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Shear Strength of Reinforced ConcreteWalls Subjected to Cyclic LoadingJ. Krolicki a , J. Maffei b & G. M. Calvi ca Arup , San Francisco, California, USAb Rutherford & Chekene , San Francisco, California, USAc Institute for Advanced Study IUSS , Pavia, ItalyPublished online: 01 Apr 2011.

To cite this article: J. Krolicki , J. Maffei & G. M. Calvi (2011) Shear Strength of ReinforcedConcrete Walls Subjected to Cyclic Loading, Journal of Earthquake Engineering, 15:S1, 30-71, DOI:10.1080/13632469.2011.562049

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Journal of Earthquake Engineering, 15(S1):3071, 2011Copyright A. S. Elnashai & N. N. AmbraseysISSN: 1363-2469 print / 1559-808X onlineDOI: 10.1080/13632469.2011.562049

Shear Strength of Reinforced Concrete WallsSubjected to Cyclic Loading

J. KROLICKI1, J. MAFFEI2, and G. M. CALVI31Arup, San Francisco, California, USA2Rutherford & Chekene, San Francisco, California, USA3Institute for Advanced Study IUSS, Pavia, Italy

This research proposes a new shear strength model, based on modifications to the UCSD shearmodel by Kowalsky and Priestley [2000], to calculate the shear capacity and predict the displacementductility of reinforced concrete walls in diagonal tension. An experimental database is created from aliterature review of experimental tests of reinforced concrete walls that exhibit strength degradationand shear failure during cyclic loading. This experimental database is used in the formulation ofthe proposed shear model and to test the accuracy of the model. The proposed model improves theaccuracy of calculating pre-emptive shear and flexure-shear strength, correctly identifies the failuremode for all collected specimens, and provides the closest prediction of the ultimate displacementductility. Based on this research, the proposed shear model is recommended for the calculation ofthe shear strength of reinforced concrete walls for either the assessment of existing buildings or thedesign of new structures.Keywords Reinforced Concrete; Walls; Shear Failure; Diagonal Tension; Shear Strength; SeismicDesign; Cyclic Loading; Ductility

1. IntroductionSignificant advances have been made over the past several decades in understanding thebehavior of reinforced concrete structures during earthquakes. Research has shown that,rather than improving strength, improving the displacement capacity of a structure willimprove the durability of the structure to resist earthquakes.

For a reinforced concrete wall to have a ductile response the more dependable flexuralyielding of the designated plastic hinge regions should control the strength, inelastic defor-mation, and energy dissipation of the structural system. All shear failure modes and otherpotential failure modes should be designed to exceed the flexural strength capacity throughall levels of ductility. Structural walls where shear failure limits the ultimate displacementhave restricted ductility capacity and are vulnerable to brittle shear failure.

In order to design a plastic hinge that meets the crucial requirement of being ductilerather than brittle, its shear capacity must exceed its flexural capacity. Thus, our ability tocalculate both the shear and flexural capacity is critical. The latter is now increasingly wellunderstood. Research has given us tools, such as moment-curvature analysis, to model andcalculate the flexural behavior of concrete elements. Shear capacity, however, is subject tomore factors and is related to a wider range of outcomes. For these reasons, current designrelies on conservative benchmarks rather than precise models.

Address correspondence to J. Krolicki, Arup, 560 Mission, Suite 700, San Francisco, CA 94105, USA;E-mail: jason.krolicki@arup.com

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Shear Strength of Reinforced Concrete Walls 31

The goal of this work is to propose an improved shear strength model to help engineersdevelop a more reliable seismic design and better understand the ultimate behavior of rein-forced concrete wall structures. It is envisioned that this improved shear model would offerengineers a methodology to calculate the shear strength, determine the ductility capacity,and identify the failure mode of structural walls with limited ductility. By meeting theseobjectives the proposed shear model could become an essential tool for displacement orperformance-based assessment and design of structural walls.

2. Experimental DatabaseAn experimental database is created from a literature review of published results of rein-forced concrete wall tests that exhibit shear strength degradation and failure during cyclicloading. A preliminary screening of the published test results eliminates wall tests that aredocumented as web-crushing, sliding shear failure or a flexural response and tests withpoor quality hysteretic data. The preliminary screening yielded three sources for reinforcedconcrete wall specimens failing in shear in diagonal tension under cyclic loading.

The first of these is a test program of reinforced concrete walls conducted by Hidalgoet al. [1998] at the Universidad Catolica de Chile in Santiago, Chile. The testing programfocused on the behavior of reinforced concrete walls that exhibit shear failure through thetest of 26 specimens subjected to cyclic loading in double curvature. The varying specimenparameters include the wall aspect ratio, the horizontal and vertical reinforcement ratio,and the compressive strength of concrete.

The second source is from the Imperial College in London where Pilakoutas andElnashai [1995a] tested six walls in single curvature and varying horizontal and verticalreinforcement, and boundary tie reinforcement. One wall is observed to fail in pre-emptiveshear and another in flexure-shear. Each of the six walls is reviewed in the detailedscreening.

Finally, Priestley et al. [1993] tested full-size bridge piers at the University ofCalifornia, San Diego. The testing program includes two rectangular piers tested in doublecurvature and fail in diagonal tension.

This preliminary literature review resulted in 34 accepted experimental test speci-mens. These specimens are further evaluated by a detailed screening process to identifythe specimen failure mode.

2.1. Flexure vs. Shear Failure

The first step to verify a flexure or shear failure mode is to plot the experimental force-displacement envelope of the hysteretic response vs. the calculated force-displacementresponse. To capture the experimental force-displacement envelope the full hystereticresponse curve for each specimen is plotted. Both the positive and negative experimen-tal hysteretic load-displacement response curves are traced to create an envelope of theexperimental response.

An average bilinear force-displacement curve, approximated as elastic-perfectly-plastic, is calculated from the two experimental response envelopes using the followingprocedure. First, the average experimental ultimate strength of the two response envelopesis calculated and plotted (VU). Second, the intersection of the bilinear curve or estimate offirst yield is calculated as 0.75 times the average experimental ultimate strength. A lineis plotted from the origin to intersect this point and terminate at the average experimen-tal ultimate strength. The intersection of the two lines identifies the experimental yield

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32 J. Krolicki, J. Maffei, and G. M. Calvi

Lateral Displacement

Late

ral F

orce

VU

Envelope of experimental response

y , Yield disp.

Bilinear response curve

0.8VU

u , Ultimate

0.75VU

FIGURE 1 Definition of experimental yield displacement and ultimate displacement.

displacement. Last, the ultimate experimental displacement is calculated when the exper-imental response has degraded to 0.8 times the average experimental ultimate strength.Figure 1 graphically demonstrates this relationship.

The calculated force-deformation response curve is calculated using the procedure out-lined by Priestley et al. [2007]. This procedure is based on a combination of the flexuralresponse calculated from moment-curvature analysis and the shear displacement response.This procedure is documented in detail later in this report.

The experimental force-displacement envelope and the calculated force-displacementresponse are plotted together to confirm that the specimen failed in shear. A shear failureis evident if the experimental envelope drops off in rapid degradation prior to reachingthe ultimate displacements predicted in the calculated force-displacement response or ifthe maximum measured experimental force does not reach the calculated ultimate flexuralcapacity. If a shear failure mode cannot clearly be determined, the specimen is discarded.

2.2. Determining the Mode of Shear FailureThe next step is to determine mode of shear failure of the specimen. There are severalpossible modes of failure for reinforced concrete walls: anchorage failure, sliding shear,diagonal compression or web-crushing, and diagonal tension.

a. Anchorage failure. To assure that an anchorage failure is not expected, the min-imum required anchorage is calculated based on the equation given by Priestleyet al. [1996]. In order to preclude anchorage rotations from occurring, the verticalreinforcement should be embedded at minimum distance, ldc, given as:

ldc = 0.3dbfyf c . (1)

b. Sliding Shear or Shear Friction. The equation by Mattock et al. [1972], given inthe ACI 318-05 Commentary R11.7.3, is used to calculate the sliding shear strengthor the shear friction capacity of the wall specimen. This alternate equation includesboth the effects of dowel action from reinforcement crossing the shear plane andfrom aggregate interlock along the crack faces. The equation gives a closer predic-tion of shear-transfer strength than the conservative codified equation. The nominal

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Shear Strength of Reinforced Concrete Walls 33

shear friction strength, where shear friction reinforcement is perpendicular to theshear plane, is given as:

Vn = 0.8Avf fy + AcK1, (2)

where K1 = 400psi for normal weight concrete.c. Diagonal compression/Web crushing. The web-crushing capacity of the speci-

men is calculated using two distinct methods to confirm that web-crushing failureof the specimen did not occur. The first method proposed by Oesterle et al. [1983]indicates that web crushing behavior is more likely at large lateral deformations andfor walls with higher axial loads.

