Physics · Thermal Properties Of Matter · NEET
Free expansion is delta L = L alpha delta T, so the longer the rod, the more it wants to grow. But the thermal strain is delta L / L = alpha delta T, and the L cancels. Stress depends only on strain: stress = Y alpha delta T. So a 1 m rod and a 5 m rod of the same material and same temperature rise develop the exact same stress. Length only matters for how much a free rod expands, not for the stress in a clamped rod.
Thermal strain is the fractional length the rod would have grown if it were free: strain = alpha delta T (no units). Thermal stress is the internal push per unit area caused by the walls stopping that growth: stress = Y times strain = Y alpha delta T (units N/m^2). First find the strain, then multiply by Young's modulus Y to get the stress.
Compressive. On heating, the rod tries to expand and pushes outward on the walls; the walls push back inward, squeezing the rod. That inward squeeze is a compressive stress. If instead you cool a clamped rod, it tries to shrink, the walls hold it stretched, and the stress becomes tensile (pulling).
Stress is force per unit area, so force = stress times area. F = (Y alpha delta T) times A = Y A alpha delta T. Here A is the area of cross-section of the rod. Do not confuse this with the length or volume; only the cross-section area is used for force.
No. delta T is a temperature difference, and a change of 1 degree Celsius equals a change of 1 Kelvin. So heating from 0 to 100 C gives delta T = 100, whether you call it 100 C or 100 K. Only use absolute Kelvin when a single temperature (not a difference) is needed, such as in gas laws.
A metallic bar of Young's modulus 0.5 x 10^11 N/m^2 and coefficient of linear thermal expansion 10^-5 per degree C, length 1 m and area of cross-section 10^-3 m^2 is heated from 0 C to 100 C without expansion or bending. The compressive force developed in it is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Thermal stress = Y alpha delta T, where Y is Young's modulus, alpha is the coefficient of linear expansion, and delta T is the temperature change. The force is F = Y A alpha delta T, with A the cross-section area.
On heating, the rod naturally tries to get longer by delta L = L alpha delta T. Rigid walls stop this, which is the same as compressing the rod back by that amount. The walls therefore exert a force, and the internal push per unit area is the thermal stress.
No. The length cancels because strain = delta L / L = alpha delta T. Stress = Y alpha delta T has no L in it, so two rods of different lengths (same material, same delta T) have equal stress.
The stress does not depend on area, but the force does. Stress = Y alpha delta T is the same for any area, while force F = stress times A = Y A alpha delta T grows with a thicker rod.
Compressive on heating (the rod is squeezed) and tensile on cooling (the rod is stretched by the walls that stop it from shrinking).
