Design of flat plate slab and its Punching Shear Reinf. for the students of B.Sc. in Civil Engg.

1. 53
CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
DESIGN OF THE FLAT PLATE SLAB BY DDM
DESIGN OF PUNCHING SHEAR REINFORCEMENT
Lecture notes for B.Sc. in Civil Engg. Students
Department of Civil Engineering
Stamford University Bangladesh.
MD. MAHBUB-UL-ALAM
ASST. PROF.

2. 54
CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
CHAPTER FIVE
DESIGN OF THE FLAT PLATE SLAB BY DDM
5.1 Design consideration
In order to design of a flat plate slab, following points are to be considered:
 Minimum thickness of flat plate slab is 5".
 In a flat plate slab, long direction +ve main reinforcement at midspan are placed
below the short direction bars. Because, most of loads are transferred in long
direction and as result, major deflection of the panel occurs in that direction.
 Effective depth of the tensile bars in long direction, dl= h-clear cover = h-1.0"
Effective depth of the tensile bars in short direction, ds= h-clear cover = h-1.5"
Concrete cover for floor slab = ¾"
Concrete cover for roof slab = 1.0"
 Generally, #4~#6 bars are used as tensile bars in flat plate slab.
 In flat plate slab, minimum bar siza is #4 because of its larger thickness compared
to other slabs. To make proper bonding with big volume of concrete, it is necessary
to provide larger bar size. Also, reinforcing bars at midspan may slightly bend in
between supporting mortar blocks due to thicker casting of concrete during
ongoing slab construction works.
 In flat plate slab, all panels have same thickness. Generally, maximum thickness of
a panel is taken as minimum required thickness of all panels in a floor. Because in
such type of slab, there is no column-connecting beams and panels are connected
directly with each other. As a result, uneven thickness of two adjacent panels may
create unpleasant outlook of the ceiling and at the same time, such uneven
thickness of the slab may induce shear failure at the face, as shown in figure below,
of the intersections of these two panels .
ds dl
Shear failure may occur at
the face where two uneven
panels meet each other

3. 55
CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
 Check for punching shear requirement
Requirement for Punching shear reinforcement is usually checked for centraly loaded
column of a flat plate slab. As it is known that punching shear failure is occurred at a
distance d/2 from all faces of a column, as shown in Figure 5.1.
The requirement of the punching shear is,
 If Vu ≤ φVc, no shear reinforcement is required.
 If Vu > φVc, hear reinforcement is required.
Where,
Here, φ = shear strength reduction factor =0.75
And bo = Perimeter of the critical section = 2(c1+d) + 2(c2+d) for rectangular column.
= 4(c+d) for square column.
Figure 5.1: Punching shear check
xx
y
y
Shaded area of
slab which gives
punching loads to
the column
Punching location of
the slab @ d/2
distance from four
faces of column
x
y
(c1+d)
(c2+d)
  





 



144
dcdc
xyw
area)unpunchedarealoadedX(totalslabofloadfactoredTotal
floortheofloadfactoredonbasedshearPunchingTotalV
21
T
u
  concwtlightfordbf
concwtnormalfordbfV
oc
occ
.475.0
..4
'
'





4. 56
CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
1 2 3 4 5 6
A
B
C
D
16'-0" 16'-0" 16'-0" 16'-0" 16'-0"
20'-0"20'-0"20'-0"
81'-0"
61'-6"
STAIR
S-1
S-2
S-1
S-3 S-3 S-3
S-1
S-2
S-1S-1 S-1
S-4 S-4 S-4
5.2 Design Problems
Design Problem-01: A four storied residential building has Flat Plate floor system having
same panel size of 16'x20' in each direction as shown in figure below. All panels are
supported by 12" square columns. The service live load is to be taken as 40 psf and the
service dead load consists of 25 psf of floor finishing and 70 psf for partition walls in
addition to the self-weight of the slab. Use f’c = 4 ksi and fy = 60 ksi.
S-1 is the Exterior corner panel.
S-2 is the Exterior panel with short direction discontinuous.
S-3 is the Interior panel.
S-4 is the Exterior panel with long direction discontinuous.
Answer the following questions:
1. Check whether the slab satisfies the DDM limitations or not.
2. Design panel S-4.

