Lecture -11 Analysis and Design of Two-Way Slab Systems (Two-Way Slab With Beams & Two Way Joist...

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Lecture-11Analysis and Design of Two-way Slab Systems

(Two-way Slab with Beams & Two Way joist Slabs)

B P f D Q i Ali

Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 1

By: Prof Dr. Qaisar Ali

Civil Engineering Department

NWFP UET [email protected]


Topics AddressedMoment Coefficient Method for Two way slab withbbeams

Introduction

Cases

Moment Coefficient Tables

R i f t R i t

Prof. Dr. Qaisar Ali

Reinforcement Requirements

Steps

Example

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Topics AddressedTwo-way Joist Slab

Introduction

Behavior

Characteristics

Basic Steps for Structural Design


Example

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Moment Coefficient Method (Introduction)

Two Way Slabs

The Moment Coefficient Method included for the first time in1963 ACI Code is applicable to two-way slabs supported onfour sides of each slab panel by walls, steel beams relativelydeep, stiff, edge beams (h = 3hf).

Although, not included in 1977 and later versions of ACI code,


its continued use is permissible under the ACI 318-08 codeprovision (13.5.1). Visit ACI 13.5.1.

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Moment Coefficient Method la

Ma,neg

Ma,pos

Two Way Slabs

Moments:Ma, neg = Ca, negwula2

Mb, neg = Cb, negwulb2

Ma, pos, (dl + ll) = M a, pos, dl + M a, pos, ll = Ca, pos, dl × wu, dl × la2 + Ca, pos, ll × wu, ll × la2

Mb, pos, (dl + ll) = Mb, pos, dl + Mb, pos, ll = Cb, pos, dl × wu, dl × lb2 + Cb, pos, ll × wu, ll × lb2

Where C C = Tabulated moment coefficients

Ma,neglb

Mb,neg Mb,negMb,pos


Where Ca, Cb = Tabulated moment coefficients

wu = Ultimate uniform load, psf

la, lb = length of clear spans in short and long directions

respectively.

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Moment Coefficient Method: Cases

Two Way Slabs

Moment Coefficient Method: CasesDepending on the support conditions, several cases are possible:

Prof. Dr. Qaisar Ali 6

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Two Way Slabs





Two Way Slabs



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Two Way Slabs




Two Way Slabs

Moment Coefficient Tables:Moment Coefficient Tables:


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Two Way Slabs




Two Way Slabs



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Two Way Slabs




Two Way Slabs



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Two Way Slabs




Two Way Slabs

Load Coefficient Table:Load Coefficient Table:


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Maximum spacing and minimum reinforcement

Two Way Slabs

requirement:

Maximum spacing (ACI 13.3.2):

smax = 2 hf in each direction.

Minimum Reinforcement (ACI 7.12.2.1):

Asmin = 0.0018 b hf for grade 60.


Asmin 0.0018 b hf for grade 60.

Asmin = 0.002 b hf for grade 40 and 50.


Special Reinforcement at exterior corner of SlabThe reinforcement at exterior ends of the slab shall be provided as per ACI

Two Way Slabs

The reinforcement at exterior ends of the slab shall be provided as per ACI13.3.6 in top and bottom layers as shown.

The positive and negative reinforcement in any case, should be of a size andspacing equivalent to that required for the maximum positive moment (per footof width) in the panel.


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Moment Coefficient Method

Two Way Slabs

Steps

Find hmin = perimeter/ 180 = 2(la + lb)/180

Calculate loads on slab (force / area)

Calculate m = la/ lb

Decide about case of slab,


Decide about case of slab,

Use table to pick moment coefficients,

Calculate moments and then design.

Apply reinforcement requirements (smax = 2hf, ACI 13.3.2)

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Moment Coefficient Method: Example

Two Way Slabs

o e t Coe c e t et od a p e

A 100′ × 60′, 3-storey commercial building is to be designed.The grids of column plan are fixed by the architect.