The web-crushing strength of a wall is calculated as:

Vwc,max = 1.8fc

1 +(

600 2000 NuAgf c)

bw(0.8lw), (3)

where is the story drift ratio to which the wall is subjected. For walls with highaxial loads, when Nu/Agf c 0.09, the web crushing strength may not exceed:

Vwc,max = 1.8fc

1 + (420) bw(0.8lw). (4)

The other method used to verify the web-crushing capacity is proposed by Paulayand Priestley [1992]. In the plastic hinge regions, walls may have significant diago-nal cracking in both directions. The diagonal compression strength of the concretestrut of the truss mechanism may be considerably reduced. It is recommended byPaulay and Priestley [1992] that the total shear stress in this region be limited to80% of that in elastic regions.

vwc,max 0.16f c 870 psi (6MPa). (5)

Paulay and Priestley [1992] reported that web crushing may occur in the plastichinge zone at lower shear stresses than the maximum limitation given in Eq. (5)after a few cycles of reverse loading involving displacement ductility ratios of fouror more. To account for the dynamic effects and expected ductility of the wall, theplastic hinge region shall have the addition limitation on shear stress, given as:

vwc,max (

0.22o,w

+ 0.03)

f c. (6)

The variable o,w is the wall flexural over strength factor, calculated as the flexuralover strength divided by the moment resulting from code forces. A typical valueassumed in the calculation is 1.4.The maximum web crushing shear strength is calculated as:

Vwc,max = (0.8lwbw)vwc,max. (7)

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34 J. Krolicki, J. Maffei, and G. M. Calvi

Specimen and Experimental Data

Shear Friction Check by Calculation

SufficientACCEPT

InsufficientREJECTED

Diagonal CompressionCheck by Calculation

Diagonal Compression

Shear Friction

InsufficientREJECTED

SufficientACCEPT

Check First Yield

Determine Flexure-shear orPre-emptive Shear Failure

Experimental F- hystereticcurves

Calculated F- Responsecurve

Plot F-Curves

Failure Mode Check

FlexuralREJECTED

ShearACCEPT

Determine Mode of Shear Failure

Pre-emptive Shear Failure

Flexure-Shear Failure

Assemble Pre-emptive Shear

Database

Assemble Flexure-Shear

Database

FIGURE 2 Shear failure screening processes for wall experimental database.

If the web-crushing capacity calculated by either equation at the measured dis-placement ductility is less than the applied maximum experimental shear force, apotential web-crushing failure is assumed and the wall specimen is discarded.

The remaining specimens are categorized as flexure-shear or pre-emptive shearfailures by plotting the experimental hysteretic envelope with the calculated flexuralresponse and comparing the experimental peak strength to the calculated first yield.

The following flow chart, Fig. 2, identifies the key steps in the detailed shearfailure screening process for the experimental database.

2.3. Pre-emptive Shear Database

Pre-emptive shear failure is defined as a member loss of strength from a shear failure indiagonal tension prior to reaching flexural yield. Indications of pre-emptive shear failureare when the maximum experimental strength of the specimen does not reach the flexural

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Shear Strength of Reinforced Concrete Walls 35

Displacement

Late

ral F

orce

Calculated First Yield

Calculated F- response curve

Idealized F- response curve

Experimental response envelope of pre-emptiveshear failure

y Calculated yield displacement

FyFexp

FIGURE 3 Definition of pre-emptive shear failure.

TABLE 1 Range of data for wall specimens exhibiting pre-emptive shear failure

Variable

Specimens withhorizontal

reinforcement datarange

Specimens withouthorizontal

reinforcement datarange

All specimensdata range

Concrete strength 24805000 (psi) 24953510 (psi) 24805000 (psi)17.134.5 (MPa) 17.224.2 (MPa) 17.134.5 (MPa)

Transverse Reinf.Ratio

0.001230.0031 0.0 0.00.0031

Longitudinal Reinf.Ratio

0.00540.0302 0.00430.0068 0.00430.0302

Shear-Span Ratio(M/VLw)

0.352.0 0.350.69 0.352.0

Axial Stress Ratio(P/fcAg)

0.00160.0645 0.00360.0057 0.00160.0645

first yield, followed by rapid strength degradation after the maximum strength is reached.An example plot of a pre-emptive shear failure is provided in Fig. 3.

The assembled database of wall specimens failing in pre-emptive shear from cyclicloading is comprised of 18 walls. The database is further subdivided into nine wall spec-imens with horizontal reinforcement and nine wall specimens without. Table 1 presentsthe range of significant variables for the collected specimens of the pre-emptive sheardatabase.

2.4. Flexure-Shear Database

Flexure-shear failure is identified by a rapid degradation of the shear resisting concretemechanism due to cyclic loading after flexural yielding has occurred. This rapid loss ofshear resistance can be seen in the hysteretic response by strength degradation on subse-quent loading cycles beyond the peak strength. Therefore, members failing in flexure-shear

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36 J. Krolicki, J. Maffei, and G. M. Calvi

Displacement

Late

ral F

orce

Calculated First Yield

Calculated F- response curve

Idealized F- response curve

Experimental response envelope of flexure-shear failure

y Calculated yield displacement

Fy

Fexp

FIGURE 4 Definition of flexure-shear failure.

TABLE 2 Range of data for wall specimens exhibiting flexure-shear failure

Variable Data range

Concrete strength 23805600 (psi)16.438.6 (MPa)

Transverse Reinforcement Ratio 0.00130.0031Longitudinal Reinforcement Ratio 0.00630.0157Shear-Span Ratio (M/VLw) 0.5 2.0Axial Stress Ratio (P/fcAg) 0.0010.007Displacement Ductility 2.45.8

have very limited displacement ductility. Figure 4 demonstrates the conditions used toclassify flexure-shear failures.

The screened database of tests that exhibit flexure-shear failure results in eight wallsidentified as exhibiting a flexure-shear failure. The collected database contains specimenswith the following range of significant variables shown in Table 2.

3. Development of the Proposed Shear Strength Model

3.1. Mechanism of Shear ResistanceSince the air force warehouse shear failure in 1955, investigated by Elstner and Hognestad[1957], there had been intensive research to understand and predict shear strength andfailure modes of reinforced concrete under static loads. Much of the current ACI codeprovisions are based on the reports from ACI-ASCE Committee 326 in 1962 and ACI-ASCE Joint Committee 426 in 1973. Elsewhere, similar research efforts to define shearresistance were underway around the same time, with significant advances in Europe andNew Zealand. In the last few decades, concrete shear failure due to earthquakes has focusedresearch on the shear capacity of members subjected to cyclic loading and produced severalresearch based shear models.

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Shear Strength of Reinforced Concrete Walls 37

FIGURE 5 Internal forces of cracked beam with stirrups [MacGregor and Wight, 2005].

The ACI-ASCE research on shear resistance of reinforced concrete identified fourof the major components of shear resistance. A section of a cracked reinforced concretebeam is represented by MacGregor and Wight [2005] in Fig. 5. The internal shear resistingmechanisms are shown between the cracked sections in equilibrium.

The internal forces resisting shear include the concrete in the compression zone Vcz,aggregate interlock along the crack face Vay, dowel action of the longitudinal reinforce-ment Vd, and transverse steel truss mechanism Vs. The distribution of shear between eachinternal mechanism is diagrammatically shown in Fig. 6 from the ACI-ASCE Committee426 [1973].

Early shear strength models reduced this complex behavior down to two main shearresisting mechanisms: a concrete contribution and a steel truss contribution. The concretemechanisms lump the shear contribution from the concrete in the compression zone Vcz,shear along the crack face trough aggregate interlock Vay, and shear resistance of dowelaction of the longitudinal reinforcement Vd into one term. The shear carried by this sim-plified concrete term is given as an average shear stress across the section calculated as aproportion of the concrete tensile strength.

The transverse steel truss analogy for the design of shear reinforcement was proposedindependently by Ritter [1899] and Morsch [1902]. A detailed description of the con-ceptual truss analogy is provided by Park and Paulay [1975]. This early model remainsthe preferred method of calculating the internal force equilibrium of reinforced concretesections. Considering a beam, the web of the analogous truss is formulated by diagonalconcrete compression struts parallel to the inclined cracks and vertical tension members ofthe transverse reinforcement. The concrete compression zone and the longitudinal flexuralreinforcement create the chords of the analogous truss. The slope of the compression diag-onals has been traditionally assumed to be 45 to the longitudinal axis for calculation ofshear strength. More recently, shear models have incorporated the observed phenomenonof variation in the angle of diagonal cracking and allow it to vary from as much as 2565depending on the loading and the location from the support.

