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Reinforced Concrete Designin Compliance with ACI 318M-08
Part 8: Design of Two-Way Slabs Shear
by
Dr. S. K. Ghosh
S. K. Ghosh Associates Inc.Palatine, IL and Aliso Viejo, CA
USA
A web seminar series in cooperation with and under sponsorship of theDepartment of Municipal Affairs, Emirate of Abu Dhabi, UAE
www.skghoshassociates.com
334 East Colfax Street, Unit E 43 Vantis DrivePalatine, IL 60067 Aliso Viejo, CA 92656Ph: (847) 991-2700 Fax: (847) 991-2702 Ph: (949) 249-3739 Fax: (949) 249-3989
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S. K. Ghosh Associates Inc.
www.skghoshassociates.com
- 1 -
REINFORCED CONCRETE DESIGN IN COMPLIANCE WITH
ACI 318M-08: DESIGN OF TWO-WAY SLABS - SHEAR
Please stay tuned. We will be starting at
8:00 am UAE Time.
You will be able to listen to the seminar
using computer speakers
If you are encountering technical difficulties, please call 00 1 847 991 2700
Visit us at: www.skghoshassociates.com
SKGA Web Seminarin cooperation with and under sponsorship of
the Department of Municipal Affairs, Emirate of Abu Dhabi, UAE
- 2 -
REINFORCED CONCRETE DESIGN IN
COMPLIANCE WITH ACI 318M-08
A web seminar series in cooperation wi th and under sponsorship of
the Department of Municipal Affairs, Emirate of Abu Dhabi, UAE.
Part 9: DESIGN OF TWO-WAY SLABS -
SHEAR
S. K. Ghosh Associates Inc.
Palatine, IL and Aliso Viejo, CA
www.skghoshassociates.com
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Shear in Two-Way Slab Systems
Shear is typically not critical where a two-way slab
system is supported on beams and/or walls.
However, shear can be critical for flat plates or flat
slabs directly supported on columns. Shear strength
at an exterior slab-column connection is especially
critical because the total exterior negative slab
moment must be transferred to the edge column.
- 4 -
Unbalanced moment
Moment distribution: gravity load
Moment distribution: lateral load
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Shear in Two-Way Slab Systems
Two types of shear are checked for flat plates or flat
slabs directly supported by columns.
- One-way shear (beam-type shear)
- Two-way shear (punching shear)
- 6 -
Shear in Two-Way Slab Systems
One-way shear (beam-type shear)
- slab acts as a wide beam spanning between
columns
- critical section is taken at a distance d (effectivedepth of slab) away from the column face
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Shear in Two-Way Slab Systems
Two-way shear (punching shear)
- failure line occurs along the surface of a truncated
cone or pyramid around a column
- critical section is taken at a distance d/2 from the
face of the column
- two-way shear is usually more critical than one-way
shear
- 8 -
Shear in Two-Way Slab Systems
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Shear in Two-Way Slab Systems
Direct shear at an interior slab-column joint
(Source: Portland Cement Association, Notes on ACI 318-08 Building Code
Requirements for Structural Concrete, Skokie, IL, 2008)
One-way shear Two-way shear
- 1 0 -
Shear in Two-Way Slab Systems
When no or insignificant moment is transferred from
slab to column, direct shear, which is distributed
uniformly around the critical shear perimeter bo,
occurs around slab-column joints.
When significant moment is transferred from slab to
column because of unbalanced gravity loads on
either side of an interior column or horizontal load,
shear transferred from unbalanced moment, in
addition to direct shear, should be considered.
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Shear in Two-Way Slab Systems
ACI 318M-08 Fig. R11.11.7.2: Assumed distribution of shear stress
- 1 2 -
Crit ical Section for Shear
ACI 11.11 Provis ions for slabs and footings
ACI 11.11.1
The shear strength of slabs and footings in the vicinity
of columns, concentrated loads, or reactions is
governed by the more severe of two conditions
ACI 11.11.1.1. Beam action where each critical
section to be investigated extends in a plane across
the entire width.
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Crit ical Section for Shear
ACI 11.11.1.2. For two-way action, each of the critical
sections to be investigated shall be located so that its
perimeter bo is a minimum but need not approach closer
than d/2 to:
(a) Edges or corners of columns, concentrated loads, or
reaction areas; and
(b) Changes in slab thickness such as edges of capitals,
drop panels, or shear caps.
