INTRODUCTION Mechanical Properties of Materials Many materials, when in service, are subjected to forces or loads, it is necessary to know the characteristics of the material and to design the member from which it is made such that any resulting deformation will not be excessive and fracture will not occur. The mechanical behaviour of a material reflects the relationship between its response or deformation to an applied load or force. Important mechanical properties are strength, hardness, ductility and stiffness. It is possible for the load to be tensile, compressive or shear and its magnitude may be constant with time or it may fluctuate continuously. Application time may be only a fraction of a second or it may extend over a period of many years. Service temperature may be an important factor. The role of structural engineers is to determine stresses and stress distributions within members that are subjected to well-defined loads. This may be accomplished by experimental testing techniques and/or by theoretical and mathematical stress analyses. These topics are treated in traditional stress analysis and strength of materials texts. Materials are frequently chosen for structural applications because they have desirable combinations of mechanical characteristics. Mechanical Properties of a material describes the response of the material to an applied force or torque. Stress:- The internally developed forces per unit area of a material due to the application of external force. Stress = = = Units N/m 2 Three types of Stress 1. Tensile stress: It occurs when equal and opposite forces are directed away from each other. 2. Compressive stress: It occurs when equal and opposite forces are directed toward each 1
other. 3. Shearing Stress: Force per unit area applied parallel to the surface of a body trying to displace the upper layers of the body. Shearing Stress or Tangential Stress = Force/Area Strain:- Fractional change in the dimensions of a material due to the application of external force. It has no unit. Strain = Three types of Strain: 1. Longitudinal Strain:- It is defined as the increase in length (Δl) per unit original length (l) when deformed by the external force. Longitudinal Strain = 2. Volumetric Strain:-It is defined as change in volume (ΔV) per unit original volume (v), when deformed by external force. Volume Strain = 3. Shear strain: It is defined as the tangent of the strain angle, θ Shear Strain = Tan θ. Hookes Law: Within the proportional limit, the strain produced in a body is directly proportional to the applied stress Stress α Strain, 2
i. e., Stress = (E) Strain. where E is proportionality constant and is called Modulus of Elasticity. Units N/m 2. Different Moduli of elasticity E: Basing on three different types of stresses and strains there are 3 types of modulii exists. 1. Young s modulus (Y): It is the ratio of tensile stress to tensile strain. Young ' s modulus = longitudinal stress Y = F / A ΔL / L Units N/m 2 = FL A ΔL longitudinal strain 2. Rigidity modulus / Shear modulus (S): It is the ratio of shear stress to shear strain. The shear modulus S is defined as the ratio of the shearing stress F/A to the shearing strain φ : S = Units N/m 2 3
Strain ϕ = 3. Bulk modulus (B): It is the ratio of volume stress to volume strain. The bulk modulus is negative because of decrease in V. IMPORTANT MECHANICAL PROPERTIES: Elasticity: The property of a material by virtue of which the material regains its original dimensions after the stress is removed. Within the elastic limit the material obeys Hooke s law, which states that stress is directly proportional to strain. Elastic deformation: Reversible: when the stress is removed, the material returns to the dimensions it had before the loading. Usually strains are small (except for the case of some plastics, e.g. rubber). 4
Plasticity: The property of the material where in the material has a permanent deformation after the stress is removed. Plastic deformation Irreversible: when the stress is removed, the material does not return to its original dimensions. In tensile tests, if the deformation is elastic, the stress-strain relationship is called Hooke's law: σ = E ε 5
Ductility: The ease with which the material can be drawn in the form of thin wires. This property exhibited by metals and alloys. Strength: The ability of a material to sustain loads without failure. It refers to the stress up to which the body obeys Hooke s law. It is load/unit area. Types of strengths: Yield strength: The strength beyond which it exhibits plasticity. Ultimate tensile strength: The strength at which the material breaks or fractures. Hardness: The ability of the material to withstand plastic deformation or indentation produced in the material. Creep: Changes that occur in the dimensions of the material as a function of time, when subjected to constant load. Fatigue: The changes in the mechanical properties of the material under the action a cyclical or repeated stresses. Fracture: The breakage of a material into separate parts under the action of stress. Brittleness: The property by virtue of which material undergoes fracture immediately after crossing the elastic limit. Toughness: The energy absorbed by the material due to application of external force up to fracture. Relationship between Engineering stress and true stress, Engineering strain and true strain: Engineering stress (σ E ) = where F is the force and A0 the original area of cross section Engineering strain (ε E ) = where L is the gauge length under the force F, and L0 is the original gauge length. 6
