In the previous section, we completed a discussion on Gravitation
• In this section we will see Mechanical properties of solids
Rigidity and Plasticity are two extreme cases. We can write about them in 3 steps:
1. In the first extreme case, the given body is perfectly rigid
• In this case, the body has a definite shape and size
♦ If we apply a force on that body, no deformation will occur. That is.,
✰ The shape of the body will not change
✰ The size of the body will not change
✰ This is shown in fig.9.1(b) below
2. In the other extreme case, the given body is perfectly plastic. This can be explained in 4 steps:
(i) In this case, the body do not have definite shape and size
♦ If we apply a force (like placing a 1 kg wt.) on that body, deformation will occur. That is.,
✰ The shape of the body will change to a ‘new shape’
✰ The size of the body will change to a ‘new size’
✰ This is shown in fig.9.1(d) above
♦ If we remove the force,
✰ There will be no return from the ‘new shape’ to the 'original shape'
✰ There will be no return from the ‘new size’ to the 'original size'
✰ This is shown in fig.9.1(e) above
3. That means, as a result of the force F, a permanent deformation takes place in a perfectly plastic body
♦ So we can write:
✰ Perfectly rigid body ⇒ No change due to force
✰ Perfectly plastic body ⇒ Permanent change due to force
• But in reality,
♦ No body is perfectly rigid
♦ No body is perfectly plastic
• So how can we describe the reality?
The answer can be written in 8 steps:
1. We will use the term ‘real-bodies’ to denote the bodies that we see in our day to day life
• The simplest definition for a real-body can be given using just two statements (a) and (b):
(a) A real-body is not perfectly rigid
(b) A real-body is not perfectly plastic
2. Let us elaborate the two statements
Statement (a): A real-body is not perfectly rigid
This statement can elaborated using 2 steps:
(i) Consider a body which appears to be perfectly rigid
• When an external force is applied on that body, it’s shape and size will definitely change
(ii) An example:
• A steel rod of small length (about 30 cm) and comparatively large diameter (about 2.5 cm) will appear to be very rigid
• But if we rivet it on one end and pull from the other end with sufficient force, it’s length will surely increase
• The ‘increase in length’ may not be visible to the naked eye. But we can measure it using precision instruments
Statement (b): A real-body is not perfectly plastic
This statement can elaborated using 2 steps:
(i) Consider a body which appears to be perfectly plastic
• When an external force is applied on that body, it’s shape and size will visibly change to ‘new shape’ and ‘new size’
• But when the force is removed, the body will definitely try to get back to the ‘original shape’and ‘original size’
(ii) An example:
• Consider a lump of putty or lump of wet clay
(See: images of putty and images of wet clay)
• Apply a force on the lump. It will easily deform and will appear to be perfectly plastic
• But if we remove the force, there will be a very small rebound
(See: Dictionary meaning of rebound)
• The ‘rebound’ may not be visible to the naked eye. But we can measure it using precision instruments
3. So we can write:
• A steel rod is very close to ‘perfectly rigid’
♦ But it is not ‘perfectly rigid’
• A lump of putty is very close to ‘perfectly plastic’
♦ But it is not ‘perfectly plastic’
4. So real-bodies also have two extremes:
♦ Close to perfectly rigid
♦ Close to perfectly plastic
• This is shown in the fig.9.2 below:
5. So now we know that, real-bodies are neither ‘perfectly rigid’ nor ‘perfectly plastic’
■ Then how can we describe real-bodies?
• For describing real-bodies, we introduce a new type of body: Elastic body
• 'Elastic' comes at the exact middle of ‘perfectly rigid’ and ‘perfectly plastic’
• This is shown in fig.9.3 below:
'Elastic body' can be described in 2 steps:
(i) Apply a force F on a elastic body
♦ The body will deform
(ii) Remove F
♦ The body will completely rebound to it’s original shape and size
• This is shown in fig.9.4 below:
6. So we can write the definition for elasticity. It can be written in 3 steps:
(i) When a force is applied on a elastic body, the body deforms
(ii) When the force is removed, the body tends to regain it’s original size and shape
(iii) This property is called elasticity
7. Elastic deformation:
• This can be explained in 3 steps:
(i) The deformation suffered by a elastic body is not permanent
(ii) When the force is removed, the deformation disappears
(iii) So this deformation is given a special name: Elastic deformation
8. Now we can compare the three types:
✰ Perfectly rigid body ⇒ No change due to force
✰ Elastic body ⇒ Change disappears when force is removed
✰ Perfectly plastic body ⇒ Permanent change due to force
♦ That is., we want to know this:
✰ Why do the deformation disappear when the external force is removed?
