Converting Roof Live Loads Down a Slope Member

[Charred]

I am trying to teach a new engineer the process of converting roof LL down a sloped member for bi-axial analysis/design. I know the code states that LL acts vertically along the horizontal projection of the member which means that this load has to be converted [cos(theta)^2] to act along the length of the member and perpendicular to the member. However he is having trouble grasping this concept so I began to sketch the process and prove it through trigonometry.

I can visually represent the conversion from the length of the member to act perpendicular through trig, but I seem to be making a mistake or not thinking about this correctly when converting the first step (from the horizontal projection to the length of the member). Does anybody know a good reference for this topic or can provide any assistance?

 

[KootK]

In the past, I've made the argument that "the total load for both the projection and the diagonal need to be the same, right?". That seems to appeal the intuition of junior engineers. And, mathematically, there's no way other than the right way to make that pan out.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[mike20793]

I think this concept is easiest learned by the classic rafter tie design example. Link

 

[Charred]

I've read that article before and it does help with understanding the loading, but the FBD's are incorrect and does not show what I am looking for. Basically, I am trying to visually prove how the LL is converted twice to act perpendicular to the member along the entire length of that member.

 

[ajh1]

You've got a vertical load to start. It may be easier to visualize as a point load rather than uniform, so convert a 1' horizontal length to a total weight. Basic vector analysis will give you a component parallel to the roof plane and a second perpendicular to the roof plane. Each of the two components could be adjusted back to a uniform load along the slope if desired by dividing the horizontal projection by the slope length to get the adjusted pounds per lineal foot acting along the length of the slope.