M E 4 14

M A C H I N E
D E S I G N 1
B Y E N G R . D E N N IS E . G A N A S
 Lecture 4b
SHAFT

B Y E N G R . D E N N IS E . G A N A S ME 414
 STRESSES IN SHAFTS (Torsion)
Example 12: Find the diameter of a solid steel shaft to transmit 20 kW at 200
rpm. The ultimate shear stress for the steel may be taken as 360 MPa and a
factor of safety as 8. If a hollow shaft is to be used in place of the solid shaft,
find the inside and outside diameter when the ratio of inside to outside
diameters is 0.5.
i) For solid shaft: 𝐀𝐧𝐬: 𝐃 = 𝟒𝟕. 𝟔 𝒎𝒎 say 𝟓𝟎𝒎𝒎
ii) For hollow shaft: 𝐀𝐧𝐬: 𝐃 = 𝟒𝟖. 𝟕 𝐦𝐦 say 𝟓𝟎𝐦𝐦, 𝐝 = 𝟎. 𝟐𝟓mm

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 STRESSES IN SHAFTS (Torsion & Bending)

B. SUBJECT TO TORSION AND BENDING LOAD:

i. For solid circular shaft Where: Ss = maximum torsional shear stress
𝟏𝟔 St = maximum tensile or compressive stress
𝑺𝒔 = 𝟑
𝑴𝟐 + 𝑻 𝟐
𝝅𝑫 T = Torsional moment
𝟏𝟔 M = Bending moment
𝑺𝒕 = 𝟑
𝑴 + 𝑴𝟐 + 𝑻 𝟐 D = Shaft diameter or outer shaft diameter
𝝅𝑫
d = inner shaft diameter
ii. For hollow circular shaft k = d/D
𝟏𝟔 Note: Equivalent torque, 𝑻𝒆 = 𝑀2 + 𝑇 2
𝑺𝒔 = 𝑴𝟐 + 𝑻 𝟐
𝝅𝑫𝟑 𝟏 − 𝒌𝟒 Equivalent bending moment, 𝑀𝑒 = 𝑀 + 𝑀2 + 𝑇 2

𝟏𝟔
𝑺𝒕 = 𝑴 + 𝑴𝟐 + 𝑻 𝟐
𝝅𝑫𝟑 𝟏 − 𝒌𝟒

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 STRESSES IN SHAFTS (Torsion & Bending)

Example 13: A solid circular shaft is subjected to a bending moment of 3000 N-m and
a torque of 10000 N-m. The shaft is made of 45 C8 steel having ultimate tensile stress
of 700 MPa and a ultimate shear stress of 500 MPa. Assuming a factor of safety as 6,
determine the diameter of the shaft.
Solution:
i) Using Shear stress, 𝑫𝒔 = 𝟖𝟔. 𝟏𝐦𝐦 ∴Use larger value: Ans: D = 86.1mm
ii) Using Tensile stress, 𝑫𝒕 = 𝟖𝟑. 𝟕𝐦𝐦

Example 14: A shaft is required to transmit a power of 25 kW at 360 rpm. The force
analysis due to attached parts results in bending moment of 830 Nm at a section
between bearings. If permissible stresses in the shaft are: 60 N/mm2 in bending and
40 N/mm2 in shear, calculate the diameter of the shaft.

i) Using Shear stress, 𝑫𝒔 = 𝟓𝟏. 𝟑𝟑 𝐦𝐦

ii) Using Tensile stress, 𝑫𝒕 = 𝟓𝟒. 𝟒 𝐦𝐦 ∴Use larger value: Ans: D = 54.4mm

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 STRESSES IN SHAFTS (Bending)
C. SUBJECT TO BENDING LOAD ONLY:
i. For solid circular shaft
Where: Sb = bending stress
𝟑𝟐𝑴 M = bending moment = W x L
𝑺𝒃 = D = diameter or outer diameter
𝝅𝑫𝟑 d = inner diameter
ii. For hollow circular shaft
𝟑𝟐𝑴
𝑺𝒃 =
𝝅𝑫𝟑 (𝟏 − 𝒌𝟒 )
Example 15: A pair of wheels of a railway wagon carries a load of 50 kN on each axle
box, acting at a distance of 100 mm outside the wheelbase. The gauge of the rails is
1.4 m. Find the diameter of the axle between the wheels, if the bending stress is not to
exceed 100 MPa.
𝑨𝒏𝒔: 𝑫 = 𝟎. 𝟎𝟕𝟗𝟖𝒎 or 𝟕𝟗. 𝟖𝒎𝒎

