# Static Analytical Approach of Moderately Thick Cylindrical Ribbed Shells Based on First-Order Shear Deformation Theory

Authors: Jinxing Lai; Chunxia Guo; Junling Qiu; Haobo Fan
Journal: Mathematical Problems in Engineering
 (2015)

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## Abstract

The classical shell theory (CST) without considering the shear deformation has been commonly used in the calculation of shells structures recently. However, the impact of theory of plates and shells subjected to the shear deformation on the calculation is increasingly pronounced along with the wide use of composite laminated structures. In this paper, based on first-order shear deformation theory (FSDT) of cylindrical shells, the displacement control differential equation of moderately thick cylindrical shells has been obtained, so has been the edge force at longitudinal of the shells. Meanwhile, a group of unit force is introduced to deduce the displacement of edge beam under the action of edge force. A join condition of moderately thick cylindrical ribbed shells is established according to the continuity of displacement as well. Most notably, the displacement analytical solution of bending problems of moderately thick cylindrical ribbed shells is obtained, which has profound theoretical significance for further improving the analytical solution of moderately thick cylindrical shells.


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## Body

## 1. Introduction

With the advent of new materials, various shell structures have experienced unprecedented development. In comparison with the beam and plate structural and the spherical shell structure, the cylindrical shell is of more reasonable structure, superior mechanical performance, simpler construction, and manufacturing. Therefore, the cylindrical shells have been widely used in tunnel, subway, and pipe jacking and other municipal engineering and machinery and aviation industries [1, 2]. Because of the engineering design requirement, the thickness of shells structures should be increased and often exceed the theoretical extent of thin shells; transverse shear deformation must be taken into account accordingly. Consequently, the effect of transverse shear deformation on the shells must be also considered; the corresponding calculated models and analytical theories should be established, which have induced increasing attention from scholars.Since Sophie Germain (1813) gained correct expression of bending of thin plate and Aaron (1874) and Love (1888) established theory of thin shells; the theory of plates and shells have been developed with the shells structure gradually application in engineering [3]. Based on Kirchhoff-Love’s classical theory of thin shells [4], the comparatively perfect result of dynamic excitation was obtained within the low radio-wave band when R/h≫1. However, it may cause 30% or even more error by putting this theory into the deformation, stress, and frequency analysis of composite shells, because the ratio of their elastic modulus and shear modulus usually reach 25–40. Therefore, the CST is not feasible for this case [5].High-order shear deformation theory (HSDT) and FSDT are adopted to analyse the effect of transverse shearing deformation in the moderately thick plates and shells theory [6–8]. The HSDT can better satisfy the boundary conditions of upper and lower surface of shells, but the equation is more complex, and some of its components of internal forces have no clear physical significance. In contrary, although the FSDT cannot always satisfy the unstressed boundary conditions of upper and lower surface of shells, it can get a simpler equation and readily establish the boundary conditions, with clear physical significance of component internal forces. Thus, the FSDT has been commonly used in actual engineering. Furthermore, the transverse shear deformation with increased equation order of shells brings tremendous difficulties on solving the equation. Therefore, whether the new equation can be solved in transverse shear deformation is another tough problem during the research of bending problem of moderately thick shells [9, 10]. By far, few analytical solutions of bending problem of moderately thick shells have been obtained based on the FSDT. This paper presents the basic equation of static forces calculation of cylindrical shells on the basis of the FSDT and derives the control equation of moderately thick cylindrical shells by introducing displacement potential functions. Furthermore, the subject of static analysis of moderately thick cylindrical ribbed shells is an interaction problem of edge beam and structure and with great complexity. Therefore, current research has been carried out mostly on thin cylindrical shells. The edge force at the longitudinal of shells has been obtained based on the derivation of control differential equation of moderately thick cylindrical shell. Meanwhile, the displacement of edge beam has been deduced under the action of edge force. In addition, the join condition of moderately thick cylindrical ribbed shells is established according to the continuity of displacement, and the displacement analytical solution of bending problems of moderately thick cylindrical ribbed shells has been obtained as well.


## 2. The Governing Equation of Moderately Thick Cylindrical Shells Based on FSDT

Take coordinate system of cylindrical shells as shown in Figure1:(1a)α1=x,α2=θ.Figure 1
Coordinate system of cylindrical shells.Then the corresponding Lamé coefficients are(1b)A1=1,A2=1.The radiuses of curvature are(1c)R1=∞,R2=R.Hence, the geometrical equations can be written as(2)ε11=∂u1∂x,ε22=∂u2R∂x+wR,w=∂u1R∂θ+∂u2∂x,κ1=∂ϕ1∂x,κ2=∂ϕ2R∂θ,τ=τ12+τ21=1R∂ϕ1∂θ+∂ϕ2∂x,ε13=ϕ1+∂w∂x,ε23=ϕ2+∂wR∂θ.And the constitutive equations are(3)T11=Kε11+νε22,T22=Kε22+νε11,T12=T21=K1-ν2ω,M11=Dκ1+νκ2,M22=Dκ2+νκ1,M12=M21=D1-ν2τ,T1n=Ghkτϕ1+∂w∂x,T2n=Ghkτϕ2+∂wR∂θ.Also, the equilibrium equations can be written as(4a)R∂T11∂x+∂T21∂θ=-Rq1,(4b)∂T22∂θ+R∂T12∂x+T2n=-Rq2,(4c)T22-R∂T1n∂x-∂T2n∂θ=Rqn,(4d)∂M22∂θ+R∂M12∂x=RT2n,(4e)R∂M11∂x+∂M21∂θ=RT1n,where J0=h, J2=h3/12, K=Eh/(1-ν2), D=Eh3/12(1-ν2), K is the tensile stiffness of the shell, and D is the bending stiffness of the shell.Substituting (2) into (3), the internal force and the bending moment, which are determined by u1, u2, w, ϕ1, and ϕ2, can be obtained. Then, substituting the internal force and the bending moment into (4a)–(4e), the displacement-based differential equation set of moderately thick cylindrical shells can be obtained as(5a)∂2u1∂x2+1-ν2∂2u1R2∂θ2+1+ν2∂2u2R∂x∂θ+νR∂w∂x=-q1K,(5b)∂2u2R2∂θ2+1-ν2∂2u2∂x2+1+ν2∂2u1R∂x∂θ+1R2∂w∂θ=-q2K,(5c)1R∂u2R∂θ+wR+ν∂u1∂x-k∇2w+∂ϕ1∂x+∂ϕ2R∂θ=qnK,(5d)∂2ϕ1∂x2+1-ν2∂2ϕ1R2∂θ2+1+ν2∂2ϕ2R∂x∂θ-k12h2(ϕ1+∂w∂x)=0,(5e)∂2ϕ2R2∂θ2+1-ν2∂2ϕ2∂x2+1+ν2∂2ϕ1R∂x∂θ-k12h2(ϕ2+∂wR∂θ)=0,where k=(1-ν)/2kτ.Bending problems of moderately thick cylindrical shells are summarized as mathematical problems that solve the displacement-based differential equation (2) under certain boundary conditions. Equations (5a)–(5e) are the differential equations composed of five second-order differential equations, whose solution is quite difficult; therefore, we introduce the displacement functions F1(x,θ), F2(x,θ), and f(x,θ), and then we have(6a)u1=-1-ν2Rν∂3∂x3-∂3R2∂x∂θ2F1,(6b)u2=-1-ν2R2+ν∂3R∂x2∂θ+∂3R3∂θ3F1,(6c)w=1-ν2∇2∇2F1.In the equation∇2=∂2/∂x2+∂2/R2∂θ2, (5a) and (5b) automatically satisfy themselves.Meanwhile, we can get(7a)ϕ1=∂F2∂x-∂fR∂θ,(7b)ϕ2=∂F2R∂θ+∂f∂x.Substituting (7a) and (7b) into (5d) and (5e) leads to(8a)∂∂x∇2F2-12kh2F2-12kh2w-∂R∂θ1-ν2∇2f-12kh2f=0,(8b)∂R∂θ∇2F2-12kh2F2-12kh2w+∂∂x1-ν2∇2f-12kh2f=0.By the Cauchy-Riemann conditions, it does not lose its generality [11]; we obtain the following equations by using (8a) and (8b):(9a)∇2F2-12kh2F2-12kh2w=0,(9b)1-ν2∇2f-12kh2f=0.Substituting (6c) into (9a), the equation that functions F1(x,θ) and F2(x,θ) should satisfy can be established as(10)∇2F2-k0F2=k01-ν2∇4F1,where k0=12k/h2. From (6a)–(6c) and (9a)-(9b), (5a)–(5e) can be satisfied. Then, substituting (6a)–(6c) and (7a)-(7b) into (5c) leads to(11)1R1-ν2R1-ν2∂4F1∂x4-k1-ν2∇6F1+∇2F2=qnK.Equations (10), (11), and (9b) are equivalent to (5a)–(5e). In order to make them a single equation, we have(12a)F1=∇2-k0F,(12b)F2=k01-ν2∇4F.Then (10) automatically satisfies itself. In (12a) and (12b), F(x,y) is another unknown function. By substituting (12a) and (12b) into (11), we obtain the equation that function F(x,θ) should satisfy; that is,(13)k1-ν2∇8F-1-ν1-ν22R2∂4∂x4∇2F+1-ν1-ν22R2k0∂4F∂x4=-qnK.To make the data easily analyzed, we introduce the nondimensional coordinateξ=x/L and make α1=R/L, α2=h/R. Thus, the governing differential equations (13) and (9b) can be, respectively, replaced by the following forms: (14a)k1-ν2∇8F~-1-ν1-ν2α142R6∂4∂ξ4∇2F~+6k1-ν1-ν2α14R8α22∂4F~∂ξ4=-qnK,(14b)1-ν2∇2f~-12kR2α22f~=0,where(15)∇8:=1R8α18∂8∂ξ8+4α16∂8∂ξ6∂θ2+6α14∂8∂ξ4∂θ4mmmm+4α12∂8∂ξ2∂θ6+∂8∂θ8(:),∇2:=1R2α12∂2∂ξ2+∂2∂θ2:.Equations (12a) and (12b) can be written as(16a)F~1=1R2α12∂2∂ξ2+∂2∂θ2-12kα22F~,(16b)F~2=6k1-νR6α22α14∂4∂ξ4+2α12∂4∂ξ2∂θ2+∂4∂θ4F~.The displacement equations can be obtained by (6a)–(6c), which are(17a)u~1=-1-ν2R4να13∂3∂ξ3-α1∂3∂ξ∂θ2F~1,(17b)u~2=-1-ν2R42+να12∂3∂ξ2∂θ+∂3∂θ3F~1,(17c)w~=1-ν2R4α14∂4∂ξ4+2α12∂4∂ξ2∂θ2+∂4∂θ4F~1.And the angles of rotation can be expressed as(18a)ϕ~1=1Rα1∂F~2∂ξ-∂f~∂θ,(18b)ϕ~2=1R∂F~2∂θ+α1∂f~∂ξ.According to (2), the deformation can be obtained as(19)ε~11=α1R∂u~1∂ξ,ε~22=1R∂u~2∂θ+w~,ω~=1R∂u~1∂θ+α1∂u~2∂ξ,κ~1=α1R∂ϕ~1∂ξ,κ~2=1R∂ϕ~2∂θ,τ~=α1R∂ϕ~2∂ξ+1R2∂ϕ~1∂θ,ε~13=ϕ~1+α1R∂w~∂ξ,ε~23=ϕ~2+1R2∂w~∂θ.According to (3), internal forces can be obtained as(20)T~11=ERα21-ν2ε~11+νε~22,T~22=ERα21-ν2ε~22+νε~11,M~11=ER3α23121-ν2κ~1+νκ~2,M~22=ER3α23121-ν2κ~2+νκ~1,M~12=M~21=ER3α2324(1+ν)τ~,T~12=T~21=ERα21-ν21-ν2ω~,T~13=GRkτα2ε~13,T~23GRkτα2ε~23.By opening up the displacement function as trigonometric series in satisfying the boundary conditions, the control equation can be simplified to a solvable eighth-order ordinary differential equation and a second-order ordinary differential equation. After solving the solution of the control equation sets, displacement, angle, and internal force can be solved too, and then set ten undetermined coefficients that it contains by using boundary conditions or join conditions to solve the solution of moderately thick cylindrical shells.


