Chapter 16 Shear Friction, Horizontal Shear Transfer, And Composite Concrete Beams

858

16-1 INTRODUCTION

This chapter considers the shear strength of interfaces between members or parts of mem-bers that can slip relative to one another. Among other cases, this includes the interfacebetween a beam and a slab cast later than the beam, but expected to act in a compositemanner.

16-2 SHEAR FRICTION

From time to time, shear must be transferred across an interface between two membersthat can slip relative to one another. The shear-carrying mechanism is known variouslyas aggregate interlock, interface shear transfer, or shear friction. The last of theseterms is used here. The interface on which the shears act is referred to as the shearplane or slip plane.

Three methods of computing shear transfer strengths have been proposed in the lit-erature. These include: (a) Shear friction models, (b) Cohesion plus friction models, and(c) Horizontal shear models as occur in composite beams.

Behavior in Shear Friction Tests

Several types of test specimens have been used to study shear transfer. The most commonare the so-called push-off specimens similar to that shown in Fig. 16-1, which were testedin one of three ways:

1. Mattock [16-1], [16-2] and his associates reported tests of push-off specimens inwhich lateral expansion or contraction and crack widths were not held constant or otherwisecontrolled during the tests except by the reinforcement perpendicular to the slip plane. Thesespecimens were tested either as constructed (uncracked) or precracked along the shear planeshown by the vertical dashed line in Fig. 16-1. Only the failure load and the amount of trans-verse reinforcement were reported.

vnh

C

T

16Shear Friction,

Horizontal ShearTransfer, and

CompositeConcrete Beams

Section 16-2 Shear Friction • 859

Fig. 16-1Shear-transfer specimen.

2. Walraven [16-3], [16-4] tested push-off specimens but held the crack width con-stant during the tests. Loads and slip were reported.

3. A third type of test, reported by Loov and Patnaik [16-5] and Hanson [16-6],measured the shear transferred between the web and the slab in a composite beam. Thisis referred to as horizontal shear.

An excellent review of tests and equations for shear transfer is presented by Ali andWhite [16-7].

Cohesion and Friction

The results of 13 tests of uncracked specimens, and 21 tests of specimens with previouslycracked shear planes are plotted in Fig. 16-2, with open circles and solid circles, respec-tively. The test results from [16-1] and [16-2] plotted in Fig. 16-2 suggest that:

(a) The strengths of cracked and uncracked specimens can be represented byequations of the form

(16-1a)

(16-1b)

where c is a cohesion-like term equal to the intercepts of the sloping lines on the verti-cal axis in Fig. 16-2, plus a friction-like term that is equal to the product of

the compressive stress on the shear plane, and the coefficient of friction, . Theplots in Fig. 16-2, can be idealized as straight sloping lines that intercept the verticalaxis at “cohesions” of and psi, for the uncracked and precrackedspecimens, respectively, and a coefficient of friction of about 0.95. The curves termi-nate at a ceiling of about 1300 psi.

(b) The initial strengths of the uncracked specimens were higher than those of theprecracked specimens. The uncracked and precracked specimens reached similarupper limits on shear transfer as shown in Fig. 16-2.

In Fig. 16-2, the shear stress at failure, is plotted against where is theratio of the area of the transverse reinforcement across the shear plane to the area of the

rvrvfy,vu,

c L 255c L 505

ms,s tan u

vn = c + msvn = c + s tan u

860 • Chapter 16 Shear Friction, Horizontal Shear Transfer, and Composite Concrete Beams

A

B

C

shear plane. The crack widths, w, were not held constant or otherwise controlled duringthese test series. The shear stress at failure increases with the value of which istaken as a measure of the clamping force due to the tension in the reinforcement cross-ing the shear plane. In Mattock’s tests of initially uncracked push-off tests, the firstcracks were a series of diagonal tension cracks, as shown in Fig. 16-3. With furtherdeformation, the compression struts between these cracks rotated at their ends (points Aand B) such that point B moved downward relative to A. At the same time, the distanceAC, measured across the crack, increased, stretching the transverse reinforcement. Thetension in the reinforcement was equilibrated by an increase in the compression in thestruts. Failure occurred when the bars yielded or the struts were crushed.

When a shear is applied to an initially cracked specimen or to a specimen formedby placing a layer of concrete on top of or against an existing layer of hardened concrete,

rvfy,

1500

1500

1000

1000

500

5000

Initially uncracked

Initially uncracked

Initially cracked

Initially cracked

v u, p

si

rv fy, psi

Fig. 16-2Variation shear strength with web reinforcementratio (From [16-1] and[16-2].)

rvfy.

vu

Fig. 16-3Diagonal tension crackingalong a previously uncrackedshear plane. (From [16-8].)


Fig. 16-4Shear-friction model. (From [16-6].)

relative slip of the layers causes a separation of the surfaces, as shown in Fig. 16-4a. If thereinforcement across the crack is developed on both sides of the crack, it is elongatedby the separation of the surfaces and hence is stressed in tension. For equilibrium, acompressive stress is needed as shown in Fig. 16-4b. Shear is transmitted across thecrack by (a) friction resulting from the compressive stresses [16-4], [16-5], and (b) inter-lock of aggregate roughness on the cracked surfaces, combined with dowel action of thereinforcement crossing the surface.

Shear-Friction Model

The original and simplest design model is the shear-friction model [16-1], [16-8], shown inFig. 16-4, which ignores the cohesion-like component, c, and assumes that the shear transferis due entirely to friction. This model is the basis of the shear-friction design procedure inACI Code Section 11.6. Because the cohesion component is ignored, unusually high valuesof the coefficient of friction must be used to fit the test data. Design is based on the shear-friction equation, given in terms of forces as

(16-2)(ACI Eq. 11-25)

or in terms of stresses,

(16-3)

where is defined above and is the appropriate value of the coefficient of friction givenin ACI Code Section 11.6.4.3.

