Walls and Shear Walls

973

18Walls and Shear

Walls

Hinge

Wall 2Wall 1

Mw 1 Mw 2

To Co

Mw1 Mw 2

Vbn

Vbi

Vb2

Vb1

(a) Hinged coupling beams. (b) Stiff coupling beams.

O

18-1 INTRODUCTION

Definitions—Walls and Wall Loadings

ACI Code Section 2.2 defines a wall as follows:“Wall—Member, usually vertical, used to enclose or separate spaces.”This definition fails to consider the structural actions of walls. ACI Section 2.1 also

defines the term “structural walls”:“Structural wall—Wall proportioned to resist combinations of shears, moments, and

axial forces. A shear wall is a structural wall.”Major factors that affect the design of structural walls include the following:

(a) The structural function of the wall relative to the rest of the structure.

• The way the wall is supported and braced by the rest of the structure.

• The way the wall supports and braces the rest of the structure.

(b) The types of loads the wall resists.(c) The location and amount of reinforcement.

Two frequent characteristics of walls are their slenderness, height to thickness ratio, whichis generally higher than for columns, and the reinforcement ratios, generally about a fifthto a tenth of those in columns.

Types of Walls

Structural walls can be classified as:

(a) Bearing walls—walls that are laterally supported and braced by the rest ofthe structure that resist primarily in-plane vertical loads acting downward on the topof the wall (see Fig. 18-1a). The vertical load may act eccentrically with respect tothe wall thickness, causing weak-axis bending.

974 • Chapter 18 Walls and Shear Walls

(b) Shear walls—walls that primarily resist lateral loads due to wind or earth-quakes acting on the building are called shear walls or structural walls. These wallsoften provide lateral bracing for the rest of the structure. (See Fig. 18-1b.) They resistgravity loads transferred to the wall by the parts of the structure tributary to the wall,plus lateral-loads (lateral shears) and moments about the strong axis of the wall.

(a) Bearing wall.

(c) Cantilever retaining wall. (d) Counterfort wall.

(b) Shear wall.

Foundation

Soilpressure

Soilpressure

Foundation

Counterfort

Soil retainingwall

(e) Compression panel ina bridge deck.

hw

h

Ow

Fig. 18-1Types of walls.

Section 18-1 Introduction • 975

(c) Nonbearing walls—walls that do not support gravity in-plane loads otherthan their own weight. These walls may resist shears and moments due to pressuresor loads acting on one or both sides of the wall. Examples are basement walls andretaining walls used to resist lateral soil pressures. (See Fig. 18-1c and d.)

(d) Tilt-up walls—are very slender walls that are cast in a horizontal positionadjacent to the structure. They are then tilted into their intended vertical position andfastened to the foundation, to the roof or floor diaphragm, and to the adjacent panels.They are designed to resist vertical and lateral loads.

(e) Although they are not walls as such, plates that resist in-plane compression,such as the compression flanges or the decks of box girder bridges, display some ofthe characteristics of walls. (See Fig. 18-1e.)

One-Way and Two-Way Walls

Walls may be supported and restrained against lateral deflections along one to four sides.Cantilever retaining walls are generally supported solely along the lower edge of the wall.Such walls act as vertical flexural cantilevers that resist lateral loads from the adjoiningsoil. Bearing walls are generally laterally supported and restrained against deflection alongtwo opposite sides, usually the top and bottom supports. Cantilever retaining walls andbearing walls transfer load in one direction: to supports at the top and bottom of the wall,for example, or to supports at the east and west edges. In the terminology used for one-wayand two-way slabs, these are referred to as one-way walls. One-way walls may be designedas wide columns spanning between the top and bottom supports, using ACI Code Chapters10 and 11, or they may be designed using ACI Code Chapter 14. Walls that transfer load inmore than one direction are called two-way walls. Walls supported on three sides mayoccur in open-topped, rectangular tanks; in storage bins for bulk materials; or in counter-fort retaining walls (see Fig. 18-1d). Walls supported on four sides are used to resist forcesor pressures applied perpendicular to the walls.

Wall Assemblies

Shear walls may be planar walls, standing in one vertical plane, or three-dimensionalassemblies of planar walls or wall segments. The latter occur as elevator shafts in buildingswhere four or more vertical walls enclose a stairwell or a group of elevators. Figure 18-2 isa photograph of a wall assembly that will enclose an elevator shaft and will brace a steel-frame building that will be built around it. The frame will be attached to the wall assemblyby welding or otherwise fastening the frame to steel plates embedded in the wall concreteduring construction of the walls.

In the design of a wall assembly, it is necessary to consider the transfer of shearforces from the wall segments serving as webs of the assemblies, to wall segments whichact as flanges.

Notation

The orientation of the walls in a vertical direction leads to ambiguity in the notation forheight and width. While most of the notation will be defined where it is first used, a fewkey symbols are defined here. Some of these are illustrated in Fig. 18-1a.

is the height of a beam

is the height of a location in a buildinghz

hb


Fig. 18-2Wall assembly in a buildingunder construction. (Photo-graph courtesy of J. K.Wight.)

is the overall height of a wall

length of compression member (column or wall) in a frame, measured from cen-ter to center of the joints in the frame

is the horizontal length of a wall

h is the thickness of a wall

18-2 BEARING WALLS

Walls used primarily to support gravity loads in buildings are referred to as bearing walls.Design is by ACI Code Section 14.5, which was derived specifically to apply to walls sub-jected to axial loads and moments due to the axial loads acting at an eccentricity of one-sixthof the thickness of the wall from the midplane of the wall (i.e., at the kern of the wall). Theresulting moments are referred to as weak-axis bending moments. ACI Code Section 14.4allows the design of bearing walls to be carried out either by:

1. Using the one-way column design and slenderness requirements in ACI CodeSections 10.2, 10.3, 10.10, 10.14, 14.2, and 14.3, or

2. The so-called empirical design method in ACI Code Section 14.5.

Walls with strong-axis moments and significant in-plane shear forces acting parallel to thewall, referred to as shear walls or structural walls, are not covered by ACI Chapter 14,although the code does not state this. Shear walls will be discussed in Section 18-5.

Axial-Load Capacity—ACI Eq. (14-1) was based on the results of 54 wall testsreported by Oberlender and Everard [18-1]. Their test results showed no effect of the rein-forcement ratios. Concentrically loaded test specimens with and eccentricallyloaded specimens with or more, exhibited buckling failures.

About the same time, Kripanarayanan [18-2] discussed the strength of slender wallsbased on analytical studies. He observed that reinforcement amounts of to 1.0percent were needed for the reinforcement to affect the failure loads of slender walls.

r = 0.75

hw>h = 16hw>h = 28,

/w

/chw

Section 18-2 Bearing Walls • 977

Equation (18-1) (ACI Eq. 14-1) was derived in a two-step procedure. First, the capacityof a short wall was derived. Then this was multiplied by a factor reflecting the effects of slen-derness on the axial-load capacity.

The largest eccentricity at which a load can be applied to a plain concrete wall with-out developing tensile stresses is at one-sixth of the wall thickness from the midthicknessof the wall (at the kern point of the section). This load case can be approximated by a rec-tangular stress block extending from the compressed face of the wall for a distance oftwo-thirds of the thickness of the wall as shown in Fig. 18-3. The force per horizontallength of wall, is

This was rounded off to and then multiplied by the term in the square bracketsin Eq. (18-1) to account for the slenderness of the wall. The slenderness term was derivedto give reasonable agreement with the slenderness effects in ACI Code Section 10.10. Theequation for the axial-load capacity of a bearing wall is

(18-1)(ACI Eq. 14-1)

where

is the clear, vertical distance between lateral supports

k is the effective length factor for a wall, taken as

0.8 if the wall is braced against translation at both ends and the top or bottom (orboth) is restrained against rotation

/c

fPn = 0.55ffcœAg c1 - a k/c

32hb2 d

0.55fcœh/w

Pn,short = 0.85fcœ *

2

3h * /w = 0.567fc

œ * h * /w

/w,

h

h /3 h /6

h /3

Pn

Pn

Midthickness

Bearing wall

Fig. 18-3Cross section through the topof a bearing wall showing theflexural compression zone.


1.0 if both ends are effectively hinged

2.0 for walls which are not effectively braced against lateral translation at the top,and therefore must be considered to be free-standing

h is the overall thickness of the wall

is the strength-reduction factor for compression-controlled sections, taken equal to0.65.

Thickness, Reinforcement, and Sustained Loads—Equation (18-1) is not affected bythe amount of wall reinforcement and does not allow for creep under sustained axial loads.This is in contrast to design by ACI Code Chapter 10.

ACI Code Section 14.5.3.1 limits the minimum thicknesses of walls designedusing the so-called empirical design method to 1/25 of the unsupported height or lengthof the wall, whichever is shorter, but not less than 4 in. ACI Code Chapter 14 does notrequire wall reinforcement to be designed for the loads on the wall. Instead, ACICode Sections 14.3.2 and 14.3.3 give the minimum vertical and horizontal reinforce-ment ratios.

These reinforcement ratios can be written in terms of the maximum spacing of thebars. Thus, for No. 5 or smaller bars with not less than 60 ksi, the maximum horizontaland vertical spacings are as follows:

Vertical steel:

(18-2a)

Horizontal steel:

(18-2b)

If the reinforcement is in two layers, Av is the total area of vertical bars within thespacing sh, and similarly for the horizontal bars.

ACI Code Section 14.3 requires more reinforcement horizontally than vertically.This reflects the greater chance that vertical cracks in walls might form as a result ofrestrained horizontal shrinkage or temperature stresses, compared with a lower chancethat horizontal cracks will form as a result of restrained vertical stresses. Generally, ifshrinkage occurs in the vertical direction, the shrinkage stresses are dissipated by verti-cal compression stresses in the wall.

EXAMPLE 18-1 Compute the Capacity of a Bearing Wall

A wall with a vertical height between lateral supports of 16 ft and a horizontal length of 25ft between intersecting walls supports a uniformly distributed factored gravity load of 41kips/ft, including the self-weight of the wall. The wall is supported on a strip footing thatprevents lateral movement of the bottom of the wall. The wall supports a wooden-frameroof deck, which acts as a diaphragm to restrain lateral displacement of the top of the wall.Is an 8-in.-thick wall adequate if If so, select reinforcement for the wall.Use the load and resistance factors from ACI Code Sections 9.2 and 9.3.

1. Check whether the Wall Thickness is Sufficient. ACI Code Section 14.5.3.1limits the thickness of walls designed by the empirical design method to the larger of 4 in.and 1/25 of the shorter of the unsupported height or the length. Thus, the minimum thick-ness is An 8-in.-thick wall satisfies the minimum thicknessgiven in ACI Code Section 14.5.3.1. Use an 8-in. wall.

116 * 122>25 in. = 7.68 in.

fcœ = 4000 psi?

sv,max = Ah /(0.0020h)

sh,max = Av /(0.0012h)

fy

f

Section 18-2 Bearing Walls • 979

2. Compute the Capacity of a 1-ft-Wide Strip of Wall.

(18-1)

ACI Code Section 14.5.2 gives for the end restraints described in the state-ment of the problem. Walls will seldom have spiral reinforcement. As a result, the wall is an“other” type of member, and ACI Code Section 9.3.2.2 specifies We thus have

This value exceeds the applied factored gravity load of 41.0 kips per ft; thus, the wallhas adequate capacity.

3. Select Reinforcement. ACI Code Sections 14.3.2 and 14.3.3 require minimumareas of and for vertical and horizontal reinforcement, respectively.ACI Code Section 14.3.4 allows the reinforcement to be placed in one layer, or “curtain,”because the wall thickness is less than 10 in. ACI Code Section 14.3.5 gives the maximumbar spacing parallel to the wall as the smaller of 3 times the wall thickness—3 times

—and an upper limit of 18 in. The maximum spacing of the reinforcement isas follows:

Horizontal spacing of vertical reinforcement—from ACI Code Section 14.3.2:

(18-2a)

If the required vertical steel is placed in a single layer of vertical No. 4 bars,and the spacing is

Vertical spacing of horizontal reinforcement—from ACI Code Section 14.3.3:

(18-2b)

Using a single layer of No. 4 bars, the spacing of the minimum horizontal reinforce-ment is . Use 12 in. on center.

Because the vertical reinforcement is not specifically used in the strength design,ACI Code Section 14.3.6 does not require the vertical bars to be tied in nonseismic regionsprovided that

(a) the area of vertical steel is less than 0.01 times the gross area of the wall, or

(b) the steel is not used as compression steel.

The vertical steel provided has times the gross areaof the wall.

It is good practice to provide a No. 5 bar vertically at each end of each curtain of wallsteel.

Use an 8-in.-thick wall with with one curtain of bars at the cen-ter of the wall with No. 4 vertical Grade-60 bars at 18 in. o.c. and No. 4 horizontalGrade-60 bars at 12-in. o.c. Add one No. 5 vertical bar at each end of the wall. ■

fcœ � 4000 psi,

As = 0.2 in.2>18 * 182 in.2 = 0.0014

= 0.20>10.0020 * 82 = 12.5 in. on centerssv,max

sv,max = Ah>10.0020h2sh,max = 0.20 in.2>10.0012 * 8 in.2 = 20.8 in. on centers. Use 18 in. on centers.

0.20 in.2,Av =

sh,max = Av>10.0012h28 in. = 24 in.

0.0020Ag0.0012Ag

= 60.1 kips>ft= 137,000 lb * [0.438] per foot of wall length

fPn = 0.55 * 0.65 * 4000 psi * 8 in. * 12 in. c1 - a1.0 * 16 * 12

32 * 8b2 d

f = 0.65.

k = 1.0

fPn = 0.55ffcœAg c1 - a k/c

32hb2 d


18-3 RETAINING WALLS

Reinforced concrete walls are used to resist the horizontal soil pressures when the sur-face of the ground is higher on one side of the wall than on the other. These are calledretaining walls. The most common type is the cantilever retaining wall shown in Fig.18-1c, which consists of a vertical cantilever wall that resists horizontal earth pressureand a footing that resists the moments at the base of the cantilever and transfers theforces to the ground. ACI Code Section 14.1.2 specifies that flexural design of retain-ing walls should be in accordance with the flexural design provisions of ACI CodeChapter 10, with minimum horizontal wall reinforcement in accordance with ACI CodeSection 14.3.3.

