A 2 m  long alloy bar of 1500 mm² cross-sectional area hangs vertically and has a collar securely fixed at its lower end. What is the stress-induced in the bar when a weight of 2 kN falls from a height of 100 mm on the collar? (Take E = 120 GPa)

TB Solution

Given Data:

Length of bar(l)= 2 m

Cross-sectional Area of bar(A)= 1500 mm²

Weight of fall on collar of bar(P)= 2 kN= 2000 N

Height from which weigh falls(h)= 100 mm

Modulus of Elasticity(E)= 120 GPa = 120 x 1000 N/mm²

Explanation:

Stress-induced due to the load is applied with impact:

\(\sigma ={P \over A}[1 \pm \sqrt(1+ {2AEh \over Pl})] \)

\(\sigma ={2000 \over 1500}[1 \pm \sqrt(1+ {2\times 1500 \times 120000 \times 100 \over 2000 \times 2000})]\)

= 127. N/mm²

Additional InformationStress-induced due to the load is applied with impact:

\(\sigma ={P \over A}[1 \pm \sqrt(1+ {2AEh \over Pl})] \)

where Length of the bar(l), Cross-sectional Area of the bar(A), Weight of fall on the collar of the bar(P)'

Height from which weight falls(h), Modulus of Elasticity(E)

Case- When Deformation is very small as compared to h, Then

  \(\sigma = \sqrt({2AEh \over Pl}) \)