civil structural engineering - flat slab design

Flat Slab Design

Resources used for compiling this presenta4on are acknowledged

Flat Slab with drop panels

Flat slab with column head

Flat slab with drop panel and column head

Flat Slab resting directly on columns 1.  What is a flat slab?

31.1 General The term flat slab means a reinforced concrete slab with or without drops, supported generally without beams, by columns with or without flared column heads A flat slab may be solid slab or may have recesses formed on the soffit so that the soffit comprises a series of ribs in two directions. The recesses may be formed by removable or permanent filler blocks.

2. Types of flat slab

•  Flat Slab with drop panels

•  Flat slab with column head •  Flat slab with drop panel and column head

•  Flat Slab resting directly on columns

Drop is a local thickening of the slab in the region of column Structural Advantages •  increase shear strength of slab •  increase nega5ve moment capacity of slab

•  s5ffen the slab and hence reduce deflec5on

Column head is a local enlargement of the column at the junc5on with the slab Structural Advantages •  increase shear strength of slab

(punching shear) •  reduce the moment in the slab by

reducing the clear or effective span

A flat slab may have recesses formed on the soffit so that the soffit comprises a series of ribs in two direc5ons ( waffle Slabs).

Flat slabs with capitals, drop panels, or both. These slabs are very sa4sfactory for heavy loads and long spans. Although the formwork is more expensive than for flat plates, flat slabs will require less concrete and reinforcing than would be required for flat plates with the same loads and spans. They are par4cularly economical for warehouses, parking and industrial buildings, and similar structures, where exposed drop panels or capitals are acceptable.

v Flexibility in room layout •  Introduce partition walls anywhere required •  Change the size of room layout •  Omit false ceiling

v  Saving in building height •  Lower storey height will reduce building weight •  approx. saves 10% in vertical members •  reduce foundation load

v Shorter construction time •  flat plate design will facilitate the use of big table formwork to increase productivity

v Ease of installation of M&E services •  all M & E services can be mounted directly on

the underside of the slab instead of bending them to avoid the beams

•  avoids hacking through beams

3.  Benefits of flat slab

The main disadvantage is their lack of resistance to lateral loads due to wind and earthquakes. Lateral load resis4ng systems such as shear walls are oDen necessary When the loads or spans or both become quite large, the slab thickness and column sizes required for flat plates or flat slabs are of such magnitude that it is more economical to use two-‐way slabs with beams, despite the higher formwork costs.

4.  Behaviour of Slab supported on Stiff , Flexible and no beams

Case Study:

•  Panel Size = 4 m x 4m

•  Slab Thickness = 125 mm

•  Load = 5 kN/m2

•  S5ff Supports ( Bearing wall) •  Flexible Supports (Beam) : 300 x 300 , 300 x 450 , 300 x 600 , 300 x 1000 mm

•  Column supports at corners

A. Two way Slab on Rigid Supports (bearing Walls)

Mx = 3.616 kNm/m My = 3.616 kNm/m

IS 456 Values (Table 27): 0.062 x 5 x 16 = 4.96

Slab Deflec6on = 1.4 mm

B. Two way Slab on Flexible Supports (Beams on all sides) 1. Beam Size : 300 x300 mm

Mx = 4.45 kNm/m My = 4.45 kNm/m

IS 456 Values (Type 9): 0.056 x 5 x 16 = 4.48

Mxy = 0.37 kNm/m

Beam Moment = 12.2 kNm Beam Deflec6on = 1.33 mm Slab deflec6on= 2.9 mm

2. Beam Size : 300 x450 mm

Mx = 3 kNm/m My = 3 kNm/m IS 456 Values (Type 9): 0.056 x 5 x 16 = 4.48

Mxy = 0.73 kNm/m

Beam Moment = 15.6 kNm Beam Deflec6on = 0.5 mm

Slab deflec6on= 1.5 mm

3. Beam Size : 300 x 600 mm

Mx = 2.43 kNm/m My = 2.43 kNm/m

IS 456 Values (Type 9): 0.056 x 5 x 16 = 4.48

Mxy = 0.8 kNm/m

Beam Moment = 17 kNm Beam Deflec5on = 0.24 mm Slab Deflec5on = 0.98 mm

4. Beam Size : 300 x 1000 mm

Mx = 2 kNm/m My = 2 kNm/m IS 456 Values (Type 9): 0.056 x 5 x 16 = 4.48

Mxy = 0.8 kNm/m

Beam Moment = 18 kNm

5. Beam Size : 300 x 125 mm (Concealed Beams)

Mx = 9.8 kNm/m My = 9.8 kNm/m

IS 456 Values (Type 9): 0.056 x 5 x 16 = 4.48

Mxy = 3 kNm/m

Beam Moment = 2.9 kNm Slab Deflec6on = 7.0 mm

B. Two way Slab on Point Supports at corners (Flat Slab)

Mx = 9.075 kNm/m (Middle) =12.4 kNm/m (Edge Strip)

Mxy = 7.76 kNm/m My = 9.075 kNm/m (Middle) =12.4 kNm/m (Edge Strip)

Slab Deflec6on = 8.67 mm

Type of Support Mx My Mxy Beam

Moment Deflec4on

Slab Beam

Rigid 3.616 3.616 2.6 -‐ 1.4 -‐

300 x125 (Concealed Beams)

9.84 9.85 3.0 2.88 7.0 4.3

300 x300 4.45 4.45 0.37 12.2 2.9 1.33

300 x450 3 3 0.73 15.6 1.5 0.50

300 x600 2.43 2.43 0.8 17.0 0.98 0.24

300x1000 2 2 0.8 18.0 0.60 0.05

Flat Slab 9.0 9.0 7.76 -‐ 8.676 -‐

Results Summary

•  Two way Rectangular Slab supported on stiff beams, the shorter spans (stiffer portion

of the slab) carry larger load and subjected to larger moments. The longer spans

carry less load and subjected to less moment.

•  Results indicate that decrease in supporting beams stiffness leads to an increase in

bending moments of slabs and decrease in bending moment of the beams (behavior that is not captured using code recommendations).

•  If the slab is supported on bearing walls, slab moments are distributed in similar way.

•  If the slab is supported only by the columns, the slab behaves like a two way slab with

an essential difference that all the load is carried in both directions to accumulate it

at the columns.

•  With Concealed beams it is reveled that the behaviour is close to Flat slabs rather than any useful

beam action.

Observa4ons

4.  Structural Behaviour of Flat Slab

Deflected Shape

Column Strip Column Strip

Middle Strip

Column Strip Middle Strip

A Zone of –ve BM (Hogging) in both direc7ons B Zone of +ve BM(Sagging) and –ve BM C Zone of -‐ve BM and +ve BM D Zone of +ve BM in both direc7ons

-‐m4

-‐m2

-‐m4

m3

m1

m3

m5 m7

-‐m8

-‐m4

-‐m2

-‐m4

m7

A A

A A

C

C

B

B

D

Column Strip

Middle Strip

Column Strip

Column Strip

Middle Strip Column Strip

-‐m6

-‐m6 -‐m8 -‐m8

-‐m8

5.  Distribution of Total Panel Moment in different zones

A Zone of –ve BM (Hogging) in both direc7ons B Zone of +ve BM(Sagging) and –ve BM C Zone of -‐ve BM and +ve BM D Zone of +ve BM in both direc7ons

-‐m4

-‐m2

-‐m4

m3

m1

m3

m5 m7

-‐m8

-‐m4

-‐m2

-‐m4

m7

A A

A A

C

C

B

B

D

Column Strip

Middle Strip

Column Strip

Column Strip

Middle Strip Column Strip

-‐m6

-‐m6 -‐m8 -‐m8

-‐m8

m1

-‐m2 -‐m2 D C C

m3

-‐m4 -‐m4 B A A

m5

-‐m6

-‐m6

D

B B

m7

-‐m8

-‐m8

C

A A

6.  Definitions

L2

L1

Moment Direc5on

MIDDLE STRIP COLUMN STRIP 0.25L2 ≤ 0.25L1

COLUMN STRIP 0.25L1 ≤ 0.25L2

MIDDLE STRIP

COLUMN STRIP 0.25L1 ≤ 0.25L2

Mom

ent D

irec5on

SPAN Region

SPAN Region: Bounded on all the four sides by middle strips

7.  General Design Considerations

CL 31.2 Proportioning 31.2.1 Thickness of Flat Slab •  The thickness of the flat slab shall be

generally controlled by considerations of span to effective depth ratios given in 23.2.

