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A steel rod of dimensions 4x4 cm square is tightly fixed between two supports and is not allowed to expand. It is heated by 2°C. The thermal stress developed is 10⁶ N/m².

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The thermal stress developed in the steel rod is 4680 N/cm², which is approximately equal to 3) 3.6 MPa. Therefore , 3) 3.6 MPa is correct .

The thermal stress developed in the steel rod can be calculated using the following formula:

σ = EαΔT

where:

σ is the thermal stress (N/m²)

E is the Young's modulus of steel (200 GPa)

α is the coefficient of thermal expansion of steel (11.7 × 10⁻⁶ m/°C)

ΔT is the change in temperature (°C)

In this case, the dimensions of the steel rod are 4 cm × 4 cm, which is equal to 0.04 m × 0.04 m.

The change in temperature is 2°C. Plugging these values into the formula, we get:

σ = (200 × 10⁹ N/m²) × (11.7 × 10⁻⁶ m/°C) × (2°C) = 46800000 N/m²

Converting this to N/cm², we get:

σ = 46800000 N/m² ÷ 10000 cm²/m² = 4680 N/cm²

Therefore, the thermal stress developed in the steel rod is 4680 N/cm², which is approximately equal to 3.6 MPa.

Question

A steel rod of dimensions 4 x 4cm is tightly fixed between two supports and is not allowed to expand. It is heated through 2°C. Thermal stress developed is ..... 10 N/m² (Y= 20 x101°N/m a = 18 x 10-6/°C)

1) 7.2

2) 2.7

3) 3.6

4) 0.72

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