Permanent Compression Force

A permanent compressive force, perpendicular to the slip plane, causes a normalstress on the slip plane in the concrete. If is compressive, the normal stress adds to the compressive stress on the concrete due to the reinforcement. If tensile forces,

Nu>AcvNu

Nu,

mrv

vn = rvfy m

Vn = Avffy m


Fig. 16-5Force components in a barinclined to the shear plane.(From [16-1].)

Nu , act on the shear plane, they must be equilibrated by tensile reinforcement providedin addition to the shear-friction reinforcement

(16-4)

Inclined Shear-Friction Reinforcement

When the shear-friction reinforcement is inclined to the slip plane such that slip along thisplane produces tensile stresses in the reinforcement, as shown in Fig. 16-5, the componentof bar force perpendicular to the slip plane causes an compressive force on the plane equalto In addition, there is a force component parallel to the crack equal to

In stress units, the nominal shear-friction resistance at failure from Eq. (16-3) is

(16-5)

which has units of stress or, in terms of forces,

(16-6)(ACI Eq. 11-26)

Shear-friction reinforcement that is inclined to the shear plane, such that the antici-pated slip along the shear plane cause compression in the inclined bars, is not desirable.This compression tends to force the crack surfaces apart, leading to a decrease in theshear that can be transferred. Such steel is not included as shear-friction reinforcement.

Other Factors Affecting Shear Transfer

Lightweight Concrete

Tests [16-2] have indicated that the resistance to slip along the shear plane is smaller forlightweight-concrete specimens than for normal-weight concrete specimens. This happensbecause the cracks penetrate the pieces of lightweight aggregate rather than following the

Vn = Avffy1m sin af + cos af2vn = mrvfy sin af + rvfy cos af

Avffy cos af.Avffy sin af.

An =Nuffy


perimeters of the aggregate. The resulting crack faces are smoother than for normal-weightconcrete [16-2]. This effect has been accounted for by multiplying the coefficient of fric-tion, , by the correction factor for lightweight concrete, given in ACI Section 11.6.4.3.Originally, was introduced in ACI Chapter 11 to account for the smaller tensile strengthof lightweight aggregate concrete. Here, it is used for convenience to account for thesmoother crack surfaces that occur in lightweight concrete.

Coefficients of Friction

ACI Code Section 11.6.4.3 gives coefficients of friction, as follows:

1. for concrete placed monolithically:

2. for concrete placed against hardened concrete with surface intentionally roughened,as specified in ACI Code Section 11.6.9:

3. for concrete placed against hardened concrete not intentionally roughened, asspecified in ACI Code Section 11.6.9:

4. for concrete anchored to as-rolled structural steel by studs or by reinforcing bars(see ACI Code Section 11.6.10):

where for normal-weight concrete and 0.75 for all-lightweight concrete. The codeallows to be determined based on volumetric proportions of lightweight and normal-weightaggregates, as specified in ACI Code Section 8.6.1, but it shall not exceed 0.85.

The first term in Eq. (16-1) represents the portion of the shear transferred by sur-face protrusions and by dowel action. Equation (16-5) with as plotted inFig. 16-6, applies only for greater than 200 psi and the radial lines in Fig. 16-6 arenot plotted below this value. For Grade-60 reinforcement, this limit requires a minimumreinforcement ratio,

Design Rules in the ACI Code

ACI Code Section 11.6 presents design rules for cases “where it is appropriate to considershear transfer across a given plane, such as an existing or a potential crack, an interfacebetween dissimilar materials, or an interface between two concretes cast at differenttimes.” Typical examples are shown in Fig. 16-7.

In design, a crack is assumed to exist along the shear plane, and reinforcement isprovided across that crack. The amount of reinforcement is computed (ACI CodeSection 11.6.4) from

(16-7)(ACI Eq. 11-1)

(16-2)(ACI Eq. 11-25)

where for designs carried out by using the load factors in ACI Code Section 9.2and where is the coefficient of friction, depending on the surfaces in contact.

In cases 2 and 3 defined previously, the surface must be clean and free of laitance (aweak layer on the top surface of a concrete placement due to bleed water collecting at thesurface). ACI Code Section 11.6.10 requires that, in case 4, the steel must be clean and freeof paint. Case 2 applies to concrete placed against hardened concrete that has been rough-ened to a “full amplitude” (wave height) of approximately but does not specify a “wavelength” for the roughened surface. This was done to allow some freedom in satisfying this

14 in.

m

f = 0.75

Vn = Avffy m

fVn Ú Vu

r = 0.0033.

rvfy

af = 90°,

l

l = 1.0

0.7l

0.6l

1.0l

1.4l

m,

l

l,m


requirement. It was intended, however, that the wave length be on the same order of magni-tude as the full amplitude, say to

Upper Limit on Shear Friction

For normal-weight concrete either cast monolithically or placed against hardened con-crete with intentionally roughened surfaces, as noted previously, ACI Code Section11.6.5 sets the upper limit on from Eqs. (16-2) and (16-6) to the smallest of

and where is the area of the concrete sectionresisting shear transfer. The limits represented by the second two expressions given hereare larger than previously permitted limits in the ACI Code and represent a reexaminationof test data as presented in references [16-10] and [16-11]. For all other cases, includinglightweight concrete and concrete placed against not intentionally roughened surfaces, shall not exceed the smaller of and In cases where a lower-strengthconcrete is cast against a higher-strength concrete, or visa versa, the value of for thelower-strength concrete shall be used to evaluate the limits given here. These upper limitson shear friction strength are necessary, because Eqs. (16-2) and (16-6) may becomeunconservative in some cases.

Cohesion-plus-Friction Model

For the usual case of transverse reinforcement perpendicular to the shear plane, Mattock[16-2] rewrote Eq. (16-2) as

(16-8)

where for normal-weight concrete, 200 psi for all-lightweight concrete, and250 psi for sand-lightweight concrete. The first term on the right-hand side of Eq. (16-8)represents the shear transferred by “cohesion,” which is caused by pieces of aggregatebearing on the surfaces of the slip plane, by the shearing off of surface protrusions, and by

K1 = 400 psi

Vn = K1Acv + 0.8Avffy

fcœ

1800 psi2Ac .0.2fcœAc

Vn

Ac11600 psi2Ac ,1480 psi + 0.08fcœ2Ac 0.2fc

œAc ,Vn

34 in.1

4

1500

1500

1000

1000

500

5002000

Initially uncracked

Initially cracked

v u, p

si

rvfy, psi

Eq. (16-5)

Eq. (16-5)

Eq. (16-5)

m � 0.6

m � 1.0

m � 1.4

ACI Code Section 11.6.5for fc� � 4000 psi

Fig. 16-6Comparison of Eq. (16-2) withtest data from [16-1], [16-2],and [16-6].


dowel action. The second term represents the “friction,” with the coefficient of frictiontaken to be 0.8 for cracked concrete sliding on cracked concrete. Equation (16-8) is calledthe modified shear-friction equation.

In 2001, Mattock [16-10] reevaluated Eq. (16-8) on the basis of 199 tests, with ranging from 2450 psi to 14,400 psi, and derived a set of equations that retained thenumerical constant 0.8 in Eq. (16-8), but presented a new family of values of forvarious types of shear planes.

Walraven Model

Walraven [16-3] idealized concrete as a series of size-graded spherical pieces of coarseaggregate embedded in a matrix of hardened concrete paste. A crack was assumed tocross the matrix between pieces of aggregate until it reached a piece of aggregate, atwhich time it followed the aggregate–matrix interface around the aggregate. The strength

K1

fœc

.

.

.

Fig. 16-7Examples of shear friction.(From [16-9].)


of the interface between the cement paste and the aggregate was assumed to be less thanthe strength of the aggregate. Such a crack is shown in Fig. 16-8a, after cracking, butbefore shearing displacement. The radius of the aggregate particle is R, the diameter is a,and the crack width is w. It is assumed that the crack width is held constant while a shear-ing load is applied. After a small slip parallel to the crack, the spherical piece of aggregatecomes into contact with the matrix in the dark shaded area, allowing shear to be trans-ferred across the crack (Fig. 16-8b). Further slip mobilizes rigid–plastic stresses at thepoint of contact, and the crack is restrained against further slip until the crack surfacesdeteriorate (see Fig. 16-8c). The maximum shearing force transferred by a single aggre-gate particle occurs at the stage shown in Fig. 16-8c, corresponding to the largest area ofplastic stresses. By assuming randomly chosen gradations of aggregate sizes, Walravendeveloped expressions for the shear forces transferred for various crack widths, w, andmaximum aggregate diameters, a.

Walraven [16-3] tested 88 push-off shear-transfer specimens similar to Fig. 16-1. Themajor difference between these tests and those by Mattock and others was that Walravenkept the crack widths, w, constant throughout each test. Vecchio and Collins [16-12] fittedEqs. (16-5) and (16-8) to Walraven’s test data and obtained

(16-9)

where is the shear stress transferred across the crack, is the compression stress requiredacross the crack in psi, and is the maximum shear stress that can be transmittedacross a given crack.

The maximum shear stress, that can be transferred across a crack when itswidth is held at w in. is given by

(16-10)

where a is the diameter of the coarse aggregate in the cracked concrete in inches. The crackwidth, w in inches, is computed from the spacing, of the inclined crack as

(16-11)

where is the principal tensile strain, which is assumed to act perpendicular to the crack,and is the spacing of the cracks, measured perpendicular to the cracks. Equation (16-11)assumes that all the strain is concentrated in the crack.

Walraven’s tests were limited to concrete strengths, from 2900 to 8200 psi. Forhigh-strength concretes with cylinder strengths in excess of 10,000 psi, the cracks tend to

fœc,

su

e1

w = e1su

su,

vcimax =2.162fœc

0.3 + a 24w

a + 0.63b

vci max,

vci max

fcivci

vci = 0.18vci max + 1.64fci - 0.82fci

2

vci max

R

a

wc

w

b

b, max

(a) No contact. (b) Growing contact. (c) Maximum contact.

c, max

Shear

Fig. 16-8Walraven crack model. (From [16-3].)


cross the individual pieces of aggregate rather than going around them. As a result, the cracksurface was smoother than for weaker concretes. For this case, the effective size of theaggregate, a, decreases, approaching zero. Angelakos, Bentz, and Collins [16-13] handledthis by arbitrarily reducing the effective aggregate size, a, in Eq. (16-10) from the nominaldiameter, a, to zero as the concrete strength increases from 8500 psi to 10,000 psi. For con-crete strengths of 10,000 psi or higher, they took a equal to zero.

Ali and White [16-7] carried out similar analyses, assuming an undulating crack path, toillustrate the force transfer that develops when the two sides of the crack come into bearing.

Loov and Patnaik—Composite Beams

From tests of composite beams, Loov and Patnaik [16-5] derived the following equation forshear transfer across cracks and for horizontal shear in composite beams (see Fig. 16-11,discussed later):

(16-12)

In this equation, accounts for lightweight concrete and k is a constant equal to 0.5 forconcrete placed against hardened concrete and 0.6 for concrete placed monolithically.The square-root term allows this equation to fit the test data more closely than do othermodels. Equation (16-12) is plotted in Fig. 16-9 for comparison with test results andwith Eq. (16-8).

Because the reinforcement is assumed to yield in order to develop the necessaryforces, the yield strength of the steel is limited to 60,000 psi. Each bar must be anchored onboth sides of the crack to develop the bar. The steel must be placed approximatelyuniformly across the shear plane, so that all parts of the crack are clamped together.

Comparison of Design Rules with Test Results

In Fig. 16-9, test data for push-off tests of initially uncracked specimens with reinforce-ment perpendicular to the shear plane [16-1] are compared with

l

vci max = lk2sfœc + rvfy cos af

1500

1500

1000

1000

500

5000

Initially uncracked

Initially cracked

Composite T-beams [16-13]

v n, p

si

rvfy, psi

Eq. 16-12, k � 0.5

Eq. 16-8

0.8 rvfy

c � 400

ACI Code Section 11.6.5for fc� � 4000 psi

Fig. 16-9Comparison of test resultsfrom [16-1] and [16-5] andequations (16-8) and (16-12).


(a) Equation (16-8), with and

(b) Equation (16-12), with for initially cracked concrete

The test data are compared with the nominal strengths, with The specimens hadconcrete strengths ranging from 3840 psi to 4510 psi.

Test data for push-off specimens with a precracked interface [16-1] and an averageconcrete strength of 4060 psi are plotted in Fig. 16-9, along with data from tests of com-posite beams with average concrete strengths of 5710 psi for the webs and 5160 psi for theflanges [16–5]. The nominal strengths computed from Eq. (16-12), with for a pre-cracked interface, and fit the data quite well.

EXAMPLE 16-1 Design of the Reinforcement in the Bearing Region of aPrecast Beam

Figure 16-10 shows the support region of a precast concrete beam. The factored beamreactions are 62 kips vertical force and a horizontal tension force of 12 kips. The horizon-tal force arises from restraint of the shrinkage of the precast beam. The cross section of thebeam is 12 in. wide by 18 in. deep. Use assume normal-density concrete,and use

1. Assume the cracked plane. The crack plane giving the maximum area, is a ver-tical crack. This will tend to overestimate the cohesion component. Assume that the support re-gion of the beam is enclosed by a structural steel angle, as shown in Fig. 16-10,and assume that the crack is at 60° to the horizontal. It intercepts the end of the beam at 10.2in. above the bottom and has a length of 12 in. We shall take Ac = 12 in. * 12 in. = 144 in.2.

2. Compute the area of steel required. Resolving the forces onto the inclinedplane gives a normal force of 20.6 kips compression and a shear force of 59.7 kips for a 60°plane. Each assumed crack angle will result in a different combination of normal and shearforces. However, it is quick and conservative to assume that the shear force is equal to thevertical reaction of 62 kips and that the normal force is equal to the horizontal reaction

(negative in tension). In addition, we shall assume that

From Eq. (16-2),

We want so

Avf ÚVuffy m

Vn Ú fVu,

Vn = Avffy m

af = 90°.Nu = -12 kips

6-in. * 6-in.

Ac,

fy = 60,000 psi.fœc = 4000 psi,

fœc = 4000 psi,k = 0.5

f = 1.0.

k = 0.5,

K1 = 400 psi,

3 No. 6

Fig. 16-10Example 16-1.

Section 16-3 Composite Concrete Beams • 869

where (the crack plane is in monolithically placed concrete), and(normal-weight concrete is used). We thus have

We must confirm that the required value of does not exceed the upper limitgiven in ACI Code Section 11.6.5. The required value of is

The value for the stress acting on the effective concrete area resisting shear transfer, islimited in ACI Code Section 11.6.5 to the smallest of

or1600 psi

Thus, the limiting stress is 800 psi, and the upper limit on is

The required value for does not exceed this value, so we do not need to increase theeffective section area resisting the factored shear force.

The tensile force must also be transferred across the crack by reinforcement, ascalculated using Eq. (16-4),

Therefore, the total steel across the crack must be

Provide three No. 6 bars across the assumed crack, . These barsmust be anchored on both sides of the crack. This is done by welding them to the bear-ing angle and by extending them as recommended in [16-14]. ■

16-3 COMPOSITE CONCRETE BEAMS

Frequently, precast beams or steel beams have a slab cast on top of them and are designedassuming that the slab and beam act as a monolithic unit to support loads. Such a beam-and-slab combination is referred to as a composite beam. This discussion will deal onlywith composite beams where the beam is precast concrete or other concrete cast at anearlier time than the slab.

Shored or Unshored Construction

When the slab concrete is placed, the precast beam can either be shored or unshored. In typ-ical shored construction, the precast beam is placed and must support its own weight. Shoresare then added to support the beam and initially resist the weight of the cast-in-place slab.When the strength of the slab concrete is high enough to resist expected stresses, the shores

1.7/d,

As = 1.32 in.210.984 + 0.267) in.2 = 1.25 in.2

An =12 kips

0.75 * 60 ksi= 0.267 in.2

Vn

Vn … 1800 psi2Ac = 800 psi * 144 in.2 = 115 kips

Vn

480 psi + 0.08fcœ = 480 + 0.08 * 4000 = 800 psi

0.2fcœ = 0.2 * 4000 psi = 800 psi

Ac ,

Vn = Vu /f = 62 kips/0.75 = 82.7 kips

Vn

Vn , Vu/f,

Avf =62 kips

0.75 * 60 ksi * 1.4= 0.984 in.2

l = 1.0m = 1.4lVu = 62 kips,


are removed and the slab dead load is resisted by the composite beam and slab. If the precastbeam is not shored when the slab is placed, the beam supports its own weight plus the weight ofthe slab and the slab forms. ACI Code Section 17.2 allows either construction process and re-quires that each element be strong enough to support all loads it supports by itself. If the beamis shored, ACI Code Section 17.3 requires that the shores be left in place until the compositesection has a strength adequate to support all loads and to limit deflections and cracking.

