Section 6-5 Analysis and Design of Reinforced Concrete Beams for
Shear-ACICode
1
Comparison of the Truss Model andthe ACI Procedure
For beams with large amounts of web reinforcement, failing at high shear
stresses, tne be
havior closely approaches that predicted by the plastic-truss model. (See Fig. 6-27.) On the
Other hand, the truss model predicts zero shear strength for beams without web
reinforcement
and tends to underestimate the shear capacity for beams with V, less than
about Vc.
The truss model emphasizes two aspects of shear behavior frequently overlooked in
the ACI procedure: the shift in the tensile force diagram, which is of great
in
detailing the longitudinal steel (as discussed in Chapter 8), and the need forimportance
adequate an
chorage of the stirrups in the top and bottom flanges.
Examples of the Design of Beams for Shear
EXAMPLE 6-2 Design of Vertical Stirrups
in a Simply Supported Beam
Figure 6-36 shows the elevation and cross section of a simply supported T beam. This
supports a uniformly distributed service (unfactored) dead load of 20 kN/m, including
its beam
own
weight, and a uniformly distributed service live load of 24 kN/m. Design vertical stirrups for this
beam. The concrete strength is 25 MPa, the yield strength of the fleXural reinforcement is
420 MPa,
D= 20 KN/m
L= 24 kN/m
b=900 mm
d=
610
mm
l = 10 m
h, = 150 mm
| y = 300
(a) Elevation.
(b) Section.
W,, = 62.4 kN/m
W,
=62.4 kN/m
WDu = 24 kN/m
312 kN
264 kN
-48 kN
168 kN
312 KN
(c) Load case 1.
"
(d) Load case 2.
312 = 416 kN
= 312 KN
0.75
Vu
WLu
Fig. 6-36
Beam and shear force
envelope--Example 6-2.
(e) Shear force envelope.
48
48 kN
0.75
() V, lo diagram.
= 64 kN
242 " Chapter 6 Shear in Beams
stirrups is 300 MPa. Use load
and the yield strength of the
and resistance factors
Sections 9.2.3 and 9.3.2.3.
Compute the design factored shear-force envelope. For
S and R all equal to zero. Total factored load:
from ACi3184
ACI (Eq. 9-2) with
W, = 1.2 X 20 kN/m + 1.6 X 24 kN/m
FIHL
= 62.4 kN/m
Factored dead load:
Wpu = 1.2 X 20 kN/m = 24.0 kN/m
Three loading cases should be considered: Fig. 6-37c, Fig. 6-37d, and
the miTOr
Fig. 6-37d. The three shear-force diagrams are superimposed in Fig. 6-37e.
of
approximate the shecar-force envelope with straight lines and design simply For
we
V,, =w,e/2 at the ends and V, = W,l/8 at
Supported
midspan,
where
Wu
1s
the
for
total
live
factored
and
load and wi, is the factored live load. From (6-14),
simplopposi
ibeamscity,te
Setting these equal, we obtain the smallest value of V,
that satisfies (6-14):
V, =
This is plotted in Fig. 6-36f.
Since this beam is loaded on the top
section is located at d = 0.61 m from the flange and supported on the bottom flange, the critical
d from the
support. From Fig. 6-36f and
support is
similar triangles, the shear at
at d = 416 kN
0.61 m
5m
-(416 64)
= 373 kN
Therefore, V,/ at d
2.
where
= 373 kN and min. V, =
373 kN.
Are stirrups required by ACI
V,=
Section 11.5.5.1? No stirrups are
required if
V, <
V2,
Vfeb,d
6
V25 MPa
(6-8)
X 300 mm X
610 mm
6 X 1000
= 153 kN
Because V, = 373 KN exceeds VJ2 = 76.3 kN,
stirrups are required.
Isthe cross section large
3
enough? ACI Section I1.5.6.9 gives the
stirrups as
maximum chear in the
(6-27a)
Thus, the maximum allowable V, is
= 5Ve
= 765 kN
(6-276)
Section6-5
Analysis and Design of Reinforced Concrete Beams for Shear-ACI Code
243
Because V, at d= 373 kN is less than V, max = 765 kN, the section is big enough.
