Engineering Materials

Transcription

1 Engineering Materials Mechanical Properties of Engineering Materials Mechanical testing of engineering materials may be carried out for a number of reasons: The tests may simulate the service conditions of a material, so that the test results may be used to predict its service performance. Mechanical testing may also be conducted in order to provide engineering design data, as well as acceptability, the main purpose of which is to check whether the material meets the specification. In the USA, the American Society for Testing Materials (ASTM) published standard specifications and methods of testing which are updated every three years. In the UK, the British Standards Institution (BSI) published an annual catalogue of all BSI Standards, and agreed European Standards (EN series). All of these organizations issue publications relating to the selection of test pieces and the conducting of mechanical tests. Tensile Test for Ductile Engineering Materials The tensile test is popular since the properties obtained could be applied to design different components. The tensile test measures the resistance of a material to a static or slowly applied force. A standard tensile specimen is shown in Figure (1). Normally, the cross section is circular, but rectangular specimens are also used. The standard diameter is approximately 12.8 mm (0.5 in.), Gauge length is used in ductility computations, the standard value is 50 mm (2.0 in.). The specimen is mounted by its ends into the holding grips of the testing apparatus Figure (2). The tensile testing machine is designed to elongate the specimen at a constant rate, and to continuously and simultaneously measure the instantaneous applied load and the 1

2 resulting elongations (using an extensometer). A stress strain test typically takes several minutes to perform and is destructive; that is, the test specimen is permanently deformed and usually fractured. Figure (1) A standard tensile specimen with circular cross section. Figure (2) Schematic representation of the apparatus used to conduct tensile stress strain tests. The specimen is elongated by the moving crosshead; load cell and extensometer measure, respectively, the magnitude of the applied load and the elongation. 2

3 The output of such a tensile test is recorded (usually on a computer) as load or force versus elongation. These load deformation characteristics are dependent on the specimen size. For example, it will require twice the load to produce the same elongation if the cross-sectional area of the specimen is doubled. To minimize these geometrical factors, load and elongation are normalized to the respective parameters of engineering stress and "engineering strain". "Engineering stress" is defined by the relationship in which F is the instantaneous load applied perpendicular to the specimen cross section, in units of newtons (N) or pounds force (lb.), and (A o ) is the original cross sectional area before any load is applied (m 2 or in. 2 ). The units of engineering stress (referred to subsequently as just stress) are mega pascals, MPa (SI) (where 1 MPa = 10 6 N/m 2 ), and pounds force per square inch, psi (Customary U.S.). "Engineering strain" is defined by the following relationship in which l is the original length before any load is applied, and l i is the instantaneous length. Sometimes the quantity (l i - l ) is denoted as Δl and is the deformation elongation or change in length at some instant, as referenced to the original length. Engineering strain (subsequently called just strain) is unitless, but meters per meter or inches per inch are often used; the value of strain is obviously independent of the unit system. Sometimes strain is also expressed as a percentage, in which the strain value is multiplied by 100. Figure (3) shows the typical stressstrain curve for low carbon steel material. 3

4 Figure (3) Example 1: Consider a cylindrical specimen of a steel alloy 10 mm in diameter and 75 mm long that in pulled in tension. Determine the engineering stress. Also, determine its elongation when a load of N is applied. The engineering strain is Example 2: 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 ( psi). (a) What is the maximum load that may be applied to a specimen with a crosssectional 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? 4

5 Properties Obtained from the Tensile Test A. Elastic Properties 1- Modulus of Elasticity, E: Figure (3) shows the stress strain for low carbon steel. Initially the graph is a straight line and the material obeys تخضع Hooke's law. With low stresses the material spring back completely to its original shape when stresses are removed, the materials being to be elastic. The modulus of elasticity, or Young's modulus (E), is the slope of the stress strain curve in the elastic region. Example: A force of N will cause a 1 cm 1 cm bar of magnesium to stretch from 10 cm to cm. Calculate the modulus of elasticity. The yield point in most ductile materials can be calculated using proof stress; as shown in figure.4. Figure (4) Schematic stress strain diagram showing non-linear elastic behavior. 5

6 Young's modulus (E) is a measure of stiffness جساءة of a component. The stiffness of a material is the ability of a material to resist bending when a strip of material is bent, one surface is tension and the opposite face in compression. A stiff component with a high modulus of elasticity will show much smaller changes in dimensions if the applied is relatively small and therefore, causes only elastic deformation. For example, the modulus of elasticity for polymeric materials (PVC and PE) is as low as 7 MPa (10 3 psi). But may run as high as 4 GPa ( psi) for some of the very stiff polymers (PS, PMMA, and PI). Modulus values for metals are larger the polymers. 2- Modulus of Resilience, Ur: Resilience is the capacity of a material to absorb energy when it is deformed elastically and then, upon unloading, to have this energy recovered. The associated property is the modulus of resilience, Ur which is the strain energy per unit volume required to stress a material from an unloaded state up to the point of yielding. Computationally, the modulus of resilience for a specimen subjected to a uniaxial tension test is just the area under the engineering stress strain curve taken to yielding Figure (5), or: For a linear elastic region: In which ε y is the strain at yielding. 6

7 Figure (5) Schematic representation showing how modulus of resilience (corresponding to the shaded area) is determined from the tensile stress strain behavior of a material. The units of resilience are the product of the units from each of the two axes of the stress strain plot. For SI units, this is joules per cubic meter (J/m 3, equivalent to Pa), whereas with Customary U.S. units it is inch-pounds force per cubic inch (in.- lbf/in. 3, equivalent to psi). Both joules and inch-pounds force are units of energy, and thus this area under the stress strain curve represents energy absorption per unit volume (in cubic meters or cubic inches) of material. Thus, resilient materials are those having high yield strengths and low modulus of elasticity; such alloys would be used in golf ball and spring applications. 3- Poisson's Ratio, υ: When a tensile stress is imposed on a metal specimen, an elastic elongation and accompanying strain ε z result in the direction of the applied stress (arbitrarily taken 7

8 to be the z direction), as indicated in Figure (6). As a result of this elongation, there will be constrictions in the lateral (x and y) directions perpendicular to the applied stress; from these contractions, the compressive strains ε x and ε y may be determined. If the applied stress is uniaxial (only in the z direction), and the material is isotropic, then ε x = ε y. A parameter termed "Poisson s ratio, υ" is defined as the ratio of the lateral and axial strains, or: Figure (6) 8

9 Axial (z) elongation (positive strain) and lateral (x and y) contractions (negative strains) in response to an imposed tensile stress. Solid lines represent dimensions after stress application; dashed lines, before. The negative sign is included in the expression so that will always be positive, since ε x and ε y will always be of opposite sign. Theoretically, Poisson s ratio for isotropic materials should be ; furthermore, the maximum value for (or that value for which there is no net volume change) is For many metals and other alloys, values of Poisson s ratio range between 0.25 and For isotropic materials, shear and elastic modulus are related to each other and to Poisson s ratio according to: E = 2G (1 + υ) In most metals G is about 0.4E; thus, if the value of one modulus is known, the other may be approximated. Example 3: A tensile stress is to be applied along the long axis of a cylindrical brass rod that has a diameter of 10 mm (0.4 in.). Determine the magnitude of the load required to produce a mm (10-4 in.) change in diameter if the deformation is entirely elastic. The Poisons' ratio of brass is (Fig. 7) Example 4: A cylindrical specimen of some metal alloy 10 mm (0.4 in.) in diameter is stressed elastically in tension. A force of N (3370 lb) produces a reduction in specimen diameter of mm ( in.). Compute Poisson s ratio for this material if its elastic modulus is 100 GPa ( psi). (Fig. 7) 9

10 Figure (7) representation of examples 3 and 4. B- Plastic deformation 1- Yield strength, σ y : As we continue to increase the applied stress the material begins to exhibit both elastic and plastic deformations. The material eventually "yields" to the applied stress. The critical stress value needed to initiate plastic deformation is defined as the "elastic limit" of the material. In metallic materials this is usually the stress required for dislocation motion or slip to be initiated. In polymer materials this stress will correspond to disentanglement of polymer molecule chains or sliding past each other. It is therefore desirable to know the stress level at which plastic deformation begins, or where the phenomenon of yielding occurs. For metals that experience this gradual elastic plastic transition, the point of yielding may be determined as 10

11 the initial departure from linearity of the stress strain curve; this is sometimes called the proportional limit. As a consequence, a convention has been established wherein a straight line is constructed parallel to the elastic portion of the stress strain curve at some specified strain offset, usually 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 σ y. 2- Tensile Strength, σ t : The stress obtained at highest applied force is the "Tensile strength, ζ t ", which is the maximum stress on the engineering stress strain curve. In many ductile materials, deformation dose not remain uniform. At some point, one region deforms more than the others and a large local decrease in the cross sectional area occurs, Figure (8). This locally deformed region is called a "necking". This phenomenon is known as "Necking phenomenon". Figure (8) Necking Phenomena 11

12 3- Ductility: Ductility is the measure of the degree of plastic deformation that has been sustained as fracture. Ductility may be expressed: A. Percent Elongation (% EL): The percent elongation (% EL) is the strain at fracture, or Where l f is the fracture length and l is the original gauge length. B. Percent Reduction in Area (% RA): Percent reduction in area is defined as: Where A is the original cross sectional area and A f is the cross sectional area at the point of fracture. Determination of the ductility of materials is important for at least two reasons. First, it indicates to a designer the degree to which a structure will deform plastically before fracture. Second, it specifies the degree of allowable deformation during fabrication operations (bars, rods, wires, plates, I beam, and fibers). Example (1): The following data were collected from a in. in diameter test specimen of magnesium (l o = 2 in.): 12

