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If the upper end of a cylinder is clamped and a torque is applied at the lower end the cylinder gets twisted by angle \[\theta \]. Simultaneously shearing strain \[\phi \] is produced in the cylinder. (i) The angle of twist \[\theta \]  is directly proportional to the distance from the fixed end of the cylinder. At fixed end \[\theta ={{0}^{o}}\] and at free end \[\theta =\]  maximum. (ii) The value of angle of shear \[\phi \] is directly proportional to the radius of the cylindrical shell. At the axis of cylinder \[\phi =0\] and at the outermost shell \[\phi =\] maximum. (iii) Relation between angle of twist \[(\theta )\] and angle of shear \[(\phi )\]            \[AB=r\theta =\phi l\] \[\therefore \]\[\varphi =\frac{r\theta }{l}\] (iv) Twisting couple per unit twist or torsional rigidity or torque required to produce unit twist. \[C=\frac{\pi \eta {{r}^{4}}}{2l}\] \[\therefore \] \[C\propto {{r}^{4}}\propto {{A}^{2}}\] (v) Work done in twisting the cylinder through an angle \[\theta \]  is \[W=\frac{1}{2}C{{\theta }^{2}}=\frac{\pi \eta {{r}^{4}}{{\theta }^{2}}}{4l}\]  

Moduli of elasticity are three, viz. Y, K and \[\eta \] while elastic constants are four, viz, Y, K, \[\eta \] and \[\sigma \]. Poisson's ratio s is not modulus of elasticity as it is the ratio of two strains and not of stress to strain. Elastic constants are found to depend on each other through the relations : \[Y=3K(1-2\sigma )\]                                                      ...(i) \[Y=2\eta (1+\sigma )\]                                                    ...(ii) Eliminating s or Y between these, we get \[Y=\frac{9K\eta }{3K+\eta }\]                                        ...(iii) \[\sigma =\frac{3K-2\eta }{6K+2\eta }\]                            ...(iv)  

If a long bar have a length L and radius r then volume \[V=\pi {{r}^{2}}L\]            Differentiating both the sides \[dV=\pi {{r}^{2}}dL+\pi 2rL\,dr\]            Dividing both the sides by volume of bar  \[\frac{dV}{V}=\frac{\pi {{r}^{2}}dL}{\pi {{r}^{2}}L}+\frac{\pi 2rL\,dr}{\pi {{r}^{2}}L}\]\[=\frac{dL}{L}+2\frac{dr}{r}\]            \[\Rightarrow \]  Volumetric strain = longitudinal strain + 2(lateral strain)                            \[\Rightarrow \]\[\frac{dV}{V}=\frac{dL}{L}+2\sigma \frac{dL}{L}\]\[=(1+2\sigma )\frac{dL}{L}\]                                                                                        \[\left[ \text{As}\,\,\sigma =\frac{dr/r}{dL/L}\,\,\,\,\Rightarrow \,\,\,\frac{dr}{r}=\sigma \frac{dL}{L} \right]\] or \[\sigma =\frac{1}{2}\left[ \frac{dV}{AdL}-1 \right]\]                                                                                          [where A = cross-section of bar] (i) If a material having \[\sigma =-0.5\] then \[\frac{dV}{V}=[1+2\sigma ]\frac{dL}{L}=0\] \[\therefore \] Volume = constant  or  \[K=\infty \] i.e. the material is incompressible. (ii) If a material having \[\sigma =0,\] then lateral strain is zero i.e. when a substance is stretched its length increases without any decrease in diameter e.g. cork. In this case change in volume is maximum. (iii) Theoretical value of Poisson?s ratio \[-1<\sigma <0.5\]. (iv) Practical value of Poisson?s ratio \[0<\sigma <0.5\]

When a long bar is stretched by a force along its length then its length increases and the radius decreases as shown in the figure.            Lateral strain : The ratio of change in radius or diameter to the original radius or diameter is called lateral strain. Longitudinal strain : The ratio of change in length to the original length is called longitudinal strain.            The ratio of lateral strain to longitudinal strain is called Poisson's ratio \[(\sigma )\]. i.e. \[\sigma =\frac{\text{Lateral strain}}{\text{Longitudinal strain}}\] \[\sigma =\frac{-dr/r}{dL/L}\] Negative sign indicates that the radius of the bar decreases when it is stretched. Poisson's ratio is a dimensionless and a unitless quantity.

Within limits of proportionality, the ratio of tangential stress to the shearing strain is called modulus of rigidity of the material of the body and is denoted by \[\eta ,\] i.e. \[\eta =\frac{\text{Shearing stress}}{\text{Shearing strain}}\] In this case the shape of a body changes but its volume remains unchanged. Consider a cube of material fixed at its lower face and acted upon by a tangential force F at its upper surface having area A. The shearing stress, then, will be Shearing stress \[=\frac{F}{A}\]   This shearing force causes the consecutive horizontal layers of the cube to be slightly displaced or sheared relative to one another, each line such as PQ or RS in the cube is rotated through an angle \[\phi \] by this shear. The shearing strain is defined as the angle \[\phi \] in radians through which a line normal to a fixed surface has turned. For small values of angle, Shearing strain\[=\varphi =\frac{QQ'}{PQ}=\frac{x}{L}\] So \[\eta =\frac{\text{shear stress}}{\text{shear strain}}=\frac{F/A}{\varphi }=\frac{F}{A\varphi }\] Only solids can exhibit a shearing as these have definite shape.

A solid sphere of radius R made of a material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area A floats on the surface of the liquid. Volume of the spherical body \[V=\frac{4}{3}\pi {{R}^{3}}\] \[\frac{\Delta V}{V}=3\frac{\Delta R}{R}\] \[\therefore \] \[\frac{\Delta R}{R}=\frac{1}{3}\frac{\Delta V}{V}\]                          ....(i) Bulk modulus \[K=-\,V\frac{\Delta P}{\Delta V}\] \[\therefore \] \[\left| \frac{\Delta V}{V} \right|=\frac{\Delta P}{K}=\frac{mg}{AK}\]                     ..(ii)                                                       \[\left[ \text{As }\Delta P=\frac{mg}{A} \right]\] Substituting the value of \[\frac{\Delta V}{V}\] from equation (ii) in equation (i) we get \[\frac{\Delta R}{R}=\frac{1}{3}\frac{mg}{AK}\]  

If a liquid of density \[\rho \], volume V and bulk modulus K is compressed, then its density increases. As density  \[\rho =\frac{m}{V}\] so  \[\frac{\Delta \rho }{\rho }=\frac{-\Delta V}{V}\]                    ...(i) But by definition of bulk modulus \[K=\frac{-V\Delta P}{\Delta V}\]\[\Rightarrow \]\[-\frac{\Delta V}{V}=\frac{\Delta P}{K}\]                                            ...(ii) From (i) and (ii) \[\frac{\Delta \rho }{\rho }=\]\[\frac{{\rho }'-\rho }{\rho }=\frac{\Delta P}{K}\]                    [As \[\Delta \rho =\rho '-\rho \]] or  \[{\rho }'=\rho \left[ 1+\frac{\Delta P}{K} \right]\]\[=\rho [1+C\Delta P]\]   \[\left[ \text{As }\frac{\text{1}}{K}=C \right]\]

When a solid or fluid (liquid or gas) is subjected to a uniform pressure all over the surface, such that the shape remains the same, then there is a change in volume. Then the ratio of normal stress to the volumetric strain within the elastic limits is called as Bulk modulus. This is denoted by K. \[K=\frac{\text{Normal stress}}{\text{volumetric strain}}\]            \[K=\frac{F/A}{-\Delta V/V}=\frac{-pV}{\Delta V}\] where p = increase in pressure; V = original volume; \[\Delta V=\] change in volume The negative sign shows that with increase in pressure p, the volume decreases by \[\Delta V\] i.e. if p is positive, \[\Delta V\] is negative. The reciprocal of bulk modulus is called compressibility. C = compressibility = \[\frac{1}{K}=\frac{\Delta V}{pV}\] S.I. unit of compressibility is \[{{N}^{1}}{{m}^{2}}\] and C.G.S. unit is \[dyn{{e}^{1}}c{{m}^{2}}\]. Gases have two bulk moduli, namely isothermal elasticity \[{{E}_{\theta }}\] and adiabatic elasticity \[{{E}_{\phi }}\]. (1) Isothermal elasticity \[({{E}_{\theta }})\]: Elasticity possess by a gas in isothermal condition is defined as isothermal elasticity. For isothermal process, PV = constant       (Boyle's law) Differentiating both sides \[PdV+VdP=0\Rightarrow PdV=VdP\] \[P=\frac{dP}{(-dV/V)}\]\[=\frac{\text{stress}}{\text{strain}}\]\[={{E}_{\theta }}\] \[\therefore \]\[{{E}_{\theta }}=P\] i.e., Isothermal elasticity is equal to pressure. (2) Adiabatic elasticity \[({{E}_{\phi }})\] : Elasticity possess by a gas in adiabatic condition is defined as adiabatic elasticity. For adiabatic process,   \[P{{V}^{\gamma }}\]= constant           (Poisson's law) Differentiating both sides, \[P\,\gamma \,{{V}^{\gamma -1}}dV+{{V}^{\gamma }}dP=0\] \[\Rightarrow \]\[\gamma \,PdV+VdP=0\] \[\gamma \,P=\frac{dP}{\left( \frac{-dV}{V} \right)}=\frac{\text{stress}}{\text{strain}}\]\[={{E}_{\varphi }}\] \[\therefore \] \[{{E}_{\phi }}=\gamma P\] i.e., adiabatic elasticity is equal to \[\gamma \] times pressure.                        [Where \[\gamma =\frac{{{C}_{p}}}{{{C}_{v}}}\]]   Note :

• Ratio of adiabatic to isothermal elasticity

\[\frac{{{E}_{\varphi }}}{{{E}_{\theta }}}=\frac{\gamma \,P}{P}=\gamma >1\] \[\therefore \] \[{{E}_{\phi }}>{{E}_{\theta }}\] i.e., adiabatic elasticity is always more than isothermal elasticity.