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Two Way Slabs

o e t Coe c e t et od a p e

Complete analysis of the slab is done by analyzing four panels

Panel I Panel IPanel III Panel III


Panel I Panel I

Panel II Panel II

Panel III Panel III

Panel IV Panel IV



Two Way Slabs

p

A 100′ × 60′, 3-storey commercial building: Sizes and Loads.Sizes:

Minimum slab thickness = perimeter/180 = 2 (20+25)/180 = 6″

However, for the purpose of comparison, take hf = 7″

Columns = 14″ × 14″ (assumed)


Beams = 14″ × 20″ (assumed)

Loads:

S.D.L = Nil ; Self Weight = 0.15 x (7/12) = 0.0875 ksf

L.L = 144 psf ; wu = 0.336 ksf

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Two Way Slabs

Panels are analyzed using Moment Coefficient MethodCase = 4m = la/lb = 0.8

Mb,neg Mb,negMb,pos

Ma,neg

Ma,pos

Ma,neg




Two Way Slabs


Ca,neg = 0.071Cb,neg = 0.029Ca,posLL = 0.048Cb,posLL = 0.020C 0 039


Ca,posDL = 0.039Cb,posDL = 0.016

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Two Way Slabs







Two Way Slabs





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Two Way Slabs







Two Way Slabs





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Two Way Slabs







Two Way Slabs

Panels are analyzed using Moment Coefficient Method

Panel I

Case = 4m = la/lb = 0.8


Mb,neg Mb,negMb,pos

Ma,neg

Ma,pos

Ma,neg



Ma,neg = 9.5 k-ftMa,pos = 6.1 k-ftMb,neg = 6.1 k-ftMb,pos = 3.9 k-ft For slab supported on Spandrals, Mneg,ext = 1/3Mpos

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Two Way SlabsMoment Coefficient Method: Example

Panel II

Case = 9m = la/lb = 0.8

Ca,neg = 0.075Cb,neg = 0.017Ca,posLL = 0.042Cb,posLL = 0.017C 0 029 M MM

Ma,neg




Ma,neg = 10.1 k-ftMa,pos = 5.1 k-ftMb,neg = 3.6 k-ftMb,pos = 3.1 k-ft

Mb,neg Mb,negMb,pos

Ma,pos

Ma,neg



Panel III

Case = 8m = la/lb = 0.8


Mb,neg Mb,negMb,pos

Ma,neg

Ma,pos

Ma,neg





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Panel IV

Case = 2m = la/lb = 0.8

Ca,neg = 0.065Cb,neg = 0.027Ca,posLL = 0.041Cb,posLL = 0.017C 0 026 M MM

Ma,neg





Mb,neg Mb,negMb,pos

Ma,pos

Ma,neg



Slab analysis summary

8.77.4

7.4

8.6 8.65.43.7

10.19.5

9.5

6.1 6.13.9

6.1


3.2

4.95.75.7

8.7

3.25.13.6

10.1

3.6

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Slab Reinforcement Details

A

C

C

C CBA

A

C

C

B BA

A= #4 @ 12″


A

BBB

C

A

BA

C

@B = #4 @ 6″C = #4 @ 4″

Two-Way Joist Slab

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Introduction

Two-Way Joist

A two-way joist system, or waffle slab, comprises evenlyspaced concrete joists spanning in both directions and areinforced concrete slab cast integrally with the joists.

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Joist

Introduction

Two-Way Joist

Like one-way joist system, a two way system will be qualifiedto be said as two-way joist system if clear spacing betweenribs (dome width) does not exceed 30 in.

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Introduction

Two-Way Joist

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Introduction

Two-Way Joist

The joists are commonly formed by using Standard Square“dome” forms and the domes are omitted around the columnsto form the solid heads.


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Introduction

Standard Dome Data

Two-Way Joist

Standard Dome Data

Generally the dome for waffle slab can be of any size. However thecommonly used standard domes are discussed as follows:

30-in × 30-in square domes with 3-inch flanges; from which 6-inchwide joist ribs at 36-inch centers are formed: these are available instandard depths of 8, 10, 12, 14, 16 and 20 inches.

