Elasticity Slides (PHY 103) PDF

# CONTENT - Elasticity - Hooke's law - Modulus of elasticity, rigidity (shear) and bulk. - Energy stored in a strain - Poisson ratio - Solved problems # LESSON OBJECTIVES After completion of this module, you should be able to: - Demonstrate your understanding of elasticity, elastic limit, stress, s...

# CONTENT - Elasticity - Hooke's law - Modulus of elasticity, rigidity (shear) and bulk. - Energy stored in a strain - Poisson ratio - Solved problems # LESSON OBJECTIVES After completion of this module, you should be able to: - Demonstrate your understanding of elasticity, elastic limit, stress, strain, and ultimate strength. - Write and apply formulae for calculating Young's modulus, shear modulus, and bulk modulus. - Solve problems involving each of the # Introduction ## Some properties of Metals - Hardness- property of a metal to resist cutting action, penetration - Brittleness -property of a metal to allow little bending or deformation - Malleability- property of a metal to allow it to be hammered, rolled - Ductility- property of a metal that permits it to be permanently bent - Density- it is the mass per unit volume of a material - Fusibility- ability of a metal to become liquid by application of heat - Conductivity- property of a metal that enables it to conduct heat and electricity. # EXTERNAL FORCE ON A SOLID - When an external force is applied to a solid it produces a change in shape, size or volume of the solid. - This change in shape produced on the solid is known as deformation. - Some deformation could be temporal or permanent - Temporal if when the applied force is removed the solid returns to its original shape and size. # Elasticity and Deformation - The ability of the solid (material) to regain its original shape and size after the deforming force is removed is known as elasticity. - The type of deformation in this case is called elastic deformation. - When the solid (material) remains permanently deformed after the removal of the external force it is the referred to as plastic deformation. - Two types of plastic deformations are: Dislocation and fracture. # Elastic and Inelastic Properties of Matter - An elastic body is one that returns to its original shape after a deformation. - An inelastic body is one that does not return to its original shape after a deformation. # Elastic and Inelastic Collision - An elastic collision loses no energy. The deformation on collision is fully restored. - In an inelastic collision, energy is lost and the deformation may be permanent. # Hooke's Law - Provided the elastic (proportional) limit is not exceeded the force applied on a material is directly proportional to the change in length or extension - The change in length is proportional to the applied force. $F = k \Delta l$ # Propotionality of f and $\Delta$l - $f \propto \Delta l$ proportionality holds until the force reaches the proportional limit. - Beyond that, the object will still return to its original shape up to the elastic limit. - Beyond the elastic limit, the material is permanently deformed; plastic region - Ultimate strength is the greatest stress a body can experience without breaking or rupturing. - Breaks at the breaking point # An Elastic Spring - A spring is an example of an elastic body that can be deformed by stretching. - A restoring force, $F$, acts in the direction opposite the displacement $x$ of the oscillating body. - $F = -kx$ # ELASTIC SPRING.. - When a spring is stretched, there is a restoring force F that is proportional to the displacement $x$. - $F = -kx$ - $F$ = force exerted by the spring. - The spring constant k is a property of the spring given by: $k = \frac{\Delta F}{\Delta x}$ - The spring constant k is a measure of the elasticity of the spring. # FACTORS AFFECTING $\Delta l$ - The change in length of a stretched object depends not only on the applied force, but also on: - length - cross-sectional area, - the material from which it is made. # Stress and Strain - $stress = \frac{force}{area} = \frac{F}{A}$ - $strain = \frac{change in length}{original length} = \frac{\Delta l}{l_0}$ - Stress refers to the cause of a deformation, and strain refers to the effect of the deformation. # Stress and Strain.. - The downward force F causes the displacement x. - Thus, the stress is the force; the strain is the elongation. # Types of Stress - **Tensile stress** occurs when equal and opposite forces are directed away from each