858
161 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 interface
between a beam and a slab cast later than the beam, but expected to act in a composite
manner.
162 SHEAR FRICTION
From time to time, shear must be transferred across an interface between two members
that can slip relative to one another. The shearcarrying mechanism is known variously
as aggregate interlock, interface shear transfer, or shear friction. The last of these
terms is used here. The interface on which the shears act is referred to as the shear
plane 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 common
are the socalled pushoff specimens similar to that shown in Fig. 161, which were tested
in one of three ways:
1.Mattock [161], [162] and his associates reported tests of pushoff specimens in
which lateral expansion or contraction and crack widths were not held constant or otherwise
controlled during the tests except by the reinforcement perpendicular to the slip plane. These
specimens were tested either as constructed (uncracked) or precracked along the shear plane
shown by the vertical dashed line in Fig. 161. Only the failure load and the amount of trans
verse reinforcement were reported.
v
nh
C
T
16
Shear Friction,
Horizontal Shear
Transfer, and
Composite
Concrete Beams
Section 162 Shear Friction •
859
Fig. 161
Sheartransfer specimen.
2.Walraven [163], [164] tested pushoff 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 [165] and Hanson [166],
measured the shear transferred between the web and the slab in a composite beam. This
is referred to as horizontal shear.
An excellent review of tests and equations for shear transfer is presented by Ali and
White [167].
Cohesion and Friction
The results of 13 tests of uncracked specimens, and 21 tests of specimens with previously
cracked shear planes are plotted in Fig.162, with open circles and solid circles, respec
tively. The test results from [161] and [162] plotted in Fig. 162 suggest that:
(a) The strengths of cracked and uncracked specimens can be represented by
equations of the form
(161a)
(161b)
where c is a cohesionlike term equal to the intercepts of the sloping lines on the verti
cal axis in Fig. 162, plus a frictionlike term
that is equal to the product of
the compressive stress on the shear plane, and the coefficient of friction, . The
plots in Fig. 162, can be idealized as straight sloping lines that intercept the vertical
axis at “cohesions” of
and psi, for the uncracked and precracked
specimens, 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 the
precracked specimens. The uncracked and precracked specimens reached similar
upper limits on shear transfer as shown in Fig. 162.
In Fig. 162, the shear stress at failure, is plotted against where is the
ratio of the area of the transverse reinforcement across the shear plane to the area of the
r
v
r
v
f
y
,v
u
,
c L 255c L 505
ms,
s tan u
v
n
= c + ms
v
n
= 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 during
these test series. The shear stress at failure increases with the value of which is
taken 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 pushoff tests, the first
cracks were a series of diagonal tension cracks, as shown in Fig. 163.With further
deformation, the compression struts between these cracks rotated at their ends (points A
and B) such that point B moved downward relative to A. At the same time, the distance
AC,measured across the crack, increased, stretching the transverse reinforcement. The
tension in the reinforcement was equilibrated by an increase in the compression in the
struts. 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 formed
by placing a layer of concrete on top of or against an existing layer of hardened concrete,
r
v
f
y
,
1500
1500
1000
1000
500
500
0
Initially uncracked
Initially uncracked
Initially cracked
Initially cracked
vu, psi
r
v
f
y
, psi
Fig. 162
Variation shear strength
with web reinforcement
ratio (From [161] and
[162].)
r
v
f
y
.
v
u
Fig. 163
Diagonal tension cracking
along a previously uncracked
shear plane. (From [168].)
Section 162 Shear Friction •
861
Fig. 1
6
4
Shearfriction model.
(From [166].)
relative slip of the layers causes a separation of the surfaces, as shown in Fig. 164a.If the
reinforcement across the crack is developed on both sides of the crack, it is elongated
by the separation of the surfaces and hence is stressed in tension. For equilibrium, a
compressive stress is needed as shown in Fig. 164b. Shear is transmitted across the
crack by (a) friction resulting from the compressive stresses [164], [165], and (b) inter
lock of aggregate roughness on the cracked surfaces, combined with dowel action of the
reinforcement crossing the surface.
ShearFriction Model
The original and simplest design model is the shearfriction model [161], [168], shown in
Fig.164, which ignores the cohesionlike component, c, and assumes that the shear transfer
is due entirely to friction. This model is the basis of the shearfriction design procedure in
ACI Code Section 11.6. Because the cohesion component is ignored, unusually high values
of 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
(162)
(ACI Eq.1125)
or in terms of stresses,
(163)
where is defined above and is the appropriate value of the coefficient of friction given
in ACI Code Section 11.6.4.3.
