First slide

Stress and strain

Question

A long thin metallic rod of length l, cross sectional area S and Young's modulus $γ$ rests on a smooth horizontal surface. A horizontal force P is applied at one end of the rod. Then the elongation produced in the rod is

Easy

A

Zero
B

$\frac{P l}{S γ}$
C

$\frac{2 P l}{S γ}$
D

$\frac{P l}{2 S γ}$

Solution

(4)
a = Acceleration of the rod, a = P/m. If we observe the rod from its frame, then pseudo force ma acts at the COM C of the rod. So under the action of P and ma, elongation produced in the rod is
$δ = \frac{P \times A C}{S γ} = \frac{P . \frac{l}{2}}{S γ} = \frac{P l}{2 S γ} .$

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