Question

A shaft arrangment as shown in which variable distributed torque is acting, the reaction generated at right support is ?

• A. $\frac{1}{6}{T}_{0}L$
• B. $\frac{1}{3}{T}_{0}L$
• C. $\frac{1}{12}{T}_{0}L$
• D. $\frac{1}{4}{T}_{0}L$

Answer 1

The correct option is D $\frac{1}{4}{T}_{0}L$

Let reaction generated at A and B are $T_A$ and $T_B$ respectively.

$T_A+T_B=\frac{1}{3}\times L\times \frac{T_0{L}^2}{L^2}=\frac{1}{3}L{T}_0...\left(i\right)$

Take a section at x from A.

$T_A+T_x=\frac{1}{3}{T}_0\frac{x^2}{L^2}\times x$

$T_x=(\frac{1}{3}{T}_0\frac{x^3}{L^2}-T_A)$


$\therefore {\theta }_{D/A}=0$ (D & A are fixed)

${\int }_0^L\frac{T_{x}dx}{GJ}=0$

$\Rightarrow \frac{1}{GJ}{\int }_0^L(\frac{1}{3}{T}_0\frac{x^3}{L^2}-T_A)dx=0$

$\Rightarrow \frac{1}{GJ}{[\frac{T_0}{3L^2}\frac{x^4}{4}-T_{A}x]}_0^L=0$

$\Rightarrow \frac{T_0{L}^4}{12L^2}-T_{A}L=0$

$T_A=\frac{T_{0}L}{12}....\left(ii\right)$

So from eq (i) and (ii)

$T_B=\frac{T_{0}L}{4}$