华中科技大学力学系板壳力学大作业 2012年12月
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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+1)
(1.4)
(xi ) =
N ∑ j =1
Wj
(n+1)
(xi )f (xj )
f (n+1) (xi ) =
d ( n) f (xi ) dx N ∑ = f (n) (xj )
j =1 N ∑ N ∑ (n) [ Wk (xj )f (xk )] j =1 k=1
=
N ∑ N ∑ (n) = [ Aij Wk (xj )]f (xk ) k=1 j =1
′′ 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
∂ 2w ∂ 2w M y = −D ( 2 + µ 2 ) ∂y ∂x The new version of the DQ method presented by [2] can be briefly summed up as follow. The key idea of the method is to take the boundary conditions into account when deciding the weighting coefficients of high-order derivatives. Since w and w′ are chosen as degrees of freedom at end points, to take the boundary conditions into account, the second-order derivative at end points take the following forms:
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)
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.
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
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
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
+
′′ BiN wN
+
N N −1 ∑ ∑ j =1 k=2
′′ Bik Bkj wj
(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
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)
(1.17)
4
Homework of Mechanics of Plates and Shells The weighting coefficients of fourth order derivatives at inner points are computed by
(4) wi
=
′′ Bi1 w1
′′ w1
=
′ A11 w1
+
′ A1N wN
+
N −2 ∑ k=2
′ A1k wk
=
N +2 ∑ Βιβλιοθήκη Baidu =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
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)
Figure 1: Rectangular plates under cosine-distributed edge compressions Thus the weighting coefficients of second-, third-, and fourth-order derivatives can be computed by N N N N ∑ ∑ ∑ ∑ Bij = Aik Akj Cij = Aik Bkj Dij = Aik Ckj = Bik Bkj (1.7)
Bo Xiao, Department of Mechanics, HUST
(n+1)
(1.4)
(xi ) =
N ∑ j =1
Wj
(n+1)
(xi )f (xj )
f (n+1) (xi ) =
d ( n) f (xi ) dx N ∑ = f (n) (xj )
j =1 N ∑ N ∑ (n) [ Wk (xj )f (xk )] j =1 k=1
=
N ∑ N ∑ (n) = [ Aij Wk (xj )]f (xk ) k=1 j =1
′′ 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
∂ 2w ∂ 2w M y = −D ( 2 + µ 2 ) ∂y ∂x The new version of the DQ method presented by [2] can be briefly summed up as follow. The key idea of the method is to take the boundary conditions into account when deciding the weighting coefficients of high-order derivatives. Since w and w′ are chosen as degrees of freedom at end points, to take the boundary conditions into account, the second-order derivative at end points take the following forms:
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)
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.
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
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
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
+
′′ BiN wN
+
N N −1 ∑ ∑ j =1 k=2
′′ Bik Bkj wj
(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
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)
(1.17)
4
Homework of Mechanics of Plates and Shells The weighting coefficients of fourth order derivatives at inner points are computed by
(4) wi
=
′′ Bi1 w1
′′ w1
=
′ A11 w1
+
′ A1N wN
+
N −2 ∑ k=2
′ A1k wk
=
N +2 ∑ Βιβλιοθήκη Baidu =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
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)
Figure 1: Rectangular plates under cosine-distributed edge compressions Thus the weighting coefficients of second-, third-, and fourth-order derivatives can be computed by N N N N ∑ ∑ ∑ ∑ Bij = Aik Akj Cij = Aik Bkj Dij = Aik Ckj = Bik Bkj (1.7)