Diese Präsentation wurde erfolgreich gemeldet.
Wir verwenden Ihre LinkedIn Profilangaben und Informationen zu Ihren Aktivitäten, um Anzeigen zu personalisieren und Ihnen relevantere Inhalte anzuzeigen. Sie können Ihre Anzeigeneinstellungen jederzeit ändern.
Theory of elasticity and plasticity
Equations sheet, Part 2
2-D formulation
Plane strain
Strain-displacement equations

...
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 ...
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
Displacem...
• 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 ca...
Nächste SlideShare
Wird geladen in …5
×

Theory of elasticity and plasticity (Equations sheet, Part 2) Att 8911

Theory of elasticity and plasticity (Equations sheet, Part 2)

  • Als Erste(r) kommentieren

  • Gehören Sie zu den Ersten, denen das gefällt!

Theory of elasticity and plasticity (Equations sheet, Part 2) Att 8911

  1. 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. 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. 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. 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

×