Ribbed Slab

RIBBED SLAB SYSTEM WITH ONE WAY SLAB
I.The given data
1.Structural plan.

Composition of the slab
Composition of the slab












2. The distance from beam axis to wall axis l1=2,3m, l2=6,05m. The thickness of load bearing wall bt=0,34m. The dimensions of the reinforced concrete column bcxhc= 0,3x0,3m.
3. Characteristics of the construction: the composition of the slab consist four layers as shown. The nominal live load: Ptc=9,25(KN/m2), reliability factor of live load n=1,2.
4. Select material: concrete grade with the compressive strength B15, main reinforcement and constructive reinforcement of slab and stirrup of beam use class CI, main reinforcement of beam use class CII.
Concrete grade B15 with Rb=8,5MPa, Rbt=0,75MPa
Steel CI: Rs=225MPa, Rsc=225MPa, Rsw=175MPa
Steel CII: Rs=280MPa, Rsc=280MPa, Rsw=225MPa
II. Design plate
1) Select the dimensions of components of slab
- Select thickness of the slab
hb= Dm.l1=1,3x230035=85.42 mm
select hb=90 mm
Where: D=1,3 ; m=35
Select the dimensions of the secondary beam
hdp=1mdpxl2=112x6050=504,17 mm
select hdp = 500mm and bdp = 220mm
Select the dimensions of the primary beam
hdc=1mdcx3l1=19,5x3x2300=726 mm
select hdc = 750 mm and bdc = 300 mm
2) Design scheme
Consider the ratio of slab: l2l1=6,052,3=2,63>2
this is one way slab


Establish the simplified model for a 1 m wide strip of the slab which is perpendicular to secondary beam and consider strip slab as a continuous beam
The design span of slab:
The outer span: l0b=l1b - bdp2-bt2+hb 2 = 2,3 - 0,5.0,22 - 0,5.0,34 + 0,5.0,09 = 2,065 m
The middle span: l0 = l1 - bdp=2,3-0,22 = 2,08 m
Span difference:
2,08-2,0652,08x100=0,72%<10%
3) The design load
Dead load is calculated as table
The layers of slab
Nominal values(KN/m2)
Realibility factor
Design values ( KN/m2)




Thickness of tile layer is 10 mm, γ =20 KN/m3
0.2
1.1
0.22
Thickness of lining mortar is 30 mm, γ=18 KN/m3
0.54
1.3
0.702
Thickness of RC layer is 90 mm, γ=25 KN/m3
2.25
1.1
2.475
Thickness of plaster mortar layer is 10 mm, γ=18 KN/m3
0.18
1.3
0.234
Total
3.17
 
3.631

Table 1: Determine dead load
Round figure: gb*=3,63 KN/m2
Live load: pb*=ptc.n=9,25.1,2=11,1 KN/m2
Total load: qb*=gb*+pb*=3,63+11,1=14,73 KN/m2
Calculate strip slab b1=1m, we have: qb=14,73.1=14,73 KN/m
4) Design internal forces
Following the plastic hinge scheme model:
-The bending moment at the outer spans and the second supports:
Mnb=Mg2=±qb.l0b211=±14,73x2,065211=±5,71 KN.m
-The bending moment at the middle spans and the middle supports:
Mng=Mgg=±qb.l0216=±14,73x2,08216=±3,98 KN.m
-The maximum shear force: QBt=0,6qb.lob=0,6.14,73.2,065 = 18,25 KN


Fig 2: the design scheme and the internal force of strip slab


5) Calculate main reinforcement
Select a = 15 mm for all section, the working height of slab h0 = hb-a = 90-15=75 mm
-At end supports and outer spans, with M = 5,71 KNm
m =MRb.b1.h02=5,71.1068,5.1000.752=0.119 < pl = 0.255
ζ=0,5(1+1-2 m) = 0,51+1-2.0,119= 0,936
As= MRs.ζh0 = 5,71.106225.0,936.75 = 362 mm2
µ%= Asb1.h0 = 3621000.75.100% = 0,483% > µmin= 0,05%
Selecting the longitudinal reinforcement which has diameter of 8 mm, as= 50,3 mm2, space of longitudinal reinforcements is
s = b1.asAs=1000.50,3362 = 139 mm
Select 8, s=130mm.
-At the middle supports and middle spans with M= 3,98 KNm.
m=MRb.b1.h02=3,98.1068,5.1000.752=0,083< pl=0,255
ζ=0,51+1-2 m=0,51+1-2.0,083=0,957
As = MRs.ζh0 = 3,98.106225.0,957.75=246 mm2
µ% = Asb1.h0=2461000.75.100%=0,328 %> µmin= 0.05%
Selecting the longitudinal reinforcement which has diameter of 6 mm, as= 28,3 mm2, space of longitudinal reinforcements is:
s=b1.asAs=1000.28,3246=115 mm
Select 6, s=120 mm.

