JEE125 Materials Technology Mechanical Properties Introduction National Centre for Maritime Engineering and Hydrodynamics – JEE125 2 What do structures need? Load bearing structures require materials with reliable and reproducible strength properties National Centre for Maritime Engineering and Hydrodynamics – JEE125 3 Unit Systems National Centre for Maritime Engineering and Hydrodynamics – JEE125 4 Unit Systems Using inputs in a single column will give the same answers National Centre for Maritime Engineering and Hydrodynamics – JEE125 5 Types of Loadings Tension Compression Shear Torsion National Centre for Maritime Engineering and Hydrodynamics – JEE125 6 Tension Tension places a stretching load on a structure. It will elongate and become thinner. National Centre for Maritime Engineering and Hydrodynamics – JEE125 7 Compression Compression places a squeezing load on the structure. It will shorten and expand. National Centre for Maritime Engineering and Hydrodynamics – JEE125 8 Shear Shear forces are unaligned forces that push in the opposite direction. National Centre for Maritime Engineering and Hydrodynamics – JEE125 9 Torsion Torsion will place a twisting load on the structure. National Centre for Maritime Engineering and Hydrodynamics – JEE125 10 Material Characterisation A standardised method is required to predict and describe how a material will behave under each of these loading scenarios National Centre for Maritime Engineering and Hydrodynamics – JEE125 11 Normal Forces Tension Compression National Centre for Maritime Engineering and Hydrodynamics – JEE125 12 Tension Tension is the most common failure condition for metals. Important parameters are force applied, cross sectional area, original length and final length. National Centre for Maritime Engineering and Hydrodynamics – JEE125 13 Compression Compression can be applied to metals but is much more commonly applied to ceramics and concrete. Same parameters as in tension. National Centre for Maritime Engineering and Hydrodynamics – JEE125 14 Engineering Stress Engineering stress is a measure of the load being applied to a structure πΉ π= π΄π π = Engineering Stress (Pa or N/m2) πΉ = Force (N) π΄π = Original Cross Section (m2) National Centre for Maritime Engineering and Hydrodynamics – JEE125 15 Engineering Strain Engineering Strain is the measure of how much a structure has deformed π= ππ −ππ ππ OR π= βπ ππ π = Engineering Strain (-) ππ = Instantaneous Length (m) ππ = Original Length (m) National Centre for Maritime Engineering and Hydrodynamics – JEE125 16 Shear Forces Force is applied to opposing faces in opposite directions National Centre for Maritime Engineering and Hydrodynamics – JEE125 17 Shear Forces πΉ π= π΄π πΎ = tan π π = Shear stress (N/m2) πΉ = Force (N) π΄π = Original Cross Section (m2) πΎ = Shear strain π = Strain Angle (rads) National Centre for Maritime Engineering and Hydrodynamics – JEE125 18 Elastic Deformation National Centre for Maritime Engineering and Hydrodynamics – JEE125 19 Elastic Deformation National Centre for Maritime Engineering and Hydrodynamics – JEE125 20 Elastic Deformation For elastic deformation the strain is totally recovered after the load is removed National Centre for Maritime Engineering and Hydrodynamics – JEE125 21 Elastic Deformation In the elastic range the stress is proportional to the strain π∝π The elastic or Youngs modulus defines this relationship π = πΈπ OR πΈ = π π National Centre for Maritime Engineering and Hydrodynamics – JEE125 22 Elastic Modulus The elastic modulus has units of Pa. For most metals it ranges between 45 and 400 GPa National Centre for Maritime Engineering and Hydrodynamics – JEE125 23 Shear Modulus Elastic deformation can also be induced by shear stresses π = πΊπΎ πΊ = Shear Modulus (Pa) For most metals: πΊ ≈ 0.4πΈ National Centre for Maritime Engineering and Hydrodynamics – JEE125 24 Question A metal rod of length 1 m and a cross section of 25 mm2 has a load of 200 kg applied to it. After applying the load the rod 0.38 mm longer. Calculate the elastic modulus. National Centre for Maritime Engineering and Hydrodynamics – JEE125 25 Answer πΉ π= π΄π 200 × 9.81 π= 25 π = 78.48 πππ National Centre for Maritime Engineering and Hydrodynamics – JEE125 26 Answer βπ π= ππ 0.38 π= 1000 π = 0.00038 National Centre for Maritime Engineering and Hydrodynamics – JEE125 27 Answer π πΈ= π 78.48 πΈ= 0.00038 πΈ = 206.5πΊππ National Centre for Maritime Engineering and Hydrodynamics – JEE125 28 Poisson's Ratio Poisson’s ratio defines the contraction in the direction perpendicular to the direction of load ππ¦ ππ₯ π£=− =− ππ§ ππ§ Typically between 0.3 and 0.45 National Centre for Maritime Engineering and Hydrodynamics – JEE125 29 Poisson's Ratio For isotropic materials the relationship between the elastic modulus, shear modulus and Poisson’s Ratio is governed by πΈ = 2πΊ(1 − π£) National Centre for Maritime Engineering and Hydrodynamics – JEE125 30 Element Analysis National Centre for Maritime Engineering and Hydrodynamics – JEE125 31 Typical Values for metals Metal alloy Modulus of Elasticity Shear Modulus Poisson’s ratio GPa GPa Aluminium 69 25 0.33 Brass 97 37 0.34 110 46 0.34 45 17 0.29 Nickel 207 76 0.31 Steel 207 83 0.30 Titanium 107 45 0.34 Tungsten 407 160 0.28 Copper Magnesium National Centre for Maritime Engineering and Hydrodynamics – JEE125 32 Material Testing Standards National Centre for Maritime Engineering and Hydrodynamics – JEE125 33 Tension Testing Tension testing is required for all metallic engineering materials National Centre for Maritime Engineering and Hydrodynamics – JEE125 34 Cylindrical Specimens National Centre for Maritime Engineering and Hydrodynamics – JEE125 35 Flat Specimens National Centre for Maritime Engineering and Hydrodynamics – JEE125 36 Tube Sections National Centre for Maritime Engineering and Hydrodynamics – JEE125 37 Gauge Length Gauge length (πΏπ ) is the length that we measure as we tension the test specimen National Centre for Maritime Engineering and Hydrodynamics – JEE125 38 Gauge Length πΏπ = 5.65 ππ for non-circular cross sections πΏπ = 5π for circular cross sections National Centre for Maritime Engineering and Hydrodynamics – JEE125 39 Typical Cylindrical Specimen Dimensions Diameter π Maximum variation of diameter of paralled length Original gauge length πΏπ = 5.65 ππ Minimum parallel length Minimum free length between grips 1 πΏπΆ = πΏπ + π 2 Minimum transition radius π πΏπ + π 20.0 ± 0.4 0.05 100 ± 2.0 110 120 20 15.0 ± 0.3 0.04 75 ± 1.5 83 90 15 10.0 ± 0.2 0.03 50 ± 1.0 55 60 10 5.0 ± 0.1 0.02 25 ± 0.5 28 30 5 National Centre for Maritime Engineering and Hydrodynamics – JEE125 40 Typical Flat Specimen Dimensions Width π Maximum variation of width within the parallel length of a test piece Minimum test piece thickness Original gauge length πΏπ Minimum parallel length 1 πΏπΆ = πΏπ + 2 π Minimum free length between grips Minimum transition radius π πΏπ + π 40 ± 2 0.2 3 200 ± 4 220 240 20 20 ± 1 0.1 0.3 200 ± 4 210 220 20 20± 1 0.1 0.1 80 ± 1.6 90 100 20 12.5 ± 0.6 0.06 - 50 ± 1.0 56 63 15 6 ± 0.3 0.04 - 24 ± 0.5 27 30 6 3 ± 0.2 0.04 - 12 ± 0.2 14 15 3 National Centre for Maritime Engineering and Hydrodynamics – JEE125 41 Tensile Testing – Key Points i. Measurement of gauge length and marking of gauge length. ii. Measurement of cross-sectional area. iii. Test grips – types iv. Strain rate is the time rate of increase of strain. The standard strain rate is 8 × 10−4 π −1 within the range 2.5 × 10−4 – 2.5 × 10−3 π −1 . v. Determination of percentage elongation and percentage reduction of area after fracture. National