2/10/2010 - civil.utm.mycivil.utm.my/iznisyahrizal/files/2013/09/chapter-7-slab-design.pdf · 11...

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Reinforced Concrete Slab

One-way

Simply supported one-

way slab

Continuous one-way slab

Two-way

Simply supported two-

way slab

Restrained two-way slab

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Step Task Standard

1 Determine design life, Exposure class & Fire resistance

EN 1990 Table 2.1

EN 1992-1-1: Table 4.1

EN 1992-1-2: Sec. 5.6

2 Determine material strengthBS 8500-1: Table A.3

EN 206-1: Table F1

3 Select size of thickness of slab EN 1992-1-1: Table 7.4N

EN 1992-1-2: Table 5.8

4 Calculate min. cover for durability , fire and bond requirements EN 1992-1-1: Sec. 4.4.1

5 Estimate actions on slabs EN 1991-1-1

6Analyze structure to obtain critical moments and

shear forcesEN 1992-1-1: Sec. 5

7 Design flexural reinforcement EN 1992-1-1: Sec. 6.1

8 Check shear EN 1992-1-1: Sec. 6.2

9 Check deflection EN 1992-1-1: Sec. 7.4

10 Check cracking EN 1992-1-1: Sec. 9.3

11 Detailing EN 1992-1-1: Sec.8 & 9.3

The selection of slab thickness from structuralviewpoint is often dictated by deflection controlcriteria. In practice, the overall depths of slabs areoften fixed in relation to their spans.

Span to overall depth ratios of 20 to 30 aregenerally found to be economical in the case ofsimply supported and continuous beams.

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BS EN 1992-1-2: 2004: Table 5.8

The analysis of reinforced concrete slab to determinebending moment and shear force may be done eitherby using elastic analysis or by considering plasticcollapse methods. Hence one of the following methodscan be use.

Elastic analysis Yield line method Hillerborg strip method Moment and shear coefficient from the Code of

Practice or design handbook

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(1) Moment Design

Calculate K = M/fckbd2

If K Kbal (0.167), compression reinforcement is NOT required:

z = d [0.5 + (0.25 – K/1.134)]

As = M0.87fykz

EN 1992-1-1: 2004: Cl. 9.3

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(2) Shear Design

Check and ensure that VEd VRd, c

whereVRd, c = [0.12k (100lfck)

1/3] bwd [0.035k3/2fck1/2]bwd

andk = (1 + 200/d) 2.0 expressed in mml = (Asl/bwd) 0.02Asl = Area of tensile reinforcement that extends (lbd + d) beyond the section consideredbw = the smallest width of the section in tensile area (mm)

EN 1992-1-1: 2004: Cl. 6.2.2

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EN 1992-1-1: 2004: Cl. 6.2.2

(3) Deflection

Check and ensure than (l/d)actual (l/d)allow

where (l/d)allow = (l/d)basic (7/leff) (As, prov/As, req)

(7/leff) if L 7 m (As, prov/As, req) 1.5

and (l/d)basic =

EN 1992-1-1: 2004: Cl. 7.4

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(4) Cracking

(i) Minimum area of principal reinforcement, As, min = 0.26fctmbtd/fyk 0.0013btdwhere bt is the mean width of the tension zone (Cl. 9.3)

(i) The minimum area of secondary reinforcement is 20% As, min. (Cl. 7.3)

(ii) For slab with less than 200 mm thickness, spacing of principal reinforcement bars 3h or 400 mm (whichever is the lesser). For secondary reinforcement, spacing 3.5h or 450 mm (whichever is the lesser). (Cl. 7.3)

EN 1992-1-1: 2004

ly/lx 2.0

Main reinforcement is design in ONE-DIRECTION only

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4250 mm

Design data

fck = 30 N/mm2

fyk = 500 N/mm2

Design life = 50 yearsFire resistance = R60Exposure class = XC3Assumed bar = 12 mm

Permanent, gk (excluding self weight) = 1.0 kN/m2

Variable, qk = 3.0 kN/m2

Slab Thickness

Minimum thickness for fire resistance = 80 mmThickness considering deflection control, h = 4250/26 (assumed)

