Calculate the instantaneous stress produced in bar 10 cm^2 in area and 3 m long by suddenly application of the tensile load of unknown magnitude, if the extension of the bar due to sudden applied load is 1.5 mm. Also determine the suddenly applied load. Take E= 2x105 N/mm2.

Data:

Area(A)=$10cm^2=10*10^2 mm^2$

$A=10^3 mm^2$

$L=3m=3000mm$

$\delta l=1.5mm$

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

Let p and P be the instantaneous stress and suddenly applied load

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

$1.5=\frac{p*3000}{2*10^5}$

$p=100 N/mm^2$

Now, for suddenly applied load,

Stress=p=$2*\frac{P}{A}$

$100=\frac{2*P}{10^3}$

$P=\frac{100*10^3}{2}$

$=50*10^3 N$

P=50KN