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Two rectangular bars of equal area 50 mm2 are subjected to two different stress of 200 N/mm2 and 400 N/mm2, respectively. Determine the total load on the composite bar.
1. 30 kN
2. 60 kN
3. 50 kN
4. 80 kN

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Correct Answer - Option 1 : 30 kN

Concept:

Stress:
 
The internal restoring force acting per unit area of the cross-section of the deformed body is called stress.

\(\rm Stress=\dfrac{Applied\;Force }{Area}=\dfrac{F}{A}\)

Its unit is N/mm2 Or MPa.

Calculation:

Given:

A = 50 mm2, σ1 = 200 N/mm2, σ2 = 400 N/mm2.

\(\rm Stress=\dfrac{Applied\;Force }{Area}=\dfrac{F}{A}\)

∴ Force = Stress × Area

F1 = σ1 × A  = 200 × 50 = 10000 N

F= 10000 N = 10 kN

F2 = σ2 × A = 400 × 50 = 20000 N 

F= 20000 N = 20 kN

∴ Total Load, F = F+ F2  = 10 + 20 = 30

Thus, Total Load = 30 kN.

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