Pearson Edexcel Physics

Unit 1: Mechanics and Materials

1.4 Materials

Pearson Edexcel Physics

Unit 1: Mechanics and Materials

1.4 Materials

Candidates will be assessed on their ability to::
23. Be able to use the equation density [math] p = \frac{m}{V} [/math]
24. Understand how to use the relationship upthrust = weight of fluid displaced
25.

a)      Be able to use the equation for viscous drag (Stokes’ Law)

[math]F = 6πηrv[/math]

b)      Understand that this equation applies only to small spherical objects moving at low speeds with laminar flow (or in the absence of turbulent flow) and that viscosity is temperature dependent
26. CORE PRACTICAL 2: Use a falling-ball method to determine the viscosity of a liquid
27. Be able to use the Hooke’s law equation, [math]∆F = k∆x[/math], where k is the stiffness of the object
28.

Understand how to use the relationships

–  (Tensile or compressive) stress = force/cross-sectional area

–  (Tensile or compressive) strain= change in length/original length young modulus = stress/strain.
29.

a)      Be able to draw and interpret force-extension and force-compression graphs

b)      Understand the terms limit of proportionality, elastic limit, yield point, elastic deformation and plastic deformation and be able to apply them to these graphs
30. Be able to draw and interpret tensile or compressive stress-strain graphs, and understand the term breaking stress
31. CORE PRACTICAL 3: Determine the Young modulus of a material
32. Be able to calculate the elastic strain energy [math]E_{el}[/math] in a deformed material sample, using the equation [math] ∆E_{el} = \frac{1}{2} F∆x [/math], and from the area under the force-extension graph The estimation of area and hence energy change for both linear and non-linear force-extension graphs is expected.

  • 23) Be able to use the equation density [math]p = \frac{m}{v}[/math]

  • ⇒ Density:
  • Density is a measure of the mass per unit volume of a substance – this is technically called ‘volumic mass’.
  • Its value depends on the mass of the particles from which the substance is made, and how closely those particles are packed:
  • [math]\begin{gather} \text{density } (\mathrm{kg\,m^{-3}}) = \frac{\text{mass } (\mathrm{kg})}{\text{volume } (\mathrm{m^3})} \\
    \rho = \frac{m}{V} \end{gather}[/math]
  • ⇒ Example:
  • A house brick is 23 cm long, 10 cm wide and 7 cm high. Its mass is 3.38 kg. What is the brick’s density?
  • ⇒ Solution:
  • – Volume =0.23 × 0.10 × 0.07
    [math]V = 1.61 × 10^{-3} m^3[/math]
  • – Mass m = 3.38 kg
  • [math]\begin{gather}
    \rho = \frac{m}{V} \\
    \rho = \frac{3.38}{1.61 \times 10^{-3}} \\
    \rho = 2100\ \mathrm{kg\,m^{-3}}
    \end{gather}[/math]
  • At 20°C, a child’s balloon filled with helium is a sphere with a radius of 20 cm. The mass of helium in the balloon is 6 grams. What is the density of helium at this temperature?
  • [math]\begin{gather}
    \rho = \frac{m}{V} \\
    r = 0.20\ \mathrm{m} \\
    V = \frac{4}{3} \pi r^3 \\
    V = \frac{4}{3} \times 3.14 \times (0.20)^3 \\
    V = 1.33 \times 3.14 \times 0.008 \\
    V = 0.0335\ \mathrm{m^3} \\
    \rho = \frac{m}{0.0335}
    \end{gather}[/math]
  • – Mass m = 0.006 kg
  • [math]\begin{gather}
    \rho = \frac{0.006}{0.0335} \\
    \rho = 0.179\ \mathrm{kg\,m^{-3}}
    \end{gather}[/math]

  • 24) Understand how to use the relationship upthrust = weight of fluid displaced

  • ⇒ Upthrust:
  • When an object is submerged in a fluid, it feels an upwards force caused by the fluid pressure.
  • The upthrust. It turns out that the size of this force is equal to the weight of the fluid that has been displaced by the object. This is known as Archimedes’ principle.
  • Figure 1 Archimedes principle
  • If the object is completely submerged, the mass of fluid displaced is equal to the volume of the object multiplied by the density of the fluid:
  • [math]m = V_p[/math]
  • The weight of fluid displaced (i.e. upthrust) is then found using the relationship:
  • [math]W = mg[/math]
  • Example:
  • ⇒ A Brick Sink:
  • If the house brick from the example calculation of density above were dropped in a pond, it would experience an upthrust equal to the weight of water it displaced. This is simply the weight of an equal volume of water.
  • As the density of water is 1000 [math]kg m^{-3}[/math], the mass of water displaced by the brick would be:
  • [math]\begin{gather}
    m = 1000\ \mathrm{kg\,m^{-3}} \times 1.61 \times 10^{-3}\ \mathrm{m^3} \\
    m = 1.61\ \mathrm{kg}
    \end{gather}[/math]
  • The water has a weight of
  • [math]\begin{gather}
    W = mg \\
    W = 1.61 \times 9.81 \\
    W = 15.8\ \mathrm{N}
    \end{gather}[/math]
  • So, there is an upward force on the brick of 15.8 N.
  • If we compare the weight of the brick with the upthrust when it is submerged, the resultant force will be downwards:
  • [math]\begin{gather}
    W = mg \\
    \text{Weight} = 3.38 \times 9.81 \\
    W = 33.2\ \mathrm{N}\ (\text{downwards}) \\
    \text{Upthrust} = 15.8\ \mathrm{N}\ (\text{upwards}) \\
    \text{Resultant force} = 33.2 – 15.8 \\
    \text{Resultant force} = 17.6\ \mathrm{N}\ (\text{downwards})
    \end{gather}[/math]
  • So, the brick will accelerate downwards within the water until it reaches the bottom of the pond, which then exerts an extra upwards force to balance the weight so the brick rests stationary on the bottom with zero resultant force.
  • Figure 2 Upthrust force apply on an object by Archimedes principle

  • 25

  • a) Be able to use the equation for viscous drag (Stokes’ Law)

  • F = 6πηrv

  • b) Understand that this equation applies only to small spherical objects moving at low speeds with laminar flow (or in the absence of turbulent flow) and that viscosity is temperature dependent

  • ⇒ Stokes Law:
  • “Drag force is directly proportional to the sphere’s velocity, its radius, and the fluid’s viscosity”.
  • [math]F = 6πηrv[/math]
  • Where r is the radius of sphere (m), v is the velocity of sphere ([math]ms^{-1}[/math]), and n is the coefficient of viscosity of the fluid (Pa s).
  • it difficult to wade through a swimming pool filled with oil because of the oil’s viscous drag. This is the friction force between a solid and a fluid. Calculating this fluid friction force can be relatively simple.
  • On the other hand, it can be very complicated for large objects, fast objects and irregularly shaped objects, as the turbulent flow creates an unpredictable situation.
  • It must be remembered that the simple slow-falling sphere of Stokes’ law is not a common situation and in most real applications the terminal velocity value is a result of more complex calculations.
  • However, the principle that larger objects generally fall faster holds true for most objects without a parachute.
Falling Object Terminal Velocity ([math]ms^{-1}[/math])
Skydiver 60
Golf ball 32
Hail Stone (0.5 cm radius) 14
Raindrop (0.2 cm radius) 9
  • We will just look at basic scenarios for the sake of simplicity, such a solid sphere moving slowly in a fluid.
  • Consider, for instance, a ball bearing descending down an oil column. Taking into account the terminal velocity of the sphere with respect to the specific forces, then:
  • [math]\begin{gather}
    \text{Weight} = \text{Upthrust} + \text{Stokes force} \\
    m_s g = \text{weight of fluid displaced} + 6\pi r \eta v_{\text{term}}
    \end{gather}[/math]
  • Where [math]m_s[/math] is the mass of the sphere and [math]v_{\text{term}}[/math] is its terminal velocity.
  • For the sphere, the mass [math]m_s[/math] is given by
  • [math]\begin{gather}
    m_s = \text{volume} \times \text{density of sphere} \\
    m_s = \frac{4}{3} \pi r^3 \times \rho_s \\
    \text{So, the weight of the sphere } W_s \text{ is given by:} \\
    W_s = m_s g \\
    W_s = \frac{4}{3} \pi r^3 \rho_s g
    \end{gather}[/math]
  • For the sphere, the upthrust is equal to the weight of fluid displaced. The mass [math]m_f[/math] of fluid displaced is given by
  • [math]\begin{gather}
    m_f = \text{volume} \times \text{density of fluid} \\
    m_f = \frac{4}{3} \pi r^3 \rho_f
    \end{gather}[/math]
  • So, the weight of fluid displaced [math]W_f[/math] is given by
  • [math]\begin{gather}
    W_f = m_f g \\
    W_f = \frac{4}{3} \pi r^3 \rho_f g
    \end{gather}[/math]
  •  Overall, then:
  • [math]\begin{gather}
    \frac{4}{3} \pi r^3 \rho_s g = \frac{4}{3} \pi r^3 \rho_f g + 6 \pi r \eta v_{\text{term}} \\
    v_{\text{term}} = \frac{2 r^2 g (\rho_s – \rho_f)}{9 \eta}
    \end{gather}[/math]
  • So terminal velocity is proportional to the square of the radius. This means that a larger sphere falls faster. Furthermore, because the radius is squared, it falls much faster.
  • ⇒ Laminar Flow:
  • Laminar flow, also known as streamline flow, de-scribes the smooth, orderly movement of a fluid where particles follow parallel paths with minimal mixing between layers.
  • Laminar flow, type of fluid (gas or liquid) flow in which the fluid travels smoothly or in regular paths, in contrast to turbulent flow, in which the fluid undergoes irregular fluctuations and mixing.
  • In laminar flow, the velocity, pressure, and other flow properties at each point in the fluid remain constant.
  • Figure 3 Laminar and Turbulent flow
  • Isaac Newton wrote extensively on the topic of fluid flow, as he did with most other scientific fields. He is specifically renowned for having created the formulas that characterize the frictional force that exists between the layers in streamline movement.
  • A liquid is referred to as Newtonian if it obeys his formulas, which is the case for the majority of liquids.

