On the Maximum Displacement and Static Buckling of a Circular Cylindrical Shell

G. E. Ozoigbo^1*, A. M. Ette²

¹Department of Mathematics/Computer Science/Statistics & Informatics, Alex Ekwueme Federal University Ndufu-Alike, Ikwo, Ebonyi State, Nigeria.
²Department of Mathematics, Federal University of Technology, Owerri, Imo State, Nigeria.

DOI: 10.4236/jamp.2019.711196                

Abstract

The static buckling load of an imperfect circular cylindrical shell is here determined asymptotically with the assumption that the normal displacement can be expanded in a double Fourier series. The buckling modes considered are the ones that are partly in the shape of imperfection, and partly in the shape of some higher buckling mode. Simply-supported boundary conditions are considered and the maximum displacement and the static buckling load are evaluated nontrivially. The results show, among other things, that generally the static buckling load, λ_s decreases with increased imperfection and that the displacement in the shape of imperfection gives rise to the least static buckling load.

Keywords

Static, Maximum Displacement, Circular, Cylindrical Shell

1. Introduction

Cylindrical shells have wide engineering applications such as in the construction and study of aircraft, spacecraft and nuclear reactor, tanks for liquid and gas storage and pressure vessels, etc. The analyses of the buckling of cylindrical shells under various loading conditions have been made in the past years and both theoretical and experimental studies have been considered just as in [1] and [2]. Earlier studies on the buckling of shells were done by [3] [4] [5] [6] and [7], while Amazigo and Frazer [8], studied the buckling under external pressure of cylindrical shells with dimple-shaped initial imperfection. It would be recalled that Budiansky and Amazigo [9], investigated the buckling of infinitely long imperfect cylindrical shells subjected to static loads and Ette and Onwuchekwa [10], equally studied the static buckling of an externally pressurized finite circular cylindrical shell using asymptotic method. In this regard, mention must be made of Lockhart and Amazigo [11], who used perturbation method to investigate the dynamic buckling of finite circular cylindrical shells with small arbitrary geometric imperfections under external step-loading. In the same way, Bich et al. [12], by using analytical approach, investigated the nonlinear static and dynamic buckling behaviour of eccentrically shallow shells and circular cylindrical shells based on Donnell shell theory. Relevant studies on the buckling analysis were investigated in [13] [14] [15] and [16] and Ette [17] [18] [19] [20] and [21].

In this study, we consider a statically loaded imperfect finite circular cylindrical shell and aim at determining the maximum displacement and the static buckling load for the case where the displacement is partly in the shape of imperfection and partly in some other buckling mode. The analysis is purely on the use of asymptotic expansions and perturbation procedures.

This analysis is organised as follows. We shall first write down the governing equations as in Amazigo and Frazer [8] and Budiansky and Amazigo [9]. Using the techniques of regular perturbation and asymptotics, we shall analytically determine a uniformly valid expression of the displacement which is followed determining by the maximum displacement. Lastly, we shall reverse the series of maximum displacement and determine the static buckling load.

2. Formulation

As in [11], the general Karman-Donnell equation of motion and the compatibility equation governing the normal deflection $W\left(X\mathrm{,}Y\right)$ and Airy stress function $F\left(X\mathrm{,}Y\right)$ for cylindrical shell, of length L, radius R, thickness h, bending stiffness $D=\frac{E{h}^3}{12\left(1-{\upsilon }^2\right)}$ (where E and $\upsilon$ are the Young’s modulus and Poisson’s ratio respectively), mass per unit area $\rho$, subjected to external pressure per unit area P, are

$\frac{1}{Eh}{\nabla }^{4}F-\frac{1}{R}{W}_{\mathrm{,}XX}=-S\left(W\mathrm{,}\frac{1}{2}W+\bar{W}\right)$ (2.1)

$D{\nabla }^{4}W+\frac{1}{R}{F}_{\mathrm{,}XX}+P\left[\frac{1}{2}\alpha {\left(W+\bar{W}\right)}_{\mathrm{,}XX}+{\left(W+\bar{W}\right)}_{\mathrm{,}YY}\right]=S\left(W+\bar{W}\mathrm{,}F\right)$ (2.2)

$W=W_{,XX}=F=F_{,XX}=0\text{at}X=0,L,0<X<\pi ,0<Y<2\pi .$ (2.3)

where, X and Y are the axial and circumferential coordinates respectively and $\bar{W}\left(X\mathrm{,}Y\right)$, is a continuously differentiable stress-free and time independent imperfection. In this work, an alphabetic subscript placed after a comma indicates partial differentiation while S is the symmetric bi-linear operator in X and Y given by

$S\left(P\mathrm{,}Q\right)=P_{\mathrm{,}XX}{Q}_{\mathrm{,}YY}+P_{\mathrm{,}YY}{Q}_{\mathrm{,}XX}-2P_{\mathrm{,}XY}{Q}_{\mathrm{,}XY}$ (2.4a)

and ${\nabla }^4$, is the two-dimensional bi-harmonic operator defined by

${\nabla }^4={\left(\frac{{\partial }^2}{\partial X^2}+\frac{{\partial }^2}{\partial Y^2}\right)}^2$ (2.4b)

here, we shall neglect both axial and circumferential inertia and shall similarly assume simply-supported boundary conditions and neglect boundary layer effect by assuming that the pre-buckling deflection is constant.

As in [11] and [22], we now introduce the following non-dimensional quantities.