From this same truss analogy, the struts closer to the face of the fixed support of a beamare known to be steeper and are an indication of arch action. According to Park and Paulay[1975], this arch action has a tendency to boost the capacity of the other shear carrying

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38 J. Krolicki, J. Maffei, and G. M. Calvi

FIGURE 6 Distribution of internal shears in beam with web reinforcement [ASCE-ACI,1973].

mechanisms. This increase in shear capacity has led to shear models that require alternatemethods for calculating the shear resistance of members with low shear span ratios.

Several improvements to the early shear models have been proposed through ongoingresearch. Research developed shear models have introduced a factor to account for theincrease in effectiveness of the concrete contribution with longitudinal web reinforcement.Research developed shear models have also worked to separate the individual contributionof key variables that have been lumped within the concrete shear resisting mechanism, suchas the benefit of axial load.

The contribution to shear resistance from axial load is only effective where cracksare closed and, therefore, only applicable to the compression zone. However, most shearstrength models do not account for the increase in shear from axial load unless post-tensioning reinforcement is used. To properly account for the axial load contribution toshear resistance, the shear contribution of the compression zone must be separated fromthe shear contribution from aggregate interlock of the remaining section.

Perhaps the most significant development in the calculation of shear capacity for theresistance of seismic loading is identifying the loss of shear strength with repeated loadreversals. Early experimental programs with cyclic loading recorded the degradation ofshear strength with increasing displacement ductility. The Applied Technology Council, inthe Seismic Guidelines for Highway Bridges [ATC-6, 1981], proposed a conceptual modelof a shear strength envelope that degrades with increasing displacement ductility. Thisconceptual model is the foundation for several research based ductility dependent shearcapacity models including the UCSD shear model by Kowalsky and Priestley [2000].

The proposed shear strength model is based on the UCSD shear model by Kowalskyand Priestley [2000] with proposed changes for walls and members with low shear spanratios. The UCSD model is selected for the development of the proposed shear model forseveral reasons.

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Shear Strength of Reinforced Concrete Walls 39

The UCSD shear model included the most significant parameters known to relate toshear strength.

Each of the identified components of shear resistance is intended to relate to physicalphenomenon, not just empirical equations best fit to data.

The research conducted by Kowalsky and Priestley [2000] showed the model to havegood agreement in predicting column shear strength.

The model identifies factors that can be used for both assessment and design. Kowalsky and Priestley [2000] suggested the model may be applicable for elements

with aspect ratio less than 1.5 but additional research is required.

The UCSD shear model was, however, developed for calculating the shear resistance ofcolumns based on cyclic loaded column tests. The shear resistance of columns and walls,and how it is calculated, is somewhat different and worthy of comparison.

3.2. Comparison of the Shear Resistance of Columns and WallsThe calculation of the shear span ratio is one of the fundamental differences betweencolumns and walls. The shear span ratio is defined as the moment to shear ratio dividedby the member depth (i.e., length of wall) parallel to the shear, given as (M/VLw). Theshear span ratio of columns is typically related to one of two conditions: columns loadedin double curvature, and columns loaded in single curvature (i.e., cantilevered). Althoughthere is some variation of the shear span ratios of columns in double curvature based on thestiffness of the end restraint condition, the shear span ratio for most columns can be calcu-lated from the geometric relationships. For double curvature, the moment is proportionalto the shear times the height divided by two. For cantilevers or single curvature columns,the moment is proportional to the shear times the height. Given this relationship, the shearspan ratio can be calculated based on the column height and width, similar to the aspectratio. These relationships are demonstrated in Fig. 7.

Compared to columns the shear span ratio of walls is more variable depending on theheight of the wall, number of floors, and coupling between walls. Structural walls sup-porting multiple floors have distributed inertial shear forces due to earthquakes over theheight of the wall. These inertial forces are dependent on the mass participation of fun-damental modes of vibration and the distribution of ductility. For coupled walls, the wallpier moments are redistributed over the height of the wall, depending on the strength of thecoupling beams. Given this complexity, the shear span ratio for walls should be estimatedbased on the earthquake demand (shear and moment) under equilibrium of the wall system.A few examples are given in Fig. 8.

V

M

M/V

hd

Shear Span Ratio = M/Vd = h/2d Column Double Curvature

V

M

M/Vhd

Shear Span Ratio = M/Vd = h/d Column Single Curvature

FIGURE 7 Columnshear span ratio, double and single curvature.

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M

M/V

M

M/V

Vi

Vi

Vi

Vi

Vi

Vi

Vi

Vi

Vi

Vi

Vi

Vi

Coupled WallMulti-Story WallShear Span Ratio = M/VLw

Lw Lw

FIGURE 8 Wallshear span ratio.

y

z

y

xV

avg = V/Ag

max = (3/2)V/AgAg

FIGURE 9 Shear stresses in a rectangular cross section.

Another comparison of the shear resistance of columns and walls is the calculation ofthe effective shear area. Based on shear flow, the shear stress varies over the depth of a sec-tion. For homogeneous elastic rectangular sections the shear stress distribution is parabolic.This relationship is presented in detail in Ugural [1991] and shown in Fig. 9. Traditionally,in calculations the shear stress has been simplified to an average shear stress across thegross area of the section. This shear stress distribution is typical of uncracked rectangularcolumn sections. In the case of cracked sections, where there is no concrete tensile capacityin the tension zone the distribution of shear stress is quite different.

Structural walls are often made up of flanged sections, creating for example L-shaped,T-shaped, C-shaped, or I-shaped cross sections. The shear stress distribution of uncrackedflanged sections is slightly different, as shown in Fig. 10. For a thin flange, the stress is verysmall as compared to the shear stress in the web. The average shear stress is calculated asacting on the effective area of the web, Aweb. In ACI 318, the effective shear area for walls,Acv, is calculated as the gross area of concrete section bounded by web thickness and lengthof section in the directions of shear force considered.

Other differences that effect the calculation of shear resistance of walls and columnsinclude the following.

Walls are more likely to be squat than columns. Some of the most common applica-tions of structural walls are in low-rise buildings, where the height to width ratio ofthe walls is typically low.

Walls have proportionally better development of reinforcement. Due to the propor-tion of the length of the wall to the diameter of the horizontal reinforcement (lw/db),

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Shear Strength of Reinforced Concrete Walls 41

y

z

y

xV

Aweb

( )max( )min

avg = V/Aweb

xy

xy

FIGURE 10 Shear stresses in flanged section.

the reinforcement can fully develop in wall sections. In addition, cyclic loading isless likely to degrade the bond of the reinforcement over the full section.

Walls may have distributed flexural reinforcement. Walls are typically designed for uniaxial bending, where columns are subjected to

biaxial demands. Columns under biaxial cyclic loading have increased degradationof the concrete and shear resistance. Unlike columns, the out of plane bending ofwalls is relatively flexible, has low strain demands, and is not considered to have asignificant contribution to lateral resistance.

There has been more testing of columns than walls. Past research and funding hasfocused on the shear resistance of columns under cyclic loading, resulting in a sub-stantial experimental database and research proposed shear models. These shearmodels have often been extrapolated to be applied to the calculation of shear strengthof walls.

These key differences between columns and walls are evaluated and considered in theproposed modifications to the UCSD shear model.

3.3. The Proposed Shear Strength Model

The form of the proposed shear strength equation is unchanged from the modified UCSDshear model by Kowalsky and Priestley [2000], given as:

Vcap = VC + VS + VP. (8)

The shear strength equation calculates the capacity as the sum of three components:

VC = Concrete shear-resisting mechanismVS = Horizontal reinforcement truss shear-resisting mechanismVP = Axial load component.

The proposed changes to each component of shear resistance are discussed in detail in thefollowing sections.

3.4. Proposed Horizontal Reinforcement Truss Shear-Resisting Mechanism (VS)The wall section illustrated in Fig. 11 demonstrates the shear strength developed by thehorizontal reinforcement truss mechanism, as suggested by Kowalsky and Priestley [2000].

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42 J. Krolicki, J. Maffei, and G. M. Calvi

FIGURE 11 Average crack angle, .

The critical inclined flexure-shear crack is shown crossing the section at an average angle, , to the vertical axis. The horizontal reinforcement crossing the crack act to transfer shearforces. The maximum shear force that can be resisted by the horizontal reinforcementequals the area of steel crossing the crack times the yield strength and the angle at whichthe load is applied.

The average cracking angle = 30 is recommended by Priestley et al. [1996] forassessment of existing structures. For design, a more conservative value of = 35 isrecommended.

In the collected experimental database, Hidalgo et al. [2002] reported an experimentalvalue for the measured horizontal reinforcement contribution to shear resistance, VS.exp, foreach wall specimen tested. Hidalgo et al. [2002] calculated the experimental shear contri-bution of horizontal reinforcement from measured strains in the reinforcement from strain-gages located close to diagonal cracks. The corresponding stresses are calculated from themeasured stress-strain curves for the reinforcement. For walls where cracking occurredaway from the strain-gages, the shear resisted by the reinforcement is calculated assumingyielding had been attained in each horizontal bar crossing the main diagonal crack.

The UCSD model calculated shear contribution from horizontal reinforcement is com-pared to the measured experimental steel contribution. Figure 12 provides the ratio of themeasured experimental horizontal reinforcement shear, Vs.exp, from Hidalgo et al. [2002],to the UCSD calculated steel contribution to shear resistance, Vs.pred. Values of the ratioVs.exp/Vs.pred of less than 1.0, suggest the UCSD model is over-predicting the steel truss

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Shear Strength of Reinforced Concrete Walls 43

0.0

1.0

2.0

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Vs.

exp

/ V

s.p

red

0.0

1.0

2.0

0 0.2 0.4 0.6 0.8 1 1.2

(b) Aspect Ratio(a) Test Number

Vs.

exp

/ V

s.p

red

FIGURE 12 Horizontal reinforcement shear strength: Vs experimental data vs. Vs UCSD model.

contribution to shear resistance. The statistical mean is equal to 0.67 for the set of data andthe coefficient of variation is 10.2%. When the ratio, Vs.exp/Vs.pred, is plotted based on spec-imen shear span ratio, as shown in Fig.12b, a trend is apparent. As the specimen shear spanratio reduces, the over-prediction of the calculated contribution horizontal reinforcementincreases.

This over-prediction may result from overestimation of the cracking angle. It wasobserved by Priestley et al. [1996] that diagonal compression struts and cracking usuallyform at 45 degree angles and then tend to flatten towards 30 as the yielding of the memberprogresses. This change in inclination of the struts from 45 to 30 is due to the interactionbetween flexure and shear. The crack angle formed may also be influenced by the ratio oflongitudinal to horizontal reinforcement. The specimens collected for this research havelow shear span ratios and limited ductility. This limited ductility after yield is thought toprevent full flexure-shear cracking from developing and result in limited cracking angles.

The reduction in the angle of the compressive struts and shear cracking with lowershear span ratio can be conceptually represented by an equivalent plastic-truss model.Walls with shear span ratios of 2.0 and 0.7 are represented in Fig. 13. The tension andcompression influence lines demonstrate a conceptual internal force path for the element.It is apparent that walls with the lower shear span ratios result in shallower compressionstruts and a larger angle of concrete cracking, .

3.4.1. Proposed Average Cracking Angle, cr. To account for this change in walls with lowshear span ratios, it is proposed that the average critical cracking angle varies as a factor ofshear span ratio (M/Vlw). For shear span ratio greater than or equal to two (M/Vlw 2), theaverage cracking angle will develop at cr = 30 degrees. For a shear span less than two, thecracking angle varies linearly to 45. This relationship is shown in Fig. 14 and is given as:

cr =(

30 452

)(M

V lw)

+ 45 30. (9)

3.4.2. Vertical Height of Inclined Crack, hcr. Diagonal tension failure of walls with lowshear span ratios may occur across tension cracks developed from corner to corner of thewall or along a steeper angle of cracking, as described by Paulay and Priestley [1992].The ultimate behavior is largely dependent on the amount of transverse reinforcement andthe horizontal axial strength of the floor structure or tie beam at the top of the wall to

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44 J. Krolicki, J. Maffei, and G. M. Calvi

a) Wall Shear Span Ratio - 2.0 b) Wall Shear Span Ratio - 0.7

Compression Influence Lines

Tension Ties

Compression Influence Lines

FIGURE 13 Plastic truss model for tall and squat cantilever walls.

FIGURE 14 Average crack angle, cr.

redistribute shear along the wall. However, the maximum vertical height of the inclinedcrack that can develop regardless of the cracking angle is limited to the height of the wall.

The projected vertical height of the critical flexure-shear inclined crack, hcr, isdependent on the average diagonal crack angle, cr, to the vertical axis, given as:

hcr = l

tan cr hw, (10)

where l is the horizontal projection of the crack length.In the compression zone the diagonal shear cracks are closed and therefore do not

engage the transverse reinforcement. The horizontal projection of the crack length isreduced by, c, the depth of the compression zone and, c0, the cover to the main bars asshown in Fig. 11. The horizontal projection of the crack length is calculated as:

l = lw c co. (11)

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Shear Strength of Reinforced Concrete Walls 45

a) Wall shear span ratio 2.0 b) Wall shear span ratio 1.0 c) Wall shear span ratio 0.5

FIGURE 15 Height of vertical cracking for walls of varying shear span ratio loaded insingle curvature.

The compression zone, c, can be calculated through moment curvature analysis orreasonably assumed for walls as 0.2lw.

Figures 15 and 16 demonstrate the relationship of the average cracking angle cr, thevertical crack height hcr, and varying shear span ratios for walls loaded in double and singlecurvature.

3.4.3. Proposed Horizontal Reinforcement Truss Shear-Resisting Mechanism. Given theproposed changes to the average cracking angle and the limit on crack height, the proposedshear contribution of the effective horizontal reinforcement crossing the inclined crack iscalculated as:

VS = t tw hcr fy = AvfyhcrS . (12)

3.4.4. Comparison to Data. Using the proposed Eq. (12), the ratio of the calculatedshear contribution from horizontal reinforcement Vs.pred and experimental shear Vs.exp fromHidalgo et al. [2002] is plotted in Fig. 17. The statistical mean is equal to 1.04 for the samedataset and the coefficient of variation is 4.9%.

The proposed revisions to the shear contribution of the horizontal reinforcement resultin improved accuracy in predicting the measured experimental values in the collecteddataset. The variation in the calculated results of the proposed model with respect to wallshear span ratio, shown in Fig. 17b, also shows significant improvement.

3.5. Proposed Concrete Shear-Resisting Mechanism (VC)The proposed concrete shear-resisting mechanism maintains the same variables as givenin the UCSD shear model. The contribution of the concrete to the shear resistance iscalculated as:

VC,proposed = P Pf c (0.8Acv). (13)

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46 J. Krolicki, J. Maffei, and G. M. Calvi

a) Wall shear span ratio 2.0 b) Wall shear span ratio 1.0 c) Wall shear span ratio 0.5

FIGURE 16 Height of vertical cracking for walls of varying shear span ratio loaded indouble curvature.

0.0

1.0

2.0

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Vs.

exp

/ V

s.p

red

0.0

1.0

2.0

0 0.25 0.5 0.75 1 1.25

Vs.

exp

/ V

s.p

red

b) Shear Span Ratioa) Test Number

FIGURE 17 Horizontal reinforcement shear strength: Vs experimental data vs. Vs proposed model.

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Shear Strength of Reinforced Concrete Walls 47

The effective shear area is the gross area of the wall web, less the compression zone.Assuming the compression zone is approximately 0.2 lw, the area is estimated as 0.8 Acv.The coefficient P accounts for the degradation of the shear resistance of concrete as afunction of displacement ductility. The coefficient P accounts for the effects of shear spanratio. The coefficient accounts for the increase in shear resistance proportional to increas-ing volumetric ratio of longitudinal reinforcement and is unchanged from the USCD shearmodel.

The coefficient is calculated as:

= 0.5 + 20g (14) 1.0, (15)

where g is the ratio of total longitudinal reinforcement over the gross cross-sectional areaof the member. This relationship is shown in Fig. 18.

3.5.1. Proposed Shear Degradation Coefficient, P. The shear resistance of concrete isknown to degrade during cyclic loading with increasing ductility demands. As displace-ments increase, the cracks widen, reducing the effectiveness of the aggregate interlockshear resistance along the crack surface. Upon reversal of the displacement, the inducedmoments in the element reverse and the yielded reinforcement must first be overcome forthe crack to close. As the crack closes there is a lack of fit along the concrete surface andprotrusions are crushed or ground away. The roughness along the crack begins to wear, thusreducing the effectiveness of this surface to resist shear upon the next displacement cycle.If vertical reinforcement in the centre of the wall or horizontal reinforcement yields, cracksmay not fully close. The coefficient P attempts to model this complex phenomenon.

In the original formulation of the UCSD model [Priestley et al., 1994] the displace-ment ductility factor, k, is given as a bilinear degradation curve with increasing ductilitydemands. The FEMA 306 [ATC, 1999a] model adjusted the UCSD model for the appli-cation to reinforced concrete walls by changing bilinear degradation to linear degradationand by reducing the high ductility limit of 8.0 down to a displacement ductility limit of 5.0.The modified UCSD model by Kowalsky and Priestley [2000] simplifies the displacementductility factor to a linear degradation. The modified UCSD model suggests for uniaxialcolumns loading, the coefficient should be taken as 0.29 MPa (3.5 psi) for low ductility( 2.0) and vary linearly to 0.05 MPa (0.6 psi) for high ductility ( 8.0).

