Module 8 French

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Chapter 12 - Diaphragms

The Reorganized ACI 318-14 Code


New Chapter

Ch. 12 - Diaphragms *



Diaphragms (Chapter 12)• New addition• Why?

– Previously not addressed for SDC A, B, and C

– Seismic design of diaphragm is required for all buildings in SDC B through F

– Special seismic requirements for diaphragms in SDC D, E, and F (ACI 18.12)

– Guidance for engineers Courtesy of NIST GCR 10-917-4 document


Diaphragms

1. Stiff thin structural member that transfers inertial forces to, or between, vertical lateral force resisting members.

2. Ties a structure together.3. Ensures continuous load path within a

building4. Essential for lateral force-resisting system5. Wall and column stability



Scope of diaphragms

• Diaphragm can be:– Cast-in-place slabs– Cast-in-place topping on

precast elements– Precast elements with end strips– Interconnected precast w/o

topping

12.1 Scope12.2 General12.3 Design limits12.4 Required strength12.5 Design strength12.6 Reinforcement limits12.7 Reinforcement detailing


Diaphragm force considerations

• In-plane forces• Transfer forces• Connection forces• Bracing forces• Out-of-plane forces




Definitions12.1 Scope12.2 General12.3 Design limits12.4 Required strength12.5 Design strength12.6 Reinforcement limits12.7 Reinforcement detailing

Collector

Fpx

Tension Chord, T

Compression Chord, C

L

B

a b a


Materials

• Concrete → Ch. 19• Reinforcing steel → Ch. 20




Design limits

• Minimum diaphragm thickness, h

– Meet:– one-way slab thickness requirements (7.3.1.1) – two-way slab thickness requirements (8.3.1.1))



Required strength

• Loads → Ch. 5• Additional load combinations for

diaphragms and collectors are (ASCE7-10 §12.4.2.3):1.2 + 0.2 + + + 0.20.9 − 0.2 +




Required strength• Analysis → Ch. 6

– Envelope analysis (Beam method)

– Finite element– Strut-and-tie → Ch. 23– ASCE 7-10

• Equivalent Lateral Load §12.8• Modal Response Spectrum

§12.9• Diaphragm §12.10


CC’

C

T


Rigid or Flexible

12.4.2.3 Any set of reasonable and consistent assumptions for diaphragm stiffness shall be permitted.

– Rigid vs flexible diaphragms• Seismic: L/B ≤ 3 (ASCE 7-10 § 12.3.1.2)• Wind: L/B ≤ 2 (ASCE 7-10 § 27.5.4)

B

L




0R=1/3 R=1/3 R=1/3

ωℓ=1

Rigid

V

M

Rigid Diaphragm

• Distribute horizontal forces to lateral force resisting system (LFRS) in direct proportion to relative stiffness

• Diaphragm deflection is irrelevant compared to LFRS

• Vertical members deflect the same• Capable to transferring torsional and

shear forces



Flexible Diaphragm

• Lateral force distribution to LFRS is independent of their relative stiffness

• Lateral loads are distributed to LFRS in proportion to tributary area

• Not capable of transferring torsional forces

• Diaphragm deflection is considerably larger than that of LFRS




Strength calculation

• Equivalent Beam Model:Treats rigid diaphragm as a horizontal beam spanning between idealized rigid supports:

= ∑ ±=


ey

ex

Center of mass

k1

k2

k3

k4k5

Center of rigidity

Diaphragm boundary

Lateral Load

Vertical element and reaction force

Fpx


Design strength

• Diaphragm, collectors, and their connections must satisfy:– φSn ≥ U– φ→ 21.2– Design strength requirements

depend on diaphragm model used• Beam model → 12.5.2 to 12.5.4• Strut and tie → 23.3• Finite element → Ch. 22




Design

• LFRS (walls and moment frame) perform in nonlinear range

• Diaphragm, collectors, and their connections to the vertical LFRS should have the strength to remain elastic during an earthquake.



