Materials Selection Case Study 1 Bases and Mechanical Properties. Professors: Anne Mertens and Davide Ruffoni Assistant: Tommaso Maurizi Enrici

Transcription

1 Materials Selection Case Study 1 Bases and Mechanical Properties Professors: Anne Mertens and Davide Ruffoni Assistant: Tommaso Maurizi Enrici Thursday, October 4, 2018

2 Mechanical Properties Case Studies Case Study 1: The Lightest STIFF Beam Case Study 2: The Lightest STIFF Tie-Rod Case Study 3: The Lightest STIFF Panel Case Study 4: Materials for Oars Case Study 5: Materials for CHEAP and Slender Oars Case Study 6: The Lightest STRONG Tie-Rod Case Study 7: The Lightest STRONG Beam Case Study 8: The Lightest STRONG Panel Case Study 9: Materials for Constructions Case Study 10: Materials for Small Springs Case Study 11: Materials for Light Springs Case Study 12: Materials for Car Body CES 2009 CES 2016 Thursday, October 4, 2018 Materials Selection 2

3 Materials selection Mechanical properties: tensile test, fatigue, hardness, toughness, creep Physical properties: density, conductivity, coefficient of thermal expansion Chemical properties : corrosion Microscopic characteristics: anisotropy of properties, hardening, microstructure, grain size, segregation, inclusions Thursday, October 4, 2018 Materials Selection 3

4 Materials selection Process linked aspects: formability, machinability, weldability, stampability Aestethic aspects: colour and surface roughness Notice: surface properties volume properties 4 poles for engineering and material science Performances Science of materials Synthesis Manufacturing Transformation Composition Structure Properties Thursday, October 4, 2018 Materials Selection 4

5 Design steps Design tools Functional analysis Methods for fonction analysing 3D modeler Simulation Optimization methods Components modelization (FEM) DFM/DFA Objectives Understand the function Define the main characteristics of the product Optimize shape Optimize realisation (manufacturing + assembling) Market Need Principle Improvement Detailing Materials Selection Choose between main classes of materials (ceramics, metals ) Choose between families inside a material classe (steel, cast iron, Al ) Choose a particular variety inside the family (6000 or 7000 alloy ) Process selection Choose between the main classes of processes (moulding, machining ) Choose between the families for a given process class (sand casting, pressure casting ) Choose between the different varieties for a given family of process (moulding, metal mould, Cosworth process ) Thursday, October 4, 2018 Materials Selection 5

6 Forecasts Evolution of materials is challenged by: mechanical properties physical and chemical properties environmental problems (manufacturing) materials ressource Key Domains : energy (nuclear, solar cells, ) transport Thursday, October 4, 2018 Materials Selection 6

7 A bit of History Thursday, October 4, 2018 Materials Selection 7

8 A bit of History 1850s, time of the Crimean War Napoleon III French military engineers had found they could control the trajectory applying a rifling or spinning in the barrels of guns (cannon) The spiraling motion added extra stresses Consequence? Need a higherstrength material Steel Cannon shatter Thursday, October 4, 2018 Materials Selection 8

9 A bit of History 1946, University of Pennsylvania Moore School of Electrical Engineering Electronic Numerical Integrator Analyser and Computer (Eniac) by John Mauchly and J. Presper Eckert 1947, Discovery of the Transistor (Semiconductors) Built from materials such as Silicon and Germanium which can either behave as an electrical insulator or conductor The first general-purpose electronic computer thermionic valves 70,000 resistors. Covered 167 square metres of floor space Weighed 30 tonnes Consumed 160 kw of electricity Companies spent tens of billion of dollars to squeeze more circuits on to a small chip of material 2010, an Intel X3370 microprocessor 820 million transistors Your computer could handle 3 billion instructions /s more than Eniac Thursday, October 4, 2018 Materials Selection 9

10 A bit of History 2012, low cost airlines company Change the material of a small pivot (46) for each seat In the air transports Weight = Costs Aluminum PE+ Glass fibers Composite 10,000,000 dollars saved each year Thursday, October 4, 2018 Materials Selection 10

11 Mike Ashby from Univesity of Cambridge Thursday, October 4, 2018 Materials Selection 11

12 Ashby Diagrams [The mechanical efficiency of natural materials, Mike Ashby, 2003] Thursday, October 4, 2018 Materials Selection 12