-At the middle spans and middle supports, we are allowed to reduce 20% amount of longitudinal reinforcement, we have As=0,8.246 = 197 mm2
µ% = 1971000.75.100%=0,26%
Selecting the longitudinal reinforcement which has diameter of 6 mm and as= 28,3 mm2, space
of longitudinal reinforcements is:
s=b1.asAs=1000.28,3197=147mm
select 6, s=140 mm, and As=202 mm2
Checking the working height h0: cover to reinforcement is 10 mm
h0= 90-10-0,5.8 = 76mm > 75mm
So the design value h0=75mm is safety
Design reinforcement resists negative moment: with pbgb=11,13,63=3.06>3
value ν=13 , the Extended segment of negative moment reinforcement:
from the edge of secondary beam: νl0= 13.2.08=0,69 m
from the axis of secondary beam: νl0+0,5bdp= 0,69+0,5.0,22= 0,8m

The longitudinal reinforcement resists negative moment which is placed alternately. Extended segment of shorter reinforcement :
From edge of secondary beam is:
16l0=16.2.08=0,35 m
From axis of secondary beam is:
16l0+0,5bdp=0,35+0,5.0.22=0,46 m
The longitudinal reinforcement resists positive moment which is placed alternately
Distance from end of shorter reinforcement to edge of secondary beam is:
18l0=18.2.08=0,26 m
Checking shear force capacity:
QBt=18,25 KN< Qbmin=0,6.Rbtb1h0=0,6.0,75.1000.75=33750 N=33,75 KN
concrete can resist completely the shearing force
6) Design constructive reinforcement
-Negative moment reinforcement is placed in the direction perpendicular to primary beam, select
6, s=200 , with area per 1m length of slab is 141mm2, It's higher than 50% area of calculating reinforcement at the middle support of slab is 0,5.246= 123 mm2, length of extended segment :
From edge of primary beam is
14l0=14.2.08=0,52 m
From axis of primary beam is:
14l0+0,5bdc=0,52+0,5.0,3 = 0,67 m
-Distribution reinforcement is placed in the direction perpendicular to main reinforcement:
Design positive moment reinforcement in the direction l2:
Select 6, s=200 with area per 1m length of slab is 141m2 to ensure higher than 20% area of calculating reinforcement at middle span ( outer span 0,2x362= 72,4 mm2, middle span 0,2x246=49,2 mm2 )
Design negative moment reinforcement of slab which is near secondary beam, primary beam
Select 6, s=250












III. Design the secondary beam.
1.Structural plan.
The secondary beam is a continuous beam with 4 symmetrical spans
Consider a left half of beam (fig.4)
The beam is directly supported on the wall at Sd= length of wall, Sd=bdp=220 mm
Cd=min(Sd/2 and l2/40); Sd/2=110 mm < l2/40= 6050/40= 151 mm
Thus Cd= 110mm
Design span of secondary beam:
-Outer span: lpb=l2-bdc2-bt2+Cd=6,05-0,5.0,3-0,5.0,34+0,11=5,84 m
-middle span: lp=l2-bdc=6,05-0,3=5,75 m
-span difference: 5,84-5,755,84x100=1,54 % <10%

Fig.4 design scheme and internal force in the secondary beam
Design scheme
Moment envelope diagram
Shear force envelop shear force diagram


2) Design loading
* Dead load:
Self-weight of beam (without slab thickness: 80mm)
g0p=bdphdp-hbγn=0,22.(0,5-0,09).25.1,1=2,481 KN/m
Dead load is transferred from the plate:
gdp1=gb*l12+l12=3,632,32+2,32=8,349 KN/m
Total dead load:
gp=g0p+gdp1=2,481+8,349=10,83KN/m
*Live load:
Live load is transferred from the plate:
pp=pbl12+l12=11,12,32+2,32=25,53 KN/m
*Total design load
qp=gp+pp=10,83+25,53=36,36 KN/m
We have ratio: ppgp=25,5310,829=2,36
3.Design internal forces
a) bending moment:
Ordinate of moment envelope (positive branch)
+ at outer span:
M+=β1qplpb2=1240,08β1 KNm
+at middle span:
M+=β1qplp2=1202,153β1 KNm
Ordinate of moment envelope (negative branch)
M-=β2qplp2=1202,153β2 KNm
Look for appendix 11 with ratio pdp/gdp= 2,36 ; k =0,2644 and coefficient β1, β2. The results are presented in the table 2


TABLE2
Span/section
value of β
ordinate of M (KNm)

β1
β2
M+
M-
outer span




support A
0

0

1
0.065

80.605

2
0.09

111.607

0.425l
0.091

112.847

3
0.072

89.286

4
0.02

24.802

support B-TD5

-0.0715

-85.954
span 2




6
0.018
-0.03216
21.639
-38.661
7
0.058
-0.01116
69.725
-13.416
0.5l
0.0625

75.135

8
0.058
-0.00816
69.725
-9.81
9
0.018
-0.02616
21.639
-31.448
support C-td 10

-0.0625

-75.135

Section which has negative moment equal to zero is far support 2 at a distance of :
x=k.lpb=0,2644.5,84=1,544 m
section which has positive moment equal to zero is far support at a distance of:
+ outer span: 0,15lpb = 0,15.5,84= 0,876m
+ middle span: 0,15lp = 0,15.5,75= 0,8625 m
b) shear force
QA=0,4qp.lpb=0,4.36,36.5,84=84,937 KN
QBt=0,6qp.lpb=0,6.36,36.5,84= 127,405 KN
QBp=QC= 0,5.qdp.lp=0,5.36,36.5,75 = 104,535 KN
The moment envelope diagram and shear force envelope diagram are presented in fig.4
4.Calculate longitudinal reinforcement
Concrete grade B15 with Rb=8,5 MPa, Rbt=0,75 MPa.
Longitudinal reinforcement class CII with Rs= 280 MPa, Rsc= 280 MPa, stirrup class CI with Rsw= 175MPa.

Negative moment
We calculate similar to a rectangular section with b=220mm, h=500mm.
Assume that a = 35mm => h0= h-a= 500-35=465 mm
-at the support B(2), with M= 85,954 KNm
m =MRbbh02=85,954.1068,5.220.4652=0,213 < αpl=0,255
ζ=1+1-2 m2=1+1-2.0,2132=0,879
As=MζRsh0=85,954.1060,879.280.465=751,05mm2
Checking μ%=Asbdph0=751,05220.465.100% =0,73 %
-at support (C )3, with M=75,135 KNm
m=MRbbh02=75,135.1068,5.220.4652=0,186< pl=0,255
ζ=1+1-2 m2=1+1-2.0,1862=0,896
As=MζRsh0=75,135.1060,896.280.465=644,06 mm2
μ%=Asbdph0=644,06220.465.100%=0,63%
Positive moment
We calculate similar to T section, the flange is in compressive zone, thickness of flange hf=90 mm
Assume that a =35mm, h0=h –a =500-35 = 465 mm
Extension of flange is Sf Which is less than the following values
16ld=16.5,75=0,958 m
A half of distance of clear spans
0,5l0= 0,5.2,08=1,04m ( because h'f >0,1h, h=500mm and space of horizontal beams is higher than that of vertical beams: 6,05m >2,3m)
Thus Sf = min(0,958;1,04) m=0,958m
Select Sf = 958mm

Width of flange is bf'=b+2Sf=220+2.958=2136 mm
We have: Mf=Rbbf'hf'h0-0,5hf'=8,5.2136.90.465-0,5.90=686,3.106 Nmm
M+max= 112,847 KNm the neutral axis goes through the flange
So we calculate the section similar to rectangular section, with b=bf = 2136mm, h = 500mm
a=35mm, h0= 465mm
-at the outer span, with M+= 112,847 KNm
m=MRbbfh02=112,847.1068,5.2136.4652=0,029< pl=0,255
ζ=1+1-2 m2=1+1-2.0,0292=0,985
As=MRsζh0=112,847.106280.0,985.465=879,92 mm2
Checking μ%=Asbdph0=879,92220.465.100%=0,86%
-at the second span, with M+= 75,135 KNm
m=MRbbfh02=75,135.1068,5.2136.4652=0,019< pl=0,255
ζ=1+1-2 m2=1+1-2.0,022=0,99
As=MRsζh0=75,135.106280.0,99.465=582,9 mm2
Checking μ%=Asbdph0=582,9220.465.100%=0,57 %
5. Select and arrange the longitudinal reinforcement

Table 3:
Section
Outer span
Support 2
The second span
Support 3
Design As
879,92
751,05
582,9
644,06
Arrange As
2 20+1 18
2 18+1 18
2 16+1 16
2 16+1 18
Area
882,8
763,4
603,2
656,6

Fig.5: arrange main reinforcement in main section of beam



Checking a:att=C0+ dmax2=25+20/2= 35 mmatt=35mm = agt= 35mm -> safety
Checking a:
att=C0+ dmax2=25+20/2= 35 mm
att=35mm = agt= 35mm -> safety



Checking a:att=C0+ dmax2=25+18/2= 34 mmatt=34mm < agt= 35mm -> safety

Checking a:
att=C0+ dmax2=25+18/2= 34 mm
att=34mm < agt= 35mm -> safety
Checking a:att=C0+ dmax2=25+18/2= 34 mmatt=34mm < agt= 35mm -> safetyChecking a:att=C0+ dmax2=25+16/2= 33 mmatt=33mm < agt= 35mm -> safety
Checking a:
att=C0+ dmax2=25+18/2= 34 mm
att=34mm < agt= 35mm -> safety
Checking a:
att=C0+ dmax2=25+16/2= 33 mm
att=33mm < agt= 35mm -> safety