Simple Stresses and Strain Notes PDF

# Simple stresses and strain ## Day: Wednesday ## Date: 7/8/2024 ### Simple Stresses and Strain * **Mechanical Properties** 1. **Elasticity**: The ability of objects to regain their original shape and size after deforming force is removed is called elasticity. 2. **Plasticity**: The abi...

# Simple stresses and strain ## Day: Wednesday ## Date: 7/8/2024 ### Simple Stresses and Strain * **Mechanical Properties** 1. **Elasticity**: The ability of objects to regain their original shape and size after deforming force is removed is called elasticity. 2. **Plasticity**: The ability of solid material to undergo permanent deformation, a non-reversible change of shape in response to applied forces. 3. **Brittleness**: It is the property of a material to get broken into pieces when impacted all of sudden ductility. 4. **Ductility**: It is the property of a material to be drawn into wire when pulled and such that material will be ductile. * **Material properties** 1. Young's modulus 2. Density 3. Poisson ratio. 4. Yield strength. * **The resistance by which material of the body opposes the deformation is known as strength of material.** * **Stress**: the force of resistance per unit area, offered by a body against deformation is known as stress. * The external force acting on the body is called Load or force. * Mathematically stress is represented by: * $(\sigma) = \frac{P}{A}$ where, * $\sigma$ is stress, * $P$ = external load * $A$ = cross section of area. * **The S.I units for stress is N/m², N/m² , 1 pascal., $(\sigma_x,\sigma_y)$ represent direction.** * **Strain**: When a body subjected to some external force, there is some change of dimension of the body. * **The ratio of change of dimension of the body to the original dimension is known as strain. Strain (E) = Change in length/original length.** * **Types of Stress** * These stresses may be divided into two: 1. **Normal stress** 2. **Shear stress** * **Normal stress**: * It is the stress which acts in the direction perpendicular to the area. It is represented by the $(\sigma)$. * The normal stress further divided into tensile stress and compressive stress. * **Tensile stress**: The stress induced in a body, when subjected to two equal and opposite pulls, as a result of which there is an increasing length is known as tensile stress. Tensile stress is called tensile load by area. * **Tensile strain**: Increasing length/original length. ## Day: Thursday ## Date: 8/8/2024 ### Compressive stress * The stress is used in a body subjected to equal and opposite push, as a result of a decreasing length of a body is known as compressive stress. * **Compressive stress = decreasing length/original length.** ### Shear stress * The Stress induced in a body subjected to two equal and opposite force which are acting together. Tangentially, across the resisting section, as a result of which the body tends to shear off across the cross is known as shear stress.. * The corresponding strain is known as shear strain. * The shear stress is the stress which acts tangential to the area. It is represented by (T). * Shear stress (T) = Shear resistance/Area = P/A = P/2XL ### Shear strain: * As the bottom face of the block is fixed, the face ABCD is distorted to ABC'D' through an angle $\phi$. As a result of force "P" as shown in the above figure. * Shear strain $(\phi) = $ Transversal displacement/distance of AD * **Types of Strain** 1. Tensile strain 2. Compressive strain 3. Change in Volume/original volume = volume strain 4. Shear strain * **Hooke's Law**: It states that when the material is loaded within elastic limit. The stress is proportional to the strain produced by the stress. * This means the ratio of the stress to the corresponding strain is a constant within the elastic limit. This constant is known as modulus of elasticity/ modulus of rigidity or elastic modulus. * **Modulus of Elasticity or Young's modulus (E)** * The ratio of tensile Stress by compressive stress to the corresponding strain is known as constant. ## Day: Friday ## Date: 9/8/2024 ### Working Stress * The stress is capable of preventing failure is called working stress. * Permissible stress or allowable stress * Working stress = ultimate stress / factor of safety. ### Factor of Safety * It is defined as the ratio of ultimate tensile stress to the working (initial, permissible) stress. * Factor of safety = ultimate tensile stress/working stress or permissible or allowable ### Elastic modulus * Stress-strain for mild steel * UTM = Universal Testing machine. * The behaviour of a material subjected to an increased tensile load is studied by testing a specimen in a tensile testing machine and plotting the stress-strain diagram.. * Stress-strain Diagrams of different materials varying widely. * **Stress ($\sigma$)** * **Strain ($\epsilon$)** * P = Proportional Limit, E = Elastic Limit * Y = Yield limit, u = ultimate strength (or U), * R = Rupture strength. * The plot between strain $\epsilon$ i corrosponding stress of a ductile material is represented graphically by a tensile test diagram. The above figure shows a stress v/s strain diagram for mild steel. ### Proportional Limit (P) * When the load is increased gradually, the strain is proportional to load or stress upto a certain value. The line OP indicates range and is known as line of proportionality. Hooke's law is applicable in this range. The stress at the end point 'P' is known as the proportional limit. ### Elastic limit (E) * If the load is increased beyond the limit of proportionality, the elongation is found to be more rapid through the material may still be in the elastic limit. I.e, on removing the load, the strain vanishes. The point E indicates the elastic limit. Hooke's law cannot be applied in this range. ### Yield point (Y) * When the load is further increased, plastic deformation occurs. I.e, on removing the load, the strain is not fully recoverable. At point 'Y' metal shows appreciable strain even without further increase of load. * **Actually, the curve props slightly at this point to Y' and the yielding goes upto the point Y". The points Y' & Y" are known as the upper & lower yield points respectively. The stress-strain curve between Y' & Y" is not steady.** ### Ultimate strength (U) * After the yield point further straining is possible only by increasing the load. The stress-strain curve raises upto the point 'U'. The stress curve at 'U' is known as the ultimate stress. ### Rupture strength (R) * If the ball is stressed further, it begins to form a neck or local reduction in cross-section occurs. The specimen ruptures at a point called rupture strength 'R'. ## Day: Monday ## Date: 12/8/2024 ### Longitudinal strain * When a body subjected to an axial tensile load, there is an increase in the length of the body but at the same time there is a decrease in other dimensions of the body at right angles to the line of action of applied load. Thus the body is having axial deformation & also deformation at right angles the line of action of the applied load(Lateral deformation). * **The ratio of axial deformation to the original length of the body is longitudinal (linear) Strain.** * Longitudinal strain = $\delta L/L$ * Let P tensile force acting on the body. * L = Length of the body * $\delta L$ = Increase in length of the body in direction of P. ### Lateral strain * **The strain at right angles to the direction of applied load is known as lateral strain.** * Let a rectangular bar of length L, breadth "b" and depth "d" is subjective to an axial tensile load "P" shown in the below figure. * The length of the bar will increase while the breadth & depth will decrease. * Let $\delta L$ = increase in length * $\delta b$ = increase in breadth * $\delta d$ = decrease in depth. * **And Lateral strain = $\delta b/b = \delta d/d$** * **Note**: 1. The longitudinal strain is tensile. The Lateral strain is compressive. 2. The longitudinal strain is compressive then the lateral strain is tensile. 