Practice Question: Solve for Twist on a Shaft

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Solution

Answer: D) 1911.8 lb-ft

Let T_a and T_e be the support reactions at points A and E respectively.

In equilibrium:

$$ T_a+T_e=590+880+750 $$

$$ T_a+T_e=2220 $$

Since the support is unyielding, the angle of twist at point A with respect to point E shall be zero. Shown below is the shafts torque diagram. 

Equilibrium equation for twist:
$$ \theta_\frac{A}{E}=\ \theta_\frac{A}{B}+\theta_\frac{B}{C}+\theta_\frac{C}{D}+\theta_\frac{D}{E}=0 $$

To compute for the angle of twist, use: $$ \theta=\ \frac{TL}{GJ} $$
where: $$ T=torsion$$ $$ L=length $$ $$ G = modulus \ of \ rigidity $$ $$ J = polar \ moment \ of \ inertia $$

The polar moment of inertia of a solid circular section is: $$\frac{\pi D^4}{32}$$

Accordingly:

$$ \theta_\frac{A}{E}=\frac{T_a(60\ in)}{\left(11.5*{10}^6\ psi\right)\left[\frac{\pi({5\ in)}^4}{32}\right]}+\ \frac{(T_a-590\ lb-ft)(40\ in)}{\left(11.5*{10}^6\ psi\right)\left[\frac{\pi({5\ in)}^4}{32}\right]}+\frac{(T_a-1470\ lb-ft)(75\ in)}{\left(11.5*{10}^6\ psi\right)\left[\frac{\pi({10\ in)}^4}{32}\right]}+\frac{(T_a-2220\ lb-ft)(75\ in)}{\left(11.5*{10}^6\ psi\right)\left[\frac{\pi({15\ in)}^4}{32}\right]}=0 $$

Simplifying:
$$ \theta_\frac{A}{E\ }= \frac{Ta(60\ in)}{{(5\ in)}^4}+\ \frac{(Ta-590\ lb-ft)(40\ in)}{{(5\ in)}^4}+\frac{(Ta-1470\ lb-ft)(75\ in)}{{(10\ in)}^4}+\frac{(Ta-2220\ lb-ft)(75\ in)}{{(15\ in)}^4}=0 $$
$$ T_a=318.16\ lb-ft $$

With this value of T_a and using the equilibrium equation: $$ T_e = 1911.84 \ lb-ft $$
 

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