Mechanical Properties of Matter Concept Page - 6

Example

Define and calculate Compressibility

Compressibility is a measure of the relative volume change of a fluid or solid as a response to a pressure (or mean stress) change.

Example: The fractional change in the volume of oil is 1 percent when a pressure of 2 x 107${}^7$ N/m2${}^2$ is applied. Find the bulk modulus and its compressibility?

Solution: P=2×107N/m2$P=2\times {10}^{7}N/m^2$

ΔVV=−0.01$\frac{\mathrm{\Delta }V}{V}=-0.01$    [Pressure is applied from outside ∴$\therefore$ Volume is decreased]

B=−P(ΔVV)$B={\displaystyle \frac{-P}{({\displaystyle \frac{\mathrm{\Delta }V}{V}})}}$

B=−2×107−0.01$B={\displaystyle \frac{-2\times {10}^7}{-0.01}}$

=2×109N/m2$=2\times {10}^{9}N/m^2$

C=1B=5×10−10m2/N$C={\displaystyle \frac{1}{B}}=5\times {10}^{-10}{m}^2/N$

Definition

Variation of bulk modulus with density

Bulk modulus is defined as B=−VdPdV$B=-V{\displaystyle \frac{dP}{dV}}$
For a liquid, dP=−ρgdh$dP=-\rho gdh$
Also, V∝1/ρ$V\propto 1/\rho$ by conservation of mass
⟹dVV=−dρρ$⟹{\displaystyle \frac{dV}{V}}=-{\displaystyle \frac{d\rho }{\rho }}$

Solving above equations, we get
dρ=−ρ2gBdh$d\rho =-{\displaystyle \frac{{\rho }^{2}g}{B}}dh$

Definition

Shear Modulus

Shear Modulus (Modulus of rigidity) is defined as the ratio of shear stress to the shear strain.
Modulus of rigidity = FLδlA${\displaystyle \frac{FL}{\delta lA}}$

Example

Problem on linear shear

Example:
A tangential force of 2100$2100$ N is applied on a surface of area 3x10$3x10$−6${}^{-6}$m2${}^2$ which is 0.1$0.1$ m from a fixed face. The force produces a shift of 7$7$ mm of upper surface with respect to bottom. Find the rigidity modulus of material.Solution:
F=2100N$F=2100N$
A=3×10−6m2$A=3\times {10}^{-6}{m}^2$
h=0.1m$h=0.1m$
Δx=0.007m$\mathrm{\Delta }x=0.007m$
Rigidity  Modulus=21003×10−60.0070.1$\text{Rigidity Modulus}={\displaystyle \frac{{\displaystyle \frac{2100}{3\times {10}^{-6}}}}{{\displaystyle \frac{0.007}{0.1}}}}$
=2100×0.1003×0.007×10−6$={\displaystyle \frac{2100\times 0.100}{3\times 0.007\times {10}^{-6}}}$
=1010N/m2$={10}^{10}N/m^2$

Example

Problem on twisting shear

Example:
A wire of length 1$1$ m and radius 4$4$ mm is clamped at upper end. The lower end is twisted by an angle of 300${}^0$. Find the angle of shear.Solution:
r=4mm=0.004m$r=4mm=0.004m$
L=1m$L=1m$
θ=300$\theta ={30}^0$
rL=ϵ0θ${\displaystyle \frac{r}{L}}={\displaystyle \frac{{ϵ}_0}{\theta }}$
0.004×3001=ϵ0${\displaystyle \frac{0.004\times {30}^0}{1}}={ϵ}_0$
ϵ0=0.120${ϵ}_0={0.12}^0$

Definition

Shear Stress

A shear stress, denoted (Greek: tau), is defined as the component of stress coplanar with a material cross section. Shear stress arises from the force vector component parallel to the cross section.

Definition

Poisson's Ratio

Definition: Poisson's ratio can be defined as a negative ratio of transverse strain to axial strain. It is a measurement of the Poisson Effect. Poisson Effect is an effect in which material expands in a perpendicular direction to the direction of compression. Inversely, material contracts in a perpendicular direction to the direction of stretching. It is denoted by Greek letter 'nu' known as Poisson Coefficient.
The range of Poisson's ratio should be between -1.0 to 0.5, because of Young's Modulus, Bulk Modulus, and Shear Modulus requirements.

Poisson's ratio of Steel ranges from 0.27 to 0.30. Poisson's ratio of Concrete ranges from 0.20 to 0.25. 
Example: When a rubber cord is stretched, the change in volume is negligible
compared to the change in its linear dimension. Then what is the poisson's ratio
for rubber?

Solution:By Lame's relation, ν=12−E6B,$\nu ={\displaystyle \frac{1}{2}}-{\displaystyle \frac{E}{6B}},$ where  B$B$ is bulk modulus.
Given, volume change is negligible, thus B$B$ tends to infinity. (B=−VdPdV)$(B=-V{\displaystyle \frac{dP}{dV}})$
 Thus, ν=12$\nu ={\displaystyle \frac{1}{2}}$

Definition

Theoretical Limits of Poisson Ratio

The Poisson's ratio of a stable, isotropic, linear elastic material cannot be less than 1.0 or greater than 0.5 because of the requirement for Young's modulus, the shear modulus and bulk modulus to have positive values. Most materials have Poisson's ratio values ranging between 0.0 and 0.5. A perfectly incompressible material deformed elastically at small strains would have a Poisson's ratio of exactly 0.5. Most steels and rigid polymers when used within their design limits (before yield) exhibit values of about 0.3, increasing to 0.5 for post-yield deformation which occurs largely at constant volume. Rubber has a Poisson ratio of nearly 0.5.

Example

Use of relation between Poisson Ratio and change in volume of wire

Example: A material has Poisson's ratio of 0.5. If a uniform rod suffers a longitudinal strain of 2 x 10−3${}^{-3}$ what is the percentage increase in its volume?

Solution:The volume change of a material due to the strain is related to the Poisson's ratio by the relation:
ΔV/V=(1−2×σ)ΔL/L$\mathrm{\Delta }V/V=(1-2\times \sigma )\mathrm{\Delta }L/L$ where, σ$\sigma$ is the Poisson's ratio of the material.
In this case the term in the bracket of R.H.S is zero since σ$\sigma$ is 0.5. So percentage volume change is 0%.

Example

Relation of Bulk Modulus and Young Modulus using Poisson Ratio

Example: Young's modulus of a metal is 15×10$15\times 10$11${}^{11}$ Pa. If its Poisson's ratio is 0.4$0.4$. What is the bulk modulus of the metal in Pa$Pa$ ?

Solution:y=15×1011$y=15\times {10}^{11}$
σ=0.4$\sigma =0.4$
y3k=1−2σ${\displaystyle \frac{y}{3k}}=1-2\sigma$
15×1011=3(1−.8)k$15\times {10}^{11}=3(1-.8)k$
25×1011=k$25\times {10}^{11}=k$

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