19 i h 19 i h d ith 2 ½ i h fl f hi h


19-inch × 19-inch square domes with 2 ½-inch flanges, from which5-inch wide joist ribs at 24-inch centers are formed. These areavailable in standard depths of 8, 10, 12, 14 and 16 inches.

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Introduction

Standard Dome Data

Two-Way Joist

Standard Dome Data


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Behavior

Two-Way Joist

The behavior of two-way joist slab is similar to a two way flatSlab system.



Characteristics

Two-Way Joist

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 ofdeepened drop panels or supported beams.

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Basic Steps for Structural Design

Step No 01 (Sizes): Sizes of all structural and non

Two-Way Joist

Step No. 01 (Sizes): Sizes of all structural and nonstructural elements are decided.

Step No. 02 (Loads): Loads on structure are determinedbased on occupational characteristics and functionality (referAppendix C of class notes).

Step No 03 (Analysis): Effect of loads are calculated on all


Step No. 03 (Analysis): Effect of loads are calculated on allstructural elements.

Step No. 04 (Design): Structural elements are designed for

the respective load effects following code provisions.

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Sizes

Minimum Joist Depth

Two-Way Joist

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 thencomputed by equalizing the moment of inertia of equivalent flat slab tothat of waffle slab


that of waffle slab.

However since this practice is time consuming, tables have beendeveloped to determine the size of waffle slab from equivalent flat slabthickness.

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Sizes

Minimum Joist Depth

Two-Way Joist

Minimum Joist Depth

Equivalent Flat Slab Thickness

ACI 318-05 – Sect. 9.5.3

Minimum thickness = ln/33


Sizes

Minimum Joist Depth

Two-Way Joist

Minimum Joist Depth

Slab and rib depth from equivalent flat slab thickness

Table 01: Waffle flat slabs (19" × 19" voids at 2'-0")-Equivalent thicknessRib + Slab Depths (in.) Equivalent Thickness te (in.)

8 + 3 8.898 + 4 ½ 10.1110 + 3 10.51

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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.

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Sizes

Minimum Joist Depth

Two-Way Joist

Minimum Joist Depth

Slab and rib depth from equivalent flat slab thickness

Table 02: Waffle flat slabs (30" × 30" voids at 3'-0")-Equivalent thicknessRib + Slab Depths (in.) Equivalent Thickness te (in.)

8 + 3 8.618 + 4 ½ 9.7910 + 3 10.18

10 + 4 ½ 11.37

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12 + 3 11.7412 + 4 ½ 12.9514 + 3 13.3

14 + 4 ½ 14.5416 + 3 14.85

16 + 4 ½ 16.1220 + 3 17.92



Sizes

Minimum Width of Rib

Two-Way Joist

Minimum Width of Rib

ACI 8.11.2 states that ribs shall be not less than 4 inch in width.

Maximum Depth of Rib

A rib shall have a depth of not more than 3 ½ times the minimumwidth of rib.

Minimum Slab Thickness

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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 in.

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LoadsFloor dead load for two-way joist with certain dome size, dome depth can

Two-Way Joist

y j , pbe calculated from the table shown for two options of slab thicknesses (3inches and 4 ½ inches).

Table 03: 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

8 3.98 71 9010 4 92 80 99

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30-in

10 4.92 80 9912 5.84 90 10914 6.74 100 11916 7.61 111 12920 9.3 132 151

19-in

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

Loads

Floor dead load (w ) for two way joist can also be

Two-Way Joist

Floor dead load (wdj) for two-way joist can also becalculated 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:

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Total Load of joists per dome:wdj = (Vsolid – Vvoid) × γconc

= ( 8.24 – 4.166) × 0.15 = 0.61 kip

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.

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Loads

At locations where solid head is present the floor dead load

Two-Way Joist

At locations where solid head is present, the floor dead loadcan be calculated as follows:

If, wdj = dead load in joist area

Wsh = dead load in solid head area

= hsolid × γconc

Wdj+sh = {wshb + wdj(l2-b)}/l2

wdjWdj+sh

ln

a a

Wdj+sh

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dj+sh { sh dj( 2 )} 2

bl2

a a

Loads

Factored loads can be calculated as:

Two-Way Joist

Factored loads can be calculated as:

If wL = live load (load/area), then

Load out of solid head region

wosh = 1.2 wdj + 1.6wL

Load in solid head region

1 2 1 6

wish wish

woshWish

ln

a a

Wish

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wish = 1.2wdj+sh+1.6wL bl2

a awosh

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Analysis

ACI code allows use of DDM for analysis of waffle slabs (ACI

Two-Way Joist

ACI code allows use of DDM for analysis of waffle slabs (ACIR13.1). In such a case, waffle slabs are considered as flatslabs, with the solid head acting as drop panels (ACI 13.1.3).



Analysis

Static moment calculation for DDM analysis:

Two-Way Joist

Static moment calculation for DDM analysis:

wosh

ln

woshWish

lna a

Wish

Mosh Mish

ln


Mosh = woshl2ln2/8 Mish = (wish-wosh)ba2/2

Mish

Mo = Mosh + Mish

b

l2

a a

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Design

Design of slab for punching shear

Two-Way Joist

Design of slab for punching shear

The solid head shall be checked against punching shear.

The critical section for punching shear is taken at a section d/2 from faceof the column, where d is the effective depth at solid head.



Design

Design of slab for

Two-Way Joist

Design of slab forpunching shear

Load on tributary area willcause punch out shear.

Within tributary area, twotypes of loads are acting:

l1


Solid head load

Joist load

Both types shall beconsidered while calculatingpunching shear demand

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l2 d/2

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Design

Design of slab for punching

Two-Way Joist

Design of slab for punchingshear

Total area = l1 × l2

Solid area = Asolid

Joist part area (Aj) = (l1×l2) -Asolid

Critical perimeter area = Acp

l1


Critical perimeter area Acp

Vu =Aj×wosh+ (Asolid – Acp) × wish

Where,

wosh = joist part load

wish = load inside solid head

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l2 d/2


DesignShear Strength of Slab in punching shear:

Two-Way Joist

Shear Strength of Slab in punching shear:

ΦVn = ΦVc + ΦVs

ΦVc is least of:

Φ4√ (fc′)bod

(2 + 4/βc) √ (fc′)bod

{(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

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DesignDesign of Joist for Beam Shear:

Two-Way Joist

Design of Joist for Beam Shear:

Beam shear Demand

Beam shear is not usually a problem in slabs including waffle slabs.However for completion of design beam shear may also bechecked. Beam shear can cause problem in case where largerspans and heavier loads with relatively shallow waffle slabs areused.


The critical section for beam shear is taken at a section d from faceof the column, where d is the effective depth at solid head.


DesignDesign of Joist for Beam Shear:

Two-Way Joist

g

Beam shear capacity of concrete joist

ΦVn = ΦVc + ΦVs

ΦVc is least of:

Φ2√ (fc′)bribd

ΦVs = ΦAvfy/bribs

Stirrup


If required, one or two single legged stirrups are provided in the rib to increase the shear capacity of waffle slab.

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Design

Design for Flexure

Two-Way Joist

Design for Flexure

The design of waffle slab is done by usual procedures.

However, certain reinforcement requirements apply discussed next.



ACI recommendations on reinforcement requirement of waffle slab:

Two-Way Joist

requirement of waffle slab:

ACI 10.6.7 states that if the effective depth d of a beam orjoist exceeds 36 in., longitudinal skin reinforcement shall beprovided as per ACI section 10.6.7.

According to ACI 13.3.2, for cellular or ribbed constructionreinforcement shall not be less than the requirements of ACI


reinforcement shall not be less than the requirements of ACI7.12.

As per ACI 7.12, Spacing of top bars cannot exceed 5h or18 inches.

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ACI recommendations on reinforcement requirement of waffle slab:

Two-Way Joist

requirement of waffle slab:



Other important points:

The amount of reinforcement and if necessary the top slab

Two-Way Joist

The amount of reinforcement and, if necessary, the top slabthickness can be changed to vary the load capacities fordifferent spans, areas, or floors of a structure.

