3 Slab Design

Chapter two Slab

design

42

3 Slab design

AB

CD

1' 2' 3' 4'

1 2 3 4

A'

B'

C'

D'

R0.28m

0.57m

0.40m0.40m

5.00m

7.00m

5.50m

2.50m

0.57mR0.28m

0.40m

0.40m

2.00m

1.10m1.65m

0.40m

0.40m

7.0 m

5.0 m 5.0 m

15.0 m

22.0 m

E E'

S1 S2 S3

S4 S5 S6

S7 S8 S9

S10 S11 S12

Chapter two Slab

design

43

Figure (2-1) plan of slab and notations.

In multistory building, there isnt rectangular reinforced concrete

beam in whole frame; all beams are T-shape or L-shape.

Figure (2-2) T-beam and L-beam.

Design of two way slabs

There are several methods for analysis and design of two way slab systems;

more useful methods include:

1. Direct Design Method DDM.

2. Equivalent Frame Method EFM.

3. Yield Line Theory YLT.

4. Coefficient method.

The ACI code 13.5.1.1 specifies two methods for designing two-way slabs for

gravity loads, its Direct Design Method and Equivalent Frame Method.

1. Direct Design Method

The code 13.6 provides a procedure with which a set of moment coefficients

can be determined. The method in effect ,involves a single cycle moment

distribution analysis of the structure based on (a) the estimated flexural

stiffness of the slabs, beams(if any) and columns (b) the torsional stiffnesses of

the slabs and beams(if any) transverse to the direction in which flexural

h

be

t

h

bw

be

t

bw

Chapter two Slab

design

44

moments are being determined. Some types of moment coefficients have been

used statistically for many years for slab design. They do not, however, give

very satisfactory results for slabs with unsymmetrical dimensions and loading

patterns.

Limitations of DDM:

1. There shall be minimum of three continuous spans in each direction.

The panels shall be rectangular, with the ratio of longer to shorter spans within

a panel not greater than 2.

3. The successive span lengths in each direction shall not differ by more than

one third the longer span.

4. Columns may be offset a maximum of 10 percent of the span in the direction

of the offset from either axis between centerlines of successive columns.

5. All loads must be due gravity and uniformly distributed, the live load shall

not exceed 2 times of dead load.

6. If beams are used on the columns lines, the relative stiffness of the beams in

the tow perpendicular directions, given by the ratio

, must be between

0.2 and 5.

2. Equivalent Frame Method

In this method a portion of the structure is taken out by itself, the same

stiffness values used for direct design method are used for equivalent frame

method. This latter method, which is very satisfactory for symmetrical frames

as well as for those with unusual dimensions or loadings.

The only difference between the DDM and the EFM is in the determination of

the longitudinal moments in the spans of the equivalent rigid frame. Whereas

the DDM involves one cycle moment distribution, the EFM involves a normal

moment distribution of several cycles. The design moments obtained by either

method are distributed to column and middle strips in the same fashion. There

are no such limitations (as in DDM) on EFM. This is a very important matter

Chapter two Slab

design

45

because so many floor systems do not meet the limitations specified for the

DDM.

3. Yield Line Theory

Although the yield line theory not included in the ACI code slab analysis by this

method may be useful in providing the needed information for understanding

the behavior of irregular or single panel with various boundary condition.

A yield line analysis uses rigid plastic theory to compute the failure loads

corresponding to given plastic moment resistances in various parts of the slab.

It does not give any information about deflections or about the loads at which

yielding first starts.

K.W Johansson 1948 developed modern yield line theory. These types of

analysis are widely used for the design in the Scandinavian countries. Although

the concepts of YLT was first presented by Ingerslev 1921-23. For solving the

yield line problems two methods are available

Figure (2-3) mode of failure of a slab.

For designing two way slabs (continuous system) all supports are assumed to be fixed and several

modes of failure are taken into consideration to find the ultimate capacity of the slab.

1.Equilibrium Method 2.Virtual Work Method

Chapter two Slab

design

46

4. Coefficient method:

(Method 3 of the ACI-code 1963 (Marcos method).

Limitation of coefficient method:

1. The method is applicable for slabs supported of the edges by required

concrete walls, steel beams or monolithic concrete beams having total

depth, h3 hf.

2. The design is based on coefficient from tables for the middle strip of the

slab.

3. The moment at the discontinuous edge is taken as 1/3 of the positive

moment.

