SSC JE Mechanical Engineering 2018 Set 3 Paper

Question : 124 of 200

Marks: +1, -0

If a load of 40 kN is applied in a compressive manner on a rod whose cross section is 10 mm × 20 mm. Then what will be the compressive stress (in MPa) on the rod?

Validate

Solution:

stress in (MPa)

={40 ×10^3}/{10×20}={40,000}/200=200

MPa

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