Question

A rod is 2m long at a temperature of 10C. Find the expansion of the rod, when

the temperature is raised to 80C. If this expansion is prevented find the stress

induced in the material of the rod. Take E = 1.0 105 MN/m2

and  = .000012/C​

Answer 1

HOPE IT HELPS U : )

Free expansion of the rod =αLΔθ

=15×10

−60

C×2m×(50−20)

0

C

=9×10

−

4m=0.9mm

If the expansion is fully prevented, then

Strain=

2

9×10

−4

=4.5×10

−4

Temperature Stress = Strain ×Y

=4.5×10

−4

×2×10

11

=9×10

7

N/m

2

If 0.4 mm expansion is allowed, then length restricted to expand

=0.9−0.4=0.5mm

Strain=

2

5×10

−4

=2.5×10

−4

Temperature stress = Strain ×Y=2.5×10

−4

×2×10

11

=5×10

7

N/m

2

Answer 2

Answer:

Therefore the stress induced in the rod is $S=8.4\times10^7 Pa$

Given:

• A rod of length $L=2.0m$ at a temperature $T_{1}=10\textdegree C$ is heated to a final temperature $T_{2} =80\textdegree}C$
• Young's modulus of the material of the rod is $E=1.0 \times 10^{5} MPa=1.0\times10^{11}Pa$
• The linear thermal expansion coefficient of the material of the rod is $\alpha =1.2\times10^{-5}\; \textdegree C^{-1}$

Explanation:

If the expansion is stopped, we must determine the stress that will be placed on the rod.

With the specified increase in temperature, the rod's length increases by $\triangle L=L\times \alpha \times(T_{2} -Tx_{1} )$

The strain produced in the rod due to the thermal expansion is

$\epsilon=\frac{\triangle L}{L}$

$\epsilon=\alpha \times(T_{2}-T_{1})$

$\epsilon=1.2\times10^{-5} \;\textdegree C^{-1}\times (80-10)\;\textdegree C$

$\epsilon=8.4\times10^{-4}$

If the rod's Young's modulus is $E$, then stress $S=E\times \epsilon$ can cause the rod to stretch, compress by length $\triangle L$, or generate strain $\epsilon$.

Therefore, if the rod's thermal expansion is stopped, the same stress $S$will be applied to it.

Therefore the stress induced in the rod is

$S=E\times \epsilon$

$S=1.0 \times 10^{11} Pa \times8.4 \times10^{-4}$

$S=8.4\times10^7 Pa$

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