Chap 6 Mech Prop

Chap 6 Mech Prop - 4 I Chapter 6 Mechanical Properties...

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Unformatted text preview: 4-. I Chapter 6: Mechanical Properties OUTLINE 6.1 Mechanical Properties of Metals 6.1.1 Nature of the Force or Stress Applied a) Tensile b) Compression c) Shear d) Torsion 6.1.2 Stress States 6.1.3 Brief Considerations of Time and Temperature 6.1.4 Elastic Defamation a) Stress-Strain Relations a1) Elastic Modulus a2) Shear Modulus a3) Bulk Modulus a4) Poisson’s ratio b) Anelasticity 6.1.5 Plastic Deformation a) Yielding and Yield Strength b) Tensile Strength c) Fracture Stress d) Ductility e) Toughness 0 Resilience 6.1 Mechanical Properties of Metals In the previous sections, we have learned most of the fundamentals necessary to understand the mechanical behavior or materials. This fundamental knowledge will allow us to understand why, for example, an aluminum alloy is used in an airplane wing, why is steel used in an automobile axle, why does a bar of steel fractures after being strained 75% at 18°C but at -253°C requires only 1%, why are stiff materials used in the construction of rocket motors, why are sharp corners avoided in the design of aircrafts and automobiles. The answer to these questions reflects the relationship between the response of a material to a deformation or, an applied load of force. This leads us to the topic of “mechanical properties of materials” which we will cover in the next lectures. 6.1.1 Nature of the F area or Stress Applied In the subsequent discussion we will assume that the applied load is static, changes slowly with time and is applied uniformly across a cross section. 103 a) Tensile Load One of the common ways of applying a load is to apply a tensile load. As shown in Figure 124, the tensile load produces an elongation and positive linear strain. In engineering, the materials can be tested in tensile machines to determine its properties under those conditions. F i I I I I I I I ‘a I I I I l i do Figure 124: Schematic illustration of how a tensile load produces an elongation and positive linear strain. 0-. f——-—-—__.—-—_,. Normally, a standard test specimen, shown in Figure 125, is mounted into the holding grips of a testing machine (Figure 126). This machine is designed to pull the specimen at a constant rate and to measure continuously the applied load and the resulting elongatious. "' Q? 7 Diameter _._i. Figure 125: A standard tensile specimen with circular cross section. Reduced section I‘—"2i"‘—’l Diameter — - —-|— 0.505‘ -— ~ =_. l~—2'—-—-l 3 Gauge length 3' Radius lea: all 5"“ using :35: mm": Figure 126: Apparatus used to conduct stress-strain tests. The load cell measures the applied load and the extensometer measure the strain. 104 The test can be recorded and presented in a plot “Load versus Elongation”. However, because the load will depend on the specimen size, the load and the elongation are nomialized to the following parameters 1) Engineering stress F o:— A where F is the instantaneous load and A is original cross-section area. The engineering stress has unit of MPa or N/m’. 2) Engineering strain where his the instantaneous length and [0 is the initial length. The engineering strain is dimensionless. b) Compressive Load In the case of a compressive applied load, the specimen will contract in the direction of the load (Figure 127). The same equations are used for engineering stress and strain but now, both 0' and e are taken as negative. F _.| Figure 127: Schematic illustration of how a compressive load produces a contraction and negative linear strain. c) Shear Load In other cases, the applied load is of the shear type as shown in Figure 128. In this case, the shear stress is given by F t:— A 105 where F is the shear force imposed parallel to the upper and lower faces and A is the area. The shear strain is given by r=Tauo where 6 is the shear angle Figure 128: Representation of a shear strain. (1) Torsion Load Torsion is a variant of pure shear, wherein a material is twisted (Figure 129). Thus, the torsion stress can be given by T t =— Iorxion A where T is torsional force and A is the area. The torsion strain can be given as = tanQ Ywm’on where (p is the angle of torsion. Torsional forces produce a rotational motion about the longitudinal axis of one end relative to the other end. Some examples of applications where this type of load is present include drive shafls and twist drills Figure 129: Torsional defamation produced by an applied torque T. 106 6.1.2 Stress States The discussion on the nature of applied stresses considers macroscopic bodies. However when you are deforming a particular material, some particular crystallographic planes will be subjected to very high stresses, others to moderate stresses and there will be planes which will not acted by an applied stress. In addition, we shall see that some planes will be subjected to shear stresses, whereas others will be subjected to tensile or compressive stresses. These aspects are important because the magnitude of the shear stresses induced by the applied load will be crucial for the deformation of the material, whereas the conditions for fracture ofien depend on the magnitude of the tensile stresses. Consequently, under a particular loading condition, we need to understand the state of stresses in all planes of the crystal. Let us start by considering a particular cylinder subjected to a tensile stress 6, as shown in Figure 130. In plane A’, the applied stress is no longer a pure tensile stress. It actually consists of a tensile stress 0’, which acts normal to the slip plane and a shear stress t’ that acts parallel to the plane. Slip Direction Figure 130: Schematic representation showing normal (0") and shear stresses (1:’) that act on a plane oriented at an angle 6 relative to a plane perpendicular to the tensile axis. 107 Both the tensile stress and the shear stress can be calculated according to the equations a'=£=m=acosz¢ and z" =—P—.—=—I-D—c—O£=O'COS(0COSH A' Aleos¢ A A/cos¢ where P’, P" and A’ can be obtained from Figure 131. P P3 Figure 131: Schematic illustration of the geometrical relationships between the applied load (P), the acting normal load (1”) and the acting shear load (P”); and also between the direction normal to the slip plane (N A), the radius of the plane perpendicular to the tensile axis (rA) and the radius of the slip plane (rAv). Let us now try to compute the maximum value of 1’. First let us pay attention to the fact that the tensile axis direction, the normal to the slip plane and the slip direction may not be coplanar and thus 6 + G can be different than 90 degrees. However, let us assume for simplification that all these directions are coplanar. In this case, we can write 6 + Z = 90 ,which means that 0 = 325-- 6, Thus we can write t = ocos Ecosg- —®J = a [email protected] which has a maximum at o = «:14. Thus, 1: m =o(o.7o7)z 5% We now realize that the maximum resolved shear stress occurs on those planes that are 45° to the tensile axis in a direction that is coplanar with the tensile axis and the plane normal. Now, we can ask: What is the value of the stress acting on the planes that are perpendicular to the tensile axis? What about the stress on the planes parallel to the tensile axis? If you go back and look at the above equation we can see that Q = 0 => 1: = O 108 and @=% :9 1: =0. In other words, there are no shear stresses acting on the planes perpendicular and parallel to the tensile axis. In general, materials are subjected to stress states considerably more complex than uniaxial tension. In most of the cases materials are subjected to both normal and shear stresses. However, regardless of the complexity of the stresses applied, we can always define the state of stresses by 3 normal stresses referred to as “principal stresses” and represented by symbols (:1, 0'2, and 03. By convention, o" 202 20,. For uniaxial tension or compression cl=o; cz=0; 63:0, and for hydrostatic compression or tension O'I=O'2=O’3, The maximum shear stress can be given by 01—0: Let me state now, and we will understand this later, that permanent plastic deformation occurs when rm, reaches a value greater than a critical value ryidd. ryicld is called the yield strength and it is actually related with the motion of dislocations. Thus, plastic flow occurs when rm 2 ryiem . Based on this discussion, we can determine the maximum load that can be applied to a structure without causing defamation if we know the relations between load and principal stresses. 6.1.3 Brief Considerations of Time and Temperature In order for you to appreciate the broader concept of an applied load, it is appropriate to present a few examples where time and temperature are of importance. One example is the application of a load, which is constant with time (ex: gas cylinder). This situation can change dramatically if the load fluctuates continuously with time, as for example with the use of an artificial bone. In addition, the load can be applied during a short-time, as for an automobile impact, or can be applied during a long period, as for example in building structures. In other applications, the load is applied at high temperatures, as for the case of turbine blades, or applied at low temperatures, as for structures used in space. 