Dept. of CE, GCE Kannur Dr.RajeshKN
Design of Two-way Slabs
Dr. Rajesh K. N.Assistant Professor in Civil Engineering
Govt. College of Engineering, Kannur
Design of Concrete Structures
Dept. of CE, GCE Kannur Dr.RajeshKN
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(Analysis and design in Module II, III and IV should be based on Limit State Method. Reinforcement detailing shall conform to SP34)
MODULE III (13 hours)
Slabs : Continuous and two way rectangular slabs (wall-supported) with different support conditions, analysis using IS 456 moment coefficients, design for flexure and torsion, reinforcement detailing –Use of SP 16 charts.
Staircases : Straight flight and dog-legged stairs – waist slab type and folded plate type, reinforcement detailing.
Dept. of CE, GCE Kannur Dr.RajeshKN
Two-Way Slabs
• Initial proportioning of the slab thickness may be done by span/effective depth ratios
• The effective span in the short span direction should be considered for this purpose
• A value of kt ≈ 1.5 ( modification factor to max l/d ratio) may be considered for preliminary design.
Dept. of CE, GCE Kannur Dr.RajeshKN
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With mild steel (Fe 250),
0 8 35
0 8 40
for simply supported slabs
for continuous slabs
.
.
x
x
l
Dl
⎧⎪⎪ ×≥ ⎨⎪⎪ ×⎩
With Fe415 steel,
35
40
for simply supported slabs
for continuous slabs
x
x
l
Dl
⎧⎪⎪≥ ⎨⎪⎪⎩
•For two-way slabs with spans up to 3.5 m and live loads not exceeding 3.0 kN/m2, span to overall depth ratio can be taken as follows, for deflection control (Cl. 24.1, Note 2):
Dept. of CE, GCE Kannur Dr.RajeshKN
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• According to the Code (Cl. 24.4), two-way slabs may be designed by any acceptable theory, using the coefficients given in Annex D.
• Code suggests design procedures (in the case of uniformly loaded two-way rectangular slabs) for:
• simply supported slabs whose corners are not restrained from lifting up [Cl. D–2].
• ‘torsionally restrained’ slabs, whose corners are restrained from lifting up and whose edges may be continuous or discontinuous [Cl. D–1].
• The flexural reinforcements in the two directions are provided to resist the maximum bending moments Mux = αx wu lx
2 (in the short span) and Muy = αy wu lx
2 (in the long span).
Dept. of CE, GCE Kannur Dr.RajeshKN
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• The moment coefficients prescribed in the Code (Cl. D–2) to estimate the maximum moments (per unit width) in the short span and long span directions are based on the Rankine-Grashoff theory.
• However, the moment coefficients recommended in the Code (Cl. D–1) are based on inelastic analysis (yield line analysis rather than elastic theory.
Dept. of CE, GCE Kannur Dr.RajeshKN
Nine different types of ‘restrained’ rectangular slab panels
lx
continuous (orfixed) edge
simply supportededge
ly
Dept. of CE, GCE Kannur Dr.RajeshKN
8
Design a simply supported slab to cover a room with internal dimensions 4.0 m × 5.0 m and 230 mm thick brick walls all around. Assume a live load of 3 kN/m2 and a finish load of 1 kN/m2. Use M 20 concrete and Fe 415 steel. Assume that the slab corners are free to lift up. Assume mild exposure conditions.
Effective short span ≈ 4150 mm
Assume an effective depth d ≈ 415020 1 5× .
