Mechanical Properties of Matter Concept Page - 7

Formula

Relation Between Different Elastic Constants

G=E2(1+ν)$G={\displaystyle \frac{E}{2(1+\nu )}}$

K=E3(1−2ν)$K={\displaystyle \frac{E}{3(1-2\nu )}}$

E=9KG3K+G$E={\displaystyle \frac{9KG}{3K+G}}$

ν=3K−2G2(3K+G)$\nu ={\displaystyle \frac{3K-2G}{2(3K+G)}}$

G: Modulus of Rigidity
ν$\nu$ : Poisson's Ratio
E: Young's Modulus
K: Bulk Modulus

Example

Various forms of equations of potential energy

Example: A uniform wire of length 4$4$ m and area of cross section 2$2$ mm2${}^2$ is subjected to longitudinal force produced an elongation of 1$1$mm.If Y=$=$0.2 x 1011${}^{11}$ NM−2${}^{-2}$, what is the elastic potential energy stored in the body?

Solution:
Workdone=∫1mm0AYLxdx$Workdone={\int }_0^{1mm}{\displaystyle \frac{AY}{L}}xdx$
=AY2L(11000)2$={\displaystyle \frac{AY}{2L}}({\displaystyle \frac{1}{1000}}{)}^2$
=2×10−6×0.2×10112×4×10−6$={\displaystyle \frac{2\times {10}^{-6}\times 0.2\times {10}^{11}}{2\times 4}}\times {10}^{-6}$
=0.005J$=0.005J$

Example 2:  A uniform wire of Youngs modulus Y$Y$ is stretched by a force within the elastic limit. If S$S$ is the stress produced in the wire and ε$\epsilon$ is the strain in it, what is the potential energy stored per unit volume?

Solution:energy=12×strain×stress×volume$energy={\displaystyle \frac{1}{2}}\times strain\times stress\times volume$
energyvolume=12×strain×stress${\displaystyle \frac{energy}{volume}}={\displaystyle \frac{1}{2}}\times strain\times stress$
=12Sε$={\displaystyle \frac{1}{2}}S\epsilon$
=12×Y×strain×strain(∴Y=stressstrain)$={\displaystyle \frac{1}{2}}\times Y\times strain\times strain(\therefore Y={\displaystyle \frac{stress}{strain}})$
=12Yε2$={\displaystyle \frac{1}{2}}Y{\epsilon }^2$
=12×stress×stressY$={\displaystyle \frac{1}{2}}\times stress\times {\displaystyle \frac{stress}{Y}}$
=12S2Y$={\displaystyle \frac{1}{2}}{\displaystyle \frac{S^2}{Y}}$

Example

Elastic Potential Energy Stored in a wire under stress

Question: A metal wire of length L, area of cross section A and Young's modulus Y$Y$ is stretched by a variable force F$F$ such that F$F$ is always slightly
greater than the elastic forces of resistance in the wire. Then the elongation of the wire is l$l$.
Find the elastic potential energy stored in the wire.

Solution:
We have
l=FLAY$l={\displaystyle \frac{FL}{AY}}$
or
 F=AYLl$F={\displaystyle \frac{AY}{L}}l$
Work done is given using the relation
dW=Fdx$dW=Fdx$
or
dW=AYLxdx$dW={\displaystyle \frac{AY}{L}}xdx$
or
W=AYL∫l0xdx$W={\displaystyle \frac{AY}{L}}{\int }_0^{l}xdx$
or
W=AYl22L$W={\displaystyle \frac{AY{l}^2}{2L}}$
The elastic potential energy is given as ΔU=W$\mathrm{\Delta }U=W$
or
ΔU=AYl22L$\mathrm{\Delta }U={\displaystyle \frac{AY{l}^2}{2L}}$ (This is the energy stored in the wire).All the work done on the wire is stored as its potential energy, thus there is no heat produced during elongation.

Example

Problems on composite wires

Example: A copper wire and a steel wire of the same length and same cross section are joined end to end to form a composite wire. The composite wire is hung from a rigid support and a load is suspended from the other end. If the increase in length of the composite wire is 2.4 mm$2.4mm$, then what is the amount of increase in lengths of steel and copper wires? (Ycu=10×1010N/m2,Ysteel=2×1011N/m2)$(Y_{cu}=10\times {10}^{10}N/m^2,Y_{steel}=2\times {10}^{11}N/m^2)$ 

Solution:Given :     Lcu=Lsteel=L$L_{cu}=L_{steel}=L$             
Acu=Asteel=A$A_{cu}=A_{steel}=A$
Let the increase in the lengths of individual wires be  Δlcu$\mathrm{\Delta }{l}_{cu}$  and Δlsteel$\mathrm{\Delta }{l}_{steel}$.
 