For two-way action, the slab or footing shall be designed
in accordance with 11.11.2 through 11.11.6
- 1 4 -
Crit ical Section for Shear
Tributary areas and critical sections for two-way shear
(Source: Portland Cement Association, Notes on ACI 318-08 Building Code
Requirements for Structural Concrete, Skokie, IL, 2008)
(Drop panel)
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Concrete Shear Strength for
Two-Way Act ion
ACI 11.11.2
The design of a slab or footing for two-way action is
based on Eq. (11-1) and (11-2). Vc shall be
computed in accordance with 11.11.2.1, 11.11.2.2,
or 11.11.3.1. Vs shall be computed in accordance
with 11.11.3. For slabs with shearheads, Vn shall be
in accordance with 11.11.4. When moment is
transferred between a slab and a column, 11.11.6
shall apply.
- 2 0 -
Concrete Shear Strength for
Two-Way Act ion
ACI 11.11.2.1. For nonprestressed slabs and footings,
Vc shall be the smallest of (a), (b), and (c):
(a)
Where is the ratio of long side to short side of thecolumn, concentrated load or reaction area;
dbf2
117.0V o'
cc
+= ACI Eq. (11-31)
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Concrete Shear Strength for
Two-Way Act ion
(b)
where s is 40 for interior columns, 30 for edgecolumns, 20 for corner columns; and
(c)
ACI Eq. (11-32)
ACI Eq. (11-33)
dbf2b
d083.0V o
'c
o
sc
+
=
dbf33.0V o'
cc =
- 2 2 -
Concrete Shear Strength for
Two-Way Act ion
ACI 318M-08 Fig. R11.11.2: Value of for a nonrectangular loaded area
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Shear Strength Provided by Bars, Wires,
and Single or Multiple-Leg Stirrups
ACI 11.11.3
Shear reinforcement consisting of bars or wires and
single- or multiple-leg stirrups shall be permitted in
slabs and footings with d greater than or equal to
150 mm, but not less than 16 times the shear
reinforcement bar diameter. Shear reinforcement
shall be in accordance with 11.11.3.1 through
11.11.3.4.
- 2 6 -
Shear Strength Provided by Bars, Wires,
and Single or Multiple-Leg Stirrups
ACI 318M-08 Fig. R11.11.3 (a) (c): Single- or multiple-leg stirrup-type slabshear reinforcement
(a) Single-leg stirrup or bar (b) Multiple-leg stirrup or bar
(c) Closed stirrups
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Shear Strength Provided by Bars, Wires,
and Single or Multiple-Leg Stirrups
ACI 11.11.3.1. Vn shall be computed by Eq. (11-2),
where Vc shall not be taken greater than
and Vs shall be calculated in accordance with 11.4.
In Eq. (11-15), Av shall be taken as the cross-
sectional area of all legs of reinforcement on one
peripheral line that is geometrically similar to the
perimeter of the column section.
dbf17.0 o'
c
scn VVV +=
s
dfAV
ytv
s =
ACI Eq. (11-2)
ACI Eq. (11-15)
- 2 8 -
Shear Strength Provided by Bars, Wires,
and Single or Multiple-Leg Stirrups
ACI 11.11.3.2. Vn shall not be taken greater than
ACI 11.11.3.3. The distance between the column face
and the first line of stirrup legs that surrounds the
column shall not exceed d/2. The spacing between
adjacent stirrup legs in the first line of shear
reinforcement shall not exceed 2d measured in a
direction parallel to the column face.
dbf5.0 o'
c
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Shear Strength Provided by Bars, Wires,
and Single or Multiple-Leg Stirrups
The spacing between successive lines of shear
reinforcement that surround the column shall not
exceed d/2 measured in a direction perpendicular to
the column face.
ACI 11.11.3.4. Slab shear reinforcement shall satisfy
the anchorage requirements of 12.13 and shallengage the longitudinal flexural reinforcement in the
direction being considered.
- 3 0 -
Shear Strength Provided by Bars, Wires,
and Single or Multiple-Leg Stirrups
Design and detailing criteria for slabs with s tirrups (interior column)(Source: Portland Cement Association, Notes on ACI 318-08 Building Code
Requirements for Structural Concrete, Skokie, IL, 2008)
dbf17.0V o'cu
dbf50.0
s/dfAdbf17.0V
o'c
ytvo'cu
+
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Shear Strength Provided by Bars, Wires,
and Single or Multiple-Leg Stirrups
Design and detailing criteria for slabs with s tirrups(edge and corner column)
(Source: Portland Cement Association, Notes on ACI 318-08 Building CodeRequirements for Structural Concrete, Skokie, IL, 2008)
- 3 2 -
Shear Strength Provided by
Headed Shear Stud Reinforcement
ACI 11.11.5
Headed shear stud reinforcement, placed
perpendicular to the plane of a slab or footing, shall
be permitted in slabs and footings in accordancewith 11.11.5.1 through 11.11.5.4.
