Lecture-06 Analysis and Design of Slab Systems

1

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

UET [email protected]

www.drqaisarali.com

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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



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.


� This topic will be covered in this lecture.

3




� 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




� 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





� This topic will be discussed in this lecture.

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� Therefore in this lecture, only the following topics will be

discussed.


� Analysis & design of two way slab system without beams (flat

plate & flat slabs)


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Analysis & design of one way

joist system

8

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Contents

� General

� Characteristics

� Basic Steps

� Serviceability Requirements

� Example

9



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

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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



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



Characteristics

� Suitable for long spans

� Economical range: 30 ft – 50 ft

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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.

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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.

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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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9




� 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.

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� 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″.

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� 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




� Analysis:

� For the purpose of analysis, a single T shaped joist is

considered.

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Uniformly distributed load (L/D ≤ 3)

≤ 1.2ln ln

Two or more spans

Prismaticmembers


� Analysis:

� ACI approximate analysis is applicable.

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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).


� Analysis:

22

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� Design:

� Design of joist for flexure:

� Design of one way joist is just like design of a beam.

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� 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.

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� Design:


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.)

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� Design:


� 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.


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� 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)

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� 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.

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� Design:


� 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.,

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� Design:


� 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.

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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

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Example

Plan View

Section A-A

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17



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.

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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″

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18



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

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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

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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

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Example

� Solution:



� 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)

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20



Example

� Solution:



� 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)

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Example

� Solution:



� 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)

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Example

� Solution:


� 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′

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Example

� Solution:


� 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.

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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″

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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

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Analysis and Design of Two-way Slab

System without Beams

(Flat Plates and Flat Slabs)

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� 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)

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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

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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

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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



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.

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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)

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Steps in DDM

� Step 01: Sizes

� ACI table 9.5 (c) are used for finding the slab thickness.

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Steps in DDM

� Step 02: Loads

� The slab load is calculated in usual manner.

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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.

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Steps in DDM


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)

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b) Marking Middle Strip (For Interior Frame)

Steps in DDM


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)

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Steps in DDM



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:

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Steps in DDM


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

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Steps in DDM


� Step III: Calculate Static Moment (Mo) for interior span of frame.

Mo =wu l2 ln

2

8 Mo

l2ln

Span of frame

59



Steps in DDM


� Step IV: Longitudinal Distribution of Static Moment (Mo).

M+

M − M −

M − = 0.65Mo

M + = 0.35Mo

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Steps in DDM


� Step V: Lateral Distribution to column and middle strips.

M − = 0.65Mo

M + = 0.35Mo

0.60M +

0.75M − 0.75M −

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Steps in DDM


� Step VI: Calculate Static Moment (Mo) for exterior span of frame.

Mo =wu l2 ln

2

8 Mo

l2ln

Span of frame

62

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Steps in DDM


� Step VII: Longitudinal distribution of static moment (Mo).

Mext+

Mext − M int−

Mext − = 0.26Mo

M ext+ = 0.52Mo

Mint- = 0.70Mo

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Steps in DDM


� 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−

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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



� 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

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� 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




� 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

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� 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




� 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




� Standard Bar Cut off Points (Practical Recommendation) for

column and middle strips both.

71




� 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



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).

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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)

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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



Contents of Section-II

� General

� Various Design Options

� Example

76

39



� Punching Shear in Flat Plates

� Punching shear occurs at column support points in flat plates and flat slabs.

General

Shear crack

Punch Out

77



� Critical Section for Punching Shear

General

78

40



� 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



� 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




bo = 2(c1,S + d/2) + (c1,L + d)

General

c1,S

c1,L

d/2

d/2

81

d/2




bo = (c1,S + d/2) + (c1,L + d/2)

General

c1,S

c1,L

d/2

d/2

82

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� 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



� 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

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� 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



� And/ or provide shear reinforcement (ΦVs) in the form of:

� Integral beams

� Bent Bars

� Shear heads

� Shear studs


86

44



� Drop Panels (ACI 9.5.3.2 and 13.3.7.1):


87



� Column Capital:


• 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

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� 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


89



� Integral Beams

� Integral Beams require the design of two main components:

i. Vertical stirrups

ii. Horizontal bars radiating outward from column faces.


