slab - rectangular - 4 edges supporeted

8/17/2019 Slab - Rectangular - 4 Edges Supporeted

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2.2.1 DESIGN INFORMATION

Material Properties

Characteristic strength of concrete, = N/mm2

Characteristic strength of steel, = N/mm2

Density of concrete, = kN/m3

Nominal maximum size of aggregate = mm

Exposure condition =

Fire resistance = hours

Base Slab Section Properties

Thickness of slab = mm

Length of slab = mm

Width of slab = mm

Reinforcement Properties

Reinforcement design width, = mm

Bottom r/f span shortway = T bars

Top r/f span shortway = T bars

Bottom r/f span longway = T bars

Top r/f span longway = T bars

2.2.2 DESIGN LOADS

Superimposed dead load per unit area = kN/m2

Live load per unit area = kN/m2

ly = 5029 mm

Type of Panel = Two adjacent edges discontinuous

Design Loading

Dead load = kN/m2

Live load = kN/m2

Ultimate design load = kN/m2

= 1.186ly/lx =

Output

ρct 24

BS8110:1985

Table 3.1

2/6/2016

PROJECT

STRUCTURE ELEMENT REF

DESIGNED BY

h 125

12

4 2 4 2 m m

CODES BS 8110 : PART 1 :1985 KUSHAN CHECKED BY

Reference Calculations

f y 460

f cu 25

Durability Requirement

ØB2

Clause 2.7.6

2.5

Discontinuous

C o n t i n u o u s

D i s c o n t i n u o u s

ØB1

ly 5029

lx 4242

bt

ØT2 12

ØT1

1000

12

12

3

P o s

i t i o n

‐ 1

Position ‐ 2

P o s i t i o n‐ 3

Position ‐ 4

gk

qk

n

ly/lx 1.19

l x

=

Continuous

5.5

3

12.5

hagg 20

Mild

2


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PROJECT


DESIGNED BYCODES BS 8110 : PART 1 :1985 KUSHAN CHECKED BY


2.2.3 DETERMINATION OF COVER

Grade of concrete =

Exposure condition =

Fire resistance = hours → Cover <

Nominal maximum size of aggregate = mm → Cover <

Maximum bar size = mm → Cover <

Minimum nominal cover = mm

Check for minimum dimension of slab for fire resistance

→ OK

= h ‐ cover ‐ ØT1/2

= 125 ‐ 25 ‐ 12/2

= mm 94mm

= h ‐ cover ‐ ØT1 ‐ ØT2/2

= 125 ‐ 25 ‐ 12 ‐ 12/2

= mm 82mm

= h ‐ cover ‐ ØB1/2

= 125 ‐ 25 ‐ 12/2

= mm 94mm

= h ‐ cover ‐ ØB1 ‐ ØB2/2

= 125 ‐ 25 ‐ 12 ‐ 12/2

= mm 82mm

Bending moment coefficients (β)

For Position ‐ 1 =






Table 3.15

Effective depth for bottom reinforcement

long span direction

dB2

82 dB2 =

82 dT2 =

Effective depth for top reinforcement short

span direction

Effective depth for top reinforcement long

span direction

Effective depth for bottom reinforcement

short span direction

dB1

94 dB1 =

Clause 3.3.1.3 20 20

Clause 3.3.1.2 12 12

25 Cover = 25mm

Figure 3.2Requirement for minimum thickness of 125mm thick

slab for 2 hours fire resistance = 125 mm

94

Table 3.4 25

→ Cover < 25Mild

Table 3.5 2 25

≤ 125 mm

OK

dT1

dT1 =

βsx

βsy

0.000

βsx

0.046

0.045

βsx

βsy

βsy

dT2

0.000

0.062

0.034

5

4

2

1 3

6


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Output

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PROJECT




Bottom Reinforcement

Shortway direction

Bending moment in shortway direction = βsxnlx2

= 0.046 x 12.5 x (4242/1000)^2

= kNm /m width

= M/bd2f cu

= 10.41x10^6/[1000x94^2x25]

=

= [0.5+(0.25‐K/0.9)1/2

]d

= [0.5+(0.25‐0.047/0.9)^0.5]d

= d

= mm

As,req = M/0.87f yZ

= (10.41x10^6)/(0.87x460x88.79)

= mm2/m width

Hence required steel % for bending moment = 100As,req/bd

= 100x293/(1000x94)

= %

→

= lesser of (3d or 750)

= mm

mm2/m width

Longway direction

Bending moment in shortway direction = βsynlx2

= 0.034 x 12.5 x (4242/1000)^2= kNm /m width

= M/bd2f cu

= 7.65x10^6/[1000x82^2x25]

=

= [0.5+(0.25‐K/0.9)1/2

]d

= [0.5+(0.25‐0.045/0.9)^0.5]d

= d

= mm

As,req = M/0.87f yZ

= (7.65x10^6)/(0.87x460x77.6210687992582)

= mm2/m width


= 100x246/(1000x82)

