Calculating Young's Modulus (E) for Metal Alloy, Low Carbon Steel, Copper

Click For Summary

Homework Help Overview

The discussion revolves around calculating Young's Modulus (E) for three materials: a metal alloy, low carbon steel, and copper, using provided force and extension data for each test piece. The test pieces share the same dimensions of 25mm in length and 4mm in diameter.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between tensile stress and tensile strain, with one suggesting the need to plot a force-extension graph to determine Young's Modulus. Questions arise regarding the definitions of tensile stress and strain, as well as how to interpret the graph for necessary values.

Discussion Status

Some participants have provided guidance on plotting the graph and interpreting the data, while others express uncertainty about how to extract relevant measurements from the graph. There is an ongoing exploration of the correct approach to use the data effectively.

Contextual Notes

Participants note confusion regarding the measurements for extension and the appropriate force to use in calculations. There is also mention of external factors, such as advertisements, affecting the clarity of the data presentation.

MathsRetard09
Messages
22
Reaction score
0

Homework Statement



Determine Young's Modulous (E) for the following materials:
Metal alloy - Test Piece 1 (TP1)
Low Carbon Steel - Test Piece 2 (TP2)
Copper - Test Piece 3 (TP3)

Each test piece has the same following dimentions - Length 25mm Diamter 4mm

Data:

TP1 = Force(kN) - 1.0 | 2.0 | 3.0 | 4.0 | 5.0 | 6.0 | 7.0 | 7.5 | 8.0 | 8.5 |
Extension (mm) - 0.012 | 0.024 | 0.040 | 0.046 | 0.60 | 0.072 | 0.084 | 0.093 | 0.11 | 0.13 |

TP2 = Force(kN) - 0.5 | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 | 3.5 | 4.0 |
Extension (mm) - 0.005 | 0.009 | 0.015 | 0.02 | 0.024 | 0.028 | 0.037 | 0.045 |

TP3 = Force(kN) - 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 | 0.7 | 0.8 | 0.9 | 1.0 |
Extension (mm) - 0.0014 | 0.003 | 0.0045 | 0.006 | 0.0075 | 0.0092 | 0.0108 | 0.0122| 0.014 | 0.018 |


Homework Equations



Don't really know but my guess is something along the lines of:

E = Tensile Stress / Tensile Strain ??

Would apreciate if i could have the correct equation for this.

The Attempt at a Solution



Just playing a guessing game and don't really know what I am doing, so not really looking for answers but mainly how to equate it all, if you could give me an example using the first one 'Metal Alloy' showing from start to finish with the answer then i'll figure out how to do the last two.

Then if i do get stuck i'll come back and let you guys know :)
 
Physics news on Phys.org
What is tensile stress? What is tensile strain? Expand the formula and you will have an equation connecting them (Young modulus, force, area, extension, length) all.
I suggest you then plot a force-extension graph. From the graph, you will find E. Hint: The graph should be a line, though all the points are not on the line. The data is not really nice.
 
Hey, I've done the graphs, i know what I am looking at, but it's what exactly am i looking for on the graph to take note of to put into the equation.

Length = 25mm

Cross-Section will be the diameter 4mm or the radius of that??

There is an extension but i don't know what the measurement for that is, the tests pieces themselves i don't have with me to measure. So for the extension how do i read from the graph to figure out what it is?

The forces are different for each subject, do i add into the equation the top force on the tale of data for each? or am i looking for the maximum force before breaking point?

As for the data, that can't be helped now that there's a massive advertisement on it. I'll try and clear it up a bit.

Thanks for your reply though much appreciated :)
 
MathsRetard09 said:
Cross-Section will be the diameter 4mm or the radius of that??
Cross-sectional diameter = 4mm as given in the problem.

There is an extension but i don't know what the measurement for that is, the tests pieces themselves i don't have with me to measure. So for the extension how do i read from the graph to figure out what it is?

The data has already given you extension of each measurement, and you use it to plot the graph, so what's the point of finding it in the graph?

The forces are different for each subject, do i add into the equation the top force on the tale of data for each? or am i looking for the maximum force before breaking point?

No. All you need to do with the data is to use it to plot a graph of force versus extension.

What is the equation you got anyway? (the one relates force, E, area, length and extension) This equation should be written with notations only; don't hastily plug the data in it as it's unnecessary. The data is only for plotting the graph.
 

Similar threads

Replies
2
Views
2K
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
1
Views
7K