other. The external force per unit area of the material resulting in the stretch of the material. - **compressive stress** occurs when equal and opposite forces are directed toward each other. # Tensile Stress - In tensile stress, forces tend to stretch the object. # Compressional Stress - Compressional stress is exactly the opposite of tensional stress. These columns are under compression. # Tensile and compressive stresses and strains - **Tensile stress** When an external force produces elongation of the body in its direction, it is termed as tensile force. - **Tensile strain** Tensile strain = Increase in length/ Original length - **Compressive stress** When an external force causes shortening of the body in the direction of force, it is termed as a compressive force. The stress developed in the body due to a compressive force is called as compressive stress. - **Compressive strain** Compressivestrain = decrease in length/ Original length # Summary of Definitions - Stress is the ratio of an applied force F to the area A over which it acts: $Stress = \frac{F}{A}$ - Units : Pa= $\frac{N}{m^2}$ or $\frac{lb}{in^2}$. - Strain is the relative change in the dimensions or shape of a body as the result of an applied stress: - Examples: Change in length per unit length; change in volume per unit volume. # Longitudinal Stress and Strain - For wires, rods, and bars, there is a longitudinal stress F/A that produces a change in length per unit length. In such cases: - Stress = $\frac{F}{A}$ - Strain = $\frac{\Delta L}{L}$ # The Elastic Limit - The elastic limit is the maximum stress a body can experience without becoming permanently deformed. - If the stress exceeds the elastic limit, the final length will be longer than the original 2 m. # The Ultimate Strength - If the stress exceeds the ultimate strength, the string breaks! # Young Modulus - Young's modulus E: modulus of elasticity. - It is a material property that describes the stiffness of a material under tensile or compressive stress. - It is named after the British scientist Thomas Young, who first introduced the concept. - It is defined as the ratio of stress (σ) to strain (ε) within the elastic (linear) region of deformation. - The formula is expressed as: $Young modulus = \frac{Tensile or compressive stress}{Tensile or compressive strain}$ ⇒ $E = \frac{\sigma}{\varepsilon}$. # SHEAR MODULUS G - It is a material property that describes the ability of a material to resist deformation under shear stress. - Shear stress occurs when a force is applied parallel to one surface of an object or material, causing the material to deform by sliding or shifting its internal layers without changing its volume. - G is a measure of the material's stiffness in response to shear stress. $shear modulus = \frac{shear stress}{shear strain}$ → $G = \frac{\tau}{\gamma}$ - The SI unit of shear modulus are typically also in pascals (Pa) # Shear Stress - Shear stress tends to deform an object. # Shear Modulus.. - A shearing stress alters only the shape of the body, leaving the volume unchanged. - For example, consider equal and opposite shearing forces F acting on the cube below: - The shearing force F produces a shearing angle φ. - The angle φ is the strain and the stress is given by F/A as before. # Calculating Shear Modulus - Stress is force per unit area: $Stress = \frac{F}{A}$. - The strain is the angle expressed in radians: $Strain = \phi = \frac{d}{l}$ - The shear modulus S is defined as the ratio of the shearing stress F/A to the shearing strain φ: - The shear modulus: $S = \frac{F/A}{\phi}$. - Units are in Pascals. # Bulk Stress, Strain, and Modulus - When you dive into water, you feel a force pressing on every part of your body from all directions: bulk stress. - Bulk stress always tends to decrease the volume enclosed by the surface of a submerged object. - The forces of this "squeezing" are always perpendicular to the submerged surface - The effect of these forces is to decrease the volume of the submerged object by an amount $\Delta V$ - This kind of deformation is called bulk strain and is described by a change in volume relative to the original volume. # Bulk Modulus K - The material property that describes the material's response to changes in volume under hydrostatic pressure. - It measures the relative change in volume with respect to a change in pressure. - Relevant when studying the compressibility of a material. - Defined as the ratio of the