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Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Lecture-06
Analysis and Design of
Slab Systems
By: Prof Dr. Qaisar Ali
Civil Engineering Department
www.drqaisarali.com
1
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Topics Addressed
� Organization of the lecture
� Analysis & design of one way slab system
� Analysis & design of one way joist system
� Analysis & design of two way slab system with beams
� Analysis & design of two way slab system without beams
(flat plate & flat slabs)
� Analysis & design of two way joist system
2
2
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Organization of the Lecture
� Analysis & design of one way slab system
� The ACI code approximate method of analysis called as strip
method is used for the analysis of one way slabs. This topic
has already been covered in BSc lectures 3 & 4, Please
download the same from the website.
� Analysis & design of one way joist system
� This topic will be covered in this lecture.
3
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Organization of the Lecture
� Analysis & design of two way slab system with beams
� The ACI code approximate method of analysis called as
moment coefficient method is generally used for the analysi s
of two way slabs supported on stiff beams or walls. This topic
has been covered in BSc lecture 4, please download the same
from the website.
4
3
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Organization of the Lecture
� Analysis & design of two way slab system without beams
(flat plate & flat slabs)
� The ACI code approximate method of analysis called as
Direct Design Method (DDM) is used for the analysis of flat
plates and flat slabs. This topic has been covered in BSc
lecture 5. The students can download the same from the
website. However this topic will also be quickly discussed i n
this lecture.
� Please note that DDM can also be used for the analysis of two wa y slabs with beams.
However as the application of this method to such systems is r elative difficult
therefore these systems are generally analyzed using moment coefficient method
instead of DDM, provided that beams are relatively stiff.
5
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Organization of the Lecture
� Analysis & design of two way joist system
� This topic will be discussed in this lecture.
6
4
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Organization of the Lecture
� Therefore in this lecture, only the following topics will be
discussed.
� Analysis & design of one way joist system
� Analysis & design of two way slab system without beams (flat
plate & flat slabs)
� Analysis & design of two way joist system
7
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Analysis & design of one way
joist system
8
5
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Contents
� General
� Characteristics
� Basic Steps
� Serviceability Requirements
� Example
9
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
General
� Joist: T-beams called joists are formed by creating void
spaces in what otherwise would be a solid slab.
� Joist Construction: Joist construction consists of a
monolithic combination of regularly spaced ribs and a top slab
(T beam or Joist) arranged to span in one direction or two
orthogonal directions.
Rib
10
6
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
General
� ACI code 8.11 contains provisions for joist construction aimed
at facilitating the analysis and design of joist system as
compared to regular slab beam system.
� A structural system will be however called as joist system if
the pan width (clear spacing between ribs) is less than or
equal to 30 inches (ACI 8.11.3).
11
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
General
� If the system does not fulfill the requirements of joist system
then it shall be designed as regular slab beam system which
means that slab shall be designed for flexure and beam shall be
designed for flexure and shear.
12
7
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Characteristics
� Suitable for long spans
� Economical range: 30 ft – 50 ft
13
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Characteristics
� Pan voids reduce dead load.
� Electrical/mechanical equipment can be placed between joists.
� Clear, unobstructed tenant space.
� Less effect of vibrations due to stiffer system.
� Standard forms for the void spaces between ribs are either 20 or
30 inches wide, and 8, 10, 12, 16, or 20 inches deep.
14
8
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Characteristics
� Economical for buildings such as apartment houses, hotels, and
hospitals, where the live loads are fairly small and the spans
comparatively long.
� Forms are tapered in cross section generally at a slope of 1 to 12
to facilitate removal.
15
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
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.
16
9
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Basic Steps for Structural Design
� Sizes:
� Depth of joist: Same table as used for one way slabs can be used to
find depth of joist.
� l = Span length (same as for solid one way slab)
� The depth of the joist includes the slab depth.
17
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Basic Steps for Structural Design
� Sizes:
� Width of joist:
� Minimum width of rib = Depth of rib/3.5 but not less than 4″.
� Depth of slab
� Minimum slab thickness = clear distance between ribs/12 but not less than 2″.
18
10
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Basic Steps for Structural Design
� Loads:
� One way Joist systems are usually designed for gravity loading
(U = 1.2D + 1.6L).
� The joist is analyzed for load (per running foot) over a width
equal to the center to center spacing between the joists as
shown:
b
b19
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Basic Steps for Structural Design
� Analysis:
� For the purpose of analysis, a single T shaped joist is
considered.
20
11
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Uniformly distributed load (L/D ≤ 3)
≤ 1.2ln ln
Two or more spans
Prismaticmembers
Basic Steps for Structural Design
� Analysis:
� ACI approximate analysis is applicable.
21
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Spandrelsupport Negative
Moment
x wuln2
1/24 1/10* 1/11 1/11 1/10* 0
*1/9 (2 spans)
* 1/12 (for all spans with ln < 10 ft)
Columnsupport 1/16
PositiveMomentx
1/14 1/16 1/11
wuln2
lnln ln
Simplesupport
Integral withsupport
wu
Note: For simply supported slab, M = wul2/8, where l = span length (ACI 8.7).
Basic Steps for Structural Design
� Analysis:
22
12
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Basic Steps for Structural Design
� Design:
� Design of joist for flexure:
� Design of one way joist is just like design of a beam.
23
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Basic Steps for Structural Design
� Design:
� Flexural Reinforcement Placement in joist
� Reinforcement for the joists usually consists of two bars in the positive
bending region, with one bar discontinued where no longer needed or bend
up to provide a part of negative steel requirement over the supporting
girder.
� One way joists are generally proportioned with the concrete providing all of
the shear strength, with no shear reinforcement used.
24
13
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Basic Steps for Structural Design
� Design:
� Flexural Reinforcement Placement in joist
For Joist positive bending: As required by flexural demand but ≥ ACI 10.5.1 (Minimum reinforcement for flexural members)
For Joist negative bending: As required by flexural demand ≥ ACI 10.5.1 (Minimum reinforcement for flexural members)
Slab Main Reinforcement (located at mid depth): As required by flexural demand but ≥ ACI 7.12 (shrinkage reinf.)
25
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Design:
� Flexural Reinforcement Placement in joist
� Skin reinforcement
� 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.
Basic Steps for Structural Design
26
14
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Basic Steps for Structural Design
� Design:
� Maximum Spacing Requirement for flexural reinforcement in slab
� If spacing is obtained by flexural demand than compare spacing with:
� Least of 3h or 18” (ACI 7.6.5)
� If shrinkage reinforcement is governing than compare spacing with:
� Least of 5h or 18” (ACI 7.12.2.2)
27
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Basic Steps for Structural Design
� Design:
� Design for Shear
� If Vu ≥ ΦVc, then shear capacity shall be increased either by
increasing rib depth or width or by providing single legged shear
reinforcement.
� If Vu < ΦVc the section is safe against shear. Even minimum
reinforcement as required for the beams with [ΦVc/2 < Vu < ΦVc] is
not required as per ACI 11.5.5.1.
28
15
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Basic Steps for Structural Design
� Design:
� Design for Shear
� ACI 11.5.5.1: Minimum shear reinforcement — A minimum area of
shear reinforcement shall be provided in all reinforced concrete
flexural members (prestressed and nonprestressed) where factored
shear force Vu exceeds one-half the shear strength provided by
concreteΦVc, except:
� (a) Slabs and footings;
� (b) Concrete joist construction defined by 8.11;
� (c) Beams with total depth not greater than 10 inches.,
29
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Basic Steps for Structural Design
� Design:
� Design for Shear
� For joist construction, contribution of concrete to shear
strength Vc shall be permitted to be 10 percent more than that
specified in Chapter 11.
� Critical shear demand section shall be at a distance “d” from
the face of the support.
30
16
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example
� Determine the required depth and reinforcement for the one-
way joist system shown next. The joists are 6 inches wide and
are spaced 36 inches c/c. The slab is 3.5 inches thick.
� fc′ = 4000 psi ; fy = 60,000 psi
� Service Live Load (LL) = 60 psf
� Superimposed Dead Load (SDL) = 64 psf
� Width of spandrel beams = 20 inches ; Width of interior beams = 36 inches
� Columns: interior = 18 × 18 inches ; exterior = 16 × 16 inches
� Story height (typical) = 13 ft
31
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example
Plan View
Section A-A
32
17
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example
� Solution:
� Step No 01: Sizes
� l/18.5 = 30 × 12/18.5 = 19.5″ ; Use 19.5″ deep joist (16 + 3.5″)
� d = 19.5 – 0.75″ clear cover – ½ bar dia. (Assuming #8 bar) = 18.25″
� Width of the joist = 6″ > (rib depth/ 3.5 = 16/3.5 = 4.57″), OK.
33
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example
� Solution:
� Step No 02: Loads
� Area of rib (A) = 6×16 = 96 in2
� Dead Load of rib = Aγc = (96/144)×0.15
= 0.10 kip/ft
� Per foot dead load of slab = 0.15 × (3.5 × 36)/144 = 0.131 kip/ft
� Per foot superimposed dead load on slab = 0.064 × 36/12 = 0.192 kip/ft
� Per foot live load on 36″ width = 0.060 × 36/12 = 0.18 kip/ft
� wu = 1.2DL+1.6LL=1.2×(0.10+0.131+0.192)+1.6×0.18 = 0.795 kip/ft
b =36″
6″
3.5″
19.5″16″
34
18
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example
� Solution:
� Step No 03: Analysis
lnln ln
Integral withsupport
wu
Spandrelsupport Negative
Moment
x wuln2
1/24 1/10 1/11 1/11 1/11 1/11
PositiveMomentx
1/14 1/16
wuln2
1/16
35
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example
� Solution:
� Step No 03: Analysis
0.795 kip/ft
10.93
12.57
12.34
12.34
12.34
25.05
42.94
60.1252.68
36.22
52.68 52.68
SFD (kip)
BMD (k-ft)
27.5 ' 27 '
Vu,ext = 9.72
Vu,int = 11.18
36
19
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example
� Solution:
� Step No 04: Design
� Design of joist for shear:
� Vu, ext = 9.72 kips ; Vu, int = 11.18 kips
� ΦVc = 1.10Φ2√ (fc′)bd
= 1.10 × 0.75 × 2 × √ (4000) × 6 × 18.25/1000 = 11.42 kips
� ΦVc > Vu, ext and Vu, int,OK
37
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example
� Solution:
� Step No 04: Design
� Design of joist for flexure:
� Design for positive moment (42.94 kip-ft)
� beff = 36″ (c/c spacing between joists)
� Check if joist is to be designed as rectangular beam or T-beam.
