Slab System
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Transcript of Slab System
![Page 1: Slab System](https://reader033.fdocuments.in/reader033/viewer/2022061614/563db912550346aa9a99bbac/html5/thumbnails/1.jpg)
Chapter 1 : Slab system
Our slabs are designed as partially prestressed , that is 80% of self weight is designed as Post-tensioned flat slab and the remaining portions are designed as reinforced concrete.
A-Slab thickness
From table 9.5 (c) , the slab thickness chosen is l33 = 303.3 mm.
Thus , use 350 mm slab thickness :
B- Load calculations: 1.Dead load:
Plastering = 1510 * 21 = 31.5 Kg
Sand = 1 * 0.125 * 1800 = 225 KgMortar = 1 * 0.03 * 2100 = 63 Kg
Tile = 2010 * 20 = 40 Kg
Total DL = 360 Kg / m2 = 3.56 KN/m2
![Page 2: Slab System](https://reader033.fdocuments.in/reader033/viewer/2022061614/563db912550346aa9a99bbac/html5/thumbnails/2.jpg)
2 .Live Load: From the ACSE , the Live load is determined referring to the occupancy category of the building ( residential building ).
From the figure ; the LL is 4.79 KN¿m2
The next step is to start the design of our slabs using csi SAFE , however ; the self load multiplier of the slab own weight is set to be 0.2 since 0.8 of the slab weight will be carried by the Strands of post-tensioning , we'll start our
design by designing the slab as reinforced concrete:
1. Defining the materials , sections , spans , design strips , loads and load combinations .
The following figures will illustrates how to input the design info. Starting from the materials until being ready for running analysis and design.
![Page 3: Slab System](https://reader033.fdocuments.in/reader033/viewer/2022061614/563db912550346aa9a99bbac/html5/thumbnails/3.jpg)
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The above figure illustrates splitting the Slab into design strips of width equal 10 m , this design strip is composed of a column strip of width 5 m , and two half middle strips with 2.5 width each.
Actually ; using the ACI Direct Design Method is not permitted here since the loads includes Non-gravity load ( earthquake loads and wind loads ) , however ; the csi SAFE will use a more accurate procedures of design , which is the Equivalent frame method.
Now defining the loads as follows :
![Page 7: Slab System](https://reader033.fdocuments.in/reader033/viewer/2022061614/563db912550346aa9a99bbac/html5/thumbnails/7.jpg)
![Page 8: Slab System](https://reader033.fdocuments.in/reader033/viewer/2022061614/563db912550346aa9a99bbac/html5/thumbnails/8.jpg)
Now after adding the loads , csi SAFE will automatically set the Design combinations and the design will be according to the
controlled one as showing in the figures below:
![Page 9: Slab System](https://reader033.fdocuments.in/reader033/viewer/2022061614/563db912550346aa9a99bbac/html5/thumbnails/9.jpg)
2- Design and analysis Results .
Now we are ready to the run the analysis and design of the slabs
Hint : moving the cursor at any point will give the value of the moment at that point ( eg. Max. moment is 445.4 KN/m
2).
![Page 10: Slab System](https://reader033.fdocuments.in/reader033/viewer/2022061614/563db912550346aa9a99bbac/html5/thumbnails/10.jpg)
As mentioned above , csi SAFE enable us to select the bar size that we will use in the slabs : the following figures will
illustrates the complete design of the strips:
Note : the blue line indicates the limit for the minimum reinforcement ; i.e : most slabs will require only minimum reinforcement except the exterior spans , that’s because a large quantity of the load is held by the STRANDS ( 80 % of self
weight )!!!!!!
In the N-S Direction:
![Page 11: Slab System](https://reader033.fdocuments.in/reader033/viewer/2022061614/563db912550346aa9a99bbac/html5/thumbnails/11.jpg)
In The E-W Direction:
Note here the top steel will require a design rather than minimum reinforcement :
Moving the cursor across the diagram ( reinforcement diagram ) will give the required amount of bars , that’s enable us easily for applying bar cut-off for more economic design.
Requirement of placing steel bars:
While applying cut off of bars in two way slabs , some bars must remain constant across the strips and must not be cut , the ACI
code showed this thing and recommended the following :
![Page 12: Slab System](https://reader033.fdocuments.in/reader033/viewer/2022061614/563db912550346aa9a99bbac/html5/thumbnails/12.jpg)
These figures is clear enough for applying cut-off providing that the requirement of the Development length is satisfied.