change in pressure ($\Delta P$) to the resulting fractional change in volume ($\Delta V/V$) within the elastic (linear) region of deformation. The formula is expressed as: $K = -\frac{V(\Delta P)}{\Delta V}$. - The negative sign is included because an increase in pressure typically leads to a decrease in volume, and vice versa. - The SI units of bulk modulus pascals (Pa) or gigapascals # Volume Elasticity - Sometimes an applied stress F/A results in a decrease of volume. - In such cases, there is a bulk modulus B of elasticity. - $B = \frac{Volume stress}{Volume strain} = \frac{-F/A}{\Delta V/V}$. - The bulk modulus is negative because of decrease in V. # The Bulk Modulus - $B = \frac{Volume stress}{Volume strain} = \frac{-F/A}{\Delta V/V}$ - Since $F/A$ is generally pressure P, we may write: $B = \frac{-P}{\Delta V/V} = \frac{-PV}{\Delta V}$. - Units remain in Pascals (Pa) since the strain is unitless. # Applications of Elastic Moduli - The shear modulus is an important parameter in material science and engineering. - In the analysis of materials' behavior under various types of loads. It is one of the elastic moduli along with Young's modulus (which describes the material's response to tensile stress) and bulk modulus (which describes the material's response to changes in volume under hydrostatic pressure). - Together, these moduli provide a comprehensive understanding of a material's mechanical properties. # ENERGY STORED IN AN ELASTIC MATERIAL - When a solid material is stretched or compressed, energy is stored in it. - This energy is released when the applied force is removed provided the elastic limit is not exceeded. - Energy is work done during deformation - But work done $w$ is given as: $w = \int_0^x Fdx $ # ENERGY STORED IN A YOUNG STRAIN - Recall: $Y = \frac{Stress}{strain} = \frac{F}{A} = \frac{FL}{\Delta L}$. - Therefore, $F = \frac{YA \Delta L}{L} $. - Where $F(x) = \frac{YA}{l_0}x$ - $W = \int_0^x Fdx = \int_0^x \frac{YA}{l_0}xdx = \frac{YA}{2l_0}x^2$, where $k = \frac{YA}{l_0}$. - Hence, $W = \frac{1}{2}kx^2$ = P.E - Thus, work is stored as potential energy in an elastic material. # Elastic Energy per volume - The energy per unit volume - $\frac{E}{V}=\frac{E}{\Delta L}$. - $\frac{E}{V} = \frac{YA}{2L\Delta L}x^2 = \frac{YA}{\Delta L}x\times \frac{x}{L}-\frac{1}{2}$ - $\frac{E}{V} = \frac{1}{2}stress \times strain$ # ENERGY STORED BY SHEAR AND BULK STRAIN - **By shear strain** $ E = \frac{1}{2}(\frac{\tau}{G})^2V$. - Where τ = shear stress. - G = shear modulus. - V= volume of the block. - Energy per volume = $\frac{E}{V} = \frac{1}{2}(\frac{τ}{G})^2$ - **By Bulk Strain** $E = \frac{1}{2}(\frac{P}{B})^2V_0$ - Where v=new volume and $V_0$ is the initial Volume - Energy per volume = $\frac{E}{V_0} = \frac{1}{2}(\frac{V}{V_0})^2B$ # Poisson's ratio σ - ratio of transverse contraction strain to longitudinal extension strain in the direction of stretching force. - Tensile deformation is considered positive. - compressive deformation is considered negative. - The definition of Poisson's ratio contains a minus sign so that normal materials have a positive ratio. - σ= - etrans / elongitudinal. # Poisson's ratio: relation to elastic moduli - Poisson's ratio is related to elastic moduli - B as the bulk modulus; S as the shear modulus; and Y, Young's modulus. - σ = (3Β – 2S)/(6B + 2S) - Y = 2S(1 + σ) - Y = 3B(1 - 2 σ) # Example 1 A steel wire 10 m long and 2 mm in diameter is attached to the ceiling and a 200-N weight is attached to the end. What is the applied stress? - First find area of wire: $A = \frac{\pi D^2}{4} = \frac{\pi (0.002m)^2}{4}$. - $A = 3.14 \times 10^{-6} m²$ - $Stress = 6.37 \times 10^7 Pa$ # Example 2 - A 10 m steel wire stretches 3.08 mm due to the 200 N load. What is the longitudinal strain? - Given: L = 10 m; $\Delta L$=3.08 mm. - $Strain = \frac{\Delta L}{L} = \frac{0.00308 m}{10 m}$ # Example 3 The elastic limit for steel is 2.48 x 10^8 Pa. What is the maximum weight that can be supported without exceeding the elastic limit? - Recall: A = 3.14 x 10^-6 m² - $Stress = \frac{F}{A} = 2.48 \times 10^8 Pa $ - F = (2.48 x 10^8 Pa) A - F = (2.48 x 10^8 Pa) (3.14 x 10^-6 m²) - F=779 N # Example 4 The ultimate strength for steel is 4.89 x 10^8 Pa. What is the maximum weight that can be supported without breaking the wire? - Recall: A = 3.14 x 10^-6 m2 - $Stress = \frac{F}{A} = 4.89 \times 10^8 Pa $ - F = (4.89 x 10^8 Pa) A - F = (4.89 x 10^8 Pa) (3.14 x 10^-6 m²) - F=1536 