A rod 13 mm in diameter is stretched by 3.5 mm under a steady load 12000 N. What stress would be induced in the bar of 800 N weight falling through 80 mm before commencing to stretch the rod if it is originally unstressed?

Note: If in question word "freely fall" found then solve by impact load formula

Solution:

$E=2.15*10^5 N/mm^2$

Case 1)

Gradually applied load=12000N

$\delta l=3.5mm$

$d=13mm$

$A=\frac{\pi}{4}*13^2$

$A=132.7322 mm^2$

Stress $p=\frac{P}{A}=\frac{12000}{132.732}=90.408 N/mm^2$

$\delta l=\frac{p*l}{E}$

$3.5=\frac{90.408*l}{2.15*10^5}=L=8323.4mm$

Case 2)

h = 80mm, P = 800N

$p=\frac{P}{A}+\sqrt {\frac({P}{A})^2+\frac{2PhE}{AL}}$

$p=\frac{800}{132.732}+\sqrt {\frac({800}{132.732})^2}+$

$\frac{2*800*80*2.15*10^5}{132.732*8323.4}$

$p=163.97 N/mm^2$