� Assume a = hf = 3.5″
As = Mu/ {Φfy(d–a/2)}=(42.94×12)/ {0.9×60×(18.25–3.5/2)}= 0.57 in2
� Re-calculate “a”:
a = Asfy/ (0.85fc′beff) = 0.57 × 60/ (0.85 × 4 × 36) = 0.27″ < hf
� Therefore design joist as rectangular beam.
After trials, As = 0.53 in2 (2 #5 bars, 0.6 in2)
38
20
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example
� Solution:
� Step No 04: Design
� Design of joist for flexure:
� Design for exterior negative moment (25.05 kip-ft)
� b = 6″ (bottom width of joist) ; h = 19.5″ ; d = 18.25″
� After trials, As = 0.31 in2
� Asmin = (3√ (fc′)/fy)bd ≥ (200/fy)bd = 0.365 in2 , governs
� Therefore, As = 0.365 in2
� Distribute bars uniformly in top slab:
� As = 0.365/ 3 = 0.121 in2/ft (#3 @ 10″ c/c)
39
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example
� Solution:
� Step No 04: Design
� Design of joist for flexure:
� Design for interior negative moment (60.12 kip-ft)
� b = 6″ (bottom width of joist) ; h = 19.5″ ; d = 18.25″
� After trials, As = 0.78 in2
� Asmin = (3√ (fc′)/fy)bd ≥ (200/fy)bd = 0.365 in2
� Therefore, As = 0.78 in2
� Distribute bars uniformly in top slab:
� As = 0.78/ 3 = 0.26 in2/ft (#3 @ 5″ c/c)
40
21
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example
� Solution:
� Step No 04: Design
� Design of slab for flexure:
� The slab reinforcement normal to the ribs is often located at mid-depth
of the slab to resist both positive and negative moments.
� wu = 1.2 (0.044+0.064)+1.6*0.06 =0.23 ksf
� Over the support, negative moment coefficient of 1/12 will be used.
� Mu = wuln2/12 = 0.23 × 2.52/12 = 0.12 ft-kip (1.44 in-kip)
� After trials, As = 0.015 in2/ft
� Asmin = 0.0018 × 12 × 3.5 = 0.08 in2/ft, governs (#3 @ 16.5″ c/c)
ln = 2.5′
41
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example
� Solution:
� Step No 04: Design
� Design of slab for flexure:
� Maximum spacing allowed for temperature steel reinforcement :
� 5hf = 5 × 3.5 = 17.5″
� 18″
� Finally provide #3 @ 16″ c/c.
42
22
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example
� Solution:
� Step No 05: Drafting
ln = 27.5′ ln = 27′
ln /4 = 6′-9″ ln /3 = 9′-3″ ln /3 = 9′-0″
A
A
#3 @ 5″ c/c#3 @ 10″ c/c
2 #5 bars in joist
#3 @ 16″ c/c
Beam
3.5″
43
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example
� Solution:
� Step No 05: Drafting
2 #5 bars (Main +ve)
#3 @ 10″ c/c (Main –ve)
#3 @ 16″ c/c, at mid depth(shrinkage & temperature)
Section A-A
44
23
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Analysis and Design of Two-way Slab
System without Beams
(Flat Plates and Flat Slabs)
45
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� There are two sections under this topic
� Section – I: Flexural Analysis of Two-Way Slab System
without Beams (Direct Design Method)
� Section – II: Shear Design for Two-Way Slab System
without Beams (Flat Plates and Flat Slabs)
46
24
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Contents of Section-1
� Introduction to Direct Design Method (DDM)
� Steps in Direct Design Method (DDM).
� Detailing of Flexural Reinforcement
� Summary of Direct Design Method (DDM)
� Example
47
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Introduction to DDM
� General:
� In the moment coefficient method of analysis used for slabs with
beams, the system is analyzed panel by panel.
� In DDM, frames rather than panels are analyzed.
Interior Frame
Exterior Frame
Interior Frame
Exterior Frame
48
25
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Introduction to DDM
� General:
� For complete analysis of slab system frames, are analyzed in E-
W and N-S directions.
E-W FramesN-S Frames
49
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Introduction to DDM
� Limitations (ACI 13.6.1):
� Though DDM is useful for analysis of slabs, specially without beams,
the method is applicable with some limitations as discussed next.
50
26
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Introduction to DDM
� Limitations (ACI 13.6.1):
Uniformly distributed loading (L/D ≤ 2)
l2
l1 l1≥2l1 /3
Three or more spans
Column offset≤ l2 /10
Rectangular slab panels (2 or less:1)
51
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Steps in DDM
� Step 01: Sizes
� ACI table 9.5 (c) are used for finding the slab thickness.
52
27
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Steps in DDM
� Step 02: Loads
� The slab load is calculated in usual manner.
53
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Steps in DDM
� Step 03: Analysis
Interior Frame
l1
l2Half width of panel on one side
Half width of panel on other side
• Step I: Marking E-W Frame (Interior Frame)
Col Centerline
Panel Centerline
Panel Centerline
According to ACI 13.6.2.3: Where the transverse span of panels on either side of the centerline of supports varies, l2 shall be taken as the average of adjacent transverse spans.
54
28
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Steps in DDM
� Step 03: Analysis
Exterior Frame
l1
l2 = Panel width/2 +h2/2
l2Half width of panel on one side
h2/2
• Step I: Marking E-W Frame (Exterior Frame)
55
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
b) Marking Middle Strip (For Interior Frame)
Steps in DDM
� Step 03: Analysis
CS/2 = Least of l1/4 or l2/4
CS/2CS/2
C.S
M.S/2
M.S/2
l2
l1
ln
Half Column strip
• Step II: a) Marking Column Strip (For Interior Frame)
56
29
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Steps in DDM
� Step 03: Analysis
CS/2 = Least of l1/4 or l2/4
l2/4 = 20/4 = 5′
5′5′
10′
5′
5′
l2
l1
ln
Half Column Strip
• Step II: a) Marking Column Strip (For Interior Frame)
b) Marking Middle Strip (For Interior Frame)
For l1 = 25′ and l2 = 20′, CS
and MS widths are given as
follows:
57
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Steps in DDM
� Step 03: Analysis
l2 = ½ (20) + ½ (14/12) = 10.58′
CS = (20/4) + ½ (14/12) = 5.58′
MS = 10.58 – 5.58 = 5′
l2
l1
ln
• Step II: a) Marking Column Strip (For Exterior Frame)
b) Marking Middle Strip (For Exterior Frame)
MS = l2 - CS
For Given Frame:
CS and MS widths are given as follows:
CS = Min. Panel Width/4 + ½ (Col. Size)
MS
Min. Panel Width/4CS
h/2
58
30
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Steps in DDM
� Step 03: Analysis
� Step III: Calculate Static Moment (Mo) for interior span of frame.
Mo =wu l2 ln
2
8 Mo
l2ln
Span of frame
59
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Steps in DDM
� Step 03: Analysis
� Step IV: Longitudinal Distribution of Static Moment (Mo).
M+
M − M −
M − = 0.65Mo
M + = 0.35Mo
60
31
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Steps in DDM
� Step 03: Analysis
� Step V: Lateral Distribution to column and middle strips.
M − = 0.65Mo
M + = 0.35Mo
0.60M +
0.75M − 0.75M −
61
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Steps in DDM
� Step 03: Analysis
� Step VI: Calculate Static Moment (Mo) for exterior span of frame.
Mo =wu l2 ln
2
8 Mo
l2ln
Span of frame
62
32
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Steps in DDM
� Step 03: Analysis
� Step VII: Longitudinal distribution of static moment (Mo).
Mext+
Mext − M int−
Mext − = 0.26Mo
M ext+ = 0.52Mo
Mint- = 0.70Mo
63
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Steps in DDM
� Step 03: Analysis
� Step VIII: Lateral Distribution to column and middle strips.
Mext- = 0.26Mo
M ext+ = 0.52Mo
Mint- = 0.70Mo
0.60Mext+
1.00Mext− 0.75Mint−
64
33
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Steps in DDM
� Step 03: Analysis (Summary)
� Distribution factors for longitudinal & Lateral Distribution to column and middle
strips.
Mext- = 0.26Mo
Mext+ = 0.52Mo
Mint- = 0.70Mo 0.60Mext+
1.00Mext− 0.75Mint−0.60M+
0.75M− 0.75M−
M- = 0.65Mo
M+ = 0.35Mo
65
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Maximum spacing and minimum reinforcement requirement for slab
systems without beams:
� Maximum spacing (ACI 13.3.2):
smax = 2 hf in each direction.
� Minimum Reinforcement (ACI 7.12.2.1):
Asmin = 0.0018 bhf for grade 60.
Asmin = 0.002 bhf for grade 40 and 50.
Detailing of flexural reinforcement
66
34
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Detailing of flexural reinforcement
� Reinforcement placement: In case of two way slabs supported on
beams, short-direction bars are normally placed closer to the top
or bottom surface of the slab, with the larger effective depth
because of greater moment in short direction.
67
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Detailing of flexural reinforcement
� Reinforcement placement: However in the case of flat
plates/slabs, the long-direction negative and positive bars, in both
middle and column strips, are placed closer to the top or bottom
surface of the slab, respectively, with the larger effective depth
because of greater moment in long direction.
68
35
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Detailing of flexural reinforcement
� Splicing: ACI 13.3.8.5 requires that all bottom bars within the column
strip in each direction be continuous or spliced with length equal to
1.0 ld. Splicing shall be provided in location where yielding is not
expected.
69
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Detailing of flexural reinforcement
� Continuity of bars: At least two of the column strip bars in each
direction must pass within the column core and must be anchored
at exterior supports (ACI 13.3.8.5).
70
36
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Detailing of flexural reinforcement
� Standard Bar Cut off Points (Practical Recommendation) for
column and middle strips both.
71
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Detailing of flexural reinforcement
� Reinforcement at Exterior Corners:
� Reinforcement should be provided at
exterior corners in both the bottom and top
of the slab, for a distance in each direction
from the corner equal to one-fifth the
longer span of the corner panel as shown
in figure.
� The positive and negative reinforcement
should be of size and spacing equivalent to
that required for maximum positive
moments (per foot of width) in the panel
(ACI 13.3.6)
l /5
l /5
l = longer clear span
72
37
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Summary of Direct Design Method
� Decide about sizes of slab and columns. The slab depth can
be calculated from ACI table 9.5 (c).
� Find Load on slab (wu = 1.2DL + 1.6LL)
� On given column plan of building, decide about location and
dimensions of all frames (exterior and interior)
� For a particular span of frame, find static moment (Mo =
wul2ln2/8).
