Consider a 250 mm $\times$ 15 mm $\times$ 10 mm steel bar which is free to expand is heated from $15^\circ$C to $40^\circ$C. What will be developed?

Question

A rectangular section has width $b$ and depth $d$ and a symmetrical triangular section has base $b$ and depth $d$. The ratio of moment of inertia of rectangular section to that of triangular section with respect to their respective centroidal axis will be:

Answer 1

Stress only appears if a body is stopped from freely reaching its natural expanded (or contracted) shape. Here, nothing stops the bar's ends from moving outward as it heats up.

Compressive stress: needs a rigid boundary pushing back on the expanding bar; no such boundary is present, so ruled out.
Tensile stress: needs the bar's ends held apart or fixed while it tries to shrink; heating causes expansion, not shrinkage, and again no restraint exists, so ruled out.
Shear stress: needs relative sliding across a section under transverse or torsional action, absent in pure axial free expansion, so ruled out.
No stress: matches the physical situation exactly, since the bar simply grows in size with the strain fully absorbing the temperature change and zero internal force is generated.

So the bar develops no stress when heated while free to expand.

Answer 2

A third way to see this is by separating total strain into a thermal part and a mechanical part, and applying the stress-strain law only to the mechanical part.

The total strain of the bar is $ \varepsilon_{\text{total}} = \varepsilon_{\text{thermal}} + \varepsilon_{\text{mechanical}} $, where $ \varepsilon_{\text{thermal}} = \alpha \Delta T $ is the strain due to the temperature rise alone, and $ \varepsilon_{\text{mechanical}} $ is the strain associated with any actual stress via $ \sigma = E\,\varepsilon_{\text{mechanical}} $. Since the bar is free to expand, its actual length change equals exactly the thermal expansion, i.e., $ \varepsilon_{\text{total}} = \varepsilon_{\text{thermal}} $, which forces $ \varepsilon_{\text{mechanical}} = \varepsilon_{\text{total}} - \varepsilon_{\text{thermal}} = 0 $.

Compressive stress: would require $ \varepsilon_{\text{mechanical}} $ to be negative, but it was just shown to be exactly zero, so this is ruled out.
Tensile stress: would require $ \varepsilon_{\text{mechanical}} $ to be positive; again this contradicts $ \varepsilon_{\text{mechanical}} = 0$, so ruled out.
Shear stress: shear stress requires a mechanical shear strain component, but the loading here is purely a uniform axial temperature change with no transverse or torsional mechanical strain at all, so ruled out.
No stress: directly follows from $ \sigma = E\,\varepsilon_{\text{mechanical}} = E \times 0 = 0 $, matching this option exactly.

Splitting the strain into thermal and mechanical parts confirms mathematically that zero mechanical strain means zero stress.

Therefore, the correct answer is no stress is developed.

Answer 3

A third way to see this is by separating total strain into a thermal part and a mechanical part, and applying the stress-strain law only to the mechanical part.

The total strain of the bar is $ \varepsilon_{\text{total}} = \varepsilon_{\text{thermal}} + \varepsilon_{\text{mechanical}} $, where $ \varepsilon_{\text{thermal}} = \alpha \Delta T $ is the strain due to the temperature rise alone, and $ \varepsilon_{\text{mechanical}} $ is the strain associated with any actual stress via $ \sigma = E\,\varepsilon_{\text{mechanical}} $. Since the bar is free to expand, its actual length change equals exactly the thermal expansion, i.e., $ \varepsilon_{\text{total}} = \varepsilon_{\text{thermal}} $, which forces $ \varepsilon_{\text{mechanical}} = \varepsilon_{\text{total}} - \varepsilon_{\text{thermal}} = 0 $.

Compressive stress: would require $ \varepsilon_{\text{mechanical}} $ to be negative, but it was just shown to be exactly zero, so this is ruled out.
Tensile stress: would require $ \varepsilon_{\text{mechanical}} $ to be positive; again this contradicts $ \varepsilon_{\text{mechanical}} = 0$, so ruled out.
Shear stress: shear stress requires a mechanical shear strain component, but the loading here is purely a uniform axial temperature change with no transverse or torsional mechanical strain at all, so ruled out.
No stress: directly follows from $ \sigma = E\,\varepsilon_{\text{mechanical}} = E \times 0 = 0 $, matching this option exactly.

Splitting the strain into thermal and mechanical parts confirms mathematically that zero mechanical strain means zero stress.

Therefore, the correct answer is no stress is developed.

Consider a 250 mm $\times$ 15 mm $\times$ 10 mm steel bar which is free to expand is heated from $15^\circ$C to $40^\circ$C. What will be developed?

Show Hint

The Correct Option is D

Approach Solution - 1

Approach Solution -2

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