2-D formulation Plane theory of elasticity Att 6672

2-D formulation Plane theory of elasticity Att 6672

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  Lecture 7 2-D formulation Planetheory of elasticity Print version Lecture on Theory of Elasticity and Plasticity of Dr. D. Dinev, Department of Structural Mechanics, UACEG 7.1 Contents 1 Plane strain 1 2 Plane stress 3 3 Plane strain vs. plane stress 5 4 Airy stress function 5 5 Polynomial solution of 2-D problem 7 6 General solution strategy 14 7.2 1 Plane strain Plane strain Introduction • Because of the complexity of the field equations analytical closed-form solutions to full 3-D problems are very difficult to accomplish • A lot of problems into the area of engineering can be approximated by 1-D or 2-D strain or stress state – rods, beams, columns, shafts etc. – Retaining walls, disks, plates, shells 7.3 Plane strain 1
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  Problem definition • Consideran infinitely long prismatic body • If the body forces and surface tractions have no components on z-direction the deformation field can be reduced into u = u(x,y) v = v(x,y) w = 0 • This deformation is called as a state of plane strain in the (x,y)-plane • Thus all cross-sections will have same displacements 7.4 Plane strain Field equations • The strain-displacement relations become εxx = ∂u ∂x , εyy = ∂v ∂y , εxy = 1 2 ∂u ∂y + ∂v ∂x εzz = εxz = εyz = 0 • In matrix form   εxx εyy 2εxy   =    ∂ ∂x 0 0 ∂ ∂y ∂ ∂y ∂ ∂x    u v • The St.-Venant’s compatibility equation is ∂2εxx ∂y2 + ∂2εyy ∂x2 = 2 ∂2εxy ∂x∂y 7.5 Plane strain Field equations • The stress-strain relations are σxx = (λ +2µ)εxx +λεyy σyy = λεxx +(λ +2µ)εyy σzz = λεxx +λεyy σxy = µ2εxy σxz = σyz = 0 2
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  • In matrixform     σxx σyy σzz σxy     =     λ +2µ λ 0 λ λ +2µ 0 λ λ 0 0 0 2µ       εxx εyy εxy   7.6 Plane strain Field equations • The equilibrium equations are reduced to ∂σxx ∂x + ∂σxy ∂y + fx = 0 ∂σxy ∂x + ∂σyy ∂y + fy = 0 • In matrix form σxx σxy σxy σyy ∂ ∂x ∂ ∂y + fx fy = 0 0 7.7 Plane strain Field equations • The Navier’s displacement equilibrium equations are µ∇2 u+(λ + µ) ∂ ∂x ∂u ∂x + ∂v ∂y + fx = 0 µ∇2 v+(λ + µ) ∂ ∂y ∂u ∂x + ∂v ∂y + fy = 0 where ∇2 = ∂2 ∂x2 + ∂2 ∂y2 - Laplacian operator • The Beltrami-Michell stress equation is ∇2 (σxx +σyy) = − 1 1−ν ∂ fx ∂x + ∂ fy ∂y • The surface tractions (stress BCs)are tx ty = σxx σxy σxy σyy nx ny 7.8 2 Plane stress Plane stress Problem definition 3
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  • Consider anarbitrary disc which thickness is small in comparison to other dimensions • Assume that there is no body forces and surface tractions in z-directions and the surface of the disc is stress free • Thus imply a stress field σxx = σxx(x,y) σyy = σyy(x,y) σxy = σxy(x,y) σzz = σxz = σyz = 0 7.9 Plane stress Field equations • The Hooke’s law     εxx εyy εzz εxy     = 1 E     1 −ν 0 −ν 1 0 −ν −ν 0 0 0 1+ν       σxx σyy σxy   • Relation between normal strains εzz = − ν 1−ν (εxx +εyy) 7.10 Plane stress Field equations • Strain-displacement equations     εxx εyy εzz 2εxy     =      ∂ ∂x 0 0 0 ∂ ∂y 0 0 0 ∂ ∂z ∂ ∂y ∂ ∂x 0        u v w   εyz = εzx = 0 • St.-Venant’s compatibility equation is ∂2εxx ∂y2 + ∂2εyy ∂x2 = 2 ∂2εxy ∂x∂y 7.11 Plane stress Field equations • Equilibrium equations - same as in plane strain σxx σxy σxy σyy ∂ ∂x ∂ ∂y + fx fy = 0 0 7.12 4
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  E ν Plane stress tostrain E 1−ν2 ν 1−ν Plane strain to stress E(1+2ν) (1+ν)2 ν 1+ν Plane stress Field equations • The Navier’s displacement equilibrium equations are µ∇2 u+ E 2(1−ν) ∂ ∂x ∂u ∂x + ∂v ∂y + fx = 0 µ∇2 v+ E 2(1−ν) ∂ ∂y ∂u ∂x + ∂v ∂y + fy = 0 • The Beltrami-Michell stress equation is ∇2 (σxx +σyy) = −(1+ν) ∂ fx ∂x + ∂ fy ∂y • The surface tractions (stress BCs)are tx ty = σxx σxy σxy σyy nx ny 7.13 3 Plane strain vs. plane stress Plane strain vs. plane stress Summary • The plane problems have identical equilibrium equations, BCs and compatibility equa- tions • The similar equations show that the differences are due to different constants involving different material constants • The field equations of plane stress can be obtained from equations of plane strain by fol- lowing substitution • When ν = 0 plane strain ≡ plane stress 7.14 4 Airy stress function Airy stress function The Method • A popular method for the solution of the plane problem is using the so called Stress func- tions • It employs the Airy stress function and reduce the general formulation to a single equation in terms of a single unknown • The general idea is to develop a stress field that satisfies equilibrium and yields a single governing equation from the compatibility equations. • The obtained equilibrium equation ca be solved analytically in closed-form 7.15 5
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  Airy stress function TheMethod • Assume that the body forces are zero • The Beltrami-Michell stress compatibility equations are ∇2 (σxx +σyy) = 0 • Equilibrium equations are ∂σxx ∂x + ∂σxy ∂y = 0 ∂σxy ∂x + ∂σyy ∂y = 0 • The stress BCs are tx ty = σxx σxy σxy σyy nx ny 7.16 Airy stress function The Method • The Beltrami-Michell equation can be expanded as ∂2σxx ∂x2 + ∂2σyy ∂x2 + ∂2σxx ∂y2 + ∂2σyy ∂y2 = 0 • The equilibrium equations are satisfied if we choose the representation σxx = ∂2φ ∂y2 , σyy = ∂2φ ∂x2 , σxy = − ∂2φ ∂x∂y where φ = φ(x,y) is an arbitrary form called an Airy stress function • Substitution of the above expressions into the Beltrami-Michell equations lied to ∂4φ ∂x4 +2 ∂4φ ∂x2∂y2 + ∂4φ ∂y4 = 0 7.17 Airy stress function The Method • George Biddell Airy (1801-1892) 7.18 6
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  Airy stress function TheMethod • The previous expression is a biharmonic equation. In short notation ∇2 ∇2 φ(x,y) = 0 • Thus all equations of the plane problem has been reduced to a single equation in terms of the Airy stress function φ(x,y). • This function is to be determined in the 2-D region R bounded by the boundary S • Appropriate BCs are necessary to complete a solution 7.19 5 Polynomial solution of 2-D problem Polynomial solution of 2-D problem The Method • The solution with polynomials is applicable in Cartesian coordinates and useful for prob- lems with rectangular domains • Based on the inverse solution concepts - we assume a form of the solution of the equation ∇2∇2φ(x,y) = 0 and then try to determine which problem may be solved by this solution 7.20 Polynomial solution of 2-D problem The Method • The assumed solution is taken to be a general polynomial and can be expressed in the power series φ(x,y) = ∞ ∑ m=0 ∞ ∑ n=0 Cmnxm yn where Cmn are constants to be determined • The method produces a polynomial stress distribution and not satisfies the general BCs • We need to modify the BCs using St.-Venant principle- with statically equivalent BCs • The solution would be accurate at points sufficiently far away from the modified boundary 7.21 Polynomial solution of 2-D problem Example 1 • Let’s use a trial solution- first order polynomial φ(x,y) = C1x+C2y+C3 • The solution satisfies the biharmonic equation • Go to the stress field σxx = ∂2φ ∂y2 = 0, σyy = ∂2φ ∂x2 = 0, σxy = − ∂2φ ∂x∂y = 0 Question • What this solution mean? 7.22 7
