nex ) yp = S yp / σ max .
From (18.54), (18.55), and (18.56), we know σ t > σ r , (σ t )max occurs @ r = a , and σ r = 0 @ r = a .
Therefore (σ t )max is uniaxial at the critical point r = a . From Table 3.4 w = 76.81 kN/m3 and form Table
3.9 ν = 0.3 . In addition, ω = 2π n / 60 = 2π (10, 000) / 60 = 1047.2 rad/sec , so from (18-57) we get
wω 2 ⎡
σ max = (σ t )max = ( 3 + ν ) b2 + (1 −ν ) a 2 ⎤⎦
4g ⎣
(76,810)(1047.2) 2 ⎡
⎣(
= 3 + 0.3) (0.3) 2 + (1 − 0.3) (0.03) 2 ⎤ ≈ 630 MPa
4(9.81) ⎦
1283
( nex ) yp = = 2.04
630
733
18-7. A disk-type flywheel, to be used in a punch press application with C f = 0.04 , is to be cut from a
1.50-inch-thick steel plate. The flywheel disk must have a central hole of 1.0-inch radius, and its mass
moment of inertia must be 50 in-lb-sec 2 .
Hint: J =
( )
wπ t ro4 − ro4 ω 2
2g
---------------------------------------------------------------------------------------------------------------------------
Solution
J=
(
wπ t ro4 − ri4 ) ⇒ ro4 − ri4 =
2 gJ
⇒ ro = 4
2 gJ
+ ri4
2g wπ t wπ t
2(386)(50)
ro = 4 + (1.0) 4 = 13.04"
0.283π (1.5)
With w = 0.283 lb/in 3 , ν = 0.3 , and ω = 2π n / 60 = 2π (3600) / 60 = 377 rad/sec , (18-57) gives
wω 2 ⎡
(σ t )max = ( 3 +ν ) b2 + (1 −ν ) a 2 ⎤⎦
4g ⎣
(0.283)(377) 2 ⎡
⎣(
= 3 + 0.3) (13.04) 2 + (1 − 0.3) (1.0)2 ⎤ = 14.635 ksi
4(386) ⎦
(σ t )max ≈ 14.64 ksi
40
n yp = = 2.73
14.64
734
18-8. A steel disk-type flywheel, to be used in a V-belt test stand operating at 3000 rpm, must have a
coefficient of speed fluctuation of C f = 0.06 . The flywheel has been analyzed in a preliminary way and it
is proposed to use a constant-thickness disk, 75 mm thick, with a central hole of 50 mm radius and an outer
radius of 250 mm. Further, it is desired to drill one small hole through the disk at a radius of 200 mm, as
shown in Figure P18.8.
a. At the location of the small hole, neglect stress concentration factors, and determine the magnitude
of the tangential and radial stresses and identify the stress as uniaxial or multiaxial and explain
you answer.
b. At the location of the small hole, take into account the stress concentration factors, and
approximate the magnitude of the tangential and radial stresses.
c. Would the state of stress at this location ( r = 200 mm ) be uniaxial or multiaxial? Why?
d. At the location of the small hole, not neglecting stress concentration, what would be a rough
approximation of the magnitude of tangential and radial stress components at the edge of the small
hole? (Hint: See Figure 5.9)
---------------------------------------------------------------------------------------------------------------------------
Solution
⎡ ( 3 + ν ) wω 2 ⎤ ⎡ 2 a 2b 2 ⎤
(σ r )hole = ⎢ ⎥ ⎢ a + b 2 − rh2 − 2 ⎥
⎢⎣ 8g ⎥⎦ ⎣⎢ rh ⎦⎥
⎡ ( 3 + ν ) wω 2 ⎤ ⎡ 2 ⎛ 1 + 3ν ⎞ 2 a b ⎤
2 2
(σ t )hole = ⎢ ⎥ ⎢a + b2 − ⎜ ⎟ r + ⎥
⎝ 3 +ν
h
⎣⎢ 8g ⎦⎥ ⎢⎣ ⎠ rh2 ⎥⎦
For steel w = 76.81 kN/m3 and ν = 0.30 . In addition, ω = 2π n / 60 = 2π (3000) / 60 = 314 res/sec and
g = 9.81 m/s 2 . Therefore
(σ r )hole =⎢
( )
⎡ ( 3 + 0.3) 76.81× 103 ( 314 )2 ⎤ ⎡
⎥ ⎢ (0.05) 2 + (0.250) 2 − (0.200) 2 − (0.05) (0.250) ⎤⎥
2 2
⎢ 8 ( 9.81) ⎥⎢ 2
⎦⎥
⎦⎥ ⎣
(0.200)
⎣⎢
(σ r )hole = 318.5 ×106 [0.065 − 0.0439] = 6.72 MPa
(σ t )hole =⎢
( )
⎡ ( 3 + 0.3) 76.81× 103 ( 314 )2 ⎤ ⎡
⎥ ⎢ (0.05) 2 + (0.250) 2 + ⎛ 1 + 3(0.3) ⎞ (0.200) 2 + (0.05) (0.250) ⎤⎥
2 2
⎢ 8 ( 9.81) ⎥⎢ ⎜ ⎟
⎝ 3 + 0.3 ⎠ (0.200) 2 ⎦⎥
⎣⎢ ⎦⎥ ⎣
735
Problem 18-8 (continued)
(b) The stress concentration factor, Kt , is determined by noting that since the hole is small, its diameter
will be d → 0 , so d / b ≈ 0 , resulting in Kt ≈ 3 . As a result
736
18-9. A proposed constant-strength flywheel is to be 1.00 inch in axial thickness at the center of rotation,
and 0.10 inch thick at its outer radius, which is 15.00 inches. If the material is AM 350 stainless steel, and
the flywheel is operating in a 400o F ambient air environment, estimate the rotational speed in rpm at
which yielding should initiate.
