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## 1.3.4 Behaviour of Springs and Materials

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**Objective**Describe how deformation is caused by a force in one direction and can be tensile or comprehensive**Deformation**Can be caused by tensile or compressive forces • Tensile • cause tension • stretching forces • Compressive • cause compression • squeezing forces**Deformation**two equal and opposite tensile forces stretching a wire two equal and opposite compressive forces squeezing a spring**Objective**Describe the behaviour of springs and wires in terms of force, extension, elasticlimit, Hooke’s Law and the force constant – i.e. force per unit extension or compression**Definitions**• Force (F) • applied to a spring or wire in tension or compression • Extension (x) • the change in length of a material when subjected to a tension, measured in metres • Elastic Limit • the point at which elastic deformation becomes plastic deformation**Definitions**• Elastic Deformation • when the deforming force is removed, the object will return to it’s original shape • eg rubber band, spring (usually) • Plastic Deformation • when the deforming force is removed, the object will not return to it’s original shape • eg Plasticine, Blutack**Hooke’s Law**• When tension is plotted against extension, a straight line graph denotes elastic deformation • This is summarised by Hooke’s Law: ‘The extension of a body is proportional to the force that causes it’ or as a formula: where F = Force F = kx x = extension k = force/spring constant**Hooke’s Law**F Tension /N x Extension /mm**Force/Spring Constant**• F = kx • Expressed in newtons per metre • How much force is required per unit of extension • eg 5 N mm-1 means a force of 5 N causes an extension of 1 mm • Can only be used when the material is undergoing elastic deformation**Objective**Determine the area under a force against extension (or compression) graph to find the work done by the force**Work Done**• Extension produced by tension F is x • Work done to reach this extension is the area under the graph work done = area of triangle = ½Fx**Objective**Select and use the equations for elastic potential energy, E = ½Fx and ½kx2**Elastic Potential Energy**• As work has been done to stretch the wire, the wire then stores Elastic Potential Energy • This also applies to compression forces • For elastic deformation, the elastic potential energy equals work done: E = ½Fx as F = kx then E = ½kx2**Objective**Define and use the terms stress, strain, Young modulus and ultimate tensile strength (breaking stress)**Stretching Materials**• One way of describing the property of a material is to compare stiffness • In order to calculate stiffness, two measurements need to be made: • strain • stress**Stretching Materials**• Strain is the fractional increase in the length of a material Strain = extension (m) original length (m) • Stress is the load per unit cross-sectional area of the material Stress (Nm-2) = force (N) cross-sectional area (m2)**Young Modulus**• To calculate stiffness, calculate the ratio of stress to strain: Young Modulus (Nm-2) = stress strain or E = stress strain**Young Modulus**Hooke’s Law Region Elastic limit Limit of proportionality stress gradient = Young modulus strain**Ultimate Tensile Stress**• Stiffness tells us about the elastic behaviour of a material (Young modulus) • Strength tells us how much stress is needed to break the material • The amount of stress supplied at the point at which the material breaks is called the ultimate tensile stress of the material**Objective**Describe an experiment to determine the Young modulus of a metal in the form of a wire**Young Modulus Practical**Young modulus practical**Objective**Define the terms elastic deformation and plastic deformation of a material**Definitions**• Elastic Deformation • when the deforming force is removed, the object will return to it’s original shape • eg rubber band, spring (usually) • Plastic Deformation • when the deforming force is removed, the object will not return to it’s original shape • eg Plasticine, Blutack**Objective**Describe the shapes of the stress against strain graphs for typical ductile, brittle and polymeric materials**Ductile**• Will stretch beyond it’s elastic limit • Will deform plastically • Can be shaped by stretching, hammering, rolling and squashing • Examples include copper, gold and pure iron**Brittle**• Will not stretch beyond it’s elastic limit • Will deform elastically • Will shatter if you apply a large stress • Examples include glass and cast iron**Polymeric**Will perform differently depending on the molecular structure and temperature • Can stretch beyond it’s elastic limit • Can deform plastically • Can be shaped by stretching, hammering, rolling and squashing • Examples include polythene**Polymeric**• Cannot stretch beyond it’s elastic limit • Can deform elastically • Can shatter if you apply a large stress • Examples include perspex**Summary**• All materials show elastic behaviour up to the elastic limit • Brittle materials break at the elastic limit • Ductile materials become permanently deformed beyond the elastic limit • Polymeric materials can show either characteristics, depending on the molecular structure and temperature**Questions**• Physics 1 – Chapter 8 • SAQ 1 to 9 • End of Chapter questions 1 to 4