6.1.4 Elastic Deformation a) Stress-Strain Relations a1) Elastic modulus When solid materials are subjected to small stresses they usually respond in an elastic fashion. This means, of course, that the strain produced by the stress is reversible (goes back to zero when the stress is removed). In this regime, stress and strain are proportional to each other through the relation 109 i=5 a where E is the elastic modulus. This equation is known as Hooke’s law. The elastic modulus is given in units of MPa or N/m2. Graphically we can represent Hooke's law by a straight line in the 0 vs 6 diagram (Figure 132). Unload Slope u modulus ol elasticity Stress O 0 Strain Figure 132: Stress~strain diagram showing linear elastic deformation. The modulus of elasticity can be thought of as stiffness, or, the materials resistance to elastic deformation. Some values of E for different materials can be seen in the Table —m— Elastic Modulus GPa -_ _IE_ ‘22;— ' Thus, the modulus of elasticity is intimately related to the atomic bonding of the material. In other words, the modulus of elasticity is a measure of the resistance to separate the atoms apart. This is shown in Figure 133, where the bond energy as a fitnction of interatomic distance is presented. Thus, a material with a higher elastic modulus will be represented by a steeper function in the diagram of Figure 133. Wit-M I Figure 133: Force vs interatomic separation for weakly and strongly bonded atoms. This force is intimately related with the modulus of elasticity exhibited by the material. es, We have assumed so far linear elastic deformation. However some materials (concrete and many polymers) do exhibit non-linear behavior. In this case, the elastic modulus is taken as the tangent to the curve Ao/Ae at a particular stress level (Figure 134). In addition, it is also important to emphasize that the elastic modulus is related to normal stresses. When dealing with shear stresses and shear strains we should consider the shear modulus. é“. - rm Mow! tn v,» , g _ Smut medulla ’ 3' [bet-m min and 0.) 5mm - Figure 134: Stress-Strain diagram showing non-linear elastic behavior. a2) Shear Modulus In pure elastic shear, the Hooke’s Law describes the relation between shear stress and shear strain, given by: E- = G (Pa or N/mz) 'Y where G is the shear modulus. Remember that a pure shear does not produce any change in volume, but does produce a change in shape. a3) Bulk Modulus Let us think about another possible situation, in which there is not a change in shape but only a change in volume. This situation is called pure dilatation. An example of this is the hydrostatic deformation of materials under a body of water. In this case, the stress can be related to the strain by the expression a = BA—If where B is called the bulk modulus. 111 a4) Poisson’s ratio During the application of a tensile or compressive stress along a particular direction, we should remember that the elongation or contraction along one axis will be followed by a change of dimensions in the two perpendicular transverse directions. In other words, as a result of elongation in the z direction (6,) there will be contractions in the x (8,) and y (8,) directions. The ratio of the induced transverse strains to the axial strain is called the Poisson’s ratio and represented by the symbol v. Thus, we can write 8 < V =—§‘-=— e, e 1 So, what does it mean to say that a material has a negative Poisson’s ratio? It means that the elongation in the z direction will produce also an elongation on the transverse directions, instead of a contraction. Some Poisson’s ratio values for different materials can be seen in the following table. —m_ “ra— ' __ The elastic parameters discussed so far can be related through the following expressions, if we assume the material to be elastically isotropic E = ZG(l+v) E = 33(1-2v) c) Anelasticity Up to now, we have assumed that elastic deformation is time independent, that is, when you apply a stress to the material, there will be an elastic strain which will remain until the stress is removed. In most materials, however, elastic deformation will continue alter the application of stress and upon load release, because some finite time is required for complete recovery. This time-dependent behavior is called anelasticity. For metals it is small and often neglected; for polymers, however, is quite important and usually an area of study called viscoelastic behavior. 112 6.1.5 Plastic Deformation a) Yielding and yield strength For most metallic materials, elastic deformation occurs only up to strains of about 0.005. Beyond this level of strain, the material starts to deform plastically. In practical terms, structures are designed to ensure that only elastic deformation will result when an applied stress is present. Therefore, it is crucial to know at which stress level plastic deformation begins. This is called yielding and represents the onset of plastic deformation. The point of yielding may be determined as the initial departure from linearity of the stress-strain curve, which is normally called the proportional limit, as indicated by point P in Figure 135. Because the yielding is many times difficult to determine (Al is a good example), a convention was established, wherein a straight line is constructed