Sabah Shawkat Cabinet of Structural Engineering 2017 3.2 Reinforced Concrete Slabs
Slabs are divided into suspended slabs. Suspended slabs may be divided into two
groups:
(1)
slabs supported on edges of beams and walls
(2)
slabs supported directly on columns without beams and known as flat slabs. Supported
slabs may be one-way slabs (slabs supported on two sides and with main reinforcement in one
direction only) and two-way slabs (slabs supported on four sides and reinforced in two
directions). In one-way slabs the main reinforcement is provided along the shorter span. In order
to distribute the load, a distribution steel is necessary and it is placed on the longer side. Oneway slabs generally consist of a series of shallow beams of unit width and depth equal to the
slab thickness, placed side by side. Such simple slabs can be supported on brick walls and can
be supported on reinforced concrete beams in which case laced bars are used to connect slabs
to beams.
Figure 3.2-1: One –way slab, two-way slab, ribbed slab, flat slab, solid flat slab with drop
panel, waffle slab
In R.C. Building construction, every floor generally has a beam/slab arrangement and
consists of fixed or continuous one-way slabs supported by main and secondary beams.
Sabah Shawkat Cabinet of Structural Engineering 2017 Figure 3.2-2: Solid flat slab, solid flat slab with drop panels
The usual arrangement of a slab and beam floor consists of slabs supported on crossbeams or secondary beams parallel to the longer side and with main reinforcement parallel to
the shorter side. The secondary beams in turn are supported on main beams or girders extending
from column to column. Part of the reinforcement in the continuous is bent up over the support,
or straight bars with bond lengths are placed over the support to give negative bending
moments.
Figure 3.2-3: Types of the reinforced concrete slab systems
Sabah Shawkat Cabinet of Structural Engineering 2017 3.2.1 Flat Slabs
Flat plate is defined as a two-way slab of uniform thickness supported by any
combination of columns, without any beams, drop panels, and column capitals. Flat plates are
most economical for spans from 4,5 to 7,5m, for relatively light loads, as experienced in
apartments or similar building.
-A flat slab is a reinforced concrete slab supported directly on and built monolithically with the
columns, the flat slab is divided into middle strips and column strips. The size of each strip is
defined using specific rules. The slab may be in uniform thickness supported on simple
columns. These flat slabs may be designed as continuous frames. However, they are normally
designed using an empirical method governed by specified coefficients for bending moments
and other requirements which include the following:
1. There should be not less than three rectangular bays in both longitudinal and transverse
directions.
2. The length of the adjacent bays should not vary by more than 10 %.
Figure 3.2.1-1: Post punching behaviour of slab- critical section
The general layout of the reinforcement is based on the both bending moments (in spans) and
bending moments in addition to direct loads (on columns).
Sabah Shawkat Cabinet of Structural Engineering Figure 3.2.1-2: Combined punching shear and transfer of moments
Figure 3.2.1-3
2017 Sabah Shawkat Cabinet of Structural Engineering Figure 3.2.1-4
2017 Sabah Shawkat Cabinet of Structural Engineering 2017 3.2.1-1 Analysis and Design of Flat Plate
To obtain the load effects on the elements of the floor system and its supporting
members using an elastic analysis, the structure may be considered as a series of equivalent
plane frames, each consisting of vertical members – columns, horizontal members - slab.
Such plane frames must be taken both longitudinally (in x-direction) and transversely (in y
direction) in the building, to assure load transfer in both directions.
For gravity load effects, these equivalent plane frames can be further simplified into
continuous beams or partial frames consisting of each floor may be analysed separately together
with the columns immediately above and below, the columns being assumed fixed at their far
ends. Such a procedure is described in the “Equivalent Frame Method”. When frame geometry
and loadings meet certain limitations, the positive and negative factored moments at critical
sections of the slab may be calculated using moment coefficients, termed “Direct Design
Method”. These two methods differ essentially in the manner of determining the longitudinal
distribution of bending moments in the horizontal member between the negative and positive
moment sections. However, the procedure for the lateral distribution of the moments is the same
for both design methods.
Figure 3.2.1.1-1: Steel shear –heads, steel plats joined by welding
Sabah Shawkat Cabinet of Structural Engineering 2017 Since the outer portions of horizontal members (slab) are less stiff than the part along the
support lines, the lateral distribution of the moment along the width of the member is not
Uniform. The procedure generally adopted is to divide the slab into column strips (along the
column lines) and middle strips and then apportion the moment between these strips and the
distribution of the moment within the width of each strip being assumed uniform.
Figure 3.2.1.1-2: Moments and frames
Sabah Shawkat Cabinet of Structural Engineering Figure: 3.2.1.1-3
Example: 3.2-1 Design and calculation of Flat Plate
Geometric Shapes
Slab thickness
hd  300mm
The geometry of the building floor plans:
l1  7.7m
lk  2.3m
Construction height of object:
l2  3.6m
ly  7.7m
kv  2.850m
Dimensions columns:
bs  400mm
hs  bs
The peripheral dimensions of the beam:
ho  0.5m
bo  0.30m
Figure: 3.2.1-1
Load calculation
Load per area
Reinforced concrete slab thickness of 300 mm
qdo  hd 25
kN
m
3
1.35
qdo  10.125
kN
m
2
2017 Sabah Shawkat Cabinet of Structural Engineering floor layer:
q1d  3
kN
2
1.4
q 1d
 4.2 
m
Live load (apartments):
vd  2.0
kN
2
1.5
qd  qdo  q1d  vd
m
vd  3
m
Total load on 1 m 2 of slab:
kN
2
kN
m
2
qd  17.325
kN
m
2
Force load
Peripheral masonry thickness of 400 mm YTONG:
F1  10
kN
kv ly 400mm1.35
3
m
Total load acting on the console:
F1  118.503kN
F1d  F1
F1d  118.503kN
Investigation replacement frame in the X-axis Frame 1:
Calculation model
Figure: 3.2.1-2
load calculation
2017 Sabah Shawkat Cabinet of Structural Engineering Load width in a direction perpendicular to the x:
zsx  ly
Load in the x-direction:
qdx  qd zsx
qdx  133.403
kN
m
Calculation of internal forces
Moment on a console:
2

