A tension test was performed on a steel specimen having an original diameter of 12.5 mm and gauge length

Question

of 50 mm. Using the data listed in the table, plot the stress–strain diagram, and determine approximately the
modulus of toughness. Use a scale of 20 mm-50 MPa and 20 mm = 0.05 mm/mm.

Load (kN)
0
11.1
31.9
37.8
40.9
43.6
53.4
62.3
64.5
62.3
58.8

Elongation (mm)
0
0.0175
0.0600
0.1020
0.1650
0.2490
1.0160
3.0480
6.3500
8.8900
11.9380

I know that I have to use the equations sigma=(P/A) and epsilon=(dL/L) to get the points for stress and strain... but how do I get the values for the equations? I tried to get the stress with the second values of the table:

sigma=(11.1e3 Pa)/(pi*(0.0175e3m)^2)
I get 11.54 and it is supposed to be 90.45 MPa. How do I do these?

Answer 1

sigma is in fact Pressure, which is force/area. you are given force, compute area from the diameter.

epsilon is change of length divided by length. You are given deltaL (elongation). L was given as 50mm.
Surely you can calculate delta L /L

For each data point given, calculate those two quantities (sigma, deltaL/L), and plot the graph