160 CHAPTER 2 Axially Loaded Members

Stress Concentrations

The problems for Section 2.10 are to be solved by considering the P P

stress-concentration factors and assuming linearly elastic behavior. b d

Problem 2.10-1 The flat bars shown in parts (a) and (b) of the figure are
subjected to tensile forces P
3.0 k. Each bar has thickness t
0.25 in.
(a) For the bar with a circular hole, determine the maximum stresses (a)
for hole diameters d
1 in. and d
2 in. if the width b
6.0 in.
(b) For the stepped bar with shoulder fillets, determine the maximum R
stresses for fillet radii R
0.25 in. and R
0.5 in. if the bar
widths are b
4.0 in. and c
2.5 in. P P
b c

Probs. 2.10-1 and 2.10-2 (b)

Solution 2.10-1 Flat bars in tension

R = radius

P P P P
b d b c

(b)

(a)

P
3.0 k t
0.25 in. (b) STEPPED BAR WITH SHOULDER FILLETS

b
4.0 in. c
2.5 in.; Obtain k from Fig. 2-64
(a) BAR WITH CIRCULAR HOLE (b
6 in.)
P 3.0 k
Obtain K from Fig. 2-63 snom


4.80 ksi
ct (2.5 in.)(0.25 in.)
FOR d
1 in.: c
b  d
5 in.
FOR R
0.25 in.: R/c
0.1 b/c
1.60
P 3.0 k
snom


2.40 ksi k
2.30
max
K
nom
11.0 ksi
ct (5 in.)(0.25 in.)
FOR R
0.5 in.: R/c
0.2 b/c
1.60
1
d/b
K
2.60
6 K
1.87
max
K
nom
9.0 ksi

max
k
nom
6.2 ksi

FOR d
2 in.: c
b  d
4 in.

P 3.0 k
snom


3.00 ksi
ct (4 in.)(0.25 in.)
1
d/b
K
2.31
3

max
K
nom
6.9 ksi
 SECTION 2.10 Stress Concentrations 161

Problem 2.10-2 The flat bars shown in parts (a) and (b) of the figure are
subjected to tensile forces P
2.5 kN. Each bar has thickness t
5.0 mm.
(a) For the bar with a circular hole, determine the maximum stresses for
hole diameters d
12 mm and d
20 mm if the width b
60 mm.
(b) For the stepped bar with shoulder fillets, determine the maximum
stresses for fillet radii R
6 mm and R
10 mm if the bar widths are
b
60 mm and c
40 mm.

Solution 2.10-2 Flat bars in tension

R = radius

P P P P
b d b c

(b)

(a)

P
2.5 kN t
5.0 mm (b) STEPPED BAR WITH SHOULDER FILLETS

b
60 mm c
40 mm;
(a) BAR WITH CIRCULAR HOLE (b
60 mm)
Obtain K from Fig. 2-64
Obtain K from Fig. 2-63
P 2.5 kN
FOR d
12 mm: c
b  d
48 mm snom


12.50 MPa
ct (40 mm)(5 mm)
P 2.5 kN
snom


10.42 MPa FOR R
6 mm: R/c
0.15 b/c
1.5
ct (48 mm)(5 mm)
1 K
2.00
max
K
nom
25 MPa
d/b
K
2.51
5 FOR R
10 mm: R/c
0.25 b/c
1.5

max
K
nom
26 MPa K
1.75
max
K
nom
22 MPa
FOR d
20 mm: c
b  d
40 mm
P 2.5 kN
snom


12.50 MPa
ct (40 mm)(5 mm)
1
d/b
K
2.31
3

max
K
nom
29 MPa
162 CHAPTER 2 Axially Loaded Members

Problem 2.10-3 A flat bar of width b and thickness t has a hole of

diameter d drilled through it (see figure). The hole may have any
diameter that will fit within the bar.
What is the maximum permissible tensile load Pmax if the P P
allowable tensile stress in the material is
t? b d

Solution 2.10-3 Flat bar in tension

d
K P*
P P b
b d
0 3.00 0.333
0.1 2.73 0.330
t
thickness 0.2 2.50 0.320

t
allowable tensile stress 0.3 2.35 0.298
Find Pmax 0.4 2.24 0.268
Find K from Fig. 2-64
smax st
Pmax
snomct
ct
(b  d)t We observe that Pmax decreases as d/b increases.
K K
Therefore, the maximum load occurs when the hole
st d becomes very small.

bt ¢ 1  ≤
K b d
( S 0
and
K S 3)
Because
t, b, and t are constants, we write: b
Pmax 1 d st bt
P*