---------------------------------------------------------------------------------------------------------------------------
Solution
From Tables 3.4 and 3.5 , w = 0.282 lb/in 3 and S yp ( )400 = 144 ksi . From (18-64)
⎡ wω 2 ro2 ⎤
zmax = zr = 0 = zo e exp ⎢ ⎥
⎣⎢ 2σ g ⎦⎥
⎡ 0.282ω 2 (15.0) 2 ⎤ 5.71×10−7 ω 2
1.0 = 0.1e exp ⎢ ⎥ ⇒ 10 = e
⎢⎣ 2(144, 000)(386) ⎥⎦
ln(10) = 5.71× 10−7 ω 2
2π n ln(10) 60(2008)
ω= = = 2008 rad/sec ⇒ n = = 19,175 rpm
60 5.71× 10 −7
2π
737
18-10. A constant-strength steel flywheel with a rim is being considered for an application in which the
allowable stress of the flywheel material is 135 MPa, the outer radius of the disk is 300 mm, the rim
loading is 780 kN per meter of circumference, and the flywheel rotates at 6000 rpm. Calculate the
thickness of the flywheel web at radii of 0, 75, 150, 225, and 300 mm, and sketch the profile of the web
cross section.
---------------------------------------------------------------------------------------------------------------------------
Solution
n = 6000 rpm , σ = σ allow = 135 MPa , ro = 300 mm , q = 780 kN/m . From (19-62) z = zo e A , where
⎡⎛ wω 2 ⎞ 2 2 ⎤
A = ⎢⎜ (
⎟ ro − r
⎢⎣⎜⎝ 2σ g ⎟⎠
)⎥⎥
⎦
From (18-65) zo = q / σ o = 780 × 103 /135 × 106 = 5.78 × 10−3 m . For steel w = 76.81 kN/m3 . In addition,
ω = 2π n / 60 = 2π (6000) / 60 = 628.3 res/sec . Therefore
( )
z = 5.78 × 10−3 e11.45(0.09 − r
2
)
r (mm) z (mm)
0 16.2
75 15.2
150 12.5
225 9.1
300 5.8
738
Chapter 19
19-1. The cylindrical bearing journal of an overhung crankshaft has been sized for wear and found to
require a diameter of 38 mm based on wear analysis. A force analysis of the journal has shown there to be a
transverse shear force of 45 kN, torsional moment of 1000 N-m and bending moment of 900 N-m at the
critical cross section of the cylindrical journal bearing. If the fatigue-based stress governs, and has been
found to be 270 MPa, calculate whether the 38-mm-diamter journal is safely designed to withstand the
fatigue loading.
----------------------------------------------------------------------------------------------------------------------------
Solution
32M 32(900)
σb = = = 167 MPa
πd 3
π (0.038)3
16T 16(1000)
τT = = = 93 MPa
πd 3
π (0.038)3
4 V 16 V 16(45 000)
τV = = = = 53 MPa
(
3 πd /4
2 3 πd 2
)3π (0.038) 2
2
167 ⎛ 167 ⎞
⎟ + ( 93) = 83.5 ± 125 ⇒ σ 1 ≈ 209 MPa , σ 3 ≈ −42 MPa
2
σ1,3 = ± ⎜
2 ⎝ 2 ⎠
(σ eq ) A = 1⎡
2 ⎢
⎣
(σ1 − σ 2 )2 + (σ 2 − σ 3 )2 + (σ 3 − σ1 )2 ⎤⎦⎥
1⎡
= ( 209 − 0 )2 + ( 0 + 42 )2 + ( −42 − 209 )2 ⎤⎥⎦ ≈ 233 MPa
2 ⎢⎣
(σ eq ) A = 233 MPa < (σ d ) fatigue = 270 MPa Pont A should not suffer fatigue failure
At point B, the state of stress is pure shear with τ = τ T + τ T = 93 + 53 = 146 MPa . The principal stresses
are σ1 = 146 MPa , σ 2 = 0 , and σ 3 = −146 MPa
(σ eq )B = 1⎡
2 ⎢⎣
(146 − 0 )2 + ( 0 + 146 )2 + ( −146 − 146 )2 ⎤⎥⎦ ≈ 253 MPa < (σ d ) fatigue = 270 MPa
Pont B should not suffer fatigue failure either, so the journal is adequately designed.
