MT-1620 al.2002 Unit 5 Engineering Constants Readings Rivello 31-35,39,3.11 Paul A Lagace, Ph. D Professor of aeronautics Astronautics and Engineering Systems Paul A Lagace @2001
MIT - 16.20 Fall, 2002 Unit 5 Engineering Constants Readings: Rivello 3.1 - 3.5, 3.9, 3.11 Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering Systems Paul A. Lagace © 2001
MT-1620 Fall 2002 We do not characterize materials by their enno the emon are mpg useful in doing transformations manipulations etc We characterize materials by their ENGINEERING CONSTANTS (or, Elastic Constants) (what we can physically measure There are 5 types 1. Longitudinal young s)(Extensional)Modulus: relates extensional strain in the direction of loading to stress in the direction of loading (3 of these) 2. Poissons Ratio: relates extensional strain in the loading direction to extensional strain in another direction (6 of these.. only 3 are independent Paul A Lagace @2001 it 5-p. 2
MIT - 16.20 Fall, 2002 We do not characterize materials by their Emnpq. The Emnpq are useful in doing transformations, manipulations, etc. We characterize materials by their “ENGINEERING CONSTANTS” (or, Elastic Constants) (what we can physically measure) There are 5 types 1. Longitudinal (Young’s) (Extensional) Modulus: relates extensional strain in the direction of loading to stress in the direction of loading. (3 of these) 2. Poisson’s Ratio: relates extensional strain in the loading direction to extensional strain in another direction. (6 of these…only 3 are independent) Paul A. Lagace © 2001 Unit 5 - p. 2
MT-1620 Fall 2002 3. Shear Modulus: relates shear strain in the plane of shear loading to that shear stress (3 of these) 4. Coefficient of mutual Influence relates shear strain due to shear stress in that plane to extensional strain or, relates extensional strain due to extensional stress to shear strain (up to 18 of these 5. Chentsoy coefficient relates shear strain due to shear stress in that plane to shear strain in another plane 6 of these Let's be more specific Longitudinal Modulus 1)E or Exx or E, or Ex: contribution of E,1 to O11 2 or Ew or E2 or E: contribution of E22 to O22 3)E33or Ez or E3 or E: contribution of e33 to O33 Paul A Lagace @2001 lt 5-p. 3
MIT - 16.20 Fall, 2002 3. Shear Modulus: relates shear strain in the plane of shear loading to that shear stress. (3 of these) 4. Coefficient of Mutual Influence: relates shear strain due to shear stress in that plane to extensional strain or, relates extensional strain due to extensional stress to shear strain. (up to 18 of these) 5. Chentsov Coefficient: relates shear strain due to shear stress in that plane to shear strain in another plane. (6 of these) Let’s be more specific: 1. Longitudinal Modulus 1) E11 or Exx or E1 or Ex: contribution of ε11 to σ11 2) E22 or Eyy or E2 or Ey: contribution of ε22 to σ22 3) E33 or Ezz or E3 or Ez: contribution of ε33 to σ33 Paul A. Lagace © 2001 Unit 5 - p. 3
MT-1620 Fall 2002 In general mm mm due to Omm applied only mm ( no summation on m) 2. Poisson's Ratios (negative ratios) 1)v12 or vx:(negative of) ratio of E22 to e,1 due to O, 2)v13 or vxz:(negative of) ratio of E33 to e11 due to 3)V23 or Vyz: (negative of) ratio of E33 to E22 due to O22 4)V21 or vox:(negative of) ratio of E, to E22 due to O22 5)V31 or Vzx:(negative of) ratio of E,1 to e33 due to 33 6)V32 or vxy:(negative of) ratio of E22 to E33 due to O33 In general: Vom Emm due to onn applied only (forn≠m Important:vm≠Vm Paul A Lagace @2001 lt 5-p. 4