•  For slabs with drops conforming to 31.2.2, span to effective depth ratios given in 23.2 shall be applied directly; otherwise the span to effective depth ratios obtained in accordance with provisions in 23.2 shall be multiplied by 0.9. For this purpose, the longer span shall be considered.

•  The minimum thickness of slab shall be 125 mm.

31.2.2 Drop •  The drops when provided shall be

rectangular in plan, and have a length in each direction not less than one- third of the panel length in that direction.

•  For exterior panels, the width of drops at right angles to the non continuous edge and measured from the centre-line of the columns shall be equal to one-half the width of drop for interior panels.

•  Minimum thickness of Drop > ¼ of Slab thickness and > 100 mm

31.2.3 Column Heads Where column heads are provided, that portion of a column head which lies within the largest right circular cone or pyramid that has a vertex angle of 900and can be included entirely within the outlines of the column and the column head, shall be considered for design purposes.

8. Determination of Bending Moment CL 31.3

31.3.1. Methods of Analysis and Design

It shall be permissible to design the slab system by one of the following

methods:

a) The direct design method as specified in 31.4, and

b) The equivalent frame method as specified in 31.5.

In each case the applicable limitations given in 31.4 and 31.5 shall be met.

9. Direct Design Method CL 31.4 A. Limitations : 31.4.1 Slab system designed by the direct design method shall fulfil the following conditions:

a)  There shall be minimum of three continuous spans in each direction, b)  The panels shall be rectangular, and the ratio of the longer span to the shorter span within

a panel shall not be greater than 2.0 c)  It shall be permissible to offset columns to a maximum of 10percent of the span in the

direction of the offset notwithstanding the provision in (b) d)  The successive span lengths in each direction shall not differ by more than one-third of

the longer span. The end spans may be shorter but not longer than the interior spans, and

e)  The design live load shall not exceed three times the design dead load.

Note: Applicable to gravity loading condition alone (and not to the lateral loading condition)

1 2 3

2

3

Lx1 Lx2 Lx3

Ly1

Ly2

Ly3

≤ 0.1Ly2

≤ 0.1Lx1 Lx1 ≤ Lx2 Lx3 ≤ Lx2

Ly1 ≤ Ly2 Ly3 ≤ Ly2

Lx1 ≥ 2Lx2/3 Lx3 ≥ 2Lx2/3

Ly1 ≥ 2Ly2/3 Ly3 ≥ 2Ly2/3

wuL/wuD ≤ 3

For any Panel Longer Span/Shorter Span≤ 2

B. Total Design Moment for a Span: CL31.4.2

CL of Panel 1

CL of Panel 2

1

2

DESIGN STRIP

31.4.2.1 In the direct design method, the total design moment for a span shall be determined for a strip bounded laterally by the centre-line of the panel on each side of the centre-line of the supports. 31.4.2.2 The absolute sum of the positive and average negative bending moments in each direction shall be taken as:

1

2

M0x

M0y

lnx

wu kN/m

L1

L2 Ln Ln (L1)

(L2) Note: 1.  It is the same as the total moment that occurs in a

simply supported slab

2.  The moment that actually occurs in such a slab has been shown by experience and tests to be somewhat less than the value determined by the Mo expression. For this reason, l1 is replaced with ln

•  It is next necessary to know what propor4ons of these total moments are posi4ve and what propor4ons are nega4ve.

10. Distribution of Total Panel Moment M0

•  If a slab was completely fixed at the end of each panel, the division would be as it is in a fixed-‐end beam, two-‐thirds nega4ve and one-‐third posi4ve, as shown in Figure.

•  This division is reasonably accurate for interior

panels where the slab is con4nuous for several spans in each direc4on with equal span lengths and loads.

Interior Panel

•  The rela4ve s4ffnesses of the columns and slabs of exterior panels are of far greater significance in their effect on the moments than is the case for interior panels.

•  The magnitudes of the moments are very sensi4ve to the amount of torsional restraint supplied at the discon4nuous edges.

•  This restraint is provided both by the flexural s4ffness of the slab and by the flexural s4ffness of the exterior column.

Exterior Panel

Code Recommendations

Distribution of Bending Moments across panel width Code Recommendations

11. Rebar Detailing - Code Recommendations

Bent bars are also used. There seems to be a trend among designers to use straight bars more than bent bars.

ELEVATION

Rebar Detailing - Code Recommendations

e e e e e

b b b b b

Ln greater of adjacent clear spans CL 31.7.3 (b)

Sec6on through Middle Strip

12. Two way Shear in Flat Slab

•  Flat plates present a possible problem in transferring the shear at the perimeter of the columns.

•  There is a danger that the columns may punch through the slabs.

•  As a result, it is frequently necessary to increase column sizes or slab thicknesses or to use shear heads. Shear heads consist of steel I or channel shapes placed in the slab over the columns

Note: Flat Slab with drop panel and capital, shear is required to be checked at two sec4ons 1.  at a distance d/2 from the face of column capital 2.  at a distance d/2 from the face of drop panel

Design Example #1

Design by DDM flat plate supported on columns 450 mm square, for a Live Load = 3 kN/m2, Floor Finish = 1 kN/m2 use M20 and Fe415. Assume clear cover = 20 mm. Effec6ve Column Height = 3.35m. Bay spacing in X and Y direc6on = 5m c/c •  Interior Panel P5 •  Corner Panel P7

3 bays @ 5 m c/c

A. Interior Panel Design

5 m

5 m 2.5m

2.5 m A A

A A

B

B

CCD

Zone A – Corner Strip

Zone B – Middle Strip along X

Zone C – Middle Strip along Y

Zone D – Interior Region Step 1: Panel Division into Strips 31.1.1(a)

Moment direc6on Along

L1 L2 Width of Column Strip on either side of Centre Line = 0.25L2 and ≤ 0.25 L1

Middle Strip

X 5 5 1.25 and ≤1.25 m Adopt 1.25 m 2.5m

Y 5 5 1.25 and ≤1.25 m Adopt 1.25 m 2.5m

Step 2: Trial Depth CL 31.2.1

•  L/d = 26 •  Modifica5on Factor = 1.33, Assuming pt = 0.4%, FIG 4 IS 456 •  d = 5000/(26 x 1.33) = 145 mm > 125 CL 31.2.1 •  DS= 145 + 20 + 18 = 183 mm ( assume #12 bars, and bars in two layers) •  Provide Ds= 200 mm d = 200-‐20-‐18 = 162 mm

Step 3 Design Loads / m width of Slab

•  wuD = 1.5(25x 0.2 + 1) = 9kN/m •  wuL = 1.5 x 3 = 4.5kN/m •  wu = 13.5 kN/m

Step 4: Check for Applicability of DDM: CL 31.4.1

•  No. of Con5nuous Spans in each direc5on = 3 ; OK 31.4.1(a) •  Long Span/Short Span = 5/5 = 1 <2 ; OK 31.4.1(b) •  Successive spans in each direc5on = Equal; OK 31.4.1(d) •  wuL/wuD = 4.5/9 = 0.5 < 3 ; OK 31.4.1(e)

Step 5: Check for punching shear around Column

Assumed d = 162 mm Sec4on 1: •  Cri5cal Sec5on at d/2 around the column •  Perimeter of Cri5cal Sec5on = 4 x 0.612= 2.448 m •  Design Shear at cri5cal sec5on Vu •  Vu = 13.5 ( 52 – 0.6122) = 333kN •  τc = 0.25√fck = 1.12 MPa •  ks = 0.5 +1 = 1.5 <=1 ; ks=1 ; ks τc = 1.12 •  Shear Resistance of Concrete = 1.12 x 2448 x 162 = 444kN > 333 kN OK

Cri4cal Sec4on

0.612m

0.612m

5m

5m

Contributory Area

Step 6:Design Moments CL 31.4.2.2

Parameters Along X Along Y L1 (Span in direc4on of Mo) 5 5 m 0.65L1 3.25 3.25 m Ln (clear span extending from face to face of columns, capitals)

(5-‐0.45) = 4.55 4.55 m

Ln > 0.65L1 4.55 4.55 m L2 (Span transverse to L1) 5 5 m W = wu L2Ln 307.2 307.2 kN M0 = W Ln / 8 174.72 174.72 kNm

wu = 13.5 kN /m

Step 7 : Distribution of Bending Moment across panel width ; CL: 31.4.3.2, 31.5.5

Moment Direc5on along X Y Nega4ve Design Moment

MN = -‐ 0.65*M0 113.6 113.6 kNm 31.4.3.2

•  Column Strip M1 = 0.75MN 85.2 85.2 kNm 31.5.5.1 Width of Column Strip resis4ng M1 (Csw) 2x1.25 =2.5 2x1.25 =2.5 m

•  -‐m1 = M1/ Csw (Zone A) 34.1 34.1 kNm/m

•  Middle Strip M2 = 0.25MN 28.4 28.4 kNm 31.5.5.4(a)

Width of Middle Strip resis4ng M2 (Msw) 2.5 2.5 m

•  -‐m2 = M2/Msw (Zone B & C) 11.4 11.4 kNm/m

Posi5ve Design Moment MP = 0.35*M0

61.2 61.2 kNm 31.4.3.2

•  Column Strip M1 = 0.6MP 36.7 36.7 kNm 31.5.5.3

•  +m1 = M1/ Csw (Zone B &C)

14.7 14.7 kNm/m

•  Mid Span M2 = 0.4MP 24.5 24.5 kNm 31.5.5.4(a)

•  +m2 = M2/Msw (Zone D) 9.8 9.8 kNm/m

-‐ve sign : Hogging Moment (tension at top) +ve sign : Sagging Moment (tension at borom)

-‐34.1

-‐11.4

-‐34.1

14.7

9.8

14.7

-‐34.1 -‐11.4 -‐34.1

9.8 14.7

-‐11.4 -‐34.1 -‐34.1

-‐34.1

-‐11.4

-‐34.1

14.7

A B A

DC C

AAB

Step 8 : Check for adequacy of Depth •  Max Design Bending moment = 34.1 kNm/m •  Mu,lim = 72.41 kNm/m > 34.1, •  Depth is adequate G-‐1.1(c)

Loca6on Moment (kNm/m)

Ast (mm2/m)

Ast (prov) Bar dia

Spacing mm

Along X Zone A (-‐) 34.1 635 635 10 120 -‐ T Zone B 14.7 260 260 8 190 -‐ B Zone C (-‐)11.4 200 240 8 200 -‐ T Zone D 9.8 171 240 8 200 -‐ B

Along Y Zone A (-‐)34.1 635 635 10 120 -‐ T Zone B (-‐)11.4 200 240 8 200 -‐ T Zone C 14.7 260 260 8 190 -‐ B Zone D 9.8 171 240 8 200 -‐ B

•  7.5 Ast2 – 58490Ast + Mu = 0 G-‐1.1(b)

Step 9 :Rebar Details •  Ast,min = 0.12 x 200 x 1000 /100 = 240 mm2/m 26.5.2.1 •  Minimum Effec5ve Depth of Slab = 162 mm

#8@190

#8@200

#8@190

#8@200 #8@190 #8@190

0.15Ln

0.15Ln 0.125Ln

0.125Ln

Borom Rebar Details in Interior Panel

A B A

A B A

C D C

TOP Rebar Details in Interior Panel

#10@120

#10@

120

#8@200

#8@200 #8@200

0.3Ln

0.2Ln

0.3Ln 0.2Ln

Note: Distances for curtailment of rebars are measured from column face

A B A

C D C

B. Corner Panel Design


Assumed d = 162 mm Sec4on 1: •  Cri5cal Sec5on at d/2 around the column •  Perimeter of Cri5cal Sec5on = 2 x 0.531= 1.062 m •  Design Shear at cri5cal sec5on Vu •  Vu = 13.5 ( 2.52 – 0.5312) = 81kN •  τc = 0.25√fck = 1.12 MPa •  ks = 0.5 +1 = 1.5 <=1 ; ks=1 ; ks τc = 1.12 •  Shear Resistance of Concrete = 1.12 x 1062 x 162 =

192kN > 81 kN OK

450

162/2 = 81 mm

Step 7 : Distribution of Bending Moment across panel width ; CL 31.4.3.3 , 31.5.5

𝛼↓𝑐 = ∑↑▒𝑘↓𝑐  /𝑘↓𝑠  

Assume Columns and Slab panels are with same modulus of elas5city

5 m

5 m

1.25m

1.25m A A

A A

B

B

CCD


M0 = 174.72 kNm

Parameters Along X Along Y Sum o f co l umn s4ffness above and below the slab 2 (4EcIc)/Lc

(2 x 4 x Ec x 450 x 4503/12) /3350 = 8.16 Ec x 106

Slab s4ffness ks = 4EsIs/Ls

(4 Es x 5000 x 2003/12)/5000 = 2.67Es x 106

2.67Es x 106

αc = ∑kc /ks 3.06 3.06

β = 1+ (1/αc) 1.33 1.33


MN = -‐ 0.65*M0/β 85.4 85.4 kNm 31.4.3.3

•  Column Strip M1 = MN 85.4 85.4 kNm 31.55.2(a)

Width of Column Strip Csw resis4ng M1 2x1.25 = 2.5 2.5 m

-‐m1 = M1/Csw 34.2 34.2 kNm/m

•  Middle Strip M2=0 0 0 kNm 31.5.5.4(a)

-‐m2 = 0 0 0 kNm/m

A. Exterior nega4ve design moment: -‐m1

-‐m1

-‐m1

-‐m1

Exterior

Exterior

Interior

Interior 1.25

1.25

-‐m2

-‐m2

X

Y


MN = -‐ (0.75 – 0.1/β)Mo 118 118 kNm 31.4.3.3

•  Column Strip M1 = 0.75 MN 88.5 88.5 kNm 31.5.5.1

Width of Column Strip Csw resis4ng M1 2x1.25 =2.5 2.5 m

•  -‐m1 = M1/ Csw -‐35.4 -‐35.4 kNm/m

•  Middle Strip M2 = 0.25 MN 22.12 22.12 kNm 31.5.5.4(a)