## 3. The Static Bending of Moderately Thick Cylindrical Shells That Considers the Effect of Edge Beam

### 3.1. Displacement of Edge Beam under Edge Force

In order to study the coworking situation of shells and edge beam, the shells and edge beam are cut and split along the edge of the shells, this section will have the internal forcesT22, T21, T23, M22, and M21 that can balance with the external force and the corresponding deformations u2, u1, w, ϕ1, and ϕ2, as shown in Figure 2.Figure 2
Force at the place where the shells link up with the edge beam.Now let us discuss the displacement state of bearing load of edge beamsT22, T21, T23, M22 and M21; the edge beam section adopts the coordinate system (x, 1, 2), among which coordinate x locates in y-direction of edge beam, coordinate axes 1 and 2 stand for main axes of section moments of inertia, the point O on the section shows the edge point on the edge beam equivalent to r=R of shells, and, in addition, it has coordinates e1, e2, as shown in Figure 3.Figure 3
Coordinate graph of edge beam.In order to calculate the deformation state of edge beam, the edge beam bears the edge load as shown in Figure4, and the rules of continuous change of a set of unit forces p-1, p-2, z-x, m-x, and m-1 on x-axis are stipulated as follows:(21)p-1=1·∑n=1∞sin⁡nπxL,p-2=1·∑n=1∞sin⁡nπxL,z-x=1·∑n=1∞sin⁡nπxL,m-x=1·∑n=1∞sin⁡nπxL,m-1=1·∑n=1∞sin⁡nπxL.Figure 4
Rules of a set of displacement and stress on shear centerT.According to the displacement differential equation of given beam and the Saint-Venant torsional equation [12, 13], by (21), we have(22)d4δ-1dx4=1EI2∑n=1∞sinnπxL,d4δ-21dx4=1EI1∑n=1∞sinnπxL,d2δ-22dx2=1EI1∑n=1∞sinnπxL,dδ-3dx=1EF∑n=1∞sinnπxL,d2δ-4dx2=-1GIt∑n=1∞sinnπxL,δ-5=dδ-2dx=dδ-21dx+dδ-22dx.The displacementsδ-1, δ-2, δ-3, δ-4, and δ-5 are the ones of shear center T corresponding to the directions of   p-1, p-2, z-x, m-x, and m-1.Defining integral to (22) and making l=L/nπ lead to(23)δ-1=∑n=1∞l4EI2sinnπxL,δ-21=∑n=1∞l4EI1sinnπxL,δ-22=-∑n=1∞l2EI1sinnπxL,δ-3=-∑n=1∞lEFcos⁡nπxL,δ-4=∑n=1∞l2GItsinnπxL.The displacement on the pointO caused by the action of a set of force p-1=1, p-2=1, z-x=1, m-x=1, and m-1=1 on point T is listed in Table 1.Table 1
Displacement on the pointO caused by unit force acting on point T.
p
-
1
=
1
p
-
2
=
1
z
-
x
=
1
m
-
x
=
1
m
-
1
=
1
δ
-
1
l
4
E
I
2
0
0
-
l
2
e
2
G
I
t
0
δ
-
2
0
l
4
E
I
1
0
l
2
e
1
G
I
t
0
δ
-
3
l
3
e
1
E
I
2
l
3
e
2
E
I
1
-
l
E
F
0
-
l
2
E
I
1
δ
-
4
0
0
0
l
2
G
I
t
0
δ
-
5
0
l
3
E
I
1
0
0
-
l
E
I
1In the table, the first column of the first row represents the displacementδ-1 of point O in the direction of axis 1 caused by the action of force p-1=1 on point T along the axis 1; the second column of the first row stands for the displacement δ-1 of point O in the direction of axis 1 caused by the action of force p-2=1 on point T along the axis 2; as axes 1 and 2 are orthogonal, δ-1=0; the third column of the first row represents the displacement δ-1 of point O in the direction of axis 1 caused by the action of force z-x=1 on point T along the axis 1; as axes 1 and x are orthogonal, δ-1=0; the fourth column of the first row stands for the displacement δ-1 of point O in the direction of axis 1 caused by the action of moment of force m-x=1 on point T when bypassing the vertical axis of point T; the fifth column of the first row represents the displacement δ-1 of point O in the direction of axis 1 caused by the action of moment of force m-1=1 on point T when rotating around axis 1. The state of the remaining rows and columns can be obtained according to this method. The displacement rules of point O of edge beam are as shown in Figure 5.Figure 5
Displacement rules of pointO.Generally, the edge force of shells often acts on the edge of beam, instead of acting on pointT directly; therefore, the edge force should be converted into a set of equivalent forces p-1, p-2, z-x, m-x, and m-1 that act on point T. When taking dx of beams during converting, the load T21=1·∑n=1∞cos⁡(nπx/L) and internal forces M1, Q2, M2, Q1, and z-x are acting on it.From the equilibrium condition [14], we have(24)dM1dx=e2·∑n=1∞cos⁡nπxL,dM2dx=e1·∑n=1∞cos⁡nπxL,dz-xdx=-∑n=1∞cos⁡⁡nπxL.Defining differential and integral to (24), respectively, we have(25)d2M1dx2=-∑n=1∞e2lsin⁡nπxL,d2M2dx2=-∑n=1∞e1lsin⁡nπxL,z-x=-∑n=1∞lsinnπxL.By comparing (25) with differential equations d2M1/dx2=p2(x) and d2M2/dx2=p1(x), it shows that the stress and displacement in the main axis 1 of moments of inertia and direction 2 caused by the action of T21=1·∑n=1∞cos⁡(nπx/L) can be replaced by equivalent axial force z-x=-∑n=1∞l·sin(nπx/L) and mathematical loads -∑n=1∞(e1/l)sin(nπx/L) and -∑n=1∞(e2/l)sin(nπx/L); therefore, the equivalent load can be calculated according to Table 2.Table 2
Equivalent load.
p
-
1
p
-
2
z
-
x
m
-
x
m
-
1
p
1 = 1
1
0
0
-
e
2
0
p
2 = 1
0
1
0
e
1
0
T
21 = 1
-
e
1
l
-
e
2
l
-
l
0
0
M
22 = 1
0
0
0
1
0
M
21 = 1
0
0
0
0
1In Table2, the first column of the first row represents the equivalent load p-1 of the corresponding point T on the axis 1 when point O acting on p1=1 along the axis 1, whose value is 1; the second column of the first row stands for the equivalent load p-2 of the corresponding point T on the axis 2 when point O acting on p1=1 along the axis 1; as axes 1 and 2 are orthogonal, its value is 0; the third column of the first row represents the equivalent load z-x of the corresponding point T on the x-axis when point O acting on p1=1 along the axis 1; as axis 1 and x-axis are orthogonal, its value is 0; the fourth column of the first row stands for the equivalent load m-x of the corresponding point T through the vertical axis of point T when point O acting on p1=1 along the axis 1, whose value is -e2; as its direction is opposite to the specified one, as shown in Figure 3, the negative sign is taken; the fifth column of the first row represents the equivalent load m-1 of the corresponding point T through the axis 1 when point O acting on p1=1 along the axis 1, whose value is 1.By using the combination of the displacement of pointO caused by the action of   p-1=1, p-2=1, z-x=1, m-x=1, and m-1=1 of point T shown in Table 1 and the equivalent load of point T caused by the action of the edge force of point O as shown in Table 2, displacements δ1, δ2, δ3, δ4, and δ5 of point O on the cross section of edge beam caused by the action of unit forces p1=1, p2=1, T21=1, M22=1, and M21=1 on the edge of point O can be calculated according to Table 3.Table 3
Displacement of pointO caused by the action of unit force.
p
1
=
1
p
2
=
1
T
21
=
1
M
22
=
1
M
21
=
1
δ
1
l
4
E
I
2
+
l
2
e
2
2
G
I
t
-
l
2
e
1
e
2
G
I
t
-
l
3
e
1
E
I
2
-
l
2
e
2
G
I
t
0
δ
2
-
l
2
e
1
e
2
G
I
t
l
4
E
I
1
+
l
2
e
1
2
G
I
t
-
l
3
e
2
E
I
1
l
2
e
1
G
I
t
-
l
2
E
I
1
δ
3
l
3
e
1
E
I
2
l
3
e
2
E
I
1
-
l
2
E
F
F
e
1
2
I
2
+
F
e
2
2
I
1
-
1
0
0
δ
4
-
l
2
e
2
G
I
t
l
2
e
1
G
I
t
0
l
2
G
I
t
0
δ
5
0
l
3
E
I
1
-
l
2
e
2
E
I
1
0
-
l
E
I
1For the convenience of data analysis, recordα3=h2/b, α4=b/L, and rewrite all values in Table 3 as the form of α3 and α4, as shown in Table 4.Table 4
Displacement of pointO caused by the action of unit force.
p
1
=
1
p
2
=
1
T
21
=
1
M
22
=
1
M
21
=
1
δ
1
12
E
α
3
(
α
4
n
π
)
4
+
3
4
G
α
3
(
α
4
n
π
)
2
-
3
4
G
α
3
2
(
α
4
n
π
)
2
-
6
E
α
3
(
α
4
n
π
)
3
-
3
2
G
α
3
2
(
α
4
n
π
)
2
b
0
δ
2
-
3
4
G
α
3
2
(
α
4
n
π
)
2
12
E
α
3
3
(
α
4
n
π
)
4
+
3
4
G
α
3
3
(
α
4
n
π
)
2
-
6
E
α
3
2
(
α
4
n
π
)
3
3
2
G
α
3
3
(
α
4
n
π
)
2
b
-
12
E
α
3
3
(
α
4
n
π
)
2
b
2
δ
3
6
E
α
3
(
α
4
n
π
)
3
6
E
α
3
2
(
α
4
n
π
)
3
-
5
E
α
3
(
α
4
n
π
)
2
0
0
δ
4
-
3
2
G
α
3
2
(
α
4
n
π
)
2
b
3
2
G
α
3
3
(
α
4
n
π
)
2
b
0
3
G
α
3
3
(
α
4
n
π
)
2
b
2
0
δ
5
0
12
E
α
3
3
(
α
4
n
π
)
3
b
-
6
E
α
3
2
(
α
4
n
π
)
2
b
0
-
12
E
α
3
3
α
4
n
π
b