Tests have shown that the ultimate strength of a composite beam is the same whetherthe member was shored or unshored during construction. For this reason, ACI CodeSection 17.2.4 allows strength computations to be made that consider only the final com-posite member.

Horizontal Shear

In the beam shown in Fig. 16-11a, there are no horizontal shear stresses transferred fromthe slab to the beam. They act as two independent members. In Fig. 16-11b, horizontal shearstresses act on the interface, and as a result, the slab and beam act in a composite manner.The ACI Code provisions for horizontal shear are given in ACI Code Section 17.5.Although the mechanism of horizontal shear transfer and that of shear friction are similar,if not identical, there is a considerable difference between the two sets of provisions. Thedifference results from the fact that Eq. (16-8) and ACI Code Sections 17.5.3 and 11.6 are allempirical attempts to fit test data. Equation (16-8) is valid for relatively short shear planeswith lengths up to several feet, but is believed to give shear strengths that are too high forlong shear-transfer regions if the maximum stress is localized. It is also unconservative forlow values of , as used in Eq. (16-3).

The term horizontal shear stress is used to describe the shear stresses acting on theinterface in Fig. 16-11b. In a normal composite beam, these stresses are horizontal. If thebeam were vertical, however, the term “horizontal shear” would still be used to distinguishbetween this shear and the orientation of shear stresses in a beam. Tests of horizontal shearin composite beams are reported in [16-3], [16-5], [16-6], [16-15], and [16-16]. The testsreported in [16-6] included members with and without shear keys along the interface. Thepresence of shear keys stiffened the connection at low slips but had no significant effect onits strength.

rvfy

(a) Noncomposite.

(b) Composite.

Fig. 16-11Horizontal shear transfer in acomposite beam.


Computation of Horizontal Shear Stress

From strength of materials, the horizontal shear stresses, on the contact surfacebetween an uncracked elastic precast beam and a slab can be computed from

(16-13)

where

force acting on the section in question

moment of the area of the slab or flange about the centroidal axis of thecomposite section

of inertia of the composite section

of the interface between the precast beam and the cast-in-place slab

Equation (16-13) applies to uncracked elastic beams and is only an approximation forcracked concrete beams.

The ACI Code gives two ways of calculating the horizontal shear stress.ACI Code Section 17.5.3 defines the nominal horizontal shear force, to be trans-

ferred as

(16-14)(ACI Eq. 17-1)

Setting and using expression for shear stress from Chapter 6 gives

(16-15a)

This is based on the observation that the shear stresses on opposite pairs of sides of anelement located at the top of the web are equal in magnitude, as shown in Fig. 6-3a, butare arranged to give couples in opposite directions. For an element taken from directlyover the beam web at the interface between the web and flange, the shear stresses on thetop and bottom sides of the element are and the shear stresses on the vertical sidesof the element are

(16-15b)

where is the factored shear force acting on the cross section of the beam as obtainedfrom a shear-force diagram for the beam. If the shear stresses on the left and right faces ofthe element form a counter-clockwise couple, those on the bottom and top of the elementmust form a clockwise couple. For equilibrium, the shear stresses on four sides of an elementmust be equal in magnitude. Thus,

(16-16)

Calculation of Horizontal Shear Stresses from C and T Forces

Alternatively, ACI Code Section 17.5.4 allows horizontal shear to be computed from thechange in compressive or tensile force in the slab in a segment of any length. Figure 16-12illustrates this clause. At midspan, the force in the compression zone is C, as shown in

vnh = vnv

Vu

vnv =Vu>fbvd

vnh,

vnh =Vu>fbvd

fVnh = Vu

fVnh Ú Vu

Vnh,

bv = width

Ic = moment

Q = first

V = shear

vuh =VQ

Icbv

vh,


Fig. 16-12a. All of this force acts above the interface. At the end of the beam, the force inthe flange is zero. Thus, the horizontal shear force to be transferred across the interfacebetween midspan and the support is

(16-17)

A similar derivation could be carried out by considering the force in the reinforcement,

(16-18)

where is the tensile force in the reinforcement related to ACI Code Section 17.5.4.1 says that when ties are provided to resist the horizontal

shear calculated using Eqs. (16-17) or (16-18), the spacing of the ties should approximatelyreflect the distribution of shear forces along the member. This is specified because the slipof the slab relative to the web at the onset of a horizontal shear failure is too small to allowmuch redistribution of horizontal shear stresses. In [16-5], the measured slip at maximumhorizontal shear stress was on the order of 0.02 in., and the slip at failure ranged from 0.08to 0.3 in. We shall calculate the horizontal shear stresses by dividing the beam into a seriesof segments and computing the value of or at both ends of each segment. Then,

(16-19)

where and are the larger and smaller tension forces on the two ends of a segment,and is the length of the segment. The tension force can be taken as If jd isassumed to be constant, varies as the diagram. For a uniformly loaded simple beam,the diagram is a parabola, and the moments at the and points of the span are 0.75and 0.438 times the maximum moment, respectively. These fractions of the moment diagramcan be calculated from the shear-force diagram because the change in moment from one sec-tion to another is equal to the area of the shear-force diagram between the two sections.

The two procedures give similar results, as will be seen in Example 16-2. The limitson from ACI Code Sections 17.5.3.1 to 17.5.3.3 are given in Table 16-1.(Vuh >f)Vnh

18

14Mu

MuTu

Tu = Mu> jd./sTu2Tu1

vuh =Tu1 - Tu2bv/s

TuCu

Mu.Tu

Vuh = Tu

Vh = C

vh

vh

C

Tu

(a)

(b)

Fig. 16-12Horizontal shear stresses in acomposite beam.