Centre for Maritime Engineering and Hydrodynamics – JEE125 42 Questions A cylindrical specimen of a nickel alloy having an elastic modulus of 207 GPa and an original 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. National Centre for Maritime Engineering and Hydrodynamics – JEE125 43 Answer Given: πΈ = 207 πΊππ π = 10.2 ππ = 10.2 π₯ 10−3 π πΉ = 8900 π Δπ = 0.25 ππ Cross-sectional area π΄ = ππ2 /4 = π π₯ (10.2 π₯ 10−3 )2 π2 = 81.71 π₯ 10−6 π2 π = πΉ = 8900 π/81.71 π₯ 10−6 π2 π΄ = 108.92 π₯ 106 ππ National Centre for Maritime Engineering and Hydrodynamics – JEE125 44 Answer π 108.92 × 106 π = = πΈ 207 × 109 = 5.26 π₯ 10 − 4 and π = π = Δπ π Δπ = 0.25 ππ/5.26 × 10−4 ππ ππ§ = 475 ππ National Centre for Maritime Engineering and Hydrodynamics – JEE125 45 Bending Forces Bending a beam induces both tension and compressive loads National Centre for Maritime Engineering and Hydrodynamics – JEE125 46 Bending Stresses π π πΈ = = π¦ πΌ π π¦ = ππ¦ π= πΌ distance from neutral axis (m) π = bending moment (n-m) πΌ = 2nd moment of area (m4) π = Radius of curvature (m) National Centre for Maritime Engineering and Hydrodynamics – JEE125 47 Moment of Inertia National Centre for Maritime Engineering and Hydrodynamics – JEE125 48 Deflection of a Beam π2 π¦ π = πΈπΌ 2 ππ₯ If solved: πΉπΏ3 πΏ= πΆπΈπΌ πΈπΌ is called the flexural rigidity National Centre for Maritime Engineering and Hydrodynamics – JEE125 49 Torsional Forces Torsion in a shaft places shear forces within the shaft National Centre for Maritime Engineering and Hydrodynamics – JEE125 50 Torsional Stresses Similar relationships can be generated for shafts π π πΊπ = = π½ π πΏ π = torque (Nm) π½ (m4) = polar second moment of area π = shear stress πΊ = shear modulus (Pa) π = angle of twist (rads) πΏ = length of shaft National Centre for Maritime Engineering and Hydrodynamics – JEE125 51 Tomorrow National Centre for Maritime Engineering and Hydrodynamics – JEE125 52 Next Class Plastic Properties Failure National Centre for Maritime Engineering and Hydrodynamics – JEE125 53 JEE125 Materials Technology Mechanical Properties Engineering Stress Engineering stress is a measure of the load being applied to a structure πΉ π= π΄π π = Engineering Stress (Pa or N/m2) πΉ = Force (N) π΄π = Original Cross Section (m2) National Centre for Maritime Engineering and Hydrodynamics – JEE125 55 Engineering Strain Engineering Strain is the measure of how much a structure has deformed π= ππ −ππ ππ OR π= βπ ππ π = Engineering Strain (-) ππ = Instantaneous Length (m) ππ = Original Length (m) National Centre for Maritime Engineering and Hydrodynamics – JEE125 56 Elastic Deformation Elastic deformation only occurs for very low values of strain, in the order of 0.005. Beyond this strain, Hooke’s law does not apply, and permanent strain results from plastic deformation. National Centre for Maritime Engineering and Hydrodynamics – JEE125 57 Elastic Deformation In the elastic range the stress is proportional to the strain π∝π The elastic or Youngs modulus defines this relationship π = πΈπ OR πΈ = π π National Centre for Maritime Engineering and Hydrodynamics – JEE125 58 Plastic Deformation by Slip National Centre for Maritime Engineering and Hydrodynamics – JEE125 59 Plastic Deformation by Slip National Centre for Maritime Engineering and Hydrodynamics – JEE125 60 Plastic Deformation by Dislocation National Centre for Maritime Engineering and Hydrodynamics – JEE125 61 Plastic Deformation by Twinning National Centre for Maritime Engineering and Hydrodynamics – JEE125 62 Plastic Deformation by Twinning Twinning only occurs on a specific crystallographic plane, in BCC and HCP crystals, but not in FCC systems. For BCC systems, the twin plane is (112), and the direction is [111]. National Centre for Maritime Engineering and Hydrodynamics – JEE125 63 Slip Lines in a Polycrystalline Metal National Centre for Maritime Engineering and Hydrodynamics – JEE125 64 Strengthening Metals Grain Size Reduction Cold working/Work Hardening Solid Solution Strengthening National Centre for Maritime Engineering and Hydrodynamics – JEE125 65 Tensile Strength National Centre for Maritime Engineering and Hydrodynamics – JEE125 66 Tensile Strength National Centre for Maritime Engineering and Hydrodynamics – JEE125 67 Ductility Ductility is the amount of plastic deformation that has been sustained at fracture. Brittle materials usually have fracture strains of <5%. National Centre for Maritime Engineering and Hydrodynamics – JEE125 68 Ductility ππ − ππ %πΈπΏ = × 100 ππ %πΈπΏ ππ ππ = the percentage elongation = final length between original gauge marks = original gauge length National Centre for Maritime Engineering and Hydrodynamics – JEE125 69 Ductility π΄0 − π΄π %π π΄ = × 100 π΄π %π π΄ π΄π π΄π = the percentage reduction of cross-sectional area. = original cross-sectional area = final cross-sectional area National Centre for Maritime Engineering and Hydrodynamics – JEE125 70 Typical Properties Metal Alloy Yield Strength MPa Tensile Strength MPa Ductility, %EL [in 50mm (2 in)] Aluminium 35 90 40 Copper 69 200 45 Brass (70Cu-30Zn) 75 300 68 Iron 130 262 45 Nickel 138 480 40 Steel (1020) 180 380 25 Titanium 450 520 25 Molybdenum 565 655 35 National Centre for Maritime Engineering and Hydrodynamics – JEE125 71 Temperature Effect Stress Strain curves for Iron National Centre for Maritime Engineering and Hydrodynamics – JEE125 72 Resilience Resilience is the capacity of a material to absorb energy when it is deformed, and then upon unloading to have this energy recovered. "Resilience = to bounce back" National Centre for Maritime Engineering and Hydrodynamics – JEE125 73 Resilience ππ¦ ππ = π΄ = ΰΆ± ππ¦ ππ 0 ππ¦ ππ¦ = yield strain = yield stress National Centre for Maritime Engineering and Hydrodynamics – JEE125 74 Resilience For a linear curve: 1 ππ = π΄ = ππ¦ ππ¦ 2 Units for resilience are Pa National Centre for Maritime Engineering and Hydrodynamics – JEE125 75 Resilience 1 ππ = π΄ = ππ¦ ππ¦ 2 OR ππ¦2 ππ = 2E National Centre for Maritime Engineering and Hydrodynamics – JEE125 76 Toughness Toughness is the ability to absorb energy up to fracture, and is thus similar to resilience. Hence toughness is equal to the area under a stress-strain curve up to the point of fracture. National Centre for Maritime Engineering and Hydrodynamics – JEE125 77 Toughness National Centre for Maritime Engineering and Hydrodynamics – JEE125 78 True Stress and Strain National Centre for Maritime Engineering and Hydrodynamics – JEE125 79 True Stress and Strain National Centre for Maritime Engineering and Hydrodynamics – JEE125 80 True Stress True Stress accounts for the reduction in crosssectional area as necking occurs πΉ ππ = π΄π π΄π = instantaneous cross-sectional area ππ = π 1 + π National Centre for Maritime Engineering and Hydrodynamics – JEE125 81 True Strain True strain accounts for the elongation of the gauge length as the specimen lengthens ππ βπΏ ππ ππ = lim ΰ· =ΰΆ± ππ ππ ππ li ππ = ln l0 ππ = ln(1 + π) National Centre for Maritime Engineering and Hydrodynamics – JEE125 82 Strain Hardening Effects The complex stresses present at necking include a strain hardening factor, in addition to a component of axial stress. Thus the axial stress is slightly less, due to the presence of the developing strain hardening component, and is shown as the corrected stress. National Centre for Maritime Engineering and Hydrodynamics – JEE125 83 Strain Hardening Effects To account for this an empirical study was performed ππ = πΎπππ πΎ = material constant accounting for heat treatment and work hardening π = material constant accounting for