AS Physics Unit 11 Materials Mr D Powell. Mr Powell 2009 Index Chapter Map Common misconceptions Don’t be overwhelmed by the large number of new terms.

AS Physics Unit 11 Materials

KS5 AS PHYSICS AQA 2450Mr D Powell

Mr Powell 2009Index

Chapter Map Common misconceptions

Don’t be overwhelmed by the large number of new terms introduced in this chapter.

The units used for stress, strain and the Young modulus can also prove to be confusing – also the conversion of mm2 (again).

Mr Powell 2009Index

11.1 Density

Specification link-up 3.2.2: Bulk properties ofSolids

How do we define density?

What is the unit of density?

How do we measure the density of an object?

v

m

http://en.wikipedia.org/wiki/Density

Mr Powell 2009Index

Consider this data.... Material ρ in kg/m3

Interstellar medium 10−25 − 10−15

Earth's atmosphere 1.2

Water 1000

Plastics 850 − 1400

The Earth 5515.3

Copper 8920 − 8960

Lead 11340

The Inner Core of the Earth ~13000

Uranium 19100

The core of the Sun ~150000

White dwarf star 1 × 109

Atomic nuclei 2.3 × 1017

Neutron star 8.4 × 1016 — 1 × 1018

Black hole 4 × 1017

We often talk about the properties of materials but if we often want to compare how materials behave we have to consider things by way a ratio.

We can look at both mass and volume as physical properties and this look at their ratio.

We call the ratio of m/v as being the density or (rho) of a material.

It basically means for the same unit volume i.e. 1m3 how much mass do I have.

Mr Powell 2009Index

Example Metal Densities...

Metal or Alloy Density (kg/m3)

Admiralty Brass 8525Aluminum 2712

Aluminum - melted 2560 - 2640Aluminum bronze (3-10% Al) 7700 - 8700

Aluminum foil 2700 -2750Brass 60/40 8520

Bronze - lead 7700 - 8700Bronze - phosphorous 8780 - 8920

Bronze (8-14% Sn) 7400 - 8900Cadmium 8640Cast iron 6800 - 7800

Iron 7850Lead 11340

White metal 7100Zinc 7135

Zirconium 6570Yellow Brass 8470

Mr Powell 2009Index

Calculations... When looking at density it is easy when you have only one substance.

However, if you are dealing with an alloy or mixture then you must take this into account. Think about two metals which are alloyed...

V

VVV

VV

V

m

VVm

Vm

Vm

BAA

BAA

BBAA

BBB

AAA

Mass of substance A is density x volume A

Mass of substance B is density x volume B

Total mass of substances as alloy

The density of the new alloy must be its mass/volume

So we have the new density

Mr Powell 2009Index

Quick Questions...

1) What are the following to densities in kg m-3.

a) 1.29 g cm-3

b) 7.6 g cm-3

c) 19.6 g cm-3

2) What is the volume of 100 g of Mercury of which the density is 13 600 kg m-3?

Answers... a) 1290 kg m-3

b) 7600 kg m-3

c) 19 600 kg m-3

Convert 100 g to kg = 0.1 kg

  V = m/ = 0.1 13 600 = 7.35 10-6 m3

Mr Powell 2009Index

Density Calc & Errors....This is a 5.00cm cube with an uncertainty of 1 mm for each dimension.. The mass of the cuboid is 4000 g with negligible uncertainty. First calculate the density of the material assuming there was no uncertainty...

Volume = 5.00 cm x 5.00 cm x 5.00 cm = 152.00 cm3

Density = 4000g/125.00cm3 = 32.00g cm-3.

Now for the uncertainties.

The percentage uncertainty of 50 mm = 1/50 x100% = 2%

Therefore the percentage uncertainty of the volume = 2% +2% + 2% to a very good approximation.

The percentage uncertainty of the density = % Uncertainty of the volume + %uncertainty of the mass. Therefore the percentage uncertainty of the density = 6 % + 0

Therefore the density......... (32 +/-1.9)g cm-3

5cm

Mr Powell 2009Index

Extension – changes in density (moving onto A2)

In general, density can be changed by changing either the pressure or the temperature.

Increasing the pressure will always increase the density of a material.

Increasing the temperature generally decreases the density

The effect of pressure and temperature on the densities of liquids and solids is small.

In contrast, the density of gases is strongly affected by pressure. The density of an ideal gas is

where R is the universal gas constant, P is the pressure, M is the molar mass, and T is the absolute temperature. This means that the density of an ideal gas can be doubled by doubling the pressure, or by halving the absolute temperature.

Mr Powell 2009Index

11.2 Springs

Specification link-up 3.2.2: Bulk properties ofSolids

Is there any limit to the linear graph of force against extension for a spring?

What is the meaning of spring constant, and in what unit is it measured?

If the extension of a spring is doubled, how much more energy does it store?

qp kkk Series....

pq kkk

111

Parallel....

Mr Powell 2009Index

Spring Extension... Hookes Law

You should already know about this exp..

We need to think about the form

F = k∆L

Where the force is related by a constant for the spring.

Mr Powell 2009Index

K – theory...

21

21

21

2211

2211

21

222

111

kkk

LkLkLk

LLLif

LkLkLk

LkLkW

FFW

FLk

FLk

When you think about it the idea of “k” is the stiffness of a spring.. F = k∆L

The higher the value of k the more force required to pull a spring to a longer length or extension... F = kx

Placing springs in series or parallel will alter the overall “k” value of a combination but as extension is same for all springs we can simplify relation to....

Mr Powell 2009Index

K – theory...

21

21

21

21

22

11

111

kkk

k

F

k

F

k

F

k

F

k

FL

LLL

FLk

FLk

We know that the springs each extend by a small amount and the force on each is the same as the force overall.

Hence the total extension is..

Then Sub for ∆L=F/k for system

Now cancel F as same everywhere to get expression in term of k only.

Mr Powell 2009Index

How does “k” change...

1. Test spring to find k.2. Test two springs in series then in

parallel to work out “k”.3. You will need to plot a graph for

each. To test the theory.4. What is the relation for each? pq

qp

kkk

kkk

111

By tabulating

k = 31.4Nm-1

By graphical methods

27.8 Nm-1

32.8 Nm-1

Average 30.3Nm-1

Mr Powell 2009Index

Results....

Parallel Results

Extension (m) Force (N)

0.000 00.005 10.017 20.035 30.050 4 0.000 0.010 0.020 0.030 0.040 0.050 0.060

00.5

11.5

22.5

33.5

44.5

f(x) = 74.3196890006859 x + 0.409558655385317

Force (N)

Extension (m)

Forc

e (N

)

Joe & Huw - Results

Series Results

Extension (m) Force (N)

0.000 0

0.035 1

0.090 2

0.175 3

0.254 4 0.000 0.050 0.100 0.150 0.200 0.250 0.3000

1

2

3

4

5

f(x) = 15.0408051472978 x + 0.333478789679408

Force (N)

Extension (m)

Forc

e (N

)

qp kkk

pq kkk

111

Mr Powell 2009Index

11.3 Deformation of solids

Specification link-up 3.2.2: Bulk properties ofsolids; the Young modulus

How is stress related to force, and strain toextension?

What is meant by tensile?

Why do we bother with stress and strain, when force and extension are more easily measured?

Stress = Force / Area

Strain = elongation/ original length

l

l

l

eStrain

A

FStress

http://en.wikipedia.org/wiki/Tensile_strength

Mr Powell 2009Index

Deformation of a Solid......

Why do we get this type of graph?

Mr Powell 2009Index

Searle's Apparatus

This is a basic experiment where you attach two wires (one a weighted control) to the equipment.

Then you load the test wire checking F, A, e, l and length variables.

As you load the wire the micrometer shows the extension of the wire for each load.

