a tapered column of modulus E and mass density rho varies linearly from a radius of r1 to r2 in a length L, and is hanging from its Broad end. Find the total deformation due to weight of the bar.
Hello,
Length of Bar = L
Radius of Broad end = r1. Hence, Diameter = d1.
Radius of smaller end = r2. Hence, Diameter = d2.
Let, the force due to self weight = P
Now, consider a small element of length dx of the bar at distance x from the broad end. Hence, diameter of bar at x from broad end = d1 - ( d1 - d2 ). x / L
Let, d1 - d2 / L = k
Hence, diameter of bar at distance x = d1 - kx
So, Cross sectional area at x = π /4 x ( d1 - kx )^2
We know that,
Stress = Force / Area
Hence, Stress = P / [ π / 4 x ( d1 - kx )^2 ]
So, Stress = 4P / π ( d1 - kx )^2
Also, Stress / Strain = E
Hence, Strain = Stress / E
So, Stress = 4P / [ π ( d1 - kx )^2 ] E
Hence, Elongation of elementary length = { 4P / [ π ( d1 - kx )^2 ] E }.dx
Let, Total elongation = dL
So, dL = lim. ( 0 to L ) ∫ { 4P / [ π ( d1 - kx )^2 ] E }.dx
Solving the above integration, we get,
dL = 4P / πEk [ ( 1/ d1 - kl ) - 1/d1 ]
Resubstituting k = d1 - d2 / L
So, dL = 4PL / πEd2.d1
Hence, the total deformation due to the self weight of bar is 4PL / πEd1d2
Best Wishes.
