Shaft Design Tutorial

(Tutorial # 1)
Example-2
A 600 mm diameter pulley driven by a horizontal belt transmits power through a
solid shaft to a 260 mm diameter pinion, which drives a mating gear. The pulley
weighs 1200 N. The pinion has a pressure angle of 20o. The arrangement of
elements, the belt tensions, and the components of the gear reactions on the
pinion are indicated in the below Figure. Both pulley and pinion is keyed to the
shaft. The shaft material is cold drawn AISI 1040. It is considered there is a minor
shock during the operation when load applied. Design the necessary diameter
along the shaft. Refer to the data given in the figure below.

Fr

Ft
Solution
Given data
Shaft material: Cold drawn ISI 1040.
Material properties: Su = 552 MPa and Sy = 490 MPa
Pulley diameter: DP = 600 mm and RP = 300 mm
Pinion diameter: dP = 260 mm and rP = 130 mm
Pulley weight: WP = 1200 N
Pulley force: T1 = 5000 N and T2 = 1500 N

Pinion : α = 20o

Load factor for minor shock: kb = 2.0 and kt = 1.5

 External torque:
Mt pulley Mt pinion Support bearings: at O and B
A B Pulley: at A
O C
Pinion: at C
250 mm 500 mm 200 mm Σ Mt = 0, Mt pulley = Mt pinion
= (T1 – T2)RP
= (5000–1500)(300) N.mm
= 1050000 N.mm

Internal torque in the shaft:

Segment OA : Mt OA = 0, no internal torque developed at this segment
Segment ABC : Mt ABC = Mt pulley = 1050000 N.mm = 1050 N.m

Tangential and radial force developed by pinion

Tangential force: Ft = Mt pulley/ rP = 1050000/130 = 8077 N

Radial force: Fr = Ft tanα = (8077)tan20o = 2940 N

 Bending moment diagrams

Vertical plane (zy-Plane) Horizontal plane (zx-Plane)

1200 N T = (T1 +T2)= 6500 N

A B A B
O C O C
R2y R2x
R1y R1x
250 mm 500 mm 200 mm Ft 250 mm 500 mm 200 mm Fr

∑ MO = 0 : ∑ MO = 0 :
Ft × 950 − R2 y × 750 − 1200 × 250 = 0
− Fr × 950 + R2 x × 750 − 6500x 250 = 0
∴ R2 y = 9831 N ∴ R2 x = 5890 N

∑ Fy = 0 : ∑ Fx = 0 :
R1 y − 1200 − 9831 + Ft = 0 R1x − 6500 + 5890 − Fr = 0
∴ R1 y = 2954 N ∴ R1x = 3550 N
 Vertical plane (zy-Plane) Horizontal plane (zx-Plane)
1200 N T = (T1 +T2)= 6500 N

A B A B
O C O C
R2y R2x
R1y R1x
0.25m 0.5 m 0.2 m Ft 0.25 m 0.5 m 0.2 m Fr

M 1615.4 N.m M 887.5N.m

B C
738.5 N.m O A
– 588 N.m
O A B C

MA’ = R1y(0.25) = (2954)(0.25) = 738.5 N.m MA’’ = R1x(0.25) =(3550)(0.25) = 887.5 N.m
MB’ = Ft(0.20) = (8077)(0.20) = 1615.4 N.m MB’’ = Fr(0.2) =( –2940)(0.2) = – 588 N.m

Bending moment at O (support): MO = 0

Bending moment at A (pulley): M A = 738.52 + 887.52 = 1154.6 N.m
Bending moment at B (support): M B = 1615.42 + ( −588) 2 = 1719 N.m
Bending moment at D (pinion): MD = 0
Shaft design A B
O C

C
O
dO dA A dAB dB B dC

250 mm 500 mm 200 mm

dO = the diameter of the segment shaft where the bearing support O is located
dA = the diameter of the segment shaft where the pulley is located
dB = the diameter of the segment shaft where the bearing support B is located

dC = the diameter of the segment shaft where the pinion is located

Solving the shaft diameter using Maximum Stress Theory
16
d3 = (kb M b )2 + (kt M t )2
πσ s
Allowable shear stress, σs
From the shaft material properties: σ s = 0.18 × Su = 0.18 × 552 = 99.36 MPa
= 0.3 × S y = 0.3 × 490 = 147 MPa
The shaft diameter dO:
The support bearing O is fixed to the shaft, we use σ s = 99.36 MPa
Bending moment and torque: Mb = MO = 0 and Mt = 0
Thus, the diameter is designed based on the analysis of bearing which will be
studied in Machine Elements Design 2 (next semester).