Permanent Compression Force
A permanent compressive force, perpendicular to the slip plane, causes a normal
stress 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,
N
u
> A
cv
N
u
N
u
,
mr
v
v
n
= r
v
f
y
m
V
n
= A
vf
f
y
m
862 •
Chapter 16 Shear Friction, Horizontal Shear Transfer, and Composite Concrete Beams
Fig. 165
Force components in a bar
inclined to the shear plane.
(From [161].)
N
u
, act on the shear plane, they must be equilibrated by tensile reinforcement provided
in addition to the shearfriction reinforcement
(164)
Inclined ShearFriction Reinforcement
When the shearfriction reinforcement is inclined to the slip plane such that slip along this
plane produces tensile stresses in the reinforcement, as shown in Fig.165, the component
of bar force perpendicular to the slip plane causes an compressive force on the plane equal
to In addition, there is a force component parallel to the crack equal to
In stress units, the nominal shearfriction resistance at failure from Eq. (163) is
(165)
which has units of stress or, in terms of forces,
(166)
(ACI Eq.1126)
Shearfriction 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 the
shear that can be transferred. Such steel is not included as shearfriction reinforcement.
Other Factors Affecting Shear Transfer
Lightweight Concrete
Tests [162] have indicated that the resistance to slip along the shear plane is smaller for
lightweightconcrete specimens than for normalweight concrete specimens. This happens
because the cracks penetrate the pieces of lightweight aggregate rather than following the
V
n
= A
vf
f
y
1m sin a
f
+ cos a
f
2
v
n
= mr
v
f
y
sin a
f
+ r
v
f
y
cos a
f
A
vf
f
y
cos a
f
.
A
vf
f
y
sin a
f
.
A
n
=
N
u
ff
y
Section 162 Shear Friction •
863
perimeters of the aggregate. The resulting crack faces are smoother than for normalweight
concrete [162]. 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 strength
of lightweight aggregate concrete. Here, it is used for convenience to account for the
smoother 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, as
specified in ACI Code Section 11.6.9:
4.for concrete anchored to asrolled structural steel by studs or by reinforcing bars
(see ACI Code Section 11.6.10):
where for normalweight concrete and 0.75 for alllightweight concrete. The code
allows to be determined based on volumetric proportions of lightweight and normalweight
aggregates, as specified in ACI Code Section 8.6.1, but it shall not exceed 0.85.
The first term in Eq. (161) represents the portion of the shear transferred by sur
face protrusions and by dowel action. Equation (165) with as plotted in
Fig. 166,applies only for greater than 200 psi and the radial lines in Fig. 166 are
not plotted below this value. For Grade60 reinforcement, this limit requires a minimum
reinforcement ratio,
Design Rules in the ACI Code
ACI Code Section 11.6 presents design rules for cases “where it is appropriate to consider
shear transfer across a given plane, such as an existing or a potential crack, an interface
between dissimilar materials, or an interface between two concretes cast at different
times.” Typical examples are shown in Fig.167.
In design, a crack is assumed to exist along the shear plane, and reinforcement is
provided across that crack. The amount of reinforcement is computed (ACI Code
Section 11.6.4) from
(167)
(ACI Eq.111)
(162)
(ACI Eq.1125)
where for designs carried out by using the load factors in ACI Code Section 9.2
and 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 (a
weak layer on the top surface of a concrete placement due to bleed water collecting at the
surface). ACI Code Section 11.6.10 requires that, in case 4, the steel must be clean and free
of 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 “wave
length” for the roughened surface. This was done to allow some freedom in satisfying this
1
4
in.
m
f = 0.75
V
n
= A
vf
f
y
m
fV
n
Ú V
u
r = 0.0033.
r
v
f
y
a
f
= 90°,
l
l = 1.0
0.7l
0.6l
1.0l
1.4l
m,
l
l,m
864 •
Chapter 16 Shear Friction, Horizontal Shear Transfer, and Composite Concrete Beams
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 normalweight concrete either cast monolithically or placed against hardened con
crete with intentionally roughened surfaces, as noted previously, ACI Code Section
11.6.5 sets the upper limit on from Eqs. (162) and (166) to the smallest of
and where is the area of the concrete section
resisting shear transfer. The limits represented by the second two expressions given here
are larger than previously permitted limits in the ACI Code and represent a reexamination
of test data as presented in references [1610] and [1611]. For all other cases, including
lightweight concrete and concrete placed against not intentionally roughened surfaces,
shall not exceed the smaller of and In cases where a lowerstrength
concrete is cast against a higherstrength concrete, or visa versa, the value of for the
lowerstrength concrete shall be used to evaluate the limits given here. These upper limits
on shear friction strength are necessary, because Eqs. (162) and (166) may become
unconservative in some cases.