4. Check anchorage of stirrups and maximum spacing. Try No. 10M double-leg stirrups,
f,= 300 MPa:
A,, = 2 X 71 = 142 mm²
(a) Check the anchorage of the stirrups. ACI Section 12.13.2.1 allows No. 25M and
smaller stirrups to be anchored by a standard 90° or 135° hook stirrup hook around a longitu
dinalbar. Provide a No. 10M or larger bar in each of the upper corners of the stirrup to anchor
them,
(b) Find the maximum stirrup spacing.
Based on the beam depth: ACISection 11.5.4.1 requires the smaller of 0.5d= 305 mm or
600 mm. ACISection l1.5,4.3requires half these spacings if V, exceeds ()Vfbd:
3
= 305 kN
Because the maximum V, is less than 305 kN, the maximum spacing is 305 mm.
Based on minimum allowed A,, ((6-24), ACI Eq. (11-13),
bys
Ay,min =
(6-24)
(ACI Eq. 11-13)
but not less than
Ap.min =
1bs
(6-25)
3 fy
Rearranging gives
16 X Afy
Smax
16 x (2 x 71) mm² x 300 MPa
V25 X 300 mm
Vfbw
= 454 mm
but not less than
Smax
3A,fy
bw
3 X (2 X 71) X 300
300
426 mm
Therefore, the maximum spacing based on the beam depth govers Maximum
S=305 mm.
5.
Compute the stirrup spacing required to resist the shear forces For verical sirups,
(6-21) applies; that is,
S=
A,fyd
Vl - V
where , = 153 kN. At dfrom the support, V,/
(6-21)
373 kN and
2 X 71 mm x 300 MPa X 610 mm = 118 mm
S
=
(373 - 153) X 1000 N
Change spacing to s = 150 mm where this ac
Try No, 10M U stirrups at s = 100mm.
The intermediate spacings selected are up to
ceptable, and then to the maximum spacing of 300 mm.
are varied in
different spacings are used and the spacings
the designer. Generally, no more than three
mm?
U-stirrups A, =2X 71 mm = 142
multiples of 50 or 75 mm. For No. 10M
244
Chapter 6 Shear in Beams
s
the factored shears is
spacing required to support
the
support
from
At d
oies
= 100 mm.
150 mm. Rearranging (6-21)
to
increased
be
can
s
where
Compute V,/d,
A,fd
118
mm.
Uge
+ V
S
(2 X 71) X 300 X 610 + 153 = 326 kN
150 × 1000
occurs at
From Fig. 6-38f and similar triangles, this shear
416 - 326
416 - 64
X 5000 mm
= 1280 mm from the end of the beam
Changes to 300 mm. Compute V,Jd, where s can be increased to 300mm. From (6-21)
V
2 X 71 X 300 X 610
+ 153
= 240kN
300 X 1000
This occurs at
416
X=
240
416 - 64
5000
= 2500 mm from the end of the beam
Stirrups must be continued to the point where V,l = V2 = 76.5 kN. This occurs at
416 - 76.5
416 - 64
5000
= 4820 mm from the end of the beam
In choosing the numbers of stirrups at each spacing, each stirrup is assumed to reinforce a
length of web extending s/2 on each side of the stirrup. For this reason, the first stirrup is placed at sy.
from the support. We shall select the following spacings: To summarize, s = one at 50, 13 at 1000
1350 mm from the support, eight at 150 to 2550 mm and eight at 300 to 4950 mm each end. IIS
leaves a space of 50 mm × 2 at midspan.
The total number of stirrups in half of the beam is 30.
30 stirrups are required ln
other half at similar spacings. The elevation and cross Another
section are shown, respectively
Figs. 6-37 and 6-36b.
2No. 13M bars in
Corners of stirrups
13@ 100 mm
8@ 150 mm
8@ 300 mm
Fig. 6-37
Stirrups in beam
Example 6-2.
No. 10M Grade-300 U
stirrups with 135° hooks on upper
ends.
Section 6-5 Analysis and Desian of Reinforced Concrete Beams for Shear-ACICode
24
CYAMPLE 6-3 Design of Stirrups in aContinuous Beam
Fngure 6-38 shows the elevation and cross section of an interior span of a continuous T beam.
lhis beam supports a floor supporting an unfactored dead load of 36 KN/m (including its own
Weight) and an unfactored live load of 35 KN/m. The concrete is of normal density, with f =
23 MPa. The yield strength of the web reinforcement is 300 MPa. Design vertical stirrups, using ACI
Sections 11.1through 11.5. Use the load and resistance factors from Sections 9.2.1 and 9.3.2 of the
2002 ACICode.
l. Compute the factored shear-force envelope. ACI Section 9.2.1 gives the following
load conmbinations for beams loaded with dead and live loads:
(ACI Eq.9-1)
U= 1.4(D + F)
U= 1.2(D+ F+ T) + 1.6(L + H) + 0.5(L, or Sor R)
(ACI Eq. 9-2)
Here, F is the load due to the weight and pressures of fluids with well-defined densities and
controllable maximum heights or due to related internal moments and forces. The beam does not
support tanks containing fluids; thus, F is zero.
T denotes the cumulative effect of temperature, creep,shrinkage, differential settlement, and
shrinkage-compensating concrete. ACI 318-02 Section 9.2.3 and ACI 318-02 Commentary Section
R9.2.3 suggest that T should be based on a realistic assessment of such effects occurring in service.
T loads tend to be dissipated by inelasticaction in the beam at high loads and will be ignored here.
Thus, T is zero.
H represents the loads due tothe weight and pressure of soil, water in soil, or other materials.
The beam being designed does not support soils; hence, H is zero.
Lr, S,and R are live load, snow load, and rain load onaroof. Since this beam supports a floor,
these loads are not applicable.
ACI 318-02 Equations (9-1) and (9-2) respectively become
U= 14D = 1.4 X 36 = 50.4 kN/m
and
U= 1.2D + 1.6L = 1.2 X 36 + 1.6 X 35 = 99.2 kN/m
800 mm
en=9 m
900 mm
(a) Elevation
400 mm
595
-504
(b) Section.
84
800 mm.
-84
(c) Shear force
Fig. 6-38
Beam and shear force envelope-Exxample 6-3.
-595
246
Chapter 6 Shear in Beams
The larger of these governs, so we have
Factored total load w, = 99.2 kN/m
and
kN/m
Factored live load W, = 56
suee
8.3.3.3 gives the shear at the face of the
This is an interior span, and ACISection
V,= w,n/2
where , = 4.5 m is the clear span.
V,, = 99.2 kN/m X9 m/2
= 446.4 kN
faces of the supnorte A
V./d= 446.4/0.75 = 595 kN at the
is the clear span of the beam and
and torsion as 0.75.
318-02 Section 9.3.2.3 gives for shear
taken the shear at midspan as
have
we
For simply supported beams,
(6-26)
V, = WL/8
= 56 kN/m X 9 m/8
= 63 kN
so V/=63/0.75 = 84 kN
continuous beam. The
the shear at midspan of a
compute
to
equation
same
the
We shall use
shear-force envelope is shown in Fig. 6-39c.
(ACIEq. l1-3)
V,=(1/6)|rb,d
=1/6X V25 X 400 ×800
= 266667 N = 266.7 kN
and
V2 = 133.3 kN
Since 504 kN
of the support is V,/ = 504 kN.
The maximum shear at d from the face Using similar triangles V,/ is equal to V/2 at
kN, web reinforcement is required.
exceeds 133.3
X=
504 kN-133.3 KIN y4s00
504 kN -84 kN
mm=3912 mm
from the faces of the supports.
with 135° hooks arouu
Anchorage of stirrups. Try No. 10M Grade-300 Ustirupssuch stirrups to be ancnoi
considers
longitudinal bar in cach upper corner. ACISection 12.13.2.1
4.
in tthe naximun stirrup spacing.
=800mm/2
d/2
of
spacing
a
maximum
11.5.4.1
gives
Section
ACI
Based on beam depih:
=A05 nmo If V, cxccsts (l/3)|fb,d= 533.3 KN, which corespondstoVw/# =(1/3
= 800 KN. ACI Secion I1.5.4.3 requires the maximum spacing to be reduced
to d/4. valueisles
kN.Since this
'he maxinunshear at dfrom the face of asupport is V,=504)
than 800 kN,the maximum spacing based on beanm depth is d/2 = 400 mm.