$x=\frac{X\pi }{L},H=\frac{h}{R},\epsilon \bar{w}=\frac{\bar{W}}{h},\lambda =\frac{L^{2}RP}{{\pi }^{2}D},\xi =\frac{L^2}{{\pi }^2{R}^2}$ (2.5a)

$y=\frac{Y}{R},w=\frac{W}{h},K\left(\xi \right)=-\frac{A^2}{{\left(1+\xi \right)}^2},A=\frac{L^2\sqrt{12\left(1-{\upsilon }^2\right)}}{{\pi }^{2}Rh}$ (2.5b)

where, $\upsilon$ is Poisson’s ratio and $\epsilon$ is a small parameter which measures the amplitude of the imperfection while L is the length of the cylindrical shell which is simply-supported at $x=0,\pi$.

We shall neglect boundary layer effect by assuming that the pre-buckling deflection is constant so that we let

$F=-\frac{1}{2}PR\left(X^2+\frac{1}{2}\alpha Y^2\right)+\left(\frac{E{h}^2{L}^2}{{\pi }^{2}R{\left(1+\xi \right)}^2}\right)f$ (2.7a)

$W=\frac{P{R}^2\left(1-\frac{1}{2}\alpha \upsilon \right)}{Eh}+hw$ (2.7b)

where, P is the applied static load and $\lambda$ is the non-dimensional load amplitude. The first terms on the right hand sides of (2.7a) and (2.7b), are pre-buckling approximations, while the parameter $\alpha$, shall take the value $\alpha =1$, if pressure contributes to axial stress through the ends, otherwise $\alpha =0$, if pressure only acts laterally.

Substituting (2.7a) and (2.7b) into (2.1 and (2.2), using (2.5a) and (2.5b), and simplifying results to

${\bar{\nabla }}^{4}f-{\left(1+\xi \right)}^2{w}_{\mathrm{,}xx}=-H{\left(1+\xi \right)}^{2}s\left(w\mathrm{,}\frac{1}{2}w+\epsilon \bar{w}\right)$ (2.8)

and

${\bar{\nabla }}^{4}w-K\left(\xi \right)f_{\mathrm{,}xx}+\lambda \left[\frac{1}{2}\alpha {\left(w+\epsilon \bar{w}\right)}_{\mathrm{,}xx}+{\left(w+\epsilon \bar{w}\right)}_{\mathrm{,}yy}\right]=-HK\left(\xi \right)s\left(w+\epsilon \bar{w}\mathrm{,}f\right)$ (2.9)

$w=w_{,xx}=f=f_{,xx}=0\text{at}x=0,\pi ,0<x<\pi ,0<y<2\pi ,0<\epsilon \ll 1,$ (2.10)

where,

$s\left(P,Q\right)=P_{,xx}{Q}_{,yy}+P_{,yy}{Q}_{,xx}-2P_{,xy}{Q}_{,xy},{\bar{\nabla }}^4={\left(\frac{{\partial }^2}{\partial x^2}+\xi \frac{{\partial }^2}{\partial y^2}\right)}^2$ (2.11)

3. Classical Buckling Load

The classical buckling load ${\lambda }_C$ is the load that is required to buckle the associated linear perfect structure and its equations, from (2.8) and (2.9) are

${\bar{\nabla }}^{4}f-{\left(1+\xi \right)}^2{w}_{,xx}=0$ (3.1)

and

${\bar{\nabla }}^{4}w-K\left(\xi \right)f_{,xx}+\lambda \left[\frac{1}{2}\alpha w_{,xx}+{\bar{w}}_{,yy}\right]=0$ (3.2)

$w=w_{,xx}=0,f=f_{,xx}=0\text{at}x=0,\pi .$ (3.3)

The solution to (3.1)-(3.3) is a superposition of the form

$\left(w,f\right)=\left(a_{mk},b_{mk}\right)\sin \left(ky+{\varphi }_{mk}\right)\sin mx$ (3.4)

where, $\left(a_{mk}\mathrm{,}{b}_{mk}\right)\ne \left(\mathrm{0,0}\right)$ and ${\varphi }_{mk}$ is an inconsequential phase.

On substituting (3.4) into (3.1), using (3.3) and after lengthy simplification, we get

$b_{mk}=-\frac{{\left(1+\xi \right)}^2{m}^2{a}_{mk}}{{\left(m^2+\xi k^2\right)}^2}$ (3.5)

Substituting (3.5) into (3.2) and simplifying, yields

$\lambda =\frac{{\left(m^2+\xi k^2\right)}^2-\frac{K\left(\xi \right)m^4{\left(1+\xi \right)}^2}{{\left(m^2+\xi k^2\right)}^2}}{\frac{1}{2}\alpha m^2+\frac{k^2{m}^4{\left(1+\xi \right)}^2\xi }{{\left(m^2+\xi k^2\right)}^2}}$ (3.6)

Thus, if n is the critical value of k that minimizes $\lambda$, then, the value of $\lambda$ at $k=n$ was taken as the classical buckling load ${\lambda }_C$. Thus, in this case, we get

$\frac{\text{d}\lambda }{\text{d}k}=0$ (3.7)

Therefore, corresponding to $k=n$,we see that (3.6) is now equivalent to

$\lambda =\frac{{\left(m^2+\xi n^2\right)}^2-\frac{K\left(\xi \right)m^4{\left(1+\xi \right)}^2}{{\left(m^2+\xi n^2\right)}^2}}{\frac{1}{2}\alpha m^2+\frac{n^2{m}^4{\left(1+\xi \right)}^2\xi }{{\left(m^2+\xi n^2\right)}^2}}$ (3.8)

Usually, m and n take the values $m=1,2,3,\cdots$ and $n=0,1,2,\cdots$

We recall that [23] had assumed that k varies continuously, and so, minimized $\lambda$ with respect to k. If $m=1$ is the nontrivial values of m and we let $\zeta =\xi n^2$, then, we have

${\lambda }_C=\frac{{\left(1+\zeta \right)}^2+\frac{A^2}{{\left(1+\zeta \right)}^2}}{\frac{1}{2}\alpha +\zeta }$ (3.9)

The corresponding displacement and Airy Stress function are

$\left(w,f\right)=\left[\left\{1,-{\left(\frac{1+\xi }{1+\zeta }\right)}^2\right\}\right]a_{1n}\sin \left(ky+{\varphi }_{1n}\right)\sin x$ (3.10)

4. Static Theory

In this section, we shall derive the equations satisfied by the displacement and Airy stress functions when the static load is applied.

