A 5 kN tensile load is applied on a steel bar of 1000 mm2 cross section and 1 m length. What will be the deflection due to external load? (Take E = 200 GPa)a.25 cmb.25 mc.25 µmd.25 mm
Question
A 5 kN tensile load is applied on a steel bar of 1000 mm2 cross section and 1 m length. What will be the deflection due to external load? (Take E = 200 GPa)a.25 cmb.25 mc.25 µmd.25 mm
Solution
To solve this problem, we need to use the formula for deflection (δ) in a material under tensile or compressive stress, which is given by:
δ = PL / AE
where: P = applied force L = length of the material A = cross-sectional area of the material E = modulus of elasticity of the material
Given in the problem: P = 5 kN = 5000 N (since 1 kN = 1000 N) L = 1 m = 1000 mm (since 1 m = 1000 mm) A = 1000 mm² E = 200 GPa = 200 * 10^3 N/mm² (since 1 GPa = 10^3 N/mm²)
Substituting these values into the formula, we get:
δ = (5000 N * 1000 mm) / (1000 mm² * 200 * 10^3 N/mm²)
After simplifying, we find that the deflection δ = 0.025 mm, or 25 µm.
So, the correct answer is c. 25 µm.
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