Question

a bar is 500 mm long and is stretched to 505mm with a force of 50KN .THE BAR IS 10mm diameter ,calculate the stress and strain​

Answer 1

Given:

A bar is 500 mm long and is stretched to 505mm with a force of 50KN . The bar is if 10mm diameter .

To find:

• Stress
• Strain

Calculation:

• Let's assume the bar to be a cylinder.

• Area of cross-section = πr² = π(10/1000)² = 0.0031 m².

$\rm \: stress = \dfrac{force}{area}$

$\rm \implies\: stress = \dfrac{50000}{0.0031}$

$\rm \implies\: stress = \dfrac{5\times {10}^{8} }{31}$

$\rm \implies\: stress = 1.6 \times {10}^{7} \: Pa$

Now, strain will be :

$\rm \: strain= \dfrac{\Delta L}{ L}$

$\rm \implies\: strain= \dfrac{505 - 500}{ 500}$

$\rm \implies\: strain= \dfrac{5}{ 500}$

$\rm \implies\: strain= 0.01$

Hope It Helps.