Each joist rib contains two bottom bars. Straight bars aresupplied over the column centerlines for negative factoredmoment.


Bottom bar

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Other important points:

For layouts that do not meet the standard 2 feet and 3 feet

Two-Way Joist

For layouts that do not meet the standard 2-feet and 3-feetmodules, it is preferable that the required additional width beobtained by increasing the width of the ribs framing into thesolid column head.

The designer should sketch out the spacing for a typical paneland correlate with the column spacing as a part of the early


p g p yplanning.

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Example: Design the slab system of hall shown in figure as waffleslab, according to ACI 318. Use Direct Design Method for slab

Two-Way Joist

analysis.fc′ = 4 ksi

fy = 60 ksi

Live load = 100 psf


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Solution:A 108′ × 144′ building divided into twelve (12) panels supported at

Two-Way Joist

A 108 × 144 building, divided into twelve (12) panels, supported attheir ends on columns. Each panel is 36′ × 36′.

The given slab system satisfies all the necessary limitations for DirectDesign Method to be applicable.



Step No 01: Sizes

Columns

Two-Way Joist

Columns

Let all columns be 18″ × 18″.

Slab

Adopt 30″ × 30″ standard dome.

Minimum equivalent flat slab thickness (hf) can be found using ACI Table9 5 (c):


9.5 (c):

Exterior panel governs. Therefore,

hf = ln/33

= [{36 – (2 × 18/2)/12}/33] × 12 = 12.45″

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Step No 01: Sizes

Slab

Two-Way Joist

Slab

The closest depth of doom that will fulfill the requirement of equivalentthickness of flat slab equal to 12.45″ is 12 in. with a slab thickness of 4 ½in. 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 9.79


10 + 3 10.1810 + 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




Step No 01: Sizes

Planning of Joist layout

Two-Way Joistl = 36′-0″ = 432″Standard module = 36″ × 36″

Planning of Joist layout No. of modules in 36′-0″:n = 432/36 = 12

Planning:First module is placed on interiorcolumn centerline and providedtowards exterior ends of panel.In this way, width of exterior joistcomes out to be 15″.


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Step No 01: Sizes

Solid Head

Two-Way Joist

Solid Head

Solid head dimension from column centerline = l/6 = 36/6 = 6′

Total length of solid head= 2 × 6 = 12′

As 3′ × 3′ module is selected, therefore 4 voids will make an interior solidhead of 12.5′ × 12.5′.

Depth of the solid head = Depth of standard module = 12 + 4.5 = 16.5′



Step No 02: LoadsFloor (joist) dead load (wdj) = 109 psf = 0 109 ksf

Two-Way Joist

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 9010 4.92 80 9912 5.84 90 10914 6.74 100 11916 61 111 129


16 7.61 111 12920 9.3 132 151

19-in

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

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Step No 02: LoadsFloor (joist) dead load (wdj) = 109 psf = 0 109 ksf

Two-Way Joist

Floor (joist) dead load (wdj) 109 psf 0.109 ksf

Solid Head dead load (wsh) = {(12 + 4.5)/12} × 0.15 = 0.206 ksf

Wdj+sh = {wshb + wdj(l2-b)}/l2

= {0.206×12.5 + 0.109 (36 – 12.5)}/36

= 0.143 ksf

wdjWdj+sh

l

a a

Wdj+sh

75

ln

b = 12.5′l2

a = 5.25′ a

Step No 02: Loads

w = 100 psf = 0 100 ksf

Two-Way Joist

wL = 100 psf = 0.100 ksf

Load out of solid head region

wosh = 1.2 wdj + 1.6wL

= 1.2×0.109 + 1.6×0.100

= 0.291 ksfwish wish

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Load in solid head region

wish = 1.2wdj+sh+1.6wL

= 1.2 × 0.143 + 1.6 × 0.100 = 0.33 ksf

bl2

a awosh


Two Way JoistStep No 03: Frame Analysis (E-W Interior Frame)