13 Ashby Diagrams Thursday, October 4, 2018 Materials Selection 13

14 Ashby Diagrams Thursday, October 4, 2018 Materials Selection 14

15 Ashby Diagrams Thursday, October 4, 2018 Materials Selection 15

16 Ashby Diagrams Thursday, October 4, 2018 Materials Selection 16

17 Ashby Diagrams Thursday, October 4, 2018 Materials Selection 17

18 Ashby Diagrams Thursday, October 4, 2018 Materials Selection 18

19 Simplification: Where is the problem? dy For a beam under flexion, the moment of inertia : I XX = bh3 12 h X Length (L): 300 mm Thickness (h) = 1 mm I XX = = 2,1 mm 4 b Width (b) = 25 mm I YY = = 1300 mm 4 Y 25 mm In the case of the mechanical properties, it is important to consider the forces applied, but it is the weakest point that determine the selection. X 1 mm It is possible to change the geometry, but if you cannot What can we do? Y X Thursday, October 4, 2018 Materials Selection 19

20 Simplification: Train Wheel (Fast Example) Thursday, October 4, 2018 Materials Selection 20

21 The Stiffness design The Stiffness design is important to avoid excessive ELASTIC deflection Thursday, October 4, 2018 Materials Selection 21

22 The Stiffness design The Stiffness design is important to avoid excessive ELASTIC deflection Your are here Thursday, October 4, 2018 Materials Selection 22

23 L The Stiffness F δ S = F δ = C 1EI L 3 δ = ε L EI = Flexural rigidity I = Second Moment of inertia E = Young s Modulus δ = Deflexion C 1 = 3 10 N Length (L): 300 mm Thickness (h)= 1 mm I XX = = 2,1 mm 4 Problem : Width (b)= 25 mm I YY = = 1300 mm 4 δ?? IF we consider that the beam is made of Stainless Steel (E = 200 GPa) Which are the consequences if I want to use Polystyrene (E = 2 GPa)? IF I can change the thickness and hold the same deflection. Thursday, October 4, 2018 Materials Selection 23

24 L The Stiffness F δ S = F δ = C 1EI L 3 EI = Flexural rigidity I = Second Moment of inertia E = Young s Modulus C 1 = 3 10 N Stainless Steel (E = 200 GPa; ρ = 7800 kg/m 3 ) Polystyrene (E = 2 GPa; ρ = 1040 kg/m 3 ) I YY = I XX = = 1300 mm 4 = 2,1 mm 4 δ = δ = 10 (0,25) ( ) FL3 C 1 EY XX = 124 mm = 0,02 mm Steel With δ = 124 mm I XX = h = 12I XX w 1/3 = (0,25) (0,124) = 210 mm4 PS 1/3 = 4,6 mm When h(steel) = 1 mm Thursday, October 4, 2018 Materials Selection 24

25 L The Stiffness C 1 = 3 F δ 10 N S = F δ = C 1EI L 3 Length: 300 mm Width = 25 mm EI = Flexural rigidity I = Second Moment of inertia E = Young s Modulus δ = Deflexion Stainless Steel (E = 200 GPa; ρ = 7800 kg/m 3 ) Polystyrene (E = 2 GPa; ρ = 1040 kg/m 3 ) Thickness = 1 mm Thickness = 4,6 mm About the weight? m SS = ,3 0,025 0,001 = 59 gr m PS = ,3 0,025 0,046 = 36 gr BIGGER Section BUT LIGHTER Depends on what you need and the conditions Thursday, October 4, 2018 Materials Selection 25

26 The Materials Selection approach Case Study 1: Find the Lightest STIFF Beam C 1 = 3 L F δ 100 N Objective Minimize the mass Constraints Stiffness specified Length L Square shape Free Variables Area (A) of the cross-section Choice of the material Length: 300 mm EI = Flexural rigidity I = Second Moment of inertia E = Young s Modulus δ = Deflexion Hypothesis: F = S S δ min F δ S min = C 1EI L 3 m = A L ρ A = m L ρ m = mass A = area of the section L = Length ρ = Density Thursday, October 4, 2018 Materials Selection 26