strain hardening National Centre for Maritime Engineering and Hydrodynamics – JEE125 84 Typical Values Material n K (MPa) Low-carbon steel (annealed) 0.21 600 4340 steel alloy (tempered @ 315°C) 0.12 2650 304 stainless steel (annealed) 0.44 1400 Copper (annealed) 0.44 530 Naval brass (annealed) 0.21 585 2024 aluminium alloy (heat treated-T3) 0.17 780 AZ-31B magnesium alloy (annealed) 0.16 450 National Centre for Maritime Engineering and Hydrodynamics – JEE125 85 Elastic Recovery Some elastic recovery will occur after the load is released Yield stress will be higher due to strain hardening National Centre for Maritime Engineering and Hydrodynamics – JEE125 86 Question 9.2 A load of 140,000 N is applied to a cylindrical specimen of a steel alloy displaying the stress-strain behaviour shown in figure 1. It has a crosssectional diameter of 10 mm. (a) Will the specimen experience elastic/plastic deformation? Why? (b) If the original specimen length is 500 mm, how much will it increase in length when this load is applied? National Centre for Maritime Engineering and Hydrodynamics – JEE125 87 Answer From the graph in Figure 1, it can be seen that the maximum stress corresponding to the elastic or linear portion of the curve is 1300 MPa. If the actual stress applied is greater than this value, then the cylindrical specimen will be experiencing plastic deformation. National Centre for Maritime Engineering and Hydrodynamics – JEE125 88 Answer – Part A πΉ π= π΄ ππ 2 π 10 π΄= = 4 4 2 = 78.54ππ2 140,000 π= 78.54 π = 1783 πππ π > ππ¦ ∴Material is undergoing plastic deformation National Centre for Maritime Engineering and Hydrodynamics – JEE125 89 Answer – Part B From the plot: National Centre for Maritime Engineering and Hydrodynamics – JEE125 90 Answer βπ π= π0 βπ 0.02 = 500 βπ = 10 ππ National Centre for Maritime Engineering and Hydrodynamics – JEE125 91 Next Class National Centre for Maritime Engineering and Hydrodynamics – JEE125 92 Next Class Impact and Fracture Hardness Flexure Strength National Centre for Maritime Engineering and Hydrodynamics – JEE125 93 JEE125 Materials Technology Mechanical Properties Today’s Class National Centre for Maritime Engineering and Hydrodynamics – JEE125 95 Introduction National Centre for Maritime Engineering and Hydrodynamics – JEE125 96 Impact and Fracture Fracture is the separation of a body into two or more pieces in response to a stress. Ductile fracture is where the materials experience plastic deformation associated with high energy absorption. Brittle materials experience little or no plastic deformation with low energy absorption. National Centre for Maritime Engineering and Hydrodynamics – JEE125 97 Fracture National Centre for Maritime Engineering and Hydrodynamics – JEE125 98 Fracture National Centre for Maritime Engineering and Hydrodynamics – JEE125 99 Ductile vs Brittle Cup and Cone Brittle National Centre for Maritime Engineering and Hydrodynamics – JEE125 100 Ductile Fracture Shear forces contribute to ductile fracture. Tensile stress creates shear stress on planes that lie at an angle to the longitudinal axis of the test piece. National Centre for Maritime Engineering and Hydrodynamics – JEE125 101 Ductile Fracture The maximum shear occurs at an angle of 45° to the applied tensile force. This matches with the cup and cone fracture pattern. National Centre for Maritime Engineering and Hydrodynamics – JEE125 102 Brittle Fracture Brittle fracture is where the surface is usually flat and shiny. It results from the repeated breaking of atomic bonds along crystallographic planes known as cleavage. National Centre for Maritime Engineering and Hydrodynamics – JEE125 103 Brittle Fracture Brittle fracture in metals is believed to take place in three stages: 1. Plastic deformation concentrates dislocations on the slip or twin planes. 