Two-Way Joist

Step 1: Marking E-W Interior Frame:

l 36′ 0″

l2 = 36′-0″

l1 = 36′-0″ln = 34′-6″




Two-Way Joist

Step 01: Marking E-W Interior Frame

Design Span of frame (c/c) = l1 = 36′

Design Length of frame = ln = 36 – (2 × 18/2)/12 = 34.5′

Width of frame = l2 = 36′

Half column strip width = (Shorter span)/ 4 = 36/4 = 9′




Two-Way Joist

Step 2: Marking Column and Middle Strips

MS/2 = 9′-0″

a = 5′-3″ CS/2 = 9′-0″

CS/2 = 9′-0″

a 5 -3

b= 12′-6″

CS/2 = Least of l1/4 or l2/4

l /4 = 36/4 = 9′

MS/2 = 9′-0″


l2/4 = 36/4 = 9



Two-Way Joist

Step 03: Static Moment Calculation

Mosh (outside head) = woshl2ln2/8

= 0.291 × 36 × 34.52/8 = 1557.56 ft-k

Mish (solid head) = (wish – wosh) ba2/2

= (0.33–0.291)×12.5×5.252/2 = 6.70 ft-k

Mo (total static moment) = Mosh + Mish = 1557.56 + 6.70 = 1564.26 ft-k

Note: Since normally, Mish is much smaller than Mosh the former can be conveniently ignored in design calculations




Two-Way Joist

Step 04: Longitudinal distribution of Total static moment (Mo).




Two-Way Joist

Step 05: Lateral Distribution of Longitudinal moment (L.M).

α INT36 =0 {no interior beams}

l2/l1 = 36/36 = 1

α INT36l2/l1 = 0



Two Way JoistStep No 03: Frame Analysis (E-W Exterior Frame)

Two-Way Joist

Step 01: Marking E-W exterior Frame

l 36′ 0″l1 = 36′-0″ln = 34′-6″

l2 = 18′-0″ + (9/12) = 18.75′


2 ( )



Two-Way Joist

Step 01: Marking E-W exterior Frame

Design Span of frame (c/c) = l1 = 36′

Design Length of frame = ln = 36 – (2 × 18/2)/12 = 34.5′

Width of frame = l2 = 9′ + 9′ + (9/12)″ = 18.75′

Half column strip width = (Shorter span)/ 4 = 36/4 = 9′




Two-Way Joist

Step 02: Marking Column and Middle Strips

l 36′ 0″l1 = 36′-0″ln = 34′-6″

CS/2 = Least of l1/4 or l2/4

l /4 = 36/4 = 9′

MS/2 = 9′-0″a = 5′ 3″


CS/2 = 9′-0″l2/4 = 36/4 = 9 a = 5 -3

b= 7′-0″



Two-Way Joist

Step 03: Static Moment Calculation

Mosh (outside head) = woshl2ln2/8

= 0.291 × 18.75 × 34.52/8 = 811.78 ft-k

Mish (solid head) = (wish – wosh) ba2/2

= (0.33–0.291)×7×5.252/2 = 3.76 ft-k

Mo (total static moment) = Mosh + Mish = 811.78 + 3.76 = 815.54 ft-k

Note: Since normally, Mish is much smaller than Mosh the former can be conveniently ignored in design calculations



Two Way JoistStep No 03: Analysis

Two-Way Joist

Step 04: Longitudinal distribution of Total static moment (Mo).




Two-Way Joist

Step 05: Lateral Distribution of Longitudinal moment (L.M)[Refer to ACI 13.6.4 to ACI 13.6.6].

α EXT36 =0 {no exterior beams}

l2/l1 = 36/36 = 1

α EXT36l2/l1 = 0




Two-Way Joist

Analysis of N-S Interior and Exterior Frame will be same as E-W respectiveframes due to square panels.