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Shear Strength Provided by
Headed Shear Stud Reinforcement
The overall height of the shear stud assembly shall
not be less than the thickness of the member less
the sum of: (1) the concrete cover on the top flexural
reinforcement; (2) the concrete cover on the base
rail; and (3) one-half the bar diameter of the tension
flexural reinforcement.
- 3 4 -
Shear Strength Provided by
Headed Shear Stud Reinforcement
Where flexural tension reinforcement is at the bottom
of the section, as in a footing, the overall height of
the shear stud assembly shall not be less than the
thickness of the member less the sum of: (1) the
concrete cover on the bottom flexural reinforcement;(2) the concrete cover on the head of the stud; and
(3) one-half the bar diameter of the bottom flexural
reinforcement.
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Shear Strength Provided by
Headed Shear Stud Reinforcement
Commentary to 11.11.5
Headed shear stud reinforcement was introduced in
the 2008 Code. Using headed stud assemblies as
shear reinforcement in slabs and footings requires
specifying the stud shank diameter, the spacing of
the studs, and the height of the assemblies for the
particular applications.
- 3 6 -
Shear Strength Provided by
Headed Shear Stud Reinforcement
Tests (Joint ACI-ASCE Committee 421) show that vertical
studs mechanically anchored as close as possible to the
top and bottom of slabs are effective in resisting
punching shear. The bounds of the overall specified
height achieve this objective while providing areasonable tolerance in specifying that height as shown
in Figure R7.7.5.
Joint ACI-ASCE Committee 421, Shear Reinforcement for
Slabs (ACI 421.1R-99) (Reapproved 2006), American
Concrete Institute, Farmington Hills, MI, 1999, 15 pp.
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Shear Strength Provided by
Headed Shear Stud Reinforcement
ACI 318M-08 Fig. R7.7.5: Concrete cover requirements for headed shearstud reinforcement
- 3 8 -
Shear Strength Provided by
Headed Shear Stud Reinforcement
ACI 318M-08 Fig. R7.7.5: Concrete cover requirements for headed shearstud reinforcement
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Shear Strength Provided by
Headed Shear Stud Reinforcement
ACI 11.11.5.1. For the critical section defined in
11.11.1.2, Vn shall be computed using Eq. (11-2),
with Vc and Vn not exceeding and
, respectively.
dbf25.0 o'
c
dbf66.0 o'
c
- 4 0 -
Shear Strength Provided by
Headed Shear Stud Reinforcement
Vs shall be calculated using Eq. (11-15) with Av equal
to the cross-sectional area of all the shear
reinforcement on one peripheral line that is
approximately parallel to the perimeter of the
column section, where s is the spacing of theperipheral lines of headed shear stud reinforcement.
Avfyt/(bos) shall not be less than'
cf.170
s
dfAV
ytv
s = ACI Eq. (11-15)
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Shear Strength Provided by
Headed Shear Stud Reinforcement
ACI 11.11.5.2. The spacing between the column face
and the first peripheral line of shear reinforcement
shall not exceed d/2. The spacing between
peripheral lines of shear reinforcement,measured in
a direction perpendicular to any face of the column,
shall be constant.
- 4 2 -
Shear Strength Provided by
Headed Shear Stud Reinforcement
For prestressed slabs or footings satisfying 11.11.2.2, this
spacing shall not exceed 0.75d; for all other slabs and
footings, the spacing shall be based on the value of the
shear stress due to factored shear force and unbalanced
moment at the critical section defined in 11.11.1.2, andshall not exceed:
(a) 0.75d where maximum shear stresses due to factored
loads are less than or equal to and
(b) 0.5d where maximum shear stresses due to factored
loads are greater than
'
cf5.0
'
cf5.0
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Shear Strength Provided by
Headed Shear Stud Reinforcement
Commentary to 11.11.5.2
The specified spacing between peripheral lines of
shear reinforcement are justified by experiments
(Joint ACI-ASCE Committee 421). The clear
spacing between the heads of the studs should be
adequate to permit placing of the flexural
reinforcement.
- 4 4 -
Shear Strength Provided by
Headed Shear Stud Reinforcement
ACI 11.11.5.3. The spacing between adjacent shear
reinforcement elements, measured on the perimeter
of the first peripheral line of shear reinforcement,
shall not exceed 2d.