Vertical Stirrups

Horizontal Bars

90

46



� 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


91



� Integral Beams (Horizontal Bars)

bo = 4R + 4c


• 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

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� 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.


X = ¾ (lv – c/2)

X =

¾ (

l v–

c/2)

R

c

Critical Perimeter, bo

¾ (lv – c/2)

93



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



Two-Way Joist System

95



Contents

� General

� Behavior

� Characteristics

� Basic Steps for Structural Design

� Some Important Points

� Example

96

49



� 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



� 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



99

General



� 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



� 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



� Standard Dome Data

102

General

52



� The behavior of two-way joist slab is similar to a two way flat

Slab system.

Behavior

103



� 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



� 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.


105



� 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


54



� Sizes


� Equivalent Flat Slab Thickness

� ACI 318-05 – Sect. 9.5.3

107




� Sizes


� 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.


55



� Sizes


� Slab and rib depth from equivalent flat slab thickness

Table 02: Waffle flat slabs (30" × 30" voids at 3'-0")-Equivalent thickness


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.




� 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.


56



� 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.




� 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


57



� 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




� 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


l1

ln

a

wsjwsj

a = b/2 – c/2 – half joist width

58



� 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


ln

wuj = 1.2 wdj + 1.6wL



� 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


59



� Analysis

� Static moment calculation for DDM analysis:

117


wuj

ln

Moj = wujl2ln2/8

wusj

lna a

wusj

Mosj = wusjba2/2

MojMosj

Mo = Moj + Mosj

bl2

ln



� Design

� Design for Flexure

� The design of waffle slab for flexure is done like solid slab design.

118


60




� 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.



� 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


61



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



� 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



� 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



� 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



� 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




� 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




� 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


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




� 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




� 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′




� 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




� 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




� 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



� Step No 03: Frame Analysis (E-W Interior Frame)

� Marking E-W Interior Frame:

l2 = 36′

133

Example




� Marking of Column and Middle Strips:


l2/4 = 36/4 = 9′

l2 = 36′CS/2 = 9′

MS/2 = 9′

MS/2 = 9′CS/2 = 9′

134

Example

68




� 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





� 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




� Static Moment Calculation.

l2 = 36′

137

Example




� 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





l2 = 36′410

820

1104 1025

552

1025 1104 410

820

Units: ft-kip

ML = Mo × (D.F)L

139

Example




� 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





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





MLat per foot = Mlat/strip width




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



� Step No 03: Frame Analysis (E-W Exterior Frame)


� 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




� Static Moment Calculation.

l2 = 18.75′

144

Example

73





0.260.52

0.70 0.650.35

0.65 0.70 0.260.52

145

Example





ML = Mo × (D.F)L

215429

578 536289

536 578 215429

Units: ft-kip

146

Example

74





0.601.00 0.75 0.75 0.75

0.60 0.601.000.75

147

Example





MLat = ML × (D.F)Lat




ML,- = 536 kip-ft

ML,+ = 289 kip-ft

215 434 402 769 215402

172172 00 144 144116134 134

257 173 257

148

Example

75








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



� 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




� 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




� 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




� 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



� 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



� 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



� 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



� 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



� 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



� 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




� Note: For the completion of design problem, the waffle slab should

also be checked for beam shear and punching shear.

160

Example

81



References

� ACI 318-02

� CRSI Design Handbook

� Design of Concrete Structures by Nilson, Darwin and Dolan [13th

Ed]

161



The End

162

Lecture-06 Analysis and Design of Slab Systems

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Transcript of Lecture-06 Analysis and Design of Slab Systems