869_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

869_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Columns with Other Support Conditions The problems for Section 11.4 are to be solved using the assumptions of ideal, slender, prismatic, linearly elastic columns (Euler buckling). Buckling occurs in the plane of the figure unless stated otherwise. Problem 11.4-1 An aluminum pipe column ( E ± 10,400 ksi) with length L ± 10.0 ft has inside and outside diameters and , respectively (see figure). The column is supported only at the ends and may buckle in any direction. Calculate the critical load P cr for the following end conditions: (1) pinned-pinned, (2) fixed-free, (3) fixed-pinned, and (4) fixed-fixed. d 2 ± 6.0 in. d 1 ± 5.0 in. SECTION 11.4 Columns with Other Support Conditions 863 Solution 11.4-1 Aluminum pipe column (1) P INNED - PINNED ± 235 k ; P cr ± p 2 EI L 2 ± p 2 (10,400 ksi) (32.94 in. 4 ) (120 in.) 2 L ± 10.0 ft ± 120 in.
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Unformatted text preview: I ± p 64 ( d 4 2 ² d 4 1 ) ± 32.94 in. 4 d 2 ± 6.0 in. d 1 ± 5.0 in. E ± 10,400 ksi (2) F IXED-FREE (3) F IXED-PINNED (4) F IXED-FIXED P cr ± 4 p 2 EI L 2 ± 939 k ; P cr ± 2.046 p 2 EI L 2 ± 480 k ; P cr ± p 2 EI 4 L 2 ± 58.7 k ; (b) F OR q ± 200 lb ft d 1 d 2 Probs.11.4-1 and 11.4-2 Problem 11.4-2 Solve the preceding problem for a steel pipe column ( E ± 210 GPa) with length L ± 1.2 m, inner diameter , and outer diameter . d 2 ± 40 mm d 1 ± 36 mm (c) F ROM (1) NUMERICALLY SOLVE FOR S WHEN 2 SOLUTIONS ARE POSSIBLE : ; s ± 0.264 ft and s ± 2.42 ft : Q ± P cr ; l b – min ± 38.5 in. 4 From (1) I b – min ± ² 1 24 sA r 8 QsL 1 ² qL 1 3 + 3 sqL 1 2 ² 3 s 2 qL 1 + s 3 q QH 11Ch11.qxd 9/27/08 3:39 PM Page 863...
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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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