Similar to (2.8) and (2.9), the structure satisfies the following equations at static loading

${\bar{\nabla }}^{4}f-{\left(1+\xi \right)}^2{w}_{\mathrm{,}xx}=-H{\left(1+\xi \right)}^{2}s\left(w\mathrm{,}\frac{1}{2}w+\epsilon \bar{w}\right)$ (4.1)

and

${\bar{\nabla }}^{4}w-K\left(\xi \right)f_{\mathrm{,}xx}+\lambda \left[\frac{1}{2}\alpha {\left(w+\epsilon \bar{w}\right)}_{\mathrm{,}xx}+\xi {\left(w+\epsilon \bar{w}\right)}_{\mathrm{,}yy}\right]=-HK\left(\xi \right)s\left(w+\epsilon \bar{w}\mathrm{,}f\right)$ (4.2)

$w=w_{,xx}=f=f_{,xx}=0\text{at}x=0,\pi$ (4.3)

We now assume the following asymptotic expansions

$\left(\begin{array}{c}w\\ f\end{array}\right)={\displaystyle \underset{i=1}{\overset{\infty }{\sum }}}\left(\begin{array}{c}{w}^{\left(i\right)}\\ f^{\left(i\right)}\end{array}\right){\epsilon }^i$ (4.4)

Substituting (4.4) into (4.1) and (4.2), and equating the coefficients of orders of ${\epsilon }^i,i=1,2,3,\cdots$, the following equations are obtained

$\mathcal{O}\left(\epsilon \right)\mathrm{<mo>\; :\; </mo>}(\begin{array}{l}{\bar{\nabla }}^4{f}^{\left(1\right)}-{\left(1+\xi \right)}^2{w}_{\mathrm{,}xx}^{\left(1\right)}=0.\\ {\bar{\nabla }}^4{w}^{\left(1\right)}-K\left(\xi \right)f_{\mathrm{,}xx}^{\left(1\right)}+\lambda \left[\frac{1}{2}\alpha {\left(w^{\left(1\right)}+\bar{w}\right)}_{\mathrm{,}xx}+\xi {\left(w^{\left(1\right)}+\bar{w}\right)}_{\mathrm{,}yy}\right]=0\end{array}$ (4.5)

$\mathcal{O}\left({\epsilon }^2\right)\mathrm{<mo>\; :\; </mo>}(\begin{array}{l}{\bar{\nabla }}^4{f}^{\left(2\right)}-{\left(1+\xi \right)}^2{w}_{\mathrm{,}xx}^{\left(2\right)}=-H{\left(1+\xi \right)}^2\left[\frac{1}{2}s\left(w^{\left(1\right)}\mathrm{,}{w}^{\left(1\right)}\right)+s\left(w^{\left(1\right)}\mathrm{,}\bar{w}\right)\right]\\ {\bar{\nabla }}^4{w}^{\left(2\right)}-K\left(\xi \right)f_{\mathrm{,}xx}^{\left(2\right)}+\lambda \left[\frac{1}{2}\alpha w_{\mathrm{,}xx}^{\left(2\right)}+\xi w_{\mathrm{,}yy}^{\left(2\right)}\right]\\ =-HK\left(\xi \right)\left[s\left(w^{\left(1\right)}\mathrm{,}{f}^{\left(1\right)}\right)+s\left(\bar{w}\mathrm{,}{w}^{\left(1\right)}\right)\right]\end{array}$ (4.6)

$\mathcal{O}\left({\epsilon }^3\right)\mathrm{<mo>\; :\; </mo>}(\begin{array}{l}{\bar{\nabla }}^4{f}^{\left(3\right)}-{\left(1+\xi \right)}^2{w}_{\mathrm{,}xx}^{\left(3\right)}=-H{\left(1+\xi \right)}^2\left[s\left(w^{\left(1\right)}\mathrm{,}{w}^{\left(2\right)}\right)+s\left(w^{\left(2\right)}\mathrm{,}\bar{w}\right)\right]\\ {\bar{\nabla }}^4{w}^{\left(3\right)}-K\left(\xi \right)f_{\mathrm{,}xx}^{\left(3\right)}+\lambda \left[\frac{1}{2}\alpha w_{\mathrm{,}xx}^{\left(3\right)}+\xi w_{\mathrm{,}yy}^{\left(3\right)}\right]\\ =-HK\left(\xi \right)\left[s\left(w^{\left(1\right)}\mathrm{,}{f}^{\left(2\right)}\right)+s\left(w^{\left(2\right)}\mathrm{,}{f}^{\left(1\right)}\right)+s\left(\bar{w}\mathrm{,}{f}^{\left(2\right)}\right)\right]\end{array}$ (4.7)

etc.