The lateral soil pressure acting on the wall and the bearing capacity of the soil underthe wall footing are obtained by consultation with a geotechnical engineer, especially if thewall resists surcharge loads acting on the upper ground surface. It is important to providedrainage through or around the wall to minimize hydrostatic pressure behind the wall.

18-4 TILT-UP WALLS

Industrial buildings and warehouses are frequently constructed from tall thin concretewalls that are cast in a horizontal position on a slab on grade that will become the floor slabfor the completed building. After the walls have gained adequate strength, a crane is usedto tilt them up to their final vertical position at which time they are connected together.Such walls and buildings are referred to as tilt-up walls and tilt-up buildings. These consti-tute a form of construction that is becoming common throughout North America. Mostaspects of the design of tilt-up buildings is covered by the report of the ACI Committee 551,Tilt-up Concrete Construction [18-3]. ACI Code Section 14.8, Alternative Design of Slen-der Walls, presents design requirements for the very slender walls used in tilt-up construc-tion. Guidance is also found in the Concrete Design Manual [18-4]. Special considerationshould be given to construction loads acting on these walls.

Nathan [18-5] analyzed thin one-way walls. His design charts are used widely in thedesign of tilt-up buildings.

18-5 SHEAR WALLS

The term shear wall is used to describe a wall that resists lateral wind or earthquake loadsacting parallel to the plane of the wall in addition to the gravity loads from the floors and roofadjacent to the wall. Such walls are referred to as structural walls in ACI Code Chapter 21.

The strength and behavior of short, one- or two-story shear walls (as shown inFig. 18-4a) generally are dominated by shear. These walls typically have a height-to-length aspect ratio of less than or equal to 2 and are called short or squat walls.Such walls can be designed by either the requirements given in ACI Code Chapter 11 orthe strut-and-tie method given in ACI Code Appendix A. If the wall is more than three orfour stories in height, lateral loads are resisted mainly by flexural action of the verticalcantilever wall (Fig. 18-4b) rather than shear action. Shear walls with greater thanor equal to 3 are referred to as slender or flexural walls. These walls typically are designedusing the provisions of ACI Code Chapters 10 and 11. Shear walls with ratiosbetween 2 and 3 exhibit a combination of shear and flexural behavior and normally wouldbe designed following the provisions of ACI Code Chapters 10 and 11.

Although shear walls may be simple planar walls, several wall segments commonlyare connected together to act as a three-dimensional unit. Such wall assemblies have

hw//w

hw//w

1hw//w2

Section 18-6 Lateral Load-Resisting Systems for Buildings • 981

(a) Squat shear wall. (b) Slender shear wall.

Tie

Fig. 18-4Shear walls.

regular or irregular C, T, L, or H-shaped cross sections with webs and flanges that may en-close spaces in buildings, such as stair wells or elevator shafts.

18-6 LATERAL LOAD-RESISTING SYSTEMS FOR BUILDINGS

Three common systems for resisting wind or earthquake lateral loads are given here.

1. Moment-resisting frames are made up of interconnected beams and columns.Lateral loads are resisted by bending of the beams and columns (see Fig. (18-5)). Suchframes undergo relatively large lateral deflections. Typically the deflected shape of amoment-resisting frame is as shown in Fig. 18-5. If all stories have beams and columnswith sizes proportional to the shear in the story, the lateral deflection of each story wouldbe similar. To simplify construction, however, the sizes of the beams and columnsselected for the lower stories are commonly used throughout, or are changed only everythird or fourth story. Hence, the beams and columns in a building tend to be oversized inthe upper stories. Moment-resisting frames are used for buildings up to 8 to 10 stories.

In a moment-resisting frame, deflections of the columns and beams both con-tribute to the sway deflections of the frame. If we isolate the shaded portion of the framein Fig. 18-5, the deflection of A relative to B is given by

(18-3)

where L is the span of the beam, center to center of supports, and is the height of the col-umn from midheight of the story above the beam to midheight of the story below thebeam. The first term on the right-hand side of Eq. (18-3) is the deflection due to the bend-ing of the column. The second term is due to the bending of the beam. Shear deformationsare not included in Eq. (18-3) because they are very small compared to bending deforma-tions in a typical frame member.

Assuming and the portion of the relative horizontal deflectiondue to bending of the beam is two-thirds of the total deflection. Frequently in tall frames, it

EcIc = EbIb,L = 2/

/

¢AB =V/3

24EcIc+V/2L

24EbIb


Fig. 18-5Moment-resisting frame.

A

�AB

BV

V

L /2

(b) Beam–column subassembly from part (a).

(a) Deflected frame.

O

VOL /2

VOL /2

may be impossible to make the beams stiff enough to prevent large deflections. In suchcases, walls or other stiffening elements are used to control lateral deflections.

2. Bearing-wall systems are used for apartment buildings or hotels. A bearing-wallbuilding has a series of parallel transverse shear walls between rooms or apartments. The wallsresist lateral loads by flexural action and deflect as vertical cantilevers.

3. Shear-wall–frame buildings, shown schematically in Fig. 18-6, are used inbuildings ranging from about 8 to about 30 stories. The lateral load is resisted in part by thewall and in part by the frame. Some of the most slender shear-wall–frame structures everbuilt are described by Grossman [18-6]. He presents some of the wind modeling rationale

Section 18-7 Shear-Wall–Frame Interaction • 983

and two case histories of buildings with heights up to 10 times the least width at groundlevel.

4. Very tall concrete buildings Structural systems for very tall concrete buildingsare discussed in [18-7]. The bottom stories of such a building are shown in Fig. 17-26. Theclosely spaced columns in the upper stories transfer vertical loads much like a continuousclosed tube would. At the top of the second floor the vertical loads are transferred to a con-tinuous deep beam called a transfer beam. It, in turn, transfers the vertical loads to the 10large columns on the perimeter of the ground floor. In this region the more-or-less uniformcompression in the tube is disrupted.

18-7 SHEAR-WALL–FRAME INTERACTION

The division of lateral load between the wall and frame in a shear-wall–frame buildingcan be analyzed by using a frame analysis and a model similar to that in Fig. 18-6. Theframe members in the model represent the sums of the stiffnesses of the columns andbeams in the building in the bays parallel to the plane of the wall. Similarly, the wall inthe model represents the sum of the walls in the structure. The wall and frame are con-nected by axially stiff link beams at every floor. The link beams in the model shown inFig. 18-6 may or may not be hinged. In computing internal forces and moments due to thefactored loads, the flexural stiffnesses, EI, from ACI Code Section 10.10.4.1 may be used.The model in Fig. 18-6 may be acceptable for buildings that are symmetrical in plan andhave rigid floor diaphragms. A three-dimensional model is required for an unsymmetricalbuilding, and where a designer wants to account for diaphragm flexibility.

Frame Hinged link beams

Wall

Fig. 18-6Analytical model of a shear-wall–frame building.


The lateral-force analysis of shear-wall–frame buildings must account for the differ-ent deformed shapes of the frame and the wall. Due to the incompatibility of the deflectedshapes of the wall and the frame, the fractions of the total lateral load resisted by the walland frame differ from story to story. Near the top of the building, the lateral deflection ofthe wall in a given story tends to be larger than that of the frame in the same story and theframe pushes back on the wall. This alters the forces acting on the frame in these stories. Atsome floors the forces change direction, as shown schematically by the range of possiblemoment diagrams in the wall as shown in Fig. 18-7. As a result, the frame resists a largerfraction of the lateral loads in the upper stories than it does in the lower stories.

Bounds on the Forces in a Shear-Wall–Frame Building

The relative portions of the lateral loads resisted by the walls and frames in a shear-wall–frame building can be estimated by considering the wall and the frame as two verti-cal cantilevers, fixed at the bottom and connected via a single extensionally rigid link beamjoining the wall and frame at the top, as shown in Fig. 18-7a [18-8]. If the frame is so stiffthat it prevents horizontal deflection of the top of the wall, the reaction of the loaded frameat the top of the wall is where w is the lateral load per foot of height and is theheight of the wall. This is equivalent to the reaction at the pinned end of a uniformly loadedbeam having a constant EI that is fixed at one end and pinned at the other. As the lateralstiffness of the frame decreases relative to the lateral stiffness of the wall, the reaction atthe top of the wall decreases, approaching zero for a very flexible frame combined with astiff wall. As a result, the shear-force and bending-moment diagrams for the wall can varybetween the approximate limits shown in Fig. 18-7b and c. The sum of the shear forces inthe frame and the wall in a given story must equal the shear due to the applied loads.

18-8 COUPLED SHEAR WALLS

Two or more shear walls in the same plane (or two wall assemblies) are sometimes con-nected at floor levels by coupling beams, so that the walls act as a unit when resisting lat-eral loads, as shown in Fig. 18-8. The coupling beams frame into the edges of the walls, as

hw3>8whw,

Shear force in wall ifframe is very flexible

Shear force in wallif frame isvery stiff

Moment in wall ifframe is very flexible

Moment in wall ifframe is very stiff

(a) Idealization of the frame from a wall–frame building as a propped cantilever.

(b) Range of shear-force diagrams for wall.

(c) Range of moment diagrams for wall.

0Fig. 18-7Effect of frame stiffness onshear and moment in theshear wall.

Section 18-8 Coupled Shear Walls • 985

shown in this figure. The discussion will be limited to the case of two walls separated by asingle vertical line of openings, which are spanned by reinforced concrete coupling beams.Walls with more than two lines of openings, as shown in Fig. 18-2, are handled similarly towhat is discussed here for two coupled walls. Other coupled wall systems may need spe-cial attention, especially if the widths and heights of the line of openings are irregular.

When the coupled wall deflects, the axes of both parts of the wall at A and deflectlaterally and rotate through an angle , as shown in Fig. 18-9. This results in shearing deflec-tions of the coupling beams that join the two walls. Localized cracking of the beam-to-walljoint reduces the angle the coupling beam must go through where it is attached to the wall. Itis customary to assume the effect of these localized deflections and reduced stiffness of thecoupling beam can be represented by moving the assumed connection point in from the faceof the wall by approximately where hb is the height of the coupling beam [18-8]. Thus,we shall assume the walls are joined by coupling beams spanning from B to The down-ward deflection of point B is

(18-4)

where is the width of the wall.The moments and axial forces in the two wall segments in Fig. 18-9 must be in equi-

librium with the axial forces, shears, and moments in the entire coupled wall system. Thesigns of the moments and shears may change over the height of the building, similar tothose shown by Fig. 18-7b and c.

bw

¢B = abw2

-hb2b tan u

B¿.hb/2,

u

A¿

Hinge

Wall 2Wall 1

Mw 1 Mw 2

To Co

Mw1 Mw 2

Vbn

Vbi

Vb2

Vb1

(a) Hinged coupling beams. (b) Stiff coupling beams.

OFig. 18-8Coupled walls.


Statical System

Figure 18-8a shows two prismatic walls, wall 1 and wall 2, connected at each floor levelby link beams hinged at each end. The moments at the bases of the two walls are equal to

(18-5)

and

(18-6)

where are the wall moments of inertia and is the moment at the base of thewall due to factored lateral loads. The total lateral load moments in the walls equal

(18-7)

Figure 18-8b shows the same two walls, except that the walls are now coupled by beams thatare continuous with the walls at every floor level and have some flexural stiffness. As the wallsdeflect laterally, the coupling beams deflect as shown in Fig. 18-9, and shears and moments aregenerated in the coupling beams. A free-body diagram through the coupling beams halfwaybetween the faces of the two walls has shear forces in each coupling beam, as shown inFig. 18-8b and Fig. 18-9. There are also axial forces in the beams. For equilibrium of verticalforces, an axial tension, must be added to the axial forces in the walls at the centroid of thebottom of wall 1 and an axial compression, at the centroid of the bottom of wall 2, where

(18-8)

Taking the distance between the lines of action of the forces and as we find that thetotal moment at the base of the coupled wall system is

(18-9)

Equation 18-9 is plotted in Fig. 18-10 for a coupled wall consisting of two walls with[18-9]. The vertical axis is the slenderness of the beam, where andhbhb>/b,EIw1 = 2EIw2

Mo = Mw1 + Mw2 + To * /

/,CoTo

To = an

i=1Vbi = Co

Co,To,

Vbi

Mw1 + Mw2 = Mo

MoIw1 and Iw2

Mw2 = MoIw2

Iw1 + Iw2

Mw1 = MoIw1

Iw1 + Iw2

LoadVbi

Vbi

Very stiffelement

Analysisnode

Axisof wall

Axis of wall 2

Axis of wall 1

A

BB �

A�

hb

bw

2hb

2� tan�

hb

2

Coupling beam

bw/2

u

u

Fig. 18-9Effect of shear wall deflec-tions on the forces in a coupling beam (couplingbeam shown thinner for clarity).

Section 18-8 Coupled Shear Walls • 987

are the height and the adjusted span of the coupling beams, respectively. The beam slen-derness is used here as a measure of the stiffness of the coupling beams. A coupling beamwith equal to zero has no flexural stiffness, and as a result, the wall moments are dividedin proportion to the ratio of the wall stiffnesses, as given by Eqs. (18-5) and (18-6). As theflexural stiffness of the coupling beams increases, the shears in them increase. As a result, thefraction of the overturning moment resisted by the axial force couple increasesasymptotically. A major effect of the coupling beams is to reduce the moments and at the bases of the two walls. This makes it easier to transmit the wall reactions to the foun-dation. The coupling beams also act to reduce the lateral deflections. If the beams are per-fectly rigid, the two walls act as one wall. Figure 18-10 illustrates trends only. The actualdivision of into and depends on more variables than just

Coupling beams may be rectangular beams, T beams, or portions of the floor slab[18-10], [18-11]. In seismic regions, short coupling beams may have diagonally placedreinforcement. This is discussed in Chapter 19.

