878_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

878_Mechanics - 10 3 mm 4 b ± 50 mm L ± 2 m P ± 60 kN e ± 25 mm Eq(11-51 Eq(11-56 M max ± Pe sec kL 2 ± 2.03 kN m d ± e a sec kL 2 ² 1 b ±

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872 CHAPTER 11 Columns Solution 11.5-1 Bar with rectangular cross section ± 1.230 I ± bh 3 12 ± 0.1667 in. 4 kL ± L A P EI P ± 2800 lb e ± 0.5 in. E ± 10 * 10 6 psi b ± 2.0 in. h ± 1.0 in. L ± 30 in. Eq.(11-51): ; ± 1710 lb-in. Eq.(11 ² 56): M max ± Pe sec kL 2 d ± e a sec kL 2 ² 1 b ± 0.112 in. ; Problem 11.5-2 A steel bar having a square cross section (50 mm ³ 50 mm 50 mm ± 50 mm) and length L ± 2.0 m is compressed by axial loads that have a resultant P ± 60 kN acting at the midpoint of one side of the cross section (see figure). Assuming that the modulus of elasticity E is equal to 210 GPa and that the ends of the bar are pinned, calculate the maximum deflection and the maximum bending moment M max . d P = 60 kN Solution 11.5-2 Bar with square cross section ± 1.481 kL ± L A P EI E ± 210 GPa I ± b 4 12 ± 520.8
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Unformatted text preview: * 10 3 mm 4 b ± 50 mm. L ± 2 m. P ± 60 kN e ± 25 mm Eq. (11-51): Eq. (11-56): M max ± Pe sec kL 2 ± 2.03 kN # m ; d ± e a sec kL 2 ² 1 b ± 8.87 mm ; Problem 11.5-3 Determine the bending moment M in the pinned-end column with eccentric axial loads shown in the figure. Then plot the bending-moment diagram for an axial load P ± 0.3 P cr . Note: Express the moment as a function of the distance x from the end of the column, and plot the diagram in nondimensional form with M/Pe as ordinate and x/L as abscissa. Probs.11.5-3, 11.5-4 and 11.5-5 B A P P P P L e e ² v x y M = Pe M = Pe (a) (b) 11Ch11.qxd 9/27/08 3:39 PM Page 872...
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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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