Question

A steel rod of area of cross-section 3.14× 10-4 m2 is stretched by a force of 100kN. Calculate the stress acting on the rod.

Answer 1

Answer :

• Stress experienced by the rod is 0.3 Pa

Explanation :

Given :

• Area of cross-section, a = 3.14 × 10⁻⁴ m².
• Force acting on the steel rod, F = 10 kN or 10000 N

To find :

• Stress experienced by the rod, σ = ?

Knowledge required :

By definition, stress acting on a body is its force per unit area.

Formula for stress acting on a body :

⠀⠀⠀⠀⠀⠀⠀⠀⠀σ = F/A⠀

Where :

• σ = Stress acting on the body.
• F = Force exerted on the body.
• A = Area of the body.

Solution :

By using the formula for stress acting on a body and substituting the values in it, we get :

⠀⠀=> σ = F/A

⠀⠀=> σ = 10000/(3.14 × 10⁻⁴)

⠀⠀=> σ = 10000/(314/100 × 10⁻⁴)

⠀⠀=> σ = 10000/(314 × 10⁻⁴ × 10⁻²)

⠀⠀=> σ = 10000/(314 × 10⁻⁶)

⠀⠀=> σ = 10⁴/(314 × 10⁻⁶)

⠀⠀=> σ = 10²/314

⠀⠀=> σ = 1/(314 × 10⁻²)

⠀⠀=> σ = 1/3.14

⠀⠀=> σ = 0.3 (approx.)

⠀⠀⠀⠀∴ σ = 0.3 Pa

Therefore,

• Stress experienced by the rod, σ = 0.3 Pa.

Answer 2

Answer:-

$\small\ \bf Longitudinal \: stress = \frac{force}{cross \: sectional \: area}$

$\ \bf Force = 100kN \: or \: 10000N$

$\bf Area = 3.14 × {10}^{ - 4} {m}^{2}$

$\ \bf Longitudinal \: stress = \frac{10000}{3.14 \times {10}^{ - 4} }$

$\ \bf = \frac{100}{314 \times {10}^{ - 4} }$

$\bf \ = \frac{ {10}^{2} }{314 \times {10}^{ - 4} }$

$\bf \ = \frac{1}{314 \times {10}^{ - 2} }$

$\bf = \frac{1}{3.14}$

$\bf \ = 0.3(approximately)$

∴ Stress acting on the rod = 0.3 Pa

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