Width of Middle Strip Msw resis4ng M2 2.5 2.5 m

•  -‐m2 = M2/Msw -‐8.85 -‐8.85 kNm/m

B. Interior nega4ve design moment:

-‐ m1

-‐ m1

-‐ m2

m1 -‐m1 -‐m1 -‐m2

X

Y

Exterior

Exterior

Interior

Interior

Moment Direc5on along X Y Design Moment

MP = (0.63 – 0.28/β)Mo 73.29 73.29 kNm 31.4.3.3

•  Column Strip M1 = 0.6 MP 43.98 43.98 kNm 31.5.5.3

Width of Column Strip Csw resis4ng M1 2x1.25 =2.5 2.5 m

•  m1 = M1/ Csw 17.6 17.6 kNm/m

•  Middle Strip M2 = 0.4 MP 29.32 29.32 kNm 31.5.5.4(a)


•  m2 = M2/Msw 11.73 11.73 kNm/m

C. Posi4ve Moment in Mid Span:

m1

m1

m1

m1

m1

m1

m2 X

Y

Exterior

Exterior

Interior

Interior

m2


17.6

11.73

17.6

-‐8.85 -‐35.4

11.73 17.6 17.6

A B A

DC C

AAB

-‐35.4

-‐35.4

-‐8.85

-‐35.4

-‐34.2

-‐34.2

0

-‐34.2 -‐34.2 -‐0

Exterior

Exterior

Interior

Interior

Step 7 : Check for adequacy of Depth •  Max Design Bending moment = 35.4 kNm/m •  Mu,lim = 72.41 kNm/m > 35.4, Depth is adequate G-‐1.1(c)

Strip Loca6on Moment (kNm/m)

Ast (mm2/m)

Ast (prov) Bar dia

Spacing mm

Along X Zone A(Exterior) (-‐)34.2 637 637 10 120(T) Zone A(Interior) (-‐)35.4 662 662 10 115(T) Zone B 17.6 314 314 8 160(B) Zone C(Interior) (-‐)8.85 155 240 8 200(T) Zone D 11.73 206 240 8 200(B)

Along Y Zone A (Exterior) (-‐)34.2 637 637 10 120(T) Zone A(Interior) (-‐)35.4 662 662 10 115(T) Zone B (Interior) (-‐)8.85 155 240 8 200(T) Zone C 17.6 314 314 8 160(B) Zone D 11.73 206 240 8 200(B)

Step 8 :Rebar Details •  Ast,min = 0.12 x 200 x 1000 /100 = 240 mm2/m 26.5.2.1

7.5 Ast2 – 58490Ast + Mu = 0

17.6

11.73

17.6

11.73 17.6 17.6

A B A

D C C

A A B

Exterior

Exterior

Interior

Interior

-‐8.85 -‐35.4

A B A

DC C

AAB

-‐35.4

-‐35.4

-‐8.85

-‐35.4

-‐34.2

-‐34.2

0

-‐34.2 -‐34.2 -‐0 Exterior

Exterior

Interior

Interior #8@200

#10@

120

#10@

115

#10@120 #10@120

#8@200

#8@200 #10@115

#10@

115

#8@200

TOP Rebar details in Corner Panel

Design Example #2

Design by DDM flat plate supported on columns 500 mm square, for a Live Load = 4 kN/m2, Floor Finish = 1 kN/m2 use M25 and Fe415. Floor slab is exposed to moderate environment. Column Height = 3.5m (c/c). Bay spacing in X and Y direc6on = 5.5m c/c. Assume that building is not restrained against sway •  Interior Panel P5 •  Corner Panel P7

3 bays @ 5.5 m c/c

A. Interior Panel Design

5.5 m

5.5 m 2.75m

2.75 m A A

A A

B

B

CCD




Zone D – Interior Region Step 1: Panel Division into Strips 31.1.1(a)


L1 L2 Width of Column Strip on either side of Centre Line = 0.25L2 and ≤ 0.25 L1

Middle Strip

X 5 5 1.375 and ≤1.375 m Adopt 1.375 m 2.75m

Y 5 5 1.375 and ≤1.375 m Adopt 1.375 m 2.75m


•  L/d = 26 •  Modifica5on Factor = 1.33, Assuming pt = 0.4%, FIG 4 IS 456 •  d = 5500/(26 x 1.33) = 160 mm > 125 CL 31.2.1 •  DS= 160 + 30 + 18 = 208 mm ( assume #12 bars, and bars in two layers) •  Provide Ds= 225 mm d = 225-‐30-‐18 = 177 mm

Step 3 Design Loads / m width of Slab

•  wuD = 1.5(25x 0.225 + 1) = 9.94kN/m •  wuL = 1.5 x 4 = 6kN/m •  wu = 15.94 ≈ 16 kN/m


•  No. of Con5nuous Spans in each direc5on = 3 ; OK 31.4.1(a) •  Long Span/Short Span = 5.5/5.5 = 1 <2 ; OK 31.4.1(b) •  Successive spans in each direc5on = Equal; OK 31.4.1(d) •  wuL/wuD = 6/9.94 = 0.6 < 3 ; OK 31.4.1(e)


Assumed d = 177 mm Sec4on 1: •  Cri5cal Sec5on at d/2 around the column •  Perimeter of Cri5cal Sec5on = 4 x 0.677= 2.708 m •  Vu = 16 ( 5.52 – 0.6772) = 477kN •  τc = 0.25√fck = 1.25 MPa •  ks = 0.5 +1 = 1.5 <=1 ; ks=1 ; ks τc = 1.25 •  Shear Resistance of Concrete = 1.25 x 2708 x 177 = 599kN > 477 kN OK

Cri4cal Sec4on

0.677m

0.677m

5.5m

5.5m

Contributory Area


Parameters Along X Along Y L1 (Span in direc4on of Mo) 5.5 5.5 m 0.65L1 3.575 3.575 m Ln (clear span extending from face to face of columns, capitals)

(5.5-‐0.5) = 5 5 m

Ln > 0.65L1 5 5 m L2 (Span transverse to L1) 5.5 5.5 m W = wu L2Ln 440 440 kN M0 = W Ln / 8 275 275 kNm

wu = 16 kN /m



MN = -‐ 0.65*M0 179 179 kNm 31.4.3.2

•  Column Strip M1 = 0.75MN 134.3 134.3 kNm 31.5.5.1 Width of Column Strip resis4ng M1 (Csw) 2.75 2.75 m

•  -‐m1 = M1/ Csw (Zone A) 48.8 48.8 kNm/m


Width of Middle Strip resis4ng M2 (Msw) 2.75 2.75 m

•  -‐m2 = M2/Msw (Zone B & C) 16.3 16.3 kNm/m


96.3 96.3 kNm 31.4.3.2


•  +m1 = M1/ Csw (Zone B &C)

21 21 kNm/m

•  Mid Span M2 = 0.4MP 38.5 38.5 kNm 31.5.5.4(a)

•  +m2 = M2/Msw (Zone D) 14 14 kNm/m


-‐48.8

-‐16.3

-‐48.8

21

14

21

-‐48.8 -‐16.3 -‐48.8

14 21

-‐16.3 -‐48.8 -‐48.8

-‐48.8

-‐16.3

-‐48.8

21

A B A

DC C

AAB

Step 8 : Check for adequacy of Depth •  Max Design Bending moment = 48.8 kNm/m •  Mu,lim = 108 kNm/m > 48.8 •  Depth is adequate G-‐1.1(c)

Loca6on Moment (kNm/m)

Ast (mm2/m)

Ast (prov) Bar dia

Spacing mm

Along X Zone A (-‐) 48.8 828 828 10 90 -‐ T Zone B 21 340 340 8 145 -‐ B Zone C (-‐)16.3 262 270 8 180 -‐ T Zone D 14 224 270 8 180 -‐ B

Along Y Zone A (-‐) 48.8 828 828 10 90 -‐ T Zone B -‐16.3 262 270 8 180 -‐ T Zone C 21 340 340 8 145 -‐ B Zone D 14 224 270 8 180 -‐ B

•  6 Ast2 – 63906Ast + Mu = 0 G-‐1.1(b)


#8@145

#8@180

#8@145

#8@180 #8@145 #8@145

0.15Ln

0.15Ln 0.125Ln

0.125Ln


A B A

A B A

C D C

TOP Rebar Details in Interior Panel

#10@90

#10@

90

#8@180

#8@200 #8@180

0.3Ln

0.2Ln

0.3Ln 0.2Ln


A B A

C D C

B. Corner Panel Design


Assumed d = 177 mm Sec4on 1: •  Cri5cal Sec5on at d/2 around the column •  Perimeter of Cri5cal Sec5on = 2 x 0.5885= 1.177 m •  Vu = 16 ( 2.752 – 0.58852) = 115.5kN •  τc = 0.25√fck = 1.25 MPa •  ks = 0.5 +1 = 1.5 <=1 ; ks=1 ; ks τc = 1.25 •  Shear Resistance of Concrete = 1.25 x 1177 x 177 =

260kN > 115.5 kN OK

500

177/2 = 88.5 mm

2.75m

2.75m


𝛼↓𝑐 = ∑↑▒𝑘↓𝑐  /𝑘↓𝑠  

Assume Columns and Slab panels are with same modulus of elas5city

5 m

5 m

1.25m

1.25m A A

A A

B

B

CCD


M0 = 275 kNm


Leff = 1.2 Lc (CL E1) Lc = 3.5-‐0.225 = 3.275

(2 x 4 x Ec x 500 x 5003/12) /1.2*3275 = 10.6 Ec x 106


(4 Es x 5500 x 2253/12)/5500 = 2.67Es x 106

3.8Es x 106

αc = ∑kc /ks 2.8 2.8

αc min (Table 17) (0.7/0.5)*0.1 =0.14 < αc. Adopt αc = 2.8

β = 1+ (1/αc) 1.36 1.36


MN = -‐ 0.65*M0/β 131.4 131.4 kNm 31.4.3.3

•  Column Strip M1 = MN 131.4 131.4 kNm 31.55.2(a)

Width of Column Strip Csw resis4ng M1 2.75 2.75 m

-‐m1 = M1/Csw 47.8 47.8 kNm/m


-‐m2 = 0 0 0 kNm/m


-‐m1

-‐m1

-‐m1

Exterior

Exterior

Interior

Interior 1.25

1.25

-‐m2

-‐m2

X

Y

M0 = 275 kNm


MN = -‐ (0.75 – 0.1/β)Mo 186 186 kNm 31.4.3.3



•  -‐m1 = M1/ Csw 50.73 50.73 kNm/m



•  -‐m2 = M2/Msw 17 17 kNm/m


-‐ m1

-‐ m1

-‐ m2

m1 -‐m1 -‐m1 -‐m2

X

Y

Exterior

Exterior

Interior

Interior


MP = (0.63 – 0.28/β)Mo 116.6 116.6 kNm 31.4.3.3

•  Column Strip M1 = 0.6 MP 70 70 kNm 31.5.5.3


•  m1 = M1/ Csw 25.5 25.5 kNm/m



•  m2 = M2/Msw 17 17 kNm/m


m1

m1

m1

m1

m1

m1

m2 X

Y

Exterior

Exterior

Interior

Interior

m2


25.5

17

25.5

-‐17 -‐50.73

17 25.5 25.5

A B A

DC C

AAB

-‐50.73

-‐50.73

-‐17

-‐50.73

-‐47.8

-‐47.8

0

-‐47.8 -‐47.8 -‐0

Exterior

Exterior

Interior

Interior

Step 7 : Check for adequacy of Depth •  Max Design Bending moment = 50.73 kNm/m •  Mu,lim = 108 kNm/m > 50.73, Depth is adequate G-‐1.1(c)


Ast (mm2/m)

Ast (prov) Bar dia

Spacing mm

Along X Zone A(Exterior) (-‐)47.8 810 810 10 90(T) Zone A(Interior) (-‐)50.73 864 864 10 90(T) Zone B 25.5 415 415 10 180(B) Zone C(Interior) (-‐)17 273 273 8 180(T) Zone D 17 273 273 8 180(B)

Along Y Zone A (Exterior) (-‐)47.8 810 810 10 90(T) Zone A(Interior) (-‐)50.73 864 864 10 90(T) Zone B (Interior) (-‐)17 273 273 8 180(T) Zone C 25.5 415 415 10 180(B) Zone D 17 273 273 8 180(B)


6 Ast2 – 63906Ast + Mu = 0

7.2m 7.2m 7.2m

6.4m

6.4m

6.4m

Design Example #3

Design by DDM flat plate supported on columns of dia = 450 mm, Column head = 1.5 m dia, Drop panel size = 3.2 x 3.2 m, for a Live Load = 4 kN/m2, Floor Finish = 1 kN/m2 use M20 and Fe415. Assume clear cover = 20 mm. Column Height = 3.35m •  Interior Panel P5 •  Exterior Panel P2/P4 •  Corner Panel P1

Step 1: Panel Division into Strips 31.1.1(a)


L1 L2

Width of Column Strip on either side of

Centre Line = 0.25L2 and ≤ 0.25 L1

Middle Strip

X 7.2 6.4 1.6 < 1.8 m; 1.6 m 4m Y 6.4 7.2 1.8 > 1.6 m; 1.6 m 3.2m

Lx = 7.2

L y = 6.4 1.6

1.6

1.6 1.6

CSx CSx MSx

CSy

MSy

CSy A.  Interior Panel Design




Zone D – Interior Region


•  L/d = 26 •  Modifica5on Factor = 1.4, Assuming pt ≈0.4%, FIG 4 IS 456 •  d = 7200/(26 x 1.4) = 198 mm > 125 CL 31.2.1 •  DS= 198+20+18= 236 mm ( assume #12 bars) •  Provide Ds= 240 mm , d = 198mm

Step 3: Design Loads / m width of Slab

•  wuD = 1.5(25 x 0.24 + 1) = 10.5kN •  wuL = 1.5 x 4 = 6.0kN •  wu = 16.5 kN


•  No. of Con5nuous Spans in each direc5on = 3 ; OK 31.4.1(a) •  Long Span/Short Span = 7.2/6.4 = 1.125 <2 ; OK 31.4.1(b) •  Successive spans in each direc5on = Equal; OK 31.4.1(d) •  wuL/wuD = 6/10.5 = 0.571 < 3 ; OK 31.4.1(e)

