876_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

876_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: 11Ch11.qxd 9/27/08 870 3:39 PM Page 870 CHAPTER 11 Columns Solution 11.4-10 E L 2.0 m Q a 72 GPa 200 kN n 1.0 m CRITICAL LOAD 3.0 d Pcr P#n 200 mm Pcr FREE-BODY DIAGRAM OF THE BEAM I Pcr 1200 kN p2EI 4L2 4PcrL2 I p2E 27.019 * 106 mm4 MOMENT OF INERTIA p cd 4 64 I 4 d t min ©Mc P 0P 2Q A (d d4 2t)4 d I 64 p 2 t min 10.0 mm ; 400 kN Problem 11.4-11 The frame ABC consists of two members AB and BC that are rigidly connected at joint B, as shown in part (a) of the figure. The frame has pin supports at A and C. A concentrated load P acts at joint B, thereby placing member AB in direct compression. To assist in determining the buckling load for member AB, we represents it as a pinned-end column, as shown in part (b) of the figure. At the top of the column, a rotational spring of stiffness b R represents the restraining action of the horizontal beam BC on the column (note that the horizontal beam provides resistance to rotation of joint B when the column buckles). Also, consider only bending effects in the analysis (i.e., disregard the effects of axial deformations). (a) By solving the differential equation of the deflection curve, derive the following buckling equation for this column: b RL (kL cot kL EI 1) k 2L2 0 in which L is the length of the column and EI is its flexural rigidity. (b) For the particular case when member BC is identical to member AB, the rotation stiffness b R equals 3EI/L (see Case 7, Table G-2, Appendix G). For this special case, determine the critical load Pcr. x P P bR C B B L L EI y A (a) A (b) ...
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This note was uploaded on 12/22/2011 for the course MEEG 310 taught by Professor Staff during the Fall '11 term at University of Delaware.

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