A brass rod 1.5 m long and 20mm diameter was found to deform by 1.9mm under tensile load of 40KN. Calculate the modulus of elasticity.
Answers
A brass rod 1.5 m long and 20mm diameter was found to deform by 1.9mm under tensile load of 40kN.
To find : The modulus of elasticity.
solution : cross sectional area of rod, A = πr²
here r = 10mm, l
= 3.14 × (10 × 10¯³)²
= 3.14 × 10¯⁴
= 3.14 × 10¯⁴ m²
strain = ∆l/l = 1.9mm/1.5 m
= (1.9 × 10¯³)/(1.5)
= 1.27 × 10¯³
now Modulus of elasticity = F/A × (∆l/l)
= (40kN)/(4.71 × 10¯⁴ × 1.27 × 10¯³)
= (40/3.14 × 1.267) × 10¹⁰
= 10.05 × 10¹⁰ N/m²
Therefore the modulus of elasticity is 10.05 × 10¹⁰ N/m² or 100.5 × 10³ N/mm².
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Answer:
A brass rod 1.5 m long and 20mm diameter was found to deform by 1.9mm under tensile load of 40KN. Calculate the modulus of elasticity.