Question

A metal cylinder has an original diameter of 1.00 cm and length of 5.00 cm. A tension test was performed on the specimen and the data are listed in the table below.

                                               Load (kN)                    Elongation (cm)

0.00                                 0.0000

1.51                                0.0006

4.61                                 0.0016

8.01                                 0.0026

11.01 0.0036

11.71 0.0051

11.81                               0.0081

12.01                                0.0201

16.61 0.0401

20.01 0.1001

21.51                                0.2801

19.51                                0.4001

18.51                                0.4601                

a) Graph the stress on the specimen versus the strain.

b) Considering the elastic region, find the slope of the best-fit straight line and determine the elastic modulus of the metal.

c) What is the likely identity of the material from which the cylinder is made?

Answer 1

From the given data

Load(kN) Elongation in (cm)

0 0

1.51 0.0006

4.61 0.0016

8.01 0.0026

11.01 0.0036

11.71 0.0051

11.81 0.0081

12.01 0.201

16.61 0.401

20.01 0.1001

21.51 0.2801

19.51 0.4001

18.51 0.4601

the strain is obtained by elongation/5 cm ( 5 cm is length of the cylinder)

hence

the following Load strain curve is obtained

the slope of the linear region from linear regression is m

m = 15182 kN

but m = E*A

hence

E = 15182/A = 15182/pi*(0.5*10^-2)^2 = 0.1933 GPa

hence this material can be PTFE (poly tetra fluoro ethylene)