N-S Exterior Frame

N-S Interior Framel2 = 36′-0″

l2 = 18′-9″



Two Way JoistStep No 04: Design

Two-Way Joist

For E-W Interior slab strip:

davg = 12 + 4.5 – 1″ (concrete cover) – 0.75 (avg. bar dia) = 14.75″g

Asmin = 0.0018bte (Where te = equivalent flat slab thickness)

Asmin = 0.0018 × 12 × 12.95 = 0.279 in2

Now, Equation used to calculate (ρ) in table below is as follows:

Mu = Φfyρbdavg2{1– 0.59ρfy/fc′} = 0.9×60×ρ×12×14.752×{1– 0.59×ρ×60/4}

After solving the above equation for ρ, we get:

ρ = [140980.5 ±√{(140980.5)2 – (4 × 1247677 × Mu′ × 12)}]/2(1247677)….(A)




Two-Way Joist

For E-W Interior slab strip:




Two-Way Joist

For E-W exterior slab strip:

davg = 12 + 4.5 – 1– 0.75 = 14.75″ g

Asmin = 0.0018bte (Where te = equivalent flat slab thickness)

Asmin = 0.0018 × 12 × 12.95 = 0.279 in2

Now, Equation used to calculate (ρ) in table below is as follows:

Mu = Φfyρbdavg2{1– 0.59ρfy/fc′} = 0.9×60×ρ×12×14.752×{1– 0.59×ρ×60/4}

After solving the above equation for ρ, we get:

ρ = [140980.5 ±√{(140980.5)2 – (4 × 1247677 × Mu′ × 12)}]/2(1247677)….(A)




Two-Way Joist

For E-W exterior slab strip:




Two-Way Joist

Design of N-S Interior and Exterior Frame will be same as E-W respective frames due to square panels and also for thereason that davg is used in design.




Two-Way Joist

Note: For the completion of design problem, the waffle slabshould also be checked for beam shear and punching shear.



Two Way JoistStep No 05: Detailing (E-W Frames)

Two-Way Joist

#6 @ 12″ #6 @ 6″ #6 @ 6″ #6 @ 12″

Prof. Dr. Qaisar Ali 96#6 @ 12″ #6 @ 6″ #6 @ 6″ #6 @ 12″

#6 @ 18″ #6 @ 18″ #6 @ 18″ #6 @ 18″


Two Way JoistStep No 05: Detailing (N-S Frames)

Two-Way Joist

#6 @ 12″#6 @ 18″#6 @ 12″

#6 @ 6″#6 @ 18″#6 @ 6″

#6 @ 6″#6 @ 18″#6 @ 6″



Two Way JoistStep No 05: Detailing (E-W Interior Frame)

Two-Way Joist

#6 @ 6″ c/c

18′-0″

Column Strip (Interior Frame); section taken over support

#6 @ 12″ c/c2 #7 Bars

#6 @ 12 c/c

Prof. Dr. Qaisar Ali 98Column Strip (Exterior Frame); section taken over support2 #7 Bars


Two Way JoistStep No 05: Detailing (E-W Interior Frame)

Two-Way Joist

#6 @ 18″ c/c

18′-0″

Middle Strip (Interior Frame); Section taken over column line

#6 @ 18″ c/c2 #7 Bars

#6 @ 18 c/c

Prof. Dr. Qaisar Ali 99Middle Strip (Exterior Frame); Section taken over column line2 #7 Bars


Two Way JoistStep No 05: Detailing (E-W Exterior Frame)

Two-Way Joist

#6 @ 6″ c/c

9′-0″

Column Strip (Interior Frame); section over support2 #7 Bars

#6 @ 12″ c/c


Column Strip (Exterior Frame); section over support2 #7 Bars


Two Way JoistStep No 05: Detailing (E-W Exterior Frame)

Two-Way Joist

#6 @ 18″ c/c

9′-0″

Middle Strip (Interior Frame) ; section over support2 #7 Bars

#6 @ 18″ c/c


Middle Strip (Exterior Frame); section over support2 #7 Bars


The End


Lecture -11 Analysis and Design of Two-Way Slab Systems (Two-Way Slab With Beams & Two Way Joist...

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Transcript of Lecture -11 Analysis and Design of Two-Way Slab Systems (Two-Way Slab With Beams & Two Way Joist...