5. 57
CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
Solution:
1. Check for limitations for using DDM
1st
Limitation: Required minimum of 3 continuous spans in each direction i.e. (3x3).
Here in this case, 5x5 panels ------satisfied.
2nd
Limitation:Panel shall be rectangular with a ratio of longer to shorter span c/c of
supports within a panel not grater then 2.
Here this value is 25.1
16
20






i.e. 1.25<2 ------satisfied.
3rd
Limitation: Successive span length c/c of supports in each direction shall not differ by
more than 1/3 of the longer span.
In this case difference of successive span length = (20-16) = 4′
1/3 of longer span = 20/3 = 6.67′>4′ ok------satisfied.
4th
Limitation: Column may be offset from the basic rectangular grid of the building by up
to 0.1 times the span parallel to the offset.
All columns are in the basic rectangular grid, so no offset is observed------
satisfied.
5th
Limitation: All loads should be gravity in nature.
Here loads are dead and live loads only and no lateral loads—satisfied.
2. Design of S-4 panel
Step-1: Calculation of Slab thickness [Table 4.1]
Since thickness of all panels of a flat plate should be the same, slab thickness for exterior
and interior panels will be calculated separately and maximum thickness will be taken.
 For an exterior panel (without edge beam):
nl = clear span in long direction = 20′- 1′=19′
For fy= 60 ksi, minimum slab thickness, "8"6.7
30
12*19
30
 nl
h
 For an interior panel:
nl = clear span in long direction = 20′- 1′=19′
For fy= 60 ksi, minimum slab thickness, "7"9.6
33
12*19
33
 nl
h
Selected slab thickness = 8″> 5″ ok.

6. 58
CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
In a flat plate slab, long direction reinforcements are placed first,
Effective depth in long direction, ld = 8-1.0 = 7.0″
Effective depth in short direction, "5.65.18 Sd
Step-2: Calculation of total load
Self-weight of slab = psf100150*
12
8

Total dead load = FF + Partition wall load + Self wt. of slab= (25+70+100) = 195 psf
Factored load, TW = 1.2DL+1.6LL= (1.2*195+1.6*40) = 298 psf
Step-3: Calculation of the Column & Middle strips
Width of column strips =
4
Sl
at both side of column center line in both directions.
"48'4
4
16

Full column strip in both directions = 2 x 4= 8′ = 96″
 Middle strip in long direction = 16′- 2*4 = 8′ = 96″
Middle strip in short direction = 20′- 4*2 = 12′ = 144″
Details of column and middle strips are shown in Figure 5.2.
Figure 5.2: Calculation of column & middle strips in long direction (left) and in short direction
(right)
4' 8' 4'
20'
16'
Long
direction
4'
12'
4'
20'
16'
short
direction

7. 59
CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
Step-4: Calculation of Moment
Short direction
 Static moment:
Total static moment
8
2
2
0
nT llW
M  will be calculated as per Figure 5.3.
167625
8
15*20*298 2
0 SM lb-ft = 167.63 k-ft
 Distribution of Moment at both Supports & Midspan:
-ve Moment at both Int. supports = 0.65 MOS = 0.65*167.65 = 108.97 k-ft
+ve Moment at mid span = 0.35 MOS = 0.35*167.65 = 58.68 k-ft
Figure 5.3: Details of MOS
Midspan +ve M
Int. Col. -ve M Int. Col. -ve M
l2 = 10'+10'
= 20'
½ panel
width, 10'
½ panel
width, 10'
20'
S-4
16'
ln = 16'-1'=15'

8. 60
CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
 Distribution of moment into Column & Middle strips from supports and midspan:
For flat plate, 01  and βt = 0
Using 0
1
21

l
l
, we can obtain distribution percentages from Table 4.2.
-ve Interior Moment for column strip from support = 0.75*108.97 = 81.73 k-ft
-ve Interior Moment for middle strip from support = 0.25*108.97 = 27.25 k-ft
+ve Moment for column strip from midspan = 0.60*58.68 = 35.21 k-ft
+ve Moment for middle strip from midspan = 0.40*58.68 = 23.47 k-ft
Long direction
 Static moment:
As per Figure 5.4, 215156
8
19*16*298 2
0 LM lb-ft = 215.16 k-ft
Figure 5.4: Details of MOL
l2 = 8'+8' = 16'
½ panel width
8'
½ panel width
8'
ln = 20'-1'=19'
S-4
20'
16'
Int. Col. –ve M
Ext. Col. –ve M
Midspan
+ve M

9. 61
CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
 Distribution of Moment at both Supports & Midspan:
-ve Moment at Ext. support = 0.26 MOL = 0.26*215.16 = 55.94 k-ft
+ve Moment at Mid span = 0.52 MOL = 0.52*215.16 =111.88 k-ft
-ve Moment at Int. support = 0.70 MOL = 0.70*215.16 = 150.61 k-ft
 Distribution of moment into Column & Middle strips from supports and midspan:
For flat plate, 01  and βt = 0
Using 0
1
21