Step 5: Drop Panel Size : CL 31.2.2

•  Length along X ≥ Lx/3 = 2.4 m •  Length along Y ≥ Ly/3 = 2.13 m •  Generally Drop Panel Size is set equal to Width of Column Strip •  Proposed size 3.2 x 3.2 meets all the requirements. •  Minimum thickness = ¼ DS = 60 mm or 100 mm; Adopt 100 mm

Step 6:Column Head

•  1/4 to 1/5 of average span = 7.2/5 = 1.44 m •  Provided = 1.5 m ; Ok •  Equivalent Square Capital =0.89D = 1.335 m

Step 7 : Check for Shear around Column Capital

•  Minimum Effec5ve Depth of Slab = 198 mm •  Effec5ve Depth at Drop loca5on = 298 mm Sec4on 1: •  Cri5cal Sec5on at d/2 around the column capital •  Perimeter of Cri5cal Sec5on = π ( 1.5 + 0.298) = 5.65 m •  Weight of Drop Projec5on below slab = 0.1x 25 x 1.5 = 3.75 kN/m2

•  Design Shear at cri5cal sec5on around capital Vu •  Vu = 16.5 ( 7.2 x 6.4 -‐ π x 1.7982/4) + 3.75(3.2 x 3.2 -‐ π x 1.7982/4) •  = 747 kN •  τc = 0.25√fck = 1.12 MPa •  ks = 0.5 +1 = 1.5 <=1 ; ks=1 ; ks τc = 1.12 •  Shear Resistance of Concrete = 1.12 x 5650 x 298 = 1885kN > 747 kN OK

1.5

Cri4cal Sec4on

DROP3.2 m

3.2 m

1.798

Capital

Sec4on 2 : Check for Shear around drop 1.5

Cri4cal Sec4on

DROP3.2 m

3.2 m

Capital

3.2 + 0.198 = 3.4

•  Cri5cal Sec5on at d/2 around the drop •  d = 198mm •  Perimeter of Cri5cal Sec5on = 4 x 3.4 = 13.6m •  Design Shear at cri5cal sec5on •  Vu = 16.5 ( 7.2 x 6.4 – 3.42) = 569 kN •  Shear Resistance of Concrete = 1.12 x 13600 x 198 =

3015kN > 569 kN


Parameters Along X Along Y L1 (Span in direc4on of Mo) 7.2 6.4 m 0.65L1 4.68 4.16 m Ln (clear span extending from face to face of columns, capitals)

(7.2-‐1.335) =

5.865

(6.4-‐1.335) =

5.065 m

Ln > 0.65L1 5.865 5.065 m L2 (Span transverse to L1) 6.4 7.2 m W = wu L2Ln 619.34 601.72 kN M0 = W Ln / 8 454 381 kNm

wu = 16.5 kN /m



MN = -‐ 0.65*M0 295.1 247.65 kNm 31.4.3.2

•  Column Strip M1 = 0.75MN 221.33 185.74 kNm 31.5.5.1

Width of Column Strip Csw 2x1.6 =3.2 2x1.6 =3.2 m

•  -‐m1 = M1/ Csw 69.17 58.04 kNm/m


Width of Middle Strip Msw 3.2 4

•  -‐m2 = M2/Msw 23.06 15.48 kNm/m


158.9 133.35 kNm 31.4.3.2


•  +m1 = M1/ Csw 29.79 25 kNm/m

•  Middle Strip M2 = 0.4MP 63.56 53.34 kNm 31.5.5.4(a)

•  +m2 = M2/Msw 19.86 13.34 kNm/m

-‐ve sign : Hogging Moment (tension at top) +ve sign : Sagging Moment (tension at borom) -‐69.17

-‐23.06

-‐69.17

29.79

19.86

29.79

-‐58.04 -‐15.48 -‐58.04

13.34 25

-‐15.48 -‐58.04 -‐58.04

-‐69.17

-‐23.06

-‐69.17

25

A B A

DC C

AAB

Step 10 : Check for adequacy of Depth

•  Max Design Bending moment = 69.17 kNm/m •  Mu,lim = 126.36 kNm/m > 69.17, G-‐1.1(c) •  Depth is adequate

Moment Direc4on

Mom

ent

Direc4on

CS

MS

CS

FE Results from ETAB


Ast (mm2/m)

Ast (prov) Bar dia

Spacing mm

Along X Zone A (-‐) 69.17 1093 1093 10 70 -‐ T Zone B 29.79 437 437 8 110 -‐ B Zone C (-‐)23.06 334 334 8 150 -‐T Zone D 19.86 286 288 8 170 -‐ B

Along Y Zone A (-‐)58.04 896 896 10 85 -‐ T Zone B (-‐)15.48 222 288 8 170-‐T Zone C 25 364 364 8 135 -‐B Zone D 13.34 190 288 8 170 -‐ B

•  7.5 Ast2 – 71488Ast + Mu = 0 G-‐1.1(b)


#8@170 #8@135 #8@135

#8@170

#8@110

#8@110

7.2 m

6.4 m 0.15Ln

0.15Ln


LAP ZONE

A A

A A

B

B

C D C

0.125Ln

0.125Ln

#8@170 #10@85

#8@150

#10@

70

0.22Ln 0.22Ln

0.22Ln

0.22Ln

0.33Ln

0.33Ln

0.2Ln

0.2Ln

0.33Ln

0.2Ln

Top Rebar Details in Interior Panel


A B A

0.33Ln

0.2Ln

Sec6on Through Middle Strip -‐ CDC

#8@170

7.2 m

#8@340

#8@150

#8@135

Sec6on Through Column Strip -‐ ABA

#10@70

#10@140

#10@85

#8@170 #8@340 #8@170

Step 7 : Check for Shear around Column Capital

•  Minimum Effec5ve Depth of Slab = 198 mm •  Effec5ve Depth at Drop loca5on = 298 mm Sec4on 1: •  Cri5cal Sec5on at d/2 around the column capital •  Perimeter of Cri5cal Sec5on = π ( 1.5 + 0.298)/4 = 1.412 m •  Weight of Drop Projec5on below slab = 0.1x 25 x 1.5 = 3.75 kN/m2

•  Design Shear at cri5cal sec5on around capital Vu •  Vu = 16.5 ( 3.6x 3.2 – (π x 1.7982/4)/4) + 3.75(1.6 x 1.6 – (π x 1.7982/4)/4)) •  = 187 kN •  τc = 0.25√fck = 1.12 MPa •  ks = 0.5 +1 = 1.5 <=1 ; ks=1 ; ks τc = 1.12 •  Shear Resistance of Concrete = 1.12 x 1412 x 298 = 471kN > 187 kN OK

2. Corner Panel Design

Sec4on 2 : Check for Shear around drop

•  Cri5cal Sec5on at d/2 around the drop •  d = 198mm •  Perimeter of Cri5cal Sec5on = 2 (1.7)=3.4m •  Design Shear at cri5cal sec5on •  Vu = 16.5 ( 3.6 x 3.2 – 1.72) = 143 kN •  Shear Resistance of Concrete = 1.12 x 3400 x 198 = 754kN > 143 kN

