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 MIT - 16.20 Fall, 2002 We do not characterize materials by their E mnpq . The E mnpq 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. Longitudina l (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 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. Longitudina l Modulus 1) E 11 or E xx or E 1 or E x : contribution of ε 11 to σ 11 2) E 22 or E yy or E 2 or E y : contribution of ε 22 to σ 22 3) E 33 or E zz or E 3 or E z : contribution of ε 33 to σ 33 Paul A. Lagace ? 2001 Unit 5 - p . 3 MIT - 16.20 Fall, 2002 In general : E mm = σ mm due to σ mm applied only ε mm (no summation on m) 2. Poisson ’ s Ratios ( negative ratios) 1) ν 12 or ν xy : (negative of) ratio of ε 22 to ε 11 due to σ 11 2) ν 13 or ν xz : (negative of) ratio of ε 33 to ε 11 due to σ 11 3) ν 23 or ν yz : (negative of) ratio of ε 33 to ε 22 due to σ 22 4) ν 21 or ν yx : (negative of) ratio of ε 11 to ε 22 due to σ 22 5) ν 31 or ν zx : (negative of) ratio of ε 11 to ε 33 due to σ 33 6) ν 32 or ν zy : (negative of) ratio of ε 22 to ε 33 due to σ 33 In general : ν nm = ? ε mm due to σ nn applied only ε nn (for n ≠ m) Important : ν nm ≠ ν mn Paul A. Lagace ? 2001 Unit 5 - p . 4 MIT - 16.20 Fall, 2002 However, these are not all independent. There are relations known as “reciprocity relations ” (3 of them) ν 21 E 11 = ν 12 E 22 ν 31 E 11 = ν 13 E 33 ν 32 E 22 = ν 23 E 33 3. Shear Moduli 1) G 12 or G xy or G 6 : contribution of (2) ε 12 to σ 12 2) G 13 or G xz or G 5 : contribution of (2) ε 13 to σ 13 3) G 23 or G yz or G 4 : contribution of (2) ε 23 to σ 23 In general : G mn = σ mn due to σ mn applied only 2 ε mn factor of 2 here since it relates physical quantities shear stress τ mn ? G mn = shear deformation ( angular ch arg e ) γ mn Paul A. Lagace ? 2001 Unit 5 - p . 5 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) η 16 : (negative of) ratio of (2) ε 12 to ε 11 due to σ 11 2) η 61 : (negative of) ratio of ε 11 to (2) ε 12 due to σ 12 3) η 26 (5) η 36 (7) η 14 (9) η 24 4) η 62 (6) η 63 (8) η 41 (10) η 42 11) η 34 (13) η 15 (15) η 25 (17) η 35 12) η 43 (14) η 51 (16) η 52 (18) η 53 5. Chentsov Coefficients ( negative ratios) 1) η 46 : (negative of) ratio of (2) ε 12 to (2) ε 23 due to σ 23 2) η 64 : (negative of) ratio of (2) ε 23 to (2) ε 12 due to σ 12 3) η 45 : (negative of) ratio of (2) ε 13 to (2) ε 23 due to σ 23 4) η 54 : (negative of) ratio of (2) ε 23 to (2) ε 13 due to σ 13 5) η 56 : (negative of) ratio of (2) ε 12 to (2) ε 13 due to σ 13 6) η 65 : (negative of) ratio of (2) ε 13 to (2) ε 12 due to σ 12 Paul A. Lagace ? 2001 Unit 5 - p . 6 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 Poisso n ’ 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: η 61 E 1 = η 16 G 6 η 51 E 1 = η 15 G 5 η 41 E 1 = η 14 G 4 η 62 E 2 = η 26 G 6 η 52 E 2 = η 25 G 5 η 42 E 2 = η 24 G 4 η 63 E 3 = η 36 G 6 η 53 E 3 = η 35 G 5 η 43 E 3 = η 34 G 4 Paul A. Lagace ? 2001 Unit 5 - p . 7 MIT - 16.20 Fall, 2002 in general: η nm E m = η m n G n (m = 1, 2, 3) no sum (n = 4, 5, 6) and η 46 G 6 = η 64 G 4 η 45 G 5 = η 54 G 4 η 56 G 6 = η 65 G 5 in general: η nm G m = η m n G n (m = 4, 5, 6) no sum (m ≠ n) Paul A. Lagace ? 2001 Unit 5 - p . 