Instantaneous stress and elongation:
What shall be the instantaneous stress & elongation of a 25 mm diameter bar, 2.6 m long, suspended vertically, if any mass of 10 kg falls through a height of 300 mm onto a collar that is rigidly associated to the bottom end of the bar? Take E = 200 GPa.
Solution
We have
= 96.24 MPa
Now,
δ= σL/ E = (96.24 × 10^{6} × 2.6 )/ (200 × 10^{9})
= 0.00125 m = 1.25 mm
Therefore, the maximums tress & elongation are 96.24 MPa & 1.25 mm, respectively.
This elongation of 1.25 mm is extremely small compared to the height of fall namely 300 mm. Let us discover the stress if this elongation were to be neglected in the calculation. While it is neglected the work strain energy equation becomes.
Wh = (σ^{2}/2E )× AL
or
It shows that the error in the stress calculation is unimportant even if the elongation is neglected.