Summary of Reinforcement :
Provide minimum reinforcement bottom steel in the Design strips , That is ρmin=0 .0018 Ag uniformly distributed across the bay width , also provide the required amount of top steel as suggested by csi SAFE uniformly distributed over the effective
band with of the support.
![Page 13: Slab System](https://reader033.fdocuments.in/reader033/viewer/2022061614/563db912550346aa9a99bbac/html5/thumbnails/13.jpg)
Manual Design of post-tensioned concrete slab to balance 80% of self weight of the slab.
MATERIALS:
-f'c = 35 Mpa-ɣ for concrete = 24 KN/m.s
-fci'= 0.75f'c= 27 Mpa-fpu= 1860 Mpa , fps not exceeding 1276 Mpa
-fpy = 1690 Mpa-fpe = 1100 Mpa-Eps = 200 Gpa
-fy = 420 Mpa -Stress-relived strands ( K 270 )
![Page 14: Slab System](https://reader033.fdocuments.in/reader033/viewer/2022061614/563db912550346aa9a99bbac/html5/thumbnails/14.jpg)
-Slab thickness = 350 mm. -Unsupported length of the column ( story height ) = 4 m
-ft = 0.25√ f ' c = 1.48 Mpa. -fc = 0.45 f'c = 15.8 Mpa.
-Self weight = 24 * 0.35 = 8.4 KN/m.s-Balance ratio = 0.8 →0.8 * 8.4 = 6.72 KN/m.s
The analysis here is based on the Equivalent-frame method as recommended by the ACI Code:
![Page 15: Slab System](https://reader033.fdocuments.in/reader033/viewer/2022061614/563db912550346aa9a99bbac/html5/thumbnails/15.jpg)
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As shown in the figure , the analysis here is based on an
equivalent frame taken in the N-S direction , since all spans
![Page 17: Slab System](https://reader033.fdocuments.in/reader033/viewer/2022061614/563db912550346aa9a99bbac/html5/thumbnails/17.jpg)
are same in magnitudes , also all columns are equal in dimension , analysis for one equivalent frame is enough for finding the required # of strands and non-bonded reinforcement for the whole slabs in the both directions.
Hmin= L / 30 = 10000/30 = 333.33 mm →Use h = 350 mm .Ww = 6.72 KN/m.sWu = 1.2 * 6.72 = 8.06 KN/m.s
→assume an average intensity of compressive stress due to balancing is fc = 1.17 Mpa , then unit F = 1.17 *350*1000 =
409.5 KN/m , so ; trying 12in diameter seven wires stress relived
strands ; Pe = Aps * fpe = 99*1100 = 108.9 KN , for L = 10 m ,
total Fe = 409.5 * 10 = 4095 KN .# of strands per bay is FePe =
4095/108.9 = 37 strands , and total Pe = Fe = 108.9 * 37 = 4029.3 KN , the actual unit force F = 4029.3/10 = 402.93 KN/m
, and the actual unit stress is fc = 402.9∗103
1000∗350 = 1.15 Mpa < 1.17
Mpa , which is satisfactory . Consequently , use fc = 1.15 Mpa due to load balancing , and assume a tendon profile as shown in
the figure below :
-a1 = 175+300
2 - 50 = 187.5 mm , a2 = 350 – 50 – 100 = 200 mm.
![Page 18: Slab System](https://reader033.fdocuments.in/reader033/viewer/2022061614/563db912550346aa9a99bbac/html5/thumbnails/18.jpg)
Wbal = 8F a
l2 =
8∗402.93∗0.1875102
= 6.044 KN/m.s
Wnet = 6.72 – 6.044 = 0.676 KN/m.s for the exterior span.
Wbal = 8F a
l2 =
8∗402.93∗0.2102
= 6.45 KN/m.s
Wnet = 6.72 – 6.45 = 0.27 KN/m.s for the interior span.
Equivalent frame characteristic
Kc = 4 I c Ec
l−2h =
4∗Ec∗400∗4003∗112
4000−(2∗350 ) = 2585858.586 Ec
C = ∑ (¿1−0.63 xy)( x3 y3 )¿
) = 1 -0.63 * 350400 ) * ( 3503∗400
3¿ =2565354167 mm4
Kt = ∑ 9C Ecs
l 2(1− c2l 2
)3 =
9∗2565354167∗Ecs
10,000∗(1− 40010000
)3 +
9∗2565354167∗Ecs
10,000∗(1− 40010000
)3
= 5219226.414Ecs
Kec = 1
1kc
+1k t
= 1
12585858.586
+1
5219226.414 = 1729152.367 Ecs
Ks = 4 I c Ecs
ln−C1
2 =
4∗10000∗3503∗Ecs
12∗(10000−4002 ) = 14583333.33 Ecs
Since all members are of the same concrete material , Ec cancels each other.