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  Polynomial solution of2-D problem Example 2 • Use a higher order polynomial φ(x,y) = C1y2 • The solution also satisfies the biharmonic equation • The stress field σxx = ∂2φ ∂y2 = 2C1, σyy = ∂2φ ∂x2 = 0, σxy = − ∂2φ ∂x∂y = 0 7.23 Polynomial solution of 2-D problem Example 2 • The solution fits with the uniaxial tension of a disc • The boundary conditions are σxx(± ,y) = t σyy(x,±h) = 0 σxy(± ,y) = σxy(x,±h) = 0 • The constant C1 can be obtained from the BCs 7.24 Polynomial solution of 2-D problem Example 3 • Pure bending of a beam - a comparison with the MoM solution 7.25 Polynomial solution of 2-D problem Review of the beam theory • á la Speedy Gonzales • Assumptions – Long beam- h – Small displacements- u h and v h – Small strains- εxx 1 – Bernoulli hypothesis- εyy ≈ 0 and σxy ≈ 0 7.26 8
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  Polynomial solution of2-D problem Review of the beam theory • Displacement field u(x,y,z) = yθ v(x,y,z) = v(x,y) • Because γxy = 0, thus θ = − ∂v ∂x • The final displacements are u(x,y,z) = −y ∂v ∂x v(x,y,z) = v(x,y) 7.27 Polynomial solution of 2-D problem Review of the beam theory • Strain field εxx = −y ∂2v ∂x2 • Compatibility equation 1 r = κ = dθ ds ≈ dθ dx = d2v dx2 • The strain field can be expressed εxx = −yκ • Hooke’s law σxx = Eεxx = −yEκ 7.28 Polynomial solution of 2-D problem Review of the beam theory 9
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  • Bending moment M= A σxxydA = −EIκ • Equilibrium equations dV dx = −q dM dx = −V 7.29 Polynomial solution of 2-D problem Review of the beam theory • Differential equation EI d4v dx4 = q • 4-th order ODE- needs of four BCs 7.30 Polynomial solution of 2-D problem Example 3- MoM solution • MoM solution 7.31 Polynomial solution of 2-D problem Example 3- Elasticity solution • Elasticity solution 7.32 Polynomial solution of 2-D problem Example 3- Elasticity solution • Strong BCs σyy(x,±c) = 0, σxy(x,±c) = 0, σxy(± ,y) = 0 7.33 10
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  Polynomial solution of2-D problem Example 3- Elasticity solution • Weak BCs- imposed in a weak form (using the St.-Venant principle) c −c σxx(± ,y)dy = 0, c −c σxx(± ,y)ydy = −M 7.34 Polynomial solution of 2-D problem Example 3- Elasticity solution • Based on the MoM solution (linear σxx distribution) we try a following solution φ(x,y) = A1y3 • The function satisfies ∇4φ(x,y) = 0 • The stress functions are σxx = 6A1y σyy = 0 → satisfies σyy(x,±c) = 0 σxy = 0 → satisfies σxy(x,±c) = 0, σxy(± ,y) = 0 • This trial solution fits with the BCs 7.35 Polynomial solution of 2-D problem Example 3- Elasticity solution • The constant A1 is obtained from the weak BC at x = ± c −c σxx(± ,y)dy ≡ 0 c −c σxx(± ,y)ydy = 4c3 A1 = −M, → A1 = − M 4c3 • Thus the Airy stress function is φ(x,y) = − M 4c3 y3 • Corresponding stresses are σxx = − 3M 2c3 y σyy = 0 σxy = 0 7.36 Polynomial solution of 2-D problem Example 3- Elasticity solution • Strain field-by the Hooke’s law εxx = − 3M 2Ec3 y εyy = 3Mν 2Ec3 y εxy = 0 • Displacement field- by strain-displacement equations u = − 3M 2Ec3 xy+ f(y) v = 3Mν 4Ec3 y2 +g(x) • The functions f(y) and g(x) have to be determined from the definition of the shear strain 7.37 11
