MT-1620 al.2002 Unit 4 Equations of Elasticity Readings R 2.3,2.6.28 T&G 84.85 B,M,P5.1-5.5,58,59 7.1-7.9 6.163,6.5-67 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 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
MT-1620 al.2002 et's first review a bit from Unified, saw that there are 3 basic considerations in elasticity Equilibrium 2. Strain-Displacement 3. Stress-Strain Relations Constitutive Relations) Consider each 1. Equilibrium (3) ΣF=0,ΣM=0 Free body diagrams Applying these to an infinitesimal eleme yields 3 equilibrium equations Figure 4.1 Representation of general infinitesimal element dy z Paul A Lagace @2001 Unit 4-p. 2
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 Fi = 0, Σ Mi = 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
MT-1620 al.2002 0031 千1=0 dy 2 22 0032十 2=0(4-2) 2 13 do 0033 53=0(43) domn f=0 ym 2. Strain-Displacement(6) Based on geometric considerations Linear considerations (1.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 ∂σ11 + ∂σ21 + ∂σ31 + f1 = 0 (4-1) ∂y1 ∂y2 ∂y3 ∂σ12 + ∂σ22 + ∂σ32 + f2 = 0 (4-2) ∂y1 ∂y2 ∂y3 ∂σ13 + ∂σ23 + ∂σ33 + f3 = 0 (4-3) ∂y1 ∂y2 ∂y3 ∂ ∂ + σmn 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
MT-1620 al.2002 (4-4) (4-5) 2 33 46) 3 12 7) 2 1 au du 31 13 十 (4-8) 3 2dy3 dv du 2 dyn dym Paul A Lagace @2001 Unit 4-p. 4
MIT - 16.20 Fall, 2002 ε11 = ∂ u1 (4-4) ∂ y1 ε22 = ∂ u2 (4-5) ∂ y2 ε33 = ∂ u3 (4-6) ∂ y3 ε21 = ε12 = 1 ∂ u1 + ∂ u2 2 ∂ y2 ∂ y1 ε31 = ε13 = 1 ∂ u1 + ∂ u3 2 ∂ y3 ∂ y1 ε32 = ε23 = 1 ∂ u2 + ∂ u3 2 ∂ y3 ∂ y2 (4-7) (4-8) (4-9) 1 ∂ um + ∂ un ε mn = 2 ∂ yn ∂ ym Paul A. Lagace © 2001 Unit 4 - p. 4
MT-1620 Fall 2002 3. Stress-Strain(6 Omn -mpg pg 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 → DYNAM|CS 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 3. Stress - Strain (6) σmn = Emnpq εpq 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
MT-1620 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=Emn 3 equilibrium (o) 6 stresses = 6fF Um 6 strain-displacements(E 3 displacements 6 stress-strain(o-e MPORTANT 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 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 = εmn 3 equilibrium (σ) 6 stresses = σmn 6 strain-displacements (ε) 3 displacements = um 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
MT-1620 Fall 2002 One other point Are all these equations/unknowns independent? NO 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 dymdy+aN, amdy dyn少。0 m 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 ets consider an example Step 1 consider how shear strain (E)is related to displacement 12 Paul A Lagace @2001 Unit 4-p. 7
∂ ∂ ∂ ∂ MIT - 16.20 Fall, 2002 One other point: Are all these equations/unknowns independent? NO 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: ∂2εnk + ∂2εml − ∂2εnl − ∂2εmk = 0 y yl ∂ ∂yk y yk ∂ ∂y m l yn m yn 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 (ε12) is related to displacement: 1 ∂u1 ∂u2 ε + 12 = 2 ∂y2 ∂y1 Paul A. Lagace © 2001 Unit 4 - p. 7
MT-1620 Fall 2002 Note that deformations(um) 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 812≠0,。y2 12 0 This is the most general case and most likely in a general structure ake derivatives on both sides 12 1(0。u 2 0yy22(叫yny2"y7y2 Step 3: rearrange slightly and recall other strain-displacement equations du 2 Paul A Lagace @2001 Unit 4-p.8
∂ ∂ ∂ ∂ ∂ ∂ MIT - 16.20 Fall, 2002 Note that deformations (um) 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 ∂ε12 ≠ 0, ∂ε12 ≠ 0 ∂y1 ∂y2 This is the most general case and most likely in a general structure Take derivatives on both sides: ∂2ε12 1 ∂3u1 ∂3u2 ⇒ = 2 + 2 y y2 2 y y2 y y2 1 1 1 Step 3: rearrange slightly and recall other strain-displacement equations ∂u1 = ε1 , ∂u2 ε = 2 ∂y1 ∂y2 Paul A. Lagace © 2001 Unit 4 - p. 8
MT-1620 al.2002 12 0yy22(-y2 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=k t spring constant(linear case Paul A Lagace @2001 Unit 4-p. 9
∂ ∂ MIT - 16.20 Fall, 2002 ∂2ε12 1 ∂2ε11 + ∂2ε22 ⇒ = 2 y y2 2 ∂y2 ∂y12 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
MT-1620 al.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 hookes law mn nnpp where e is the "elasticity tensor mpg How many components does this appear to have? m,n,p,q=1,2,3 3x 3X3x3=81 components But there are several symmetries 1. sin (energy considerations (symmetry in switching first two indices 2. Since (geometrical considerations) → Paul A Lagace @2001 Unit 4 -p 10
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: σmn = Emnpq εpq where Emnpq 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 σmn = σnm (energy considerations) ⇒ Emnpq = Enmpq (symmetry in switching first two indices) 2. Since εpq = εqp (geometrical considerations) ⇒ Emnpq = Emnqp Paul A. Lagace © 2001 Unit 4 - p. 10