MIT - 16.20 Fall, 2002 Unit 4 Equations of Elasticity Readings : R 2.3, 2.6, 2.8 T & G 84, 85 B, M, P 5.1-5.5, 5.8, 5.9 7.1-7.9 6.1-6.3, 6.5-6.7 Jones (as background on composites) Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering Systems Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 Let ’s first review a bit … from Unified, saw that there are 3 basic considerations in elasticity: 1. Equilibrium 2. Strain - Displacement 3. Stress - Strain Relations ( Constitutive Relations) Consider each: 1. Equilibrium (3) ? Σ F i = 0, Σ M i = 0 ? Free body diagrams ? Applying these to an infinitesimal element yields 3 equilibrium equations Figure 4.1 Representation of general infinitesimal element Paul A. Lagace ? 2001 Unit 4 - p . 2 MIT - 16.20 Fall, 2002 ? σ 11 + ? σ 21 + ? σ 31 + f 1 = 0 (4-1) ? y 1 ? y 2 ? y 3 ? σ 12 + ? σ 22 + ? σ 32 + f 2 = 0 (4-2) ? y 1 ? y 2 ? y 3 ? σ 13 + ? σ 23 + ? σ 33 + f 3 = 0 (4-3) ? y 1 ? y 2 ? y 3 ? ? + σ mn m n y f 0 = 2. Strain - Displacement (6) ? Based on geometric considerations ? Linear considerations ( I.e., small strains only -- we will talk about large strains later ) (and infinitesimal displacements only) Paul A. Lagace ? 2001 Unit 4 - p . 3 MIT - 16.20 Fall, 2002 ε 11 = ? u 1 (4-4) ? y 1 ε 22 = ? u 2 (4-5) ? y 2 ε 33 = ? u 3 (4-6) ? y 3 ε 21 = ε 12 = 1 ?? ? u 1 + ? u 2 ?? 2 ? ? ? y 2 ? y 1 ? ? ε 31 = ε 13 = 1 ?? ? u 1 + ? u 3 ?? 2 ? ? ? y 3 ? y 1 ? ? ε 32 = ε 23 = 1 ?? ? u 2 + ? u 3 ?? 2 ? ? ? y 3 ? y 2 ? ? (4-7) (4-8) (4-9) 1 ?? ? u m + ? u n ?? ε mn = 2 ? ? ? y n ? y m ? ? Paul A. Lagace ? 2001 Unit 4 - p . 4 MIT - 16.20 Fall, 2002 3. Stress - Strain (6) σ mn = E mnpq ε pq we’ ll come back to this … Let ’s review the “ 4th important concept ”: Static Determinance There are there possibilities (as noted in U.E.) a. A structure is not sufficiently restrained (fewer reactions than d.o.f.) degrees of freedom ? DYNAMICS b. Structure is exactly (or “ simply ” ) restrained (# of reactions = # of d.o.f.) ? STATICS (statically determinate) Implication : can calculate stresses via equilibrium ( as done in Unified ) Paul A. Lagace ? 2001 Unit 4 - p . 5 MIT - 16.20 Fall, 2002 c. Structure is overrestrained (# reactions > # of d.o.f.) ? STATICALLY INDETERMINATE …must solve for reactions simultaneously with stresses, strains, etc. in this case, you must employ the stress-strain equations --> Overall, this yields for elasticity: 15 unknowns and 15 equations 6 strains = ε mn 3 equilibrium ( σ ) 6 stresses = σ mn 6 strain-displacements ( ε ) 3 displacements = u m 6 stress-strain ( σ - ε ) IMPORTANT POINT: The first two sets of equations are “universal ” (independent of the material) as they depend on geometry (strain-displacement) and equilibrium (equilibrium). Only the stress-strain equations are dependent on the material. Paul A. Lagace ? 2001 Unit 4 - p . 6 ?? ?? MIT - 16.20 Fall, 2002 One other point : Are all these equations/unknowns independent? N O Why? --> Relations between the strains and displacements (due to geometrical considerations result in the Strain Compatibility Equations (as you saw in Unified) General form is: ? 