Simple Stress and Strain

Elasticity

It is the property by virtue of which certain materials return back to their original position after the removal of the external force.

Plasticity

It is the property of the solid material by virtue of which it tends to retain its deformed shape even after the removal of external load.

   Normal Stress or Direct Stress

The internal resistance per unit area, offered by a body against deformation is known as Normal stress or direct stress or axial stress.

where 

P = External force or load applied normal to the cross section. 

R = Internal resistance developed at the cross section as shown in fig,
A = Cross-sectional area of bar

But to satisfy the equilibrium of the bar, ∑ Fx = 0

R - P = 0

R = P ( The internal resistance R is always equal to normal force P)

The internal resistance per unit area is called normal stress, and is given by

s  =R/A =P/A. 

The unit of Stress is expressed as

1 N/m2 ,kN/m2 ,  MN/m2 , 

MPa = Megapascal = 10⁶ N/m2 = 1 N/mm2

Normal Strain or Axial Strain
The deformation for unit length is called Strain.
We can also define, it is a ratio of change of dimension of the body to the original dimension is known as strain.

Let us take a bar of length (L) and it is elongated to length (L+ dL) by the action of a normal tensile force P,

Length of the bar = L

Increase in length due to Tensile Force = dL

Tensile strain = increase in length / Original length
(+ Ve) (e ) = dL /L
Compressive strain = decrease in length/Original length
(- Ve) (e ) = dL/L
Note : Strain is dimensionless quantity.

Hooke’s law : 

It states that within elastic limit, the stress is proportional to the corresponding strain.

      s µ e 

or s = e E 

Young’s modulus: 

The ratio of tensile stress (or compressive stress) to the corresponding strain is known as Young’s modulus or modulus of elasticity and is denoted by E.

E = s/e

The Normal Stress required to produce one unit normal strain within the elastic limit is also called Young's Modulus.

Change in length of a bar

Normal stress s = P/A
and Normal strain e = dL / L
According to Hook’s Law, Up to elastic limit, Stress and strain are related by the equation,

                         s = e E 

substitute stress and strain values in the above equation, 

P/A = E (dL / L)

Therefore Change in length of the bar =

 dL =PL /AE

The Total Change in length of a bar, when it subjected to an axial load of 'P' is

dL = PL / AE

Example 1.

A short hollow, cast iron cylinder with wall thickness of 10 mm is to carry compressive load of 100 kN. Compute the required outside diameter `D’, if the allowable working stress in compression is 80 N/mm2.  

Solution: 

The safe Working stress s = 80N/mm2; 

Load acting on cylinder P= 100 kN = 100000 N 

Let the outer diameter of hollow cylinder is =D

Thickness of wall t = 10 mm

The inner diameter of cylinder is 

Di = D - 2t = D - 2 x 10 = (D - 20)

The area of cylinder A =(p/4) *{D² - (D-20)²}

The normal compressive stress s= P/A 

substituting P and A values in above eq. and solving.

The external diameter of cylinder, 

         D = 49.8 mm 

Example:2

An aluminium bar of 1.8 meters long and has a 25 mm  square cross sectional area.  How much will the bar elongate under a tensile load of P=17.5 kN, if the young's modulus of material E = 75000 Mpa.

Solution: 

The tensile load acting on the bar = 

P =17.5 kN = 17.5 x 1000 N

Length of the bar L = 1.8 m = 1.8 x 1000 mm

Cross sectional area of the bar =

A = 25 x 25 = 625 mm2

young's modulus of material E = 75000 Mpa.

The elongation of the bar due to tensile load is =

 d = PL/AE

Substitute the values in the above equation, 

the elongation in the bar dL =

  d = 17500x1800 / (625x75000)  = 0. 672 mm

Assignment:

1. A steel rod 500mm long and 20mm10mm in cross-section is subjected to axial pull of 300 KN. If modulus of elasticity is 2x10⁵ N/mm2 Calculate the elongation of the rod. Also calculate strain induced in the bar.
(Ans: 3.75mm,e=7.5x10³)
2. A hollow cylinder 2m long has an outside diameter of 50mm and inside diameter of 30mm. If the cylinder is carrying a load of 25 KN. FInd the stress in cylinder, also find deformation of the cylinder E = 100 Gpa.
(Ans:Stress=20.0N/mm2,Deformation= 0.4mm)
3. A load of 5 kN is to be raised with the help of a steel wire. FInd the minimum diameter of wire if stress is not to exceed 100 MPa.
(Ans: d=7.98mm)
4.In an experiment a steel specimen of 13mm diameter was found to elongate 0.2mm in a 200 mm gauge length when it was subjected to a force of 26.8 KN. If specimen was tested within elastic range, Calculate Young's modulus for the steel. (Ans: E=2.019x10⁵ N/mm2 )