Chapter 9.2 - Young's Modulus

In the previous section, we saw the basic details about stress-strain diagram. In this section we will see Young's modulus

• We have already seen the basic details about the portion OA of the stress-strain curve
• Here we will discuss a few more details about this portion. It can be written in 8 steps:
1. We know that, 'point A' is the point of proportional limit
■ OA is a straight line because, Hooke’s law is satisfied in this portion
    ♦ That is., $\mathbf\small{\rm{stress = k \times strain}}$ is satisfied in this portion
    ♦ That is., $\mathbf\small{\rm{\frac{stress}{strain}=k}}$ is satisfied in this portion
2. We have seen three different types of stresses and strains. They are:
(i) Compressive stress which causes compressive strain
and Tensile stress which causes tensile strain
• These two can be combined as:
Longitudinal stress which causes longitudinal strain
(ii) Shear stress which causes shear strain
(iii) Hydraulic stress which causes volumetric strain
3. All items mentioned in (2) are stresses and strains
■ We can draw a separate stress-strain curve for each of them
That is.,
(i) We can draw a longitudinal stress - longitudinal strain curve
    ♦ We have already seen this curve
(ii) We can draw a shear stress- shear strain curve
    ♦ Some examples can be seen here
(iii) We can draw a hydraulic stress - hydraulic strain curve
    ♦ Some examples can be seen here 
4. Each of the curves mentioned in (3), will have a proportional limit
• We have seen that, this 'proportional limit' is marked as point A
• Let us elaborate this a little more. It can be done in 3 steps:
(i) Consider the longitudinal stress - longitudinal strain curve
    ♦ The initial portion OA will be a straight line
    ♦ Within this OA, we will get: $\mathbf\small{\rm{\frac{Longitudinal\;stress}{Longitudinal\;strain}=k_1}}$
          ✰ Where k₁ is a constant
(ii) Consider the shear stress - shear strain curve
    ♦ The initial portion OA will be a straight line
    ♦ Within this OA, we will get: $\mathbf\small{\rm{\frac{Shear\;stress}{Shear\;strain}=k_2}}$
          ✰ Where k₂ is a constant
(iii) Consider the hydraulic stress - volumetric strain curve
    ♦ The initial portion OA will be a straight line
    ♦ Within this OA, we will get: $\mathbf\small{\rm{\frac{Hydraulic\;stress}{Volumetric\;strain}=k_3}}$
          ✰ Where k₃ is a constant
5. So we have to learn about three constants: k₁, k₂ and k₃
■ These constants are commonly known as moduli of elasticity or elastic moduli
    ♦ Modulus is singular
    ♦ Moduli is plural
6. So we can write:
    ♦ k₁ is the Elastic modulus related to longitudinal stress and longitudinal strain
    ♦ k₂ is the Elastic modulus related to shear stress and shear strain
    ♦ k₃ is the Elastic modulus related to hydraulic stress and volumetric strain
7. For differentiating between the three elastic moduli, each one is given a special name:
    ♦ The constant k₁ is called Young's Modulus
    ♦ The constant k₂ is called Shear Modulus   
    ♦ The constant k₃ is called Bulk Modulus
• So we can write:
    ♦ $\mathbf\small{\rm{\frac{Longitudinal\;stress}{Longitudinal\;strain}}}$ = Young's modulus
    ♦ $\mathbf\small{\rm{\frac{Shear\;stress}{Shear\;strain}}}$ = Shear modulus
    ♦ $\mathbf\small{\rm{\frac{Hydraulic\;stress}{Volumetric\;strain}}}$ = Bulk modulus
8. So the items given in (7) are the three elastic moduli
■ Note that:
• The word Elastic or Elasticity is specially mentioned along with moduli
• This is because, these moduli are applicable only when the material is within the proportional limit (That is., within the portion OA)
• At higher parts of the curve, the material is no longer elastic and so, the moduli are not applicable at those parts

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Young's modulus

• First we will discuss about Young’s modulus. It is represented by the letter Y
• We have seen that,Young’s modulus is related to longitudinal stress and strain
• Let us see an interesting feature related to longitudinal stress and strain. We will write it in 7 steps:
1. Fig.9.17(a) below shows the original length L of a steel cylinder