N # Example 5. If the stress applied to the steel wire was 6.37 x 10^7 Pa and the strain was 3.08 x 10^-4. Find the modulus of elasticity for steel. - $Modulus = \frac{Stress}{Strain} = \frac{6.37 \times 10^7Pa}{3.08 \times 10^{-4}}$ - $Modulus = 207 \times 10^9 Pa$ - This longitudinal modulus of elasticity is called Young's Modulus and is denoted by the symbol Y # Example 6: Young's modulus for brass is 8.96 x 10^11Pa. A 120-N weight is attached to an 8-m length of brass wire; find the increase in length. The diameter is 1.5 mm. - First find area of wire: $A = \frac{\pi D^2}{4} = \frac{\pi (0.0015 m)²}{4}$ - A= 1.77 x 10^-6 m² - $Y = \frac{FL}{\Delta L}$ or $\Delta L = \frac{FL}{AY}$ # Example 7: (Continued) - Y = 8.96 x 10^11 Pa; F = 120 N; L=8m; A= 1.77 x 10^-6m² - F = 120 N; $\Delta L$ = ? - $Y = \frac{FL}{\Delta L}$ or $\Delta L = \frac{FL}{AY}$ - $\Delta L = \frac{FL}{AY} = \frac{(120N)(8.00 m)}{(1.77 \times 10^{-6}m²)(8.96 \times 10^{11}Pa)}$ - Increase in length: $\Delta L$ = 0.605 mm # Example 8 A steel stud (G = 8.27 x 10^10Pa) 1 cm in diameter projects 4 cm from the wall. A 36,000 N shearing force is applied to the end. What is the defection d of the stud? - $A = \frac{\pi D^2}{4}=\frac{\pi (0.01m)^2}{4}$ - Area: A = 7.85 x 10^-5 m² - $S = \frac{F/A}{\phi} = \frac{F/A}{d/l} = \frac{Fl}{Ad}$ - $d = \frac{Fl}{AS}$ - d = 0.222 mm - $\frac{(36,000 N)(0.04m)}{(7.85 \times 10^{-5}m²) (8.27 \times 10^{10}Pa)}$ # Example 9 A hydrostatic press contains 5 liters of oil. Find the decrease in volume of the oil if it is subjected to a pressure of 3000 kPa. (Assume that B = 1700 MPa.) - $B= \frac{-P}{\Delta V/V} = \frac{-PV}{\Delta V}$ - $\Delta V = \frac{-PV}{B} = \frac{-(3 x 10^6Pa)(5 L)}{(1.70 x 10^9Pa)}$ - Decrease in V; milliliters (mL): $\Delta V = -8.82 mL$ # Example 10 A load of 50 N attached to a spring hanging vertically stretches the spring 5.0 cm. The spring is now placed horizontally on a table and stretched 11.0 cm. What force is required to stretch the spring this amount? - $F_s = kx $ - 50 = k(0.05). - $k = \frac{1000 N}{m}$ - $F_s = kx$ - $F_s = (1000)(0.11)$ - $F_s = 110 N$ # Example 11 - A slingshot consists of a light leather cup, containing a stone, that is pulled back against 2 rubber bands. It takes a force of 30 N to stretch the bands 1.0 cm (a) What is the potential energy stored in the bands when a 50.0 g stone is placed in the cup and pulled back 0.20 m from the equilibrium position? (b) With what speed does it leave the slingshot? - a) $F_s = kx$ 30=k(0.01) $k = \frac{3000 N}{m}$ - b) $U_i = \frac{1}{2}kx^2 = 0.5(k)(.20) = 300 J $ - c) $E_B = E_A$ ,$U_i = K$ - $U_i = \frac{1}{2}mv^2 = \frac{1}{2}(0.050)v^2$ - $v = 109.54 m/s$ # Example 12 An aluminium (B = 70 x 10^9 N / m) ball with a radius of 0.5m falls to the bottom of the sea where the pressure is 150atm. (a) What is the original volume of the ball (b) what is the change in volume (c) what is the new volume of the ball (d) what is the bulk stress (e) Calculate the bulk strain in the ball. - Solution - (a) $V = \frac{4}{3}\pi r^3$ - $V= \frac{4}{3}\pi(0.5)^3$ . - $V = 0.5236m³$. - (b) $\Delta V = \frac{-PV}{B}$. - P=150atm=150×101,325=1.52×10^7 pa - $\Delta V = \frac{-1.52×10^7×0.5236}{70×10^9}$ - $\Delta V = -1.137×10^{-3}m²$. - (c) $V_f = V - \Delta V$. - $V_f = -1.137×10^{-3} + 0.5236$. - $V_f = 0.5235m³,$. - (d) Bulk Stress = Pressure = 1.52×10^7 Nm². - (e) Bulk Strain = $\frac{\Delta V }{V_0} = \frac{-1.137×10^{-3}}{0.5236} = -2.17×10^{-3}$. # Example 13 - A steel wire 10m long and 2mm in diameter is attached to the ceiling and a 200N weight is attached to the end. What is attached to the end. What is the applied stress. What is the strain if the wire is strained 3.08 mm - Solution - $A = \frac{\pi D^2}{4} = \frac{\pi(0.002)^2}{4} = 3.14×10^{-6} m²$. - *Stress = $\frac{F}{A}$* - $Stress = \frac{200}{3.14×10^{-6}}= 6.37×10^7 pa$. - (b) $Strain = \frac{\Delta L }{L_0}$. - $Strain = \frac{3.08×10^{-3}}{10} = 3.08×10^{-4}$ longitudinal strain # Example 14 - The elastic limit for steel of diameter 2mm is 2.48×10^8 pa. (a) What is the maximum weight that can be supported without exceeding the elastic limit. (b) If the ultimate strength is 4089×10^8 pa. What is the maximum weight that can be supported without breaking the wire? - Solution - (a) A=3.14×10^-6 m² - $Stress = \frac{F}{A} = 2.48×10^8 pa$ - F=2.48×10^8×3.14×10^-6 N - F=779N - (b) $Stress = \frac{F}{A} = 4089×10^8 pa$ - F=4089×10^8×3.14×10^-6 N - F=1536N