Tuesday, December 20, 2016

Mechanical Properties of Solids Problems with Solutions Two

We are solving series of problems  based on mechanical properties of solids and one of the major property of the solids is elasticity due to which the solid materiel comes back to its original position, the body comes back  to its original position. It cannot come back to its original shape by hundred percent and to know by how much the body is able to recover its original position, we have a physical quantity called modulus of elasticity. It can be measured along the length using Young's modulus, along shape using Shearing modulus and along the volume it is measured with bulk modulus.

Problem

When a force is applied on a body of known surface area on the upper surface of the body keeping its lower surface fixed, the shift of the layer is given to us. Using the data given in the problem, we need to measure the rigidity modulus of the elasticity.


Solution

We know that rigidity modulus is the ability of a body to regain its original shape when the applied force is done on the upper surface and it is withdrawn. We can define it as the ratio of shearing stress to the shearing strain. Data can be applied and the problem cab be solved as shown in the diagram below.


Problem

Young's modulus of the wire is given to us and we need to measure the work done in the process per unit volume when the applied force is able to produce a known strain and the data is as shown in  the diagram below.


Solution

We know that when ever we apply force on a wire and it has produced some strain, we must have used some of our energy. That used energy can not disappear and it has satisfy the law of conservation of the energy. This energy is stored in the form of potential energy and it can be measured as shown in the diagram below.


Problem

A copper wires initial length and increase in the length is given to us and poisson's ratio is also given to us in the problem. We need to measure the lateral stain in the problem and the problem is as shown in the diagram below.


Solution

We know that poission's ratio is defined as the ratio of lateral strain to the longitudinal strain. When a force is applied along the length, its change in the length happens in the same direction and change in its width happens in the perpendicular direction. This change in the breadth to its original breadth is called lateral strain. For an ideal body, poission's ratio has to be equal to half and for all practical bodies, it is always less than that half. We can solve the problem by applying the definition as shown in the diagram below.


Problem

A steel bar of length one meter at zero degree centigrade is fixed between two rigid supports so that its length cannot be changed and its Young's modulus is given to us in the data as shown below. If coefficient of linear expansion of the material is also given to us, we need to measure the force developed in the wire when the temperature is raised.


Solution

We can rewrite the force in the Young's modulus in terms of the terms given in the data like coefficient of linear expansion. We can write the increase in the length from the definition and coefficient of linear expansion and the formula can be obtained as shown in the diagram below.


Problem

A stone of mass two kilogram is attached to one end of a wire of known area of cross section and length. If the breaking stress of the wire is given to us, we need to measure the number of revolutions that the wire can be revolved with out breaking the wire and the problem is as shown in the diagram below.


Solution

Breaking stress is the maximum stress that the wire can experience with out breaking itself. We can find the force as the product of breaking stress and area of cross section of the wire. This force acts like centripetal force and it can be expressed in terms of angular velocity. Problem can be solved as shown in the diagram below.


Problem

Length of the wire when applied force is four newton and five newton is given to us. We need to find the length of the wire when applied force is nine newton. Problem is as shown in the diagram below.


Solution

Both the lengths given in the problem are under the influence of force and hence its initial length is less than that. We can  express the increase in the length as the final length and initial length and it can be expressed in terms of the definition as solved in the diagram below.


Related Posts

Mechanical Properties of Solids Problems with Solutions One


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