= 138 mm
With a clear cover of 20 mm and say, 10 φ bars, overall thickness of slab D ≈ 138 + 20 + 5 = 163 mm
Provide D = 165 mmdx = 165 – 20 – 5 = 140 mmdy = 140– 10 = 130 mm
Design Problem 1
1. Effective span and trial depths
Dept. of CE, GCE Kannur Dr.RajeshKN
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Effective spans ⎩⎨⎧
=+==+=
mm mm
5130130500041401404000
y
x
ll
51304140
y
x
lr
l≡ = = 1.239
self weight @ 25 kN/m3 × 0.165m = 4.13 kN/m2
finishes (given) = 1.0 kN/m2
live loads (given) = 3.0 kN/m2
Total w = 8.13 kN/m2
Factored load wu = 8.13 × 1.5 = 12.20 kN/m2
2. Loads:
Dept. of CE, GCE Kannur Dr.RajeshKN
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3. Design Moments (for strips at midspan, 1 m wide in each direction)
As the slab corners are torsionally unrestrained, Table 27 gives moment coefficients:
αx = 0.0878αy = 0.0571
short span: Mux = αx wulx2 = 0.0878 × 12.20 × 4.1402 = 18.36 kNm/m
long span: Muy = αy wulx2 = 0.0571 × 12.20 × 4.1402 = 11.94 kNm/m
Required spacing of 10 φ bars =385
5.781000× = 204 mm
4. Design of Reinforcement6
2 3 218 36 1010 140
.ux
x
Mbd
×=
×= 0.9367 MPa
(Ast)x, reqd = (0.275 × 10–2) × 1000 × 140 = 385 mm2/m
a. Shorter span
[Table 3, Page 49, SP: 16]0 275,( ) .t x reqdp =
Dept. of CE, GCE Kannur Dr.RajeshKN
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(Ast)y, reqd = (0.204 × 10–2) × 1000 × 130 = 265.7 mm2/m
Required spacing of 10 φ bars =7.265
5.781000×= 295 mm
3 3 1403 3 130
(short span)(long span)v
ds
d= ×⎧
≤ ⎨ = ×⎩Maximum spacing (Cl.26.3.3 b)
10 200 392 510 290 270 7
2
2
(short span) mm m (long span) mm m
,
,
@ .@ .
st x
st y
c c Ac c A
ϕϕ
⎧ ⇒ =⎪⎨ ⇒ =⎪⎩
Provide
6
2 3 211 94 1010 130
.uy
y
Mbd
×=
×= 0.7065 MPa
b. Longer span
Dept. of CE, GCE Kannur Dr.RajeshKN
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5. Check for deflection control
3392 5 100
10 140,.
t xp = ×× = 0.280
fs = 0.58 × 415 × 385/392.5 = 236 MPa
Modification factor kt = 1.5 (Fig. 3 of Code)
(l/d)max = 20 × 1.5 = 30
(l/d)provided = 1404140
= 29.6 < 30 — OK.
Dept. of CE, GCE Kannur Dr.RajeshKN
13
6. Check for shear
Average effective depth d = (140 + 130)/2 = 135 mm
Vu = wu(0.5lxn – d)
uv
Vbd
τ = = 22.75 × 103/(1000 × 135) = 0.169 MPa
For pt = 0.28 ,
k c vτ τ> — Hence, OK.
= 0.376 MPaτ c
where lxn is the clear span in the short span direction
• The critical section for shear is to be considered d away from the face of the support.
•An average effective depth d = (dx + dy)/2 may be considered in the calculations.
= 12.20 (0.5 × 4.0 – 0.135) = 22.75 kN/m
(Table 19, Page 73)
1 3.k = (Cl. 40.2.1.1)
Dept. of CE, GCE Kannur Dr.RajeshKN
4000
230
8 φbars
165
525
SECTION AA
PLAN OF FLOOR SLAB
A
165 mm thick
A
10 φ@ 200 c/c
10 φ@ 290 c/c
10 φ@ 290 c/c
10 φ@ 200 c/c
5000230
230 230
525
425
7. Detailing
Dept. of CE, GCE Kannur Dr.RajeshKN
Repeat Design Problem 1, assuming that the slab corners are prevented from lifting up.
