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Two similar bars of Steel and Aluminium are heated to a same temperature. Forces are applied at the ends of the bars to maintain their lengths unaltered. If the ratio of Young’s moduli of Steel and Aluminium is 3, and the ratio of the coefficients of thermal expansion of Steel to that of Aluminium is 0.5, what is the stress on the Aluminium bar if the stress on the Steel bar is 100 MPa?

Two similar bars of Steel and Aluminium are heated to a same temperature. Forces are applied at the ends of the bars to maintain their lengths unaltered. If the ratio of Young’s moduli of Steel and Aluminium is 3, and the ratio of the coefficients of thermal expansion of Steel to that of Aluminium is 0.5, what is the stress on the Aluminium bar if the stress on the Steel bar is 100 MPa? Correct Answer 66.7 MPa

Concept:

Ordinary materials expand when heated and contract when cooled, hence, an increase in temperature produce a positive thermal strain. Thermal strains usually are reversible in a sense that the member returns to its original shape when the temperature return to its original value.

dt = α .L.t or εt  = αt or σt = Eαt

Where,

dt = Elongation or contraction of bar and εt = Thermal strain

α = coefficient of linear expansion for the material

L = original Length

t = temp. Change

σt = Stress due to temperature change

E = Modulus of elasticity of the material

If however, the free expansion of the material is prevented by some external force, then a stress is set up in the material. They stress is equal in magnitude to that which would be produced in the bar by initially allowing the bar to its free length and then applying sufficient force to return the bar to its original length.

The value of stress is given above.

Calculation:

Given,

 Ratio of Young’s moduli of Steel and Aluminium (Es/EAl) = 3

Ratio of the coefficients of thermal expansion of Steel to that of Aluminium (αs/αAl) = 0.5

Stress on the Steel bar (σs) = 100 MPa

So σs = Esαst

σAl = EALαAlt

The temperature difference for both the bar is the same.

So (σs/σAl) = (Es/EAl)(αs/αAl)

(100/σAl) = 3 x 0.5 = 1.5

∴ σAl = 100/1.5 = 66.67 MPa

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