In composite beams, the contact surfaces must be clean and free from laitance. Thewords “intentionally roughened” imply that the surface has been roughened with a “fullamplitude” of as discussed in connection with shear friction. When the factored shearforce, at the section exceeds ACI Code Section 17.5.3.4requires design be based on shear friction, in accordance with ACI Code Section 11.6.4.This limit reflects the range of test data used to derive ACI Code Section 17.5.3.3.

ACI Code Section 17.6 requires that the ties provided for horizontal shear be not lessthan the minimum stirrups required for shear, given by

(16-20)(ACI Eq. 11-13)

The tie spacing shall not exceed four times the least dimension of the supported element,which is usually the thickness of the slab, but not more than 24 in. The ties must be fullyanchored both in the beam stem and in the slab.

Deflections

The beam cross section considered when calculating deflections depends on whether thebeam was shored or unshored when the composite slab is placed. If it is shored so that thefull dead load of both the precast beam and the slab is carried by the composite section,ACI Code Section 9.5.5.1 allows the designer to consider the loads to be carried by the fullcomposite section when computing deflections. The modulus of elasticity should be basedon the strength of the concrete in the compression zone, while the modulus of ruptureshould be based on the strength of the concrete in the tension zone. For nonprestressedbeams constructed with shores, it is not necessary to check deflections if the overall heightof the composite section satisfies ACI Code Table 9.5(a).

ACI Code Section 9.5.5.2 covers the calculation of deflections for unshored con-struction of nonprestressed beams. If the thickness of the precast member satisfies ACITable 9.5(a), it is not necessary to consider deflections. If the thickness of the compositesection satisfies the table, but the thickness of the precast member does not, it is not neces-sary to compute deflections occurring after the section becomes composite, but it is neces-sary to compute the instantaneous deflections and that part of the sustained loaddeflections occurring prior to the beginning of effective composite action. The latter can beassumed to occur when the modulus of elasticity of the slab reaches 70 to 80 percent of its28-day value, usually about 4 to 7 days after the slab is placed.

ACI Code Section 9.5.5.1 states that if deflections are computed, they shouldaccount for the curvatures induced by the differential shrinkage between the slab and theprecast beam. Shrinkage of the slab relative to the beam causes the slab to shorten relativeto the beam. Because the slab and beam are joined together, this relative shortening causes

Av =0.752fœc bws

fyt, and Ú

50 bws

fyt

f1500bvd2 psi,Vu = fVnh,14 in.,

TABLE 16-1 Calculation of

ACI Section Contact Surfaces Ties

17.5.3.1 Intentionally roughened None17.5.3.2 Not roughened Minimum ties

from ACI Code Section 17.6

17.5.3.3 Intentionally roughened

but not more than 500bvd

a260 +0.6Avfyt

bvsb lbvdAvfyt

80bvd80bvd

Vnh

Vnh


the beam to deflect downward, adding to the deflections due to loads. Some of the shrink-age of the concrete in the beam will have occurred before the beam is erected in the struc-ture. All of the slab shrinkage occurs after the slab is cast. As the slab shrinks relative to thebeam, tensile stresses are induced in the slab and compressive stresses in the beam. These areredistributed to some degree by creep of the concrete in the slab and beam. This effect canbe modeled by using an age-adjusted effective modulus, and an age-adjusted trans-formed section in the calculations, as discussed in Section 3-6 and in [16-17] and [16-18].

EXAMPLE 16-2 Design of a Composite Beam

Precast, simply supported beams that span 24 ft and are spaced 10 ft on centers arecomposite with a slab that supports an unfactored live load of 100 psf, a partition loadof 20 psf, and a superimposed dead load of 10 psf. Design the beams and the compositebeam and slab. Use for the slab, 5000 psi for the precast beams, and

Use load factors from ACI Code Section 9.2.1.

1. Select the trial dimensions. For the end span of the slab, ACI Code Table 9.5(a)gives the minimum thickness of a one-way slab as For asimply supported beam, the table gives Deflectionsof the composite beam may be a problem if the overall depth is less than 18 in. However,for unshored construction, ACI Code Section 9.5.5.2 requires that deflections of the precastmember be considered if its overall depth is less than that given by Table 9.5(a). To avoidthis, we shall try an 18-in.-deep precast beam 12 in. wide to allow the steel to be in onelayer, plus a 5-in. slab. For the precast beam,

2. Compute the factored loads on the precast beam. Because the floor will beconstructed in an unshored fashion, the precast beam must support its own dead load, thedead load of the slab, the weight of the forms for the slab, assumed to be 10 psf, and someconstruction live load, assumed to be 50 psf. Thus, we have the following data:

Dead loads:

Live load:

The ACI Code does not specifically address the load factors for this construction-load case. We shall take giving

3. Compute the size of the precast member required for flexure.

We shall select a steel percentage close to, but less than, the tension-controlled limit in theprecast beam, so that and later check whether that provides enough steel forf = 0.9,

Mu =1.94 * 242

8= 140 kip-ft

wu = 1.2 * 0.950 + 1.6 * 0.500 = 1.94 kips>ftU = 1.2D + 1.6L,

w = 10 ft * 0.050 ksf = 0.500 kip>ftTotal = 0.950 kip>ftForms w = 10 ft * 0.010 ksf = 0.100 kip>ftSlab w = 5 in.>12 * 10 ft * 0.150 kcf = 0.625 kip>ftBeam stem w =12 * 18

144* 0.150 kcf = 0.225 kip>ft

d = 18 - 2.5 = 15.5 in.

h = />16 = 124 * 122>16 = 18 in.h = />24 = 120 in.>24 = 5 in.

fy = 60,000 psi.fcœ = 3000 psi

Ecaa,


flexure in the composite section. From Table A-3 for 5000 psi, the highest value of cor-responding to tension-controlled behavior is 0.021. We shall try For this valueof , and from Eq. (5-23a),

For this requires that and Therefore,the size chosen for deflection control is adequate.

