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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.

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 [

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 [

As in [

1 E h ∇ 4 F − 1 R W , X X = − S ( W , 1 2 W + W ¯ ) (2.1)

D ∇ 4 W + 1 R F , X X + P [ 1 2 α ( W + W ¯ ) , X X + ( W + W ¯ ) , Y Y ] = S ( W + W ¯ , F ) (2.2)

W = W , X X = F = F , X X = 0 at X = 0 , L , 0 < X < π , 0 < Y < 2 π . (2.3)

where, X and Y are the axial and circumferential coordinates respectively and W ¯ ( X , Y ) , 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 ( P , Q ) = P , X X Q , Y Y + P , Y Y Q , X X − 2 P , X Y Q , X Y (2.4a)

and ∇ 4 , is the two-dimensional bi-harmonic operator defined by

∇ 4 = ( ∂ 2 ∂ X 2 + ∂ 2 ∂ Y 2 ) 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 [

x = X π L , H = h R , ε w ¯ = W ¯ h , λ = L 2 R P π 2 D , ξ = L 2 π 2 R 2 (2.5a)

y = Y R , w = W h , K ( ξ ) = − A 2 ( 1 + ξ ) 2 , A = L 2 12 ( 1 − υ 2 ) π 2 R h (2.5b)

where, υ is Poisson’s ratio and ε 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 , π .

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

F = − 1 2 P R ( X 2 + 1 2 α Y 2 ) + ( E h 2 L 2 π 2 R ( 1 + ξ ) 2 ) f (2.7a)

W = P R 2 ( 1 − 1 2 α υ ) E h + h w (2.7b)

where, P is the applied static load and λ 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 α , shall take the value α = 1 , if pressure contributes to axial stress through the ends, otherwise α = 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

∇ ¯ 4 f − ( 1 + ξ ) 2 w , x x = − H ( 1 + ξ ) 2 s ( w , 1 2 w + ε w ¯ ) (2.8)

and

∇ ¯ 4 w − K ( ξ ) f , x x + λ [ 1 2 α ( w + ε w ¯ ) , x x + ( w + ε w ¯ ) , y y ] = − H K ( ξ ) s ( w + ε w ¯ , f ) (2.9)

w = w , x x = f = f , x x = 0 at x = 0 , π , 0 < x < π , 0 < y < 2 π , 0 < ε ≪ 1 , (2.10)

where,

s ( P , Q ) = P , x x Q , y y + P , y y Q , x x − 2 P , x y Q , x y , ∇ ¯ 4 = ( ∂ 2 ∂ x 2 + ξ ∂ 2 ∂ y 2 ) 2 (2.11)

The classical buckling load λ C is the load that is required to buckle the associated linear perfect structure and its equations, from (2.8) and (2.9) are

∇ ¯ 4 f − ( 1 + ξ ) 2 w , x x = 0 (3.1)

and

∇ ¯ 4 w − K ( ξ ) f , x x + λ [ 1 2 α w , x x + w ¯ , y y ] = 0 (3.2)

w = w , x x = 0 , f = f , x x = 0 at x = 0 , π . (3.3)

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

( w , f ) = ( a m k , b m k ) sin ( k y + ϕ m k ) sin m x (3.4)

where, ( a m k , b m k ) ≠ ( 0,0 ) and ϕ m k is an inconsequential phase.

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

b m k = − ( 1 + ξ ) 2 m 2 a m k ( m 2 + ξ k 2 ) 2 (3.5)

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

λ = ( m 2 + ξ k 2 ) 2 − K ( ξ ) m 4 ( 1 + ξ ) 2 ( m 2 + ξ k 2 ) 2 1 2 α m 2 + k 2 m 4 ( 1 + ξ ) 2 ξ ( m 2 + ξ k 2 ) 2 (3.6)

Thus, if n is the critical value of k that minimizes λ , then, the value of λ at k = n was taken as the classical buckling load λ C . Thus, in this case, we get

d λ d k = 0 (3.7)

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

λ = ( m 2 + ξ n 2 ) 2 − K ( ξ ) m 4 ( 1 + ξ ) 2 ( m 2 + ξ n 2 ) 2 1 2 α m 2 + n 2 m 4 ( 1 + ξ ) 2 ξ ( m 2 + ξ n 2 ) 2 (3.8)

Usually, m and n take the values m = 1 , 2 , 3 , ⋯ and n = 0 , 1 , 2 , ⋯

We recall that [

λ C = ( 1 + ζ ) 2 + A 2 ( 1 + ζ ) 2 1 2 α + ζ (3.9)

The corresponding displacement and Airy Stress function are

( w , f ) = [ { 1 , − ( 1 + ξ 1 + ζ ) 2 } ] a 1 n sin ( k y + ϕ 1 n ) sin x (3.10)

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

∇ ¯ 4 f − ( 1 + ξ ) 2 w , x x = − H ( 1 + ξ ) 2 s ( w , 1 2 w + ε w ¯ ) (4.1)

and

∇ ¯ 4 w − K ( ξ ) f , x x + λ [ 1 2 α ( w + ε w ¯ ) , x x + ξ ( w + ε w ¯ ) , y y ] = − H K ( ξ ) s ( w + ε w ¯ , f ) (4.2)

w = w , x x = f = f , x x = 0 at x = 0 , π (4.3)

We now assume the following asymptotic expansions

( w f ) = ∑ i = 1 ∞ ( w ( i ) f ( i ) ) ε i (4.4)

Substituting (4.4) into (4.1) and (4.2), and equating the coefficients of orders of ε i , i = 1 , 2 , 3 , ⋯ , the following equations are obtained

O ( ε ) : ( ∇ ¯ 4 f ( 1 ) − ( 1 + ξ ) 2 w , x x ( 1 ) = 0. ∇ ¯ 4 w ( 1 ) − K ( ξ ) f , x x ( 1 ) + λ [ 1 2 α ( w ( 1 ) + w ¯ ) , x x + ξ ( w ( 1 ) + w ¯ ) , y y ] = 0 (4.5)

O ( ε 2 ) : ( ∇ ¯ 4 f ( 2 ) − ( 1 + ξ ) 2 w , x x ( 2 ) = − H ( 1 + ξ ) 2 [ 1 2 s ( w ( 1 ) , w ( 1 ) ) + s ( w ( 1 ) , w ¯ ) ] ∇ ¯ 4 w ( 2 ) − K ( ξ ) f , x x ( 2 ) + λ [ 1 2 α w , x x ( 2 ) + ξ w , y y ( 2 ) ] = − H K ( ξ ) [ s ( w ( 1 ) , f ( 1 ) ) + s ( w ¯ , w ( 1 ) ) ] (4.6)

O ( ε 3 ) : ( ∇ ¯ 4 f ( 3 ) − ( 1 + ξ ) 2 w , x x ( 3 ) = − H ( 1 + ξ ) 2 [ s ( w ( 1 ) , w ( 2 ) ) + s ( w ( 2 ) , w ¯ ) ] ∇ ¯ 4 w ( 3 ) − K ( ξ ) f , x x ( 3 ) + λ [ 1 2 α w , x x ( 3 ) + ξ w , y y ( 3 ) ] = − H K ( ξ ) [ s ( w ( 1 ) , f ( 2 ) ) + s ( w ( 2 ) , f ( 1 ) ) + s ( w ¯ , f ( 2 ) ) ] (4.7)

etc.

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

( f ( i ) w ( i ) ) = ∑ ( k = 1 ) , ( p = 0 ) ∞ [ ( f 1 ( i ) w 1 ( i ) ) cos p y + ( f 2 ( i ) w 2 ( i ) ) sin p y ] sin k x (4.8)

and now assume

w ¯ ( x , y ) = a ¯ sin m x sin n y (4.9)

As earlier obtained, we shall need the following simplifications

∇ ¯ 4 = ( ∂ 2 ∂ x 2 + ξ ∂ 2 ∂ y 2 ) 2 = ( ∂ 4 ∂ x 4 + 2 ξ ∂ 4 ∂ x 2 ∂ y 2 + ξ 2 ∂ 4 ∂ y 4 ) (4.10a)

so that, if

f ( 1 ) = f Γ 1 ( 1 ) cos p y sin k x

then, we have

∇ ¯ 4 f ( 1 ) = ( k 2 + ξ p 2 ) 2 f Γ 1 ( 1 ) cos p y sin k x , Γ 1 = 1 , 2 (4.10b)

and, if

f ( 2 ) = f Γ 2 ( 2 ) sin p y sin k x

we get

∇ ¯ 4 f ( 2 ) = ( k 2 + ξ p 2 ) 2 f Γ 2 ( 2 ) sin p y sin k x , Γ 2 = 1 , 2 (4.10c)

We shall use the fact that

K ( ξ ) = − A 2 ( 1 + ξ ) 2 (4.10d)

Solution of Equations of First Order Perturbation

The equations necessary here, from (4.5), are

∇ ¯ 4 f ( 1 ) − ( 1 + ξ ) 2 w , x x ( 1 ) = 0 (4.11)

and

∇ ¯ 4 w ( 1 ) − K ( ξ ) f , x x ( 1 ) + λ [ 1 2 α ( w ( 1 ) + w ¯ ) , x x + ξ ( w ( 1 ) + w ¯ ) , y y ] = 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 n y sin m x 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 ( 1 ) = − m 2 ( 1 + ξ ) 2 w 1 ( 1 ) ( m 2 + n 2 ξ ) 2 (4.13)

Similarly, by multiplying the resultant equation through by sin n y sin m x , 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 ( 1 ) = − m 2 ( 1 + ξ ) 2 w 2 ( 1 ) ( m 2 + n 2 ξ ) 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 n y sin m x and integrating with respect to x from 0 to π and y from 0 to 2π, and for p = n , k = m , we get

[ ( m 2 + n 2 ξ ) 2 − λ ( 1 2 α m 2 + n 2 ξ ) ] w 2 ( 1 ) + K ( ξ ) m 2 V 2 ( 1 ) = λ a ¯ ( 1 2 α m 2 + n 2 ξ ) (4.15)

On substituting for f 2 ( 1 ) from (4.14) and K ( ξ ) from (4.10d) in (4.15) and simplifying, yields

w 2 ( 1 ) = B 0 (4.16)

where,

B 0 = λ a ¯ ( 1 2 α m 2 + n 2 ξ ) φ 0 2 , φ 0 2 = ( m 2 + n 2 ξ ) 2 + ( m 2 A m 2 + n 2 ξ ) 2 − λ ( 1 2 α m 2 + n 2 ξ ) (4.17)

Next, multiplying the resultant equation by cos n y cos m x 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 ( 1 ) = 0 (4.18)

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

w ( 1 ) = B 0 sin m x sin n y ; f ( 1 ) = − Φ 0 B 0 sin m x sin n y , Φ 0 = m 2 ( 1 + ξ ) 2 ( m 2 + n 2 ξ ) 2 (4.19)

Solution of Equations of Second Order Perturbation

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

∇ ¯ 4 f ( 2 ) − ( 1 + ξ ) 2 w , x x ( 2 ) = − H ( 1 + ξ ) 2 [ 1 2 s ( w ( 1 ) , w ( 1 ) ) + s ( w ( 1 ) , w ¯ ) ] (4.20)

and

∇ ¯ 4 w ( 2 ) − K ( ξ ) f , x x ( 2 ) + λ [ 1 2 α w , x x ( 2 ) + ξ w , y y ( 2 ) ] = − H K ( ξ ) [ s ( w ( 1 ) , w ( 1 ) ) + s ( w ¯ , w ( 1 ) ) ] (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)

∇ ¯ 4 f ( 2 ) − ( 1 + ξ ) 2 w , x x ( 2 ) = − H ( 1 + ξ ) 2 ( m n ) 2 [ ( 1 2 B 0 2 + B 0 a ¯ ) ( cos 2 m x + cos 2 n y ) ] (4.22)

and

∇ ¯ 4 w ( 2 ) − K ( ξ ) f , x x ( 2 ) + λ [ 1 2 α w , x x ( 2 ) + ξ w , y y ( 2 ) ] = − H K ( ξ ) ( m n ) 2 [ Φ 0 B 0 2 + a ¯ B 0 ] ( cos 2 m x + cos 2 n y ) (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 2 n y sin m x and integrate with respect to x and y, to get

f 1 ( 2 ) = Φ 1 ( 1 2 B 0 2 + B 0 a ¯ ) − Φ 2 w 1 ( 2 ) , Φ 1 = 4 H m n 2 ( 1 + ξ ) 2 π ( m 2 + 4 n 2 ξ ) 2 , Φ 2 = m 2 ( 1 + ξ ) 2 π ( m 2 + 4 n 2 ξ ) 2 , m = o d d (4.24)

Similarly, we next multiply the resultant equation by sin n y sin m x and integrate as usual, and for p = n , k = m , we get

f 2 ( 2 ) = − m 2 ( 1 + ξ ) 2 w 2 ( 2 ) ( m 2 + n 2 ξ ) 2 . (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 2 n y sin m x and integrate, as usual for p = 2 n , k = m to get

w 1 ( 2 ) = Φ 5 , Φ 5 = Φ 3 ( B 0 2 Φ 0 + B 0 Φ 0 a ¯ ) + Φ 4 ( 1 2 B 0 2 + B 0 a ¯ ) (4.26)

Φ 3 = ( m n ) 2 H K ( ξ ) ( m 2 + 4 n 2 ξ ) 2 + ( m 2 A m 2 + 4 n 2 ξ ) 2 − λ ( 1 2 α m 2 + 4 n 2 ξ )

Φ 4 = 4 m 3 n 2 A 2 ( m 2 + 4 n 2 ξ ) 2 [ ( m 2 + 4 n 2 ξ ) 2 + ( m 2 A m 2 + 4 n 2 ξ ) 2 − λ ( 1 2 α m 2 + 4 n 2 ξ ) ] (4.27)

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

w 2 ( 2 ) = 0 , f 2 ( 2 ) = 0 (4.28)

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

w ( 2 ) = w 1 ( 2 ) cos 2 n y sin m x ; f ( 2 ) = f 1 ( 2 ) cos 2 n y sin m x , (4.29)

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

w ( 2 ) = Φ 5 cos 2 n y sin m x ; f ( 2 ) = Φ 10 cos 2 n y sin m x , Φ 10 = [ Φ 1 ( 1 2 B 0 2 + B 0 a ¯ ) − Φ 2 Φ 5 ] (4.30)

Solution of Equations of Third Order Perturbation

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

∇ ¯ 4 f ( 3 ) − ( 1 + ξ ) 2 w , x x ( 3 ) = − H ( 1 + ξ ) 2 [ s ( w ( 1 ) , w ( 2 ) ) + s ( w ( 2 ) , w ¯ ) ] (4.31)

and

∇ ¯ 4 w ( 3 ) − K ( ξ ) f , x x ( 3 ) + λ [ 1 2 α w , x x ( 3 ) + ξ w , y y ( 3 ) ] = − H K ( ξ ) [ s ( w ( 1 ) , f ( 2 ) ) + s ( w ( 2 ) , f ( 1 ) ) + s ( w ¯ , f ( 2 ) ) ] (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

∇ ¯ 4 f ( 3 ) − ( 1 + ξ ) 2 w , x x ( 3 ) = − 1 4 H ( 1 + ξ ) 2 ( m n ) 2 ( Φ 5 B 0 + Φ 5 a ¯ ) [ 9 sin 3 n y − sin n y − cos 2 m x sin 3 n y + 9 cos 2 m x sin n y ] (4.33)

and

∇ ¯ 4 w ( 3 ) − K ( ξ ) f , x x ( 3 ) + λ [ 1 2 α w , x x ( 3 ) + ξ w , y y ( 3 ) ] = − 1 4 H K ( ξ ) ( m n ) 2 [ − Φ 0 Φ 5 B 0 + B 0 { Φ 1 ( 1 2 B 0 2 + B 0 a ¯ ) − Φ 2 Φ 5 } + a ¯ { Φ 1 ( 1 2 B 0 2 + B 0 a ¯ ) − Φ 2 Φ 5 } ] [ 9 sin 3 n y − sin n y − cos 2 m x sin 3 n y + 9 cos 2 m x sin n y ] (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 ( r , p ) ( 3 ) with their respective Airy stress functions f i ( r , p ) ( 3 ) . These buckling modes correspond to the following terms on the right hand sides of (4.33) and (4.34) : sin 3 n y sin m x , sin n y sin m x , cos 2 m x sin 3 n y and cos 2 m x sin n y .

However, of the four modes, it is only the mode in the shape of sin n y sin m x that is in the shape of the imperfection. We shall consider this mode and the additional mode in the shape of sin 3 n y sin m x .

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 3 n y sin m x and integrate and for k = m , p = 3 n , to get

f 2 ( m ,3 n ) ( 3 ) = 1 ( m 2 + 9 n 2 ξ ) 2 { 9 π H ( 1 + ξ ) 2 m n 2 A 0 − m 2 ( 1 + ξ ) 2 w 2 ( m ,3 n ) ( 3 ) } (4.35)

In the same way, we multiply the resultant equation by sin m x sin n y , integrate and for k = m , p = n , to get

f 2 ( m , n ) ( 3 ) = 1 ( m 2 + n 2 ξ ) 2 { 1 π H ( 1 + ξ ) 2 m n 2 A 0 − m 2 ( 1 + ξ ) 2 w 2 ( m , n ) ( 3 ) } , A 0 = B 0 3 l 0 , l 0 = Φ 5 ( 1 B 0 2 + 1 B 0 3 a ¯ ) (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 3 n y sin m x , integrate and note that for, k = m , p = 3 n , we get

w 2 ( m , 3 n ) ( 3 ) = Θ 0 A 0 − Θ 1 A 01 ( m 2 + 9 n 2 ξ ) 2 + ( m 2 A ( m 2 + 9 n 2 ξ ) ) 2 − λ ( 1 2 α m 2 + 9 n 2 ξ ) , Θ 0 = 9 π ( m 2 + 9 n 2 ξ ) 2 H m 3 n 2 A 2 , Θ 1 = 9 π H K ( ξ ) m n 2 , A 01 = B 0 3 l 01 , l 01 = 1 2 Φ 1 − 1 B 0 2 Φ 0 Φ 5 + 1 B 0 a ¯ Φ 1 − 1 B 0 2 Φ 2 Φ 5 (4.37)

Similarly, we multiply through by sin n y sin m x and note that for k = m , p = k , we get

w 2 ( m , n ) ( 3 ) = Θ 2 A 01 − Θ 3 A 0 ( m 2 + n 2 ξ ) 2 + ( m 2 A ( m 2 + n 2 ξ ) ) 2 − λ ( 1 2 α m 2 + n 2 ξ ) , Θ 3 = 1 π ( m 2 + n 2 ξ ) 2 H m 3 n 2 A 2 , Θ 2 = 1 π H K ( ξ ) m n 2 (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

w 2 ( m ,3 n ) ( 3 ) sin 3 n y sin m x , w 2 ( m , n ) ( 3 ) sin n y sin m x , f 2 ( m ,3 n ) ( 3 ) sin 3 n y sin m x and f 2 ( m , n ) ( 3 ) sin n y sin m x (4.39)

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

( w f ) = ε ( w 2 ( m , n ) ( 1 ) f 2 ( m , n ) ( 1 ) ) sin m x sin n y + ε 2 ( w 1 ( m , 2 n ) ( 2 ) f 1 ( m , 2 n ) ( 2 ) ) cos n y sin m x + ε 3 [ ( w 2 ( m ,3 n ) ( 3 ) f 2 ( m ,3 n ) ( 3 ) ) sin m x sin 3 n y + ( w 2 ( m , n ) ( 3 ) f 2 ( m , n ) ( 3 ) ) sin m x sin n y ] + ⋯ (4.40)