The answer can be written in steps:
1. Consider an elastic body
• Let us apply a compressive force on that body
♦ A compressive force will try to decrease the size of the body
♦ So the molecules and atoms in the body will get closer and closer to each other
2. But there are forces acting in between those molecules and atoms
• These forces are called:
♦ Inter-molecular forces
♦ Inter-atomic forces
3. These forces will act against the external compressive force. That is.,
♦ These forces will try to resist the external compressive force
4. When the external compressive force is removed, the forces mentioned in (2) will push the molecules and atoms back to their original positions
■ Thus the deformation disappears
5. Consider an elastic body
• Let us apply a stretching force on that body
♦ A stretching force will try to increase the size of the body
♦ So the molecules and atoms in the body will get further and further away from each other
6. When the external stretching force is removed, the forces mentioned in (2) will pull the molecules and atoms back to their original positions
■ Thus the deformation disappears
Stress can be explained in 8 steps:
1. When a force tries to deform a body, a restoring force develops in the body
2. We know that, any force has both magnitude and direction
♦ The magnitude of the restoring force is
✰ Equal to the magnitude of the applied force
♦ The direction of the restoring force is
✰ Opposite to the direction of the applied force
3. When we divide the restoring force by ‘area of cross section of the body’, we get ‘stress’
• So, if F is the magnitude of the restoring force and A the area of cross section, we have:
$\mathbf\small{\rm{Stress\,=\,\frac{F}{A}}}$
4. F is in the numerator and A is in the denominator. So we can write:
■ Stress is the restoring force per unit area
5. Obviously,
♦ The unit of stress will be: N m-2
✰ Another name for N m-2 is pascal
✰ The symbol for pascal is Pa
♦ The dimensional formula of stress will be: [ML-1T-2]
✰ The reader may write the steps to derive the dimensional formula in his/her own notebooks
6. Let us see an example to demonstrate the calculation of stress. It can be explained in 4 steps:
(i) In fig.9.5(a) below, a force F is applied on the top end of a steel cylinder
• The cylinder is resting on a rigid platform
(ii) We see that, the top surface of the cylinder is a circle
♦ If 'd' is the diameter of the cylinder, the area of the top circle will be $\mathbf\small{\frac{\pi\,d^2}{4}}$
♦ Then, the stress experienced by the cylinder = $\mathbf\small{\frac{F}{\frac{\pi\,d^2}{4}}=\frac{4F}{\pi\,d^2}}$
(iii) We see that, the 'cylinder is compressed' by the action of F
♦ So the stress experienced by the cylinder is called compressive stress
(iv) It is important to note that, in fig.9.5(a), F is perpendicular to the top surface of the cylinder
7. Let us see another example. This can be explained in 4 steps:
(i) Consider fig,9.4(b) above. The same cylinder is suspended from a rigid support (The support can be the strong and rigid ceiling of a special laboratory room)
• This time the F is trying to stretch the cylinder
(ii) The stress experienced is same as before: $\mathbf\small{\frac{4F}{\pi\,d^2}}$
(iii) We see that, the cylinder is stretched by the action of F
♦ So the stress experienced by the cylinder is called tensile stress
(iv) It is important to note that, in fig.9.5(b), F is perpendicular to the bottom surface of the cylinder
8. There is a common name which can be used for both compressive stress and tensile stress
■ The common name is: Longitudinal stress
• So when we see the term ‘longitudinal stress’, it can be either compressive stress or tensile stress
• The symbol for longitudinal stress is 𝜎 (Greek small letter 'sigma')
1. When a force is applied on a body, it deforms
2. We must be able to measure the deformation
♦ If the deformation is large, we can measure it using simple instruments
♦ If the deformation is small, we will need precision instruments to measure it
3. An experienced scientist or engineer can predict whether the deformation will be small or large
• The deformation will be small or large depending on two factors:
(i) Material with which the body is made
♦ For example, a wire may be made of steel or copper
(ii) Magnitude of the force which causes the deformation
♦ For example, a large force will cause a large deformation
4. Once a deformation occurs, we will want to measure the 'value of that deformation'
• To measure deformation, 2D figs are more effective. We will write the calculations in 4 steps:
(i) Fig.9.6(a) below shows the original length L of the cylinder
• The cylinder is resting on a rigid platform
(ii) We see that, due to the compressive force F, the length of the cylinder decreases by ΔL
♦ So we can write: Change in length = ΔL
(iii) When we divide the 'change in length' by ‘original length’, we get ‘strain’
• So in fig.9.6(a), it is clear that:
♦ Strain experienced by the cylinder = $\mathbf\small{\rm{\frac{\Delta L}{L}}}$
(iv) We see that, the 'cylinder is compressed' by the action of F
♦ So the strain experienced by the cylinder is called compressive strain
5. Let us see another example. it can be written in 4 steps:
(i) Fig.9.6(b) above shows the original length L of the cylinder
• The cylinder is suspended from a rigid ceiling
(ii) We see that, due to the stretching force F, the length of the cylinder increases by ΔL
♦ So we can write: Change in length = ΔL
(iii) When we divide the 'change in length' by ‘original length’, we get ‘strain’
• So in fig.9.6(b), it is clear that:
♦ Strain experienced by the cylinder = $\mathbf\small{\rm{\frac{\Delta L}{L}}}$
(iv) We see that, the 'cylinder is stretched' by the action of F
♦ So the strain experienced by the cylinder is called tensile strain
6. There is a common name which can be used for both compressive strain and tensile strain
■ The common name is: Longitudinal strain
• So when we see the term ‘longitudinal strain’, it can be either compressive strain or tensile strain
• The symbol for longitudinal strain is 𝜀 (Greek small letter 'epsilon')
Shearing stress can be explained in 6 steps:
1. Consider the body in fig.9.7(a) below
• It has the shape of a rectangular box (A box is hollow inside. But the body in fig.a is solid)
• The body is resting on a rigid platform
• The bottom face of the body is fixed to the platform
• The height of the body is L
2. A force F is applied on the top face of the body
• This F is parallel to the top face
• So naturally, F will be parallel to the bottom face also
3. This force F has a few more peculiarities
• To explain those peculiarities, we will use 2D figs.
♦ Fig.9.8(a) below shows the original body
♦ Fig.9.8(b) shows the deformed shape
(i) The force F is
♦ Parallel to the top face
♦ Parallel to the bottom face
(ii) The force F passes only through 'points on the top surface of the body'
(iii) The force F does not pass through any 'point in the interior of the body'
(iv) In fig.c, the force F passes through some 'points in the interior of the body'
♦ For our present discussion, this is not acceptable
(v) In fig.d, the force F passes through some 'points in the interior of the body'
♦ For our present discussion, this is not acceptable
(vi) In short, for our present discussion,
♦ F must be parallel to the top and bottom faces
♦ F must pass only through 'points on the top surface of the body'
4. Due to the application of F, a restoring force develops inside the body
♦ This restoring force is equal in magnitude to F
♦ This restoring force is opposite in direction to F
5. When we divide the 'restoring force' by the 'area of top face', we get: shearing stress
■ So shearing stress is: Restoring force per unit area
6. Another name for shearing stress is: Tangential stress
• The force F also can be called by any of the two names:
♦ Shearing force
♦ Tangential force
1. Consider fig.9.8(b) above
• The bottom face of the body is fixed
2. Due to the shearing force F, the top face is displaced
3. But there is no displacement for the bottom face
4. When compared to the bottom face, the displacement of the top face is Δx
♦ Since Δx is obtained in comparison with the bottom face, we say that:
✰ Δx is the 'relative displacement' between the top and bottom faces
5. Now we take ratio of the two items:
♦ Relative displacement Δx
♦ Height of the body L
• This ratio is called shearing strain
■ So we can write: Shearing strain = $\mathbf\small{\rm{\frac{\Delta x}{L}}}$
6. But in fig.9.8(b), we see that:
• Δx and L forms the perpendicular sides of a right triangle
• From that same fig.b, we get: $\mathbf\small{\rm{\tan \theta=\frac{\Delta x}{L}}}$
♦ Where 𝜽 is the angular displacement of the body from the initial position
7. We have seen in math classes that:
• If 𝜽 is small, tan 𝜽 = 𝜽
♦ Where 𝜽 on the right side is expressed in radians
♦ (𝜽 on the left side can be in degrees or radians because, whatever be the unit, 'tan 𝜽' will be the same)
■ Let us see an example. It can be written in 4 steps:
(i) Let 𝜽 = 10o
♦ Then 𝜽 in radians = 0.1745
(ii) We have: tan 𝜽 = tan 10o = tan 0.1745c = 0.1763
(iii) What is the error if we use '𝜽' instead of 'tan 𝜽' ?