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  Polynomial solution of2-D problem Example 3- Elasticity solution • The definition of the shear strain gives εxy = − 3M 4Ec3 x+ 1 2 ∂ f(y) ∂y + 1 2 ∂g(x) ∂x • This result can be compared with the shear strain obtained from the constitutive relations εxy = 0 − 3M 4Ec3 x+ 1 2 ∂g(x) ∂x + 1 2 ∂ f(y) ∂y = 0 7.38 Polynomial solution of 2-D problem Example 3- Elasticity solution • The equation can be partitioned into ∂ f(y) ∂y = ω0 ∂g(x) ∂x = 3M 2Ec3 x+ω0 where ω0 is an arbitrary constant • Integration of the above equation gives f(y) = uo +yω0 g(x) = 3M 4c3E x2 +xω0 +v0 • The constants u0, v0 and ω0 express the rigid-body motion of the beam 7.39 Polynomial solution of 2-D problem Example 3- Elasticity solution • The back substitution into the displacement field gives u(x,y) = u0 +yω0 − 3M 2c3E xy v(x,y) = v0 +xω0 + 3M 4c3E x2 + 3Mν 4c3E y2 • The constants can be found from the essential BCs 7.40 Polynomial solution of 2-D problem Example 3- Elasticity solution • The essential BCs are concentrated at points of beam ends u(− ,0) = 0 → u0 = 0 v(− ,0) = 0 → 3M 2 4c3E +v0 − ω0 = 0 v( ,0) = 0 → 3M 2 4c3E +v0 + ω0 = 0 7.41 12
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  Polynomial solution of2-D problem Example 3- Elasticity solution • The constants are u0 = 0 ω0 = 0 v0 = − 3M 2 4c3E • Displacement field can be completed as u(x,y) = − 3M 2c3E xy v(x,y) = 3M 4c3E (x2 +νy2 − 2 ) 7.42 Polynomial solution of 2-D problem 3 2 1 0 1 2 3 2 1 0 1 2 Example 3- Elasticity solution • Vector plot of the displacement field 7.43 Polynomial solution of 2-D problem Example 3- Elasticity solution • Elasticity solution u(x,y) = − M EI xy v(x,y) = M 2EI (x2 +νy2 − 2 ) • MoM solution u(x) = − M EI xy v(x) = M 2EI (x2 − 2 ) where I = 2 3 c3 Note It is convenient to use a computer algebra system for the mathematics (Maple, Mathematica etc.) 7.44 13
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  Polynomial solution of2-D problem Example 3 • General conclusion 7.45 6 General solution strategy General solution strategy Selection of the polynomial order • Step 1- Determine the maximum order of polynomial using MoM arguments Example 1 • Normal loading-q(x) → xn • Shear force- V(x) → xn+1 • Bending moment- M(x) → xn+2 • Stress- σxx → xn+2y • Airy function- xn+2y3 • Maximum order= n+5 7.46 General solution strategy Example 2 • Shear loading-n(x) → xm • Shear force- V(x) → xm • Bending moment- M(x) → xm+1 • Stress- σxx → xm+1y • Airy function- xm+1y3 • Maximum order= m+4 7.47 14
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  General solution strategy Selectionof the polynomial order • Step 2- Write down a polynomial function φ(x,y) that contains all terms up to order max(m+4,n+5) φ(x,y) = C1x2 +C2xy+C3y2 +C4x3 +... 7.48 General solution strategy Selection of the polynomial order • May use the Pascal’s triangle for the polynomial 1 x y x2 xy y2 x3 x2y xy2 y3 x4 x3y x2y2 xy3 y4 x5 x4y x3y2 x2y3 xy4 y5 • And constants C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 C12 C13 C14 C15 C16 C17 C18 • The first three terms have no physical meaning (zero stress field) 7.49 General solution strategy Selection of the polynomial order • Step 3 Compatibility condition ∇2 ∇2 φ(x,y) = 0 • Step 4 Boundary conditions- strong and weak. Lead to a set of equations for Ci • Step 5 Solve all equations and determine Ci Other types of solution • Fourier series method • ...... 7.50 General solution strategy The End • Any questions, opinions, discussions? 7.51 15