4. The moments at the column strip are taken as 2/3 of that at the middle

strip.

Slab design calculation:

The slabs in the our project are designed by coefficient method.

For slab(1) with panels (7*5).

1. slab thickness:

i. Perimeter= 2*(5+7)=24 m.

t primary=

=133.33 mm.

t provided by architecture =180 mm not economic.

ii. finding t by DDM.

600

150

1300 mm

400mm

850mm

150mm

400mm

h=6

00 mm

Chapter two Slab

design

47

Figure (2-3) dimension of beam cross-sections.

beT-section=bw+2(h-t)bw+8t

beL-section=bw+ (h-t)bw+4t

kf1= (

) (

t

hf ) (

t

hf (

t

hf)

) (

t

hf )

(

) (

t

hf )

KfL-beam=1.388

KfT-beam=1.642

Ib T-beam=

=11.822*109 mm4

Ib L-beam=

=9.993*109 mm4

Is:

Sec 1 Is=2500*

=0.703*109 mm4

Sec 2 Is=3500*

=0.984*109 mm4

Sec 3 Is=5000*

=1.406*109 mm4

Sec 4 Is=7000*

=1.968*109 mm4

Sec 5 Is=2500*

=0.703*109 mm4

Sec 6 Is=6250*

=1.758*109 mm4

=

1=14.214

2=10.155

1From Chu-kia Wang , Charles G. Salmon, Jose A. Pincheira, Reinforced concrete design,7th ed.,2007,John Wiley & sons inc., page 632.

Chapter two Slab

design

48

3=8.408

4=6.0

5=14.214

6=6.724

m for each panel:

panel A m=

=9.694

panel B m=

=8.242

panel C m=

=8.836

panel D m=

=7.385

panel E m=

=9.0175

panel F m=

=7.566

*For all panels m >2.0 Use ACI eq.9.13

tmin=

c=

=1.434

for critical panel A,B ln=7000-400=6600mm

tmin=

=141.68 mm use 150mm

Chapter two Slab

design

49

2. Weights:

L.L= 4 KPa

Selfweight=([0.05+0.02]*23)+(0.15*

24)+(0.02*14)

Self weight= 5.49 KPa use 5.5 KPa.

Wu= 1.2 Wud.l+1.6 Wul.l2

Wud=1.2*5.5=6.6 KPa.

Wul=1.6*4=6.4 KPa.

Wu=6.6+6.4=13.0 KPa.

3. Using coefficient method to determining factored moments:

la and lb are clear spans.

la=shorter span length, m.

lb=longer span length, m.

la=5.0-0.4=4.6 m

lb=7-0.4=6.6 m

m=la

l =

=0.6969 0.7

from appendix(B), table (B-1,2,3,4) case 4 for one end continuous in each

direction may be critical case for slabs (multi-bays).

Negative moments:

For m=0.7

Ca=0.081

Cb=0.019

Ma=Ca Wu la2=0.081*13.0*4.62

=22.28 kN.m

Mb=Cb Wu lb2

2 From ASCE 7-05, page 5.

Tile

Plastering (gypsum) R.C slab

Mortar

15 cm

2

5

2

figure(2-4)slab layers. 7.0m

5.0m

Mb

Ma

These slab moments

becomes torsion (tu) on the beams.

Chapter two Slab

design

50

=0.019*13.0*6.62=10.76 kN.m

Positive moments:

Dead positive moments:

Ca=0.046

Cb=0.011

Ma=0.046*6.6*4.62=6.42 kN.m

Mb=0.011*6.6*6.62=3.16 kN.m

Live load positive moments:

Ca=0.057

Cb=0.014

Ma=0.057*6.4*4.62=7.72 kN.m

Mb=0.014*6.4*6.62=3.90 kN.m

Total positive moments:

Ma total=MaL+MaD

Ma total=6.42+7.72=14.14 kN.m

Mb total=MbL+MbD

Mb total=3.16+3.9=7.06 kN.m

4. Determining Reinforcement:

In two way slab the main reinforcement placed in two directions.

Reinforcement can be determining according to (7.12) ACI 318M-08:

i. Slab with grade 50 (350 MPa) or less min=0.002.

ii. Slab with grade 60 (420 MPa) min=0.0018.

iii. Slab with grade more than 60 min=0.0018*

f but min should not

less than 0.0014.

Use fy=350 MPa, fc=21 MPa.

min=0.002

The down main reinforcement is along shorter span.