• This can be calculated as follows:
♦ Difference between the two = (tan 𝜽 - 𝜽) = (0.1763 - 0.1745) = 0.0018
♦ So % error = $\mathbf\small{\rm{\frac{0.0018}{0.1745}\times 100=1.03}}$ %
(iv) So it is clear that, when 𝜽 is small, we can take tan 𝜽 = 𝜽
8. The 𝜽 shown in fig.9.8(b) will be very small in most practical cases
• So we can take 'tan 𝜽' equal to 𝜽
• Thus, comparing the results in (5) and (6), we get:
♦ Shearing strain = $\mathbf\small{\rm{\frac{\Delta x}{L}}}$ = tan 𝜽 = 𝜽
■ That is., Shearing strain = 𝜽
Hydraulic stress can be explained in 5 steps:
1. Consider a solid sphere immersed in a fluid
• The fluid will exert force from all directions on the sphere
• This is shown in fig.9.9(a) below:
2. In the above fig.9.9(a), only a very few forces are shown. This is to avoid congestion
• If there is enough space, we can draw infinite number of forces in the ‘infinite number of directions’ available
■ That means:
♦ Every point on the surface of the sphere will experience a force
• The force experienced by all those points will be equal in magnitude
3. Consider any one of those point
■ The force acting at that point will be perpendicular to the surface of the sphere at that point
• But the surface of the sphere is not a flat surface. It is a curved surface
• Then how can we draw the perpendicular?
4. The answer is simple. It can be written in 5 steps:
(i) For curved surfaces, we make use of tangents
(ii) Suppose that, we want a perpendicular at the point P on the surface of our sphere
•We first draw a tangent through P
♦ This is shown in fig.b
(iii) Then we draw a perpendicular to this tangent
♦ This perpendicular should be drawn in such a way that, it passes through P
♦ This is shown in fig.c
(iv) In our present case, the body has a spherical shape. The 'spherical shape' is a 'regular shape'
• If the body has a irregular shape, it will be more difficult to draw the tangents. We will see such cases in higher classes
(v) Note that, in the case of spheres and circles, the perpendicular can be drawn even more easily. Because, all the perpendiculars on the surface will pass through the center of the sphere/circle (Details here)
5. Due to all the external forces, an internal restoring force will develop inside the sphere
■ When we divide this restoring force by the surface area of the sphere, we get the hydraulic stress
• In other words, hydraulic stress is the restoring force per unit area
■ It’s units and dimensional formula are same as those of longitudinal stress that we saw before
1. We saw that, the sphere in fig.9.9(a) is subjected to forces from all directions
♦ As a result, the sphere will shrink in all directions
♦ This is shown in fig,9.10 below:
2. So the volume will decrease
♦ Since the shrinking occurs in all directions, the volume will be preserved
♦ That means, the sphere will remain a sphere, but with lesser diameter
3. So we have a new volume
• The difference between the initial and final volumes will give the change in volume ΔV
4. When we divide this ΔV by the initial volume V, we get the volume strain
• So we can write: Volume strain = $\mathbf\small{\rm{\frac{\Delta V}{V}}}$
1. We have seen three types of strains:
Longitudinal strain, Shear strain, Volume strain
2. Consider the longitudinal strain:
• We have: Longitudinal strain = $\mathbf\small{\rm{\frac{\Delta L}{L}}}$
♦ The numerator is length
♦ The denominator is length
• So longitudinal strain will have no units
♦ Also, it will have no dimensional formula
3. Consider the shear strain:
• We have: Shear strain = $\mathbf\small{\rm{\frac{\Delta x}{L}}}$
♦ The numerator is length
♦ The denominator is length
• So shear strain will have no units
♦ Also, it will have no dimensional formula
4. Consider the volume strain:
• We have: Volume strain = $\mathbf\small{\rm{\frac{\Delta V}{V}}}$
♦ The numerator is volume
♦ The denominator is volume
• So volume strain will have no units
♦ Also, it will have no dimensional formula
5. It is clear that, strain is just a ratio. It has no units. Also it has no dimensional formula
• We know that, it is possible to convert any ratio into percentage format
• For example:
♦ $\mathbf\small{\rm{\frac{1}{2}=50}}$%
♦ $\mathbf\small{\rm{\frac{1}{3}=33.33}}$%
♦ $\mathbf\small{\rm{\frac{2}{5}=40}}$%
• So it is clear that, strain can be expressed in percentage format also
• In problems, if we are given strain in percentage format, we can convert it back into ratio format
Following solved examples demonstrate the applications of longitudinal stress and strain:
Solved examples 9.1 to 9.4
• In this section we will see Mechanical properties of solids
Rigidity and Plasticity are two extreme cases. We can write about them in 3 steps:
1. In the first extreme case, the given body is perfectly rigid
• In this case, the body has a definite shape and size
♦ If we apply a force on that body, no deformation will occur. That is.,
✰ The shape of the body will not change
✰ The size of the body will not change
✰ This is shown in fig.9.1(b) below
 |
| Fig.9.1 |
(i) In this case, the body do not have definite shape and size
♦ If we apply a force (like placing a 1 kg wt.) on that body, deformation will occur. That is.,
✰ The shape of the body will change to a ‘new shape’
✰ The size of the body will change to a ‘new size’
✰ This is shown in fig.9.1(d) above
♦ If we remove the force,
✰ There will be no return from the ‘new shape’ to the 'original shape'
✰ There will be no return from the ‘new size’ to the 'original size'
✰ This is shown in fig.9.1(e) above
3. That means, as a result of the force F, a permanent deformation takes place in a perfectly plastic body
♦ So we can write:
✰ Perfectly rigid body ⇒ No change due to force
✰ Perfectly plastic body ⇒ Permanent change due to force
• So we have seen the two extremes
♦ No body is perfectly rigid
♦ No body is perfectly plastic
• So how can we describe the reality?
