04 tributary

Tributary load Copyright Prof Schierle 2011 1

Tributary Load and Load PathChildhorsepost

rotating platform

supporting gear

Assume 1’ tributary


Load Path andTributary Load• Load path is the path load

travels from where it acts to where it is resisted

• Tributary load is the load acting on a member(needed to design it)

It is convenient to visualize andcompute load on a strip of unit width (1 foot or 1 meter)For example:• 1’ slab, resting on • 1’ wall, resting on• 1’ footing, resting on• 1’ soil


Lateral wind load

Load path: A > B > C• A wind wall • B floor and roof diaphragms • C shear walls

Tributary load:A Wind wall resists wind pressureB Floor/roof diaphragms resist wind wall load

(½ of wall above & ½ of wall below) C Shear walls resist ½ each (2 walls) of

floor and roof diaphragms


Load Path1 Slab / wall

Slab rests on walls

2 Deck / joist / wallDeck rests on joistsJoists rest on walls

3 Slab / beam / wallSlab rests on beamsBeams rest on walls

4 Deck / joist / beam / wallDeck rests on joistsJoists rest on beamsBeams rest on walls

5 Deck / joist / beam / girder / postDeck rests on joistsJoists rest on beamsBeams rest on girdersGirders rest on post (column)All supported by footing


Tributary load

Uniform loadw = 100 psf (pounds per square foot)Post reactionsRa = Rb = Rc = RR = 100 x 12’ x 10’ / 4 = 3000 #R = 3000 # / 1000 R = 3.0 kNote:# = poundk = kip (1 kip = 1000 pounds)

Point loadP = 8kPost reactionsRa = Rb = Rc = RR = 8 / 4 R = 2.0 k


Tributary loadSimple beam on 2 columnsAssume:w = 200 plf (pounds per linear foot)

ReactionsRa = Rb = R = w L/2R = 200 x 30 / 2 = 3000 #R = 3000 #/ 1000 R = 3.0 k

Two simple beams on three columns Assume:w = 2 klfReactionsRa = 2 x 10 / 2 Ra = 10 kRb = 2 x (10+20) / 2 Rb = 30 kRc = 2x20 / 2 Rc = 20 k


Tributary load: deck / joist / beam / columnAssume Uniform load w = 80 psfJoist spacing e = 2’Joist span L1 = 12’Beam spans L2 = 10’

L3 = 20’Find load path and tributary loadLoad path: plywood deck > joist > beam > columnsTributary loads:Uniform joist loadwj = w e = 80 psf x 2’ wj = 160 plfBeam load (assume uniform load due to narrow joist spacing)wb = 80 psf L1/2 = 80 psf x 12’ /2 wb = 480 plfPost reactionsRa = wb L2 / 2 = 480 plf x 10 /2 Ra = 2,400 #Rb = wb (L2+L3)/2 = 480 (10+20) / 2 Rb = 7,200 #Rc = wb L3 / 2 = 480 x 20 / 2 Rc = 4,800 #


Tributary loadThree-story building1 Isometric view2 Exploded visualization3 DimensionsWind wall > diaphragms > shear wallsAssume Wind pressure P = 20 psfShear wall shear (2 walls)Third floorV3 = 20 psf x 100’ x 5’/1000 V3 = 10 kSecond floorV2 = 20 psf x 100’ x 15’/1000 V2 = 30 kFirst floor = base shear VV = 20 psf x 100’ x 25’/1000 V = 50 kNote:Each diaphragm resists wind pressure from half thewall above and below. Lower half of 1st floor resistedby footing; hence shear walls don’t resist lower half.


Tributary load1 Concrete slab > wall

Concrete slab t = 8”, span L = 20’LL = 50 psfDL =120 psf (150 pcf) =170 psf

Slab load on wall (per linear foot of wall)w = 170 psf x 20’/2 w =1700 plf

2 Deck > joist > wallPlywood roof deck2x12 wood joists at 24”, span L = 18’LL = 30 psfDL= 20 psf = 50 psf

Roof load on wall (per linear foot of wall)w = 50 psf x 18’/2 w = 450 plf


Beam

P = 16 k

Tributary load3 Concrete slab / beam / wall

Slab span L = 10’, t = 5” Beam span L = 30’LL = 20 psfDL = 70 psf (including beam DL) = 90 psfBeam load w = 90 psf x10’ / 1000 w = 0.9 klfWall reaction R = 0.9 klf x 30’ / 2 R = 13.5 k

4 Concrete slab on metal deck / joist/ beamDeck span L = 8’Joist span L = 20’Beam span L = 40’LL = 40 psfDL = 60 psf (including joist and beam DL) = 100 psfJoist load w = 100 psf x 8’ / 1000 w = 0.8 klf

Beam point loads P = 0.8 klf x 20’ P = 16 kBeam reaction R = 4 P /2 = 4 x 16 k / 2 R = 32 kUniform wall load w = 100 psf x 4’ / 1000 w = 0.4 klf

Note:Wall requires pilaster to support beams


P=35k P=35k

Girder

Beam

7 x P=10k

Tributary loadConcrete slab on metal deck / joist/ beam / girderAssume:Spans: Deck L = 5’Joist L = 20’Beam L = 40’Girder L = 60’Loads:LL = 50 psfDL = 50 psf (combined framing and deck load) = 100 psfUniform joist loadw = 100 psf x 5’/1000 w = 0.5 klfBeam point loads (from joists) P = 0.5 klf x 20’ P = 10 kGirder point loads (from beams) P = 7 x 10 k/2 P = 35Girder uniform loadw = 100 psf x 2.5’ / 1000 w = 0.25 klf Column reaction R=(100 psf/1000)x40’x60’/4 R = 60 k


One-story concrete structureLoads:Roofing 3 psfCeiling 2 psf10” concrete slab 125 psf (150 pcf x 10” / 12”)DL 130 psfLL 20 psf 150 psfLx = 30’Lxc = 34Ly = 25’

Columns, 12”x12” (t=12”, t/2 = 6” = 0.5’)Column reactions A, B, C, DRa = 150 psf (30+34)/2 (25) Ra = 120,000 #Rb = 150 (30+34)/2 (25/2+0.5) Rb = 62,400 #Rc = 150 (30/2+0.5) (25) Rc = 58,125 #Rd = 150 (30/2+0.5) (25/2+0.5) Rd = 30,225 #


Level 2 column reactions w = 150 psfRa = 150 psf (30+34)/2 (25) = 150 (800) Ra = 120,000 #Rb = 150 (30+34)/2 (25/2+1) = 150 (432) Rb = 64,800 #Rc = 150 (30/2+1) (25) = 150 (400) Rc = 60,000 #Rd = 150 (30/2+1) (25/2+1) = 150 (216) Rd = 32,400 #Level 1 reactions w=150+200 w = 350 psfRa = 350 (800) Ra = 280,000 #Rb = 350 (432) Rb = 151,200 #Rc = 350 (400) Rc = 140,000 #Rd = 350 (216) Rd = 75,600 #Level 0 reactions w=150+200+200 w = 550 psfRa = 550 (800) Ra = 440,000 #Rb = 550 (432) Rb = 237,600 #Rc = 550 (400) Rc = 220,000 #Rd = 550 (216) Rd = 118,800 #

Three-story concrete structureRoof DL 130 psfRoof LL 20 psfRoof 150 psfFloor DL 150 psf (includes columns, etc.)Floor LL 50 psf (Office LL)Floor 200 psfColumns, 2’x2’ (t =2’, t/2 =1’)


happy end

04 tributary

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