Originally Posted by nealtw
I don't know Bridgeman: The structure would be changed with new bearing poinst for stairs, dormers and whatever, I would want to know where that weight is going, what type of foundation I have and the condition it is in. . . . . . .We have drilled lots of holes in basement floors to determan the size of footing.
OK, nealtw, I'll play along. So you've "drilled lots of holes in basement floors," in order to accomplish exactly what? In case you didn't know, most residential foundation footings extend outside of basement walls as well as inside of them (or do you also drill holes, typically 10' deep, through the walls' backfill to get down to and verify the presence of external footings?). And the practice of drilling random holes in basement floors does little, if anything, to determine the actual load-bearing capability of footings beneath those floors. An accurate determination of concrete strength is required, along with knowing the amount and size of any steel reinforcement present. As a minimum, concrete core samples need to be taken and broken by a testing lab to determine compressive strength, and the entire footing needs to be exposed to enable using a Pachometer or other means to determine the size and location of rebar.
So perhaps you could explain exactly what your basement floor hole-drilling accomplishes, other than considerably weakening those floors while verifying that footings are present.
To explain my earlier comments regarding the small amount of increased loads placed on a typical residential building undergoing an attic remodel, I've crunched some numbers:
For the sake of argument, let's say a typical, yet simple remodel will include a walled area approximately 20' by 20', with 8' ceilings, no internal walls. The total dead load weight of drywall (including fasteners, texturing and paint) and wall studs/top and bottom plates and fasteners for such a room comes out to 3000 pounds (rounded up from an actual number of 2932 lb.), based on 1/2" drywall at 2 lb./s.f., and seasoned Doug fir density of 32 lb./c.f. Dividing that total by the total number of studs for such a room (I used 70, based on 16" spacing and triple-stud corners, doubler at a door), results in slightly less than 43 pounds additional compressive load on each stud. And this amount, less any distribution factors for more wall members below, could be expected to carry on down through the lower level walls and into the foundation. Using the Doug fir, No. 1 stud-grade are good for 1250 psi, resulting in (total of 1250 x 5.25 sq. in.) each stud being capable of resisting a total load of 6563 pounds. The 43 pounds we're adding to each stud having a total working capacity of more than 3 tons of resisting force, doesn't really amount to much. Doing the arithmetic, actually just a bit more than 6/10 of 1% of total stud capacity.
Feel free to correct me if you think I'm wrong, but I think the foregoing numbers are good. And I agree with you that anyone attempting to do a major remodel of any kind, needs to verify what's there as best he/she can, making every effort possible to make sure the existing structure isn't being compromised. Please keep in mind that I am not a design engineer, so I suspect astute designers may question some of my reasoning, having more experience with load-distribution theory.