Design strength• Diaphragm beam assumption

– Moment and axial → 12.5.2• Moment strength → 22.3• Axial Strength → 22.4• Tension reinforcement within h/4 of tension edge




Diaphragm

fpx=γfx

RRRL


Collector

Fpx

Tension Chord, T


L

B


Chord calculation example

• Assume:– Fpx = 460 kip– L = 180 ft– B = 80 ft– a = 25 ft– lw = 25 ft


lw

Collector

Fpx

Tension Chord, T


L

B

a b a



Example

= 260kip= 200kip=0= 0

= 1.8kip/ft= 3.3kip/ftFpx

L

B

q1

q2

RL RR

a



Example

Shear diagram (kip)

Moment diagram (ft-kip)

-48

152

-180

80

590

4637

932




Chord−Example§12.5.1.3= = = 4637ft − kip(0.95)(80ft) = 61kip

Moment diagram (ft-kip)

Design tension strength of reinforcement:= ; = 61kip(0.9)(60ksi) = 1.1in.Provide:Four No. 5 bars within h/4 of the slab edge, or space at 6 in. o.c.

590

4637

932



Chord−Example

• Post-tensioned:– Effective stress: = 160psi

Tension stress in diaphragm:= = ( )( , )( .) ( ) = 52psi>

Therefore, reinforcement is not required.




Design strength

• Diaphragm beam assumption– Shear → 12.5.3

• φ = 0.75 or as required in 21.2.4 (special moment frames and walls)

• CIP - = (2 + ) (12.5.3.3)where is the reinforcement perpendicular to diaphragm flexural reinforcement (parallel to shear force)

• CIP - ≤ 8 (12.5.3.4)• Transfer of shear from collectors to walls

– Shear friction may apply → 22.9



Shear

Vu@E = 180 kip

Check diaphragm shear strength:= (2 + ) ignoring reinforcement, ρt = 0:= 713kip > Vu@E = 180 kip

Shear diagram (kip)

-48

152

-180

-80




t=tpc+ttop

t=ttop

Shear (12.5.3.5)

• For precast elements with CIP topping (a) and (b) must be satisfied :

(a) as for cast-in-placehas a thickness:

Composite topping: =+Noncomposite topping: =

(b) cannot exceed shear friction at connections



Shear (12.5.3.6)

• Diaphragms with:– Interconnected precast elements without concrete

topping, and for– Precast elements with end strips (CIP topping slab or

beam)– Satisfy (a), (b), or both:

(a) Nominal strength of grouted joints ≤ 80 psi; shear friction reinforcement in addition to moment and axial forces

(b) Mechanical connectors crossing joints between precast elements.




Shear (12.5.3.7)

• For any diaphragm, where shear is transferred to a collector or vertical element, (a) or (b) must apply:

(a)Satisfy shear friction of 22.9(b)Mechanical connector or dowels must

consider uplift and rotation of vertical elements or LFRS


Shear (12.5.3.6) (Continue)



Design strengthDiaphragm beam assumption• Collectors → 12.5.4• Continuous across diaphragm

depth • Tension or compression members → 22.4• Extend along vertical element the greater

of:– ℓd in tension– Length required to transfer collector

force through shear friction



Collector

• Collectors within the wall width:– Tension and compression

forces are transferred into the wall at wall boundary

(b) Collector tension and compression forces

(a) Collector reinforcement

Shear

Tension Compression

Collector reinforcement

Shear wall




Collector Example• Axial force distribution in collector:= @@ = 260kip80ft = 3.25kip/ft

@ = @@ = 260kip25ft = 10.4kip/ft


Collector

Fpx

Tension Chord, T


260 kip


Collector Example

Reinforcement required to resist Collector Tension:= = 89.4(0.9)(60 ) = 1.66 .Provide four No. 6 bars in addition to the reinforcement required for the gravity load. Bars can fit within the wall width.

Collector

Fpx

Tension Chord, T


3.25

kip

/ft 10.4

kip

/ft

3.25

kip

/ft

3.25

kip

/ft

7.15

kip

/ft

-89.

4 ki

p

89.4

kip




Collector

If collector is wider than the wall width, seismic load is resisted:1- Part by bars directly in line with shear wall.2- Balance by bars placed eccentric to the wall and uses slab shear-friction capacity at the wall-to-slab interface to transfer seismic forces to the wall.



Collector

R12.5.4 states:Where a collector width extends into the slab, the collector width on each side of the vertical element should not exceed approximately one-half the contact length between the collector and the vertical element.

lwlw/2


beff



Collector

ec

M

Ve

• Moment due to eccentricity between collector and wall.

