Properties of solid materials Hardness Resistant to cutting, indentation and abrasion. The Mohs scale of hardness grades ten minerals from the softest (talc) rated 1, to the hardest (diamond) rated 10 Cutting tools (chisels) need to be harder than the stuff they’re cutting…made of harden steel.

In engineering, the hardness of a metal is measured using the Brinell Hardness Number (BHN), which is a ratio of the load applied to a small steel sphere to the area of the indentation it makes in the surface of the metal being tested.

Stiffness High resistance to bending and stretching Compare retort stand with polythene meter rule The stiffness of a material is measured in terms of its modulus of elasticity, higher value stiffer the material. Outer protective casing of safety helmets and boots need to be stiff to keep their shape, not to crush onto your body.

Toughness Material difficult to break = absorb a lot of energy, before breaking. Tough metals usually undergo considerable plastic deformation in order to absorb the energy Car tyres : made from tough rubber/steel compositions…absorbed energy is transferred to the tyres as internal energy…get hot

Brittleness A brittle object will shatter or crack when subjected to dynamic shocks or impacts. Brittle materials undergo little or no plastic deformation before breaking Glass for car windscreens are brittle, concrete and biscuits and chocolate!

Malleability Materials change shape but may lost their strength. Malleable materials can be hammered out into thin sheets. Gold is very malleable – gold leaf to decorate picture frames, pottery etc

Ductility Materials that are ductile can be drawn into a wire without losing their strength. Copper used for electrical connections are produced by drawing out cylinders to the desired thickness Although most ductile materials are also malleable, the reverse is not always true Many malleable materials will shred or break when extended.

Strength An object is strong if it can withstand large force before it breaks. Depends on size – eg/thick cotton thread required a bigger breaking force than thin copper wire Strength of a material defined in terms of its breaking stress, F/A

Extension is Proportional to Force Hooke’s Law Extension is Proportional to Force

If you measure how a spring stretches (extends its length) as you apply increasing force and plot extension (x) against force (F); the graph will be a straight line.

The elastic limit can be seen on the graph. This is where it stops obeying Hookes law. Elastic limit can be seen on the graph. Anything before the limit and the spring will behave elastically. This is where the graph stops being a straight line. If you stretch the spring beyond this point it will not return to its original size or shape.

You can write Hooke's law as an equation: F = k ∆ x Where: F is the applied force (in newtons, N), x is the extension (in metres, m) and k is the spring constant (in N/m). The extension ∆x (delta-x) is sometimes written e or ∆l. You find the extension from: ∆x = stretched length – original length.

Hold on a minute, K? Spring Constant?! The spring constant measures how stiff the spring is. The larger the spring constant the stiffer the spring. You may be able to see this by looking at the graphs below: The spring constant k is measured in Nm-1 because it is the force per unit extension. The value of k does not change unless you change the shape of the spring or the material that the spring is made of. k is measured in units of newtons per metre (Nm -1).

Example F = k ∆ x = 50 N m-1 2.0N = k x 0.04m A spring is 0.38m long. When it is pulled by a force of 2.0 N, it stretches to 0.42 m. What is the spring constant? Assume the spring behaves elastically. Extension, ∆x = Stretched length – Original length = . 0.42m – 0.38m = 0.04 m F = k ∆ x So, k = 2.0 N 0.04 m = 50 N m-1 2.0N = k x 0.04m

Limitations Hooke’s Law stops working when the load is great enough! P is the elastic limit If you increase the load past P, the material will be permanently stretched. Remove the force material will be longer than at the start

Questions 1.A metal guitar string stretches 4mm when a 10N force is applied. a)If the string obeys Hooke’s law how far will the string stretch with a 15N force? b)calculate the stiffness constant for this string. c)the string is tightened beyond its elastic limit. What would be noticed about the string?

2. A rubber band is 6cm long. When it is loaded with 2 2.A rubber band is 6cm long. When it is loaded with 2.5N, its length becomes 10.4cm. Further loading increases the length to 16.2cm when the force is 5N. Does the rubber band obey Hooke’s law when the force on it is 5N? Justify your answer with a suitable calculation. 3.A vertical spring stretches 10cm from its original position under 200g load, determine the spring load.

Stress and Strain

Tensile stress = tensile force (N) cross-section area (m2) A stretching force is also called a tensile force. Tensile stress = tensile force (N) cross-section area (m2) σ = F / A unit – Pa (pascal) or Nm-2 Note: 1 Pa = 1 Nm-2

Qu Nylon has an ultimate tensile stress of 7x107Pa. The cross sectional area of the nylon rope in the diagram is 3x10-5m2. If the Bassem weighs 600N. Will it hold him? Answer: 2x107Pa This is less than the ultimate tensile stress, he’ll live to see another physics example! Dammit! : )

Breaking stress This is the stress required to cause a material to break. Sometimes known as the ultimate tensile stress. A steel wire of length 2m and diameter 0.4mm is extended by 4.0mm when a stretching force of 50N is applied. Calculate The applied stress Answer: 4.0x108 Pa

Tensile strain (ε) Tensile strain = extension original length unit – none (it’s a ratio like pi)

Qu A steel wire of length 2.3m stretches by 1.5mm. What is the tensile strain? Answer: 6.5x10-4 Qu: A steel wire of length 2m and diameter 0.4mm is extended by 4.0mm when a stretching force of 50N is applied. Calculate b) The strain on the wire Answer: 2x10-3 = 20%

Young Modulus (E ) This is a measure of the stiffness of a material. Definition: The Young modulus, E = tensile stress, σ tensile strain, ε Show that E = F L A ΔL What are the units?

Qu Qu: A steel wire of length 2m and diameter 0.4mm is extended by 4.0mm when a stretching force of 50N is applied. Calculate b) The Young Modulus, E Answer: 2.0x1011Pa

Question: A metal wire of original length 1.6m, cross sectional area 0.8 mm2 extends by 4mm when stretched by a tensile force of 200N. Calculate the wire’s (a) strain, (b) stress & (c) Young Modulus strain: 2.5x10-3 c) E: 2.5x108Pa/2.5x10-3Pa b) stress: 200/8x107=2.5x108Pa