Correct Answer - Option 1 : 126.5 MPa
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
Stress-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
\(\sigma = \sqrt({2AEh \over Pl}) \)