• Resolve through:– Shear forces ⊥ to

collector ( )– Bending in plane of

diaphragm= + − ℎ


eten T2

h

Slab width to resist compression

Slab width to resist tension

Ve

Ve

Mu

Cc


Collector

• = ( ℎ)• =As2 is supplemental reinforcement to resist Mu

Refer to “Design of Concrete Slabs as Seismic Collectors,” SEAOC Seismology and Structural Standards Committee, May 2005, 15 pp.


h

Dowels

jh

As2



Opening in Diaphragm

b1 = 24 ftb2 = 14 ftc = 22 ft


q2

L

B

a b a

b1

b2

c

2c

q1



L

B

a b a

T1

1.8 kip/ft 3.3 kip/ft

2.45 kip/ft 2.65 kip/ft

Ta

Tb

Cb

Ca

C1





• Primary chord forces T1 = |C1|= 61 kip • Secondary chord forces (Ta, Ca) and (Tb, Cb)• qoW = 2.45 kip/ft and qoE = 2.65 kip/ft• Mass of segment above opening = ½ mass of

segment below opening • The segment above opening will resist one-third of

the total diaphragm load over this segment.• Segment below opening will resist 2/3 of total

diaphragm load over this segment.



Opening in Diaphragm• Segment above opening:

. kip/ft =0.82 kip/ft. kip/ft =0.88 kip/ft

40.5 ft-kip 41 ft-kip

21 ft-kip

The secondary chord force near midspan:= = 0.95 = 1.0





( ) . kip/ft =1.63 kip/ft( ) . kip/ft =1.77 kip/ft

81 ft-kip 82 ft-kip

41 ft-kip

The secondary chord force near midspan:= = 0.95(2 ) = 0.98

12.1 Scope12.2 General12.3 Design limits12.4 Required strength12.5 Design strength12.6 Reinforcement limits12.7 Reinforcement detailingSegment below opening:



Total chord force:= + = 61kip + 1kip = 62kipThe required tension reinforcement along slab edge:= = 1.15in.Use Four No.5 chord bars at midheight of slab.





Tensile chord forces at opening corners:= . =1.96 kip

Tension reinforcement:= = 0.04in.Use one No.5 chord bar along the slab edge adjacent to the opening. Provide one No.5 along each side of slab along opening.



Reinforcement limits

• Minimum reinforcement– One-way slabs → 7.6 – Two-way slabs → 8.6– Shrinkage and temperature → 24.4– Slab reinforcement designed to resist other load

effects is not allowed to also resist in-plane shear• Temperature and shrinkage steel may be used for in-

plane shear• Steel to resist in-plane shear is not required if φVc ≥ Vu




Reinforcement detailing

• Detail for one and two-way slabs• Maximum bar spacing

– Lesser of 5h,18 in.• Develop tensile or compressive force on

each side of section• Extend tension reinforcement ℓd past the

point it is no longer required– Exception at edges and expansion joints



Chapter 18 – Earthquake resistant structures

• Beams• Columns

• Beams• Columns• Two-way slabs• Precast structural

walls

• Beams• Columns• Beam-column joints• Moment frames

using precast concrete

• Diaphragms • Members not part

of seismic force resisting system

• Structural walls• Foundations

Ordinary Systems (Min. for SDC B)

Intermediate Systems (Min. for SDC C)

Special Systems (SDC D, E, F)



Design forces → 18.12.2

• Check governing building code for additional diaphragm force requirements

• Design of collectors for overstrength factor Ωo may be required– Avoid brittle behavior

• Elastic diaphragm behavior is very desirable


Diaphragms → 18.12

• Design forces Chapter 5– When load combinations with overstrength factor, Ωo are required (ASCE 7-10 §12.4.3.2):1.2 + 0.2 + Ω + + 0.20.9 − 0.2 + ΩWhere Ωo = 2.5

• Reduced φ for shear in 21.2.4 may apply



Axially loaded elements

• Elements primarily carrying axial load that transfer shear around diaphragm discontinuities must meet collector requirements of 18.12.7.5 and 18.12.7.6


Collector elements w/ compression > 0.2f’c

• Amount of transverse reinforcement per Table 18.12.7.5

• Transverse reinforcement detailing per special moment frame columns

• Continue trans. reinforcement until compression < 0.15f’c

• If overstrength factors are used for vertical elements– 0.2f’c → 0.5f’c– 0.15f’c → 0.4f’c

A long collector with confinement reinforcement-source NISTGCR10-917-4



Splices and anchorage

• At collector splices and anchorage zones:– Bar spacing ≥ larger of 3db and

1.5 in.– Cover ≥ larger of 2.5db and 2 in.– Transverse reinforcement, Av ≥

larger of 0.75 (bws/fyt) and 50 bws/fyt

Collector connection to shear wall boundary zone-Source NISTGCR10-917-4

Module 8 French

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