2 ε nk + ? 2 ε m l ? ? 2 ε n l ? ? 2 ε mk = 0 yy l ?? y k yy k ?? y l m y n m y n This results in 6 strain-compatibility (in 3-D). What a mess!!! What do these really tell us??? The strains must be compatible , they cannot be prescribed in an arbitrary fashion. Let ’s consider an example: Step 1 : consider how shear strain ( ε 12 ) is related to displacement: 1 ? ? u 1 ? u 2 ? ? + ? ε 12 = 2 ? ? ? y 2 ? y 1 ? ? Paul A. Lagace ? 2001 Unit 4 - p . 7 ?? ?? ?? MIT - 16.20 Fall, 2002 Note that deformations (u m ) must be continuous single-valued functions for continuity. (or it doesn ’ t make physical sense !) Step 2 : Now consider the case where there are gradients in the strain field ? ε 12 ≠ 0 , ? ε 12 ≠ 0 ? y 1 ? y 2 This is the most general case and most likely in a general structure Take derivatives on both sides: ? 2 ε 12 1 ? ? 3 u 1 ? 3 u 2 ? ? = 2 + 2 yy 2 2 ? ? yy 2 yy 2 ? ? 1 1 1 Step 3 : rearrange slightly and recall other strain-displacement equations ? u 1 = ε 1 , ? u 2 ε 2 = ? y 1 ? y 2 Paul A. Lagace ? 2001 Unit 4 - p . 8 ?? MIT - 16.20 Fall, 2002 ? 2 ε 12 1 ? ? 2 ε 11 + ? 2 ε 22 ? ? = 2 yy 2 2 ? ? ? y 2 ? y 1 2 ?? 1 So, the gradients in strain are related in certain ways since they are all related to the 3 displacements. Same for other 5 cases … Let ’s now go back and spend time with the … Stress-Strain Relations and the Elasticity Tensor In Unified, you saw particular examples of this, but we now want to generalize it to encompass all cases. The basic relation between force and displacement (recall 8.01) is Hooke ’s Law: F = kx spring constant (linear case) Paul A. Lagace ? 2001 Unit 4 - p . 9 MIT - 16.20 Fall, 2002 If this is extended to the three-dimensional case and applied over infinitesimal areas and lengths, we get the relation between stress and strain known as: Generalized Hooke ’ s law: σ mn = E mnpq ε pq where E mnpq is the “elasticity tensor ” How many components does this appear to have? m, n, p, q = 1, 2, 3 ? 3 x 3 x 3 x 3 = 81 components But there are several symmetries: 1. Since σ mn = σ nm (energy considerations) ? E mnpq = E nmpq (symmetry in switching first two indices ) 2. Since ε pq = ε qp (geometrical considerations) ? E mnpq = E mnqp Paul A. Lagace ? 2001 Unit 4 - p . 1 0 2212 MIT - 16.20 Fall, 2002 (symmetry in switching last two indices ) 3. From thermodynamic considerations (1st law of thermo) ? E mnpq = E pqmn (symmetry in switching pairs of indices ) Also note that: Since σ mn = σ nm are only 6! , the apparent 9 equations for stress With these symmetrics, the resulting equations are: ? σ 11 ? ? E 1111 E 1122 E 1133 2 E 1123 2 E 1113 2 E 1112 ? ? ε 11 ? ? ? ? ? ? ?? E 1122 E 2222 E 2233 2 E 2223 2 E 2213 2 E 2212 ? ? ε 22 ? ? σ 22 ? ? σ 33 ? ? E 1133 E 2233 E 3333 2 E 3323 2 E 3313 2 E 3312 ? ? ε 33 ? ? ? = ? ? ? ? ? σ 23 ? ? E 1123 E 2223 E 3323 2 E 2323 2 E 1323 2 E 1223 ? ? ε 23 ? ? σ 13 ? ? E 1113 E 2213 E 3313 2 E 1323 2 E 1313 2 E 1213 ? ? ε 13 ? ? ? ? ? ? ? ? σ 12 ? ? E 1112 E 2212 E 3312 2 E 1223 2 E 1213 2 E 1212 ? ? ε 12 ? Paul A. Lagace ? 2001 Unit 4 - p . 