Rigid Body and Elasticity

A body that can not be deformed is called rigid body. In fact rigid body is an ideal concept. No body is rigid; some bodies are also elastic and can be deformed. The interesting thing of elastic nature of bodies is that they come back to their original state after being deformed considering that the force applied is small enough.

Here we first discuss the force applied on a spring and what work the force does on the spring. You may be quiet familiar with springs that they come to their equilibrium condition after being deformed. One limitation for this is the force applied should be small enough.

Spring Force and Work

A box is attached to one end of a spring in Figure 1 and the position of the box is at the origin of our coordinate system without any force being applied. Now an external net force is applied on the box towards positive x-direction. And let the magnitude of the force applied on the spring is $F$.

Figure 1 A spring stretched towards positive x-direction.

The force applied stretches the spring a distance $x$ at a particular point and it is found that the force on the spring is directly proportional to the elongation $x$ of the spring also called Hooke's law, that is, $F \propto x$ and

\[F = kx \tag{1} \label{1}\]

In \eqref{1}, $k$ is the proportionality constant. Note that the Eq. \eqref{1} is ture for small elongation $x$ or the force is small enough so that the value of $k = \frac{F}{x}$ remains constant. We introduced the direct proportionality of the force on the spring and the elongation as Hooke's law but you should also consider the limitation of the validity of this law which is "the force applied should be small enough".

The force exerted by the spring is equal and opposite to the force applied on the spring according to Newton's third law. The same magnitude of force should be applied on the spring on both of its ends to stretch it. You saw the direct proportionality of the force and the elongation in Eq. \eqref{1} and know that the force is not constant during the displacement or elongation. If you suppose an infinitesimally small elongation $dx$ towards positive x-axis, the force during this elongation or displacement can be taken as constant and we can use the definition of work done $dW$ to find the work done for the small displacement which is,

\[dW = Fdx\]

Now the total work done within the limits $x_1$ and $x_2$ is obtained by integrating the above expression from $x_1$ to $x_2$.

\[W_{\text{force}} = \int\limits_{{x_1}}^{{x_2}} {Fdx} = \int\limits_{{x_1}}^{{x_2}} {k{\kern 1pt} xdx} = \frac{1}{2}kx_2^2 - \frac{1}{2}kx_1^2 \tag{2} \label{2}\]

The Eq. \eqref{2} is the work done on the spring by the force of magnitude $F$, so the work done by the spring is the negative of the work done in Eq. \eqref{2} which is

\[{W_{{\rm{spring}}}} = \frac{1}{2}kx_1^2 - \frac{1}{2}kx_2^2 \tag{3} \label{3}\]

The work done by a conservative force is the negative of the change in potential energy that is, $W = {U_1} - {U_2} = - ({U_2} - {U_1}) = - \Delta U$. If you compare Eq. \eqref{3} with this, you'll find the potential energy due to the spring force at a particular elongation to be

\[U = \frac{1}{2}kx_{}^2 \tag{4} \label{4}\]

If a stretched spring does work on a block and the block moves in the direction of the force, the work done should be positive and this requires that the initial potential energy should be greater than the final potential energy which means $x_1$ should be greater than $x_2$ in Eq. \eqref{3}.

Stress and Strain

Here we discuss the relationship between stress and strain. The idea of stress and strain comes from the force that causes deformation. The main idea of stress and strain is that the stress is defined in terms of the force per unit area and the strain in terms of the change in dimension per unit initial dimension. We discuss three types of stress and strain below.

Tensile Stress and Strain

In Figure 2 you can see a rod of cross-sectional area $A$ and a force of magnitude $F$ acts perpendicular to the cross-section on the right end of the rod. The initial length of the rod without any force is $L_i$. The force on the rod stretches the rod and the final length is $L_i + \Delta L$.

Figure 2 A force of magnitude F acts perpendicular to the cross-section of a bar and the bar is stretched due to the force.

The tensile stress is ratio of the force on the rod perpendicular to the cross-sectional area to the cross-sectional area $A$ that is,

\[\text{Tensile Stress}=\frac{{F}}{A}\]

The force on the rod changes the length of the rod to the final length $L_i + \Delta L$, so the tensile strain is the ratio of the change in length $\Delta L$ to the initial length $L_i$. Therefore,

\[{\text{Tensile Strain}} = \frac{\Delta L}{L_i}\]

The ratio of the tensile stress to the tensile strain is constant if the force is small enough which is called Young's Modulus denoted by $Y$.

\[Y = \frac{{{\text{Tensile Stress}}}}{{{\text{Tensile Strain}}}} = \frac{{FL}}{{A{\kern 1pt} \Delta L}} \tag{5} \label{5}\]

On the other hand we can also consider the compression of the rod instead of stretching it. If the rod is compressed, the stress is compressive stress and corresponding strain is compressive strain. Note that in Figure 2 the same magnitude of force should be applied on the left end to apply the force on the right end.

