+
′′ BiN wN
+
N N −1 ∑ ∑ j =1 k=2
′′ Bik Bkj wj
1
Homework of Mechanics of Plates and Shells
1 Preliminary works
First, the differential quadrature (DQ) method for one dimensional problem can be interpreted as below. For generality, consider a continuously differentiable function f (x) with one single variable x defined within [a,b]. Setting N points within [a, b], the function can be assumed as f (x) =
Next consider the buckling of the plate. The governing equation is written as D( ∂ 4w ∂ 4w ∂ 4w ∂ 2w ∂ 2w ∂ 2w + 2 + ) = σ h + 2 σ h + σ h x xy y ∂x4 ∂x2 ∂y 2 ∂y 4 ∂x2 ∂x∂y ∂y 2 3 (1.14)
A reproduction of DQ analysis of buckling of thin rectangular plates with cosine-distributed compressive loads on two opposite sides
Bo Xiao, Department of Mechanics, HUST
N ∑ j =1
Wj (x)f (xj )
(1.1)
where N , Wj (x), and f (xj ) are the number of grid points, the interpolation function, and the values at grid point j , respectively. Consider the k -th order of f (xj ) gives f
′′ wi
=
Bij wj =
Aik Akj wj =
¯ij δj B
(1.16)
(i = 2, 3, · · · , N − 1) ¯ is consistent with the B in [1], and can be computed by where the denotation of B { ∑N if j = 1, 2, · · · , N k=1 Aik Akj ¯ij = (i = 2 , 3, · · · , N ) B 0 if j = N + 1, N + 2 ∑N −1 k=2 Aik Akj if j = 1, 2, · · · , N ¯ Bij = A11 (i = 1, N ) if j = N + 1 A1N if j = N + 2
where Airy stress function should satisfy the compatibility equation: ∂ 4φ ∂ 4φ ∂ 4φ + 2 + =0 ∂x4 ∂x2 ∂y 2 ∂y 4 The boundary conditions are b y=− , 2 a , x= 2 b y= , 2 a x=− , 2 ∂φ =0 ∂y σ0 b πy φ = 2 [2b cos( ) − 2πy − πb], 2π b σ 0 b2 ∂φ 2σ0 b φ=− , =− π ∂y π σ0 b πy φ = 2 [2b cos( ) − 2πy − πb], 2π b φ= (1.10) ∂φ =0 ∂x (1.11) (1.12) ∂φ =0 ∂x (1.13) (1.9)
′′ w1
=
′ A11 w1
+
′ A1N wN
+
N −2 ∑ k=2
′ A1k wk
=
N +2 ∑ j =1
¯1j δj B ¯ N j δj B
′′ ′ ′ wN = AN 1 w1 + AN N wN +
N −2 ∑ k=2
′ A N k wk =
N +2 ∑ j =1
(1.15)
′ ′ where {δ }T = {w1 , w2 , · · · , wN , w1 , wN }. And the weighting coefficients of second order derivatives at inner points are computed by N ∑ j =1 N ∑ N ∑ j =1 k=1 N +2 ∑ j =1
Homework of Mechanics of Plates and Shells where D, h, and w are the flexural rigidity, plate thickness, and deflection, respectively. For the plate buckling analysis, the boundary conditions are 1) Simply supported (S): either w = Mx = 0 at x = ±a/2, or w = My = 0 at y = ±b/2; 2) Clamped (C): either w = wx = 0 at x = ±a/2, or w = wy = 0 at y = ±b/2; where the bending moments are defined as M x = −D ( ∂ 2w ∂2w + µ ) ∂x2 ∂y 2
k=1 k=1 k=1 k=1
Now the present problem is an isotropic thin rectangular plate under uni-axial cosine-distributed in-plane compressions shown in Fig.1. Additional complexity arises when having to first solve the problem in plane-stress elasticity for obtaining the internal pre-stress distribution, and then the buckling problem. Method based on stress function can be used for obtaining in-plane stressed, since all boundary conditions are in terms of stresses. Applying Airy stress function φ = φ(x, y ) without body forces, the stresses take the following forms: σx = ∂ 2φ ∂y 2 σy = ∂ 2φ ∂x2 τxy = − ∂2φ ∂x∂y (1.8)
Comparing the two formulas above gives
(n+1) Wj (xi )
=
N ∑ k=1
Aik Wj (xk )
(n)
(1.5)
Or in a simpler form,
(n+1) Wij
=
N ∑ k=1
Aik Wkj
(n)
(1.6)
2
Homework of Mechanics of Plates and Shells
(k)
(xi ) =
N ∑ j =1
Wj (xi )f (xj )
(k)
(1.2)
Denote the summation coefficient as Aij when k = 1, namely, f (xi ) =
′ N ∑ j =1
Aij f (xj )
(1.3)
Where using Lagrange interpolation function, Aij can be explicitly computed by ∏N N ∑ 1 k=1,k̸=i,j (xi − xk ) (i ̸= j ) Aij = (i = j ) Aij = ∏N x j − xk k=1,k̸=j (xj − xk ) k=1,k̸=j Now the relation of coefficients of neighboring derivatives will be derived. f
Abstract The problem of buckling of thin rectangular plates with cosine-distributed compressive loads on two opposite sides is solved again using differential quadrature method. First the plane elasticity problem is solved to obtain the in-plane stresses, and then the buckling problem is solved. And the results obtained using Matlab codes are appropriately the same as in the literature.
(1.1பைடு நூலகம்)