lk 
Mk   F1d lk  qdx 
2 

Mk  625.407kN m
Moment of inertia:
Transverse replacement frame:
Ip 
ly hd
3
Ip  0.017m
12
4
Central girders replacement of frame:
Ist  Ip
column:
Is 
1
12
bs hs
Is  5.208  10 3 m
3
4
Bending stiffness:
Transverse replacement frame:
Kp 
Ip
l1
1000
kN
Kp  2.25kN 
2
rad m
Central girders replacement frame:
Kst 
Ist
 l2
1000
kN
rad m
2
m
rad
Kst  4.813kN 
m
rad
Column
Ks 
Is
kv
1000
kN
rad m
2
Ks  1.827kN 
m
rad
2017 Sabah Shawkat Cabinet of Structural Engineering 2017 Figure: 3.2.1-3
 1
 10  1
M1010'  1
M910  1
M10'10  1
9
M108  1
M911  1
M109  1
M97  1
M1012  1
Primary moments in node 9:
M910o  
M97o  0kN m
1
12
qdxl1
2
M911o  0kN m
Primary moments in node 10:
M109o 
1
12
M1010ò  
qdxl1
1
12
2
qdxl2
M108o  0kN m
2
M1012o  0kN m
M10'10o  M1010ò
Given
M97kN m M97o  Ks  3 9rad 
M911kN m M911o  Ks  2 9rad 
M910kN m M910o  Kp  2 9rad   10rad 
M108kN m M108o  Ks  3 10rad  M109kN m M109o  Kp  2 10rad   9rad 
M1012kN m M1012o  Ks  2 10rad 
M1010'kN m M1010ò  Kst   10rad 
M10'10kN m M10'10o  Kst   10rad 
Equilibrium conditions:
Node 9
Mk  M97kN m  M910kN m  M911kN m
Node 10:
0 kN m
Sabah Shawkat Cabinet of Structural Engineering M109kN m  M1012kN m  M1010'kN m  M108kN m 0kN m
v  Find M97 M910 M911 M109 M1012M1010'M108  9  10 M10'10
The calculated moments of individual members of equilibrium conditions:
M910  v ( 1.0) kN m
M10'10  v ( 9 0) kN m
M1010'  v ( 5.0) kN m
M911  v ( 2.0) kN m
M109  v ( 3.0) kN m
M910  691.408kN m
M10'10  282.659kN m
M1010'  282.659kN m
M911  26.401kN m
M109  545.787kN m
The computation of shear forces in the individual members:
V910o  qdx
0
l1
2
V109o  V910o
V910  V910o 
M910  M109
l1
V910  532.511kN
M910  M109
V109  V109o 
l1
V109  494.688kN
 l1 
V1010ò  qdx 
0
39.601
1
-691.408
2
26.401
3
545.787
v  4
-105.251
5
-282.659
6
-157.877
7
7.223
8
-28.797
9
282.659
 4
V1010'  V1010ò
V1010'  256.8kN
Maximum moment between 9-10 Mmax
a  V910
l1
a  3.992m
V910  V109
2
Mmax   V910a  M910  qdx
2
a
Mmax  371.423kN m
Maximum moment between 10-10 Mstr
Mstr
 l2 
 
