two-way Lecture24DesignofTwo
WayFloorSlabSystem
August4,2003
CVEN444
LectureGoals
ShearStrengthofSlabs
ShearExample
DirectDesignMethod
ShearStrengthofSlabs
In floor systems, the slab must have adequate
thickness to resist both bending moments and shear
forces at critical section. There are three cases to look at
for shear.
1. Two-way Slabs supported on beams
2. Two-Way Slabs without beams
3. Shear Reinforcement in two-way slabs
without beams.
ShearStrengthofSlabs
Two-way slabs supported on beams
The critical location is found at d distance from the
column, where
(
Vc = 2 f c bd
)
The supporting beams are stiff and are capable of
transmitting floor loads to the columns.
ShearStrengthofSlabs
The shear force is calculated using the triangular and
trapezoidal areas. If no shear reinforcement is provided,
the shear force at a distance d from the beam must equal
(
Vud Vc 2 f c bd
)
where,
l2
Vud = wu d
2
ShearStrengthofSlabs
Two-Way Slabs without beams
There are two types of shear that need to be addressed
1. One-way shear or beam shear at distance d
from the column
2. Two-way or punch out shear which occurs
along a truncated cone.
ShearStrengthofSlabs
1. One-way shear or beam shear at distance d from
the column
2. Two-way or punch out shear which occurs along a
truncated cone.
ShearStrengthofSlabs
One-way shear considers critical section a distance d
from the column and the slab is considered as a wide
beam spanning between supports.
(
Vud Vc = 2 f c bd
)
ShearStrengthofSlabs
Two-way shear fails along a a truncated cone or pyramid
around the column. The critical section is located d/2 from
the column face, column capital, or drop panel.
ShearStrengthofSlabs
If shear reinforcement is not provided, the shear strength
of concrete is the smaller of:
2 + 4 f b d 4 f b d
Vc =
co
co
c
(
)
bo = perimeter of the critical section
c = ratio of long side of column to short side
ShearStrengthofSlabs
If shear reinforcement is not provided, the shear
strength of concrete is the smaller of:
d
s + 2 f b d
Vc =
co
bo
s is 40 for interior columns, 30 for edge
columns, and 20 for corner columns.
ShearStrengthofSlabs
Shear Reinforcement in two-way slabs without beams.
For plates and flat slabs, which do not meet the condition
for shear, one can either
- Increase slab thickness
- Add reinforcement
Reinforcement can be done by shearheads, anchor bars,
conventional stirrup cages and studded steel strips.
ShearStrengthofSlabs
Shearhead consists of steel I-beams or channel welded
into four cross arms to be placed in slab above
a column. Does not apply to external columns
due to lateral loads and torsion.
ShearStrengthofSlabs
Anchor bars consists of steel reinforcement rods or
bent bar reinforcement
ShearStrengthofSlabs
Conventional stirrup cages
ShearStrengthofSlabs
Studded steel strips
ShearStrengthofSlabs
The reinforced slab follows section 11.12.4 in the
ACI Code, where Vn can not
Vn = Vc + Vs 6 f c bo d
Vc = 4 f c bo d
Vs =
The spacing, s, can not exceed d/2.
If a shearhead reinforcement is provided
Vn 7 f c bo d
Av f y d
s
ExampleProblem
Determine the shear
reinforcement required for an
interior flat panel considering
the following: Vu= 195k, slab
thickness = 9 in., d = 7.5 in.,
fc = 3 ksi, fy= 60 ksi, and
column is 20 x 20 in.
ExampleProblem
Compute the shear terms find b0 for
Vc = 4 f c b0 d
column
b0 = 4
+ d = 4 ( 20 in. + 7.5 in.)
width
= 110 in.
ExampleProblem
Compute the maximum allowable shear
Vc = 4 f c b0 d
= 0.75 ( 4 )
1k
3000 ( 110 in.) ( 7.5 in.)
1000 lbs
= 135.6 k
Vu =195 k > 135.6 k Shear reinforcement is need!
ExampleProblem
Compute the maximum allowable shear
Vc = 6 f c b0 d
= 0.75 ( 6 )
1k
3000 ( 110 in.) ( 7.5 in.)
1000 lbs
= 203.3 k
So Vn >Vu Can use shear reinforcement
ExampleProblem
Use a shear head or studs as
in inexpensive spacing.
Determine the a for
Vc = 2 f c b0 d
column
b0 = 4
+ 2a
width
ExampleProblem
Determine the a for
Vu = 2 f c b0 d
((
19500 lb = 0.75 ( 2 ) 3000 4 20 in. + 2a
) ) ( 7.5 in.)
a = 41.8 in.
The depth = a+d
= 41.8 in. +7.5 in. = 49.3 in. 50 in.
ExampleProblem
Determine shear reinforcement
Vs = Vu Vc
= 195 k 135.6 k
= 59.4 k
The Vs per side is Vs / 4 = 14.85 k
ExampleProblem
Determine shear reinforcement
14.85 k
Vs =
= 19.8 k
0.75
Use a #3 stirrup Av = 2(0.11 in2) = 0.22 in2
Vs =
Av f y d
s
s=
Av f y d
Vs
ExampleProblem
Determine shear reinforcement spacing
s=
Av f y d
Vs
=
0.22 in 2 ) ( 60 ksi ) ( 7.5 in.)
(
19.8 k
= 5.0 in.
Maximum allowable spacing is
d 7.5 in.
=
= 3.75 in.
2
2
ExampleProblem
Use s = 3.5 in.
50 in.
# of stirrups =
= 14.3 Use 15 stirrups
3.5 in.
The total distance is 15(3.5 in.)= 52.5 in.
ExampleProblem
The final result:
15 stirrups at total distance of
52.5 in. So that a = 45 in. and
c = 20 in.
DirectDesignMethodforTwoway
Slab
Method of dividing total static moment Mo into
positive and negative moments.
Limitations on use of Direct Design method
1.Minimumof3continuousspansineachdirection.
(3x3panel)
2.Rectangularpanelswithlongspan/shortspan2
DirectDesignMethodforTwoway
Slab
Limitations on use of Direct Design method
3. Successive span in each direction shall not differ by
more than 1/3 the longer span.
4. Columns may be offset from
the basic rectangular grid of
the building by up to 0.1
times the span parallel to the
offset.
DirectDesignMethodforTwoway
Slab
Limitations on use of Direct Design method
5. All loads must be due to gravity only (N/A to
unbraced laterally loaded frames, from mats or
pre-stressed slabs)
6. Service (unfactored) live load 2 service dead
load
DirectDesignMethodforTwoway
Slab
Limitations on use of Direct Design method
7. Forpanelswithbeamsbetweensupportsonall
sides,relativestiffnessofthebeamsinthe2
perpendiculardirections.
l
2
12
l
2
21
Shallnotbelessthan0.2norgreaterthan5.0
DefinitionofBeamtoSlabStiffness
Ratio,
Accounts for stiffness effect of beams located along
slab edge
reduces deflections of panel
adjacent to beams.
=
flexural stiffness of beam
flexural stiffness of slab
DefinitionofBeamtoSlabStiffness
Ratio,
=
4E cb I b / l
=
4E cs I s / l
4E cb I b
4E cs I s
E cb = Modulus of elasticity of beam concrete
E sb = Modulus of elasticity of slab concrete
I b = Moment of inertia of uncracked beam
I s = Moment of inertia of uncracked slab
With width bounded laterally by centerline of adjacent
panels on each side of the beam.
TwoWaySlabDesign
Static Equilibrium of Two-Way Slabs
Analogy of two-way slab to plank and beam floor
Section A-A:
Moment per ft width in planks M =
Total Moment
M f = ( wl2 )
l12
8
wl12
k - ft 8
k - ft/ft
TwoWaySlabDesign
Static Equilibrium of Two-Way Slabs
Analogy of two-way slab to and plank beam floor
wl1
k/ft
Uniform load on each beam
2
2
wl1 l2
k - ft
M lb =
Moment in one beam (Sec: B-B)
2 8
TwoWaySlabDesign
Static Equilibrium of Two-Way Slabs
Total Moment in both beams
M = ( wl1 )
2
l2
k - ft
8
Full load was transferred east-west by the planks and then was
transferred north-south by the beams;
The same is true for a two-way slab or any other floor system.
BasicStepsinTwowaySlab
Design
1. Choose layout and type of slab.
2. Choose slab thickness to control deflection. Also,
check if thickness is adequate for shear.
3. Choose Design method
Equivalent Frame Method- use elastic frame
analysis to compute positive and negative
moments
Direct Design Method - uses coefficients to
compute positive and negative slab moments
BasicStepsinTwowaySlab
Design
4. Calculate positive and negative moments in the slab.
5. Determine distribution of moments across the width of
the slab. - Based on geometry and beam stiffness.
6. Assign a portion of moment to beams, if present.
7. Design reinforcement for moments from steps 5 and 6.
8. Check shear strengths at the columns
MinimumSlabThicknessfortwo
wayconstruction
Maximum Spacing of Reinforcement
s 2t ( ACI 13.3.2 )
At points of max. +/- M:
and s 18 in. ( ACI 7.12.3)
Min Reinforcement Requirements
As( min ) = As( T&S) from ACI 7.12
( ACI
13.3.1)
DistributionofMoments
Slab is considered to be a series of frames in two
directions:
DistributionofMoments
Slab is considered to be a series of frames in two
directions:
DistributionofMoments
Total static Moment, Mo
M0 =
2
wu l2ln
( ACI 13 - 3)
8
where wu = factored load per unit area
l2 = transverse width of the strip
ln = clear span between columns
( for circular columns, calc. ln using h = 0.886d c )
ColumnStripsandMiddle
Strips
Moments vary across width of slab panel
Design moments are averaged over the
width of column strips over the
columns & middle strips between
column strips.
ColumnStripsandMiddle
Strips
Column strips Design w/
width on either side of a
column centerline equal to
smaller of
0.25 l2
0.25 l1
l1= length of span in
direction moments are
being determined.
l2= length of span
transverse to l1
ColumnStripsandMiddle
Strips
Middle strips: Design
strip bounded by two
column strips.
PositiveandNegativeMomentsin
Panels
M0 is divided into + M and -M Rules given in ACI
sec. 13.6.3
MomentDistribution
PositiveandNegativeMomentsin
Panels
M0 is divided into + M and -M Rules given in ACI
sec. 13.6.3
M u + M u ( avg ) M 0 =
2
wu l2ln
8
LongitudinalDistributionof
MomentsinSlabs
For a typical interior panel, the total static moment is
divided into positive moment 0.35 Mo and negative
moment of 0.65 Mo.
For an exterior panel, the total static moment is
dependent on the type of reinforcement at the outside
edge.
DistributionofM0
MomentDistribution
The factored components
of the moment for the
beam.
TransverseDistributionof
Moments
The longitudinal moment values mentioned are for the
entire width of the equivalent building frame. The
width of two half column strips and two half-middle
stripes of adjacent panels.
TransverseDistributionof
Moments
Transverse distribution
of the longitudinal
moments to middle and
column strips is a
function of the ratio of
length l2/l1,1, and t.
TransverseDistributionof
Moments
Transverse distribution of the longitudinal moments to
middle and column strips is a function of the ratio of
length l2/l1,1, and t.
1 =
Ecb I b
Ecs I s
t =
EcbC
2 Ecs I s
0.63 x x 3 y
1
C=
torsional constant
y 3
DistributionofM0
ACI Sec 13.6.3.4
For spans framing into a common support negative
moment sections shall be designed to resist the larger
of the 2 interior Mus
ACI Sec. 13.6.3.5
Edge beams or edges of slab shall be proportioned to
resist in torsion their share of exterior negative
factored moments
FactoredMomentinColumn
Strip
1 = Ratio of flexural stiffness of beam to stiffness of
slab in direction l1.
t= Ratio of torsional stiffness of edge beam to
flexural stiffness of slab(width= to beam length)
FactoredMomentinan
InteriorStrip
FactoredMomentinan
ExteriorPanel
FactoredMomentinan
ExteriorPanel
FactoredMomentinColumn
Strip
1 = Ratio of flexural stiffness of beam to stiffness of
slab in direction l1.
t= Ratio of torsional stiffness of edge beam to
flexural stiffness of slab(width= to beam length)
FactoredMomentinColumn
Strip
1 = Ratio of flexural stiffness of beam to stiffness of
slab in direction l1.
t= Ratio of torsional stiffness of edge beam to
flexural stiffness of slab(width= to beam length)
FactoredMomentinColumn
Strip
1 = Ratio of flexural stiffness of beam to stiffness of
slab in direction l1.
t= Ratio of torsional stiffness of edge beam to
flexural stiffness of slab(width= to beam length)
FactoredMoments
Factored Moments in beams (ACI Sec. 13.6.3)
Resist a percentage of column strip moment plus
moments due to loads applied directly to beams.
FactoredMoments
Factored Moments in Middle strips (ACI Sec. 13.6.3)
The portion of the + Mu and - Mu not resisted
by column strips shall be proportionately
assigned to corresponding half middle strips.
Each middle strip shall be proportioned to
resist the sum of the moments assigned to its 2
half middle strips.
ACIProvisionsforEffectsof
PatternLoads
The maximum and minimum bending moments at
the critical sections are obtained by placing the live
load in specific patterns to produce the extreme
values. Placing the live load on all spans will not
produce either the maximum positive or negative
bending moments.
ACIProvisionsforEffectsof
PatternLoads
1. The ratio of live to dead load. A high ratio will
increase the effect of pattern loadings.
2. The ratio of column to beam stiffness. A low ratio
will increase the effect of pattern loadings.
3. Pattern loadings. Maximum positive moments
within the spans are less affected by pattern loadings.
ReinforcementDetailsLoads
After all percentages of the static moments in the
column and middle strip are determined, the steel
reinforcement can be calculated for negative and
positive moments in each strip.
a
d = Ru bd 2
M u = As f y
2
ReinforcementDetailsLoads
Calculate Ru and determine the steel ratio , where
=0.9. As = bd. Calculate the minimum As from
ACI codes. Figure 13.3.8 is used to determine the
minimum development length of the bars.
Ru = wu f c (1 0.59 wu )
wu =
fy
fc
Minimumextensionfor
reinforcementinslabswithout
beams(Fig.13.3.8)
MomentDistribution
The factored components
of the moment for the
beam.
TransverseDistributionof
Moments
The longitudinal moment values mentioned are for the
entire width of the equivalent building frame. The
width of two half column strips and two half-middle
stripes of adjacent panels.