A vertically hung bar is 2 m long and has a diameter of 25 mm. A weight of 600 N is dropped from a height (h) on a collar attached to the lower end of the bar. Find the height of drop, if the stress in the bar is not to exceed 100 MPa. Also, find the maximum weight that can be dropped from this height without causing any permanent deformation. The elastic limit is 220 MPa & E= 200 GN/m2.

Data: L=2m,

L=2000mm

$\phi=25mm$, P=600N,

p=100MPa=$100 N/mm^2$

Elastic element=$220 MPa=220 N/mm^2$

$E=200 GN/mm^2=200*10^3 N/mm^2$

Case 1: Calculating the height of drop(h) if stress doesnt exceed $p=100 N/mm^2$

Now, equating strain energy to the loss of potential energy.

$\frac{p^2*AL}{2E}=P(h+\delta l)$

$\delta l=\frac{p*l}{E}=\frac{100*2000}{200*10^3}=1mm$

$\frac{100^2*\frac{\pi}{4}*25^2*2000}{2*200*10^3}=600(h+1)$

$h=39.90mm$

Case 2: Calculating maximum weight (p') for elastic limit=$220 N/mm^2$

$i.e p'=220 N/mm^2$

Strain energy stored corresponding elastic limit stress=

$=\frac{p'^2*AL}{2E}=\frac{200}{2*200*10^3}*\frac{\pi}{4}*25^2*2000$

$=118.79*10^3 Nmm$

Now, Let P" be the maximum weight that can be dropped from height of 39.90mm

Equating strain energy to the loss of potential energy,

$\frac{p'*AL}{2E}=P"(h+\delta l)$

$=118.97*10^3=p'(39.9+1)$

$p'=2904.43 N$