ACI 11.11.5.4. Shear stress due to factored shear
force and moment shall not exceed at
the critical section located d/2 outside the outermost
peripheral line of shear reinforcement.
dbf17.0 o'
c
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- 4 7 -
QuestionandAnswerSession
If you are encountering technical difficulties, please call 00 1 847 991 2700
- 4 8 -
Openings in Slabs
ACI 11.11.6 Openings in slabs
When openings in slabs are located at a distance less
than 10 times the slab thickness from a
concentrated load or reaction area, or whenopenings in flat slabs are located within column
strips as defined in Chapter 13, the critical slab
sections for shear defined in 11.11.1.2 and
11.11.4.7 shall be modified as follows:
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Openings in Slabs
ACI 11.11.6.1. For slabs without shearheads, that part
of the perimeter of the critical section that is
enclosed by straight lines projecting from the
centroid of the column, concentrated load, or
reaction area and tangent to the boundaries of the
openings shall be considered ineffective.
ACI 11.11.6.2. For slabs with shearheads, theineffective portion of the perimeter shall be one-half
of that defined in 11.11.6.1.
- 5 0 -
Openings in Slabs
Effect of openings in slabs on shear strength(Source: Portland Cement Association, Notes on ACI 318-08 Building Code
Requirements for Structural Concrete, Skokie, IL, 2008)
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Transfer of Moment in Slab-
Column Connections
ACI considers transfer of moment between a slab and
a column in 13.5.3 and 11.11.7.
Critical section for shear due to moment transfer is
d(effective depth)/2 from the face of the column,
which is the same critical section for direct two-way
shear.
- 5 2 -
Transfer of Moment in Slab-
Column Connections
The total shear stress due to direct shear and shear
caused by moment transfer is
where
Vu = direct shear at the critical section
Ac = area of critical section
J
cM
A
Vv uv
c
uu
+=
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Transfer of Moment in Slab-
Column Connections
v = factor used to determine the unbalanced momenttransferred by eccentricity of shear at slab-column
connections
Mu = unbalanced moment
c = distance from centroid of critical section to
face of section where stress is being computed
J = property of critical section analogous to polar
moment of inertia
- 5 4 -
Transfer of Moment in Slab-
Column Connections
Shear stress dist ribution due to moment-shear transfer at slab-columnconnections
(Source: Portland Cement Association, Notes on ACI 318-08 Building CodeRequirements for Structural Concrete, Skokie, IL, 2008)
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Transfer of Moment in Slab-
Column Connections
ACI 13.5.3
When gravity load, wind, earthquake, or other lateral
forces cause transfer of moment between slab and
column, a fraction of the unbalanced moment shall
be transferred by flexure in accordance with
13.5.3.2 through 13.5.3.4.
ACI 13.5.3.1. The fraction of unbalanced moment nottransferred by flexure shall be transferred by
eccentricity of shear in accordance with 11.11.7.
- 5 6 -
Transfer of Moment in Slab-
Column Connections
ACI 13.5.3.2.A fraction of the unbalanced moment
given by fMu shall be considered to be transferredby flexure within an effective slab width between
lines that are one and one-half slab or drop panel
thickness (1.5h) outside opposite faces of thecolumn or capital, where Mu is the factored moment
to be transferred and
( ) 21
321
1
bbf
+= ACI Eq. (13-1)
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Transfer of Moment in Slab-
Column Connections
(b) For unbalanced moments at interior support, and for
edge columns unbalanced moments about an axis
perpendicular to the edge, increase f to as much as1.25 times the value from Eq. (13-1), but not more than
f= 1.0, provided that Vu at the support does not exceed0.4Vc. The net tensile strain t calculated for theeffective slab width defined in 13.5.3.2 (c2+3h) shall not
be less than 0.010.
The value of Vc in item (a) and (b) shall be calculated in
accordance with 11.11.2.1 (Eqs. 11-31, 11-32, 11-33).
- 6 0 -
Transfer of Moment in Slab-
Column Connections
Commentary to 13.5.3.3
At exterior supports, for unbalanced moments about an
axis parallel to the edge, the portion of moment
transferred by eccentricity of shear vMu may bereduced provided that the factored shear at the support(excluding the shear produced by moment transfer)
does not exceed 75 percent of the shear strength Vc asdefined in 11.12.2.1 for edge columns or 50 percent for
corner columns. Tests indicate that there is no
significant interaction between shear and unbalanced
moment at the exterior support in such cases.
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Transfer of Moment in Slab-
Column Connections
Evaluation of tests of interior supports indicate that
some flexibility in distributing unbalanced moments
transferred by shear and flexure is possible, but with
more severe limitations than for exterior supports.
For interior supports, the unbalanced moment
transferred by flexure is permitted to be increased
up to 25 percent provided that the factored shear
(excluding the shear caused by the momenttransfer) at the interior supports does not exceed 40
percent of the shear strength Vc as defined in11.11.2.1.
- 6 2 -
Transfer of Moment in Slab-
Column Connections
ACI 13.5.3.4. Concentration of reinforcement over the
column by closer spacing or additional
reinforcement shall be used to resist moment on the
effective slab width defined in 13.5.3.2.
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Transfer of moment in slab-
column connections
Often column strip reinforcement is concentrated near
the column to accommodate this unbalanced
moment. Available test data (Hanson and Hanson
1968) seem to indicate that this practice does not
increase shear strength but may be desirable to
increase the stiffness of the slab-column junction.
Hanson, N.W., and Hanson J.M., Shear and MomentTransfer between Concrete Slabs and Columns,
Journal, PCA Research and Development
Labortories, V.10, No.1, Jan. 1968, pp. 2-16.
- 6 4 -
Transfer of moment in slab-
column connections
ACI 11.11.7 Transfer of moment in slab-column
connections
ACI 11.11.7.1. Where gravity load, wind, earthquake,or other lateral forces cause transfer of unbalanced
moment Mu between a slab and column, fMu shallbe transferred by flexure in accordance with 13.5.3.
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Transfer of moment in slab-
column connections
The remainder of the unbalanced moment, vMu, shallbe considered to be transferred by eccentricity of
shear about the centroid of the critical section
defined in 11.11.1.2 where
v = (1 f) ACI Eq. (11-37)
- 6 6 -
Transfer of moment in slab-
column connections
Graphical solut ion of ACI Eqs. (13-1) and (11-37)(Source: Portland Cement Association, Notes on ACI 318-08 Building Code
Requirements for Structural Concrete, Skokie, IL, 2008)
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Transfer of moment in slab-
column connections
The shear stress due to factored shear force and
moment shall not exceed at the critical
section located d/2 outside the outermost line of
stirrup legs that surround the column.
)f.( 'c 170
- 7 0 -
Transfer of moment in slab-
column connections
ACI 318M-08 Fig. R11.11.7.2: Assumed distribution of shear stress
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- 7 3 -
Transfer of moment in slab-
column connections
Ac = area of concrete of assumed critical section
= 2d (c1 + c2 + 2d)
Jc = property of assumed critical section analogous to
polar moment of inertia
=
Similar equations may be developed for Ac and Jc, for
columns located at the edge or corner of a slab.
( ) ( ) ( )( )2
dcdcd
6
ddc
6
dcd2
123
1
3
1 ++++
++
- 7 4 -
Transfer of moment in slab-
column connections
Section properties for shear stress computations(Source: Portland Cement Association, Notes on ACI 318-08 Building Code
Requirements for Structural Concrete, Skokie, IL, 2008)
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- 7 5 -
Transfer of moment in slab-
column connections
Section properties for shear stress computations(Source: Portland Cement Association, Notes on ACI 318-08 Building Code
Requirements for Structural Concrete, Skokie, IL, 2008)
- 7 6 -
Transfer of moment in slab-
column connections
Section properties for shear stress computations(Source: Portland Cement Association, Notes on ACI 318-08 Building Code
Requirements for Structural Concrete, Skokie, IL, 2008)
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Transfer of moment in slab-
column connections
Section properties for shear stress computations(Source: Portland Cement Association, Notes on ACI 318-08 Building Code
Requirements for Structural Concrete, Skokie, IL, 2008)
- 7 8 -
Factored Shear in Slab Systems
with Beams
ACI 13.6.8 Factored shear in slab systems wi th
beams
ACI 13.6.8.1. Beams with f1l2/l1 equal to or greaterthan 1.0 shall be proportioned to resist shear
caused by factored loads on tributary areas which
are bounded by 45-degree lines drawn from the
corners of the panels and the centerlines of the
adjacent panels parallel to the long sides.
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Factored Shear in Slab Systems
with Beams
f1 = [(EcbIb)/(EcsIs)] in direction l1l1 = length of span in direction that moments are
being determined
l2 = length of span in direction perpendicular to l1.
EcbIb = Flexural stiffness of beam
Ecs
Is
= Flexural stiffness of slab
- 8 0 -
Factored Shear in Slab Systems
with Beams
ACI 318M-08 Fig. R13.6.8: Tributary area for shear on an interior beam
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Factored Shear in Slab Systems
with Beams
ACI 13.6.8.2. In proportioning beams with f1l2/l1 lessthan 1.0 to resist shear, linear interpolation,
assuming beams carry no load at f1 = 0, shall bepermitted.