3. Hence, every longitudinal strain in the direction of accompanied by lateral strains of the opposite kind in all directions perpendicular to the load. ### Poisson's ratio * **Poisson's ratio is the ratio of lateral strain to the longitudinal strain is a constant for a given material when the material is stressed within the elastic limit. This ratio is called poisson's ratio.** * It is denoted by $\mu$. * Poisson's ratio = Lateral Strain/Longitudinal Strain * (or) * Lateral strain = Poisson's ratio x Longitudinal Strain. * As Lateral strain is opposite sign to longitudinal strain. Hence, lateral strain = -$\mu$xlongitudinal strain. ### Volumetric strain * **It is the ratio of increase in volume of a body to its original volume when it is acted upon by three mutually perpendicular stresses.** * Volumetric strain = ($\sigma1 + \sigma 2 + \sigma 3 )(1 - 2\mu$) * ~E1 + E2 + E3 ## Day: Tuesday ## Date: 13/8/2024 ### Elastic Constants * **The factors to determine the deformations produced by a stress system acting on a material within elastic limits are constant and termed as elastic constants.** * Two elastic constants (modulus of elasticity (E) & modulus of rigidity) have already being define in previous section. * A third elastic constant is being defined in this section. * **If three mutually perpendicular stresses of equally intensity are applied to a body of initial volume of volume (V) as shown in figure (a).** * **Then the ratio of the direct stress to the volumetric strain is known as bulk modulus (K).** * **Usually, bulk modulus is applicable mainly to fluid problems with pressure intensity (P) in all directions.** * **Thus, K = Direct stress / Volumetric strain = P/E.** ### Volumetric strain: * Ev = ($\sigma1 + \sigma 2 + \sigma 3) (1 - 2\mu$) * $\sigma$ = 3(1-2$\mu$) (P/V) * For three perpendicular of equally intensity, P (compressive) * K = P/3(1-2$\mu$)(P/v) => E = 3K(1-2$\mu$) * K = E/3(1-2$\mu$) ### Relationship between elastic constants: * Consider a square element ABCD, under the action of simple shear stress (T), shown in the figure. * In the figure, the resultant distortion of the element, showing the total change in the corner angles +/-$\phi$. * However, for convenience, take the side AB may be considered to be fixed, show in the figure. * As angle $\phi$ is extremely small, CC', DD', can be assumed to be arise let 'C' be perpendicular on the diagonal AC'. * **Linear strain of Diagonal AC can be taken as:** * E = AC'/ AC = EC'/AC * E = EC'/AC * E = EC'/AC * From triangle ECC * Cos 45^o = EC/EC'= Cos 45^o CC'/CC * From triangle ABC'SA * Cos 45^o = AB/AC = AB/Cos 45^o * E = Ed = CC'Cos45^o = CC'Cos² 45^o * E = Ed = AB/Cos45^o = AB/Cos² 45^o * CC' = BC $\phi$ * E = CC'Cos² 45^o = L/2 BC $\phi$/AB * AB = BC + CD = DA * E = 1/2 BC $\phi$/BC = 1/2 $\phi$ = strain * From $\phi$ = Transversal Distance = CC' / Total Distance = CC * Shear strain $\phi$ = CD/BC => CC = BC$\phi$ * The modulus of rigidity G = T/$\phi$ * From this $\phi$ = T/G * From eq (5) we have 2E = T/G => E = T/2G * **In the state of simple sheer on two perpendicular plains the plains at 45^o are subjected to tensile stress(magnitudes, equal to the that of shear stress)** * **While the plains at 135^o are subjected to a compressive stress of the same magnitude with no shear stress on this plains.** * The plans AC & BC are subjected to tensile and compressive stress are respectively, each equal to ''T'' as shown in the figure. * Hence, linear strain of diagonal "s" is, * E = s'/s = (1 + v)/s * T = (1+v) * From eq 4 & 8 * T = 2G(1+v) * 2G = E(1+v) * => E = 2G(1 + v) * From E = 3K(1-2v) * => **E = 2G(1+v) = 3K(1-2v)**. * This equation relates the elastic constant. * Also, from above * 2(1 + v) = E/G => E = 2G + 2Gv * 3K (1 - 2v) => 1 - 2v = E/3K. * **By adding ea 5 & 6 we have:** * 2 + 2v = E/G * 1 - 2v = E/3K * 3 = E [1/G + 1/3K] * 3 = E [3K + G/3KG] * **=> E = 9KG/3K + G** ## Day: Wednesday ## Date: 14/8/2024 ### A bar of 24mm diameter and 400mm length is acted upon. By an axial load of 38 Kg/N. The elongation of the bar and change in diameter are measured as 0.165mm & 0.0031 mm respectively. Determine Poisson's ration, the values of three modulus. * Given data of circular bar: * d = 24mm * l = 400mm * P = 38 kN * $\delta L$ = 0.165 * $\delta d$ = 0.0031 = 3e x 10^-3 * P = 38000 kN * Area (A) = $\pi$$d^2$ = $\pi$$(24)^2$ = 452.3 mm² * $\sigma$ = 38000/452.3 = 84.01 N/mm² (pascal) * **Poisson ration ($\mu$) = v. Longitudinal strain/Lateral strain = v. $\delta L/L$/$\delta d/d$** * 0.0031 = v* 0.165/400 * v = 0.0031*400/24*0.165 * v = 0.313 * **Three elastic modulus** * E = $\sigma$$\epsilon$ = 84/0.165/400 * **E = 203.636 mpa** * E = 2G(1 + v) * G = E/2(1+v) = 203.636/2(1+0.313) * **G = 77.546 mpa** * E = 3K(1-2v) => K = E/3(1-2v) * K = 203.636/3(1-2(0.313)) * K = 18/493 mpm ## Day: Thursday ## Date: 22/8/2024 ### Bars of Varying Section * **An expression for the elongation of the bar of length "l" and cross sectional area "A", under the action of force "P" is: ** * E = E * As = E = E * E = $\delta L/L$ * $\delta L/L = \sigma/E$ * $\delta L = \sigma L/E$ * $\delta L = P L/AE$ * $\delta L = P L/AE$ * **$\delta L$ = $PL/AE$** ### Analysis of Bars of Varying Section: * A bar of different lengths and diameters (Different cross-sectional areas). Let this bar subjected to an axial load P. * $\delta l1 = PL1/A1E$ * $\delta l2 = PL2/A2E$ * $\delta l3 = PL3/A3E$ * $\delta l$ = $\delta l1 + \delta l2 + \delta l3$ * $\delta l$ = $PL1/A1E + PL2/A2E + PL3/A3E$ * **Young's modulus of different sections is different * An axial pull of 35kN is acting on a bar consisting of a bar, 3 lengths is shown in figure. If the Young's modulus E = 2.1 x 10⁵ N/mm² and P = 35000N. Determine (a) stress in each section, (b) total extension of bar. * Given that E = 2.1 x 10⁵ N/mm² * P = 35000N * l1 = 20cm = 20 x 10 mm = 200 mm * d1 = 2cm = 2 x 10 mm = 20 mm * A1 = $\pi$$d^2$ = $\pi$$(20)^2$ = 314 mm² * P = 35000N * Stress $\sigma$ = P/A1 * $\sigma$1 = 35000/314 = 111.46 N/mm² * Section 2: * d2 = 3cm = 3 x 10 mm = 30 mm * l2 = 2.5cm = 25 x 10 mm = 250 mm * A2 = $\pi$$d^2$/4 = $\pi$$(30)^2$ = 706.85 * Stress $\sigma$ = 35000/706.85 = 49.515 * Section 3 * d3 = 5cm = 5 x 10 mm = 50 mm * l3 = 22 cm = 22 x 10 mm = 220mm * A3 = $\pi$$d^2$/4 = $\pi$$(50)^2$ = 1963.49 * Stress $\sigma3$ = 35000/1963.49 = 17.82 * **Total extension of bar is:** * $\delta L $ = $PL1/A1E + PL2/A2E + PL3/A3E$ * $\delta L$ = (35000 x 200)/(314 x 21 x 10 ^5) + (35000 x 250)/(706.83 x 2.1 x 10^5) + (35000 x 220)/(1963.49 x 2.1 x 10^5) * $\delta L$ = (35 x 2)/(314 x 2.1) + (35 x 2.5)/(706.83 x 2.1) + (3.5 x 2.2)/(1963.49 x 2.1) * $\delta L$ = 50 + 0.0589 + 0.0186 * $\delta L$ = 0.1836 * **Total expansion is $\delta L$ = 0.1836 mm. ** ## Day: Wednesday ## Date: 21/8/2024 ### Analysis of Composite bars * A bar made up of 2 or more bars of equal length and of different materials are rigidly fixed with each other and behaving as for has 1 unit for extension, when surface to an axial and compressive load is called composite bar. * For the composite bar, two points are important: 1. The extension or compressive in each bar is equal. Hence deformation per unit length that is strain in equal bar is equal. 2. The total external load on the composite bar is equal to the sum of the loads carried by each different material. * The above figure shows a Composite bar is made up of two different materials. * **P = total load on the composite bar** * **L = length of bar's of different materials(**or**)** * **L = Length of composite bar** * **A1 = Area of cross-section of bar 1** * **A2 = Area of cross-section of bar 2** * **E1 = Young's modulus of bar 1** * **E2 = Young's modulus of bar 2** * **P1 = Load share by bar 1** * **P2 = Load share by bar 2** * **$\sigma$1 = stress induced in bar 1** * **$\sigma$2 = stress induced in bar 2** * **Now, the total load on Compasite bar is equal to the sum of the load carried by the two bar's.** * **P = P1 + P2** * **The stress in bar 1 = load carried by bar 1/Area of cross section of bar 1.** * $\sigma$1 = P1/A1 or, P1 = $\sigma$1A1 * Strain bar 2: $\sigma$2 = P2/A2 * **For bars of different modulus, the ratio of E1/E2 is called modular ratio of the 1st material and the 2nd material.