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Torsion_Strength of Materials, Papers of Law of Torts

Strength of Materials: Lesson about torsion

Typology: Papers

2023/2024

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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.
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
where J is the polar moment of inertia of the section and r is the outer radius.
For solid cylindrical shaft:
For hollow cylindrical shaft:
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff
pf12
pf13
pf14
pf15

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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.

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

where J is the polar moment of inertia of the section and r is the outer radius.

For solid cylindrical shaft:

For hollow cylindrical shaft:

ANGLE OF TWIST

The angle θ through which the bar length L will twist is

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.

POWER TRANSMITTED BY THE SHAFT

A shaft rotating with a constant angular velocity ω (in radians per second) is being acted by a twisting moment T. The power transmitted by the shaft is

where T is the torque in N·m, f is the number of revolutions per second, and P is the power in watts.

Solved Problems in Torsion

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 × 10^6 psi.

Solution 304

ksi, determine the maximum horsepower that can be transmitted.

Problem 309 A steel propeller shaft is to transmit 4.5 MW at 3 Hz without exceeding a shearing stress of 50 MPa or twisting through more than 1° in a length of 26 diameters. Compute the proper diameter if G = 83 GPa.

Solution 309

Solution 311

Problem 312 A flexible shaft consists of a 0.20-in-diameter steel wire encased in a stationary tube that fits closely enough to impose a frictional torque of 0.50 lb·in/in. Determine the maximum length of the shaft if the shearing stress is not to exceed 20 ksi. What will be the angular deformation of one end relative to the other end? G = 12 × 10^6 psi.

Solution 312

Problem 313 Determine the maximum torque that can be applied to a hollow circular steel shaft of 100-mm outside diameter and an 80-mm inside diameter without exceeding a shearing stress of 60 MPa or a twist of 0.5 deg/m. Use G = 83 GPa.

Solution 313

Problem 314 The steel shaft shown in Fig. P-314 rotates at 4 Hz with 35 kW taken off at A, 20 kW removed at B, and 55 kW applied at C. Using G = 83 GPa, find the maximum shearing stress and the angle of rotation of gear A relative to gear C.

Problem 316 A compound shaft consisting of a steel segment and an aluminum segment is acted upon by two torques as shown in Fig. P-316. Determine the maximum permissible value of T subject to the following conditions: τst = 83 MPa, τal = 55 MPa, and the angle of rotation of the free end is limited to 6°. For steel, G = 83 GPa and for aluminum, G = 28 GPa.

Solution 316

Problem 317 A hollow bronze shaft of 3 in. outer diameter and 2 in. inner diameter is slipped over a solid steel shaft 2 in. in diameter and of the same length as the hollow shaft. The two shafts are then fastened rigidly together at their ends. For bronze, G = 6 × 10^6 psi, and for steel, G = 12 × 10^6 psi. What torque can be applied to the composite shaft without exceeding a shearing stress of 8000 psi in the bronze or 12 ksi in the steel?

Problem 318 A solid aluminum shaft 2 in. in diameter is subjected to two torques as shown in Fig. P-

  1. Determine the maximum shearing stress in each segment and the angle of rotation of the free end. Use G = 4 × 10^6 psi.

Solution 318

Problem 319 The compound shaft shown in Fig. P-319 is attached to rigid supports. For the bronze segment AB, the diameter is 75 mm, τ ≤ 60 MPa, and G = 35 GPa. For the steel segment BC, the diameter is 50 mm, τ ≤ 80 MPa, and G = 83 GPa. If a = 2 m and b = 1.5 m, compute the maximum torque T that can be applied.

Solution 321

Problem 322 A solid steel shaft is loaded as shown in Fig. P-322. Using G = 83 GPa, determine the required diameter of the shaft if the shearing stress is limited to 60 MPa and the angle of rotation at the free end is not to exceed 4 deg.

Solution 322

Problem 323 A shaft composed of segments AC, CD, and DB is fastened to rigid supports and loaded as shown in Fig. P-323. For bronze, G = 35 GPa; aluminum, G = 28 GPa, and for steel, G = 83 GPa. Determine the maximum shearing stress developed in each segment.

Problem 324 The compound shaft shown in Fig. P-324 is attached to rigid supports. For the bronze segment AB, the maximum shearing stress is limited to 8000 psi and for the steel segment BC, it is limited to 12 ksi. Determine the diameters of each segment so that each material will be simultaneously stressed to its permissible limit when a torque T = 12 kip·ft is applied. For bronze, G = 6 × 10^6 psi and for steel, G = 12 × 10 6

psi.

Solution 324