### 3.2. Solution Method

Recently, open cylindrical shells, especially the reinforced concrete open cylindrical shells which are usually used to be large-span roofs, have been applied successfully in many projects. The curvy edges of cylindrical shells are braced by the reinforced concrete circular-arc thin plate or thin arch circle, which can be regarded as simply supported edges with large stiffness. Thus, we can introduce(26)F~=∑n=1∞F~nθsinnπξ,f~=∑n=1∞f~n(θ)cos⁡(nπξ),where ξ=0 and ξ=1 are the boundary conditions of simply supported edges.Substituting (26) into (14a), we have(27a)d8F~ndθ8+B1d6F~ndθ6+B2d4F~ndθ4+B3F~n′′+B4F~n=-2R8qnKk1-ν.Then, substituting (26) into (14b), we have(27b)-1-ν2α1nπ2-12kα22f~n+1-ν2d2f~ndθ2=0,where(28)B1=-4α1nπ2,B2=6α1nπ4,B3=-4α1nπ6-1-ν2kα1nπ4,B4=α1nπ8+1-ν2kα1nπ6+12(1-ν2)α22α1nπ4.The particular solution of (27a) is(29)F~np=-2R8qnKk1-νB4.To obtain the solution of (27a) and (27b), substituting F~nh=eλmθ and f~n=eλmθ into (27a) and (27b), we have(30a)λm8+B1λm6+B2λm4+B3λm2+B4=0,mmmmmmlmmm=1,2,3,…,8,(30b)λm2=α1nπ2+24k1-vα22m=9,10.Equation (30a) can be solved using MATLAB, which is composed of 8 complex roots(31)λ1,2=d1±d2i,λ3,4=-d1±d2i,λ5,6=d3±d4i,λ7,8=-d3±d4i.The homogeneous solution of (27a) is(32)F~nh=ed1θC1·cos⁡⁡d2θ+C2·sin⁡d2θ+e-d1θC3·cos⁡⁡d2θ+C4·sin⁡d2θ+ed3θC5·cos⁡⁡d4θ+C6·sin⁡d4θ+e-d3θC7·cos⁡d4θ+C8·sind4θ.The general solution of (27a) is(33)F~n=F~nh+F~np.And the solution of (27b) is (34)f~n=C9eλ9θ+C10eλ10θ,where C1~C10 are undetermined coefficients, which can be determined by boundary conditions or join conditions.To satisfy boundary conditions or join conditions, displacement components, stress components, and bending moment components are expanded according to trigonometric series.According to (16a) and (16b), the expansions of displacement functions can be expressed as(35a)F~1=∑n=1∞1R2d2F~ndθ2-12kα22+α1nπ2F~nsin⁡(nπξ)=∑n=1∞F~1nθsinnπξ,(35b)F~2=∑n=1∞6k1-νR6α22α1nπ4F~n-2α1nπ2d2F~ndθ2mmmmmmmmm·d2F~ndθ2+d4F~ndθ4sin⁡(nπξ)=∑n=1∞F~2nθsinnπξ.According to (17a)–(17c), the expansions of displacement components can be expressed as (36a)u~1=∑n=1∞1-ν2R4να1nπ3F~1n+α1nπd2F~1ndθ2cos⁡⁡nπξ=∑n=1∞u~1ncos⁡nπξ,(36b)u~2=∑n=1∞1-ν2R42+να1nπ2dF~1ndθ-d3F~1ndθ3sin⁡nπξ=∑n=1∞u~2nsinnπξ,(36c)w~=∑n=1∞1-ν2R4α1nπ4F~1n-2α1nπ2d2F~1ndθ2mmmmmm·d2F~1ndθ2+d4F~1ndθ4sin(nπξ)=∑n=1∞w~nsin⁡nπξ.According to (18a) and (18b), the expansions of angles of rotation can be expressed as(37a)ϕ~1=∑n=1∞1Rα1nπF~2n-df~dθcos⁡(nπξ)=∑n=1∞u~4ncos⁡⁡nπξ,(37b)ϕ~2=∑n=1∞1RdF~2ndθ-α1nπf~nsinnπξ=∑n=1∞u~5nsin⁡nπξ.According to (19), the expansions of strain components can be expressed as(38)ε~11=∑n=1∞-α1nπRu~1nsinnπξ=∑n=1∞ε~1nsinnπξ,ε~22=∑n=1∞1Rdu~2ndθ+w~nsinnπξ=∑n=1∞ε~2nsinnπξ,ω~=∑n=1∞1Rdu~1ndθ+α1nπu~2cos⁡nπξ=∑n=1∞ω~ncos⁡nπξ,κ~1=∑n=1∞-α1nπRu~4nsinnπξ=∑n=1∞κ~1nsinnπξ,κ~2=∑n=1∞1Rdu~5ndθsinnπξ=∑n=1∞κ~2nsinnπξ,τ~1=∑n=1∞α1nπRu~5ncos⁡nπξ=∑n=1∞τ~1ncos⁡nπξ,τ~2=∑n=1∞1Rdu~4ndθcos⁡nπξ=∑n=1∞τ~2ncos⁡nπξ,ε~13=∑n=1∞u~4n+α1nπRu~3ncos⁡nπξ=∑n=1∞ε~13ncos⁡nπξ,ε~23=∑n=1∞u~5n+1Rdu~3ndθsinnπξ=∑n=1∞ε~23nsinnπξ.According to (20), the expansions of internal force and bending moment components can be expressed as (39)T~11=ERα21-ν2∑n=1∞(ε~1n+vε~2n)sinnπξ=∑n=1∞T~1nsinnπξ,T~22=ERα21-ν2∑n=1∞(ε~2n+vε~1n)sinnπξ=∑n=1∞T~2nsinnπξ,M~11=ER3α23121-ν2∑n=1∞(κ~1n+vκ~2n)sinnπξ=∑n=1∞M~1nsinnπξ,M~22=ER3α23121-ν2∑n=1∞(κ~2n+vκ~1n)sin⁡(nπξ)=∑n=1∞M~2nsinnπξ,T~12=T~21=ERα221+ν∑n=1∞ω~ncos⁡nπξ=∑n=1∞T~12ncos⁡nπξ,M~12=M~21=ER3α23241+ν∑n=1∞(τ~1n+vτ~2n)cos⁡⁡(nπξ)=∑n=1∞M~12ncos⁡nπξ,T~13=GRkτα2∑n=1∞ε~13ncos⁡nπξ=∑n=1∞T~13ncos⁡nπξ,T~23=GRkτα2∑n=1∞ε~23nsinnπξ=∑n=1∞T~23nsinnπξ.