When the wire is loaded beyond the elastic limit, then strain increases much more rapidly. The maximum stress corresponding to B (see stress-strain curve) after which the wire begin to flow and breaks, is called breaking stress or tensile strength and the force by application of which the wire breaks is called the breaking force. (i) Breaking force depends upon the area of cross-section of the wire i.e., Breaking force \[\propto \,A\] \[\therefore \]  Breaking force \[=P\times A\] Here P is a constant of proportionality and known as breaking stress. (ii) Breaking stress is a constant for a given material and it does not depend upon the dimension (length or thickness) of wire.            (iii) If a wire of length L is cut into two or more parts, then again it's each part can hold the same weight. Since breaking force is independent of the length of wire. (iv) If a wire can bear maximum force F, then wire of same material but double thickness can bear maximum force 4F            (v) The working stress is always kept lower than that of a breaking stress.            So that safety factor \[=\frac{\text{breaking stress}}{\text{working stress}},\] may have large value.            (vi) Breaking of wire under its own weight.            Breaking force = Breaking stress \[\times \] Area of cross section Weight of wire = Mg = ALdg = PA      [P =Breaking stress]                               [As mass = volume \[\times \] density = ALd] \[\Rightarrow \] \[Ldg=P\] \[\therefore \]\[L=\frac{P}{dg}\]            This is the length of wire if it breaks by its own weight.  

In stretching a wire work is done against internal restoring forces. This work is stored in the wire as elastic potential energy or strain energy. If a force F acts along the length L of the wire of cross-section A and stretches it by \[x\] then \[Y=\frac{\text{stress}}{\text{strain}}=\frac{F/A}{x/L}=\frac{FL}{Ax}\]\[\Rightarrow \] \[F=\frac{YA}{L}x\] So the work done for an additional small increase dx in length, \[dW=Fdx=\frac{YA}{L}x\,.\,dx\] Hence the total work done in increasing the length by l,       \[W=\int_{0}^{l}{dW}=\int_{0}^{l}{Fdx}=\int_{0}^{l}{\frac{YA}{L}.x\,dx}=\frac{1}{2}\frac{YA}{L}{{l}^{2}}\] This work done is stored in the wire. \[\therefore \] Energy stored in wire \[U=\frac{1}{2}\frac{YA{{l}^{2}}}{L}=\frac{1}{2}Fl\]  \[\left[ \text{As }F=\frac{YAl}{L} \right]\] Dividing both sides by volume of the wire we get energy stored in unit volume of wire.   \[{{U}_{V}}=\frac{1}{2}\times \frac{F}{A}\times \frac{l}{L}\]\[=\frac{1}{2}\times \text{stress}\times \text{strain}=\frac{\text{1}}{\text{2}}\times Y\times {{(\text{strain})}^{2}}\] \[=\frac{1}{2Y}{{(\text{stress})}^{2}}\]      [As AL = volume of wire]

Total energy stored in wire (U)	Energy stored in per unit volume of wire (UV)
\[\frac{1}{2}Fl\]	\[\frac{1}{2}\frac{Fl}{\text{volume}}\]
\[\frac{1}{2}\times \text{stress}\times \text{strain}\times \text{volume}\]	\[\frac{1}{2}\times \text{stress}\times \text{strain}\]
\[\frac{1}{2}\times Y\times {{(\text{strain})}^{2}}\times \text{volume}\]	\[\frac{1}{2}\times Y\times {{(\text{strain})}^{2}}\]
\[\frac{1}{2Y}\times {{(\text{stress)}}^{\text{2}}}\times \text{volume}\]	\[\frac{1}{2Y}\times {{(\text{stress})}^{2}}\]

Note :

• If the force on the wire is increased from F1 to F2 and the elongation in wire is l then energy stored in the wire \[U=\frac{1}{2}\frac{({{F}_{1}}+{{F}_{2}})}{2}l\]
• Thermal energy density = Thermal energy per unit volume =\[\frac{1}{2}\] \[\times \] Thermal stress \[\times \] strain

= \[\frac{1}{2}\frac{F}{A}\frac{l}{L}\] = \[\frac{1}{2}(Y\alpha \Delta \theta )(\alpha \Delta \theta )\] = \[\frac{1}{2}Y{{\alpha }^{2}}{{(\Delta \theta )}^{2}}\]