{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

878_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

878_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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 56): M max Pe sec kL 2 d e a sec kL 2 1 b 0.112 in. ; Problem 11.5-2 A steel bar having a square cross section (50 mm 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
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern