lecture 10 analysis and design of waffle slabs_2011
TRANSCRIPT
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Lecture-10
Analysis and Design of
Two-way Slab Systems
(Two Way Joist Slabs & Two-way Slab
with Beams)
By: Prof Dr. Qaisar Ali
Civil Engineering Department
UET Peshawar
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Topics Addressed
Two-Way Joist Slab
Introduction
Behavior
Characteristics
Basic Steps for Structural Design
Example
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Topics Addressed
Moment Coefficient Method for Two Way Slab with
Beams
Introduction
Cases
Moment Coefficient Tables
Reinforcement Requirements
Steps
Example
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011 4
Two-Way Joist Slab
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011 5
Introduction
A two-way joist system, or waffle slab, comprises evenly
spaced concrete joists spanning in both directions and a
reinforced concrete slab cast integrally with the joists.
Two-Way Joist
Joist
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011 6
Introduction
Like one-way joist system, a two way system will be called
as two-way joist system if clear spacing between ribs (dome
width) does not exceed 30 inches.
Two-Way Joist
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011 7
Introduction
Two-Way Joist
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Introduction
The joists are commonly formed by using standard square
dome forms and the domes are omitted around the columns
to form the solid heads.
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Two-Way Joist
Solid Head
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two-Way Joist
Introduction
Standard Dome Data
Generally the dome for waffle slab can be of any size. However the
commonly used standard domes are discussed as follows:
30-inch 30-inch square domes with 3-inch flanges; from which 6-
inch wide joist ribs at 36-inch centers are formed: these are
available in standard depths of 8, 10, 12, 14, 16 and 20 inches.
19-inch 19-inch square domes with 2 -inch flanges, from which
5-inch wide joist ribs at 24-inch centers are formed. These are
available in standard depths of 8, 10, 12, 14 and 16 inches.
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two-Way Joist
Introduction
Standard Dome Data
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Behavior
The behavior of two-way joist slab is similar to a two way flat
Slab system.
Two-Way Joist
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Characteristics
Dome voids reduce dead load.
Attractive ceiling (waffle like appearance).
Electrical fixtures can be placed in the voids.
Particularly advantageous where the use of longer spans
and/or heavier loads are desired without the use of
deepened drop panels or supported beams.
Two-Way Joist
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Basic Steps for Structural Design
Step No. 01 (Sizes): Sizes of all structural and non
structural elements are decided.
Step No. 02 (Loads): Loads on structure are determined
based on occupational characteristics and functionality.
Step No. 03 (Analysis): Effect of loads are calculated on all
structural elements.
Step No. 04 (Design): Structural elements are designed for
the respective load effects following code provisions.
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Two-Way Joist
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Sizes
Minimum Joist Depth
For Joist depth determination, waffle slabs are considered as flat slab
(ACI 13.1.3, 13.1.4 & 9.5.3).
The thickness of equivalent flat slab is taken from table 9.5 (c).
The thickness of slab and depth of rib of waffle slab can be then
computed by equalizing the moment of inertia of equivalent flat slab to
that of waffle slab.
However since this practice is time consuming, tables have been
developed to determine the size of waffle slab from equivalent flat slab
thickness.
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Two-Way Joist
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Sizes
Minimum Joist Depth
Equivalent Flat Slab Thickness
ACI 318-05 Sect. 9.5.3
Minimum thickness = ln/33
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Two-Way Joist
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011 16
Sizes
Minimum Joist Depth
Slab and rib depth from equivalent flat slab thickness
Two-Way Joist
Table 01: Waffle flat slabs (19" 19" voids at 2'-0")-Equivalent thickness
Rib + Slab Depths (in.) Equivalent Thickness te (in.)
8 + 3 8.89
8 + 4 10.11
10 + 3 10.51
10 + 4 11.75
12 + 3 12.12
12 + 4 13.38
14 + 3 13.72
14 + 4 15.02
16 + 3 15.31
16 + 4 16.64
Reference: Table 11-2 of CRSI Design Handbook 2002.
Note: Only first two columns of the table are reproduced here.
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011 17
Sizes
Minimum Joist Depth
Slab and rib depth from equivalent flat slab thickness
Two-Way Joist
Table 02: Waffle flat slabs (30" 30" voids at 3'-0")-Equivalent thickness
Rib + Slab Depths (in.) Equivalent Thickness te (in.)
8 + 3 8.61
8 + 4 9.79
10 + 3 10.18
10 + 4 11.37
12 + 3 11.74
12 + 4 12.95
14 + 3 13.3
14 + 4 14.54
16 + 3 14.85
16 + 4 16.12
20 + 3 17.92
20 + 4 19.26
Reference: Table 11-2 of CRSI Design Handbook 2002.
Note: Only first two columns of the table are reproduced here.
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011 18
Sizes
Minimum Width of Rib
ACI 8.11.2 states that ribs shall be not less than 4 inches in width.
Maximum Depth of Rib
ACI 8.11.2 also states that ribs shall have a depth of not more than 3
times the minimum width of rib.
Minimum Slab Thickness
ACI 8.11.6.1 states that slab thickness shall be not less than one-
twelfth the clear distance between ribs, nor less than 2 inch.
Two-Way Joist
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011 19
Sizes
Solid Head
Dimension of solid head on either side of column centerline is equal to
l/6.
The depth of the solid head is equal to the depth of the combined
depth of ribs and top slab.
Two-Way Joist
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011 20
Loads
Floor dead load for two-way joist with certain dome size, dome depth can
be calculated from the table shown for two options of slab thicknesses (3
inches and 4 inches).
Two-Way Joist
Table 03: Standard Dome Dimensions and other Data
Dome Size Dome Depth
(inches)
Volume of Void
(ft3)
Floor Dead Load (psf) per slab
thickness
3 inches 4 inches
30 inches
8 3.98 71 90
10 4.92 80 99
12 5.84 90 109
14 6.74 100 119
16 7.61 111 129
20 9.3 132 151
19 inches
8 1.56 79 98
10 1.91 91 110
12 2.25 103 122
14 2.58 116 134
16 2.9 129 148 Reference: Table 11-1, CRSI Design Handbook 2002
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011 21
Loads
Floor dead load (wdj) for two-way joist can also be
calculated as follows:
Two-Way Joist
36
8
3
30
Volume of solid:
Vsolid = (36 36 11)/1728 = 8.24 ft3
Volume of void:
Vvoid = (30 30 8)/1728 = 4.166 ft3
Total Load of joists per dome:
wdj = (Vsolid Vvoid) conc = ( 8.24 4.166) 0.15 = 0.61 kips/ dome
Total Load of joists per sq. ft:
wdj/ (dome area) = 0.61/ (3 3) = 0.0679 ksf
= 68 psf 71 psf (from table 03)
The difference is because sloped ribs are not considered. Plan
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011 22
Loads
Taking a panel out of the system:
Two-Way Joist
l2
l1 wdj
wdj + wsh
l1
a a
b l2
wdj = Dead load at joist (load/ area)
wsh = Dead load of solid head (load/ area)
wdj + wsh
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011 23
Loads
If the complete area l1 l2 is assumed to occupy joists
alone, then the dead load in the area l1 l2 will be wdj.
However since there are solid head regions present,
therefore additional dead load due to solid head region shall
be:
wdsh = hsolid conc - wdj
Two-Way Joist
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011 24
Loads
Factored loads can be calculated as:
If wL = live load (load/area)
wdj = dead load at joists, then
Factored load due to joists (wj)
wuj = 1.2 wdj + 1.6wL
Factored load due to solid head (wsh)
wush = 1.2(wdsh wdj)
Two-Way Joist
wj wsh
l1
a a
b
wsh
l2
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Analysis
ACI code allows use of DDM for analysis of waffle slabs (ACI
R13.1). In such a case, waffle slabs are considered as flat
slabs, with the solid head acting as drop panels (ACI 13.1.3).
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Two-Way Joist
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Analysis
Static moment calculation for DDM analysis:
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Two-Way Joist
wuj
ln
Moj = wujl2ln2/8
wush
ln a a
wush
Mosh = wushba2/2
Moj Mosh
Mo = Moj + Mosh
b l2
ln
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Design
Design of slab for punching shear
The solid head shall be checked against punching shear.
The critical section for punching shear is taken at a section d/2 from face
of the column, where d is the effective depth at solid head.
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Two-Way Joist
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Design
Design of slab for
punching shear
Load on tributary area will
cause punch out shear.
Within tributary area, two
types of loads are acting:
Solid head load
Joist load
Both types shall be
considered while calculating
punching shear demand
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Two-Way Joist
l2 d/2
l1
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Design
Design of slab for punching
shear
Total area (At) = l1 l2
Solid head area = Asolid
Critical perimeter area = Acp
Vu =At wuj + Asolid (wish wuj)
Where,
wuj = Factored load considering
joist alone
wsh = Factored solid head dead
load
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Two-Way Joist
l2 d/2
l1
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011 30
Design
Shear Strength of Slab in Punching Shear:
Vn = Vc + Vs
Vc is least of:
4 (fc)bod
(2 + 4/c) (fc)bod
{(sd/bo +2} (fc)bod
c = longer side of column/shorter side of column
s = 40 for interior column, 30 for edge column, 20 for corner columns
Two-Way Joist
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011 31
Design
Design of Joist for Beam Shear:
Beam Shear Demand
Beam shear is not usually a problem in slabs including waffle slabs.
However for completion of design beam shear may also be
checked. Beam shear can cause problem in case where larger
spans and heavier loads with relatively shallow waffle slabs are
used.
The critical section for beam shear is taken at a section d from face
of the column, where d is the effective depth at solid head.
Two-Way Joist
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011 32
Design
Design of Joist for Beam Shear:
Beam shear capacity of concrete joist:
Vn = Vc + Vs
Vc is least of:
2 (fc)bribd
Vs = Avfy/bribs
For joist construction, contribution of concrete to shear strength Vc shall be
permitted to be 10 percent more than that specified in Chapter 11.
If required, one or two single legged stirrups are provided in the rib to
increase the shear capacity of waffle slab.
Two-Way Joist
Stirrup
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Design
Design for Flexure
The design of waffle slab for flexure is done by usual procedures.
However, certain reinforcement requirements apply discussed next.
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Two-Way Joist
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
ACI Recommendations on Reinforcement
Requirement of Waffle Slab:
Recommendations for Ribs:
ACI 10.6.7 states that if the effective depth d of a beam or joist
exceeds 36 inches, longitudinal skin reinforcement shall be provided as
per ACI section 10.6.7.
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Two-Way Joist
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
ACI Recommendations on Reinforcement
Requirement of Waffle Slab:
Recommendations for Slab:
According to ACI 13.3.2, for cellular or ribbed construction reinforcement
shall not be less than the requirements of ACI 7.12.
As per ACI 7.12, Spacing of top bars cannot exceed 5h or 18 inches.
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Two-Way Joist
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Other Important Points:
The amount of reinforcement and, if necessary, the top slab
thickness can be changed to vary the load capacities for
different spans, areas, or floors of a structure.
Each joist rib contains two bottom bars. Straight bars are
supplied over the column centerlines for negative factored
moment.
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Two-Way Joist
Bottom bar
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Other Important Points:
For layouts that do not meet the standard 2-feet and 3-feet
modules, it is preferable that the required additional width be
obtained by increasing the width of the ribs framing into the
solid column head.
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Two-Way Joist
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Other Important Points:
The designer should sketch out the spacing for a typical panel
and correlate with the column spacing as a part of the early
planning.
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Two-Way Joist
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Example: Design the slab system of hall shown in figure as waffle
slab, according to ACI 318. Use Direct Design Method for slab
analysis.
fc = 4 ksi
fy = 60 ksi
Live load = 100 psf
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Two-Way Joist
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Solution:
A 108 144 building, divided into twelve (12) panels, supported at
their ends on columns. Each panel is 36 36.
The given slab system satisfies all the necessary limitations for Direct
Design Method to be applicable.
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Two-Way Joist
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 01: Sizes
Columns
Let all columns be 18 18.
Slab
Adopt 30 30 standard dome.
Minimum equivalent flat slab thickness (hf) can be found using ACI Table
9.5 (c):
Exterior panel governs. Therefore,
hf = ln/33
ln = 36 (2 18/2)/12 = 34.5
hf = (34.5/33) 12 = 12.45
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Two-Way Joist
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 01: Sizes
Slab
The closest depth of doom that will fulfill the requirement of equivalent
thickness of flat slab equal to 12.45 is 12 in. with a slab thickness of 4
in. for a dome size of 30-in.
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Table: Waffle flat slabs (30" 30" voids at 3'-0")-Equivalent thickness
Rib + Slab Depths (in.) Equivalent Thickness te (in.)
8 + 3 8.61
8 + 4 9.79
10 + 3 10.18
10 + 4 11.37
12 + 3 11.74
12 + 4 12.95
14 + 3 13.3
14 + 4 14.54
16 + 3 14.85
16 + 4 16.12
20 + 3 17.92
20 + 4 19.26
Two-Way Joist
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 01: Sizes
Planning of Joist layout
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Two-Way Joist
l = 36-0 = 432
Standard module = 36 36
No. of modules in 36-0:
n = 432/36 = 12
Planning:
First joist is placed on interior column
centerline with progressive placing of
other joists towards exterior ends of
panel. To flush the last joist with
external column, the width of exterior
joist comes out to be 15 (6+Column
size /2) as shown in plan view.
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 01: Sizes
Solid Head
Solid head dimension from column centerline = l/6 = 36/6 = 6
Total required length of solid head= 2 6 = 12
As 3 3 module is selected, therefore 4 voids including joist witdh will
make an interior solid head of 12.5 12.5. (Length of solid head = c/c
distance between rib + rib width )
Depth of the solid head = Depth of standard module = 12 + 4.5 = 16.5
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Two-Way Joist
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 02: Loads
Floor (joist) dead load (wdj) = 109 psf = 0.109 ksf
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Table: Standard Dome Dimensions and other Data
Dome Size Dome Depth (in.) Volume of Void
(ft3)
Floor Dead Load (psf) per slab
thickness
3 inches 4 inches
30-in
8 3.98 71 90
10 4.92 80 99
12 5.84 90 109
14 6.74 100 119
16 7.61 111 129
20 9.3 132 151
19-in
8 1.56 79 98
10 1.91 91 110
12 2.25 103 122
14 2.58 116 134
16 2.9 129 148 Reference: Table 11-1, CRSI Design Handbook 2002
Two-Way Joist
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011 46
Step No 02: Loads
Floor (joist) dead load (wdj) = 109 psf = 0.109 ksf
Solid Head dead load (wsh) = chsh wdj
= 0.15 {(12 + 4.5)/12} 0.109 = 0.097 ksf
Two-Way Joist
wdj Wdj+sh
l1
a a
b
Wdj+sh
l2 a = 5.25 ft
b = 12.5 ft
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011 47
Step No 02: Loads
wL = 100 psf = 0.100 ksf
Load due to joists plus LL (wuj)
wuj = 1.2 wdj + 1.6wL
= 1.2 0.109 + 1.60.100
= 0.291 ksf
Load due to solid head dead load (wush)
wush = 1.2wsh
= 1.2 0.097 = 0.1164 ksf
Two-Way Joist
wush
l1
a a
b l2
wuj
wush
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 03: Frame Analysis (E-W Interior Frame)
Step 1: Marking E-W Interior Frame:
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Two-Way Joist
l2 = 36
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 03: Frame Analysis (E-W Interior Frame)
Step 2: Marking of Column and Middle Strips:
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Two-Way Joist
l2 = 36 CS/2 = Least of l1/4 or l2/4
l2/4 = 36/4 = 9
CS/2 = 9
MS/2 = 9
MS/2 = 9 CS/2 = 9
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 03: Frame Analysis (E-W Interior Frame)
Step 03: Static Moment Calculation
Moj (due to joists) = wojl2ln2/8
= 0.291 36 34.52/8 = 1557.56 ft-kip
Mosh (due to solid head excluding joists) = wush ba2/2
= 0.116412.55.252/2 = 20 ft-kip
Mo (total static moment) = Moj + Mosh = 1557.56 + 20 = 1577.56 ft-kip
Note: Since normally, Mosh is much smaller than Moj the former can be
conveniently ignored in design calculations.