** * A steel rod of 3cm diameter is enclosed centrally in a hollow copper tube of external diameter 5cm, and internal diameter of 4cm. The composite bar is then subjected to an axial pull of 45000N. if the length of each bar is equal to 15cm, determine: * (I) The stresses in the rod and tube; and * (II) Load carried by each bar. * Take E for steel = 2.1 x 10⁵ N/mm² and for copper = 1.1 x 10⁵ N/mm² * Given: * Dia. of steel rod = 3cm = 30mm. * Area of steel rod As = $\pi$$(30)^2$/4 = 706.86 mm² * External dia. of copper tube = 5cm = 50mm * Internal dia. of copper tube = 4cm = 40mm. * Area of copper tube, Ac = $\pi$$(50 - 40)^2$/4 = 706.86mm² * Axial pull on composite, P = 45000N * Length of each bar, L = 15 cm. = 150mm * Young's modulus for steel, Es = 2.1 x 10⁵ N/mm² * Young's modulus for copper, Ec = 1.1 x 10⁵ N/mm² * **(1) The stress in the rod and tube, let:** * $\sigma$s = stress in steel. * Ps = Load carried by steel rod. * $\sigma$c = stress in copper, and * Pc = Load carried by copper tube. * **Now, strain in steel = strain in copper** * $\sigma$s/Es = $\sigma$c/Ec * $\sigma$s = $\sigma$c*Es/Ec * $\sigma$s = $\sigma$c*2.1 x 10⁵/1.1 x 10⁵ * **$\sigma$s = 1.909 $\sigma$c (1)** * **Now, stress = Load/Area** * Load = stress x Area * Load on steel + load on copper = Total load * $\sigma$s x As + $\sigma$c x Ac = P * 1.909 x 706.86 + $\sigma$c x 706.86 = 45000 * $\sigma$c( 1.909 x 706.86 + 706.86) = 45000 * **$\sigma$c*2056.256 = 45000** * **$\sigma$c = 45000/2056.25 = 21.88 N/mm²** * Substituting the value $\sigma$c in eq (1), * $\sigma$s = 1.909 x 21.88 N/mm² * **$\sigma$s = 41.77 N/mm²** * **(II) Load carried by each bar** * As, load = stress x area. * Load carried by steel rod, * Ps = $\sigma$s x As * Ps = 41.77 x 706.86 = 29525.5 N. * Ps = 29525.5 N. * Load carried by copper tube, * Pc = 45000 - 29525.5 * **Pc = 15474.5 N.** ## Day: Friday ## Date: 23/8/2024 ### Temperature Stress (Thermal Stress) * Thermal stresses are the stresses induced in a body due to change in temperature. * Thermal stresses are setup in a body, then the temperature of the body is raised or lowered, and the body is not allowed to expand or contract freely. But if the body is allowed to expand or contract freely, no stresses be setup in the body. * Consider a body, which is heated to a certain temperature: * a) A ----B B' * b) A----B B' * c) A B * Let: * L = Original length of the body * T = raise in K * E = young's modulus * $\alpha$ = coeff of linear expension or linear enter expension, of rod due to rise at temperature. * $\delta l$ = Extension of rod due to rise at temperature. * If the rod is free to expand then extension the rod is $\delta l$ = $\alpha .T.L$. * If Fig (a), AB, represents original length, then BB' represents the increase in length due to temperature rise. * Now suppose that when the external compressive load "P" is applied at (B') so that the load is decreased in its length from (L + $\alpha$.T.L) to L, shown in fig B and D is original length. * Compressive strain = Decreasing length/original length * Then compressive strain = $\alpha$T.L/L = $\alpha$ T * But stress = E x strain * From this stress = E x $\alpha$T * **Load overall or thrust on the load** * **Load = stress x area**, **Load = $\alpha$T.E.A**, **Load = $\alpha$T.A.E** * If the end of the body are firmed to rigid supports, so the its expansion prevented, then compressive stress or strain will be setup in the rod. This stress and strain are known as Thermal stress and thermal strain. * **Thermal strain = extension prevented/original length = $\alpha$T.L/L = \alpha T** * **E = $\delta L/L$ = $\alpha TL/L = \alpha T$** * **Thermal stress = Thermal strain x E = $\alpha$T.E** * **Thermal stress is also known as temperature stress. ** * **Thermal strain is also know as Temperature strain** * **Stress & Strain when the supports yield.