![Page 19: Slab System](https://reader033.fdocuments.in/reader033/viewer/2022061614/563db912550346aa9a99bbac/html5/thumbnails/19.jpg)
Exterior column :
DF = Ks
∑ K
=Ks
Kec+Ks ¿+Ks¿=
14583333.33
14583333.33+1729152.367 = 0.894
Interior columns:
DF = Ks
∑ K=
Ks
Kec+Ks ¿+Ks¿=
14583333.33
14583333.33+14583333.33+1729152.367 = 0.472
EXT. Span : FEM = w l2
12 = 0.676∗10
2
12 = 5.633 KN.m
Int . Span : FEM = w l2
12 = 0.27∗10
2
12 = 2.25 KN.m
DCBAPointDC DE
CB CD
BA BC
ABMember
0.47 0.470.47 0.470.47 0.470.894DF+2.25- 2.25
0 0+2.25- 2.25
0 0+5.6-
2.25-1.6- 1.6
-5.64.98
FEMDist.
-0.70.37 0.37
+ 2.5 0 -1.17- 0.95
-0.790.71
CODist.
![Page 20: Slab System](https://reader033.fdocuments.in/reader033/viewer/2022061614/563db912550346aa9a99bbac/html5/thumbnails/20.jpg)
5.36- 5.01
-0.734Final Mnet
Mnet = 5.36 KN.m
Slab concrete tensile stress at support.
Maximum Mnet = 5.36 – ( 0.27∗102 *
0.43 ) = 5.18 KN.m
Slab section modulus , S = b h2
6 = 10000∗350
2
6 = 204166666.7mm3.
Ftop = −P
A + MS = -1.15 +
5.18∗106
204166666.67 = -1.13 Mpa , OK.
Slab concrete tensile stress at Midspan.
Maximum Mnet = 0.27∗102
8 - 5.36 = -1.805 KN.m
Ftop = -1.15 + −1.805∗106
204166666.67 = -1.158 Mpa , OK .
Ultimate flexural strength analysis
1. Balanced Moment
-Exterior span :
FEMbal=W b l2
12 = 6.044∗102
12 = 50.4 KN.m
-Interior span:
FEMbal=W b l2
12 = 6.45∗102
12 = 53.8 KN.m
![Page 21: Slab System](https://reader033.fdocuments.in/reader033/viewer/2022061614/563db912550346aa9a99bbac/html5/thumbnails/21.jpg)
Now Running a moment distribution will determine the maximum balanced moment for the exterior column joints.
DCBAPointDC DE
CB CD
BA BC
ABMember
0.47 0.470.47 0.470.47 0.470.894DF+53.8- 53.8
0 0+53.8- 53.8
0 050.4-
53.8+1.6- 1.6
-50.444.8
FEMDist.
-0.80.37 0.37
22.4 0 -10.5- 10.5
0.80.71-
CODist.
63.9- 65.9
-5.46Final Mnet
2. Secondary Moments ( Ms ) and factored load moment ( Mu )
-Exterior span :
From tendon profile , e = 0.0 mm.
M1 = Primary moment = Pe.e = 0.0 KN.m
M bal- = 5.46 KN.m
Ms = M bal - M1 = -5.46 KN.m
Factored FEM = W u l2
12 = 8.064∗10
2
12 = 67.2 KN.m
-Interior span :
From tendon profile , e = 175-50 = 125 mm
M1 = Pe . e = 402.93 * 125 = 50.4 KN.m
M bal = 63.9 KN.m
Ms = 50.4 – 63.9 = 13.5 KN.m
![Page 22: Slab System](https://reader033.fdocuments.in/reader033/viewer/2022061614/563db912550346aa9a99bbac/html5/thumbnails/22.jpg)
Factored FEM = 67.2 KN.m
And now running a moment distribution for the factored moments , here analysis of pattern loading should be made to determine the worst case of the service and factored loads.
DCBAPointDC DE
CB CD
BA BC
ABMember
0.47 0.470.47 0.470.47 0.470.894DF+67.2- 67.2
0 067.2-
67.20 0
-67.259.81
FEMDist.
