DIAPHRAGM ACTION WITH HOLLOW CORE SLABS

CHAPTER 4 DIAPHRAGM ACTION WITH HOLLOW CORE SLABS

satisfy Section 7.13 in precast concrete structures. When hollow core slabs are used as floor or

4.1 General Information

For large panel bearing wall structures, minimum roof decks to support vertical loads, the natural

forces are specified to provide ties throughout the extension is to use the slabs as a diaphragm to re-

structure. For other types of precast structures, sist and transmit lateral loads. Lateral loads will

only general detailing philosophies are specified.

be applied to building structures in the form of lat- In either case, the fundamental requirement is to eral earth pressures, wind loads or seismic loads.

provide a complete load path from any point in a The function of a diaphragm is to receive these

structure to the foundation. Clearly, a diaphragm loads from the building elements to which they

is a significant element in this load path. A tie sys- have been applied and transmit the loads to the lat-

tem that satisfies the strength and force transfer eral-resisting elements which carry the lateral

demands on a diaphragm will generally satisfy the loads to the foundation. The design issues in a

detailing requirements for structural integrity. hollow core diaphragm are the design of connec- tions to get loads into the diaphragm, the strength

4.2 Design Loads

and ductility of the slab system to transmit these Lateral loads imposed on hollow core dia- loads to the lateral-resisting elements and the de-

phragms can include lateral earth pressures, wind sign of the connections required to unload the lat-

loads or seismic loads. Lateral earth pressures eral forces from the diaphragm to the lateral-res-

will be established by the characteristics of the isting elements.

soil being retained. Wind and seismic loads will Clear communication is required between the

be dictated by the applicable building code for the building designer and the hollow core slab suppli-

structure. Soil and wind loads are forces actually er when the hollow core system is to be used as a

applied to the structure. Seismic forces are gener- diaphragm. Some elements of the diaphragm de-

ated from within the structure as inertial forces sign may be delegated to the hollow core slab sup-

due to lateral displacement from ground motions. plier. However, only the building designer is in

While soil and wind loads can be safely treated as the position to know all the parameters involved in

static loads, seismic loads must be considered as generating the applied lateral loads. Because of

dynamic loads. In all cases, the same elements many design issues, only the building designer

will comprise a complete diaphragm, but the duc- can determine the location and relative stiffnesses

tility demands on a seismic resistant system are of the lateral-resisting elements. These parame-

significantly more important. ters dictate the distribution of forces in the dia-

The balance of the discussion in this chapter phragm. If any design responsibility will be dele-

will be concerned with lateral loads from wind gated to the hollow core supplier, the location and

and seismic. This is not intended to slight the im- magnitude of the lateral loads applied to the dia-

portance of considering unbalanced soil pressures phragm and the location and magnitude of forces

which can commonly be a significant consider- to be transmitted to lateral-resisting elements

ation in many projects using hollow core slabs. must be specified. Where hollow core slabs must

The basic principles of hollow core diaphragms connect to other building materials, or where de-

which will be discussed are equally applicable to mands on connections go beyond simple strength

lateral soil pressures.

demands, the connection details should be shown There are many documents which cover design in the contract documents.

for wind and seismic loads. The references used An additional consideration in detailing dia-

for this chapter are the 1994 UBC code 32 and the phragms is the need for structural integrity. ACI

1996 BOCA code 33 . For wind load, both codes Section 16.5 provides minimum requirements to

are similar in that a basic wind speed is selected are similar in that a basic wind speed is selected

The magnitude of F px need not exceed an importance factor is selected based on the oc-

0.75ZIw px but shall not be less than 0.35 ZIw px . cupancy of the building, modifying factors are de-

Many other requirements are included in the UBC termined for the geometry of the building and the

code which are not restated in this summary. design positive and negative wind pressures are

The BOCA code prescribes a seismic design calculated.

procedure. A very important note is that the For seismic loads, the two codes take different

BOCA provisions result in forces that are already approaches. The UBC code allows an equivalent

factored and are intended to be used with ultimate static load approach for many building types. For

strength design methods with no additional load others, where certain heights or irregularities are

factors. The base shear is calculated as: present, a dynamic lateral force procedure is re-

(Eq. 4.2.3.) quired. The static force procedure allows design

V=C s W

for a base shear of:

where

V = ZIC W

C s = coefficient related to peak velocity-related R w

(Eq. 4.2.1)

acceleration, soil profile, structural system where

type and building fundamental period Z = seismic zone factor

W = total dead load plus other applicable loads

I = importance factor The base shear is distributed over the height of

C = factor dependent on site and structure fun- the building in proportion to the distribution of the damental period

building mass with consideration of the building period. A minimum eccentricity of 5% of the per-

