Solution to Problem 331 | Flanged bolt couplings

$T = (PRn)_{st} + (PRn)_{al}$

$T = (A\tau Rn)_{st} + (A\tau Rn)_{al}$

$T = \frac{1}{4}\pi (1/2)^2 \tau_{st}(6)(6) + \frac{1}{4}\pi (3/4)^2 \tau_{al}(4)(4)$

$T = 2.25\pi \tau_{st} + 2.25\pi \tau_{al}$

$T = 2.25\pi (\tau_{st} + \tau_{al})$       → Equation (1)
 

$\left( \dfrac{\tau}{GR} \right)_{st} = \left( \dfrac{\tau}{GR} \right)_{al}$

$\dfrac{\tau_{st}}{(12 \times 10^6 )(6)} = \dfrac{\tau_{al}}{(4 \times 10^6 )(4)}$

$\tau_{st} = \frac{9}{2}\tau_{al}$       → Equation (2a)

$\tau_{al} = \frac{2}{9}\tau_{at}$       → Equation (2b)
 

Equations (1) and (2a)
$T = 2.25\pi(\frac{9}{2}\tau_{al} + \tau_{al}) = 12.375\pi \tau_{al}$

$T = 12.375\pi(6000) = 74\,250\pi \, \text{lb}\cdot\text{in}$

$T = 233.26 \, \text{kip}\cdot\text{in}$
 

Equations (1) and (2b)
$T = 2.25\pi(\tau_{st} + \frac{2}{9}\tau_{st}) = 2.75\pi \tau_{st}$

$T = 2.25\pi(9000) = 24\,750\pi \, \text{lb}\cdot\text{in}$

$T = 77.75 \, \text{kip}\cdot\text{in}$
 

Use T = 77.75 kip·in           answer