0.0

0.3

0.5

0.8

1.0

1.3

1.5

0 0.01 0.02 0.03 0.04 0.05 0.06Longitudinal Reinforcement Ratio

-

fact

or

FIGURE 18 Longitudinal reinforcement ratio coefficient, .

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48 J. Krolicki, J. Maffei, and G. M. Calvi

Considering that the degradation of the concrete contribution is affected by the crackwidth, the factor maybe better expressed in terms of curvature ductility. However, for thisresearch, comparison of results was made with experimental force-displacement hystereticdata where curvature was not recorded. Therefore, displacement ductility was consideredto provide a more straight forward comparison to the data.

As the shear span ratio of a section is reduced, the influence of shear displacementand the effects of shear cracking become increasingly prominent in the force-displacementresponse. Therefore, it is less likely that a section with a low shear span ratio would reachhigh displacement ductility before the shear failure occurs. For this reason the proposedmodel limits the displacement ductility factor P to 6.0 for walls with aspect ratios lessthan 2.0. The proposed displacement ductility factor, P, shown in Fig. 19, accounts forthe degradation of the shear resistance of concrete as a function of displacement ductility,given as:

P = 0.29MPa (3.5 psi) for low ductility 2.0P = 0.05 MPa (0.6 psi) for high ductility 6.0P is calculated by linear interpolation for values of displacement ductility between the

above limits.

For members protected against plastic hinging by capacity design and sections awayfrom the plastic hinge region, the displacement ductility is taken as 1.0, P = 0.29 MPa(3.5 psi).

For walls with shear span ratios greater that 2.0, the displacement ductility factor isexpected to match the UCSD model recommendations of a high displacement ductilitylimit of 8.0. However, the maximum shear span ratio of the experimental database forthis study is limited to 2.0 and therefore could not be studied. In addition, the limits ondisplacement ductility may vary based on shear span ratio but there is insufficient evidenceto support this.

3.5.2. Proposed Member Shear Span Ratio Coefficient, P. The coefficient P accountsfor the well documented strengthening effects of elements with lower shear span ratios.Kowalsky and Priestley [2000] suggested that it is probable that the value of in the UCSDmodel continues to increase with M/Vlw less than 1.5, but data is not available to confirm.

FIGURE 19 Proposed displacement ductility factor, P.

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Shear Strength of Reinforced Concrete Walls 49

Therefore, particular attention is focused on understanding the relationship of the parameterP, shear span ratio, and concrete contribution to shear strength.

The experimental value of concrete contribution to shear resistance, VC,exp, is estimatedby subtracting the calculated contribution from the proposed transverse reinforcementmechanism, VS, and the contribution from axial loads, VP, from the measured experimentalshear capacity VU,exp,

(VC,exp = VU,exp (VS + VP)

). To solve the experimental data val-

ues of P the experimental concrete contribution Vc,exp is divided by the remaining factorsof Eq. (13). Figure 20 represents the experimental database normalized to P plotted vs.specimen shear span ratio.

For members with a shear span ratio of less than about 2.0 the behavior transitionsfrom beam action to arch action as loads are redistributed to compression struts and tensionties. The data trend in Fig. 20 represents this increasing concrete contribution with smallershear span ratios attributed to arch action.

The experimental database is assembled for wall tests that are reported to fail in shear.However, this includes specimens that fail in both pre-emptive shear and shear after flexuralyielding has occurred. As previously discussed, the coefficient P reduces for displacementductility greater than 2.0. Therefore, to calibrate the model for the coefficient P, indepen-dent from ductility, the experimental database should be filtered for specimens that fail witha displacement ductility less than 2.0 or in pre-emptive shear.

Another source of variation in the dataset is from the cyclic tests of specimens with notransverse reinforcement. These tests are observed to have a broad dispersion of results.Since the focus of the proposed shear model is intended to be a tool for the assess-ment of existing wall structures and the design of new structures that require horizontalreinforcement, it seems unlikely to encounter a wall with no horizontal reinforcement.The specimens with no horizontal reinforcement might unnecessarily cloud the results.Therefore, specimens that failed in pre-emptive shear with no horizontal reinforcementwhere also removed.

The remaining experimental data that fails in pre-emptive shear with horizontalreinforcement normalized to P is plotted in Fig. 21. Several variations of empiricalcurves were statistically matched to the data. Because of the limited remaining data,the calculated results did not prove to be significantly more accurate than a linearincrease in with reducing shear span ratio following the same slope as proposed byKowalsky and Priestley [2000] for aspect ratios less than 1.5. This linear extrapolation pro-duced sufficiently accurate results without introducing unnecessary complications to thecalculation.

The proposed model calculates p as linearly increasing for wall shear span ratios lessthan 2.0. The proposed coefficient is calculated as:

0.0

1.0

2.0

3.0

4.0

5.0

0 0.5 1 1.5 2 2.5 3

P

Shear Span Ratio

FIGURE 20 Experimental database normalized to vs. shear span ratio.

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50 J. Krolicki, J. Maffei, and G. M. Calvi

FIGURE 21 Proposed shear span ratio coefficient, P.

P = 3 MVlw 1.0. (16)

3.6. Axial Load Component (Vp)The axial load component accounts for the enhancement of shear strength with increasedaxial compression. The axial load component Vp is used directly in the proposed model asoriginally formulated in UCSD shear model. The axial load component is a based on a sim-plified assumption of a linear compression strut [Priestley et al., 1996]. The actual behavioris a combination of applied axial compression force and flexural compression force. Asshown in Fig. 22 for a member in double curvature, the end regions of the curved com-pression strut have higher magnitudes, but acting at a lower angle to the member than themiddle section. Priestley et al. [1996] suggested that the difference of magnitude and incli-nation have a similar effect on shear resistance over the member height and that the simplelinear assumption of this complex behavior appears to provide a reasonable fit to test data.

The axial load contribution to shear resistance from the simplified linear compressionstrut is shown in Fig. 23, from Priestley et al. [2007], for walls loaded in single curvature

FIGURE 22 Linear representation of curved compression strut [Priestley et al., 1996].

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Shear Strength of Reinforced Concrete Walls 51

FIGURE 23 Axial load contribution Vp for walls loaded in double and single curvature.

and double curvature. The applied shear force is resisted by the horizontal component ofthe compression strut.

The axial load contribution to shear resistance is calculated as [Priestley et al., 2007]:

Vp = P tan . (17)

For cantilever wall tests that are loaded in single curvature, the equation can berewritten as:

Vp = lw c2M/V P =lw c2hw

P. (18)

For wall tests loaded in double curvature, the equation becomes:

Vp = lw c2M/V P =lw c

hwP. (19)

3.7. Recommended Procedure for the Proposed Shear ModelOutlined below is the recommended calculation procedure for the proposed shear model.The proposed equations are repeated here for clarity. As previously stated, the shearstrength capacity of the section is calculated as sum of the concrete shear-resisting mech-anism, the transverse reinforcement truss shear-resisting mechanism, and the axial loadcomponent, given as:

Vcap = (VC + VS + VP). (20)

3.8. Concrete Shear-Resisting Mechanism (VC)The contribution of the concrete to the shear resistance is calculated as:

VC,proposed = P Pf c (0.8Acv). (21)

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52 J. Krolicki, J. Maffei, and G. M. Calvi

Proposed member shear span ratio coefficient, P:

P = 3 MVlw 1.0. (22)

The coefficient is calculated as:

= 0.5 + 20g (23)

1.0. (24)

The proposed coefficient P is a function of displacement ductility, given as:

p = 0.29 MPa (3.5 psi) for low ductility 2.0p = 0.05 MPa (0.6 psi) for high ductility 6.0p is calculated by linear interpolation for values of displacement ductility between the

above limits.

3.9. Proposed Transverse Reinforcement Truss Shear-Resisting Mechanism (Vs )

VS = t tw hcr fy = Avfyhcrs

. (25)

Vertical height of inclined crack, hcr:

hcr = l

tan cr hw. (26)

Proposed average cracking angle, cr:

cr =(

30 452

)(M

V lw)

+ 45 30. (27)

The horizontal length of the crack is calculated as:

l = lw c co. (28)

3.10. Axial Load Component (Vp)

Vp = P tan , (29)

where the angle is the simplified linear compression strut measured from the verticalaxis of the member.