Negative moments:

Short direction reinforcement:

Figure (2-5) panel (1,3)

Chapter two Slab

design

51

d= 150-20-

=123.65 mm.

Ma=22.28 kN.m

=

m(1-

mR

f )

m=f

f c ; R=

u

d

m=

=19.6 ; R=

=1.619140

=

(1-

)=0.004858 > min=0.002 ok.

As= bd=0.004858*1000*123.65=600.609 mm2/m

Use 13 Ab=129 mm2

No. of bars=

=4.65588 bars/m

Spacing=

=214.78 mm

Smax=2h=2*150=300 mm Sprovided=214.78mm Ok.

Use S=200 mm

Use No. 13@200 mm.

Long direction reinforcement:

d= 150-20-

-12.7=110.95 mm.

Mb=10.76 kN.m

=

m(1-

mR

f )

m=f

f c ; R=

u

d

m=

=19.6; R=

=0.971153

Chapter two Slab

design

52

=

(1-

)=0.002855 > min=0.002 ok.

As= bd=0.002855*1000*110.95=316.719 mm2/m

Use 13 Ab=129 mm2

No. of bars=

=2.455 bars/m

Spacing=

=407.33 mm

Smax=2h=2*150=300 mm Sprovided=407.33 mm not Ok.

Use S=300 mm

Use No. 13@300 mm.

Positive moment reinforcements (mid span reinforcement):


d= 150-20-

=123.65 mm.

Ma=14.14 kN.m

=

m(1-

mR

f )

m=f

f c ; R=

u

d

m=

=19.6 ; R=

=1.02759

=

(1-

)=0.003026 > min=0.002 ok.

As= bd=0.003026*1000*123.65=374.125 mm2/m

Use 13 Ab=129 mm2

No. of bars=

=2.9 bars/m

Spacing=

=344.8 mm

Chapter two Slab

design

53

Smax=2h=2*150=300 mm Sprovided=344.8mm not ok.

Use S=300 mm

Use No. 13@300 mm.


d= 150-20-

-12.7=110.95 mm.

Mb=7.06 kN.m

=

m(1-

mR

f )

m=f

f c ; R=

u

d

m=

=19.6 ; R=

=0.0.6372

=

(1-

)=0.0018544 < min=0.002 use min.

As= bd=0.002*1000*110.95=221.9 mm2/m

Use 13 Ab=129 mm2

No. of bars=

=1.720 bars/m

Spacing=

=581.343 mm


Use S=300 mm

Use No. 13@300 mm.

5. Check for shear:

Also by coefficient method [appendix (B-4)].

m=la

l =0.7

Wa=0.81

Wb=0.19

Chapter two Slab

design

54

Wa,Wb Both direction is checked.

Vushort=Wa

=

=24.22 kN

Vulong=W

=

=8.15 kN

Strength of slab:

=0.75

Vc=f c

*b*d

Vc=

*1000*123.65

Vc=94.439 kN

Vc=70.83 kN

Vc=70.83 >24. 22 Ok. for shear.

Design of slab (7) with panel 5.5*5.0:

la=5.0-0.4=4.6 m

lb=5.5-0.4=5.1 m

m=la

l =

=0.9

from appendix(B), table (B-1,2,3,4) case 9 for three end continuous may be

critical case for slabs (multi-bays).

Negative moments:

For m=0.9

Ca=0.068

Cb=0.025

Ma=Ca Wu la2

=0.068*13.0*4.62=18.71 kN.m

Mb=Cb Wulb2

=0.025*13.0*5.12=8.453 kN.m

Positive moments:

Dead positive moments:

Ca=0.026

Cb=0.015

Ma=0.026*6.6*4.62=3.63 kN.m

5.0 m

5.5

m

Figure (2-6) panel (7).

Chapter two Slab

design

55

Mb=0.015*6.6*5.12=2.58 kN.m

Live load positive moments:

Ca=0.036

Cb=0.022

Ma=0.036*6.4*4.62=4.88 kN.m

Mb=0.022*6.4*5.12=3.66 kN.m

Total positive moments:

Ma total=MaL+MaD

Ma total=3.63+4.88=8.51 kN.m

Mb total=MbL+MbD

Mb total=2.58+3.66=6.24 kN.m

6. Determining Reinforcement:


Negative reinforcement:

d= 150-20-

=123.65 mm.