The shaft diameter dA:

The pulley A is keyed to the shaft, we use σ s = 0.75x 99.36 MPa = 74.52 MPa
Bending moment and torque: Mb = MA = 1154.56 N.m and Mt = Mt ABC = 1050 N.m
1
 16 2 3
dA =  6
(2 . 0 × 1154. 56 )2
+ (1.5 × 1050 ) 
π × 74.52 × 10 
⇒ d A = 0.0576 m = 57.6 mm
Thus, use a standard shaft of dA = 60 mm
The shaft diameter dB:
The bearing support B is fixed to the shaft, we use σ s = 99.36 MPa
Bending moment and torque: Mb = MC = 1719 N.m and Mt = Mt ABC = 1050 N.m
1
 16 2 3
dB =  6
(2 .0 × 1719 )2
+ (1.5 × 1050 ) 
π × 99.36 × 10 
⇒ d B = 0.0579 m = 57.9 mm
Thus, use a standard shaft of dB = 60 mm

The shaft diameter dC:

The pinion C is keyed to the shaft, we use σ s = 0.75x 99.36 MPa = 74.52 MPa
Bending moment and torque: Mb = MD = 0 and Mt = Mt ABC = 1050 N.m

1
 16 2 3
dC =  6
0 + (1.5 × 1050) 
π × 74.52 × 10 
⇒ d C = 0.0475 m = 47.5 mm
Thus, use a standard shaft of dC = 50 mm
 C
O
dO dA A dAB dB B dC

250 mm 500 mm 200 mm

Results:
dO = determined by bearing analysis
dA = 60 mm
dB = 60 mm
dC = 50 mm
NOTE: diameter of segment AB should be made dAB > 60 mm
Solving the shaft diameter using Distortion Energy Theory

Endurance strength (S‘e )

For cold drawn AISI 1040 having

the tensile strength of Su = 552
MPa, the estimated endurance
strength is

S‘e = 210 MPa

Actual endurance strength (Se ):
’
Se = kb ke Se’
The value of size factor is estimated as kb = 0.75
For the desired reliability of 0.99, from the related Table we have the value of
approximated reliability factor is ke = 0.81
Thus, Se = 210x0.75x0.81 = 127.6 MPa
1/3
 2
 
2

 32nd  K t M b  3 M
Shaft diameter: d =     +  t  
 π   S e  4  S y  
   
Considering the shaft undergoes reversed bending and steady torsion with a minor
shock during operation, we apply the design factor or safety factor nd =2.0.

The shaft diameter dO:

The support bearing O is fixed to the shaft, so no stress concentration or Kb = 1

Bending moment and torque: Mb = MO = 0 and Mt = 0

Thus, the diameter is designed based on the analysis of bearing which will be
studied in Machine Elements Design 2 (next semester).
The shaft diameter dA:
The pulley A is keyed to the shaft with the type of profile keyseat, so we consider
the stress concentration factor is Kb = 2.0
Bending moment and torque: Mb = MA = 1154.56 N.m and Mb = Mb ABC = 1050 N.m
1/3
 32x2.0 3  1050  
2 2
   2.0x1154.56 
Thus, d A =    6 
+  6

 π   127.6x10  4  490x10  
 
dA = 0.0718 m = 71.8 mm
Thus, use a standard shaft of dA = 75 mm

The shaft diameter dB:

The support bearing B is fixed to the shaft, but there is a well-rounded fillet at
section B, so we consider the stress concentration factor is Kb = 1.5
Bending moment and torque: Mb = MB = 1719 N.m and Mt = MtABC = 1050 N.m
1/3
 32x2.0 3  1050  
2 2
   1.5x1719 
Thus, d B =    6
+  6

