Recent content by ErLupo

Ah. Yes. This makes sense. That was my mistake. When I use the changed distances to the moment arms I get Dy = 1.65L and Ay = 1.34L. When the truss is cut through Q and the vertical forces are summed I got: Ay - BEsin(θ) - L = 0 BE = (1.34L - L)/(sin(26.1°)) BE = .787L Which matches the...

Yes, FE = BC - 2BEx, correct? So that would make FE = 12.29 But if I were to use a right triangle with hypotenuse BE wouldn't I just add BEx which would be 14.14? Even if I only use the actual length of FE (12.29) and a combination of law of sines and cosines to get the unknowns I still end...

Wouldn't that just be; Let x = vertical distance. 12/sin30 = y/sin60 y = 20.78. That would make the hypotenuse = 24 (Pythagorean Theorem). Then to solve for the hypotenuse BF of the small 15 degree triangle: BF/sin15 = 24/sin120 BF = 7.17 Then to solve for the two legs joined at a right...

Homework Statement Determine the force in member BE of the loaded truss. See the attached picture. Homework Equations Sum of the Moments = 0 Sum of the Forces = 0 The Attempt at a Solution [/B] Sum moments about A to get: -12L - 56L +40Dy = 0 Dy=1.7L Sum the forces in the y...