A bar suspended freely from a fixed end is subjected to an elongation dl which is caused by its own weight. Here, the value of elongation of a freely suspended bar is determined.

Elongation of a Bar by its Self Weight |

### Elongation of a Bar by its Self Weight

Consider a bar AB that is freely suspended, with one end fixed. The length of the bar be 'L', area of cross-section be 'A', 'E' be the modulus of elasticity and 'w' is the weight per unit volume of the bar material.

Consider a small strip of thickness'dx' at a distance 'x' from the free end 'B'.

#### Step 1: Weight of the bar of the length 'x'

Let the weight of the bar of length 'x' be P. Then,

P = Specific weight x Volume = Specific weight x area x thickness = w . A.x

This implies, on the strip of length 'x' a weight is acting downwards whose value is P = wAx.

#### Step 2: Stress on the element 'dx'

The stress on the element 'dx' is given by = Weight acting on the element / Area of Cross-section

i.e Stress = wAx/A = wx;

This means that, the stress developed due to self weight of the bar is not uniform. It varies with the value of 'x'.

#### Step 3: Strain in the element 'dx'

Strain in the element = Stress/E = wx/E

#### Step 4: Elongation of the Element

Elongation of the element = Strain x length of element = ( wx .dx ) E

#### Step 5: Elongation of the Bar

The elongation of the bar is obtained by integrating the above equation from 0 to L, and we get

**dl**

**= WL/2E**

## Comments