09 A. K. M. B. Rashid Professor, Department of MME BUET, Dhaka Introduction to Properties and The Elastic Modulus Topics to Cover Introduction to properties Introduction to mechanical properties The elastic modulus References: 1. Ashby and Jones. Engineering Materials 1, 3 rd Ed, Ch01 (pp.1-8), Ch03 (pp.29-42), Ch06 (pp.83-87). 2. Callister & Rethwisch. Materials Science and Engineering: An Introduction, 9 th Ed, Ch06 (pp.169-174, 177-179).
Introduction to Properties An engineer has a vast range of materials at his disposal: metals and alloys, polymers, glasses and ceramics, composites How does he go about selecting the material, or combination of materials, which best suit his purpose? by selecting properties Mistakes can cause disasters. SS John P. Gaines split in two (1943) 3/34 Case Study 1: Plastic-handled screwdriver Shaft and blade high-carbon steel (metal) High modulus (of elasticity), measuring the resistance of the material to elastic deflection or bending (a.k.a. stiffness). A polymer, if used, would twist far too much. High yield strength, measuring the resistance to cause permanent deformation, so that it would not bend plastically or permanently if twist it hard. Why not use glass? High hardness, measuring the resistance to cause localized deformation, otherwise it would be indented by the material of the head of the screw and damaged. Must have high fracture toughness (energy required to cause brittle fracture) to resist brittle fracture, so that it gives or bends before it breaks. 4/34
Handle Perspex (polymer/plastic) Have large section, so its twisting or modulus is less important. Cannot be made using rubber, since its modulus is too low. Cannot be used metal either, because it is too heavy. Other commonly used material is wood, a rather complicated but cheap polymer. Perspex (Poly-methly-meth-acrylate, PMMA) is preferred because of its ease of fabrication soft and moulded easily when hot aesthetic value appearance, feel, texture are right low density not unnecessarily tiring to hold Low cost 5/34 Case Study 2: Rolls-Royce RB211 turbofan aero-engine Air is compressed into the engine by turbofan, which also provides the aerodynamic thrust around the outside of the casing. Air is further compressed by the compressor blades, then mixed with fuel and burnt in the combustion chamber. The expanding gases drives the turbine blades, which provide power to the turbofan and the compressor blade, and finally pass out of the rear of the engine, contributing to the thrust. Turbine blades Compressor blades Turbofan 6/34
Turbofan blade Titanium alloy (metal) Good modulus, yield strength, fracture toughness. Must also resist fatigue failure (due to rapid fluctuating loads) surface wear (from striking water droplets at high speed) corrosion (important when taking off over the sea) inally, it must have low density. Why not use composite blades of CRP (carbon fibre-reinforced plastics)? Engine blade Complex nickel-base superalloys (metal) Requirements additional to those used for turbofan blade must be used. or economy, fuel must be burnt at as high temperature as possible. The first row of engine blades nowadays runs at about 950 C. This adds high creep and oxidation resistance to the requirement. Although it has a density less than one-half of that of titanium alloy, it is simply not tough enough even a bird strike would instantly demolish a CRP turbofan. 7/34 Case Study 3: Sparking plug for IC engine Spark electrode Tungsten alloy Resistant to thermal fatigue (from rapidly fluctuating temperature) wear (from spark erosion) oxidation and corrosion (from hot gases containing S and Pb) Insulation Alumina (ceramic) Good electrical insulation Good resistance to thermal fatigue, corrosion and oxidation 8/34
Summaries of Properties Bulk Mechanical Properties Density Modulus Yield strength, tensile strength, and ductility Hardness Impact strength and fracture toughness atigue strength and thermal fatigue resistance Creep strength Bulk Non-mechanical Properties Physical properties melting point Thermal properties thermal stability, heat capacity, thermal expansion, thermal conductivity Magnetic properties para-, dia-, and ferromagnetic properties Chemical properties corrosion and oxidation Optical properties refractive index, colour, light absorption, reflection and transmission Electrical properties electrical conductivity, dielectric properties Production properties Ease of manufacture Castability Heat treatability Economic properties Price and availability Surface properties Hardenability ormability Machinability Weldability Oxidation and corrosion riction, abrasion, and wear Aesthetic properties Appearance, texture, feel 9/34 Price and Availability of Materials Table 1.3: Breakdown of Material Prices Class of Use Material Price per ton Basic construction Wood, concrete, structural steel US$200 $500 Medium and light Metals, alloys and polymers for aircraft, US$500 $30,000 Engineering automobiles, appliances, etc. Special materials Turbine-blade alloys, advanced composites. US$30,000 $100,000 (CRP, BRP), etc Precious metals, etc. Sapphire bearings, silver contacts, gold microcircuits, US$100,000 $60m industrial diamond cutting and polishing tools Availability of Materials Depends upon 3 factors: Degree of localization The size of the reserves (the resource base) The energy required to mine and process Critically important in materials selection process Materials for large-scale structural use Lowest price Many materials have all property requirements but their use is eliminated by their price Materials for light and mediumengineering work Larger value addition Lesser economic constraint High labour and production / fabrication cost Severe competition Materials for high-performance applications Regime of high materials technology Actively under research Intense competition from new materials 10/34
