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