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Two-Way Joist
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 03: Frame Analysis (E-W Interior Frame)
Step 03: Static Moment Calculation.
51
Two-Way Joist
l2 = 36
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 03: Frame Analysis (E-W Interior Frame)
Step 04: Longitudinal distribution of Total static moment (Mo).
52
Two-Way Joist
l2 = 36 0.26
0.52
0.70 0.65
0.35
0.65 0.70 0.26
0.52 Longitudinal
distribution
factors (D.F)L
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 03: Frame Analysis (E-W Interior Frame)
Step 04: Longitudinal distribution of Total static moment (Mo).
53
Two-Way Joist
l2 = 36 410
820
1104 1025
552
1025 1104 410
820
Units: ft-kip
ML = Mo (D.F)L
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 03: Frame Analysis (E-W Interior Frame)
Step 05: Lateral Distribution of Longitudinal moment (L.M).
54
Two-Way Joist
l2 = 36 0.60
Lateral
distribution
factors (D.F)Lat 1.00 0.75 0.75 0.75
0.60 0.60
1.00 0.75
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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 03: Frame Analysis (E-W Interior Frame)
Step 05: Lateral Distribution of Longitudinal moment (L.M).
55
Two-Way Joist
l2 = 36 492
410 828 769 769
331 492
410 828 MLat = ML (D.F)Lat
ML,ext- = 410 kip-ft
ML,ext+ = 820 kip-ft
ML,int- = 1104 kip-ft
ML,- = 1025 kip-ft
ML,+ = 552 kip-ft
328/2 328/2
328/2 328/2
0 0
0 0
276/2
276/2
276/2
276/2
221/2
221/2
256/2
256/2
256/2
256/2
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 03: Frame Analysis (E-W Interior Frame)
Step 05: Lateral Distribution of Longitudinal moment (L.M).
56
Two-Way Joist
l2 = 36 27.3
22.8 46 42.7 42.7
18.38 27.3
22.8 46 MLat per foot = Mlat/strip width
ML,ext- = 410 kip-ft
ML,ext+ = 820 kip-ft
ML,int- = 1104 kip-ft
ML,- = 1025 kip-ft
ML,+ = 552 kip-ft
18.22 18.22
18.22 18.22
0 0
0 0
15.33
15.33
15.33
15.33
12.3
12.3
14.22
14.22
14.22
14.22
-
29
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 03: Frame Analysis (E-W Exterior Frame)
Step 03: Static Moment Calculation
Moj (due to joists) = wojl2ln2/8
= 0.291 18.75 34.52/8 = 811.78 ft-kip
Mosh (due to solid head excluding joists) = wush ba2/2
= 0.116475.252/2 = 12.83 ft-kip
Mo (total static moment) = Moj + Mosh = 811.78 + 12.83 = 825 ft-kip
Note: Since normally, Mosh is much smaller than Moj the former can be
conveniently ignored in design calculations.
57
Two-Way Joist
l2 = 18 + c/2
= 18 + (18/12)/2
= 18.75
c = column dimension
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 03: Frame Analysis (E-W Exterior Frame)
Step 03: Static Moment Calculation.
58
Two-Way Joist
l2 = 18.75
-
30
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 03: Frame Analysis (E-W Exterior Frame)
Step 04: Longitudinal distribution of Total static moment (Mo).
59
Two-Way Joist
0.26
0.52 0.70 0.65
0.35 0.65 0.70 0.26
0.52
Longitudinal
distribution
factors (D.F)L
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 03: Frame Analysis (E-W Exterior Frame)
Step 04: Longitudinal distribution of Total static moment (Mo).
60
Two-Way Joist
215
429 578 536
289 536 578 215
429
Units: ft-kip
ML = Mo (D.F)L
-
31
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 03: Frame Analysis (E-W Exterior Frame)
Step 05: Lateral Distribution of Longitudinal moment (L.M).
61
Two-Way Joist
0.60
Lateral
distribution
factors (D.F)Lat
1.00 0.75 0.75 0.75
0.60 0.60
1.00 0.75
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 03: Frame Analysis (E-W Exterior Frame)
Step 05: Lateral Distribution of Longitudinal moment (L.M).
62
Two-Way Joist
215 434 402 769 215 402
MLat = ML (D.F)Lat
ML,ext- = 215 kip-ft
ML,ext+ = 429 kip-ft
ML,int- = 578 kip-ft
ML,- = 536 kip-ft
ML,+ = 289 kip-ft
172 172 0 0 144 144 116 134 134
257 173 257
-
32
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 03: Frame Analysis (E-W Exterior Frame)
Step 05: Lateral Distribution of Longitudinal moment (L.M).
