technical notes on brickwall construct

8/14/2019 Technical notes on brickwall construct

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For additional information on earthquake forces anddesign, the designer may refer to the following publica-tions:

1. Uniform Building Code, 1970 edition, InternationalConference of Building Officials.2. Reinforced Brick Masonry and Lateral ForceDesign, Harry C. Plummer and John C. Blume,Structural Clay Products Institute, 1953.3. Seismic Design for Buildings, (Department of theArmy, TM 5-809-10; Department of the Navy, NAV-

DOCKS P-355; Department of the Air Force, AFM 88-3, Chapter 13), March 1966.4. Earthquake Engineering Research, The Committeeon Earthquake Engineering Research - Division ofEngineering - National Research Council, NationalAcademy of Engineering, National Academy ofSciences, 1969.

DISTRIBUTION OF LATERAL FORCESDiaphragms. Horizontal distribution of lateral forces

to shear walls is achieved by the floor and roof systems

acting as diaphragms (see Fig. 2).To qualify as a diaphragm, a floor and roof system

must be able to transmit the lateral forces to the shearwalls without exceeding a deflection which would causedistress to any vertical element. The successful action ofa diaphragm also requires that it be properly tied into thesupporting shear walls. The designer should insure thisaction by appropriate detailing at the juncture betweenhorizontal and vertical structural elements of the building.

Diaphragms may be considered as analogous to hori-

zontal (or inclined, in the case of some roofs) plate gird-ers. The roof or floor slab constitutes the web; the joists,beams and girders function as stiffeners; and the walls orbond beams act as flanges.

Diaphragms may be constructed of materials such asconcrete, wood or metal in various forms. Combinationsof such materials are also possible. Where a diaphragmis made up of units such as plywood, precast concreteplanks or steel deck units, its characteristics are, to alarge degree, dependent upon the attachments of one unitto another and to the supporting members. Such attach-ments must resist shearing stresses due to internal trans-lational and rotational actions.

The stiffness of a horizontal diaphragm affects the dis-tribution of the lateral forces into the shear walls. Nodiaphragm is infinitely rigid or flexible. However, for thepurpose of analysis, diaphragms may be classified intothree groups: rigid, semirigid or semiflexible, and flexible.

A rigid diaphragm is assumed to distribute horizontalforces to the vertical resisting elements in proportion totheir relative rigidities (see Fig. 3).

Semirigidor semiflexible diaphragms are those whichhave significant deflections under load, but which alsohave sufficient stiffness to distribute a portion of the loadto the vertical elements in proportion to the rigidities of thevertical resisting elements. The action is analogous to acontinuous beam system of appreciable stiffness on yield-ing supports (see Fig. 4). The support reactions aredependent upon the relative stiffness of both diaphragmand the vertical resisting element.

A flexible diaphragm is analogous to a shear deflect-ing continuous beam or series of beams spanningbetween supports. The supports are considered non-yielding, and the relative stiffness of the vertical resistingelements compared to that of the diaphragm is great.Thus, a flexible diaphragm is considered to distribute thelateral forces to the vertical resisting elements on a tribu-tary area basis (see Fig. 5).

Where the center of rigidity of a shear wall systemdoes not coincide with the center of application of the lat-eral force, the distribution of the rotational forces due to atorsional moment on the system must also be considered.Where rigid or semirigid diaphragms are used, it may beassumed that the torsional forces are distributed to theshear walls in direct proportion to their relative rigiditiesand their distance from the center of rigidity (see Fig. 6).In the design provisions for earthquake forces of the 1970Uniform Building Code, shear resisting elements arerequired to resist an arbitrary torsional moment equivalentto the story shear acting with an eccentricity of not less

TABLE 1Horizontal Force Factor K for Buildings or Other Structures

Type or Arrangement of Resisting Elements

All building framing systems except ashereinafter classified

Buildings with a box system

Buildings with a dual bracing systemconsisting of a ductile moment resistingspace frame and shear walls using thefollowing design criteria:(1) The frames and shear walls shall

resist the total lateral force in accordancewith their relative rigidities considering theinteraction of the shear walls and frames.(2) The shear walls acting independentlyof the ductile moment resisting portions of

the space frame shall resist the totalrequired lateral forces.(3) The ductile moment resisting spaceframe shall have the capacity to resist not

less than 25 percent of the required later-al force

Buildings with a ductile moment resistingspace frame designed in accordance withthe following criteria: The ductile moment

resisting space frame shall have thecapacity to resist the total required lateralforce

Elevated tanks plus full contents, on four or

more cross-braced legs and not support-ed by a building

Structures other than buildings

Value of K

1.00

1.33

0.80

0.67

3.00

2.00

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than five per cent of the maximum building dimension atthat level. A flexible diaphragm is not considered capableof distributing torsional stresses.

When dealing with a rigid diaphragm and distributingthe horizontal forces to vertical resisting elements in pro-portion to the relative rigidities, the relative rigidity of theshear wall is dependent upon the shear and flexuraldeflections. However, for the proportions of the shearwalls in most high-rise buildings, the flexural deflection

greatly exceeds the shear deflection, in which case onlyflexural rigidity need be considered in determining the rel-ative stiffness of the shear walls. For determination of rel-ative shear and flexural deflections, see Fig. 7.

A rigorous analysis of the lateral load distribution tothe shear wall is sometimes very time-consuming and fre-quently unjustified by the results. Therefore, in manycases a design based on reasonable limits may be used.For example, the load may be distributed by first consider-

FIG. 2Diaphragm Action

FIG. 3

FIG. 4

FIG. 5

Relative Rigidities

Wall A 0.33

Wall B 0.22

Wall C 0.45

x = (0.22)(16) + (0.45)(40) = 21.52 ft.Torsional moment = (20,000)(21.52 - 20)

= 30,400 ft-lb

If F were applied at center of rigidity, it would be distributed as follows:

Wall A: (0.33)(20,000) = 6600 lbWall B: (0.22)(20,000) = 4400 lb.Wall C: (0.45)(20,000) = 9000 lb.

Since F is not applied at center of rigidity, forces due to torsionalmoment are distributed to walls in proportion to their relative rigiditiesand the distance from the center of rigidity.

FB = FA 0.22 5.52 = 0.17 FA

0.33 21.52

FC = FA 0.45 18.48 = 1.17 FA

0.33 21.52

21.52FA + 5.52FB + 18.48FC = 30,400 ft-lb

21.52FA + 5.52(0.17FA) + 18.48(1.17FA) = 30,400

44.06FA = 30,400

FA = 690 lb

FB = 117 lb

FC = 808 lb

Due to torsional moment, forces on walls A and B are increased, while

force acting on wall C is decreased. However, to be on safe side in designingthe walls, no reduction should be made. Therefore, the walls should be

designed to resist the following forces:Wall A= 6600 +690 = 7290 lb

Wall B = 4400 + 117 = 4517 lbWall C = 9000 lb

(((()) )

)

FIG. 6

Torsional Moment on a Sheer Wall System

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stress in the masonry, assuming a reduced cross sectionof wall. The effect of reinforcement which may be presentin a reinforced brick masonry wall or as a tie to the floorsystem in a non-reinforced or partially reinforced masonrywall is not considered. It should also be pointed out that

the limit on deflection is actually a limit on differentialdeflection between two successive floor or diaphragm lev-els.

Maximum span-to-width or depth ratios fordiaphragms are usually used to indirectly controldiaphragm deflection. Normally, if the diaphragm isdesigned with the proper ratio, the diaphragm deflectionwill not be critical. As a guide for horizontal diaphragmproportions, Table 2, taken from the State of CaliforniaAdministrative Code, Title 21, Public Works, may proveuseful.

Rigidity of Shear Walls. Where shear walls are con-nected by a rigid diaphragm so that they must deflectequally under horizontal load, the proportion of total hori-zontal load at any story or level carried by a perpendicularshear wall is based on its relative rigidity or stiffness. Therigidity of a shear wall is inversely proportional to itsdeflection under unit horizontal load. The total deflectionof the shear wall can be determined from the sum of theshear and moment deflections. Equations for the deflec-

tion of fixed and cantilevered walls or piers are shown inFig. 9.

Where a shear wall contains no openings, the compu-tations for deflection and rigidity are quite simple. In Fig.10(a), the shear walls are of equal length and rigidity, andeach takes one half of the total load. In Fig. 10(b), wall Cis one half the length of wall D and it, therefore, receivesless than one eighth of the total load. Where shear wallscontain openings such as doors and windows, the compu-tations for deflection and rigidity are more complex.However, approximate methods have been developedwhich may be used. See Fig. 11.

To increase the stiffness of shear walls as well as

their resistance to bending, intersecting walls or flangesmay be used. Very often in the design of buildings, Z, T,U, and I-shape sections develop as natural parts of thedesign. See Figs. 12 and 13. Shear walls with theseshapes, of course, have better flexural resistance. The1969 BIA Standard, Building Code Requirements forEngineered Brick Masonry, (Sec. 4.7.12A) limits the effec-tive flange width that may be used in calculating flexuralstresses. In the case of symmetrical T or I sections, theeffective flange width may not exceed one sixth of thetotal wall height above the level being analyzed. In thecase of unsymmetrical L or C sections, the width consid-ered effective may not exceed one sixteenth of the totalwall height above the level being analyzed. In eithercase, the overhang for any section may not exceed sixtimes the flange thickness (see Figs. 14 and 15). It is, ofcourse, necessary to insure that the shear stress at theintersection of the walls does not exceed the permissibleshear stress. This will depend on the method used inbonding the two walls together.

Coupled Shear Walls. Another method that may beused to increase the stiffness of a bearing wall structureand reduce the possibility of tension developing in shearwalls due to wind parallel to the wall is the coupling ofcollinear shear walls. Figures 16 and 17 indicate theeffect of coupling on the stress distribution in the wall dueto parallel forces. A flexible connection between the wallsis assumed in Figs. 16(a) and 17(a), so that the walls actas independent vertical cantilevers in resisting the lateralloads. Figures 16(b) and 17(b) assume the walls to beconnected with a more rigid member which is capable ofshear and moment transfer so that a frame-type actionresults. This can be accomplished with a steel, reinforcedconcrete or reinforced brick masonry section. The platetype action, which is indicated in Figs. 16(c) and 17(c),assumes an extremely rigid connection between walls,such as full story height walls or deep rigid spandrels.

FIG. 8Diaphragm Deflection Limitation

TABLE 2

Roof or Floor Diaphragms -- MaximumSpan-Width Ratios

Diaphragm Construction

Concrete

Steel deck (continuous sheetin a single plane)Steel deck (without continu-

ous sheet)

Poured reinforced gypsumroofsPlywood (nailed all edges)Plywood (nailed to supports

only--blocking may be omit-ted between joints)Diagonal sheathing (special)Diagonal sheathing (conven-tional construction)

Maximum Span-Width Ratiol/b

Masonry and

ConcreteWalls

Limited by

Deflection

4:1

2:1

3:1

3:1

2 1/2 :11

3:11

2:11

Wood and

Light SteelWalls

5:1

2 1/2 : 1

4:1

4:1

3 1/2 : 1

3 1/2 : 1

2 1/2 : 1

1 The use of diagonal sheathed or unblocked plywood diaphragms for buildings

having masonry or reinforced concrete walls shall be limited to one-story buildingsor to the roof of a top story.

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In this method of test, horizontal movement of the wallspecimen (8 by 8 ft), due to the horizontal racking load atthe top of one end, is prevented by a stop block at thebottom of the other end. To counteract rotation of thespecimen due to this overturning couple, tie rods are usednear the loaded edge of the wall specimen. Under rack-ing load these rods superimpose an indeterminate com-pressive force which suppresses the critical diagonal ten-sile stresses and increases the load required to rack thespecimen. A summary of the racking data for non-rein-

forced brick masonry walls tested in accordance withASTM E 72 is given in Table 3. A typical mode of failurefor a 4-in. brick wall subject to racking is shown in Fig. 18.

FIG. 12Shear Walls of Equivalent Stiffness

FIG. 13Shear Walls With Flanges

FIG. 14Effective Flanae Width

FIG. 15Effective Flange Width

FIG. 16End Shear Walls

FIG. 17Interior Shear Walls

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Circular Shear Specimens. Based on experimentalwork done at the Balcones Research Center of theUniversity of Texas by Professors Neils Thompson andFrank Johnson, the Research Division of the StructuralClay Products Institute conducted a series of diagonaltensile tests on circular brick masonry specimens. Inthese tests, a 15-in. diameter specimen is tested in com-pression with the line of load at 45 deg to the bed joints.As shown in Fig. 19 the diametrical stresses are largelytensile over the central 80 per cent of the specimen. The

tensile stress is approximately constant for about 60 percent of the diameter and may be calculated by the follow-ing equation:

ft = 2P

Dt

where:P = load at rupture, in poundsD = diameter of specimen, in inchest = thickness of specimen, in inchesThe test results for 133 specimens built with 27 types

of brick and type S portland cement-lime mortar of ASTM

C 270, Specifications for Mortar for Unit Masonry, aresummarized in Table 4. A typical mode of failure for a cir-cular brick specimen is shown in Fig. 20. While there wasno consistent relationship between diagonal tensilestrength and brick properties, such as initial rate ofabsorption, it appeared that brick with the weakest bondcharacteristics as shown by flexural strength values alsoyielded the lowest diagonal tensile strengths.

The test results for 20 circular specimens built withone type of brick and four types of mortars are summa-

rized in Table 5. As indicated, the mortar type had amarked effect on the diagonal tensile strength of circularback masonry specimens.

Square Shear Specimens. The Brick Institute ofAmerica has continued to study the diagonal tensile orshear strength of brick masonry in an effort to developboth test methods and design information. Working withthe National Bureau of Standards and the ResearchDivision of John A. Blume and Associates (SanFrancisco), a test method has been developed that hasseveral advantages over the ASTM E 72 racking test pro-cedure. As previously stated, in the E 72 procedure thehold-down tie rods required to prevent overturning of thespecimen under load produce an indeterminate bearingcondition at the bottom edge of the specimen, thus pre-venting an analytic determination of the stress within thewall specimen itself.

In the alternate test procedure, which will be submittedto ASTM as a proposed alternate to the ASTM E 72 proce-dure, the specimens are nominally 4 ft by 4 ft as opposed tothe 8-ft square specimen in the E 72 procedure. In thismethod, the test results are susceptible to stress analysis.In addition, they are more reproducible and thus more reli-able for comparison and design data purposes.

FIG. 18Typical Mode of Failure in Racking

FIG. 19Stress Distribution in Tensile Splitting Test

FIG. 20Small-Scale Diagonal Tension Test

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The square specimen is placed in the testing frame soas to be loaded in compression along a diagonal, thusproducing a diagonal tension failure with the specimensplitting apart along the loaded diagonal (see Figs. 21 and22). The results of both small scale (2 ft square) and fullscale (4 ft square) tests of 4, 6 and 8-in. thick brickmasonry specimens are summarized in Table 6 and typi-cal shear stress-strain curies are displayed in Fig. 23. Allspecimens were constructed with type S mortar. The

shearing stress Vm is determined by the equation

vm = 0.707 F

tl

where:F = diagonal compressive force or load, in poundst = thickness of wall specimen, in inches

l= length of a side of a square specimen, in inches

TABLE 3

Racking Strength Data--Non-Reinforced Brick

Masonry Walls1

Sourceof

Data

3

4

5

6

Construction

8-in. brick wall;headers inevery 6th

course

10-in. cavitywall; 4-in. brick,2-in. space, 4-

in.brick

6-in. brick wall

4-in. brick wall

MortarType

M

N

N

N

S

UltimateRacking

Load,kips per

ft

6.25

2

6.252

5.66

8.14

12.5

ShearStress,

psi

64

2

642

64

123

285

No. ofWallsTested

2

5

3

3

5

1

Tested in accordance with ASTM E 72.2

Walls not loaded to failure.3

Structural Properties of Six Masonry Wall Constructions, H. L. Whittemore, A. H.Stang and D. E. Parsons, National Bureau of Standards Report BMS5, 1938.4

Structural Properties of a Brick Cavity-Wall Construction, H. L. Whittemore, A.H.

Stang and D.E. Parsons, National Bureau of Standards Report BMS23, 1939.5

SCR brick* Wall Tests, C. B. Monk, Jr., Structural Clay Products ResearchFoundation, Research Report No.1, June 1953.6

Compressive, Transverse and Racking Strength Tests of Four-Inch Brick Walls,Structural Clay Products Research Foundation, Research Report No. 9, August

1965.* Reg. U.S. Pat. Off., SCPI

TABLE 4

Summary of Diagonal Tensile Tests

On Circular Brick Masonry Specimens1

Brick

Initial Rate ofAbsorption,

g per min per30 sq in.

Less than 55 to 30

More than 30

Brick Masonry Specimens2

n

307825

Diagonal Tensile Strength,psi

LowAverage

177131167

HighAverage

341354287

x

235256231

1

Small Scale Specimen Testing, Progress Report No. 1, SCPI-SCPRF, October1964.2

All specimens built with type S mortar.

TABLE 5

Diagonal Tensile Strength of BrickMasonry Specimens as a Function of

Mortar Type1

Mortar Brick Masonry Specimens4

Type

M

SNO

n

5

555

Proportions

by Volume2

1C:1/4L:3S

1C:1/2L:4

1/2

S1C:1L:6S

1C:2L:9S

Diagonal

TensileStrength, psi

313

25217693

Tensile

Strength,

psi3

308

1488764

StrengthRatio

1.24

1.000.700.37

1

Small Scale Specimen Testing, Progress Report No. 1, SCPI-SCPRF, October1964.2

C=portland cement; L=hydrated lime (type S); S=sand.3

28-day briquets.4

All specimens built with 3/8-in. joints and brick having an average compressivestrength of 11,771 psi and an initial rate of absorption of 10.6 g per min per 30 sqin.

FiG. 21Wallette Shear Test Setup

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Effects of Normal Loads. The Brick Institute ofAmerica has also investigated the effects of compressiveloads normal to the bed joints on the shearing strength of

plain (non-reinforced) brick masonry walls. Table 7 sum-marizes the results of these tests with the normal com-pressive load on the wall varying from 0 (unloaded) to 375psi. The specimens were built with one type of brick uti-lizing type S mortar and inspected workmanship. Allspecimens were two wythes, 8 in. in thickness and bond-ed with metal ties.

Allowable Stresses. The allowable shear stressesfor non-reinforced masonry provided in the 1969 SCPIStandard, Building Code Requirements for EngineeredBrick Masonry, are shown in Table 8, and for reinforcedmasonry in Table 9.

The Standard also provides a basis for the design ofbiaxially loaded shear walls. Section 4.7.12.1 states:"In non-reinforced shear walls, the virtual eccentricity

(el) about the principal axis which is normal to the length

(l) of the shear wall shall not exceed an amount which willproduce tension. In non-reinforced shear walls subject tobending about both principal axes, (etl + e lt) shall not

exceed (tl/ 3) where e t = virtual eccentricity about the prin-

cipal axis which is normal to the thickness (t) of the shearwall. Where the virtual eccentricity exceeds the values

given in this section, shear walls shall be designed inaccordance with Section 4.7.9 or 4.7.11". (Reinforced orpartially reinforced walls.)