27 The Materials Selection approach Case Study 1: Find the Lightest STIFF Beam L F Beam: Square Section b=h dy F δ = C 1EI L 3 A = m L ρ S min δ h X Since A = b 2 C 1 = 3 A = m L ρ 100 N b Y The Area will be the Free Variable I = bh3 12 = A2 12 Just remember: m 12 S C 1 L 1/2 L 3 ρ E 1/2 m E 1/2 ρ Constraints Stiffness specified Length L Square shape Thursday, October 4, 2018 Materials Selection 27

28 The Material Index (M) Case Study 1: Find the Lightest STIFF Beam M = A B Log(M) = Log A Log(B) For instance E1/2 ρ A Log(A) = Log(B) + Log M 1 1 Slope Selection Direction (+) B Thursday, October 4, 2018 Materials Selection 28

29 Ashby Diagrams Thursday, October 4, 2018 Materials Selection 29

30 The Material Index (M) Case Study 1: Find the Lightest STIFF Beam m E 1/2 ρ 2 Slope 1 Stainless Steel (E = 200 GPa; ρ = 7800 kg/m 3 ) Polystyrene (E = 2 GPa; ρ = 1040 kg/m 3 ) Thursday, October 4, 2018 Materials Selection 30

31 Lightest Beam (Bending conditions) Case Study 1: Find the Lightest STIFF Beam Length: 300 mm Thickness = 1 mm? Width and thickness F = 100 N Width = 25 mm δ = 0,34 mm S min = N/m Stainless Steel (E = 200 GPa; ρ = 7800 kg/m 3 ) Polystyrene (E = 2 GPa; ρ = 1040 kg/m 3 ) m A = 12 S C 1 L m L ρ 1/2 L 3 ρ E 1/2 Material Weight (kg) A (mm 2 ) Stainless Steel Width and Thickness (mm) 0, Polystyrene 1, Thursday, October 4, 2018 Materials Selection 31

32 CES Case Study 1: Find the Lightest STIFF Beam To Minimize mass 1 2 Slope Thursday, October 4, 2018 Materials Selection 32

33 To Minimize Stiffness CES 1 2 Slope Depends on what you need It is not in flexion Thursday, October 4, 2018 Materials Selection 33

34 Ok, slow down.. Case Study 2: Find the Lightest STIFF Tie-Rod Tie-Rod = TRACTION CONDITIONS Objective Minimize the mass Constraints Stiffness specified Length L Free Variables Area (A) of the cross-section Choice of the material DATA F =1000 N Dimensions: Length: 300 mm Thickness = 1 mm Width = 25 mm In Traction, the shape of the cross-section is not important m = A L ρ A = m L ρ From material σ ε = E From definition: δ = ε L F = σ A F δ S min = S Thursday, October 4, 2018 Materials Selection 34

35 Lightest Tie-Rod (Traction conditions) Case Study 2: Find the Lightest STIFF Tie-Rod F δ S min = S F = 1000 N δ = 3, mm S min = 264, N/m Dimensions: Length: 300 mm Thickness = 1 mm Width = 25 mm σ A ε L S min E A L A = S min m L ρ m S L 2 ρ E m (264, ) ( ) 2 ρ E ρ E m E ρ Thursday, October 4, 2018 Materials Selection 35

36 Case Study 2: Find the Lightest STIFF Tie-Rod CES m E ρ Thursday, October 4, 2018 Materials Selection 36

37 Lightest Tie-Rod (Traction conditions) Case Study 2: Find the Lightest STIFF Tie-Rod F = 1000 N δ = 3, mm S min = 264, N/m Dimensions: Length: 300 mm Thickness = 1 mm Width = 25 mm E A L A = S min m L ρ m S L 2 ρ E Stainless Steel (E = 200 GPa; ρ = 7800 kg/m 3 ) Silicon carbide (E = 430 GPa; ρ = 3150 kg/m 3 ) Al Alloys (E = 75 Gpa; ρ = 2700 kg/m 3 ) m E ρ Material Weight (kg) A (mm 2 ) Width and Thickness (mm) Silicon Carbide 0, ,8 13,4 Al Alloys 0, ,4 Thursday, October 4, 2018 Materials Selection 37

