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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.

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Presentation on theme: "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."— Presentation transcript:

1 AS Physics Unit 11 Materials Mr D Powell

2 Mr Powell 2009 Index 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 mm 2 (again).

3 Mr Powell 2009 Index 11.1 Density Specification link-up 3.2.2: Bulk properties of Solids How do we define density? What is the unit of density? How do we measure the density of an object? Specification link-up 3.2.2: Bulk properties of Solids How do we define density? What is the unit of density? How do we measure the density of an object? http://en.wikipedia.org/wiki/Density

4 Mr Powell 2009 Index Consider this data.... Material ρ in kg/m 3 Interstellar medium10 −25 − 10 −15 Earth's atmosphere1.2 Water1000 Plastics850 − 1400 The EarthEarth5515.3 Copper8920 − 8960 Lead11340 The Inner Core of the EarthInner Core ~13000 Uranium19100 The core of the SunSun~150000 White dwarfWhite dwarf star1 × 10 9 Atomic nuclei2.3 × 10 17 Neutron star8.4 × 10 16 — 1 × 10 18 Black hole4 × 10 17  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. 1m 3 how much mass do I have.

5 Mr Powell 2009 Index Example Metal Densities... Metal or Alloy Density (kg/m 3 ) Admiralty Brass8525 Aluminum2712 Aluminum - melted2560 - 2640 Aluminum bronze (3-10% Al)7700 - 8700 Aluminum foil2700 -2750 Brass 60/408520 Bronze - lead7700 - 8700 Bronze - phosphorous8780 - 8920 Bronze (8-14% Sn)7400 - 8900 Cadmium8640 Cast iron6800 - 7800 Iron7850 Lead11340 White metal7100 Zinc7135 Zirconium6570 Yellow Brass8470

6 Mr Powell 2009 Index 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... 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

7 Mr Powell 2009 Index 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 m 3

8 Mr Powell 2009 Index 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 cm 3 Density = 4000g/125.00cm 3 = 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

9 Mr Powell 2009 Index Extension – changes in density (moving onto A2) In general, density can be changed by changing either the pressure or the temperature.pressure 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 ideal gas 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.universal gas constantmolar massabsolute temperature

10 Mr Powell 2009 Index 11.2 Springs Specification link-up 3.2.2: Bulk properties of Solids 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? Specification link-up 3.2.2: Bulk properties of Solids 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? Series.... Parallel....

11 Mr Powell 2009 Index 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.

12 Mr Powell 2009 Index K – theory... 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....

13 Mr Powell 2009 Index K – theory...  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.

14 Mr Powell 2009 Index 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? By tabulating k = 31.4Nm -1 By graphical methods 27.8 Nm -1 32.8 Nm -1 Average 30.3Nm -1

15 Mr Powell 2009 Index Results.... Parallel Results Extension (m)Force (N) 0.0000 0.0051 0.0172 0.0353 0.0504 Joe & Huw - Results Series Results Extension (m)Force (N) 0.0000 0.0351 0.0902 0.1753 0.2544

16 Mr Powell 2009 Index 11.3 Deformation of solids Specification link-up 3.2.2: Bulk properties of solids; the Young modulus How is stress related to force, and strain to extension? What is meant by tensile? Why do we bother with stress and strain, when force and extension are more easily measured? Specification link-up 3.2.2: Bulk properties of solids; the Young modulus How is stress related to force, and strain to extension? 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 http://en.wikipedia.org/wiki/Tensile_strength

17 Mr Powell 2009 Index Deformation of a Solid...... Why do we get this type of graph?

18 Mr Powell 2009 Index 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

19 Mr Powell 2009 Index 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 10 11 Pa.

20 Mr Powell 2009 Index Stress – Strain graphs. P = limit of proportionality E = elastic limit Y 1 = yield point where it is weakened temporarily Y 2 = yield point beyond which plastic flow occurs. UTS = ultimate tensile stress – loses strength and thins B – catastrophic failure!  Evidently all materials behave differently.  You should know these basic graphs from memory!