CohesionplusFriction Model
For the usual case of transverse reinforcement perpendicular to the shear plane, Mattock
[162] rewrote Eq. (162) as
(168)
where for normalweight concrete, 200 psi for alllightweight concrete, and
250 psi for sandlightweight concrete. The first term on the righthand side of Eq. (168)
represents the shear transferred by “cohesion,” which is caused by pieces of aggregate
bearing on the surfaces of the slip plane, by the shearing off of surface protrusions, and by
K
1
= 400 psi
V
n
= K
1
A
cv
+ 0.8A
vf
f
y
f
c
œ
1800 psi2A
c
.0.2f
c
œ
A
c
V
n
A
c
11600 psi2A
c
,1480 psi + 0.08f
c
œ
2A
c
0.2f
c
œ
A
c
,V
n
3
4
in.
1
4
1500
1500
1000
1000
500
500200
0
Initially uncracked
Initially cracked
vu, psi
r
v
f
y
, psi
Eq. (165)
Eq. (165)
Eq. (165)
m 0.6
m 1.0
m 1.4
ACI Code Section 11.6.5
for f
c
4000 psi
Fig. 166
Comparison of Eq. (162) with
test data from [161], [162],
and [166].
Section 162 Shear Friction •
865
dowel action. The second term represents the “friction,” with the coefficient of friction
taken to be 0.8 for cracked concrete sliding on cracked concrete. Equation (168) is called
the modified shearfriction equation.
In 2001, Mattock [1610] reevaluated Eq. (168) on the basis of 199 tests, with
ranging from 2450 psi to 14,400 psi, and derived a set of equations that retained the
numerical constant 0.8 in Eq. (168), but presented a new family of values of for
various types of shear planes.
Walraven Model
Walraven [163] idealized concrete as a series of sizegraded spherical pieces of coarse
aggregate embedded in a matrix of hardened concrete paste. A crack was assumed to
cross the matrix between pieces of aggregate until it reached a piece of aggregate, at
which time it followed the aggregate–matrix interface around the aggregate. The strength
K
1
f
œ
c
.
.
.
Fig. 167
Examples of shear friction.
(From [169].)
866 •
Chapter 16 Shear Friction, Horizontal Shear Transfer, and Composite Concrete Beams
of the interface between the cement paste and the aggregate was assumed to be less than
the strength of the aggregate. Such a crack is shown in Fig.168a, after cracking, but
before 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 aggregate
comes into contact with the matrix in the dark shaded area, allowing shear to be trans
ferred across the crack (Fig. 168b). Further slip mobilizes rigid–plastic stresses at the
point of contact, and the crack is restrained against further slip until the crack surfaces
deteriorate (see Fig. 168c).The maximum shearing force transferred by a single aggre
gate particle occurs at the stage shown in Fig. 168c, corresponding to the largest area of
plastic stresses. By assuming randomly chosen gradations of aggregate sizes, Walraven
developed expressions for the shear forces transferred for various crack widths, w, and
maximum aggregate diameters, a.
Walraven [163] tested 88 pushoff sheartransfer specimens similar to Fig. 161. The
major difference between these tests and those by Mattock and others was that Walraven
kept the crack widths, w, constant throughout each test. Vecchio and Collins [1612] fitted
Eqs. (165) and (168) to Walraven’s test data and obtained
(169)
where is the shear stress transferred across the crack, is the compression stress required
across the crack in psi, and is the maximum shear stress that can be transmitted
across a given crack.
The maximum shear stress, that can be transferred across a crack when its
width is held at win. is given by
(1610)
where a is the diameter of the coarse aggregate in the cracked concrete in inches. The crack
width,win inches, is computed from the spacing, of the inclined crack as
(1611)
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 (1611)
assumes that all the strain is concentrated in the crack.
Walraven’s tests were limited to concrete strengths, from 2900 to 8200 psi. For
highstrength concretes with cylinder strengths in excess of 10,000 psi, the cracks tend to
f
œ
c
,
s
u
e
1
w = e
1
s
u
s
u
,
v
ci max
=
2.162f
œ
c
0.3 + a
24w
a + 0.63
b
v
ci max
,
v
ci max
f
ci
v
ci
v
ci
= 0.18v
ci max
+ 1.64f
ci
 0.82
f
ci
2
v
ci max
R
a
w
c
w
b
b, max
(a) No contact.(b) Growing contact.(c) Maximum contact.
c, max
Shear
Fig. 168
Walraven crack model.
(From [163].)
Section 162 Shear Friction •
867
cross the individual pieces of aggregate rather than going around them. As a result, the crack
surface was smoother than for weaker concretes. For this case, the effective size of the
aggregate, a, decreases, approaching zero. Angelakos, Bentz, and Collins [1613] handled
this by arbitrarily reducing the effective aggregate size, a, in Eq. (1610) from the nominal
diameter, 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 [167] carried out similar analyses, assuming an undulating crack path, to
illustrate 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 [165] derived the following equation for
shear transfer across cracks and for horizontal shear in composite beams (see Fig. 1611,
discussed later):
(1612)
In this equation, accounts for lightweight concrete and k is a constant equal to 0.5 for
concrete placed against hardened concrete and 0.6 for concrete placed monolithically.
The squareroot term allows this equation to fit the test data more closely than do other
models. Equation (1612) is plotted in Fig.169 for comparison with test results and
with Eq. (168).
Because the reinforcement is assumed to yield in order to develop the necessary
forces, the yield strength of the steel is limited to 60,000 psi. Each bar must be anchored on
both sides of the crack to develop the bar. The steel must be placed approximately
uniformly across the shear plane, so that all parts of the crack are clamped together.
Comparison of Design Rules with Test Results
In Fig. 169, test data for pushoff tests of initially uncracked specimens with reinforce
ment perpendicular to the shear plane [161] are compared with
l
v
ci max
= lk2sf
œ
c
+ r
v
f
y
cos a
f
1500
1500
1000
1000
500
500
0
Initially uncracked
Initially cracked
Composite Tbeams [1613]
vn
, psi
r
v
f
y
, psi
Eq. 1612, k 0.5
Eq. 168
0.8
r
v
f
y
c 400
ACI Code Section 11.6.5
for f
c
4000 psi
Fig. 169
Comparison of test results
from [161] and [165] and
equations (168)
and (1612).
868 •
Chapter 16 Shear Friction, Horizontal Shear Transfer, and Composite Concrete Beams
(a) Equation (168), with and
(b) Equation (1612), with for initially cracked concrete
The test data are compared with the nominal strengths, with The specimens had
concrete strengths ranging from 3840 psi to 4510 psi.
Test data for pushoff specimens with a precracked interface [161] and an average
concrete strength of 4060 psi are plotted in Fig. 169, along with data from tests of com
posite beams with average concrete strengths of 5710 psi for the webs and 5160 psi for the
flanges [16–5]. The nominal strengths computed from Eq. (1612), with for a pre
cracked interface, and fit the data quite well.
EXAMPLE 161 Design of the Reinforcement in the Bearing Region of a
Precast Beam
Figure 1610 shows the support region of a precast concrete beam. The factored beam
reactions 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 the
beam is 12 in. wide by 18 in. deep. Use assume normaldensity 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. 1610,
and assume that the crack is at 60° to the horizontal. It intercepts the end of the beam at 10.2
in. above the bottom and has a length of 12 in. We shall take A
c
= 12 in. * 12 in. = 144 in.
2
.
2.Compute the area of steel required.Resolving the forces onto the inclined
plane 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 shear
forces. However, it is quick and conservative to assume that the shear force is equal to the
vertical 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. (162),
We want so
A
vf
Ú
V
u
ff
y
m
V
n
Ú fV
u
,
V
n
= A
vf
f
y
m
a
f
= 90°.N
u
= 12 kips
6in.* 6in.
A
c
,
f
y
= 60,000 psi.
f
œ
c
= 4000 psi,
f
œ
c
= 4000 psi,
k = 0.5
f = 1.0.
k = 0.5,
K
1
= 400 psi,
3 No. 6
Fig. 1610
Example 161.
Section 163 Composite Concrete Beams •
869
where (the crack plane is in monolithically placed concrete), and
(normalweight concrete is used). We thus have
We must confirm that the required value of does not exceed the upper limit
given 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, is
limited in ACI Code Section 11.6.5 to the smallest of
or
1600 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 the
effective section area resisting the factored shear force.
The tensile force must also be transferred across the crack by reinforcement, as
calculated using Eq. (164),
Therefore, the total steel across the crack must be
Provide three No. 6 bars across the assumed crack, .These bars
must 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 [1614].
■
163 COMPOSITE CONCRETE BEAMS
Frequently, precast beams or steel beams have a slab cast on top of them and are designed
assuming that the slab and beam act as a monolithic unit to support loads. Such a beam
andslab combination is referred to as a composite beam. This discussion will deal only
with composite beams where the beam is precast
concrete or other concrete cast at an
earlier 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. Shores
are then added to support the beam and initially resist the weight of the castinplace slab.
When the strength of the slab concrete i
s high enough to resist expected stresses, the shores
1.7/
d
,
A
s
= 1.32 in.
2
10.984 + 0.267) in.
2
= 1.25 in.
2
A
n
=
12 kips
0.75 * 60 ksi
= 0.267 in.
2
V
n
V
n
… 1800 psi2A
c
= 800 psi * 144 in.
