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 

TYPES OF STRESSES

Stresses are two types they are

               1. Normal stresses, and 

               2. Shear stresses

          Normal stress is the stress which acts in a way perpendicular to he area. It is represented by 'σ' (SIGMA). Normal stress is further divided into Tensile stress and Compressive stress.

Let us discuss each of them briefly

1. Tensile stress:

          The stress induced in a body, when subjected to two equal and opposite pull forces as shown fig., as a result of which there is an increase in length, is known as Tensile stress. The ratio of increase in length to the initial (or) original length is known as Tensile strain. The tensile stress acts normal to the area and it pulls on the area. 

              Let P = Pull force acting on the body,

                     A = Cross-sectional area of the body, 

                     L = Original length of the body, 

                   dL = Change in length 

                    σ  = Stress induced in the body, and 

                     e = Strain 

          Below fig., shows a bar subjected to a tensile force 'P' at its ends. Consider a cross section x-x which divides the bar into two parts. The part left to the section x-x, will be in equilibrium if Tensile force (P)=Resisting force (R). Similarly the part which is right to the cross section x-x, will be in equilibrium if P= Resisting force. This resisting force per unit area is known as Tensile stress or Intensity of stress.

2. Compressive stress: The stress induced in a body, when it is subjected to two equal and opposite push forces as shown in fig., as a result of which there is a decrease in length of the body, is known as Compressive Stress. The ratio of decrease in length to the initial (or) original length is known as Compressive strain. The compressive stress acts normal to the area and it pushes on the area. 

  

3. Shear stress: The stress induced in a body, when it is subjected to two equal and opposite push forces which are acting tangentially across the resisting section as shown in fig., as a result of which the body tends to shear off across the section, is known as Shear Stress. The corresponding strain is known as Shear strain. The shear stress which acts tangential to the area. Which is represented by 'τ' (Tau).

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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.

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STRESS

          Consider some external force acting on a body, the force of resistance per unit area offered by a body against deformation is known as Stress. This external force which is acting on body is called the Load (or) Force. When the load is applied on the body while the stress is induced in the material of the body. The loaded body remains in equilibrium when the resistance offered by the body against deformation and the applied load are equal. 

Stress is denoted by a letter "σ" (SIGMA)

Mathematically stress is written as σ = P/A

          Where σ = Stress (or  Intensity of stress)

                       P = External force (or) Load

                       A = Cross sectional area.

Units of Stress:

          Units of stress depends upon two factors, they are units of Load and Units of area. 

In MKS units, the force is expressed in kgf and area in square meters, so stress units are kgf/m^2.
In SI units, the force is expressed in Newtons and area in meter squares, hence units of stress becomes N/m^2.

SIMPLE STRESSES AND STRAINS

INTRODUCTION:

          When an external force acts on a body, the body tends to undergo some deformation. The body resists the deformation due to cohesion between the molecules. The resistance by which the body tends to oppose the deformation is termed as "Strength of Materials". The resistance that the material offered within certain limit is proportional to the deformation of the material by the external force. Also within this limit resistance is equal to the deformation. But beyond the elastic limit, the resistance offered by the material is less than the applied load. In this case, the deformation increases until the failure takes place.

          Within the elastic stage, the resisting force equals to the external force. This resisting force per unit area is called Stress (or) Intensity of Stress. 

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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