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
1 Comments
Beautifully explained
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