How Do You Calculate Distributed Load on a Roof Truss Due to Snow?

AI Thread Summary
The discussion focuses on calculating the distributed load of snow on a roof truss, specifically addressing the confusion over whether to use horizontal or diagonal lengths for the load calculation. The distributed load is given as 20 lb/ft along the slope of the roof, and participants emphasize that this measurement should be applied directly as stated. One participant argues for using horizontal lengths, while others clarify that the load must be treated as acting along the slope, leading to a higher resultant load. The method of joints and equilibrium equations are referenced as essential tools for solving the problem. Ultimately, the consensus is that the load must be calculated based on its specified orientation along the roof slope.
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Homework Statement


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.

Homework Equations


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 length 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?
 

Attachments

  • SNOW DISTRIBUTED LOAD.jpg
    SNOW DISTRIBUTED LOAD.jpg
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arestes said:
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 length as they do
It clearly states
arestes said:
measured along the roof
 
haruspex said:
It clearly states
Hi, yeah but looking at the direction of the load, it is vertical, therefore, I can imagine that the "area" of the distributed force would be computed by using a horizontal span. I can also imagine that I can transmit each differential element dF downward, leveling the height and getting a horizontal rectangle.
Why can't I do this?
 
arestes said:
I can imagine that the "area" of the distributed force would be computed by using a horizontal span.
Imagine what you like, but it tells you, very clearly, that the given weight is 20lb per foot, measuring along the slope of the roof. As you go up the slope, the weight of snow on each foot you cover is 20lb.
If you prefer, you can use that to calculate the total weight and then calculate the weight per horizontal foot, but it will be more than 20lb.
 
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