Section 6-5 Analysis and Desian of Reinforced Concrete Beams for Shear-ACI Code2
Based on minimum A,
(6-24)
A,,min
(ACI Eq. l1-13)
but not more than
A,min =
(6-25)
3f,
Rearranging (6-24) gives
16A,f,
max
16x 142x 300
25 X400
=340.8 mm
and, from (6-25),
3A,f_3 X142 X300 319.5mm
400
Smax
b,,
Therefore, the maximum spacing s based on Amin gOverns. Accordingly, we try s = 300 mm.
6.
Compute the stirrup spacing required to resist the factored shear force.
S =
Afyd
(6-21)
where, from step 3, V, = 266.7 kN.
At d from support,
'u = 504 kN and
s =
2X71mm X300 MPa X800 mm
(504 kN - 266.7 kN) X 1000 N
= 144 mm
At d from support, use s = 100 mm.
Change s to 150 mm:
V, Afyd +
V
for s = 150 mm=
V
(6-21)
2X71X 300 X800
150 X 1000 N
+ 266.7
= 494 kN
595
This occurs at x
494
595- 84
X4500 = 889 mm from the faces of the supports.
Change s to 300mm:
Vu for
=300 mm =
= 380 kN
2X71X 300 X800
300 X 1000
+ 266.7
248
Chapter 6 Shear in Beams
595
This occurs at x= 595
from
380 X4500=J893 mm
the faces of
the supports.
84
Select stirrups:
face
support
of the support)
from the
fromfce of
= 950 mm
100
One at 50 mm1
X
9
+
the
(extending to 50
face of the
2000 mm from
nine at 100 mm
950 + 1050 =
suppo)
the
to
(extending
4400 mm from
seven at 150 mm
+ 2400 =
support)
(extending to2000
4400) = 200 mm without any stirrups near midspan. This is just
eight at 300mm
(4500
This leaves a space of 2 X
12.10.3 requires that flexural
fine.
reinforcement. ACI Section
for flexure., except at reinflexural
required
Anchorage of
no longer
is
it
7.
where
flexural
point
reinforcement athe suppast the
Extend all
12d,
dor
cantilevers.
extend
forcement
ends of
spans and atthe free
ports of simple
of inflection.
points
flexural
d= 800 mm past the
hooks around bars in the top
stirrups with 135°
corGrade-300U
10M
at l50 mm, and eight at 300 mm,
No.
seven
Use
mm,
100
at
Summary.
one at 50 mm, nine half the beam is 25. Another 25 stirrups are required in
ners of the section
in
stirrups
each half. The total number of
the other half at similar spacings.
distance
6-6
MODERN SHEAR DESIGN METHODS
concrete codes,
In most current
(6-9),
shear design is based on
(6-9)
V, = V t V,
was empiri
concrete." The currently used value of V,
the
by
carried
'shear
the
is
where V,
studied, a me
since reinforced concrete began to be
cally derived from test results. Ever
Today, the
researchers.
Grail sought by
chanical explanation of V. has been the Holyunderstood enough to allow shear design
underlying mechanisms of shear transfer are
equations extrapolated from test results.An
based at least partially on theory, instead of on
design methods is given in [6-2).
shear
excellent review of the current status of modern
slender beams are compared in
The details of six methods of designing for shear in
traditional truss analogy, discussed in
Table 6-1. The first and second methods are the
Section 6-5. Three of the re
Section 6-4, and the traditionalACI method, examined in
field theory
maining four methods are successively improved versions of the compression
Canadian concreie
(a) the compression field theory (CFT-84), from the 1984
code [6-25] [6-14],
[0-20]
(b) the modified compression field theorv, from the 1994 Canadian Code
andthe AASHTO Limit States Design Specification, and
(c) the revision of MCFT currently being considered for inclusion in
Canadian code (MCFT-04). See [6-27], [6-28]. [6-29).
the 2004
The sixth procedure listed is based on an application of shear friction to s.cnder
ds
methods
gn
desi
the
method,
failing in shear. Except for the truss analogy andthe CFT-84
calculating V,, as
are based on (6-9), although each of the methods has a unique way of
shown in Table 6-1.
Definitions and Design Methods
Design procedures for shear fall intotwoclassifications:
1. Whole-member design
methods, such as the truss
3s a truss or a strut-and-tie model that
represents the
for one load case.
member
analogy,
idealize themechanisam
complete load-resisting