### 3.3. Join Conditions

There are five edge forces at the place where the shells is connected with the edge beam, namely,T~21, T~22, T~23, M~22, and M~21. By opening them up with trigonometric series, we have(40)T~21=∑n=1∞T~12ncos⁡nπξ,T~22=∑n=1∞T~2nsinnπξ,T~23=∑n=1∞T~23nsinnπξ,M~21=∑n=1∞M~12ncos⁡nπξ,M~22=∑n=1∞M~2nsinnπξ,where T~21, T~22, T~23, M~22, and M~21 are functions of θ and contain ten undetermined coefficients; then the edge forces are as follows:on the edgeI:(41)T~22θ=-θ0=∑n=1∞T~2nθ=-θ0sinnπξ=∑n=1∞r1nsinnπξ=r1,T~23θ=-θ0=∑n=1∞T~23nθ=-θ0sinnπξ=∑n=1∞r2nsinnπξ=r2,T~21θ=-θ0=∑n=1∞T~12nθ=-θ0cos⁡nπξ=∑n=1∞r3nsin(nπξ)=r3,M~22θ=-θ0=∑n=1∞M~2nθ=-θ0sin(nπξ)=∑n=1∞r4nsin(nπξ)=r4,M~21θ=-θ0=∑n=1∞M~12nθ=-θ0sin(nπξ)=∑n=1∞r9nsin(nπξ)=r9;on the edgeII:(42)T~22θ=θ0=∑n=1∞T~2nθ=θ0sin(nπξ)=∑n=1∞r5nsin(nπξ)=r5,T~23θ=θ0=∑n=1∞T~23nθ=θ0sin(nπξ)=∑n=1∞r6nsin(nπξ)=r6,T~21θ=θ0=∑n=1∞T~12nθ=θ0cos⁡(nπξ)=∑n=1∞r7nsin(nπξ)=r7,M~22θ=θ0=∑n=1∞M~2nθ=θ0sin(nπξ)=∑n=1∞r8nsin(nπξ)=r8,M~21θ=θ0=∑n=1∞M~12nθ=θ0sin(nπξ)=∑n=1∞r10nsin(nπξ)=r10.The edge forces are as shown in Figure6.Figure 6
Edge forces distribution at the place where the shells are connected with the edge beam.In order to use the results in Table4, we decompose the edge forces r1~r10 along axes 1 and 2.Edge
I
Decomposer1, r2 along the axis 1: (43a)r1cos⁡θ-r2sinθ=∑n=1∞(r1ncos⁡θ-r2nsinθ)sin(nπξ).
Decomposer1, r2 along the axis 2:(43b)r1sinθ+r2cos⁡θ=∑n=1∞(r1nsinθ+r2ncos⁡θ)sinnπξ.According to Table4, the displacement generated by the edge beam I can be denoted by(44)δ1I=∑n=1∞δ1nIsinnπξ,δ2I=∑n=1∞δ2nIsinnπξ,δ3I=∑n=1∞δ3nIcos⁡nπξ,δ4I=∑n=1∞δ4nIsinnπξ,δ5I=∑n=1∞δ5nIsinnπξ,where(45)b0=12Eα3α4nπ4+34Gα3α4nπ2,b1=-34Gα32α4nπ2,b2=-6Eα3α4nπ3,b3=-32Gα32α4nπ2b,c1=12Eα33α4nπ4+34Gα33α4nπ2,c2=-6Eα32α4nπ3,c3=32Gα33α4nπ2b,c4=12Eα33α4nπ2b2,d0=-5Eα3α4nπ2,f0=3Gα33α4nπ2b2g0=12Eα33α4nπ3b,g1=-6Eα32α4nπ2b,g2=-12Eα33α4nπb.Thus, the subdisplacement and angle of rotation of edge beam I inu, v, and w directions are(46a)u~1I=δ3I=∑n=1∞δ3nIcos⁡nπξ,(46b)u~2I=∑n=1∞δ2nIsinθ+δ1nIcos⁡θsinnπξ,(46c)w~I=∑n=1∞-δ1nIsinθ+δ2nIcos⁡θsinnπξ,(46d)u~4I=δ5I=∑n=1∞δ5nIsinnπξ,(46e)u~5I=δ4I=∑n=1∞δ4nIsinnπξ.Edge II
Decomposer5, r6 along the axis 1:(47a)-r5cos⁡θ-r6sinθ=∑n=1∞(-r5ncos⁡θ-r6nsinθ)sin(nπξ).
Decomposer5, r6 along the axis 2:(47b)r5sinθ-r6cos⁡θ=∑n=1∞r5nsinθ-r6ncos⁡θsinnπξ.According to Table3, the displacement generated by the edge beam II can be denoted by(48)δ1II=∑n=1∞δ1nIIsinnπξ,δ2II=∑n=1∞δ2nIIsinnπξ,δ3II=∑n=1∞δ3nIIcos⁡nπξδ4II=∑n=1∞δ4nIIsinnπξ,δ5II=∑n=1∞δ5nIIsinnπξ.Thus, the subdisplacement and angle of rotation of edge beam II inu, v, and w directions are(49a)u~1II=δ3II=∑n=1∞δ3nIIcos⁡nπξ,(49b)u~2II=∑n=1∞-δ2nIIsinθ+δ1nIIcos⁡θsinnπξ,(49c)w~II=∑n=1∞δ1nIIsinθ+δ2nIIcos⁡θsinnπξ,(49d)u~4II=δ5II=∑n=1∞δ5nIIsinnπξ,(49e)u~5II=δ4II=∑n=1∞δ4nIIsinnπξ.The expression for shell’s displacements is(50)u~1=∑n=1∞u~1ncos⁡nπξ,u~2=∑n=1∞u~2nsinnπξ,w~=∑n=1∞w~nsinnπξ,u~4=∑n=1∞u~4ncos⁡nπξ,u~5=∑n=1∞u~5nsinnπξ.Then, shell’s displacement on the edge is as follows.Edge I is(51a)u~1qIθ=-θ0=u~1θ=-θ0=∑n=1∞u~1nθ=-θ0cos⁡nπξ,(51b)u~2qIθ=-θ0=u~2θ=-θ0=∑n=1∞u~2nθ=-θ0sinnπξ,(51c)w~qIθ=-θ0=w~θ=-θ0=∑n=1∞w~nθ=-θ0sinnπξ,(51d)u~4qIθ=-θ0=u~4θ=-θ0=∑n=1∞u~4nθ=-θ0cos⁡nπξ,(51e)u~5qIθ=-θ0=u~5θ=-θ0=∑n=1∞u~5nθ=-θ0sinnπξ.Edge II is(52a)u~1qIIθ=θ0=u~1θ=θ0=∑n=1∞u~1nθ=θ0cos⁡nπξ,(52b)u~2qIIθ=θ0=u~2θ=θ0=∑n=1∞u~2nθ=θ0sinnπξ,(52c)w~qIIθ=θ0=w~θ=θ0=∑n=1∞w~nθ=θ0sinnπξ,(52d)u~4qIIθ=θ0=u~4θ=θ0=∑n=1∞u~4nθ=θ0cos⁡nπξ,(52e)u~5qIIθ=θ0=u~5θ=θ0=∑n=1∞u~5nθ=θ0sinnπξ.By using the continuous conditions of displacements to compare (46a)–(46e), (51a)–(51e), (49a)–(49e), and (52a)–(52e), respectively, the displacement boundary conditions of shells can be obtained as follows.Edge I is(53)u1nθ=-θ0=δ3nI,u2nθ=-θ0=δ2nIsinθ+δ1nIcos⁡θ,wnθ=-θ0=-δ1nIsinθ+δ2nIcos⁡θ,u4nθ=-θ0=δ5nIu5nθ=-θ0=δ4nI.Edge II is(54)u1nθ=θ0=δ3nII,u2nθ=θ0=-δ2nIIsinθ+δ1nIIcos⁡θ,wnθ=θ0=-δ1nIIsinθ+δ2nIIcos⁡θ,u4nθ=θ0=δ5nII,u5nθ=θ0=δ4nII.According to the join condition, the ten undetermined coefficients can be solved, and trigonometric series expressions of displacements, angles of rotation, internal forces, and moments can be further obtained, whose changing curves can be drawn with MATLAB program.


