Stress/Strain with two rods joined together

  • Thread starter Thread starter aal0315
  • Start date Start date
Click For Summary

Homework Help Overview

The problem involves a rod composed of two sections made of different materials, steel and brass, which are joined end to end. The scenario describes a fixed end with the other end being pulled, resulting in a total change in length of 1.20 mm. Participants are exploring how much each section of the rod lengthens under this condition.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relationship between stress, strain, and the Young's modulus for each material. There is an exploration of how the total change in length relates to the individual changes in length of the steel and brass sections. Some participants question the assumptions regarding the deformation of each material and how to express the total change in length mathematically.

Discussion Status

There is an ongoing exploration of the deformation characteristics of the two materials, with some participants suggesting that the brass deforms more than the steel. A specific equation relating the changes in length of the two sections has been proposed, and there is acknowledgment of the reasoning behind it, though no consensus has been reached on the final values.

Contextual Notes

Participants are working under the assumption that the force and cross-sectional area are constant across both sections of the rod. There is also a noted uncertainty regarding the total length of the rod and how it affects the calculations.

aal0315
Messages
41
Reaction score
0

Homework Statement


A rod is made of two sections joined end to end. The sections are identical, except the one is steel and the other is brass. While one end is held fixed, the other is pulled to result in change in length of 1.20 mm. By how much does the length of each section increase?


Homework Equations


Strain = stress / E
Stress = F/A
delta L = 1/E x L


The Attempt at a Solution


I know that E(steel) = 200x10^9 and E(brass) = 100x10^9
I know that the total length is L= L1 +L2, where L1 is the length of the steel and L2 is the length of the brass
I know that the force and cross-sectional area is the same, so those don't matter.
I feel that the delta L = 1/(E(steel)+E(brass))xL would be right, but i don't know where to find the value of L.
or is it that delta L = 1/E(steel)xL1, where delta L = 1.20mm? but i feel that this is wrong because it says in the question that the whole rod lengthens that much.
any suggestions?
 
Physics news on Phys.org
the total change in length of the rod is equal to the sum of the change in lengths of its component parts. If E_B = 1/2E_S, which piece deforms more, and how much more?
 
the brass deforms and it is double the amount of the steel, so is this right:
1.2 = x+2x , where x is the length of steel
x = 0.4mm and the length of brass increases by 0.8mm?
 
aal0315 said:
the brass deforms and it is double the amount of the steel, so is this right:
1.2 = x+2x , where x is the length of steel
x = 0.4mm and the length of brass increases by 0.8mm?
Yes, very good.
 
thank you :o)
 

Similar threads

Find the electric field at a point away from two charged rods
  • · Replies 11 ·
Replies
11
Views
4K
Basic Stress and Strain question -- Rock on top of a vertical column
  • · Replies 19 ·
Replies
19
Views
2K
Length change for a rod of two sections
  • · Replies 14 ·
Replies
14
Views
2K
Temperature of a bar to produce certain force on wall
  • · Replies 6 ·
Replies
6
Views
2K
Finding the net elongation in a rod due to its own weight
  • · Replies 13 ·
Replies
13
Views
3K
Stress, Strain, and Sound in a Projectile Steel Rod
  • · Replies 2 ·
Replies
2
Views
2K
Platinum resistance thermometer and Kirchoff's Laws
  • · Replies 1 ·
Replies
1
Views
3K
Calculate the stress in the duralumin rod
  • · Replies 4 ·
Replies
4
Views
2K
Confusion Calculating Young's Modulus
  • · Replies 5 ·
Replies
5
Views
2K
Temperature at the junction of 2 wires
  • · Replies 8 ·
Replies
8
Views
2K