1. 1.3.4 Behaviour of Springs and Materials

2. Objective
Describe how deformation is caused by a force in one direction and can be tensile or comprehensive

3. Deformation Can be caused by tensile or compressive forces Tensile
cause tension
stretching forces
Compressive
cause compression
squeezing forces

4. Deformation two equal and opposite tensile forces stretching a wire
two equal and opposite compressive
forces squeezing a spring

5. 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

6. 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

7. 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

8. 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

9. Hooke’s Law F
Tension /N x
Extension /mm

10. 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

11. Objective
Determine the area under a force against extension (or compression) graph to find the work done by the force

12. 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

13. Objective
Select and use the equations for elastic potential energy, E = ½Fx and ½kx2

14. 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

15. Objective
Define and use the terms stress, strain, Young modulus and ultimate tensile strength (breaking stress)

16. 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

17. 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)

18. Young Modulus
To calculate stiffness, calculate the ratio of stress to strain:
Young Modulus (Nm-2) = stress
strain
or E = stress

19. Young Modulus Hooke’s Law Region Elastic limit
Limit of proportionality
stress
gradient = Young modulus
strain

20. 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

21. Objective
Describe an experiment to determine the Young modulus of a metal in the form of a wire

22. Young Modulus Practical

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

24. 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

25. Objective
Describe the shapes of the stress against strain graphs for typical ductile, brittle and polymeric materials

26. 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

27. 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

28. 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

29. Polymeric Cannot stretch beyond it’s elastic limit
Can deform elastically
Can shatter if you apply a large stress
Examples include perspex

30. 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

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