MIT - 16.20 Fall, 2002 In general: Emm = σmm due to σmm applied only εmm (no summation on m) 2. Poisson’s Ratios (negative ratios) 1) ν12 or νxy: (negative of) ratio of ε22 to ε11 due to σ11 2) ν13 or νxz: (negative of) ratio of ε33 to ε11 due to σ11 3) ν23 or νyz: (negative of) ratio of ε33 to ε22 due to σ22 4) ν21 or νyx: (negative of) ratio of ε11 to ε22 due to σ22 5) ν31 or νzx: (negative of) ratio of ε11 to ε33 due to σ33 6) ν32 or νzy: (negative of) ratio of ε22 to ε33 due to σ33 In general: νnm = − εmm due to σnn applied only εnn (for n ≠ m) Important: νnm ≠ νmn Paul A. Lagace © 2001 Unit 5 - p. 4
MT-1620 al.2002 However, these are not all independent. There are relations known as "reciprocity relations"(3 of them) 21-11 12-22 31-11 13-33 v E 2-22 23-33 3. Shear modul 1)G12 or Gxy or Gg: contribution of (2)E12 to O12 2)G13or Gx or G: contribution of (2)E13 to O13 3)G23 or Gvz or G4: contribution of (2)E23 to O23 general:Gmm=mm due to Omn applied only factor of 2 here since it relates physical quantities shear stress shear deformation(angular charge Paul A Lagace @2001 lt 5-p. 5
MIT - 16.20 Fall, 2002 However, these are not all independent. There are relations known as “reciprocity relations” (3 of them) ν21 E11 = ν12 E22 ν31 E11 = ν13 E33 ν32 E22 = ν23 E33 3. Shear Moduli 1) G12 or Gxy or G6: contribution of (2)ε12 to σ12 2) G13 or Gxz or G5: contribution of (2)ε13 to σ13 3) G23 or Gyz or G4: contribution of (2)ε23 to σ23 In general: Gmn = σmn due to σmn applied only 2εmn factor of 2 here since it relates physical quantities shear stress ⇒ G τmn mn = shear deformation (angular charge) γ mn Paul A. Lagace © 2001 Unit 5 - p. 5
MT-1620 al.2002 4. Coefficients of Mutual Influence (negative ratios (also known as"coupling coefficients") Note: need to use contracted notation here 1)n16:(negative of) ratio of (2)E12 to E,1 due to o, 2)m:(negative of) ratio of E to(2)E12 due to O12 3)126 (5) (7)m14 24 (6) 41 (10)n4 m34 (13)n 7)n5 12)n43 (14)n51 (16) (18)n53 5. Chentsov Coefficients (negative ratios) 1) m46.(negative o of)ratio of (2 )E,2 to(2)&23 due to o23 2)n64:(negative of) ratio of(2)E23 to(2)E,2 due to O12 3)n4:(negative of) ratio of(2)E13 to(2)E23 due to 23 4)n54:(negative of ratio of (2)E23 to(2)E13 due to O13 5)n56:(negative of) ratio of(2)E12 to(2)E13 due to 013 6)ma5:(negative of ratio of(2)E13 to (2)E12 due to O12 Paul A Lagace @2001 Unit 5-p
MIT - 16.20 Fall, 2002 4. Coefficients of Mutual Influence (negative ratios) (also known as “coupling coefficients”) Note: need to use contracted notation here: 1) η16: (negative of) ratio of (2)ε12 to ε11 due to σ11 2) η61: (negative of) ratio of ε11 to (2)ε12 due to σ12 3) η26 (5) η36 (7) η14 (9) η24 4) η62 (6) η63 (8) η41 (10) η42 11) η34 (13) η15 (15) η25 (17) η35 12) η43 (14) η51 (16) η52 (18) η53 5. Chentsov Coefficients (negative ratios) 1) η46: (negative of) ratio of (2)ε12 to (2)ε23 due to σ23 2) η64: (negative of) ratio of (2)ε23 to (2)ε12 due to σ12 3) η45: (negative of) ratio of (2)ε13 to (2)ε23 due to σ23 4) η54: (negative of) ratio of (2)ε23 to (2)ε13 due to σ13 5) η56: (negative of) ratio of (2)ε12 to (2)ε13 due to σ13 6) η65: (negative of) ratio of (2)ε13 to (2)ε12 due to σ12 Paul A. Lagace © 2001 Unit 5 - p. 6
