45 Degree balcony supports
Here in Vermont there are no building inspectors except for commercial building.
I am pretty handy but I am unsure about my current project. I need to build a 4x4 landing, 8' above ground for a side door exit. Due to the grounds slope, retaining wall etc. there is no way to place posts and get a proper frost free foundation. I decided to have two 1/4" steel plates built which will allow me to attach supports off of the concrete foundation. I have built two 2x6 bolted together supports that are angled off of the concrete portion of the house wall at a 45 degree angle up to doubled 2x8 deck rim joists. The remaining 2x8 joists are spaced at 12" oc. The ledger board is lag bolted to the houses rim joist. I also have to attach a set of stairs however they will come off of the side of the landing (Balcony). which will put additional side forces on the angled 2x6' supports. I can however attach one of the stair stringers to the house. The other will have no mid point support. Suggestions? Will the 45 degree supports off of the side wall provide enough support. I am bolting , not nailing everything together?
Thanks for any sound advice.
as long as the plate on the wall can support almost double the weight of the deck in a vertical load it would be good there but you also need to be certain the deck can't be pushed away from the building cause now the beams will give horizontal pressure where there is normally only vertical load. the steps might actually help you in this case to give better ancor to side of building.
Looking at it from an engineering standpoint, the design of your balcony is fine. Even if (4) 240 lb. guys were standing on it with a beer in each hand, it would equate to a design live load of 60 lbs/sq.ft. which meets most building code requirements for balconys. Regardless of the remote possibility of it ever seeing these loads, it has to be designed to sustain them and even then it doesn’t mean it’s at the point of collapsing. The 2 x 8’s over 4 ft. spans are not even in question.
The compression loads on the diagonals are calculated as follows:
4 x 4 area = 16 sq.ft. 60 lbs./sq.ft. = a total design live load of 960 lbs. The load is equally supported by the house and the diagonals, so the load at the front end = 480 lbs. The vertical load taken up by the diagonals is then 240 lbs. each. At a 45 degree angle going back to the foundation, the compression on each diagonal works out to 340 lbs.
I’ll explain why I went through all of the above. The connection method of the diagonals to the joists is the critical factor here. If the diagonals go up to the underside of the joists and are held in place by say a bolted plate, the 340 lb. load is of no concern as it is being taken up completely by the member. However, if the diagonals go on to the side of the joists and are bolted through, it needs a closer look.
Here’s why. That 340 lb. load is now being transferred to the diagonals via the shank of the bolt and is parallel to the grain. I can only assume that you are using pressure treated spruce or a similar grade of lumber. The shear value parallel to the grain is 135 lbs./sq.in. If you used (1) half inch diameter bolt, the bearing area of the bolt through the (2) 2 x 6’s is 2.35 sq.in. Now, going back to the 340 lb. load, bearing on 2.35 sq.in. results in 144 lbs./sq.in which exceeds the shear vale of 135 lbs./sq.in. It means the diagonals could split.
Using (2) ˝” bolts, the bearing load becomes 72 lbs./sq.in….well below the 135 value.
Using (1) 5/8” bolt, the bearing load becomes 115 lbs./sq.in…again well below 135.
To calculate the bearing area of the bolt:
Bolt dia. * 3.142 / 2 * 3 (for the thickness of the (2) 2 x 6’s)
I don’t mean to blind you with science or baffle you with bull****, but this is handy to know and is often overlooked as a critical factor in wood construction. Make sure that you lag to the house with at least (4) ˝” – 5/8” lag bolts. As chain088, pointed out, there is also a pull out load created.
NICE VERY NICE !
Thanks for the info !
Wish I could get answers like that on every question.
Andy, I like your numbers but down here we have something called a human safety factor and the minimum is 3. Now your guys weigh in at 720 lbs. and those little 12 oz. beers are now over a qt. You may want to revise your numbers. I generally work to an HSF of 5 to avoid lawsuits.
The mechanical properties and design values for wood that are established and set out by the USDA, US Forestry Dept and a host of other agencies, already have a safety factor and are published as “allowable” loads in the design of wood structures. These values are set for various species and grades of lumber and are used by the designers. As long as the design of a structure shows that the values for bending, shear etc., were not exceeded, you don’t have to fear a lawsuit. Well, not one that you wouldn’t win that is. Again, I’m looking at this from an engineering perspective and as a contractor yourself, you will be well aware that on a sizable project, over designing could cost you the job.
I don’t mean to under rate the importance of safety, but if you design and build any structure within these guidelines, you can rest with a clear conscience knowing that you haven’t compromised public safety. Then again, I don’t know if your local codes call for a factor of 3 over and above.
However, I don’t disagree that on a balcony or deck for example, if one 5/8” bolt would do the job, what the h*** , use two. I over designed my own deck for the additional few dollars.
The only instance where bigger is better is in the matter of the beers. Yes sir, quarts over pints anytime. http://www.diychatroom.com/images/smilies/biggrin.gif
|All times are GMT -5. The time now is 07:27 PM.|
User Alert System provided by Advanced User Tagging (Pro) - vBulletin Mods & Addons Copyright © 2015 DragonByte Technologies Ltd.