We seek solutions to (4.5)-(4.7) in the form

$\left(\begin{array}{c}{f}^{\left(i\right)}\\ w^{\left(i\right)}\end{array}\right)={\displaystyle \underset{\left(k=1\right),\left(p=0\right)}{\overset{\infty }{\sum }}}\left[\left(\begin{array}{c}{f}_1^{\left(i\right)}\\ w_1^{\left(i\right)}\end{array}\right)\cos py+\left(\begin{array}{c}{f}_2^{\left(i\right)}\\ w_2^{\left(i\right)}\end{array}\right)\sin py\right]\sin kx$ (4.8)

and now assume

$\bar{w}\left(x\mathrm{,}y\right)=\bar{a}\sin mx\sin ny$ (4.9)

As earlier obtained, we shall need the following simplifications

${\bar{\nabla }}^4={\left(\frac{{\partial }^2}{\partial x^2}+\xi \frac{{\partial }^2}{\partial y^2}\right)}^2=\left(\frac{{\partial }^4}{\partial x^4}+2\xi \frac{{\partial }^4}{\partial x^2\partial y^2}+{\xi }^2\frac{{\partial }^4}{\partial y^4}\right)$ (4.10a)

so that, if

$f^{\left(1\right)}=f_{{\Gamma }_1}^{\left(1\right)}\cos py\sin kx$

then, we have

${\bar{\nabla }}^4{f}^{\left(1\right)}={\left(k^2+\xi p^2\right)}^2{f}_{{\Gamma }_1}^{\left(1\right)}\cos py\sin kx,{\Gamma }_1=1,2$ (4.10b)

and, if

$f^{\left(2\right)}=f_{{\Gamma }_2}^{\left(2\right)}\sin py\sin kx$

we get

${\bar{\nabla }}^4{f}^{\left(2\right)}={\left(k^2+\xi p^2\right)}^2{f}_{{\Gamma }_2}^{\left(2\right)}\sin py\sin kx,{\Gamma }_2=1,2$ (4.10c)

We shall use the fact that

$K\left(\xi \right)=-\frac{A^2}{{\left(1+\xi \right)}^2}$ (4.10d)

Solution of Equations of First Order Perturbation

The equations necessary here, from (4.5), are

${\bar{\nabla }}^4{f}^{\left(1\right)}-{\left(1+\xi \right)}^2{w}_{\mathrm{,}xx}^{\left(1\right)}=0$ (4.11)

and

${\bar{\nabla }}^4{w}^{\left(1\right)}-K\left(\xi \right)f_{\mathrm{,}xx}^{\left(1\right)}+\lambda \left[\frac{1}{2}\alpha {\left(w^{\left(1\right)}+\bar{w}\right)}_{\mathrm{,}xx}+\xi {\left(w^{\left(1\right)}+\bar{w}\right)}_{\mathrm{,}yy}\right]=0$ (4.12)

Substituting (4.8) and (4.9) into (4.11), using (4.10a), (4.10b) and (4.10c), multiplying the resultant equation through by $\cos ny\sin mx$ and integrating with respect to y from 0 to 2π and with respect to x from 0 to π, we note that for $p=n,k=m$, we easily get

$f_1^{\left(1\right)}=-\frac{m^2{\left(1+\xi \right)}^2{w}_1^{\left(1\right)}}{{\left(m^2+n^2\xi \right)}^2}$ (4.13)

Similarly, by multiplying the resultant equation through by $\sin ny\sin mx$, and integrating with respect to y from 0 to 2π and with respect to x from 0 to π, and for $p=n,k=m$, we obtain

$f_2^{\left(1\right)}=-\frac{m^2{\left(1+\xi \right)}^2{w}_2^{\left(1\right)}}{{\left(m^2+n^2\xi \right)}^2}$ (4.14)

In the same manner, substituting (4.8) and (4.9) into (4.12), assuming (4.10a), (4.10b) and (4.10c), thereafter multiplying the resultant equation by $\sin ny\sin mx$ and integrating with respect to x from 0 to π and y from 0 to 2π, and for $p=n,k=m$, we get

$\left[{\left(m^2+n^2\xi \right)}^2-\lambda \left(\frac{1}{2}\alpha m^2+n^2\xi \right)\right]w_2^{\left(1\right)}+K\left(\xi \right)m^2{V}_2^{\left(1\right)}=\lambda \bar{a}\left(\frac{1}{2}\alpha m^2+n^2\xi \right)$ (4.15)

On substituting for $f_2^{\left(1\right)}$ from (4.14) and $K\left(\xi \right)$ from (4.10d) in (4.15) and simplifying, yields

$w_2^{\left(1\right)}=B_0$ (4.16)

where,

$\begin{array}{l}{B}_0=\frac{\lambda \bar{a}\left(\frac{1}{2}\alpha m^2+n^2\xi \right)}{{\phi }_0^2},\\ {\phi }_0^2={\left(m^2+n^2\xi \right)}^2+{\left(\frac{m^{2}A}{m^2+n^2\xi }\right)}^2-\lambda \left(\frac{1}{2}\alpha m^2+n^2\xi \right)\end{array}$ (4.17)

Next, multiplying the resultant equation by $\cos ny\cos mx$ and integrating with respect to x and y from 0 to π and 0 to 2π, respectively for $p=n,k=m$, and simplifying, we get

$w_1^{\left(1\right)}=0$ (4.18)

We therefore expect from (4.8) that for $i=1$

$w^{\left(1\right)}=B_0\sin mx\sin ny;f^{\left(1\right)}=-{\Phi }_0{B}_0\sin mx\sin ny,{\Phi }_0=\frac{m^2{\left(1+\xi \right)}^2}{{\left(m^2+n^2\xi \right)}^2}$ (4.19)