2. Shear stresses build up where dislocation movement is blocked, and micro-cracks are nucleated at these sites. 3. Further stress will cause the microcracks to coalesce into macro-cracks, which then propagate through the metal and result in failure. National Centre for Maritime Engineering and Hydrodynamics – JEE125 104 Impact Fracture Testing These tests measure the energy absorbed during the impact of a notched specimen of standardized geometry. This parameter is useful in assessing the ductile to brittle transition. National Centre for Maritime Engineering and Hydrodynamics – JEE125 105 Impact Fracture Testing National Centre for Maritime Engineering and Hydrodynamics – JEE125 106 Impact Fracture Testing The hammer of the impact tester is raised to a set height, β, and then released. Impact energy is absorbed by the specimen at fracture, such that the hammer loses energy. National Centre for Maritime Engineering and Hydrodynamics – JEE125 107 Impact Fracture Testing The height after impact is β′. The dial on the tester will record the energy lost by the pendulum, which is proportional to β – β′. National Centre for Maritime Engineering and Hydrodynamics – JEE125 108 Ductile to Brittle Transition At lower temperatures, brittle fracture is more likely to occur, and is accompanied by a reduction in impact energy. National Centre for Maritime Engineering and Hydrodynamics – JEE125 109 Ductile to Brittle Transition National Centre for Maritime Engineering and Hydrodynamics – JEE125 110 Ductile to Brittle Transition The carbon content of the steel vastly impacts the ductile to brittle transition National Centre for Maritime Engineering and Hydrodynamics – JEE125 111 Assessing Fracture National Centre for Maritime Engineering and Hydrodynamics – JEE125 112 Fracture Mechanics Charpy testing provides a good test comparative analysis, but difficult to extrapolate to actual failure modes. Stress concentrations such as notches or brittle welds can be introduced to start failure. National Centre for Maritime Engineering and Hydrodynamics – JEE125 113 Fracture Mechanics ππ = 2π0 π ππ‘ 1 2 National Centre for Maritime Engineering and Hydrodynamics – JEE125 114 Fracture Toughness 1 - Tensile 2 - Sliding 3 - Shearing Fracture toughness πΎπΆ is defined as the measure of a material’s resistance to fracture when a crack is present. National Centre for Maritime Engineering and Hydrodynamics – JEE125 115 Fracture Toughness 1 - Tensile 2 - Sliding 3 - Shearing πΎπ = πππ ππ π ππΆ π = dimensionless constant = stress at crack tip (MPa) = crack size (m) National Centre for Maritime Engineering and Hydrodynamics – JEE125 116 Fracture Toughness 1. That crack dimensions are small relative to the bulk size of the component 2. That the mode of crack surface displacement is specified as being 1 tensile, 2 sliding or 3 tearing. National Centre for Maritime Engineering and Hydrodynamics – JEE125 117 Fracture Toughness National Centre for Maritime Engineering and Hydrodynamics – JEE125 118 Application of Fracture Toughness With the previous equations limits if the loading condition is known the vessel can be surveyed to identify cracks above a size governed by: 1 πΎ1πΆ ππ = π ππ 2 National Centre for Maritime Engineering and Hydrodynamics – JEE125 119 Hardness Hardness is a measure of a materials resistance to localized plastic deformation. Moh’s hardness test is used for minerals. A mineral is selected from the above, which just scratches the unknown, thus the hardness number is assigned. National Centre for Maritime Engineering and Hydrodynamics – JEE125 120 Mohs Scale for Minerals Mineral Moh hardness Talc 1 Gypsum 2 Calcite 3 Fluorite 4 Apatite 5 Orthoclase feldspar 6 Quartz 7 Topaz 8 Sapphire (corundum) 9 Diamond 10 National Centre for Maritime