 π   127.5x10  4  490x10  
 
dB = 0.0745 m = 74.5 mm
Thus, use a standard shaft of dB = 80 mm
The shaft diameter dC:
The pinion C is keyed to the shaft with the type of profile keyseat, so we consider
the stress concentration factor is Kb = 2.0
Bending moment and torque: Mb = MC = 0 and Mt = Mt ABC = 1050 N.m
1/3
 32x2.0 3  1050  
2
  
Thus, d C =   0+  6
 π  4  490x10  
 
dC = 0.0336 m = 33.6 mm

Thus, use a standard shaft of dC = 36 mm

 C
O
dO dA A dAB dB B dC

250 mm 500 mm 200 mm

Results from maximum shear theory Results from distorsion energy theory

dO = determined by bearing analysis dO = determined by bearing analysis

dA = 60 mm dA = 75 mm
dB = 60 mm dB = 80 mm
dC = 50 mm dC = 36 mm

NOTE: diameter of segment AB should NOTE: diameter of segment AB should

be made dAB > 60 mm be made dAB > 80 mm
Example-3

The Figure below shows an arrangement for a motor and exciter with a pinion on
the same shaft. The pinion drives a gear with the gear directly below the pinion.
The motor develops 55 kW at 200 rpm. The exciter absorbs 5 kW, the remainder
going to the pinion. The motor and the exciter are assembled to the shaft by
means of a force fit while the pinion is keyed to the shaft. What is the required
diameters of the shaft, if the shaft material is made of hot rolled AISI 1050
having an ultimate strength of 620 MPa and a yield point of 338 MPa. Use kb =
kt = 1.5, the pressure angle of the gear is 20°, and the desired reliability is 0.99.
*Given:

Power: @ Motor , PMR = 55 kW @ N = 200 r.p.m.

@ Exciter, PER = 5 kW
@ Pinion, PP = 50 kW
Shaft Material:
Su = 620 MPa.
Sy = 338 MPa.
Pinion Pressure angle (α) = 20°.

Load factors: kb = kt = 1.5

*Solution
External torques:
The torque @ Pinion is calculated by:

9550 × Power in kW 9550 × 50 ⇒ M t P = 2387.5 N.m.

MtP = =
Speed in r.p.m. 200

The torque @ the excited rotor is:

9550 × Power in kW 9550 × 5 ⇒ M t ER = 238.75 N.m.
M t ER = =
Speed in r.p.m. 200

The torque @ the motor rotor is:

9550 × Power in kW 9550 × 55 ⇒ M t MR = 2626.25 N.m.
M t MR = =
Speed in r.p.m. 200
 Internal torque in each segment of the shaft

y
Segment OAB:
MtMR Mt ER Mt P
O C D Mt OAB = Mt MR = 2626.25 N.m
A B
0,5 m 0,5 m 0,5 m 0,25 m
Segment BCD:
Mt BCD = Mt P = 2387.5 N.m

Tangential and radial force developed by pinion

For the pinion  Mt P = Ft × Rg , ( )

where Rg = 200 × 10-3 2 = 0.1 m

The tangential force: ⇒ Ft = Mt P Rg = 2387.5 0.1 = 23875 N .

The radial force: Fr = Ft x tanα = 23875 x tan(20°) = 8690 N.

 Constructing the bending moment diagrams to know the location of
the maximum bending moment.

Vertical plane (zy-Plane) Horizontal plane (zx-Plane)

16 kN 4 kN y x

O C D O C D
A B
0,5 m 0,5 m 0,5 m 0,25 m 1,5 m 0,25 m

R1y R2y Fr R1x R2x Ft

∑ MO = 0 : ∑ MO = 0 :
Fr × 1.75 − R2 y × 1.5 − 4 × 1 − 16 × 0.5 = 0 Ft × 1.75 − R2 x × 1.5 = 0
∴ R2 y = 2.138 kN ∴ R2 x = 27.854 kN

