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EFFECT OF MATERIAL PROPERTIES ON DESIGN. Factors affecting the behavior of materials in components Component design : Function, Material properties Manufacturing processes. Behavior of component material : Properties of stock material Geometry & external forces

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## EFFECT OF MATERIAL PROPERTIES ON DESIGN

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**EFFECT OF MATERIAL PROPERTIES ON DESIGN**• Factors affecting the behavior of materials in components • Component design : • Function, • Material properties • Manufacturing processes. • Behavior of component material : • Properties of stock material • Geometry & external forces • Effect of fabrication method.**FUNCTION &**REQUIREMENTS COMPONENT DESIGN MANUFACTURING PROCESSES MATERIAL PROPERTIES Factors that should be consider in component design**PROPERTIES OF**STOCK MATERIAL BEHAVIOR OF MATERIAL IN THE COMPONENT GEOMETRY & EXTERNAL FORCES EFFECT OF FABRICATION METHOD Factors that should be considered in anticipating the behavior of material in the component.**Statistical variation of material properties**• In practice, material properties are seldom homogenous, as they are sensitive to variations in parameter such as : • Composition, heat treatment & processing condition. • Others that influence material behavior • Surface roughness, internal stress, sharp corners and stress raiser. • Variation in material properties can be solved by choosing factor of safety. • Factor of safety accommodates unknown influences that affect the life of component under service condition.**Statistical variation of material properties**• Material properties can be statistically described by mean value, standard derivation and coefficient of variation. • Experiment data obtained from large number of samples, material properties follow a normal distribution. • It is possible to estimate statistical data from non-statistical sources that give range or tolerance limits. • In this case the standard deviation S is approx. given by : • S = (max. value of property – min. value)/6 …… ( 1.1 )**Statistical variation of material properties**• Example • If the range of strength of an alloy is given as 800 – 1200 MPa, and the mean value can be taken as 1000 MPa • The standard deviation S can be estimate as • S = ( 1200 – 800 )/6 = 66.67 MPa • The coefficient of variation v is then • v = 66.67/1000 = 0.0667 • If sample is about 25 test eq. 1.1 is to divide by 4 • A sample of about 5 to divide by 2**Stress Concentration**• In almost engineering components and machine have to incorporate design features which introduce changes in thin cross-section. • Changes in cross section causes localized stress concentrations and severity depends on the geometry of the discontinuity and nature of the material. • Stress concentration factor, • Kt = Smax/Sav • Smax, maximum stress at discontinuity and Sav, nominal stress. • Kt, value depends only on geometry of the part. • Applies to brittle & notch sensitive materials. Should consider when using high-strength, low ductility, case-hardened & / or heavily cold work materials.**Stress Concentration**• Stress concentration in fatigue. • Kf, fatigue stress concentration factor, • Kf = endurance limit of notch free • endurance limit of notched part • Notch sensitivity factor, q = Kf - 1 • Kt - 1 • Generally , q 1 as material strength increases and sensitive to notches. • Kf = Kt • q also depend on component size. • Stress-raiser, dangerous in larger masses.**Stress Concentration**• Guidelines for design. • Abrupt changes in cross-section should be avoided. • Fillet radii or stress-relieving groove should be provided. • Fig. 11.3(d) • Slot and grooves should be provided with generous run-out radii and with fillet radii in all corners. Fig. 11.3(b) • Stress relieving grooves or undercut should be provided at the end of threads and splines. Fig. 11.3(c) • Sharp internal corners and external edges should be avoided • Weakening features like bolt and oil holes, identification marks, and part number should not be located in highly stressed areas. • Weakening features should be staggered to avoid the addition of their stress concentration effects, Fig. 11.3(d)**DESIGN ASPECT – Loading Condition**• Static loading – load is applied gradually and remains applied throughout part’s life. • Repeated loading – applied and removed several times (repetitive) during life. Fail by fatigue at stress lower than yield strength. Higher design factor is needed. • Impact – require large design factor. (i) sudden load causes stresses much higher than computed. (ii) require part to absorb energy of the impact. • Static loading but at high T – consider creep, microstructural changes, oxidation & corrosion & influence of method of fabrication on creep.**Designing For Static Strength**• Static strength • Ability to resist short-term steady load at moderate T. • Measured in terms as yield strength, UTS, compressive strength & hardness. • Aimed at avoiding yielding of the component and / or fracture. • Component must be strong enough to support the load & may require stiffness to ensure deflections do not exceed certain limits. • Stiffness , important to avoid misalignment and maintain dimensional accuracy. • Elasticity (Young’s M) • important when designing struts, columns & thin-walled cylinders subjected to compressive axial loading where failure can take place by buckling.