8 MIT - 16.20 Fall, 2002 This gives 21 independent (at most) engineering constants: Total 3 E n 6 ν nm 3 G m 18 η nm 6 η nm ↓ ↓ ↓ ↓ ↓ 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: ε mn = S mnpq σ pq and consider only the first equation: ε 11 = S 1111 σ 11 + 2 S 1123 ε 1 = S 1 1 σ 1 + Paul A. Lagace ? 2001 + S 1122 σ 22 + S 1133 σ 33 σ 23 + 2 S 1113 σ 13 + 2 S 1112 σ 12 (we’ ll have to use contracted notation, so … ) S 1 2 σ 2 + S 1 3 σ 3 + S 14 σ 4 + S 1 5 σ 5 + S 16 σ 6 ( Note : 2’ s disappear!) Unit 5 - p . 9 MIT - 16.20 Fall, 2002 Consider each of the compliance terms separately: Case 1 : Only σ 11 applied ε 1 = S 11 σ 1 and we know E 1 = σ 1 due to σ 1 only ε 1 1 ? S 11 = E 1 Case 2 : Only σ 22 applied ε 1 = S 12 σ 2 We need two steps here. The direct relation to σ 2 is from ε 2 : E 2 = σ 2 due to σ 2 only ε 2 and we know ν 21 = ? ε 1 due to σ 2 only ε 2 ? σ 2 = ? E 2 due to σ 2 only ε 1 ν 21 Paul A. Lagace ? 2001 Unit 5 - p . 1 0 MIT - 16.20 Fall, 2002 Thus: S 12 = ? ν 21 (due to σ 2 only) E 2 --> to make this a bit simpler (for later purposes), we recall the reciprocity relation: ν 12 E 2 = ν 2 1 E 1 ν 21 ν 12 ? = E 2 E 1 Thus: S 12 = ? ν 12 E 1 Case 3 : Only σ 3 applied In a similar manner we get: S 13 = ? ν 13 E 1 Case 4 : Only σ 4 ( σ 23 ) applied ε 1 = S 14 σ 4 Again, two steps are needed. Paul A. Lagace ? 2001 Unit 5 - p . 1 1 MIT - 16.20 Fall, 2002 First the direct relation of ε 4 to σ 4 : G 4 = σ 4 due to σ 4 only ε 4 This is engineering strain! and then: η 41 = ? ε 1 due to σ 4 only ε 4 σ 4 = ? G 4 ? ε 1 η 41 ? S 14 = ? η 41 due to σ 4 only G 4 Again, we use a reciprocity relation to get: η 41 η 14 = G 4 E 1 ? S 14 = ? η 14 due to σ 4 only E 1 We do the same for Case 5 of only σ 5 applied, and Case 6 of only σ 6 applied to Paul A. Lagace ? 2001 Unit 5 - p . 1 2 MIT - 16.20 Fall, 2002 get: S 15 = ? η 15 E 1 and S 16 = ? η 16 E 1 With all this we finally get: 1 1 44 ? ησ 5 ? ησ 6 ] ε 1 = E 1 [ σ 1 ? ν 1 2 σ 2 ? ν 1 3 σ 3 ? ησ 1 5 1 6 we used the reciprocity relations so we could “pull out ” this common factor. We can do this for all the other cases. In general we write: 1 6 n m = 1 Note : ν nn = -1 If we let η ’s be ν ’s Paul A. Lagace ? 2001 Unit 5 - p . 1 3 ε n = ? E ∑ ν nm σ m MIT - 16.20 Fall, 2002 “Engineering ” Stress-Strain Equations General Form 1 ε 1 = E 1 [ σ 1 ? ν 1 2 σ 2 ? ν 1 3 σ 3 ? ησ 1 5 1 6 1 44 ? ησ 5 ? ησ 6 ] 1 ε 2 = E 2 [ ? ν 21 σ 1 + σ 2 ? ν 2 3 σ 3 ? ησ 4 ? ησ 5 ? ησ 6 ] 24 2 5 26 1 ε 3 = E 3 [ ? ν 31 σ 1 ? νσ 2 + σ 3 ? η 3 4 σ 4 ? η 35 σ 5 ? η 3 6 σ 6 ] ν 32 1 γ 4 = ε 4 = G 4 [ ? η 41 σ 1 ? η 4 2 σ 2 ? η 43 σ 3 + σ 4 ? η 4 5 σ 5 ? η 46 σ 6 ] 1 γ 5 = ε 5 = G 5 [ ? η 51 σ 1 ? η 5 2 σ 2 ? η 53 σ 3 ? η 5 4 σ 4 + σ 5 ? η 56 σ 6 ] 1 γ 6 = ε 6 = G 6 [ ? ησ 1 ? η 6 2 σ 2 ? η 63 σ 3 ? η 6 4 σ 4 ? η 65 σ 5 + σ 6 ] 61 Paul A. Lagace ? 2001 Unit 5 - p . 1 4 MIT - 16.20 Fall, 2002 In general: 1 6 ε n = ? E ∑ ν nm σ m n m = 1 Note : ν nn = -1 η ’s --> ν ’s We have developed these for a fully anisotropic material. Again, there are currently no useful engineering materials of this nature. Thus, these would need to be reduced accordingly. Paul A. Lagace ? 2001 Unit 5 - p . 