13 After fracture, the gage length is in. and the diameter is in.. Plot the data and calculate (a) the 0.2% offset yield strength, (b) the tensile strength, (c) the modulus of elasticity, (d) the %Elongation, (e) the %Reduction in area, (f) the engineering stress at fracture, and (g) the modulus of resilience. 4- True Stress, σ T, and True Strain, ε T : The decrease in engineering stress beyond the tensile strength point on an engineering stress strain curve is related to the definition of engineering stress. We used the original area A º in our calculations, but this is not precise because the area continually changes. We define true stress and true strain as: "True stress, ζ T " is defined as the load divided by the instantaneous cross sectional area, A i over which deformation is occurring (until the neck, past the tensile point), or: 13

14 Furthermore, it is occasionally more convenient to represent as "True strain, ε T ", defined by: 5-Relationships between true stress and engineering stress and between true strain and engineering strain: True and engineering stress and strain are related according to: σ T = σ (1 + ε) and ε T = ln ( 1 + ε) Example: Prove that 1: ζ T = ζ (1 + ε) 2: ε T = ln ( 1 + ε) 14

15 For some metals and alloys the region of the true stress strain curve from the onset of plastic deformation to the point at which necking begins may be approximated by: In this expression, K and n are constants; these values will vary from alloy to alloy, and will also depend on the condition of the material (i.e., whether it has been plastically deformed, heat treated, etc.). The parameter n is often termed the strain hardening exponent and has a value less than unity. Example : Compute the strain-hardening exponent n for an alloy in which a true stress of 415 MPa produces a true strain of 0.10; assume a value of 1035 MPa for K. 15

16 The Bend Test for Brittle Materials In ductile materials, the engineering stress strain curve typically goes through a maximum; thus maximum stress is the tensile strength of the material. Failure occurs at a lower stress after necking has reduced the cross sectional area supporting the load. In more brittle materials, failure occurs at the maximum load, where the tensile strength and breaking strength are the same. In brittle materials, including many ceramics and thermosetting polymers, yield strength, tensile strength, and breaking strength are all the same Figure (9). Figure (9) The stress strain behavior of brittle materials compared with that of more ductile materials. In many brittle materials, the normal tensile test cannot easily be performed because of the presence of flaws at the surface. Often just placing a brittle material in the grips of tensile testing machine cause cracking. These materials may be tested using bend test Figure (10). By applying the load at three points and causing bending, a tensile force acts on the material opposite the midpoint. Fracture begins at this location. The flexural strength (also called the bending strength, or fracture strength) describes the material strength: 16

17 General relation: for rectangular beam for circular beam Figure (10) A three-point loading scheme for measuring the stress strain behavior and flexural strength of brittle ceramics, including expressions for computing stress for rectangular and circular cross sections. 17

18 Where F is the fracture load, L is the distance between the two outer points, b is the width of the specimen, d is the thickness of specimen, and R is the specimen radius. The flexural strength has the units of stress and is designated by (ζ bend ). The results of the bend test are similar to the stress strain curve, however, the stress is plotted versus deflection rather than versus strain. Figure (11). Figure (11) Stress deflection curve for MgO obtained from a bend test. The modulus of elasticity in bending, or the flexural modulus (E bend ), is calculated in the elastic region of Figure (11). Where δ is the deflection of the beam when a force F is applied. 18

19 Example (5): A circular specimen of MgO is loaded using a three-point bending mode. Compute the minimum possible radius of the specimen without fracture, given that the applied load is 425 N, the flexural strength is 105 MPa, and the separation between load points is 50 mm. Example (6): The flexural strength of a composite material reinforced with glass fiber is 310 MPa and the flexural modulus is 125 GPa. A sample, which is 12.5 mm wide, 10 mm high or thick, and 200 mm long, is supported between two rods125 mm a part. Determine the force required to fracture material and the max deflection of the sample at fracture. The force is applied in the midpoint between supporting rods. Example (7): A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 3.5 mm; the specimen fractured at a load of 950 N when the distance between the support points was 50 mm. Another test is to be performed on a specimen of this same material, but one that has a square cross section of 12 mm on each edge. At what load would you expect this specimen to fracture if the support point separation is 40 mm? 19