∑ Fy = 0 : ∑ Fx = 0 :
R1 y − 16 − 4 − 2.138 + 8.69 = 0 R1x − 27.854 + 23.875 = 0
∴ R1 y = 13.448 kN ∴ R1x = 3.979 kN
 16 kN 4 kN y x

O C D O C D
A B
0,5 m 0,5 m 0,5 m 0,25 m 1,5 m 0,25 m

R1y R2y Fr R1x R2x Ft

M 6.724 kN.m M
5.968 kN.m
5.448 kN.m
3.979 kN.m

2.172 kN.m 1.989 kN.m

O A B C D O A B C D

MA’ = R1y(0.5) = 6.724 kN.m MA’’ = R1x(0.5) = 1.989 kN.m

MB’ = R1y(1.0) – 16(0.5) = 5.448 kN.m MB’’ = R1x(1.0) = 3.979 kN.m
MC’ = Fr(0.25) = 2.172 kN.m MC’’ = Ft(0.25) = 5.968 kN.m

Bending moment at O (support): MO = 0

Bending moment at A (motor rotor): M A = 6.7242 + 1.9892 = 7.012 kN.m = 7012 N.m
Bending moment at B (excited rotor): M B = 5.4482 + 3.9792 = 6.746 kN.m = 6746 N.m
Bending moment at C (support: M C = 2.1722 + 5.9682 = 6.351 kN.m = 6351 N.m
Bending moment at D (pinion): MD = 0
Shaft design

O D
dO dA A d B dC C
AB dB dD

0,5 m 0,5 m 0,5 m 0,25 m

dO = the diameter of the segment shaft where the bearing support O is located
dA = the diameter of the segment shaft where the motor rotor is located

dB = the diameter of the segment shaft where the excited rotor is located

dC = the diameter of the segment shaft where the bearing support C is located

dD = the diameter of the segment shaft where the pinion is located

Solving the shaft diameter using Maximum Stress Theory
16
d3 = (kb M b )2 + (kt M t )2
πσ s
Allowable shear stress, σs
From the shaft material properties: σ s = 0.18 × Su = 0.18 × 620 = 111.6 MPa
= 0.3 × S y = 0.3 × 338 = 101.4 MPa
The shaft diameter dO:
The support bearing O is fixed to the shaft, we use σ s = 101.4 MPa
Bending moment and torque: Mb = MO = 0 and Mt = 0
Thus, the diameter is designed based on the analysis of bearing which will be
studied in Machine Elements Design 2 (next semester).

The shaft diameter dA:

The motor rotor A is fixed to the shaft, we use σ s = 101.4 MPa
Bending moment and torque: Mb = MA = 7012 N.m and Mt = Mt OAB = 2626.25 N.m
1
 16 2 3
dA =  6
(1 . 5 × 7012 )2
+ (1.5 × 2626.25) 
π × 101.4 × 10 
⇒ d A = 0.0826 m = 82.6 mm
Thus, use a standard shaft of dA = 84 mm
The shaft diameter dB:
The excited rotor B is fixed to the shaft, we use σ s = 101.4 MPa
Bending moment and torque: Mb = MB = 6746 N.m and Mt = Mt BCD = 2387.5 N.m
1
 16 2 3
dB =  6
(1 .5 × 6746 )2
+ (1.5 × 2387.5) 
π × 101.4 × 10 
⇒ d B = 0.0814 m = 81.4 mm
Thus, use a standard shaft of dB = 82 mm

The shaft diameter dC:

The bearing support C is fixed to the shaft, we use σ s = 101.4 MPa
Bending moment and torque: Mb = MC = 6351 N.m and Mt = Mt BCD = 2387.5 N.m
1
 16 2 3
dC =  6
(1 .5 × 6351)2
+ (1.5 × 2387.5) 
π × 101.4 × 10 
⇒ d C = 0.0799 m = 79.9 mm
Thus, use a standard shaft of dC = 80 mm
 The shaft diameter dD:
The pinion Dis keyed to the shaft, we use σ s = 0.75x101.4 MPa = 76.05 MPa
Bending moment and torque: Mb = MD = 0 and Mt = Mt BCD = 2387.5 N.m