739
19-2. The overhung crankshaft shown in Figure P19.2A is supported by bearings at R1 and R2 , and
loaded by P on the crankpin, vertically, as shown. The crank position shown may be regarded as the most
critical position. In this critical position, load P ranges form 900 lb up to 1800 lb down. The material
properties are given in Figure P19.2B. Based on wear estimates, all cylindrical bearing diameters should be
0.875 inch. Neglecting stress concentration effects, and using a safety factor of 1.5, determine whether the
diameter of 0.875 inch at R1 is adequate if infinite life is desired.
----------------------------------------------------------------------------------------------------------------------------
Solution
From specifications: Pmax = 1800 lb (down) , Pmin = −900 lb (up) , db = 0.875" , nd = 1.5 ,
S f = 70 ksi (Figure P19.2) , and N desired = ∞
The potential failure modes are fatigue and yielding. The specified loading produces non zero-mean cyclic
stresses. From (5-??)
SN
Smax − N = for σ m ≥ 0 ; Smax − N ≤ S yp
1 − mt Rt
For N desired = ∞
Sf
S max − f =
1 − mt Rt
where mt =
Su − S f
=
140 − 70 σ P
= 0.5 and Rt = m = m =
(1800 + (−900)) / 2 )
= 0.25
Su 140 σ max Pmax 1800
So
70
Smax − f = = 80 ksi
1 − 0.5(0.25)
Smax − f 80
And σ d = = = 53.333 ksi
nd 1.5
740
Problem 19-2 (continued)
4⎛ F ⎞ 4 ⎛ 4(1800) ⎞
τ ts = ⎜ ts ⎟ = ⎜⎜ ⎟ = 3990 psi
3 ⎝ A ⎠ 3 ⎝ π (0.875) 2 ⎟⎠
The shear stress due to torsion and the normal stress due to bending are
Two critical point are identified on this section, as shown in the figure below
At point A, σ 2 = 0 and
2
σb ⎛σ ⎞
+ ⎜ b ⎟ + (τ T )
2
σ1 =
2 ⎝ 2 ⎠
(σ eq ) A = 1⎡
2 ⎢⎣
(σ1 − σ 2 )2 + (σ 2 − σ 3 )2 + (σ 3 − σ1 )2 ⎤⎥⎦
1⎡
= ⎢
⎣
( 59.645 − 0 )2 + ( 0 + 4.905)2 + ( −4.905 − 59.645)2 ⎤⎦⎥ ≈ 62.24 ksi
2
At point B, σ 2 = 0 and
(σ eq )B = 1⎡
2 ⎢⎣
(σ1 − σ 2 )2 + (σ 2 − σ 3 )2 + (σ 3 − σ1 )2 ⎤⎥⎦
1⎡
= ⎢
⎣
( 21.14 − 0 )2 + ( 0 + 21.14 )2 + ( −21.14 − 21.14 )2 ⎤⎦⎥ ≈ 36.62 ksi
2
741
19-3. The overhung crankshaft shown in Figure P19.3 is supported on bearings R1 and R2 and loaded by
force P on the crank pin. This is taken to be the most critical crank position. The load P ranges from 6.25
kN up to 6.25 kN down. The crankshaft is made from forged carburized AISI 4620 steel
( Su = 696 MPa , S yp = 586 MPa , S 'f ≈ 348 MPa ). It has been determined that wear is the governing
failure mode for the bearing journal, and the allowable bearing design pressure, based on wear, is 5 MPa. A
5-mm-radius fillet is desired where cylindrical journal A blends into rectangular cheek B, which has a width
to height ratio of w / h = 0.5 . A factor of safety of 3 is to be used for all failure modes except wear, for hich
the safety factor has already been included in allowable stress for wear (5 MPa).
----------------------------------------------------------------------------------------------------------------------------
Solution
Su = 696 MPa , S yp = 586 MPa , S 'f ≈ 348 MPa , Pmax = 6.25 kN ↓ , Pmin = 6.25 kN ↑ , σ w− allow = 5 MPa ,
rAB = 5 mm , w / h = 0.5 , nd = 3
(a) (σ d ) f = S 'f / nd = 348 / 3 = 116 MPa . Since S yp = 586 MPa , (σ d ) f = 116 MPa is the governing
design stress for cheek B.
