Design of Flat Plate (2)
Transcript of Design of Flat Plate (2)
IN THE NAME OF ALLAH, THE MOST BENEFICENT,
THE MOST MERCIFUL
DESIGN OF FLAT PLATE
DESIGN OF FLAT PLATE
A five storey building has a line plan as shown below.
The floor consist of reinforced concrete flat plate with no edge
beam and has a ceiling height of 10 ft. The building is
subjected to gravity loads only. The dead load consist of 2 ½”
F.F, ½” ceiling plaster, 10 psf for mechanical fixtures and 25
psf for partition load. The live load = 60 psf. The external wall
weighs 350 Ib/ft. f’c = 4 ksi and fy = 60 ksi. Design the end
panel Q of the floor system. Check the conditions of DDM.Q
S
W
N
E
16'
16'
16'
18' 18' 18'
P
RR S
Q
16“ x 12"
16“ x 16“ COL
16“ x 12" COL
LINE PLAN
SOLUTIONSlab Thickness Refer to table 9.5 (c) of ACI Code.
h = ln/30 = 200/30 = 6.66" say 7.0"ln= 18x12 – 16 = 200"
Check for Geometry and Loading Condition of DDMACI 13.6.1 Refers
Three or more spans is each direction Panels are rectangular and 18/16 = 1.125 <2.0 Successive span don't differ No column offset Loads are due to gravity only
wd= 7x12.5+30+6+10+25= 158.5 psfwl= 60 psf2wd > wl ok
No beam present
Check for Shear LOADS
wu = 1.4(158.5) + 1.7x60 = 324 psf
Assuming ¾" clear cover and # 4 bar being used.
d= 7- 0.75 - 0.5/2 = 6"
Interior Column Critical section for punching shear is at a
distance d/2 from face of support.
Vu = [18x16 – (22/12)2]x324
= 92220 Ib
bo = 22x4 = 88"
According to ACI 11.12.2.1, Vc is smallest of the following
16"
22"
22" 16"
16"
22"
22" 16’
18’
Assumed Loaded Area for Interior Column
16"
• Vc= (2+4/βc)x √fc' bod βc = 1.0
=(2+4/1.0) √4000x88x6 =200362 lb
• Vc = (αsd/bo+2)√fc'bod
=(40x6/88+2) √4000x88x6 =157860 Ib.
αs = 40 for interior column
• Vc = 4√fc' bod = 4x√4000x88x6 = 133574 Ib.
Vc is the lowest of above three values i.e. 133574
Ib.
ΦVc = 0.85x133574= 113538 Ib.
ΦVc > Vu Safe
Exterior Column
bo = 15x2 + 22 = 52"
22"
15"
15”
22”
18’
8.5’
Assumed loaded area for exterior column
Shear is caused by floor load and weight of exterior wall.
Vu= [18x (8+0.5) – 22x15/144] 324 +[(18-16/12) 350 x 1.4]
= 57000 Ib.
Vc is smallest of the following
Vc = (2+4/βc)√ fc‘ bod = (2+4/1.33)√4000x52x6 = 98678 Ib
βc= 16/12 = 1.33
Vc = (αsd/bo+2) √fc' bod αs = 30 for exterior column
=(30x6/52+2) √4000x52x6 =107770 lb
Vc =4 √fc’ bo d =4x√4000x52x6 = 78930 lb
ΦVc = 0.85x78930 = 67090 Ibs.
ΦVc > Vu Safe
Total Factored Static Moment in E-W Dir and its Distr
Equivalent Rigid Frame on Inner Column Line
Mo = wul2ln2 /8 = 0.324x16(16.67)2/8 = 180.07 kft
ln = 18 - 16/12 = 16.67 ft
D.F ACI 13.6.3.2
- ve moment = 0.65Mo= 117.05 k'
+ ve moment = 0.35Mo= 63.02 k'
Moment in Column Strip ACI 13.6.4
l2/l1 = 16/18= 0.89, αl2/l1 = 0
- ve moment in C.S = 75 %
+ ve moment in C.S = 60 %
0.65 0.35 0.65
S
W
N
E
18' 18' 18'
P
RR S
Q
16“ x 12“ COL
16“ x 16“ COL
16“ x 12" COL
LINE PLAN
16’
16'
16’
Distribution of Moment
Location Total C.S Moment (k') M.S.Moment (k')
E-W Dir 117.05 117.05x0.75 = 87.78 29.26
- ve moment
E-W Dir 63.02 63.02x0.6 = 37.81 25.21+ ve moment
Equivalent Rigid Frame on Outer Column Line
Mo= 0.324(8+0.5)x (16.67)2/8+ 0.35 (16.67)2/8 x 1.4 = 112.68 kftD.F. For interior span
- ve moment = 0.65Mo = 0.65x112.68 = 73.24 kft
+ ve moment = 0.35Mo = 0.35x112.68 = 39.44 kft Percentage moment in C.S.=Same as for inner column line.