l
l
, we can obtain distribution percentages from Table 4.2.
-ve Exterior moment for column strip from Ext. support = 1*55.94 = 55.94 k-ft
-ve Exterior moment for middle strip from Ext. support = 0 k-ft
+ve Moment for column strip from Mid span = 0.60*111.88 = 67.13 k-ft
+ve Moment for middle strip from Mid span = 0.40*111.88 = 44.75 k-ft
-ve Interior moment for column strip from Int. support = 0.75*150.61 = 112.96 k-ft
-ve Interior moment for middle strip from Int. support = 0.25*150.61 = 37.65 k-ft
Step-5: Check for‘d’
Maximum moment is found in –ve Interior column strip of long direction.
Mu = 112.96 k-ft
Column strip width, b = 96″
Let, steel ratio, ρ = ρmax= 0.75ρb
Step-6: Calculation of Reinforcement
Reinforcement requirement is shown in Table 5.1.
Sample calculation (short direction):
Interior –ve, column strip moment, Mu = 81.73 k-ft Column strip width, b = 96″
Total depth, h =8.0" Effective depth, d = 6.5″
okd
x
xdxxxx
MMNow nu
''0.7''88.3
4
600214.0
59.0196600214.090.01296.112
,
2






 
0214.0
87000
87000
85.075.0
'
1 










yy
c
ff
f


10. 62
CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
 Selected bar is #4 and Ab= 0.20 in2
 Minimum steel requirement for 60-grade bar, Ast = 0.0018bh = 0.0018x96x8 = 1.38 in2
 Total required reinforcement,
Asreq= ρbd
So provided steel area, As =2.91 in2
 Maximum spacing, Smax = Smaller of 2h or 18″= Smaller of 2x8″ or 18″= 16″
 No. of bars required, n =
 Spacing, S =
Sample calculation (long direction):
Interior –ve, column strip moment, Mu = 112.96 k-ft Column strip width, b = 96″
Total depth, h =8.0" Effective depth, d = 7.0″
 Selected bar is #4 and Ab= 0.20 in2
 Minimum steel requirement for 60-grade bar, Ast = 0.0018bh = 0.0018x96x8 = 1.38 in2
 Total required reinforcement,
Asreq= ρbd
So provided steel area, As =3.77 in2
 Maximum spacing, Smax = Smaller of 2h or 18″= Smaller of 2x8″ or 18″= 16″
 No. of bars required, n =
 Spacing, S =
22
2
2'
'
38.191.25.696
5.696490.085.0
1273.812
11
60
485.0
85.0
2
11
85.0
ininxx
xxxx
xxx
bd
bdf
M
f
f
c
U
y
c

















.1657.14
20.0
91.2
nos
A
A
b
s

."16"0.6''4.6
116
96
1
max okSoS
n
b




22
2
2'
'
38.177.3796
796490.085.0
1296.1122
11
60
485.0
85.0
2
11
85.0
ininxx
xxxx
xxx
bd
bdf
M
f
f
c
U
y
c



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











.1985.18
20.0
77.3
nos
A
A
b
s

."16''0.5"33.5
119
96
1
max okS
n
b





11. 63
CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
Table 5.1: Details of slab reinforcement arrangement of panel (S-4)
+ve bar arrangement of column and middle strips for short and long directions are shown in Figure 5.5.
-ve bar arrangement of column and middle strips for short and long directions are shown in Figure 5.6.
Cross sectional view of A-A is shown in Figure 5.7.
Cross sectional view of B-B is shown in Figure 5.8.
Direction Location Mu
k-ft
b
in
d
in
Minimum
Ast=ρstbh
in2
Required
As=ρbd
in2
Provided
As
in2
Use # 4 as main bar
Spacing & no. of bars
long
direction
column
strip
-ve Ext. 55.94 96" 7.0 1.38 1.82 1.82 09 @ 12.0″c/c
+ve Midspan 67.13 96" 7.0 1.38 2.19 2.19 11 @ 9.5″c/c
-ve Int. 112.96 96" 7.0 1.38 3.77 3.77 19 @ 5.0″c/c
middle
strip
-ve Ext. 0 96" 7.0 1.38 0 1.38 07 @ 16.0″c/c
+ve Midspan 44.75 96" 7.0 1.38 1.45 1.45 08 @ 13.5″c/c
-ve Int. 37.65 96" 7.0 1.38 1.22 1.38 07 @ 16.0″c/c
short
direction
column
strip
-ve Int. 81.73 96" 6.5 1.38 2.91 2.91 16 @ 6.0″c/c
+ve Midspan 35.21 96" 6.5 1.38 1.23 1.38 07 @ 16.0″c/c
middle
strip
-ve Int. 27.25 144" 6.5 2.07 0.94 2.07 11 @ 14.0″c/c
+ve Midspan 23.47 144" 6.5 2.07 0.81 2.07 11 @ 14.0″c/c

12. 64
CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
16'
20'
Figure 5.5: +ve bars arrangement in both directions Figure 5.6: -ve bars arrangement in both directions
16'
20'
A A
B B