True Stress (σt) = Instantaneous load / Instantaneous cross-sectional area = True strain: = = ln = = ln Where dl is the infinitesimal elongation, li is the instantaneous length and lo is the original length. Assuming that the density of the material does not change, the volume has to remain the same. Hence we have the relation, l o A o = l i A i = ---------(1) ln = ln = True strain From eq. (1) = = = 1 + ------ (2) ln = ln (1 + ) = ln (1 + ) This is the relation between true stress and true strain. And From eq. (2) = -----(3) and since = ------(4) From eq.n s (3) & (4) = (1 + ) = (1 + ) 7
Where and are the engineering stress and strain values at a particular load. The distinction between the plots of true stress-strain and engineering stress-strain curves is shown in below figure. Note that the true stress continuously increases up to the fracture point while the engineering stress slightly decreases near the fracture point. Figure: Comparison of engineering stress-engineering strain curve and true stress- true strain curve. Poisson s Ratio: Virtually all common materials undergo a transverse contraction when stretched in one direction and a transverse expansion when compressed. The extension of the wire is not the only change that takes place under the action of a tensile stress. There are also dimensional changes in the perpendicular direction. The diameter of the wire decreases with tensile stress and increases with compressive stress. If we consider a cube, with its edges parallel to the x, y and z-axes, the tensile stress which produces an increase in length along l x, leads to a decrease in l y and l z. This behaviour is quantitatively described by the Poisson s ratio (ν) which is defined as the ratio of lateral strain to the longitudinal strain. 8
The minus sign indicates that the change in length in the directions perpendicular to the applied stress is of the opposite sense to that in the direction parallel to the stress i.e. the tensile stress which produces an increase in length along l x leads to a decrease in length along l y and l z. Either transverse strain or longitudional strain is negative then ν is positive. Since the value of the Poisson s ratio (ν) is always found to be less than 0.5, the volume of the solid always increases under a tensile stress. Stress Strain behaviour of Ductile Materials: The below figure shows the typical relation between stress and strain exhibited by ductile materials. For a substantial range of applied stress, the stress verses strain relation is linear. This linearity between stress and strain was discovered by Hooke and is referred to as the Hooke s law. The slope of this curve gives the particular modulus of elasticity. For ductile materials, the point A marks the limit of Hooke s law, where the stress is linearly proportional to strain. Region AB corresponds to the nonlinear relation between stress and strain. The behaviour is still within the elastic limits. Strain is reversible up to point B. After point B material starts to deform plastically. From an atomic point of view, plastic deformation corresponds to breaking of bonds with 9
adjacent atoms and formation of new bonds. At this point stress is removed the atoms do not return to their original positions. In the figure point C is called upper yield point and D is called lower yield point. Usually yield strength is defined for 0.002 amount of permanent deformation or the stress corresponding to the intersection of this line and the stress-strain curve as it bends over in the plastic region is defined as the yield strength. At the upper yield point, plastic deformation is initiated with an actual decrease in stress. Continued deformation fluctuates slightly about some constant stress value, called the lower yield point, stress subsequently rises with increasing strain. In this process raising the yield strength by permanently straining the material is called strain hardening. After yielding, the stress necessary to continue plastic deformation in metals increases to a maximum till E. This is termed as the tensile strength. From this point non uniform plastic deformation takes place. This process of the material is called necking. Point F marks the fracture point. The stress corresponding to point F is called the ultimate strength or breaking strength. The ability of a material to sustain loads without failure is called strength. It refers to the stress up to which the body obeys Hooke s law. It is load/unit area. It is of two types: 1. Yeild Strength: A yield strength or yield point of a material is defined as stress at which a material begins to deform plastically. 2. Ultimate Strength: It is the maximum stress that a material can withstand while being stretched or pulled before failing or breaking. Work hardening: It also known as strain hardening or cold working. It is strengthening of a metal by plastic deformation. This strengthening occurs because of dislocation movements and dislocation generation within the crystal structure of material. Ductile materials exhibit plastic deformation while brittle materials do not. Brittle materials fracture immediately after the elastic limit is reached. 10
Figure: Stress - Strain curve for ductile materials. Tensile Properties: Ductility: Ductility is a measure of the deformation at fracture. It is defined by percent of elongation %EL = (l f - l 0 )X100/ l 0 The yield and tensile strengths and modulus of elasticity decrease with increasing temperature, ductility increases with temperature. Percentage reduction in area %RA = ( A 0 A f ) x100/ A 0 Effect of temperature: The yield and tensile strengths and modulus of elasticity decrease with increasing temperature, ductility increases with temperature. Toughness: The ability to absorb energy up to fracture = the total area under the strain-stress curve up to fracture. Units: the energy per unit volume, e.g. J/m 3. Figure 3: Stress-strain curve for ductile and brittle materials. 11