This enables you to create a stress strain graph

Mr Powell 2009Index

Example...A wire of length 2m and diameter 0.4mm is hung from the ceiling. Find the extension caused in the wire when a weight of 100N is hung on it. Young Modulus (E) for the wire is 2.0 x 1011 Pa.

Mr Powell 2009Index

Stress – Strain graphs.

P = limit of proportionalityE = elastic limitY1 = yield point where it is weakened

temporarilyY2 = yield point beyond which plastic flow

occurs.UTS = ultimate tensile stress – loses strength

and thinsB – catastrophic failure!

Evidently all materials behave differently.

You should know these basic graphs from memory!

Mr Powell 2009Index

Breaking Strain of Glass

This is a simple experiment which will enable you to estimate the breaking strain of glass.

1. Take a glass rod and heat and bend both ends to produce a swan shape (don’t burn your hands)

2. Heat the middle of the swan neck and pull out to a thin strand about 0.5mm across (don’t burn your hands)

3. Take readings for width and then load until it breaks. (Eye Protection)

4. This will give you the UTS – Ultimate Breaking Stress of Glass

Mr Powell 2009Index

Glass Results.Diameter in mm Area m2 Breaking Force (Nm-2) Pressure Pa MPa

1.15 1.03869E-06 25 24068800.47 24.1

1.12 9.85203E-07 28 28420525.55 28.4

1.15 1.03869E-06 36 34659072.67 34.7

1.05 8.65901E-07 14 16168121.2 16.2

0.84 5.54177E-07 50 90223890.64 90.2

1.14 1.0207E-06 14 13716030.8 13.7

0.73 4.18539E-07 31 74067228.16 74.1

0.86 5.8088E-07 33.8 58187529.22 58.2

0.63 3.11725E-07 10.31 33074073.33 33.1

0.5 1.9635E-07 16.19 82454992.92 82.5

0.15 1.76715E-08 3.42 193532410.8 193.5

Quoted Value of Ultimate Tensile Stress is 33MPa

http://en.wikipedia.org/wiki/Ultimate_tensile_strength

http://en.wikipedia.org/wiki/Glass

Mr Powell 2009Index

YM of Copper...

1.Set up the apparatus as shown in the diagram . Use 32SWG wire.

2.Fix the marker then Initially add the mass hanger to tension the wire. Consider the original length of the wire to be from the clamped end to the tape marker. The extension will be measured using the tape marker.

3. Prepare a table with columns for length, extension, force (weight) and cross-sectional area.

4. Measure and tabulate the diameter of the wire, taking two perpendicular measurements at several places. Hence calculate the average cross-sectional area of the wire.

5. Add a 100g mass to the mass hanger and take measurements of the extension and diameter. Repeat until the wire breaks. If the wire stretches (creeps), wait for it to stop before taking measurements.

6. Estimate the uncertainty in the measurement of the extension, the diameter of the wire and its length

Aims

In this experiment you will use a graph to calculate the Young modulus of copper. You will analyse data from your experiment and make suitable calculations in order to plot a graph of stress against strain.

Mr Powell 2009Index

YM of Copper...

Questions1. What are the units of the Young modulus?2. Write down an equation for the extension ΔL of

the wire in terms of the load W, the area of cross section of the wire A, the wire’s length L and the Young modulus E of the wire material.

ResultsTabulate your results and then plot a graph of stress against strain. Measure the gradient of your graph in order to determine the Young modulus of the wire.

DiscussionWhat is the greatest source of error in this experiment?Estimate the percentage uncertainty in your result.

Aims

In this experiment you will use a graph to calculate the Young modulus of copper. You will analyse data from your experiment and make suitable calculations in order to plot a graph of stress against strain.

Mr Powell 2009Index

YM of Copper Wire - Results

Original Length (m)

Extension (m) Strain Force (N) Diameter (m) Area (m^2) Stress

(N/m^2)YM (stress /

strain) in MPa

1.655 0 0 0 0.00035 9.62113E-08 0 #DIV/0!