= 163 mm Use h or hf = 175 mm

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Durability, Fire & Bond Requirements (Concrete Cover)

Minimum concrete cover regard to bond, Cmin, b = 12 mm (Table 4.2)Minimum concrete cover regard to durability, Cmin, dur = 20 mm (Table 4.4N)Minimum required axis distance for R60 fire resistance, a = 20 mm (Table 5.8: BS EN 1992-1-2: 2004)Minimum concrete cover regard to fire, Cmin, fire = a – bar/2

= 20 – 12/2 = 14 mm

Allowance in design for deviation, Cdev = 10 mm

Nominal cover, Cnom = Cmin + Cdev = 20 + 10 = 30 mm

Use Cnom = 30 mm

Analysis

Slab self weight = 25.0 x 0.175 = 4.38 kN/m2

Finishing etc. = 1.00 kN/m2

Total permanent action, gk = 4.38 + 1.00 = 5.38 kN/m2

Total variable action, qk = 3.00 kN/m2

Design action, nd = 1.35gk + 1.5qk = 11.76 kN/m2

Consider 1 m width, wd = nd 1m = 11.76 kN/m/m width

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Mmax = wL2/8 = 26.50 kNm/m widthVmax = wL/2 = 25.0 kN/m width

4.25 m

wd = 11.76 kN/m/m width

Main Reinforcement

Effective depth, d = 175 – 30 – 12/2 = 139 mm

K = M/fckbd2 = 26.50 106/(30 1000 1392) = 0.046 Kbal = 0.167

Compression reinforcement is NOT required

z = d[0.5 + 0.25 – (K/1.134)] = 0.96d 0.95d Use z = 0.95d

As = M/0.87fykz= 26.50 x 106 / (0.87 500 0.95 139)= 462 mm2/m width

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Check Area of ReinforcementAs, min = 0.26(fctm/fyk) bd

= 0.26 (2.90/500) bd = 0.0015 bd 0.0013 bd As, min = 0.0015bd = 209 mm2/m width

As, max = 0.04Ac = 0.04bh= 7000 mm2/m width

Main bar H12-225 (As = 503 mm2/m width)and

Secondary bar H12-450 (As = 251 mm2/m width)

Shear

Maximum design shear force, VEd = 25.0 kN/m width

k = 1 + (200/d)1/2 2.0= 1 + (200/139)1/2 = 2.20 2.0 Use k = 2.0

l = Asl/bd 0.02= 503 / (1000 139) = 0.0036

VRd, c = [0.12k(1000lfck)1/3] bd

= [0.12 2.0 (1000 0.0036 30)1/3] 1000 139 10-3

= 73.9 kN/m widthVmin = [0.0035k3/2fck

1/2]bd= [0.0035 2.03/2 301/2] 1000 139 10-3 = 75.4 kN/m width

VEd (25 kN/m width) 75.4 kN/m width OK

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Deflection

Percentage of required tension reinforcement; = As, req / bd

=462/ 1000 139 = 0.0033

Reference reinforcement ratio;o = (fck)

1/2 10-3 = (30)1/2 10-3 = 0.0055

Since o Use Eq. (7.16a) in EC 2 Cl. 7.4.2

For structural system, K = 1.0

(l/d)basic = 1.0 (11 + 13.5 + 9.13) = 33.70

Modification factor for span less than 7 m = 1.00Modification for steel area provided = As, prov / As, req = 503/462 = 1.09 1.50

(l/d)allow = 33.70 1.00 1.09 = 36.63

(l/d)actual = 4250/139 = 30.6 (l/d)allow

Deflection OK

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Cracking

hf = 175 mm 200 mm

Main bar:Smax, slab = 3hf 400 mm = 400 mmMax bar spacing = 225 mm Smax, slab OK

Secondary bar:Smax, slab = 3.5hf 450 mm = 450 mmMax bar spacing = 450 mm Smax, slab OK

