A bar is subjected to axial forces as shown. | vedclass

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(D) To find the total elongation,we analyze the internal force in each part of the bar.$1$. For the $I^{st}$ part (left section of length $l$): The fo

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$\frac{4Fl}{AE}$

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(D) To find the total elongation,we analyze the internal force in each part of the bar.$1$. For the $I^{st}$ part (left section of length $l$): The force acting on the left end is $3F$ (to the left). To maintain equilibrium,the internal tensile force in this section must be $3F$. Thus,the elongation is $\Delta x_1 = \frac{(3F)l}{AE} = \frac{3Fl}{AE}$.$2$. For the $II^{nd}$ part (right section of length $l$): The force acting on the right end is $F$ (to the right). The internal tensile force in this section is $F$. Thus,the elongation is $\Delta x_2 = \frac{Fl}{AE}$.$3$. Total elongation: Since both parts are being stretched,the total elongation is $\Delta x = \Delta x_1 + \Delta x_2 = \frac{3Fl}{AE} + \frac{Fl}{AE} = \frac{4Fl}{AE}$.

$A$ bar is subjected to axial forces as shown. If $E$ is the modulus of elasticity of the bar and $A$ is its cross-section area,its total elongation will be:

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