2
= 115 kips
V
n
480 psi + 0.08f
c
œ
= 480 + 0.08 * 4000 = 800 psi
0.2f
c
œ
= 0.2 * 4000 psi = 800 psi
A
c
,
V
n
= V
u
/f = 62 kips/0.75 = 82.7 kips
V
n
V
n
,V
u
/f,
A
vf
=
62 kips
0.75 * 60 ksi * 1.4
= 0.984 in.
2
l = 1.0
m = 1.4lV
u
= 62 kips,
870 •
Chapter 16 Shear Friction, Horizontal Shear Transfer, and Composite Concrete Beams
are removed and the slab dead load is resisted by the composite beam and slab. If the precast
beam is not shored when the slab is placed, the beam supports its own weight plus the weight of
the 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 beam
is shored, ACI Code Section 17.3 requires that the shores be left in place until the composite
section 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 whether
the member was shored or unshored during construction. For this reason, ACI Code
Section 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.1611a, there are no horizontal shear stresses transferred from
the slab to the beam. They act as two independent members. In Fig. 1611b, horizontal shear
stresses 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. The
difference results from the fact that Eq. (168) and ACI Code Sections 17.5.3 and 11.6 are all
empirical attempts to fit test data. Equation (168) is valid for relatively short shear planes
with lengths up to several feet, but is believed to give shear strengths that are too high for
long sheartransfer regions if the maximum stress is localized. It is also unconservative for
low values of , as used in Eq. (163).
The term horizontal shear stress is used to describe the shear stresses acting on the
interface in Fig. 1611b. In a normal composite beam, these stresses are horizontal. If the
beam were vertical, however, the term “horizontal shear” would still be used to distinguish
between this shear and the orientation of shear stresses in a beam. Tests of horizontal shear
in composite beams are reported in [163], [165], [166], [1615], and [1616]. The tests
reported in [166] included members with and without shear keys along the interface. The
presence of shear keys stiffened the connection at low slips but had no significant effect on
its strength.
r
v
f
y
(a) Noncomposite.
(b) Composite.
Fig. 1611
Horizontal shear transfer in a
composite beam.
Section 163 Composite Concrete Beams •
871
Computation of Horizontal Shear Stress
From strength of materials, the horizontal shear stresses, on the contact surface
between an uncracked elastic precast beam and a slab can be computed from
(1613)
where
force acting on the section in question
moment of the area of the slab or flange about the centroidal axis of the
composite section
of inertia of the composite section
of the interface between the precast beam and the castinplace slab
Equation (1613) applies to uncracked elastic beams and is only an approximation for
cracked 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
(1614)
(ACI Eq.171)
Setting and using expression for shear stress from Chapter 6 gives
(1615a)
This is based on the observation that the shear stresses on opposite pairs of sides of an
element located at the top of the web are equal in magnitude, as shown in Fig. 63a, but
are arranged to give couples in opposite directions. For an element taken from directly
over the beam web at the interface between the web and flange, the shear stresses on the
top and bottom sides of the element are and the shear stresses on the vertical sides
of the element are
(1615b)
where is the factored shear force acting on the cross section of the beam as obtained
from a shearforce diagram for the beam. If the shear stresses on the left and right faces of
the element form a counterclockwise couple, those on the bottom and top of the element
must form a clockwise couple. For equilibrium, the shear stresses on four sides of an element
must be equal in magnitude. Thus,
(1616)
Calculation of Horizontal Shear Stresses from C and T Forces
Alternatively, ACI Code Section 17.5.4 allows horizontal shear to be computed from the
change in compressive or tensile force in the slab in a segment of any length. Figure 1612
illustrates this clause. At midspan, the force in the compression zone is C, as shown in
v
nh
= v
nv
V
u
v
nv
=
V
u
> f
b
v
d
v
nh
,
v
nh
=
V
u
> f
b
v
d
fV
nh
= V
u
fV
nh
Ú V
u
V
nh
,
b
v
= width
I
c
= moment
Q = first
V = shear
v
uh
=
VQ
I
c
b
v
v
h
,
872 •
Chapter 16 Shear Friction, Horizontal Shear Transfer, and Composite Concrete Beams
Fig. 1612a. All of this force acts above the interface. At the end of the beam, the force in
the flange is zero. Thus, the horizontal shear force to be transferred across the interface
between midspan and the support is
(1617)
A similar derivation could be carried out by considering the force in the reinforcement,
(1618)
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. (1617) or (1618), the spacing of the ties should approximately
reflect the distribution of shear forces along the member. This is specified because the slip
of the slab relative to the web at the onset of a horizontal shear failure is too small to allow
much redistribution of horizontal shear stresses. In [165], the measured slip at maximum
horizontal shear stress was on the order of 0.02 in., and the slip at failure ranged from 0.08
to 0.3 in. We shall calculate the horizontal shear stresses by dividing the beam into a series
of segments and computing the value of or at both ends of each segment. Then,
(1619)
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 is
assumed 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 0.75
and 0.438 times the maximum moment, respectively. These fractions of the moment diagram
can be calculated from the shearforce diagram because the change in moment from one sec
tion to another is equal to the area of the shearforce diagram between the two sections.
The two procedures give similar results, as will be seen in Example 162. The limits
on from ACI Code Sections 17.5.3.1 to 17.5.3.3 are given in Table 161.(V
uh
>f)V
nh
1
8
1
4
M
u
M
u
T
u
T
u
= M
u
> jd./
s
T
u2
T
u1
v
uh
=
T
u1
 T
u2
b
v
/
s
T
u
C
u
M
u
.T
u
V
uh
= T
u
V
h
= C
v
h
v
h
C
T
u
(a)
(b)
Fig. 1612
Horizontal shear stresses in a
composite beam.
Section 163 Composite Concrete Beams •
873
In composite beams, the contact surfaces must be clean and free from laitance. The
words “intentionally roughened” imply that the surface has been roughened with a “full
amplitude” of as discussed in connection with shear friction. When the factored shear
force, at the section exceeds ACI Code Section 17.5.3.4
requires 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 less
than the minimum stirrups required for shear, given by
(1620)
(ACI Eq.1113)
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 fully
anchored both in the beam stem and in the slab.
Deflections
The beam cross section considered when calculating deflections depends on whether the
beam was shored or unshored when the composite slab is placed. If it is shored so that the
full 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 full
composite section when computing deflections. The modulus of elasticity should be based
on the strength of the concrete in the compression zone, while the modulus of rupture
should be based on the strength of the concrete in the tension zone. For nonprestressed
beams constructed with shores, it is not necessary to check deflections if the overall height
of 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 ACI
Table 9.5(a), it is not necessary to consider deflections. If the thickness of the composite
section 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 load
deflections occurring prior to the beginning of effective composite action. The latter can be
assumed to occur when the modulus of elasticity of the slab reaches 70 to 80 percent of its
28day 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 should
account for the curvatures induced by the differential shrinkage between the slab and the
precast beam. Shrinkage of the slab relative to the beam causes the slab to shorten relative
to the beam. Because the slab and beam are joined together, this relative shortening causes
A
v
=
0.752f
œ
c
b
w
s
f
yt
, and Ú
50 b
w
s
f
yt
f1500b
v
d2 psi,V
u
= fV
nh
,
1
4
in.,
TABLE 161 Calculation of
ACI Section Contact Surfaces Ties
17.5.3.1 Intentionally roughened None
17.5.3.2 Not roughened Minimum ties
from ACI Code Section 17.6
17.5.3.3 Intentionally roughened
but not more than 500b
v
d
a260 +
0.6A
v
f
yt
b
v
s
b lb
v
dA
v
f
yt
80b
v
d
80b
v
d
V
nh
V
nh
874 •
Chapter 16 Shear Friction, Horizontal Shear Transfer, and Composite Concrete Beams
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 the
beam, tensile stresses are induced in the slab and compressive stresses in the beam. These are
redistributed to some degree by creep of the concrete in the slab and beam. This effect can
be modeled by using an ageadjusted effective modulus, and an ageadjusted trans
formed section in the calculations, as discussed in Section 36 and in [1617] and [1618].
EXAMPLE 162 Design of a Composite Beam
Precast, simply supported beams that span 24 ft and are spaced 10 ft on centers are
composite with a slab that supports an unfactored live load of 100 psf, a partition load
of 20 psf, and a superimposed dead load of 10 psf. Design the beams and the composite
beam 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 oneway slab as For a
simply supported beam, the table gives Deflections
of 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 precast
member be considered if its overall depth is less than that given by Table 9.5(a). To avoid
this, we shall try an 18in.deep precast beam 12 in. wide to allow the steel to be in one
layer, plus a 5in. slab. For the precast beam,
2.Compute the factored loads on the precast beam.Because the floor will be
constructed in an unshored fashion, the precast beam must support its own dead load, the
dead load of the slab, the weight of the forms for the slab, assumed to be 10 psf, and some
construction 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 tensioncontrolled limit in the
precast beam, so that and later check whether that provides enough steel forf = 0.9,
M
u
=
1.94 * 24
2
8
= 140 kipft
w
u
= 1.2 * 0.950 + 1.6 * 0.500 = 1.94 kips>ft
U = 1.2D + 1.6L,
w = 10 ft * 0.050 ksf = 0.500 kip>ft
Total = 0.950 kip>ft
Forms w = 10 ft * 0.010 ksf = 0.100 kip>ft
Slab w = 5 in.>12 * 10 ft * 0.150 kcf = 0.625 kip>ft
Beam 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.
f
y
= 60,000 psi.
f
c
œ
= 3000 psi
E
caa
,
Section 163 Composite Concrete Beams •
875
flexure in the composite section. From Table A3 for 5000 psi, the highest value of cor
responding to tensioncontrolled behavior is 0.021. We shall try For this value
of , and from Eq. (523a),
For this requires that and Therefore,
the size chosen for deflection control is adequate.