Ma=18.71 kN.m

=

m(1-

mR

f )

m=f

f c ; R=

u

d

m=

=19.6 ; R=

=1.3597

=

(1-

)=0.004045 > min=0.002 ok.

As= bd=0.004045*1000*123.65=500.191 mm2/m

Use 13 Ab=129 mm2

No. of bars=

=3.878 bars/m

Spacing=

=257.9 mm

Chapter two Slab

design

56

Smax=2h=2*150=300 mm Sprovided=257.9mm ok.

Use S=250 mm

Use No. 13@250 mm.


d= 150-20-

-12.7=110.95 mm.

Mb=8.45 kN.m

=

m(1-

mR

f )

m=f

f c ; R=

u

d

m=

=19.6 ; R=

=0.7708

=

(1-

)=0.002252 > min=0.002 ok.

As= bd=0.002242*1000*110.95=249.87 mm2/m

Use 13 Ab=129 mm2

No. of bars=

=1.937 bars/m

Spacing=

=516.27 mm

Smax=2h=2*150=300 mm Sprovided=516.27 mm not ok.

Use S=300 mm

Use No. 13@300 mm.

Positive moment reinforcements (mid span reinforcement):


d= 150-20-

=123.65 mm.

Chapter two Slab

design

57

Ma=8.51 kN.m

=

m(1-

mR

f )

m=f

f c ; R=

u

d

m=

=19.6 ; R=

=0.61844

=

(1-

)=0.001799 < min=0.002 not ok.

As= bd=0.002*1000*123.65=247.3 mm2/m

Use 13 Ab=129 mm2

No. of bars=

=1.9171 bars/m

Spacing=

=521.63 mm


Use S=300 mm

Use No. 13@300 mm.


d= 150-20-

-12.7=110.95 mm.

Mb=6.24 kN.m

=

m(1-

mR

f )

m=f

f c ; R=

u

d

m=

=19.6 ; R=

=0.5632

=

(1-

)=0.001635 < min=0.002 not ok.

Chapter two Slab

design

58

As= bd=0.002*1000*110.95=221.9 mm2/m

Use 13 Ab=129 mm2

No. of bars=

=1.72 bars/m

Spacing=

=581.34 mm


Use S=300 mm

Use No. 13@300 mm.

7. Check for shear:

Also by coefficient method [appendix (-)].

m=la

l =0.9

Wa=0.75

Wb=0.25


Vushort=Wa

= 7

=22.43 kN

Vulong=W

=

=8.29 kN

Strength of slab:

=0.75

Vc=f c

*b*d

Vc=

*1000*123.65

Vc=94.44 kN

Vc=70.83 kN

Vc=70.83 >22.43 Ok. for shear.

Design of cantilever slab (12):

All slabs are (5*2.5)

la=2.5-.2=2.3 m

lb=5-0.4=4.6 m

Chapter two Slab

design

59

m=la

l =

=0.5

from appendix(B), table(B-1) case 4 for one end continuous in each

direction may be critical case for slabs (multi-bays).

Negative moments:

For m=0.5

Ca=0.094

Cb=0.060

Ma=Ca Wu la2=0.094*13*2.32

=6.46 kN.m

Mb=Cb Wu lb2

=0.006*13.0*4.62=1.65 kN.m

*Cantilever slab there isnt positive moment.

Finding reinforcement:


d= 150-20-

=123.65 mm.

Ma=6.46 kN.m

=

m(1-

mR

f )

m=f

f c ; R=

u

d

m=

=19.6 ; R=

=0.4694

=

(1-

)=0.00136 < min=0.002 use =0.002 .

As= bd=0.002*1000*123.65=247.3 mm2/m

Use 13 Ab=129 mm2

No. of bars=

=1.917 bars/m

2.5

m

5.0 m

figure (2-7) panel(10,12).

Chapter two Slab

design

60

Spacing=

=521.63 mm


Use S=300 mm

Use No. 13@300 mm.


d= 150-20-

-12.7=110.95 mm.

Mb=1.65 kN.m

=

m(1-

mR

f )

m=f

f c ; R=

u

d

m=

=19.6 ; R=

=0.1489

=

(1-

)=0.0004273 < min=0.002 NOT

As= bd=0.002*1000*110.95=221.9 mm2/m

Use 13 Ab=129 mm2

No. of bars=

=1.72 bars/m

Spacing=

=581.34 mm

Smax=2h=2*150=300 mm Sprovided=581.34 mm OK.