For seismic design, the Canadian concrete code [18-10] defines a coupled wall as onein which is at least 66 percent of In Fig. 18-10, this occurs when is about0.2. If is less than 66 percent of the wall is called a partially coupled wall.

Analysis of Coupled Walls

Before modern structural analysis programs, the individual coupling beams shown inFigs. 18-8b and 18-9 were replaced for analysis by a series of closely spaced laminae,each having a unit height and stiffness where is the story height. This allowed ahsIb>hs,

Mo,To * /hb>/bMo.To * /

hb>/b.To * /Mw1,Mw2,Mo

Mw2Mw1

To * /

h>/b/b

Mo

Mw1 � 2/3 Mo Mw 2 � 1/3 Mo

Mw1/Mo Mw 2/Mo

00.

20.

40.

60.

81.

0

h b/O

b

ToO/Mo

ToO � 0.66 Mo

Fig. 18-10Effect of the stiffness of thecoupling beams on themoments in walls 1 and 2,

(From [18-9].)Iw1 = 2Iw2.


closed-form solution of the forces in the laminae and the walls. Such analyses were limit-ed to a few uniform wall layouts [18-9]. Modern structural-analysis programs have madeit possible to model coupled structural walls, as shown in Fig. 18-8b, and to compute theforces and moments in the coupling beams directly. As a result, laminar analyses are sel-dom used now.

In a frame analysis to determine factored moments for design, the member stiff-nesses may be based on ACI Code Section 10.10.4.1. The coupling beams are joinedto hypothetical members with high values of the moment of inertia, I, between the faceof the wall and the centerline of the wall, as shown by the dark shaded regions inFig. 18-9. Short, deep coupling beams develop both flexural and shear deflections. Theshear deflections can be included by replacing the of the coupling beam between thewalls with

(18-10)

This equation comes from adding the moment deflections and shear deflections of thebeam, where and are the depth and the span of the coupling beam from face to faceof the walls. The second term in the denominator of Eq. (18-10) accounts for the shear de-flections of the beam.

Floor slabs may serve as soft coupling beams. Their stiffness can be based on a slabwith a width perpendicular to the wall equal to the wall thickness plus half of the widthof the opening, between the walls, added on each side of the opening [18-11], [18-12],[18-13]. In tests of shear walls coupled by slabs, the specimens failed by punching-shearfailures in the slab around the ends of the walls. Under cyclic loads, the stiffness of slabsserving as coupling beams decreased rapidly.

Figure 18-11a, b, and c shows the distributions of moments in the walls, the axial-force couple, and the shears in the coupling beams for a typical coupled wall [18-4] where

Typically, the maximum shear in the coupling beams occurs at about one-thirdof the height above the base. The sawtooth shape of Fig. 18-11a results from the endmoments in the coupling beams.

Iw1 = Iw2.

/b>2,

/bhb

Ieff =Ib

1 + 2.8ahb/bb2

Ib

Penthouse121110987654321

Ground

Levels

Max Vb

(a) Wall moments, Mw1 and Mw 2. (b) Axial couple, TO. (c) Shear in coupling beams, Vb.

Fig. 18-11Typical distribution of wall moments, axial-force couple, and shear in the coupling beams,

(From [18-4].)Iw1 = Iw2.

Section 18-9 Design of Structural Walls—General • 989

18-9 DESIGN OF STRUCTURAL WALLS—GENERAL

Layout of Building

Major considerations in selecting a structural system for a multistory building with struc-tural walls are the following:

(a) The building must have enough rigidity to withstand the service loads with-out excessive deflections or vibrations.

(b) It is desirable that the wall be loaded with enough vertical load to resist anyuplift of parts of the wall foundations due to lateral loads.

(c) The locations of frames and walls should minimize torsional deformationsof the building about the vertical axis of the building. (See also Section 19-3.)

(d) The walls must have adequate strength in shear, and in combined flexureand axial loads.

(e) The wall thickness or cover on the reinforcement may be governed by thefire code.

Diaphragms

Lateral loads are transferred to the lateral-force resisting system by the floor and roof slabsserving as horizontal diaphragms that act as wide, flat beams in the plane of the floor or roofsystem. The diaphragms must have a tension flange, a compression flange, and a web capableof transmitting the lateral loads. Because the direction of the lateral load changes back and forthduring wind or earthquake cycles, the compression and tension flanges of the diaphragm mustbe able to accommodate changes in the sign of the loading. Notches or other discontinuities inthe tension and compression flanges of the diaphragm should be avoided or reinforced to trans-mit the forces around the discontinuities. Diaphragms are also discussed in Section 19-10.

Distribution of Walls in a Building Floor Plan

A common design recommendation is to minimize the separation, commonly referred toas the eccentricity, between the center of mass (geometric centroid of the floor plate) andthe center of lateral resistance (CR) provided by the shear walls and moment resistingframes in the lateral-load system. Because lateral loads are assumed to act through thecenter of mass (CM), any eccentricity between the CM and CR will result in the generationof torsional moments. A central-core wall system, similar to that shown in Fig. 18-12,commonly is used to minimize eccentricity between the CM and CR.

When a building structure is subjected to large lateral displacements due to earth-quake ground motions, the stiffnesses of the lateral-load resisting members are likely tochange in a nonuniform fashion. As a result, the CR is likely to be relocated and the eccen-tricity between the CM and CR may increase. To account for this, the International Build-ing Code [18-14] specifies a minimum eccentricity in the two principal directions thatmust be added to any calculated eccentricity. For structures where substantial torsionalmoments may be generated, a wide distribution of shear walls around the perimeter of thefloor plan would be most efficient for resisting that torsion.

Distribution of Story Forces (Story Shears) to Walls

In the following analysis of the distribution of lateral loads to structural members, onlyisolated shear walls will be considered as lateral-load resisting elements. It should be


clear that this analysis could be expanded to include moment-resisting frames and wallassemblies, such as T-shapes, L-shapes, U-shapes, etc.

Consider the floor plan and isolated shear walls shown in Fig. 18-13. We will assumethat the slab (diaphragm) connecting these walls is stiff in-plane but has a low flexuralstiffness. Thus, the walls are not coupled, but they all should have the same lateral dis-placement under the acting lateral loads. We will assume the walls have a very low stiff-ness when bent about their weak axis, and thus, we will only consider the stiffness of thewalls when they are bent about their strong axes. We initially will assume that the walls areslender. Thus, only flexural stiffnesses will be considered. However, modifications to ac-count for shear deformations in short walls will be discussed at the end of this subsection.Finally, for this analysis of the distribution of lateral forces, we initially will assume thatthe walls are uncracked and thus will use the gross moment of inertia for the walls. Theeffect of flexural cracking can be accounted for in a second interation of this analysis byassuming that the moment of inertia for a flexurally cracked wall is equal to one-half of thegross moment of inertia. If lateral displacements are to be calculated, the moment of iner-tia values should be reduced by 30 percent to correspond to those recommended in ACICode Section 10.10.4.1:

Walls uncracked:Walls cracked:

Returning to Fig. 18-13 and using the assumptions discussed here, we now want tocalculate the shear forces induced in each wall due to the lateral forces, and If therewas no eccentricity between the CM and CR, the total lateral force would be distributedto the four walls along the north and south edges of the floor plate in proportion to their

Vx

Vy .Vx

0.35 Ig

0.7 Ig

RC beam

RC column

21 ft

RC wall

Couplingbeam

21 ft

30 ft

21 ft

21 ft

114 ft

21 ft 21 ft 21 ft 21 ft 21 ft 21 ft

126 ftFig. 18-12General building floor plan.


moments of inertia about their strong axis (i.e., their y-axis), as shown in Fig. 18-13. Thus,the lateral force resisted by wall j would be

(18-11a)

where n is the number of walls resisting in bending about their strong axis. Similarly,would be resisted by the two walls along the east and west edges of the floor plate, and thelateral force resisted by wall i would be

(18-11b)

where m is the number of walls resisting in bending about their strong axis (x-axis).If there is an eccentricity between the CM and CR or a minimum eccentricity is

specified by a design code, then the effects of torsion must be considered. To find the CRfor the floor plate in Fig. 18-13, we initially will assume an origin at the southwest cornerof the plate and measure distances from that origin in terms of X and Y. To find the locationof the CR in the Y-direction, we will consider only the walls resisting through bendingabout their y-axis. Following this procedure, the value for is

(18-12a)Yr =ajIyjYj

ajIyj

Yr

Vx

Vy

Vyiœ = C Ixi

amIxm SVy

VyVx

Vxjœ = C Iyja

nIyn SVx

CM

CR

L1/2

y

x

ex

Wall j

N

L1/2

xi

L2/2

L2/2

Vx

Yr

Xr

Vy

yj

lxi

lyj

Wall i

ey

Fig. 18-13Eccentricity of center ofresistance (CR) with respectto center of mass (CM).


Similarly, to find the location of CR in the X-direction, we will consider only the wallsresisting by bending about their x-axis. Thus,

(18-12b)

The location of the CM is given in Fig. 18-13, so the eccentricities from CM to CR are

(18-13a)

(18-13b)

These eccentricities or an increased value of eccentricity required to satisfy a governing build-ing code requirement can be used to calculate the torsion caused by the lateral loads as follows:

(18-14a)

(18-14b)

The torsion resisted by each wall in the floor plan will be related to the lateral stiffnessof the wall for bending about its strong axis multiplied by the distance in the x- or y-directionfrom the wall to the CR, as measured perpendicular to the weak axis of the wall. Thus, asstated previously, if a significant torsion is to be resisted, the use of widely distributed wallsis more effective in resisting torsion. An equivalent torsional stiffness for all of the walls inthe floor system (acting as a unit) can be calculated as the sum of the torsional resistancefrom each wall multiplied by their respective perpendicular distance to the CR. This torsionalstiffness can be expressed as

(18-15)

With this equivalent torsional stiffness for the walls acting as a combined system, wecan determine how much shear is induced in each wall when resisting the torsionalmoments. However, there may be some question regarding what torsional moments shouldbe used when determining the total shear force in each wall. Typically, a structure is ana-lyzed for lateral loads (wind or especially seismic) acting in only one principal directionand then reanalyzed for the lateral loads acting in the other principal direction. These tworesults normally are considered separately in the design. In this case, the torsion that isgenerated by either of the lateral loads acting in one principal direction is resisted by all ofthe walls—not just those with their principal axes perpendicular to the lateral load. Thus, itis not clear how much of the torsion generated due to lateral loading in the second princi-pal direction should be included when considering the effect of lateral loading in the firstprincipal direction. Assuming that some percentage of the effect of lateral loading in thesecond principal direction should be considered, the author presents the following expres-sions. The first expression is for the shear induced in walls that have their strong axisperpendicular to the lateral force

(18-16a)Vxjfl = B Iyjyj

KtR1Tx + aTy2

Vx .

Kt = aiIxixi

2 + ajIyj # yj2

Ty = Vyex

Tx = Vxey

ex =L1

2- Xr

ey =L2

2- Yr

Xr =aiIxiXi

aiIxi

Vy


Similarly, for the walls that have their strong axis perpendicular to the lateral force

(18-16b)

For both expressions, the author recommends using to reflect the low probabilityof having the maximum lateral forces acting simultaneously in both principal directions.

We now can combine the results from Eqs. (18-11) and (18-16) to obtain the totalshear resisted by walls with their strong axis perpendicular to the lateral force as

(18-17a)

For walls with their strong axis perpendicular to the lateral force we have

(18-17b)

For all of the analysis results given up to this point, only flexural stiffnesses of thewalls have been considered. For walls with aspect ratios less than 3, the effect ofshear deformations starts to become more significant. For such walls, it is recommended touse a modified moment of inertia to reflect the increased importance of shear deforma-tions. A value recommended in the PCA Design Handbook [18-15] is

(18-18)

where is the effective flexural moment of inertia that has been selected based on theprior discussions, and is the wall area,

Wall Foundations

The foundations at the base of the wall must be able to transfer the shear, moment, andaxial force as required from the base of the wall to the supporting soil or rock. If theaxial gravity forces in the wall are small, the moments at the base of the wall maycause uplift under one side of the shear-wall footing, as shown in Fig. 18-14a. Both anaxial load and a moment are resisted by the soil, and as a result, the soil pressure willvary across the width of the wall footing. For a vertical load acting at the kern of thefooting area, or for a vertical load and moment acting on the footing, the soil pressurewill range from smaller than average (as low as zero) on one side of the footing tohigher than the average stress (up to twice as high) at the other side. Because tensileuplift stresses are difficult to resist, they should be avoided. If the footing size becomesexcessive, possible solutions are:

(a) Replace the rectangular footing with an H-shaped footing, to increase the ra-

dius of gyration of the footing.

(b) Use pile or caisson foundations.

(c) Use coupled shear walls to widen the footprint of the wall and divide themoment to be transferred into the three components shown in Fig. 18-8b.

(d) Attach the base of the wall to horizontal outrigger beams in the basementwhich extend the foundation to receive the vertical loads from adjacent framecolumns so that there is no uplift; or the uplift is counteracted (see Fig. 18-14b).

ar = AInertia

Areab

Pu

h * /w .Aw

Ie

Imod =Ie

1 +24IeAwhw

2

1hw//w2Vyi = Vyiœ + Vyifl

Vy ,

Vxj = Vxjœ + VxjflVx

a = 0.25

Vyifl = B Ixixi

KtR1Ty + aTx2

Vy ,


(e) Attach the base of the wall to the ground-floor diaphragm and basement floorto provide a horizontal force couple to react to the moment at the base of the wall.

(f) Use a mat foundation.

Wyllie [18-16] and Paulay and Priestly [18-17] both discuss shear-wall foundations.

Required Size of Wall

In choosing a structural wall section for a given building, the wall must

(a) have enough strength to resist the factored moments, shears, and axial loadsacting on it; and

(b) have enough stiffness to limit the lateral deflections.

Column load

Basementfloor–diaphragm

Slab–groundfloor diaphragm

Outrigger beam or wall

(b) Combined footing.

(a) Isolated wall footing.

Uplift

Fig. 18-14Wall foundations.


