A steel flat of thickness 10 mm tapers uniformly from 60 mm at one end to 40 mm at other end in a length of 600 mm . If the bar is subjected to a load of 80 KN , find its extension. Take $\text{E}=2\times105\text{MPa}.$

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This question was previously asked in

JKSSB JE (Civil) Official Paper (Held On: 19 Nov, 2023)

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Answer 1

Concept:

For a tapered flat bar under axial load, the extension $\Delta L$ is given by:

$\Delta L=\frac{4PL}{Et(b_1+b_2)}$

where, P = axial load, L = length, E = modulus of elasticity, t = thickness,

b₁ b₂= widths at ends

Calculation:

Given:

P = 80 kN

$\text{E}=2\times105\text{MPa}.$

t = 10 mm

b₁ = 60 mm; b₂ = 40 mm

$\Delta L=\frac{4\times80\times10^3\times600}{2\times10^5\times10(60+40)}$

$\Delta L=0.386mm$