6 v and m diagrams for dam v m f equations f h f 12

Info icon This preview shows pages 6–9. Sign up to view the full content.

View Full Document Right Arrow Icon
6 V and M Diagrams for Dam V M F Equations F h F = 1/2 γ h 2 acting at centroid M = F x lever arm = F x1/3h = 1/6 γ h 3 Typical Problem: f r = M x c/I I = bh t 3 /12 Solve for the thickness h t of the dam
Image of page 6

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
7 CE207 Review Beam Diagrams and Formulas for BRAE433 a) Simple Beam – uniform load (w) Loading Vx = w( L/2 - x) Vmax = wL/2 x L wL Given: w = 300 lb/ft L = 22 ft Shear Mx = wx/2( L - x) Mmax = wL 2 /8 Moment Find: R1,Mmax Solution: R1 = Vmax = wL/2 3,300 lbs Mmax = wL 2 /8 18,150 ft-lbs
Image of page 7
8 CE207 Review b) Simple Beam – concentrated load (P) Loading Vx = P/2 Vmax = P/2 x L P Given: P = 10,000 lb L = 25 ft Shear Mx = Px/2 Mmax = PL/4 Moment Find: R1,Mmax CE207 Review b) Simple Beam – concentrated load (P) L Loading Vx = P/2 Vmax = P/2 Shear Mx = Px/2 Mmax = PL/4 x P Given: P = 10,000 lb L = 25 ft Find: Moment R1,Mmax Solution: R1 = Vmax = P/2 5,000 lbs Mmax = PL/4 62,500 ft-lbs
Image of page 8

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
9 CE207 Review c) Cantilevered Beam – increasing load Loading L Given: γ = 62.4 lb/cuft h = 5 ft Find: Shear Moment Mmax = γ h 3 /6 F,Mmax •F = 1/2 γ h 2 acting at centroid •M = F x lever arm = F x1/3h = 1/6 γ h 3 F h max pressure = γ h h CE207 Review c) Cantilevered Beam – increasing load Loading L Given: γ = 62.4 lb/cuft h = 5 ft Find: Shear Moment Mmax = γ h 3 /6 F h h F, Mmax •F = 1/2 γ h 2 acting at centroid •M = F x lever arm = F x1/3h = 1/6 γ h 3 max pressure = γ h Solution: F = 1/2 x specific weight x h 2 780 lbs Mmax = 1/6 x specific weight x h 3 1300 ft-lbs
Image of page 9
This is the end of the preview. Sign up to access the rest of the 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