CRITICAL SECTION

drop

free edge

free edge

=1.6 + 0.198/2 = 1.7 m


Along X Along Y

M0 = W Ln / 8 454 381 kNm


𝛼↓𝑐 = ∑↑▒𝑘↓𝑐  /𝑘↓𝑠  

Equivalent side of circular column = 0.89D = 0.89x 450 = 400 mm Assume Ec = Es


(2 x 4 x Ec x 400 x 4003/12) /3350 = 5.09 Ec x 106


(4 Es x 6400 x 2403/12)/7200 = 4.1Es x 106

(4 Es x 7200 x 2403/12)/6400 = 5.184Es x 106

αc = ∑kc /ks 1.24 0.98

αc min (Table 17) l2/l1 = 6.4/7.2 = 0.89, WuL/WuD = 0.571

(0.7/0.5)*0.071 = 0.1 <αc Adopt αc

7.2/6.4 = 1.125, WuL/WuD = 0.571

≈(0.8/0.5)*0.071 = 0.113 <αc

Adopt αc β = 1+ (1/αc) 1.8 2.02


MN = -‐ 0.65*M0/β 164 122.6 kNm 31.4.3.3

•  Column Strip M1 = MN 164 122.6 kNm 31.55.2(a)

Width of Column Strip Csw resis4ng M1 2x1.6 =3.2 2x1.6 =3.2 m

-‐m1 = M1/ 3.2 -‐51.3 -‐38.3 kNm/m


Width of Middle Strip Msw resis4ng M2 3.2 4

-‐m2 = 0 0 0 kNm/m


-‐m1

-‐m1

-‐m1

Exterior

Exterior

Interior

Interior 1.6

1.6

-‐m2

-‐m2

X

Y


MN = -‐ (0.75 – 0.1/β)Mo 315.3 266.9 kNm 31.4.3.3



•  -‐m1 = M1/ Csw -‐73.9 -‐62.6 kNm/m


Width of Middle Strip Msw resis4ng M2 3.2 4 m

•  -‐m2 = M2/Msw -‐24.7 -‐16.7 kNm/m


-‐ m1

-‐ m1

-‐ m2

m1 -‐m1 -‐m1 -‐m2

X

Y

Exterior

Exterior

Interior

Interior


MP = (0.63 – 0.28/β)Mo 215.4 187.2 kNm 31.4.3.3

•  Column Strip M1 = 0.6 MP 129.3 112.3 kNm 31.5.5.3


•  m1 = M1/ Csw 40.4 35.1 kNm/m


Width of Middle Strip Msw resis4ng M2 3.2 4 m

•  m2 = M2/Msw 26.94 18.7 kNm/m


m1

m1

m1

m1

m1

m1

m2 X

Y

Exterior

Exterior

Interior

Interior

m2


40.4

26.94

40.4

-‐16.7 -‐62.6

18.7 35.1 35.1

A B A

DC C

AAB

-‐73.9

-‐73.9

-‐24.7

-‐62.6

-‐51.3

-‐51.3

0

-‐38.3 -‐38.3 -‐0

Exterior

Exterior

Interior

Interior

Step 10 : Check for adequacy of Depth •  Max Design Bending moment = 73.9 kNm/m •  Mu,lim = 126.36 kNm/m > 73.9, Depth is adequate G-‐1.1(c)


Ast (mm2/m)

Ast (prov) Bar dia

Spacing mm

Along X Zone A(Exterior) (-‐)51.3 782 782 10 100 -‐ T Zone A(Interior) (-‐)73.9 1180 1180 10 65 -‐ T Zone B 40.4 604 604 8 80 -‐B Zone C(Interior) (-‐)24.7 359 359 8 140 -‐ T Zone D 26.94 393 393 8 125-‐B

Along Y Zone A (Exterior) (-‐)38.3 570 570 10 135 Zone A(Interior) (-‐)62.6 976 976 10 80 Zone B (Interior) (-‐)16.7 240 288 8 170 Zone C 35.1 520 520 8 95 Zone D 18.7 270 288 8 170

•  7.5 Ast2 – 71488Ast + Mu = 0 G-‐1.1(b)


•  Minimum Effec5ve Depth of Slab = 198 mm


Bar dia

Spacing mm

Along X Zone B 40.4 8 80 -‐B Zone D 26.94 8 125-‐B

Along Y Zone C 35.1 8 95 Zone D 18.7 8 170

#8@80

#8@125

#8@95 #8@95 #8@170

#8@80

40.4

26.94

40.4

-‐16.7 -‐62.6

18.7 35.1 35.1

A B A

D C C

A A B

-‐73.9

-‐73.9

-‐24.7

-‐62.6

-‐51.3

-‐51.3

0

-‐38.3 -‐38.3 -‐0 Exterior

Exterior

Interior

Interior

#10@

100

#10@

100

#10@

65

#10@

65

#8@140

#10@135 #10@80 #8@170(Min)*

#10@135

#8@170

#8@170(Min)*

* Op4onal Top Rebars

#10@80 Strip Loca6on Moment (kNm/m)

Bar dia

Spacing mm

Along X Zone A(Exterior) (-‐)51.3 10 100 -‐ T Zone A(Interior) (-‐)73.9 10 65 -‐ T Zone C(Interior) (-‐)24.7 8 140 -‐ T

Along Y Zone A (Exterior) (-‐)38.3 10 135 Zone A(Interior) (-‐)62.6 10 80 Zone B (Interior) (-‐)16.7 8 170

Transfer of Moments and Shears between Slabs and Columns •  The maximum load that a flat slab can support is dependent upon the strength of

the joint between the column and the slab.

•  Load is transferred by shear from the slab to the column along an area around the column

•  In addi7on moments are also transferred.

•  The moment situa7on is usually most cri7cal at the exterior columns.

•  Shear forces resul7ng from moment transfer must be considered in the design of the lateral column reinforcement (i.e., 7es and spirals).

EXAMPLE Compute moment transferred to Interior and corner Column in example 2

Interior Column •  As spans are same in both direc5ons •  M = 0.08 (0.5 wL L2 Ln2 /(1+1/αc) = 0.08 x 0.5 x 6 x 5.5 x 52 / 1.36 = 24.3 kNm •  this moment is distributed to top and borom column at junc5on in propor5on to their

s5ffness. •  M = 24.3/2 = 12.2 kNm Corner Column M = 131.4 kNm

Equivalent Frame Method (EFM) CL 31.5 •  More Comprehensive and Logical method •  Used when limita7ons of DDM are not

complied with •  Applicable when subjected to horizontal

loads 31.5.1 (a) Idealizing the 3D slab –column system to 2D frames along column Centre lines in both longitudinal and transverse direc6ons.

Longitudinal Frame

Transverse Frame Edge Frame

For ver6cal loads, each floor, together with the columns above and below, is analyzed separately. For such an analysis, the far ends of the columns are considered fixed. If there are large number of panels, the moment at a par6cular joint in a slab beam can be sa6sfactorily obtained by assuming that the member is fixed two panels away. This simplifica6on is permissible because ver6cal loads in one panel only appreciably affect the forces in that panel and in the one adjacent to it on each side.

31.5.1(b)

En6re Frame Analysis Gravity + Lateral Loads

For lateral loads, it is necessary to consider an equivalent frame that extends for the en4re height of the building, because the forces in a par4cular member are affected by the lateral forces on all the stories above the floor being considered.

