In material science and engineering, we always try to know more about the behaviour of a rigid body when it goes through stretching, twisting and squeezing. We make our effort to understand the behaviour of rigid bodies by establishing a relation between applied force and deformation. So to understand the mechanical properties of a substance, we have first to understand stress and types of stress.
Stress is defined as the internal restoring force per unit area. Let us take an example. If we apply a force on a rubber wire, it elongates. After a while, when we remove the force, then the rubber regains its original shape and size. Now the question is, what is the responsible force which drives back the rubber to its original position? It can be explained like this: An internal force is developed when we apply the external force. The magnitude of this restoring force increases as the elongation of wire increases. When we remove the external force, the rubber wire returns to its original shape and size due to that restoring force. So stress is the restoring force per unit area that acts inside the body.
As we discussed, stress is restoring force per unit area, So mathematically,
Stress =FA
where,
F is restoring force
A is the surface area to which F is perpendicular
Some important points on stress
In the SI unit, the unit of stress is Nm-2 or Pascal. However, this unit of stress is minimal, and thus, in engineering applications, some larger units are used. They are MPa and GPa-
1MPa=106Pa=1N/mm2
1GPa=1000MPa=1KN/mm2
In border sense, various types of stress can classify as-
Let us understand these stresses in more detail.
Given figure forces on each cross-section are shown. The surface area of cross-sections is like the area of AB=Area of CD=2mm2, Area of EF= Area of GH=1mm2. Then find the stress on the surface AB, GH, EF and CD.
Hence total load on either side of AB=40N (positive sign means tensile force )
Surface area of AB= 2mm2
Hence stress on AB= 402N/mm2= 20MPa
As the rod is in equilibrium, then we can say that the Total force on either side of the surface GH= 60N
Surface area GH= 1mm2
Hence stress on GH= 601N/mm2=60 MPa
Surface area of CD= 2mm2
Hence stress on CD= 602N/mm2= 30MPa
Surface area EF= 1mm2
Hence stress on EF= 501N/mm2=50 MPa
A balloon is filled with air having an internal pressure of 3Pa. The diameter of the balloon is 4cm. Then find the volumetric stress.
As the balloon is in equilibrium, its volume is constant and the total force on either side of the surface is the same. Hence pressure is equal to the volumetric stress here.
Hence volumetric stress = 3Pa
There are various studies and applications of stress in material engineering and science. By the information of stress on a body, we can calculate the elongation produced on the body. When the stress on a material increases, various changes that are described by various limits can be explained based on stress-strain diagrams. By using Hooke’s law, we can derive Elastic constants from stress information.
In this study material, we discussed stress, types of stress and solved some stress-related questions.