1 1 MIT - 16.20 Fall, 2002 Results in 21 independent components of the elasticity tensor ? Along diagonal (6) ? Upper right half of matrix (15) [don ’ t worry about 2 ’s] Also note : 2 ’s come out automatically … don ’ t put them in ε~ For example: σ 12 = … E 1212 ε 12 + E 1221 ε 21 … = … 2E 1212 ε 12 … These E mnpq can be placed into 3 groups : ? Extensional strains to extensional stresses E 1111 E 1122 E 2222 E 1133 E 3333 E 2233 e.g., σ 11 = … E 1122 ε 22 … ? Shear strains to shear stresses E 1212 E 1213 E 1313 E 1323 E 2323 E 2312 Paul A. Lagace ? 2001 Unit 4 - p . 1 2 MIT - 16.20 Fall, 2002 e.g., σ 12 = … 2E 1223 ε 23 … ? Coupling term : extensional strains to shear stress or shear strains to extensional stresses E 1112 E 2212 E 3312 E 1113 E 2213 E 3313 E 1123 E 2223 E 3323 e.g., σ 12 = …E 1211 ε 11 … 11 = …2E 1123 ε 23 … σ A material which behaves in this manner is “fully” anisotropic However, there are currently no useful engineering materials which have 21 different and independent components of E mnpq The “ type ” of material (with regard to elastic behavior) dictates the number of independent components of E mnpq : Paul A. Lagace ? 2001 Unit 4 - p . 1 3 MIT - 16.20 Fall, 2002 2 Isotropic 3 Cubic 5 “Transversely Isotropic”* 6 Tetragonal 9 Orthotropic 13 Monoclinic 21 Anisotropic # of Independent Components of E mnpq Material Type Useful Engineering Materials Composite Laminates Basic Composite Ply Metals (on average) Good Reference : BMP, Ch. 7 *not in BMP For orthotropic material s (which is as complicated as we usually get), there are no coupling terms in the principal axes of the material Paul A. Lagace ? 2001 Unit 4 - p . 1 4 MIT - 16.20 Fall, 2002 ? When you apply an extensional stress, no shear strains arise e.g., E 1112 = 0 (total of 9 terms are now zero) ? When you apply a shear stress, no extensional strains arise (some terms become zero as for previous condition) ? Shear strains (stresses) in one plane do not cause shear strains (stresses) in another plane ( E 1223 , E 1213 , E 1323 = 0) With these additional terms zero, we end up with 9 independent components: (21 - 9 - 3 = 9) and the equations are: Paul A. Lagace ? 2001 Unit 4 - p . 1 5 ??? 13 MIT - 16.20 Fall, 2002 ? σ 11 ? ? E 1111 E 1122 E 1133 0 0 0 ? ? ε 11 ? ? ? ?? E 1122 E 2222 E 2233 0 0 0 ? ? ? ? σ 22 ? ? ? ε 22 ? ? σ 33 ? ? E 1133 E 2233 E 3333 0 0 0 ? ? ε 33 ? ? ? = ? ? ? ? σ 23 ? ? ? 0 0 0 2 E 2323 0 0 ? ? ε 23 ? ? σ 13 ? ? 0 0 0 0 2 E 1313 0 ? ? ε 13 ? ? ? ? ? ? ? σ 12 ? ? ? 0 0 0 0 0 2 E 1212 ? ? ε 12 ? For other cases, no more terms become zero , but the terms are not Independent. For example, for isotropic materials: ? E 1111 = E 2222 = E 3333 ? E 1122 = E 1133 = E 2233 ? E 2323 = E 1313 = E 1212 ? And there is one other equation relating E 1111 , E 1122 and E 2323 ? 2 independent components of E mnpq (we’ll see this more when we do engineering constants) Paul A. Lagace ? 2001 Unit 4 - p . 