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Shear Strength of Reinforced Concrete Walls 53

3.11. Conservatism for Design vs. AssessmentPriestley et al. [2007] distinguishes between shear resistance calculated for the assessmentof existing structures and design of new members. The assessment model is formulatedto calculate expected or mean shear strength, which is valuable in evaluating existingstructures, test data, or earthquake damage, as discussed in FEMA 306 [ATC, 1999a].

In the design of new structures providing an extra margin of shear strength typicallycosts little and improves the reliability of performance, thus some conservative factors canbe applied. The recommended design strength is calculated as:

Vdes = sVcap, (30)

where the shear strength reduction factor s = 0.85 is recommended for design. Thisstrength reduction factor is taken in addition to the conservatism applied in each of theshear resisting components for design listed below.

For the horizontal reinforcement shear resisting component, the calculation of averagecrack angle is increased by five degrees to produce more conservative results for designrather than assessment. The proposed average crack angle, cr, for design of walls is takenas 35 for walls with a shear span ratio greater or equal to 2.0 and varies linearly up to 50for walls with lower shear span ratios. The equation for design is given as:

cr,design = 50 7.5(

MV lw

) 35. (31)

To provide a consistent level of conservatism for design of walls where the projected verti-cal height of the crack exceed the height of the wall, the projected crack height to the heightof the wall should be reduced proportionally to the five degree increase of the average crackangle. The vertical height of the crack for design is given as:

hcr,design = hcr tan (cr)tan (cr + 5) 0.83 hcr. (32)

For design, the concrete contribution to shear resistance should also be reduced to assurethe shear strength of a wall is not over-predicted. The UCSD model recommends reducingthe displacement ductility factor by (0.85P) as a conservative value for design.

The axial load contribution to shear resistance is similarly reduced for design of newstructures by 0.85Vp.

4. Flexural ResponseTo determine the shear strength of a reinforced concrete wall using the proposed shearmodel, the calculation of the shear capacity envelop and the force-displacement responseis required. In addition, the shear strength calculated by ductility dependent shear modelsrely on calculation of the displacement ductility factor, . The level of refinement in theflexural model used for predicting the yield displacement can affect the accuracy of theshear capacity calculation, especially the estimate of ultimate displacements.

The force-displacement response maybe calculated using the results from a moment-curvature program. However, for walls with low shear-span ratios, the effect of sheardeformations and loss of stiffness resulting from shear cracking can contribute significantly

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54 J. Krolicki, J. Maffei, and G. M. Calvi

to the ultimate displacements of the element. It is therefore recommended to include theadditional displacements from shear deformations in the total force-displacement response.

Detailed finite element analysis models such as the modified compression field theoryproposed by Vecchio and Collins [1986] include shear deformations. However, if detailedmember analysis is not available, the member shear deformations can be estimated by theprocedure outlined by Priestley et al. [2007] or that by Miranda et al. [2005]. The totalforce-deformation response curve is a combination of an idealized flexure response curveand an idealized shear response curve corresponding to the member limit states.

The procedure of Priestley et al. [2007] is used in this study to estimate force-deformation response and the specimen failure mode. The procedure includes the followingsteps to calculate the combined force-displacement response.

1. Calculate the moment-curvature flexure response.2. Translate moment-curvature to force displacement.3. Calculate shear displacement response.4. Superimpose the flexural and shear response to calculate the total force-

displacement response.

4.1. Moment-Curvature Analysis Model

The moment-curvature assumptions and calculations as proposed by Priestley et al. [1996]are used to generate the moment-curvature curve for the specimen. The moment-curvatureanalysis assumes:

Plane sections remain plane Steel and concrete strains are proportional to the distance from neutral axis Concrete tension is ignored Axial force is applied the section centroid Concrete and reinforcement nonlinear stress-strain relationships are as defined in

the following Unconfined concrete stress-strain relationship used outside of the centerline of the

boundary ties

4.2. Concrete Properties for Moment-Curvature AnalysisThe stress strain relationship by Mander et al. [1988] is used for both confined and uncon-fined concrete in the moment-curvature analysis. The effective lateral confining stress forrectangular walls sections uses the recommendations from Priestley et al. [2007], whereasthe ultimate confined concrete compression strain assumed uses the Priestley et al. [1996]recommendations.

4.3. Steel Properties for Moment-Curvature AnalysisThe stress-strain characteristics for reinforcement under monotonic loading are given byPriestley et al. [2007]. For grade 60 reinforcement ASTM 706, Priestley et al. [2007] pro-vided typical values for strain hardening starting around sh = 0.088, ultimate strain is aboutsu = 0.10 to 0.12, and the ratio of ultimate yield stress is typically fu/fy = 1.35 to 1.50.Reinforcement stress-strain test data provided in the research reports is used in lieu of typ-ical values when available. To account for the effects due to cyclic loading, the momentcurvature analysis is based on an ultimate strain limit of s = 0.6su, as recommended byPriestley et al. [2007].

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Shear Strength of Reinforced Concrete Walls 55

4.4. Identification of Member Limit StatesThe moment-curvature analysis calculates the member limit states proposed by Priestlyet al. [1996] outlined in Table 3. Figure 24 graphically represents each limit state and thebilinear idealization of the moment-curvature response.

TABLE 3 Member limit states

Flexural cracking Mcr, cr

Mcr =(Ig/

xt) (fct + P/Ag) Where xt is the distance to the extreme

tension fibrecr = Mcr

/(EcIc) Curvature at flexural cracking of concrete

cr = crlw (fct/

Ec) Strain at flexural cracking of concreteFirst yield My, y is defined as the first occurrence of either:

s = y = fy/

Es Strain at yielding of reinforcementc = 0.002 Strain at peak stress of unconfined concrete

Nominal capacity Mn is defined as the first occurrence of either:s = 0.015 Reinforcement strain after yielding (onset of

1mm crack widths)c = 0.004 Ultimate strain of unconfined concrete

(onset of spalling)Yield curvature y is defined as a linear extrapolation from first yield to nominal capacity:

y = y(Mn

/My

)Yield curvature

Ultimate Capacity Mu, u is defined as the first occurrence of either:s = 0.6su Ultimate strain of reinforcement, su = 0.10

to 0.12c = cu Ultimate strain of confined concrete

Bilinear Idealization

Curvature

Mom

ent My

MnMu

Moment-Curvature Response

Mcr

ucr yy

FIGURE 24 Bilinear idealization of moment curvature.

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56 J. Krolicki, J. Maffei, and G. M. Calvi

4.5. Force-Displacement Response from Moment-CurvatureThe member force-displacement response can be estimated using the simplified approachbased on the concept of a plastic hinge. The plastic hinge is a fictitious zone of lengthLp, typically located at the base of a wall, over which strain and curvature are assumedto be equal to the maximum value at the base of the wall (see, e.g., Priestley et al., 2007).The plastic hinge method also accounts for strain penetration and assumes that outsidethe plastic hinge curvatures are distributed linearly over the shear span according to thebilinear approximation of the moment-curvature response. The length of the plastic hingeis calculated as:

LP = k (LSS) + 0.1lw + LSP 2LSP (33)

k = 0.2( fu

fy 1)

0.08. (34)

The strain penetration length recommended by Priestley et al. [2007] is given as:

LSP = 0.15fyedbl, (35)

where fye is the yield strength (ksi) and dbl is the diameter of longitudinal reinforcement.The member shear span LSS is the distance for the point of maximum moment to the

point of contra-flexure. Shear span can be calculated as the moment to shear ratio at thecritical section:

LSS = MV =hwkT

. (36)

For cantilever members in single curvature, the coefficient kT = 1 and the shear span equalsthe wall height. For members in double curvature, kT = 2 and the shear span is equal to onehalf the wall height.

The force-displacement curve is calculated by converting moments into lateral forceand curvature into displacement using the equations provided in Table 4. A sample force-displacement curve is plotted in Fig. 25.

4.6. Shear Flexibility of Concrete MembersFor members with a low shear span ratio, shear deformations become a significant por-tion of the member displacements. To provide more accurate prediction of ultimatedisplacements the effects of shear deformation and loss of stiffness, after shear cracking,should be included in the force-displacement response.

The original procedure to calculate shear deformations was presented by Park andPaulay [1975]. Miranda et al. [2005] provided further development and study on reinforcedconcrete columns and most recently the procedure was updated by Priestley et al. [2007].

The procedure used in this research for calculation of shear deformations followsthe recommendations from Priestley et al. [2007] with a few modifications. One suchmodification is to calculate the shear response prior to flexural cracking to estimate theshear deformations of an uncracked section. The shear stiffness of an uncracked section iscalculated using the recommendations from Park and Paulay [1975]. The modifiedprocedure used to calculate shear deformations is summarized in Table 5.