  • 26) CORE PRACTICAL 2: Use a falling-ball method to determine the viscosity of a liquid

  • The falling-ball method is a technique used to measure the viscosity of a liquid.
  • ⇒ Steps:
  • Prepare the setup: Fill a tall, clear tube with the liquid of interest.
  • Release the ball: Release a small ball (usually made of a dense material like steel or glass) into the liquid.
  • Measure the terminal velocity: Measure the terminal velocity of the ball as it falls through the liquid.
  • Calculate viscosity: Use Stokes’ Law to calculate the viscosity of the liquid based on the terminal velocity, ball density, and liquid density.
  • Stokes’ Law:
  • [math]\begin{gather}
    \eta = \frac{2}{9} \frac{(\rho_{\text{ball}} – \rho_{\text{liquid}}) g r^2}{v}
    \end{gather}[/math]
  • Variables:
  • – η: Viscosity of the liquid
  • – [math]\rho_{\text{ball}}[/math]: Density of the ball
  • – [math]\rho_{\text{liquid}}[/math]: Density of the liquid
  • – g: Acceleration due to gravity
  • – r: Radius of the ball
  • – v: Terminal velocity of the ball
  • Figure 4 Small sphere falling in a long glass cylindrical jar
  • This method provides a practical way to measure the viscosity of a liquid by observing the behavior of a falling object within it.

  • 27) Be able to use the Hooke’s law equation, ∆F = k∆x , where k is the stiffness of the object:

  • ⇒ Hook’s Law:
  • Isaac Newton’s adversary, Robert Hooke, was a superb experimental scientist. In addition to having a wide range of interests, he made several discoveries that go beyond Hooke’s law.
  • According to Hooke’s law, the force required to lengthen a spring increase with its length. Only when a substance has not beyond what been known as the limit of proportionality does it comply with Hooke’s law.
  • Hooke’s law is best described mathematically with the equation:
  • [math]\begin{gather}
    \text{Force applied (N)} = \text{Stiffness constant (N\,m}^{-1}\text{)} \times \text{Extension (m)} \\
    \Delta F = k \Delta x
    \end{gather}[/math]
  • Figure 5 Hook’s Law
  • The force required to prolong a spring is proportional to its extension up to a particular point in proportionality. An item will undergo elastic deformation when only a slight force is applied to it.
  • As long as the elastic limit is not exceeded, it returns to its initial size and form when the force is withdrawn. We will examine elastic and plastic deformation in greater depth.
  • ⇒ Example:
  • A spring has a stiffness constant of 50 [math]Nm^{-1}[/math] and is 3.0 cm long. How long would it be if a 200 g mass were hung from it?
  • ⇒ Solution:
  • Weight Force:
  • [math]\begin{gather}
    W = mg \\
    W = 0.200 \times 9.81 \\
    W = 1.962\ \mathrm{N} \\[6pt]
    \Delta F = k \Delta x \\[6pt]
    \Delta x = \frac{\Delta F}{k} \\
    \Delta x = \frac{1.962}{50} \\
    \Delta x = 0.0392\ \mathrm{m} = 3.92\ \mathrm{cm} \\[6pt]
    \text{Final length} = \text{original length} + \text{extension} \\
    L = L_0 + \Delta x \\
    L = 3 + 3.92 \\
    L = 6.92\ \mathrm{cm}
    \end{gather}[/math]