There is no widely accepted way of doing this. A rough estimate of the minimum wall stiff-ness, EI, required to limit the lateral deflections to an acceptable value can be made byconsidering the walls as a vertical cantilever with a constant EI over the height, loaded witha constant wind load over the height, as shown in Fig. 18-14a. Wall thicknesses, and thusthe structure’s EI would generally decrease as the moments and shears decrease near thetop of the structure, we shall assume it is constant over the building height and equal to thesum of the EI values of the walls in the bottom story. This cantilever will be loaded with aconstant, uniform service wind load of lb per foot of height. The wind load is the prod-uct of the length and height of the building and the wind pressure per foot of height, takenequal to the algebraic sum of the windward wind pressures and leeward wind suctionsevaluated at the top of the building. It is customary to limit the relative horizontal deflec-tion of the top and bottom of a story to a fraction of the height of the story, expressed as

(200 to 500), where is called the story drift. This limit is expressed in termsof the slope of the wall relative to the vertical in any story throughout the height of thebuilding. The largest slope in a shear wall or shear-wall–frame building is generally at thetop of the structure and is given by [18-6]

(18-19)

If the base is fixed against rotation and assuming that is constant over theheight of the building, , the following gives the slope at the top relative to the unde-flected position:

(18-20)

Here,

is the height of the wall

is the modulus of elasticity of concrete, psi (because all other units are in poundsand feet, the modulus of elasticity in psi is multiplied by to change thepsi units to psf)

is the uncracked moment of inertia of the wall or walls at the bottom of the struc-ture (an uncracked wall is taken to be representative of service-load levels, becausecodes limit service-load deflections and the walls are generally not severely crackedunder service loads); ACI Code Section 10.10.4.1 gives for an un-cracked wall

is the unfactored wind load per vertical foot of wall, evaluated at the top of thewall, in lb/ft of height

Limits on Story Drift

ASCE/SEI Committee 7 [18-18] does not limit the lateral deflection or the story drift.The National Building Code of Canada [18-19] limits the maximum story drift under un-factored loads to less than 1/500 of the story height. The maximum acceptable drift de-pends on the ability for occupants of the building to perceive the motion of the buildingduring a major windstorm. This storm may be chosen as a rare event, such as the maxi-mum annual windstorm or the maximum windstorm in one tenancy of the building—say,every 3 to 8 years. Grossman discusses design for wind induced movement of slenderconcrete buildings [18-6].

ws

I = 0.70Ig ft4

Ig

1144 in.2>ft22Ec

hw

Slope at top =wshw

3

6EcIg

hw

EI = EIbase

1slope at top2 = 1rotation of base of wall2 + 1area of M>EI diagram from base to top2¢>hs¢ = hs>

ws


If a wall having height and constant wall stiffness EI is fixed at the base, themaximum drift will be in the top story. Setting the service-load story drift from Eq. (18-20)equal to 1/500, we can compute the minimum total for the walls parallel to the directionof the wind load as

(18-21)

Once the minimum has been estimated, walls can be selected to give equal to orgreater than what is required by Eq. (18-21). Although this analysis is intentionally simpli-fied, it gives an order-of-magnitude estimate of story drift. It does not consider torsionalloadings on the structure, and it assumes that the wind pressure is constant over the heightof the building. At the stage in design where the sizes of shear walls are initially chosen,these details are not yet known. A better estimate of can be obtained by calculating thesecond term on the right-hand side of Eq. (18-19) more accurately based on a betterestimate of the EI and moment.

Minimum Wall Thickness

The minimum thicknesses given in ACI Code Section 14.5.3 are intended to applyonly to bearing walls designed via the empirical design method from ACI CodeSection 14.5. The author would limit the thickness of walls with rectangular cross sec-tions to the absolute minimum of 1/20 of the unsupported height of the wall. Prefer-ably, the wall thickness should not be less than 1/15 of the unsupported height. Thewall thickness must be large enough to allow the concrete to be placed withouthoneycombing.

Reinforcement in Shear Walls

Distributed and Concentrated Reinforcement

The reinforcement in a shear wall is generally made up of:

(a) Distributed horizontal and vertical reinforcement spread uniformly over thelength between the boundary elements and over the height of the wall.

(b) Concentrated vertical reinforcement is located in boundary elements ator near the edges of the wall and is tied in much the same way that column cagesare.

Minimum Wall Reinforcement

In addition to the distributed reinforcement required by ACI Code Section 14.3, severalother ACI sections require minimum amounts of distributed horizontal and vertical steelthat may apply to walls. These are given in Table 18-1.

For deformed bars not larger than No. 5, ACI Code Section 14.3 requires mini-mum areas for distributed vertical reinforcement in walls to be equal to andthe minimum area of distributed horizontal reinforcement in walls to be equal to

This steel can be placed in two layers or curtains of distributed vertical andhorizontal reinforcement with the bars in the two curtains tied together at intervals, butnot enclosed in ties. A single reinforcement curtain is allowed at the midplane of wallshaving thicknesses of 10 in. or less (ACI Code Section 14.3.4). This steel cannot be tied

0.0020Ag.

0.0012Ag

¢>h©Ig©Ig

©10.70Ig2 =500wshw

3

6Ec

Ig

hw


because there is only one curtain. Thicker walls require two curtains of reinforcement,each consisting of not less than half of the total minimum reinforcement required ineach direction. Each of these two layers is placed not more than one-third of the wallthickness from the surface. It is desirable to have the steel in two layers close to thesides of the wall because the internal lever arm, for weak-axis bending is muchsmaller if the reinforcement is at the center of the wall. ACI Code Section 14.3.5 re-quires that distributed vertical and horizontal bars be spaced not further apart than threetimes the wall thickness or 18 in., whichever is less. ACI Code Section 11.9.9.3 allowsthe maximum spacing of horizontal shear reinforcement to be the smallest of and 18 in. ACI Section 11.9.9.5 allows the maximum spacing of vertical shear rein-forcement to be the smallest of and 18 in./w>3, 3h,

/w>5, 3h,

jd,

TABLE 18-1 Minimum Reinforcement in Walls Compared to Other Membersa

Reason ACI Code Requirement Maximum SpacingSection

Shrinkage and 7.12.2.1 (b) Slabs where Grade-60 deformed Five times the slab thickness,temperature bars or welded-wire fabric no farther apart than 18 in.

(plain or deformed) are used: 0.0018

Minimum flexural 10.5.4 The minimum area of tensile reinforcement Three times the slab thickness, no farthersteel in one-way slabs in the direction of the slab span is the same apart than 18 in.

as by 7.12.2.1

Deep beams 11.7.4.1 The area of shear reinforcement perpendicular s shall not exceed d/5 or 12 in.to the span shall not be less than 0.0025bws

11.7.4.2 The area of shear reinforcement parallel s2 shall not exceed d/5 or 12 in.to the span not be less than 0.0025bws2

Walls 11.9.9.2 Ratio of horizontal shear reinforcement Spacing of horizontal shear area to gross concrete area shall not be reinforcement shall not exceedless than 0.0025 or 18 in.

11.9.9.4 Ratio of vertical shear reinforcement Spacing of vertical shear reinforcement shall not be less than shall not exceed or 18 in.

nor 0.0025

Two-way 13.3.1 Area of reinforcement in each direction Spacing of reinforcement at critical slab reinforcement shall be determined from moments at sections shall not exceed two times

critical sections, but shall not be less the slab thicknessthan required by 7.12.2.1

Minimum 14.3.2 Minimum ratio of vertical reinforcement 14.3.5 Vertical and horizontal reinforcement— area to gross concrete area shall be: reinforcement shall not be spaced Walls (a) 0.0012 for deformed bars not larger farther apart than three times the wall

than No. 5 with a specified yield strength thickness, nor farther apart than 18 in.not less than 60,000 psi

14.3.3 Minimum ratio of horizontal reinforcement area to gross concrete area shall be:(a) 0.0020 for deformed bars not larger than No. 5 with a specified yield strength not less than 60,000 psi

aIf more than one of these sections apply, the sections requiring the largest minimum area and the smallest spacing shall govern.

(rt - 0.0025)r/ = 0.0025 + 0.5(2.5 - hw//w) *

/w/3, 3h,r/

/w/5, 3h,

rt


Shear walls subjected to large moment reversals, as likely would occur duringstrong earthquake ground motions, may require larger minimum vertical-reinforcementareas than given by the previous requirements in order to prevent possible fracture of thevertical reinforcement [18-20].

Ties for Vertical Reinforcement

ACI Code Section 14.3.6 specifies that the distributed vertical reinforcement need not beenclosed by lateral ties if

(a) the vertical reinforcement area is not greater than 0.01 times the gross con-crete area or

(b) the vertical reinforcement is not required as compression reinforcement.

Many designers interpret part (b) of ACI Code Section 14.3.6 to mean that wall steel satis-fying (a) in this list that is not enclosed in transverse ties should be ignored in strength cal-culations if

• it is stressed in compression under static loads, or

• it is stressed by cyclic loads that cause compression in the bars.

Distributed wall steel placed in two separate curtains can be tied through the thick-ness of the wall using stirrups or through-the-wall cross-ties engaging reinforcement onboth faces of the wall. Although the ACI Code does not specify what fraction of the barsshould be tied in this manner, it is customary for such ties to engage every second or thirdbar each way on both faces [18-21].

In many cases the distributed vertical reinforcement has enough moment capacityto resist the wind load moments. If not, concentrated reinforcement is provided inboundary elements at the ends of the walls or at the intersections of walls. These ele-ments should contain vertical steel satisfying the minimum reinforcement requirementsfrom ACI Code Section 10.9.1 based on the area of the boundary elements, rather thanon the gross area of the wall. This steel is to be enclosed by ties satisfying ACI CodeSection 7.10.5. As a result, the steel is assumed to be restrained against buckling, andthus able to resist compressive bar forces. The boundary elements may be the same thick-ness as the rest of the wall, or they may be thicker.

Although ACI Code Chapter 14 does not require concentrated reinforcement atthe ends of the walls, it is good practice to at least use larger bars at the extreme ends ofthe wall.

Transfer of Wall Load through Floor Systems

In tall buildings, the strength of the concrete in the walls may be higher than the strength ofthe concrete in the floor system. Walls may be of high-strength concrete to reduce the lat-eral deflections and the wind-induced sway vibrations of the building. On the basis of testsof column–floor joints, ACI Code Section 10.12 allows an increase in the effective strengthof the floor concrete in locations where the floor concrete is confined on all sides. Thiseffect is smaller in edge and corner column joints than in interior joints, because there isless confinement of the joint concrete at edge or corner columns. The lack of confinementis even more pronounced in wall–floor joints, because a greater length of joint concrete isunconfined. In the absence of tests, it is recommended that the strength of wall-to-floorjoints be based on the lower of the wall and floor concrete strengths.

Section 18-10 Flexural Strength of Shear Walls • 999

18-10 FLEXURAL STRENGTH OF SHEAR WALLS

Flexure—Nominal Strength, Factored Loads, and Resistance Factors

Cross sections in a wall are designed to satisfy

(18-22)

(18-23)

(18-24)

where is the nominal resistance based on the specified material strengths, is the re-quired resistance computed from the factored loads, and so on. The strength-reduction fac-tor, comes from ACI Code Section 9.3.2 for flexure and axial loads and from ACI CodeSection 9.3.2.3 for shear. The factored loads are from ACI Code Section 9.2.1.

Strength-Reduction Factors for Flexure and Axial Load—ACICode Section 9.3.2

The strength-reduction factors for combined flexure and axial loads for a shear wall vary,depending on the maximum steel strains anticipated at ultimate load. As explained in Chapters5 and 11, the calculation of strength-reduction factors, is based on the strain, in the layerof steel at the depth, which is located farthest from the extreme-compression fiber.

Tension-controlled limit for a rectangular wall. The strength-reduction factor, can becomputed directly from the computed strain in the extreme-tension layer of reinforcement, .ACI Code Section 10.3.4 defines the tension-controlled limit load as the load causing a straindistribution having a maximum strain of 0.003 in compression on the most compressed face ofthe member, when the steel strain in the extreme layer of tension reinforcement, reaches0.005 in tension. From similar triangles, the neutral axis will be located at from the compressed face, where is the distance from the extreme-compression fiber to thecentroid of the layer of bars farthest from the compression face of the member. Thus, when cis less than or equal to , the wall is tension-controlled, and .

Compression-controlled limit for a rectangular wall. The compression-controlledlimit corresponds to a strain distribution with the neutral axis at . So, when c is greaterthan or equal to , the wall is compression-controlled, and

Flexural Strength of Rectangular Walls with UniformCurtains of Vertical Distributed Reinforcement

Code guidance on the use of vertical distributed reinforcement in walls loaded cyclically isambiguous. In seismic regions, the loads resisted by vertical bars that are not tied, are ignored.ACI Code Section 14.3.6 is more lenient. It requires that vertical bars be tied (a) if the totaldistributed vertical reinforcement, exceeds or (b) if the vertical reinforcement isincluded as compression reinforcement in the calculations. The following strength analysisapplies to walls with two curtains of reinforcement with ties through the wall to the other cur-tain of bars. The equations in this subsection also apply if the area of steel is less than 0.01.

ACI Code Section 21.6.4.2 suggests that, in seismic regions, column reinforcementwould be adequately tied if the center-to-center spacing of cross-ties or the legs of hoops did

0.01Ag,As,

f = 0.65.0.6dt

0.6dt

f = 0.90.375dt

dt

c>dt = 0.375dt

et,

et

f,

dt,et,f,

1f2

f,

MuMn

fVn Ú Vu

fNn Ú Nu

fMn Ú Mu


not exceed 14 in. Given this guidance, the author believes that the vertical reinforcement in anonseismic wall can be taken as “tied” if at least every second bar in a curtain of reinforcementis tied through the wall to a bar in the other curtain of steel, near the opposite face.