1 5 MIT - 16.20 Fall, 2002 Orthotropic Case: In material principal axes, there is no coupling between extension and shear and no coupling between planes of shear, so: all η mn = 0 Thus, only the following constants remain: E 1 ν 12 , ν 21 G 12 E 2 ν 13 , ν 31 G 13 E 3 ν 23 , ν 32 G 23 3 + 3 + 3 = 9 (same as E mnpq , better be ! ) Paul A. Lagace ? 2001 Unit 5 - p . 1 6 σ MIT - 16.20 Fall, 2002 So the six equations become: 1 ε 1 = E 1 [ σ 1 ? ν 1 2 σ 2 ? ν 13 σ 3 ] 1 ε 2 = E 2 [ ? ν 21 σ 1 + σ 2 ? ν 2 3 σ 3 ] 1 ε 3 = E 3 [ ? ν 31 σ 1 ? ν 3 2 σ 2 + σ 3 ] 1 ε 4 = σ 4 G 23 1 ε 5 = σ 5 G 13 1 ε 6 = σ 6 G 12 Paul A. Lagace ? 2001 Unit 5 - p . 1 7 ? ? 0 MIT - 16.20 Fall, 2002 matrix form: ? ε 1 ? ? 1 ? ν 12 ? ν 13 0 0 0 ? ? σ 1 ? ? ? ?? E 1 E 1 E 1 ? ? ? ? ? ε 2 ? ? ? ? ? ? ? ν 21 E 1 2 ? ν 23 0 0 0 ? ? ? σ 2 ? ? ? ? E 2 E 2 ? ? ? ε 3 ? ?? ? ν 31 ? ν 32 1 0 0 0 ? ? ? σ 3 ? ? ? ? ? ? ? ? ? = ? E 3 E 3 E 3 ? ? ? ? ε 4 ? ? 0 0 0 1 0 0 ? ? σ 4 ? ? ? ? G 23 ? ? ? ? ? ? 1 ? ? ? ? ε 5 ? ? 0 0 0 0 G 13 0 ? ? σ 5 ? ? ? ? ? ? ? ? 1 ? ? ? ? ? ? 0 0 0 0 0 G 12 ?? ? σ 6 ? ? ε 6 ? ? ε = S σ ~ ~ ~ this is, in fact, the compliance matrix Paul A. Lagace ? 2001 Unit 5 - p . 1 8 MIT - 16.20 Fall, 2002 Thus: ? If we know the engineering constants (through tests -- this is upcoming) ? Relate engineering constants to S mnpq ? Get E mnpq by inversion of S mnpq matrix (combine steps to directly get relationships between E mnpq and the engineering constants) Isotropic Case As we noted in the last unit, as we get to materials with less elastic constants (< 9) than an orthotropic material, we no longer have any more zero terms in the elasticity or compliance matrix, but more nonzero terms are related. For the isotropic case: ? All extensional moduli are the same: E 1 = E 2 = E 3 = E ? All Poisson ’ s ratios are the same: ν 12 = ν 21 = ν 13 = ν 31 = ν 23 = ν 32 = ν ? All shear moduli are the same: G 4 = G 5 = G 6 = G Paul A. Lagace ? 2001 Unit 5 - p . 1 9 MIT - 16.20 Fall, 2002 ? And, there is a relationship between E, ν and G (from Unified): E G = 21 + ν ) ( Thus, there are only 2 independent constants. We now have all the relationships to do the manipulations, but we need to measure the basic properties. We must therefore talk about … Testing Testing is used for a variety of purposes. Depending on the purpose, the technique and “ care” will vary. The “fidelity ” needed in the testing depends on the use. Basically, apply a load (stress) condition and measure appropriate responses: Paul A. Lagace ? 2001 Unit 5 - p . 2 0 MIT - 16.20 Fall, 2002 ? Strain ? Displacements ?F ailure (or maybe vice versa) Concerns in test specimens ? Boundary conditions and introduction of load ? Stress concentrations ? Achievement of desired stress state ? Cost and ease of use also important (again, depends on use of test) Many of these concerns will depend on the material / configuration and load condition. This generally involves issues of scale. “Properties ” of a material / structure depends on the “ scale ” at which you look at it. Scale ≈ level of homogenization (average behavior over a certain size) Paul A. Lagace ? 2001 Unit 5 - p . 2 1 MIT - 16.20 Fall, 2002 Example 1: Atoms make up a material Figure 5.1 Representation of atomic bonding as springs The behavior of the “materials ” is some combination of the atoms and their bonds. Example 2 : Composite Figure 5.2 Representation of unidirectional composite fibers in a matrix Fibers and matrix respond differently. “ Averag e ” their response to get “composite ” properties Paul A. Lagace ? 