38 Change of the section Panel: b fixed h free Beam: Square Section b=h Panel: h fixed b free Thursday, October 4, 2018 Materials Selection 38

39 Lightest Panel (Bending conditions) Case Study 3: Find the Lightest STIFF Panel L F Objective Minimize the mass Constraints Stiffness specified Length L and b specified Free Variables h (thickness) of the cross-section Choice of the material C 1 = 3 Y Z δ 100 N Y X b = 25 mm h Hypothesis: F S = S δ min Length: 300 mm Width: 25 mm F δ C 1EI L 3 m = A L ρ = S min A = m L ρ m = mass A = area of the section L = Length ρ = Density Thursday, October 4, 2018 Materials Selection 39

40 Lightest Panel (Bending conditions) Case Study 3: Find the Lightest STIFF Panel Panel: b fixed h free Since A = bh h = A b I = A b b 12 3 = A3 12 b 2 S min C 1EI A = L 3 m L ρ Y h X b = 25 mm A = m m L ρ 12 S b2 C 1 The Area will be the Free Variable, but all the consequences of the selection are on the thickness 1/3 L 2 ρ E 1/3 m E 1/3 ρ h = A b Thursday, October 4, 2018 Materials Selection 40

41 CES Case Study 3: Find the Lightest STIFF Panel m E 1/3 ρ Thursday, October 4, 2018 Materials Selection 41

42 Bending conditions m E 1/2 ρ m E 1/3 ρ Stainless Steel (E = 200 GPa; ρ = 7800 kg/m 3 ) Polystyrene (E = 2 GPa; ρ = 1040 kg/m 3 ) Thursday, October 4, 2018 Materials Selection 42

43 Bending conditions F = 100 N δ = 0,34 mm S min = N/m Beam Panel (b= 25 mm) Stainless Steel (E = 200 GPa; ρ = 7800 kg/m 3 ) Polystyrene (E = 2 GPa; ρ = 1040 kg/m 3 ) Material Weight (kg) A (mm 2 ) Stainless Steel Thickness h (mm) 0, Polystyrene 1, Stainless Steel 1, ,59 Polystyrene 0, ,34 Thursday, October 4, 2018 Materials Selection 43

44 Stiffness Summary E E 1/2 E 1/3 To m At fixed S min ρ ρ ρ S min δ max At fixed m Stiffness Traction : Stiffness Bending : Stiffness Bending : Thursday, October 4, 2018 Materials Selection 44

45 Case Study 4: Materials for Oars Thursday, October 4, 2018 Materials Selection 45

46 Case Study 4: Materials for Oars Objective Minimize the mass Constraints Stiffness specified Length L Circular shape (beam) Free Variables Area (A) of the cross-section Choice of the material L (Outboard) = 2 m Thursday, October 4, 2018 Materials Selection 46

47 Case Study 4: Materials for Light Oars We assume solid section A = π r 2 r I = πr4 4 = A2 4π Y Y Z X F δ S min = C 1EI A = m L 3 A = m L ρ m L ρ 4 π S L5 3 1/2 ρ E 1/2 m E 1/2 ρ Thursday, October 4, 2018 Materials Selection 47

48 Case Study 4: Materials for Light Oars m E 1/2 ρ Thursday, October 4, 2018 Materials Selection 48

49 Case Study 4: Materials for Light and Slender Oars We assume solid section A = π r 2 r I = πr4 4 = A2 4π Y Y Z X F δ S min = C 1EI L 3 I = πr4 4 = A2 4π A 4 π S L3 3 1/2 1 A E E 1/2 Place LIMITS to a single Property Evaluating the Properties Chart 10 Gpa < E < 200 GPa Thursday, October 4, 2018 Materials Selection 49

50 Case Study 4: Materials for Light and OarsSlender Oars m E 1/2 ρ CFRP - best material with more control of the properties Bamboo Traditional material for oars for canoes Woods Traditional, but with natural variabilities Ceramics Low toughness and high cost 10 Gpa < E < 200 GPa Thursday, October 4, 2018 Materials Selection 50

51 Case Study 4: Materials for Light and OarsSlender Oars Solid I = πr4 4 = A2 4π Tube I = πr 3 t S 3 m2 1,58 kg Probably Tube shape Assume 2,5 kg for a Solid Oar (exagerated) 4 π L 5 E ρ 2 CFRP (E = 110 GPa; ρ = 1550 kg/m 3 ) Bamboo (E = 17,5 GPa; ρ = 700 kg/m 3 ) m S min δ max At fixed S min At fixed m S CFRP = 853,94 N/m S Bamboo = 666,1 N/m CFRP good for Competition Oar Thursday, October 4, 2018 Materials Selection 51

52 Case Study 5: Materials for CHEAP and Slender Oars Objective Minimize the cost Constraints Stiffness specified Length L Circular shape (beam) Free Variables Area (A) of the cross-section Choice of the material L (Outboard) = 2 m Thursday, October 4, 2018 Materials Selection 52

53 Case Study 5: Materials for CHEAP and Slender Oars m C = m C m 4 π S L5 3 1/2 ρ E 1/2 m = C C m C 4 π S L5 3 1/2 ρ C m E 1/2 C Cost C m Cost per unit of mass C E 1/2 ρ C m Better to consider cost always as a function of mass Thursday, October 4, 2018 Materials Selection 53

54 Case Study 5: Materials for CHEAP and Slender Oars C E 1/2 ρ C m Bamboo Traditional material for oars for canoes Woods Traditional, but with natural variabilities Stone and Concrete Low toughness and difficult to manufacture 10 Gpa < E < 200 GPa Thursday, October 4, 2018 Materials Selection 54

55 Case Study 5: Materials for CHEAP and Slender Oars m 4 π S L5 3 1/2 ρ E 1/2 C 4 π S L5 3 1/2 ρ C m E 1/2 C = m C m m = C C m C Cost C m Cost per unit of mass C E 1/2 ρ C m Better to consider cost always as a function of mass Woods good for Commercial Oar Thursday, October 4, 2018 Materials Selection 55

56 The Strength design The Strength design is important to avoid plastic collapse or maybe not Thursday, October 4, 2018 Materials Selection 56

57 The Strength design The Strength design is important to avoid plastic collapse Your are here Thursday, October 4, 2018 Materials Selection 57

58 Lightest Tie-Rod (Traction conditions) Case Study 6: Find the Lightest STRONG Tie-Rod L DATA F =1000 N Dimensions: Length: 300 mm Thickness = 1 mm Width = 25 mm F Objective Minimize the mass Constraints Support tensile load F without yielding Length L Free Variables Area (A) of the crosssection Choice of the material In Traction, the shape of the cross-section is not important m = A L ρ A = m L ρ From material F A σ y m F L ρ σ y Thursday, October 4, 2018 Materials Selection 58

59 Ashby Diagrams Thursday, October 4, 2018 Materials Selection 59

60 CES Case Study 6: Find the Lightest STRONG Tie-Rod m σ y ρ Thursday, October 4, 2018 Materials Selection 60

61 Lightest Tie-Rod (Traction conditions) Case Study 6: Find the Lightest STRONG Tie-Rod m F L ρ σ y m σ y ρ It is possible to do as before, but let s calculate the maximum F on the precedent Tie-Rod Material Weight (kg) Width and Thickness (mm) Al Alloys 1,25 63 Stainless Steel (σ y = 600 MPa; ρ = 7800 kg/m 3 ) Wood (σ y = 50 MPa; ρ = 700 kg/m 3 ) Al Alloys (σ y = 270 Mpa; ρ = 75 kg/m 3 ) F m L σ y 0 kn = 416 kn ρ X kn Elastic Throughout Plastic deformation/ Collapse Thursday, October 4, 2018 Materials Selection 61

62 Lightest Beam (Bending conditions) Case Study 7: Find the Lightest STRONG Beam L M f Objective Minimize the mass Constraints Stiffness specified Length L Square shape Free Variables Area (A) of the cross-section Choice of the material C 1 = 3 Length: 300 mm Hypothesis: y max = h/2 σ max σ 100 N σ max = M f y max I M = Moment σ = Stress y max = max distance from the neutral axis σ max σ f m = A L ρ A = m L ρ Beam: Square Section b=h Since A = b 2 I = bh3 12 A3/2 6 Thursday, October 4, 2018 Materials Selection 62