Solution of Equations of Second Order Perturbation

Equations of the second order to be solved are from (4.6), namely

${\bar{\nabla }}^4{f}^{\left(2\right)}-{\left(1+\xi \right)}^2{w}_{,xx}^{\left(2\right)}=-H{\left(1+\xi \right)}^2\left[\frac{1}{2}s\left(w^{\left(1\right)},w^{\left(1\right)}\right)+s\left(w^{\left(1\right)},\bar{w}\right)\right]$ (4.20)

and

$\begin{array}{l}{\bar{\nabla }}^4{w}^{\left(2\right)}-K\left(\xi \right)f_{,xx}^{\left(2\right)}+\lambda \left[\frac{1}{2}\alpha w_{,xx}^{\left(2\right)}+\xi w_{,yy}^{\left(2\right)}\right]\\ =-HK\left(\xi \right)\left[s\left(w^{\left(1\right)},w^{\left(1\right)}\right)+s\left(\bar{w},w^{\left(1\right)}\right)\right]\end{array}$ (4.21)

Evaluating the symmetric bi-linear functions on the right hand sides of (4.20) and (4.21), substituting the same and simplifying, we get (after simplifying trigonometric terms)

${\bar{\nabla }}^4{f}^{\left(2\right)}-{\left(1+\xi \right)}^2{w}_{,xx}^{\left(2\right)}=-H{\left(1+\xi \right)}^2{\left(mn\right)}^2\left[\left(\frac{1}{2}{B}_0^2+B_0\bar{a}\right)\left(\cos 2mx+\cos 2ny\right)\right]$ (4.22)

and

$\begin{array}{l}{\bar{\nabla }}^4{w}^{\left(2\right)}-K\left(\xi \right)f_{,xx}^{\left(2\right)}+\lambda \left[\frac{1}{2}\alpha w_{,xx}^{\left(2\right)}+\xi w_{,yy}^{\left(2\right)}\right]\\ =-HK\left(\xi \right){\left(mn\right)}^2\left[{\Phi }_0{B}_0^2+\bar{a}{B}_0\right]\left(\cos 2mx+\cos 2ny\right)\end{array}$ (4.23)

Next we substitute (4.8) and (4.9) into (4.22), assuming (4.10a), (4.10b) and (4.10c), for $i=2$. Thereafter, we multiply the resultant equation through by $\cos 2ny\sin mx$ and integrate with respect to x and y, to get

$\begin{array}{l}{f}_1^{\left(2\right)}={\Phi }_1\left(\frac{1}{2}{B}_0^2+B_0\bar{a}\right)-{\Phi }_2{w}_1^{\left(2\right)},\\ {\Phi }_1=\frac{4Hm{n}^2{\left(1+\xi \right)}^2}{\pi {\left(m^2+4n^2\xi \right)}^2},{\Phi }_2=\frac{m^2{\left(1+\xi \right)}^2}{\pi {\left(m^2+4n^2\xi \right)}^2},m=odd\end{array}$ (4.24)

Similarly, we next multiply the resultant equation by $\sin ny\sin mx$ and integrate as usual, and for $p=n,k=m$, we get

$f_2^{\left(2\right)}=-\frac{m^2{\left(1+\xi \right)}^2{w}_2^{\left(2\right)}}{{\left(m^2+n^2\xi \right)}^2}\mathrm{.}$ (4.25)

Next, we substitute (4.8) and (4.9) into (4.23), using (4.10a), (4.10b) and (4.10c), for $i=2$, and then multiply the resultant equation through by $\cos 2ny\sin mx$ and integrate, as usual for $p=2n,k=m$ to get

$w_1^{\left(2\right)}={\Phi }_5,{\Phi }_5={\Phi }_3\left(B_0^2{\Phi }_0+B_0{\Phi }_0\bar{a}\right)+{\Phi }_4\left(\frac{1}{2}{B}_0^2+B_0\bar{a}\right)$ (4.26)

${\Phi }_3=\frac{{\left(mn\right)}^{2}HK(\xi )}{{\left(m^2+4n^2\xi \right)}^2+{\left(\frac{m^{2}A}{m^2+4n^2\xi }\right)}^2-\lambda \left(\frac{1}{2}\alpha m^2+4n^2\xi \right)}$

${\Phi }_4=\frac{4m^3{n}^2{A}^2}{{\left(m^2+4n^2\xi \right)}^2\left[{\left(m^2+4n^2\xi \right)}^2+{\left(\frac{m^{2}A}{m^2+4n^2\xi }\right)}^2-\lambda \left(\frac{1}{2}\alpha m^2+4n^2\xi \right)\right]}$ (4.27)

In the same manner, multiply the resultant equation by $\sin ny\sin mx$ and integrate with respect to x and y, for $p=n,k=m$, and simplify to get

$w_2^{\left(2\right)}=0,f_2^{\left(2\right)}=0$ (4.28)

Therefore, we observe from (4.8), and for $i=2$, that

$w^{\left(2\right)}=w_1^{\left(2\right)}\cos 2ny\sin mx;f^{\left(2\right)}=f_1^{\left(2\right)}\cos 2ny\sin mx,$ (4.29)

On substitution in (4.29) using (4.24) and in the first part of (4.26), we get

$\begin{array}{l}{w}^{\left(2\right)}={\Phi }_5\cos 2ny\sin mx;f^{\left(2\right)}={\Phi }_{10}\cos 2ny\sin mx,\\ {\Phi }_{10}=\left[{\Phi }_1\left(\frac{1}{2}{B}_0^2+B_0\bar{a}\right)-{\Phi }_2{\Phi }_5\right]\end{array}$ (4.30)

Solution of Equations of Third Order Perturbation

The actual equations of the third order are from (4.7), namely

${\bar{\nabla }}^4{f}^{\left(3\right)}-{\left(1+\xi \right)}^2{w}_{,xx}^{\left(3\right)}=-H{\left(1+\xi \right)}^2\left[s\left(w^{\left(1\right)},w^{\left(2\right)}\right)+s\left(w^{\left(2\right)},\bar{w}\right)\right]$ (4.31)

and

$\begin{array}{l}{\bar{\nabla }}^4{w}^{\left(3\right)}-K\left(\xi \right)f_{,xx}^{\left(3\right)}+\lambda \left[\frac{1}{2}\alpha w_{,xx}^{\left(3\right)}+\xi w_{,yy}^{\left(3\right)}\right]\\ =-HK\left(\xi \right)\left[s\left(w^{\left(1\right)},f^{\left(2\right)}\right)+s\left(w^{\left(2\right)},f^{\left(1\right)}\right)+s\left(\bar{w},f^{\left(2\right)}\right)\right]\end{array}$ (4.32)

Evaluating the symmetric bi-linear functions on the right sides of (4.31) and (4.32) and substituting the same and simplifying, yields

$\begin{array}{l}{\bar{\nabla }}^4{f}^{\left(3\right)}-{\left(1+\xi \right)}^2{w}_{\mathrm{,}xx}^{\left(3\right)}\\ =-\frac{1}{4}H{\left(1+\xi \right)}^2{\left(mn\right)}^2\left({\Phi }_5{B}_0+{\Phi }_5\bar{a}\right)[9\sin 3ny-\sin ny\\ -\cos 2mx\sin 3ny+9\cos 2mx\sin ny]\end{array}$ (4.33)

and

$\begin{array}{l}{\bar{\nabla }}^4{w}^{\left(3\right)}-K\left(\xi \right)f_{\mathrm{,}xx}^{\left(3\right)}+\lambda \left[\frac{1}{2}\alpha w_{\mathrm{,}xx}^{\left(3\right)}+\xi w_{\mathrm{,}yy}^{\left(3\right)}\right]\\ =-\frac{1}{4}HK\left(\xi \right){\left(mn\right)}^2[-{\Phi }_0{\Phi }_5{B}_0+B_0\left\{{\Phi }_1\left(\frac{1}{2}{B}_0^2+B_0\bar{a}\right)-{\Phi }_2{\Phi }_5\right\}\\ +\bar{a}\left\{{\Phi }_1\left(\frac{1}{2}{B}_0^2+B_0\bar{a}\right)-{\Phi }_2{\Phi }_5\right\}][9\sin 3ny-\sin ny\\ -\cos 2mx\sin 3ny+9\cos 2mx\sin ny]\end{array}$ (4.34)

We observe from the simplifications on the right hand sides of (4.33) and (4.34) that there will be four buckling modes generated $w_{i\left(r\mathrm{,}p\right)}^{\left(3\right)}$ with their respective Airy stress functions $f_{i\left(r\mathrm{,}p\right)}^{\left(3\right)}$. These buckling modes correspond to the following terms on the right hand sides of (4.33) and (4.34) : $\sin 3ny\sin mx$, $\sin ny\sin mx$, $\cos 2mx\sin 3ny$ and $\cos 2mx\sin ny$.

However, of the four modes, it is only the mode in the shape of $\sin ny\sin mx$ that is in the shape of the imperfection. We shall consider this mode and the additional mode in the shape of $\sin 3ny\sin mx$.

We now substitute (4.8) and (4.9) into (4.33), using (4.10a), (4.10b) and (4.10c), for $i=3$, and thereafter, multiply the resultant equation through by $\sin 3ny\sin mx$ and integrate and for $k=m$, $p=3n$, to get

$f_{2\left(m\mathrm{,3}n\right)}^{\left(3\right)}=\frac{1}{{\left(m^2+9n^2\xi \right)}^2}\left\{\frac{9}{\pi }H{\left(1+\xi \right)}^{2}m{n}^2{A}_0-m^2{\left(1+\xi \right)}^2{w}_{2\left(m\mathrm{,3}n\right)}^{\left(3\right)}\right\}$ (4.35)

In the same way, we multiply the resultant equation by $\sin mx\sin ny$, integrate and for $k=m$, $p=n$, to get

$\begin{array}{l}{f}_{2\left(m,n\right)}^{\left(3\right)}=\frac{1}{{\left(m^2+n^2\xi \right)}^2}\left\{\frac{1}{\pi }H{\left(1+\xi \right)}^{2}m{n}^2{A}_0-m^2{\left(1+\xi \right)}^2{w}_{2\left(m,n\right)}^{\left(3\right)}\right\},\\ A_0=B_0^3{l}_0,l_0={\Phi }_5\left(\frac{1}{B_0^2}+\frac{1}{B_0^3}\bar{a}\right)\end{array}$ (4.36)

We next substitute (4.8) and (4.9) into (4.34), using (4.10a), (4.10b) and (4.10c), for $i=3$. Thereafter, we multiply the resultant equation by $\sin 3ny\sin mx$, integrate and note that for, $k=m$, $p=3n$, we get

