Question

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.

Answer 1

Answer:

a) the maximum shear stress τ$_{max}$ the bar is 16T$_{max}$ /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$_{max}$ = 16T$_{max}$ /πd³

Therefore, the maximum shear stress τ$_{max}$ the bar is 16T$_{max}$ /πd³

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

let theta ($\theta$) be the angle of twist

polar moment of inertia $I_p}$ = 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$\theta$ = T(x)dx / G$I_{p}$

so

d$\theta$ = tx.dx / G(πd⁴/32)

d$\theta$ = 32tx.dx / πGd⁴

so total angle of twist $\theta$ will be;

$\theta$ =  $\int\limits^L_0 \, d\theta$

$\theta$ =  $\int\limits^L_0 \,$ 32tx.dx / πGd⁴

$\theta$ = 32t / πGd⁴  $\int\limits^L_0 \, xdx$

$\theta$ = 32t / πGd⁴ [ L²/2]

$\theta$ = 16tL² / πGd⁴  

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