R w = coefficient dependent on structural system pendicular building dimension is also required by type

BOCA when distributing forces to the lateral-re- W = total dead load plus other applicable loads

sisting elements. For Seismic Performance Cate- This base shear is then distributed over the

gories B and greater, each floor or roof diaphragm height of the structure in proportion to the dis-

shall be designed for a minimum load equal to tribution of weights over the height. Additionally,

50% of the effective peak velocity-related accel-

a minimum eccentricity of 5% of the building di- eration times the weight attributable to the level mension perpendicular to the direction being con-

under consideration. The peak velocity-related sidered shall be included when determining the

acceleration is determined by the project location. distribution of forces to the lateral-resisting ele-

Again, there are many provisions in the BOCA ments when the diaphragm is not flexible. Specif-

code which are not covered in this summary. ic to diaphragms, for Zones 2, 3, and 4, the UBC

In light of the performance of some diaphragms requires that a floor or roof diaphragm resist a

in recent earthquakes, the seismic demand on dia- force equal to:

phragms is an area of new focus. Preliminary in-

dications are that diaphragms should remain elas-

F t +Σ F i tic during a seismic event to ensure that post-elas-

F px =

tic behavior can be achieved in the Σ w i

lateral-resisting elements. By designing a dia-

phragm to remain elastic, several things are ac- where

i=x

complished. Diaphragm flexibility, discussed in

F px = force applied to diaphragm at level under Section 4.3 will be less significant. The ductility consideration

requirements for connection details will be of less

F = additional portion of base shear applied at concern. The horizontal distribution of forces to

top level lateral-resisting elements can be maintained. The building code provisions summarized

F i = portion of base shear applied at level i above are based on achieving post-elastic perfor- w i = portion of W at level i

mance. To keep a diaphragm compatible with

Fig. 4.3.1 Diaphragm bending moments

Flexible diaphram on

Rigid diaphram on

Rigid supports

flexible supports

(a)

(b)

post-elastic performance in the lateral-resisting to be considered as a flexible diaphragm. Analy- system system, an analysis can be done to evaluate

sis considering flexible diaphragms is much more the total potential post-elastic capacity of the lat-

complex than for rigid diaphragms and should be eral-resisting elements. Providing a diaphragm

considered in light of the project complexity and with strength beyond this capacity will achieve

seismicity for the project location. For most low compatibility, but will involve significant analy-

and mid-rise structures in low seismic risk areas, sis. Alternatively, the diaphragm design forces

an assumption of a rigid diaphragm will be rea- prescribed by the building codes can be increased

sonable.

by a factor of 2R/5 to keep the diaphragm elastic The difference in behavior of flexible and rigid and minimize required analysis. Whether build-

diaphragms is illustrated in Figure 4.3.1. In (a), ing code provisions are based on service or fac-

the flexible diaphragm with rigid supports be- tored load levels, use of 2R/5 loads will result in

haves as a continuous beam. Shears and moments factored loads for design.

in the diaphragm are a function of the plan geome- try. In (b), the deflections of the flexible supports

must be the same because of the rigid diaphragm. Once the lateral forces to be applied to the dia-

4.3 Distribution of Lateral Forces

The diaphragm shears and moments will be a phragm have been determined, the next problem

function of the relative stiffnesses of the supports. is to determine the distribution of those lateral

The differences between (a) and (b) can be consid- forces to the lateral-resisting elements which will

erable. Actual behavior will fall between the two carry the forces to the foundation. This problem is

cases tending toward one or the other as a function usually structurally indeterminate which means

of the diaphragm stiffness.

that deformation compatibilities must be consid- In seismic areas, the topic of diaphragm flexi- ered for establishing equilibrium. The stiffnesses

bility has become a more significant issue. UBC to be considered are those of the diaphragm and

requires consideration of the diaphragm flexibil- the lateral-resisting elements. Concrete dia-

ity for the horizontal distribution of forces. A phragms are normally considered to be rigid when

flexible diaphragm is defined by the UBC as one compared to the lateral-resisting elements. De-

having a maximum lateral deformation more than pending on the type and magnitude of lateral

twice the average story drift for the level under forces applied, a hollow core diaphragm may need

consideration. It may be inferred from UBC Sec- consideration. It may be inferred from UBC Sec-

