Axial Deformation

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

dδ =

From similar triangles AEE’ and ADD’, x = b1 + (b2 –b1) /L .Y