20 Hardness Test of Engineering Materials The hardness of a material is a measure of the resistance of a material to abrasion or indentation. Indentation hardness tests: There are two types of indentation hardness test. The first type (Brinell, Vickers, and Knoop) measures the size of the impression left by an indenter of prescribed geometry under a known load whereas the second type (Rockwell) measures depth of penetration of an indenter under specified conditions. 1- The Brinell Test: With the Brinell test, a hardened steel ball is pressed for a time of 10 to 15 sec. into the surface of the material by a standard force (fig. 12). After the load and ball have been removed, the diameter of the indentation is measured. The Brinell hardness number signified by HB, is obtained by dividing the size of the force applied by the surface area of the spherical indentation: The units used for the area are mm 2 and for the force kgf (1 kgf = 9.8 N, and is the gravitational force exerted by 1 kg). The area can be obtained from the measured diameter of the indentation and ball diameter, either calculation or the use of tables: Where D is the diameter of the ball and d that of the indentation. 20

21 Table (1) shows the materials and standard measuring in Brinell test. The F/D 2 values is thus chosen to fit the materials concerned with size of test ball and weigh used in test, the harder the material the higher the used value. Table (1) shows the materials and standard measuring in Brinell test. Material Steel Cu and Al alloys Pure Cu and Al Pure lead, tin and its alloys F/D Vickers Test: The Vickers hardness test involves a diamond indenter, in the form of a square based pyramid with an apex angle of 136, being pressed under load for 10 to 15 sec. into the surface of the material under test (fig. 12). The result is a square shaped impression. After the load and indenter are removed the diagonals d of the indentation are measured. The Vickers hardness number (HV) is obtained by dividing the size of the force F, in units of kgf, applied by the surface area in mm 2 of the indentation. The surface area can be calculated from the mean diagonal value, the indentation being assumed to be a right pyramid with a square base and an apex angle of 136, or obtained by using tables: 21

22 Thus the Vickers hardness HV is given by: The Vickers test has advantage over the Brinell test of the increased accuracy that is possible in determining the diagonals of a square as opposed to the diameter of a pyramid. Up to a hardness value of about 300 HV, the hardness value number given by the Vickers test is the same as that given by the Brinell test. Table (2) shows the materials and standard measuring in Vickers test. Table (2) The materials and standard measuring in Vickers test. Material Load (kg) Steel Cu alloys Pure Cu and Al alloys Pure Al Pure lead, tin and its alloys Example (8): With the Vickers hardness test a 30 kg load gave for a sample of steel an indention with diagonal having mean length of mm. What is the hardness? 22

23 3-The Knoop Test: The hardness of ceramic materials is measured by Knoop test. The Knoop test used a diamond pyramid as shown in Figure (12). Indenter which is designed to give a long thin impression. The length (l) being seven times greater than the width (b). The surface area of the indentation is given by: Where l/b = 7.11 and b/t = 4 And thus, the Knoop hardness, HK by: 4-The Rockwell Test: The Rockwell hardness test is differs from the Brinell, Vickers, and Knoop hardness tests is not obtaining a value for the hardness in terms of the area of an indentation but using the depth of indentation, this depth being directly indicated by a pointer on a calibrated scale. The test uses either a diamond cone or a hardened steel ball as the indenter (fig. 12). A preliminary force is applied to press the indenter into contact with the surface. A further force is then applied, and causes the indenter to penetrate into the material. The additional force is then removed and there is some reduction in the depth of the indenter due to the deformation of the material not being entirely plastic. The difference in the final 23

24 depth of the indenter and the initial depth before the additional force was applied is determined. This is the permanent increase in penetration e due to the additional force. The Rockwell hardness number (HR) is then given by: The Rockwell hardness number (HR) = E e Where E is a constant determined by the form of the indenter. For the diamond cone indenter E is 100, for steel ball 130. There is a number of Rockwell scales, the scale being determined by the indenter and the additional force used. In any reference to the results of a Rockwell test the scale letter must be quoted. For metals the B and C scales are probably the most commonly used ones. Correlation between Hardness and Tensile Strength: Both tensile strength and hardness are indicators of a metal s resistance to plastic deformation. Consequently, they are roughly proportional, as shown in figure (13), for tensile strength as a function of the HB for cast iron, steel, and brass. The same proportionality relationship does not hold for all metals, as figure (13) indicates. As a rule of thumb for most steels, brass, and aluminum alloys the HB and the tensile strength are related according to: Tensile Strength (MPa) = k Hardness Where k is a constant depends on type of hardness test and the tested metal. In Brinell test, for steel k = 3.45, for brass k = 5.6, and for aluminum alloys k = 4.2, while in Vickers test, for steel k =

25 Figure (12) Hardness test techniques 25

26 Example (9): An aluminum alloy has a hardness of 45 HB when annealed and 100 HB when hardened. Estimate the tensile strength in MPa of the alloys in these two conditions. Figure (13) Relationships between hardness and tensile strength for steel, brass, and cast iron 26