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

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sics Deformation

Can be caused by tensile or compressive forces

Tensile cause tension stretching forces

Compressive cause compression squeezing forces

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sics Deformation

two equal and opposite tensileforces stretching a wire

two equal and opposite compressiveforces 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

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sics 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

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sics 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

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sics 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 = extensionk = force/spring constant

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sics Hooke’s Law

F

xExtension /mm

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sion

/N

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sics 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

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sics 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

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sics 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)

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sics 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

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sics 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)

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sics Young Modulus

To calculate stiffness, calculate the ratio of stress to strain:

Young Modulus (Nm-2) = stress

strain

or E = stress

strain

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sics Young Modulus

Limit of proportionality

strain

stress

Hooke’s Law Region

Elastic limit

gradient = Young modulus

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sics 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

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sics Young Modulus Practical

markerwire

rule

load

wooden blocks

cardboard bridges

Young modulus practical

Objective

Define the terms elastic deformation and plastic deformation of a material

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sics 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

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sics 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

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sics 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

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sics 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

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sics Polymeric

Cannot stretch beyond it’s elastic limit

Can deform elastically

Can shatter if you apply a large stress

Examples include perspex

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sics 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

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sics Questions

Physics 1 – Chapter 8 SAQ 1 to 9 End of Chapter questions 1 to 4