¢1  ≤ Pmax

st bt K b 3

Problem 2.10-4 A round brass bar of diameter d1
20 mm

has upset ends of diameter d2
26 mm (see figure). The lengths
of the segments of the bar are L1
0.3 m and L2
0.1 m. P
d2 d1 d2
P
Quarter-circular fillets are used at the shoulders of the bar,
and the modulus of elasticity of the brass is E
100 GPa.
If the bar lengthens by 0.12 mm under a tensile load P, L2 L1 L2
what is the maximum stress
max in the bar?
Probs. 2.10-4 and 2.10-5
 SECTION 2.10 Stress Concentrations 163

Solution 2.10-4 Round brass bar with upset ends

d2 = 26 mm d1 = 20 mm
P P Use Fig. 2-65 for the stress-concentration factor:
P EA2 E
snom



L2 L1 L2 A1 2L2 A1  L1 A2 2L2 ( A2 )  L1
A1

E
E
100 GPa
d
2L2 ( d12 ) 2  L1

0.12 mm
SUBSTITUTE NUMERICAL VALUES:
L2
0.1 m
(0.12 mm)(100 GPa)
L1
0.3 m snom

28.68 MPa
26 )  0.3 m
2(0.1 m)( 20 2

26 mm  20 mm
R
radius of fillets

3 mm R 3 mm
2

0.15
D1 20 mm
PL2 PL1

2¢ ≤ Use the dashed curve in Fig. 2-65. K
1.6
EA2 EA1
smax
Ksnom
(1.6)(28.68 MPa)
EA1 A2
Solve for P:
P

46 MPa
2L2 A1  L1 A2

Problem 2.10-5 Solve the preceding problem for a bar of monel metal
having the following properties: d1
1.0 in., d2
1.4 in., L1
20.0 in.,
L2
5.0 in., and E
25  106 psi. Also, the bar lengthens by 0.0040 in.
when the tensile load is applied.

Solution 2.10-5 Round bar with upset ends

d2 = 1.4 in. d1 = 1.0 in Use Fig. 2-65 for the stress-concentration factor.
P P
P EA2 E
snom



L2 L1 L2 A1 2L2A1  L1A2 2L2 ( AA12 )  L1
E

d
2L2 ( d12 ) 2  L1
E
25  106 psi

0.0040 in. SUBSTITUTE NUMERICAL VALUES:
L1
20 in. (0.0040 in.)(25  106 psi)
snom

3,984 psi
1.4 )  20 in.
2(5 in.)( 1.0 2
L2
5 in.
1.4 in.  1.0 in. R 0.2 in.
R
radius of fillets
R


0.2
2 D1 1.0 in.

0.2 in. Use the dashed curve in Fig. 2-65. K
1.53

PL2 PL1 smax
Ksnom
(1.53)(3984 psi)


2¢ ≤
EA2 EA1
6100 psi
EA1A2
Solve for P: P

2L2A1  L1A2
164 CHAPTER 2 Axially Loaded Members

Problem 2.10-6 A prismatic bar of diameter d0
20 mm is being

compared with a stepped bar of the same diameter (d1
20 mm) that
is enlarged in the middle region to a diameter d2
25 mm (see figure). P1
The radius of the fillets in the stepped bar is 2.0 mm.
(a) Does enlarging the bar in the middle region make it stronger
than the prismatic bar? Demonstrate your answer by determining
P2
the maximum permissible load P1 for the prismatic bar and the
maximum permissible load P2 for the enlarged bar, assuming
that the allowable stress for the material is 80 MPa. d1 d0
P1
(b) What should be the diameter d0 of the prismatic bar if it is to
have the same maximum permissible load as does the
stepped bar?
d2
d1
P2

Soluton 2.10-6 Prismatic bar and stepped bar

Stepped bar: See Fig. 2-65 for the stress-

P1
concentration factor.
R
2.0 mm D1
20 mm D2
25 mm
RD
0.10
D2 D
1.25
K
1.75
1 1
P2
P2 P2 smax
snom
 2

snom

d1 d0 4 d1 A1 K
P1
smax st
P2
snom A1
A
A
K 1 K 1
80 MPa 
d2
¢ ≤¢ ≤ (20 mm) 2
1.75 4
d1
P2
14.4 kN
d0
20 mm Enlarging the bar makes it weaker, not stronger. The
ratio of loads is P1P2
K
1.75
d1
20 mm
d2
25 mm (b) DIAMETER OF PRISMATIC BAR FOR THE SAME
ALLOWABLE LOAD
Fillet radius: R
2 mm
Allowable stress:
t
80 MPa d02 st d12 d12
P1
P2
st ¢ ≤
¢ ≤
d0

