Sd i-module3- rajesh sir

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

2
(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.

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.

4
With mild steel (Fe 250),
0 8 35
0 8 40
for simply supported slabs
for continuous slabs
.
.
x
x
l
D
l
⎧
⎪⎪ ×
≥ ⎨
⎪
⎪ ×⎩
With Fe415 steel,
35
40
for simply supported slabs
for continuous slabs
x
x
l
D
l
⎧
⎪⎪
≥ ⎨
⎪
⎪⎩
•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):

5
• 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).

6
• 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.

Nine different types of ‘restrained’ rectangular slab panels
lx
continuous (or
fixed) edge
simply supported
edge
ly

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 ≈ 4150
20 15× .
= 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 mm
dx = 165 – 20 – 5 = 140 mm
dy = 140– 10 = 130 mm
Design Problem 1
1. Effective span and trial depths

9
Effective spans
⎩
⎨
⎧
=+=
=+=
mm
mm
51301305000
41401404000
y
x
l
l
5130
4140
y
x
l
r
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:

10
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 = 0.0878
αy = 0.0571
short span: Mux = αx wulx
2 = 0.0878 × 12.20 × 4.1402 = 18.36 kNm/m
long span: Muy = αy wulx
2 = 0.0571 × 12.20 × 4.1402 = 11.94 kNm/m
Required spacing of 10 φ bars =
385
5.781000× = 204 mm
4. Design of Reinforcement
6
2 3 2
18 36 10
10 140
.ux
x
M
bd
×
=
×
= 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 =

11
(Ast)y, reqd = (0.204 × 10–2) × 1000 × 130 = 265.7 mm2/m
7.265
5.781000×
= 295 mm
3 3 140
3 3 130
(short span)
(long span)v
d
s
d
= ×⎧
≤ ⎨
= ×⎩
Maximum spacing (Cl.26.3.3 b)
10 200 392 5
10 290 270 7
2
2
(short span) mm m
(long span) mm m
,
,
@ .
@ .
st x
st y
c c A
c c A
ϕ
ϕ
⎧ ⇒ =⎪
⎨
⇒ =⎪⎩
Provide
6
2 3 2
11 94 10
10 130
.uy
y
M
bd
×
=
×
= 0.7065 MPa
b. Longer span

12
5. Check for deflection control
3
392 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 =
140
4140
= 29.6 < 30 — OK.

13
6. Check for shear
Average effective depth d = (140 + 130)/2 = 135 mm
Vu = wu(0.5lxn – d)
u
v
V
bd
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)

4000
230
8 φ bars
165
525
SECTION AA
PLAN OF FLOOR SLAB
A
165mm
thick
A
10 φ@ 200 c/c
10 φ@ 290 c/c
10 φ@ 290 c/c
10 φ @ 200 c/c
5000
230
230 230
525
425
7. Detailing

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 4140
5000 130 5130
mm
mm
x
y
l
l
= + =⎧
⎨
= + =⎩
1 24.y
x
l
l
=
Factored load wu = 12.20 kN/m2
Design Problem 2
1. Effective span and trial depths
2. Loads

( )
1 240 1 2
0 072 0 079 0 072
1 3 1 2
. .
. . – .
. .
−
= + ×
−
= 0.0748
Mux = αx wu lx
2
= 0.0748 × 12.20 × 4.142 = 15.61 kNm/m
Short span: αx
= 0.056
Mux = αy wu lx
2 = 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:

4. Design of reinforcement
6
2 3 2
15 61 10
10 140
.ux
x
M
bd
×
=
×
= 0.844 MPa
(Ast)x, reqd = (0.246 × 10–2) × 1000 × 140 = 334 mm2/m
334
3.501000×
= 150.7 mm
Maximum spacing permitted = 3 × 140 = 420 mm, but < 300 mm.
a. Shorter span

6
2 3 2
11 69 10
10 130
.uy
y
M
bd
×
=
×
= 0.714 MPa
(Ast)x, reqd = (0.206 × 10–2) × 1000 × 130 = 264 mm2/m
1000 50 3
264
.×
= 191 mm
Maximum spacing permitted = 3 × 130 = 375 mm, but < 300 mm.
b. Longer span
⎩
⎨
⎧
span)(long
span)(short
cc
cc
190@8
150@8
φ
φ
Provide

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 =
4140
140
= 30.4 < 31 — Hence, OK.

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.
( )2
830 8 4
0 75 334.
π× ×
×
160 c/c≅

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

830
5 nos 8φ
U–shaped
bars
160
SECTION BB
525 8 φ@ 190 c/c
8 φ@ 150 c/c
160
SECTION AA

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.

Sd i-module3- rajesh sir

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Sd i-module3- rajesh sir