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


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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.


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.


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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):


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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).


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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.


Nine different types of ‘restrained’ rectangular slab panels

lx

continuous (orfixed) edge

simply supportededge

ly


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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


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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:


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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 =


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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

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2 3 211 94 1010 130

.uy

y

Mbd

×=

×= 0.7065 MPa

b. Longer span


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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.


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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)


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


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


( ) 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:


4. Design of reinforcement


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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



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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


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.


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≅


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 8 φU–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.