# 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 !

## Answers and Replies

Mentor
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?

Homework Helper
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?

Mentor
Why is stress zero?
Because the bar is unconstrained while it is expanding. There are no forces acting on it.

Hamza Abbasi
Hamza Abbasi
Thank you for guiding . Problem solved !