Fig. 4.4.1 Tie forces in bearing wall buildings

ply in seismic zones 2, 3, and 4. The BOCA code simply states that the horizontal distribution of forces consider the relative stiffnesses of the later- al-resisting system and the diaphragm. This pro-

vision would apply to Seismic Performance Cate-

gories B and greater. By code then, diaphragm

flexibility need not be considered when designing

for wind or for seismic loads in Zones 0 and 1 un- T 3 T 1

der the UBC or Seismic Performance Category A TT

under BOCA. 1

When diaphragm flexibility must be consid- T 2 ered, a cracked moment of inertia calculation is suggested in Reference 34 and a Virendeel truss

T 2 model in suggested in Reference 35. Since the

analysis of a structure with a flexible diaphragm is dependent on so many factors beyond the dia- phragm itself, such analysis is beyond the scope of

For seismic loading, it is preferable to use con- this manual.

ventional reinforcing steel for these types of ties to limit the elongations and deformations. When structural integrity requirements control in non-

seismic areas, untensioned prestressing strands As noted in the introduction to this chapter, the

4.4 Structural Integrity

may be used to satisfy the strength requirements. ACI code requires consideration of structural in-

tegrity for all precast concrete structures. While proper detailing for lateral loads will satisfy the 4.5 Elements of a Diaphragm

Figure 4.5.1 illustrates the various elements complete load path philosophy of structural integ- which comprise a complete diaphragm. The fol- rity, there are some minimum provisions in ACI lowing definitions will be used to describe the var- Section 16.5 which must be met. With specific re-

ious elements:

gard to diaphragms, provisions to be aware of in- Boundary Element :Edge member around the pe- clude: rimeter of a diaphragm or

1. For buildings other than large panel bearing the perimeter of an opening wall buildings, the connection to the dia-

in a diaphragm which ties phragm of members being laterally braced by

the diaphragm together. The the diaphragm shall have a minimum nominal

boundary element may func- tensile strength of 300 lb per lin ft (4.4KN/m).

tion as a chord or a drag strut.

2. For large panel bearing wall structures, a sum- Collector : Elements which transfer mary of the tie forces is given in Figure 4.4.1

shear from the diaphragm to and are required to have the following mini-

a lateral-resisting element. mum nominal strengths:

Chord : Tension or compression ele- ment creating a flange for

T 1 = nominal strength of 1500 lb per lin ft the diaphragm to develop (21.9 KN/m) of floor or roof span

flexural integrity in the dia- T 2 = nominal strength of 16,000 lb (71 KN)

phragm. Drag Strut : Element used to “drag” lat- T 3 = nominal strength of 1500 lb per lin ft

eral loads into the lateral-re- (21.9 KN/m) of wall

sisting elements and to dis- These minimum strengths shall not control if

tribute shears over a greater the actual forces in the diaphragm are greater.

length of the diaphragm length of the diaphragm

Drag struts are not required for structural integrity Longitudinal joint : Joint oriented parallel to the

as long as the diaphragm is connected directly to slab span.

the lateral-resisting elements. Drag struts simply Transverse joint : Joint oriented perpendicular

spread out shears that might otherwise be highly to the slab span.

localized. Under the UBC code, it is implied that To satisfy structural integrity, all diaphragms

drag struts are required elements in Zones 2, 3, should have boundary elements of some type.

and 4. The BOCA code is silent on the use of drag The boundary elements are essential to ensure that

struts, but it can be implied that they are required

a diaphragm will have the strength to transfer lat- for Seismic Performance Categories B and great- eral loads to the lateral-resisting system. As a

er.

chord, tension reinforcement is placed in the When a bonded structural topping is used with boundary element to allow the diaphragm to act as

a hollow core slab diaphragm, these elements can

be provided directly by reinforcement in the top- forcement can also provide shear friction steel for

a deep horizontal beam or tied arch. This rein-

ping. When no topping is provided, these ele- shear transfer along the longitudinal joints.

ments are developed as grouted or concrete ele- Collectors are required in all diaphragms to

ments external to the hollow core slabs. As a sim- transfer forces from the diaphragm to the lateral-

ple example, Figure 4.5.2 depicts two common resisting elements. Such connectors are also re-

boundary conditions. In (a), the boundary rein- quired for structural integrity to provide a com-

forcement is placed in a masonry bond beam and plete load path for lateral forces to the foundation.

the collector reinforcement is placed in the key- Collectors may also function to get forces into a

ways between slabs. In (b), the boundary rein- diaphragm.

