Solution to Problem 320 Torsion

From the solution of Problem 319:
Maximum T_br = 4.970 kN·m
Maximum T_st = 1.963 kN·m
 

$\theta_{br} = \theta_{st}$

$\left( \dfrac{TL}{JG} \right)_{br} = \left( \dfrac{TL}{JG} \right)_{st}$

$\dfrac{4.973a(1000^2)}{\frac{1}{32}\pi (75^4)(35\,000)} = \dfrac{1.963b (1000^2)}{\frac{1}{32}\pi (50^4)(83\,000)}$

$b/a = 1.187$
 

$T = T_{br \, max} + T_{st \, max}$

$T = 4.970 + 1.963$

$T = 6.933 \, \text{kN}\cdot\text{m}$           answer