Fig.9.17

• The cylinder is resting on a rigid platform
• We see that, due to the compressive force F, the length of the cylinder decreases by ΔL_c
    ♦ So we can write: Change in length due to compression = ΔL_c
    ♦ Then, strain (𝜺_c) due to compression will be given by: $\mathbf\small{\rm{\epsilon_c=\frac{\Delta L_c}{L}}}$
2. Fig.9.17(b) above shows the same steel cylinder mentioned in (1)
• This time, the cylinder is suspended from a rigid ceiling
• We see that, due to the stretching force F, the length of the cylinder increases by ΔL_t
[This 'F' is same in magnitude as the 'F' in (1)]
    ♦ So we can write: Change in length due to tension = ΔL_t
    ♦ Then, strain (𝜺_t) due to tension will be given by: $\mathbf\small{\rm{\epsilon_t=\frac{\Delta L_t}{L}}}$
3. Remember that:
• The same steel cylinder is used in both the cases
    ♦ So the 'cross sectional area on which force is applied' will be the same in both cases
• The same force F is applied in both the cases
• Since forces and areas are the same, we get:
Compressive stress (𝜎_c) applied in (1) = Tensile stress (𝜎_t) applied in (2)
• Also, since the same steel cylinder is used in both cases, the initial length L is same
4. Experiments show that:
If all the conditions in (3) are satisfied, ΔL_c will be equal to ΔL_t 
• From this we get:
$\mathbf\small{\rm{\frac{\Delta L_c}{L}=\frac{\Delta L_t}{L}}}$
⇒ $\mathbf\small{\rm{\epsilon_c=\epsilon_t}}$
• That is: Compressive strain (𝜺_c) = Tensile strain (𝜺_t)
• Also, from (3), we have: Compressive stress (𝜎_c) = Tensile stress (𝜎_t)
5. Now let us calculate Y:
• Let us calculate Y using the compressive test
    ♦ We have: $\mathbf\small{\rm{Y=\frac{\sigma_c}{\epsilon_c}}}$
• Next, let us calculate Y using the tensile test
    ♦ We have: $\mathbf\small{\rm{Y=\frac{\sigma_t}{\epsilon_t}}}$
6. Equality of the two fractions:
    ♦ Both the numerators in (5) are equal
    ♦ Both the denominators in (5) are also equal
• So both the results in (5) are the same
7. So we can write:
■ If the body is made of a material like steel and if all the conditions in (3) are satisfied, we can determine Y using either compressive method or tensile method. Both methods will give the same result

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Next we will derive a formula to find Y. It can be written in 3 steps:
1. We have: $\mathbf\small{\rm{Y=\frac{Longitudinal\;stress\;(\sigma)}{Longitudinal\;strain\;(\epsilon)}}}$
2. But $\mathbf\small{\rm{\sigma=\frac{F}{A}\;\;and\;\;\epsilon=\frac{\Delta L}{L}}}$
3. So the result in (1) becomes: $\mathbf\small{\rm{Y=\frac{\sigma}{\epsilon}=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}}}$
■ Thus we get: $\mathbf\small{\rm{Y=\frac{F \times L}{A \times \Delta L}}}$
■ Note that, since strain has no unit, the 'unit of Y' is same as that of stress, which is: Nm^-2

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Now we will see two important properties related to Y. They can be written in 7 steps:
1. We have: $\mathbf\small{\rm{Y=\frac{\sigma}{\epsilon}}}$
• This can be rearranged as: $\mathbf\small{\rm{\sigma=Y \times \epsilon}}$
2. The '=' sign implies that '𝜎' and '(Y × 𝜺)' are equal
• So if Y is large, 𝜺 will become small. Then only the equality between '𝜎' and '(Y×𝜺)' can be maintained
3. So we can write:
■ If the 'material of large Y' is used to make a cylinder, even a large 𝜎 will produce only a small strain
• In other words:
■ A 'material having large Y' will be stronger than a 'material having smaller Y'
4. Let us see a comparison:
• From the data book, we have:
    ♦ Y_Aluminium = 70 ×10⁹ N m^-2
    ♦ Y_Copper = 120 ×10⁹ N m^-2
    ♦ Y_Steel = 200 ×10⁹ N m^-2
5. Let us apply a same load on three different cylinders, made of aluminium, copper and steel
• Assume that the cylinders have the same length and diameter 
• Since the loads and diameters are the same, the stress (𝜎) will be the same
• Then we get:
𝜎 = (Y_Aluminium × 𝜺_Aluminium) = (Y_Copper × 𝜺_Copper) = (Y_Steel × 𝜺_Steel)
• Comparing the values in (4), we can write:
    ♦ 𝜺_Steel will be the smallest
    ♦ 𝜺_Aluminium will be the largest
6. We see that, large forces are required to cause large strains in steel
• That means, very large forces will be required to make steel elongate like plastic
• That means, if we do not apply such very large forces, the steel will always return to it's original length
• That means, steel is more elastic than aluminium and copper
■ That is why, steel is used in the construction of buildings, transmission towers etc.,
• The engineer will specify the maximum load that is allowed on a building or tower
    ♦ When we apply that load, the pillars and beams will deform a little
    ♦ When that load is removed, the pillars and beams will regain their original dimensions
    ♦ This is because, steel is elastic even when large loads are applied
7. Thus we see two important properties related to Y:
(i) A material having larger Y will be stronger
(ii) A material having larger Y will be more elastic

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Let us now write a useful information that can be obtained from graphs. It can be written in 7 steps:

1. We have seen that:
• Stress = k × Strain (or Longitudinal stress = Y × Longitudinal strain)
• is similar to
• the equation y = mx 
    ♦ We often this equation in coordinate geometry classes
■ So the slope m corresponds to the young's modulus Y
2. Consider two graphs shown in the fig.9.18 below:

Fig.9.18

• The graph in fig.a has a steep slope
• The graph in fig.b has a gentle slope
3. Forming a right triangle:
(i) Mark any two convenient points P and Q in the graph in fig.a
(ii) Draw horizontal and vertical dashed lines through P and Q
    ♦ Let them meet at R
(iii) We will get a right triangle PQR with PQ as the hypotenuse
(iv) We see that:
The height (QR) of the triangle is large and the base (PR) is small
4. Slope of the graph is obtained as: $\mathbf\small{\rm{Slope=\frac{Height}{Base}}}$
    ♦ The numerator (Height) is larger
    ♦ The denominator (Base) is smaller
    ♦ So the slope of the graph in fig.a is high
5. But the slope corresponds to Y
■ So we can write:
If the portion OA in the stress-strain curve is steep (high slope), the material will have a high value for Y
6. The opposite happens in fig.b
• We see that:
(i) The height of the triangle is small and the base is large
(ii) That is:
    ♦The numerator (Height) is smaller
    ♦ The denominator (Base) is larger 
(iii) So the slope of the graph in fig.a is low
■ So we can write:
If the portion OA in the stress-strain curve has a gentle slope, the material will have a lower value for Y
7. Note that, if we want to compare two materials in this way, their graphs must be drawn to the same scale

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Let us see some solved examples:

Solved examples 9.5 to 9.10

Solved examples 9.11 to 9.16

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Now we will see an experiment to determine the Young's modulus. It can be written in 12 steps:

1. We will be determining the ‘Y of the material’ with which, a wire is made. Consider fig.9.19 below:

Fig.9.19

• Two wires are suspended from a rigid support

• The two wires:

    ♦ Are long and straight

    ♦ Have the same length

    ♦ Have the same diameter

2. The wire on the left is called the reference wire

    ♦ The main scale is attached to this wire

3. The wire on the right is called the experimental wire

    ♦ The vernier scale is attached to this wire

4. The reference wire carries a pan at it’s bottom

    ♦ A fixed weight is placed in this pan

5. The experimental wire also carries a pan at it’s bottom

    ♦ The weight in this pan can be increased or decreased

6. An initial small mass is placed on both the pans

    ♦ Due to the weights of those masses, both the wires become straight

7. The initial reading (r₁) is noted from the scale

8. The load in the experimental wire is gradually increased

    ♦ The additional mass (m) is noted

    ♦ Then the additional weight = mg

9. Due to the additional weight, the wire will experience tensile stress and will elongate

    ♦ The new reading (r₂) is noted

    ♦ The difference (r₂-r₁) will give the elongation ΔL

• Let L be initial length of the experimental wire

    ♦ Then strain is given by: $\mathbf\small{\rm{\epsilon=\frac{\Delta L}{L}}}$ 

10. Let r be the radius of the experimental wire

• Then area of cross section of the experimental wire is given by: $\mathbf\small{\rm{A=\pi r^2}}$ 

• Then stress experienced by the experimental wire is given by: $\mathbf\small{\rm{\sigma=\frac{mg}{\pi r^2}}}$ 

11. So we get:

• Y of the material of the wire = $\mathbf\small{\rm{\frac{\sigma}{\epsilon}=\frac{\frac{mg}{\pi r^2}}{\frac{\Delta L}{L}}}}$

$\mathbf\small{\rm{\Rightarrow Y=\frac{mg\;L}{\pi r^2\;\Delta L}}}$

12. An advantage of using the reference wire can be written in 3 steps:

(i) Even when the room temperature is normal, a wire made of steel may elongate a little

(ii) So the elongation that we measure after applying the weight, will not be the 'true elongation due to tensile stress'

    ♦ It will include the 'elongation due to temperature increase' also

(iii) But if a reference wire is present, such a problem will not arise because, both wires will be 'expanding to the same amount' due to temperature increase 

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In the next section, we will see shear modulus

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Chapter 9 - Mechanical Properties of Solids

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

Fig.9.1

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

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• So we have seen the two extremes

• 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:

Fig.9.2

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:

Fig.9.3

'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:

Fig.9.4

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

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• Next, we want to know the ‘cause of elasticity’

    ♦ 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

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Stress experienced by a body

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

Fig.9.5

• 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')

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

• 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 experienced by a body

Shearing stress can be explained in 6 steps:
1. Consider the body in fig.9.7(a) below

Fig.9.7

• 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

Fig.9.8

(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

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 = 10^o
    ♦ Then 𝜽 in radians = 0.1745 
(ii) We have: tan 𝜽 = tan 10^o = tan 0.1745^c = 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:

Fig.9.9

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

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:

Fig.9.10

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 = $\mathbf\small{\rm{\frac{\Delta V}{V}}}$

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Strain as a percentage

This can be explained in 5 steps:
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

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Following solved examples demonstrate the applications of longitudinal stress and strain:
Solved examples 9.1 to 9.4

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In the next section, we will see Hooke's Law

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