2
4 K 4 K
(a) COMPARISON OF BARS
d02 d1 20 mm
Prismatic bar:
P1
st A0
st ¢ ≤ d0


15.1 mm
4 K 1.75


(80 MPa) ¢ ≤ (20 mm) 2
25.1 kN
4
 SECTION 2.10 Stress Concentrations 165

Problem 2.10-7 A stepped bar with a hole (see figure) has widths
b
2.4 in. and c
1.6 in. The fillets have radii equal to 0.2 in. P P
d b c
What is the diameter dmax of the largest hole that can be drilled
through the bar without reducing the load-carrying capacity?

Solution 10-7 Stepped bar with a hole

P P
d b c

b
2.4 in. BASED UPON HOLE (Use Fig. 2-63)

c
1.6 in. b
2.4 in. d
diameter of the hole (in.) c1
b  d
Fillet radius: R
0.2 in. smax
Pmax
snom c1t
(b  d)t
Find dmax K
1 d
BASED UPON FILLETS (Use Fig. 2-64)
¢ 1  ≤ btsmax
K b
b
2.4 in. c
1.6 in. R
0.2 in. R/c
0.125
b/c
1.5 K
2.10 d (in.) d/b K Pmax btsmax
smax smax c 0.3 0.125 2.66 0.329
Pmax
snomct
ct
¢ ≤ (bt)
K K b 0.4 0.167 2.57 0.324

0.317 bt smax 0.5 0.208 2.49 0.318
0.6 0.250 2.41 0.311
0.7 0.292 2.37 0.299

Based upon hole

0.33
Based upon fillets
0.32 0.317

0.31 dmax ≈ 0.51 in.

Pmax
bt
max
0.30 d (in.)
0.3 0.4 0.5 0.6 0.7 0.8
166 CHAPTER 2 Axially Loaded Members

Nonlinear Behavior (Changes in Lengths of Bars)

Problem 2.11-1 A bar AB of length L and weight density  hangs

vertically under its own weight (see figure). The stress-strain relation
for the material is given by the Ramberg-Osgood equation (Eq. 2-71): A



 
m

  0 
E E
0
Derive the following formula L
L2
L L
 
m

  0  
2E (m 1)E
0
for the elongation of the bar.
B

Solution 2.11-1 Bar hanging under its own weight

STRAIN AT DISTANCE x
s s0 s m gx s0 gx m
Let A
cross-sectional area e
 ¢ ≤
 ¢ ≤
E E s0 E E s0
dx Let N
axial force at distance x
L N
Ax ELONGATION OF BAR
L
s0 L

  
L
N gx gx m
x s

gx 
e dx
dx  ¢ ≤ dx
E E s0
A 0 0 0

gL2 s0L gL m

 ¢ ≤

Q.E.D.
2E (m  1)E s0

Problem 2.11-2 A prismatic bar of length L
1.8 m and cross-sectional

area A
480 mm2 is loaded by forces P1
30 kN and P2
60 kN (see
figure). The bar is constructed of magnesium alloy having a stress-strain
A B P1 C P2
curve described by the following Ramberg-Osgood equation:


 1
70 
1 10

    (

MPa) 2L L
45,000 618 — —
3 3
in which
has units of megapascals.
(a) Calculate the displacement C of the end of the bar when the load
P1 acts alone.
(b) Calculate the displacement when the load P2 acts alone.
(c) Calculate the displacement when both loads act simultaneously.

Solution 2.11-2 Axially loaded bar L
1.8 m A
480 mm2

P1
30 kN P2
60 kN
A B P1 C P2
RambergOsgood Equation:
2L L s 1 s 10
—
3
—
3 e
 ¢ ≤ (s
MPa)
45,000 618 170
Find displacement at end of bar.
 SECTION 2.11 Nonlinear Behavior 167

(a) P1 ACTS ALONE

(c) BOTH P1 AND P2 ARE ACTING
P1 30 kN
AB: s


62.5 MPa P1  P2 90 kN
A 480 mm2 AB: s


187.5 MPa
A 480 mm2
e
0.001389 e
0.008477
2L 2L
c
e ¢ ≤
1.67 mm AB
e ¢ ≤
10.17 mm
3 3
P2 60 kN
(b) P2 ACTS ALONE BC: s


125 MPa
A 480 mm2
P2 60 kN
ABC: s


125 MPa e
0.002853
A 480 mm2
L
e
0.002853 BC
e ¢ ≤
1.71 mm
3
c
eL
5.13 mm C
AB  BC
11.88 mm
(Note that the displacement when both loads act
simultaneously is not equal to the sum of the
displacements when the loads act separately.)