## 3.1. Displacement of Edge Beam under Edge Force

In order to study the coworking situation of shells and edge beam, the shells and edge beam are cut and split along the edge of the shells, this section will have the internal forcesT22, T21, T23, M22, and M21 that can balance with the external force and the corresponding deformations u2, u1, w, ϕ1, and ϕ2, as shown in Figure 2.Figure 2
Force at the place where the shells link up with the edge beam.Now let us discuss the displacement state of bearing load of edge beamsT22, T21, T23, M22 and M21; the edge beam section adopts the coordinate system (x, 1, 2), among which coordinate x locates in y-direction of edge beam, coordinate axes 1 and 2 stand for main axes of section moments of inertia, the point O on the section shows the edge point on the edge beam equivalent to r=R of shells, and, in addition, it has coordinates e1, e2, as shown in Figure 3.Figure 3
Coordinate graph of edge beam.In order to calculate the deformation state of edge beam, the edge beam bears the edge load as shown in Figure4, and the rules of continuous change of a set of unit forces p-1, p-2, z-x, m-x, and m-1 on x-axis are stipulated as follows:(21)p-1=1·∑n=1∞sin⁡nπxL,p-2=1·∑n=1∞sin⁡nπxL,z-x=1·∑n=1∞sin⁡nπxL,m-x=1·∑n=1∞sin⁡nπxL,m-1=1·∑n=1∞sin⁡nπxL.Figure 4
Rules of a set of displacement and stress on shear centerT.According to the displacement differential equation of given beam and the Saint-Venant torsional equation [12, 13], by (21), we have(22)d4δ-1dx4=1EI2∑n=1∞sinnπxL,d4δ-21dx4=1EI1∑n=1∞sinnπxL,d2δ-22dx2=1EI1∑n=1∞sinnπxL,dδ-3dx=1EF∑n=1∞sinnπxL,d2δ-4dx2=-1GIt∑n=1∞sinnπxL,δ-5=dδ-2dx=dδ-21dx+dδ-22dx.The displacementsδ-1, δ-2, δ-3, δ-4, and δ-5 are the ones of shear center T corresponding to the directions of   p-1, p-2, z-x, m-x, and m-1.Defining integral to (22) and making l=L/nπ lead to(23)δ-1=∑n=1∞l4EI2sinnπxL,δ-21=∑n=1∞l4EI1sinnπxL,δ-22=-∑n=1∞l2EI1sinnπxL,δ-3=-∑n=1∞lEFcos⁡nπxL,δ-4=∑n=1∞l2GItsinnπxL.The displacement on the pointO caused by the action of a set of force p-1=1, p-2=1, z-x=1, m-x=1, and m-1=1 on point T is listed in Table 1.Table 1
Displacement on the pointO caused by unit force acting on point T.
p
-
1
=
1
p
-
2
=
1
z
-
x
=
1
m
-
x
=
1
m
-
1
=
1
δ
-
1
l
4
E
I
2
0
0
-
l
2
e
2
G
I
t
0
δ
-
2
0
l
4
E
I
1
0
l
2
e
1
G
I
t
0
δ
-
3
l
3
e
1
E
I
2
l
3
e
2
E
I
1
-
l
E
F
0
-
l
2
E
I
1
δ
-
4
0
0
0
l
2
G
I
t
0
δ
-
5
0
l
3
E
I
1
0
0
-
l
E
I
1In the table, the first column of the first row represents the displacementδ-1 of point O in the direction of axis 1 caused by the action of force p-1=1 on point T along the axis 1; the second column of the first row stands for the displacement δ-1 of point O in the direction of axis 1 caused by the action of force p-2=1 on point T along the axis 2; as axes 1 and 2 are orthogonal, δ-1=0; the third column of the first row represents the displacement δ-1 of point O in the direction of axis 1 caused by the action of force z-x=1 on point T along the axis 1; as axes 1 and x are orthogonal, δ-1=0; the fourth column of the first row stands for the displacement δ-1 of point O in the direction of axis 1 caused by the action of moment of force m-x=1 on point T when bypassing the vertical axis of point T; the fifth column of the first row represents the displacement δ-1 of point O in the direction of axis 1 caused by the action of moment of force m-1=1 on point T when rotating around axis 1. The state of the remaining rows and columns can be obtained according to this method. The displacement rules of point O of edge beam are as shown in Figure 5.Figure 5
Displacement rules of pointO.Generally, the edge force of shells often acts on the edge of beam, instead of acting on pointT directly; therefore, the edge force should be converted into a set of equivalent forces p-1, p-2, z-x, m-x, and m-1 that act on point T. When taking dx of beams during converting, the load T21=1·∑n=1∞cos⁡(nπx/L) and internal forces M1, Q2, M2, Q1, and z-x are acting on it.From the equilibrium condition [14], we have(24)dM1dx=e2·∑n=1∞cos⁡nπxL,dM2dx=e1·∑n=1∞cos⁡nπxL,dz-xdx=-∑n=1∞cos⁡⁡nπxL.Defining differential and integral to (24), respectively, we have(25)d2M1dx2=-∑n=1∞e2lsin⁡nπxL,d2M2dx2=-∑n=1∞e1lsin⁡nπxL,z-x=-∑n=1∞lsinnπxL.By comparing (25) with differential equations d2M1/dx2=p2(x) and d2M2/dx2=p1(x), it shows that the stress and displacement in the main axis 1 of moments of inertia and direction 2 caused by the action of T21=1·∑n=1∞cos⁡(nπx/L) can be replaced by equivalent axial force z-x=-∑n=1∞l·sin(nπx/L) and mathematical loads -∑n=1∞(e1/l)sin(nπx/L) and -∑n=1∞(e2/l)sin(nπx/L); therefore, the equivalent load can be calculated according to Table 2.Table 2
Equivalent load.
p
-
1
p
-
2
z
-
x
m
-
x
m
-
1
p
1 = 1
1
0
0
-
e
2
0
p
2 = 1
0
1
0
e
1
0
T
21 = 1
-
e
1
l
-
e
2
l
-
l
0
0
M
22 = 1
0
0
0
1
0
M
21 = 1
0
0
0
0
1In Table2, the first column of the first row represents the equivalent load p-1 of the corresponding point T on the axis 1 when point O acting on p1=1 along the axis 1, whose value is 1; the second column of the first row stands for the equivalent load p-2 of the corresponding point T on the axis 2 when point O acting on p1=1 along the axis 1; as axes 1 and 2 are orthogonal, its value is 0; the third column of the first row represents the equivalent load z-x of the corresponding point T on the x-axis when point O acting on p1=1 along the axis 1; as axis 1 and x-axis are orthogonal, its value is 0; the fourth column of the first row stands for the equivalent load m-x of the corresponding point T through the vertical axis of point T when point O acting on p1=1 along the axis 1, whose value is -e2; as its direction is opposite to the specified one, as shown in Figure 3, the negative sign is taken; the fifth column of the first row represents the equivalent load m-1 of the corresponding point T through the axis 1 when point O acting on p1=1 along the axis 1, whose value is 1.By using the combination of the displacement of pointO caused by the action of   p-1=1, p-2=1, z-x=1, m-x=1, and m-1=1 of point T shown in Table 1 and the equivalent load of point T caused by the action of the edge force of point O as shown in Table 2, displacements δ1, δ2, δ3, δ4, and δ5 of point O on the cross section of edge beam caused by the action of unit forces p1=1, p2=1, T21=1, M22=1, and M21=1 on the edge of point O can be calculated according to Table 3.Table 3
Displacement of pointO caused by the action of unit force.
p
1
=
1
p
2
=
1
T
21
=
1
M
22
=
1
M
21
=
1
δ
1
l
4
E
I
2
+
l
2
e
2
2
G
I
t
-
l
2
e
1
e
2
G
I
t
-
l
3
e
1
E
I
2
-
l
2
e
2
G
I
t
0
δ
2
-
l
2
e
1
e
2
G
I
t
l
4
E
I
1
+
l
2
e
1
2
G
I
t
-
l
3
e
2
E
I
1
l
2
e
1
G
I
t
-
l
2
E
I
1
δ
3
l
3
e
1
E
I
2
l
3
e
2
E
I
1
-
l
2
E
F
F
e
1
2
I
2
+
F
e
2
2
I
1
-
1
0
0
δ
4
-
l
2
e
2
G
I
t
l
2
e
1
G
I
t
0
l
2
G
I
t
0
δ
5
0
l
3
E
I
1
-
l
2
e
2
E
I
1
0
-
l
E
I
1For the convenience of data analysis, recordα3=h2/b, α4=b/L, and rewrite all values in Table 3 as the form of α3 and α4, as shown in Table 4.Table 4
Displacement of pointO caused by the action of unit force.
p
1
=
1
p
2
=
1
T
21
=
1
M
22
=
1
M
21
=
1
δ
1
12
E
α
3
(
α
4
n
π
)
4
+
3
4
G
α
3
(
α
4
n
π
)
2
-
3
4
G
α
3
2
(
α
4
n
π
)
2
-
6
E
α
3
(
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## 3.2. Solution Method

