04 tributary

7/18/2019 04 Tributary

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Tributary load Prof Schierle 1

Tributary Load and Load Path

Child horse post

rotating platform

supporting gear

Assume 1’ tributary




Load Path and

Tributary 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 and

compute 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 pressure

B 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 / wall

Deck rests on joists

Joists rest on walls

3 Slab / beam / wallSlab rests on beams

Beams rest on walls

4 Deck / joist / beam / wallDeck rests on joists

Joists rest on beams

Beams rest on walls

5 Deck / joist / beam / girder / postDeck rests on joists

Joists rest on beams

Beams rest on girders

Girders rest on post (column) All supported by footing




Tributary load

Uniform loadw = 100 psf (pounds per square foot)

Post reactions

Ra = Rb = Rc = R

R = 100 x 12’ x 10’ / 4 = 3000 #R = 3000 # / 1000 R = 3.0 k

Note:

# = pound

k = kip (1 kip = 1000 pounds)

Point load

P = 8kPost reactions

Ra = Rb = Rc = R

R = 8 / 4 R = 2.0 k


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Tributary load

Simple beam on 2 columns

Assume:w = 200 plf (pounds per linear foot)

Reactions

Ra = 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 klf

ReactionsRa = 2 x 10 / 2 Ra = 10 k

Rb = 2 x (10+20) / 2 Rb = 30 k

Rc = 2x20 / 2 Rc = 20 k



Tributary load: deck / joist / beam / column

Assume

Uniform load w = 80 psf

Joist spacing e = 2’Joist span L1 = 12’

Beam spans L2 = 10’

L3 = 20’

Find load path and tributary load

Load path: plywood deck > joist > beam > columns

Tributary loads:

Uniform joist loadw j = w e = 80 psf x 2’ w j = 160 plf

Beam load

(assume uniform load due to narrow joist spacing)

wb = 80 psf L1/2 = 80 psf x 12’ /2 wb = 480 plf

Post reactions

Ra = wbL2 / 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 load

Three-story building

1 Isometric view2 Exploded visualization

3 Dimensions

Wind wall > diaphragms > shear walls

AssumeWind pressure P = 20 psf

Shear wall shear (2 walls)

Third floor

V3 = 20 psf x 100’ x 5’/1000 V3 = 10 kSecond floor

V2 = 20 psf x 100’ x 15’/1000 V2 = 30 k

First floor = base shear V

V = 20 psf x 100’ x 25’/1000 V = 50 kNote:

Each diaphragm resists wind pressure from half the

wall above and below. Lower half of 1st floor resisted

by footing; hence shear walls don’t resist lower half.



Tributary load

1 Concrete slab > wall

Concrete slab t = 8”, span L = 20’

LL = 50 psf DL =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 > wall

Plywood roof deck2x12 wood joists at 24”, span L = 18’

LL =30 psf

DL= 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 load

3 Concrete slab / beam / wall

Slab span L = 10’, t = 5”

Beam span L = 30’

LL = 20 psf

DL = 70 psf (including beam DL)

=90 psf

Beam load w = 90 psf x10’ / 1000 w = 0.9 klf

Wall reaction R = 0.9 klf x 30’ / 2 R = 13.5 k

4 Concrete slab on metal deck / joist/ beam

Deck span L = 8’

Joist span L = 20’

Beam span L = 40’LL = 40 psf

DL = 60 psf (including joist and beam DL)

=100 psf

Joist load w = 100 psf x 8’ / 1000 w = 0.8 klf Beam point loads P = 0.8 klf x 20’ P = 16 k

Beam reaction R = 4 P /2 = 4 x 16 k / 2 R = 32 k

Uniform 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 load

Concrete slab on metal deck / joist/ beam / girder

Assume:

Spans:Deck L = 5’

Joist L = 20’

Beam L = 40’

Girder L = 60’

Loads:LL = 50 psf

DL = 50 psf (combined framing and deck load)

=100 psf

Uniform joist loadw = 100 psf x 5’/1000 w = 0.5 klf

Beam point loads (from joists)

P = 0.5 klf x 20’ P = 10 k

Girder point loads (from beams)P = 7 x 10 k/2 P = 35

Girder uniform load

w = 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 structure

Loads:

Roofing 3 psf

Ceiling 2 psf

10” concrete slab 125 psf (150 pcf x 10” / 12”)

DL 130 psf

LL 20 psf

150 psf

Lx = 30’

Lxc = 34

Ly = 25’

Columns, 12”x12” (t=12”, t/2 = 6” = 0.5’)

Column reactions A, B, C, D

Ra = 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 psf

Ra = 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 psf

Ra = 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 psf

Ra = 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 structure

Roof DL 130 psf

Roof LL 20 psf

Roof 150 psf

Floor DL 150 psf (includes columns, etc.)

Floor LL 50 psf (Office LL)

Floor 200 psf

Columns, 2’x2’ (t =2’, t/2 =1’)



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04 tributary

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