63
Two-Way Joist
23.8 48.2 44.7 44.7 23.8 48.2
ML,ext- = 215 kip-ft
ML,ext+ = 429 kip-ft
ML,int- = 578 kip-ft
ML,- = 536 kip-ft
ML,+ = 289 kip-ft
17.64 17.64 0 0 14.76 14.76 11.89 13.74 13.74
28.6 19.2 28.6
MLat per foot = Mlat/strip width
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 03: Frame Analysis
Analysis of N-S Interior and Exterior Frame will be same as E-W respective
frames due to square panels.
64
N-S
Interior
Frame
l2 = 36-0
N-S
Exterior
Frame
l2 = 18-9
Two-Way Joist
-
33
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 04: Design
For E-W Interior Slab Strip:
65
Two-Way Joist
l2 = 36 27.3
22.8 46 42.7 42.7
18.38 27.3
22.8 46
18.22 18.22
18.22 18.22
0 0
0 0
15.33
15.33
15.33
15.33
12.3
12.3
14.22
14.22
14.22
14.22
Selective values
are used for design.
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 04: Design
E-W Exterior Frame.
66
Two-Way Joist
23.8 48.2 23.8 48.2
17.64 17.64 0 0 14.76 14.76 11.89
28.6 19.2 28.6
Selective values
are used for
design
-
34
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 04: Design
Design of N-S Interior and Exterior Frame will be same as E-
W respective frames due to square panels and also for the
reason that davg is used in design.
davg = 16.5 (0.75 inch (cover) + inch (Assumed bar
diameter) = 15 inch
This will be used for both directions positive as well as
negative reinforcement.
67
Two-Way Joist
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 05: Detailing (E-W Frames)
68
#6 @ 12 #6 @ 6 #6 @ 6 #6 @ 12
#6 @ 12 #6 @ 6 #6 @ 6 #6 @ 12
#6 @ 18 #6 @ 18 #6 @ 18 #6 @ 18
Two-Way Joist
Negative Reinforcement in
E-W direction
For M = 46 ft-k =46 x12
= 552 in-k
d = 15
fy = 60 ksi
As = 0.7 in2
-
35
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 05: Detailing (N-S Frames)
69
#6 @ 6
#6 @ 12
#6 @ 6
#6 @ 18
#6 @ 18
#6 @ 18
#6 @ 12
#6 @ 6
#6 @ 6
Two-Way Joist
Negative Reinforcement
in N-S direction
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 05: Detailing (N-S Frames)
Positive reinforcement
For M = 27 ft-k =27x12 = 324in-k
d = 15
fy = 60 ksi
As = 0.373 in2 This is per foot reinforcement. For 18 feet col strip, this will
be equal to 0.373 x 18 = 6.714 in2
There are 6 joists in 18 feet with. Therefore per rib reinforcement = 1.12
Using # 7 bars, 2 bars per joist rib will be provided.
70
Two-Way Joist
-
36
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 05: Detailing (E-W Interior Frame)
71
Column Strip (Interior Frame); section taken over support
Column Strip (Exterior Frame); section taken over support
#6 @ 12 c/c
#6 @ 6 c/c
2 #7 Bars
18-0
2 #7 Bars
Two-Way Joist
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 05: Detailing (E-W Interior Frame)
72
Middle Strip (Interior Frame); Section taken over column line
Middle Strip (Exterior Frame); Section taken over column line
#6 @ 18 c/c
#6 @ 18 c/c
2 #7 Bars
18-0
2 #7 Bars
Two-Way Joist
-
37
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 05: Detailing (E-W Exterior Frame)
73
Column Strip (Interior Frame); section over support
#6 @ 6 c/c
2 #7 Bars
9-0
Column Strip (Exterior Frame); section over support
#6 @ 12 c/c
2 #7 Bars
Two-Way Joist
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 05: Detailing (E-W Exterior Frame)
74
Middle Strip (Interior Frame) ; section over support
#6 @ 18 c/c
2 #7 Bars
9-0
Middle Strip (Exterior Frame); section over support
#6 @ 18 c/c
2 #7 Bars
Two-Way Joist
-
38
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Step No 04: Design
Note: For the completion of design problem, the waffle slab
should also be checked for beam shear and punching shear.
75
Two-Way Joist
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011 76
Moment Coefficient
Method
-
39
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Method (Introduction)
The Moment Coefficient Method included for the first time in
1963 ACI Code is applicable to two-way slabs supported on
four sides of each slab panel by walls, steel beams relatively
deep, stiff, edge beams (h = 3hf).
Although, not included in 1977 and later versions of ACI code,
its continued use is permissible under the ACI 318-08 code
provision (13.5.1). Visit ACI 13.5.1.
77
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Method
Moments:
Ma, neg = Ca, negwula2
Mb, neg = Cb, negwulb2
Ma, pos, (dl + ll) = M a, pos, dl + M a, pos, ll = Ca, pos, dl wu, dl la2 + Ca, pos, ll wu, ll la
2
Mb, pos, (dl + ll) = Mb, pos, dl + Mb, pos, ll = Cb, pos, dl wu, dl lb2 + Cb, pos, ll wu, ll lb
2
Where Ca, Cb = Tabulated moment coefficients
wu = Ultimate uniform load, psf
la, lb = length of clear spans in short and long directions respectively.
