Strain produced in a rod after expansion

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Homework Help Overview

The discussion revolves around the strain produced in a rod due to thermal expansion when subjected to a temperature increase. The rod is described as being on a frictionless surface, and participants are analyzing the implications of this setup on strain and stress.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants explore the relationship between temperature change, linear expansion, and the resulting strain. There are attempts to clarify the correct equations for calculating strain and to address the discrepancy between the original poster's calculations and the book's answer.

Discussion Status

Multiple interpretations of the problem are being explored, particularly regarding the definitions of strain and stress in the context of a frictionless surface. Some participants have offered corrections to the equations presented, while others question the assumptions regarding stress and strain in this scenario.

Contextual Notes

There is an ongoing debate about the accuracy of the book's answer and the implications of the frictionless condition on stress and strain. Participants note that the approximation used may not hold for larger temperature changes.

Hamza Abbasi
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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 !
 
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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?
 
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!
 
Chestermiller said:
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/
 
mjc123 said:
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 :smile::smile:
 
mjc123 said:
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?
 
Hamza Abbasi said:
Why is stress zero?
Because the bar is unconstrained while it is expanding. There are no forces acting on it.
 
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Thank you for guiding :smile: . Problem solved !
 

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