B Y E N G R . D E N N IS E . G A N A S M E4 14
 STRESSES IN SHAFTS (Torsion, Bending & Axial)
D. SUBJECT TO TORSIONAL, BENDING AND AXIAL LOADS:
𝟐
𝟏𝟔 𝜶𝑭𝒂 𝑫𝒐 (𝟏+𝑲𝟐 𝟏
𝑺𝒔 𝒎𝒂𝒙 = 𝑴+ + 𝑻𝟐
𝝅𝑫𝒐 𝟑 𝟖 𝟏−𝑲𝟒

𝟐
𝟏𝟔 𝜶𝑭𝒂 𝒅𝒐 𝟏 + 𝑲𝟐 𝜶𝑭𝒂 𝒅𝒐 𝟏 + 𝑲𝟐 𝟏
𝑺𝒕𝒎𝒂𝒙 = 𝑴+ + 𝑴+ + 𝑻𝟐
𝝅𝑫𝟑𝒐 𝟖 𝟖 𝟏 − 𝑲𝟒

Where: Ss = maximum shear stress

St = maximum tensile or compressive stress
M = bending moment
Fa = axial tension or compression load
Do= outer diameter
K = ratio of inside to outside diameter of hollow shafts
= 0 for solid shafts
α = ratio of the maximum intensity of stress resulting from the
axial load to the average axial stress
𝟏
α= (for slenderness ratio of ≤115) where Sy = yield stress in compression, psi
𝟏−𝟎.𝟎𝟎𝟒𝟒 𝑳/𝒌
L = length between supporting bearings
k = radius of gyration of the shaft
𝑺𝒚 𝑳 𝟐 n = constant for the type of column end support
α= Euler’s Eqn. (for slenderness ratio of >115)
𝒏𝝅𝟐 𝑬 𝒌 = 1 for hinged ends, 2.25 for fixed ends, and 1.6 for ends
partly restrained, as in bearings
E = modulus of elasticity, psi

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 STRESSES IN SHAFTS (Torsion, Bending & Axial)
Example 16: A hollow steel shaft is to transmit 19.8hp at 250 rpm. The loading
is such that the maximum bending moment is 10,000 in.-lb, the maximum
torsional moment 5,000 in.-lb, and the axial compressive load 4,000 lb. The
shaft is supported on rigid bearings 6 ft apart and is subjected to minor
torsional loads suddenly applied. The maximum allowable shear stress is
3,100 psi. The ratio of the inside diameter to the outside diameter is 0.75. Find
the inner diameter of the shaft.
Given: P = 19.8hp, N = 250 rpm, Mb = 10,000 in.-lb, Mt = 5,000 in.-lb, Fa = 4,000lb, L = 6ft (72in),
Ssmax = 3,100 psi, K = 0.75
Solution:
Since Do is unknown, the values of k and α cannot be determined so it is necessary
to assume a trial value of Do. Note that α ranges from 1 to 2.02 for slenderness ratios
from 0 to 115 so we’ll assume α as 2.00 for which the slenderness ratio is 115.
2
16 𝛼𝐹𝑎 𝐷𝑜 (1 + 𝐾 2 1
𝑆𝑠 𝑚𝑎𝑥 = 3 𝑀+ + 𝑇2
𝜋𝐷𝑜 8 1 − 𝐾4

2
16 2.00 𝐷𝑜 4000(1 + 0.752 ) 1
3100 = 3 10,000 + + 50002
𝜋𝐷𝑜 8 1 − 0.754

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 STRESSES IN SHAFTS (Torsion, Bending & Axial)