1 6 MIT - 16.20 Fall, 2002 Why, then, do we bother with anisotropy? Two reasons : 1. Someday, we may have useful fully anisotropic materials (certain crystals now behave that way) Also, 40-50 years ago, people only worried about isotropy 2. It may not always be convenient to describe a structure (i.e., write the governing equations) along the principal material axes. How else? Loading axes Examples Figure 4-2 wing rocket case fuselage Paul A. Lagace ? 2001 Unit 4 - p . 1 7 MIT - 16.20 Fall, 2002 In these other axis systems, the material may have “more ” elastic components. But it really does n ’ t . (you can’ t “ create ” elastic components just by describing a material in a different axis system, the inherent properties of the material stay the same ). Figure 4-3 Example : Unidirectional composite (transversely isotropic) No shear / extension coupling Shears with regard to loading axis but still no inherent shear/extension coupling In order to describe full behavior, need to do …TRANSFORMATIONS (we’ll review this/expand on it later) Paul A. Lagace ? 2001 Unit 4 - p . 1 8 MIT - 16.20 Fall, 2002 --> It is often useful to consider the relationship between stress and strain (opposite way). For this we use COMPLIANCE ε mn = S mnpq σ pq where: S mnpq = compliance tensor Paul A. Lagace ? 2001 Unit 4 - p . 1 9 MIT - 16.20 Fall, 2002 Using matrix notation: σ = E ε ~~ ~ and E -1 σ ~ ~ = ε ~ inverse with ε = S σ ~ ~~ this means E -1 = S ~ ~ ? E S = I ~~ ~ ? The compliance matrix is the inverse of the elasticity matrix Note : the same symmetries apply to S mnpq as to E mnpq Paul A. Lagace ? 2001 Unit 4 - p . 2 0 MIT - 16.20 Fall, 2002 Meaning of each: ? Elasticity term E mnpq : amount of stress ( σ mn ) related to the deformation/strain ( ε pq ) ? Compliance term S mnpq : amount of strain ( ε mn ) the stress ( σ pq ) causes These are useful in defining/ determining the “engineering constants ” All of this presentation on elasticity (and what you had in Unified ) is based on assumptions which limit their applicability: which we will review / introduce / expand on in the next lecture. CAUTION ? Small strain ? Small displacement / infinitesimal (linear) strain Fortunately, most engineering structures are such that these assumptions cause negligible error. Paul A. Lagace ? 2001 Unit 4 - p . 2 1 MIT - 16.20 Fall, 2002 However, there are cases where this is not true: ? Manufacturing (important to be able to convince) ? Compliant materials ? Structural examples: dirigibles, … So let ’s explore: Large strain and the formal definition of strain What we defined before are the physical manifestation of strain / deformation ? Relative elongation ? Angular rotation Strain is formally defined by considering the diagonal length of a cube: Figure 4-4 undeformed x 3 (small letters) x 2 Paul A. Lagace ? 2001 x 1 Unit 4 - p . 2 2 MIT - 16.20 Fall, 2002 and looking at the change in length under general (and possibly large) deformation: Figure 4-5 deformed (capital letters) x 3 x 2 x 1 The formal definition of the strain tensor is: 2 2 2 γ mn dx m dx n = ( dS ) - ( ds ) ? 2 γ 11 dx 1 dx 1 + 2 γ 2 2 dx 2 dx 2 + 2 d γ 3 3 d x 3 d x 3 + 2 ( γ 12 + γ 21 ) dx 1 dx 2 + 2 ( γ 13 + γ 31 ) dx 1 dx 3 2 2 + 2 ( γ 23 + γ 32 ) dx 2 dx 3 = ( dS ) ? ( ds ) Paul A. Lagace ? 