Solution to Problem 319 Torsion

$\Sigma M = 0$

$T = T_{br} + T_{st}$       → Equation (1)
 

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

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

$\dfrac{T_{br}(2)(1000)}{\frac{1}{32}\pi (75^4)(35\,000)} = \dfrac{T_{st}(1.5)(1000)}{\frac{1}{32}\pi (50^4)(83\,000)}$

$T_{br} = 1.6011T_{st}$       → Equation (2a)

$T_{st} = 0.6246T_{br}$       → Equation (2b)
 

 

$\tau_{max} = \dfrac{16T}{\pi D^3}$
 

Based on τ_br ≤ 60 MPa

$60 = \dfrac{16T_{br}}{\pi (75^3)}$

$T_{br} = 4\,970\,097.75 \, \text{N}\cdot\text{mm}$

$T_{br} = 4.970 \, \text{ kN}\cdot\text{m}$       → Maximum allowable torque for bronze
 

From Equation (2b)
$T_{st} = 0.6246(4.970)$

$T_{st} = 3.104 \, \text{ kN}\cdot\text{m}$

 

Based on τ_br ≤ 80 MPa

$80 = \dfrac{16T_{st}}{\pi (50^3)}$

$T_{st} = 1\,963\,495.41 \, \text{N}\cdot\text{mm}$

$T_{st} = 1.963 \, \text{ kN}\cdot\text{m}$       → Maximum allowable torque for steel
 

From Equation (2a)
$T_{br} = 1.6011(1.963)$

$T_{br} = 3.142 \, \text{ kN}\cdot\text{m}$

 

From Equation (1), use T_br = 3.142 kN·m and T_st = 1.963 kN·m
$T = 3.142 + 1.963$

$T = 5.105 \, \text{kN}\cdot\text{m}$           answer