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

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 m x sin 3 n y . In this respect, we neglect the displacements of order O ( ε 2 ) in (4.40) so that the displacement becomes

w = ε w 2 ( m , n ) ( 1 ) sin m x sin n y + ε 3 { w 2 ( m , n ) ( 3 ) sin m x sin n y + Ω w 2 ( m ,3 n ) ( 3 ) sin m x sin 3 n y } + O ( ε 4 ) (5.1)

where,

Ω = 0 or 1

When Ω = 0 , we get the exact displacement that is purely in the shape of imperfection, but when Ω = 1 , we get the resultant displacement incorporating the modes sin m x sin 3 n y and sin m x sin n y .

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 = π 2 m ; y a = π 2 n (5.3)

where (5.3) are the least nontrivial values of x a and y a .

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 = ε w 2 ( m , n ) ( 1 ) ( λ ) + ε 3 { w 2 ( m , n ) ( 3 ) ( λ ) − Ω w 2 ( m ,3 n ) ( 3 ) ( λ ) } + ⋯ (6.1)

Meanwhile, (6.1) can be recast as

w a = ε c 1 + ε 3 c 3 + O ( ε 4 ) (6.2)

where,

c 1 = w 2 ( m , n ) ( 1 ) ( λ ) , c 3 = w 2 ( m , n ) ( 3 ) ( λ ) − Ω w 2 ( m , 3 n ) ( 3 ) ( λ ) (6.3)

The static buckling load, λ S according to [

d λ d w a = 0 (7.1)

The usual procedure (as in [

ε = d 1 w a + d 3 w a 3 + ⋯ (7.2)

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

d 1 = 1 c 1 , d 3 = − 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,

ε = 2 3 3 c 1 c 3 (7.4)

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

{ ( m 2 + n 2 ξ ) 2 + ( m 2 A m 2 + n 2 ξ ) 2 − λ S ( 1 2 α m 2 + n 2 ξ ) } 3 2 = 3 3 2 λ S ( ε a ¯ ) ( 1 2 α m 2 + n 2 ξ ) Ψ 0 (7.5)

This determines the static buckling load λ S of the circular cylindrical shell structure, and the determination is implicit in the load parameter λ S ,

where,

Ψ 0 = ( Θ 2 l 01 − Θ 3 l 0 ) 1 − Ω Q 02 Q 01 , Q 01 = ( Θ 2 l 01 − Θ 3 l 0 ) φ 0 2 , Q 02 = ( Θ 0 l 0 − Θ 1 l 01 ) φ 0 2 , φ 0 2 = [ ( m 2 + n 2 ξ ) 2 + ( m 2 A m 2 + n 2 ξ ) 2 − λ S ( 1 2 α m 2 + n 2 ξ ) ] (7.6)

The result (7.5) is asymptotic in nature. The results of the classical buckling load λ 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 λ S , is as shown in (7.5). A computer program in MATLAB gives the relationship between the static buckling load λ S , and the imperfection parameter ε , at Ω = 0 or Ω = 1 , and where we have fixed the following as α = 1 , A = 0.2 , H = 0.2 , a ¯ = 0.02 , ξ ¯ = 0.8 , m = 1 and n = 1 is as shown in

A careful appraisal of the graph of

ε | λ S | |
---|---|---|

Ω = 0 | Ω = 1 | |

0.0100 | 2.4945 | 2.4954 |

0.0200 | 2.4944 | 2.4953 |

0.0300 | 2.4943 | 2.4952 |

0.0400 | 2.4942 | 2.4951 |

0.0500 | 2.4941 | 2.4949 |

0.0600 | 2.4940 | 2.4948 |

0.0700 | 2.4939 | 2.4946 |

0.0800 | 2.4937 | 2.4944 |

0.0900 | 2.4936 | 2.4942 |

0.1000 | 2.4933 | 2.4938 |

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.

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

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

Ozoigbo, G.E. and Ette, A.M. (2019) On the Maximum Displacement and Static Buckling of a Circular Cylindrical Shell. Journal of Applied Mathematics and Physics, 7, 2868-2882. https://doi.org/10.4236/jamp.2019.711196