forcement is placed in a grouted or concrete filled Drag struts act to engage a longer length of dia-

space at the end of the slabs. The collector rein- phragm web for transferring diaphragm shears

forcement is again placed in the keyways between into the lateral-resisting elements. A drag strut is

slabs. The primary difference between the details parallel to the applied load, receives load from the

is that the boundary reinforcement in (a) is eccen- diaphragm and transfers load to the lateral-resist-

tric from the diaphragm web while it is concentric

Fig. 4.5.1 Diaphragm elements

Lateral - Resisting Element

Boundary Element

Drag Strut

Lateral Resisting Element

Transverse Joint

Collector

Lateral - Resisting

Load

Boundary Element

Longitudinal

Element

(chord)

Joint

Fig. 4.5.2 Boundary elements

(a)

(b)

in (b). The concentric boundary element will ex-

Fig. 4.6.1 Shear friction steel in butt joint

hibit better performance in a seismic situation and should be used in Zones 3 and 4 under the UBC or Seismic Performance Categories C, D and E un- der the BOCA code.

4.6 Diaphragm Strength

The diaphragm must have the strength to trans- fer imposed lateral loads from the point of ap- plication to the point of resistance. The dia- phragm spans between lateral-resisting elements as a deep beam or tied arch. Shears and tensions will develop and must be resisted in the dia- phragm to have a complete system.

When the grout strength is exceeded or ductile The grouted keyways between slabs do have

4.6.1 Longitudinal Joints

behavior is required, shear friction principles may capacity to transfer longitudinal shear from one

be used to design reinforcement to be placed per- slab to the next. Using a shear stress of 80 psi 36 pendicular to the longitudinal joints. This rein-

(0.55 MPa), the useable ultimate strength for lon- forcement may be placed in the transverse joints at gitudinal shear is:

the slab ends rather than being distributed along the length of the joints. Placed as shown in Figure

φ V n = φ(0.08)h n ℓ

(Eq. 4.6.1)

4.6.1, the area of steel is calculated as: where

A vf =V u

(Eq. 4.6.2)

ℓ = length of joint under consideration (in)

Ôf y μ

h n = net height of grout key (in)

where

V u = factored applied shear

Fig. 4.6.2 Alternate longitudinal shear

Fig. 4.6.3 Collector detail

connections

Bar Spacing by Design

Intermittent slab cut-outs

Reinforcement across

Welded connection

grout keyway (a)

(b)

µ Boundary = 1.0 for shear parallel to longitudinal joints

Reinforcement

= 1.4 for shear parallel to transverse joints where concrete can flow into cores

φ = 0.85 While the detail shown in Figure 4.6.1 is the

most economical means of providing a mechani- cal connection across the longitudinal joints, al-

Fig. 4.6.4 Potential effects of rigid lap

ternate connections are available which may be

connection

desirable in certain circumstances. Figure 4.6.2(a) shows reinforcing steel placed across the Vertically rigid

connection

longitudinal joint and grouted into the cores. This detail might be considered when the amount of re- inforcement required in the transverse joints is great enough to cause congestion. Figure 4.6.2(b)

Potential joint

shows weld anchors in the slabs and a loose plate

cracking and

welded across the longitudinal joint. Use of this grinding detail should be carefully coordinated with the

hollow core slab supplier to ensure that proper an- chorage of the weld plates in the slabs can be ac- complished.

deflection to occur without distress at the connec- Where the diaphragm must unload shear into a

tion. Figure 4.6.4 shows potential damage at the lateral-resisting element, boundary element or in-

first interior longitudinal joint when a vertically terior drag strut, a condition similar to the longitu-

rigid connection is used. The potential for distress dinal joint exists. For longitudinal shear, again

is dependent on the slab span and the real applied shear friction can be used to design reinforcement

loads. Short, lightly loaded spans may experience as the collector to cross potential crack planes and

no problems.

transfer the shear. Figure 4.6.3 depicts an exam- The effect of different vertical stiffnesses may ple of such a collector detail. While drag struts

be accounted for by:

and boundary elements may have a vertical stiff- ness similar to the slab deck, the lateral-resisting

1. Determining that distress will not affect the elements will usually have a significantly higher

strength or performance of the system, vertical stiffness. The collectors connecting di-

rectly to the lateral-resisting elements will tend to

2. Locating vertically rigid connections near the

be rigid vertically. While strength and toughness slab supports where vertical movement is at such collectors is certainly important, it is

minimized, or

equally important to consider every day perfor- mance of the structure. At rigid vertical elements,

3. Providing allowance for vertical movement in it may be desirable to allow slab camber growth or

the connection detail.

