A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant intensity t per unit distance . (a) Determine the maximum shear stress tmax in the bar
(b) Determine the angle of twist between the ends of the bar.

a) the maximum shear stress τ the bar is 16T /πd³

b) the angle of twist between the ends of the bar is 16tL² / πGd⁴  

Explanation:

Given the data in the question, as illustrated in the image below;

d is the diameter of the prismatic bar of length AB

t is the intensity of distributed torque

(a) Determine the maximum shear stress tmax in the bar

Maximum Applied torque  T_max = tL

we know that;

shear stress τ = 16T/πd³

where d is the diameter

so

τ = 16T /πd³

Therefore, the maximum shear stress τ the bar is 16T /πd³

(b) Determine the angle of twist between the ends of the bar.

let theta () be the angle of twist

polar moment of inertia = πd⁴/32

now from the second image;

lets length dx which is at distance of "x" from "B"

Torque distance x

T(x) = tx

Elemental angle twist = d = T(x)dx / G

so

d = tx.dx / G(πd⁴/32)

d = 32tx.dx / πGd⁴

so total angle of twist will be;

=  

=   32tx.dx / πGd⁴

= 32t / πGd⁴  

= 32t / πGd⁴ [ L²/2]

= 16tL² / πGd⁴  

Therefore,  the angle of twist between the ends of the bar is 16tL² / πGd⁴  

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