Chapter 3 – Torsion
Torsion
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TORSIONAL SHEARING STRESS, τ
For a solid or hollow circular shaft subject to a twisting moment T, the
torsional shearing stress τ at a distance ρ from the center of the shaft is
angle of twist
𝜏=
power transmitted by a shaft
𝑇𝜌
𝐽
and
𝜏𝑚𝑎𝑥 =
𝑇𝑟
𝐽
torsion
torsional shearing stress
where J is the polar moment of inertia of the section and r is the outer radius.
TORSION
Consider a bar to be rigidly attached at one end and twisted at the other end
by a torque or twisting moment T equivalent to F × d, which is applied
perpendicular to the axis of the bar, as shown in the figure. Such a bar is said
to be in torsion.
For solid cylindrical shaft:
𝐽=
𝜋
32
𝐷4
For hollow cylindrical shaft:
𝐽=
𝜋
32
𝜏𝑚𝑎𝑥
16𝑇
𝜏𝑚𝑎𝑥 = 𝜋𝐷3
(𝐷 4 − 𝑑 4 )
16𝑇𝐷
𝜋 (𝐷 4 − 𝑑 4 )
ANGLE OF TWIST
The angle θ through which the bar length L will twist is
𝜃=
𝑇𝐿
𝐽𝐺
Problem 305
What is the minimum diameter of a solid steel shaft that will not twist through
more than 3° in a 6-m length when subjected to a torque of 12 kN·m? What
maximum shearing stress is developed? Use G = 83 GPa.
in radians
Solution
𝜃=
where T is the torque in N·mm, L is the length of shaft in mm, G is shear
modulus in MPa, J is the polar moment of inertia in mm 4, D and d are
diameter in mm, and r is the radius in mm.
𝑇𝐿
𝐽𝐺
3𝑜 (
𝜋
)=
180𝑜
12(6)(10003 )
1
𝜋𝑑 4 (83000)
32
𝑑 = 113.98 𝑚𝑚
POWER TRANSMITTED BY THE SHAFT
A shaft rotating with a constant angular velocity ω (in radians per second) is
16𝑇 16(12)(10003 )
=
= 41.27 𝑀𝑃𝑎
𝜋𝑑 3
𝜋(113.983 )
being acted by a twisting moment T. The power transmitted by the shaft is
𝜏𝑚𝑎𝑥 =
𝑷 = 𝑻𝝎 = 𝟐𝝅𝑻𝒇
Problem 306
A steel marine propeller shaft 14 in. in diameter and 18 ft long is used to
where T is the torque in N·m, f is the number of revolutions per second, and
P is the power in watts.
Problem 304
A steel shaft 3 ft long that has a diameter of 4 in is subjected to a torque of
15 kip·ft. Determine the maximum shearing stress and the angle of twist. Use
G = 12 × 106 psi.
Solution 304
16𝑇 16(15)(1000)(12)
𝜏𝑚𝑎𝑥 =
=
= 14324 𝑝𝑠𝑖 = 14.3 𝑘𝑠𝑖
𝜋𝐷3
𝜋(43 )
𝜃=
𝑇𝐿 15(3)(1000)(122 )
= 1
= 0.0215 𝑟𝑎𝑑 = 1.23𝑜
4
6
𝐽𝐺
𝜋(4 )(12 𝑥 10 )
32
transmit 5000 hp at 189 rpm. If G = 12 × 106 psi, determine the maximum
shearing stress.
Solution 306
𝑃
5000(396000)
𝑇=
=
= 1667337.5 𝑙𝑏 ∙ 𝑖𝑛
2𝜋𝑓
2𝜋(189)
𝜏𝑚𝑎𝑥 =
16𝑇 16(1667337.5)
=
= 3094.6 𝑝𝑠𝑖
𝜋𝑑 3
16(14)3
Torsion of thin-walled tube
Solution 337
𝑇 = 2𝐴𝑡𝜏
The torque applied to thin-walled tubes is expressed as
Where: 𝑇 = 600 𝑁 ∙ 𝑚, 𝐴 = 30(80) = 2400 𝑚𝑚2 , 𝜏 = 80 𝑀𝑃𝑎
600000 𝑁 ∙ 𝑚𝑚 = 2(2400 𝑚𝑚2 (𝑡) (80
𝑁
)
𝑚𝑚2
𝑡 = 1.5625 𝑚𝑚
where T is the torque in N·mm, A is the area enclosed by the center line of
the tube (as shown in the stripe-filled portion) in mm 2, and q is the shear flow
in N/mm.
At any convenient center O within the section, the farthest side is the
shortest side, thus, it is induced with the maximum allowable shear stress of
80 MPa.
The average shearing stress across any thickness t is:
𝑞
𝑇
𝜏= =
𝑡 2𝐴𝑡
Problem 338
A tube 0.10 in. thick has an elliptical shape shown in Fig. P-338. What torque
will cause a shearing stress of 8000 psi?
Thus, torque T can also be expressed as:
𝑇 = 2𝐴𝑡𝜏
Problem 337
A torque of 600 N·m is applied to the rectangular section shown in Fig. P-337.
Determine the wall thickness t so as not to exceed a shear stress of 80 MPa.
What is the shear stress in the short sides? Neglect stress concentration at
So
the corners.
Where: 𝐴
= 𝜋𝑎𝑏 = 𝜋(3)(1.5) = 4.5𝜋 𝑖𝑛2
𝑇 = 2𝐴𝑡𝜏
𝑡 = 0.10 𝑖𝑛, 𝜏 = 8000 𝑝𝑠𝑖
𝑇 = 2(4.5𝜋)(0.10)(8000) = 22619.47 𝑙𝑏 ∙ 𝑖𝑛
𝑇 = 22.62 𝑘𝑖𝑝 ∙ 𝑖𝑛 answer
Problem 339
Problem 340
A torque of 450 lb·ft is applied to the square section shown in Fig. P-339.
A tube 2 mm thick has the shape shown in Fig. P-340. Find the shearing
Determine the smallest permissible dimension a if the shearing stress is
limited to 6000 psi.
stress caused by a torque of 600 N·m.
𝑇 = 2𝐴𝑡𝜏
𝑇 = 2𝐴𝑡𝜏
𝑇 = 450 𝑙𝑏 ∙ 𝑓𝑡 = 450 (12)𝑙𝑏 ∙ 𝑖𝑛 = 5400 𝑙𝑏 ∙ 𝑖𝑛
𝐴 = 𝜋(10)2 + 80(20) = 1914.16 𝑚𝑚2
𝐴 = 𝑎2
600 000 𝑁 ∙ 𝑚𝑚 = 2 (1914.16 𝑚𝑚2 )(2 𝑚𝑚)𝜏
Solution:
𝜏 = 78.36 𝑀𝑃𝑎 answer
5400 𝑙𝑏 ∙ 𝑖𝑛 = 2 (𝑎2 )(0.10 𝑖𝑛) (6000
𝑎 = 2.12 𝑖𝑛 answer
𝑙𝑏
)
𝑖𝑛2
Flanged bolt couplings
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flanged bolt coupling
For rigid flanges, the shear deformations in the bolts are proportional to their
radial distances from the shaft axis. The shearing strains are related by
𝛾1 𝛾2
=
𝑅1 𝑅2
torque
torsional shearing stress
In shaft connection called flanged bolt couplings (see figure), the torque is
transmitted by the shearing force P created in he bolts that is assumed to be
Using Hooke's law for shear,𝐺
=
𝜏
𝛾
, we have
𝑃1
uniformly distributed. For any number of bolts n, the torque capacity of the
coupling is
𝑃2
𝜏1
𝜏2
𝐴1
𝐴2
=
𝑜𝑟
=
𝐺1 𝑅1 𝐺2 𝑅2
𝐺1 𝑅1
𝐺2 𝑅2
If the bolts on the two circles have the same area, 𝐴1 = 𝐴2 , and if the bolts
are made of the same material,𝐺1 = 𝐺2 , the relation between 𝑃1 and 𝑃2
reduces to:
𝑃1
𝑃2
=
𝑅1
𝑅2
𝜋𝑑2
𝑇 = 𝑃𝑅𝑛 =
𝜏𝑅𝑛
4
If a coupling has two concentric rows of bolts, the torque capacity is
𝑇 = 𝑃1 𝑅1 𝑛1 + 𝑃2 𝑅2 𝑛2
Where:
the subscript 1 refer to bolts
on the outer circle and
subscript 2 refer to bolts on
the inner circle. See figure