Problem 2.11-3 A circular bar of length L
32 in. and diameter d
0.75

in. is subjected to tension by forces P (see figure). The wire is made of a
copper alloy having the following hyperbolic stress-strain relationship: d
P P
18,000


 0    0.03 (

ksi)
1  300
(a) Draw a stress-strain diagram for the material. L
(b) If the elongation of the wire is limited to 0.25 in. and the maximum
stress is limited to 40 ksi, what is the allowable load P?

Solution 2.11-3 Copper bar in tension

d (b) ALLOWABLE LOAD P
P P
Max. elongation max
0.25 in.
L Max. stress
max
40 ksi

L
32 in. d
0.75 in. Based upon elongation:

d2 max 0.25 in.

A

0.4418 in.2 emax


0.007813
4 L 32 in.
18,000 emax
(a) STRESS-STRAIN DIAGRAM smax

42.06 ksi
1  300 emax
18,000e
s

0  e  0.03
(s
ksi)
1  300e BASED UPON STRESS:
Slope = 18,000 ksi

60 smax
40 ksi
(ksi) Stress governs. P

max A
(40 ksi)(0.4418 in.2)
Asymptote
40
17.7 k
equals 60 ksi

20


0 0.01 0.02 0.03
168 CHAPTER 2 Axially Loaded Members

Problem 2.11-4 A prismatic bar in tension has length L
2.0 m 200

and cross-sectional area A
249 mm2. The material of the bar has the
stress-strain curve shown in the figure.

(MPa)
Determine the elongation  of the bar for each of the following axial
loads: P
10 kN, 20 kN, 30 kN, 40 kN, and 45 kN. From these results,
plot a diagram of load P versus elongation  (load-displacement diagram). 100

0
0 0.005 0.010


Solution 2.11-4 Bar in tension

P P

50
L
40
L
2.0 m 30
A
249 mm2 P (kN)
20
STRESS-STRAIN DIAGRAM 10
(See the problem statement for the diagram)
0 5 10 15 20  (mm)
LOAD-DISPLACEMENT DIAGRAM

P

P/A e 
eL NOTE: The load-displacement curve has the same

(kN) (MPa) (from diagram) (mm) shape as the stress-strain curve.
10 40 0.0009 1.8
20 80 0.0018 3.6
30 120 0.0031 6.2
40 161 0.0060 12.0
45 181 0.0081 16.2

Problem 2.11-5 An aluminum bar subjected to tensile forces P has
length L
150 in. and cross-sectional area A
2.0 in.2 The stress-strain
behavior of the aluminum may be represented approximately by the
bilinear stress-strain diagram shown in the figure.
E2 = 2.4 × 106 psi
Calculate the elongation  of the bar for each of the following axial 12,000
loads: P
8 k, 16 k, 24 k, 32 k, and 40 k. From these results, plot a psi
diagram of load P versus elongation  (load-displacement diagram).
E1 = 10 × 106 psi

0 
 SECTION 2.11 Nonlinear Behavior 169

Solution 2.11-5 Aluminum bar in tension

P P LOAD-DISPLACEMENT DIAGRAM

L P

P/A e 
eL
(k) (psi) (from Eq. 1 or Eq. 2) (in.)
L
150 in. 8 4,000 0.00040 0.060
A
2.0 in.2 16 8,000 0.00080 0.120

STRESS-STRAIN DIAGRAM 24 12,000 0.00120 0.180

32 16,000 0.00287 0.430


40 20,000 0.00453 0.680
E2

1
40 k
40
E1
30 24 k

0 1  P (k) 20
0.68 in.

10 0.18 in.
E1
10  106 psi
E2
2.4  106 psi  (in.)
0 0.2 0.4 0.6 0.8

1
12,000 psi
s1 12,000 psi
e1


E1 10  106 psi

0.0012
For 0  s  s1:

s s
e

(s
psi) Eq. (1)
E2 10  106 psi
For s  s1:
s  s1 s  12,000
e
e1 
0.0012 
E2 2.4  106
s

 0.0038
(s
psi) Eq. (2)
2.4  106
170 CHAPTER 2 Axially Loaded Members

Problem 2.11-6 A rigid bar AB, pinned at end A, is supported by a wire

CD and loaded by a force P at end B (see figure). The wire is made of
high-strength steel having modulus of elasticity E
210 GPa and yield C
stress
Y
820 MPa. The length of the wire is L
1.0 m and its
diameter is d
3 mm. The stress-strain diagram for the steel is defined L
by the modified power law, as follows:
A D B


E 0 

Y

 E
 
n



Y 

Y
Y P

(a) Assuming n
0.2, calculate the displacement B at the end of the 2b b

bar due to the load P. Take values of P from 2.4 kN to 5.6 kN in
increments of 0.8 kN.
(b) Plot a load-displacement diagram showing P versus B.

