Question

A 20 cm long rod of uniform rectangular section, 8 mm wide $\times$ 1.2 mm thick is bent into the form of a circular arc resulting in a central displacement of 0.8 cm. Neglecting second-order quantities in computations, what is the longitudinal surface strain (approximate) in the rod ?

• A. $7.2\times {10}^{-4}$
• B. $8.4\times {10}^{-4}$
• C. $9.6\times {10}^{-4}$
• D. $10.8\times {10}^{-4}$

Answer 1

The correct option is C $9.6\times {10}^{-4}$

The arrangement can be shown as in the figure below:

The straight rod AC is bent into a circular arc ABC.
From the property of circle, we have
$BE\times ED=AE\times EC$

$\Rightarrow \delta \times (2R-\delta )=\frac{L}{2}\times \frac{L}{2}$

$\Rightarrow 2\delta R-{\delta }^2=\frac{L^2}{4}$

Neglecting 2nd quantities, we get

$\Rightarrow R=\frac{L^2}{8\delta }$

Here, $\delta =0.8cm,L=20cm$

$\therefore R=\frac{(20{)}^2}{8\times 0.8}=62.5cm$

Also, from the theory of pure bending, we have

$\frac{f}{y}=\frac{E}{R}$

$\Rightarrow f=\frac{E}{R}\times y$

$\Rightarrow \epsilon \times E=\frac{E}{R}\times y$

$\Rightarrow \epsilon =\frac{y}{R}$

$\Rightarrow \epsilon =\frac{1.2\times {10}^{-1}}{2\times 62.5}=9.6\times {10}^{-4}$

Alternatively

$\epsilon =\frac{4t\delta }{L^2}=\frac{4\times 1.2\times 0.8\times 10}{(28\times 10{)}^2}$

$=9.6\times {10}^{-4}$