Theory of elasticity and plasticity (Equations sheet, Part 2) Att 8911
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Theory of elasticity and plasticity (Equations sheet, Part 2)
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Theory of elasticity and plasticity (Equations sheet, Part 2) Att 8911
- 1. Theory of elasticity and plasticity Equations sheet, Part 2 2-D formulation Plane strain Strain-displacement equations εxx εyy 2εxy = ∂ ∂x 0 0 ∂ ∂y ∂ ∂y ∂ ∂x u v Hooke’s law σxx σyy σzz σxy = λ + 2µ λ 0 λ λ + 2µ 0 λ λ 0 0 0 2µ εxx εyy εxy Plane stress Strain-displacement equations εxx εyy εzz 2εxy = ∂ ∂x 0 0 0 ∂ ∂y 0 0 0 ∂ ∂z ∂ ∂y ∂ ∂x 0 u v w 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 Equilibrium equations σxx σxy σxy σyy ∂ ∂x ∂ ∂y + fx fy = 0 0 Stress BCs tx ty = σxx σyx σxy σyy nx ny 1
- 2. Airy stress function σxx = ∂2 φ ∂y2 , σyy = ∂2 φ ∂x2 , σxy = − ∂2 φ ∂x∂y where φ = φ(x, y) is an arbitrary form called an Airy stress function Biharmonic equation ∂4 φ ∂x4 + 2 ∂4 φ ∂x2∂y2 + ∂4 φ ∂y4 = 0 Polynomial solution of 2-D problem Main steps • Select the polynomial function φ(x, y) = C1x2 + C2xy + C3y2 + C4x3 + . . . • Check the compatibility condition (biharmonic equation) 2 2 φ(x, y) = 0 • Use the Strong and Weak BCs to obtain a set of equations for Ci • Solve all equations and determine Ci FEM in plane elasticity- Constant Strain Triangle Displacement interpolation ux uy = ζ1 0 ζ2 0 ζ3 0 0 ζ1 0 ζ2 0 ζ3 ux1 uy1 ux2 uy2 ux3 uy3 Strain-displacement matrix B = 1 2A y23 0 y31 0 y12 0 0 x32 0 x13 0 x21 x32 y23 x13 y31 x21 y12 where 2A = det 1 1 1 x1 x2 x3 y1 y2 y3 and xjk = xj − xk, yjk = yj − yk 2
- 3. Element stiffness matrix k = A hBT E BdA where E = E 1 − ν2 1 ν 0 ν 1 0 0 0 1−ν 2 Classical plate theory Displacement, curvature and strain fields u(x, y, z) v(x, y, z) w(x, y, z) = −z ∂w0 ∂x −z ∂w0 ∂y w0(x, y) , κxx κyy κxy = −∂w2 ∂x2 −∂w2 ∂y2 −2 ∂w2 ∂x∂y , εxx εyy εxy = z κxx κyy κxy Stresses, stress resultants σxx σyy σxy = E 1 − ν2 1 ν 0 ν 1 0 0 0 1−ν 2 z κxx κyy κxy , Mxx Myy Mxy = D 1 ν 0 ν 1 0 0 0 1−ν 2 κxx κyy κxy where the flexural rigidity is D = Et3 12(1−ν2) Equilibrium equation ∂4 w ∂x4 + 2 ∂4 w ∂x2∂y2 + ∂4 w ∂y4 = q D Boundary conditions (edge, orthogonal of the x-axis) • Fixed edge- w = 0, dw dx = 0 • Free edge- Mx = 0, Vx = 0, where Vx = Qx + ∂Mxy ∂y • Simply supported edge- Mx = 0, w = 0 Navier’s solution • Furier series coefficient of the loading qmn = 4 ab b 0 a 0 q(x, y) sin mπ a x sin nπ b ydxdy 3
- 4. • Solution for the displacement w(x, y) = 1 Dπ4 ∞ m=1 ∞ n=1 qmn m2 a2 + n2 b2 2 sin mπ a x sin nπ b y Marcus Method BCs cases Lx Ly Case 1 Lx Ly Case 2 Lx Ly Case 3 Lx Ly Case 4 Lx Ly Case 5 Lx Ly Case 6 Cx = 1 Cx = 2 Cx = 1 Cx = 1 Cx = 1 Cx = 1 Cy = 1 Cy = 5 Cx = 5 Cx = 1 Cx = 2 Cx = 1 Directional loads qx = Cy 4 y Cx 4 x + Cy 4 y q, qy = q − qx Bending moments • Maximum span moments Mx = ¯Mx 1 − 5 6 2 x 2 y ¯Mx M0 x , My = ¯My 1 − 5 6 2 y 2 x ¯My M0 y where ¯Mx and ¯My are maximum span moments in strips loaded by the corresponding directional load and M0 x = q 2 x 8 , M0 y = q 2 y 8 • Edge moments- the edge moments are calculated as a strip supported with the same type of supports as a plate and loaded with directional load qx or qy 4