For the 12-by-18-in. beam, , and , so

Try a 12-by-18-in. precast beam with three No. 9 bars; Then

This is less than , which is the tension-controlled limit. Hence, and we have

Therefore, o.k.

4. Check the capacity of the composite member in flexure. The factored loadson the composite member are as follows:

Dead load:

Superimposed dead load

Live load:

Mu =3.06 * 242

8= 220 kip-ft

Total factored load = 3.06 kips>ftTotal w = 1.20 kips>ft Factored = 1.6 * 1.20 = 1.92 kips>ft Partitions w = 10 ft * 0.020 ksf = 0.2 kip>ft Floor load w = 10 ft * 0.100 ksf = 1.0 kip>ft Total w = 0.950 kip>ft Factored = 1.2 * 0.950 = 1.14 kips>ftw = 0.100 kip>ftSlab w = 0.625 kip>ft Precast stem w = 0.225 kip>ft

fMn = fAsfyad -a

2b = 185 k-ft

f = 0.9,3/8 d

c = a/b1 = 3.53/0.8 = 4.41 in.

a =3.00 * 60

0.85 * 5 * 12= 3.53 in.

As = 3.00 in.2.

= 2.23 in.2

As =Muffyjd

=140 * 12

0.9 * 60 * 0.9 * 15.5

jd � 0.9dd = 18 - 2.5 = 15.5

h = 12.3 + 2.5 = 14.8 in.d = 12.3 in.b = 12 in.,

bd2 ÚMufR

=140 k-ft * 12 in./ft

0.9 * 1.03 ksi= 1810 in.3

r, R = 1030 psi = 1.03 ksir = 0.02.

r

Compute From ACI Code Section 8.12.2, the effective flange width is 72 in.The overall height is Assuming rectangularbeam action, where in the slab is 3000 psi, we have

Because a is less than the flange thickness, rectangular beam action exists. Also, ismuch less than for the tension-controlled limit; thus, for flexure, and we have

Therefore, the steel chosen is adequate to resist the moments acting on the composite sec-tion. Use a 12-by-18-in. precast section with three No. 9 longitudinal bars and a 5-in.cast-in-place slab.

5. Check vertical shear. at d from the support = 3.06 kips/ft * (12 - 20.5/12)

(where we shall use the smaller of the two concrete strengths)(ACI Eq. 11-3)

Because we need stirrups. Thus,

Try Grade-60, No. 3 U stirrups, and let Then

(from ACI Eq. 11-15)

(from ACI Eq. 11-13)

We will select the stirrups after considering horizontal shear.

6. Compute the horizontal shear. Horizontal shear may be computed accordingto ACI Code Section 17.5.3 or Section 17.5.4. We shall do the calculations both ways andcompare the results. We shall assume that the interface is clean, free of laitance, and inten-tionally roughened.

ACI Code Section 17.5.3: From Eq. (16-14) (ACI Eq. (17-1)), at d from the support.

From ACI Code Sec. 17.5.3.2, an intentionally roughened surface without ties is ad-equate for

Therefore, ties are required.From ACI Code Section 17.5.3.3, if minimum ties are provided according to ACI

Code Section 17.6 and the interface is intentionally roughened,

fVnh = f1260 + 0.6rvfyt2lbvdfVnh = f80bvd = 0.75 * 80 * 12 * 20.5 = 14.8 kips

fVnh Ú Vu = 31.5 kips

= 22.0 in. on centers

Minimum stirrups (50 psi governs): s =Avfyt

50bw=

0.22 * 60,000

50 * 12

Maximum spacing = d>2 = 10.25 in., say, 10 in.

= 17.9 in. on centers

s =Avfytd

Vs=

0.22 * 60 * 20.5

15.1

Av = 0.22 in.2.

= 15.1 kips at d from the support

Vs =Vuf

- Vc =31.5

0.75- 26.9

Vu 7 fVc,= 2 * 123000 * 12 * 20.5 = 26.9 kips fVc = 20.2 kips

Vc = 2l2fcœ bwdft = 31.5 kips.

Vu

= 270 kip-ft

fMn = fAsfyad -a

2bf = 0.93/8

c>da =3.00 * 60

0.85 * 3 * 72= 0.98 in.

fcœ18 + 5 = 23 in.; d = 23 - 2.5 = 20.5 in.

fMn.



The maximum tie spacing allowed by ACI Code Section 17.6.1 is but notmore than 24 in. The maximum spacing for vertical shear governs. Assume that the ties areNo. 3 two-leg stirrups at the maximum spacing allowed for shear, 10 in. Then

and

Thus, minimum stirrups provide more than enough ties for horizontal shear. We shall useclosed stirrups to better anchor them into the top slab. Use No. 3, Grade-60 closed stir-rups at 10 in. on centers throughout the length of the beam.

ACI Code Section 17.5.4: From Eq. (16-19),

(16-19)

Consider the section of the flange between the midspan and the quarter point of the span.At midspan, The distance from the support to thequarter point is The moment at the quarter point is

The average horizontal shear stress between the midspan and the quarter point is

Because this is less than 80 psi, minimum ties are sufficient between the midspan and thequarter points.