V =M h u (Eq. 4.6.5) The transverse joints serve many functions. As

4.6.2 Transverse Joints

jh

described in Section 4.6.1, reinforcement in the In the first case, a unit shear is calculated and shear transverse joints may provide the shear friction re-

friction reinforcement is distributed according to inforcement for shear in the longitudinal joints.

the shear diagram. In the second case, the total The transverse joint may also have to act as a drag

shear is calculated as the tension or compression strut with axial tension or compression to carry di-

of the internal couple. In this case, shear friction aphragm loads to the lateral-resisting elements. A

reinforcement is uniformly distributed over the transverse joint may also be the chord member

length between zero moment and maximum mo- where flexural tension is resisted. Finally, an inte-

ment. It is suggested that the shear friction rein- rior transverse joint disrupts the web of the hori-

forcement be distributed according to the shear zontal beam where horizontal shear would have to

diagram in UBC zones 3 and 4 and BOCA Seis-

be transferred to maintain the composite depth of mic Performance Categories C, D and E to mini- the diaphragm.

mize the force redistribution required with a uni- The design of shear friction reinforcement for

form spacing.

longitudinal joint shear is covered in Section Because of the orientation of the joints and the

4.6.1. Drag strut reinforcement is calculated sim- loading directions considered, the reinforcement ply as:

in the transverse joint discussed above is not all additive. Typically, the chord tension and longitu-

A =T s u

Ôf y dinal joint shear will be concurrent. The drag strut tension will typically occur with loads applied in

(Eq. 4.6.3)

Chord tension is resisted by reinforcement to pro-

the perpendicular direction.

vide flexural strength to the diaphragm. It is sug- gested that the effective depth of the reinforce-

4.7 Collectors

ment from the compression side of the diaphragm Collectors function as connections to transfer

be limited to 0.8 times the depth of the diaphragm. forces into diaphragms and from diaphragms to Hence, the chord reinforcement is calculated as:

boundary elements, drag struts or lateral-resisting M

elements. The preceding discussion has indicated

(Eq. 4.6.4)

Ô0.8hf y that reinforcing bars may be used as collectors us- ing shear friction design procedures. As shear

where friction reinforcement, the steel is used in tension to resist a shear force. In detailing the steel, a

h = depth of the diaphragm crack plane is defined and the bars must be an-

φ = 0.9 chored for full strength on each side of the crack Because diaphragms tend to act as tied arches

plane. For anchorage at a transverse boundary rather than beams, tension in the chord reinforce-

element, the bars may be grouted into the keyways ment does not go to zero at the ends of the dia-

or into slab cores where the top of the core is cut phragm. The chord reinforcement must be an-

away. Concrete is then used to fill the cores for the chored at the ends of the diaphragm where a stan-

length of the bar embedment. Based on a review dard hook at the corner will suffice. For

of the literature, it is not clear when anchorage of horizontal shear in the web of the diaphragm, a

collector bars in keyways is sufficient and when shear parallel to the transverse joint is developed.

the collector bars should be placed in slab cores. Shear friction reinforcement perpendicular to the

There is a concern that as the boundary element transverse joint and embedded in the slab key-

and keyway crack, anchorage for a collector bar in ways can be used to reinforce for this shear. The

a keyway may be lost. Deformations and revers- applied shear can be calculated as:

ible loading in a seismic event would suggest that anchoring collector bars in slab cores would be

h =V preferable in more intense seismic areas. In keep-

I ing with code philosophy, it is suggested that bars

or

be anchored in slab cores in UBC zones 3 and 4

Fig. 4.9.1 Example Problem

and BOCA Seismic Performance Categories C design forces. It is suggested that a topping be and greater.

considered in high seismic zones in buildings with In non-seismic and low seismic design situa-

plan irregularities or large diaphragm span to tions, the collectors need not be reinforcing bars.

depth ratios.

Particularly for direct connections to lateral-re- Untopped hollow core diaphragms are sug- sisting elements, welded and bolted connections

gested when the diaphragm force system is will suffice for the collector connections when

straightforward and the in-plane diaphragm they are compatible with the slab system used.

deflections are acceptable. An example at the end of this chapter illustrates a procedure for deter-

mining diaphragm deflections. In high seismic When a composite structural topping is pro-

4.8 Topped vs. Untopped Diaphragms

areas, local codes may limit the use of untopped, vided, it should have a minimum thickness of 2 to

hollow core diaphragms.