Figure 18-15a shows a rectangular wall section with a uniform distribution of verti-cal steel. We will assume that the wall is subjected to a factored axial load, and wewant to find the nominal flexural strength, for this wall using the assumed straindistribution in Fig. 18-15b. We will use a procedure developed by A. E. Cardenas and hiscolleagues [18-22] and [18-23]. They made the following assumptions at nominal strengthconditions for shear wall sections similar to that in Fig. 18-15a.

1. All steel in the tension zone yields in tension.

2. All steel in the compression zone yields in compression.

3. The tension force acts at middepth of the tension zone.

Mn ,Nu ,

Mn

(a) Typical wall section.

� 0.005

(b) Assumed strain distribution.

c /2

c

External forces

Internal forces

(c) Resultant external and internal forces acting on wall section.

Nu

Ow

n.a. 0.003

Mn

Nu

(Ow�c)/2 Ow /2

Ow /2

T C

Fig. 18-15Wall with uniform distribu-tion of vertical reinforcementsubjected to axial load andbending.


4. The total compression force (sum of steel and concrete contributions) acts atmiddepth of the compression zone.

From those assumptions and using to represent the total area of longitudinal (vertical)reinforcement, we can obtain the following expressions for the vector forces in Fig. 18-15c.

(18-25a)

(18-25b)

(18-25c)

and

(18-25d)

The percentage of total longitudinal reinforcement is

(18-26a)

and the longitudinal reinforcement index is

(18-26b)

Finally, Cardenas et al. defined an axial stress parameter as,

(18-27)

where represents the factored axial load, positive in compression.Enforcing section equilibrium leads to

Combining some terms and dividing both sides of this force equilibrium expression byresults in

Substituting the definitions from Eqs. (18-26) and (18-27), we can solve for the distance tothe neutral axis from the compression edge of the wall,

(18-28)

With this value for c, we can use Eq. (18-25) to find all of the section forces. Then, sum-ming moment about the compression force, C, in Fig. 18-15c, we get the followingexpression for the nominal moment strength of the wall section.

c = ¢ a + v0.85b1 + 2v

≤/w0.85b1

c

/w- ¢1 -

2c

/w≤ Asth/w

fy

fcœ =

Nuh/wfcœ

h/wfcœ

0.85fcœ hb1c + Astfy¢ c/w ≤ - Astfy¢ /w - c

/w≤ = Nu

Cc + Cs - T = Nu

Nu

a =Nuh/wfcœ

v = r/fy

fcœ

r/ =Asth/w

C = Cs + Cc

Cc = 0.85fcœ hb1c

Cs = Astfy¢ c/w ≤T = Astfy¢ /w - c

/w≤Ast


(18-29)

Cardenas and his colleagues were able to show good agreement between measuredmoment strengths from tests of shear walls and strengths calculated using Eq. (18-29).

The critical load case for evaluating the nominal moment strength of a structural wallnormally corresponds to ACI Eq. (9-6) or (9-7) in ACI Code Section 9.2.1.

(ACI Eq. 9-6)

(ACI Eq. 9-7)

If service-level wind forces are specified by the governing building code, use a load factorof 1.6 for W in ACI Eq. (9-6).

Either of these load cases will minimize the factored wall axial load, and thusminimize the wall nominal moment strength. Also, it can be shown for essentially all struc-tural walls that Thus, the wall section is tension-controlled, and

Moment Resistance of Wall Assemblies, Walls withFlanges, and Walls with Boundary Elements

Frequently, shear walls have webs and flanges that act together to form H-, C-, T-, andL-shaped wall cross sections referred to as wall assemblies. The effective flange widthscan be taken from ACI Code Sections 8.12.2 and 8.12.3. In regions subject to earthquakes,ACI Code Section 21.9.5.2 limits the flange widths to the smaller of

(a) half the distance to an adjacent web or

(b) 25 percent of the total height of the wall.

We shall use the limiting flange thicknesses from ACI Code Section 21.9.5.2 for both seis-mic and nonseismic walls.

Frequently, the thickness is increased at the ends of a wall to give room for concen-trated vertical reinforcement that is tied like a tied column (see ACI Code Section 7.10.5).The increased thickness also helps to prevent buckling of the flanges. Regions containingconcentrated and tied reinforcement are known as boundary elements, regardless ofwhether or not they are thicker than the rest of the wall.

Nominal Moment Strength of Walls with BoundaryElements or Flanges

In this subsection, we will discuss structural walls that have longitudinal reinforcementconcentrated at the edges to increase their nominal moment strength. A few typical exam-ples of such walls are shown in Fig. 18-16. The longitudinal reinforcement in boundaryelements similar to those in Fig. 18-16a and b will need to be tied with transverse rein-forcement that satisfies ACI Code Section 7.10.5 if the walls are in regions of low or noseismic risk. The confinement requirements are more stringent for boundary elements ofstructural walls in regions of high seismic risk, as will be discussed in Chapter 19.

When calculating the nominal moment strength, for walls similar to those inFig. 18-16, the contribution from the vertical reinforcement in the web is usually ignoredbecause its contribution to is quite small compared to the contribution from the verticalreinforcement concentrated at the edges of the wall. For some flanged sections or wallassemblies, as shown in Fig. 18-16c, ignoring the vertical reinforcement in the web may be

Mn

Mn ,

f = 0.9.c 6 0.375d.

Nu

U = 0.9D + 1.0E

U = 0.9D + 1.0W

Mn = T¢ /w2≤ + Nu¢ /w - c

2≤


too conservative. An alternative procedure for analyzing such wall assemblies in flexure isgiven in the next subsection.

The model used to analyze the nominal moment strength of a structural wall with bound-ary elements is shown in Fig. 18-17. For the boundary element in tension, the tension force is

(18-30)

where is the total area of longitudinal steel in the boundary element. The longitudinalsteel in the compression boundary element is ignored. Using the compression stress-blockmodel from Chapter 4, the compression force is

(18-31)

where b is the width of the boundary element. Enforcing section equilibrium for the verti-cal forces in Fig. 18-17 results in

(18-32)

Normally, the compression stress block is contained within the boundary element, as shown inFig. 18-17. If the compression force required for section equilibrium is large enough to causethe value of a to exceed the length of the boundary element shown as in Fig. 18-17, then asection analysis similar to that discussed in Section 4-8 for flanged sections will be required.

As discussed in the prior subsection, the critical load case for evaluating the nominalmoment strength of a structural wall normally corresponds to ACI Eq. (9-6) or (9-7) in ACICode Section 9.2.1. Either of these load cases will minimize the factored axial load,and thus minimize the wall nominal moment strength. Summing the moment about thecompression force in Fig. 18-17 leads to the following expression for

(18-33)Mn = Tad -a

2b + Nu¢ /w - a

2≤ Mn .

Nu ,

b¿

a =T + Nu0.85fc

œ b

C = 0.85fcœ ba

As

T = Asfy

(a) Boundary element within dimensions of wall.

(b) Wall with enlarged boundary element.

(c) Wall with reinforcement concentrated in flanges.

Fig. 18-16Structural walls with concen-trated reinforcement at theiredges.


For essentially all structural walls, it can be shown that the neutral-axis depth, c, is wellless than 0.375d, so the section is tension-controlled, and

Nominal Moment Strength of Wall Assemblies

Paulay and Priestley [18-17] present the following method of computing the required rein-forcement in a wall assembly. Figure 18-18 shows a plan of a wall consisting of a web andtwo flanges. The wall is loaded with a factored axial load, and a moment, that causescompression in flange 1; and a shear, parallel to the web. The axial load and the momentact through the centroid of the area of the wall. The moment can be replaced by an eccentricaxial load located at

(18-34)ea =MuaNu

Vu,Mua,Nu,

f = 0.9.

Nu

Mua

Centroidal axis

2

13

Nu

ea eb

C1 T3T2

xa x1 x2 xb

Nu

Fig. 18-18Reinforcement forces actingon a wall segment. (FromSeismic Design of ReinforcedConcrete and Masonry Build-ings, Paulay and Priestley,Copyright John Wylie &Sons, 1992. Reprinted by permission of John Wylie & Sons, Inc.)

b�

Mn

T C

a

a /2

b

Ow

Nu

d

Ow/2Fig. 18-17Flexural strength model forwall with boundary elements.

Section 18-11 Shear Strength of Shear Walls • 1005

from the centroid, which is equivalent to it acting at from the centroid of theflange that is in compression.

Because of the shear, the vertical reinforcement in the web will be assumed to bethe minimum steel required by ACI Code Section 11.9.9.4. The calculations will be sim-plified by assuming that all of the web steel yields in tension. We can assume this becauseunder cyclic loads, the neutral axis will alternately be close to the left and the right ends ofthe web. The steel in the web resists an axial force of

(18-35)

where is the area of distributed steel in wall element 2, the web. Summing momentsabout the centroid of the compression flange gives the following equation for the tension,

in the right-hand flange:

(18-36)

In a similar manner, the force in the tension reinforcement, required in the left-hand flangefor the axial load and the moment causing compression in the right-hand flange canbe computed as

(18-37)

The required area of concentrated vertical reinforcement in the flanges can be computedby dividing the tension forces from Eqs. (18-36) and (18-37) by the product,

Biaxially loaded walls. A wall is said to be biaxially loaded if it resists axial loadplus moments about two axes. One method of computing the strength of such walls is theequivalent eccentricity method presented in Section 11-7. In this method, a fraction between0.4 and 0.8 times the weak-axis moment is added to the strong-axis moment. The wall is thendesigned for the axial load and the combined biaxial moment treated as a case of uniaxialbending and compression.

Strictly speaking, the elastic moment resistance of an unsymmetrical wall shouldbe computed allowing for moments about both principal axes of the cross section. Thisis not widely done in practice. It is generally assumed that cracking of the walls and theproportioning of the vertical wall reinforcement can be done considering momentsabout one orthogonal axis at a time.

Shear Transfer between Wall Segments in Wall Assemblies

For the flanges to work with the rest of the cross section of a wall assembly, so-called “ver-tical shear stresses” must exist on the interface between the flange and web, even when thewall and the wall segments are constructed monolithically. The stresses to be transferred arecalculated in the same manner as for a composite beam, by using Eqs. (16-13) and (16-15).The reinforcement should satisfy ACI Code Section 11.6, Shear Friction.

18-11 SHEAR STRENGTH OF SHEAR WALLS

The design of structural walls for shear in nonseismic regions is covered in ACI CodeSection 11.9, Provisions for Walls. The basis for this code section was a series of testsof one-third-size, planar, shear walls, reported in [18-22], [18-23], [18-24]. The test

f * fy.

T1 LNuxb - T2x2

x1 + x2

Mub,

T3 LNuxa - T2x1

x1 + x2

T3,

As2

T2 = As2fy

Vu,

xa = ea - x1


specimens were divided between flexural shear walls with ratios of of 1.0,2.0, and higher and short shear walls with The basic shear-designequations are similar to those for the shear design of prestressed concrete beams:

(18-38)

(ACI Eq. 11-1)

(18-39)

(ACI Eq. 11-2)

(18-40)

ACI Code Section 11.9.3 limits to a maximum value of where d shall betaken as unless a strain–compatibility analysis is used to define the centroid of thetension force in bending. For walls with concentrated vertical reinforcement in boundaryelements at the edges of the walls, d may be measured from the extreme compressionedge to the centroid of the concentrated vertical reinforcement near the tension edge.

for Shear Walls

As with beam design, the concrete contribution to shear strength is set approximately equalto the value of shear that causes shear (inclined) cracking in a structural wall. For wallssubjected to axial compression, a designer is permitted to use Eq. (18-41), unless a moredetailed analysis is made, as will be discussed in the next paragraph,

(18-41)

where is the factor for lightweight aggregate concrete. It is taken as 1.0 for normal-weight concrete and shall be used as defined in ACI Code Section 8.6.1 for lightweightconcrete. For walls subjected to axial tension, a designer must use ACI Code Eq. (11-8)with h substituted for ,

(18-42)

where is negative for tension and shall be expressed in psi units.For structural walls subjected to axial compression, ACI Code Section 11.9.6

permits to be taken as the smaller of

(18-43)

(ACI Eq. 11-27)

or,

(18-44)

(ACI Eq. 11-28)

Vc = E0.6l2fcœ +/w¢1.25l2fcœ + 0.2

Nu/wh≤

MuVu

-/w2

UhdVc = 3.3l2fcœ hd +

Nud

4/w

Vc

Nu/AgNu

Vc = 2¢1 +Nu

500Ag≤l2fcœ hdbw

l

Vc = 2l2fcœ hd

Vc

0.8/w102fcœ hd,Vn

Vs Ú aVuf

- VcbVn = Vc + Vs

fVn Ú Vu

Mu>(Vu/w) = 0.50.Mu>(Vu/w)


Equation (18-43) corresponds to the shear force at the initiation of web-shear crackingand normally will govern for short walls. Equation (18-44) normally will govern for slen-der walls and corresponds to the shear force at the initiation of flexural-shear cracking at asection approximately above the base of the wall. If the quantity in Eq.(18-44) is negative, then Eq. (18-44) does not apply to the wall being analyzed. The valuefor is to be evaluated at a section above the base of the wall, and the distance to thatsection is to be taken as the smallest of and one story height (Fig. 18-19). Thevalue of computed at that section may be used throughout the height of the shear wall.

Shear Reinforcement for Structural Walls

Shear reinforcement for structural walls always consists of evenly distributed vertical andhorizontal reinforcement. In many cases, shear cracks in walls are relatively shallow (i.e.,their inclination with respect to a horizontal line is less than 45°), so vertical reinforcementwill be just as effective—if not more effective—as horizontal reinforcement in controllingthe width and growth of such cracks. However, the shear-strength contribution from wallreinforcement is based on the size and spacing of the horizontal reinforcement.