78
Ma,neg lb
la
Ma,neg
Ma,pos Mb,neg Mb,neg
Mb,pos
-
40
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
4 spans @ 25-0
3 sp
ans @
20
-0
Two Way Slabs
Moment Coefficient Method: Cases
Depending on the support conditions, several cases are possible:
79
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Method: Cases
Depending on the support conditions, several cases are possible:
80
4 spans @ 25-0
3 sp
ans @
20
-0
-
41
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Method: Cases
Depending on the support conditions, several cases are possible:
81
4 spans @ 25-0
3 sp
ans @
20
-0
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
4 spans @ 25-0
3 sp
ans @
20
-0
Two Way Slabs
Moment Coefficient Method: Cases
Depending on the support conditions, several cases are possible:
82
-
42
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Tables:
83
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Tables:
84
-
43
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Tables:
85
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Tables:
86
-
44
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Tables:
87
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Tables:
88
-
45
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Load Coefficient Table:
89
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
90
Maximum Spacing and Minimum Reinforcement
Requirement:
Maximum spacing (ACI 13.3.2):
smax = 2 hf in each direction.
Minimum Reinforcement (ACI 7.12.2.1):
Asmin = 0.0018 b hf for grade 60.
Asmin = 0.002 b hf for grade 40 and 50.
-
46
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Special Reinforcement at Exterior Corner of Slab
The reinforcement at exterior ends of the slab shall be provided as per ACI
13.3.6 in top and bottom layers as shown.
The positive and negative reinforcement in any case, should be of a size and
spacing equivalent to that required for the maximum positive moment (per foot
of width) in the panel.
91
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Method
Steps
Find hmin = perimeter/ 180 = 2(la + lb)/180
Calculate loads on slab (force / area)
Calculate m = la/ lb
Decide about case of slab,
Use table to pick moment coefficients,
Calculate moments and then design.
Apply reinforcement requirements (smax = 2hf, ACI 13.3.2)
92
-
47
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Method: Example 1
A 100 60, 3-storey commercial building is to be designed.
The grids of column plan are fixed by the architect.
93
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Method: Example 1
Complete analysis of the slab is done by analyzing four panels
94
Panel I Panel I
Panel I Panel I
Panel II Panel II
Panel III Panel III
Panel III Panel III
Panel IV Panel IV
4 spans @ 25-0
3 sp
ans @
20
-0
-
48
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Method: Example 1
A 100 60, 3-storey commercial building: Sizes and Loads.
Sizes:
Minimum slab thickness = perimeter/180 = 2 (20+25)/180 = 6
However, for the purpose of comparison, take hf = 7
Columns = 14 14 (assumed)
Beams = 14 20 (assumed)
Loads:
S.D.L = Nil ; Self Weight = 0.15 x (7/12) = 0.0875 ksf
L.L = 144 psf ; wu = 0.336 ksf
95
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
4 spans @ 25-0
3 sp
ans @
20
-0
Two Way Slabs
Moment Coefficient Method: Example 1
Panels are analyzed using Moment Coefficient Method
96
Case = 4
m = la/lb = 0.8
Mb,neg Mb,neg Mb,pos Ma,neg
Ma,pos
Ma,neg
-
49
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Method: Example 1
Panels are analyzed using Moment Coefficient Method
97
Case = 4
m = la/lb = 0.8
Ca,neg = 0.071
Cb,neg = 0.029
Ca,posLL = 0.048
Cb,posLL = 0.020
Ca,posDL = 0.039
Cb,posDL = 0.016
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Method: Example 1
Panels are analyzed using Moment Coefficient Method
98
Case = 4
m = la/lb = 0.8
Ca,neg = 0.071
Cb,neg = 0.029
Ca,posLL = 0.048
Cb,posLL = 0.020
Ca,posDL = 0.039
Cb,posDL = 0.016
-
50
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Method: Example 1
Panels are analyzed using Moment Coefficient Method
99
Case = 4
m = la/lb = 0.8
Ca,neg = 0.071
Cb,neg = 0.029
Ca,posLL = 0.048
Cb,posLL = 0.020
Ca,posDL = 0.039
Cb,posDL = 0.016
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Method: Example 1
