Since I have this huge suply of free 8 foot 2x6's I was wondering if any of you knew how to figure the angles for gambrel roofs for differant widths of buildings. Taking a stab in the dark here but I'm guessing I could cover an 18 to 20 foot span using 8 foot stock
Thanks Jon, I've got a general Idea of what to do And if I was energetic enough I coult just lay out the 2X6's on the barn floor and figure out the angles that way but I don't fell like moving all the equipment to get that much floor space to work
I was just sort of aimlessly wandering around the 'net and found these plans for gambrel roof trusses. I recalled your post, so here's the link. There seem to be plans with detailed drawings for almost every span you could imagine. I hope it helps
**Here's a quick layout. Of course all the angles are adjustable to achive various combinations of height vs. span, but I've used 135 degrees as the inside angle for each joint. The figures shown assume 8' framing members AFTER the ends are cut. Just cut both ends of each member to 22.5 degrees. The pitch of the top section is 4'7" in 12".
That's a good idea saving on plywood cuts. Less work that way. Here's how I arrived at the calculations. NOTE: This technique will work for gambrel designs where all angles are equal and all framing members are equal in length. In the figures below, all numbers in [font color="green"]GREEN[/font] are angles.
The "outside" angles (this may not be the right term) of any polygon must add up to 360.
See Fig. 1: 90*4=360.
What really is a gambrel roof except half of a big octagon? See Fig. 2: 360 divided by 8 sides = 45 degrees for the outside angle at each joint. 180 - 45 = 135, the inside angle.
Transferring the 135 to Fig. 3 and using 8 ft as the length of each leg, we can form a triangle by connecting the leg ends and call that "h". Find "h" by using the Law of Cosines = 14.782.
Notice that doing the same for the other side of the gambrel forms two legs of a new triangle. By closing this new triangle across the bottom, we find the SPAN. In Fig. 4, use the Law of Cosines or the Pythagorean Theorem to find 20.905.
Fig. 5 just shows that you cut the end of each board at 22.5 degrees to make a 45 degree turn. And, Fig. 6 is the span converted from decimal to feet & inches. Also a little more trig. yields the overall height, but I'm too tired to draw that out right now.
Quart, I won't be online over the weekend, but there are plenty of others who know a lot more than me on this stuff. Look forward to hearing of your progress.
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