2 1 / 2 in (50-65 mm). The topping can then be de- signed as the diaphragm without consideration of the hollow core slabs. When the topping provides

4.9 Design Example

the strength and stiffness for the diaphragm but Given the building plan shown in Figure 4.9.1, the connections are made in the hollow core slabs,

design and detail the untopped hollow core dia- shear stresses will be present at the interface of the

phragm assuming:

topping and the hollow core slabs. These stresses will generally be well distributed throughout the

a. wind design per UBC

interface, but may be more highly localized near

b. Zone 2A seismic per UBC. the connections. As discussed in Chapter 2, hori- zontal shear stresses should be kept below a nomi-

Building data

nal strength of 80 psi (0.55 MPa).

6 stories

The primary benefits of a composite structural

14 ft floor to floor

topping are to increase stiffness and to allow easi-

8 in hollow core floors wt = 53.5 psf er continuous ties in plans with irregular shapes or

partitions & mechanical wt = 20 psf large openings. However, in seismic areas, the

precast framing system wt = 32 psf additional topping weight increases the seismic

exterior wall system (avg.) wt = 35 psf

Solutions: M u = 1.3(1739) = 2261 ft-k

a. Seismic Zone 0; Basic wind speed 80 mph

• Chord Forces:

Use Exposure C Using the perimeter beams as chords:

Design wind pressure = P = C e C q q s I w M u

where

Ô0.8h

C e = 1.53

C q = +0.8, -- 0.5 = 39.3k

q s = 16.4 Connect beams through columns for this force

I w = 1.0 plus forces due to volume change and gravity P = 1.53(0.8)(16.4)(1.0) = 20.1

loads. (Fig. 4.9.2 Det. C) = 1.53(0.5)(16.4)(1.0) = 12.5

The chord must continue through the center

32.6 psf wall.

Wind to diaphragm = w = 14(0.0326) = 0.46k/ft

A = s u (φ was included in T u ) • Consider load applied parallel to the slabs

Total V = 200(0.46) = 92k

Assuming a rigid diaphragm, the shear dis-

= 0.66 in 2

tribution to the walls based on their flexural stiff- ness is:

Use 2 - #6 (Fig. 4.9.2 Det. F) • Connect diaphragm web to chords

30 ft walls: V = 40k

V uh =M u

20 ft wall: V = 12k

jh

The diaphragm equilibrium is:

j ≅ 0.8

V uh = 2261

0.46 k/ft

0.8(80) = 35.3k

Distribute over length from zero moment to maximum moment

40 6 V uh = 35.3 = 0.41k/ft

V 87ft

6 40 Additionally, this connection must resist the negative wind pressure from the exterior wall

system.

w u = 1.3(0.0125)(14)

= 0.23k/ft Use 300 lb/ft for structural integrity

1739 ft-k

The factored design forces are then:

(Fig. 4.9.2 Det. A) The same forces must be resisted at the trans-

V u30 = 1.3(40) = 52k verse joints. Use shear friction for the shear

V u20 = 1.3(6) = 7.8k with bars placed in the keyways perpendicular V u20 = 1.3(6) = 7.8k with bars placed in the keyways perpendicular

A = 1.02 in vf 2 (from above) or

A s = 3(0.3) + 0.9(60) 0.85(60)(1.4)

6 = 0.11 in 2 does not control

= 0.034 in 2 /keyway Use 4 - #5 located near slab ends

Use #3 at every 2nd keyway

(Fig. 4.9.2 Det. D)

(Fig. 4.9.2 Det. B) Alternatively, mechanical connections of slab

• Longitudinal shear to wall could be used to transfer the same forces.

The maximum longitudinal joint shear is at the • Shear at center 20 ft wall:

first slab joint from the 30 ft shear wall. Since con- nections will be made directly from the center bay

With the rigid diaphragm assumption: to the shear wall, only the center bay joint length

should be considered.

