Thursday

March 26, 2015

March 26, 2015

Posted by **Catherine** on Saturday, September 10, 2011 at 9:19pm.

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?

- Mechanics -
**bobpursley**, Saturday, September 10, 2011 at 9:25pmsigma 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

**Answer this Question**

**Related Questions**

mechanical properties of materials - following data was obtained from a tensile ...

mechanical properties of materials - following data was obtained from a tensile ...

mechanical properties of materials - following data was obtained from a tensile ...

mechanics of materials - The following data was obtained from a tensile test of ...

mechanics of materials - The following data was obtained from a tensile test of...

college- strength of material - i have initial and final gauge length and ...

mechanics - Sketch a stress-strain curve for which the stress is calculated ...

physics - A strip of rubber originally 75 mm long is stretched until it is 100 ...

college - I have initial and final gauge length and diameter and also have max ...

mechanics - The change in length of a steel wire under an applied load can be ...