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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
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 = 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
Step 6:Design Moments CL 31.4.2.2
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
Moment Direc5on along X Y Nega4ve Design Moment
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
Moment Direc5on along X Y Nega4ve Design Moment
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)
Width of Middle Strip Msw resis4ng M2 2.5 2.5 m
• 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
-‐ve sign : Hogging Moment (tension at top) +ve sign : Sagging Moment (tension at borom)
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 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 1.375 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
Step 2: Trial Depth CL 31.2.1
• 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
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 = 5.5/5.5 = 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)
Step 5: Check for punching shear around Column
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
Step 6:Design Moments CL 31.4.2.2
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
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 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
• Middle Strip M2 = 0.25MN 44.8 44.8 kNm 31.5.5.4(a)
Width of Middle Strip resis4ng M2 (Msw) 2.75 2.75 m
• -‐m2 = M2/Msw (Zone B & C) 16.3 16.3 kNm/m
Posi5ve Design Moment MP = 0.35*M0
96.3 96.3 kNm 31.4.3.2
• Column Strip M1 = 0.6MP 57.8 57.8 kNm 31.5.5.3
• +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
-‐ve sign : Hogging Moment (tension at top) +ve sign : Sagging Moment (tension at borom)
-‐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)
Step 9 :Rebar Details • Ast,min = 0.12 x 225 x 1000 /100 = 270 mm2/m 26.5.2.1 • Minimum Effec5ve Depth of Slab = 177 mm
#8@145
#8@180
#8@145
#8@180 #8@145 #8@145
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@90
#10@
90
#8@180
#8@200 #8@180
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
Step 5: Check for punching shear around Column
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
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
Step 6:Design Moments CL 31.4.2.2
M0 = 275 kNm
Parameters Along X Along Y Sum o f co l umn s4ffness above and below the slab 2 (4EcIc)/Lc
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
Slab s4ffness ks = 4EsIs/Ls
(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
Moment Direc5on along X Y Nega4ve Design Moment
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
• 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
M0 = 275 kNm
Moment Direc5on along X Y Nega4ve Design Moment
MN = -‐ (0.75 – 0.1/β)Mo 186 186 kNm 31.4.3.3
• Column Strip M1 = 0.75 MN 139.5 139.5 kNm 31.5.5.1
Width of Column Strip Csw resis4ng M1 2.75 2.75 m
• -‐m1 = M1/ Csw 50.73 50.73 kNm/m
• Middle Strip M2 = 0.25 MN 46.5 46.5 kNm 31.5.5.4(a)
Width of Middle Strip Msw resis4ng M2 2.75 2.75 m
• -‐m2 = M2/Msw 17 17 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 116.6 116.6 kNm 31.4.3.3
• Column Strip M1 = 0.6 MP 70 70 kNm 31.5.5.3
Width of Column Strip Csw resis4ng M1 2.75 2.75 m
• m1 = M1/ Csw 25.5 25.5 kNm/m
• Middle Strip M2 = 0.4 MP 46.7 46.7 kNm 31.5.5.4(a)
Width of Middle Strip Msw resis4ng M2 2.75 2.75 m
• m2 = M2/Msw 17 17 kNm/m
C. Posi4ve Moment in Mid Span:
m1
m1
m1
m1
m1
m1
m2 X
Y
Exterior
Exterior
Interior
Interior
m2
-‐ve sign : Hogging Moment (tension at top) +ve sign : Sagging Moment (tension at borom)
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)
Strip Loca6on Moment (kNm/m)
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)
Step 8 :Rebar Details • Ast,min = 0.12 x 225 x 1000 /100 = 270 mm2/m 26.5.2.1
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)
Moment direc6on Along
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 A – Corner Strip