32M
(σ b )nom =
π d3
742
Problem 19-3 (continued)
32(241.6)
(σ b )nom = = 55.5 MPa
π (0.0354)3
Checking for stress concentrations, with D / d ≈ h / d = 45.4 / 35.4 ≈ 1.3 and r / d = 5 / 35.4 ≈ 0.14 we
determine Kt ≈ 1.7 Using q = 0.85
3V 3⎛ 6250 ⎞
(g) τ V = = ⎜ ⎟ ≈ 9.1 MPa
2 bh 2 ⎝ 0.0227(0.0454) ⎠
6M 6(212.5)
σb = 2 = ≈ 27.3 MPa
bh 0.0227(0.0454) 2
T
τ T = (τ max )T =
Q
where
241.6
τ T = (τ max )T = = 40.33 MPa
5.99 × 10−6
743
Problem 19-3 (continued)
2
σb ⎛σ ⎞
σ1,3 = ± ⎜ b ⎟ + τ T2 = 13.65 ± 42.6 ⇒ σ 1 = 56.25 MPa , σ 3 = −28.95 MPa
2 ⎝ 2 ⎠
(σ eq )1 = 1⎡
2 ⎣⎢
( 56.25 − 0 )2 + ( −28.95 − 0 )2 + ( −28.95 − 56.25)2 ⎤⎦⎥ ≈ 75 MPa
Point 4: Pure shear with τ = τ T + τ V = 40.33 + 9.1 = 49.43 MPa , therefore σ 2 = 0 and
(σ eq )4 = 1⎡
2 ⎢
⎣
( 49.43 − 0 )2 + ( −49.43 − 0 )2 + ( −49.43 − 49.43)2 ⎤⎦⎥ ≈ 85.5 MPa
( )1 = 75 MPa < (σ d ) f
(i) Since σ eq = 116 MPa and σ eq ( )4 = 85.5 MPa < (σ d ) f = 116 MPa , we
conclude that the design is acceptable.
744
19-4. The cylindrical bearing journal of a straddle-mounted crankshaft has been tentatively sized based on
wear requirements, and it has been found that a diameter of 1.38 inches is required for this purpose. A force
analysis of the journal at the critical cross section has shown there to be a transverse shear force of 8500 lb,
torsional moment of 7500 in-lb , and bending moment of 6500 in-lb. If the governing failure mode is
fatigue, and the design stress based on fatigue has been found to be 40,000 psi, calculate whether the 1.38-
inch-diameter journal is properly designed to safely withstand the fatigue loading.
----------------------------------------------------------------------------------------------------------------------------
Solution
From Specifications d = 1.38 in , Fts = 8500 lb , T = 7500 in-lb , M b = 6500 in-lb , (σ d ) f = 40 ksi
32M 32(6500)
σb = = = 25,193 psi
πd 3
π (1.38)3
16T 16(7500)
τT = = = 14,534 psi
π d3 π (1.38)3
2
25.193 ⎛ 25.193 ⎞
⎟ + (14.534 ) = 12.6 ± 19.2 ⇒ σ1 ≈ 31.8 ksi , σ 3 ≈ −6.6 ksi
2
σ1,3 = ± ⎜
2 ⎝ 2 ⎠
(σ eq ) A =
1⎡
2 ⎣⎢
( 31.8 − 0 )2 + ( 0 + 6.6 )2 + ( −6.6 − 31.8)2 ⎤⎦⎥ ≈ 35.6 ksi
At point B, the state of stress is pure shear with τ = τ T + τ ts = 22.1 ksi . The principal stresses are
σ 1 = 22.1 ksi , σ 2 = 0 , and σ 2 = −22.1 ksi , which results in
(σ eq )B =
1⎡
2 ⎢⎣
( 22.1 − 0 )2 + ( 0 + 22.1)2 + ( −22.1 − 22.1)2 ⎤⎥⎦ ≈ 38.3 ksi
Point B is therefore critical, and since σ eq ( )B = 38.3 ksi < (σ d ) f = 40 ksi , we conclude that the bearing is
designed correctly.
745
19-5. A straddle-mounted crankshaft for a belt-driven single-cylinder refrigeration compressor is to be
designed to meet the following specifications:
a. The force on the connecting rod bearing ranges from 3500 lb down to 1300 lb up.
b. The belt tension ratio is T1 = 8T2 .
c. The crank throw is 2.2 inches.
d. Allowable maximum bearing pressure is 700 psi.
e. The pulley pitch diameter = 8.00 inches.
f. All bearings are to be made alike
g. Main bearings are 12 inches apart center-to-center, with the connecting rod bearing halfway
between.
h. The pulley overhangs the main bearing 4 inches.
Design a suitable chrankshaft and construct good engineering sketches of the final design.
------------------------------------------------------------------------------------------------------------------------
Solution
1. A preliminary conceptual
sketch may be drawn as Figure 1.
