design of flat plate (2)
TRANSCRIPT
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