In many cases, only minimum amounts of shear reinforcement are required in struc-tural walls, and those minimum amounts are a function of the amount of shear being re-sisted by the structural wall. If the factored shear force, is less than then thedistributed vertical and horizontal wall reinforcement must satisfy the requirements of ACICode Section 14.3, as summarized in Table 18-1. If a designer uses Grade-60 reinforce-ment and bar sizes not larger than No. 5, then the minimum percentage of vertical steel is0.0012, and the minimum percentage of horizontal steel is 0.0020. Referring to Fig. 18-20for notation definitions, the percentage of vertical (longitudinal) steel is

(18-45a)

And the percentage of horizontal (transverse) steel is

(18-45b)

For both the horizontal and vertical steel, ACI Code Section 14.3.5 limits the bar spacingto the smaller of 3h and 18 in.

rt =Av,horiz

hs2

r/ =Av,vert

hs1

fVc/2,Vu ,

Vc

/w/2, hw/2,Mu/Vu

1Mu/Vu - /w/22/w/2

hwCritical section

�story height

Ow

�Ow /2

�hw /2

Fig. 18-19Location of critical sectionfor checking flexural-shearstrength.


For walls resisting a higher factored shear force (i.e., ), the verticaland horizontal steel in the wall must satisfy the minimum percentages and the maximumspacing requirements of ACI Code Section 11.9.9, as summarized in Table 18-1. The mini-mum percentage of horizontal (transverse) reinforcement, is 0.0025 and is to be placedat a spacing that does not exceed the smallest of and 18 in. Vertical reinforcementis to be placed at a spacing that does not exceed the smallest of and 18 in. The per-centage of vertical (longitudinal) steel, shall not be less than the larger of 0.0025 andthat calculated using ACI Code Eq. (11-30):

(18-46)

(ACI Eq. 11-30)

For walls with this equation will not govern. In shorter walls where thehorizontal reinforcement percentage, exceeds 0.0025, the value of calculated inEq. (18-46) does not need to exceed the amount (percentage) of horizontal reinforce-ment required for shear strength, as given next.

If walls are required to resist a factored shear force that exceeds then hori-zontal reinforcement must be provided to satisfy the strength requirement expressed inEq. (18-40). The shear strength, provided by the horizontal reinforcement is given byACI Code Eq. (11-29):

(18-47)

(ACI Code Eq. 11-29)

In this equation, is the same as shown in Fig. 18-20, and s is the same as in thatfigure. As was done for shear-strength requirements in Chapter 11, the designer has the optionto select a bar size and spacing to satisfy the strength requirement in Eq. (18-40). In addition tosatisfying the shear-strength requirement, a designer must check that both the horizontal andvertical wall reinforcement satisfy the minimum reinforcement percentages and maximumspacing limits given in ACI Code Section 11.9.9, which were discussed in the prior paragraph.

Shear Strength of Structural Walls Resisting Seismic Loads

Chapter 19 discusses Design for Earthquake Effects, but the author thought it was essentialto include here the modified shear-strength requirements given in ACI Code Chapter 21 forstructural walls. Although the equation that defines the nominal shear strength in ACI Code

s2Av,horizAv

Vs =Avfyd

s

Vs ,

fVc ,

r/rt ,hw//w Ú 2.5,

r/ = 0.0025 + 0.5¢2.5 -hw/w≤1rt - 0.00252

r/ ,/w/3, 3h,

/w/5, 3h,rt ,

fVc/2 6 Vu … fVc

s1 Av,vert

s2

Av,horiz

Fig. 18-20Distribution of vertical andhorizontal shear reinforcementin the web of a shear wall.


Chapter 21 appears to be significantly different from that given by the equations in ACICode Chapter 11, the final values for are not substantially different.

The nominal shear strength, for a wall designed to resist shear forces due toearthquake ground motions is given by ACI Code Eq. (21-7):

(18-48)

(ACI Eq. 21-7)

In this equation, is taken as the width of the web of the wall, h, multiplied by the totallength of the wall, This area is larger than the effective shear area, hd, used for theequations in ACI Code Chapter 11.

The first term inside the parenthesis of Eq. (18-48) represents the concrete contribu-tion to shear strength, The coefficient, represents the difference between theexpected occurrence of flexure-shear cracking in slender walls and web-shear cracking inshort walls. The value of is taken as 2.0 for walls with and as 3.0 for wallswith A linear variation for the value of is to be used for walls with ratios between 1.5 and 2.0.

The second term inside the parenthesis of Eq. (18-48) represents the shear-strengthcontribution from the horizontal wall reinforcement, Multiplying the definition of given in Eq. (18-45b) by the definition of and the yield strength, results inthe following equivalent value for

(18-49)

This equation is similar to Eq. (18-47), which gives the value for used in ACI CodeChapter 11. However, because the value of used in ACI Code Chapter 21exceeds the value of used in ACI Code Chapter 11.

The maximum allowable value for the nominal shear strength, from Eq. (18-48)is limited to in ACI Code Section 21.9.4.4. Again, this value is similar, but notthe same as the upper limit of given in ACI Code Section 11.9.3.

The requirements for minimum vertical and horizontal reinforcement percent-ages and for maximum permissible spacing are the same as those given in ACI CodeSection 11.9.9, which were discussed previously. The only modification is given inACI Code Section 21.9.4.3, which states that for walls with the valueof (vertical steel) shall not be less that the value of This requirement reflectsthe fact that in short walls, the vertical reinforcement is equal to or more efficient thanthe horizontal reinforcement in controlling the width and growth of inclined shearcracks.

The discrepancies noted here between the shear-strength equations given in ACICode Chapters 11 and 21 can create a dilemma for structural designers. The shearstrength of walls designed to resist lateral wind forces should be determined from theequations in ACI Code Chapter 11. However, if the same wall is designed to also resistequivalent lateral forces due to earthquake ground motions, the shear strength must bechecked using the equations in ACI Code Chapter 21. A unification of these differentshear-strength equations is a major goal of the Code Committee as it works toward futureeditions of the ACI Code.

rt .r/

hw//w … 2.0,

102fcœ hd8Acv2fcœ

Vn ,Vs

Vs/w 7 d,Vs

Vs,equiv =Av,horizfy/ws2

Vs,equiv =Av,horiz

hs2fyh/w

Vs .fy ,Acv1h/w2 rtVs .

hw//wachw//w … 1.5.hw//w Ú 2.0ac

ac ,Vc .

/w .Acv

Vn = Acv 1acl2fcœ + rtfy2Vn ,Vn


Shear Transfer across Construction Joints

The shear force transferred across construction joints can be designed using shear friction.The clamping force is the sum of the tensile reinforcement force components of all the ten-sile bar forces and permanent compressive forces acting on the joint. This includes all thereinforcement perpendicular to the joint, regardless of the primary use of this steel.

EXAMPLE 18-2 Structural Wall Subjected to Lateral Wind Loads

The moment and shear strengths of the structural wall shown in Fig. 18-21 are to beevaluated for the combined gravity and lateral loads applied to the wall. The wall is 18 ftlong and is 10 in. thick. A uniform distribution of vertical and horizontal reinforcement isused in two layers, one near the front face of the wall and the other near the back face. Thevertical reinforcement consists of No. 5 bars at 18 in. on centers in each face, and the hor-izontal reinforcement consists of No. 4 bars at 16 in. on centers in each face. Normal-weight concrete with a compressive strength of 4000 psi is used in the wall. All of the wallreinforcement is Grade-60 steel

The gravity loads applied at each floor level, as shown in Fig. 18-21, are due to deadload. The live loads are not shown but are assumed to be equal to approximately one-halfof the dead loads. The lateral wind loads are based on service-level wind forces and didinclude the directionality factor.

1. Make an Initial Check of Wall Reinforcement. The percentage and spacing ofthe vertical and horizontal reinforcement will be checked before we do the strength calcu-lations. The percentage of horizontal reinforcement is found using Eq. (18-45b):

This satisfies the minimum requirement in ACI Code Section 11.9.9.2, so it should be ac-ceptable unless a larger amount is required to satisfy shear strength requirements. The maxi-mum center-to-center spacing for the horizontal reinforcement is the smallest of /w /5

rt =Av,horiz

hs2=

2 * 0.20 in.2

10 in. * 16 in.= 0.0025

1fy = 60 ksi2.

Critical sectionfor flexural-shear

9 ft

10.5 ft

10.5 ft

10.5 ft

10.5 ft

12.0 ft

30 k

6 k

11 k

16 k

20 k

22 k

50 k

50 k

50 k

50 k

18 ftFig. 18-21Structural wall for Example 18-2.


3h (30 in.), and 18 in. Thus, the provided spacing for the horizontal rein-forcement is o.k.

Although it is good practice to use larger vertical bars at the edges of the wall, sayNo. 6 or No. 7 bars, we will calculate the percentage of vertical reinforcement assumingonly No. 5 bars are used. From Eq. (18-45a),

Because the wall has an aspect ratio, Eq. (18-46) will not governfor the minimum percentage of vertical reinforcement. Thus, the minimum percentage of ver-tical reinforcement is 0.0025 (ACI Code Section 11.9.9.4), which is less than what is pro-vided. The spacing limit for the vertical reinforcement is the smallest of 3h (30 in.), and 18 in. Thus, the provided spacing of vertical reinforcement is o.k.

2. Check Moment Strength. The moment at the base of the wall is equal to the sumof the products of the lateral forces times their respective distances to the base of the wall:

The appropriate load factor for service-level wind forces is 1.6. Thus, the factored momentat the base of the wall is

The analysis given in [18-23] will be used to evaluate the moment strength of the wall.Using ACI Code Eq. (9-6), the factored axial load is

For 4000-psi concrete, Other required parameters are

(18-26b)

and

(18-27)

With these parameters, we can use Eq. (18-28) to find the depth to the neutral axis:

If we assume that the effective flexural depth, d, is approximately equal to it is clear that c is significantly less than 0.375d. Thus, this is a tension-controlled section,and we will use a strength reduction factor, equal to 0.9.

To calculate the nominal moment strength, we first must determine the tension forceat nominal strength conditions. The value of for the vertical steel can be calculated as

Ast = 2Ab/ws1

= 2 * 0.31 in.2 *216 in.

18 in.= 7.44 in.2

Ast

f,

173 in.,0.8/w =

c = a 0.0240 + 0.0516

0.85 * 0.85 + 2 * 0.0516b216 in. = 19.8 in.

c = ¢ a + v0.85b1 + 2v

≤/wa =

Nuh/wfcœ

=207 kip

10 in. * 216 in. * 4 ksi= 0.0240

v = r/fy

fcœ = 0.00344

60 ksi

4 ksi= 0.0516

b1 = 0.85.

Nu = 0.9ND = 0.9 * 230 kips = 207 kips

Mu1base2 = 1.6 * 2910 = 4660 kip-ft

= 2910 kip-ft

M1base2 = 22 * 54 + 20 * 43.5 + 16 * 33 + 11 * 22.5 + 6 * 12

/w/3 16 ft = 72 in.2,hw//w = 3.0 7 2.5,

r/ =Av,vert

hs1=

2 * 0.31 in.2

10 in. * 18 in.= 0.00344

(3.6 ft = 43.2 in.2,


Then, from Eq. (18-25a),

Now, referring to the vector forces in Fig. 18-15c, we can use Eq. (18-29) to calculate thenominal moment strength of the structural wall.

Using the strength reduction factor, we get

Because is larger than (4660 kip-ft), the wall has adequate flexural strength.

3. Check Shear Strength. The factored shear at the base of the wall is

Because this wall is slender, we can assume that Eq. (18-44) will govern forthe shear-strength contribution from the concrete, However, we will check both Eqs.(18-43) and (18-44) to demonstrate how each is used. For normal-weight concrete,

Using (173 in.) and putting all of the quantities into units of poundsand inches, the value of from Eq. (18-43) is

For Eq. (18-44), we need to evaluate the ratio of at the critical section abovethe base of the wall. The distance to that section is the smallest of (9 ft), (27 ft),and one story height (12 ft). In this case, governs, as shown in Fig. 18-21. The factoredmoment at that section can be found from the following as

Thus, the ratio of Using this value in Eq. (18-44) andexpressing all of the quantities in pounds and inches, we have

Mu/Vu = 3580/120 = 29.8 ft.

= 4660 kip-ft - 120 kip * 9 ft = 3580 kip-ft

Mu1crit. sect.2 = Mu 1base2 - Vu 1base2 /w2

/w/2hw/2/w/2

Mu/Vu

= 1361,000 + 41,4002 lbs = 402 kips

= 3.3 * 124000 * 10 * 173 +207,000 * 173

4 * 216

Vc = 3.3l2fcœ hd +Nud

4/w

Vc

d = 0.8/wl = 1.0.

Vc .hw//w = 3.0,

Vu = 1.616 + 11 + 16 + 20 + 222 = 120 kips

MufMn

fMn = 0.9 * 5340 = 4800 kip-ft

f,

= 64,000 kip-in. = 5340 kip-ft

= 405 ka216 in.

2b + 207 ka216 in. - 19.8 in.

2b

Mn = T¢ /w2≤ + Nu¢ /w - c

2≤

T = 7.44 in.2 * 60 ksia216 in. - 19.8 in.

216 in.b = 405 kips

T = Astfy¢ /w - c/w

≤


This value governs for as expected. Using as defined in ACI Code Section9.3.2.3 for shear and torsion, This exceeds so nocalculation of the value of for the provided horizontal (transverse) reinforcement is required.It is clear that the value of is less than so the requirements for horizontal and verti-cal reinforcement in ACI Code Section 11.9.9 will govern for this wall. Those requirementswere checked in step 1 and were found to be at or above the ACI Code required values.

Thus, the 10-in. thick structural wall with the indicated vertical and horizontalreinforcement has adequate moment and shear strength and the reinforcement satis-fies all of the ACI Code requirements. ■

EXAMPLE 18-3 Structural Wall Subjected to Equivalent Lateral Earthquake Loads

The structural wall shown in Fig. 18-22 is to be analyzed for a combined gravity load andequivalent lateral load due to earthquake ground motions. The dimensions of the wall andthe section reinforcement (Fig. 18-22b) are taken from a paper by Wallace and Thomsen[18-25], in which the authors analyzed the need for special confinement reinforcement forthe boundary elements at the edges of the wall. The details for special confinement rein-forcement will be discussed in Chapter 19.

This wall is one of several walls resisting lateral loads for the structure that was ana-lyzed by Wallace and Thomsen [18-25]. The lateral earthquake force, E, assigned to thiswall is 205 kips. The dead and reduced live load at the base of the wall are and Use (normal-weight concrete) and for allreinforcement.