** * If the supports yield by an amount equal to "S" then actual expansion is equal expansion due (raised in temp - S). * **Actual expansion = $\alpha$T.L - S** * **Actual strain = actual expansion/original length** * **Actual strain = $\alpha$T.L - S/L** * **Actual stress = actual strain x E** * **Actual stress = $\alpha$T.L- S/L x E** * **A rod is 2mt long, at temp of 10^oc. Find the expansion of the rod, when the temperature is raised to 80^oc. if this expansion is prevented, find the stressed induced in the material of the rod. ** * **E = 1.0 x 10⁵ MN/m²**, **\alpha = 0.000012 per degree centigrate** * Given that: * L = 2mt = 2000 mm * T1 = 10^oC * T2 = 80^oC * Raise in temp 'T' = T2 - T1 = 80 - 10 = 70^oC * E = 1.0 x 10⁵ MN/m² * E = 1.0 x 10⁵ N/mm² * \alpha = 0.000012 * (1) The expansion of rod due to temp rise is given by eqn. * **Expansion of the rod = \alpha.T.L** * **= 0.000012 x 70 x 200** * **= 0.168cm** * **(II) The stress in the material of the rod if expansion is prevented is given by eqn.** * Thermal stress, $\sigma$ = $\alpha$T.E * $\sigma$ = 0.000012 x 70 x 1.0 x 10¹¹ N/m² * $\sigma$ = 84 x 10⁶ N/m² = 84 N/mm² ## Day: Saturday ## Date: 24/8/2024 ### Strain energy * Whenever a body is strain, the energy observed in the body due to straining effect is known as strain energy. * The straining effect may be due to gradually applied load or suddenly applied load or load with impact. ### Resilience * The total strain energy stored in a body is knon as resilience. Whenever the straining force is removed from the strained body. The body is capable of doing work. Hence, the resilience is also defined as the capacity of strained body for doing work on the removal of the straining force. ### Proof resilience * The maximum strain energy stored in a body is known as proof resilience. * The strain energy stored in a body will be maximum when the body is stressed upto elastic limit. * **Modulus of Resilience = Proof resilience/Volume of body** ### Expression for Strain energy stored in a body when the load is applied gradually: * The load extension diagram of a body under tensile test upto elastic limit. The tensile load os increases gradually from D to value of P. On the extension of the body, increase from 0 to the "x" value. * **The load "P" perform's work in stretching like body this work will be stored in the body as strain energy, which is recovered after the load "P" is removed.** * Let: * P = gradually applied load * x = extension of the body * A = cross section area * l = length of the body * V = Volume of the body * U = strain energy stored in the body * $\sigma$ = stressed induced in the body, Load/Area * The work done by the Load = Area of Load extension curve (shaded areas) * Area of triangle ONM = 1/2Px * But Load, P = Stress x Area, Stress = $\sigma$/Area * Extension, x = Strain x length. * Strain = Extension/length * x = $\sigma$L/E * P = $\sigma$xA * **Substituting the values of P & x in eq(1) we get work done by the load.** * U = 1/2Px = 1/2 $\sigma$A. $\sigma$L/E * U = 1/2 E²AL * U = V$\sigma$²/2E * **The energy stored in the body is U = V$\sigma$²/2E** * In the variation proof resilience is as max. strain energy stored in a body. The max. strain energy stored in a body without permanent deformation (up to elastic limit) is known as proof resilience. Hence, if eq(2) the stress ($\sigma$) is taken at the elastic limit, we will get proof resilience. * **Proof resilience = E²/2E x V** * **Modulus of Resilience = strain energy per unit volume = Total Strain Energy/Volume.** * **Modulus of Resilience = E²/2E x V / V = $\sigma$²/2E** ### Expression for Strain energy stored in a body when the load is applied suddenly: * **When the load is applied suddenly to a body, the load is constant throughout the process of the deformation of the body.** * Consider a bar subjected to a sudden load. * P = Load applied suddenly. * L = Length of the bar. * A = Area of the cross-section. * V = Volume of the bar = A x L * E = Young's modulus * x = Extension of the bar. * $\sigma$ = stress induced by the suddenly applied load, and * U = Strain energy stored. * As the load is applied suddenly, the load P is constant when the extension of the bar takes place. * **Work done by the load = Load x Extension = P x x.** * **The maximum strain energy stored (i.e. energy stored upto elastic limit) in a body is given by:**