The constrained buckling problem of geometrically imperfect beams: a mathematical approach for the determination of the critical instability points

Abstract

The constrained buckling problem of geometrically imperfect beams with intermediate unilateral supports is studied in the present paper. The proposed methodology offers the ability to calculate analytically the critical loads and the buckling shape of beams with arbitrary initial geometric imperfections, for a variety of different initial contact conditions in the framework of elastic stability theory. The proposed mathematical approach is based on the formulation of the equilibrium equations in the deformed position, in which the function of the unilateral constraints is appropriately taken into account. The analytical solution is obtained after the splitting of the initial constrained non-homogeneous boundary value problem (BVP) into constrained subproblems and the utilization of a classical mathematical theorem from the field of ordinary non-homogeneous BVPs. The implementation of the presented technique is demonstrated through characteristic examples. In order to validate the proposed mathematical method, the obtained results are compared with the respective numerical ones. The latter are obtained through the utilization of geometric nonlinear finite element analysis. The paper ends with the presentation of an investigation on the variation of the critical load with respect to different positions of the unilateral constraints.

Appendices

Appendix 1: The fundamental theorem for the solution of non-homogeneous BVPs

In the mathematical field of ordinary differential equations and Boundary Value Problems it is well known that the solution of non homogeneous BVPs is strongly connected with the solution of the corresponding homogeneous BVPs. More specifically, the solution of any non homogeneous BVP with ordinary differential equation can be obtained after the utilization of a mathematical theorem which can be found in any classical mathematical text book concerning differential equations and BVPs (see e.g. [20, 21]). In order to facilitate the reader, the latter is briefly presented here, in a form that is compatible to the formulation presented in Sect. 2.

Let us consider the general form of a non-homogeneous BVP [20, 21] constituted by the governing ordinary differential equation together with the boundary conditions:

$$ M(w) - \lambda N(w) = f(x) \ne 0 $$

(40)

$$ \begin{aligned} & a_{i,0} w(a) + b_{i,0} w(b) + a_{i,1} w^{\prime}(a) + b_{i,1} w^{\prime}(b) + \cdots + \\ \, &+ a_{i,2m - 1} w^{(2m - 1)} (a) + b_{i,2m - 1} w^{(2m - 1)} (b) = 0, \quad i = 1,2,3, \ldots ,2m \\ \end{aligned} $$

(41)

where, a, bare the boundaries of a closed interval [a, b], a _i,o , b _i,0,…. are real constants and M(w), N(w)are self-adjoint expressions with respect to the differential operators M, N having orders 2m and 2n respectively, with m > n.

The previous differential operators are defined in the domain D _A which is a subset of the space of all continuous functions including their derivatives up to the order of 2 m in a closed region $\bar{\varOmega }$. In Eqs. (40) and (41) $f(x):{\mathbb{R}} \to {\mathbb{R}}$ describes the non-homogeneous term and w ^(2m−1)denotes the 2 m – 1 derivative of function w. Then, the following theorem concerns the solvability of the former non-homogeneous problem:

Theorem

Let a real number λ satisfying (41) be given. Then:

• If this value λ is not an eigenvalue of the corresponding homogeneous problem (f(x) = 0), then the given non-homogeneous problem has exactly one solution for every arbitrary right-hand side function f(x).

• If this value λ is an eigenvalue of the corresponding homogeneous problem, then the given non-homogeneous problem is in general not solvable. It is solvable (but not uniquely) if and only if the function f(x) is orthogonal to every function ϕ(x)corresponding to that λ, thus if the following equation holds for every such eigenfunction:

$$ (f,\phi ) = \int\limits_{a}^{b} {f(x)\phi (x)dx = 0} $$

(42)

Appendix 2: Formulas for the determination of the unknown coefficients in the case of uniquely solvable problems

The following formulas can be used for the calculation of the unknown coefficients in the case of uniquely solvable problems for each contact case of Table 1:

• Contact Case CC1

  $$ B_{2} = \frac{{\Pi _{2} +\Pi _{3} +\Pi _{4} }}{{\Pi _{1} }} $$

  (43)
  $$ \begin{aligned} B_{2} &= B_{1} \left[ {{ \cos }\,kaL - \frac{a + b}{kabL}{ \sin }\,kaL} \right] - \frac{a + b}{kabL}\sum\limits_{r = 1}^{n} {g_{r} } F_{r} { \sin }\,r\pi a \\ & \quad + B_{3} \frac{{{ \sin }\,kcL}}{kbL} - \frac{1}{kbL}\sum\limits_{r = 1}^{n} {g_{r} } F_{r} { \sin }\,r\pi c( - 1)^{r} \\ \end{aligned} $$

  (44)
  $$ B_{3} = - B_{1}\Lambda _{1} +\Lambda _{2} +\Lambda _{3} $$

  (45)
  $$ C_{1} = \frac{{ - B_{1} { \sin }\,kaL - \sum\nolimits_{r = 1}^{n} {g_{r} F_{r} { \sin }r\pi a} }}{aL} $$

  (46)
  $$ C_{2} = \frac{{ - B_{3} { \sin }\,kcL + B_{1} { \sin }\,kaL + \sum\nolimits_{r = 1}^{n} {g_{r} F_{r} { \sin }\,r\pi a} + \sum\nolimits_{r = 1}^{n} {g_{r} F_{r} { \sin }\,r\pi c( - 1)^{r} } }}{bL} $$

  (47)
  $$ C_{3} = \frac{{ - B_{3} { \sin }\,kcL + \sum\nolimits_{r = 1}^{n} {g_{r} F_{r} { \sin }\,r\pi c( - 1)^{r} } }}{cL} $$

  (48)
  $$ A_{2} = B_{1} { \sin }\,kaL $$

  (49)
  $$ D_{2} = - B_{1} { \sin }\,kaL - \sum\limits_{r = 1}^{n} {g_{r} } F_{r} { \sin }\,r\pi a $$

  (50)

In the above relations the terms Π₁, Π₂, Π₃, Π₄ and Λ₁, Λ₂, Λ₃ are given by the following relations:

$$ \Pi _{1} = - { \cos }[k(a + b)L] + \frac{a + b}{kabL}{ \sin }\,kaL\,{ \cos }\,kbl - \frac{{{ \sin }\,kaL}}{kbL} -\Lambda _{1} {\rm K} $$

(51)

$$ \Pi _{2} = -\Lambda _{2} {\rm K} $$

(52)

$$ \Pi _{3} = -\Lambda _{3} {\rm K} $$

(53)

$$ \begin{aligned}\Pi _{4} &= \frac{b + c}{kbcL}\sum\limits_{r = 1}^{n} {g_{r} } F_{r} { \sin }(r\pi c)( - 1)^{r} + \frac{1}{kbL}\sum\limits_{r = 1}^{n} {g_{r} } F_{r} { \sin }\,r\pi a - \\ &\quad - \frac{a + b}{kabL}{ \cos }\,kbL\sum\limits_{r = 1}^{n} {g_{r} } F_{r} { \sin }\,r\pi a - \frac{{{ \cos }\,kbL}}{kbL}\sum\limits_{r = 1}^{n} {g_{r} } F_{r} { \sin }(r\pi c)( - 1)^{r} \\ \end{aligned} $$

(54)

$$ \Lambda _{1} = \frac{{{ \sin }[k(a + b)L] - \frac{a + b}{kabL}{ \sin }\,kaL{ \sin}\,kbL}}{{\frac{1}{kbL}{ \sin }\,kcL\,{ \sin }\,kbL - { \sin}\,kcL}} $$

(55)

$$ \Lambda _{2} = \frac{{\frac{a + b}{kabL}{ \sin }\,kbL\sum\nolimits_{r = 1}^{n} {g_{r} F_{r} { \sin }\,r\pi a} }}{{\frac{1}{kbL}{ \sin }\,kcL\,{ \sin }\,kbL - { \sin }\,kcL}} $$

(56)

$$ \Lambda _{3} = \frac{{\frac{1}{kbL}\sin kbL\sum\nolimits_{r = 1}^{n} {g_{r} F_{r} \sin \,(r\pi c)( - 1)^{r} } }}{{\frac{1}{kbL}\sin kcL\sin kbL - \sin kcL}} $$

(57)

where:

$$ {\rm K} = - { \cos }\,kcL + \frac{b + c}{kbcL}{ \sin}\,kcL - \frac{{{ \sin}\,kcL{ \cos }\,kbL}}{kbL} $$

(58)

• Contact Case CC2

  $$ B_{1} = B_{2} = B_{3} = C_{1} = C_{2} = C_{3} = A_{2} = D_{2} = 0 $$

  (59)

  • Contact Case CC3

$$ B_{1} = \frac{{\left[ {\sum\nolimits_{r = 1}^{n} {g_{r} F_{r} { \sin}\,r\pi a} } \right]\frac{{{ \sin }\,\left[ {k(b + c)L} \right]}}{ka(b + c)L}}}{{{ \sin }\,kL - \frac{{{ \sin }\,kaL\,{ \sin }\,\,\left[ {k(b + c)L} \right]}}{ka(b + c)L}}} $$

(60)

$$ B_{2} = B_{1} { \cos }\,kaL - \frac{{\sum\nolimits_{r = 1}^{n} {g_{r} F_{r} { \sin }\,r\pi a + B_{1} { \sin }\,kaL} }}{ka(b + c)L} $$

(61)

$$ B_{3} = \frac{{\left[ {\sum\nolimits_{r = 1}^{n} {g_{r} F_{r} { \sin }\,r\pi a} } \right]\frac{{{ \sin }\,kaL}}{ka(b + c)L}}}{{{ \sin }\,kL - \frac{{{ \sin }\,kaL\,{ \sin }\left[ {k(b + c)L} \right]}}{ka(b + c)L}}} $$

(62)

$$ C_{1} = \frac{{ - B_{1} { \sin }\,kaL - \sum\nolimits_{r = 1}^{n} {g_{r} F_{r} { \sin }\,r\pi a} }}{aL} $$

(63)

$$ C_{2} = \frac{{B_{1} { \sin }\,kaL + \sum\nolimits_{r = 1}^{n} {g_{r} F_{r} { \sin }\,r\pi a} }}{(b + c)L} $$

(64)

$$ C_{3} = - C_{2} $$

(65)

$$ A_{2} = B_{1} { \sin }\,kaL $$

(66)

$$ D_{2} = - B_{1} { \sin }\,kaL - \sum\limits_{r = 1}^{n} {g_{r} } F_{r} { \sin }\,r\pi a $$

(67)

• Contact Case CC4

  $$ B_{1} = \frac{{ - \left[ {\sum\nolimits_{r = 1}^{n} {g_{r} F_{r} { \sin }\,r\pi c( - 1)^{r} } } \right]\frac{{{ \sin }\,kcL}}{c(a + b)L}}}{{k\,{ \sin }\,kL - \frac{{{ \sin }\,kcL\,{ \sin }\left[ {k(a + b)L} \right]}}{c(a + b)L}}} $$

  (68)
  $$ B_{2} = B_{1} { \cos }\,kaL $$

  (69)
  $$ B_{3} = \frac{{ - \left[ {\sum\nolimits_{r = 1}^{n} {g_{r} F_{r} { \sin }\,r\pi c( - 1)^{r} } } \right]\frac{{{ \sin }[k(a + b)L]}}{c(a + b)L}}}{{k\,{ \sin }\,kL - \frac{{{ \sin }\,kcL\,{ \sin }\left[ {k(a + b)L} \right]}}{c(a + b)L}}} $$

  (70)
  $$ C_{1} = \frac{{\sum\nolimits_{r = 1}^{n} {g_{r} F_{r} { \sin }\,r\pi c( - 1)^{r} - B_{3} { \sin }\,kcL} }}{(a + b)L} $$

