A steel bolt of 20mm diameter passes centrally through a copper tube of internal diameter 25mm and thickness 10mm. The tube is 600mm long and is closed by rigid washers of negligible thickness

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.

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Mumbai university > MECH > SEM 3 > Strength Of Materials

Marks: 10M

Year: Dec 2013


Thank you so much for solving this but in copper tubing, the value of area is typed wrong ---- 10099.56 ----


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.

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