= %

→


= mm

mm2/m width

equation 14 msy

7.65

Clause 3.4.4.4 K

0.045

Clause 3.4.4.4 Z

Bottom r/f in longway direction, provide T12 @ 200 mm C/C Providing

0.95

Z 78

OK

246

565.5

Clause 3.4.4.4 Required steel area to carry bending

moment,

246

0.30

Required minimum percentage of tension

reinforcement= ≤ 0.30 %0.13 %

%

OK

754.0Bottom r/f in shortway direction, provide T12 @ 150 mm C/C Providing

282

0.31


reinforcement= 0.13 % ≤ 0.31

Required steel area to carry bending

moment,

293

Clause 3.4.4.4

Clause 3.4.4.4

Clause 3.4.4.4

equation 14

K

0.047

Z

0.94

Z 89

msx

10.41

The maximum spacing of reinforcement for slabs with

less than 200mm thick



Clause

3.12.11.2.6

Clause

3.12.11.2.6

K < 0.156, and compression r/f are not required



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Output

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PROJECT




Top Reinforcement

Shortway direction

Bending moment in shortway direction = βsxnlx2

= 0.062 x 12.5 x (4242/1000)^2

= kNm /m width

= M/bd2f cu

= 13.94x10^6/[1000x94^2x25]

=

= [0.5+(0.25‐K/0.9)1/2

]d

= [0.5+(0.25‐0.063/0.9)^0.5]d

= d

= mm

As,req = M/0.87f yZ

= (13.94x10^6)/(0.87x460x86.87)

= mm2/m width


= 100x401/(1000x94)

= %

→


= mm

mm2/m width

Longway direction

Bending moment in shortway direction = βsynlx2

= 0.045 x 12.5 x (4242/1000)^2= kNm /m width

= M/bd2f cu

= 10.12x10^6/[1000x82^2x25]

=

= [0.5+(0.25‐K/0.9)1/2

]d

= [0.5+(0.25‐0.06/0.9)^0.5]d

= d

= mm

As,req = M/0.87f yZ

= (10.12x10^6)/(0.87x460x76.087546223696)

= mm2/m width


= 100x332/(1000x82)

= %

→


= mm

mm2/m widthTop r/f in longway direction, provide T12 @ 200 mm C/C Providing 565.5


moment,

332

0.40


reinforcement= 0.13 % ≤ 0.40 %

0.060

Clause 3.4.4.4 Z

0.93

Z 76

OK

246

Top r/f in shortway direction, provide T12 @ 250 mm C/C Providing 452.4

equation 14 msy

10.12

Clause 3.4.4.4 K


reinforcement= 0.13 % ≤ 0.43 %

OK

282

0.92

Z 87


moment,

401

0.43

equation 14 msx

13.94

Clause 3.4.4.4 K

0.063

Clause 3.4.4.4 Z

Table 3.27





Clause

3.12.11.2.6

Clause

3.12.11.2.6




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PROJECT




Nominal reinforcement for discontinuous edge

Shortway direction

= lesser of (3d or 750)= mm

Top r/f in shortway direction for discontinuous edge, provide T12 @ 250 mm C/C providing

mm2/m width

Longway direction


= mm

Top r/f in longway direction for discontinuous edge, provide T12 @ 200 mm C/C providing

mm2/m width

Shear resistance of slab

Shear force coefficients (β)





Maximum design shear force in slab = βv,maxnlx

= 0.47 x 12.5 x 4.242

= kN/m width

Minimum effective depth in slab = mm

= Vmax/bdmin

= (24.69x10^3)/(1000x82)

= N/mm2

Minimum of (0.8√f cu or 5 N/mm2) = N/mm

3 >

→

Table 3.27 Required minimum percentage of tension

reinforcement= 0.13 %

452.4

Clause3.12.11.2.6

The maximum spacing of reinforcement for slabs withless than 200mm thick 282

Clause

3.12.11.2.6


less than 200mm thick 246

565.5

βvx 0.47

Vmax

Table 3.16

βvy 0.40

βvx 0.31

βvy 0.26

Maximum possible design shear stress in

slab

4.00

Ok

vmax

0.301

24.69

v

Ok

dmin 82

2

3

4

1


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Output

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PROJECT




→ = (100x452.39)/(1000x94)

= < 3

→

= 400/82

= > 1

→ Ok

vc, min = 0.79[100As/bd]1/3

[400/d]1/4

[f cu/25]1/3

/γm

= {0.79[0.5]^(1/3)x[4.9]^(1/4)x[25/25]^(1/3)}/1.25

= N/mm2

> → OK

Deflection

Actual (Span/effective depth) ratio le/d =

Basic (Span/effective depth) ratio =

1

βo

1

1

= N/mm2

Allowable (Span/effective depth) ratio = 26x1.78= ≥ le/d

2.2.4 SUMMARY OF REINFORCEMENT

Span shortway r/f = @ 150 mm C/C

Span longway r/f = @ 200 mm C/C

Continuous edge shortway r/f = @ 250 mm C/C

Continuous edge longway r/f = @ 200 mm C/C

If necessary provide

Discontinuous edge shortway r/f = @ 250 mm C/C

Discontinuous edge longway r/f = @ 200 mm C/C

equation 7 Design service stress f s =

5f yAs, reqx8As, prov

BS8110:1985

Part ‐ 1 Table

3.9 Conditions to

Satisfy

0.481265

Ok

400/dmin

4.878049

Table 3.9

0.74

(100As/bd)min 100As,max/bdmin

Minimum possible design shear

stress capacity of slab,

Table 3.10

T12

T12

T12

T12

2.0120[0.9 + M/bd

2]

= 0.55 + 477 ‐ 111.72

120[0.9 + (13.94x10^6)/(1000x94^2)]

= 1.78

= 5x460x293

x8x452.39

111.7235

Table 3.11 Modification factor for tension

reinforcements, = 0.55

T12

T12

46.2

Deflection of slab is adequate OK

+[477 ‐ f s]

≤

vmax OK

45.13

26.00

slab - rectangular - 4 edges supporeted

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