parallel to the elastic portion of the stress-strain curve at some specified strain offset, usually 0.002. The stress at this point is called yield strength and the strain is called yield strain (Figure 135). For materials exhibiting a non-linear elastic region this method is not possible and typically the practice is to define the yield at a particular strain, e. g. 0.005. (mm Plat-c ~o—-*—-O I I -J L— Sun-n 0.002 Figure 135: Stress-Strain behavior and the proportional limit P, which separates the elastic regime from the plastic regime. c) Tensile strength After yielding, the stress necessary to continue plastic deformation in metals increases to a maximum (Figure 136). This stress is called tensile strength and is the maximum stress that can be sustained by a structure in tension. An interesting observation of Figure 136 shows that before reaching point M, we see that the stress required in producing an increase in strain increases. Why is that so? How is it possible that a material can get stronger despite the fact that it is subjected to an increasing applied load? The reason is called “strain hardening" (to be discussed later), in which dislocations interact between each other. The more a material is plastically deformed, the more the dislocations interact and the more difficult it becomes to deform it. So what happens in point M? Why there is all of sudden a decrease in stress beyond point M? The reason is that before 113 point M as plastic deformation increases, the cross-sectional decreases but due to strain hardening, the load increases. All the defamation is uniform throughout the narrow region of the tensile specimen. However, eventually, an elongation is reached, such that the increase in loading due to “strain hardening" is less than the decrease in load capacity carried by the material because of a decreasing load~bearing area. So the specimen cannot withstand further increases in load. In short, it is creating a plastic instability called “necking” where any additional elongation will be confined to this region. Because deformation is now very localized, cross-section decreases rapidly and the load starts to decrease even though the material in the neck continues to strain harden. In the following table, values of tensile strength for different materials can be seen. Stress 5mm Figure 136: Engineering stress-strain behavior up to fracture (point P). The point M is the point at which the stress is maximum. (1) Fracture stress The fracture stress is simply the stress at fracture and is represented by point F in Figure 136. e) Ductility The ductility of a material measures the strain required to cause fracture. Therefore a material is called ductile if it exhibits a large plastic deformation upon fracture, whereas if it exhibits little or no plastic deformation upon fracture, the material will be called brittle. This difference in behavior can be depicted in Figure 138, for which stress-strain curves for a ductile a brittle material can be seen. 114 Brittle Ductile Stress Q ——--—-—-— Strain Figure 138: Differences in stress-strain behavior for ductile and brittle materials. Ductilin can be expressed as either percent elongation (%EL) or percent reduction in area (%RA) according to the following expressions 3 —e %EL= ’ °*100 0 A —A %RA=°—£*ioo D Ductility values for different materials can be seen in the following table. Ductility is of crucial importance in the design of materials because, even when wrong stress calculations are made for a particular application, a ductile material will deform before fracture, and thus enables some preventive measures. On the other hand, a brittle material will fracture without any kind of warning. —I_ __ __ t) Toughness The toughness of a material measures the work/unit volume required to cause fracture. For a simple tensile test this work can be calculated as follows Work = J F - are Work/uni! volume = Iidf A£' = fade 115 In other words, this integral represents the area under the stress-strain curve, up to fi'acture. Generally speaking, tough materials have moderately high strengths and ductility, whereas materials with high strengths and low ductility have low toughness. g) Resilience The resilience of a material measures the ability of the material to absorb energy when deformed elastically and recovered. For a simple tensile test this work can be calculated as follows Work/ unit volume = fads Assuming a linear elastic region, as shown in Figure 139, we can write 1 Wk =3°y8r We can now insert 5,, = ay / E from Hooke’s law and obtain Thus, we see that materials having high yield strengths and low modulus of elasticity are resilient materials, such as for spring applications. 116 ...
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Chap 6 Mech Prop - 4 I Chapter 6 Mechanical Properties...

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