73
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Summary of Direct Design Method
� Find longitudinal distribution of static moment:
� Exterior span (Mext - = 0.26Mo; M ext + = 0.52Mo; Mint - = 0.70Mo)
� Interior span (Mint - = 0.65Mo; M int + = 0.35Mo)
� Find lateral Distribution of each longitudinal moment:
� 100 % of Mext – goes to column strip
� 60 % of Mext + and Mint+ goes to column strip
� 75 % of Mint – goes to column strip
� The remaining moments goes to middle strips
� Design and apply reinforcement requirements (smax = 2hf)
74
38
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example
� Analyze the flat plate shown below using DDM. The slab supports a live
load of 144 psf. All columns are 14″ square. Take fc′ = 4 ksi and fy = 60
ksi.
� See BSc Lecture 5 for solution of this example.
25′
20′
25′ 25′ 25′
20′
20′
75
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Contents of Section-II
� General
� Various Design Options
� Example
76
39
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Punching Shear in Flat Plates
� Punching shear occurs at column support points in flat plates and flat slabs.
General
Shear crack
Punch Out
77
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Critical Section for Punching Shear
General
78
40
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Critical Section for Shear Design
� In shear design of flat plates, the critical section is an area taken at a
distance “d/2” from all face of the support.
Slab thickness (h)
Critical perimeter
d/2d/2
d = h − cover
Tributary Area, At
Column
Slab
General
79
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Punching Shear: Critical Perimeter, bo
bo = 2(c1,S + d) + 2(c1,L + d)
General
c1,S
c1,L
d/2
d/2
80
41
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Punching Shear: Critical Perimeter, bo
bo = 2(c1,S + d/2) + (c1,L + d)
General
c1,S
c1,L
d/2
d/2
81
d/2
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Punching Shear: Critical Perimeter, bo
bo = (c1,S + d/2) + (c1,L + d/2)
General
c1,S
c1,L
d/2
d/2
82
42
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Punching Shear Demand (Vu): For Square Column
General
l1
l2
Critical Perimeter, bo:
bo = 4(c + d)
Area Contributing to Load (Excluding Area of bo), At :
At = (l1 × l2) – (c + d)2 / 144
[l1 & l2 are in ft. units and c & d are in inches]
Punching Shear Demand:
Vu = Wu × At
83
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Capacity of Slab in Punching Shear:
� ΦVn = ΦVc + ΦVs
� ΦVc is least of:
� Φ4√ (fc′)bod
� Φ(2 + 4/βc) √ (fc′)bod
� Φ{(αsd/bo +2} √ (fc′)bod
βc = longer side of column/shorter side of column
αs = 40 for interior column, 30 for edge column, 20 for corner columns
General
84
43
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� When ΦVc ≥ Vu (Φ = 0.75): Nothing is required.
� When ΦVc < Vu, then we need to increase the punching shear capacity of the
slab.
� Punching shear capacity of the flat plates can be increased by either of the
following ways:
i. Increasing d ,depth of slab: This can be done by increasing the slab depth as a
whole or in the vicinity of column (Drop Panel)
ii. Increasing bo, critical shear perimeter: This can be done by increasing column size
as a whole or by increasing size of column head (Column capital)
iii. Increasing fc′ (high Strength Concrete)
Various Design Options
85
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� And/ or provide shear reinforcement (ΦVs) in the form of:
� Integral beams
� Bent Bars
� Shear heads
� Shear studs
Various Design Options
86
44
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Drop Panels (ACI 9.5.3.2 and 13.3.7.1):
Various Design Options
87
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Column Capital:
Various Design Options
• ACI 13.1.2 requires the column capital should be oriented no greater than 450 to the axis of the column.
• ACI 6.4.6 requires that the capital concrete be placed at the same time as the slab concrete. As a result, the floor forming becomes considerably more complicated and expensive.
• The increased perimeter can be computed by equating Vu to ΦVc and simplifying the resulting equation for b0
88
45
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Integral Beam and Bent Bars:
� In case of integral beam or bent bar reinforcement following must be
satisfied.
� ACI 11.12.3 requires the slab effective depth d to be at least 6 in., but not
less than 16 times the diameter of the shear reinforcement.
� When bent bars and integral beams are to be used, ACI 11.12.3 reduces
ΦVc by 2
Various Design Options
89
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Integral Beams
� Integral Beams require the design of two main components:
i. Vertical stirrups
ii. Horizontal bars radiating outward from column faces.
Various Design Options
Vertical Stirrups
Horizontal Bars
90
46
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Integral Beams (Vertical Stirrups)
Vertical stirrups are used in
conjunction with supplementary
horizontal bars radiating outward
in two perpendicular directions
from the support to form what are
termed integral beams contained
entirely within the slab thickness.
In such a way, critical perimeter
is increased
Vertical stirrups
For 4 sides, total stirrup area is 4 times
individual 2 legged stirrup area
Increased
Critical
Perimeter, b o
Various Design Options
91
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Integral Beams (Horizontal Bars)
bo = 4R + 4c
Various Design Options
• How much should be the length of the horizontal bars, lv
lv
• lv can be determined using the critical perimeter bo¾ (lv – c/2)
• Distance from the face of column to the boundary of critical perimeter = ¾ (lv – c/2)
X = ¾ (lv – c/2)
X =
¾ (
l v–
c/2)
R
c
Critical Perimeter, b o
92
47
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Integral Beams
� For Square Column of Size “c”:
� bo = 4R + 4c ........ (1)
� R = √ (X 2 + X 2 ) = √ (2) X
� Eq (1) => bo = 4√ (2) X + 4c
putting value of lv : bo = 4√ (2){(3/4)(lv – c1/2)} + 4c
after simplification, we get: bo = 4.24 lv + 1.88c
� The above equation can be used for determining the
length up to which the horizontal bars should be extended
beyond the face of column.
