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Question

A bar is subjected to an axial forces as shown in figure. Find the total elongation in the bar. (E is the modulus of elasticity of the bar and A is its cross-section)

A

\frac{F I}{A E}

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B

\frac{2 F I}{A E}

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C

\frac{3 F I}{A E}

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D

\frac{4 F I}{A E}

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Solution

The correct option is D $\frac{4 F I}{A E}$
In given figure,
Considering the two halves separately.
Since, this heavy need to be in equilibrium.
An additional force facts towards light.
So, $Δ h = \frac{F}{A} \times \frac{l}{E} = \frac{3 F l}{A E} = \frac{3 F l}{A E}$
The other haly becomes.
$Δ h_{2} = \frac{F l}{A E}$
Net elongation $= \frac{4 F l}{A E}$

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