Hardness test methods: HARDNESS Hardness is defined as "Resistance of metal to plastic deformation, usually by indentation. However, the term may also refer to stiffness or temper or to resistance to scratching, abrasion, or cutting. The greater the hardness of the metal, the greater resistance it has to deformation. In metallurgy hardness is defined as the ability of a material to resist plastic deformation. This is the usual type of hardness test, in which a pointed or rounded indenter is pressed into a surface under a substantially static load. 2. HARDNESS MEASUREMENT Hardness measurement can be defined as macro-, micro- or nano- scale according to the forces applied and displacements obtained. Microhardness is the hardness of a material as determined by forcing an indenter such as a Vick-ers or Knoop indenter into the surface of the material under 15 to 1000 gf load; usually, the in-dentations are so small that they must be measured with a microscope. Micro-indenters works by pressing a tip into a sample and continuously measuring: applied load, penetration depth and cycle time. HARDNESS TESTS Hardness tests may be classified as (i) (ii) (iii) Scratch hardness test (Mohs hardness scale) Indentation test (Brinell, Rockwell and Vickers hardness tests) Rebound (Dynamic hardness) test (Shore scleroscope test) Most of the hardness tests are based on resistance to indentation. The indentation tests consists of forcing a indentor into the specimen and measuring the effects. Brinell Hardness Test Brinell hardness is determined by forcing a hard steel or carbide sphere of a specified diameter 12
under a specified load into the surface of a material and measuring the diameter of the indentation left after the test. The Brinell hardness number, or simply the Brinell number, is obtained by dividing the load used, in kilograms, by the actual surface area of the indentation, in square millimeters. The result is a pressure measurement, but the units are rarely stated. The Brinell hardness test uses a desk top machine to press a 10mm diameter, hardened steel ball into the surface of the test specimen. The machine applies a load of 500 kilograms for soft metals such as copper, brass and thin stock. A 1500 kilogram load is used for aluminum castings, and a 3000 kilogram load is used for materials such as iron and steel. The load is usually applied for 10 to 15 seconds. After the impression is made, a measurement of the diameter of the resulting round impression is taken. It is measured to plus or minus 0.05mm using a low-magnification portable microscope. The hardness is calculated by dividing the load by the area of the curved surface of the indention. The Brinell hardness test was one of the most widely used hardness tests during World War II. For measuring armour plate hardness the test is usually conducted by pressing a tungsten carbide sphere 10mm in diameter into the test surface for 10 seconds with a load of 3,000kg, then measuring the diameter of the resulting depression. The BHN is calculated according to the following formula: Figure: Brinell's hardness test method. 13
where BHN = the Brinell hardness number F = the imposed load in kg D = the diameter of the spherical indenter in mm Di = diameter of the resulting indenter impression in mm Several BHN tests are usually carried out over an area of armour plate. On a typical plate each test would result in a slightly different number. This is due not only to minor variations in quality of the armour plate (even homogenous armour is not absolutely uniform) but also because the test relies on careful measurement of the diameter of the depression. Small errors in this measurement will lead to small variations in BHN values. As a result, BHN is usually quoted as a range of values (e.g. 210 to 245, or 210-245) rather than as a single value. Rockwell Hardness Test Rockwell hardness tests use different loads and penetrators (indentors) and depth of indentation is calibrated on a dial to a hardness number. To eliminate the effect of slackness and surface roughness, the indentor is loaded with minor load (10kg) and the major load (100kg for soft material and 140 kg for hard material) is applied. Depth of indentation due to major load is used as a measure of hardness. Hardened steel ball with 100 kg total load is used for B-scale; while diamond tipped cone of 120 o angle and 0.2 mm tip radius called Brale with 150 kg total load is used for c-scale. A minor load of 10 kg is first applied, which causes an initial penetration and holds the indenter in place. Then, the dial is set to zero and the major load is applied. Upon removal of the major load, the depth reading is taken while the minor load is still on. The hardness number may then be read directly from the scale. This test measures the difference in depth caused by two different forces, using a dial gauge. Using standard hardness conversion tables, the Rockwell hardness value is determined for the load applied, the diameter of the indentor, and the indentation depth. 14
