Now we are going further to start our discussion to understand the concept of “Elongation of bar due to its self weight, with the help of this post.

### Let us see here the of elongation of bar due to its self weight

As we have already discussed that when an external force will be applied on a body, there will be deformation or changes in the size and shape of the body. But molecular forces which will be acting between the molecules of the body will resist the deformation in the body due to the external force and this is termed as internal resistance developed by the body in order to resist the deformation in the body due to the external force.

This internal resistance force per unit area will be termed as stress.

Same concept will also be applied here when we will analyze stress and strain for a bar of uniform section due to its self weight.

If a bar of uniform cross-section will be fixed at one end and hanging freely under its self weight, there will also be induced stress and strain due to self weight of the bar.

Let us consider one bar AB as shown in following figure. Let us assume that bar is fixed at one end and hanging freely under its self weight. Let us assume that length of the bar is L, cross sectional area of the bar is A, young’s modulus of elasticity of the bar is E and specific weight of the bar i.e. weight per unit volume is ω.
Let us consider one small strip of thickness dy and at a distance y from the lower end as displayed in above figure.

Now we will consider the weight of the bar for a length of y in order to understand the load acting over the small strip of thickness dy towards downward direction.

Therefore weight of the bar for a length of y = ω x A x y

As weight of the bar for a length of y is acting on the small strip of thickness dy towards downward direction, there will be some stress and some changes in length of the bar due to self weight of the bar.

Therefore let us see, stress induced in the bar due to the load acting over the small strip of thickness dy towards downward direction.

Stress, σy = (ω x A x y)/A
Stress, σy = ω x y
And from here we can easily conclude that stress induced in the bar due to the self weight of the bar will be directionally proportional to the y.

Now we will determine the strain and elongation in small strip

Strain = Stress/E
Strain = ω x y /E

Elongation in small strip = Strain x length of small strip
Elongation in small strip = (ω x y /E) x dy

Elongation of the bar due to its self weight will be determined by integrating the above equation from 0 to L.

Elongation of the bar, δL will be determined as displayed here
Do you have any suggestions? Please write in comment box

### Reference:

Strength of material, By R. K. Bansal

We will see another important topic i.e. Thermal stresses in composite bars, in the category of strength of material, in our next post.

1. extremely amazing that i wanted that only was displayed

2. Hi, can you tell me how to select a material if the aim is to minimize elongation under its own weight?
and how to select material if the aim is to minimize the elongation of the rope carrying extra mass M at its end where M >> (ro)(A)(L)

3. thank u