diaphragm design

DIAPHRAGM DESIGN

Bruce Burr P.E. & Gargi Talati P.E.Burr and Cole Consulting Engineers, Inc.Burr and Cole Consulting Engineers, Inc.

CB& URR

1. Resourcesa.) Steel Deck Institute – Diaphragm Design Manual

b.) Seismic Design for Buildings – TI-809-04 (Corps of Engineers)

c.) Standard Building Code & Commentary – 1999

d.) American Plywood Association Publications

1) Report 138 – Plywood Diaphragms

2) Diaphragms & Shear walls – Design/Construction Guide

3) Plywood Design Specifications

4) Panel Design Specification

e.) NEHRP – Recommended Provisions for Seismic Regulations for New

Buildings & Other Structures, and Commentary – 1997

f.) International Building Code – 2003

g.) SEAOC Seismology Committee

h.) Precast & Prestressed Concrete – PCI Design Handbook

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2. Usual Classification of Diaphragms:Flexible Rigid

a.) Examples: Untopped Precast Concrete Precast Concrete with Topping

Steel Deck Conc. Slab on Steel Deck

Plywood Cast-in-place Concrete

b.) Force Distribution: Tributary Areas Rigidity of Lateral Elements

3. Test for Classification:∆D > 2 * Story Drift ∆D < 2 * Story Drift

∆D < Permissible ∆D < Permissible

a.) Examples: Conc. or Masonry Shear Walls Conc. Slab or Topping with Steel or Wood Deck & Steel Rigid Frames

b.) Test Required: Wood Diaphragms & Shear Walls Conc. Slab & Steel Bracing

Conc. Slab & Conc. Or Masonry Shear walls

Per some resources: ∆D > 2 * Story Drift - Flexible∆D > 0.5 * Story Drift; ∆D < 2 * Story Drift – Semi-Rigid

∆D < 0.5 * Story Drift - Rigid3.

4. How to Play It Safe – Enveloping Analysis a.) Check chords, collectors, attachments for worst case

b.) Diaphragm frequently has the least reserve strength of the lateral system elements; for

instance concrete tilt-up walls with steel deck diaphragm

5. Stiffness of Diaphragm Significantly Greater If:a.) Steel deck welded @ supports @ 6” o.c. ~ five times the stiffness of 12” o.c.

b.) Blocked wood diaphragm > 2 ½ times stiffness of unblocked diaphragm

6. See Code Provisions for Limitations On:a.) Span – Width ratios for diaphragms

b.) Flexible diaphragm limitations for transferring torsion in open ended buildings,

and where masonry or concrete walls cannot withstand the large movements

c.) Limitations on particle board and gypsum board shear walls in higher

seismicity areas

d.) Special attention required at diaphragm chord splices, corners, reentrant corners,

openings, collectors, and connection to walls, bracing, or frames. 4.

Diaphragm Design of Two Story Building

Design Criteria: SBC’ 1999

Peak velocity related acceleration Av =0.18Peak Acceleration Aa =0.16Seismic Hazard Exposure Group ISeismic Performance Category CSoil Profile S = 1.2Basic Structural System Building frame systemSeismic resisting system Reinf. masonry shear wallsResponse modification Factor R = 4.5Deflection amplification Factor Cd = 4Seismic base shear V = Cs * W (Cs = 0.09)Story Height 12’

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Diaphragm Design Forces (N-S Direction)Seismic DL:

Roof (1 ½” wide rib steel deck):Roof seismic DL= 35 psf (100’ * 300’) = 1050k2nd floor (3” NW concrete slab):Floor seismic DL = 85 psf (100’ * 300’) = 2550k

Total DL for seismic design, W = 3600k

Base Shear:

V (total)= 0.09 * W= 0.09 * (3600) = 324k

*Dead load includes contributing walls, partitions and columns.*Assumed floor and wall loads are distributed uniformly.

8.

Diaphragm design forces (N-S Direction)

Shear based on vertical distribution (1607.4.2)Fx= Cvx * V Cvx=

Minimum force= 50% Av * Wi + shear required to be transferred because of offsets or changes in stiffness of seismic resisting elements above and below the diaphragm (1607.3.6.2.7)

Diaphragm shear at each story:V (roof)= 146k >0.5 * 0.18 * 1050 = 95kV (2nd flr)= 178k <0.5 * 0.18 * 2550 = 230k

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Flexible Diaphragm Design

Flexible diaphragm:

Max. lateral deformation of the diaphragm > 2 * story drift

Distribution of story shear based on tributary area

Shear distribution based on direct shear only

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Max. Diaphragm Shear = 51.1 K / 100’ = 0.511 K/ft

Max. Collector Force = (0.511 + 0.219) * 20’ = 15 K

1 ½” Wide Rib 22 Ga. steel deck span 5’-0”, Support fasteners @ 6” o.c., Side Lap fasteners @ 12” o.c.Allow. Diaph. Shear Strength = 0.421 K/ft > 0.511 K/ft / 1.4 = 0.365 K/ft (Working Load = E/ 1.4)

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Max. Chord Force = C=T= M / d = 2685 / 100 = 27 K

Ast = T / (Ф * Fy) = 27 / (0.9 * 36) = 0.83 in2 (3 x 3 x 1/4 Contin. L , As = 1.44 in2 )

*Compression Chord Force to be Resisted By Steel Beams & Continuous Angle.

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* VULCRAFT TABLE


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Calculate diaphragm deflection of roof deck

Moment deflection= 5qL4/384EsI = 0.1”Shear deflection = qL2/8BG’ = 0.48”Total diaphragm deflection = 0.58” >2 * story drift

where:q= diaphragm shear (0.511 K/ft)L= diaphragm span (210’)B= diaphragm depth (100’)Es= Modulus of elasticity (29000 Ksi) I= Moment of Inertia= 2(A)(B/2)2= 7.2 * 106 in4 (conservative)A= Area of perimeter beam= 10 in2

G’= Effective shear modulus (58.4 K/in support fasteners @ 6”o.c.)

*Roof Diaph. can become rigid if story drift increases. For example, momentframe with drift 3/8”, given diaph. will be rigid

*For support fastener layout 36/5, G’ = 16.4 K/in - shear deflection= 1.72” >> 0.48”

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Rigid Diaphragm Design

Rigid diaphragm:

Max. lateral deformation of the diaphragm < 2 * story drift

Distribution of story shear based on rigidity ofseismic resisting system

Shear distribution based on direct shear and torsional shear (based on calculated and accidental torsional moment)

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East Wall: Diaphragm Shear = 100 K / 100’ = 1.0 K/ft

Diaphragm Shear w/o collector = 100 K / 80’ = 1.25 K/ft

Center Wall: Diaphragm Shear = 82 K / (100’-20’) = 1.0 K/ft

Shear Capacity w/o Shear Reinf., Φ Vc = Φ * 2 (sqrt. (fc’)) * (bw * d) = 0.75 (2 * (sqrt. 3000) * 12 * 2.5)= 2.46 K/ft > 1.25 K/ft

3” NW Concrete Slab (9/16” 28 Ga. Steel deck span 2’-0”), support fasteners @ 10” o.c.

Allow. Diaph. Shear Strength = 1.782 k/ft > 1.25 / 1.4 = 0.89 K/ft (Working Load E / 1.4) 16.


Max. Chord Force = C=T= M / d = 5770 / (100’ –1’) = 58 K

Ast = T / (Ф * Fy) = 58 / (0.9 * 60) = 1.10 in2 (4 #5 Contin. As = 1.24 in2 )

*Compression Chord Force to be Resisted By Steel Beam & Concrete Slab

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* VULCRAFT TABLE

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Calculate diaphragm deflection of floor slab

Moment deflection= 5qL4/384EcI = 0.04”Shear deflection = qL2/8BG’ = 0.02”Total diaphragm deflection = 0.06” < 2 * story drift

where:q = diaphragm shear (1.0 K/ft)L = diaphragm span (210’)B = diaphragm depth (100’)Ec = Modulus of elasticity = 33 * w1.5 * sqrt. (fc’) = 3156 ksiI = Moment of Inertia= t (B * 12)3 / 12= 3.6 x 108 in.4t = slab thickness= 2.5” (above form deck)G’ = Effective shear modulus (2444 K/in support fasteners @ 10”o.c.)


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Chord Reinforcement at Parking Garage (Topping Slab Over Precast Double Tee)

*Part of chord reinforcement continuous through columns20.

Wood Diaphragm Design of One Story Building

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Wood Diaphragm Design of One Story Building

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Roof Loads:

Built-up roof & Insulation 10 psfBeams, Joists, & Deck 5Ceiling & Misc. 5Interior Partitions 5

Dead Load 25Live Load 20

Total Load 45 psf

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Walls

Brick 40Studs, Plywood, Gypsum Board 8

Total 48 psf

Glass or Curtain Wall 15 psf

Story Height = 12’

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Design Criteria (1999 SBC)

Seismic CriteriaAa = Av = 0.2Seismic Hazard Exposure Group = ISeismic Performance Category = CSoil Profile Type = S2Basic Structural System = Frame

Light framed walls with shear panelsResponse modification factor = 6.5Deflection amplification factor = 4

Equivalent lateral force procedure

25.