MT-1620 al.2002 Again, since these are physical ratios, engineering shear strain factor of 2 is used Again, these are not all independent. Just as for the Poisson's ratios there are reciprocity relations. these involve the longitudinal and shear moduli(since these couple extensional and shear or shear to shear). There are 12 of them M61 E,= m16 G6 n51E1=n15 n14G4 n62E2=126G6m52E2=n2565 2E2 44 163E3=n36G6n53E3=n35G5 n43E3=n34(4 Paul A Lagace @2001 Unit 5-p. 7
MIT - 16.20 Fall, 2002 Again, since these are physical ratios, engineering shear strain factor of 2 is used. Again, these are not all independent. Just as for the Poisson’s ratios, there are reciprocity relations. These involve the longitudinal and shear moduli (since these couple extensional and shear or shear to shear). There are 12 of them: η61 E1 = η16 G6 η51 E1 = η15 G5 η41 E1 = η14 G4 η62 E2 = η26 G6 η52 E2 = η25 G5 η42 E2 = η24 G4 η63 E3 = η36 G6 η53 E3 = η35 G5 η43 E3 = η34 G4 Paul A. Lagace © 2001 Unit 5 - p. 7
MT-1620 al.2002 in general nm -m mn n (m=1, 2, 3) no sum and n46G6 n45G5=n54G G n6565 general nm m (m =4, 5, 6 no sum Paul A Lagace @2001 Unit 5-p. 8
MIT - 16.20 Fall, 2002 in general: ηnm Em = ηmn Gn (m = 1, 2, 3) no sum (n = 4, 5, 6) and η46 G6 = η64 G4 η45 G5 = η54 G4 η56 G6 = η65 G5 in general: ηnm Gm = ηmn Gn (m = 4, 5, 6) no sum (m ≠ n) Paul A. Lagace © 2001 Unit 5 - p. 8
MT-1620 al.2002 This gives 21 independent (at most) engineering constants Total 3 En 6vnm 3 Gm 18n nm Indp't. 3 21 Now that we have defined the terms we wish to write the engineering stress-strain equations Recall compliances inp pq and consider only the first equation 11110511+511022+3113033 +2S123023+2 113013+2S 1112012 (we'lI have to use contracted notation, so.) E1=S101+S1202+S133+S1404+S1505+S1606 (Note: 2's disappear!) Paul A Lagace @2001 it 5-p. 9
MIT - 16.20 Fall, 2002 This gives 21 independent (at most) engineering constants: Total 3 En 6νnm 3 Gm 18ηnm 6 ηnm ↓ ↓ ↓ ↓ ↓ Indp’t.: 3 3 3 9 3 = 21 --> Now that we have defined the terms, we wish to write the “engineering stress-strain equations” Recall compliances: εmn = Smnpq σpq and consider only the first equation: ε11 = S1111 σ11 + 2S1123 ε1 = S11 σ1 + Paul A. Lagace © 2001 + S1122 σ22 + S1133 σ33 σ23 + 2S1113 σ13 + 2S1112 σ12 (we’ll have to use contracted notation, so…) S12 σ2 + S13 σ3 + S14 σ4 + S15 σ5 + S16 σ6 (Note: 2’s disappear!) Unit 5 - p. 9
MT-1620 al.2002 Consider each of the compliance terms separately Case 1: Only O,1 applied 1101 and we know E due to O, only Case 2: Only O22 applied E,=S We need two steps here The direct relation to o2 is from e2 due to o2 only and we know due to o2 only due to o2 only Paul A Lagace @2001 Unit 5-p. 10
MIT - 16.20 Fall, 2002 Consider each of the compliance terms separately: Case 1: Only σ11 applied ε1 = S11 σ1 and we know E1 = σ1 due to σ1 only ε1 1 ⇒ S11 = E1 Case 2: Only σ22 applied ε1 = S12 σ2 We need two steps here. The direct relation to σ2 is from ε2: E2 = σ2 due to σ2 only ε2 and we know ν21 = − ε1 due to σ2 only ε2 ⇒ σ2 = − E2 due to σ2 only ε1 ν21 Paul A. Lagace © 2001 Unit 5 - p. 10