Assume D = 165 mm dx = 165 – 20 – 5 = 140 mm, dy = 140 – 10 = 130 mm
4000 140 41405000 130 5130
mm mm
x
y
ll= + =⎧
⎨ = + =⎩1 24.y
x
ll=
Factored load wu = 12.20 kN/m2
Design Problem 2
1. Effective span and trial depths
2. Loads
Dept. of CE, GCE Kannur Dr.RajeshKN
( ) 1 240 1 20 072 0 079 0 0721 3 1 2. .. . – .. .
−= + ×
−= 0.0748
Mux = αx wu lx2 = 0.0748 × 12.20 × 4.142 = 15.61 kNm/m
Short span: αx
= 0.056
Mux = αy wu lx2 = 0.056 × 12.20 × 4.142 = 11.69 kNm/m
Long span: αy
3. Design Moments
As the slab corners are to be designed as torsionally restrained, from Table 26 (Cl. D–1), the moment coefficients for ly/lx = 1.240 are:
Dept. of CE, GCE Kannur Dr.RajeshKN
4. Design of reinforcement
[Table 3, Page 49, SP: 16]0 2465,( ) .t x reqdp =
6
2 3 215 61 1010 140
.ux
x
Mbd
×=
×= 0.844 MPa
(Ast)x, reqd = (0.246 × 10–2) × 1000 × 140 = 334 mm2/m
Required spacing of 8 φ bars =334
3.501000× = 150.7 mm
Maximum spacing permitted = 3 × 140 = 420 mm, but < 300 mm.
a. Shorter span
Dept. of CE, GCE Kannur Dr.RajeshKN
[Table 3, Page 49, SP: 16]0 206,( ) .t x reqdp =
6
2 3 211 69 1010 130
.uy
y
Mbd
×=
×= 0.714 MPa
(Ast)x, reqd = (0.206 × 10–2) × 1000 × 130 = 264 mm2/m
Required spacing of 8 φ bars =1000 50 3264
.× = 191 mm
Maximum spacing permitted = 3 × 130 = 375 mm, but < 300 mm.
b. Longer span
⎩⎨⎧
span) (long span) (short
cccc
190@8150@8
φφ
Provide
Dept. of CE, GCE Kannur Dr.RajeshKN
5. Check for deflection control
0 2465, .t xp =
fs = 0.58 × 415 × 334/335 = 240 MPa
Modification factor kt = 1.55 (Fig. 3 of Code)
(l/d)max = 20 × 1.55 = 31
(l/d)provided = 4140140
= 30.4 < 31 — Hence, OK.
Dept. of CE, GCE Kannur Dr.RajeshKN
6. Corner Reinforcement [as per Cl. D–1.8]
As the slab is ‘torsionally restrained’ at the corners, corner reinforcement has to be provided at top and bottom (four layers),
• over a distance lx/5 = 830 mm in both directions • each layer comprising 0.75 Ast, x.
spacing of 8 φ bars
Provide 8 φ @ 160 c/c both ways at top and bottom at each corner over an area 830 mm × 830 mm.
( )2830 8 40 75 334.
π× ×
×160 c/c≅
Dept. of CE, GCE Kannur Dr.RajeshKN
PLAN
830
830
525
425
5000
AA
B B
230 230
4000
230
230
5 nos 8 φbars (U–shaped)
both ways (typ) at each corner
8 φ @ 150 c/c
8 φ @ 190 c /c
Dept. of CE, GCE Kannur Dr.RajeshKN
830
5 nos 8 φU–shaped
bars
160
SECTION BB
525 8 φ@ 190 c/c
8 φ@ 150 c/c
160
SECTION AA
Dept. of CE, GCE Kannur Dr.RajeshKN
Summary
Slabs : Continuous and two way rectangular slabs (wall-supported) with different support conditions, analysis using IS 456 moment coefficients, design for flexure and torsion, reinforcement detailing –Use of SP 16 charts.
Staircases : Straight flight and dog-legged stairs – waist slab type and folded plate type, reinforcement detailing.