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1800-102-2727We have played a lot with the erasers in our school times. We have seen that when we apply a stretch or compress the eraser, it gets deformed. Either it will elongate (due to stretching) or it will compress (due to compression). This change in the shape and size of the eraser produces stress and strain in the eraser. And when we stop stretching or compressing the eraser, it regains its original shape and size. This ability to regain its shape and size is nothing but the elasticity of the eraser.
A general law was proposed in the 19th century which showed the relationship between stress and strain. It was observed that the stress and strain will vary linearly up to a certain point known as the proportional limit. We will discuss Hooke’s law in detail within this topic.
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It is the law that gives the relationship between stress and strain produced in the body when a deforming force is applied to it. But before going deeper into the concept of Hooke’s law, we need to have a better understanding of stress and strain, deforming force and restoring force. Therefore below stress and strain are explained in detail.
Stress: Stress is defined as the restoring force per unit area of the body when the body is subjected to an external force. This external force is also known as the deforming force. When the deforming force is applied to a body, it tries to get deformed but this deformation is resisted by the internal resistance due to net intermolecular force which is known as restoring force. The restoring force opposes the applied force and tries to bring back the body to its original state. Restoring force is equal to the deforming force but in the opposite direction. Stress is mathematically denoted by (sigma) which is a tensor quantity. Mathematically it is expressed as,
$Stress=\frac{RestoringForce}{CrossSectionalArea}$
If a block is subjected to a deforming force Fd and Fr restoring force is generated in the block then,
$\left|{F}_{d}\right|=\left|{F}_{r}\right|...\left(i\right)$
If the cross-sectional area of the block where the deforming force is applied is A, then the stress can be written as,
$\sigma =\frac{{F}_{r}}{A}$
As from the equation (i), we can write,
$\sigma =\frac{{F}_{d}}{A}$
Types of stress: It has mainly three types that are categorised by the type of deforming force applied to a body.
${\sigma}_{l}=\frac{{F}_{d}}{A}$
Where, Fd= Deforming force
A= Cross-sectional area on which the deforming force is acting
In the above figure, a block is subjected to a tangential deforming force that causes the
deformation of the block tangentially. Then the shear stress produced in the block is,
$ShearStress=\frac{Tangentialforce}{Areaofcrosssection}$
If the tangential force is denoted by Ft and the cross-sectional area of the surface on
which the force is acting is A then,
${\sigma}_{s}=\frac{{F}_{t}}{A}$
In the figure, a ball is dropped in a liquid. If the force acting on the surface of the ball is F. And the surface area of the ball is A then the volumetric stress is,
${\sigma}_{v}=\frac{F}{A}$
Strain: When a body is subjected to a deformation force the shape and size of the body are changed. The ratio of this change in the size of the body to its original size is known as the strain produced in the body. Strain is a dimensionless quantity denoted by ε.
$\epsilon =\frac{Changeinsize}{Originalsize}$
There are mainly three types of strain corresponding to the stress produced in the body as follows,
$Longitudinalstrain\left({\epsilon}_{l}\right)=\frac{ChangeinLength}{OriginalLength}$
As seen from the above figure, the strain produced in the rod will be,
${\epsilon}_{l}=\frac{\mathrm{\Delta L}}{L}$
If the distance between the moving layer and the fixed layer is l and the displacement of the moving layer is x then the shear strain will be
${\epsilon}_{s}=\frac{x}{l}=tan\theta \approx \theta $ (if is small)
If the change in volume of the ball in the above figure is V and the original volume is V.
Then the volumetric strain,
${\epsilon}_{v}=\frac{\mathrm{\Delta V}}{V}$
Stress and strain will generate in a body when it is subjected to an external force or deforming force. Stress and strain may cause changes in the shape and size of the body. In the 19th century, English scientist Robert Hooke proposed a relationship between stress and strain which states that,
“In a body, the strain produced due to stress is directly proportional to the stress within the elastic limit ”
Consider a block is subjected to a deforming force F. Let the stress in the block is and strain in the block be ε, then Hooke's law can be mathematically represented as,
$\sigma \propto \epsilon $
In the above relationship, proportionally is replaced by a proportionality constant which is known as the modulus of elasticity and is represented by Y.
$\sigma =Y\epsilon $
In the statement, it is mentioned that the law is applicable within the elastic limit. It means that, within the elastic limit, the body will regain its original shape and size when removing the deforming force. The plot between stress and strain will be a straight line (OA) within the elastic limit as shown below. The slope of the straight line gives the modulus of elasticity.