-0.80.37 0.37
29.9-14.1- 14.1
00
CODist.
82.2- 81.3
-5.46Final Mnet
3. Design Moments Mu
The design moments are the difference between the factored load moments and the secondary moments.
Exterior span:
Mu = 7.4 – 5.46 = 1.94 KN.m
Interior span BA :
Mu = 82.2 -13.5 = 68.7 KN.m
Interior span BC :
Mu = 81.3 – 15.5 = 65.8 KN.m
Moment reduction to the column face of support = V . c3
V = W u .l
2 -
∆ Mln
![Page 23: Slab System](https://reader033.fdocuments.in/reader033/viewer/2022061614/563db912550346aa9a99bbac/html5/thumbnails/23.jpg)
For the exterior span :
V = 8.064∗10
2 – 82.2−7.410 = 32.84 KN.m
V . c3
= 32.84∗0.42
= 4.4 KN.m
Centerline moment = 6.6 – 1.94 = 2.5 KN.m
Required Mn = Mu∅ =
4.70.9 = 5.2 KN.m
For the interior span BA:
V = 8.064∗10
2 – 7.4−82.210 = 47.8 KN.m
V . c3
= 47.8∗0.43
= 6.4 KN.m
Mu = 68.7 – 6.4 = 62.3 KN.m
Required Mn = Mu∅ =
62.30.9 = 69.2 KN.m
For the interior span BC:
Since the factored moment is almost the same →the required moment is smaller than the previous span , thus ignore it.
Span AB maximum positive moment
Assume point of zero shear ( or max. moment ) is at ( X ) m from face A :
X = V AB
W AB =
32.848.064 = 4.07 m
Maximum positive moment = V AB . x−W u x2
2−M u+M s
![Page 24: Slab System](https://reader033.fdocuments.in/reader033/viewer/2022061614/563db912550346aa9a99bbac/html5/thumbnails/24.jpg)
=4.07*32.84−8.06∗4.072
2−¿ 7.4 + 5.46
=64.93 KN.m
Required M n=M u
∅ = 64.930.9 = 72.14 KN.m
Span AB maximum positive moment
Unadjusted Max. +ve moment = W u .l2
8 = 8.064∗10
2
8 = 100.1 KN.m
+ve Moment = 100.1 – 82.2 + 15.5 = 33.4 KN.m
3 . Flexural strength ( Nominal moment strength )
1-Interior supports ( top face of the support )
As required by the ACI Code :
As = 0.00075 Acf
![Page 25: Slab System](https://reader033.fdocuments.in/reader033/viewer/2022061614/563db912550346aa9a99bbac/html5/thumbnails/25.jpg)
= 0.00075 * 350 * 1000 = 262.5 mm2
Thus use 5 ∅ 10 → As provided = 71*5 = 355 mm2
Spacing = 1000– (5∗10)
4 = 238 mm ≤ 300 mm , Ok.
ρp = A ps
bd =
36∗9910000∗300 = 0.0014
f pe = 1100 Mpa.
f ps =f pe + 70 + f c
'
100 ρp
= 100 + 70 + 35
100∗.0012 = 1462 Mpa.
F ps = 1462∗99∗36
10 = 521.1 KN per meter width of the slab.
F s = 420 * 355 = 149.1 KN per meter width of the slab.
Now , the total force provided by both reinforcement types
( bonded and non bonded ) is:
F t = 521.1 + 149.1 = 670.2 KN.
a=A s . f y+ Aps . f ps
0.85 f c' b
= 670.2∗103
0.85∗35∗1000 = 22.53 mm
ad p
= 22.53300
= 0.08 < 0.319 → T.C , Ok
.
M n=( A s . f y+ Aps . f ps )(d−a2 )
![Page 26: Slab System](https://reader033.fdocuments.in/reader033/viewer/2022061614/563db912550346aa9a99bbac/html5/thumbnails/26.jpg)
= 670.2 ) * 300 – 22.532 = ( 193.5 KN.m < 69.2 KN.m , Ok.
2- Mid-span section ( +ve moment )
F ps = 521.1 KN
a= 521.1∗103
0.85∗35∗1000 =17.5 mm
ad
< 0.319 , Ok ,,,,, TC!
M n=521.1∗(300−17.52 )=136.6KN >72.14KN .m
Summary :
Use 37 Strands uniformly distributed across the bay width with 5 ∅ 10 at the top fiber at the support section .
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