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A solid steel shaft of diameter d=1/2 in is built in as its ends A and B and car

ID: 1824151 • Letter: A

Question

A solid steel shaft of diameter d=1/2 in is built in as its ends A and B and carries a disk at C as shown in figure. If the working stress in shear for the shaft is 10,000 psi, what is the maximum safe angle of rotation that can be given to the disk which is rigidly attached to the shaft? What is the torsion applied at C?

A solid steel shaft of diameter d=1/2 in is built in as its ends A and B and carries a disk at C as shown in figure. If the working stress in shear for the shaft is 10,000 psi, what is the maximum safe angle of rotation that can be given to the disk which is rigidly attached to the shaft? What is the torsion applied at C?

Explanation / Answer

maximum safe angle of twist, max = (max*L)/(G*R)

this angle of twist is controlled by lower length of the shaft because torque acting on disc is inversely proportional to the length of the shaft. that means lesser in length, maximum will be torque.

therefore for maximum torque, we should consider length of the shaft, L = 8 inch

max = 10000*8/(12*106*0.25) = 0.02667 radian = 1.530

Tmax = J*max/R = J*G*max/L

J = *D4/32 = *0.54/32 = 6.136*10-3 in4

put all the values in the formula, we get;

Tmax = J*max/R = 6.136*10-3 *104/0.25 = 245.43 lb-in

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