21 Mr Powell 2009 Index 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

22 Mr Powell 2009 Index Glass Results. Diameter in mmArea m 2 Breaking Force (Nm -2 )Pressure PaMPa 1.151.03869E-062524068800.47 24.1 1.129.85203E-072828420525.55 28.4 1.151.03869E-063634659072.67 34.7 1.058.65901E-071416168121.2 16.2 0.845.54177E-075090223890.64 90.2 1.141.0207E-061413716030.8 13.7 0.734.18539E-073174067228.16 74.1 0.865.8088E-0733.858187529.22 58.2 0.633.11725E-0710.3133074073.33 33.1 0.51.9635E-0716.1982454992.92 82.5 0.151.76715E-083.42193532410.8 193.5 Quoted Value of Ultimate Tensile Stress is 33MPa http://en.wikipedia.org/wiki/Ultimate_tensile_strength http://en.wikipedia.org/wiki/Glass

23 Mr Powell 2009 Index 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.

24 Mr Powell 2009 Index YM of Copper... Questions 1.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. Results Tabulate 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. Discussion What 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.

25 Mr Powell 2009 Index YM of Copper Wire - Results Original Length (m) Extension (m) StrainForce (N)Diameter (m)Area (m^2) Stress (N/m^2) YM (stress / strain) in MPa 1.6550000.000359.62113E-080#DIV/0! 1.6550.0020.0012080.9810.000359.62113E-0810196310.158437.446649 1.6550.0020.0012081.9620.000359.62113E-0820392620.316874.8933 1.6550.0020.0012082.9430.000359.62113E-0830588930.4525312.33995 1.6550.0030.0018138.8290.000359.62113E-0891766791.3550624.6799 1.6550.0060.00362513.7340.000349.0792E-08151268788.141724.97406 1.6550.0220.01329319.620.000349.0792E-08216098268.816256.4834 1.655 029.43 0#DIV/0!

26 Mr Powell 2009 Index Example Results....

27 Mr Powell 2009 Index Exam Question

28 Mr Powell 2009 Index Exam Question

29 Mr Powell 2009 Index 11.4 More about stress and strain Specification link-up 3.2.2: Bulk properties of solids; the Young modulus If a metal wire is stretched to a point below its elastic limit and then unloaded, does it return to its original length? What happens when a metal wire is stretched beyond its elastic limit and then unloaded? How does the deformation of other materials such as rubber and polythene compare with a metal wire? Specification link-up 3.2.2: Bulk properties of solids; the Young modulus If a metal wire is stretched to a point below its elastic limit and then unloaded, does it return to its original length? What happens when a metal wire is stretched beyond its elastic limit and then unloaded? How does the deformation of other materials such as rubber and polythene compare with a metal wire?

30 Mr Powell 2009 Index 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 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 rubber band is loaded and behaves...... Then it........ The polythene is loaded and behaves...... Then it........ The polythene is loaded and behaves...... Then it........

31 Mr Powell 2009 Index Summary Quick Questions... Which material, A or B, has the larger Young modulus and how can you tell? tensile strain tensile stress (Nm –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.

32 Mr Powell 2009 Index 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 metre 3 ). 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

33 Mr Powell 2009 Index 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 =  r 2 =   (0.5  10 -3 ) 2 = 7.85  10 -7 m 2 Stress = F/A = 500 N  7.85  10 -7 m 2 = 6.37  10 8 Pa Strain = e/l = 0.008  2.5 = 0.0032 Young’s Modulus = stress/strain = 6.37  10 8 Pa  0.0032 = 2.0  10 11 Pa

34 Mr Powell 2009 Index Exam Question...

35 Mr Powell 2009 Index Exam Question....

36 Mr Powell 2009 Index Exam Question....

37 Mr Powell 2009 Index Quick Reference.... PropertyDefinition 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


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