Zone B – Middle Strip along X
Zone C – Middle Strip along Y
Zone D – Interior Region
Step 2: Trial Depth CL 31.2.1
• 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
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 = 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
Step 8:Design Moments CL 31.4.2.2
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
Step 9 : 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 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
• Middle Strip M2 = 0.25MN 73.78 61.91 kNm 31.5.5.4(a)
Width of Middle Strip Msw 3.2 4
• -‐m2 = M2/Msw 23.06 15.48 kNm/m
Posi5ve Design Moment MP = 0.35*M0
158.9 133.35 kNm 31.4.3.2
• Column Strip M1 = 0.6MP 95.34 80.01 kNm 31.5.5.3
• +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
Strip Loca6on Moment (kNm/m)
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)
Step 11 :Rebar Details • Ast,min = 0.12 x 240 x 1000 /100 = 288 mm2/m 26.5.2.1 • Minimum Effec5ve Depth of Slab = 198 mm
#8@170 #8@135 #8@135
#8@170
#8@110
#8@110
7.2 m
6.4 m 0.15Ln
0.15Ln
Borom Rebar Details in Interior Panel
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
Note: Distances for curtailment of rebars are measured from column face
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
Step 8:Design Moments CL 31.4.2.2
Along X Along Y
M0 = W Ln / 8 454 381 kNm
Step 9 : Distribution of Bending Moment across panel width ; CL 31.4.3.3 , 31.5.5
𝛼↓𝑐 = ∑↑▒𝑘↓𝑐 /𝑘↓𝑠
Equivalent side of circular column = 0.89D = 0.89x 450 = 400 mm Assume Ec = Es
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 400 x 4003/12) /3350 = 5.09 Ec x 106
Slab s4ffness ks = 4EsIs/Ls
(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
Moment Direc5on along X Y Nega4ve Design Moment
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
• Middle Strip M2=0 0 0 kNm 31.5.5.4(a)
Width of Middle Strip Msw resis4ng M2 3.2 4
-‐m2 = 0 0 0 kNm/m
A. Exterior nega4ve design moment: -‐m1
-‐m1
-‐m1
-‐m1
Exterior
Exterior
Interior
Interior 1.6
1.6
-‐m2
-‐m2
X
Y
Moment Direc5on along X Y Nega4ve Design Moment
MN = -‐ (0.75 – 0.1/β)Mo 315.3 266.9 kNm 31.4.3.3
• Column Strip M1 = 0.75 MN 236.5 200.2 kNm 31.5.5.1
Width of Column Strip Csw resis4ng M1 2x1.6 =3.2 2x1.6 =3.2 m
• -‐m1 = M1/ Csw -‐73.9 -‐62.6 kNm/m
• Middle Strip M2 = 0.25 MN 78.83 66.7 kNm 31.5.5.4(a)
Width of Middle Strip Msw resis4ng M2 3.2 4 m
• -‐m2 = M2/Msw -‐24.7 -‐16.7 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 215.4 187.2 kNm 31.4.3.3
• Column Strip M1 = 0.6 MP 129.3 112.3 kNm 31.5.5.3
Width of Column Strip Csw resis4ng M1 2x1.6 =3.2 2x1.6 =3.2 m
• m1 = M1/ Csw 40.4 35.1 kNm/m
• Middle Strip M2 = 0.4 MP 86.2 74.9 kNm 31.5.5.4(a)
Width of Middle Strip Msw resis4ng M2 3.2 4 m
• m2 = M2/Msw 26.94 18.7 kNm/m
C. Posi4ve Moment in Mid Span:
m1
m1
m1
m1
m1
m1
m2 X
Y
Exterior
Exterior
Interior
Interior
m2
-‐ve sign : Hogging Moment (tension at top) +ve sign : Sagging Moment (tension at borom)
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)
Strip Loca6on Moment (kNm/m)
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)
Step 11 :Rebar Details • Ast,min = 0.12 x 240 x 1000 /100 = 288 mm2/m 26.5.2.1
• Minimum Effec5ve Depth of Slab = 198 mm
Strip Loca6on Moment (kNm/m)
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
A. For bomom steel ( Sagging Moment +ve, Hogging Moment –ve)
Compute : Mx* = Mx +|Mxy| and My
* = My +|Mxy|
Case 1: If Mx* ≥ 0 and MY
* ≥ 0 then no change in computed values of Mx* and My
*
Case 2: If Mx* < 0 then Mx
* = 0 and MY* = MY + | Mxy
2/Mx|
Case 3: If My* < 0 then My
* = 0 and Mx* = Mx + | Mxy
2/My|
B. For Top steel ( Sagging Moment +ve, Hogging Moment –ve)
Compute : Mx* = Mx -‐|Mxy| and My
* = My -‐|Mxy|
Case 1: If Mx* ≤ 0 and MY
* ≤ 0 then no change in computed values of Mx* and My
*
Case 2: If Mx* > 0 then Mx
* = 0 and MY* = MY -‐| Mxy
2/Mx|
Case 3: If My* > 0 then My
* = 0 and Mx* = Mx -‐| Mxy
2/My|
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