Int. J. Engng Ed. Vol. 14, No. 3, p. 204±216, 1998 0949-149X/91 $3.00+0.00 Printed in Great Britain. # 1998 TEMPUS Publications. Interesting Instructional Problems in Column Buckling for the Strength of Materials and Mechanics of Solids Courses* J. NEURINGER and I. ELISHAKOFF Department of Mechanical Engineering, Florida Atlantic University, Boca Raton, FL 33431-0991, USA. E-mail: [email protected] In this study the following problem is addressed: a uniform column is subjected to a compressive load; an additional support is placed to increase the buckling load. The following question is posed: where to place the support location so as to maximize and evaluate the resulting buckling load? It turns out that this question can be effectively dealt with in the standard courses of Strength of Materials, Mechanics of Solids, or Mechanics of Materials, since all the necessary tools needed are presently uniformly taught in these existing courses. Including this interesting case into the curriculum may enhance students' grasp of the subject, sharpen their mind, and trigger an additional interest in the exciting subject of theoretical and applied mechanics. Topics covered in sections 1±3 can be taught in one or two 50-minute lectures, whereas the exposition of the entire material may take between two to three 50-minute lectures, depending on the interest of students. SUMMARY OF EDUCATIONAL ASPECTS report results on optimization of columns, plates OF THE PAPER and shells under buckling conditions [11, 12]. It appeared to the authors, that the motivation of 1. The paper discusses materials for a course in the students can be enhanced, if some model Strength of Materials and Mechanics of Solids. problems can be included on more advanced 2. Students of second-year Mechanical Engineer- topics in the undergraduate courses. This would ing, Ocean Engineering, Civil Engineering and provide students with some `raisins to look for'. As Aerospace Engineering are taught this course. Budiansky and Hutchinson [13] note, `Everyone 3. The mode of presentation is by lecture and is loves a buckling problem'. We propose some run as a regular course. problems the solutions to which hopefully will 4. Hours required to cover the material is 4 to 5 enlarge the number of `buckling lovers,' to include with 2 to 3 revision hours. more undergraduate students. Maybe even every 5. The novel aspects presented in this paper hope- undergraduate textbook should include a section fully help to sharpen the minds of the students on `Some Interesting Problems' providing, gastro- and increase interest in applied mechanics. nomically speaking, the dessert after the main 6. The standard text recommended for the course serving of essential material is taught. We deal is Hibbeler, Mechanics of Materials [2]. here with the problem of locating the intermediate support in a uniform column with ideal boundary conditions so as to maximize the buckling load. The analysis is conducted via the straightforward 1. INTRODUCTION optimization. It turns out that with the knowledge acquired by the undergraduate students in the NATURALLY THERE IS a big gap between standard course, some interesting problems can the exposition of the topic of buckling in the be solved, even those with seemingly un-intuitive undergraduate textbooks devoted to Mechanics solutions. of Solids, Strength of Materials or Mechanics of Note that the beams and columns with intermedi- Materials [1±5] and more specialized monographs, ate supports have been dealt with by Olhoff and suitable for the elective or graduate courses [6±10]. Taylor [14], Rozvany and Mroz [15], Wang, et al. The undergraduate textbooks invariably report [16±19], Liu, et al. [20, 21], and Hou [22] in various buckling of uniform columns with various bound- optimization contexts. The above studies used ary conditions, whereas the advanced texts include Bernoulli-Euler theory. Ari-Gur and Elishakoff numerous results of uniform or non-uniform [23] studied the effect of shear deformation on columns, plates and shells, pertinent to engineering buckling of columns with intermediate support. practice. Among other topics, specialized books Here we deal with the instructional side of this problem for uniform columns with different * Accepted 1 November 1997. boundary conditions. 