Various Design Options
X = ¾ (lv – c/2)
X =
¾ (
l v–
c/2)
R
c
Critical Perimeter, bo
¾ (lv – c/2)
93
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Example
� Design the flat plate as shown below for punching shear. The slab
supports a live load of 144 psf. All columns are 14″ square. Take fc′ = 4
ksi and fy = 60 ksi.
� See BSc Lecture 5 for solution of this example.
25′
20′
25′ 25′ 25′
20′
20′
94
48
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Joist System
95
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Contents
� General
� Behavior
� Characteristics
� Basic Steps for Structural Design
� Some Important Points
� Example
96
49
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� 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.
General
Joist
97
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� 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.
General
98
50
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
99
General
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� The joists are commonly formed by using standard square “dome”
forms and the domes are omitted around the columns to form the
solid heads.
Solid Head
100
General
51
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Standard Dome Data: 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 ½-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.
101
General
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Standard Dome Data
102
General
52
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� The behavior of two-way joist slab is similar to a two way flat
Slab system.
Behavior
103
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� 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.
Characteristics
104
53
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� 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.
Basic Steps for Structural Design
105
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� 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.
106
Basic Steps for Structural Design
54
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Sizes
� Minimum Joist Depth
� Equivalent Flat Slab Thickness
� ACI 318-05 – Sect. 9.5.3
107
Basic Steps for Structural Design
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Sizes
� Minimum Joist Depth
� Slab and rib depth from equivalent flat slab thickness
Table 01: Waffle flat slabs ( 19" × 19" voids at 2'-0")-Equivalent thickness
Rib + Slab Depths (in.) Equivalent Thickness te (in.)
8 + 3 8.898 + 4 ½ 10.1110 + 3 10.51
10 + 4 ½ 11.7512 + 3 12.12
12 + 4 ½ 13.3814 + 3 13.72
14 + 4 ½ 15.0216 + 3 15.31
16 + 4 ½ 16.64Reference: Table 11-2 of CRSI Design Handbook 2002.
Note: Only first two columns of the table are reproduced here.
Basic Steps for Structural Design
55
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Sizes
� Minimum Joist Depth
� Slab and rib depth from equivalent flat slab thickness
Table 02: Waffle flat slabs ( 30" × 30" voids at 3'-0")-Equivalent thickness
Rib + Slab Depths (in.) Equivalent Thickness te (in.)
8 + 3 8.618 + 4 ½ 9.7910 + 3 10.18
10 + 4 ½ 11.3712 + 3 11.74
12 + 4 ½ 12.9514 + 3 13.3
14 + 4 ½ 14.5416 + 3 14.85
16 + 4 ½ 16.1220 + 3 17.92
20 + 4 ½ 19.26Reference: Table 11-2 of CRSI Design Handbook 2002.
Note: Only first two columns of the table are reproduced here.
Basic Steps for Structural Design
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� 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 ½ 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.
Basic Steps for Structural Design
56
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� 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.
Basic Steps for Structural Design
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� 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 ½
inches).
Table 03: Standard Dome Dimensions and other Data
Dome SizeDome Depth
(inches)Volume of Void
(ft3)
Floor Dead Load (psf) per slab thickness
3 inches 4 ½ inches
30 inches
8 3.98 71 9010 4.92 80 9912 5.84 90 10914 6.74 100 11916 7.61 111 12920 9.3 132 151
19 inches
8 1.56 79 9810 1.91 91 11012 2.25 103 12214 2.58 116 13416 2.9 129 148
Reference: Table 11-1, CRSI Design Handbook 2002
Basic Steps for Structural Design
57
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Loads
� Floor dead load (wdj) for two-way joist can
also be calculated as follows:
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) × γconc
= ( 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
Basic Steps for Structural Design
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Loads
� Loads in an interior span of a two way joist system can be
calculated as follows:
Wsh = Wsj + Wdj
Wsj = Wsh - Wdj
Wsh = Dead load of solid head
Wdj = Dead load of joist
Wsj = Dead load of solid head excluding joist
wdj
ln
a abl2
a
Basic Steps for Structural Design
l1
ln
a
wsjwsj
a = b/2 – c/2 – half joist width
58
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Loads
� Factored loads can be calculated as:
� If wL = live load (load/area)
� wdj = dead load of joists, then
� Factored load due to joists (wj)
wuj = 1.2 wdj + 1.6wL
� Factored load due to wsj
wusj = 1.2 wsj
Where, wsj = (wdsh –wdj)
1.2 wsj
lna a
b
1.2 wsj
l2
Basic Steps for Structural Design
ln
wuj = 1.2 wdj + 1.6wL
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� 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).
116
Basic Steps for Structural Design
59
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Analysis
� Static moment calculation for DDM analysis:
117
Basic Steps for Structural Design
wuj
ln
Moj = wujl2ln2/8
wusj
lna a
wusj
Mosj = wusjba2/2
MojMosj
Mo = Moj + Mosj
bl2
ln
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Design
� Design for Flexure
� The design of waffle slab for flexure is done like solid slab design.
118
Basic Steps for Structural Design
60
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
Basic Steps for Structural Design
� Design
� Placement of Flexural Reinforcement
For Main Positive Reinforcement: Asrequired by flexural demand but ≥ ACI7.12 (shrinkage reinf.)