1. Check Moment Strength at Base of Wall. ACI Code Eq. (9-7) will be used tofind the factored loads for this analysis.

(ACI Eq. 9-7)

Seismic lateral loads, as defined in [18-18], typically have an inverted triangular distributionover the height of the building. Therefore, we will assume that the variable x shown in Fig.18-22a, which is taken as the assumed distance from the base of the wall to the centroid ofthe lateral force, has a value of Thus, the factored moment at the base of the wall is

The nominal moment strength of the wall will be analyzed by the procedure shown inFig. 18-17 for a wall with a vertical reinforcement concentrated in a boundary element. Thetension strength for the concentrated reinforcement is

T = Asfy = 10 * 1.27 in.2 * 60 ksi = 762 kips

= 16,400 kip-ft

Mu = 1.0 * 205 k * 2�3 * 120 ft

2�3 hw .

U = 0.9D + 1.0E

fy = 60 ksifcœ = 4000 psiNL = 450 kips.

ND = 1000 kips

Vu ,0.5fVc

Vs

Vu ,fVc = 0.75 * 212 = 159 kips.f = 0.75,Vc ,

= 212,000 lbs = 212 kips

= c37.9 psi +216 in.179.1 psi + 19.2 psi2

358 in. - 108 in.d * 10 in. * 173 in.

Vc = ≥0.6l2fcœ +/w¢1.25l2fcœ + 0.2

Nu/wh≤

MuVu

-/w2

¥hd


The distance from the compression edge to the centroid of the tension reinforcement is

The factored axial load for this load case is

Now, we will use Eq. (18-32) to determine the depth of the compression stress block. Inthis case, giving

The distance to the neutral axis is, which is well lessthan Therefore, this is a tension-controlled section, and

Eq. (18-33) will be multiplied by to find

fMn = f sTad -a

2b + Nu¢ /w - a

2≤ tfMn:ff = 0.9.

0.375d = 0.375 * 273 = 102 in.c = a/b1 = 40.7/0.85 = 47.9 in.,

a =T + Nu0.85fc

œ b=

762 k + 900 k

0.85 * 4 ksi * 12 in.= 40.7 in.

b = h = 12 in.,

Nu = 0.9ND = 0.9 * 1000 k = 900 kips

d = /w - 13 in. + 2 * 6 in.2 = 288 - 15 = 273 in.

24 ft

x

120 ft

EG

6 in.3 in.

3 in.

9 in. 3 in.

(a) Wall elevation.

(b) Wall section.

12 in.

3 in. 4 at 6 in. 12 at 18 in. 9 in. 4 at 6 in.

No.4 hoop & cross tie at 12 in.Ab � 0.20 in2

No.5 bars at 18 in.Ab � 0.31 in2

10 No.10 barsAb � 1.27 in2

Fig. 18-22Structural wall for Example 18-3.


Because exceeds the wall has adequate moment strength.

2. Capacity-Based Design Shear. In seismic design, the design shear force, isnot often based on the factored loads but rather on the probable flexural strength of themember in question. This is referred to as the capacity-based design procedure and is in-tended to ensure a ductile flexural response in the member—as opposed to a brittle shearfailure—if the member is loaded beyond its elastic range of behavior during an earthquake.

For the wall in this example, the probable flexural strength should be based on aprobable axial load that the wall will be carrying at the time of an earthquake. A value forsuch an axial load is not defined in either the ACI Code or ASCE/SEI 7-10 [18-18]. The au-thor will assume that a reasonable value for the probable axial load is the sum of the un-factored dead load plus the unfactored reduced live load.

With this axial load, the wall moment strength will be reevaluated and referred to as theprobable moment strength, First, the depth of the compression stress block is

Although this is larger than calculated previously, it still is clear that the tension steel in theboundary element will be yielding. Note, we do not need to show that this is a tension-controlled section. We only need to show that this is an underreinforced section and thatthe tension steel will be yielding when the concrete in the compression zone reaches themaximum useable strain of 0.003. With this calculated value of a, we are now ready to cal-culate the moment strength.

The final step is to find the lateral shear force required to develop this moment at thebase of the wall. As originally discussed by Bertero et al. [18-26], the lateral load distribu-tion over the height of a shear wall that is part of a complete structural system will tend tobe closer to a uniform distribution at peak response of the structure when it is subjected tolarge earthquake motions. Therefore, to more conservatively predict the capacity-basedshear force acting on the wall when the base moment reaches the author recommendsusing a value of in Fig. 18-22a. Using that value, the capacity-based designshear is

For this particular structural wall, the capacity-based design shear is approximately 2.5times the design shear used to check the flexural strength.

Vu1cap-based2 =Mpr

0.5hw=

29,700 k-ft

0.5 * 120 ft= 495 kips

x = 0.5hw

Mpr ,

= 357,000 kip-in. = 29,700 kip-ft

= 187,000 k-in. + 169,000 k-in.

Mpr = Tad -a

2b + Npr ¢ /w - a

2≤

a =T + Nu0.85fc

œ b= 54.2 in.

Mpr .

Npr = ND + NL = 1000 k + 450 k = 1450 kips

Vu ,

Mu ,fMn

= 273,000 kip-in. = 22,800 kip-ft

= 0.93193,000 k-in. + 111,000 k-in.4= 0.9 s762 ka273 -40.7

2b in. + 900 ka288 - 40.7

2b in. t


3. Check Shear Strength. Eq. (18-48), which is the same as ACI Code Eq. (21-7),will be used to check the shear strength of the wall. For this wall, the value of in thatequation is equal to the gross wall area, so

Because this is a slender wall, Also, because the wall is constructed with normal-weight concrete, Eq. (18-45b) will be used to determine for the distributed hor-izontal reinforcement in Fig. 18-22b as

This is greater than the minimum required percentage of 0.0025 from ACI Code Section11.9.9.2, and the 18-in. spacing satisfies the limits in ACI Code Section 11.9.9.3. Using thevalues calculated here and units of pounds and inches, the nominal shear strength of thewall from Eq. (18-48) is

ACI Code Section 21.9.4.4 limits the value of to Thus, we canuse the value calculated here. ACI Code Section 9.3.2.3 states that for shear and torsion,

So, the reduced, nominal shear strength of the wall is

Because (cap-based), the wall shear strength is o.k.

Thus, the 12-in. thick structural wall with the indicated vertical and horizontalreinforcement has adequate moment and shear strength, and the reinforcement satis-fies all of the ACI Code requirements. ■

18-12 CRITICAL LOADS FOR AXIALLY LOADED WALLS

Buckling of Compressed Walls

The critical stress for buckling of a hinged column or a one-way wall, with a rectangularcross sectional area is

(18-50)

where b is the width of the wall, and is the effective length of the wall. The flexural stiff-ness of a wall section of width b and thickness h is.

EI =Ebh3

12

k/

scr =Pcrbh

=p2EI1k/22 a 1

bhb

1b * h2

fVn 7 Vu

fVn = 0.75 * 1030 k = 773 kips

f = 0.75.

8Acv2fcœ = 1750 kips.Vn

= 1.03 * 106 lbs = 1030 kips

= 3460 in.2 1126 psi + 172 psi2= 3460 in.2 12 * 124000 + 0.00287 * 60,0002Vn = Acv1acl2fcœ + rtfy2

rt =Av,horiz

hs2=

2 * 0.31 in.2

12 in. * 18 in.= 0.00287

rtl = 1.0.ac = 2.0.

Acv = h/w = 12 in. * 288 in. = 3460 in.2

Acv

Section 18-12 Critical Loads for Axially Loaded Walls • 1017

Buckling of Compressed Walls

The corresponding stiffness of a slab or a two-way wall, per unit of width of wall, is

(18-51)

For an axially loaded and uncracked concrete column with little or no reinforcement thecritical stress, is a function of the tangent modulus of elasticity, evaluated at thecritical stress, Substituting into Eq. (18-50), replacing with replacing EI withthe plate stiffness, D, and including an appropriate edge restraint factor, K, gives the criti-cal stress for a two-way wall:

(18-52)

In this equation, b is the effective width of the panel, taken equal to the smaller of thewidth and height of the wall between lateral supports, h is the thickness of the wall,is Poisson’s ratio, and K is an edge restraint factor that varies depending on thedegree of fixity of the edges of the plate. It is similar to the effective length term inEq. (18-50). The term K is plotted in Fig. 18-23. Values of K at buckling for other typesof compressed plates are given in books on structural stability [18-27]. Equations (18-50)and (18-52) have been included to show the similarity between the critical loads forcolumns and walls.

k2

m

scr =Pcrbh

= Kp2ET

1211 - m22 ahb b2

b2,/2ETscr.ET,scr,

D =Eh3

1211 - m22

Fig. 18-23Wall buckling factor, K, forbuckling of a compressed plateanalogous to a compressedwall in a hollow rectangularbridge pier. (From [18-4].)

8

6

4

2

01 2 3 4

hw /b � 1

hw /b

K

Values of wall buckling constant, K.


Properties of Concrete Affecting the Stability of Concrete Walls

Compressive Stress–Strain Curve for the Concrete

The tangent modulus of the concrete in the wall will be used to estimate the tangent mod-ulus buckling loads. A good approximation of the stress–strain curve for the wall is thesecond-degree parabola given by Eq. (18-53) and shown in Fig. 18-24a, with a peak stressof The apex of the parabola will be taken at a strain of and:

(18-53)

where is the compressive stress in the most compressed wall of the box pier; is thecorresponding strain, assumed constant over the thickness of the wall. The strain at anylevel of stress in Eq. (18-53) is given by

(18-54)

The tangent modulus of elasticity, is the slope of the tangent to the stress–strain curveat some particular stress, in this case of point A in Fig. 18-24. It is computed using

evaluated at the critical stress, For example, the tangent modulus at point A in Fig. 18-24bis calculated as

(18-55)

The 2 in Eq. (18-55) comes from the fact that the tangent to a second-order parabola inter-sects the vertical axis through at 2 times the stress increment between and asshown in Fig. 18-24b. For example, the initial tangent modulus at for thestress–strain curve given by Eq. (18-55) is for :

ET =2 * 3600 psi

0.0020= 3,600,000 psi

fcœ = 4000 psi

f2 = 0f2,max,f2eco

ET =2¢f2¢e2

scr.

ET = ds>deET,

ec = eco a1 - A1 -f2f2, max

be2f2

f2 = f2, max c2a e2eco b - a e2ecob2 d

eco = 0.002f2,max = 0.90fcœ .

00 0.002 0.003 Strain

(a) Stress–strain curve.

0 0.002 0.003 Strain

(b) Tangent modulus at point A.

A

�e2

�f2�f2

Str

ess

0.9fc�0.9fc�

Fig. 18-24Second-order parabolicstress–strain curve for com-pressed concrete .


This is very close to the value obtained from ACI Code Section 8.5:

The amount of reinforcement in a wall is generally so small that it can be ignored whendetermining .

Post Buckling Behavior of Concrete plates

After a uniaxially loaded wall panel hinged on four edges buckles, the strips of wall along theedges parallel to the load continue to resist a portion of the compressive in-plane load asshown in Fig. 18-25. At the same time, the load resisted by the center portion of the buckledplate decreases. Failure generally comes from excess bending of the buckled parts. The neteffect is that the plate does not suddenly fail. Instead, strips along the edges of the wall con-tinue to support a stress that may approach the buckling stress. This is offset by a drop in thestress resisted by the center of the plate.

EXAMPLE 18-4 Compute the Buckling Load of a Wall

A bridge pier consists of a hollow, rectangular concrete tube similar to the one shown inFig. 18-26. Side A–B–C–D–A has the largest slenderness ratios of any of the sides of thebox pier. It has The concrete strength is 4000 psi. The applied loads causeaxial load and bending about the weak axis of the box as shown in Fig. 18-26. Thestress–strain curve of the concrete is assumed to be given by Eqs. (18-53) and (18-54),with a peak compressive stress, equal to

Compute the buckling load for a side wall having a slenderness ratio equal to 25.hw>h0.9fcœ .f2,max,

hw>h = 25.

ET

E = 57,0002fcœ = 3,600,000 psi

Buckle

Fig. 18-25Stresses in a plate after buckling.


1. Assume a Stress–Strain Curve for the Concrete. We will use the second-degree parabolic stress–strain curve given by Eq. (18-53) with a peak stress and strain of

and

The strain at any level of stress in Eq. (18-54) is given by

Assume the wall thickness is small compared to the overall distance from theextreme-compression fiber to the neutral axis in the pier. This allows us to assume a uni-form stress distribution across the section at the middle of the compressed wall. We shallignore the effect of reinforcement on the stiffness of the concrete.

2. Compute the Tangent Modulus of Elasticity for Various Values of Weshall take the maximum stress at a crushing failure of the wall equal to

Values of will be computed for a series of values of and, hence, of

(18-55)ET = 21¢f22>1¢e22 f2>f2,max.f2ec3600 psi.0.90 * 4000 psi =

f2>f2,max.

ec = eco a1 - A1 -f2f2 max

bf2 = f2, max c2a e2

ecob - a e2

ecob2 deco = 0.002:f2,max = 0.90fc

œ

2.9 in.

A

F

H

E

B

D

51 in.C

72 in. � hw

P

Axis ofbending

P

P

Pb � 102 in.

Fig. 18-26Geometry, dimensions, andbuckled shape of the compres-sion flange of a hollow piertest specimen. (Note (a) thatthe concrete end blocks are notshown and (b) the deflectionsare plotted to an enlarged scaling). (From [18-21].


where is the increase in stress from to and is the increase in strain as thestress increases from to and the strain increases from to

3. Compute the Critical Wall Slenderness Ratios for Buckling at VariousStress Ratios, Use to ratios of 0.85, 0.90, 0.925, 0.95, and 0.975.