Distribution of momentsLocation Total Moment C.S.(kft) M.S .(kft)E-W Dir 73.24 73.24x0.75= 54.93 18.31- ve momentE-W Dir 39.44 39.44x0.60 = 23.66 15.78+ ve moment
0.65 0.35 0.65
Total Factored Static Moment in N-S Dir and its Distr
Mo= wu l2 ln2/8 = 0.324x18(14.83)2/8 = 160.40 kft
ln = 16 - (6+8)/12 = 14.83 ftD.F. ACI 13.6.3.3
Ext –ve moment = 0.26Mo= 41.70 k'
+ve moment = 0.52Mo= 83.41 k'
Int –ve moment = 0.70Mo= 112.28 k’Percentage Moment in C.S. ACI 13.6.4
l2/l1 = 18/16, = 1.13
α l2/l1 = 0 βt = 0Ext –ve moment in C.S = 100 %
+ve moment in C.S = 60 % Int –ve moment in C.S = 75 %
0.26
0.52
0.70
S
W
N
E
18' 18' 18'
P
RR S
Q
16“ x 12"
16“ x 16"
16“ x 12" Col
LINE PLAN
16’
16'
16’
Distribution of Moments.
Location Total moment C.S kft MS kft
N-S Dir 41.7 41.7 0.0
Ext -ve
N-S Dir 83.41 0.6x83.41=50.05 33.36
+ ve moment
N-S Dir 112.28 0.75x112.28=84.2 28.07
Int -ve moment
Design of Slab Reinforcement Panel QStrip Loc Muk’ b ft Mu/ft
kft
d” ρ As
in2
No of
bars
Remarks
E-W Dir
2x1/2 C.S
-ve
+ve
87.78
37.81
8
8
10.97
4.72
6
6
0.006
0.00258
3.46
1.49
18
8
E-W Dir
2x1/2 M.S
-ve
+ve
29.26
25.21
8
8
3.66
3.15
6
6
0.00208
0.00208
1.2
1.2
7
7
Use ρmin
Use ρmin
E-W Dir
1/2 C.S
-ve
+ve
54.93
23.66
4.5
4.5
12.2
5.26
6
6
0.00669
0.00288
2.17
0.93
12
5 Use ρmin
E-W Dir
1/2 M.S
-ve
+ve
18.31
15.78
4
4
4.58
3.95
6
6
0.0025
0.00216
0.72
0.63
4
4
N-S Dir
2x1/2 C.S
Ext-ve
+ve
Int-ve
41.7
50.05
84.21
8
8
8
5.21
6.26
10.53
5.5
5.5
5.5
0.0034
0.00408
0.00686
1.8
2.15
3.62
10
11
19
N-S Dir
2x1/2 M.S
Ext-ve
+ve
Int-ve
0
33.36
28.07
10
10
10
0
3.34
2.81
5.5
5.5
5.5
0.00227
0.00227
0.00227
1.5
1.5
1.5
8/9
8/9
8/9
Use ρmin
Use ρmin
Use ρmin
Asmin = 0.0018 bxh = 0.0018x12x7 = 0.15 in2
ρmin in E-W direction = 0.15/(12x6) = 0.00208
ρmin in N-S direction = 0.15/(12x5.5)= 0.00227
Area of steel can be calculated from flexural formula.