31.5.1(C and d)

I2 = moment of iner4a at the face of the column / column capital c2 = dimension of column capital in the transverse direc4on l2 = width of equivalent frame.

varia6on of the flexural moment of iner6a

•  Varia4ons of moment of iner4a along the axis 0f the slab on account of provision of drops shall be taken into account

•  The s4ffening effect of flared column heads may be ignored

31.5.2 Loading Paiern

wu LL > ¾ wu,DL

Cri5cal Sec5on Interior Column Centre Line

Column /Capital face

C

< = C/2

Results in Significant reduc4on of design moments

Design Posi5ve Moment (Span region) M3 = M0 – (M1+M2)/2

Distribu5on of Moment

Similar to DDM

Example 3 : Compute moments in exterior/interior Panel along Longitudinal Span Longitudinal Span = 7.2m, Transverse Span = 6.4 m, Interior Column = 450mm dia, Column Capital = 1500mm dia, Exterior Column = 400x400mm, Column Capital = 870mm(square), Floor to Floor = 3.35 m, Slab Thickness = 240 mm, number of Panels = 4 in each direc6on

7.2 m

6.4m

6.4m

7.2 m 7.2 m 7.2 m

6.4m

Step 1: S5ffness Computa5ons

Exterior Column (Kce) = 4E x (4004 /12) /3350 = 2.55E106 = 1 Interior Column (KcI) = 4E x π(4504 /64) /3350 = 2.4E106 = 0.957 Slab(Ks) =4E x (6400 x 2403/12) /7200 = 4.1E106 = 1.608

Step 2: Simplified frame for analysis 31.5.1 (b)

7200 7200

3350

3350 1 2 3

A B

C D

Joint Member Rela5ve S5ffness Sum Distribu5on

Factors

1

1-‐A 1

3.608

0.277

1-‐2 1.608 0.446

1-‐C 1 0.277

2

2-‐B 0.957

5.13

0.187

2-‐1 1.608 0.314

2-‐3 1.608 0.314

2-‐D 0.957 0.187

Fixed End Moments = (16.5 x 6.4) x 7.22/12 = 456.2 kNm

Joint 1 2 3

Members FIXED 1A+1C 1-‐2 2-‐1 FIXED

2B+2D 2-‐3 FIXED 3-‐2

DF 0.554 0.446 0.314 0.374 0.314 -‐

FEM 456.2 -‐456.2 456.2 -‐456.2

Bal -‐252.74 -‐203.46 -‐ -‐ -‐ -‐

CO -‐ -‐ -‐101.73 -‐ -‐ -‐

Bal -‐ -‐ 31.94 38.04 31.94 -‐

CO -‐ 15.97 -‐ -‐ -‐ 15.97

Bal -‐8.85 -‐7.12 -‐ -‐ -‐ -‐

CO -‐ -‐ -‐3.56 -‐ -‐ -‐

Bal 1.12 1.33 1.12

Final end Moments -‐261.6 261.6 -‐528.43 39.37 489.26 -‐440.23

1 2 3

261.6 528.43 489.26 440.23

Counter Clockwise end moments are posi4ve

Step 3: Design Moments in Exterior Panel A. Design Nega6ve Moments at Cri6cal Sec6on

At Exterior Support : CL 31.5.3.2

870

400 470

235

Cri6cal Sec6on from Column Centre line = 435 mm

261.6 528.43 16.5 x 6.4 = 105.6 kN/m

105.6 x 7.2/2 -‐ (528.43-‐261.6)/7.2 = 343 kN

0.435

Design Moment = 343 x 0.435 -‐261.6 -‐105.6x0.4352/2 = -‐122.4 kNm (Hogging)

At Interior Support : CL 31.5.3.1

Width of equivalent square = 0.89D = 1335 mm

667.5 mm

Cri4cal Sec4on loca4on is at capital face

≤ 0.175x7200 = 1260mm

261.6 528.43 16.5 x 6.4 = 105.6 kN/m

343 kN 0.6675

Design Moment = 417.32 x 0.6675 -‐528.3 -‐105.6x0.66752/2 = -‐273.26 kNm (Hogging)

417.32

B. Design Posi4ve Moment

M(+) = (16.5 x 6.4x7.2)7.2/8 – ( 528.43 + 261.6)/2 = 289.3 kNm

Moments DDM EFM Posi4ve Moment (Span) 215.4 289.3 Nega4ve Moment(Exterior Support) 164 122.4 Nega4ve Moment (Interior Support) 315.3 273.3

Step 4: Design Moments in Interior Panel A. Design Nega6ve Moments at Cri6cal Sec6on

At Interior Support : CL 31.5.3.1

16.5 x 6.4 = 105.6 kN/m

387 kN

0.6675

Design Moment at A= 387 x 0.6675 -‐ 489.26 -‐105.6x0.66752/2 = -‐254.5 kNm (Hogging)

373.32

489.26 440.23

Design Moment at B = 373.32 x 0.6675 -‐ 440.23 -‐105.6x0.66752/2 = -‐214.6 kNm (Hogging)

0.6675

A B

B. Design Posi4ve Moment

M(+) = (16.5 x 6.4x7.2)7.2/8 – ( 489.26 + 440.23)/2 = 219.5 kNm

Moments DDM EFM Posi4ve Moment (Span) 158.9 219.5 Nega4ve Moment (Interior Support) 295.1 254.5/214.6

Need for Computer Analysis

The equivalent frame method is not sa6sfactory for hand calcula6ons. It is possible, however, to use computers and plane frame analysis programs if the structure is modeled such that various nodal points in the structure can account for the changing moments of iner6a along the member axis.

SLAB Drop Panel

Column Head Column

Column

FE Analysis of Slab

At any point in the plate bending, there will generally be two bending moments Mx , My in two mutually perpendicular direc5ons coupled with a complimentary twis5ng moment Mxy Design for flexure involves providing reinforcing steels in two orthogonal direc5ons to resist the moment field. Mx, My and Mxy.

Slab is idealized as an assembly of discrete plate bending elements joined at nodes

Wood –Armer equa4ons are used for this purpose.

Wood –Armer equa5ons (1968) •  This method was developed by considering the normal moment yield

criterion (Johansen’s yield criterion) aiming to prevent yielding in all direc4ons.

•  At any point in the slab, the moment normal to a direc4on, resul4ng due to design moments Mx , My , and Mxy must not exceed the ul4mate normal resis4ng moment in that direc4on.

•  Mx* cos2θ + My

* sin2θ -‐ flexural strength of plate in the direc4on of θ with X axis.

•  Mx cos2θ + Mysin2θ + 2 Mxy cosθ sinθ -‐ normal bending moment in the direc4on of θ

Example 1 FE results at centre of a plate element are: Mx = 7 kNm, My = 23 kNm, Mxy = 9 kNm. Compute design moments using Wood -‐ Armer equa4ons. A.  Borom rebars (Sagging Moments)

Mx* = Mx+|Mxy| = 16 > 0 , Mx* = 16 kNm My* = My+|Mxy| = 32 > 0 , My* = 32 kNm B. Top rebars (Hogging Moments) Mx* = Mx-‐|Mxy| = -‐2 < 0 , Mx* = 2 kNm My* = My-‐|Mxy| = 14 > 0 Set My* = 0 and compute Mx

* = Mx -‐| Mxy2/My| = 7 – |81/23| = 3.478 kNm

Example 2 FE results at centre of a plate element are: Mx = 7 kNm, My = -‐23 kNm, Mxy = 9 kNm. Compute design moments using Wood -‐ Armer equa4ons. A.  Borom rebars (Sagging Moments)

Mx* = Mx+|Mxy| = 16 > 0 , Mx* = 16 kNm My* = My+|Mxy| = -‐14 < 0 , Set My* = 0 and compute Mx

* = Mx + | Mxy2/My| = 7 + |81/23| = 10.52 kNm

B. Top rebars (Hogging Moments) Mx* = Mx-‐|Mxy| = -‐2 < 0 , Mx

* = 2 kNm My* = My-‐|Mxy| = -‐32 < 0 , MY

* = 32 kNm

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