1
 16 2 3
dD =  6
0 + (1.5 × 2387.5) 
π × 76.05 × 10 
⇒ d D = 0.0621 m = 62.1 mm
Thus, use a standard shaft of dD = 64 mm

Results:
O dO = determined by bearing analysis
A d B D
dO dA AB dB dC C dD dA = 84 mm
dB = 82 mm
0,5 m 0,5 m 0,5 m 0,25 m dC = 80 mm
dD = 64 mm

NOTE: diameter of segment AB should be made dAB > 84 mm

Solving the shaft diameter using Distortion Energy Theory

Endurance strength (S‘e )

For hot rolled AISI 1050 having

the tensile strength of Su = 620
MPa, the estimated endurance
strength is

S‘e = 180 MPa

Actual endurance strength (Se ): Se = kb ke Se’
The value of size factor is estimated as kb = 0.75
For the desired reliability of 0.99, from the related Table we have the value of
approximated reliability factor is ke = 0.81
Thus, Se = 180x0.75x0.81 = 109.35 MPa
1/3
 2
 
2

 32nd  K b M b  3 M
Shaft diameter: d =     +  t  
 π   S e  4  S y  
   
n
Considering the shaft undergoes reversedd bending and steady torsion, we apply
the design factor or safety factor nd =1.5.

The shaft diameter dO:

The support bearing O is fixed to the shaft, so no stress concentration or Kb = 1

Bending moment and torque: M b = MO = 0 and Mt = 0

Thus, the diameter is designed based on the analysis of bearing which will be
studied in Machine Elements Design 2 (next semester).
The shaft diameter dA:
The motor rotor A is fixed to the shaft, but there is a well-rounded fillet at section
A, so we consider the stress concentration factor is Kb = 1.5
Bending moment and torque: Mb = MA = 7012 N.m and Mt = Mt OAB = 2626.25 N.m
1/3
 32x1.5 2 2
   1.5x7012  3  2626.25  
Thus, d A =    6
+  6
 π   109.35x10  4  338x10  
 
dA = 0.1138 m = 113.8 mm
Thus, use a standard shaft of dA = 115 mm

The shaft diameter dB:

The excited rotor B is fixed to the shaft, but there is a well-rounded fillet at section
A, so we consider the stress concentration factor is Kb = 1.5
Bending moment and torque: Mb = MB = 6746 N.m and Mt = Mt BCD = 2387.5 N.m
1/3
 32x1.5 1.5x6746  3  2387.5  
2 2
   
Thus, d B =    6
+  6
 π   109.35x10  4  338x10  
 
dB = 0.1123 m = 112.4 mm

Thus, use a standard shaft of dB = 115 mm

The shaft diameter dC:
The support bearing C is fixed to the shaft, but there is a well-rounded fillet at
section C, so we consider the stress concentration factor is Kb = 1.5
Bending moment and torque: Mb = MC = 6351 N.m and Mt = Mt BCD = 2387.5 N.m
1/3
 32x1.5 2 2
   1.5x6351  3  2387.5 

Thus, d C =    6  +  6
 π   109.35x10  4  338x10  
 
dC = 0.110 m = 110 mm
Thus, use a standard shaft of dC = 110 mm

The shaft diameter dD:

The pinion D is keyed to the shaft with the type of profile keyseat, so we consider
the stress concentration factor is Kb = 2.0
Bending moment and torque: Mb = MD = 0 and Mt = Mt BCD = 2387.5 N.m
1/3
 32x1.5 3  2387.5  
2
  
Thus, d D =   0+  6
 π  4  338x10  
 
dD = 0.0454 m = 45.4 mm

Thus, use a standard shaft of dD = 48 mm

 O D
dO dA A d B dC C
AB dB dD

0,5 m 0,5 m 0,5 m 0,25 m

Results from maximum shear theory Results from distorsion energy theory
dO = determined by bearing analysis
dO = determined by bearing analysis
dA = 115 mm
dA = 84 mm
dB = 115 mm
dB = 82 mm
dC = 110 mm
dC = 80 mm
dD = 48 mm
dD = 64 mm

NOTE: diameter of segment AB should NOTE: diameter of segment AB should

be made dAB > 84 mm be made dAB > 115 mm