美国麻省理工大学:《结构力学》英文版 Unit 8 Solution Procedures

MT-1620 al.2002 Unit 8 Solution Procedures Readings R Ch. 4 T&G 17,Ch.3(18-26) Ch.4(27-46) Ch.6(54-73) Paul A Lagace Ph D Professor of aeronautics Astronautics and engineering systems Paul A Lagace @2001
MIT - 16.20 Fall, 2002 Unit 8 Solution Procedures Readings: R Ch. 4 T & G 17, Ch. 3 (18-26) Ch. 4 (27-46) Ch. 6 (54-73) Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering Systems Paul A. Lagace © 2001

MT-1620 al.2002 Summarizing what we've looked at in elasticity, we have 15 equations In 15 unknowns 3 equilibrium 6 strains 6 strain-displacement 3 displacements 6 stress-strain 6 stresses These must be solved for a generic body under some generic loading subject to the prescribed boundary conditions(BCs There are two types of boundary conditions 1. Normal(stress prescribed) 2. Geometric(displacement prescribed) you must have one or the other To solve this system of equations subject to such constraints over the continuum of a generic body is, in general, quite a challenge. There are basically two solution procedures 1. Exact--satisfy all the equations and the B.C.s 2. Numerical --come as close as possible"(energy methods, etc. Paul A Lagace @2001 Unit 8-p. 2
MIT - 16.20 Fall, 2002 Summarizing what we’ve looked at in elasticity, we have: 15 equations in 15 unknowns - 3 equilibrium - 6 strains - 6 strain-displacement - 3 displacements - 6 stress-strain - 6 stresses These must be solved for a generic body under some generic loading subject to the prescribed boundary conditions (B.C.’s) There are two types of boundary conditions: 1. Normal (stress prescribed) 2. Geometric (displacement prescribed) --> you must have one or the other To solve this system of equations subject to such constraints over the continuum of a generic body is, in general, quite a challenge. There are basically two solution procedures: 1. Exact -- satisfy all the equations and the B.C.’s 2. Numerical -- come as “close as possible” (energy methods, etc.) Paul A. Lagace © 2001 Unit 8 - p. 2

MT-1620 Fall 2002 Let's consider"exact techniques. a common and classic, one is Stress functions Relate six stresses to (fewer) functions defined in such a manner at they identically satisfy the equilibrium conditon Can be done for 3-d case Can be done for anisotropic(most often orthotropic)case See: Lekhnitskii, Anisotropic Plates, Gordan breach. 1968 >Lets consider plane stress (eventually) isotropic 8 equations in 8 unknowns 2 equilibrium 3 strains 3 strain-displacement 2 displacements 3 stress-strain 3 stresses Paul A Lagace @2001 Unit 8-p.3
MIT - 16.20 Fall, 2002 Let’s consider “exact” techniques. A common, and classic, one is: Stress Functions • Relate six stresses to (fewer) functions defined in such a manner that they identically satisfy the equilibrium conditon • Can be done for 3-D case • Can be done for anisotropic (most often orthotropic) case --> See: Lekhnitskii, Anisotropic Plates, Gordan & Breach, 1968. --> Let’s consider • plane stress • (eventually) isotropic 8 equations in 8 unknowns - 2 equilibrium - 3 strains - 3 strain-displacement - 2 displacements - 3 stress-strain - 3 stresses Paul A. Lagace © 2001 Unit 8 - p. 3

MT-1620 al.2002 Define the" Airy"Stress Function = o(x, y) English a scalar mathematician + v 8-2) yy v (8-3) aXd where: V=potential function for body forces fx and ty dV V exists if VXf=0 (CI that is of Paul A Lagace @2001 Unit 8-p. 4
MIT - 16.20 Fall, 2002 Define the “Airy” Stress Function = φ(x, y) English a scalar mathematician ∂2φ σxx = 2 + V (8 -1) ∂y ∂2φ σyy = ∂x2 + V (8 - 2) 2φ σxy = − ∂ (8 - 3) ∂ ∂x y where: V = potential function for body forces fx and fy fx = − ∂V fy = − ∂V ∂x ∂y V exists if ∇ x f = 0 (curl) that is: ∂fx = ∂fy ∂y ∂x Paul A. Lagace © 2001 Unit 8 - p. 4

MT-1620 al.2002 Recall that curl f=0→“ conservative" field gravity forces spring forces etc What does that compare to in fluids? rrotational flow Look at how p has been defined and what happens if we place these equations(8-1-8-3)into the plane stress equilibrium equations do 0 f=0 (E1) yy (E2) aX we then get Paul A Lagace @2001 Unit 8-p. 5
MIT - 16.20 Fall, 2002 Recall that curl f = 0 ⇒ “conservative” field - gravity forces - spring forces - etc. What does that compare to in fluids? Irrotational flow Look at how φ has been defined and what happens if we place these equations (8-1 - 8-3) into the plane stress equilibrium equations: ∂σxx + ∂σxy + fx = 0 (E1) ∂x ∂y ∂σxy ∂σyy ∂x + ∂y + fy = 0 (E2) we then get: Paul A. Lagace © 2001 Unit 8 - p. 5

MT-1620 al.2002 E1) 0(0p +V|+ dy axd ao dv d'o dv es) 0(0 E2) 0=0? dx dxdy dy( dx dφ0p,dv a_0 (yes =Equilibrium automatically satisfied using Airy stress function Does that mean that any function we pick for o(x, y) will be valid? No, it will satisfy equilibrium, but we still have the strain-displacement and stress-strain equations If we use these we can get to the governing equation Paul A Lagace @2001 Unit 8-p. 6
∂ ∂ ∂ ∂ MIT - 16.20 Fall, 2002 2 ∂ ∂2φ ( ) E1 : 2 +V + ∂ − ∂ φ − ∂V = 0 ? ∂x ∂y ∂y ∂ ∂x y ∂x ∂3φ + ∂V − ∂3φ − ∂ V = 0 ⇒ (yes) ∂ ∂x y x y 2 ∂x ∂ ∂ 2 ∂x 2 ∂ − ∂ φ ∂ ∂2φ (E2) : ∂x ∂ ∂y + ∂y ∂x 2 + V − ∂V = 0 ? x ∂y ∂3φ + ∂3φ + ∂ V − ∂ V ⇒ = 0 (yes) 2 2 xy xy ∂y ∂y ⇒Equilibrium automatically satisfied using Airy stress function! Does that mean that any function we pick for φ (x, y) will be valid? No, it will satisfy equilibrium, but we still have the strain-displacement and stress-strain equations. If we use these, we can get to the governing equation: Paul A. Lagace © 2001 Unit 8 - p. 6

MT-1620 al.2002 Step 1 Introduce o into the stress-strain equations(compliance form) E3) E (E4) 2(1+ E (E5) So 十 V E E3) oX E y EaX E4) E 2(1+y)a2φ (E5) e day Paul A Lagace @2001 Unit 8-p. 7
(E MIT - 16.20 Fall, 2002 Step 1: Introduce φ into the stress-strain equations (compliance form): 1 εxx = E3 E (σxx − νσyy ) () 1 εyy = (−νσxx + σyy ) (E4) E εxy = 2 (1 E + ν) σxy (E5) So: εxx = 1 ∂2φ − ν ∂2φ + (1 − ν) V (E3′) E ∂y2 ∂x2 E εyy = 1 ∂2φ − ν ∂2φ + (1 − ν) V (E4′) E ∂x2 ∂y2 E εxy = − 2 (1 + ν) ∂2φ (E5′) E ∂ ∂x y Paul A. Lagace © 2001 Unit 8 - p. 7

MT-1620 al.2002 Step 2: Use these in the plane stress compatibility equation E6) dX we get quite a mess! After some rearranging and manipulation this results in 02(△T)02(△T) Eat ax temperature term we a= coefficient of thermal expansion haven't yet considered AT= temperature differential This is the basic equation for isotropic plane stress in stress function form Reca:φ is a scalar Paul A Lagace @2001 Unit 8-p. 8
MIT - 16.20 Fall, 2002 Step 2: Use these in the plane stress compatibility equation: ∂2εxx ∂2εyy ∂2εxy + = ∂y x y 2 ∂x2 ∂ ∂ (E6) ⇒ we get quite a mess! After some rearranging and manipulation, this results in: V V ∂ ∂ + ∂ ∂ ∂ + ∂ ∂ = − ∂ ( ) ∂ + ∂ ( ) ∂ − ( ) ∂∂ + ∂∂ 4 4 4 2 2 4 2 2 2 2 2 2 2 2 1 φ φ α x y y E T x T y y 2 ∆ − 2 φ ν x x ∆ (*) temperature term we haven’t yet considered α = coefficient of thermal expansion ∆T = temperature differential This is the basic equation for isotropic plane stress in Stress Function form Recall: φ is a scalar Paul A. Lagace © 2001 Unit 8 - p. 8

MT-1620 al.2002 If we recall a little mathematics, the Laplace Operator in 2-D is x4+20 This is the biharmonic operator(also used in fluids) So the f)equation can be written Ey4AT)-(1=y2v() Finally, in the absence of temperature effects and body forces this becomes v4=0 homogeneous form What happened to E, v?? Paul A Lagace @2001 Unit 8-p. 9
MIT - 16.20 Fall, 2002 If we recall a little mathematics, the Laplace Operator in 2-D is: ∇2 ∂2 ∂2 = + ∂x2 ∂y2 ∂4 ∂4 ∂4 ⇒ ∇2∇2 = ∇4 = ∂x4 + 2 2 2 + ∂y4 ∂ ∂ x y This is the biharmonic operator (also used in fluids) So the (*) equation can be written: ∇4φ = −Eα∇2 (∆T) − (1 − ν) ∇2 V (*) Finally, in the absence of temperature effects and body forces this becomes: ∇4φ = 0 homogeneous form What happened to E, ν?? Paul A. Lagace © 2001 Unit 8 - p. 9

MT-1620 Fall 2002 this function, and accompanying governing equation, could be defined in any curvilinear system(we'l see one such example later) and in plane strain as well But. what's this all useful for??? This may all seem like"magic". Why were the os assumed as they were? This is not a direct solution to a posed problem, per se, but is known as The Inverse method In general, for cases of plane stress without body force or temp (V#o =O) A stress function o(x, y) is assumed that satisfies the biharmonic equation 2. The stresses are determined from the stress function as defined in equations (8-1)-(8-3) 3. Satisfy the boundary conditions (of applied tractions ☆☆4. Find the( structural problem that this satisfies Paul A Lagace @2001 Unit 8-p. 10
MIT - 16.20 Fall, 2002 ⇒ this function, and accompanying governing equation, could be defined in any curvilinear system (we’ll see one such example later) and in plane strain as well. But…what’s this all useful for??? This may all seem like “magic”. Why were the σ’s assumed as they were? This is not a direct solution to a posed problem, per se, but is known as… The Inverse Method In general, for cases of plane stress without body force or temp (∇4φ = 0): 1. A stress function φ (x, y) is assumed that satisfies the biharmonic equation 2. The stresses are determined from the stress function as defined in equations (8-1) - (8-3) 3. Satisfy the boundary conditions (of applied tractions) 4. Find the (structural) problem that this satisfies Paul A. Lagace © 2001 Unit 8 - p. 10
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