ACI 13.6.8.3. In addition to shears calculated
according to 13.6.8.1 and 13.6.8.2, beams shall be
proportioned to resist shears caused by factoredloads applied directly on beams.
- 8 2 -
Factored Shear in Slab Systems
with Beams
ACI 13.6.8.4. Computation of slab shear strength on
the assumption that load is distributed to supporting
beams in accordance with 13.6.8.1 or 13.6.8.2 shall
be permitted. Resistance to total shear occurring on
a panel shall be provided.
ACI 13.6.8.5. Shear strength shall satisfy the
requirements of Chapter 11.
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- 8 3 -
BREAK!
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- 8 4 -
Question
and
Answer
Session
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Example 1: Two-Way Slab Shear
Design
Check shear for the shaded design strip as shownbelow
5.5 m
5.5 m
5.5 m
4.3 m 4.3 m 4.3 m
Design Direction
- 8 6 -
Example 1: Two-Way Slab Shear
Design
Structural system
Two-way slab systems without beams (flat plate
slab system); no edge beams
Material properties
Concrete: fc = 40 MPa wc = 2400 kg/m3
Reinforcement: fy = 420 MPa
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Example 1: Two-Way Slab Shear
Design
Member dimension
Column: 400 mm 400 mm
Loading condition
Live load = 2.0 kN/m2
Partition load = 1.0 kN/m2
- 8 8 -
Example 1: Two-Way Slab Shear
Design
Determine slab thickness
From ACI Table 9-5 (c)
h = ln / 30
= (5500 400) / 30 = 170 mm
Use 200 mm
According to ACI 9.5.3.2, allowable minimum
thickness is 125 mm < 200 mm O.K.
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Example 1: Two-Way Slab Shear
Design
Effective depth d = 200 32 (20 mm cover and
diameter of one No.12 bar = 20 +12 = 32 mm) =
168 mm
- 9 0 -
Example 1: Two-Way Slab Shear
Design
Determine factored load quDead load (qD):
qD = 0.2 2400 9.8/1000 = 4.7 kN/m2
Live load (qL):
Reducible live load (2009 IBC Section 1607.9.2)
R = 0.861(A 13.94)
= 0.861(23.65 13.94) = 8.4 %
where A = 4.3 5.5 = 23.65 m2
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Example 1: Two-Way Slab Shear
Design
Reducible live load = (1-0.084)2.0 = 1.83 kN/m2
Non-reducible live load (Partition) = 1.0 kN/m2
qL = 1.83 + 1.0 = 2.83 kN/m2
qu = 1.4qD = 1.44.7 = 6.58 kN/m2
= 1.2qD+1.6qL = 1.24.7+1.62.83= 10.2 kN/m2 (Governs)
- 9 2 -
Example 1: Two-Way Slab Shear
Design
Check one-way shear (interior column)
Panel centerline
4.3 m
5.5 m
Critical section
for one-way shear
0.4 m
0.4 m
Slab effective depth
d= 0.168 m
Design
Direction
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Example 1: Two-Way Slab Shear
Design
Factored shear at critical section:
Vu = 10.2(5.5/2 0.4/2 0.168)4.3
= 104 kN
Vc= 0.750.17(fc)0.5bwd
= 0.750.171.0(40)0.543001681/1000
= 583 kN > Vu (= 104 kN) O.K.
- 9 4 -
Example 1: Two-Way Slab Shear
Design
Check one-way shear (edge column)
Panel
centerline
4.3 m
5.5/2 = 2.75 mCritical section
for one-way shear0.4 m
0.4 m
Slab effective depth
d= 0.168 m
Design Direction
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Example 1: Two-Way Slab Shear
Design
Factored shear at critical section:
Vu = 10.2(2.75 0.4/2 0.168)4.3
= 104 kN
Vc= 0.750.17(fc)0.5bwd
= 0.750.171.0(40)0.543001681/1000= 583 kN > Vu (= 104 kN) O.K.