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Shear Strength of Reinforced Concrete Walls 57

TABLE 4 Calculation of the force-displacement response

Member limit state Displacement Force

Flexural cracking cr = kT crL2SS

3 Fcr = McrLss (37)First yield y = kT

y (LSS+LSP)23 Fy = MyLss (38)

Yield Displacement y = kT y(LSS+LSP)2

3 Fn = MnLss (39)Plastic Displacement p = kT

(u y

)LP

[LSS

( LP2 LSP

)] (40)Ultimate Displacement u = y + p Fu = MuLss (41)The full force-displacement response curve can be solved for all values of moment-curvature response greater than first yield by the relationship:

F- Response Curve i = ykT(i y

)LP

[LSS

( LP2 LSP

)]Fi = MiLss (42)

Displacement

For

ce

Fy

Fn

Fu

Force-DisplacementResponse

Bilinear Idealization

Fcr

unycr

FIGURE 25 Bilinear idealization of force-displacement response [Priestley et al., 1996].

Another modification is that the calculation of shear cracking in this study is performedusing the proposed shear model. Shear cracking is calculated as the sum of the concreteand axial load contribution to shear resistance. The contribution from the horizontal trussreinforcement mechanism is not included, since at this level of shear force the concrete hasnot cracked. In addition, the shear degradation coefficient, p, is set to 3.5. The axial loadcontribution, Vp, is unchanged.

The proposed shear model equation for shear cracking is given as:

Vcr,proposed = p 3.5 f c (0.8Ag) + P tan . (50)

4.7. Construction of the Combined Force-Displacement Response CurveThe total force-displacement response curve is calculated as the sum of deformations fromthe flexural and shear response, based on the procedure outlined by Miranda et al. [2005].

Table 6 provides a summary of the calculation of the total force-displacement responsecurve. Figure 26 graphs the relationship between the flexural and shear response curves.

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58 J. Krolicki, J. Maffei, and G. M. Calvi

TABLE 5 Calculation of shear deformation

Phase 1: Shear response prior to flexural cracking

Shear stiffness of an uncracked beam, as given by Parkand Paulay [1975]:

kse = GAvhw (43)

Shear displacement at flexural cracking: se = Fcrkse (44)Phase 2: Shear response after flexural cracking

Shear stiffness prior to shear cracking is approximatelyproportional to the reduction in flexural stiffness,taken as:

ksf = GAvhw EIeffEIg

(45)

Shear displacement after flexural cracking to onset ofshear cracking:

sf = VcrFcrksf (46)

Phase 3: Shear response after shear cracking

Shear stiffness after shear cracking, where n = 10: ks,sc = t(0.25)0.25+nt Esbwd (47)Shear displacement from shear cracking to flexural first

yield is given as:s,sc = FyVc,scks,sc (48)

Phase 4: Shear response after flexural first yield

After flexural first yield the shear deformation beincreased proportional to the flexural deformation,given as:

s,d = f s,scf ,y (49)

TABLE 6 Summary of force-displacement components for the calculation of thetotal response curve

Responsephase

Displacement

Flexure + Shear = Total ForcePhase 1: Elastic response f,e + s,e = T,fcr FcrPhase 2: Flexural cracked response f,f + s,f = T,sc VcrPhase 3: Shear cracked response f,f + s,sc = T,fy FyPhase 4: Response after first yield f,y + s,d = T,fu Fu

Figure 27 graphs the combined force-displacement response curve and the proposedshear capacity envelope. The proposed shear capacity envelop is scaled such that displace-ment ductility of 1.0 equals the adjusted nominal yield displacement based on the combinedforce-displacement response, shown in the figure as y Total. Shear failure is identified asthe intersection of the force-displacement curve and the shear capacity curve. The point ofintersection of the two curves corresponds to the predicted ultimate displacement and shearcapacity of the section.

Measuring the accuracy of the calculated total flexure-shear response was not afocus of this research. However, the calculated total response was observed to produce

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Shear Strength of Reinforced Concrete Walls 59

First FlexuralYield

Shear Cracking

Flexural Cracking Phase I

Phase II

Phase III

Phase IV

UltimateFlexuralStrength

Shear Response

V

Flexural Cracking

Shear Cracking

First FlexuralYield

Phase I

Phase II

Phase III

Phase IV

Ultimate Flexural Strength

F

Flexural Response

f,cr f,sc f,fy f,fu s,e s,f s,sc s,d

FIGURE 26 Flexural force-displacement response and shear displacement responsecurves.

Flexural Cracking

Shear Cracking

First Flexural Yield

Ultimate Flexural Strength

Shear Capacity Envelope

U

y Total

T,fy T,fuT,scT,fcr

Shear Capacity

Combined Total (Flexure-Shear) Response

Adjusted Bilinear Response

F

Phase III

Phase IV

Phase II

Phase I

FIGURE 27 Combined force-displacement response and shear capacity envelope.

a reasonable match of the measured experimental envelop which was sufficiently accu-rate for calculation of the nominal yield displacement used in scaling the displacementductility.

Deviation from the procedure used to calculate the nominal yield displacement canimpact the scaling of the shear capacity curve and ultimately lead to differing results.A calculated force-displacement response based on flexure only would shift the shearcapacity envelope. Although this approach loses accuracy and correlation with the actualtotal displacements, it would result in a conservative prediction of the shear strength andductility capacity.

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60 J. Krolicki, J. Maffei, and G. M. Calvi

4.8. Calculation Flow Chart

The process for calculating the shear strength, ultimate displacement, and failure mode issummarized in the flow chart shown in Fig. 28. The flow chart provides an outline of the rec-ommended procedure for calculating the shear capacity envelope and force-displacementresponse curve.

5. Results

5.1. Calculation of Pre-Emptive Shear FailureThe calculation of pre-emptive shear failure for ductility dependent shear modes is notdependent on the flexural response and therefore can be used to measure the accuracy ofthe shear models calculation of the ultimate shear capacity. The calculated shear capacityfrom the proposed and UCSD shear model is compared to the recorded strength of thetest specimen in the pre-emptive shear database. Although a more thorough examinationof the formulation of the proposed shear model would be to use independent test data fromthat which was used in the formulation of the model, it was not possible do to the limitedsources of test data available.

Table 7 gives the mean value and coefficient of variation of the measured experi-mental strength divided by the predicted strength for each shear model. The table alsoincludes results from the building codes ACI 318-05, CSA A23.3-04, and Eurocode 8 [CEN2003] for comparison. A mean value greater than 1.0 represents an average value that isunder-predicting shear strength. The coefficient of variation measures the dispersion of theresults.

Figure 29 represents the ratio of experimental shear strength to predicted shear strengthfrom the UCSD model (on the left) and the proposed model (on the right) for the spec-imens of the pre-emptive shear database. The ratios of experimental shear strength topredicted shear strength are plotted by test number, longitudinal reinforcement ratio, trans-verse reinforcement ratio, and shear span ratio. By presenting the calculated ratios for thesesignificant variables, trends of the shear model related to the variables can be identified.

For members with transverse reinforcement the UCSD model results in a reasonableprediction of shear strength considering the model was developed based on columns. Theproposed shear model results in a closer mean ratio of experimental to predicted shearstrength and improves the dispersion of the results from the UCSD model in calculatingthe shear capacity of specimens with very low shear span ratios. The proposed shear modelproduced nearly equivalent accuracy in predicting the shear strength of the walls in thepre-emptive shear database with horizontal reinforcement as the UCSD shear model. Thereis a notable improvement over the results from the UCSD model for members with lowhorizontal reinforcement, suggesting the proposed model has improved the estimate of theconcrete contribution.

The model appears to remain somewhat conservative for wall tests 16, 17, and 18.These specimens have a shear span ratio of 0.35 and no horizontal shear reinforcement. Forthese three wall specimens, each wall had an initial shear failure in one direction, resultingin unbalanced hysteresis loops for the positive and negative cyclic response.

The recorded experimental shear strength of the specimen for this research is takenas the average strength of the positive and negative envelope hysteretic response curves.Typically, for symmetric sections, both positive and negative hysteretic responses havesimilar strength. For sections with brittle failures one side of the specimen can have asignificant loss of strength across a single diagonal crack. As the load cycle is reversed,

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Shear Strength of Reinforced Concrete Walls 61

Section Properties and MemberCharacteristics

Input Variables Required:Lw, t, H, M/VLw, fc, fy, t, l

START

Calculate Shear Strength, as a function of Displacement Ductility:

Construct Shear Capacity Envelope

Construct Force-Displacement Response Curve

ConcreteMechanism

VC

Horizontal Reinforcement

Vs

Axial Force Mechanism

Vp

Calculate Lateral Displacement for each Limit State

Overlay Plots Find Intersection

END

Output: Shear Strength

Ultimate Displacement Failure Mode

Calculate Member ResponseCalculate Member Limit States

Flexural CrackingFcr

Shear CrackingVcr

Flexural First YieldM- Analysis

Ultimate Flexural StrengthM- Analysis

Phase 2: Flexural Cracked Response

T,sc

Phase 3: Shear Cracked Response

T,fy

Phase 4: Response After First Yield

T,fu

Phase 1: Elastic ResponseT,fcr

Plot Force-Displacement Response CurvePlot Shear Capacity Envelope

Convert Shear Capacity Envelope from Displacement Ductility to Lateral

Displacement

FIGURE 28 Calculation flow chart, proposed shear model.

