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Material Selection Tutorial Selecting an appropriate material is a critical part ofalmost all engineering designs There are many factors to consider– Strength, stiffness, durability, corrosion, cost,formability, etc Methods– Experience: how do you get it? limiting– Ashby selection T.html)– Quantitative ranking of options (described here)E MCH 213DMaterial Selection - 1
Ashby Material Selection orial/non IE/selchart.htmlE MCH 213DMaterial Selection - 2
Quantitative Ranking of Options forMaterial Selection* Objective: develop a rational method to select the bestmaterial for an application based upon known materialparameters and the requirements of the applicationUse a 5-step method1. Select a quantity, Q, to minimize or maximize2. Classify the variables3. Determine the relationship between the geometry variable, therequirements, and material properties4. Determine Q as a function of requirements and materialproperties5. Rank candidate materials based upon function f2* Based on N.E. Dowling, Mechanical Behavior of Materials, section 3.8E MCH 213DMaterial Selection - 3
Step 1: Select a quantity, Q, tominimize or maximize Mass (weight), m Cost, Care the most common and the only ones that we willconsiderE MCH 213DMaterial Selection - 4
Step 2: Classify the variables Requirements – variables that have prescribed valuesthat will not change Geometry – variables that define the dimensions of thecomponent and depend implicitly upon the materialproperties Material Properties – variables used to define thematerial in terms of physical behavior, mechanicalbehavior, and costE MCH 213DMaterial Selection - 5
Step 3: Determine the relationshipbetween the geometry variable, therequirements, and materialproperties Strength– Bar, axial stress– Beam, flexural stress Stiffness– Bar, deformation– Beam, deflectionE MCH 213DMaterial Selection - 6
Step 4: Determine Q as a functionof requirements and materialproperties Q f1(requirements)* f2(material props)E MCH 213DMaterial Selection - 7
Step 5: Rank candidate materialsbased upon function f2 If both weight and cost are important then separaterankings can be generated and results combined Calculate geometry variable after ranking materials– Adjustments may be necessary if calculateddimensions are impractical (either too large or toosmall) There may be multiple requirements such as strengthand serviceability– Often material can be selected based on strength andthen the serviceability requirements checkedE MCH 213DMaterial Selection - 8
Sample Problem We must bridge a gap of L 8’The bridge must have a width of b 4”A load P 300 lb can be applied at any pointThere must be a safety factor X 1.5 for strengthThe deflection, v, must not exceed 1”Weight (mass) and cost have equal importanceOBJECTIVE: select the best candidate material from AISI 1020 steelAISI 4340 steel7075-T6 aluminumTi-6Al-4V (titanium alloy)PolycarbonateLoblolly pineGFRP (glass fiber reinforced polymer)CFRP (carbon fiber reinforced polymer)E MCH 213DMaterial Selection - 9
Step 1: Select a quantity, Q, tominimizeHere, mass and cost have equal importance Mass, m Cost, CSelect Q to be the sum of the normalized mass and cost Q m/min(m) C/min(C)E MCH 213DMaterial Selection - 10
Step 2: Classify the variables Requirements: L 8’, b 4”, P 300 lb, X 1.5, v 1” Geometry: restrict analysis to a rectangular crosssection, h height Material Properties (need step 3 & 4 results here):ρ mass density, E Young’s modulus, S strength,Cm cost indexE MCH 213DMaterial Selection - 11
Step 3: Determine the relationshipbetween the geometry variable, therequirements, and materialproperties We have a simply supported beam with a rectangularcross-section The worst case occurs when the concentrated load, P, isapplied at the centerE MCH 213DMaterial Selection - 12
bPhLStrength – elastic flexural formula shows the maximum stressoccurs at the extreme fibers of the beam at midspanMc, M PL / 4, c h / 2, I bh 3 / 12IPL h 123PL( 4 2 bh 3 2bh 2( 3PLXS X( 2bh 2& 3PLX #' h % 2bS !"12Deflection – from integration, is found to be maximumat midspan333PL12 PLPLv 48 EI 48 Ebh 3 4 Ebh 33& PL #' h !% 4 Ebv "E MCH 213D13Material Selection - 13
Step 4: Determine Q as a functionof requirements and materialproperties – strength basisTry using strength as the basis for material selectionand then check deflection& 3PLX #m 'bhL 'bL % 2bS !"121&3 3 # 2 & ' # PL Xb! 1 !%2" %S 2 "1& 3 3 # 2 & 'Cm #C Cm m PL Xb! 1 !%2" %S 2 "f2E MCH 213DMaterial Selection - 14
Step 5: Rank materials basedupon function f2 – strength basisRequirementsL (in) P (lb) b (in) X v (in) Use spreadsheet todetermine rankings9630041.51Selection Table - strength basis (check si)IndexAISI 1020 steel8.87E-03377081AISI 4340 steel8.87E-0315997137075-T6 ycarbonate1.35E-0389925Loblolly 3134880200MaterialAISI 1020 steelAISI 4340 steel7075-T6 aluminumTi-6Al-4VPolycarbonateLoblolly .67CombinedRank64273158Depth, h(in)0.6550.3180.4880.3071.3421.1270.5420.347f2for 1.961.00MassRank87456231f2for NGNGNGNGNGNGE MCH 213DCostRank23475168Material Selection - 15
Step 4: Determine Q as a functionof requirements and materialproperties – deflection basisTry using deflection as the basis for material selectionand then check strength3136 21& PL #& PL b # 3 & ' #m 'bhL 'bL ! ! 13 !4Ebv4v%"%" %E "1& PL6b 2 # 3 & 'Cm #C Cm m ! 13 !4v%" %E "f2E MCH 213DMaterial Selection - 16
Step 5B: Rank materials basedupon function f2 – deflection basisRequirementsL (in) P (lb) b (in) X v (in) 9630041.51Use spreadsheet todetermine rankingsSelection Table - deflection basis (check )IndexAISI 1020 steel8.87E-03377081AISI 4340 steel8.87E-0315997137075-T6 ycarbonate1.35E-0389925Loblolly 3134880200MaterialAISI 1020 steelAISI 4340 steel7075-T6 aluminumTi-6Al-4VPolycarbonateLoblolly 9.67CombinedRank25374168Depth, h(in)0.8260.8211.1720.9923.6262.1031.7591.146f2for 3.281.71MassRank87465132f2for OKOKOKE MCH 213DCostRank24375168Material Selection - 17
Sample Problem Results Material selection based only on strength results in thedeflection criterion being violated Material selection based only on deflection results in thestrength criterion being satisfied We can say that deflection governs this design Pine is best, 1020 steel is second best, CFRP is worstE MCH 213DMaterial Selection - 18
E MCH 213D Material Selection - 3 Quantitative Ranking of Options for Material Selection* Objective: develop a rational method to select the best material for an application based upon known material parameters and the requirements of the application Use a 5-step method 1.Select a quantity, Q, to minimize or maximize 2.Classify the variables