Shear Stress and Strain

In this case a force acts on a surface being parallel to the surface (not perpendicular). See in Figure 3 that a force of magnitude $F$ is applied to the top surface of a box of area $A$ in such a way that the surface and the force are both parallel. The same magnitude of force should be there at the bottom of the box parallel to the surface.

Figure 3 A force of magnitude $F$ acts at the top and the bottom surface of a box parallel to the surface.

The shear stress is the ratio of the force parallel to the surface to the area of the surface which is,

\[\text{Shear Stress}=\frac{F}{A}\]

The parallel force on the surface changes the shape of the box. The ratio of $x$ to the height $h$ in Figure 3 is the shear strain. Thus,

\[{\text{Shear Strain}} = \frac{x}{h}\]

The ratio of the shear stress to the shear strain is constant for forces that are small enough called shear modulus denoted by $S$ which is,

\[S = \frac{{{\text{Shear Stress}}}}{{{\text{Shear Strain}}}} = \frac{{Fx}}{{A{\kern 1pt} h}} \tag{6} \label{6} \]

Bulk Stress and Strain

A box is immersed in a fluid in Figure 4. Suppose the box experiences a uniform force at every point on the surface perpendicular to the surface. If the initial pressure before immersing the box is ${P_i} = \frac{{{F_i}}}{A}$ and the final pressure after immersing the box is ${P_f} = \frac{{{F_f}}}{A}$, then the change in pressure is $\Delta P = \frac{{{F_f}}}{A} - \frac{{{F_i}}}{A} = \frac{{\Delta F}}{A}$. If the initial pressure on the box is $P_i$, its final pressure after immersing in the fluid is $P_i + \Delta P$ where $\Delta P$ is the change in pressure. The resulting change in pressure decreases the volume of the box. Here the change in pressure $\Delta P = \frac{{\Delta F}}{A}$ is the bulk stress.

\[{\text{Bulk Stress}} = \frac{{\Delta F}}{A} = \Delta P\]

Note that the pressure acts on the entire surface of the box perpendicular to the surface. The change in pressure causes the deformation of the box that is, the volume is decreased in Figure 4.

Figure 4 A box is deformed after immersing the box in a fluid.

The initial volume of the box is $V_i$ and the final volume is $V_i + \Delta V$, so the change in volume is $\Delta V$. Notice here that the volume has decreased due to the increase in pressure, hence $\Delta V$ is negative and $\Delta P$ is positive. Now the bulk strain is the ratio of the change in volume to the initial volume which is,

\[{\text{Bulk Strain}} = \frac{{\Delta V}}{{{V_i}}}\]

The final step is to know the bulk modulus, the ratio of the bulk stress to the bulk strain. The bulk modulus is constant for small deformations or forces. The limitation of small forces or deformations is to make the ratio of stress to strain constant. The bulk modulus denoted by $B$ is

\[B = - \frac{{\Delta P\,{V_i}}}{{\Delta V}} \tag{7} \label{7}\]

The negative sign is put to make the bulk modulus a positive quantity because $\Delta V$ and $\Delta P$ always have opposite signs, that is if the pressure increases ($\Delta P$ is positive), the volume decreases ($\Delta V$ is negative). The reciprocal of the bulk modulus is called compressibility denoted by $k$,

\[k = -\frac{{\Delta V}}{{\Delta P\,{V_i}}} \tag{8} \label{8}\]

Note that stress is not a vector quantity and its SI unit is $N/{m^2}$. The Young's modulus, shear modulus and bulk modulus are called elastic moduli. The unit of elastic modulus is the same as the unit of stress because strain does not have any unit. The direct proportionality of stress and strain is another form of Hooke's law which is valid for small forces that do not exceed the proportionality limit and elastic modulus remains constant.

\[{\text{Elastic Modulus}} = \frac{{{\text{Stress}}}}{{{\text{Strain}}}}\]

Mechanics

Was this article helpful?

Physics Key uses third party cookies to show you relevent ads, analyze traffic and for the security. If you continue to use this website, you acknowledge that you have read and understand the Terms and Privacy Policy.