Introduction to Mechanical Properties rom an application standpoint, one of the most important topics within MME is the study of the mechanical behavior of material i.e. how materials respond or deform in relation to an applied load or force. Most components, even if used primarily for other property (electronic properties, for example) have to fulfill certain mechanical functions as well. It is necessary to know the characteristics of the material so that any resulting deformation will not be excessive and fracture will not occur. Laboratory testing to measure mechanical properties attempts to replicate the service conditions. Consistency is accomplished by using standardized test, so that people are measuring same thing in the same way. Many international bodies like American Society for Testing Materials (ASTM) maintains and updates standards for mechanical properties. Several other standards organizations include BDS, IS, SAE, ANSI, BS, ISO, JS, DIN, etc. Important mechanical properties are: strength, ductility, stiffness, toughness and hardness. 11/34 What happens to material when it is loaded with a mechanical force? A 0 A 0 τ θ τ l l 0 l 0 l θ A 0 pulling or stretching squeezing or squashing sliding twisting Material deforms, either elastically (i.e. temporarily), or, plastically (i.e. permanently) depending on the magnitude of the force applied. 12/34
Elastic Deformation Initial state Small load applied Load removed bonds stretched bonds return to initial state d Linear elastic Elastic means reversible!! Non-linear elastic This happens when strains (i.e. deformations) are small (~0.5%) (except for the case of polymers) d 13/34 Plastic Deformation 1. Initial state 2. Large load applied 3. Unload Load removed bonds stretched and planes sheared bond un-stretched, but planes still sheared d e+p Plastic deformation, d p Elastic deformation, d e Plastic means permanent!! linear elastic d p d e d
The Elastic Modulus Elastic modulus measures the resistance of the material to elastic or springy deformation. Low modulus materials are floppy, and stretch a lot when they are pulled (squash down a lot when pushed). High modulus materials are the opposite they stretch very little when pulled (squash down very little when pushed). loppy materials like rubber are ideal for things like bungee-jumping ropes Steels are ideally used for making bridges precisely because it is stiff in tension and doesn t give Stretching of a rubber band Sydney Harbour Bridge it is very easy to stretch a rubber band it would be useless if you couldn t but there is no way you could stretch a strip of steel of this cross-section using you bare hands. What will happen if someone uses steel wire rope for climbing protection, or some designer uses rubber to build a bridge? 15/34 Concept of Stress and Strain Stress - orce or load per unit area of cross-section over which the force or load is acting t s Stress is considered positive for tensile loads, negative for compressive loads A A s Important point magnitude of a stress is always equal to the magnitude of a force divided by the area of the face on which it acts Tensile stress σ = A t Tensile stress Shear stress σ = t A τ = s A
Common states of stress Common states of stress: tension, compression, hydrostatic pressure, and shear 17/34 orce, : Newton (N), kilogramme-force (kg f ), pound-force (lb f ) Area A : sq. metre (m 2 ), sq. centimeter (cm 2 ), sq. inch (in 2 ) Unit of Stress Newton per sq. metre (N m 2 ) Kilogramme-force per sq. centimeter (kg f cm 2 ) Pound-force per sq. inch or psi (lb f in 2 ) SI Unit of Stress Pascal (Pa) 1 Pa = 1 kg m 1 s 2 (Since, = ma = kg m s 2 ) Unit Conversion of Stress 1 Pa = N m 2 1 MPa = 1 Mega Pascal = 10 6 Pa 1 GPa = 1 Giga Pascal = 10 9 Pa 1 MPa = 1 MN m 2 = 1 N mm 2 = 145 psi Stresses/Pressure Convention Stress: positive when pull (in tension), negative when push (compression) Pressure: positive when it pushes, negative when pulls 18/34
Strain - Change in dimension (elongation) per unit length Strain is considered positive for tensile loads, negative for compressive loads w q l t w Engineering shear strain γ = w l = tanθ q for small strains P t p u/2 Nominal tensile strain ε n = u l V w Nominal lateral strain ε l = v w p V DV p l Poisson s ratio, ν = lateral strain tensile strain v/2 P v/2 u/2 Important Point Since strains are the ratios of two lengths or of two volumes, they are dimensionless. p Dilation (volume strain) Δ = ΔV V 19/34 Hooke s Law The elastic moduli are defined through Hooke s law. This is a description of the experimental observation that when strains are small the strain is very nearly proportional to the applied stress Such behavior of the solid is called linear elastic. Thus, for simple tension for example, the nominal tensile strain is proportional to the tensile stress s = E e n (Hooke s Law) Here E is called Young s modulus (of elasticity). Materials having a high value of E are termed as stiff, while those having a low value of E are called pliable, or floppy Hooke s law also holds good for stresses and strains in simple compression. In the same way, the shear strain is proportional to the shear stress, with t = G g G = shear modulus Also, the negative of the dilatation is proportional to the pressure (because positive pressure causes a shrinkage of volume) so that Important Point p = KD K = bulk modulus Because strain is dimensionless, the moduli have the same dimensions as those of stress.