For Main Negative Reinforcement:As required by flexural demand but ≥ACI 7.12 (shrinkage reinf.), with maxspacing least of 2h or 18”
Slab Reinforcement (located atmid depth): As required by flexuraldemand but ≥ ACI 7.12 (shrinkagereinf.), with max spacing least of 5hor 18”
119
Each joist rib contains two bottom bars.Straight bars are supplied over the columncenterlines for negative factored moment.
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Design
� Design 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.
120
Basic Steps for Structural Design
61
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
121
� Design
� Design for Beam Shear
� 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.
� 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
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� 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.
122
Some Important Points
62
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� The designer should sketch out the spacing for a typical panel
and correlate with the column spacing as a part of the early
planning.
123
Some Important Points
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� 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
Example
124
63
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� 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.
125
Example
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� 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″
126
Example
64
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� 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 ½ in. for a dome size of
30-in.
Table: Waffle flat slabs (30" × 30" voids at 3'-0")-Equivalent thickness
Rib + Slab Depths (in.) Equivalent Thickness te (in.)
8 + 3 8.618 + 4 ½ 9.7910 + 3 10.18
10 + 4 ½ 11.3712 + 3 11.74
12 + 4 ½ 12.9514 + 3 13.3
14 + 4 ½ 14.5416 + 3 14.85
16 + 4 ½ 16.1220 + 3 17.92
20 + 4 ½ 19.26
127
Example
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 01: Sizes
� Planning of Joist layout
l = 36′-0″ = 432″Standard module = 36″ × 36″
No. of modules in 36′-0″:n = 432/36 = 12
Joist width = 6”
Planning:First joist is placed on interior columncenterline with progressive placing ofother joists towards exterior ends ofpanel. To flush the last joist with externalcolumn, the width of exterior joist comesout to be 15″ (6″+Column size /2) asshown in plan view.
128
Example
65
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� 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 width 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′′
129
Example
c/c between ribs= 4x3 = 12′
Rib width= 3+3 inches
Total = 12′ +0.5′=12.5′
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 02: Loads
� Floor (joist) dead load (wdj) = 109 psf = 0.109 ksf
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 ½ 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 148Reference: Table 11-1, CRSI Design Handbook 2002
130
Example
66
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 02: Loads
� Floor (joist) dead load (wdj) = 109 psf = 0.109 ksf
� Solid head dead load wsh = γchsh = 0.15 ×{(12 + 4.5)/12 = 0.20625
� Solid Head dead load excluding joist (wsj) = wsh – wdj
= 0.20625 – 0.109 = 0.097 ksf
Example
wdj
ln
a a
wsj wsj
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� 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 + 1.6×0.100
= 0.291 ksf
� Load due to solid head (wush)
wush = 1.2wsj
= 1.2 × 0.097 = 0.1164 ksf
Example
1.2 wsj
lna a
1.2 wsj
wuj = 1.2 wdj + 1.6wL
67
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 03: Frame Analysis (E-W Interior Frame)
� Marking E-W Interior Frame:
l2 = 36′
133
Example
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 03: Frame Analysis (E-W Interior Frame)
� Marking of Column and Middle Strips:
CS/2 = Least of l1/4 or l2/4
l2/4 = 36/4 = 9′
l2 = 36′CS/2 = 9′
MS/2 = 9′
MS/2 = 9′CS/2 = 9′
134
Example
68
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 03: Frame Analysis (E-W Interior Frame)
� Static Moment Calculation
135
Example
wuj
ln
Moj = wujl2ln2/8
wusj
lna a
wusj
Mosj = wusjba2/2
MojMosj
Mo = Moj + Mosj
a = 12.5/2 - 1.5/2 - 0.25 = 5.25 ft
b = 12.5 ft
bl2
l1
ln
a a
b
a
b
Solid head
a = b/2 – c/2 – half joist width
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 03: Frame Analysis (E-W Interior Frame)
� Static Moment Calculation
� Moj (due to joists) = wojl2ln2/8
= 0.291 × 36 × 34.52/8 = 1557.56 ft-kip
Mosj (moment due to solid head excluding joists) = wusj ba2/2
= 0.1164×12.5×5.252/2 = 20 ft-kip
Mo (total static moment) = Moj + Mosj = 1557.56 + 20 = 1577.56 ft-kip
Note: Since normally Mosj is much smaller than Moj , the former can be
conveniently ignored in design calculations.
136
Example
69
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 03: Frame Analysis (E-W Interior Frame)
� Static Moment Calculation.
l2 = 36′
137
Example
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 03: Frame Analysis (E-W Interior Frame)
� Longitudinal distribution of Total static moment (Mo).
l2 = 36′0.26
0.52
0.70 0.65
0.35
0.65 0.70 0.26
0.52
138
Example
70
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 03: Frame Analysis (E-W Interior Frame)
� Longitudinal distribution of Total static moment (Mo).
l2 = 36′410
820
1104 1025
552
1025 1104 410
820
Units: ft-kip
ML = Mo × (D.F)L
139
Example
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 03: Frame Analysis (E-W Interior Frame)
� Lateral Distribution of Longitudinal moment (L.M).
l2 = 36′0.60
1.00 0.75 0.75 0.75
0.60 0.60
1.000.75
140
Example
71
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 03: Frame Analysis (E-W Interior Frame)
� Lateral Distribution of Longitudinal moment (L.M).
