The problem is to find out an expression for axial deformation 8, as a function of P, L, A, E
Normal stress = σ = p /A
Normal strain = e = δ /L
Hook’s Law : Hook’s law states that the stress is proportional to strain.
σ = E .e where E is a constant called modulus of elasticity.
(2.1) => P /A = E x δ / L
δ = PL / AE
Equation (2.2) is applied when P, A, E are constants throughout the bar. The weight
of the bar is neglected.
2.3.1 Varying P (Case-1)
When a bar is suspended freely and has got its weight itself, we can consider a case Of varying P, The total weight acts only in its C.G, but other than C.G it is different in every point. Let us take a differential length dy. For this differential length dy, we can assume that P, A, E does not vary. Therefore, we can apply the formula δ = PL / AE for the differential length dy of the bar.
Y=spec fic weight
Dδ = (Yy A) dy / AE
= YL2 / 2E
= WL / 2 AE
δ= WL / 2 AE
2.3.2 Varying A (Case-2)
BC = b2 – b1 / 2
ED = y – b1 / 2
AC = L
From similar triangles AED and ABC, we have
BC / ED = AC / AD
(b2 – b1)2 / 2 (y-b1) = L / x
y = (b2 – b1) / L x + b1
Applying the formula δ = PL / AE in the elemental area, we have
Dδ = p dx / (yt) E
2.3.3 Varying P and A (Case-3)
Volume of ABCD solid = 1/2 (x + b1) ty
Weight of the ABCD solid = 1/2 ( x + b1) yt x y
From similar triangles AEE’ and ADD’, x = b1 + (b2 –b1) /L .Y