Solution 2.11-6 Rigid bar supported by a wire

sY s 1n
From Eq. (2):
e
¢ ≤ (5)
C E sY
3P
L Axial force in wire: F

2
A D B
F 3P
Stress in wire: s

(6)
A 2A
P PROCEDURE: Assume a value of P
2b b Calculate
from Eq. (6)
Calculate e from Eq. (4) or (5)
Wire: E
210 GPa Calculate B from Eq. (3)

Y
820 MPa P
(MPa) e B (mm)

(kN) Eq. (6) Eq. (4) or (5) Eq. (3)
L
1.0 m
2.4 509.3 0.002425 3.64
d
3 mm
3.2 679.1 0.003234 4.85
d2
A

7.0686 mm2 4.0 848.8 0.004640 6.96
4
4.8 1018.6 0.01155 17.3
STRESS-STRAIN DIAGRAM
5.6 1188.4 0.02497 37.5
s
Ee
(0  s  sY ) (1)
For


Y
820 MPa:
Ee n
s
sY ¢ ≤
(s  sY )
(n
0.2) (2)
sY e
0.0039048 P
3.864 kN B
5.86 mm

(a) DISPLACEMENT B AT END OF BAR (b) LOAD-DISPLACEMENT DIAGRAM

3 3

elongation of wire
B

eL (3) 8
2 2
Obtain e from stress-strain equations: 6
P P = 3.86 kN
s (kN) 4
From Eq. (1):
e

(0  s  sY ) (4)
Y = 820 MPa
E 2
B = 5.86 mm

0 20 40 60 B (mm)
 SECTION 2.12 Elastoplastic Analysis 171

Elastoplastic Analysis

The problems for Section 2.12 are to be solved assuming that the
material is elastoplastic with yield stress
Y , yield strain Y , and
modulus of elasticity E in the linearly elastic region (see Fig. 2-70).
A   C
Problem 2.12-1 Two identical bars AB and BC support a vertical
load P (see figure). The bars are made of steel having a stress-strain
curve that may be idealized as elastoplastic with yield stress Y.
Each bar has cross-sectional area A.
B
Determine the yield load PY and the plastic load PP.
P

Solution 2.12-1 Two bars supporting a load P

A   C

YA
YA


B B

P P

Structure is statically determinate. The yield load PY JOINT B

and the plastic lead PP occur at the same time,
Fvert
0
namely, when both bars reach the yield stress.
(2
Y A) sin 
P
PY
PP
2sY A sin u

Problem 2.12-2 A stepped bar ACB with circular cross sections

is held between rigid supports and loaded by an axial force P at
d1 P d2 B
midlength (see figure). The diameters for the two parts of the bar A C
are d1
20 mm and d2
25 mm, and the material is elastoplastic
with yield stress
Y
250 MPa.
Determine the plastic load PP. L L
— —
2 2
172 CHAPTER 2 Axially Loaded Members

Solution 2.12-2 Bar between rigid supports

d1 d2 B SUBSTITUTE NUMERICAL VALUES:
A C
P
 2
PP
(250 MPa) ¢ ≤ (d1  d2 )
2
4
L L
— —
2 2 

(250 MPa) ¢ ≤ [ (20 mm) 2  (25 mm) 2 ]
d1
20 mm d2
25 mm
Y
250 MPa 4

201 kN
DETERMINE THE PLASTIC LOAD PP:
At the plastic load, all parts of the bar are stressed to
the yield stress.
Point C: P
FAC FCB

FAC

Y A1 FCB

Y A2
P
FAC  FCB
PP
sYA1  sYA2
sY (A1  A2 )

Problem 2.12-3 A horizontal rigid bar AB supporting a load P is hung

from five symmetrically placed wires, each of cross-sectional area A
R
(see figure). The wires are fastened to a curved surface of radius R.
(a) Determine the plastic load PP if the material of the wires is
elastoplastic with yield stress
Y.
(b) How is PP changed if bar AB is flexible instead of rigid?
(c) How is PP changed if the radius R is increased?
A B