- 9 6 -
Example 1: Two-Way Slab Shear
Design
Check two-way shear (Interior column)
Panel centerline
4.3 m
5.5 m
Critical section
for two-way shear
0.4 m
0.4 m
d/2= 0.084 md/2= 0.084 mDesign
Direction
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Example 1: Two-Way Slab Shear
Design
Side dimension of square critical section = 400 + 168
= 568 mm
Factored shear at critical section:
Vu = 10.2(5.54.3 0.5680.568)
= 238 kN
- 9 8 -
Example 1: Two-Way Slab Shear
Design
Shear strength Vc for two-way slab (ACI 11.11.2.1)
= 0.4 / 0.4 = 1.0
s = 40 (Interior)
bo = 4568 = 2272 mm
dbf2
117.0V o'
cc
+=
dbf2b
d
083.0V o'
co
s
c
+
=
dbf33.0V o'
cc =
ACI Eq. (11-31)
ACI Eq. (11-32)
ACI Eq. (11-33)
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Example 1: Two-Way Slab Shear
Design
Vc = 0.75 797 = 598 kN > Vu (= 238 kN) O.K.
kN12311000/1682272401
2117.0
dbf2
117.0V o'
cc
=
+=
+=
kN9931000/16822724022272
16840083.0
dbf2b
d083.0V o
'
c
o
sc
=
+
=
+
=
kN7971000/16822724033.0dbf33.0V o'cc === (Governs)
- 100 -
Example 1: Two-Way Slab Shear
Design
Check two-way shear (edge column)
Panel
centerline
4.3 m
5.5/2 = 2.75 mCritical section
for two-way shear0.4 m
0.4 m
d/2= 0.084 m
Design Direction
d/2= 0.084 m
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Example 1: Two-Way Slab Shear
Design
Dimension of critical section, b1 = 400 + 168 = 568 mm
Dimension of critical section, b2 = 400 + 168/2 = 484 mm
Factored shear at critical section:
Vu = 10.2[(5.5/2+0.2)4.3 0.5680.484]
= 127 kN
- 102 -
Example 1: Two-Way Slab Shear
Design
Shear strength Vc for two-way slab (ACI 11.11.2.1)
= 0.4 / 0.4 = 1.0
s = 30 (Edge column)
bo = 568 + 484 + 484 = 1536 mm
dbf2
117.0V o'
cc
+=
dbf2b
d
083.0V o'
co
s
c
+
=
dbf33.0V o'
cc =
ACI Eq. (11-31)
ACI Eq. (11-32)
ACI Eq. (11-33)
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Example 1: Two-Way Slab Shear
Design
Vc = 0.75 539 = 404 kN > Vu (= 127 kN) O.K.
Slab thickness is determined as 200 mm.
kN8321000/1681536401
2117.0
dbf2
117.0V o'
cc
=
+=
+=
kN7151000/16815364021536
16830083.0
dbf2b
d083.0V o
'
c
o
sc
=
+
=
+
=
kN5391000/16815364033.0dbf33.0V o'
cc === (Governs)
- 104 -
Example 1: Two-Way Slab Shear
Design
Direct Design Method can be used for this example
(13.6.1). Positive and negative moments at column
and middle strips were determined using Direct
Design Method.
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Example 1: Two-Way Slab Shear
Design
Total factored moment (Mo)
where
qu = factored load (10.2 kN/m2)l2 = length perpendicular to design direction (4.3m)
ln = clear span length in design direction (5.1m)
mkN6.1428
1.53.42.10
8
qM
2
2
n2uo
=
=
= ll
- 106 -
Example 1: Two-Way Slab Shear
Design
Table: Moments and provided reinforcementLocation Mu (kN-m) Width
(mm)
Required
As (mm2)
Provided
reinforcement
End Span Column
strip
Ext. neg. 0.26Mo = 37.1 2150 583 9 No.12
Positive 0.31Mo = 44.2 2150 696 7 No.12
Int. neg. 0.53Mo =75.6 2150 1201 11 No.12
Middle strip Ext. neg. 0.0Mo = 0 2150 0 7 No.12
Positive 0.21Mo = 30.0 2150 471 7 No.12
Int. neg. 0.17Mo = 44.2 2150 696 7 No.12
Interior
Span
Column
strip
Positive 0.21Mo = 30.0 2150 471 7 No.12
Negative 0.49Mo = 69.9 2150 1109 11 No.12
Middle strip Positive 0.14Mo = 20.0 2150 313 7 No.12
Negative 0.16Mo = 22.8 2150 357 7 No.12
Minimum As = 0.0018bh = 0.00182150200 = 774 mm2, Maximum spacing = 2h = 400 mm, Use 7-No.12 (A s = 791 mm
2) to
satisfy minimum As and maximum spacing
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Example 1: Two-Way Slab Shear
Design
Shear Design
Total shear stress is the sum of the direct shear stress
plus the shear stress due to the fraction of
unbalanced moment transferred by eccentricity of
shear. Assume shear stress due to moment transfer
by eccentricity of shear varies linearly about thecentroid of the section.
- 108 -
Example 1: Two-Way Slab Shear
Design
Case 1: Shear check at end support
Panel
centerline
4.3 m
5.5/2 = 2.75 mCritical section
for two-way shear0.4 m
0.4 m
d/2= 0.084 m
Design Direction
d/2= 0.084 m
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Example 1: Two-Way Slab Shear
Design
Direct shear Vu can be computed as below
A = area tributary to column
= (5.5/2+0.2)4.3 = 12.69 m2
b1 = dimension of critical section in direction
of analysis = 0.4 + 0.168/2 = 0.484 m
( )n
2121uu
MMbbAqV
l
=
- 110 -
Example 1: Two-Way Slab Shear
Design
b2 = dimension of critical section perpendicular to
direction of analysis
= 0.4 + 0.168 = 0.568 m
M1 = total negative design strip moment at interiorsupport determined from Direct Design Method
= 0.70Mo = 99.8 kN-m
M2 = total negative design strip moment at exterior
support determined from Direct Design Method
= 0.26Mo = 37.1 kN-m
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Example 1: Two-Way Slab Shear
Design
0.4Vc (= 0.4404 = 162 kN) > Vu (= 114.3 kN)
Use 5-No.12 bars over effective width equal to c2 + 3h = 400
+ 3200 = 1000 mm. From section analysis, a (the depthof rectangular stress block) is 7.0 mm.
c (distance from top to neutral axis) = a/1= 7/0.76 = 9.2 mm
t (net tensile strain) = 0.003(d-c)/c= 0.003(168-9.2)/9.2 = 0.052 > 0.01
Thus, f= 1.250.62 = 0.78; v = 1-f= 1 0.78 = 0.22
Mu = 0.3Mo = 42.8 kN-m (ACI 13.6.3.6)
- 114 -
Example 1: Two-Way Slab Shear
Design
Therefore,
( ) ( )
( ) ( )
3
32
1
213
2121
m0464.0
484.06
568.0484.02168.0568.02484.0168.0484.02
b6
bb2db2bdb2
c
J
=
+++
=
+++=
2
uv
c
uu
m/kN6430464.0
8.4222.0
26.0
3.114
J
cM
A
Vv
=
+=
+=
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Example 1: Two-Way Slab Shear
Design
Shear strength
vc = 0.750.33(40)0.5 = 1.56 N/mm2 = 1560 kN/m2
> Vu (= 643 kN/m2) O.K.
- 116 -
Example 1: Two-Way Slab Shear
Design
Case 2: Shear check at interior support
Panel centerline
4.3 m
5.5 m
Critical section
for two-way shear
0.4 m
0.4 m
d/2= 0.084 md/2= 0.084 mDesign
Direction
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Example 1: Two-Way Slab Shear
Design
where
A = 5.54.3 = 23.65 m2
b1 = b2 = 0.4 + 0.168 = 0.568 m
M1 = 0.70Mo = 99.8 kN-m
M2 = 0.26Mo = 37.1 kN-m
( )
[ ] kN3.2503.120.2381.5
1.378.99568.065.232.10
MMbbAqV
2
n
2121uu
=+=
+=
+=
l
- 118 -
Example 1: Two-Way Slab Shear
Design
Total shear stress is computed.
Ac = bod = 2(0.568+0.568)0.168
= 0.38 m2
( )
( ) 33
3
211
m0738.03
168.0568.03568.0168.0568.0
3
db3bdb
c
J
=++
=
++=
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Example 1: Two-Way Slab Shear
Design
The difference between the slab moments acting on
opposite faces of the interior support needs to be
transferred by shear to the first interior column.
Exterior moment at the face of the support is 0.70Mo =
0.70142.6 = 99.8 kN-m, and the interior moment atthe face of the support is 0.65Mo = 0.65142.6 =92.7 kN-m. Therefore, the unbalanced moment M
u
=
99.8 92.7 = 7.1 kN-m
- 120 -
Example 1: Two-Way Slab Shear
Design
0.4Vc
(= 0.4598 = 239 kN) < Vu
(= 250.3 kN)
Thus, there is no increase in f(= 0.6).
v = 1 0.6 = 0.4
( )60.0
568.0568.0321
1f =
+=
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Example 1: Two-Way Slab Shear
Design
vc = 0.750.33(40)0.5 = 1.56 N/mm2 = 1560 kN/m2
> Vu (= 697 kN/m
2
) O.K.
2
uv
c
uu
m/kN6970738.0
1.740.0
38.0
3.250
J
cM
A
Vv
=
+=
+=
- 122 -
Question
and
Answer
Session
If you are encountering technical difficulties, please call 00 1 847 991 2700
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