2001 Unit 5 - p . 2 2 ??? MIT - 16.20 Fall, 2002 ? Must look at responses at length of several fiber diameters Example 3: A Truss Figure 5.3 Representation of a generic truss Each truss member responds in a certain way, but we can characterize the truss behavior on a large scale by looking at the displacement between A and B ( ? l ) and get a “ truss stiffness”. Final Example : put all of these together ? Truss ? Members are composites ? Composites have fibers and matrix ? Fibers and matrix are composed of atoms ? a continuum! Paul A. Lagace ? 2001 Unit 5 - p . 2 3 MIT - 16.20 Fall, 2002 Any material is really a structure. We are making an engineering approximation by characterizing it at a certain level. ? that is the limit / assumption on “material properties ” and the tests we use to measure them. (so elastic constants are an engineering representation of “micromechanical ” behavior) Thus, let ’ s keep in mind such issues of scale as we consider test methods. Must measure at or above pertinent scale of homogenization / averaging. There are many different types of tests, but in dealing with elastic constants, there are 3 basic load conditions (tension, compression, shear) The test specimen will depend on the material and the load condition. 1. Tension (easiest to do) Paul A. Lagace ? 2001 Unit 5 - p . 2 4 MIT - 16.20 Fall, 2002 Figure 5.4 Tension specimens Tapered Dogbone Straight-edged coupon Bar Specimen (composites) …stress concentration problem 2. Compression Similar specimens can be used, but must beware of buckling (global and local instabilities) Possibility of local reinforcement to prevent buckling. Paul A. Lagace ? 2001 Unit 5 - p . 2 5 MIT - 16.20 Fall, 2002 3. Shear Very hard to apply pure shear. Figure 5.5 Possible shear specimens Tube Iosipescu (beam theory shows area of pure shear -- test section) Refer to : ASTM (American Society for Testing and Materials) Annual Book of Standards Voluntary test standards are contained there. Paul A. Lagace ? 2001 Unit 5 - p . 2 6 MIT - 16.20 Fall, 2002 ? what to do with test data Test data must be “ reduced ” to give the engineering constants. Figure 5.6 Typical stress-strain data (for ductile material) ? The engineering constants are defined (somewhat arbitrary) as various parts of this curve. Generally within the “ initial linear region ” . ? Often use linear regression Paul A. Lagace ? 2001 Unit 5 - p . 2 7 MIT - 16.20 Fall, 2002 Key ? stress-strain relations with engineering constants attained in this manner are valid only in the linear region. Thus, one must report: ? slope (engineering constant) ? region of applicability Note : we have not dealt with temperature effects. We will consider this later. Also note : strength / failure properties are much harder to measure. (recall Unified: average behavior vs. local / weakest link) Paul A. Lagace ? 2001 Unit 5 - p . 2 8 MIT - 16.20 Fall, 2002 We have now developed the general 3-D stress-strain relations. But we often deal with a problem where we can simplify (model as) to a 2-D system. Two important cases to next consider: Plane Stress Plane Strain Paul A. Lagace ? 2001 Unit 5 - p . 2 9