The answer can be written in 8 steps:
1. We will use the term ‘real-bodies’ to denote the bodies that we see in our day to day life
• The simplest definition for a real-body can be given using just two statements (a) and (b):
(a) A real-body is not perfectly rigid
(b) A real-body is not perfectly plastic
2. Let us elaborate the two statements
Statement (a): A real-body is not perfectly rigid
This statement can elaborated using 2 steps:
(i) Consider a body which appears to be perfectly rigid
• When an external force is applied on that body, it’s shape and size will definitely change
(ii) An example:
• A steel rod of small length (about 30 cm) and comparatively large diameter (about 2.5 cm) will appear to be very rigid
• But if we rivet it on one end and pull from the other end with sufficient force, it’s length will surely increase
• The ‘increase in length’ may not be visible to the naked eye. But we can measure it using precision instruments
Statement (b): A real-body is not perfectly plastic
This statement can elaborated using 2 steps:
(i) Consider a body which appears to be perfectly plastic
• When an external force is applied on that body, it’s shape and size will visibly change to ‘new shape’ and ‘new size’
• But when the force is removed, the body will definitely try to get back to the ‘original shape’and ‘original size’
(ii) An example:
• Consider a lump of putty or lump of wet clay
(See: images of putty and images of wet clay)
• Apply a force on the lump. It will easily deform and will appear to be perfectly plastic
• But if we remove the force, there will be a very small rebound
(See: Dictionary meaning of rebound)
• The ‘rebound’ may not be visible to the naked eye. But we can measure it using precision instruments
3. So we can write:
• A steel rod is very close to ‘perfectly rigid’
♦ But it is not ‘perfectly rigid’
• A lump of putty is very close to ‘perfectly plastic’
♦ But it is not ‘perfectly plastic’
4. So real-bodies also have two extremes:
♦ Close to perfectly rigid
♦ Close to perfectly plastic
• This is shown in the fig.9.2 below:
 |
| Fig.9.2 |
■ Then how can we describe real-bodies?
• For describing real-bodies, we introduce a new type of body: Elastic body
• 'Elastic' comes at the exact middle of ‘perfectly rigid’ and ‘perfectly plastic’
• This is shown in fig.9.3 below:
 |
| Fig.9.3 |
(i) Apply a force F on a elastic body
♦ The body will deform
(ii) Remove F
♦ The body will completely rebound to it’s original shape and size
• This is shown in fig.9.4 below:
 |
| Fig.9.4 |
(i) When a force is applied on a elastic body, the body deforms
(ii) When the force is removed, the body tends to regain it’s original size and shape
(iii) This property is called elasticity
7. Elastic deformation:
• This can be explained in 3 steps:
(i) The deformation suffered by a elastic body is not permanent
(ii) When the force is removed, the deformation disappears
(iii) So this deformation is given a special name: Elastic deformation
8. Now we can compare the three types:
✰ Perfectly rigid body ⇒ No change due to force
✰ Elastic body ⇒ Change disappears when force is removed
✰ Perfectly plastic body ⇒ Permanent change due to force
• Next, we want to know the ‘cause of elasticity’
✰ Why do the deformation disappear when the external force is removed?