The equation for the critical stress of a wall hinged on all four sides and subjected toin-plane load in one direction is:

(18-52)

A plate with height (parallel to the load) and width will buckle in a seriesof square bulges, alternately on one side of the plate, and then the other. One bulge willform if two bulges if and so on. At integer values of the buck-ling constant,

Substituting and assuming thatthe critical slenderness ratios, are computed in Table 18-2 for the most highly

stressed side of the box (side A–B–C–D–A) in Fig. 18-26. This is done for a series of val-ues of stress ratios given in Table 18-2.

The calculated values of in Table 18-2 indicate that a side wall with would buckle at a stress of 94 percent of the computed flexural capacity.

In the bridge pier tests described in [18-21] and [18-28], the failure strengths ofpiers with wall thickness ratios between 15 and 35 were reduced by plate buckling. For

Table 18-2 predicts, a buckling stress of 94 percent of the flexural capacity.This suggests that plates, this thin or thinner, should not be used in situations where suchwalls are highly stressed in compression. It is difficult to apply this directly to the stabilityof an assemblage of plates such as those that might enclose an elevator shaft, becausethe floor slabs and elevator door sills brace the wall at various points along the height ofthe wall. ■

EXAMPLE 18-5 Compute the Buckling Load of a Compressed Wall Flange

Stability of Compressed Flanges.

Figure 18-27a shows a portion of the projecting flange A–C–D–F–A of an H-shapedwall. When such a flange is loaded in compression, it may buckle in a series of up-and-down bulges at the free edge of the flange, as shown in Fig. 18-27b. For analysis, theflange can be idealized as a rectangular plate, connected to the web by a hinge A–B–Calong one long edge of length free along the other long edge D–E–F, and hinged alongthe two short edges, A–F and C–D. The flange projects a distance b from the web andhas a thickness t. The buckling load of such a flange can be computed from Eq. (18-52).

/,

hw>h = 25,

hw>h = 25hw>hf2>f2,max

hw >h,K = 4,scr = 0.85fc

œ 0.90fcœ = 3600 psi;m = 0.20;

K = 4.hw>/,hw>/ = 2hw>/ = 1,

b = /hw

scr =Pcrbh

= Kp2ET

1211 - m22 ahb b2

f2,maxf2f2>f2,max.

eco.e2f2,max,f2

¢e2f2,maxf2¢f2

TABLE 18-2 Critical Slenderness Ratios for a Wall Subjected to Various of

0.85 0.90 0.925 0.942 0.95 0.975

0.00095 0.00113 0.00124 0.00133 0.00138 0.001561,025,100 825,900 710,600 622,100 576,400 405,00034 30 27 25 24 20

Origin: Equation 18-52, Example 18-4.

hw >hET psi =ec =

f2>f2,max =

f2>f2,max


Hinged support

Free edge

E

D

F

Hinged supports

b

b

b

b

b

A

B

C

A

F

D

C

(a) Location of portion of the flange considered.

(b) Deformed shape of buckled flange.

O

Fig. 18-27Buckling of an outstandingflange of a wall assemblage.

In this case, books on elastic stability show that K is nearly constant and is equal to 0.5when [18-27].

Substituting into Eq. (18-52), the critical slenderness ratio for an out-standing flange is found to range from 7 to 12 as the average compressive stress in theflange, increases from 0.85 to 0.975 times . The calculations are summarizedin Table 18-3. The values of the critical slenderness ratios, b/t for and0.925 have been rounded off to the nearest integer, in this case to 10. The analysis givenin Table 18-3 predicts that an outstanding flange with a width-to-thickness ratio of

is in danger of buckling at flange stresses of 92.5 percent of the maximumstress in the stress–strain curve. This may limit the flexural capacity of flangedshear walls. These are similar to the limits on the flange widths allowed in codes for thedesign of steel structures. ■

f2, max

b>t = 10

f2>f2, max = 0.90f2, maxf2,

K = 0.5b>t Ú 4

TABLE 18-3 Critical Slenderness Ratios, b/t, for a Wall FlangeSubjected to Various Compression Stress Ratios,

.85 0.90 0.925 0.95 0.975

0.000946 0.001130 0.001240 0.001380 0.0015561,025,000 825,900 710,600 578,000 405,000

12 10 10 8 7

Origin: Equation 18-52, Example 18-5.

b>t =ET, psi =ec =

f2>f2,max

f2>f2,max

Problems • 1023

PROBLEMS

Horizontal reinforcement:No. 4 at 16 in., EF

Vertical reinforcement:No. 4 at 12 in., EF

Typical wall section

20 ft

15 ft

11 ft

11 ft

11 ft

11 ft

30 k20 k

22 k

18 k

12 k

7 k

10 in.

50 k

50 k

50 k

50 k

Fig. P18-1

18-1. Check the moment and shear strength at the base ofthe structural wall shown in Fig. P18-1. Also, showthat the given horizontal and vertical reinforcementsatisfies all of the ACI Code requirements regardingminimum reinforcement percentage and maximumspacing. The given lateral loads are equivalent windforces, considering both direct lateral forces and the

effects of any torsion. Use a load factor of 1.6 for thewind load effects. The given vertical loads representdead loads, and you can assume that the vertical liveloads are equal to 60 percent of the dead loads.Assume the wall is constructed with normal-weightconcrete that has a compressive strength of 4500psi. Assume all of the steel is Grade 60.

18-2. Design a uniform distribution of vertical and hori-zontal reinforcement for the structural wall shownin Fig. P18-2. Your design must satisfy all of theACI Code strength requirements, as well as therequirements for minimum reinforcement percent-age and maximum spacing. The given lateral loadsare equivalent wind forces, considering both directlateral forces and the effects of any torsion. Use aload factor of 1.6 for the wind load effects. Assumethe wall is constructed with normal-weight con-crete that has a compressive strength of 4000 psi.Assume all of the steel is Grade 60.

D � 40 k, L � 25 k

D � 70 k, L � 60 k

h � 10 in.

20 ft

75 k

100 k

12

12

Fig. P18-2


E

D, L

Wall thickness:10 in.

15 ft

20 ft

Horizontal reinforcement:No. 4 at 16 in., EF

Vertical reinforcement:No. 5 at 16 in., EF

Fig. P18-4

18-3. The structural wall shown in Fig. P18-3 is subjectedto gravity loads and an equivalent, static, lateral earthquake load,

(torsional effects included). Checkthe moment strength at the base of the structuralwall assuming that the distance, x, from the base ofthe structure to the lateral force, E, is equal to two-thirds of the wall height. Use a capacity-design ap-proach to check the shear strength of the wall andassume the distance, x, is equal to one-half of thewall height for this check. Also, show that the given

E = 150 kips

L = 100 kips21D = 150 kips and

Boundary element: 20 in. 20 in. with 12 No.8 bars

Web: 10 in. thick with No. 4 bars at 16 in., EF, EW

E

X A

D, L

A

60 ft

15 ft

15 ft

Wall section A–A

Fig. P18-3

horizontal and vertical reinforcement in the web ofthe structural wall satisfies all of the ACI Code re-quirements regarding minimum reinforcement per-centage and maximum spacing. Assume the wall isconstructed with normal-weight concrete that has acompressive strength of 5000 psi. Assume all of thesteel is Grade 60.

18-4. The structural wall shown in Fig. P18-4 is sub-jected to gravity loads

and a single, equivalent, static, lat-eral earthquake load at the top ofthe wall (at the roof level). For the given uniformdistribution of vertical and horizontal reinforce-ment, check the moment strength at the base of thewall and use a capacity-design approach to checkthe shear strength of the wall. Also, show that thegiven horizontal and vertical reinforcement in thestructural wall satisfies all of the ACI Code require-ments regarding minimum reinforcement percent-age and maximum spacing. Assume the wall isconstructed with normal-weight concrete that has acompressive strength of 4000 psi. Assume all of thesteel is Grade 60.

18-5. Design a uniform distribution of vertical and hori-zontal reinforcement for the first story of the struc-tural wall shown in Fig. P18-5. The given horizontalloads are strength-level static-equivalent earthquakeforces, so use a load factor of 1.0. The vertical deadand live loads are from the tributary area adjacent tothe wall. Assume the given lateral loads include thedirect shear force and any torsional effects that needto be considered. Use a compressive strength of4000 psi and Grade 60 reinforcement. Your final de-sign should satisfy all ACI Code requirements forminimum reinforcement percentages and maximumspacing.

1E = 220 kips2L = 40 kips2 1D = 80 kips and

References • 1025

REFERENCES

18-1 G. D. Oberlender and N. J. Everard, “Investigation of Reinforced Concrete Walls,” ACI Journal,Proceedings, Vol. 74, No. 6, June 1977, pp. 256–263.

18-2 K.M. Kripanarayanan, “Interesting Aspects of the Empirical Wall Design Equation,” ACI Journal,Proceedings, Vol. 74, No. 5, May 1997, pp. 204–207.

18-3 ACI Committee 551, “Tilt-Up Concrete Construction Guide,” (ACI 551.1R-05), ACI Manual ofConcrete Practice, American Concrete Institute, Farmington Hills, MI.

18-4 Concrete Design Manual, Cement Association of Canada, Ottawa, Ontario, January 2006.18-5 Noel D. Nathan, “Slenderness of Prestressed Concrete Columns,” PCI Journal, Vol. 28, No. 2,

March–April 1983, pp. 50–77.18-6 Jacob S. Grossman, “Slender Concrete Structures—The New Edge,” ACI Structural Journal, Vol. 87,

No. 1, January–February 1990, pp. 39–52.18-7 (Former) ACI Committee 442, “Response of Concrete Buildings to Lateral Forces,” (ACI 442R-88),

ACI Manual of Concrete Practice, American Concrete Institute, Farmington Hills, MI, 1989.18-8 Iain A. MacLeod, Shear Wall-Frame Interaction, Special Publication SP011.01D, Portland Cement

Association, Skokie, IL, 1971, 62 pp.18-9 A. R. Santhakumar and Thomas Paulay (Supervisor), Ductility of Coupled Shearwalls, Ph.D. Thesis,

University of Canterbury, Christchurch, New Zealand, October 1974.18-10 CSA Technical Committee on Reinforced Concrete Design, A23.3-04, Design of Concrete Structures,

Canadian Standards Association, Mississauga, Ontario, January 2006.18-11 Alexander Coull and J. R. Choudhury, “Analysis of Coupled Shear Walls,” ACI Journal, Proceedings,

Vol. 64, No. 9, September 1967, American Concrete Institute, Detroit, pp. 587–593.18-12 Joseph Schwaighofer and Michael P. Collins, “Experimental Study of the Behavior of Reinforced

Concrete Coupling Slabs,” ACI Journal, Proceedings, Vol. 74, No. 3, March 1977, pp. 123–127.18-13 Thomas Paulay and R. G. Taylor, “Slab Coupling of Earthquake-Resisting Shearwalls,” ACI Journal,

Proceedings, Vol. 78, No. 2, March–April 1981, pp. 130–140.18-14 International Building Code, International Code Council, Washington, D.C. 2009.18-15 Notes on ACI 318-05 Building Code Requirements for Structural Concrete, Portland Cement

Association, Skokie, IL, 2005.18-16 Loring A. Wyllie, Jr., “Chapter 7, Structural Walls and Diaphragms—How They Function,” Building

Structural Design Handbook, Wiley–Interscience, New York, 1987, pp. 188–215.18-17 Thomas Paulay and M.J. Nigel Priestley, Seismic Design of Reinforced Concrete and Masonry

Buildings, Wiley Interscience, New York, 1992, 744 pp.18-18 Minimum Design Loads for Buildings and Other Structures (ASCE/SEI 7–10), American Society of Civil

Engineers, Reston, VA, 2010, 608 pp.18-19 National Building Code of Canada, 2005, Canadian Commission on Building and Fire Codes, National

Research Council of Canada, Ottawa, Ontario, Canada.

16 ft

12 ft

12 ft

16 ft

22 k

24 k

20 k

D � 35 k, L � 25 k

D � 50 k, L � 40 k

D � 70 k, L � 60 k

h � 10 in.

Fig. P18-5


18-20 Sharon Wood, “Minimum Tensile Reinforcement Requirements in Walls,” ACI Structural Journal, Vol.86, No. 4, September–October 1989, pp. 582–591.

18-21 Andrew W. Taylor, Randall B. Rowell, and John E. Breen, “Behavior of Thin-Walled Concrete BoxPiers,” ACI Structural Journal, Vol. 92, No. 3, May–June 1995, pp. 319–333.

18-22 Alex E. Cardenas and Donald D. Magura, “Strength of High-Rise Shear Walls—Rectangular CrossSection,” Response of Multistory Concrete Structures to Lateral Forces, SP-36, American ConcreteInstitute, Detroit, 1973, pp. 119–150.

18-23 A. E. Cardenas, J. M. Hanson, W. G. Corley, and E. Hognestad, “Design Provisions for Shear Walls,”ACI Journal, Proceedings, Vol. 70, No. 3, March 1973, pp. 221–230

18-24 B. Felix, J. M. Hanson, and W. G. Corley, “Shear Strength of Low-Rise Walls with Boundary Elements,”Reinforced Concrete Structures in Seismic Zones, SP-53, American Concrete Institute, Farmington Hills,MI, 1977, pp. 149–202.

18-25 J. W. Wallace and J. H. Thomsen IV, “Seismic Design of RC Structural Walls, Part II: Applications,”Journal of Structural Engineering, ASCE, Vol. 121, No. 1, January 1995, pp. 88–101.

18-26 V. Bertero, A. E. Aktan, F. Charney and R. Sause, “Earthquake Simulator Tests and AssociatedExperimental, Analytical, and Correlation Studies of One-Fifth Scale Model,” Earthquake Effects onReinforced Concrete Structures: U.S.-Japan Research, SP-84, J.K. Wight, Editor, American ConcreteInstitute, Farmington Hills, MI, 1985, pp. 375–424.

18-27 Alexander Chajes, Principles of Structural Stability Theory, Prentice-Hall, Englewood Cliffs, NJ, 1974,336 pp.

18-28 Andrew W. Taylor and John E. Breen, Design Recommendations for Thin-Walled Box Piers and Pylons,”Concrete International, American Concrete Institute, Vol. 16, No. 12, December 1994, pp. 36–41.

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