  • 28) Understand how to use the relationships

  • – (Tensile or compressive) stress = force/cross-sectional area

  • – (Tensile or compressive) strain= change in length/original length young modulus = stress/strain.

  • (Tensile or compressive) Stress = force/cross-sectional area
  • ⇒ Stress:
  • A material sample’s internal force is measured by tensile stress, also known as compressive stress, which also considers the sample’s cross-sectional area. This makes it possible to compare forces between samples of various sizes, in order for them to be measured under similar circumstances.
  • [math]\begin{gather}
    \text{stress (pascals, Pa or N\,m}^{-2}\text{)} = \frac{\text{force (N)}}{\text{cross-sectional area (m}^2\text{)}} \\
    \sigma = \frac{F}{A}
    \end{gather}[/math]
  • Example:
  • A cylindrical stone column has a diameter of 60 cm and supports a weight of 2500 N (figure 6). What is the compressive stress in the column?


    Figure 6 This support column is under a compressive stress of 8850 Pa.

  • Solution:
  • [math]\begin{gather}
    A = \pi r^2 \\
    A = 3.14 \times (0.30)^2 \\
    A = 0.2826\ \mathrm{m^2} \\[6pt]
    \sigma = \frac{F}{A} \\
    \sigma = \frac{2500}{0.2826} \\
    \sigma = 8850\ \mathrm{Pa}
    \end{gather}
    [/math]
  • ⇒ Strain:
  • A material sample’s extension or compression is measured by tensile (or compressive) strain, which accounts for the sample’s initial length.
  • This makes it possible to analyze samples of various sizes under similar circumstances and compare extension comparisons between them.
  • [math]\begin{gather}
    \text{strain (no unit)} = \frac{\text{extension (m)}}{\text{original length (m)}} \\
    \varepsilon = \frac{\Delta x}{x}
    \end{gather}[/math]
  • As strain is a ratio, it has no units. However, it is often expressed as a percentage by multiplying the ratio by 100%..
  • Example:
  • A copper wire of length 1.76 m is stretched by a force to a length of 1.80 m. What is the tensile strain in the wire?
  • Figure 7 This copper wire has a tensile strain of 2.3%.
  • Solution:
  • [math]\begin{gather}
    \varepsilon = \frac{\Delta x}{x} \\
    \varepsilon = \frac{1.80 – 1.76}{1.76} \\
    \varepsilon = \frac{0.04}{1.76} \\
    \varepsilon = 0.023 \\
    \varepsilon = 2.3\%
    \end{gather}[/math]
  • ⇒ Young Modulus:
  • Stress and strain will be proportionate if a material is deformed elastically. The proportionality constant is a measurement of the material’s stiffness, or how much it deforms under a given stress. The Young modulus is the name given to the stiffness constant.
  • Thus, the Young modulus is a measurement of a material’s stiffness that accounts for the sample’s size and shape, meaning that many samples of the same material will all have the same Young modulus value.
  • The amount that a material deforms when forces are applied to it is known as its stiffness.
  • [math]\begin{gather}
    \text{Young modulus (Pa)} = \frac{\text{Stress (Pa)}}{\text{Strain (no units)}} \\
    E = \frac{\sigma}{\varepsilon}
    \end{gather}[/math]
  • – The definition for the Young modulus also includes the fact that the material must be undergoing elastic deformation.
  • – Beyond the limit of proportionality, this equation will no longer work to calculate the stiffness of the material.
  • Example:
  • The copper wire from above has a diameter of 0.22 mm and was stretched using a force of 100 N. What is the Young modulus of copper?
  • Solution:
  • [math]\begin{gather}
    A = \pi r^2 \\
    A = 3.14 \times \left(1.1 \times 10^{-4} \right)^2 \\
    A = 3.8 \times 10^{-8}\ \mathrm{m^2} \\[6pt]
    \sigma = \frac{F}{A} \\
    \sigma = \frac{100}{3.8 \times 10^{-8}} \\
    \sigma = 2.63 \times 10^9\ \mathrm{Pa} \\[6pt]
    E = \frac{\sigma}{\varepsilon} \\
    E = \frac{2.63 \times 10^9}{0.023} \\
    E = 1.16 \times 10^{11}\ \mathrm{Pa}
    \end{gather}[/math]
  • Figure 8 Copper has a Young modulus,

