# A rod made of Aluminium alloy (E = 72 GPa) has length 0.5 m and diameter 10 mm. The tensile stiffness (N/m) of this rod is

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A rod made of Aluminium alloy (E = 72 GPa) has length 0.5 m and diameter 10 mm. The tensile stiffness (N/m) of this rod is
1. 10π × 105
2. 36π × 105
3. 9π ×105
4. 12π × 105
5. 36π ×106

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Correct Answer - Option 2 : 36π × 105

Concept:

If the material of the spring is linearly elastic, the load P and elongation δ are proportional, or P = k δ.

k = P / δ is the stiffness (or “spring constant”) with units N/m.

f = δ / P is the flexibility (or “compliance”) with units m/N.

For a prismatic bar,

$\begin{array}{l} \sigma = E\varepsilon \\ \left( {\frac{P}{{{A_o}}}} \right) = E\left( {\frac{\delta }{{{L_o}}}} \right) \Rightarrow \delta = \frac{{P{L_o}}}{{E{A_o}}}\\ k = \frac{P}{\delta } = \frac{{E{A_o}}}{{{L_o}}}\\ f = \frac{\delta }{P} = \frac{{E{A_o}}}{{{L_o}}} \end{array}$

Calculation:

Given,

E = 72 GPa = 72000 MPa

(1 GPa = 1000 MPa) and (1 MPa = 1 N/mm2)

Lo = 0.5 m = 500 mm

diameter, d = 10 mm

${A_o} = \frac{π }{4}{d^2} = \frac{π }{4} × {10^2} = 25π$

$k = \frac{{E{A_o}}}{{{L_o}}} = \frac{{72000 × 25π }}{{500}} = 3600π\ N/mm$

(1 m = 1000 mm)

Stiffness, k = 36 π × 105 N/m