For the 12by18in. beam,
, and
, so
Try a 12by18in. precast beam with three No. 9 bars; Then
This is less than , which is the tensioncontrolled limit. Hence, and we have
Therefore, o.k.
4.Check the capacity of the composite member in flexure.The factored loads
on the composite member are as follows:
Dead load:
Superimposed dead load
Live load:
M
u
=
3.06 * 24
2
8
= 220 kipft
Total factored load = 3.06 kips>ft
Total 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>ft
w = 0.100 kip>ft
Slab w = 0.625 kip>ft
Precast stem w = 0.225 kip>ft
fM
n
= fA
s
f
y
ad 
a
2
b = 185 kft
f = 0.9,3/8 d
c = a/b
1
= 3.53/0.8 = 4.41 in.
a =
3.00 * 60
0.85 * 5 * 12
= 3.53 in.
A
s
= 3.00 in.
2
.
= 2.23 in.
2
A
s
=
M
u
ff
y
jd
=
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.,
bd
2
Ú
M
u
fR
=
140 kft * 12 in./ft
0.9 * 1.03 ksi
= 1810 in.
3
r,R = 1030 psi = 1.03 ksi
r = 0.02.
r
Compute From ACI Code Section 8.12.2, the effective flange width is 72 in.
The overall height is Assuming rectangular
beam action, where in the slab is 3000 psi, we have
Because a is less than the flange thickness, rectangular beam action exists. Also, is
much less than for the tensioncontrolled 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 12by18in. precast section with three No. 9 longitudinal bars and a 5in.
castinplace 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. 113)
Because we need stirrups. Thus,
Try Grade60, No. 3 U stirrups, and let Then
(from ACI Eq. 1115)
(from ACI Eq. 1113)
We will select the stirrups after considering horizontal shear.
6.Compute the horizontal shear.Horizontal shear may be computed according
to ACI Code Section 17.5.3 or Section 17.5.4. We shall do the calculations both ways and
compare 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. (1614) (ACI Eq. (171)),
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,
fV
nh
= f1260 + 0.6r
v
f
yt
2lb
v
d
fV
nh
= f80b
v
d = 0.75 * 80 * 12 * 20.5 = 14.8 kips
fV
nh
Ú V
u
= 31.5 kips
= 22.0 in. on centers
Minimum stirrups (50 psi governs): s =
A
v
f
yt
50b
w
=
0.22 * 60,000
50 * 12
Maximum spacing = d>2 = 10.25 in., say, 10 in.
= 17.9 in. on centers
s =
A
v
f
yt
d
V
s
=
0.22 * 60 * 20.5
15.1
A
v
= 0.22 in.
2
.
= 15.1 kips at d from the support
V
s
=
V
u
f
 V
c
=
31.5
0.75
 26.9
V
u
7 fV
c
,
= 2 * 123000
* 12 * 20.5 = 26.9 kips fV
c
= 20.2 kips
V
c
= 2l2f
c
œ
b
w
d
ft = 31.5 kips.
V
u
= 270 kipft
fM
n
= fA
s
f
y
ad 
a
2
b
f = 0.93/8
c>d
a =
3.00 * 60
0.85 * 3 * 72
= 0.98 in.
f
c
œ
18 + 5 = 23 in.; d = 23  2.5 = 20.5 in.
fM
n
.
876 •
Chapter 16 Shear Friction, Horizontal Shear Transfer, and Composite Concrete Beams
Section 163 Composite Concrete Beams •
877
The maximum tie spacing allowed by ACI Code Section 17.6.1 is but not
more than 24 in.The maximum spacing for vertical shear governs. Assume that the ties are
No. 3 twoleg 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 use
closed stirrups to better anchor them into the top slab. Use No. 3, Grade60 closed stir
rups at 10 in. on centers throughout the length of the beam.
ACI Code Section 17.5.4:From Eq. (1619),
(1619)
Consider the section of the flange between the midspan and the quarter point of the span.
