Shear Modulus Formula
Shear Modulus Formula
A rigid body model is an idealised representation of an object that does not deform under external forces. This is very useful for evaluating mechanical systems. Also, many objects are very rigid bodies. The degree to which an object is considered rigid depends on the physical properties of the material from which it is made. A plastic ping-pong ball becomes brittle under pressure, while a rubber tennis ball is resilient. However, in other situations, both a ping-pong ball and a tennis ball can bounce nicely as rigid bodies. Similarly, prosthesis designers may be able to mimic the dynamics of human limbs by modelling them as rigid bodies. Nevertheless, a true mixture of bone and tissue is an elastic medium. This article will discuss Shear Modulus Formula which is available in detail within the learning resources provided by Extramarks.
Deformation is the change in shape caused by the application of a force. It is known that even relatively small stresses can cause some degree of deformation. Deformation occurs when an object or physical medium is subjected to external pressure such as crushing, squeezing, tearing, twisting, shearing or pulling apart. The forces on a body undergoing deformation are described in Physics by two words pressure and force. Learners are advised to learn about Shear Modulus Formula from the Extramarks website which is an authentic source for information related to Shear Modulus Formula
SI unit of pressure: Pascal (Pa). A force of 1 Newton applied to a unit area of 1 square meter results in a pressure of 1 Pascal.
Stress: When an object or medium is stressed, it deforms. Elongation is the term that describes this deformation. Elongation is expressed as a percentage change in length (under tension), volume (under bulk stress), or shape (under shear) and as a result, force is a dimensionless quantity. Students must learn about Shear Modulus Formula as it is an important formula from an examination point of view.
Tensile strain is the strain caused by tensile stress. Bulk strain (or volumetric strain) is the shear stress and Shear strain is the strain caused by shear stress. The more stress, the greater the burden. However, the relationship between strain and stress need not be linear. Deformation occurs in direct proportion to the stress value only if the stress is low enough. Elastic modulus is the constant of proportionality. The general relationship between stress and strain is indicated by the linear limit at low-stress levels.
stress = (Young’s modulus) × strain
Concept Of Shear Modulus
The Shear Modulus Formula is used to describe how a material resists lateral deformation. However, this is only practical for small deformations, after which it can return to its original state. Due to the high shear forces, permanent deformation occurs, i.e. the body loses its elasticity.
The Formula For Shear Modulus
All the formulas relate to Shear Modulus are available on Extramarks.
Derivation Of The Shear Modulus Formula
For the derivation of the Shear Modulus Formula, learners can go to Extramarks which is a student-friendly website that provides the best solutions.
1] Shear Stress
The internal restoring force causes the elastic body to return to its original shape. The restoring force per unit area acting on this deformed object is called stress. If the force acting on the surface is parallel to it and therefore the stress acting on the surface is also tangential. Stress is referred to herein as shear stress or tangential stress. This stress is expressed in Newtons per square meter. Students can peruse more such study materials on Shear Modulus Formula on Extramarks.
2] Shear Strain
Strain is a measure of the deformation experienced by a body in the direction of an applied force. It is also divided by the initial measurement of the body.
Shear Modulus Unit And Dimension
Students can learn about the unit and dimension of the Shear Modulus Formula on the Extramarks website.
Solved Examples For Shear Modulus Formula
Q.1: The thickness of the metal plate is 0.3 inches. Drill a 0.6-inch radius hole in the plate. If the shear strength is FA = 4 × 104 lb sq in, then determine the force required to make a hole.
Solution: Shear stress is applied over the cylindrical surface. Apply Shear Modulus Formula
Hence the area of the cylindrical surface,
=2πrh=2×3.14×0.06×0.30
= 0.11304 square inches
Given, FA = 4 × 104 pounds squared
Therefore, the force required to pierce = 4 × 104 × 0.11304
Force = 4521.6 lbs
For more such examples on the Shear Modulus Formula, students can visit the Extramarks website.