Mu = ɸρbd2fy(1-.59ρfy/fc’)
C.S8' - 0”
M.S10' - 0”
C.S8' - 0”
C.S4' - 6”
M.S8' - 0”
C.S8' - 0”
REINFORCEMENT PLAN
12#4T
10#4T
8#4T
11#4B
18#4T
19#4T
12#4T
10#4T
8#4T
11#4B
18#4T
19#4T
5#4B9#4T
8#4B9#4B
8#4B
9#4T
DESIGN THE INTERIOR PANEL OF THE ABOVE FLOOR SYSTEM
Solution
1. Slab Thickness Same as for exterior panel i.e. 7"
3. Total Factored Static Moment in E-W Dir and its Distribution Same as for exterior panel on interior column line
4. Total Factored Static Moment in N-S Dir and its Distr
Mo = wul2ln2/8 = 0.324 x 18 (16-16/12)2/8 = 156.82 k’
0.65
0.35
0.65D.F
S
ACI Code 13.6.3.2
S
W
N
E
18' 18' 18'
P
RR S
Q
16“ x 12" Col
LINE PLAN
16’
16'
16’
- ve Moment = 0.65 Mo = 101.93 k'
+ve Moment = 0.35 Mo = 54.89 k'
Percentage Moment in C.S.
l2/l1 = 18/16 = 1.13 αl2 / l1 = 0
+ve moment in C.S = 60%
- ve moment in C.S = 75%
Distribution of Moment
Location Total Moment C.S moment M.S moment
N-S Dir 0.75x101.93=
-ve moment 101.93 76.45 25.48
N-S Dir 0.6x54.89=
+ve moment 54.89 32.93 21.96
5.Design of Slab Reinforcement - Panel
Strip Loc M kft b ft Mu/ft
k'
d
in
ρ As
in2
No of
#4 Bar
Remark
E-W Dir
2x1/2 C.S
-ve
+ve
87.78
37.81
8
8
10.97
4.72
6
6
0.006
0.0025
8
3.46
1.49
18
8
E-W Dir
2x1/2 M.S
-ve
+ve
29.26
25.21
8
8
3.66
3.15
6
6
0.0020
8
0.0020
8
1.2
1.2
7
7
ρmin
“
N-S Dir
2x1/2 C.S
- ve
+ve
76.45
32.93
8
8
9.56
4.12
5.5
5.5
0.0061
9
0.0026
7
3.27
1.41
17
8
N-S Dir
2x1/2 M.S
- ve
+ve
25.48
21.96
10
10
2.55
2.20
5.5
5.5
0.0022
7
0.0022
7
1.5
1.5
8/9
8/9
ρmin
“
S
Asmin =0.0018 bxh= 0.15 in2
ρmin in E-W direction = 0.15/(12x6) = 0.00208
ρmin in N-S direction =0.15/(12x5.5) = 0.00227
Area of steel is calculated using flexural formula.
Mu = ɸρbd2fy(1- 0.59ρfy/fc’)
For example for moment of 10.97 kft, As is calculated as fol
12x10.97 = 0.9 ρ 12 (6)2 60 (1- 0.59 ρ 60/4)
8.85ρ2 – ρ + Mu/1944 = 0
ρ = 0.0060
As = 0.006x8x12x6 = 3.46 in2
C.S8' - 0”
M.S10' - 0”
C.S8' - 0”
C.S8' - 0”
M.S8' - 0”
C.S8' - 0”
Reinforcement Plan
6 Sketch
18#4T
17#4T
7#4T
8#4B
18#4T
17#4T
8#4B
9#4T
7#4B
9#4B
8#4B
9#4T
18#4T
17#4T
7#4T
8#4B
18#4T
17#4T
6#4T
6#4T
C.S M.S C.S M.S C.S
8#4T
7#4B
C.S
M.S
C.S
M.S
C.S
10#4T
12#4T
7#4T
5#4B
10#4T
11#4T
8#4B
9#4T
10#4B
10#4B
12#4B
9#4T
8#4B
9#4B
12#4B
9#4T
12#4T
10#4T
8#4T
11#4B
19#4T
19#4T
7#4T
8#4B
19#4T
17#4T
5#4B
9#4T
8#4B
9#4B
8#4B
9#4T
7#4B
9#4B
8#4B
9#4T
12#4T
10#4T
8#4T
11#4B
18#4T
19#4T
7#4T
8#4B
18#4T
17#4T
P
R S
Q
ANY QUESTION ?
ThanksThanks