Solution 2.12-3 Rigid bar supported by five wires

F F F F F

A B

P
A B F

Y A
P (b) BAR AB IS FLEXIBLE
(a) PLASTIC LOAD PP At the plastic load, each wire is stressed to the yield
stress, so the plastic load is not changed.
At the plastic load, each wire is stressed to the yield
stress. ∴ PP
5sY A (c) RADIUS R IS INCREASED
Again, the forces in the wires are not changed, so the
plastic load is not changed.
 SECTION 2.12 Elastoplastic Analysis 173

Problem 2.12-4 A load P acts on a horizontal beam that is supported

by four rods arranged in the symmetrical pattern shown in the figure.
Each rod has cross-sectional area A and the material is elastoplastic
with yield stress
Y.
Determine the plastic load PP.

 

Solution 2.12-4 Beam supported by four rods

F F F F

 
P
F

Y A
Sum forces in the vertical direction and solve for the
P
load:
At the plastic load, all four rods are stressed to the PP
2F  2F sin 
yield stress.
PP
2sY A (1  sin )

Problem 2.12-5 The symmetric truss ABCDE shown in the figure is 21 in. 54 in. 21 in.
constructed of four bars and supports a load P at joint E. Each of the
two outer bars has a cross-sectional area of 0.307 in.2, and each of A B C D
the two inner bars has an area of 0.601 in.2 The material is elastoplastic
with yield stress
Y
36 ksi.
Determine the plastic load PP. 36 in.

P
174 CHAPTER 2 Axially Loaded Members

Solution 2.12-5 Truss with four bars

21 in. 27 in. 27 in. 21 in. PLASTIC LOAD PP
At the plastic load, all bars are stressed to the yield
A B C D stress.
3

5 4 5 FAE

Y AAE FBE

Y ABE

3 36 in.
6 8
4 PP
s A  s A
5 Y AE 5 Y BE
E SUBSTITUTE NUMERICAL VALUES:
P AAE
0.307 in.2 ABE
0.601 in.2
LAE
60 in. LBE
45 in.
Y
36 ksi
6 8
JOINT E PP
(36 ksi)(0.307 in.2 )  (36 ksi)(0.601 in.2 )
5 5
FBE Equilibrium:
FAE 3 4
13.26 k  34.62 k
47.9 k
2FAE ¢ ≤  2FBE ¢ ≤
P
5 5
E or
P 6 8
P
FAE  FBE
5 5

b b b b
Problem 2.12-6 Five bars, each having a diameter of 10 mm,
support a load P as shown in the figure. Determine the plastic
load PP if the material is elastoplastic with yield stress

Y
250 MPa.
2b

Solution 2.12-6 Truss consisting of five bars

F F F F F At the plastic load, all five bars are
b b b b
stressed to the yield stress
F

Y A
Sum forces in the vertical direction
2b
and solve for the load:
P
1 2
PP
2F ¢ ≤  2F ¢ ≤F
2 5
sY A
d
10 mm P

(52  45  5)
5
d2
4.2031 sY A
A

78.54 mm2
4 Substitute numerical values:

Y
250 MPa PP
(4.2031)(250 MPa)(78.54 mm2)

82.5 kN
 SECTION 2.12 Elastoplastic Analysis 175

Problem 2.12-7 A circular steel rod AB of diameter d
0.60 in. is

stretched tightly between two supports so that initially the tensile stress A B
in the rod is 10 ksi (see figure). An axial force P is then applied to the
rod at an intermediate location C. d
(a) Determine the plastic load PP if the material is elastoplastic
with yield stress
Y
36 ksi. A P B
(b) How is PP changed if the initial tensile stress is doubled to
20 ksi? C

Solution 2.12-7 Bar held between rigid supports

A B POINT C:
P

C
A P
A
d
C
d
0.6 in. 
PP
2sY A
(2)(36 ksi) ¢ ≤ (0.60 in.) 2

Y
36 ksi 4
Initial tensile stress
10 ksi
20.4 k

(a) PLASTIC LOAD PP (B) INITIAL TENSILE STRESS IS DOUBLED

The presence of the initial tensile stress does not PP is not changed.
affect the plastic load. Both parts of the bar must
yield in order to reach the plastic load.