Recently, open cylindrical shells, especially the reinforced concrete open cylindrical shells which are usually used to be large-span roofs, have been applied successfully in many projects. The curvy edges of cylindrical shells are braced by the reinforced concrete circular-arc thin plate or thin arch circle, which can be regarded as simply supported edges with large stiffness. Thus, we can introduce(26)F~=∑n=1∞F~nθsinnπξ,f~=∑n=1∞f~n(θ)cos⁡(nπξ),where ξ=0 and ξ=1 are the boundary conditions of simply supported edges.Substituting (26) into (14a), we have(27a)d8F~ndθ8+B1d6F~ndθ6+B2d4F~ndθ4+B3F~n′′+B4F~n=-2R8qnKk1-ν.Then, substituting (26) into (14b), we have(27b)-1-ν2α1nπ2-12kα22f~n+1-ν2d2f~ndθ2=0,where(28)B1=-4α1nπ2,B2=6α1nπ4,B3=-4α1nπ6-1-ν2kα1nπ4,B4=α1nπ8+1-ν2kα1nπ6+12(1-ν2)α22α1nπ4.The particular solution of (27a) is(29)F~np=-2R8qnKk1-νB4.To obtain the solution of (27a) and (27b), substituting F~nh=eλmθ and f~n=eλmθ into (27a) and (27b), we have(30a)λm8+B1λm6+B2λm4+B3λm2+B4=0,mmmmmmlmmm=1,2,3,…,8,(30b)λm2=α1nπ2+24k1-vα22m=9,10.Equation (30a) can be solved using MATLAB, which is composed of 8 complex roots(31)λ1,2=d1±d2i,λ3,4=-d1±d2i,λ5,6=d3±d4i,λ7,8=-d3±d4i.The homogeneous solution of (27a) is(32)F~nh=ed1θC1·cos⁡⁡d2θ+C2·sin⁡d2θ+e-d1θC3·cos⁡⁡d2θ+C4·sin⁡d2θ+ed3θC5·cos⁡⁡d4θ+C6·sin⁡d4θ+e-d3θC7·cos⁡d4θ+C8·sind4θ.The general solution of (27a) is(33)F~n=F~nh+F~np.And the solution of (27b) is (34)f~n=C9eλ9θ+C10eλ10θ,where C1~C10 are undetermined coefficients, which can be determined by boundary conditions or join conditions.To satisfy boundary conditions or join conditions, displacement components, stress components, and bending moment components are expanded according to trigonometric series.According to (16a) and (16b), the expansions of displacement functions can be expressed as(35a)F~1=∑n=1∞1R2d2F~ndθ2-12kα22+α1nπ2F~nsin⁡(nπξ)=∑n=1∞F~1nθsinnπξ,(35b)F~2=∑n=1∞6k1-νR6α22α1nπ4F~n-2α1nπ2d2F~ndθ2mmmmmmmmm·d2F~ndθ2+d4F~ndθ4sin⁡(nπξ)=∑n=1∞F~2nθsinnπξ.According to (17a)–(17c), the expansions of displacement components can be expressed as (36a)u~1=∑n=1∞1-ν2R4να1nπ3F~1n+α1nπd2F~1ndθ2cos⁡⁡nπξ=∑n=1∞u~1ncos⁡nπξ,(36b)u~2=∑n=1∞1-ν2R42+να1nπ2dF~1ndθ-d3F~1ndθ3sin⁡nπξ=∑n=1∞u~2nsinnπξ,(36c)w~=∑n=1∞1-ν2R4α1nπ4F~1n-2α1nπ2d2F~1ndθ2mmmmmm·d2F~1ndθ2+d4F~1ndθ4sin(nπξ)=∑n=1∞w~nsin⁡nπξ.According to (18a) and (18b), the expansions of angles of rotation can be expressed as(37a)ϕ~1=∑n=1∞1Rα1nπF~2n-df~dθcos⁡(nπξ)=∑n=1∞u~4ncos⁡⁡nπξ,(37b)ϕ~2=∑n=1∞1RdF~2ndθ-α1nπf~nsinnπξ=∑n=1∞u~5nsin⁡nπξ.According to (19), the expansions of strain components can be expressed as(38)ε~11=∑n=1∞-α1nπRu~1nsinnπξ=∑n=1∞ε~1nsinnπξ,ε~22=∑n=1∞1Rdu~2ndθ+w~nsinnπξ=∑n=1∞ε~2nsinnπξ,ω~=∑n=1∞1Rdu~1ndθ+α1nπu~2cos⁡nπξ=∑n=1∞ω~ncos⁡nπξ,κ~1=∑n=1∞-α1nπRu~4nsinnπξ=∑n=1∞κ~1nsinnπξ,κ~2=∑n=1∞1Rdu~5ndθsinnπξ=∑n=1∞κ~2nsinnπξ,τ~1=∑n=1∞α1nπRu~5ncos⁡nπξ=∑n=1∞τ~1ncos⁡nπξ,τ~2=∑n=1∞1Rdu~4ndθcos⁡nπξ=∑n=1∞τ~2ncos⁡nπξ,ε~13=∑n=1∞u~4n+α1nπRu~3ncos⁡nπξ=∑n=1∞ε~13ncos⁡nπξ,ε~23=∑n=1∞u~5n+1Rdu~3ndθsinnπξ=∑n=1∞ε~23nsinnπξ.According to (20), the expansions of internal force and bending moment components can be expressed as (39)T~11=ERα21-ν2∑n=1∞(ε~1n+vε~2n)sinnπξ=∑n=1∞T~1nsinnπξ,T~22=ERα21-ν2∑n=1∞(ε~2n+vε~1n)sinnπξ=∑n=1∞T~2nsinnπξ,M~11=ER3α23121-ν2∑n=1∞(κ~1n+vκ~2n)sinnπξ=∑n=1∞M~1nsinnπξ,M~22=ER3α23121-ν2∑n=1∞(κ~2n+vκ~1n)sin⁡(nπξ)=∑n=1∞M~2nsinnπξ,T~12=T~21=ERα221+ν∑n=1∞ω~ncos⁡nπξ=∑n=1∞T~12ncos⁡nπξ,M~12=M~21=ER3α23241+ν∑n=1∞(τ~1n+vτ~2n)cos⁡⁡(nπξ)=∑n=1∞M~12ncos⁡nπξ,T~13=GRkτα2∑n=1∞ε~13ncos⁡nπξ=∑n=1∞T~13ncos⁡nπξ,T~23=GRkτα2∑n=1∞ε~23nsinnπξ=∑n=1∞T~23nsinnπξ.


