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A solid circular steel shaft having an outside diameter of 2.00 in. is subjected to a pure torque of T = 8380 lb-in. The shear modulus of the steel is G = 12,000 ksi. Determine the magnitude of the angle of twist in a 5.00-ft length of shaft.

Respuesta :

Answer:

0.02667 rad (counterclockwise)

Explanation:

Parameters given:

Diameter of shaft = 2 in

Torque = 8380 lbin

Shear modulus = 12000 ksi = 12000 * 1000 pounds force per inch

Length of shaft = 5 ft = 60 in

The angle of twist is given as

A = (T*L) / (P*G)

Where T = Torque

L = length of shaft

P = polar moment of the shaft

G = Shear modulus

Polar moment of a cylindrical shaft is given as

P =( pi * R⁴) / 2

R = radius

P = (3.142 * 1⁴) / 2

P = 1.571 m⁴

A = (8380 * 60) / (12000 * 1000 * 1.571)

A = 0.02667 rad (counterclockwise)

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