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62 J. Krolicki, J. Maffei, and G. M. Calvi

TABLE 7 Comparison of proposed and UCSD shear model: measured shear strength tocalculated strength for specimens failing in pre-emptive shear

Results Vexp/Vpred ACI 318 CSA A23.3 Eurocode 8 Proposed UCSD

With HorizontalReinf.

Mean 1.52 1.41 1.75 1.07 1.10CoV 0.207 0.140 0.152 0.065 0.060

Without HorizontalReinf.

Mean 1.94 4.02 2.08 1.31 2.09CoV 0.287 0.411 0.226 0.223 0.260

All Specimens Mean 1.71 2.71 1.89 1.18 1.60CoV 0.289 0.651 0.226 0.211 0.398

the crack closes and the section regains strength. In this case the average of the hystereticresponse loses practical meaning and does not represent the primary or governing fail-ure strength of the section. To demonstrate this relationship an example plot of Wall 17(Specimen H-30 from Hidalgo et al., 2002) is provided in Fig. 30.

The predicted shear strength from the proposed model for these walls is found to becloser to the recorded peak strength in the direction of the first shear failure. Therefore, theproposed model provides a closer prediction of the ultimate strength of the specimen thanis reflected in the plotted results of Fig. 29.

5.2. Calculation of flexure-shear failureFor ductility dependent shear models, the calculation of nominal yield displacementfrom the force-displacement response is required to set the ductility scale of the shearcapacity curve. The calculated nominal yield displacement by definition correspondswith a displacement ductility of one. The procedure used in this research to calculate thecombined force-displacement response includes the effects of shear deformation and lossof stiffness after shear cracking. Accounting for this additional displacement shifts theshear capacity curve.

When measuring the accuracy of the shear capacity envelope using test specimensfailing in flexure-shear, different interpretations of the predicted strength are possible.Kowalsky and Priestley [2000] provided an example of the different interpretations pos-sible and recommended a methodology to examine the accuracy of the predicted ultimatestrength of a test specimen using ductility dependent shear models.

The method used by Kowalsky and Priestley [2000] is employed to calculate theexperimental/predicted ratios for flexure-shear failures independent of flexural response.The predicted strength is calculated using the measured ultimate experimental displacementand the shear capacity envelope. First, the ultimate experimental displacement, defined asthe displacement corresponding to a rapid loss in strength limited to 0.8 times the peakstrength (see Fig. 1), is converted to displacement ductility. Second, the calculated shearcapacity envelope is plotted with the ultimate experimental displacement ductility. Thepredicted strength is read from the point of intersection. An example of this procedure isplotted in Fig. 31.

The calculated shear capacity from the proposed model and the UCSD shear modelare compared to the recorded strength of the test specimen in the flexure-shear database.Table 8 gives the mean value and coefficient of variation of the proposed and UCSDmodel of the measured experimental strength divided by the predicted strength. Included inTable 8 are the results from the building codes ACI 318-05, CSA A23.3-04, and Eurocode8 for comparison.

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Shear Strength of Reinforced Concrete Walls 63

UCSD Shear Model

0

1

2

3

0 5 10 15 20

A) Test Number

Vex

p /

Vp

red

Vex

p /

Vp

red

Vex

p /

Vp

red

Vex

p /

Vp

red

0

1

2

3

0 0.01 0.02 0.03 0.04C) Longitudinal Reinforcement Ratio

0

1

2

3

0 0.001 0.002 0.003 0.004

D) Transverse Reinforcement Ratio

0

1

2

3

0 0.5 1 1.5 2 2.5

F) Shear Span Ratio (M/Vlw)

Vex

p /

Vp

red

Vex

p /

Vp

red

Vex

p /

Vp

red

Vex

p /

Vp

red

0 0.5 1 1.5 2 2.5

F) Shear Span Ratio (M/Vlw)

Proposed Shear Model

0

1

2

3

0 5 10 15 20A) Test Number

0

1

2

3

0 0.01 0.02 0.03 0.04C) Longitudinal Reinforcement Ratio

0

1

2

3

0 0.001 0.002 0.003 0.004D) Transverse Reinforcement Ratio

0

1

2

3

FIGURE 29 Calculation of flexure-shear failure: Comparison of UCSD model andproposed model (results organized by significant variables).

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64 J. Krolicki, J. Maffei, and G. M. Calvi

FIGURE 30 Example plot of Specimen H-30 (Wall 17) from Hidalgo et al. [2002].

exp

Experimental Displacement Ductility

Str

engt

h

Ductility

Vpred Calculated ShearCapacity Envelope

FIGURE 31 Shear strength assessment based on displacement ductility.

TABLE 8 Comparison of proposed and UCSD shear model: measured shearstrength to calculated strength for specimens failing in flexure-shear

Results Vexp/Vpred ACI 318 CSA A23.3 EC 8 Proposed UCSD

Mean 1.13 1.10 1.43 0.99 0.98CoV 0.106 0.061 0.088 0.063 0.061

Figure 32 represents the ratio of experimental shear strength to predicted shear strengthfrom the UCSD model and proposed model for the specimens of the flexure-shear database.The ratios of experimental to predicted shear strength are plotted by test number in the firstgraph, Fig. 32a. The remaining figures present the ratios of experimental shear strength topredicted shear strength arranged by variable.

Both models produce fairly accurate results in predicting the shear strength for wallsgoverned by flexure-shear failure. Both models also shows consistent dispersion when thedata is arranged by each significant variable.

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Shear Strength of Reinforced Concrete Walls 65

UCSD Shear Model

0

1

2

0 2 4 6 8 10

A) Test Number

Vex

p /

Vp

red

Vex

p /

Vp

red

Vex

p /

Vp

red

Vex

p /

Vp

red

0

1

2

0 1 2 3 4

B) Displacement Ductility at Failure

0

1

2

0 0.005 0.01 0.015 0.02

C) Longitudinal Reinforcement Ratio

0

1

2

0 0.001 0.002 0.003 0.004

D) Transverse Reinforcement Ratio

Vex

p /

Vp

red

Vex

p /

Vp

red

Vex

p /

Vp

red

Vex

p /

Vp

red

Proposed Shear Model

0

1

2

0 2 4 6 8 10

A) Test Number

0

1

2

0 1 2 3 4B) Displacement Ductility at Failure

0

1

2

0 0.005 0.01 0.015 0.02

C) Longitudinal Reinforcement Ratio

0

1

2

0 0.001 0.002 0.003 0.004D) Transverse Reinforcement Ratio

FIGURE 32 Calculation of pre-emptive shear failure: Comparison of UCSD model andProposed model (Results organized by significant variables).

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66 J. Krolicki, J. Maffei, and G. M. Calvi

5.3. Identification of Failure ModeThe identification of failure mode can help engineers determine the limiting behavior ofa wall or lateral system and help determine the course of action, whether it retrofit ornew design. To determine the accuracy of each of these models in identifying the shearfailure mode, the calculated shear strength envelopes are plotted with the calculated flex-ural response for all specimens of the collected database. Based on the criteria shownin Figure 33, the intersection of the curves yields three possible failure modes, flexure,flexure-shear, and pre-emptive shear.

The proposed and UCSD shear model are used to identify the failure modes of allwalls in the pre-emptive and flexure-shear database. The UCSD model, even with the underprediction of shear strength of specimens without horizontal reinforcement, correctly iden-tified all but one specimen in the pre-emptive shear database. The proposed shear modelis found to correctly predict pre-emptive shear failures and flexure-shear failure for allspecimens of the collected database.

5.4. Displacement Ductility

The calculation of the ductility capacity of primary lateral load resisting elements is a fun-damental component in understanding structural seismic behavior and displacement baseddesign. For flexure controlled sections, the calculation of ductility capacity is well cali-brated using moment-curvature analysis. On the other hand, the calculation of displacementductility for flexure-shear controlled elements has a higher degree of variability. This vari-ability is due to several factors, including the calculation of the nominal yield displacement,the shear capacity envelope, and the shape of the shear capacity envelope relative to dis-placement ductility. However, even as such, a well-calibrated ductility dependent shearmodel can help engineers determine the limitations of the structure i