Problem 2.12-8 A rigid bar ACB is supported on a fulcrum at

C and loaded by a force P at end B (see figure). Three identical
wires made of an elastoplastic material (yield stress
Y and
modulus of elasticity E) resist the load P. Each wire has
cross-sectional area A and length L. L
(a) Determine the yield load PY and the corresponding yield A C B
displacement Y at point B.
(b) Determine the plastic load PP and the corresponding L P
displacement P at point B when the load just reaches a a a a
the value PP.
(c) Draw a load-displacement diagram with the load P as
ordinate and the displacement B of point B as abscissa.
176 CHAPTER 2 Axially Loaded Members

Solution 2.12-8 Rigid bar supported by wires

(b) PLASTIC LOAD PP
L
A C B

Y A
Y A

L P
a a a a
A C B

(a) YIELD LOAD PY

Yielding occurs when the most highly stressed wire
Y A PP
reaches the yield stress
Y.
At the plastic load, all wires reach the yield stress.

Y A
2
Y A ©MC
0
4sY A
PP

3
A C B
At point A:
PY L sYL
A
(sY A) ¢ ≤

EA E

Y A
2 At point B:
MC
0 3sY L
B
3A
P

PY
sY A E

At point A: (c) LOAD-DISPLACEMENT DIAGRAM

sY A L sY L P 4
A
¢ ≤¢ ≤
PP
P
2 EA 2E
PP 3 Y
At point B: PY P
2Y
3sYL
B
3A
Y

2E
0 Y P B

Problem 2.12-9 The structure shown in the figure consists of a

horizontal rigid bar ABCD supported by two steel wires, one of length
L and the other of length 3L/4. Both wires have cross-sectional area A L 3L
and are made of elastoplastic material with yield stress
Y and modulus 4
of elasticity E. A vertical load P acts at end D of the bar. A B C D

(a) Determine the yield load PY and the corresponding yield

displacement Y at point D. P
(b) Determine the plastic load PP and the corresponding displacement 2b b b
P at point D when the load just reaches the value PP.
(c) Draw a load-displacement diagram with the load P as ordinate and
the displacement D of point D as abscissa.
 SECTION 2.12 Elastoplastic Analysis 177

Solution 2.12-9 Rigid bar supported by two wires

STRESSES
FB FC
L sB

sC

∴ sC
2sB (7)
3L A A
4
A B C D Wire C has the larger stress. Therefore, it will yield first.

(a) YIELD LOAD

P sC sY
2b b b sC
sY
sB


(From Eq. 7)
2 2
1
A
cross-sectional area FC
sY A
FB
s A
2 Y

Y
yield stress
From Eq. (3):
E
modulus of elasticity
1
2 ¢ sY A ≤  3(sY A)
4P
DISPLACEMENT DIAGRAM 2
A B C D P
PY
sY A
From Eq. (4):
B C D
FBL sY L
B


EA 2E
From Eq. (2):
COMPATIBILITY: sY L
D
Y
2B

3 E
C
B (1)
2
(b) PLASTIC LOAD
D
2B (2) At the plastic load, both wires yield.
FREE-BODY DIAGRAM
B

Y

C FB
FC

Y A
FB FC From Eq. (3):
2(
Y A)  3(
Y A)
4P
A B C D 5
P
PP
sY A
4
P From Eq. (4):
2b b b FBL sY L
B


EA E
EQUILIBRIUM:
From Eq. (2):
©MA
0

FB (2b)  FC (3b)
P(4b)
2sY L
2FB  3FC
4P (3) D
P
2B

E
FORCE-DISPLACEMENT RELATIONS (c) LOAD-DISPLACEMENT DIAGRAM
3
FC ¢ L ≤ P
FBL 4
B

C
(4, 5) PP
EA EA 5
PY PP
P
Substitute into Eq. (1): 4 Y
3FCL 3FBL P
2Y


4EA 2EA
0 Y P D
FC
2FB (6)
178 CHAPTER 2 Axially Loaded Members

Problem 2.12-10 Two cables, each having a length L of approximately 40 m, support a

loaded container of weight W (see figure). The cables, which have effective cross-sectional
area A
48.0 mm2 and effective modulus of elasticity E
160 GPa, are identical except
that one cable is longer than the other when they are hanging separately and unloaded. The
difference in lengths is d
100 mm. The cables are made of steel having an elastoplastic
stress-strain diagram with
Y
500 MPa. Assume that the weight W is initially zero and is
slowly increased by the addition of material to the container. L
(a) Determine the weight WY that first produces yielding of the shorter cable. Also,
determine the corresponding elongation Y of the shorter cable.
(b) Determine the weight WP that produces yielding of both cables. Also, determine the
elongation P of the shorter cable when the weight W just reaches the value WP.
(c) Construct a load-displacement diagram showing the weight W as ordinate and the
elongation  of the shorter cable as abscissa. (Hint: The load displacement diagram
is not a single straight line in the region 0  W  WY.)
W