204 Interesting Instructional Problems in Column Buckling 205 2. UNIFORM COLUMN SIMPLY- The continuity conditions read: SUPPORTED AT BOTH ENDS, WITH AN INTERMEDIATE SUPPORT w1 a w2 a 0 8 w 0 a w 0 a 9 The buckling of a uniform column is governed 1 2 by the following differential equation: 00 00 w1 a w2 a 10 d 4w d2w Equation (9) signifies the continuity of the slopes, EI P 0 1 dx4 dx2 whereas equation (10) indicates the continuity of bending moments. where w x is the transverse displacement, The general solution w1 x in the first region is: x axial coordinate, E modulus of elasticity, w x A sin kx A cos kx A x A 11 I moment of inertia, P compressive loading. 1 1 2 3 4 We divide equation (1) by flexural stiffness EI, and Satisfying the conditions of equation (6) and the denote: first of the conditions in equation (8) we get: P 2 x k 2 w x A sin kx sin ka 12 EI 1 1 a to obtain: The general solution w2 x in the second region is: 4 2 d w 2 d w k 0 3 w2 x B1 sin kx B2 cos kx B3 x B4 13 dx4 dx2 where j's are the constants of integration. After At the cross-section x a, an additional support is satisfaction of the conditions in equation (7) and placed (Fig. 1a). Thus, two regions are created. the second condition, equation (8), equation (13) For the purpose of identification the displacement reduces to: in the first region is denoted by w1 x, whereas the displacement in the second region a < x L, is L x w2 x B1 sin k L x sin k L a denoted by w2 x. Thus, instead of equation (3) we L a have the following two equations: 14 4 2 Equation (9) results in: d w1 2 d w1 k 0; for 0 x < a 4 dx4 dx2 sin ka A k cos ka 4 2 1 a d w2 2 d w2 k 0; for a x L 5 dx4 dx2 sin k L a B1 k cos k L a 0 The boundary conditions read: L a 15 w 0 w 00 0 0 6 1 1 Equation (10) yields: 00 w2 L w2 L 0 7 A1 sin ka B1 sin k L a 0 16 Non-triviality of A1 and B1 yields the transcen- dental equation: sin k L a sin ka k cos k L a L a 1 sin k L a k cos ka sin ka 0 a 17 Introducing non-dimensional quantities: a u ; kL; 18 L and using trigonometric identities we arrive at the following characteristic equation: Fig. 1. (a) Uniform column, simply supported at its ends 1 with an intermediate support. (b) Optimal location of an 2 sin cos cos 2u 1 0 19 intermediate support is in column's middle. u u2 206 J. Neuringer and I. Elishakoff Now we turn to determining a location a aà of We conclude that the best location for the the intermediate support, such that the buckling intermediate support, if the column is simply load will attain a maximum at a aà corresponds supported at both ends is the middle cross section to u aÃ=L uÃ. We first note that one of the (Fig. 1b), with the buckling load given in equation arguments on the left hand side of equation (19), (29). Let us consider a limiting case. denoted by f: f u; u 2 sin 3. BUCKLING LOAD OF A SIMPLY- 1 cos cos 2u 1 20 SUPPORTED COLUMN WITH u u2 INTERMEDIATE SUPPORT'S LOCATION APPROACHING ZERO namely u, is an implicit function of u. We wish to determine the derivative d =du and set it equal Taking into account the expressions for the to zero. We proceed as follows. Differentiating transverse displacement w1 x and w2 x given in equation (19) with respect to u, we get: equations (12) and (14), respectively, we obtain the @ f d @ f following formulas for the slopes at the intermediate 0 21 support location: @ du @u Thus, 0 sin ka d @ f @ f w a A1 k cos ka 30 22 1 a du @u @ 1 On the assumption that @ f =@ differs from zero, w 0 a B k cos k L a sin k L a 2 1 L a we see that the condition: 31 d 0 23 du Now, when a approaches zero (Fig. 2a), we get: is equivalent to: @ f 0 24 0 1 lim w1 a A1 lim k cos ka sin ka @ u a ! 0 a ! 0 a The latter equation yields: 1 A1 k lim cos ka lim sin ka 1 2u a ! 0 a ! 0 a cos cos 2u 1 u u22 A1 k k 0 32 2 sin 2u 1 0 25 u u2 0 0 Since condition (9) requires w1 a w2 a for all a, we thus also require that: We are interested in the smallest non-zero solution for of the coupled set of equations (19) and (25), 0 and call it cr; the corresponding value of u is lim w2 a 0 33 a ! 0 called ucr. Let us examine equation (25). It is satisfied by uà aÃ=L 1=2 independent of the value of . Hence, by putting u uà 1=2 into equation (19) we get: sin cos 2 sin 0 26 2 2 2 Equation (26) yields two equations.