Engineering and Hydrodynamics – JEE125 121 Brinell Hardness (HB) In this test, a 10mm ball is impressed, using loads ranging from 500-3000 kg. The diameter of the indentation is measured with a hand microscope, and the Brinell hardness (HB) calculated using a chart for the load employed. National Centre for Maritime Engineering and Hydrodynamics – JEE125 122 Brinell Hardness (HB) π»π΅ = π π· π 2π ππ· π· − π·2 − π 2 = load applied (kg) = diameter of sphere (mm) = diameter of indentation (mm) National Centre for Maritime Engineering and Hydrodynamics – JEE125 123 Brinell Hardness (HB) National Centre for Maritime Engineering and Hydrodynamics – JEE125 124 Hardness Conversion Charts National Centre for Maritime Engineering and Hydrodynamics – JEE125 125 Hardness and Tensile Strength Since both properties, hardness and tensile strength, are a function of the metals resistance to plastic deformation, they are roughly proportional ππ ≅ 3.45 × π»π΅ National Centre for Maritime Engineering and Hydrodynamics – JEE125 126 Question 9.3 A 10mm Brinell hardness indenter produced an indentation 2.50 mm in diameter in a steel alloy when a load of 1000 kg was used. (a) Compute the HB of this material. (b) What will be the diameter of an indentation to yield a hardness of 300 HB when a 500 kg load is used? (c) Estimate the tensile strength of the steels tested above. National Centre for Maritime Engineering and Hydrodynamics – JEE125 127 Question 9.3 a) Given: π· = 10ππ, π = 2.5 ππ, πΏ = 1000ππ π»π΅ = 2π ππ· π· − π·2 − π 2 π»π΅ = 200.5 National Centre for Maritime Engineering and Hydrodynamics – JEE125 128 Question 9.3 b) Given: π»π΅ = 300, π· = 10 ππ, π = 500ππ π»π΅ = 2π ππ· π· − π·2 − π2 π = 1.45 ππ National Centre for Maritime Engineering and Hydrodynamics – JEE125 129 Question 9.3 c) Given: ππ πππ = 3.45 π»π΅ π»π΅ = 200.5 ππ πππ = 1035 πππ National Centre for Maritime Engineering and Hydrodynamics – JEE125 130 Flexural Strength Flexural strength is a materials resistance to bending. For isotropic materials this can be calculated based on tensile properties. For composites and polymers, a more applied test is required. National Centre for Maritime Engineering and Hydrodynamics – JEE125 131 Three-Point Bend Test National Centre for Maritime Engineering and Hydrodynamics – JEE125 132 Bending Forces Bending a beam induces both tension and compressive loads National Centre for Maritime Engineering and Hydrodynamics – JEE125 133 Bending Stresses π π πΈ = = π¦ πΌ π π¦ π πΌ π = = = = ππ¦ π= πΌ distance from neutral axis (m) bending moment (n-m) 2nd moment of area (m4) Radius of curvature (m) National Centre for Maritime Engineering and Hydrodynamics – JEE125 134 Flexure Stress ππ = Flexure stress is the maximum stress that will occur in the beam. π πΏ π d ππ (MPa) 3ππΏ 2ππ 2 = applied load (N) = span (mm) = 16π = width (mm) = thickness (mm) = flexural stress National Centre for Maritime Engineering and Hydrodynamics – JEE125 135 Flexure Strength Maximum flexure stress sustained National Centre for Maritime Engineering and Hydrodynamics – JEE125 136 Flexural Strain Flexural Strain ππ , is the nominal fractional change in the length of an element of the outer surface of the test specimen at midspan, where the maximum strain occurs 6ππ· ππ = 2 πΏ National Centre for Maritime Engineering and Hydrodynamics – JEE125 137 Flexural Modulus Flexural Modulus, (modulus of bending, modulus of rupture), πΈπ΅ is defined as the ratio, within the elastic limit, of stress to corresponding strain. ππΏ3 πΈπ΅ = 4ππ 3 π = slope of the tangent to the initial straight line portion of the load-deflection curve National Centre for Maritime Engineering and Hydrodynamics – JEE125 138 Flexural Modulus National Centre for Maritime Engineering and Hydrodynamics – JEE125 139 Question 9.7 The