Now consider the portion of the beam between the quarter point and the eighth point:

The average horizontal shear stress between the quarter and the eighth point of the span is

Because this value exceeds 80 psi, stirrups are required to satisfy ACI Code Section17.5.3.1. Minimum ties, according to ACI Code Section 17.6, are equivalent to No. 3 two-legged stirrups at 10 in. on centers, with For minimum ties, ACI CodeSection 17.5.3.3 gives

= 326 psi

vnh = 1260 + 0.6rvfyt2 psi = 1260 + 0.6 * 0.00183 * 60,0002rv = 0.00183.

vnh =(107 - 62.7) * 1000

12 * 13 * 122 = 103 psi

Tu1 1at the quarter point2 = 107 kips, and Tu2 1at the eighth point2 = 62.7 kips

Mu at the eighth point = 96.4 kip-ft

vnh =1143 - 1072 kips * 1000

12 in. * 16 * 122 in.= 41.7 psi

Tu2 = Mu>0.9d = 107 kips

= 220 - 55.1 = 165 kip-ft

Mu = Rx - wx2>2 = 13.06 kip>ft * 12 ft2 * 6 ft - 3.06 * 62>224>4 = 6 ft.Tu1 = Mu>jd = Mu>0.9d = 143 kips.

vuh =Tu1 - Tu2bv/s

= 60.1 kips

fVnh = 0.751260 + 0.6 * 0.00183 * 60,0002 * 1.0 * 12 * 20.5

= 0.00183

rv =Avbvs

=0.22

12 * 10

4 * 5 in. = 20 in.


Minimum stirrups are more than enough. Consider the portion of beam between the sup-port and the eighth point. At the eighth point, At the support, The average horizontal shear stress on this segment is

Thus, minimum ties are satisfactory. We shall use closed stirrups to better anchor theminto the top slab. Use No. 3, Grade-60 closed stirrups at 10 in. on centers throughoutthe length of the beam. ■

REFERENCES

16-1 J. A. Hofbeck, I. A. Ibrahim, and Alan H. Mattock, “Shear Transfer in Reinforced Concrete,” ACIJournal, Proceedings, Vol. 66, No. 2, February 1969, pp. 119–128.

16-2 Alan H. Mattock and Neil M. Hawkins, “Shear Transfer in Reinforced Concrete—Recent Research,”Journal of the Prestressed Concrete Institute, Vol. 17, No. 2, March–April 1972, pp. 55–75.

16-3 J. C. Walraven, “Fundamental Analysis of Aggregate Interlock,” Journal of the Structural Division,Proceedings of the American Society of Civil Engineers, Vol. 107, No. ST11, November 1981, pp. 2245–2271.

16-4 H. W. Reinhardt and J. C. Walraven, “Cracks in Concrete Subject to Shear,” Journal of the StructuralDivision, Proceedings of the American Society of Civil Engineers, Vol. 108, No. ST1, January 1982, pp. 225–244.

16-5 Robert E. Loov and A. K. Patnaik, “Horizontal Shear Strength of Composite Concrete Beams with aRough Interface,” PCI Journal, Vol. 39, No. 1, January–February 1994, pp. 48–69.

16-6 Norman W. Hanson, “Precast-Prestressed Concrete Bridges,” Part 2, “Horizontal Shear Connections,”Journal of the Research and Development Laboratories, Portland Cement Association, Vol. 2, No. 2,May 1960, pp. 38–58.

16-7 M. A. Ali and R. N. White, “Enhanced Contact Model for Shear Friction of Normal and High-StrengthConcrete,” ACI Structural Journal, May–June 1999, Vol. 96, No. 3, pp. 348–360.

16-8 ACI-ASCE Committee 426, “The Shear Strength of Reinforced Concrete Members,” ProceedingsASCE, Journal of the Structural Division, Vol. 99, No. ST6, June 1973, pp. 1091–1187.

16-9 Mast, R. F., “Auxiliary Reinforcement in Precast Connections,” Proceedings, Journal of the StructuralDivision ASCE, Vol. 94, ST6, June 1968, pp. 1485–1504.

16-10 A. H. Mattock, “Shear Friction and High-Strength Concrete,” ACI Structural Journal, Vol. 98, No. 1,January–February 2001, pp. 50–59.

16-11 L. F. Kahan and A. D. Mitchell, “Shear Friction Tests with High-Strength Concrete,” ACI StructuralJournal, Vol. 99, No. 1 January–February 2002, pp. 98–103.

16-12 F. J. Vecchio and M. P. Collins, “The Modified Compression Field Theory,” ACI Journal, Proceedings,Vol. 83, No. 2, March–April 1986, pp. 219–231.

16-13 Dino Angelakos, Evan C. Bentz, and Michael P. Collins, “The Effect of Concrete Strength andMinimum Stirrups on the Shear Strength of Large Members,” ACI Structural Journal, Vol. 98, No. 3,May–June 2001, pp. 290–296.

16-14 PCI Design Handbook—Precast and Prestressed Concrete, Sixth edition, Prestressed Concrete Institute,Chicago, IL, 2004, 740 pp.

16-15 Paul H. Kaar, L. B. Kriz, and Eivind Hognestad, “Precast-Prestressed Bridges: (1) Pilot Tests ofContinuous Girders,” Journal of the Research and Development Laboratories, Portland CementAssociation, Vol. 2, No. 2, May 1960, pp. 21–37.

16-16 J. C. Saemann and George W. Washa, “Horizontal Shear Connections Between Precast Beams and Cast-in-Place Slabs,” ACI Journal, Proceedings, Vol. 61, No. 11, November 1964, pp. 1383–1409.

16-17 Amin Ghali and Rene Favre, Concrete Structures: Stresses and Deformations, Chapman & Hall, NewYork, 1986, 348 pp.

16-18 Walter H. Dilger, “Creep Analysis of Prestressed Concrete Structures Using Creep-Transformed SectionProperties,” PCI Journal, Vol. 27, No. 1, January–February 1982, pp. 99–118.

vnh =162.7 - 02 * 1000

12 * 13 * 122 = 145 psi

Tu2 = zero.Tu1 = 62.7 kips.

Documents

Chapter 16 Shear Friction, Horizontal Shear Transfer, And Composite Concrete Beams