Provision is also made in the standard for increasingthe shear capacity of the wall by taking into considerationcompressive loads on the wall. Section 4.7.12.3 states:

"The allowable shearing stresses in non-reinforcedand reinforced shear walls shall be taken as the allowablestresses given in Tables 3 and 4 (Tables 8 and 9 of thisTechnical Notes) ,respectively, plus one fifth of the aver-

age compressive stress due to dead load at the levelbeing analyzed. In no case, however, shall the allowableshear stresses exceed the maximum values given inTables 3 and 4.

"In computing the shear resistance of the wall, onlythe web shall be considered."

REFERENCES1. Technical Notes on Brick Construction, BIA, (amonthly series).2. Gross, J. G.; Dikkers, R. D.; and Grogan, J. C.;Recommended Practice for Engineered Brick

Masonry, SCPI, November 1969.

3. Allen M. H. and Watstein, D.; Compressive,Transverse and Shear Strength Tests of Six and

Eight-lnch Single-Wythe Walls Built with Solid and

Heavy-Duty Hollow Clay Masonry Units, ResearchReport No. 16, SCPI, September 1969.4. Blume, J. A. and Prolux, J.; Shear in GroutedBrick Masonry Wall Elements, Western States ClayProducts Association, August 1968.

FIG. 22Full Scale Shear Test Setup

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TABLE 6Shear Strength of SIngle-Wythe 4, 6 and 8-in. Clay Masonry Walls

Specimen

Series Thickness, t,in.

4S 3.63

6S 5.50

8S 7.50

Wallettes1

Ultimate Unit Average v,

Load, F, Stress, Unit Stress,kips vm, psi psi %

30 35927 325 341 5

28 34065 353

62 333 352 569 371

76 29868 269 280 669 272

Walls2

Ultimate Unit Average v,

Load, F, kips Stress Unit Stress,vm, psi psi %

113 456123 496 447 12

96 388

151 404154 413 383 11124 332

190 373

187 367 376 3198 388

1

2 ft square2

4 ft square

TABLE 7

Effect of Normal Load on Shear Strength

of 8-in. Brick Masonry1

NormalLoad

psi

0

250

375

Load

kips

97.595.088.5

201215201

227226244

v

mpsi

183178166

378404378

426424457

x

psi

176

387

436

v,

%

5.0

3.9

4.3

1

Built with type S mortar, inspected workmanship and metal-tied.

fb = 11,100 psi IRA= 20.7 g/min/30 in.2

TABLE 8

Allowable Shear Stresses in Non-Reinforced Brick Masonry

Description

ShearM or S mortar vm

N mortar vm

Allowable Stresses, psi

Without

Inspection

0.5 fm, but not to

exceed 40

0.5 fm, but not to

exceed 28

With

Inspection

0.5 fm, but not to

exceed 80

0.5 fm, but not to

exceed 56

TABLE 9Allowable Shear Stresses in Reinforced Brick Masonry

Description

ShearNo shear

reinforcementFlexuralMembers vm

Shear Walls vm

With shear

reinforcementtaking entireshear

FlexuralMembers v

Shear Walls v

Allowable Stresses, psi

WithoutInspection

0.7 fm, but not

to exceed 25

0.5 fm, but not

to exceed 50

2.0 fm, but not

to exceed 60

1.5 fm, but not

to exceed 75

WithInspection

0.7 fm, but not

to exceed 50

0.5 fm , but not

to exceed 100

2.0 fm, but not

to exceed 120

1.5 fm, but not

to exceed 150

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FIG. 23Typical Shear Stress-Strain Curves for 6 and 8-in. Walls

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technical notes on brickwall construct

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