63 Lightest Beam (Bending conditions) Case Study 7: Find the Lightest STRONG Beam I = bh2 6 A3/2 6 σ max = M f y max I = M f h 2 bh 3 12 m = A L ρ A = m L ρ = M f I σ f σ f M f 6 A 3/2 = M f 6 L 3/2 ρ 3/2 m 3/2 m (M f 6) 2/3 L ρ σ f 2/3 m σ y 2/3 ρ Thursday, October 4, 2018 Materials Selection 63

64 CES Case Study 7: Find the Lightest STRONG Beam m σ y 2/3 ρ Thursday, October 4, 2018 Materials Selection 64

65 Lightest Beam (Bending conditions) Case Study 7: Find the Lightest STRONG Beam I = bh2 6 A3/2 6 σ max = M f y max I = M f h 2 bh 3 12 m = A L ρ A = m L ρ It is always better to choose a shape that uses less material to provide the same strength TO SUPPORT BENDING = M f I σ f m σ f M f 6 A 3/2 = M f 6 L 3/2 ρ 3/2 m 3/2 m (M f 6) 2/3 L σ y 2/3 ρ ρ σ f 2/3 Thursday, October 4, 2018 Materials Selection 65

66 Lightest Panel (Bending conditions) Case Study 8: Find the Lightest STRONG Panel L M f Objective Minimize the mass Constraints Stiffness specified Length L and b specified Free Variables h (thickness) of the cross-section Choice of the material C 1 = 3 Length: 300 mm Hypothesis: y max = h/2 σ max σ 100 N σ max = M f y max I σ f m = A L ρ A = m L ρ Y X b = 25 mm I = bh2 6 A2 b 6 h Since A = bh h = A b Thursday, October 4, 2018 Materials Selection 66

67 Lightest Panel (Bending conditions) Case Study 8: Find the Lightest STRONG Panel I = bh2 6 A2 b 6 σ max = M f y max I = M f b 6 A 2 m = A L ρ A = m L ρ = M f I σ f σ f M f b 6 A 2 = M f 6 b L 2 ρ 2 m 2 m (M f 6 b) 1/2 L ρ σ f 1/2 m σ y 1/2 ρ Thursday, October 4, 2018 Materials Selection 67

68 CES Case Study 8: Find the Lightest STRONG Panel m σ y 1/2 ρ Thursday, October 4, 2018 Materials Selection 68

69 Lightest Panel (Bending conditions) Case Study 8: Find the Lightest STRONG Panel I = bh2 6 A3/2 6 σ max = M f y max I = M f h 2 bh 3 12 m = A L ρ A = m L ρ It is always better to choose a shape that uses less material to provide the same strength TO SUPPORT BENDING = M f I σ f m σ f M f 6 A 3/2 = M f 6 L 3/2 ρ 3/2 m 3/2 m (M f 6 b) 1/2 L σ y 1/2 ρ ρ σ f 1/2 Thursday, October 4, 2018 Materials Selection 69

70 Summary (to minimize the mass) Stiffness Traction : Stiffness Bending (Beam): Stiffness Bending (Panel): Strength Traction : E ρ σ y ρ E 1/2 ρ σ y 2/3 ρ Strength Bending : E 1/3 ρ σ y 1/2 ρ Strength Bending (Panel): Thursday, October 4, 2018 Materials Selection 70

71 Case Study 9: Materials for Constructions Some data : Nowadays, half the expense of building a house is the cost of the materials Family house : 200 tons Large apartment block : 20,000 tons Thursday, October 4, 2018 Materials Selection 71

72 Case Study 9: Materials for Constructions Mr. Pincopallo asks a new cover Understand the problem and translate it in selection criteria, thus properties, A cover has 3 BROAD ROLES: Cladding: Protection from the environment The frames: Mechanical support Internal surfacing: Control heat, light and sound Thursday, October 4, 2018 Materials Selection 72

73 Case Study 9: Materials for Constructions (Structural Frame) Objective Minimize the cost Constraints Length L specified Stiffness: must not deflect too much under loads Strength: must not fall under design loads Free Variables Area (A) of the cross-section Choice of the material Hypothesis: F S δ min = S Floor joints are beams, loaded in bending. F δ S min = C 1EI L 3 m = A L ρ A = m L ρ C = m C m m = C C m C 4 π S L5 C 3 E 1/2 1/2 ρ C m ρ C m E 1/2 Thursday, October 4, 2018 Materials Selection 73

74 Case Study 9: Materials for Constructions (Structural Frame) Objective Minimize the cost Constraints Length L specified Stiffness: must not deflect too much under loads Strength: must not fall under design loads Free Variables Area (A) of the cross-section Choice of the material I = bh3 12 A3/2 6 Floor joints are beams, loaded in bending. σ max = M f y max I σ f m = A L ρ A = m L ρ C = m C m m = C C m C (M f 6) 2/3 L ρ C m σ f 2/3 C σ f 2/3 ρ C m Thursday, October 4, 2018 Materials Selection 74

75 Case Study 9: Materials for Constructions ATTENTION!!! Selection with the cost/kg and with the cost/m 3 is DIFFERENT Thursday, October 4, 2018 Materials Selection 75

76 Case Study 9: Materials for Constructions C E 1/2 ρ C m Thursday, October 4, 2018 Materials Selection 76

77 Case Study 9: Materials for Constructions C σ f 2/3 ρ C m Thursday, October 4, 2018 Materials Selection 77

78 Case Study : Materials for Springs Thursday, October 4, 2018 Materials Selection 78

79 Case Study 10: Materials for Small Springs Objective Maximize stored elastic energy Constraints No failure σ < σ f throughout the spring Free Variables Choice of the material Condition of elasticity σ y σ σ = E ε ε = σ E SMALL?? V FREE VARIABLE!! dv = dxdydz [Solid Mechanics Part I Kelly] Thursday, October 4, 2018 Materials Selection 79

80 Case Study 10: Materials for Small Springs σ y σ Strain energy density ε = σ E dv = dxdydz m = V ρ W el = 1 2 න σ ε dv = 1 σ ε V 2 Total strain energy in the piece considered W el = σ y 2 2E = M 1 Total strain energy PER UNIT OF VOLUME [Solid Mechanics Part I Kelly] Thursday, October 4, 2018 Materials Selection 80

81 Case Study 10: Materials for Small Springs CFRP = Comparable in performance with steel; expensive [TRUCK SPRINGS] STEEL = The traditional choice: easily formed and heat treated TITANIUM = Expensive, corrosion resistant RUBBERS = have Excellent M 1 but low tensile strength high loss factor NYLON = Inexpensive and easily shaped, but high loss factor [CHILDREN S TOYS] Thursday, October 4, 2018 Materials Selection 81

82 Case Study Materials for Springs PAY ATTENTION W el = σ y 2 3E W el = σ y 2 2E Valid for axial springs Because much of the material is not fully loaded For torsion springs (less efficient) Thursday, October 4, 2018 Materials Selection 82

83 Case Study Materials for Springs W el = σ y 2 4E For leaf springs (less efficient) Thursday, October 4, 2018 Materials Selection 83

84 Case Study 9: Materials for Light Springs σ y σ Strain energy density ε = σ E dv = dxdydz m = V ρ W el = 1 2 න σ ε dv = 1 σ ε V 2 Total strain energy in the piece considered LIGHT?? V FREE VARIABLE!! ρ W el ρ = σ f 2 E ρ = M 2 Total strain energy PER UNIT OF MASS 2 [Solid Mechanics Part I Kelly] Thursday, October 4, 2018 Materials Selection 84

85 Case Study 11: Materials for Light Springs CFRP = Comparable in performance with steel; expensive [TRUCK SPRINGS] RUBBERS = 20 times better than Steel; but low tensile strength high loss factor NYLON = Inexpensive and easily shaped, but high loss factor [CHILDREN S TOYS] METALS Almost DISAPPEARED Thursday, October 4, 2018 Materials Selection 85

86 Case Study 12: Materials for Car Body Some context Car Evolution 1932 Ford Model B 1934 Bonnie and Clyde car Thursday, October 4, 2018 Materials Selection 86

87 Case Study 12: Materials for Car Body Some context Car Evolution 1932 Ford Model B 1970 Buick GSX 2010 Ferrari 458 Italia Km/h 325 km/h 184 km/h 80 km/h Thursday, October 4, 2018 Materials Selection 87

