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
• 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
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: $\mathbf\small{\rm{A=\frac{\pi d^2}{4}}}$
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:
• 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
(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
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:
• 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 (𝜎y)
• 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
• 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:
(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 (𝜎u)
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'
• 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
• 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:
• 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
• 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
♦ 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
• We have:
⇒ 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 |
♦ 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 |
♦ 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 |
(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
• 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 |
♦ 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 (𝜎y)
• 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
19. We have completed a discussion on portion AB
♦ 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 |
(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 (𝜎u)
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'
25. We have completed a discussion on portion BD
• 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
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 |
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
In the next section, we will see Young's modulus
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