Rockwell hardness number on B-scale, = 130 - And on C-scale = 100 - Where e = depth of indentation corresponding to major load. The Rockwell hardness tester to measure the hardness of metal measures resistance to penetration like the Brinell test, but in the Rockwell case, the depth of the impression is measured rather than the diametric area. With the Rockwell tester, the hardness is indicated directly on the scale attached to the machine. This dial like scale is really a depth gauge, graduated in special units. The Rockwell hardness test is the most used and versatile of the hardness tests. Vickers Hardness Test Figure: Rock-well hardness test method It is the standard method for measuring the hardness of metals, particularly those with extremely hard surfaces: the surface is subjected to a standard pressure for a standard length of time by means of a pyramid-shaped diamond. The diagonal of the resulting indention is measured under a microscope and the Vickers Hardness value read from a conversion table. Vickers hardness is a measure of the hardness of a material, calculated from the size of an impression produced under load by a pyramid-shaped diamond indenter. Devised in the 1920s by engineers at Vickers, Ltd., in the United Kingdom, the diamond pyramid hardness test, as it also became known, permitted the establishment of a continuous scale of comparable numbers that accurately reflected the wide range of hardnesses found in steels. 15
The indenter employed in the Vickers test is a square-based pyramid whose opposite sides meet at the apex at an angle of 136º. The diamond is pressed into the surface of the material at loads ranging up to approximately 120 kilograms-force, and the size of the impression (usually no more than 0.5 mm) is measured with the aid of a calibrated microscope. The Vickers number (HV) is calculated using the following formula: HV = 1.854 with F being the applied load (measured in kilograms force) and the area of the indentation (measured in square millimetres). The applied load is usually specified when HV is cited. The Vickers machine uses a penetrator that is square in shape, but tipped on one corner so it has the appear ance of a playing card "diamond". The Vickers indenter is a 136 degrees square-based diamond cone, the diamond material of the indenter has an advantage over other indenters because it does not deform over time and use. The impression left by the Vickers penetrator is a dark square on a light background. The Vickers impression is more easily "read" for area size than the circular impression of the Brinell method. Like the Brinell test, the Vickers number is determined by dividing the load by the surface area of the indentation (H = P/A). The load varies from 1 to 120 kilo-grams. To perform the Vickers test, the specimen is placed on an anvil that has a screw threaded base. The anvil is turned raising it by the screw threads until it is close to the point of the inden - ter. With start lever activated, the load is slowly applied to the indenter. The load is released and the anvil with the specimen is lowered. The operation of applying and removing the load is con-trolled automatically. Vickers hardness test are that extremely accurate readings can be taken, and just one type of indenter is used for all types of metals and surface treatments. Although thoroughly adaptable and very precise for testing the softest and hardest of materials, under varying loads, the Vickers machine is a floor standing unit that is rather more expensive than the Brinell or Rockwell machines. 16
Figure: Vickers hardness test method. 17
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Problems 1. A piece of copper originally 305 mm (12 in.) long is pulled in tension with a stress of 276 MPa (40,000 psi). If the deformation is entirely elastic, what will be the resultant elongation? 2. A specimen of copper having a rectangular cross section 15.2 mm X 19.1 mm (0.60 in.x 0.75in.) is pulled in tension with 44,500 N (10,000 lb ) force, producing only elastic deformation. Calculate the resulting strain. 3. A cylindrical specimen of nickel alloy having an elastic modulus of 207 Gpa and an original cross diameter of 10.2 mm will experience only elastic deformation when a tensile load of 8900 N is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is 0.25 mm. 4. A piece of copper originally 305 mm long is pulled in a tension with a stress of 276 M Pa. If the deformation is entirely elastic, what will be resultant elongation? Given 1 M Pa is 10 6 N/m 2. 5. A specimen of copper having a rectangular cross section 15.2 mm 19.1 mm (0.60 in.0.75 in.) is pulled in tension with 44,500 N (10,000 lb ) force, producing only elastic deformation. Calculate the resulting strain. 6. A cylindrical specimen of a nickel alloy having an elastic modulus of 207 GPa (psi) and an original diameter of 10.2 mm (0.40 in.) will experience only elastic deformation when a tensile load of 8900 N (2000 lb) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is 0.25 mm (0.010 in.). 7. An aluminum bar 125 mm (5.0 in.) long and having a square cross section 16.5 mm (0.65 in) on an edge is pulled in tension with a load of 66,700 N (15,000 lb), and experiences an elongation of 0.43 mm (in.). Assuming that the deformation is entirely elastic, calculate the modulus of elasticity of the aluminum. 8. Consider a cylindrical nickel wire 2.0 mm (0.08 in.) in diameter and mm (1200 in.) long. Calculate its elongation when a load of300 N (67 lb) is applied. Assume that the deformations totally elastic. 9. For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103 GPa( 15X10 6 psi). (a) What is the maximum load that may be applied to a specimen with a cross-sectional area of 130 mm 2 (0.2 in 2 ) without plastic deformation? (b) If the original specimen length is 76 mm (3.0 in.), what is the maximum length to which it may be stretched without causing plastic deformation? 10. A cylindrical rod of steel ( E=207GPa, 30X10 6 psi) having a yield strength of 310 MPa (45,000 psi) is to be subjected to a load of 11,100 N (2500 lb ). If the length of the rod is 500 mm (20.0 in.), what must be the diameter to allow an elongation of 0.38 mm (0.015 in.)? 19