88 Case Study 12: Materials for Car Body Deformation?? ENERGY CONSUMPTION At first, automotive industry move to too deformable cars and then move to have a mix FOR PEOPLE SAFETY Sometimes exaggerate Thursday, October 4, 2018 Materials Selection 88

89 Case Study 12: Materials for Car Body (Car Hood or Car Door) Objective Maximize plastic deformation at high load Constraints Geometry High σ y Division for price Consider manufacture Free Variables Choice of the material W el LIGHT?? m ρ = σ f 2 E ρ = M 2 Total strain energy PER UNIT OF MASS 2 Thursday, October 4, 2018 Materials Selection 89

90 Case Study 12: Materials for Car Body (Car Hood or Car Door) W el LIGHT?? m ρ = σ f 2 E ρ = M 2 Total strain energy PER UNIT OF MASS 2 Objective Maximize plastic deformation at high load Constraints Geometry High σ y Division for price Consider manufacture Free Variables Choice of the material Steps: Stiffness selection (Take off flexible materials) Yield strength selection to minimize the costs (Automotive) Minimum Yield Strength Maximization of stored energy Thursday, October 4, 2018 Materials Selection 90

91 Case Study 12: Materials for Car Body (Car Hood or Car Door) m E ρ Thursday, October 4, 2018 Materials Selection 91

92 Case Study 12: Materials for Car Body (Car Hood or Car Door) Thursday, October 4, 2018 Materials Selection 92

93 Case Study 12: Materials for Car Body (Car Hood or Car Door) + minimum σ y (200 MPa) Thursday, October 4, 2018 Materials Selection 93

94 Case Study 12: Materials for Car Body (Car Hood or Car Door) W el = σ f 2 2E = M 1 Total strain energy PER UNIT OF VOLUME Thursday, October 4, 2018 Materials Selection 94

95 Case Study 12: Materials for Car Body (Car Hood or Car Door) W el ρ = σ f 2 E ρ = M 2 Total strain energy PER UNIT OF MASS 2 Thursday, October 4, 2018 Materials Selection 95

96 Case Study 12: Materials for Car Body (Car Hood or Car Door) W el ρ = σ f 2 E ρ = M 2 Total strain energy PER UNIT OF MASS 2 Lamborghini Huracan 2015 Thursday, October 4, 2018 Materials Selection 96

97 Case Study 12: Materials for Car Body (Car Hood or Car Door) Manufacturing Consideration [ Thursday, October 4, 2018 Materials Selection 97

98 Case Study 12: Materials for Car Body (Car Hood or Car Door) Vacuum and pressure bag molding Not adapt for high productions No way for a commercial car Thursday, October 4, 2018 Materials Selection 98

99 Case Study 12: Materials for Car Body (Car Hood or Car Door) And which Metal? W el ρ = σ f 2 E ρ = M 2 Total strain energy PER UNIT OF MASS 2 Thursday, October 4, 2018 Materials Selection 99

100 Case Study 12: Materials for Car Body (Car Hood or Car Door) Stamping Adapt for high productions Thursday, October 4, 2018 Materials Selection 100

101 Case Study 12: Materials for Car Body (Car Hood or Car Door) PROCESSABILITY Metals easy to stamp Thursday, October 4, 2018 Materials Selection 101

102 Case Study 12: Materials for Car Body (Car Hood or Car Door) W el ρ = σ f 2 E ρ = M 2 Total strain energy PER UNIT OF MASS 2 + Minimum Formability (4) Thursday, October 4, 2018 Materials Selection 102

103 Case Study 12: Materials for Car Body (Car Hood or Car Door) PROCESSABILITY Metals easy to stamp Thursday, October 4, 2018 Materials Selection 103

104 Case Study 12: Materials for Car Body (Car Hood or Car Door) W el ρ = σ f 2 E ρ = M 2 Total strain energy PER UNIT OF MASS 2 + Minimum Formability (4) + Minimum Brazability (3) Thursday, October 4, 2018 Materials Selection 104

105 Case Study 12: Materials for Car Body (Car Hood or Car Door) BMW M3 Low alloy steel Audi A8 Al-alloys Deeper selection? LEVEL 3 [ Thursday, October 4, 2018 Materials Selection 105

106 Materials Selection Steps Optional: Thursday, October 4, 2018 Materials Selection 106