But proportionality has its own limitations that is limited by the strength of the material up to which it regains its original shape. When the limit is reached, the material will not follow Hooke's law and permanent deformation occurs. This limit is the elastic limit of the material. A detailed discussion about the relationship between stress and strain is explained later in this topic (Stress-strain curve).
Modulus of elasticity is a property of the material that describes the elastic behaviour of that material. It does not depend on either the stress or strain. The value of the modulus of elasticity for different materials is calculated by experiments and is used for the calculations.
From Hooke's law of elasticity, the modulus of elasticity can be written as,
$Y=\frac{\sigma}{\epsilon}$
On the basis of type of strain, the modulus of elasticity has three types,
Poisson's ratio is a dimensionless quantity which shows how much strain is produced in a material in the transverse direction for a unit strain in the longitudinal direction. Therefore Poisson's ratio is the ratio of strain in the transverse direction (lateral) to the strain in longitudinal direction (axial).
$Poisson\text{'}sRatio=-\frac{TransverseStrain\left({\epsilon}_{t}\right)}{LongitudinalStrain\left({\epsilon}_{l}\right)}$
Figure shows a rectangular block of length l and the lateral dimension is b. The block is subjected to a deforming force of F due to which elongation occurs. After the elongation, let the length of the block is l1 and the lateral dimension is decreased to b1.
Then the change in length will be, l=l1-l
And the change in lateral dimension, b=b1-b
Then the Poisson's ratio will be,
$Poisson\text{'}sRatio\left(\gamma \right)=-\frac{{\epsilon}_{t}}{{\epsilon}_{l}}=-\frac{\frac{{b}_{1}-b}{b}}{\frac{{l}_{1}-l}{l}}=-\frac{\frac{\mathrm{\Delta b}}{b}}{\frac{\mathrm{\Delta l}}{l}}$
The stress-strain curve shows how the stress and strain are related to each other when an external force is applied and hence elongation in the material occurs. The strain is shown on the horizontal axis in terms of percentage elongation and the stress is shown on the vertical axis. A typical stress-strain curve for a material such as copper or soft iron is shown below,
Between points O to a , the plot will be a straight line which shows that when we apply the load on the material it elongates. Then the stress is directly proportional to the strain which is nothing but Hooke's law. And for the given material, this proportionality relationship is valid for the strain less than 1%. Point A is said to be the proportional limit of the material and till this point, the material will regain its original shape and size when we remove the applying load. After point A the material will not follow Hooke's law. It means that the stress will not be proportional to the corresponding strain and the slope of the curve decreases. But, even if the material is not following Hooke's law after point A, it will still come back to its original shape and size if we remove the applied load gradually till point B. This point B is said to be the elastic limit or yield point of the material, till which the material will show its elastic behaviour.
Now when again we apply the load and the material is stretched beyond point B then the material will not come to its original shape and size when we remove the load. And till point C, there will be a small change in stress corresponding to a significant increase in elongation. As we further increase the load on the material then a situation will reach (point D) when the material will fracture. The region from points B to D is known as the plastic region for a material in which the deformation is irreversible.
Q. A small rod is subjected to a longitudinal force of magnitude 50 N. The diameter of the rod is measured as 5 mm. Calculate the stress in the rod corresponding to the 50 N force.
A. Given, diameter of the rod, d=5 mm=510-3 m
Applied load, F=50 N
Then the area of the rod, $A=\frac{\pi}{4}{d}^{2}=\frac{\pi}{4}\times {\left(5\times {10}^{-3}\right)}^{2}=1.96\times {10}^{-5}{m}^{2}$
Stress produced in the rod will be normal stress,
${\sigma}_{l}=\frac{F}{A}=\frac{50}{1.96\times {10}^{-5}}$
${\sigma}_{l}=2.54\times {10}^{6}=2.54MPa$
Q. A ball of radius 5 mm is subjected to a load of 120 N on its surface. Due to this load, the volume of the ball is reduced till the radius 4.5 mm. Find the strain produced in the ball.
A. Applied load, F=120 N
Initial radius of the ball, r1=5 mm=510-3 m
Final radius of the ball, r2=4.5 mm=4.510-3 m
Then the initial and final volume of the ball is calculated as,
${V}_{1}=\frac{4}{3}\pi {{r}_{1}}^{3}=\frac{4}{3}\times \pi \times {(5\times {10}^{-3})}^{3}=5.23\times {10}^{-7}{m}^{3}$
${V}_{2}=\frac{4}{3}\pi {{r}_{2}}^{3}=\frac{4}{3}\times \pi \times {(4.5\times {10}^{-3})}^{3}=3.82\times {10}^{-7}{m}^{3}$
Then the change in the volume of the ball will be,
$\mathrm{\Delta V}={V}_{2}-{V}_{1}=(3.82\times {10}^{-7})-(5.23\times {10}^{-7})=-1.41\times {10}^{-7}{m}^{3}$
Now the strain of the ball will be the volumetric strain and it is calculated as,
${\epsilon}_{v}=\frac{\mathrm{\Delta V}}{V}=\frac{{V}_{2}-{V}_{1}}{{V}_{1}}$
${\epsilon}_{v}=\frac{-1.41\times {10}^{-7}}{5.23\times {10}^{-7}}=-0.27$
Hence the strain in the ball is 27% and the -ve sign shows that the volume is decreased by applying the load to it.
Q. Find stress in a block when it is subjected to a tensile load of 150 N as shown in the figure. The length of the block is 15 mm and length is increased to 15.7 mm when the load is applied. Young’s modulus of elasticity of the material is 210 MPa.
A. Given, Tensile load, F=150 N
Initial length of the block, l1=15 mm=1510-3 m
Final length of the block, l2=15.7 mm=15.710-3 m
Young’s modulus, Y=210 MPa=210106 N/m2
From Hooke's law of elasticity,
=Yε
Where, = Stress in the block and ε= Strain in the block
Then the stress in the block can be written as,
$\sigma =Y\times \frac{\mathrm{\Delta l}}{{l}_{1}},\sigma =Y\times \frac{{l}_{2}-{l}_{1}}{{l}_{1}}$
$\sigma =210\times {10}^{6}\times \frac{(15.7\times {10}^{-3})-(15\times {10}^{-3})}{15\times {10}^{-3}}=210\times {10}^{6}\times \frac{0.7}{15}$
$\sigma =9.8\times {10}^{6}N/{m}^{2}=9.8MPa$
Q. A block of length 25 mm and lateral dimension 10 mm is subjected to a tensile load F. The elongation due to this load is 0.8 mm. If the change in its lateral dimension due to the elongation is 0.2 mm then find the Poisson's ratio.
A. Given, Length of the block, L=25 mm=2510-3 m
Change in length, L=0.8 mm=0.810-3 m
Lateral dimension of the block, b=10 mm=1010-3 m
Change in lateral dimension, b=-0.2 mm=0.210-3 m
Then the Poisson's ratio will be, $Poisson\text{'}sRatio\left(\gamma \right)=-\frac{{\epsilon}_{t}}{{\epsilon}_{l}}=-\frac{\frac{\mathrm{\Delta b}}{b}}{\frac{\mathrm{\Delta l}}{l}}$
$\gamma =\frac{\frac{0.2\times {10}^{-3}}{10\times {10}^{-3}}}{\frac{0.8\times {10}^{-3}}{25\times {10}^{-3}}}=0.625$
Q. What are the limitations of Hooke’s law?
A. This law is valid only for elastic materials and even in the elastic material, there is a limit up to which it is valid, known as the elastic limit. Beyond the elastic limit, permanent deformation occurs in the material. Also, there is no material which can be compressed beyond its minimum limit or there is no material which can be stretched beyond its maximum limit. It means that, if we keep applying and increasing the load on the material, it will break eventually.
Q. What kind of stress is produced between layers of the water in the river?
A. In the river, the water flows with a decreasing velocity in the vertical direction (with depth). It is due to the tangential resistance offered by each layer of the water on one another. Therefore the top layer will have a maximum velocity and the bottom most layer lateral velocity will have zero value. Since the resistive force applied between the layers is tangential, the stress produced during the flow of water in the river is shear stress.
Q. What is Poisson's ratio of cork and what is its significance?
A. Poisson's ratio of the cork is zero. It means that there will be no change in its lateral dimension (diameter) when it is stretched or compressed. Cork is generally used in wine bottles. Since the diameter of the bottle’s mouth is fixed and when the bottle has to be closed for storage for a long time (sometimes years), the cap of the bottle must not expand from its diameter while compressing the cap in order to fill the bottle mouth. If the diameter of the cap increases, the bottle may break. Therefore we use cork as a cap for the wine bottle.
Q. What is Poisson's effect?
A. When a material is stretched in from a certain dimension, it will increase the dimension of the material in that direction. At the same time, there will be a decrease in the lateral dimension of the material as vise-versa. This effect is known as Poisson's effect.