# Stress vs. Strain

1. Sep 13, 2007

### Link-

1. Find modulus of toughness
Given Data :
Load (kip)
0.00
1.50
4.60
8.00
11.00
11.80
11.80
12.00
16.60
20.00
21.50
19.50
18.50

Elongation
0.0000
0.0005
0.0015
0.0025
0.0035
0.0050
0.0080
0.0200
0.0400
0.1000
0.2800
0.4000
0.4600

Lenght= 2 in
Diameter= 0.501 in

2. Relevant equations:
$$\epsilon$$=$$\delta$$/L
$$\sigma$$=F/A

3. The attempt at a solution
Since the Prof. want the graph from excel I made the calculations on the sheet to obtain the Stress vs. Strain graph.

ε (in/in)
0
0.00025
0.00075
0.00125
0.00175
0.0025
0.004
0.01
0.02
0.05
0.14
0.2
0.23

σ (ksi)
0
7.608970949
23.33417758
40.58117839
55.79912029
59.85723813
59.85723813
60.87176759
84.20594517
101.452946
109.0619169
98.91662233
93.84397503

Now I obtain the graph with the chart wizard.

4. Problem: How could I obtain the modulus of toughness with this graph? There's way in excel to obtain the are under the graph? (Not using any numerical method, let excel do it)

Thanks

2. Sep 13, 2007

### Astronuc

Staff Emeritus
In the elastic region, $\sigma$ = E $\epsilon$, so one simply needs to find the slope.

In Excel, there is an option to fit the data and then one can select the option to display the equation, which will give y as f(x). However, fit only the data up to yield point, which means just plotting the data separately up to the yield point and then selecting a linear curve fit. Select a straight line fit.

Alternatively, one can simply take the stress at or just before yield and divide by the corresponding strain, since the curve starts at (0,0).

3. Sep 13, 2007

### Link-

I was thinking on taking the area under the graph with the triangle formula uisng $$\sigma$$ (proportional limit) as h and the strain corresponding to it. Like this

A= $$\sigma$$$$_{pl}$$*$$\epsilon$$$$_{pl}$$*.5, that's how I found the resilence modulus. But the rest of the graph is not linear and is not easily describe with a function.

About the fit using excel... How can I do a fit to the graph? If it can make a fit I could integrate to find the area very easily with the function.

Thanks

Last edited: Sep 13, 2007
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