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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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