Solution to Problem 350 | Helical Springs

$\Sigma F_V = 0$

$P_1 + P_2 + P_3 = 490.5$       → Equation (1)
 

$\Sigma M_1 = 0$

$P_2(1) + P_3(3) = 490.5(1.5)$

$P_2 + 3P_3 = 735.75$       → Equation (2)
 

 

$\dfrac{\delta_2 - \delta_1}{1} = \dfrac{\delta_3 - \delta_1}{3}$

$\delta_2 = \frac{1}{3}\delta_3 + \frac{2}{3}\delta_1$

$\dfrac{64P_2(50^3)(24)}{83\,000(10^4)} = \dfrac{1}{3} \left[ \dfrac{64P_3(75^3)(48)}{42\,000(20^4)} \right] + \dfrac{2}{3} \left[ \dfrac{64P_1(50^3)(24)}{83\,000(10^4)} \right]$

$\frac{3}{830}P_2 = \frac{9}{8960}P_3 + \frac{1}{415}P_1$

$\frac{3}{166}P_2 = \frac{9}{1792}P_3 + \frac{1}{83}P_1$       → Equation (3)
 

From Equation (1)
$P_1 = 490.5 - P_2 - P_3$
 

Substitute P₁ to Equation (3)
$\frac{3}{166}P_2 = \frac{9}{1792}P_3 + \frac{1}{83}(490.5 - P_2 - P_3)$

$\frac{3}{166}P_2 = \frac{9}{1792}P_3 + \frac{981}{166} - \frac{1}{83}P_2 - \frac{1}{83}P_3$

$\frac{5}{166}P_2 = \frac{981}{166} - \frac{1045}{148\,736}P_3$       → Equation (4)
 

From Equation (2)
$P_2 = 735.75 - 3P_3 = \frac{2943}{4} - 3P_3$
 

Substitute P₂ to Equation (4)
$\frac{5}{166}(\frac{2943}{4} - 3P_3) = \frac{981}{166} - \frac{1045}{148\,736}P_3)$

$(\frac{1045}{148\,736} - \frac{5}{166})P_3 = \frac{981}{166} - \frac{14\,715}{664}$

$P_3 = 195.01 \, \text{N}$

$P_2 = 735.75 - 3(195.01) = 150.72 \, \text{N}$

$P_1 = 490.5 - 150.72 - 195.01 = 144.77 \, \text{N}$
 

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

For steel at left:
$\tau_{max1} = \dfrac{16(144.77)(50)}{\pi (10^3)} \left[ 1 + \dfrac{10}{4(50)} \right] = 38.709 \, \text{ MPa}$       answer
 

For steel at right:
$\tau_{max2} = \dfrac{16(150.72)(50)}{\pi (10^3)} \left[ 1 + \dfrac{10}{4(50)} \right] = 40.300 \, \text{ MPa}$       answer
 

For phosphor bronze:
$\tau_{max3} = \dfrac{16(195.01)(75)}{\pi (20^3)} \left[ 1 + \dfrac{20}{4(75)} \right] = 9.932 \, \text{ MPa}$       answer