Diaphragm Design

DIAPHRAGM DESIGN

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

1. Resources a.) Steel Deck Insti Institute tute – Diaph Diaphragm ragm Desig Design n Manual b.) Seismic Seismic Design Design for Buildings Buildings – TI-809TI-809-04 04 (Corps of Engineers) Engineers) c.) Standard Standard Building Building Code Code & Commentary Commentary – 1999 d.) American Plywood Association Publications 1) Report Report 138 – Plyw Plywood ood Diaphra Diaphragms gms 2) Diaphragms Diaphragms & Shear walls walls – Desi Design/Cons gn/Construct truction ion Guide 3) Plywood Design Specifications 4) Panel Design Specification e.) NEHRP – Recomm Recommended ended Provisions Provisions for Seismic Seismic Regulations Regulations for New Buildings Buildi ngs & Other Other Structures, Structures, and and Commentary Commentary – 1997 f.) Internat International ional Buildi Building ng Code – 2003 g.) SEAOC Seismology Committee h.) Preca Precast st & Prestre Prestressed ssed Concre Concrete te – PCI Desig Design n Handbook Handbook

2. Usu Usual al Cla Classi ssific ficati ation on of Dia Diaphr phragm agms: s: Flexible a.) Exampl Examples: es:

Untopped Untopp ed Preca Precast st Concre Concrete te

b.) Force Distribution:

Rigid Precast Preca st Concre Concrete te with Topping

Steel Deck

Conc. Slab on Steel Deck

Plywood

Cast-in-place Concrete

Tributary Areas

Rigidity of Lateral Elements

3. Tes Testt for Cla Classi ssific ficati ation: on:

a.) Examples:

∆D >

2 * Story Drift

∆D <

2 * Story Drift

∆D <

Permissible

∆D <

Permissible

Conc. or Masonry Shear Walls with Steel or Wood Deck

b.) Test Required: Wood Diaphragms & Shear Walls

Conc. Slab or Topping & Steel Rigid Frames Conc. Slab & Steel Bracing Conc. Slab & Conc. Or Masonry Shear walls

Per some resources:

∆D >

2 * Story Story Drift Drift - Flexi Flexible ble ∆D > 0.5 * Story Drift; ∆D < 2 * Story Story Drift Drift – Semi Semi-Rigi -Rigid d ∆

< 0.5 * Story Drift Rigid

4. How to Play It Safe – Envelo Enveloping ping Analy Analysis sis 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. Stiffn Stiffness ess of Diaph Diaphragm ragm Signif Significantl icantly y Great Greater er If: a.) Steel deck welde welded d @ supports supports @ 6” o.c. ~ five five times times the stiffne stiffness ss of 12” o.c. b.) Blocked Blocked wood diaphragm diaphragm > 2 ½ times stiffne stiffness ss of unblocked unblocked diaphragm diaphragm

6. See Code Provisions for Limitations On: a.) Span Span – Width ratios for diaphragm diaphragmss b.) Flexible diaphragm limitations limitations for transferring torsion in open ended buildings, and where masonry or concrete walls cannot withstand the large movements c.) Limitations Limitations on particle particle board and gypsum gypsum board shear walls walls in higher seismicity areas d.) Special attention required at diaphragm chord splices, corners, reentrant corners,

Diaphragm Design of Two Story Building Design Criteria:

SBC’ 1999

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

Diaphragm Design of Two Story Building

Diaphragm Design of Two Story Building

Diaphragm Design Forces (N-S Direction) Seismic DL: Roof (1 ½” wide rib steel deck): Roof seismic DL= 35 psf (100’ * 300’) = 1050k 2nd 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.

Diaphragm design forces (N-S Direction) Shear based on vertical distribution (1607.4.2) Fx= Cvx * V

Cvx=

Minimum force= Minimum force= 50% Av Av * Wi + shear require required d to be transferr transferred ed 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. 5 * 0.18 0.18 * 105 1050 0 = 95k 95k

V (2nd flr)= 178k

<0.5 * 0.18 * 2550 = 230k

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

Flexible Diaphragm Design

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.


Max. Chord Force = C=T= M / d = 2685 / 100 = 27 K Ast = T / (Ф * Fy) = 27 / (0.9 * 36) = 0.83 in 2 (3 x 3 x 1/4 Contin. L , As = 1.44 in2 ) *Compression Chord Force to be Resisted By Steel Beams & Continuous Angle.


* VULCRAFT TABLE

Flexible Diaphragm Design 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, moment frame with with drift 3/8”, given diaph. will be rigid *For support support fastener fastener layout 36/5, 36/5, G’ = 16.4 K/in K/in - shear deflecti deflection= on= 1.72” >> 0.48”

Rigid Diaphragm Design

Rigid diaphragm: Max. lateral deformation of the diaphragm < 2 * story drift Distribution of story shear based on rigidity of seismic resisting system Shear distribution based on direct shear and torsional shear (based on calculated and accidental torsional moment)


East Wall:

Diaphragm Shear = 100 K / 100’

= 1.0 K/ft

Diaphragm Shear w/o collector = 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 support fasteners @ 10” o.c.


