$\tau_{max} = \dfrac{16PR}{\pi d^3} \left( 1 + \dfrac{d}{4R} \right)$ → Equation (3-9)

$20\,000 = \dfrac{16P(3)}{\pi(3/4)^3} \left[ 1 + \dfrac{3/4}{4(3)} \right]$

$P = 519.75 \, \text{ lb}$

For this problem, the critical spring is the one subjected to tension. Use P_{2} = 519.75 lb.

$\dfrac{\delta_1}{2} = \dfrac{\delta_2}{4}$

$\delta_1 = \frac{1}{2} \delta_2$

$\dfrac{64P_1R^3n}{Gd^4} = \dfrac{1}{2} \left( \dfrac{64P_2R^3n}{Gd^4} \right)$

$P_1 = \frac{1}{2}P_2 = \frac{1}{2}(519.75)$

$P_1 = 259.875 \, \text{ lb}$

$\Sigma M_O = 0$

$7W = 2P_1 + 4P_2$

$7W = 2(259.875) + 4(519.75)$

$W = 371.25 \, \text{ lb}$ *answer*

## Comments

## why is the light spring

Why is the light spring equation used here instead of the heavy spring equation? d/4R = 0.0625 (d/4R >= 0.05, hence heavy spring). If heavy spring equation is used m=8, P2 = 466.41lb and W = 333.15 lb. Is this not the safer load also?