1.655 0.002 0.001208 0.981 0.00035 9.62113E-08 10196310.15 8437.446649

1.655 0.002 0.001208 1.962 0.00035 9.62113E-08 20392620.3 16874.8933

1.655 0.002 0.001208 2.943 0.00035 9.62113E-08 30588930.45 25312.33995

1.655 0.003 0.001813 8.829 0.00035 9.62113E-08 91766791.35 50624.6799

1.655 0.006 0.003625 13.734 0.00034 9.0792E-08 151268788.1 41724.97406

1.655 0.022 0.013293 19.62 0.00034 9.0792E-08 216098268.8 16256.4834

1.655 0 29.43 0 #DIV/0! #DIV/0!

Mr Powell 2009Index

Example Results....

Mr Powell 2009Index

Exam Question

Mr Powell 2009Index

Exam Question

Mr Powell 2009Index

11.4 More about stress and strain

Specification link-up 3.2.2: Bulk properties ofsolids; the Young modulus

If a metal wire is stretched to a point below itselastic limit and then unloaded, does it return to its original length?

What happens when a metal wire is stretchedbeyond its elastic limit and then unloaded?

How does the deformation of other materials such as rubber and polythene compare with a metal wire?

Mr Powell 2009Index

Loading & Unloading...

Each graph shows how a substance behaves as it is loaded and then unloaded.

The metal wire is loaded and behaves......

Then it reaches the elastic limit and........

Then it........

The rubber band is loaded and behaves......

Then it........

The polythene is loaded and behaves......

Then it........

Mr Powell 2009Index

Summary Quick Questions...

Which material, A or B, has the larger Young modulus and how can you tell?

tensile strain

tens

ile s

tres

s (N

m–2

)

A

B

Answer...

Material A has the larger Young modulus because its stress–strain graph has a steeper gradient.

This means that the value for stress/strain (which is equal to the Young modulus) is greater for A.

Mr Powell 2009Index

Young Modulus

The Young Modulus is the gradient of the stress-strain graph for the region that obeys Hooke’s Law.

This is why we have the stress on the vertical axis when we would expect the stress to be on the horizontal axis.

The area under the stress strain graph is the strain energy per unit volume (joules per metre3).

Strain energy per unit volume = 1/2 stress x strain.

The units arise because stress is in Nm-2 and strain is mm-1 (NOTE:

This unit here is not "millimetres to the minus one", but metres per metre which mean no units).

Nm-2 x mm-1 = Nm m-3. Nm is joules, hence Jm-3

strain

Mr Powell 2009Index

Young Modulus – Quick Question

Example

A wire made of a particular material is loaded with a load of 500 N. The diameter of the wire is 1.0 mm. The length of the wire is 2.5 m, and it stretches 8 mm when under load. What is the Young Modulus of this material?

  First we need to work out the area:   A = r2 = (0.5 10-3)2 = 7.85 10-7 m2   Stress = F/A = 500 N 7.85 10-7 m2 = 6.37 108 Pa   Strain = e/l = 0.008 2.5 = 0.0032   Young’s Modulus = stress/strain = 6.37 108 Pa 0.0032   = 2.0 1011 Pa

Mr Powell 2009Index

Exam Question...

Mr Powell 2009Index

Exam Question....

Mr Powell 2009Index

Exam Question....

Mr Powell 2009Index

Quick Reference....

Property Definition

Strength

How much force is needed to break something. Not always a fair comparison. Something that is thick will be stronger than a thin section. A fairer test is the breaking stress.

Breaking Stress

Breaking stress = breaking force area Force is applied at 90 o to the area.

Stiffness

How difficult it is to change the shape of the object. If we load wires of the same length and diameter with the same tension, the stiffest is the one that stretches the least.

Brittle Stiff, but not strong

Elastic Ability of a material to regain its original shape after it is distorted.

Plastic A material that does NOT regain its