  (71)
  $$ C_{2} = C_{1} $$

  (72)
  $$ C_{3} = C_{1} \frac{a + b}{c} $$

  (73)
  $$ A_{2} = B_{1} { \sin }\,kaL $$

  (74)
  $$ D_{2} = C_{1} aL $$

  (75)

  • Contact Case CC5

$$ B_{1} = \frac{{\left[ {\sum\nolimits_{r = 1}^{n} {g_{r} F_{r} { \sin }\,r\pi a} } \right]z}}{{ka(b + c)L\left[ {{ \sin }(k(a + b)L + { \cos }\left[ {k(a + b)L} \right]} \right.{ \tan }\,kcL - \frac{{{ \sin }(kaL)z}}{ka(b + c)L}}} $$

(76)

$$ B_{2} = B_{1} { \cos }\,kaL + \frac{{C_{1} }}{k(b + c)} $$

(77)

$$ B_{3} = - B_{1} \frac{{{ \cos }\left[ {k(a + b)L} \right]}}{{{ \cos }kcL}} + \frac{{B_{1} { \sin }\,kaL + \sum\nolimits_{r = 1}^{n} {g_{r} F_{r} { \sin }\,r\pi a} }}{k(b + c)aL}\frac{{{ \cos }\,kbL}}{{{ \cos }\,kcL}} $$

(78)

$$ C_{1} = \frac{{ - B_{1} { \sin }\,kaL - \sum\nolimits_{r = 1}^{n} {g_{r} F_{r} { \sin }\,r\pi a} }}{aL} $$

(79)

$$ C_{2} = - C_{1} \frac{a}{b + c} $$

(80)

$$ C_{3} = - C_{2} $$

(81)

$$ A_{2} = B_{1} { \sin }\,kaL $$

(82)

$$ D_{2} = C_{1} aL $$

(83)

where:

$$ z = { \sin }\,kbL + { \cos }\,kbL\,{ \tan }\,kcL $$

(84)

• Contact Case CC6

  $$ B_{1} = B_{2} = B_{3} = C_{1} = C_{2} = C_{3} = A_{2} = D_{2} = 0 $$

  (85)

• Contact Case CC7

  $$ B_{1} = \frac{{\frac{{{ \sin }\left[ {kcL} \right]}}{caL}\sum\nolimits_{r = 1}^{n} {g_{r} F_{r} { \sin }\,r\pi a} }}{{k\,{ \sin }\,kL - \frac{{{ \sin }\,kaL\,{ \sin }\,kcL}}{caL}}} $$

  (86)
  $$ B_{2} = B_{1} { \cos }\,kaL $$

  (87)
  $$ B_{3} = \frac{{\frac{{{ \sin }\left[ {k(a + b)L} \right]}}{caL}\sum\nolimits_{r = 1}^{n} {g_{r} F_{r} { \sin }\,r\pi a} }}{{k\,{ \sin }\,kL - \frac{{{ \sin }\,kaL\,{ \sin }\,kcL}}{caL}}} $$

  (88)
  $$ C_{1} = \frac{{ - B_{1} { \sin }\,kaL - \sum\nolimits_{r = 1}^{n} {g_{r} F_{r} { \sin }\,r\pi a} }}{aL} $$

  (89)
  $$ C_{2} = C_{1} $$

  (90)
  $$ C_{3} = C_{1} \frac{a + b}{c} $$

  (91)
  $$ A_{2} = B_{1} { \sin }\,kaL $$

  (92)
  $$ D_{2} = C_{1} aL $$

  (93)

• Contact Case CC8

  $$ B_{1} = B_{2} = B_{3} = C_{1} = C_{2} = C_{3} = A_{2} = D_{2} = 0 $$

  (94)

• Contact Case CC9

  $$ B_{1} = B_{2} = B_{3} = C_{1} = C_{2} = C_{3} = A_{2} = D_{2} = 0 $$

  (95)