 # 1.3.4 Behaviour of Springs and Materials

## Presentation on theme: "1.3.4 Behaviour of Springs and Materials"— Presentation transcript:

1.3.4 Behaviour of Springs and Materials

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, elastic limit, Hooke’s Law and the force constant – i.e. force per unit extension or compression

Definitions Force (F) Extension (x) Elastic Limit
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 Plastic 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

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

Objective Define the terms elastic deformation and plastic deformation of a material

Definitions Elastic Deformation Plastic 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