Solving the equation results to a trial value of Do = 2.93in. Using this value of
Do , we can solve for α and k.
2
16 𝛼 2.93 (4000)(1 + 0.752 ) 1
3100 = 10,000 + + 50002  𝜶 = 1.76
𝜋(2.93)3 8 1 − 0.754

1 1
α=  1.76 = 1−0.0044 72)/𝑘  𝒌 = 𝟎. 𝟕𝟑
1−0.0044 𝐿/𝑘

Using this value of α = 1.76, we can solve for the real value of Do .
2
16 1.76𝐷𝑜 (4000)(1 + 0.752 ) 1
3100 = 10,000 + + 50002
𝜋𝐷𝑜 3 8 1 − 0.754

𝟏𝟓
Do = 2.94 in or 𝟐 𝟏𝟔 in Answer:
15
Since K = 0.75 and Do= 2 16 in. then
𝟑
Di = 𝟐. 𝟐𝟎𝟓𝐢𝐧, 𝒐𝒓 𝒖𝒔𝒆 𝟐 𝟏𝟔 in

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 STRENGTH OF SHAFT (PSME CODE P.18)
𝑫𝟑 𝑵 𝟑 𝟖𝟎𝑷
A. FOR MAIN POWER TRANSMITTING SHAFTS: 𝑷= or 𝑫 =
𝟖𝟎 𝑵
Example 17: Determine the power transmitted by a main power transmitting steel shaft
with 2 7/8 inches in outside diameter, SAE 1040 driving conveyor head pulleys at a shaft
speed of 150 rpm.
𝟕 𝟑
𝑫𝟑 𝑵 𝟐
𝟖
(𝟏𝟓𝟎)
𝑷=
𝟖𝟎  𝑷= 𝟖𝟎
= 𝟒𝟒. 𝟔 𝑯𝒑
𝑫𝟑 𝑵 𝟑 𝟓𝟑.𝟓𝑷
B. FOR LINESHAFT CARRYING PULLEYS: 𝑷= or 𝑫 =
𝟓𝟑.𝟓 𝑵

Example 18: Compute the power in hp of a lineshaft having a diameter of 1 7/8 inch with
a speed of 200 rpm.
𝟕 𝟑
𝟏 𝟐𝟎𝟎
𝟖
𝑷= = 𝟐𝟒. 𝟔 𝐇𝐩
𝟓𝟑. 𝟓
𝑫𝟑 𝑵 𝟑 𝟑𝟖𝑷
C. FOR SMALL SHORT SHAFTS : 𝑷= or 𝑫 =
𝟑𝟖 𝑵

Example 19: A short 61mm shaft transmit 120 Hp. Compute the linear speed of a pulley
55cm mounted on the shaft.
(𝟐. 𝟒)𝟑 𝑵
𝟏𝟐𝟎 =  𝑵 = 𝟑𝟑𝟎 𝒓𝒑𝒎
𝟑𝟖

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 CRITICAL SPEED OF SHAFTS
− 𝒕𝒉𝒆 𝒔𝒑𝒆𝒆𝒅 𝒂𝒕 𝒘𝒉𝒊𝒄𝒉 𝒂 𝒔𝒉𝒂𝒇𝒕 𝒃𝒆𝒄𝒐𝒎𝒆𝒔 𝒅𝒚𝒏𝒂𝒎𝒊𝒄𝒂𝒍𝒍𝒚 𝒖𝒏𝒔𝒕𝒂𝒃𝒍𝒆

Relation between deflection in any shaft and the force producing that deflection:
𝑭𝑳𝟑 Where: F = Force producing the deflection, lb.
𝒚=𝑪 C = constant depending on type of beam& method of loading
𝑬𝑰
I = rectangular moment of inertia, in4
y = deflection, in.
Critical Speed of Shafts M = bending moment, in.-lb
E = modulus of elasticity, psi
𝑬𝑰 g = acceleration due to gravity, in/s2
𝝎𝒄𝒓 =
𝑪𝑴𝑳𝟑

i. For a central disk of mass M of negligible weight, rotating on antifriction

bearings, and shaft is simply supported:

