Solution to Problem 323 Torsion

Stress developed in each segment with respect to T_A:
 

 

The rotation of B relative to A is zero.
$\theta_{A/B} = 0$

$\left( {\Large \Sigma} \dfrac{TL}{JG} \right)_{A/B} = 0$

$\dfrac{T_A (2)(1000^2)}{\frac{1}{32}\pi (25^4)(35\,000)} + \dfrac{(T_A - 300) (2)(1000^2)}{\frac{1}{32}\pi (50^4)(28\,000)} + \dfrac{(T_A - 1000) (2.5)(1000^2)}{\frac{1}{32}\pi (25^4)(83\,000)} = 0$

$\dfrac{2T_A}{(25^4)(35)} + \dfrac{2(T_A - 300)}{(50^4)(28)} + \dfrac{2.5(T_A - 1000)}{(25^4)(83)} = 0$

$\dfrac{16T_A}{35} + \dfrac{T_A - 300}{28} + \dfrac{20(T_A - 1000)}{83} = 0$

$\frac{16}{25}T_A + \frac{1}{28}T_A - \frac{75}{7} + \frac{20}{83}T_A - \frac{20\,000}{83} = 0$

$\frac{8527}{11\,620}T_A = 251.678$

$T_A = 342.97 \, \text{N}\cdot\text{m}$
 

$\Sigma M = 0$

$T_A + T_B = 300 + 700$

$342.97 + T_B = 1000$

$T_B = 657.03 \, \text{N}\cdot\text{m}$
 

$T_{br} = 342.97 \, \text{N}\cdot\text{m}$

$T_{al} = 342.97 - 300 = 42.97 \, \text{N}\cdot\text{m}$

$T_{st} = 342.97 - 1000 = -657.03 \, \text{N}\cdot\text{m} = -T_B$       (okay!)
 

$\tau_{max} = \dfrac{16T}{\pi D^3}$

$\tau_{br} = \dfrac{16(342.97)(1000)}{\pi (25^3)} = 111.79 \, \text{ MPa}$           answer

$\tau_{al} = \dfrac{16(42.97)(1000)}{\pi (50^3)} = 1.75 \, \text{ MPa}$           answer

$\tau_{st} = \dfrac{16(657.03)(1000)}{\pi (25^3)} = 214.16 \, \text{ MPa}$           answer