Question

A steel bar AB of uniform thickness 2 cm, tapers uniformly from 100 mm to 50 mm in a length of 400 mm. From first principles determine the elongation of plate; if an, axial tensile force of 50 kN is applied on it. [E = 2 x 10⁵ N/mm²]

Answer 1

Consider a small element of length dx of the plate, at a distance x from the larger end. Then at this section, 

Width W_x = 100 (100 50)( / 400) = \(\left(100-\frac x8\right)\)mm

Cross section area Ax = thickness x width =10 \(\left(100-\frac x8\right)\)mm²

 

\(\therefore\) Elongation of the elementary length

The total change in length of the plate can be worked out by integrating the above identity between the limits x = 0 and x = 400 mm.