Max. Chord Force = C=T= M / d = 5770 / (100’ –1’) = 58 K Ast = T / (Ф * Fy) = 58 / (0.9 * 60) = 1.10 in 2 (4 #5 Contin. As = 1.24 in2 ) *Compression Chord Force to be be Resisted By Steel Beam & Concrete Slab


* VULCRAFT TABLE

Rigid Diaphragm Design 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. (f c’) = 3156 ksi I = Moment of Inertia= t (B * 12)3 / 12= 3.6 x 108 in.4 t = slab thickness= 2.5” (above form deck) G’ = Effective shear modulus (2444 K/in support fasteners @ 10”o.c.)

Chord Reinforcement at Parking Garage (Toppi (To pping ng Slab Slab Over Prec Precast ast Dou Double ble Tee) Tee)

*Part of chord reinforcement continuous through columns

Wood Diaphragm Design of One Story Building

Wood Diaphragm Design of One Story Building

Roof Loads:

Built-up roof & Insulation

10 psf

Beams, Joists, & Deck

5

Ceiling & Misc.

5

Interior Partitions

5

Dead Load

25

Live Load

20

Total Load

45 psf

Walls

Brick

40

Studs, Plywood, Gypsum Board Total Glass or Curtain Wall Story Height = 12’

8 48 psf 15 psf

Design Criteria (1999 SBC) Seismic Criteria Aa = Av = 0.2 Seismic Hazard Exposure Group = I Seismic Performance Category = C Soil Profile Type = S2 Basic Structural System = Frame Light framed walls with shear panels Response modification factor = 6.5 Deflection amplification factor = 4 Equivalent lateral force procedure

Design Criteria (1999 SBC) Wind Criteria Velocity = 70 mph Veloci Vel ocity ty Pre Pressu ssure re = 10 10 psf GCp = 1.2 Horizo Hor izonta ntall Wind Wind Load Load = 12 psf

Seismic Diaphragm Forces Cs = 2.5 Aa = 2.5 2.5 (. (.2) 2) = .07 .077 7 R

6.5

Minimum Diaphragm Force = .5AvW = .5(.2)W = .1 W

North-South Diaphragm Design West Side

Dead Load

x

Wx

Roof = 90(100)(.025) =

225k ,

x 45 =

10,125

W1 = (7+2)(90)(.048) =

39k ,

x 45 =

1,750

W5 = (7+2)(60)(.048) =

26k ,

x 30 =

778

W6 = (7+2)(30)(.015) =

4k ,

x 75 =

304

Total

294k

x1 = 12,957 / 294 = 44.1 e1 = (90 / 2) - 44.1 = 0.9 ft (Ig (Ignor noree Tor Torsio sion) n) Vmax = .1 (294/2) = 14.7k Vwind = .012(9)(90/2) = 4.9k (Seismic Controls)

12,957

North-South Diaphragm Design East Side

Dead Load

Roof = 210(100)(.025) = W2 = (7+2)(30)(.048) = W3 = (7+2)(60)(.015) = W4 = (7+2)(120)(.015) = W7 = (7+2)(150)(.015) = W8 = (7+2)(60)(.015) = Total

525k , 13k , 8k , 16k , 20k , 8k , 590k

x x 105 = x 15 = x 60 = x 150 = x 75 = x 180 =

x2 = 61,212 / 590 = 103.7 e2 = (210 / 2) - 103. 103.7 7 = 1.3 ft (Ig (Ignore nore Tor Torsio sion) n) Vmax = .1 (590/2) = 29.5k Vwind = .012(9)(210/2) = 11.3k (Seismic Controls)

Wx 55,125 194 486 2430 1519 1458 61,212

Diaphragm Shears Assuming Uniform Loads

29.5

14.7 14.7 29.5

v = 14.7/10 14.7/100 0 = .15 k/ft

v = 29.5/10 29.5/100 0 = .3 k/ft

.3 k/ft / 1.4 = .21 k/ft

(Wor (W orki king ng Lo Load ad– – E/ E/1. 1.4) 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.

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)

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.

Shear Wall Deflection (APA Diaphragms & Shear Walls, Design/Construction Guide) Max shear in intermediate wall = (29.8+14.7) / (2x40) = 0.56 k/ft ASD, v = .56/ 1.4 = .40 k/ft, use 7/16” PS1 each side Capacity = .24(2) = .48 k/ft, with 8d nails @ 6” oc ∆sw

= 8 vh3 + vh Eab =

+ .75 h en + h da

Gt 8 (.56)(12)3

b +

.56(12)

1600(2x1.5x5.5)(40)

(90)(.298)

(2 - 2x6)

(dry fir/pine)

= .0073

+ .75(12)(.0114)(1.2) + h da

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

Compare Diaphragm Deflection To Shear Wall Deflection ∆D

= .91”,

∆SW

= .38”, .91/.38 = 2.39, > 2, flexible

b

Diaphragm Design

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