The answer can be written in steps:
1. Consider an elastic body
• Let us apply a compressive force on that body
♦ A compressive force will try to decrease the size of the body
♦ So the molecules and atoms in the body will get closer and closer to each other
2. But there are forces acting in between those molecules and atoms
• These forces are called:
♦ Inter-molecular forces
♦ Inter-atomic forces
3. These forces will act against the external compressive force. That is.,
♦ These forces will try to resist the external compressive force
4. When the external compressive force is removed, the forces mentioned in (2) will push the molecules and atoms back to their original positions
■ Thus the deformation disappears
5. Consider an elastic body
• Let us apply a stretching force on that body
♦ A stretching force will try to increase the size of the body
♦ So the molecules and atoms in the body will get further and further away from each other
6. When the external stretching force is removed, the forces mentioned in (2) will pull the molecules and atoms back to their original positions
■ Thus the deformation disappears
Stress experienced by a body
1. When a force tries to deform a body, a restoring force develops in the body
2. We know that, any force has both magnitude and direction
♦ The magnitude of the restoring force is
✰ Equal to the magnitude of the applied force
♦ The direction of the restoring force is
✰ Opposite to the direction of the applied force
3. When we divide the restoring force by ‘area of cross section of the body’, we get ‘stress’
• So, if F is the magnitude of the restoring force and A the area of cross section, we have:
$\mathbf\small{\rm{Stress\,=\,\frac{F}{A}}}$
4. F is in the numerator and A is in the denominator. So we can write:
■ Stress is the restoring force per unit area
5. Obviously,
♦ The unit of stress will be: N m-2
✰ Another name for N m-2 is pascal
✰ The symbol for pascal is Pa
♦ The dimensional formula of stress will be: [ML-1T-2]
✰ The reader may write the steps to derive the dimensional formula in his/her own notebooks
6. Let us see an example to demonstrate the calculation of stress. It can be explained in 4 steps:
(i) In fig.9.5(a) below, a force F is applied on the top end of a steel cylinder
 |
| Fig.9.5 |
(ii) We see that, the top surface of the cylinder is a circle
♦ If 'd' is the diameter of the cylinder, the area of the top circle will be $\mathbf\small{\frac{\pi\,d^2}{4}}$
♦ Then, the stress experienced by the cylinder = $\mathbf\small{\frac{F}{\frac{\pi\,d^2}{4}}=\frac{4F}{\pi\,d^2}}$
(iii) We see that, the 'cylinder is compressed' by the action of F
♦ So the stress experienced by the cylinder is called compressive stress
(iv) It is important to note that, in fig.9.5(a), F is perpendicular to the top surface of the cylinder
7. Let us see another example. This can be explained in 4 steps:
(i) Consider fig,9.4(b) above. The same cylinder is suspended from a rigid support (The support can be the strong and rigid ceiling of a special laboratory room)
• This time the F is trying to stretch the cylinder
(ii) The stress experienced is same as before: $\mathbf\small{\frac{4F}{\pi\,d^2}}$
(iii) We see that, the cylinder is stretched by the action of F
♦ So the stress experienced by the cylinder is called tensile stress
(iv) It is important to note that, in fig.9.5(b), F is perpendicular to the bottom surface of the cylinder
8. There is a common name which can be used for both compressive stress and tensile stress
■ The common name is: Longitudinal stress
• So when we see the term ‘longitudinal stress’, it can be either compressive stress or tensile stress
• The symbol for longitudinal stress is 𝜎 (Greek small letter 'sigma')
Strain experienced by a body
Strain can be explained in 6 steps:1. When a force is applied on a body, it deforms
2. We must be able to measure the deformation
♦ If the deformation is large, we can measure it using simple instruments
♦ If the deformation is small, we will need precision instruments to measure it
3. An experienced scientist or engineer can predict whether the deformation will be small or large
• The deformation will be small or large depending on two factors:
(i) Material with which the body is made
♦ For example, a wire may be made of steel or copper
(ii) Magnitude of the force which causes the deformation
♦ For example, a large force will cause a large deformation
4. Once a deformation occurs, we will want to measure the 'value of that deformation'
• To measure deformation, 2D figs are more effective. We will write the calculations in 4 steps:
(i) Fig.9.6(a) below shows the original length L of the cylinder
 |
| Fig.9.6 |
(ii) We see that, due to the compressive force F, the length of the cylinder decreases by ΔL
♦ So we can write: Change in length = ΔL
(iii) When we divide the 'change in length' by ‘original length’, we get ‘strain’
• So in fig.9.6(a), it is clear that:
♦ Strain experienced by the cylinder = $\mathbf\small{\rm{\frac{\Delta L}{L}}}$
(iv) We see that, the 'cylinder is compressed' by the action of F
♦ So the strain experienced by the cylinder is called compressive strain
5. Let us see another example. it can be written in 4 steps:
(i) Fig.9.6(b) above shows the original length L of the cylinder
• The cylinder is suspended from a rigid ceiling
(ii) We see that, due to the stretching force F, the length of the cylinder increases by ΔL
♦ So we can write: Change in length = ΔL
(iii) When we divide the 'change in length' by ‘original length’, we get ‘strain’
• So in fig.9.6(b), it is clear that:
♦ Strain experienced by the cylinder = $\mathbf\small{\rm{\frac{\Delta L}{L}}}$
(iv) We see that, the 'cylinder is stretched' by the action of F
♦ So the strain experienced by the cylinder is called tensile strain
6. There is a common name which can be used for both compressive strain and tensile strain
■ The common name is: Longitudinal strain
• So when we see the term ‘longitudinal strain’, it can be either compressive strain or tensile strain
• The symbol for longitudinal strain is 𝜀 (Greek small letter 'epsilon')
Shearing stress experienced by a body
Shearing stress can be explained in 6 steps:
1. Consider the body in fig.9.7(a) below
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| Fig.9.7 |
• The body is resting on a rigid platform
• The bottom face of the body is fixed to the platform
• The height of the body is L
2. A force F is applied on the top face of the body
• This F is parallel to the top face
• So naturally, F will be parallel to the bottom face also
3. This force F has a few more peculiarities
• To explain those peculiarities, we will use 2D figs.
♦ Fig.9.8(a) below shows the original body
♦ Fig.9.8(b) shows the deformed shape
 |
| Fig.9.8 |
♦ Parallel to the top face
♦ Parallel to the bottom face
(ii) The force F passes only through 'points on the top surface of the body'
(iii) The force F does not pass through any 'point in the interior of the body'
(iv) In fig.c, the force F passes through some 'points in the interior of the body'
♦ For our present discussion, this is not acceptable
(v) In fig.d, the force F passes through some 'points in the interior of the body'
♦ For our present discussion, this is not acceptable
(vi) In short, for our present discussion,
♦ F must be parallel to the top and bottom faces
♦ F must pass only through 'points on the top surface of the body'
4. Due to the application of F, a restoring force develops inside the body
♦ This restoring force is equal in magnitude to F
♦ This restoring force is opposite in direction to F
5. When we divide the 'restoring force' by the 'area of top face', we get: shearing stress
■ So shearing stress is: Restoring force per unit area
6. Another name for shearing stress is: Tangential stress
• The force F also can be called by any of the two names:
♦ Shearing force
♦ Tangential force
Shearing strain experienced by a body
Shearing strain can be explained in 8 steps:1. Consider fig.9.8(b) above
• The bottom face of the body is fixed
2. Due to the shearing force F, the top face is displaced
3. But there is no displacement for the bottom face
4. When compared to the bottom face, the displacement of the top face is Δx
♦ Since Δx is obtained in comparison with the bottom face, we say that:
✰ Δx is the 'relative displacement' between the top and bottom faces
5. Now we take ratio of the two items:
♦ Relative displacement Δx
♦ Height of the body L
• This ratio is called shearing strain
■ So we can write: Shearing strain = $\mathbf\small{\rm{\frac{\Delta x}{L}}}$
6. But in fig.9.8(b), we see that:
• Δx and L forms the perpendicular sides of a right triangle
• From that same fig.b, we get: $\mathbf\small{\rm{\tan \theta=\frac{\Delta x}{L}}}$
♦ Where 𝜽 is the angular displacement of the body from the initial position
7. We have seen in math classes that:
• If 𝜽 is small, tan 𝜽 = 𝜽
♦ Where 𝜽 on the right side is expressed in radians
♦ (𝜽 on the left side can be in degrees or radians because, whatever be the unit, 'tan 𝜽' will be the same)
■ Let us see an example. It can be written in 4 steps:
(i) Let 𝜽 = 10o
♦ Then 𝜽 in radians = 0.1745
(ii) We have: tan 𝜽 = tan 10o = tan 0.1745c = 0.1763
(iii) What is the error if we use '𝜽' instead of 'tan 𝜽' ?
• This can be calculated as follows:
♦ Difference between the two = (tan 𝜽 - 𝜽) = (0.1763 - 0.1745) = 0.0018
♦ So % error = $\mathbf\small{\rm{\frac{0.0018}{0.1745}\times 100=1.03}}$ %
(iv) So it is clear that, when 𝜽 is small, we can take tan 𝜽 = 𝜽
8. The 𝜽 shown in fig.9.8(b) will be very small in most practical cases
• So we can take 'tan 𝜽' equal to 𝜽
• Thus, comparing the results in (5) and (6), we get:
♦ Shearing strain = $\mathbf\small{\rm{\frac{\Delta x}{L}}}$ = tan 𝜽 = 𝜽
■ That is., Shearing strain = 𝜽
Hydraulic stress experienced by a body
Hydraulic stress can be explained in 5 steps:
1. Consider a solid sphere immersed in a fluid
• The fluid will exert force from all directions on the sphere
• This is shown in fig.9.9(a) below:
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| Fig.9.9 |
• If there is enough space, we can draw infinite number of forces in the ‘infinite number of directions’ available
■ That means:
♦ Every point on the surface of the sphere will experience a force
• The force experienced by all those points will be equal in magnitude
3. Consider any one of those point
■ The force acting at that point will be perpendicular to the surface of the sphere at that point
• But the surface of the sphere is not a flat surface. It is a curved surface
• Then how can we draw the perpendicular?
4. The answer is simple. It can be written in 5 steps:
(i) For curved surfaces, we make use of tangents
(ii) Suppose that, we want a perpendicular at the point P on the surface of our sphere
•We first draw a tangent through P
♦ This is shown in fig.b
(iii) Then we draw a perpendicular to this tangent
♦ This perpendicular should be drawn in such a way that, it passes through P
♦ This is shown in fig.c
(iv) In our present case, the body has a spherical shape. The 'spherical shape' is a 'regular shape'
• If the body has a irregular shape, it will be more difficult to draw the tangents. We will see such cases in higher classes
(v) Note that, in the case of spheres and circles, the perpendicular can be drawn even more easily. Because, all the perpendiculars on the surface will pass through the center of the sphere/circle (Details here)
5. Due to all the external forces, an internal restoring force will develop inside the sphere
■ When we divide this restoring force by the surface area of the sphere, we get the hydraulic stress
• In other words, hydraulic stress is the restoring force per unit area
■ It’s units and dimensional formula are same as those of longitudinal stress that we saw before
Volume strain experienced by a body
Volume strain can be explained in 4 steps:1. We saw that, the sphere in fig.9.9(a) is subjected to forces from all directions
♦ As a result, the sphere will shrink in all directions
♦ This is shown in fig,9.10 below:
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| Fig.9.10 |
♦ Since the shrinking occurs in all directions, the volume will be preserved
♦ That means, the sphere will remain a sphere, but with lesser diameter
3. So we have a new volume
• The difference between the initial and final volumes will give the change in volume ΔV
4. When we divide this ΔV by the initial volume V, we get the volume strain
• So we can write: Volume strain = $\mathbf\small{\rm{\frac{\Delta V}{V}}}$
Strain as a percentage
This can be explained in 5 steps:1. We have seen three types of strains:
2. Consider the longitudinal strain:
• We have: Longitudinal strain = $\mathbf\small{\rm{\frac{\Delta L}{L}}}$
♦ The numerator is length
♦ The denominator is length
• So longitudinal strain will have no units
♦ Also, it will have no dimensional formula
3. Consider the shear strain:
• We have: Shear strain = $\mathbf\small{\rm{\frac{\Delta x}{L}}}$
♦ The numerator is length
♦ The denominator is length
• So shear strain will have no units
♦ Also, it will have no dimensional formula
4. Consider the volume strain:
• We have: Volume strain = $\mathbf\small{\rm{\frac{\Delta V}{V}}}$
♦ The numerator is volume
♦ The denominator is volume
• So volume strain will have no units
♦ Also, it will have no dimensional formula
5. It is clear that, strain is just a ratio. It has no units. Also it has no dimensional formula
• We know that, it is possible to convert any ratio into percentage format
• For example:
♦ $\mathbf\small{\rm{\frac{1}{2}=50}}$%
♦ $\mathbf\small{\rm{\frac{1}{3}=33.33}}$%
♦ $\mathbf\small{\rm{\frac{2}{5}=40}}$%
• So it is clear that, strain can be expressed in percentage format also
• In problems, if we are given strain in percentage format, we can convert it back into ratio format
Solved examples 9.1 to 9.4
In the next section, we will see Hooke's Law
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