PROBLEM 1.1

          A rod 300 cm long and of diameter 4.0 cm is subjected to an axial pull of 20 KN. If the modulus of Elasticity of the material of the rod is 4.0 x 10⁵ N/mm²; Determine:

(i) The Stress, 

(ii) The Strain, and 

(iii) The Elongation of the rod.

Sol:

Given,  Length of the rod, L = 300 cm,

              Diameter,               D = 4.0 cm = 40 mm, 

              Area,                       A = (π/4) D² = (π/4) 40²  = 400 π mm²

              ^{Axial pull,              P = 20 KN = 20,000 N}

             Modulus of Elasticity, E  = 4.0 x 10⁵ N/mm²

(i) The Stress (σ) is given by equation as

                         σ = P/A = 20,000/400 π  = 15.92 N/mm² 

                 

(ii) Using the equation, the strain is obtained as 

                         E = σ/e

Strain              e = σ/E = 15.92 / 4.0 x 10⁵ = 0.0000398

(iii) Elongation is obtained by using equation as

                          e = dL/L

Elongation,     dL = e x L  = 0.0000398 x 300 = 0.01194 cm

                   

              

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RELATION BETWEEN STRESS AND STRAIN

FOR ONE-DIMENSIONAL STRESS SYSTEM:

          The relationship between stress and strain for one directional stress (i.e., Normal stress in one direction only) is given by Hook's Law. This law states that when a material is loaded within elastic limit, the developed normal stress is proportional to the strain produced. This means that the ratio of the normal stress to the corresponding strain is a constant within the elastic limit. This constant is represented by ‘E’ and is known as Modulus of Elasticity or Young’s modulus of Elasticity.

FOR TWO-DIMENSIONAL STRESS SYSTEM:

          Before going to learn about Two dimensional stress system, we have to know about Longitudinal strain, Lateral strain and Poison's ratio. 
1. Longitudinal Strain:
          When a body is subjected to an axial tensile load, then there is an increase in length of the body. But at the same time there is a decrease in other dimensions of the body at right angles to the line of action of the load applied. Thus the body having axial deformation and also deformation at right angles to the line of action of the applied load (i.e., Lateral deformation).
          The ratio of axial deformation to the original length of the body is known as Longitudinal strain (Linear). The Longitudinal strain is also defined as the deformation of the body per unit length in the direction of applied load.    
2. Lateral Strain:
          The strain which is produced right angles to the direction of applied load is known as Lateral strain.   

          Let us take an example, A rectangular bar of length 'L', breadth 'B' and depth 'D' is subjected to an axial tensile load 'P' as shown in below fig., . The length of the bar will increase while applying load, the breadth and the depth will decrease.

Let     δl = Increase in length
          δb = Decrease in breadth and, 
          δd = Decrease in depth.

Then Longitudinal strain = 
and Lateral strain =

Note: 
1. If longitudinal stain is tensile, the lateral strain will be compressive, vice-versa,
2. Hence for every longitudinal strain in the direction of the load applied is accompanied by lateral strain of the opposite kind in all directions perpendicular to the load.

3. Poisson's Ratio:
          When the material is subjected to stress within the elastic limit, The ratio of lateral strain to longitudinal strain is a constant. This ratio is called Poisson's ratio and it is denoted by a symbol 'μ' (MU).

Mathematically, 

as lateral strain is opposite in sign to longitudinal strain, hence algebraically, lateral strain is written as,  

4. Relation between Stress and Strain:
          Consider a two dimensional figure ABCD, subjected to two mutually perpendicular stresses 


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MODULUS OF RIGIDITY (or) SHEAR MODULUS

          The ratio of Shear Stress to the corresponding Shear Strain within the elastic limit is known as Modulus of Rigidity (or) Shear Modulus and is denoted by ‘C’ , ‘G’, or ‘N’.

Formula : MODULUS OF RIGIDITY (or) SHEAR MODULUS 

WHAT IS HOOK’S LAW?

          Hook’s law (or) Law of Elasticity was discovered in 1660 by the English scientist Robert Hook. Hook's law states that "when a material is loaded within elastic limit, the stress is proportional to the strain produced". This means that the ratio of the stress to the corresponding strain is constant within the elastic limit. This constant is called Modulus of Elasticity (or) Modulus of Rigidity (or) Elastic Modulii. 

Hook’s law (or) Law of Elasticity

Credits: Wkipedia

STRAIN

          When a body is subjected to some external force, we observe some change in dimension of the body. The ratio of change in dimension to the original dimension of the body is known as "Strain". Strain is Dimensionless quantity.

           Strain is classified into:

                    1. Tensile strain

                    2. Compressive strain

                    3. Volumetric strain

                    4. Shear strain

          If there is some increase in length of a body due to some external force, then the ratio of increase in length to the original length of the body is known as Tensile strain. But if there is some decrease in length of the body, then the ratio of decrease in length to the original length of the body is called as Compressive strain.

          The ratio of change in volume to the original volume of the body is called Volumetric strain.

          The strain produced by the shear stress is known as Shear strain.

Credits: Mechanical engineers

PROBLEM 1.2

Find the minimum diameter of a steel wire, which is used to raise a load of 6000 N, if the stress in the rod is not to exceed 105 MN/m² .

Sol:

As per given data:

            Load           P = 6000 N

            Stress        σ = 105 MN/m²    = 105 x 10 ⁶ N/m²            (1 MN = 10⁶ N)

                                  = 105 N/mm²                                        ( 10⁶ N/m² = 1 N/mm²)

 Let       D = The diameter of the rod in mm

Area      A = (π/4) D²

We know that, Stress  σ = Load/Area = P/A

                                    105 = 6000/(π/4) D²

                                     D² = 6000 x 4 / π x 105

                                     D   = 72.75