Solution 2.12-10 Two cables supporting a load

L
40 m A
48.0 mm2 (b) PLASTIC LOAD WP
E
160 GPa F1
sY A
F2
sY A
d
difference in length
100 mm WP
2sY A
48 kN
1 2

Y
500 MPa 2P
elongation of cable 2
L sYL
L
F2 ¢ ≤

0.125 mm
125 mm
INITIAL STRETCHING OF CABLE 1 EA E
Initially, cable 1 supports all of the load. 1P
2P  d
225 mm
Let W1
load required to stretch cable 1 P
1P
225 mm
to the same length as cable 2
EA (c) LOAD-DISPLACEMENT DIAGRAM
W1
d
19.2 kN
L WP
W W 50
1
100 mm (elongation of cable 1 ) (kN)
W1 Ed 40
s1


400 MPa (s1 6 sY ∴ OK)
A L WY
30
(a) YIELD LOAD WY W1
20
Cable 1 yields first. F1

Y A
24 kN
10
1Y
total elongation of cable 1
1 Y P
F1L sYL 0
1Y


0.125 m
125 mm 100 200 300  (mm)
EA E
WY Y
Y
1Y
125 mm
1.5

1.25
W1 1
2Y
elongation of cable 2 WP P

1.667

1.8

1Y  d
25 mm WY Y
EA 0
W
W1: slope
192,000 N/m
F2

4.8 kN
L 2Y
W1
W
WY: slope
384,000 N/m
WY
F1  F2
24 kN  4.8 kN
WY
W
WP: slope
192,000 N/m

28.8 kN
 SECTION 2.12 Elastoplastic Analysis 179

Problem 2.12-11 A hollow circular tube T of length L
15 in. P

is uniformly compressed by a force P acting through a rigid plate
(see figure). The outside and inside diameters of the tube are 3.0
and 2.75 in., repectively. A concentric solid circular bar B of 1.5 in.
diameter is mounted inside the tube. When no load is present, there
is a clearance c
0.010 in. between the bar B and the rigid plate. Both
bar and tube are made of steel having an elastoplastic stress-strain c
diagram with E
29  103 ksi and
Y
36 ksi. T

(a) Determine the yield load PY and the corresponding shortening

Y of the tube. L
T B T B
(b) Determine the plastic load PP and the corresponding shortening
P of the tube.
(c) Construct a load-displacement diagram showing the load P
as ordinate and the shortening  of the tube as abscissa.
(Hint: The load-displacement diagram is not a single straight
line in the region 0  P  PY.)

Solution 2.12-11 Tube and bar supporting a load

Clearance = c

T B T L B

BAR:
d
1.5 in.
d2
L
15 in. AB

1.7671 in.2
4
c
0.010 in.
INITIAL SHORTENING OF TUBE T
E
29  103 ksi
Initially, the tube supports all of the load.

Y
36 ksi
Let P1
load required to close the clearance
TUBE: EAT
P1
c
21,827 lb
d2
3.0 in. L
d1
2.75 in. Let 1
shortening of tube 1
c
0.010 in.
 2 P1
AT
(d  d21 )
1.1290 in.2 s1

19,330 psi (s1 6 sY ∴ OK)
4 2 AT
(Continued)
180 CHAPTER 2 Axially Loaded Members

(a) YIELD LOAD PY (c) LOAD-DISPLACEMENT DIAGRAM

Because the tube and bar are made of the same PP
material, and because the strain in the tube is larger P 100
(kips)
than the strain in the bar, the tube will yield first.
80 PY
FT

Y AT
40,644 lb
TY
shortening of tube at the yield stress 60

FTL sYL 40
TY


0.018621 in.
EAT E P1
20
Y
TY
0.01862 in.
1 Y P
BY
shortening of bar 0
0.01 0.02 0.03  (in.)

TY  c
0.008621 in.
PY Y
EAB
3.21

1.86
FB

29,453 lb P1 1
L BY
PP P
PY
FT  FB
40,644 lb  29,453 lb
1.49

1.54
PY Y

70,097 lb
0
P
P1: slope
2180 k/in.
PY
70,100 lb
P1
P
PY: slope
5600 k/in.
(b) PLASTIC LOAD PP PY
P
PP: slope
3420 k/in.
FT

Y AT FB

Y AB
PP
FT  FB
sY (AT  AB )

104,300 lb
BP
shortening of bar
L sYL

FB ¢ ≤

0.018621 in.
EAB E
TP
BP  c
0.028621 in.
P
TP
0.02862 in.