$\begin{array}{l}{w}_{2\left(m,3n\right)}^{\left(3\right)}=\frac{{\Theta }_0{A}_0-{\Theta }_1{A}_{01}}{{\left(m^2+9n^2\xi \right)}^2+{\left(\frac{m^{2}A}{\left(m^2+9n^2\xi \right)}\right)}^2-\lambda \left(\frac{1}{2}\alpha m^2+9n^2\xi \right)},\\ {\Theta }_0=\frac{9}{\pi {\left(m^2+9n^2\xi \right)}^2}H{m}^3{n}^2{A}^2,{\Theta }_1=\frac{9}{\pi }HK\left(\xi \right)m{n}^2,\\ A_{01}=B_0^3{l}_{01},l_{01}=\frac{1}{2}{\Phi }_1-\frac{1}{B_0^2}{\Phi }_0{\Phi }_5+\frac{1}{B_0}\bar{a}{\Phi }_1-\frac{1}{B_0^2}{\Phi }_2{\Phi }_5\end{array}$ (4.37)

Similarly, we multiply through by $\sin ny\sin mx$ and note that for $k=m,p=k$, we get

$\begin{array}{l}{w}_{2\left(m\mathrm{,}n\right)}^{\left(3\right)}=\frac{{\Theta }_2{A}_{01}-{\Theta }_3{A}_0}{{\left(m^2+n^2\xi \right)}^2+{\left(\frac{m^{2}A}{\left(m^2+n^2\xi \right)}\right)}^2-\lambda \left(\frac{1}{2}\alpha m^2+n^2\xi \right)},\\ {\Theta }_3=\frac{1}{\pi {\left(m^2+n^2\xi \right)}^2}H{m}^3{n}^2{A}^2,{\Theta }_2=\frac{1}{\pi }HK\left(\xi \right)m{n}^2\end{array}$ (4.38)

Thus, of the four non-zero buckling modes of this order and their respective Airy stress function, the ones we shall consider are

$\begin{array}{l}{w}_{2\left(m\mathrm{,3}n\right)}^{\left(3\right)}\sin 3ny\sin mx,w_{2\left(m\mathrm{,}n\right)}^{\left(3\right)}\sin ny\sin mx,\\ f_{2\left(m\mathrm{,3}n\right)}^{\left(3\right)}\sin 3ny\sin mx\text{and}{f}_{2\left(m\mathrm{,}n\right)}^{\left(3\right)}\sin ny\sin mx\end{array}$ (4.39)

As a summary so far, we can write the displacement and its respective Airy stress functions as

$\begin{array}{c}\left(\begin{array}{c}w\\ f\end{array}\right)=\epsilon \left(\begin{array}{c}{w}_{2\left(m\mathrm{,}n\right)}^{\left(1\right)}\\ f_{2\left(m\mathrm{,}n\right)}^{\left(1\right)}\end{array}\right)\sin mx\sin ny+{\epsilon }^2\left(\begin{array}{c}{w}_{1\left(m,2n\right)}^{\left(2\right)}\\ f_{1\left(m,2n\right)}^{\left(2\right)}\end{array}\right)\cos ny\sin mx\\ +{\epsilon }^3\left[\left(\begin{array}{c}{w}_{2\left(m\mathrm{,3}n\right)}^{\left(3\right)}\\ f_{2\left(m\mathrm{,3}n\right)}^{\left(3\right)}\end{array}\right)\sin mx\sin 3ny+\left(\begin{array}{c}{w}_{2\left(m\mathrm{,}n\right)}^{\left(3\right)}\\ f_{2\left(m\mathrm{,}n\right)}^{\left(3\right)}\end{array}\right)\sin mx\sin ny\right]+\cdots \end{array}$ (4.40)

Equation (4.40) determines the displacement and the corresponding Airy stress functions.

5. Values of Independent Variables at Maximum Displacement

The analysis henceforth will be concerned with the displacement components that are partly in the shape of the imperfection or partly in the shape of $\sin mx\sin 3ny$. In this respect, we neglect the displacements of order $\mathcal{O}\left({\epsilon }^2\right)$ in (4.40) so that the displacement becomes

$\begin{array}{c}w=\epsilon w_{2\left(m\mathrm{,}n\right)}^{\left(1\right)}\sin mx\sin ny+{\epsilon }^3\{w_{2\left(m\mathrm{,}n\right)}^{\left(3\right)}\sin mx\sin ny\\ +\Omega w_{2\left(m\mathrm{,3}n\right)}^{\left(3\right)}\sin mx\sin 3ny\}+\mathcal{O}\left({\epsilon }^4\right)\end{array}$ (5.1)

where,

$\Omega =0\text{or}1$

When $\Omega =0$, we get the exact displacement that is purely in the shape of imperfection, but when $\Omega =1$, we get the resultant displacement incorporating the modes $\sin mx\sin 3ny$ and $\sin mx\sin ny$.

Since the displacement w in (5.1) depends on x and y then, the conditions for maximum displacement are as follows

$w_{,x}=w_{,y}=0$ (5.2)

We now let $x_a$ and $y_a$ be critical values of x and y respectively at maximum displacement.

From (5.1), using (5.2), we see that for maximum displacement,

$x_a=\frac{\pi }{2m};y_a=\frac{\pi }{2n}$ (5.3)

where (5.3) are the least nontrivial values of $x_a$ and $y_a$.