Cracking OK

Max bar spacing

4250 mm

225 mm

H12

450 mm

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4250 mm

H12-2-450B

H12

-1-2

25

B

1

2 2175 mm

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BS 8110: Part 1: 1997

F is the total ultimate load = (1.35Gk + 1.5Qk)l is the effective span

BS 8110: Part 1: 1997

Rules & Regulations

(i) Area of bay 30 m2

(ii) qk/gk 1.25(iii) qk 5.0 kN/m2

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Design data

fck = 25 N/mm2

fyk = 500 N/mm2


6 @ 4 m

9 mly/lx = 2.25 2.0

Permanent, gk (excluding self weight) = 1.0 kN/m2

Variable, qk = 3.5 kN/m2

Slab Thickness, hf

Minimum thickness for fire resistance = 100 mmEstimated thickness considering deflection control = hf = 4000/30 = 133 mm

Therefore, try hf = 150 mm

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= 20 – 10/2 = 15 mm



Use Cnom = 25 mm

Actions

Slab self weight = 25.0 x 0.150 = 3.75 kN/m2

Permanent action (excluding self weight) = 1.00 kN/m2

Characteristics permanent action, gk = 4.75 kN/m2

Characteristics variable action, qk = 3.50 kN/m2


Consider 1 m width, wd = 1.0 m x 11.66 kN/m2 = 11.66 kN/m/m width

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Bay area = 6(9 x 4) = 216 m2 30 m2 OK

qk/gk = 3.50/4.75 = 0.74 1.25 OKqk = 3.50 kN/m2 5.0 kN/m2

OK

Use Table 3.12: BS 8110: Part 1Bending Moment

At near middle of end span;M = 0.075FL = 0.075 x (11.66 x 4) x 4 = 14.0 kNm/m width

At middle interior spans and interior supports;M = 0.063FL = 0.063 x (11.66 x 4) x 4 = 11.8 kNm/m width

At outer support;M = 0.04FL = 0.04 x (11.66 x 4) x 4 = 7.5 kNm/m width

At first interior support;M = 0.086FL = 0.086 x (11.66 x 4) x 4 = 16.0 kNm/m width

16.0

14.011.8

11.816.0

14.0

4 m 4 m 4 m 4 m 4 m 4 m

Bending Moment Diagram

7.5 11.8 11.8

11.8 11.8 11.8

7.5

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Main Reinforcement

Effective depth, d = 150 – 25 – 10/2 = 120 mm

Minimum and Maximum Reinforcement

As, min = 0.26(fctm/fyk) bd= 0.26 (2.56/500) bd = 0.0013 bd 0.0013 bd

As, min = 0.0013bd = 160 mm2/m width

As, max = 0.04Ac = 0.04bh = 0.04 1000 150= 6000 mm2/m width

Provide H10-425 (As = 185 mm2/m) for Secondary Reinforcements

(a) At near middle of end span

K = M/fckbd2 = 14.0 x 106/(25 x 1000 x 1202) = 0.039 Kbal = 0.167Compression reinforcement NOT required

z = 0.97d 0.95d Use z = 0.95d

As = M/0.87fykz= 14.0 x 106 / (0.87 x 500 x 0.95 x 120)= 282 mm2/m width As, min

Provide H10-275 bottom (As = 286 mm2/m width)

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(b) At middle interior spans and interior supports


z = 0.97d 0.95d Use z = 0.95d



(c) At outer supports


z = 0.98d 0.95d Use z = 0.95d


Use As, min = 160 mm2/m width


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(d) At first interior support


z = 0.96d 0.95d Take z = 0.95d


Provide H10-225 (As = 349 mm2/m width)

Shear

Maximum design shear force, VEd = 0.6F = 0.6 11.66 4 = 28.0 kN/m width

k = 1 + (200/d)1/2 2.0= 1 + (200/120)1/2 = 2.29 2.0 Use k = 2.0

l = Asl/bd 0.02= 286 / (1000 120) = 0.0024

VRd, c = [0.12k(1000lfck)1/3] bd

= [0.12 2.0 (1000 0.0024 25)1/3] 1000 120 10-3 = 52.2 kN/m widthVmin = [0.0035k3/2fck

1/2]bd= [0.0035 2.03/2 251/2] 1000 120 10-3 = 59.4 kN/m width

VEd (28.0 kN/m width) 59.4 kN/m width OK

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Deflection

* Check at the near middle of end span. WHY?