2001 Unit 4 - p . 2 3 MIT - 16.20 Fall, 2002 where γ mn = formal strain tensor. This is a definition . The physical interpretation is related to this but not directly in the general case. One can show (see BMP 5.1 - 5.4) that the formal strain tensor is related to relative elongation (the familiar ? l ) via: l relative elongation in m-direction: E m = m 1 γ m + 2 ? 1 (no summation on m) and is related to angular change via: 2 γ mn sin φ = mn ( 1 + E m ) ( 1 + E n ) Thus, it also involves the relative elongations! Most structural cases deal with relatively small strain. If the relative elongation is small (<<100%) ? E m <<1 Paul A. Lagace ? 2001 Unit 4 - p . 2 4 MIT - 16.20 Fall, 2002 look at: E m = 1 + 2 γ m m ? 1 2 ? ( E m + 1 ) = 1 + 2 γ m m E 2 + 2 E m = 2 γ m m m but if E m << 1, then E 2 m ≈ 0 ? E m = γ mm Relative elongation = strain ? l = ε small strain approximation! l Can assess this effect by comparing 2E m and E m (2 + E m ) relative elongation = E m 2E m E m (2 + E m ) % error 0.01 0.02 0.0201 0.5% 0.02 0.04 0.0404 1.0% 0.05 0.10 0.1025 2.4% 0.10 0.20 0.2100 4.8% Paul A. Lagace ? 2001 Unit 4 - p . 2 5 MIT - 16.20 Fall, 2002 Similarly, consider the general expression for rotation: 2 γ mn sin φ = mn ( 1 + E m ) ( 1 + E n ) for small elongations ( E m << 1, E n << 1) ? sin φ mn = 2 γ mn and , if the rotation is small: sin φ mn ≈ φ mn ? φ mn = 2 γ mn = 2 ε mn small strain approximation! (as before) Note : factor of 2 ! Even for a balloon, the small strain approximation may be good enough So : from now on, small strain assumed, but ? understand limitations ? be prepared to deal with large strain ? know difference between formal definition and the engineering approximation which relates dire ctly to physical reality. Paul A. Lagace ? 2001 Unit 4 - p . 2 6 MIT - 16.20 Fall, 2002 What is the other limitation? It deals with displacement, so consider Large Displacement and Non-Infinitesimal (Non-linear) Strain See BMP 5.8 and 5.9 The general strain-displacement relation is: 1 ? ? u ? u ? u ? u ? γ mn = ? m + n + r s δ rs ? 2 ? ? x n ? x m ? x m ? x n ? Where: δ rs = Kronecker delta The latter terms are important for larger displacements but are higher order for small displacements and can then be ignored to arrive back at: 1 ?? ? u m + ? u n ?? ε mn = 2 ? ? ? y n ? y m ? ? Paul A. Lagace ? 2001 Unit 4 - p . 2 7 MIT - 16.20 Fall, 2002 How to assess? Look at ? u m ? u r ? u s vs. δ rs ? x n ? x m ? x n and compare magnitudes Small v s . large and linear vs. nonlinear will depend on: ? material(s) ? structural configuration ? mode of behavior ? the loading Examples ? Rubber in inflated structures ? Large strain ( Note : generally means larger displacement) ? Diving board of plastic or wood ? Small strain but possibly large displacement (will look at this more when we deal with beams) Paul A. Lagace ? 2001 Unit 4 - p . 2 8 MIT - 16.20 Fall, 2002 ? Floor beam of steel ? Small strain and linear strain ( Note : linear strain must also be ? small) Next…back to constitutive constants …now their physical reality Paul A. Lagace ? 2001 Unit 4 - p . 2 9