Hooke s law (i.e. the linear relation between stress and strain) is very useful when calculating the response of a solid to stress. But many solids are elastic only to very small strains: up to about 0.002. We defined Poisson s ratio as the ratio of the lateral shrinkage strain to the tensile strain. This quantity is also an elastic constant So altogether we have four elastic constants: Beyond that some break and some become plastic. A few solids, such as rubber, are elastic up to very much larger strains of order 4 or 5 E, G, K, u or many materials it is useful to know that but they cease to be linearly elastic (that is, the stress is no longer proportional to the strain) after a strain of about 0.01. K E G 3 E/8 u 0:33 21/34 Measurement of Young s Modulus To measure Young s modulus, both stress and strain need to be measured with enough accuracy. In the case of metals, because they are stiff (and thus, strain data is very small), either the strain needs to be measured very accurately, or there needs to be some way of magnifying it. So we could load a bar of material in tension, having first glued strain gauges to its surface, and use the amplified electrical signal from them to measure the strain. Or, we could load a bar in bending and measure deflection for a given applied load. 22/34
3-Point bend test Compression test Especially easy geometry to adopt. Good for brittle materials (brittle metals, ceramics, polymers, composites), because if loaded in tension they may break where they are gripped by the testing machine. It is also good for natural composites, like wood or bamboo. loppy materials, like the lower modulus thermoplastics, rubbers and foamed polymers can be tested in compression strain is read directly from the movement of the testing machine. for such materials, the rate at which the specimen is strained has a significant effect on the modulus values calculated from the test. Three-point bend test 23/34 Velocity of sound method An electronic pulser-receiver is placed in contact with one end face of a short solid cylinder of the material. The times of travel of longitudinal and shear waves over the known distance are measured electronically, and used to determine the velocity of longitudinal waves, V L and the velocity of transverse waves, V T. V L = V T = ν = E 1 ν ρ 1 + ν 1 2ν G ρ 1/2 2 V 1 2 T VL 2 V 2 2 T VL 1/2 24/34
Worked Example 1 An aluminium rod is to withstand an applied force of 45,000 pounds. To assure a sufficient safety, the maximum allowable stress on the rod is limited to 25,000 psi. The rod must be at least 150 in. long but must deform elastically no more than 0.25 in. when the force is applied. Design an appropriate rod. Young s modulus for aluminium is 10x10 6 psi. Using the definition of stress, the required minimum cross-sectional area of the rod A = σ = 45000 lb 25000 lb/in 2 = 1.80 in2 or, d = 1.51 in However, the minimum length of rod is specified as 150 in. To produce a longer rod, we might make the cross-sectional area of the rod larger. The minimum strain allowed for the 150 in rod is ε = Δl l = 0.25 in 150 in = 0.001667 Using the Hook s law σ = E ε = 10x10 6 lb/in 2 0.001667 = 16670 lb/in 2 Then, the area required to withstand this stress A = σ = 45000 lb 16670 lb/in 2 = 2.70 in2 or, d = 1.85 in Thus, in order to satisfy both the maximum stress and the minimum elongation requirements, cross-sectional area of the rod must be at least 2.70 in 2, or a minimum diameter of 1.85 in. 25/34 Worked Example 2 Your friend wants to keep her large collection of books in a bookshelf in her sitting room. The bookshelf is 1600 mm wide and you offer to install a set of pine shelves, each 1600 mm long, supported at their ends on small brackets screwed into the wall. You guess that shelves measuring 20 mm deep (or thick) by 250 mm wide will work. However, professional integrity demands that you calculate the maximum stress and deflection first. or pine, E =10 GN m -2 and s y = 60 MN m -2. A book 40 mm thick weighs about 1 kg. 26/34
