6 CHAPTER 1. PROPERTIES OF MATTER
Worked out Example 1.2.
The heels on a pair of women’s shoes have a radius of 0.5 cm at the bottom. If 30%
of the weight (480 N) of a woman is supported by each heel, find the stress on each
heel.
Solution:
Force on one heel=(0.3)◊(480N)
Area of a heel=fi◊r^2 =(3.14)(0.005m)^2
Stress =‡=
Force
Area
=
(0.3)◊(480N)
(3.14)(0.005m)^2
=1. 83 ◊ 106 N/m^2
A stress–strain test takes several minutes to perform and is destructive; that is, the test
specimen is permanently deformed and usually fractured. The output of such a tensile
test is recorded (usually on a computer) as load or force versus elongation to obtain the
engineering stress-strain curve.
Learning Resource : Apparatus and Procedure for Tensile Test
(1). Tensile testing using an automated Universal Testing
Machine - Youtube channelMaterialsScience
https://youtu.be/D8U4G5kcpcM
(2). Tensile testing and plotting of stress-strain graph -
Youtube channelToolNotes
https://youtu.be/uWgnBNOy-rA
When we look at the stress-strain graph in Figure1.5(a), we can see that the graph
has two regions, namelyelasticandplastic, separated by the point P as shown in Figure
1.5(b). In the elastic region (from the origin to the point P), the stress-strain graph is
linear, that is stress is proportional to strain and also the strain is very small for this
region. In the plastic region (beyond the point P), the graph is nonlinear. The point P is
called the proportionality limit.
Let us now make a careful examination of the stress-strain graph and see how we can
use the graph to understand the mechanical behaviour of the material.
1.3 Engineering Stress-Strain Diagram
Deformation in which stress and strain are proportional is calledelastic deformation.For
elastic deformation, a plot of stress (ordinate) versus strain (abscissa) results in a linear
relationship, called Hooke’s law, as shown in Figure1.6. The slope of this linear segment
corresponds to the modulus of elasticity E. This modulus may be thought of as sti ness,
or a material’s resistance to elastic deformation. The greater the modulus, the sti er