wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A steel bar 2.5 cm in diameter is rigidly attached to two parallel supports which are 5 m apart. Find the stress and the change in the diameter of the bar when the temperature is increased by 100C.
Take α=12×106/C,Y=210 GPa.

A
252 MPa, 0.03 mm
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
150 MPa, 0.13 mm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
59 MPa, 0.33 mm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
212 MPa, 0.13 mm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 252 MPa, 0.03 mm

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,
αΔTLσLY=0σ=YαΔT
From the data given in the question ,
σ=210×109×12×106×(1000)σ=252 MPa
Change in the diameter (d) of the bar is given by
Δd=dαΔT =(0.025×12×106×(1000))×103 mm
Δd=0.03 mm
Hence, option (a) is the correct answer.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon