Youngs Modulus of copper wire experiment


by jokiemay
Tags: experiment, youngs modulus
jokiemay
jokiemay is offline
#1
Jan18-12, 12:32 PM
P: 18
1. The problem statement, all variables and given/known data
I’m doing some basic young’s modulus homework and I think I’ve solved it but I want other people to give their answers to see if the match my own. Its young are of copper wire.



2. Relevant equations
E = stress over strain
Stress = applied force (F) over Cross sectional area (A)
Strain = Extension (e) over Original length (Lo)
A = 0.27mm
F = mxg
Lo = 1.788m
e = variable

My table of results are:
M / kg e / mm
0.100 0.45
0.200 0.94
0.300 1.40
0.400 1.97
0.500 2.46
0.600 3.29
0.700 4.12
0.800 5.00
0.900 6.36
1.000 8.22


3. The attempt at a solution

I’ve worked it out with 0.100kg

M (0.100) x g (9.81) = 0.981 so F = 0.981
Pi x 0.27 squared over 4 = 0.05725566 so A = 0.05725566

So stress = 17.13371733038794867967

e = 0.45
Lo = 1.788

So strain = 0.25167785234899

Young’s Modulus = 68.0779701927423389726993185

I know I’m going wrong somewhere so any help would be appreciated
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LawrenceC
LawrenceC is offline
#2
Jan18-12, 02:13 PM
P: 1,195
You have an error in your strain computation. You seem to be dividing mm/m.
jokiemay
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#3
Jan18-12, 02:21 PM
P: 18
Ohh yes i see that - i know im going wrong somehwere and i cant get my head fully around that.

Why would the strain be such a low figure but the stress be so high

Im sure 00025167785235 is to low for strain

jokiemay
jokiemay is offline
#4
Jan18-12, 02:22 PM
P: 18

Youngs Modulus of copper wire experiment


Ohh yes i see that - i know im going wrong somewhere and i cant get my head fully around that.

Why would the strain be such a low figure but the stress be so high

Im sure 00025167785235 is to low for strain
jokiemay
jokiemay is offline
#5
Jan18-12, 02:34 PM
P: 18
im getting 68077.97019246913753978559011651

couldnt be right
LawrenceC
LawrenceC is offline
#6
Jan18-12, 02:38 PM
P: 1,195
Because

sigma = e * E

where E is Young's Modulus, a large number. Strain, e, is small. For steel, the modulus is 30X10^6 psi. sigma is stress. So as an example, for a strain of 0.00025, the stress is 7500 psi. Reasonable stress example.
jokiemay
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#7
Jan18-12, 02:42 PM
P: 18
so would be answer of 68077.97019246913753978559011651 be right ?
LawrenceC
LawrenceC is offline
#8
Jan18-12, 02:51 PM
P: 1,195
Units would be N/mm^2. Watch your significant figures. Based on your first set of data, you did it correctly but consider the comment below.

You say A is area and its value is 0.27 mm. Is that a typo? Should 0.27 be the diameter? That is how you used it.

I looked up the modulus for Cu. It's somewhat higher than what you calculate.
jokiemay
jokiemay is offline
#9
Jan18-12, 03:02 PM
P: 18
Yes the 0.27 is the diameter which is the cross sectional area (A)

I calculated that as Pi x 0.27squared / 4

Thanks very much for your help so far.
jokiemay
jokiemay is offline
#10
Jan19-12, 04:05 AM
P: 18
can anybody else hint to me where im going wrong as i see Cu is suppost to be 117GPa
LawrenceC
LawrenceC is offline
#11
Jan19-12, 08:05 AM
P: 1,195
If you plot your data, the curve looks ok. Are you certain you measured the diameter and wire length correctly?
jokiemay
jokiemay is offline
#12
Jan22-12, 07:40 AM
P: 18
Hi there - My mistake was the diameter - it should have been calculated as 0.000573mm or 5.73x10-3

so the result in getting now is 6.03 x 10(10) Pa as the youngs modulus.

Im also working out the percentage error;

ive worked out the length as 1.788m +/- 0.01m so 0.01/1.788 x 100 = +/- 0.56%

does this look correct?

My next task is to work out the percentage uncertainty of the cross section area and the gradient of the graph.


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