Question

A bar 3m long and 5 cm diameter hands vertically and has a collar securely attached at the lower end. Find the maximum stress induced when; 

(i) a weight 2.5KN falls from 12 cm on the collar 

(ii) a weight of 25KN falls from 1 cm on the collar Take E = 2.0 × 10⁵ N/mm^2.

Answer 1

Cross-sectional area of the bar = A = π/4.d² = π/4(50)² = 1962.5 mm² 

 (i)Instantaneous elongation of bar ;

δL = WL/AE = (2500 × 3000)/(1962.5 × 2 × 10⁵) = 0.01911mm

This elongation is very small as compared to 120mm height of fall and as such can be neglected

Accordingly stress induced in the bar can be worked out by using the relation. 

 (ii) Instantaneous elongation

δL = WL/AE = (25000 × 3000)/(1962.5 × 2 × 10⁵) = 0.1911mm

This elongation is comparable to 10 mm height of fall. Further the falling weight is large and hence extension of the bar cannot be neglected. Accordingly stress induced in the bar is worked out from the relation.