Q.

steel bar 10 mm x 10 mm cross section is subjected to an axial tensile load of 20kN. If the length of bar is 1 m and E = 200 GPa, then elongation of the bar is:

A. 1 mm
B. 0.5 mm
C. 0.75 mm
D. 1.5 mm
Answer» A. 1 mm
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Discussion

NOOR MAZNI BINTI ISMAIL
1 year ago

The correct answer is A. 1 mm.
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NOOR MAZNI BINTI ISMAIL
1 year ago

cross sectional area, A = 10 mm x 10 mm = 100 mm^2
σ = F/A = 20 000 N /100 mm^2 = 200 N/mm^2 = 200 x 10^6 Pa
E = σ / strain
strain = σ / E = 200 x 10^6 Pa / 200 x 10^9 Pa = 0.001
strain = dL / L
elongation, dL = 0.001 x 1 000 mm = 1 mm (Ans.)
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McqMate
1 year ago

You're correct. I've fixed the answer.
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