Total increase in the length of composite wire     Δltotal=2.4 mm$\mathrm{\Delta }{l}_{total}=2.4mm$∴$\therefore$   Δlcu+Δlsteel=2.4$\mathrm{\Delta }{l}_{cu}+\mathrm{\Delta }{l}_{steel}=2.4$                       ...........(1)

Also    Y=FLAΔl$Y={\displaystyle \frac{FL}{A\mathrm{\Delta }l}}$       
 ⟹Δl=FLAY$⟹\mathrm{\Delta }l={\displaystyle \frac{FL}{AY}}$
∴$\therefore$    Δlcu=FLcuAcuYcu=FLAYcu$\mathrm{\Delta }{l}_{cu}={\displaystyle \frac{F{L}_{cu}}{A_{cu}{Y}_{cu}}}={\displaystyle \frac{FL}{A{Y}_{cu}}}$

Similarly     Δlsteel=FLsteelAsteelYsteel=FLAYsteel$\mathrm{\Delta }{l}_{steel}={\displaystyle \frac{F{L}_{steel}}{A_{steel}{Y}_{steel}}}={\displaystyle \frac{FL}{A{Y}_{steel}}}$
Dividing these two equations we get           ΔlcuΔlsteel=YsteelYcu${\displaystyle \frac{\mathrm{\Delta }{l}_{cu}}{\mathrm{\Delta }{l}_{steel}}}={\displaystyle \frac{Y_{steel}}{Y_{cu}}}$∴$\therefore$          ΔlcuΔlsteel=2×101110×1010=2${\displaystyle \frac{\mathrm{\Delta }{l}_{cu}}{\mathrm{\Delta }{l}_{steel}}}={\displaystyle \frac{2\times {10}^{11}}{10\times {10}^{10}}}=2$                   ⟹$⟹$    Δlcu=2Δlsteel$\mathrm{\Delta }{l}_{cu}=2\mathrm{\Delta }{l}_{steel}$                          ............(2)

From (1) and (2),  
2Δlsteel+Δlsteel=2.4$2\mathrm{\Delta }{l}_{steel}+\mathrm{\Delta }{l}_{steel}=2.4$                 ⟹$⟹$   Δlsteel=0.8 mm$\mathrm{\Delta }{l}_{steel}=0.8mm$∴$\therefore$   Δlcu=2×0.8=1.6 mm$\mathrm{\Delta }{l}_{cu}=2\times 0.8=1.6mm$

Definition

Elastic Potential Energy Stored in a wire

Force exerted by spring:
F=−kx$F=-kx$
Potential energy = dU =  F. dx = -kx.dx
or for a stretch of x$x$ in the spring
U=kx22$U={\displaystyle \frac{k{x}^2}{2}}$

Definition

Application of Elasticity

Mechanical properties like strength, stiffness (Rigidity), ductility, malleability and brittleness have to be carefully studied to select a material for a particular job

• The metallic parts of machines should not be subjected to stress beyond the elastic limit otherwise they will be deformed.

• Beams are the simplest and most common parts of large structures. When beams are

Definition

Conservation of energy in the process of elastic deformation

A temporary shape change that is self-reversing after the force is removed, so that the object returns to its original shape, is called elastic deformation. In other words, elastic deformation is a change in shape of a material at low stress that is recoverable after the stress is removed.
Elastic deformation is a change in shape of a material at low stress that is recoverable after the stress is removed. This type of deformation involves stretching of the bonds, but the atoms do not slip past each other.
When the material restores its shape the converted form of work to potential energy transforms again to do work. Meanwhile in elastic deformation there isn't any energy loss.

Example

Finding Elongation using Young's modulus

Example: Elongation of a wire under its own weight is function of which quantities?

Solution:
Let's assume for wire,
 mass =$=$ M$M$
Force =ρ$=\rho$ALg$ALg$
Let weight of wire is W$W$.
consider a small length dx$dx$ of the rod at distance x$x$ from the fixed end. The tension T$T$ in element equals the weight of rod below it.
T=(L−x)WL$T=(L-x){\displaystyle \frac{W}{L}}$

elongation=L×StressAy=(L−x)WdxLAy$elongation={\displaystyle \frac{L\times Stress}{Ay}}={\displaystyle \frac{(L-x)Wdx}{LAy}}$

Total elongation =∫L0(L−x)WdxLAy$={\int }_0^L{\displaystyle \frac{(L-x)Wdx}{LAy}}$

=WLAy(Lx−x22)L0=WL2Ay$={\displaystyle \frac{W}{LAy}}(Lx-{\displaystyle \frac{x^2}{2}}{)}_0^L={\displaystyle \frac{WL}{2Ay}}$

now  W=ρ$W=\rho$  ALg$ALg$.

Putting it in there
ΔL=ρAL2g2Ay$\mathrm{\Delta }L={\displaystyle \frac{\rho A{L}^{2}g}{2Ay}}$

=ρL2g2y$={\displaystyle \frac{\rho L^{2}g}{2y}}$

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