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
l2 = 36′492
410 828 769 769
331 492
410828
328/2 328/2
328/2328/2
0 0
00
276/2
276/2
276/2
276/2
221/2
221/2
256/2
256/2
256/2
256/2
141
Example
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 03: Frame Analysis (E-W Interior Frame)
� Lateral Distribution of Longitudinal moment (L.M).
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
l2 = 36′27.3
22.8 46 42.7 42.7
18.38 27.3
22.846
18.22 18.22
18.2218.22
0 0
00
15.33
15.33
15.33
15.33
12.3
12.3
14.22
14.22
14.22
14.22
142
Example
72
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 03: Frame Analysis (E-W Exterior Frame)
� Static Moment Calculation
� Moj (due to joists) = wojl2ln2/8
= 0.291 × 18.75 × 34.52/8 = 811.78 ft-kip
Mosj (moment due to solid head excluding joists) = wusj ba2/2
= 0.1164×7×5.252/2 = 12.83 ft-kip
Mo (total static moment) = Moj + Mosj = 811.78 + 12.83 = 825 ft-kip
l2 = 18 + c/2= 18 + (18/12)/2= 18.75′
c = column dimension
143
Example
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 03: Frame Analysis (E-W Exterior Frame)
� Static Moment Calculation.
l2 = 18.75′
144
Example
73
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 03: Frame Analysis (E-W Exterior Frame)
� Longitudinal distribution of Total static moment (Mo).
0.260.52
0.70 0.650.35
0.65 0.70 0.260.52
145
Example
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 03: Frame Analysis (E-W Exterior Frame)
� Longitudinal distribution of Total static moment (Mo).
ML = Mo × (D.F)L
215429
578 536289
536 578 215429
Units: ft-kip
146
Example
74
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 03: Frame Analysis (E-W Exterior Frame)
� Lateral Distribution of Longitudinal moment (L.M).
0.601.00 0.75 0.75 0.75
0.60 0.601.000.75
147
Example
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 03: Frame Analysis (E-W Exterior Frame)
� Lateral Distribution of Longitudinal moment (L.M).
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
215 434 402 769 215402
172172 00 144 144116134 134
257 173 257
148
Example
75
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 03: Frame Analysis (E-W Exterior Frame)
� Lateral Distribution of Longitudinal moment (L.M).
ML,ext- = 215 kip-ft
ML,ext+ = 429 kip-ft
ML,int- = 578 kip-ft
ML,- = 536 kip-ft
ML,+ = 289 kip-ft
23.8 48.2 44.7 44.7 23.848.2
17.6417.64 00 14.76 14.7611.8913.74 13.74
28.6 19.2 28.6
MLat per foot = Mlat/strip width
149
Example
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� 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.
150
Example
76
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 04: Design
� E-W Interior Slab Strip:
l2 = 36′27.3
22.8 46 42.7 42.7
18.38 27.3
22.846
18.22 18.22
18.2218.22
0 0
00
15.33
15.33
15.33
15.33
12.3
12.3
14.22
14.22
14.22
14.22
151
Example
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 04: Design
� E-W Exterior Frame.
23.8 48.2 23.848.2
17.6417.64 00 14.76 14.7611.89
28.6 19.2 28.6
152
Example
77
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� 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) + ¾ inch (Assumed bar diameter) =
15 inch
� This will be used for both directions positive as well as negative
reinforcement.
153
Example
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 05: Detailing (E-W Frames)
#6 @ 12″ #6 @ 6″ #6 @ 6″ #6 @ 12″
#6 @ 12″ #6 @ 6″ #6 @ 6″ #6 @ 12″
#6 @ 18″ #6 @ 18″ #6 @ 18″ #6 @ 18″
• Negative Reinforcement
• M = 46 ft-k =46 x12 = 552 in-k
• d = 15
• fy = 60 ksi
• As = 0.7 in2
• S= 0.44/0.7 x 12 = 7.2 inch
• Finally using #6 @ 6” c/c & 12” c/c in
the column strips at all interior &
exterior supports respectively.
• in the middle strip use #6 @ 18” c/c.
154
Example
78
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 05: Detailing (E-W 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 in the column as well as middle strips.
155
Example
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 05: Detailing (E-W Interior Frame)
2 #7 Bars
Column Strip at interior support
Column Strip at exterior support
18′-0″
#6 @ 12″ c/c
#6 @ 6″ c/c
2 #7 Bars
156
Example
79
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 05: Detailing (E-W Interior Frame)
Middle Strip at Exterior support
Middle Strip at interior support
#6 @ 18″ c/c
#6 @ 18″ c/c
2 #7 Bars
18′-0″
2 #7 Bars157
Example
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 05: Detailing (E-W Exterior Frame)
9′-0″
Column Strip at interior support
#6 @ 6″ c/c
2 #7 Bars
Column Strip at exterior support
#6 @ 12″ c/c
2 #7 Bars
158
Example
80
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 05: Detailing (E-W Exterior Frame)
9′-0″
Middle Strip at interior support
#6 @ 18″ c/c
2 #7 Bars
Middle Strip at exterior support
#6 @ 18″ c/c
2 #7 Bars
159
Example
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
� Step No 04: Design
� Note: For the completion of design problem, the waffle slab should
also be checked for beam shear and punching shear.
160
Example
81
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
References
� ACI 318-02
� CRSI Design Handbook
� Design of Concrete Structures by Nilson, Darwin and Dolan [13th
Ed]
161
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures
Department of Civil Engineering, University of Engineering and Technology Peshawar
The End
162