derive the formula for deformation of tapered bar with rectangular cross sectional area under its own weight?
Hello,
Let,
A = Uniform cross sectional area of the bar
E = Young’s modulus for the bar
L = Length of the bar
w = Specific wt. of bar material
W = weight of the bar
Consider an element of length ‘dy’ at a distance of ‘y’ from the bottom of the bar being elongated due to the force ‘P’
So, weight of bar for length y is w.A.y....... ( Since, Sp. Wt. = Weight / Area )
As elongation is taking place due to own weight, P = W
We know that,
Elongation = PL / AE
So, dL = Wdy / AE
So, dL = ( w.A.y ) x dy / A.E
So, dL = ( w / E ) x y.dy
Now, for total elongation, Intergrating above eqn with respect to y
So, dL = ( w / E ) x ∫ y.dy
So, dL = ( w / E ) x [ ( lim 0 to L ) y^2 / 2 ]
So, dL = ( wL ^ 2 ) / 2E
Since, W = wAL for full length of bar,
w = W / AL
So, dL = WL / 2AE
Hence, the elongation is WL / 2AE .
Best Wishes.