=282/ 1000 120 = 0.0024


1/2 10-3 = (25)1/2 10-3 = 0.0050



(l/d)basic = 1.0 (11 + 15.9 + 19.12) = 59.9


(l/d)allow = 59.9 1.00 1.01 = 60.61


Deflection OK

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Cracking

hf = 150 mm 200 mm


Secondary bar:Smax, slab = 3.5hf 450 mm = 450 mmMax bar spacing = 425 mm Smax, slab OK

Cracking OK

Max bar spacing

4000 mm

H10-4-425B

H10

-1-2

75B

4000 mm

H10

-2-2

25T

H10

-3-4

00T

H10-4-425T

H10-4-425B

H10

-1-3

00B

H10-4-425T

H10

-2-3

00T

H10-4-425T

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Main reinforcement in BOTH DIRECTION

Slab is supported at all sides

ly/lx 2.0

Simply Supported Two-way Slab

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Simply Supported Two-way Slab – Bending Moment

msx = sxnlx2

msy = synlx2

n = total ultimate load per unit arealx = length of the shorter sidely = length of the longer sidesx and sy are the moment coefficient from Table 3.13: BS 8110: Part 1: 1997

Design data

Slab area = 5 m 7.5 m and supported on steel beams at ALL sidesSlab depth, hf = 200 mm

fck = 25 N/mm2

fyk = 500 N/mm2


Permanent action (excluding self weight), gk = 1.20 kN/m2

Variable action, qk = 2.5 kN/m2

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= 15 – 10/2 = 10 mm



Use Cnom = 25 mm

Action

Slab self weight = 25.0 kN/m3 0.200 m = 5.00 kN/m2

Permanent (excluding self weight) = 1.20 kN/m2





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Bending Moment

ly/lx = 7.5/5.0 = 1.5 2.0 Two-way slab

(a) In the x-direction

msx = sx.n.lx2 = 0.104 12.12 52 = 31.51 kNm/m width

(b) In the y-direction

msy = sy.n.lx2 = 0.046 12.12 52 = 13.94 kNm/m width

Main Reinforcement

dx = 200 – 25 – 10/2 = 170 mmdy = 200 – 25 – 10 – 10/2 = 160 mm



As, min = 0.0013bd = 221 mm2/m width


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(a) In the Short Span

K = M/fckbd2 = 31.51 106/(25 1000 1702) = 0.044 Kbal = 0.167Compression reinforcement NOT required

z = 0.96d 0.95d Use z = 0.95d

As = M/0.87fykz= 31.51 106 / (0.87 500 0.95 170)= 449 mm2/m width As min


(b) In the Long Span


z = 0.98d 0.95d Use z = 0.95d

As = M/0.87fykz= 13.94 106 / (0.87 500 0.95 160)= 211 mm2/m width As min

Use As, min = 221 mm2/m width


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Shear

Maximum design shear force, VEd = nlx/2= 12.12 5.0/2 = 30.3 kN/m width

k = 1 + (200/d)1/2 2.0= 1 + (200/170)1/2 = 2.08 2.0 Use k = 2.0

l = Asl/bd 0.02= 449 / (1000 170) = 0.0026

VRd, c = [0.12k(1000lfck)1/3] bd

= [0.12 2.0 (1000 0.0026 25)1/3] 1000 170 10-3 = 76.5 kN/m widthVmin = [0.0035k3/2fck

1/2]bd= [0.0035 2.03/2 251/2] 1000 170 10-3 = 84.1 kN/m width


Deflection


=449/ 1000 170 = 0.0026


1/2 10-3 = (25)1/2 10-3 = 0.0050


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(l/d)basic = 1.0 (11 + 14.2 + 13.55) = 38.8


(l/d)allow = 59.9 1.00 1.00 = 59.9


Deflection OK

Cracking

hf = 200 mm 200 mm


Cracking OK

Max bar spacing

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ly =7500 mm

H10-1-175B

H10

-2-3

50

B

l x=5

00

0 m

m

Plan View

2 1 1

200 mm

Cross Sectional View

ly =7500 mm

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Restrained Two-way Slab

Restrained Two-way Slab – Bending Moment

msx = sxnlx2

msy = synlx2

n = total ultimate load per unit arealx = length of the shorter sidely = length of the longer sidesx and sy are the moment coefficient from Table 3.14: BS 8110: Part 1: 1997

