Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations The Obturator on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Live Load Reduction for Bearing Wall & Strip Footing 1

Status
Not open for further replies.

SteveGregory

Structural
Jul 18, 2006
554
I have a 5-story condo with reinforced CMU bearing walls and hollow-core plank floors. The bearing wall is 72' long, bays are 32' wide, floor-floor is 12'. IBC 2009 / ASCE 7-05

1. Can I use the entire length of the wall to figure At? If so, At=72(5x32)=11,520 sf. I used Kll=2. This allows for the maximum reduction down to L=0.4Lo. Or you could argue that the wall is a one-way slab and the width is limited to 1.5 x the floor-floor height. Then At=18(5x32)=2,880 sf. Now, L=0.448Lo. Your interpretation?

2. Even though I can't get a reduction of the snow load on the roof, I used the roof area as part of the tributary area being supported by the wall and footing. Does this make sense?

The strip footing load is getting very large. I think piles and a grade beam would not work for this project for budget reasons.
 
Replies continue below

Recommended for you

1. ASCE 7's commentary doesn't address this that I can see but your use of the 1.5 x span (i.e. 1.5 x wall height) to get the width of the floor participating in the footing load sounds right.

2. No - you don't use the roof area to reduce floor live loads.

 
I do not include roof area in the LL reduction calculation. In my area snow controls the roof framing, and the load combinations themselves take care of the live and snow loads not being present at the same time.

I also wouldn't use the full length of the wall. At most I would only a length equal to how far the wall could clear span without a footing, but would more likely just use a typical bay width
 
See ASCE 7-05 section 4.8.5 that states: "The tributary area, AT ,
for one-way slabs shall not exceed an area defined by the slab
span times a width normal to the span of 1.5 times the slab span."
 
@SteveGregory- This is how we incorporate reduction in LL in the design of the exterior load bearing masonry walls with openings. (typical scenario in condos)
Let us consider masonry pier 2'-8" wide with 4'-0" wide openings on each side. For simplicity assume the openings align on all the floors. Tributary area for masonry pier / floor = 6'-8" x (32' / 2) = 107 sft. AT = 4(floors) x 107 = 428 sft.
KLL for exterior column without cantilever slabs = 4.
AL = KLL x AT = 4 X 428 = 1712 sft.
L = 0.61 Lo.
 
DST,
This is similar to what I have on the exterior walls. There are garage door openings with piers.

However, the interior walls have long solid sections. Do you use KLL=2 [Ai/At similar to a beam] and use full length of the wall for AT or use 1.5 times the flr-flr height span of the wall for the AT length?

Everyone,
This may be a little picky. Roof live load is a separate load from roof snow load. Roof live load is reducible but snow is not. You could argue that the roof area could be used in the total tributary area calculation for AT even though snow controls and you take no reduction on the roof. Your thoughts?
 
I've always treated the roof live load reduction as a separate statistical calc from the floor live load reduction. So the AT of the roof is different from that of the floors.
 
Jae,
I am not calculating the AT for the design of the roof members. I am calculating the AT for the bearing wall and footing that supports all of the floors and the roof.
 
Roof live loads are a different load case than floor live load (Lr instead of just L). The load combinations in ASCE7 do not apply full floor live and full roof live in the same combination. I do not (and would not) mix tributary areas from multiple load cases to determine your LL reduction
 
@SteveGregory - For interior walls we use KLL = 2 - similar to beams. For calculating AT we typically use floor to floor height but not more than 10 feet. This is to accommodate any openings in the future, if required. (we had to accommodate relocated doors in the cellar on few occasions)
But no matter what the calculations show, we always design walls on the lower storys for the following minimum live loads.
Exterior loadbearing walls L = 0.60 Lo
Interior loadbearing walls L = 0.50 Lo
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor