A steel bar 2.5cm in diameter is rigidly attached to two parallel supports which are 5m apart. Find the stress and the change in the diameter of the bar when the temperature is increased by 100∘C.
Take α=12×10−6/∘C,Y=210GPa.
A
252MPa, 0.03mm
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B
150MPa, 0.13mm
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C
59MPa, 0.33mm
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D
212MPa, 0.13mm
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Solution
The correct option is A252MPa, 0.03mm
Let us assume σ be the thermal stress induced by the temperature increment.
Since supports are rigid , there is zero net deflection in bar AB ∴δAB=0 ⇒δfe+δth=0......(1)
where, δfe= Thermal free expansion in bar, in the absence of rigid support. δth= contraction due to thermal stress.
We know that, when the temperature of a rod is changed by ΔT , From the concept of linear expansion ,the Change in length of the rod is given by ΔL=δfe=LαΔT and contraction due to thermal stress is given by δth=−σLY∵Y=stressstrain
Using the above formulae in (1) we get, αΔTL−σLY=0⇒σ=YαΔT
From the data given in the question , σ=210×109×12×10−6×(100−0)⇒σ=252MPa
Change in the diameter (d) of the bar is given by Δd=dαΔT=(0.025×12×10−6×(100−0))×103mm ⇒Δd=0.03mm
Hence, option (a) is the correct answer.