Question

Q1.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

Answer ✍️...

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:StressStrainCalculation:Let's assume the bar to be a cylinder.Area of cross-section = πr² = π(10/1000)² = 0.0031 m².Stress= Force/Area.

stree = 50000/ 0.0031

stree = 50000/ 0.0031stree = 5× 10⁸ / 31

stree = 50000/ 0.0031stree = 5× 10⁸ / 31stree = 1.6 × 10⁷ pa

stree = 50000/ 0.0031stree = 5× 10⁸ / 31stree = 1.6 × 10⁷ pa Now, strain will be :

stree = 50000/ 0.0031stree = 5× 10⁸ / 31stree = 1.6 × 10⁷ pa Now, strain will be :strain= ∆L / L

strain = 505 - 500 / 500

=5/500

=5/500⟹strain=0.01

=5/500⟹strain=0.01Hope It Helps you ...

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Answer 2

Answer:

The stress and strain are 0.636GPa and 0.01 respectively

Step-by-step explanation:

Given that,

The length of the bar is 500mm and it is strecged upto a length of 505mm.

Also given that,

The force with which it was stretched is 50kN

The diameter of the bar is 10mm.

First we write the given data properly

$l=500mm\\∆l=505-500=5mm\\F=50kN=5\times10^{4}N \\D=10mm$

The strain is defined as the ratio of change in length to original length.Therefore,

$\epsilon=\frac{∆l}{l}$

Substituting the values in above term,we get

$\epsilon=\frac{5}{500} = 0.01$

Strain is found.Now,Stress is defined as ratio of force to the Area of application.Area is calculated in the following way

$A=\frac{π}{4}D^{2} = \frac{\pi}{4 } \times (10 \times 10 {}^{ - 3} ) {}^{2} \\ = 7.85 \times 10 {}^{ - 5} {m}^{2}$

Hence the stress is

$\sigma=\frac{F}{A}$

Substituting the values in above relation we get

$\sigma=\frac{5 \times {10}^{4} }{7.85 \times {10}^{ - 5} } =0. 636 \times 10 {}^{9} Pa \\ = 0.636GPa$

Therefore,The stress and strain are 0.636GPa and 0.01 respectively

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