6. Maximum Displacement

The maximum displacement is obtained from (5.1) at the critical values of x and y where w has a maximum value. Hence, the value of w at these values becomes

$w_a=\epsilon w_{2\left(m\mathrm{,}n\right)}^{\left(1\right)}\left(\lambda \right)+{\epsilon }^3\left\{w_{2\left(m\mathrm{,}n\right)}^{\left(3\right)}\left(\lambda \right)-\Omega w_{2\left(m\mathrm{,3}n\right)}^{\left(3\right)}\left(\lambda \right)\right\}+\cdots$ (6.1)

Meanwhile, (6.1) can be recast as

$w_a=\epsilon c_1+{\epsilon }^3{c}_3+\mathcal{O}\left({\epsilon }^4\right)$ (6.2)

where,

$c_1=w_{2\left(m,n\right)}^{\left(1\right)}\left(\lambda \right),c_3=w_{2\left(m,n\right)}^{\left(3\right)}\left(\lambda \right)-\Omega w_{2\left(m,3n\right)}^{\left(3\right)}\left(\lambda \right)$ (6.3)

7. Static Buckling Load

The static buckling load, ${\lambda }_S$ according to [24] [25] and [26], is obtained from the maximization

$\frac{\text{d}\lambda }{\text{d}{w}_a}=0$ (7.1)

The usual procedure (as in [22] [26] and [27]), is to first reverse the series (6.2) in the form

$\epsilon =d_1{w}_a+d_3{w}_a^3+\cdots$ (7.2)

By substituting (6.1) into (7.2) and equating the coefficients of powers of orders of $\epsilon$, we get

$d_1=\frac{1}{c_1},d_3=-\frac{c_3}{c_1^4}$ (7.3)

The maximization in (7.1) easily follows from (7.2) where $w_a$ is now being substituted for w to yield, after some simplifications,

$\epsilon =\frac{2}{3\sqrt{3}}\sqrt{\frac{c_1}{c_3}}$ (7.4)

On substituting into (7.4), using (6.3) and simplifying, we get

$\begin{array}{l}{\left\{{\left(m^2+n^2\xi \right)}^2+{\left(\frac{m^{2}A}{m^2+n^2\xi }\right)}^2-{\lambda }_S\left(\frac{1}{2}\alpha m^2+n^2\xi \right)\right\}}^{\frac{3}{2}}\\ =\frac{3\sqrt{3}}{2}{\lambda }_S\left(\epsilon \bar{a}\right)\left(\frac{1}{2}\alpha m^2+n^2\xi \right){\Psi }_0\end{array}$ (7.5)

This determines the static buckling load ${\lambda }_S$ of the circular cylindrical shell structure, and the determination is implicit in the load parameter ${\lambda }_S$,

where,

$\begin{array}{l}{\Psi }_0=\left({\Theta }_2{l}_{01}-{\Theta }_3{l}_0\right)\sqrt{1-\frac{\Omega Q_{02}}{Q_{01}}},Q_{01}=\frac{\left({\Theta }_2{l}_{01}-{\Theta }_3{l}_0\right)}{{\phi }_0^2},\\ Q_{02}=\frac{\left({\Theta }_0{l}_0-{\Theta }_1{l}_{01}\right)}{{\phi }_0^2},{\phi }_0^2=\left[{\left(m^2+n^2\xi \right)}^2+{\left(\frac{m^{2}A}{m^2+n^2\xi }\right)}^2-{\lambda }_S\left(\frac{1}{2}\alpha m^2+n^2\xi \right)\right]\end{array}$ (7.6)

8. Results and Discussion

The result (7.5) is asymptotic in nature. The results of the classical buckling load ${\lambda }_C$, and that of the cylindrical shell structure are as seen in (3.9), whereas, the corresponding displacement and Airy Stress function of the structure are as in (3.10). Similarly, the static buckling load ${\lambda }_S$, is as shown in (7.5). A computer program in MATLAB gives the relationship between the static buckling load ${\lambda }_S$, and the imperfection parameter $\epsilon$, at $\Omega =0$ or $\Omega =1$, and where we have fixed the following as $\alpha =1$, $A=0.2$, $H=0.2$, $\bar{a}=0.02$, $\bar{\xi }=0.8$, $m=1$ and $n=1$ is as shown in Table 1.

A careful appraisal of the graph of Figure 1, shows that static buckling load ${\lambda }_S$, decreases with increased imperfection parameter $\epsilon$. This is expected. In other words, static buckling load ${\lambda }_S$ increases with less imperfection. The value of static buckling load ${\lambda }_S$ is higher when the buckling mode is a combination of the modes in the shape of imperfection ( $\sin mx\sin ny$ ) and shape of other geometric form ( $\sin mx\sin 3ny$ ), i.e. ( $\Omega =1$ ) compared to the case when the buckling mode is in the shape of imperfection ( $\sin mx\sin ny$ ) (i.e. $\Omega =0$ ).

Table 1. Relationship between static buckling load, ${\lambda }_S$ and imperfection parameter, $\epsilon$ for some values of $\Omega$ of the circular cylindrical shell structure using Equation (7.5).

Figure 1. Graph of Static buckling load ${\lambda }_S$, as a function of imperfection parameter, $\epsilon$ for some values of $\Omega$ of the circular cylindrical shell structure using Table 1.

9. Conclusion

This analysis has analytically determined the maximum of the out-of-plane normal displacement of a finite imperfect cylindrical shell trapped by a static load. We have used the techniques of perturbation and asymptotics to derive an implicit formula for determining the static buckling load of the cylindrical shell investigated. The formulation contains a small parameter depicting the amplitude of the imperfection and on which all asymptotic series are expanded. Such an analytical approach can be duplicated for other structures including toroidal shell segments and plates.

Acknowledgements

This paper is dedicated to the memory of late EZIKE OGBUEHI FELIX IKENNA OZOIGBO, who died on 20^thday of July, 2019 and to be buried on 22^nd day of November, 2019.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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