∑ Fx = 0 : (identically satisfied)
∑ Fy = 0 : (identically satisfied)
∑ Fz = 0 : RL + P + RR + Ts + Tt = 0
∑ M z = 0 : (identically satisfied)
∑ M x = 0 : 2.2 P − 4Ts + 4Tt = 0
∑ M y = 0 : 12 RR + 6 P + 16 (Tt + Ts ) = 0
By specification; T1 = Tt = 8T2 = 8Ts , Pmax = −3500 lb (down) , Pmin = 1300 lb (up) . Therefore
746
Problem 19-5 (continued)
Table A
Critical Section
Section Element Shape Fz (lb) T (in-lb) M (in-lb)
1 A 2575 0 0
2 A 2575 0 7725
2 B 2575 7725 0
747
Problem 19-5 (continued)
Since all bearings are to be the same, we see form Table A that sections 4C, 6E, and 7E are more critical
than other circular sections. Similarly, sections 3B and 6D are the most critical rectangular sections.
Therefore we have 5 sections to check (3B, 4C, 6D, 6E, and 7E).
7. Based on the discussions of 19.3, the failure modes to be investigated should include wear, fatigue, and
yielding.
8. Based on the methods of Example 3.1, the tentative material selection will be forged 1020 steel, case
hardened at the bearing sites. The basic properties of 1020 steel are Su = 61 ksi , S yp = 51 ksi , and
e(2") = 15% .
(10 + t )2 8
nd = 1 + ; t ≥ −6 , where t = ∑ ( RN )i
100 i =1
The rating numbers (RN) for the 8 rating factors may be chosen as shown in Table B.
Table B
Selected Rating
Rating Factor Number (RN)
1. Accuracy of loads knowledge -1
2. Accuracy of stress calculations 0
3. Accuracy of strength knowledge 0
4. Need to conserve -2
5. Seriousness of failure consequences +2
6. Quality of manufacture 0
7. Condition of operation -2
8. Quality of inspection/maintenance 0
Summation , t = -2
(10 − 3)2
nd = 1 + = 1.49 ≈ 1.5
100
10. Based on steps 7, 8, and 9, the design stresses using the specified “d” are
1 ⎛ Se / K f ⎞
(σ max − N )d = ⎜ ⎟
nd ⎝ 1 − mt Rt ⎠
Assuming K f = 2
Su − S e / K f 61, 000 − 30,500 / 2
mt = = = 0.75
Su 61, 000
748
Problem 19-5 (continued)
Rt =
σm P
= m =
[3500 + (−1300)] / 2 = 0.314
σ max Pmax 3500
So
1 ⎛ 30,500 / 2 ⎞
(σ max − N )d = ⎜ ⎟ = 13, 298 ≈ 13,300 psi
1.5 ⎝ 1 − (0.75)(0.314) ⎠
For yielding
(σ yp )d =
S yp 51, 000
= = 34, 000 psi
nd 1.5
(σ wear )d = σ w− max =
Pmax
Ap
, so ( Ap )req ' d = (σPmax ) =
3500
700
= 5 in 2
wear d
(b) Since (σ max − N )d < σ yp ( )d , fatigue is more critical that yielding, so designing for fatigue will
automatically satisfy the yielding requirement.. Dimensions for the circular (4C, 6E, and 7E) and
rectangular (3B and 6D) sections is now determined.
Circular Sections - The critical points for the circular sections are
subjected to bending and shear. Two points are of interest, point i and
point ii as shown in Figure 4. Point I experienced a normal stress due
to bending, ( σ b ) and point ii experiences a shear stress due to torsion
( τ T ) and a shear stress due to transverse shear ( τ ts ). These stresses
are defined by
Section 4C:
4 4(2575) 4372 16(5665) 28,852 32(15, 450) 157,372
τ ts = = 2 , τT = = , and σ b = =
3 πd 2
d πd 3
d 3
π d3 d3
749
Problem 19-5 (continued)
2 2
157,372 ⎛ 157,372 ⎞ ⎛ 28,852 ⎞ 78, 686 83,809 162, 495 5123
σ1,3 = ± ⎜ ⎟ +⎜ ⎟ = ± = ,− 3
2d 3 ⎝ 2d 3
⎠ ⎝ d 3
⎠ d3 d3 d3 d
1 ⎡⎛ 162, 495
2 2 2⎤
⎞ ⎛ 5123 ⎞ ⎛ 5123 162, 495 ⎞
(σ eq )i = ⎢
2 ⎢⎣⎜⎝ d 3
− 0⎟ + ⎜0 + 3 ⎟ + ⎜ − 3 −
⎠ ⎝ d ⎠ ⎝ d d3 ⎠
⎟ ⎥=
⎥⎦
201,128
d3
201,128
= 13,300 ⇒ d = 2.47
d3
Using d = 2.5" we find that σ eq ( )i = 12,872 < 13,300 , so point i at 4C satisfies the criteria. Next we turn
to point ii, which is in a state of pure shear. The shear stress due to transverse shear at this point is
4372
τ ts = ≈ 700
(2.5)2
Since τ ii = τ T + τ ts = 1846 + 700 = 2546 psi , we get σ1 = 2546 psi , σ 2 = 0 , σ 3 = −2546 psi , which
produces
(σ eq )ii = 1⎡
2 ⎣⎢
( 2546 − 0 )2 + ( 0 + 2546 )2 + ( −2546 − 2546 )2 ⎤⎦⎥ = 4410 psi
Section 6E: This section has a smaller bending stress, but larger torque. Using d = 2.5" , at 6E
2
8623 ⎛ 8623 ⎞
+ ( 2510 ) = 4312 ± 4989 = 9301, −677
2
At point i, this results in σ 2 = 0 and σ1,3 = ± ⎜ ⎟
2 ⎝ 2 ⎠
750
Problem 19-5 (continued)
Section 7E: This section has a smaller bending stress, but larger torque. Using d = 2.5" , at 7E
2
6454 ⎛ 6454 ⎞
⎟ + ( 2510 ) = 3227 ± 4088 = 7315, −861
2
At point i, this results in σ 2 = 0 and σ1,3 = ± ⎜
2 ⎝ 2 ⎠
Based on this, we tentatively assume d = 2.5" will be used for all bearing cross sections.