𝟒𝟖𝑬𝑰𝒈 𝑭𝑳𝟑
𝝎𝒄𝒓 = and 𝒚 =
𝑳𝟑 𝑾 𝟒𝟖𝑬𝑰

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 CRITICAL SPEED OF SHAFTS
ii. For a central disk of mass M of negligible weight, rotating on sleeve bearings,
and shaft is fixed-ended:
𝑭𝑳𝟑 𝟏𝟗𝟐 𝑬𝑰𝒈
𝒚= and 𝝎𝒄𝒓 =
𝟏𝟗𝟐𝑬𝑰 𝑾𝑳𝟑

iii. For non-central disk of mass M at a distance a from left support and at a
distance b from right support, shaft of negligible weight, rotating in rolling-
element bearings:
𝒂𝟐 𝒃𝟐 𝑭 𝟑𝑬𝑰𝑳𝒈
𝒚= and 𝝎𝒄𝒓 = 𝑾𝒂𝟐 𝒃𝟐
𝟑𝑬𝑰𝑳
iv. For same conditions as (iii) except using sleeve bearings:

𝒂𝟐 𝒃𝟐 𝑭 𝟑𝑬𝑰𝑳𝟑 𝒈
𝒚= and 𝝎𝒄𝒓 =
𝟑𝑬𝑰𝑳𝟑 𝑾𝒂𝟐 𝒃𝟐

v. For a shaft with uniform load or supports no load except its own weight:

𝝅𝟒 𝑬𝑰𝒈 𝟏. 𝟐𝟕𝟓𝒈
𝝎𝒄𝒓 = =
𝑾𝑳𝟑 𝒚

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 CRITICAL SPEED OF SHAFTS
Critical Speed in terms of weight and deflection:
i. For a shaft with single disk:
𝟔𝟎 𝟑𝟎 𝑭 𝟑𝟎 𝒈
𝑵𝒄𝒓 = 𝝎𝒄𝒓 = =
𝟐𝝅 𝝅 𝑴𝒚 𝝅 𝒚

ii. For a shaft supporting n disk:

𝟔𝟎 𝟑𝟎 𝒈(𝑾𝟏𝒚𝟏 + 𝑾𝟐𝒚𝟐 + ⋯ . +𝑾𝒏𝒚𝒏) Where: W = weights of disks
𝑵𝒄𝒓 = 𝝎 = y = deflection of disks
𝟐𝝅 𝒄𝒓 𝝅 𝑾𝟏𝒚𝟐𝟏 + 𝑾𝟐𝒚𝟐𝟐 + ⋯ . +𝑾𝒏𝒚𝟐𝒏

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 CRITICAL SPEED OF SHAFTS
Example 20: The shaft shown below has attached to it a gear m1, weighing
50 lb. and a flywheel m2 weighing 100 lb. Static deflections y1 and y2 have
been found to be 0.0012 in. and 0.0003 in. respectively. Determine the first
critical speed, ignoring the mass of the shaft itself.
60 30 𝑔(𝑊1𝑦1 + 𝑊2𝑦2 + ⋯ . +𝑊𝑛𝑦𝑛)
𝑁𝑐𝑟 = 𝜔𝑐𝑟 =
2𝜋 𝜋 𝑊1𝑦12 + 𝑊2𝑦22 + ⋯ . +𝑊𝑛𝑦𝑛2
𝑖𝑛
30 386 (50(0.0012) + (100)(0.0003)𝑙𝑏 ⋅ 𝑖𝑛)
𝑠2
𝑁𝑐𝑟 =
𝜋 50 0.00122 + 100 0.00032 𝑙𝑏 ⋅ 𝑖𝑛2

𝑵𝒄𝒓 = 𝟔, 𝟐𝟓𝟒 𝒓𝒑𝒎

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REMINDER:

MARCH 18, 2024

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