  • 29)

  • a) Be able to draw and interpret force-extension and force-compression graphs

  • b) Understand the terms limit of proportionality, elastic limit, yield point, elastic deformation and plastic deformation and be able to apply them to these graphs

  • 30) Be able to draw and interpret tensile or compressive stress-strain graphs, and understand the term breaking stress

  • ⇒ Stress – Strain analysis:
  • According to the Young modulus definition, if a material is experiencing elastic deformation, the stress and strain should be proportionate.
  • Plotting stress versus strain should thus result in a straight-line graph. We do discover a straight-line association for minor tensions after doing this.
  • The material’s internal structure begins to act differently after the proportionality limit is exceeded.
  • As a result, the graph begins to curve. The graph will move through different phases depending on the substance being tested because each material’s molecular structure dictates how it reacts to increased stress. The material will eventually shatter due to excessive stress.
  • Since no further information on the material can be gathered at this time, the graph’s line must terminate. The graph cannot be drawn farther because it cannot tolerate higher stresses.
  • Figure 9 The stress-strain graph for a metal gives detailed information about how the material behaves under different levels of stress. The gradient of the straight-line portion of the graph will be equal to the value of the Young modulus.
  • In the straight-line portion from the origin to point A, the metal extends elastically, and will return to its original size and shape when the force is removed. The gradient of the straight-line portion of the graph is equal to the Young modulus for the metal.
  • Point A is the limit of proportionality. Slightly beyond this point, the metal may still behave elastically, but it cannot be relied upon to increase strain in proportion to the stress.
  • Point B is the elastic limit. Beyond this point, the material is permanently deformed and will not return to its original size and shape, even when the stress is completely released.
  • Point C is the yield point, beyond which the material undergoes a sudden increase in extension as its atomic substructure is significantly re-organised. The metal ‘gives’ just beyond its yield point as the metal’s atoms slip past each other to new positions where the stress is reduced.
  • Point D represents the highest possible stress within this material. It is called the Ultimate Tensile Stress, or UTS, [math]σ_U[/math]
  • Point E is the fracture stress, or breaking stress. It is the value that the stress will be in the material when the sample breaks.

  • 31) CORE PRACTICAL 3: Determine the Young modulus of a material

  • ⇒ Aim:
  • To ascertain the material’s Young’s modulus of elasticity for a certain wire.
  • ⇒ Materials Required:
  • – Searle’s apparatus
  • – Two long steel wires of the same length and diameter
  • – A meter scales
  • – A screw gauges
  • – Eight 0.5 kg slotted weights
  • – 1 kg hanger
  • ⇒ Theory:
  • When weight Mg is applied to a wire of length L and radius r, where l is the length increase, the normal stress is as follows:
  • [math]\begin{gather}
    \text{Stress} = \frac{Mg}{\pi r^2} \\[6pt]
    \text{Strain} = \frac{l}{L}
    \end{gather}[/math]
  • Hance,
  • [math]\begin{gather}
    \text{Young’s Modulus} = \frac{\text{Stress}}{\text{Strain}} \\
    Y = \frac{\dfrac{Mg}{\pi r^2}}{\dfrac{l}{L}} \\
    Y = \frac{MgL}{\pi r^2 l}
    \end{gather}[/math]
  • ⇒ Diagram:
  • Figure 10 Searle’s Apparatus
  • ⇒ Observation:
  • Length of experimental wire AB, L = ……. cm = …… m
  • Measurement of diameter of the wire
  • Pitch of the screw gauge (p) = 0.1 cm
  • No of divisions on the circular scale = 100
  • Least count of screw gauge (L.C) = 0.1/100 = 0.001 cm
  • Zero error of screw gauge (e) = …….cm
  • Zero error of screw gauge (e) = -e = ……cm
Sr No Linear Scale reading N (cm) Circular Scale reading Total reading