13. 65
CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
Figure 5.7: Cross sectional view of A-A of short direction
d=6.5" h=8.0"
12" 12"
l =16'
0.15x16'≡2.5'0.15x16'≡2.5'
0.20x15'= 3' 0.20x15'= 3'
ln=15'
Half col. strip=4'Full middle strip=8'Half col. strip=4'
Short direction
Ext. -ve middle
strip bars
Long direction
+ve ½ col. strip
bars
Long direction
+ve middle strip
bars
Long direction
+ve ½ col. strip
bars
Short direction
Int. -ve middle
strip bars
Short direction +ve
middle strip alternate
cut bars @ 2.5' from
column centerline
binder bar

14. 66
CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
Figure 5.8: Cross sectional view of B-B of short direction
d=6.5" h=8.0"
12" 12"
l =16'
0.15x16'≡2.5'0.15x16'≡2.5'
0.20x15'= 3' 0.20x15'= 3'
ln=15'
Half col. strip=4'Full middle strip=8'Half col. strip=4'
Short direction
Ext. -ve col.
strip bars
Long direction
+ve ½ col. strip
bars
Long direction
+ve middle strip
bars
Long direction
+ve ½ col. strip
bars
Short direction
Int. -ve column
strip bars
Short direction +ve ½
col. strip alternate
cut bars @ 2.5' from
column centerline
Long direction
Int. -ve ½ col.
Strip bars
Long direction
Int. -ve ½ col.
Strip bars
Long direction
Int. -ve middle
Strip bars
binder
bar

15. 67
CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
Design Problem-02: A flat plate slab, as shown in figure below, having each panel size
20′x16′ is designed with its total thickness 8″. It supports service dead loads which include
floor finish 25 psf, partition wall loads 70 psf in addition to its self-weight and standard
live load 100 psf. Material properties are fy = 60,000 psi, and f’c = 2,800 psi. Check the
requirement of punching shear reinforcement of the slab and design for shear if required.
Assume all columns are 12 square.
Solution:
Slab thickness, "8h
ld = 7″ and "5.6Sd
Self-weight of slab, w = psf100150*
12
8

Total dead load = (100+25+70) = 195 psf
Factored load, TW = (1.2*195+1.6*100) = 394 psf
Step-1: Check for requirement of punching shear reinforcement
Punching shear is checked at
2
d
from all faces of the column and for safety, we will
consider smaller effective depth, ds= 6.5".
As shown in Figure 5.9, shaded area of slab for central column = 20′x16′ = 320 sft
Punching perimeter, bo = 4(c + ds) = 4(12+6.5) = 74.0"

16. 68
CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
Punching shear at critical section,
When no shear reinforcement is used, the shear strength of normal weight concrete,
5.6*74*800,24*75.0'4 0  dbfV cc  = 76,356.38 lb = 76.36 k
Since ,uc VV  so punching reinforcement is required.
Step-2: Determination of punching shear reinforcement
Use beam stirrup case as punching shear reinforcement as shown in Figure below.
The punching shear perimeter considering 12" square column,
 24 acbo 
When shear reinforcement is used, the shear strength of concrete will be reduced to
dbfV cc 0'2 
For preventing punching failure,
  
''40.34
5.6*2124*28002*75.0100014.125
'2 0



a
axor
dbfVV ccu 
Figure 5.9: Punching shear check
20'20'
16'
16'
Shaded area of
slab which gives
punching loads to
the column
Punching location of
the slab @ d/2
distance from four
faces of column
20'
16'
(c+d)
(c+d)
   klbxx
144
dcdc
xywV Tu 14.12557.143,125
144
)5.612)(5.612(
2016394 


 





 


17. 69
CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
The minimum length to be provided for placing punching steel at each side of the column,
= a + d = 34.40 + 6.5 = 40.9 = 41"
Shear to be resisted by the punching reinforcement,
kVV cu 78.4836.7614.125  
Shear per side of the column, Vs = k2.12
4
78.48

Use #3 U-legged stirrups and Av = 2 x 0.11 = 0.22 in2
Spacing of the shear reinforcement, "0.5"3.5
2.12
5.66022.075.0

xxx
V
dfA
s
s
yv


But maximum spacing, s = d/2= 6.5/2 = 3.25" = 3.0" c/c
No. of stirrups, n = 41/3 =13.67 = 14 nos. and total distance = 3 x 14 = 42".
Figure 5.10: Beam stirrup case punching shear reinforcement

18. 70
CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
First stirrup will be placed at s/2 = 3/2=1.5" from all column faces.
Use 4#5 straight bars to hold up the stirrups at all sides of the column.
Details of the punching shear reinforcement are shown in Figure 5.11 below.
4#5 straight
bars
a + d +s/2= 43.5
14#3 punching
steel @ 3 c/c
Figure 5.11: Details of punching shear reinforcement