V u = 7.8k on each side of wall

φ V n = φ(0.08)h n ℓ

= 0.15 in 2

= 0.85(0.08)(8 -- 2)(20 x 12) Use 2 - #4 located near slab ends or use

= 97.9k

mechanical connections

With concerns for shrinkage cracking in joints,

(Fig. 4.9.2 Det. E)

transverse shear friction reinforcement can be • Consider load applied perpendicular to the provided in the transverse joints at each end of the

slabs

center bay. Total V = 80(0.46) = 36.8k

A vf u =V Ôf y μ

Distribution to walls is:

= 1.02 in 2 / 2 transverse joints The diaphragm equilibrium is:

= 0.51 in 2 per joint

0.46 k/ft

Use 1 - #7 in transverse joint (Fig. 4.9.2 Det. B)

18.4 Shear connection to 30 ft wall k

Additionally, negative wind pressure must be

resisted across this joint, but would not be con-

current with shear. Structural integrity ties will control for this case.

T u = (0.3)(20) = 6k for bay

368 ft-k

The factored design forces are then: #3 at every 2nd keyway will be adequate

V u = 1.3(18.4) = 23.9k

(Fig. 4.9.2 Det. B)

M u = 1.3(368) = 478 ft-k

b. Seismic Zone 2A

The building weight attributable to each floor • Chord force:

is:

M u w i = 80(200)(0.0535 + 0.020 + 0.032) T u = Ô0.8h

A = 3.3 = 0.06 in s 2

= 11772k

• Base shear

The #3 bars across the transverse joints will be adequate for the chord force. (Fig. 4.9.2 Det. B)

V = ZIC W

Longitudinal shear

= 3.0k will not control

V = 0.15(1.0)(2.75) (11772)

• Shear connection to walls

= 607k

Using shear friction reinforcement

• Vertical Distribution

V u = 23.9k/30 ft wall

F t = 0.07TV

= 0.8k/ft controls over parallel wind with a site coefficient of 1.2 and C = 2.75 With bars in keyways at 3 ft on center

A vf = 0.85(60)(1.4)

F t )w x h x x = n

(V − F

= 0.034 in 2 per keyway

i=1 i

Use #3 in every 2nd keyway

F x (Fig. 4.9.2 Det. F)

173 • Shear in transverse joint

V u = 1.3(18.4 -- 0.46 × 30)

A vf 6 = 0.85(60)(1.0)

29 = 0.12 in 2

Fig. 4.9.2 Wind design summary

#4 near each 0.3k/ft

end of wall 0.41k/ft

39.3k Chord Force

Intermittent slab cut-outs

#3 @ every 2nd keyway

#7 cont.

2-#5 near each end of bay

#3 every 2nd keyway

2-#6 Intermittent slab cut-outs

• Diaphragm load The diaphragm equilibrium is:

F t +Σ F

2.77 k/ft n

241 w k px F t+ Σ F i Σ w i F px 1962

Minimum diaphragm load

F px = 0.35ZIw px

• Chord forces:

= 0.35(0.15)(1.0)(1962) Using reinforcement in a perimeter boundary = 103k

element

Maximum diaphragm load

Ô0.8hf y

F px = 0.75ZIw px

To keep diaphragm in the elastic range, multi-

(Fig. 4.9.3 Det. A)

ply the diaphragm loads by 2R/5. At the roof

F pxu = 173(2)(8)/5 • Connect diaphragm web to chord = 554k

V uh =M u

jd

The factored roof diaphragm load by code pro- visions is

F pxu = (1.1)(1.3)(173)

= 164k

= 248k Distribute over length from zero moment to

Design roof diaphragm for a factored load of

maximum moment

554k to keep in elastic range.

V uh = 164 = 1.89k/ft

• For shear parallel to slabs Additionally, this connection must resist the Using a rigid diaphragm, the shear distribution

outward force from the exterior wall system. to the walls is:

Conservatively, this force will be:

30 ft walls: V = 241k

T = 0.75ZIw w

20 ft wall: V = 72k = 0.75(0.15)(1.0)(14 × 0.035)

= 0.055k/ft

A vf = 0.85(60)(1.0)

T u = 0.055(2)(8)/5 = 5.27 in 2 / 4 joints = 0.176k/ft

= 1.32 in 2 per joint

A s = + Ôf y Ôf y μ

In boundary elements, add chord requirement At first joint

M = 241(3) -- 3 2 u (2.77)/2

= 711 ft-k

= 0.033 in /ft Use #3 at 3 ft on center grouted into cores

A s = 1.32 +

(Fig. 4.9.3 Det. A)

= 1.53 in 2

At the transverse joint, the same shear parallel

4 - #8 ok

to the transverse joint as at the chord must be transferred. However, the tension should con-

(Fig. 4.9.3 Det. A)

sider the inertial force from the weight of the

In transverse joints

exterior bay. Conservatively

A = 1.32 in s 2

T = 0.75ZIw p

Use 2 - #8

w p = 14(0.035) + 30(0.0535 + 0.020 + 0.032)

