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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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 

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