load-extension curve below was obtained from a three-point bend test carried out in accordance with ASTM 790-10. The test piece was taken from a composite panel manufactured from a 5 mm PVC foam core and clad with Kevlar/glass/epoxy. The dimensions of the test piece were width = 11.1 mm, depth = 11.1 mm and span = 179 mm. National Centre for Maritime Engineering and Hydrodynamics – JEE125 140 Question 9.7 Determine: (a) the flexural strength at the yield point (b) the maximum flexural strength (c) the flexural modulus (d) the flexural strain at the maximum load (e) the mode of fracture. National Centre for Maritime Engineering and Hydrodynamics – JEE125 141 Question 9.7 National Centre for Maritime Engineering and Hydrodynamics – JEE125 142 Question 9.7 a) π = 9000 π = 9.00 ππ × 9.81 π/π 2 = 88.3 π 3ππΏ ππ = 2ππ2 ππ = 17.3 πππ National Centre for Maritime Engineering and Hydrodynamics – JEE125 143 Question 9.7 b) ππππ₯ = 11,000 π = 9.00 ππ × 9.81 π/π 2 = 107.9 π 3ππΏ ππ = 2ππ2 ππ = 21.2 πππ National Centre for Maritime Engineering and Hydrodynamics – JEE125 144 Question 9.7 c) ππΏ3 πΈπ΅ = 4ππ3 π = 30.4 π/ππ πΈπ΅ = 2871 πππ National Centre for Maritime Engineering and Hydrodynamics – JEE125 145 Question 9.7 d) 6ππ· ππ = 2 πΏ ππ = 0.0083 National Centre for Maritime Engineering and Hydrodynamics – JEE125 146 Question 9.7 e) The test sample neither yields or breaks before the 5 % strain limit National Centre for Maritime Engineering and Hydrodynamics – JEE125 147 Fatigue Fatigue is defined as failure of structures at relatively low stress levels, when they are subjected to cyclic and fluctuating stresses, often over long periods. National Centre for Maritime Engineering and Hydrodynamics – JEE125 148 Fatigue Testing National Centre for Maritime Engineering and Hydrodynamics – JEE125 149 SN Curves A plot of the load amplitude S against the logarithm of the number of cycles N, yields what is known as an S-N curve National Centre for Maritime Engineering and Hydrodynamics – JEE125 150 SN Curves Some materials have a fatigue limit below which fatigue failure will not occur National Centre for Maritime Engineering and Hydrodynamics – JEE125 151 Stages of Fatigue 1. Crack initiation, whereby a crack begins at a surface imperfection, which acts as a stress raiser because of its higher level of surface energy. 2. Crack propagation, whereby the crack advances incrementally with each cycle of stress. 3. Final failure, where the crack reaches a critical size, and rapidly advances causing sudden failure. National Centre for Maritime Engineering and Hydrodynamics – JEE125 152 Mechanical Properties Summary Tension Compression Shear Torsion National Centre for Maritime Engineering and Hydrodynamics – JEE125 153 Mechanical Properties Summary Engineering stress is a measure of the load being applied to a structure πΉ π= π΄π Engineering Strain is the measure of how much a structure has deformed π= ππ −ππ ππ OR π= βπ ππ National Centre for Maritime Engineering and Hydrodynamics – JEE125 154 Mechanical Properties Summary National Centre for Maritime Engineering and Hydrodynamics – JEE125 155 Mechanical Properties Summary National Centre for Maritime Engineering and Hydrodynamics – JEE125 156 Mechanical Properties Summary National Centre for Maritime Engineering and Hydrodynamics – JEE125 157 Mechanical Properties Summary National Centre for Maritime Engineering and Hydrodynamics – JEE125 158 Mechanical Properties Summary National Centre for Maritime Engineering and Hydrodynamics – JEE125 159 Mechanical Properties Summary National Centre for Maritime Engineering and Hydrodynamics – JEE125 160 Mechanical Properties Summary National Centre for Maritime Engineering and Hydrodynamics – JEE125 161 Next Topic National Centre for Maritime Engineering and Hydrodynamics – JEE125 162
