Answer to Question #230027 in Mechanical Engineering for richard

Question

A helical spring is provided with a rigid plate at each end. A bolt passes through the spring and the end plates. By turning the nut, the spring is compressed until the coils touch. The height of the unloaded spring is 700 mm, its mean coil diameter is 200 mm, and the diameter of the spring wire is 30 mm. The spring has 20 coils. If the diameter of the bolt is 26 mm and each plate is 26 mm thick. Calculate: a. The elongation of the bolt when the spring is fully compressed b. The stress in the spring wire. (E=200 GPa and G= 80 GPa) 

Answer 1

Part a

The spring is compressed until the coils touch

$l= d *h \\ l= 30*20\\ l=600 mm$

Deflection of the spring

$\delta = L-l\\ \delta = 700-600\\ \delta = 100mm$

We know that

$\delta = \frac{8 PD^3 n}{Gd^4}\\ \implies P = \frac{Gd^4 \delta}{8D^3n}\\ P = \frac{80*10^9*30^4 *100}{8*200^3*20}= 5.0625*10^{10}N$

Elongation of the bolt

$\rho_B = \frac{P l_b}{AE}\\ \implies A= \frac{\pi}{4}d_b^2= \frac{\pi}{4}*26^2=530.92 mm^2\\ \rho_B = \frac{5.0625*10^{10}*l_b}{530.92*200*10^9} \\ \frac{\rho_B}{l_b}=4.76*10^{-4}$

As the length of the bolt is not provided we calculate the elongation per unit length of the bolt

Part b

Torsional shear stress

$Z_1= \frac{16 T_1}{\pi d^2}= \frac{16 (0.5PD)}{\pi d^3}= \frac{8*5.0625*10^{10}*200}{\pi *30^2}=954.92 MPa$

Direct shear stress

$Z_2= \frac{T}{\pi d^2}= \frac{8PD}{\pi d^3}= \frac{4*5.0625*10^{10}}{\pi *30^2}=71.61 MPa$

Total stress

$Z_1+Z_2= 954.92+71.61=1026.53 MPa$