**Designing for simple axial loading**• Component and structure made from ductile material are usually designed, so that no yield take place under static loading condition • But, when the component is subjected to uniaxial stress, yielding take place • When local stress reaches the yield strength of the material • Critical cross-sectional area, A ; • A = KtnsL • YS • Kt = stress concentration factor • L = applied load • ns = factor of safety • YS = yield strength**Designing for simple axial loading**• Factor of safety, ns. • Is applied in designing component to ensure it will satisfactory perform its intended function • To get the strength of material at allowable stress. • The definition, strength of material depends on the type of material and loading condition • The factor of safety, ns. • ns = S / Sa • S = Nominal strength, Sa = Allowable strength / Design strength • It is important to define which type of service condition will the material work on before calculating the ns. • Normal working condition • Limit working condition**Designing for Static Strength**• Brittle material such as gray cast iron do not exhibit yielding, so design based on ultimate strength. • General - % of elongation in a 2-in gage length is less than 5% • Brittle. e.g Al – except for castings, all is ductile. • Ductile material – large plastic deformation, so unfit for intended use. Design based on yield strength. • Design stress guidelines : • Ductile material, δd = Sy / 2 • Brittle material, δd = UTS / 6**Determine Dimensions & Shape of Component**• Given – (1) Magnitude & type of loading & (2) material condition. • Determine yield & ultimate strength & % elongation of material. Decide ductile or brittle. • Specify design factor (factor of safety). • Compute design stress. • Write equation for expected max stress. For direct normal stress, δmax = F/A • Set δmax = δd & solve for required cross-sectional area. • Determine minimum required dimension.**Contoh**Soalan : Satu struktur penyokong mesin akan dibebankan dengan beban tegangan statik 16kN. Dicadangkan, dihasilkan dalam bentuk rod empat segi daripada keluli gelek panas, AISI 1020. Berikan dimensi yang sesuai untuk keratan rentas rod tersebut. Analisis : Objektif – tentu dimensi keratan rentas rod. Diberi – F = 16 kN = 16 000 N beban statik. Bahan – AISI 1020 HR, Sy = 331 MPa, 36% elongation**Jawapan :**Analisis – Biar δ = δd = Sy / 2 bahan mulur. Analisis tegasan – δ = F/A, maka A diperlukan A = F / δd Tetapi A = a2 (a = dimensi sisi square) Dimensi minimum yang dibenarkan a = A Maka Tegasan, δd = Sy/2 = 331 MPa/2 = 165.5 MPa (N/mm2) Keratan rentas, A = F/ δd = (16 000 N) = 96.7mm2 (165.5 N/mm2) Dimensi minimum a = A = 96.7 mm2 = 9.83 mm. Dimensi minimum a = 10 mm.**Designing for Torsional Loading**• Torsional – loading of a component / part that tends to cause it to rotate or twist. • When torque is applied, shearing stress is developed & torsional deformation occurs, resulting in an angle of twist of one end of part relative to the other. • Material must have sufficient rigidity for the part to perform properly in service. • Torque = T = F x d where F = applied force & d = distance from action of force to axis of the part. • Power = torque x rotational speed (n in rad/s). • Torsional shear stress, ζmax = Td / 2Ip where T = applied torque, d = diameter & Ip = polar moment of inertia of the cross section.**Designing for Torsional Loading**• Critical cross sectional area can be calculated, • for circular shaft, 2Ip = KtnsT • d ζmax • where Kt = stress concentration factor • ns = factor of safety • Moment, Ip = πd4 / 32 for solid circular shaft & • Ip = π(d04 – di4) / 32 for hollow circular shaft of inner di & outer d0. • ASTME code of practice ; allowable value of shear stress of 0.3 yield or 0.18 UTS. • For ductile material, design shear stress = yield / 2N (steady torsion, N = 2, so ζd = yield / 4)**Designing for Torsional Loading**• Torsional rigidity of component is usually measured by the angle of twist, θ, per unit length. For circular shaft, θ is given by, • θ = T / GIp where G = modulus of elasticity in shear. • G = E / (2(1 + ν)) where ν = Poisson’s ratio. • Usual practice is limit the angular deflection in shafts to about 1 degree, i.e π/180 rad, in length of 20 times the diameter. • Stiffness of part differ depending on shape of the cross-section – circular section has higher rigidity compared to other structural shapes, i.e I-beams,wide-flange beams**Designing for Bending**• Beam – component that carries load transversely, that is, perpendicular to its long axis. • Loading – normal concentrated load, inclined concentrated load, uniformly distributed load, varying distributed load & concentrated moments. • Moment – an action that tends to cause rotation of an object. Can be produced by a pair of parallel forces acting in opposite directions, called couple. • Beam types ; simple, overhanging, cantilever, compound & continuous. • Bending moments – internal moments cause bending.**Load**M Cantilever – one fixed end, provide support & moment produced by load Designing for Bending Overhanging – bend downward, negative bending**F**F Compound – two or more parts extending in different directions. Continuous – extra support or both ends fixed, require different approaches to analyze forces & moments. Designing for Bending**Designing for Bending**• Relation between bending moment, max allowable stress & dimensions given by ; • Z = nsM • YS • where Z = section modulus = I/c • c = distance from center of gravity of cross section to the outermost fiber/beam. • I = moment of inertia of cross section with respect to neutral axis normal to direction of load. • M = bending moment & YS = max allowable stress. • ns = factor of safety.**Designing for Bending**• When load is placed on a beam, the beam is bent and every portion of it is moved in a direction parallel to the direction of the load. • The distance that a point on the beam moves/ deflection depends • Its position in the beam • Type of beam • Type of support**Example**Question : Determine the diameter of a cantilever beam of length 1 m and rectangular cross section of depth-to-width ratio 2:1. The cantilever is expected not to deflect more than 50 mm for every 1000 N increment of load at its tip. The material used in making the beam is steel AISI 4340 with a yield strength of 1420 MPa and UTS 1800 MPa. What is the max permissible load ? Assume a suitable factor of safety.**Example**Solution : The deflection ( y ) is given by relationship : y = ( Ll3 ) / ( 3 EI ) where l = the length of cantilever E = elastic modulus of the cantilever material = 210 Gpa I = the second moment of area of the cross-section From fig 4.3 I = b x ( 2b )3 = 1000 x 1 x 1000 12 50 x 3 x 210 x 109 Where b is the width of the beam; b = 14.77 mm**Example**Taking a factor of safety n = 1.5 and using eq. Z = 2148 mm3 = ( nM ) / YS = ( 1.5 x L x l ) / 1420 The safe value of L = 2033 N

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