Poisson’s ratio, Strain energy, Thermal Stress
This post is the 3rd post of the Elasticity series. Already in our last 2 posts we have discussed the meaning of Elasticity and related concepts like Stress, Strain, Hooke’s Law and modulus of elasticity.Today we will briefly discuss couple of very important terms and concepts of the same stream. Those are Poisson’s ratio, Strain energy, Thermal Stress etc. We will also derive expressions for these.
When a wire is stretched, its length increases but diameter is reduced. In other words, both shape and volume change under Longitudinal Stress. Here comes the Poisson’s ratio to measure 2 resulting strains because of this longitudinal stress. These 2 strains are known as Lateral Strain and Longitudinal Strain.
Lateral Strain = (change in diameter) / (Original diameter) = ΔD/D
Longitudinal Strain = (change in length) / (original length) = ΔL/L
Now, the Poisson’s ratio (σ) is derived from the ratio of lateral strain and longitudinal strain.
σ = Lateral Strain / Longitudinal Strain = (ΔD/D)/(ΔL/L) = (ΔD . L ) / ( D. ΔL) ……………. (1)
This ratio is dimensionless and unitless as well.
Strain Energy is the energy stored in a strained wire because of longitudinal stress.
say F is the force applied on the cross sectional surface of area A. This causes a length change of ΔL for a wire of original length L.
So work done on the wire = Energy stored in the wire = Average force X displacement = (F/2) X ΔL
Now, let’s find out the expression of the energy stored in per unit volume of the wire.
Volume of the wire = A L
Energy stored per unit volume = (F/2) X ΔL = (1/2) (F X ΔL )/(A X L) = (1/2) (F/A) (ΔL/L) = (1/2) X (Stress) X (Strain) ……….(2)
As temperature changes, it affects the length of a solid wire, and eventually causes Thermal stress. Let’s find out the expression of the Thermal Stress.F is the applied force on a surface with area A.
If initial length is L and α be the coefficient of linear expansion, then the change in length ΔL due to Δt is as follows.
ΔL = α L Δt …………….(3)
Young Modulus =Y = (FL)/(ΔL.A) …… (4)
From (3) and (4)
So F =( Y ΔL.A )/L = (Y α L Δt .A )/L = Y α Δt .A
F = Y α Δt .A …………… (5)
So, Thermal Stress = F/A = Y α Δt .A/A= Y α Δt …………………..(6)