: The shaft is made of L2 tool steel, has a diameter of 40 mm, and is fixed at its ends A and B. If it is subjected to the torque, determine the maximum shear stress in the bar. If the shear modulus of the steel is 75 GPa, determine the twisting angle at C.

the maximum shear stress in the bar = 9.55 MPa

the twisting angle =

Explanation:

From the diagram below;

Torque (T) = 2 kN × 5 mm +  2 kN × 5 mm

=

= 200 N.m

Now; If we divide the shaft into two parts AC & BC

Then :

         ------------ equation (1)

At the junction :

⇒

replace into above equation (1)

Again:

Now ; we can deduce that the maximum shear comes from since

So;

where;

R = 20 mm

D = 40 mm = 0.04 m

Then;

Thus ; the maximum shear stress in the bar = 9.55 MPa

b)

Again:

Thus, the twisting angle =

i don't understand is there any other way you can rephrase it plz let me know in the comments and i will do it.

explanation: