Analysis of Bars due to Axial loads

Determination of stress, strain and deformations in the bars subjected to axial loads.

Principle of Superposition

The resultant deformation of the body is equal to the algebraic sum of the deformation of the individual section. Such principle is called as principle of superposition

In the analysis of axial loaded bars we use the principle of superposition. By draw the free body diagrams of each section of the bar for finding out the required stresses and strains or deformations. we can understand the concept with the following example.

Example:

Q.1. Determine the stress in each section of the bar shown in the figure.The PQ and RS segments are having a length of 1 meter and QR segment is 1.5 m.  The cross section of bar is 10 mm square. What will be the total extension of the bar? The Young's modulus of bar material E = 210000 MPa.

Solution :

Check the equilibrium of the bar, the sum of total forces in X- axis is ∑ Fx = 0

                          - 63 + 35 + 49 - 21 = 0

Write the free body diagram of the each portion of the bar

The each part of the bar should be in equilibrium by introducing the internal forces as shown in the above fig.

1.Segment- PQ

The area of the bar = A = 10x10 =100 mm2

The axial tensile force acting in left side of segment  P1 = + 63 kN

For equilibrium the right side of the segment is also subjected to same force P1 = + 63 kN

The stress in segment PQ =

s1 = P1/A = 63000/100 =63N/mm2 (Tensile stress)

The elongation in segment PQ =ΔL1 =PL/AE

ΔL1 = (63000x1000)/ (100x2100000) = 0.3 mm

Segment-2, QR

The area of the bar = A = 10x10 =100 mm2

The axial tensile force in segment QR =

The right side net internal force P2 =

     (+P1) - 35 = 63 - 35 = + 28kN 

The axial tensile stress in segment 2 =

s2 = P2/A = 28000/100 = 280N/mm2

The elongation in segment QR =ΔL2 =PL/AE

ΔL2 = (28000x1500)/ (100x2100000) = 0.2 mm

3.Segment- RS

The area of the bar = A = 10x10 =100 mm2

The axial compressive force acting left side of  segment RS =

 P3 = - 21kN =-21000 N

Check: The internal force on right side of the segment is = p2 - 49 = 28 - 49 = -21kN

The stress in segment 3 =

s3 =  P3/A = - 21000/100 

               = -210 N/mm2 (compressive stress)

The shortening in segment 3 =ΔL3= PL/AE

ΔL3 = (-21000x1000)/ (100x2100000) = -0.1 mm

The total extension of the bar is equal to sum of the extensions of all three segments PQ,QR,RS

ΔL= ΔL1+ ΔL2+ΔL3

ΔL = 0.3 + 0.2 - 0.1 = 0.4 mm 

The Deformations of bars of varying sections

Let us see the following figure, where we can see a bar of having different length and different cross-sectional area and bar is subjected with an axial load P. 

As we can see here that length and cross-sectional areas of each section of bar is different and therefore stress induced, strain and change in length too will be different for each section of bar.

Young’s modulus of elasticity of each section might be same or different depending on the material of the each section of bar.

Axial load for each section will be same i.e. P. When we will go to determine the total change in length of the bar of varying sections, we will have to add change in length of each section of bar.

P = Bar is subjected here with an axial Load

A1, A2 and A3 = Area of cross section of section 1, section 2 and section 3 respectively

L1, L2 and L3 = Length of section 1, section 2 and section 3 respectively

σ1, σ2 and σ3 = Stress induced for the section 1, section 2 and section 3 respectively

ε1, ε2 and ε3 = Strain developed for the section 1, section 2 and section 3 respectively

E = Young’s Modulus of the bar

Let us see here stress and strain produced for the section 1

Stress, σ1= P / A1

Strain, ε1 = σ1/E = (P/AE)1

Similarly stress and strain produced in the section 2
Stress, σ2= P / A2
Strain, ε2 = σ2/E =(P/AE)2
Similarly stress and strain produced for the section 3
Stress, σ3= P / A3
Strain, ε3 = σ3/E = (P/AE)3
Now we will determine change in length for each section with the help of definition of strain

ΔL1 = (PL/AE)1

ΔL2 = (PL/AE)2

ΔL3  = (PL/AE)3

Total change in length of the bar
To determine the total change in length of the bar of varying sections, we will have to add change in length of each section of the bar.

ΔL= ΔL1+ ΔL2+ΔL3

Assignment :

1.A copper bar shown in figure is subjected to a tensile load of 30 KN. Determine elongation of the bar if E=100 GPa. Also find maximum stress induced.

2.A bar of uniform cross sectional area 100mm2 is subjected to forces as shown below. Calculate the change in the length of the bar.Young's modulus of material is 200 GPa