(Fig. 4.9.3 Det. B)

= 3.66k/ft • Shear connection to 30 ft wall:

T = 0.75(0.15)(1.0)(3.66) Transfer shear to wall and drag strut

= 0.41k/ft T u = 0.41(2)(8)/5

V u = 253

= 1.31k/ft

= 3.16k/ft

A s 1.89

0.85(60)(1.4)) vf 0.9(60) 0.85(60)(1.0) = 0.051 in 2 2 /ft = 0.062 in /ft

Use #4 at 3 ft on center in keyways Use #4 hairpins at 3 ft on center (Fig. 4.9.3 Det. B)

(Fig. 4.9.3 Det. D)

• Longitudinal shear

Drag strut reinforcement

The maximum longitudinal shear is at the first

slab joint from the 30 ft wall. Provide shear

friction reinforcement in the two transverse

= 79k

joints and the two boundary elements for shear

A s = resistance. Conservatively consider 5% mini- 79

mum eccentricity being resisted only in end = 1.46 in walls. 2

V u = 241 + (0.05 x 200)(554)/200

Use 2 - #8

= 269k

(Fig. 4.9.3 Det. C)

• Shear connection at 20 ft wall

nA s = 6.74(3.16)

V u = 36k

= 21.3 in 2

over building width

V u = 36 = 0.45k/ft

80 2 57(x -- 4) + 4.3(x -- 8) /2 = 21.3(956 -- x)

A vf =

find x = 87.9 in

= 0.009 in 2 /ft Use #4 dowels at 8 ft on center

I cr = 57(87.9 -- 4) 2 + 4.3(87.9 -- 8) 3 /3 + 21.3(956 -- 87.9) 2

(Fig. 4.9.3 Det. F) Drag strut reinforcement

A s = 27 0.9(60)

As a rigid diaphragm, the factored load deflec-

tion between end shear walls is: = 0.5 in

Use 2 - #5 4 (2.77)(200) 72(200) 3

384 − (4300)(829)(12) 48(4300)(829)(12) (Fig. 4.9.3 Det. E)

• In-plane deflection of diaphragm = 1.07 in (ignoring shear deflections) Idealize the diaphragm section as

As a flexible diaphragm with rigid supports the deflection will be substantially smaller. The

diaphragm deflection plus the deflection of the

lateral-resisting system is used to evaluate the gravity load support members for integrity

when deformed.

• Consider load applied perpendicular to the with 4000 psi concrete in chord

slabs

E c = 3835 with 5000 psi concrete in slab

Total V u = 554k

E c = 4300 normalize on slab concrete

Distribution to walls is

n chord = 0.89

A Tchord = 0.89(64)

V u = 554/2

= 57 in 2 n steel = 6.74

= 277k

The diaphragm equilibrium is:

Drag strut reinforcement

6.93 k/ft

2 = 129.2k

A s = 129.2 = 2.39 in 2

277 k

277 k

0.9(60) Chord reinforcement from load parallel to

slabs controls.

• Shear in transverse joint

In center bay

w p = 20(200)(0.0535 + 0.020 + 0.032) + 20(14)(0.035)(2)

54 ft-k

= 442k

• Chord force

Conservatively use

V = 0.75ZIw

V u = 49.7(0.55)(2)(8)/5 = 87.5k per joint including

A s = 38.5 = 0.64 in 2

5% eccentricity

The #4 bars across the transverse joints at 3 ft Load parallel to slabs will control on center will be adequate. (Fig. 4.9.3 Det. B) See Figure 4.9.3 for summary Longitudinal shear

V =M uh u jd

= 34.7k will not control • Shear connection to walls With 5% eccentricity

V u = 1.1(277) = 304.7k Transfer shear to wall and drag strut

V u = 304.7/200 ft = 1.52k/ft Loading parallel to slabs controls

Fig. 4.9.3 Seismic design summary

#3 @ 3ft. o.c.

#3 @3ft o.c. grout in cores

2-#8 cont.

2-#5 cont.

#4 @ 8ft. o.c.

in cores

#4 @ 3ft. o.c.

4 - #8 cont.

Intermittent slab

4 - #8 cont.

cut-outs

#4 @8ft. o.c. 2 - #8 cont.

#4 @ 3ft. o.c.

#4 @ 3ft. o.c. in keyways

2 - #8 cont. Intermittent slab cut-outs

2 - #5 cont.