## 3.3. Join Conditions

There are five edge forces at the place where the shells is connected with the edge beam, namely,T~21, T~22, T~23, M~22, and M~21. By opening them up with trigonometric series, we have(40)T~21=∑n=1∞T~12ncos⁡nπξ,T~22=∑n=1∞T~2nsinnπξ,T~23=∑n=1∞T~23nsinnπξ,M~21=∑n=1∞M~12ncos⁡nπξ,M~22=∑n=1∞M~2nsinnπξ,where T~21, T~22, T~23, M~22, and M~21 are functions of θ and contain ten undetermined coefficients; then the edge forces are as follows:on the edgeI:(41)T~22θ=-θ0=∑n=1∞T~2nθ=-θ0sinnπξ=∑n=1∞r1nsinnπξ=r1,T~23θ=-θ0=∑n=1∞T~23nθ=-θ0sinnπξ=∑n=1∞r2nsinnπξ=r2,T~21θ=-θ0=∑n=1∞T~12nθ=-θ0cos⁡nπξ=∑n=1∞r3nsin(nπξ)=r3,M~22θ=-θ0=∑n=1∞M~2nθ=-θ0sin(nπξ)=∑n=1∞r4nsin(nπξ)=r4,M~21θ=-θ0=∑n=1∞M~12nθ=-θ0sin(nπξ)=∑n=1∞r9nsin(nπξ)=r9;on the edgeII:(42)T~22θ=θ0=∑n=1∞T~2nθ=θ0sin(nπξ)=∑n=1∞r5nsin(nπξ)=r5,T~23θ=θ0=∑n=1∞T~23nθ=θ0sin(nπξ)=∑n=1∞r6nsin(nπξ)=r6,T~21θ=θ0=∑n=1∞T~12nθ=θ0cos⁡(nπξ)=∑n=1∞r7nsin(nπξ)=r7,M~22θ=θ0=∑n=1∞M~2nθ=θ0sin(nπξ)=∑n=1∞r8nsin(nπξ)=r8,M~21θ=θ0=∑n=1∞M~12nθ=θ0sin(nπξ)=∑n=1∞r10nsin(nπξ)=r10.The edge forces are as shown in Figure6.Figure 6
Edge forces distribution at the place where the shells are connected with the edge beam.In order to use the results in Table4, we decompose the edge forces r1~r10 along axes 1 and 2.Edge
I
Decomposer1, r2 along the axis 1: (43a)r1cos⁡θ-r2sinθ=∑n=1∞(r1ncos⁡θ-r2nsinθ)sin(nπξ).
Decomposer1, r2 along the axis 2:(43b)r1sinθ+r2cos⁡θ=∑n=1∞(r1nsinθ+r2ncos⁡θ)sinnπξ.According to Table4, the displacement generated by the edge beam I can be denoted by(44)δ1I=∑n=1∞δ1nIsinnπξ,δ2I=∑n=1∞δ2nIsinnπξ,δ3I=∑n=1∞δ3nIcos⁡nπξ,δ4I=∑n=1∞δ4nIsinnπξ,δ5I=∑n=1∞δ5nIsinnπξ,where(45)b0=12Eα3α4nπ4+34Gα3α4nπ2,b1=-34Gα32α4nπ2,b2=-6Eα3α4nπ3,b3=-32Gα32α4nπ2b,c1=12Eα33α4nπ4+34Gα33α4nπ2,c2=-6Eα32α4nπ3,c3=32Gα33α4nπ2b,c4=12Eα33α4nπ2b2,d0=-5Eα3α4nπ2,f0=3Gα33α4nπ2b2g0=12Eα33α4nπ3b,g1=-6Eα32α4nπ2b,g2=-12Eα33α4nπb.Thus, the subdisplacement and angle of rotation of edge beam I inu, v, and w directions are(46a)u~1I=δ3I=∑n=1∞δ3nIcos⁡nπξ,(46b)u~2I=∑n=1∞δ2nIsinθ+δ1nIcos⁡θsinnπξ,(46c)w~I=∑n=1∞-δ1nIsinθ+δ2nIcos⁡θsinnπξ,(46d)u~4I=δ5I=∑n=1∞δ5nIsinnπξ,(46e)u~5I=δ4I=∑n=1∞δ4nIsinnπξ.Edge II
Decomposer5, r6 along the axis 1:(47a)-r5cos⁡θ-r6sinθ=∑n=1∞(-r5ncos⁡θ-r6nsinθ)sin(nπξ).
Decomposer5, r6 along the axis 2:(47b)r5sinθ-r6cos⁡θ=∑n=1∞r5nsinθ-r6ncos⁡θsinnπξ.According to Table3, the displacement generated by the edge beam II can be denoted by(48)δ1II=∑n=1∞δ1nIIsinnπξ,δ2II=∑n=1∞δ2nIIsinnπξ,δ3II=∑n=1∞δ3nIIcos⁡nπξδ4II=∑n=1∞δ4nIIsinnπξ,δ5II=∑n=1∞δ5nIIsinnπξ.Thus, the subdisplacement and angle of rotation of edge beam II inu, v, and w directions are(49a)u~1II=δ3II=∑n=1∞δ3nIIcos⁡nπξ,(49b)u~2II=∑n=1∞-δ2nIIsinθ+δ1nIIcos⁡θsinnπξ,(49c)w~II=∑n=1∞δ1nIIsinθ+δ2nIIcos⁡θsinnπξ,(49d)u~4II=δ5II=∑n=1∞δ5nIIsinnπξ,(49e)u~5II=δ4II=∑n=1∞δ4nIIsinnπξ.The expression for shell’s displacements is(50)u~1=∑n=1∞u~1ncos⁡nπξ,u~2=∑n=1∞u~2nsinnπξ,w~=∑n=1∞w~nsinnπξ,u~4=∑n=1∞u~4ncos⁡nπξ,u~5=∑n=1∞u~5nsinnπξ.Then, shell’s displacement on the edge is as follows.Edge I is(51a)u~1qIθ=-θ0=u~1θ=-θ0=∑n=1∞u~1nθ=-θ0cos⁡nπξ,(51b)u~2qIθ=-θ0=u~2θ=-θ0=∑n=1∞u~2nθ=-θ0sinnπξ,(51c)w~qIθ=-θ0=w~θ=-θ0=∑n=1∞w~nθ=-θ0sinnπξ,(51d)u~4qIθ=-θ0=u~4θ=-θ0=∑n=1∞u~4nθ=-θ0cos⁡nπξ,(51e)u~5qIθ=-θ0=u~5θ=-θ0=∑n=1∞u~5nθ=-θ0sinnπξ.Edge II is(52a)u~1qIIθ=θ0=u~1θ=θ0=∑n=1∞u~1nθ=θ0cos⁡nπξ,(52b)u~2qIIθ=θ0=u~2θ=θ0=∑n=1∞u~2nθ=θ0sinnπξ,(52c)w~qIIθ=θ0=w~θ=θ0=∑n=1∞w~nθ=θ0sinnπξ,(52d)u~4qIIθ=θ0=u~4θ=θ0=∑n=1∞u~4nθ=θ0cos⁡nπξ,(52e)u~5qIIθ=θ0=u~5θ=θ0=∑n=1∞u~5nθ=θ0sinnπξ.By using the continuous conditions of displacements to compare (46a)–(46e), (51a)–(51e), (49a)–(49e), and (52a)–(52e), respectively, the displacement boundary conditions of shells can be obtained as follows.Edge I is(53)u1nθ=-θ0=δ3nI,u2nθ=-θ0=δ2nIsinθ+δ1nIcos⁡θ,wnθ=-θ0=-δ1nIsinθ+δ2nIcos⁡θ,u4nθ=-θ0=δ5nIu5nθ=-θ0=δ4nI.Edge II is(54)u1nθ=θ0=δ3nII,u2nθ=θ0=-δ2nIIsinθ+δ1nIIcos⁡θ,wnθ=θ0=-δ1nIIsinθ+δ2nIIcos⁡θ,u4nθ=θ0=δ5nII,u5nθ=θ0=δ4nII.According to the join condition, the ten undetermined coefficients can be solved, and trigonometric series expressions of displacements, angles of rotation, internal forces, and moments can be further obtained, whose changing curves can be drawn with MATLAB program.


## 4. Numerical Examples

Example 1.
With the cylindrical shells roof that has been widely used in practical engineering as an example, assume the shells and edge beams are steel reinforcement and concrete materials. The calculating parameters are as follows: elastic modulus isE=2.6×1010 Pa; Poisson’s ratio is v= 0.167; shells radius is R=10 m, α1=5/6, and α2=0.06; open angle is 2θ0=π/3; widths of edge beam are b=1.2 m, α3=1.5, and α4=0.1; and the load is q=-1×103N/m2. Through MATLAB program [15], the displacement and internal force of cylindrical shells are calculated, and by comparing the calculated results with the corresponding values of fixed edges and simply supported edges calculated by literatures [16], we get the following results as shown in Figure 7.
It can be seen from Figure7 that, in the same load and dimensions of shells, the displacement of ribbed shells is bigger than that of fixed edges while it is smaller than that of simply supported edges; namely, the displacement of ribbed shells is between these of fixed edges and simply supported edges; the bending moments M1 and M2 attain their maximum value when they are near the ends, while the value is bigger than that of fixed edge and is smaller than that of simply supported edge when it is near the middle position. This change conforms to the basic mechanical laws.Figure 7
Comparison of the displacement and internal force under three boundary conditions.
(a)
(b)
(c)
(d)Example 2.
With the same calculating parameters (including dimensions of shells and edge beam and the applied loads) of Example1, the displacement with different values of α1 and α2 is obtained, as shown in Figures 8 and 9.
Generally speaking, as seen in Figures8 and 9, considering the shells radius is unchanged, the influence of transverse shear deformation on displacement increases gradually with the increase of α1 and α2.Figure 8
Comparison of the displacement with different values ofα1.Figure 9
Comparison of the displacement with different values ofα2.


## 5. Conclusions

Based on FSDT considering the effect of transverse shear, the displacement-based differential equation set for general problems of moderately thick cylindrical shells is obtained; the problem is simplified into a solvable eighth-order ordinary differential equation and a second-order ordinary differential equation by introducing four displacement functions.In consideration of the effect of edge beams, the displacement of edge beams under edge force is obtained by using the given displacement differential equation of beams, and the displacement of edge beams is converted into the shells coordinate system. The join condition of moderately thick cylindrical ribbed shells is established according to the continuity of displacement. By using this join condition, the ten undetermined constants the displacement function contains are solved, and thus the solution of bending problem of moderately thick cylindrical ribbed shells is obtained. In addition, by comparing the displacement, internal force of ribbed shells with the displacement, internal force of fixed edges, and simply supported edges, the numerical results show that the displacement and internal force of ribbed shells are between these of fixed edges and simply supported edges, which explains that the solving methods proposed and the join condition established in this paper are correct.For ease of the calculation, the first-order model considering the effect of transverse shear is adopted in this paper. The basic equations derived by using the above model are simpler, but they cannot fully satisfy the edge condition that the shear stress is 0 on the free surface, which is inconsistent with the actual situation and is also the defect-existing in the first-order model itself. To remedy this defect, the higher-order model can be adopted for calculation in the follow-up study, so that the calculation results can get closer to the actual.



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*Source: 274091-2015-03-10.xml*