and fastened by nuts threaded on the bolt. Find the stresses in the bolt and tube when one of the nuts is tightened by one quarter of a turn, -
relative to the other.The pitch of the thread is 2mm. Take $E_s$ = 200GPa and $E_c$ = 100GPa. -

The steel bolt is under tension and the copper tubing under compression.

The nut is fastened by one quarter of a turn.

Allowed Deformation = $\frac{Pithch of Nut}{4}$

Allowed Deformation = $\frac{2}{4} = 0.5mm$

The elongation for the bolt and the tube is the same.

$\dbinom{PI}{AE}_{bolt} = \dbinom{PI}{AE}_{tubing} = 0.5mm ........(i)$

Steel Bolt – Tension

$d = 20mm \ \ \ \ \ dl = 0.5mm$

$l = 600mm \ \ \ \ \ E_s = 200GPa$

$A_s = \frac{\pi}{4}(20)^2 = 314.159mm^2$

From (I),

$\frac{P_s(600)}{(314.159)(200 × 10^3)} = 0.5$

$P_s = 52.36kN$

Stress Induced $(δ)_s = \frac{P_s}{A_s} = \frac{52.36 × 10^3}{314.159} = 166.67N/mm^2$

This is the value of the tensile stress induced in the steel bolt.

Copper Tubing – Compression

$d = 25mm \ \ \ \ t = 10mm$

$D = d + 2t = 45mm \ \ \ \ E_c= 100GPa$

$A_c = \frac{\pi}{4}[45^2 - 25^2] = 109.56mm^2$

From (I),

$\frac{P_c(600)}{(10099.56)(100 × 10^3)} = 0.5$

$P_c = 91.63kN$

Stress Induced $δ_c = \frac{P_c}{A_c} = \frac{91.63 × 10^3}{1099.56} = 83.33N/mm^2$

This is the value of the compressive stress induced in the copper tubing.