6M T T (1.5h + 0.9b )
σb = 2
, τT = = , and
bh Q 0.5h 2b 2
3 Fts 3 Fts
τ ts = =
2 A 2 bh
For the circular sections we found that d = 2.5" was a good size.
For the rectangular cross sections we arbitrarily assume that
h = 3.0" , and select an aspect ration of b / h = 0.5 , which results in
b = 1.5" . This produces the stress relationships
σ b = 0.444(5665) = 2517 psi , τ ts = 0.333(2575) = 858 psi , and τ T = 0.5778(7725) = 4464 psi
At point i, σ 2 = 0 and
751
Problem 19-5 (continued)
2
2517 ⎛ 2517 ⎞
+ ( 4464 ) = 1259 ± 4638 = 5897, −3379
2
σ1,3 = ± ⎜ ⎟
2 ⎝ 2 ⎠
(σ eq )i = 1⎡
2 ⎢⎣
( 5897 − 0 )2 + ( 0 + 3379 )2 + ( −3379 − 5897 )2 ⎤⎥⎦ = 8131 psi
At point ii , τ ii = τ T + τ ts = 4464 + 858 = 8322 psi , which produces σ1 = 5322 psi , σ 2 = 0 , and
σ 3 = −5322 psi . These result in
(σ eq )ii = 1⎡
2 ⎢
⎣
( 5322 − 0 )2 + ( 0 + 5322 )2 + ( −5322 − 5322 )2 ⎤⎦⎥ = 9218 psi
Both are less than (σ max − N )d = 13,300 psi , so these two point are safe from fatigue failure.
σ b = 0.444(7700) = 3419 psi , τ ts = 0.333(925) = 308 psi , and τ T = 0.5778(12, 675) = 7324 psi
At point i, σ 2 = 0 and
2
3419 ⎛ 3419 ⎞
⎟ + ( 7324 ) = 1710 ± 7521 = 9231, −5804
2
σ1,3 = ± ⎜
2 ⎝ 2 ⎠
(σ eq )i = 1⎡
2 ⎣⎢
( 9321 − 0 )2 + ( 0 + 5804 )2 + ( −5804 − 9321)2 ⎤⎦⎥ = 13, 216 psi
These result in
(σ eq )ii = 1⎡
2 ⎢
⎣
( 7632 − 0 )2 + ( 0 + 7632 )2 + ( −7632 − 7632 )2 ⎤⎦⎥ = 13, 219 psi
Both are less than (σ max − N )d = 13,300 psi , so these two point are safe from fatigue failure.
All critical points are adequately designed for wear, fatigue and yielding. The circular cross sections have a
diameter of d = 2.5" , and the rectangular sections have dimensions of b = 1.5" and h = 3.0" .
752
Chapter 20
20-1. Splined connections are widely used throughout industry, but little research has been done to provide
the designer with either analytical tools or good experimental data for spline strength or compliance
estimates. In question are such matters as basic spline-tooth strength, shaft strength, notch effect, and spline
geometry effects, as well as spline-compliance effects on the torsional spring rate of a system containing
one or more splined connections.
It is desired to construct a splined-joint testing setup versatile enough to facilitate both strength and
life testing of various splined connections, as well as to perform torsional compliance testing on such joints.
The testing setup is to accommodate in-line splined connections, offset parallel shafts connected by double
universal joints, and angular shaft connections. Parallel shaft offsets up to 250 mm must be accommodated,
and angular shaft centerline displacements up to 45o may be required. Splines up to 75 mm in diameter
may need to be tested in the device, and shafting samples, including splined connections, may be up to 1
meter in length. Rotating speeds up to 3600 rpm may be required.
