If a load of 40 kN is applied in a compressive manner on a rod wh
| 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?
A. 0.2
B. 2
C. 20
D. 200
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Right Answer is: D
SOLUTION
\(\sigma = \frac{P}{A} = \frac{P}{{bd}} = \frac{{40 \times 1000}}{{10 \times 20}} = 200\frac{N}{{m{m^2}}} = 200\;MPa\)