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Restrained Two-way Slab – Shear Force

vsx = vxnlxvsy = vynlx

n = total ultimate load per unit arealx = length of the shorter sidely = length of the longer sidevx and vy are the moment coefficient from Table 3.15: BS 8110: Part 1: 1997

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Reinforcement – Torsional

Lx/5

0.75 x As @ mid-span

Continuous

Co

nti

nu

ou

s

4000

4000

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Design dataSlab thickness, hf = 125 mm

fck = 25 N/mm2

fyk = 500 N/mm2


Permanent action (excluding self weight), gk = 1.5 kN/m2

Variable action, qk = 4.0 kN/m2



= 15 – 10/2 = 10 mm



Use Cnom = 25 mm

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Action

Slab self weight = 25.0 kN/m3 0.125 m = 3.13 kN/m2

Permanent (excluding self weight) = 1.50 kN/m2





Bending Moment

Ly/Lx = 7.0/4.0 = 1.75 2.0 Two-way slab

Case 4: Two Adjacent Edges Discontinuous

Staircase

0.065

0.034-0.045

-0.087

1B 1C

2B 2C

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(a) In the x-direction

Mid-span: msx = sx.n.lx2 = 0.065 12.24 42 = 12.7 kNm/m width

Support: msx = sx.n.lx2 = 0.087 12.24 42 = 17.0 kNm/m width

(b) In the y-direction

Mid-span: msy = sy.n.lx2 = 0.034 12.24 42 = 6.66kNm/m width

Support: msy = sy.n.lx2 = 0.045 12.08 42 = 8.82 kNm/m width

Main ReinforcementEffective Depth

dx = 125 – 25 – 10/2 = 95 mmdy = 125 – 25 – 10 – 10/2 = 85 mm



As, min = 0.0013bd= 0.0013 1000 95 = 127 mm2/m width


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(a) In the x-direction (Short Span)Mid-span


z = 0.95d

As = M/0.87fykz= 12.7 106 / (0.87 500 0.95 95)= 325 mm2/m width As, min

Provide H10-200 (393 mm2/m width)

Support


z = 0.93d 0.95d



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(b) In the y-direction (Long Span)Mid-span


z = 0.97d 0.95d Use z = 0.95d



Support


z = 0.96d 0.95d Use z = 0.95d



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Torsional Reinforcement

1B 1C

2C

75% As @ mid-span= 0.75 x 325 = 244 mm2/m As min

Provide H10-300 (262 mm2/m)

Void

2B





0.57

0.260.40

0.38

Void

Shear

1B 1C

2C2B

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Shear

Maximum design shear force, VEd = 0.57 12.24 4 = 27.9 kN/m width

k = 1 + (200/d)1/2 2.0= 1 + (200/95)1/2 = 2.45 2.0 Use k = 2.0

l = Asl/bd 0.02= 449 / (1000 95) = 0.0047 Why choose Asl = 449 mm2 (at support)?

VRd, c = [0.12k(1000lfck)1/3] bd

= [0.12 2.0 (1000 0.004725)1/3] 1000 95 10-3 = 51.9 kN/m widthVmin = [0.0035k3/2fck

1/2]bd= [0.0035 2.03/2 251/2] 1000 95 10-3 = 47.0 kN/m width


Deflection


= 325/ 1000 95 = 0.0034


1/2 10-3 = (25)1/2 10-3 = 0.0050


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(l/d)basic = 1.3(11 + 11.0 + 5.00) = 35.0


(l/d)allow = 59.9 1.00 1.21 = 42.32


Deflection OK

Cracking

hf = 125 mm 200 mm


Secondary bar:Smax, slab = 3.5hf 450 mm = 437.5 mmMax bar spacing = 425 mm Smax, slab OK

Cracking OK

Max bar spacing

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2/10/2010 - civil.utm.mycivil.utm.my/iznisyahrizal/files/2013/09/chapter-7-slab-design.pdf · 11...

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