Use S=300 mm

Use No. 13@300 mm.

Check for shear:

Also by coefficient method [appendix (-)].

Chapter two Slab

design

61

m=la

l =0.5

Wa=0.94

Wb=0.06


Vushort=Wa

=

=14.05 kN

Vulong=W

=

=1.79 kN

Strength of slab:

=0.75

Vc=f c

*b*d

Vc=

*1000*123.65

Vc=94.439 kN

Vc=70.83 kN

Vc=70.83 >14.05 Ok. for shear.

Design of slab 6:

Panel 6 can be divided into three one-way slab systems.

Slab 1

la

l =

=0.27 < 0.5 one way slab

Slab 2

la

l =


Slab 3

la

l =


7.0 m

5.4 m

1.6m

5.0 m

1.5m 3.5 m

3.5m

1.5m

2

1

3

Chapter two Slab

design

62

Design of slab1 @ panel 6:

Wu=13.0 kN.m

1. Check depth for moment and shear:

a. Check for moment:

Mucantilever =Wu l

Mu= fy b d (1-

7 f c)

Mucantilever =

=14.63 kN.m

14.63*10=0.9*350*0.01*1000*d *(1-

)

d=71.75 mm < 123.65 ok.

b. Check for shear:

Vu=Wu.l= 13.0* 1.5=19.5 kN

VuVc

Vc=

bd

19.5=0.75*

*

d=34.04 mm < 123.65 mm ok.

2. Finding reinforcement:

Mu= As fy z

Assume z=0.925 d

d=123.65 mm

14.63*10=0.9*As*350*0.925*123.65

As=406.067 mm

Number of bars=As

A =

=3.1478 bars

Spacing=

o of ars=

7 =317.68 mm

Smax=3h=3*150=450 mm

Chapter two Slab

design

63

Smax=450 mm

Smax=

f -2h=

-2*20=412.38 mm

Smax=

=

=360 mm

S provided=317.68 < all Smax

Use No. 13@300 mm


Wu=13.0 kN.m


a. Check for moment:

Mucantilever =Wu l

Mu= fy b d (1-

7 f c)

Mucantilever =

=16.64 kN.m

16.64*10=0.9*350*0.01*1000*d *(1-

)

d=76.53 mm < 123.65 ok.

b. Check for shear:

Vu=Wu.l= 13.0* 1.6=20.8 kN

VuVc

Vc=

bd

20.8=0.75*

*

d=36.31 mm < 123.65 mm ok.


Mu= As fy z

Assume z=0.925 d

Chapter two Slab

design

64

d=123.65 mm

16.64*10=0.9*As*350*0.925*123.65

As=461.856 mm

Number of bars=As

A =

=3.58 bars

Spacing=

o of ars=

=279.31 mm

Smax=3h=3*150=450 mm

Smax=450 mm

Smax=

f -2h=

-2*20=412.38 mm

Smax=

=

=360 mm


Use No. 13@275 mm


Wu=13.0 kN.m


c. Check for moment:

Mucantilever =Wu l

Mu= fy b d (1-

7 f c)

Mucantilever =

=14.63 kN.m

14.63*10=0.9*350*0.01*1000*d *(1-

)

d=71.75 mm < 123.65 ok.

d. Check for shear:

Vu=Wu.l= 13.0* 1.5=19.5 kN

VuVc

Vc=

bd

Chapter two Slab

design

65

19.5=0.75*

*

d=34.04 mm < 123.65 mm ok.


Mu= As fy z

Assume z=0.925 d

d=123.65 mm

14.63*10=0.9*As*350*0.925*123.65

As=406.067 mm

Number of bars=As

A =

=3.1478 bars

Spacing=

o of ars=

7 =317.68 mm

Smax=3h=3*150=450 mm

Smax=450 mm

Smax=

f -2h=

-2*20=412.38 mm

Smax=

=

=360 mm


Use No. 13@300 mm

Smax=

=

=360 mm

S provided=255 < all Smax

Use No. 13@250 mm

Other holes and openings can be neglected because all panels remain two-

way system.