b l d I = 250 x 203 12 = 1.67 10 5 mm 4 = 1600 mm = 40 kgf 40 mm/kgf M = 40 kgf 9.81 N/kgf 1600 mm 8 = 7.85 10 4 N mm σ max = 7.85 104 N mm 20 mm/2 1.67 10 5 mm 4 = 4.7 N mm 2 = 4.7 MPa δ = 5 40 kgf 9.81 N/kgf 1600 mm 3 384 10 4 N mm 2 1.67 10 5 mm 4 = 12.5 mm δ = 5 384 l 2 E I I = b d3 12 σ max = M (d/2) I M = l 8 The shelves will not break but they will deflect noticeably. You are better advised to increase the thickness of the shelves. 27/34 Data for Young s Modulus Diamond, E = 1000 GPa Soft rubbers and foamed polymers, E < 0.001 GPa Practical engineering materials, E = 10 3 to 10 3 GPa Ceramics and metals even the floppiest of them, like lead lie near the top of this range. Polymers and elastomers are much more compliant, the common ones (PE, PVC and PP) lying several decades lower. Composites span the range between polymers and ceramics. E (GPa) Why polymers are much less stiff than metals, and what we can do about it? What is the origin of the modulus? To answer these questions, we have to examine the structure of materials, and the nature of the forces holding the atoms together. 28/34
Physical Basis of Young s Modulus Two things are very important in determining the modulus: 1. The forces that hold atoms together (the interatomic bonds) which act like little springs, linking one atom to the next in the solid state. 2. The ways in which atoms pack together (the atom packing), since this determines how many little springs there are per unit area, and the angle at which they are pulled. The spring-like bond between two atoms Atom packing and bond-angle 29/34 orce required for any separation of the atoms rom the force/distance curve that we get = du dr 1. = 0 at the equilibrium point r = r 0 ; however, if the atoms are pulled apart by distance (r r 0 ), a resisting force appears. or small (r r 0 ), the resisting force is proportional to (r r 0 ) for all materials, in both tension and compression. 2. The stiffness, S, of the bond is given by S = d dr = d2 U dr 2 When the stretching is small, S is constant and equal to S 0 = d2 U dr 2 r=r 0 The bond behaves in linear elastic manner this is the physical origin of Hooke s law. the shape of the force/distance curve that we get from a typical energy/distance curve 30/34
S 0 = d2 U dr 2 r=r 0 S 0 is sometimes called the spring constant of the bond The force between a pair of atoms, stretched apart to a distance r (r r 0 ), is = S 0 r r 0 The total force exerted across unit area, if the two planes are pulled apart a distance (r r 0 ) σ = NS 0 r r 0 = 1 r 0 2 S 0 r r 0 = S 0 r 0 ε n N = number of bonds/unit area r 0 2 = average area per atom ε n = strain = r r 0 /r 0 Young s modulus is The method of calculating Young s modulus from the stiffnesses of individual bonds E = σ ε n = S 0 r 0 S 0 can be calculated from the theoretically derived U(r) curves 31/34 Table 6.1: Hierarchy of Bond Stiffnesses Bond Type S 0 (N m 1 ) E (GPa); from Equation (with r 0 = 2.5x10 10 m) Covalent, e.g., C C 50 180 200 1000 Metallic, e.g., Cu Cu 15 75 60 300 Ionic, e.g., Na Cl 8 24 32 96 H-bond, e.g., H2O H2O 2 3 8 12 Van der Waals, e.g., polymers 0.5 1 2 4 or metals and ceramics, the values of E we calculate are about right: the bond-stretching idea explains the stiffness of these solids. There exists a whole range of polymers and rubbers that have moduli that are lower by up to a factor of 100 than the lowest we have calculated. Why is this? What determines the moduli of these floppy polymers if it is not the springs between the atoms?
All polymers, if really solid, should have moduli above the lowest level we have calculated about 2 GPa since they are held together partly by Van der Waals and partly by covalent bonds. If you take ordinary rubber and cool it down in liquid nitrogen, it becomes stiff its modulus rises rather suddenly from around 10 2 GPa to a proper value of 2 GPa. But if you warm it up again, its modulus drops back to 10 2 GPa The covalent bonds, with occasional cross-links, are very stiff, but they contribute very little to the overall modulus because when you load the structure it is the flabby Van der Waals bonds between the chains that stretch, and it is these that determine the modulus. At low T, rubber has a proper modulus of a few GPa and behaves like a true solid. As rubber warms up above the glass transition temperature, Tg, the Van der Waals bonds melt. Rubber remains solid because of the cross-links, which form a sort of skeleton, but becomes leathery and rubbery. But when you load it, the chains slide over each other in places where there are no cross-linking bonds. This, of course, gives extra strain, and the modulus goes down. Young s modulus increases with increasing density of covalent cross-links in polymers, including rubbers above, Tg. Below T g, the modulus of rubbers increases markedly because the Van der Waals bonds kick in. Above T g they melt, and the modulus drops. Next Class 10 Yield Strength, Tensile Strength and Ductility