Definition and Moduli of Elasticity
This note provides information about Moduli of Elasticity. On the basis of them, this not describes the note shows relation between Relation between Y, B and \(\sigma \).
Summary
This note provides information about Moduli of Elasticity. On the basis of them, this not describes the note shows relation between Relation between Y, B and \(\sigma \).
Things to Remember
- \(\frac{stress}{strain}=E(constant)\)
- Young's modulus, Y=\(\frac{Normal\space stress}{Longitudinal\space strain}\)=\(\frac{\frac{F}{A}}{\frac{\Delta l}{l}}\)=\(\frac{F.l}{A.\Delta l}\)
- Bulk modulus, B=\(\frac{Normal\space stress}{Volumetric\space strain}\)=\(\frac{\frac{F}{A}}{\frac{\Delta v}{v}}\)=\(\frac{P.v}{\Delta v}\)
- -ve sign indicates the relation between P and v.
- Modulus of Rigidity, \(\eta =\frac{Tangential\space stress}{Shearing\space strain}\)=\(
- $$\frac{F/A}{\theta }=\frac{F/A}{x/l}=\frac{Fl}{Ax}$$
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Definition and Moduli of Elasticity
Elasticity: It is the property of a body by the virtue of which it regains its original configuration after deforming force is removed. An elastic body can regain its configuration upto certain maximum value of deforming force called the elastic limit. Deforming force and reforming force are equal in magnitude but opposite to each other within elastic limit.
Stress: Internal resistance of the body to the deforming force is called stress. Types of stress:
1. Normal stress(Tensile stress or compression stress): Stress produced on a body due to the perpendicular force is called Normal stress. It causes the change in size of the body without change in shape.
2. Tangential stress(Shearing stress): Stress produced on a body due to the tangential force is called tangential stress or shearing stress. It causes change in shape of the body without change in size.
Strain: The change in dimension per unit original dimension of the body due to application deforming force is called strain. Types of strain:
1. Longitudinal strain(Tangential strain)\(=\frac{change\space in\space length}{original\space length}=\frac{\Delta l}{l}\)
2. Volumetric strain\(=\frac{change\space in\space volume}{Original volume}=\frac{\Delta v}{v}\)
3. Shear strain: The angle in radian through which the face of a body which was originally perpendicular to the fixed surface, is turned when under the action of tangential stress.
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From figure, \(tan\theta =\frac{x}{L}\)
for small deformation(i.e. within elastic limit)\(tan\theta \approx \theta\)
\(\therefore shear\space strain,\theta =\frac{x}{L} \)
Elastic Constants(Coefficient of Elasticity)
Hooke's law states that within elastic limit, stress\(\prec \)strain
i.e.\(\frac{stress}{strain}=E(constant)\), E=modulus of elasticity
1. Young's modulus, Y=\(\frac{Normal\space stress}{Longitudinal\space strain}\)=\(\frac{\frac{F}{A}}{\frac{\Delta l}{l}}\)=\(\frac{F.l}{A.\Delta l}\)
2. Bulk modulus, B=\(\frac{Normal\space stress}{Volumetric\space strain}\)=\(\frac{\frac{F}{A}}{\frac{\Delta v}{v}}\)=\(\frac{P.v}{\Delta v}\)
-ve sign indicates the relation between P and v.
Compressibility, C=\(\frac{1}{B}\)
Modulus of Rigidity, \(\eta =\frac{Tangential\space stress}{Shearing\space strain}\)=\(
\frac{F/A}{\theta }=\frac{F/A}{x/l}=\frac{Fl}{Ax}\)
Poisson's ratio, \(\sigma =\frac{d/D}{l/L}=\frac{lateral \space strain}{longitudinal \space strain}\)
Relation between Y, B and \(\sigma \)
Consider a unit cube having faces parallel to the co-ordinate axes. Let the cube be acted upon by force F normally outward on the surfaces along the axes=\(\frac{F}{A}=f \)
Longitudinal strain, \(\alpha =\frac{l}{L}=l\)[\(\because L=1\)]
If Y be the Young's Modulus of elasticity of material of the cube, then
Y=\(\frac{\frac{F}{A}}{\frac{l}{L}}\)=\(\frac{f}{l}\space \implies l=\frac{f}{Y}\)
\(\therefore Longitudinal\space strain, \alpha =\frac{f}{Y}\)
\(\therefore Poisson's\space ratio, \sigma =\frac{lateral\space strain}{longitudinal\space strain}\)\begin{align*}or,\space \sigma = \frac{\beta}{\alpha } \end{align*}\begin{align*}\implies \beta =\sigma \alpha \end{align*}\begin{align*}\implies \beta = \sigma\frac{f}{Y} \end{align*}If we apply the stress simultaneously on different faces, then the corresponding strain produced on different faces can be tabulated as
