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 9.1 - The Stress-Strain Curve

In the previous section, we saw various stresses and strains. In this section we will see Hooke's law

• The English Physicist Robert Hooke discovered that strain is directly proportional to stress
• That means:
    ♦ When stress increases, strain also increases
    ♦ When stress decreases, strain also decreases
■ So we can write: stress ∝ strain
• This is known as Hooke’s law

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• Note the difference between ‘directly proportional’ and ‘inversely proportional’
    ♦ When x is directly proportional to y, we write: $\mathbf\small{\rm{x \propto y}}$
    ♦ When x is inversely proportional to y, we write: $\mathbf\small{\rm{x \propto \frac{1}{y}}}$ 
• In our present case, we have a direct proportionality
• That is why we write: stress ∝ strain

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• We have: 

stress ∝ strain
⇒ stress = k × strain
    ♦ Where k is the constant of proportionality
• We know that k is a constant. It’s value is fixed. So our next aim is to find the value of k
• We do an experiment to find k. The basics about the experiment can be written in 29 steps:
1. In fig.9.11(a₁) below, a thin wire is suspended from a rigid ceiling
    ♦ Length of the wire is L
    ♦ Diameter of the wire is d
          ✰ So area of cross section (A) of the wire can be obtained as: $\mathbf\small{\rm{A=\frac{\pi d^2}{4}}}$

Fig.9.11

  2. In fig.9.11(a), a mass of 0.5 kg is attached to the bottom of the wire
    ♦ Due to the weight of the 0.5 kg mass, the length of the wire increases by ΔL₁
    ♦ Due to the weight of the 0.5 kg mass, the wire experiences a stress 𝜎₁
          ✰ This stress can be obtained as: $\mathbf\small{\rm{\sigma_1=\frac{F_1}{A}=\frac{0.5(kg)\times g(ms^{-2})}{A(m^2)}}}$
    ♦ Due to the weight of the 0.5 kg mass, the wire experiences a strain 𝜺₁
          ✰ This strain can be obtained as: $\mathbf\small{\rm{\epsilon_1=\frac{\Delta L_1}{L}}}$
• So due to the weight of the 0.5 kg mass, we get (𝜺₁, 𝜎₁)
3. In fig.b, an additional mass of 0.5 kg is attached to the bottom of the wire
    ♦ Due to the weight of the 1.0 kg mass, the length of the wire increases by ΔL₂
          ✰ Note that, this ΔL₂ is measured from the bottom tip of the original length L
    ♦ Due to the weight of the 0.5 kg mass, the wire experiences a stress 𝜎₂
          ✰ This stress can be obtained as: $\mathbf\small{\rm{\sigma_2=\frac{F_2}{A}=\frac{1.0(kg)\times g(ms^{-2})}{A(m^2)}}}$
    ♦ Due to the weight of the 1.0 kg mass, the wire experiences a strain 𝜺₂
          ✰ This strain can be obtained as: $\mathbf\small{\rm{\epsilon_2=\frac{\Delta L_2}{L}}}$
• So due to the weight of the 1.0 kg mass, we get (𝜺₂, 𝜎₂)
4. In fig.c, an additional mass of 0.5 kg is attached to the bottom of the wire
    ♦ Due to the weight of the 1.5 kg mass, the length of the wire increases by ΔL₃
          ✰ Note that, this ΔL₃ is measured from the bottom tip of the original length L
    ♦ Due to the weight of the 1.5 kg mass, the wire experiences a stress 𝜎₃
          ✰ This stress can be obtained as: $\mathbf\small{\rm{\sigma_3=\frac{F_3}{A}=\frac{1.5(kg)\times g(ms^{-2})}{A(m^2)}}}$
    ♦ Due to the weight of the 1.5 kg mass, the wire experiences a strain 𝜺₃
          ✰ This strain can be obtained as: $\mathbf\small{\rm{\epsilon_3=\frac{\Delta L_3}{L}}}$
• So due to the weight of the 1.5 kg mass, we get (𝜺₃, 𝜎₃)
5. This procedure is repeated several times. Each time, the mass is increased by 0.5 kg
• Thus we get several points: (𝜺₁, 𝜎₁), (𝜺₂, 𝜎₂), (𝜺₃, 𝜎₃), (𝜺₄, 𝜎₄), . . .
• These points are plotted in a graph paper
    ♦ 𝜺 is plotted along the x-axis
    ♦ 𝜎 is plotted along the y-axis
6. If the material used for making the wire in the fig.9.11 is steel, the graph will have a shape as shown in fig.9.12(a) below:

Fig.9.12

• This graph has many salient points on it. They are marked as A, B, C etc., in fig.9.12(b)
    ♦ OA is shown in red color
    ♦ AB is shown in yellow color
    ♦ BD is shown in green color
    ♦ DE is shown in magenta color
• We see that only OA is straight. All others (AB, BD and DE) are curves
• Our present discussion on Hooke's Law is related to OA. So we will discuss it first
• The following steps from (7) to (12) are related to OA
7. The straight line character of OA needs to be analysed in detail. It can be written in 3 steps:
(i) From our earlier math classes, we know that:
    ♦ Equation of a straight line passing through the origin is y = mx
          ✰ 'm' is a constant. It is the 'slope' of the line. It will not change (Details here)
(ii) At the beginning of this section, we wrote:
Stress = k × Strain
(iii) Let us compare (i) and (ii)
    ♦ In the place of y, we have stress (𝜎)
    ♦ In the place of x, we have strain (𝜺)
• So it is obvious that, m corresponds to k
■ That is., 'slope of the line OA' is equal to 'k'
8. So our next aim is to find the slope of OA. It can be written in 5 steps:
(i) In fig.9.13(a) below, The portion OA is shown separately

Fig.9.13

(ii) Mark two convenient points (P and Q) on OA
(iii) To determine the slope, we need horizontal and vertical dashed lines:
    ♦ Draw a horizontal dashed line through P   
    ♦ Draw a vertical dashed line through Q
(iv) The two dashed lines will meet at R
    ♦ Measure the lengths of PR and QR
(v) Then the slope of line OA will be equal to $\mathbf\small{\rm{\frac{QR}{PR}}}$ 
■ So we can write:
The constant of proportionality 'k' in (stress = k×strain) is given by: $\mathbf\small{\rm{k=\frac{QR}{PR}}}$
9. We want the wire to obey Hooke's law
    ♦ That is., we want stress and strain to be proportional to each other
    ♦ That is., we want the relation (stress = k×strain) to be valid
• If the relation (stress = k×strain) is valid, the graph will be a straight line
10. But we see that, beyond the point A in fig.9.12(b), the graph is not linear
    ♦ That means, beyond the the point A, Hooke's law is not obeyed
11. We see that, beyond the point A, the stresses are higher
    ♦ That means, we are applying higher loads
■ So we can write:
If we want the wire to obey Hooke's law, we must not apply higher loads
(If we apply higher loads, we will be creating higher stresses. Because, stress = ^load⁄_area)
• Thus the importance of the portion OA becomes obvious
■ Since point A is the limiting point which shows proportionality, that point is given a special name:
Proportional limit
12. Another interesting character related to OA can be written in 7 steps:
(i) While doing the experiment, stop adding the loads. Stop just before reaching point A
(ii) Remove one 0.5 kg mass
Note down the new (𝜺, 𝜎) value
(iii) Remove another 0.5 kg mass
Note down the new (𝜺, 𝜎) value
(iv) Continue the procedure till all the masses are removed
(v) We will see that, all the new (𝜺, 𝜎) values will lie exactly on OA
(vi) Finally, when the load is zero, the strain will also be zero
• That is., all the extensions which were produced while moving up along OA are now neutralized
• The wire returns to the original length L
• This is obvious from the relation: Stress = k × Strain
    ♦ When the left side is zero, right side will also be zero
(vii) So we can write:
If before reaching A, we gradually remove the loads one by one, we will back track exactly along OA

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13. We have completed a discussion on portion OA
• From this present step (13) up to the step (18), we will be discussing about portion AB
• First we discuss the stress-strain relation in AB. It can be written in 2 steps:
(i) Portion AB is shown in fig.9.14(a) below:

Fig.9.14

• Imagine that, we are moving from point A to point B
    ♦ As we move from A to B, we are moving horizontally to the right
          ✰ This means, the strain is increasing
    ♦ As we move from A to B, we are moving vertically upwards
          ✰ This means, the stress is increasing
(ii) So we can write:
■ As we move from A to B, both stress and strain increases
• But this 'increase' is not proportional
    ♦ If it was proportional, AB would have been a straight line
14. So we have a situation: A non-proportional increase 
• Let us see an example of a ‘non-proportional increase’. It can be written in 5 steps:
(i) Bill amount at a grocery store:
Bill amount for the purchase of sugar = Price of one kg of sugar × Quantity of sugar purchased (in kg)
• Here, 'price of one kg of sugar' is a constant. Let it be Rs 32
(ii) Then, based on proportionality, we get:
    ♦ If 2 kg of sugar is purchased, the bill amount would be Rs 64
    ♦ If 5 kg of sugar is purchased, the bill amount would be Rs 160
    ♦ If 7 kg of sugar is purchased, the bill amount would be Rs 224
(iii) But if the shop keeper issues a bill of Rs 250 for 7 kg, we will say that:
• The ‘increase in bill amount’ is non-proportional
• The shop keeper may say that, due to scarcity, proportionality cannot be maintained
(iv) Graph of the values:
• All the values (Quantity, Bill amount) in (ii) will lie in a same straight line
• The value (Quantity, Bill amount) in (iii) will not lie in that line   
(v) This is a simple example for non-proportional increase
• We will see more complicated examples in higher classes
• However, now we have a basic idea of what 'non-proportional increase' is
15. The increase of strain from A to B is also non-proportional to stress
■ But there is an interesting fact:
While in the portion AB, if a 0.5 kg mass is removed, the strain will return to the just previous value
■ That means, while in the portion AB, the wire is still elastic
16. So, while in the portion AB, if we remove the 0.5 kg masses one by one, we will back track exactly along BAO
17. However, point B is the last point where we can expect elastic behaviour
• Once B is crossed, the wire will never return to the original length even if all loads are removed
• We say that: Once the point P is crossed, the ‘material of the wire’ begins to yield
• We will see details of ‘yielding’ when we discuss the next portion BD
■ Since B is the point at which yielding begins, it is an important point. It is given a special name: yield point
18. The strength corresponding to the 'yield point' is also important. That strength is called: yield strength (𝜎_y)
• The yield strength can be determined in 3 steps:
(i) Draw a horizontal dashed line through B
(ii) Mark the point at which this dashed horizontal line meets the y-axis
(iii) The stress at this meeting point will be the yield strength

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19. We have completed a discussion on portion AB

• From this present step (19) up to the step (24), we will be discussing about portion BD
    ♦ Note that, the name of the portion is BD. It is not BC
    ♦ Point C is just a 'sample point' within the portion BD
          ✰ We will soon see the reason for taking such a sample point 
• First we discuss the stress-strain relation in BD. It can be written in 2 steps:
(i) Portion BD is shown in fig.9.14(b) above
• Imagine that, we are moving from point B to point D
    ♦ As we move from B to D, we are moving horizontally to the right
          ✰ This means, the strain is increasing
    ♦ As we move from B to D, we are moving vertically upwards
          ✰ This means, the stress is increasing
(ii) So we can write:
■ As we move from B to D, both stress and strain increases
• But this 'increase' is non-proportional
    ♦ If it was proportional, BD would have been a straight line
20. So, just like AB, the portion BD also has a non-proportional increase
• In AB, we saw that:
When loads are gradually removed one by one, we will back track exactly along BAO
• But in BD, this is not the case. It can be explained using an example in 6 steps:
(i) Consider a point C within BD
(ii) If we remove the loads gradually from C, the back tracking will not be along CBAO
• Instead, the back tracking will be along the dashed magenta line CC'
• This dashed magenta line will be parallel to OA
• This is shown in fig.9.15(a) below:

Fig.9.15

(iii) Since the back tracking is not along CBAO, we will not reach the origin O when all the loads are removed
(iv) Since the back tracking is along CC', we will reach C' when all the loads are removed
■ So when all the loads are removed, a strain of CC' will remain
• That means, when all the loads are removed, the wire will not return to it's original length 
(v) When all the loads are removed, we know that stress is zero
• So we can write:
Once the yield point B is passed, the wire will never return to it's original length, even stress is reduced to zero
(vi) The remaining strain is called permanent set 
21. All the points within the portion BD, will behave like point C
• The deformation caused beyond point B is called plastic deformation
22. The last point D of the portion BD has much significance. it can be explained in 3 steps:
(i) To understand the significance, we must consider the nest portion DE
    ♦ This is shown in fig.9.15(b) above
(ii) We see that, DE is sloping downwards
    ♦ So it is clear that, point D corresponds to the maximum stress that the wire can take
(iii) The stress at D is called ultimate strength (𝜎_u)
23. The ultimate strength can be determined in 3 steps:
(i) Draw a horizontal dashed line through D
(ii) Mark the point at which this dashed horizontal line meets the y-axis
(iii) The stress at this meeting point will be the ultimate strength
24. Once the point B (yield point) is passed, the wire elongates very easily
• That., the wire will put up only a small amount of resistance against elongation
• This can be proved in steps:
(i) Mark two convenient points P and Q, in the portion BD
    ♦ One near B and the other near D
    ♦ This is shown in fig.9.15(b) above
(ii) Draw horizontal and vertical dashed lines through P and Q
    ♦ We get a right triangle with PQ as hypotenuse
(iii) Consider the base and height of this triangle
• We see that: the base is larger than the height
    ♦ Base corresponds to 'increase in strain'
    ♦ Height corresponds to 'increase in stress'
(iv) So we can write:
In the portion BD, only a small 'increase in stress' is required to bring about a large 'increase in strain'

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25. We have completed a discussion on portion BD

• From this present step (25) up to the step (28), we will be discussing about portion DE
• First we discuss the stress-strain relation in DE. It can be written in 2 steps:
(i) Portion DE is shown in fig.9.15(b) above
• Imagine that, we are moving from point D to point E
    ♦ As we move from D to E, we are moving horizontally to the right
          ✰ This means, the strain is increasing
    ♦ As we move from D to E, we are moving vertically downwards
          ✰ This means, the stress is decreasing
(ii) So we can write:
■ As we move from D to E, stress decreases but strain increases
• But this 'decrease-increase relation' is not proportional
    ♦ If it was proportional, DE would have been a straight line, dipping downwards
26. We see that point D corresponds to the ultimate point.
• That is., D is the point of 'maximum possible stress'
• So, after D, we may wish to reduce the load and thus decrease the stress. But it is of no use
    ♦ Once the ultimate point D is passed, the wire has become useless
    ♦ Even if we reduce the load, the wire will continue to elongate
27. After 'continuing to elongate' for some time, fracture will occur at point E
■ So point E is called fracture point
28. From the positions of D and E, we can say whether a material is brittle or ductile
(i) If E is close to D, the material is brittle
    ♦ Because, fracture occurs suddenly, without much elongation
(ii) If E is far away from D, the material is ductile
    ♦ Because, fracture occurs after much elongation
29. An important note:
This can be written in steps:
(i) In step (25) above, we mentioned that:
Once the point D is passed, the wire becomes useless
(ii) In fact, for engineering and scientific purposes, the wire is considered useless even when point B is passed
• This is because, after B, the wire begins to yield
(iii) Even the point B is not taken as such
• The required safety factors are applied
• We will see them in higher classes

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Elastomers

• What we discussed above is the stress-strain curve of a metallic substance like steel
• The stress-strain curves of materials like rubber, have different shapes
• Consider the curve shown in fig.9.16(a) below:

Fig.9.16

• Details of this curve can be written in 4 steps
1. Proportionality between stress and strain
This can be written in 2 steps:
(i) We see that:
    ♦ Only a ‘small portion OA in the initial stage’ is linear
          ✰ This is shown in fig.b
    ♦ Rest of portions are curves
(ii) So we can write:
    ♦ The material obeys Hooke's law only in the initial small portion
    ♦ In this small portion, stress and strain will be proportional to each other
2. We do not see a well defined plastic region
The ‘absence of a well defined plastic region’ can be proved in 6 steps:
(i) Mark two points P and Q in the 'portion after A'
(We do not have to consider the 'portion before A' for this proof because, OA obeys Hooke’s law. A portion which obeys Hooke’s law cannot be plastic)
(ii) Draw horizontal and vertical dashed lines through P and Q
    ♦ We get a right triangle with PQ as hypotenuse
(iii) Consider the base and height of this triangle
• We see that: the base is smaller than the height
    ♦ Base corresponds to 'increase in strain'
    ♦ Height corresponds to 'increase in stress'
(iv) So we can write:

In the portion after A, a large 'increase in stress' is required to bring about a small 'increase in strain'
(v) If it was a plastic region, it would elongate easily
That is., only a small increase in stress could bring about a large increase in strain
(vi) So it is clear that, there is no plastic region in this curve
• Since there is no well defined plastic region, we can write:
■ The material is elastic
    ♦ That is., even after producing large strains, it will still return to it's original size
3. 'Large strains' can be explained as follows:
    ♦ If the length increases from L to 2L, we get: $\mathbf\small{\rm{\epsilon=\frac{2L-L}{L}=\frac{L}{L}=1}}$
    ♦ If the length increases from L to 3L, we get: $\mathbf\small{\rm{\epsilon=\frac{3L-L}{L}=\frac{2L}{L}=2}}$
    ♦ If the length increases from L to 4L, we get: $\mathbf\small{\rm{\epsilon=\frac{4L-L}{L}=\frac{3L}{L}=3}}$
• Note that, we are getting whole numbers for strains
• For a metallic substance, the strain values will be fractions only
4. The above three steps can be considered as 'three conditions'
• Materials which satisfy the three conditions are called elastomers
    ♦ Rubber, tissue of Aorta etc., are examples

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In the next section, we will see Young's modulus

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