Panels are analyzed using Moment Coefficient Method
100
Case = 4
m = la/lb = 0.8
Ca,neg = 0.071
Cb,neg = 0.029
Ca,posLL = 0.048
Cb,posLL = 0.020
Ca,posDL = 0.039
Cb,posDL = 0.016
-
51
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Method: Example 1
Panels are analyzed using Moment Coefficient Method
101
Case = 4
m = la/lb = 0.8
Ca,neg = 0.071
Cb,neg = 0.029
Ca,posLL = 0.048
Cb,posLL = 0.020
Ca,posDL = 0.039
Cb,posDL = 0.016
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Method: Example 1
Panels are analyzed using Moment Coefficient Method
102
Case = 4
m = la/lb = 0.8
Ca,neg = 0.071
Cb,neg = 0.029
Ca,posLL = 0.048
Cb,posLL = 0.020
Ca,posDL = 0.039
Cb,posDL = 0.016
-
52
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
4 spans @ 25-0
3 sp
ans @
20
-0
Two Way Slabs
Moment Coefficient Method: Example 1
Panels are analyzed using Moment Coefficient Method
103
Panel I
Case = 4
m = la/lb = 0.8
Ca,neg = 0.071
Cb,neg = 0.029
Ca,posLL = 0.048
Cb,posLL = 0.020
Ca,posDL = 0.039
Cb,posDL = 0.016
Ma,neg = 9.5 k-ft
Ma,pos = 6.1 k-ft
Mb,neg = 6.1 k-ft
Mb,pos = 3.9 k-ft
Mb,neg Mb,neg Mb,pos Ma,neg
Ma,pos
Ma,neg
For slab supported on Spandrals, Mneg,ext = 1/3Mpos
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
4 spans @ 25-0
3 sp
ans @
20
-0
Two Way Slabs
Moment Coefficient Method: Example 1
Panels are analyzed using Moment Coefficient Method
104
Panel II
Case = 9
m = la/lb = 0.8
Ca,neg = 0.075
Cb,neg = 0.017
Ca,posLL = 0.042
Cb,posLL = 0.017
Ca,posDL = 0.029
Cb,posDL = 0.010
Ma,neg = 10.1 k-ft
Ma,pos = 5.1 k-ft
Mb,neg = 3.6 k-ft
Mb,pos = 3.1 k-ft
Mb,neg Mb,neg Mb,pos
Ma,neg
Ma,pos
Ma,neg
-
53
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
4 spans @ 25-0
3 sp
ans @
20
-0
Two Way Slabs
Moment Coefficient Method: Example 1
Panels are analyzed using Moment Coefficient Method
105
Panel III
Case = 8
m = la/lb = 0.8
Ca,neg = 0.055
Cb,neg = 0.041
Ca,posLL = 0.044
Cb,posLL = 0.019
Ca,posDL = 0.032
Cb,posDL = 0.015
Ma,neg = 7.4 k-ft
Ma,pos = 5.4 k-ft
Mb,neg = 8.6 k-ft
Mb,pos = 3.7 k-ft
Mb,neg Mb,neg Mb,pos
Ma,neg
Ma,pos
Ma,neg
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
4 spans @ 25-0
3 sp
ans @
20
-0
Two Way Slabs
Moment Coefficient Method: Example 1
Panels are analyzed using Moment Coefficient Method
106
Panel IV
Case = 2
m = la/lb = 0.8
Ca,neg = 0.065
Cb,neg = 0.027
Ca,posLL = 0.041
Cb,posLL = 0.017
Ca,posDL = 0.026
Cb,posDL = 0.011
Ma,neg = 8.7 k-ft
Ma,pos = 4.9 k-ft
Mb,neg = 5.7 k-ft
Mb,pos = 3.2 k-ft
Mb,neg Mb,neg Mb,pos
Ma,neg
Ma,pos
Ma,neg
-
54
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
4 spans @ 25-0
3 sp
ans @
20
-0
Two Way Slabs
Moment Coefficient Method: Example 1
Slab analysis summary
107
3.2
4.9 5.7 5.7
8.7
8.7
7.4
7.4
8.6 8.6 5.4
3.7
3.2
5.1 3.6
10.1
10.1
9.5
9.5
6.1 6.1
3.9
6.1
3.6
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Method: Example 1
Slab Reinforcement Details
108
A
B B B
C
C
C
C C B
A
A
B A
C
C
C
B B
A
A= #4 @ 12
B = #4 @ 6
C = #4 @ 4
4 spans @ 25-0
3 sp
ans @
20
-0
-
55
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Method: Example 1
Load On Beams from coefficient tables
109
Table: Load on beam in panel I, using
Coefficients
(wu = 0.336 ksf)
Bea
m
Length
(ft)
Width
(bs) of
slab
panel
support
ed by
beam
Wa Wb
Load
due to
slab,
Wwubs
(k/ft)
B1 25 10 0.71 - 2.39
B2 25 10 0.71 - 2.39
B3 20 12.5 - 0.29 1.22
B4 20 12.5 - 0.29 1.22
Panel I
4 spans @ 25-0
3 sp
ans @
20
-0
B1
B1
B2
B2
B3 B3 B3 B4 B4
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
B1
B1
B2
B2
B3 B3 B3 B4 B4
4 spans @ 25-0
3 sp
ans @
20
-0
Panel I
Two Way Slabs
Moment Coefficient Method: Example 1
Load On Beams from coefficient tables
110
Table: Load on beam in panel I, using
Coefficients
(wu = 0.336 ksf)
Bea
m
Length
(ft)
Width
(bs) of
slab
panel
support
ed by
beam
Wa Wb
Load
due to
slab,
Wwubs
(k/ft)
B1 25 10 0.71 - 2.39
B2 25 10 0.71 - 2.39
B3 20 12.5 - 0.29 1.22
B4 20 12.5 - 0.29 1.22
-
56
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Method: Example 1
Load On Beams from coefficient tables
111
Table: Load on beam in panel I, using
Coefficients
(wu = 0.336 ksf)
Bea
m
Length
(ft)
Width
(bs) of
slab
panel
support
ed by
beam
Wa Wb
Load
due to
slab,
Wwubs
(k/ft)
B1 25 10 0.83 - 2.78
B3 20 12.5 - 0.17 0.714
B4 20 12.5 - 0.17 0.714