Value of stress along | Value of strain produced along | ||||
OX | OY | OZ | OX | OY | OZ |
f | 0 | 0 | f/Y | -\(\sigma \)f/Y | -\(\sigma \)f/Y |
0 | f | 0 | -\(\sigma \)f/Y | f/Y | -\(\sigma \)f/Y |
0 | 0 | f | -\(\sigma \)f/Y | -\(\sigma \)f/Y | f/Y |
f | f | f | f/Y(1-2\(\sigma \)) | f/Y(1-2\(\sigma \)) | f/Y(1-2\(\sigma \)) |
It is seen that when equal forces are applied simultaneously along all the faces, the strain produced in each direction is the same. Let us take the produced strain as \begin{align*}x= f/Y(1-2\sigma )\end{align*}\begin{align*}\therefore dimension\space of\space the\space cube =1+x \end{align*}\begin{align*}\therefore volume\space of\space the\space cube =(1+x)^3=1+x^3+3x^2+3x \end{align*}For very small 'x' the terms containing the higher powers of x can be neglected.\begin{align*}\therefore volume\space of\space the\space cube =1+3x \end{align*}\begin{align*}chane\space in\space volume( \Delta V)=(1+3x)-1=3x \end{align*}\begin{align*}\end{align*}Now, Bulk modulus of elasticity,\begin{align*}B=\frac{\frac{F}{A}}{\frac{\Delta V}{V}} \end{align*}\begin{align*}B=\frac{f}{\Delta V}=\frac{f}{\frac{3f}{Y}(1-2\sigma )} \end{align*}\begin{align*}=\frac{Y}{ 3(1-2\sigma )} \end{align*}\begin{align*}\implies Y=3B(1-2\sigma ) \end{align*}
Relation between \( Y, \eta \space and\space \sigma\)
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Consider a unit cube on which let us apply compressional and expansional forces having equal magnitude F along Y and X-axes respectively by keeping the face along Z-axis fixed. Then the strain produced can be summarized as follows:
Stress along | Strain along | ||||
OX | OY | OZ | OX | OY | OZ |
F | 0 | 0 | F/Y | -\(\sigma \)F/Y | -\(\sigma \)F/Y |
0 | -F | 0 | \(\sigma \)F/Y | -F/Y | \(\sigma \)F/Y |
F | -F | 0 | (\(1+\sigma \))F/Y | -(\(1+\sigma \))F/Y | 0 |
This table shows that the starin produces on the faces along X and Y-axes are equal but the net effect on the face along Z is zero which keeps the face along Z-direction fixed. In this case, shear strain is produced on the cube is given by \begin{align*}\theta = |\frac{F}{Y}(1+\sigma ) |+|-\frac{F}{Y}(1+\sigma ) |=\frac{2F}{Y}(1+\sigma ) \end{align*}If \(\eta \) be the modulus of rigidity of the cube,\begin{align*}\eta =\frac{\frac{F}{A}}{\theta }=\frac{F}{\theta }=\frac{F}{\frac{2F}{Y}(1+\sigma ) } \end{align*}\begin{align*}\implies Y=2\eta (1+\sigma ) \end{align*}we have,\begin{align*}Y=3B(1-2\sigma) \rightarrow 1\end{align*}\begin{align*}Y=2\eta (1+\sigma ) \rightarrow 2 \end{align*}from 2,\begin{align*}\sigma =\frac{Y}{2\eta }-1 \end{align*}substituting \(\sigma \) in 1,\begin{align*}\frac{3}{Y}=\frac{1}{3B}+\frac{1}{\eta } \end{align*}In terms of \(\sigma \),
Equating 1 and 2,\begin{align*}3B(1-2\sigma)=2\eta (1+\sigma ) \end{align*}\begin{align*}\implies 6B\sigma +2\eta \sigma =3B-2\eta \end{align*}\begin{align*}\implies \sigma = \frac{3B-2\eta }{6B+2\eta } \end{align*}
Limiting value of \(\sigma \)
we know,\begin{align*}3B(1-2\sigma)=2\eta (1+\sigma ) \end{align*}In this equation, B and \(\eta \) are +ve.
Depending upon the value of \(\sigma \), we have two cases,
1. If \(\sigma \) is +ve. RHS of above equation becomes +ve. Thus for LHS to be +ve,\begin{align*}(1-2\sigma)>0\implies \sigma
2. If \(\sigma \) is -ve. The LHS of above equation becomes +ve. Thus, RHS to be +ve,\begin{align*}1+\sigma >0\implies \sigma >-1 \end{align*}\begin{align*}\implies -1<\sigma <0.5\end{align*}
References
Adhikari, Pitri Bhakta. A Textbook of Physics Volume-I. Kathmandu: Sukunda Pustak Bhawan, 2015.
Feynman, Richard P. The Feynman Lectures on Physics Volume 1. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.
Mathur, D S. Mechanics. New Delhi: S. Chand & Company Pvt. Ltd., 2015.
Young, Hugh D, Roger A Freedman and A Lewis Ford. University Physics. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.
Lesson
Elasticity
Subject
Physics
Grade
Bachelor of Science
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