BRAE433 - CE207 Review

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

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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

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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
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

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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
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