Panel II
4 spans @ 25-0
3 sp
ans @
20
-0
B1
B1
B2
B2
B3 B3 B3 B4 B4
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
B1
B1
B2
B2
B3 B3 B3 B4 B4
Panel II
4 spans @ 25-0
3 sp
ans @
20
-0
Two Way Slabs
Moment Coefficient Method: Example 1
Load On Beams from coefficient tables
112
Table: Load on beam in panel I, using
Coefficients
(wu = 0.336 ksf)
Bea
m
Length
(ft)
Width
(bs) of
slab
panel
support
ed by
beam
Wa Wb
Load
due to
slab,
Wwubs
(k/ft)
B1 25 10 0.83 - 2.78
B3 20 12.5 - 0.17 0.714
B4 20 12.5 - 0.17 0.714
-
57
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Method: Example 1
Load On Beams from coefficient tables
113
Table: Load on beam in panel I, using
Coefficients
(wu = 0.336 ksf)
Bea
m
Length
(ft)
Width
(bs) of
slab
panel
support
ed by
beam
Wa Wb
Load
due to
slab,
Wwubs
(k/ft)
B1 25 10 0.55 - 1.84
B3 20 12.5 - 0.45 1.89
Panel III
4 spans @ 25-0
3 sp
ans @
20
-0
B1
B1
B2
B2
B3 B3 B3 B4 B4
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
B1
B1
B2
B2
B3 B3 B3 B4 B4
Panel III
4 spans @ 25-0
3 sp
ans @
20
-0
Two Way Slabs
Moment Coefficient Method: Example 1
Load On Beams from coefficient tables
114
Table: Load on beam in panel I, using
Coefficients
(wu = 0.336 ksf)
Bea
m
Length
(ft)
Width
(bs) of
slab
panel
support
ed by
beam
Wa Wb
Load
due to
slab,
Wwubs
(k/ft)
B1 25 10 0.55 - 1.84
B3 20 12.5 - 0.45 1.89
-
58
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Method: Example 1
Load On Beams from coefficient tables
115
Table: Load on beam in panel I, using
Coefficients
(wu = 0.336 ksf)
Bea
m
Length
(ft)
Width
(bs) of
slab
panel
support
ed by
beam
Wa Wb
Load
due to
slab,
Wwubs
(k/ft)
B1 25 10 0.71 - 2.39
B3 20 12.5 - 0.29 1.22
Panel IV
4 spans @ 25-0
3 sp
ans @
20
-0
B1
B1
B2
B2
B3 B3 B3 B4 B4
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
B1
B1
B2
B2
B3 B3 B3 B4 B4
Panel IV
4 spans @ 25-0
3 sp
ans @
20
-0
Two Way Slabs
Moment Coefficient Method: Example 1
Load On Beams from coefficient tables
116
Table: Load on beam in panel I, using
Coefficients
(wu = 0.336 ksf)
Bea
m
Length
(ft)
Width
(bs) of
slab
panel
support
ed by
beam
Wa Wb
Load
due to
slab,
Wwubs
(k/ft)
B1 25 10 0.71 - 2.39
B3 20 12.5 - 0.29 1.22
-
59
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Method: Example 1
Load On Beams
117
2.39 k/ft
1.22 k/ft
2.39 k/ft
1.22 k/ft
2.77 k/ft
2.77 k/ft
0.71 k/ft 0.71 k/ft
1.84 k/ft
1.89 k/ft
1.84 k/ft
1.89 k/ft
2.39 k/ft
1.22 k/ft
2.39 k/ft
1.22 k/ft
4 spans @ 25-0
3 sp
ans @
20
-0
B1
B1
B2
B2
B3 B3 B3 B4 B4
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Method: Example 1
Load On Beams
118
1.45 k/ft
2.62 k/ft
5.39 k/ft
0.94 k/ft
3.34 k/ft
2.07 k/ft
2.16 k/ft
4.46 k/ft
4 spans @ 25-0
3 sp
ans @
20
-0
B1
B1
B2
B2
B3 B3 B3 B4 B4
Loads
including self
weight of
beams 0.23
kip/ft
-
60
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Method: Example 2
Similarly, slab analysis results of 25 20 structure.
119
Panel I Panel I
Panel I Panel I
Panel II Panel II
Panel III Panel III
Panel III Panel III
Panel IV Panel IV
4 spans @ 25-0
3 sp
ans @
20
-0
Slab thickness = 6
SDL = 40 psf
LL = 60 psf
fc =3 ksi
fy = 40 ksi
wu = 0.234 ksf
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
4 spans @ 25-0
3 sp
ans @
20
-0
Two Way Slabs
Moment Coefficient Method: Example 2
Slab analysis summary
120
2.0
3.0 3.9 3.9
6.1
6.1
5.1
5.1
6.0 6.0 3.5
2.4
1.9
3.2 2.5
7
7
6.6
6.6
4.2 4.0
2.6
4.2
2.5
-
61
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
Two Way Slabs
Moment Coefficient Method: Example 3
Similarly, slab analysis results of 20 15 structure.
121
Panel I Panel I
Panel I Panel I
Panel II Panel II
Panel III Panel III
Panel III Panel III
Panel IV Panel IV
4 spans @ 20-0
3 sp
ans @
15
-0
Slab thickness = 6
SDL = 40 psf
LL = 60 psf
fc =3 ksi
fy = 40 ksi
wu = 0.234 ksf
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
4 spans @ 25-0
3 sp
ans @
20
-0
Two Way Slabs
Moment Coefficient Method: Example 3
Slab analysis summary
122
1.0
1.8 2.1 2.1
3.6
3.6
3.2
3.2
3.4 3.4 2.2
1.3
0.9
2.0 1.3
4.1
4.1
4
4
2.2 2.5
1.3
2.2
1.3
-
62
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
References
CRSI Design Handbook
ACI 318
Design of Concrete Structures 13th Ed. by Nilson, Darwin and Dolan.
123
Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011
The End
124