Snow on a roof supported by the Howe truss of the figure can be approximated as a distributed load of 20 lb/ft (measured along the roof). Treat the distributed load as you would the weight of the members; that is, replace the total load on each of the upper members as a vertical force, half applied to the joint at each end of the member. Determine the force in members BC, BG, CG, and GH.
joint equilibrium, method of joints and method of sections.
Method of joints: Sum of vertical forces = 0 and Sum of horizontal forces =0.
The Attempt at a Solution
The only part I'm not sure about is the distributed load reduction. The book shows the figure I'm attaching. I also happen to have (supposedly) the solutions with which I contrasted my solution. The solutions manual reduces the distributed load differently than what I did: I thought that because the load was uniform in "magnitude" the load would be just 20 lb/ft multiplied by the HORIZONTAL length, not the diagonal lenght as they do in the solution (where they use Pythagoras' theorem).
Am I right or wrong? shouldn't the resultant load on members AB, BC, CD and DE be equal to 8ft * 20 lb/ft = 160 lb. And then I can proceed and put half of this onto the joints?
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