At midspan, The distance from the support to the
quarter 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 the
quarter 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 Section
17.5.3.1. 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 Code
Section 17.5.3.3 gives
= 326 psi
v
nh
= 1260 + 0.6r
v
f
yt
2 psi = 1260 + 0.6 * 0.00183 * 60,0002
r
v
= 0.00183.
v
nh
=
(107  62.7) * 1000
12 * 13 * 122
= 103 psi
T
u1
1at the quarter point2 = 107 kips, and T
u2
1at the eighth point2 = 62.7 kips
M
u
at the eighth point = 96.4 kipft
v
nh
=
1143  1072 kips * 1000
12 in.* 16 * 122 in.
= 41.7 psi
T
u2
= M
u
>0.9d = 107 kips
= 220  55.1 = 165 kipft
M
u
= Rx  wx
2
>2 = 13.06 kip>ft * 12 ft2 * 6 ft  3.06 * 6
2
>2
24>4 = 6 ft.
T
u1
= M
u
>jd = M
u
>0.9d = 143 kips.
v
uh
=
T
u1
 T
u2
b
v
/
s
= 60.1 kips
fV
nh
= 0.751260 + 0.6 * 0.00183 * 60,0002 * 1.0 * 12 * 20.5
= 0.00183
r
v
=
A
v
b
v
s
=
0.22
12 * 10
4 * 5 in.= 20 in.
878 •
Chapter 16 Shear Friction, Horizontal Shear Transfer, and Composite Concrete Beams
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 them
into the top slab. Use No. 3, Grade60 closed stirrups at 10 in. on centers throughout
the length of the beam.
■
REFERENCES
161 J. A. Hofbeck, I. A. Ibrahim, and Alan H. Mattock, “Shear Transfer in Reinforced Concrete,” ACI
Journal, Proceedings, Vol. 66, No. 2, February 1969, pp. 119–128.
162 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.
163 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.
164 H. W. Reinhardt and J. C. Walraven, “Cracks in Concrete Subject to Shear,” Journal of the Structural
Division,Proceedings of the American Society of Civil Engineers, Vol. 108, No. ST1, January 1982,
pp. 225–244.
165 Robert E. Loov and A. K. Patnaik, “Horizontal Shear Strength of Composite Concrete Beams with a
Rough Interface,” PCI Journal, Vol. 39, No. 1, January–February 1994, pp. 48–69.
166 Norman W. Hanson, “PrecastPrestressed 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.
167 M. A. Ali and R. N. White, “Enhanced Contact Model for Shear Friction of Normal and HighStrength
Concrete,” ACI Structural Journal, May–June 1999, Vol. 96, No. 3, pp. 348–360.
168 ACIASCE Committee 426, “The Shear Strength of Reinforced Concrete Members,” Proceedings
ASCE, Journal of the Structural Division, Vol. 99, No. ST6, June 1973, pp. 1091–1187.
169 Mast, R. F., “Auxiliary Reinforcement in Precast Connections,” Proceedings,Journal of the Structural
Division ASCE, Vol. 94, ST6, June 1968, pp. 1485–1504.
1610 A. H. Mattock, “Shear Friction and HighStrength Concrete,” ACI Structural Journal, Vol. 98, No. 1,
January–February 2001, pp. 50–59.
1611 L. F. Kahan and A. D. Mitchell, “Shear Friction Tests with HighStrength Concrete,” ACI Structural
Journal, Vol. 99, No. 1 January–February 2002, pp. 98–103.
1612 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.
1613 Dino Angelakos, Evan C. Bentz, and Michael P. Collins, “The Effect of Concrete Strength and
Minimum Stirrups on the Shear Strength of Large Members,” ACI Structural Journal, Vol. 98, No. 3,
May–June 2001, pp. 290–296.
1614 PCI Design Handbook—Precast and Prestressed Concrete, Sixth edition, Prestressed Concrete Institute,
Chicago, IL, 2004, 740 pp.
1615 Paul H. Kaar, L. B. Kriz, and Eivind Hognestad, “PrecastPrestressed Bridges: (1) Pilot Tests of
Continuous Girders,” Journal of the Research and Development Laboratories, Portland Cement
Association, Vol. 2, No. 2, May 1960, pp. 21–37.
1616 J. C. Saemann and George W. Washa, “Horizontal Shear Connections Between Precast Beams and Cast
inPlace Slabs,” ACI Journal, Proceedings, Vol. 61, No. 11, November 1964, pp. 1383–1409.
1617 Amin Ghali and Rene Favre, Concrete Structures: Stresses and Deformations, Chapman & Hall, New
York, 1986, 348 pp.
1618 Walter H. Dilger, “Creep Analysis of Prestressed Concrete Structures Using CreepTransformed Section
Properties,” PCI Journal, Vol. 27, No. 1, January–February 1982, pp. 99–118.
v
nh
=
162.7  02 * 1000
12 * 13 * 122
= 145 psi
T
u2
= zero.T
u1
= 62.7 kips.
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