Chapter two Slab

design

66

Table of moments, ,As, and spacing of the floor panels:

Negative parameters:

fc=21 MPa; fy=350 Mpa

by using 13mm(Ab=129 mm2)

min=0.002 for fy

Chapter two Slab

design

67

Positive parameters:

fc=21 MPa; fy=350 Mpa

by using 13mm(Ab=129 mm2)

min=0.002 for fy

Chapter two Slab

design

68

Design of over roof slab:

7.00

5.00

0.30

0.30

PARAPET

SLAB

0.30

? 1

d

d

? 2

By using yield line method we can find moment of the slab:

W=Wu*[(

*5*

*

)+(

*(

)2 *

*2 )+( 2*

*

*)]

=Wu*(2.0833+2.0833+2.5)*

W=6.667**Wu ..(1)

1=2=

M=(Mn+Mp)*5*1+(Mn+Mp)*7*2

M=(Mn+Mp)*5*

+(Mn+Mp)*7*

M=4.8*(Mn+Mp)*

Assume Mn=2Mp

M=14.4*Mp* (2)

By equating (1) and (2) we get the slab moments:

M=W

2

1

Figure (3-9) detail of parapet slab.

Chapter two Slab

design

69

Wu=(1.2D.L+1.6L.L)

LL=1.5 KPa for parapet.

Wu=(1.2*5.5)+(1.6*1.5)=9.0 KN/m2

6.667*9=14.4*Mp

Mp=4.166 KN.m

Mn=2*4.166=8.33KN.m

Design for negative reinforcement:

Using fy=350 MPa, fc=21 MPa.

min=0.002

d= 150 -20-

=123.65 mm (using 13mm).

=

m(1-

mR

f )

m=f

f c ; R=

u

d

m=

=19.6 ; R=

=0.60536

=

(1-

)=0.0017599 < min=0.002

Use min=0.002

As= bd=0.002*1000*123.65=247.3 mm2/m

Use 13 Ab=129 mm2

No. of bars=

=1.917bars/m

Spacing=

=521.63 mm

Smax=2h=2*150=300 mm Sprovided=521.63mm NOT OK.

Chapter two Slab

design

70

Use S=300 mm ;Use No. 13@300 mm.Other direction and positive

reinforcement also minimum control.

Use 13mm@300mm c/c.

Table (2-1) distribution of the reinforcement for the floor panels.

Slab No. Short Long

Top Bottom Top Bottom

Right 13@200 mm c/c 13@600 mm c/c 13@300 mm c/c 13@600 mm c/c

S1 Mid .. 13@300 mm c/c .. 13@300 mm c/c

Left 13@200 mm c/c 13@600 mm c/c 13@300 mm c/c 13@600 mm c/c


S2 Mid ..... 13@300 mm c/c .. 13@300 mm c/c



S3 Mid .. 13@300 mm c/c .. 13@300 mm c/c



S4 Mid .. 13@300 mm c/c .. 13@300 mm c/c



S5 Mid .. 13@300 mm c/c .. 13@300 mm c/c


S6-1 13@300mm c/c 13@300 mm c/c

S6 S6-2 13@275mm c/c 13@300mm c/c

S6-3 13@300 mm c/c 13@300 mm c/c


S7 Mid .. 13@300 mm c/c .. 13@300 mm c/c



S8 Mid .. 13@300 mm c/c .. 13@300 mm c/c



S9 Mid .. 13@300 mm c/c .. 13@300 mm c/c



S10 Mid .. 13@300 mm c/c .. 13@300 mm c/c



S11 Mid .. 13@300 mm c/c .. 13@300 mm c/c



Chapter two Slab

design

71

S12 Mid .. 13@300 mm c/c .. 13@300 mm c/c


Chapter two Slab

design

72

12 3 4

A

B

C

D

E

5.005.005.00

7.00

7.00

5.50

2.50

22.20

15.46

4.60

1.30

4.60

1.30

2.90

1.28

3.40

1.30

a

bc

d

e

f

a:13mm@600mmc/cBOT

b:13mm@600mmc/c BOT

c:13mm@200mmc/c TOP

d:13mm@600mmc/cBOT

e:13mm@600mmc/c BOT

f:13mm@250mmc/c TOP

g

g:213mm@bottom&top

@ all open corners

90

h

h:313mm@bottom&top

@ all corners

Chapter two Slab

design

73

Cutoff point for bars:

L1 L2

L1/7 L1/5 L2/4 L2/4

L2/3 L2/3L1/3L1/5

15cm

minimum

L1 L2

L1/8 L2/8 L2/8

L2/3 L2/3L1/3L1/4

15cm

minimum

3 Slab Design

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