Strain produced in a rod after expansion

[Hamza Abbasi]

Homework Statement

A rod of length $L_o$ is kept on a friction-less surface. The coefficient of linear expansion for the material of the rod is $\alpha$. The the temperature of the rod is increased by $\Delta T$ the strain developed in the rod will be?

Homework Equations

1. $$ \Delta L= L_o(1+\alpha \Delta T) $$
2. $$Strain (Linear ) = \frac{\Delta L}{ L_o}$$

The Attempt at a Solution

$$ Strain= \frac{ L_o(1+\alpha \Delta T)}{L_o} $$
$$ Strain =(1+\alpha \Delta T)$$

Whereas the answer in book is zero !

 

[Chestermiller]

Your answer is incorrect, and so is the answer in the book. The first equation should read $$\Delta L=L_0(1+\alpha \Delta T-L_0=L_0\alpha \Delta T$$So the strain is just $\alpha \Delta T$. Are you sure they weren't asking for the stress?

 

[mjc123]

The book answer is wrong. The stress developed in the rod will be zero if the surface is frictionless, but the strain won't.
Your answer is also wrong. Equation 1 should be L = L₀(1 + αΔT). Then ΔL = L - L₀.
(Note this is an approximation that applies when ΔT is small. What if it is large?)

Edit: Beat me to it!

 

[Hamza Abbasi]

Your answer is incorrect, and so is the answer in the book. The first equation should read $$\Delta L=L_0(1+\alpha \Delta T-L_0=L_0\alpha \Delta T$$So the strain is just $\alpha \Delta T$. Are you sure they weren't asking for the stress?

Yes , I am sure . Question was about strain/

 

[Hamza Abbasi]

The book answer is wrong. The stress developed in the rod will be zero if the surface is frictionless, but the strain won't.
Your answer is also wrong. Equation 1 should be L = L₀(1 + αΔT). Then ΔL = L - L₀.
(Note this is an approximation that applies when ΔT is small. What if it is large?)

Edit: Beat me to it!

Oh yes! I wrote equation 1 wrong !
Got it

 

[Hamza Abbasi]

The book answer is wrong. The stress developed in the rod will be zero if the surface is frictionless, but the strain won't.
Your answer is also wrong. Equation 1 should be L = L₀(1 + αΔT). Then ΔL = L - L₀.
(Note this is an approximation that applies when ΔT is small. What if it is large?)

Edit: Beat me to it!

Why is stress zero?

 

[Chestermiller]

Why is stress zero?

Because the bar is unconstrained while it is expanding. There are no forces acting on it.

 

[Hamza Abbasi]

Thank you for guiding . Problem solved !