A bar is subjected to axial forces as shown. If E is the modulus of elastically of the bar and A is its cross-section area. Its elongation will be.

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Solution

For left part, force in any cross-section is $3 F$ and $F$ in right part for equilibrium.

$Δ l = \frac{F l}{A E}$

Total elongation $Δ l = Δ_{1} + Δ_{2}$

$⟹ Δ l = \frac{3 F l}{A E} + \frac{F L}{A E}$

$⟹ Δ l = \frac{4 F l}{A E}$

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A bar is subjected to axial forces as shown. If E is the modulus of elastically of the bar and A is its cross-section area. Its elongation will be.

For left part, force in any cross-section is 3F and F in right part for equilibrium.Δl=FlAETotal elongation Δl=Δ1+Δ2⟹Δl=3FlAE+FLAE⟹Δl=4FlAE

For left part, force in any cross-section is $3 F$ and $F$ in right part for equilibrium.

$Δ l = \frac{F l}{A E}$

Total elongation $Δ l = Δ_{1} + Δ_{2}$

$⟹ Δ l = \frac{3 F l}{A E} + \frac{F L}{A E}$

$⟹ Δ l = \frac{4 F l}{A E}$