[math] N + n × (LC)d[/math]
1 [math]d_1[/math]
2 [math]d_2[/math]
3 [math]d_3[/math]
4 [math]d_4[/math]
5 [math]d_5[/math]
6 [math]d_6[/math]
  • ⇒ Measurement for extension of the wire:
  • Breaking stress for steel (from a table), B = ……N.m-2
  • Area of a cross section of a wire, [math]πr^2 = ⋯ cm^2 = ⋯ m^2[/math]
  • Breaking load = [math]Bπr^2 = ⋯N[/math]
  • [math]\begin{gather}
    \frac{B \pi r^2}{9.8} = \ldots\ \mathrm{kg} \\
    1\ \mathrm{kg} = 9.8\ \mathrm{N}
    \end{gather}[/math]
  • [math]1/2^{rd}[/math] of breaking load = ….. kg
  • Pitch of spherometer screw, (p) = 0.1 cm
  • No of divisions in the disc = 100
  • Least count of spherometer (LC) = 0.1/100 = 0.001 cm
  • ⇒ Load and Extension:
Sr No Load on hanger M (kg) Spherometer screw reading Extension for load 2.5 kg/cm
Load increasing x (cm) Load decreasing y (cm) Mean

[math]Z = (x + y) / 2 (cm)[/math]
1 0.0 [math]Z_1[/math]
2 0.5 [math]Z_2[/math]
3 1.0 [math]Z_3[/math]
4 1.5 [math]Z_4[/math]
5 2.0 [math]Z_5[/math]
6 2.5 [math]Z_6[/math]
  • ⇒ Calculation:
  • Mean observed diameter of the wire, [math]\begin{gather}
    d_0 = \frac{d_1 + d_2 + \ldots + d_6}{6} = \ldots\ \mathrm{cm}
    \end{gather}[/math]
  • Mean corrected diameter of the wire, [math]d = (d_0 + c) = ⋯ .cm = ⋯ .m[/math]
  • Mean radius of wire, [math]\begin{gather}
    r = \frac{d}{2}
    \end{gather}[/math]
  • ⇒ Result:
  • The Young’s modulus for steel as determined by Searle’s apparatus = …….. Nm-2
  • Straight-line graph between load and extension shows that stress ∝ This verifies Hooke’s law.

  • 32) Be able to calculate the elastic strain energy in a deformed material sample, using the equation [math] \Delta E_{\text{el}} = \frac{1}{2} F \Delta x [/math], and from the area under the force-extension graph. The estimation of area and hence energy change for both linear and non-linear force-extension graphs is expected.

  • ⇒ Elastic strain energy:
  • Elastic strain energy ([math]\Delta E_{\text{el}}[/math]) is the energy stored in a deformed material.
  • ⇒ Calculation:
  • – Using the equation:
  • [math]\begin{gather}
    \Delta E_{\text{el}} = \frac{1}{2} F \Delta x
    \end{gather}[/math]
  • F: Force applied, ∆x: Extension or deformation
  • – From the force-extension graph:
  • [math]\begin{gather}
    \Delta E_{\text{el}} = \text{Area under the force-extension graph}
    \end{gather}[/math]
  • ⇒ Graph Interpretation:
  • – Linear graph: For materials following Hooke’s Law (linear relationship), the area under the graph is a triangle.
  • [math]\begin{gather}
    \Delta E_{\text{el}} = \frac{1}{2} F \Delta x
    \end{gather}
    [/math]
  • Figure 11 Elastic strain energy graph
  • – Non-linear graph: For materials with non-linear force-extension relationships, estimate the area under the curve to find [math]\Delta E_{\text{el}}[/math]
  • – Energy storage: Elastic strain energy is stored in the material when it’s deformed.
  • – Work done: The energy stored is equal to the work done on the material.
  • – Material behavior: Understanding elastic strain energy helps analyze material behavior under different loads.
  • By calculating the elastic strain energy, you can determine the energy stored in a deformed material and understand its behavior under various loading conditions.
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