The basic setup, sketched in Figure P20.1, is to utilize a variable-speed drive motor to supply power
to the input shaft of the testing setup, and a dynamometer (device for measuring mechanical power) used to
dissipate from the output shaft of the test setup.
a. Select an appropriate type of frame or supporting structure for integrating the drive motor, testing
setup, and dynamometer into a laboratory test stand for investigating splined-connection behavior, as
just discussed.
b. Sketch the frame.
c. Identify potential safety issues that should , in your opinion, be addressed before putting the test
stand into use.
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Solution
(a) Since this is a design task, there is no “one” correct solution. The approach suggested is simply one
possible proposal. The following observations have pertinence to conception of the supporting frame
structure:
1. The drive motor shaft centerline should be at the same elevation as the dynamometer shaft
centerline.
2. Some vertical “adjustment” of the drive motor shaft centerline should be provided.
3. The drive motor support structure must be “moveable” to any location in the plane of the base so
as to accommodate a 250-mm shaft centerline offset relative to the dynamometer shaft centerline, and
to accommodate test sample lengths up to 1 meter.
4. The drive motor support structure should provide a means to accommodate rotation of the drive
motor about a vertical centerline up to 45o .
1. Use a simple T-Slotted base plate mounted on a concrete foundation of floor as the support
structure for the dynamometer and motor.
2. Use an open truss frame, mounted to the T-slotted base plate, to support the motor. Provide a
series of mounting holes so motor location can be changed.
3. Use a rotary table on top of the motor mount frame so motor can be rotated about a vertical axis.
4. Use shims between motor mounting feet and rotary table to provide vertical adjustment of the
motor shaft centerline.
753
Problem 20-1 (continued)
(c) Rotating machinery is always dangerous. Guards should be used and protective barriers and signs
should be used .
754
20-2. You have been given the task of designing a special hydraulic press for removing and replacing
bearings in small to medium-size electric motors. You are to utilize a commercially available 1 ½ ton (3000
lb) capacity hydraulic actuator, mounted vertically, as sketched in Figure P20.2. As shown, the actuator
body incorporates a 2-inch-diameter mounting boss. A lower plate for supporting the motor bearing
packages is to be incorporated, as sketched. A minimum vertical clearance of 3 inches is required, as
indicated, and a minimum unobstructed horizontal clearance of 5 inches between the vertical centerline of
the hydraulic press and the closest structural member is also required. The operator’s intended position is
indicated in Figure P20.2, as well.
a. Select an appropriate type of frame or supporting structure for integrating the hydraulic actuator
and support platen into a compact, stand-alone assembly, giving reasons for your selection.
b. Make a neat sketch of the frame, as you envision it.
c. Design the frame that you have sketched in (b) so that it may be expected to operate for 20 years
in an industrial setting without failure. It is estimated that the press will be operated on average, once a
minute, 8 hours each day, 250 days a year.
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Solution
(a) Since this is a design task, there is no “one” correct solution. The approach presented here is only one of
many possible proposals. The following observations have pertinence to the conception of a supporting
frame structure:
1. The frame should incorporate a means of clamping the 2-inch diameter mounting boss to
support the hydraulic actuator.
2. The frame should incorporate a means of securing the lower platen in line with the
hydraulic actuator.
3. The frame should provide open access to the operator for loading, positioning, and
unloading workpieces (motor bearing support plates).
4. The frame should provide specified minimum clearances (5 inches from vertical
centerline to closest structure and 3 inches minimum between bottom end of the actuator
stem (and any special pressure fitting) and top of the lower platen.
Based on these observations, it would seem that the best choice for the supporting frame would be a basic
C-frame (Figure 20.1e).
755
Problem 20-2 (continued)
4. Reviewing Chapter 3, cast iron or steel are reasonable candidate materials to be evaluated.
5. For cast iron, it is probably redundant to say casting is the process of choice. For steel, casting or
welding would be reasonable choices.
6. To select critical points, review section 4.4 and Chapter 6.
7. To select appropriate equations of mechanics, review section 4.4.
8. To determine dimensions at each critical section, review section 4.4. Review Chapter 2 to
determine an appropriate safety factor. For this application, an appropriate value would be nd ≈ 2 .
9. Review all aspects of the above 8 steps for compatibility, with attention to maintenance and
inspection requirements.