Design Criteria (1999 SBC)Wind Criteria

Velocity = 70 mphVelocity Pressure = 10 psfGCp = 1.2Horizontal Wind Load = 12 psf

Seismic Diaphragm ForcesCs = 2.5 Aa = 2.5 (.2) = .077

R 6.5Minimum Diaphragm Force

= .5AvW = .5(.2)W = .1 W

26.

North-South Diaphragm Design

West Side Dead Load x WxRoof = 90(100)(.025) = 225k, x 45 = 10,125W1 = (7+2)(90)(.048) = 39k, x 45 = 1,750W5 = (7+2)(60)(.048) = 26k, x 30 = 778W6 = (7+2)(30)(.015) = 4k, x 75 = 304

Total 294k 12,957

x1 = 12,957 / 294 = 44.1e1 = (90 / 2) - 44.1 = 0.9 ft (Ignore Torsion)Vmax = .1 (294/2) = 14.7k

Vwind = .012(9)(90/2) = 4.9k (Seismic Controls)

27.

North-South Diaphragm Design

East Side Dead Load x Wx Roof = 210(100)(.025) = 525k, x 105 = 55,125W2 = (7+2)(30)(.048) = 13k, x 15 = 194W3 = (7+2)(60)(.015) = 8k, x 60 = 486W4 = (7+2)(120)(.015) = 16k, x 150 = 2430W7 = (7+2)(150)(.015) = 20k, x 75 = 1519W8 = (7+2)(60)(.015) = 8k, x 180 = 1458

Total 590k 61,212

x2 = 61,212 / 590 = 103.7e2 = (210 / 2) - 103.7 = 1.3 ft (Ignore Torsion)Vmax = .1 (590/2) = 29.5k

Vwind = .012(9)(210/2) = 11.3k (Seismic Controls)

28.

Diaphragm Shears Assuming Uniform Loads

14.729.5

14.729.5

v = 14.7/100 = .15 k/ft v = 29.5/100 = .3 k/ft

.3 k/ft / 1.4 = .21 k/ft (Working Load– E/1.4)

Try unblocked 5/8” PS1 sheathing w/ 10d nails @ 6” oc @ all panel edges and @ 12” oc elsewhere. Allow .215 k/ft. If blocked edges, allow shear = .43 k/ft and ∆ = ~ 40%. Use 5/8” PS1 with blocked edges and same nailing.

29.

Moment to check chord forces

WL/8 = 29.4(90) / 8 = 331 k-ft WL/8 = 59.0(210) / 8 = 1549 k-ft

Chord Force = 1549/100 = 15.5 k

Area of steel required = 15.5 k/ (.9)(36) = .5 sq. in. minimum,

Use a minimum of 2” x 1/4” steel bar continuous @ edge

(or end plate steel beam connection)

30.

Diaphragm Deflection(per APA Research Report #138)

∆ = Σ (bending defl. + shear defl. + nail slip + chord splice slip)= 5vL3 + vL + 0.188 Len + ∆ (∆cx) (ignore last term if steel chord)

8EAb 4Gt 2b= 5 (.3) (210)3 + .3(210) + .188(210)(.006) (dry/dry)

8(29000)(.5)100 4(90)(.319)= 0.12” + 0.55” + .24” = 0.91” ; L/600 = 210(12) /600 = 4.2”

∆ * Cd = .91(4) = 3.6” 2 * h = 2 * (144) = 1.60” 180 180

Brick wall should survive with this deflection.

31.

Shear Wall Deflection(APA Diaphragms & Shear Walls, Design/Construction Guide)

Max shear in intermediate wall = (29.8+14.7) / (2x40) = 0.56 k/ftASD, v = .56/ 1.4 = .40 k/ft, use 7/16” PS1 each sideCapacity = .24(2) = .48 k/ft, with 8d nails @ 6” oc∆sw = 8 vh3 + vh + .75 h en + h da

Eab Gt b= 8 (.56)(12)3 + .56(12) + .75(12)(.0114)(1.2) + h da

1600(2x1.5x5.5)(40) (90)(.298) b(2 - 2x6) (dry fir/pine)

= .0073 + .25” + .12” + hold-down slip (ignore) = .38”

Compare Diaphragm Deflection To Shear Wall Deflection∆D = .91”, ∆SW = .38”, .91/.38 = 2.39, > 2, flexible

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diaphragm design

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