written 3.6 years ago by
Pooja Joshi • 870

•
modified 3.6 years ago

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