10. Sketch the design proposal, including dimensions and specifications.
756
20-3. Cutting firewood is popular with “do-it-yourselfers” in many parts of the world, but hand-splitting the
logs is a less-popular task. You are being asked to design a compact, “portable” moderately priced
firewood splitting machine for “home” use. The device should be capable of handling logs up to 400 mm in
diameter and 600 mm long, splitting them into fireplace-size pieces. A cord of wood (a stack of wood
1.2 m × 1.2 m × 4.4 m ) should take no longer than a hour to split. Management has decided that a power-
screw-driven splitting wedge should be investigated as a first choice. The concept is sketched crudely in
Figure P20.3. Safety is to be considered, as well as compactness and portability.
a. Select an appropriate type of frame or supporting structure for integrating the power-screw-driven
wedge and adjustable log-support arm into a compact, portable, stand-alone assembly, giving reasons
for your selection.
b. Make a sketch of the frame as you envision it.
c. From your sketch, identify each coherent subassembly, and give each subassembly a descriptive
name.
d. Make a neat sketch of each coherent subassembly, and, treating each subassembly as a free body,
qualitatively indicate all significant forces on each subassembly.
e. Design the power-screw subassembly. Preliminary estimates indicated that with a properly shaped
splitting wedge, the wedge travel need not exceed half the length of the log, and that the “splitting
force” required of the power screw need not exceed about 38 kN in the direction of the screw axis.
f. Design the adjustable log-support unit.
g. Design any other subassembly that you have named in (c).
h. Design the frame that you have sketched in (b).
i. Discuss any potential safety issues that you envision to be important.
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Solution
Since this is a design task, there is no “one” correct solution. The general nature of this project would
require a comprehensive effort from several persons for multiple weeks. Clearly, a “solution” can not be
presented here. Many possible configurations are possible. The following observations have pertinence to
conception of the supporting frame structure:
TR
38 000 =
⎡ cos θ n sin α + µt cos α ⎤
rp ⎢ ⎥
⎣ cos θ n cos α − µt sin α ⎦
757
20-4. The input shaft of a rotary coal-grinding mill is to be driven by a gear reducer through a flexible
shaft-coupling, as shown in Figure P20-4. The output shaft of the gear reducer is to be supported on two
bearings mounted 10 inches apart at A and C, as shown. A 1:3 spur gear mesh is being proposed to drive
the gear-reducer output shaft. A spur gear is mounted on the output shaft at midspan between the bearings,
as shown, and is to have a pitch diameter of 9 inches. The pitch diameter of the drive pinion is to be 3
inches. The coal-grinding mill is to be operated at 600 rpm, and requires 100 horsepower continuously at its
input shaft.
An 1800-rpm electric motor is to supply power to the pinion input shaft. Concentrating attention
on the spur gear speed reducer sketched in Figure P20.4, do the following:
a. Select an appropriate type frame or supporting structure for integrating the gears, shafts, and
bearings into a compact stand-alone subassembly, giving reasons for your selection.
b. Make a neat sketch of the frame, as you envision it.
c. Design or select a spur gear set.
d. Design the gear-reducer output shaft.
e. Design the pinion input shaft.
f. Select appropriate bearings for the gear-reducer output shaft.
g. Select appropriate bearings for the pinion input shaft.
h. Specify appropriate lubrication for the gears and bearings.
i. Design the frame that you have sketched in (b).
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Solution
This problem is in the nature of a “project”, requiring the cooperative efforts of perhaps three persons
working together over a period of 10 -15 weeks to produce a reasonable result. Clearly, a “solution” can not
be presented here. Many acceptable configurations can be conceived to fulfill specifications. The following
suggestions are offered as guidelines:
1. A “housing” or “case” (see Figure 20.1g) would seem to be a good choice for the supporting
structure, since it can provide bearing supports for the two shafts, a sump for lubrication, and an
enclosure to protect the gear mesh from environmental contamination as well as provide safety
shielding.
2. In designing the spur gear set, see Chapter 15 for procedures and details.
3. In designing the gear-reducing output shaft and the pinion input shaft, see Chapter 8 for
procedures and details.
4. For selection of bearings for the gear reducer output shaft and the pinion input shaft, see details of
Chapter 10 if plain bearings are chosen, or Chapter 11 if roller bearings are chosen.
5. For selecting and specifying appropriate lubrication for bearings and gears, see details of Chapter
10 or Chapter 11, and Chapter 15, as appropriate.
758
20-5. a. In the context of mechanical design, define the terms safety, danger, hazard, and risk.
b. List the actions a designer might take to provide proper safeguards before releasing a machine to
customers in the marketplace.
c. Make a list of safeguarding devices that have been developed to help reduce to an acceptable level
the risks associated with engineered products.
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Solution
(a) Definitions;
(c) From Table 20.2, a list of safeguarding devices may be made as follows;
1. Photoelectric sensors
2. RF capacitance devices
3. Electromechanical devices
4. Pullback devices
5. Holdback devices
6. Safety trip controls
7. Pressure sensitive body bars
8. Safety tripods
9. Safety tripwires
10. Two-hand controls
11. Two-hand trips
12. Gates
13. Automatic feeders
14. Semiautomatic feeders
15. Automatic ejectors
16. Semiautomatic ejectors
17. Robots
759