l1
 2
 V1010'  M1010'  qdx
4
2
2
Mstr  4.431kN m
2017 Sabah Shawkat Cabinet of Structural Engineering 2017 Figure: 3.2.1-4
Transformation moments for the part columned strip and between the columns
Ma  M910
Mb  M109
Ma  691.408m kN Mb  545.787m kN
Mc'  Mmax1.25 Mc  Mstr 1.25 Mc'  464.278kN m Mc  5.539kN m
Moments over support:
 0.75
M1a   p Ma
M2a  172.852kN m
M2b   1   p Mb
p
M1a  518.556kN m
M1b   p Mb
M2b  136.447kN m
M2a   1   p  Ma
M1b  409.34kN m
Positively support moments:
 m  0.60
M3c  Mc'  m
M3c  278.567kN m
M4c  Mc'  1   m
M4c  185.711kN m
Sabah Shawkat Cabinet of Structural Engineering 2017 Figure: 3.2.1-5
Dimensioning of the reinforcement:
Material characteristic of concrete f ckcyl and steel fyk
fyd  375MPa
fcd  12MPa
The top reinforcement for moments:
effective height:
d  hd  3cm
width, which act the moment
b 
ly
b  3.85m
2
Column strip M 1a:
M1a  518.556kN m
fcd  12MPa
d  0.27m


M1a  0.518MN m
M1a

2
b  3.85m
 0.0475

 0.154
b d fcd
Ast   b d fcd
Ast 
  b d fcd 
MN
2
100cm
2
Ast  59.252cm
Ast
3.85
2
 15.39cm
Sabah Shawkat Cabinet of Structural Engineering 2017 Among the columned strip M 2a:
M2a  172.852kN m
M2a  0.172MN m
fcd  12MPa
d  0.27m


M2a

2
b  3.85m
 0.01439

 0.051
b d fcd
Ast 
  b d fcd 
MN
2
Ast
2
100cm
Ast  17.95cm
2
3.85
 4.662cm
Column strip M 1b:
M1b  409.34kN m
M1b  0.409MN m
fcd  12MPa
d  0.27m


M1b

2
b  3.85m
 0.03596

 0.121
b d fcd
Ast 
  b d fcd 
MN
2
2
100cm
Ast  44.857cm
Ast
3.85
2
 11.651cm
Among the columned strip M 2b:
M2b  136.447kN m
M2b  0.136MN m
fcd  12MPa


b  3.85m
M2b

2
d  0.27m
 0.01145

 0.04
b d fcd
Ast 
  b d fcd 
MN
2
100cm
2
Ast  14.283cm
The lower reinforcement for moments:
Ast
3.85
2
 3.71cm
Sabah Shawkat Cabinet of Structural Engineering 2017 Column strip M 3c:
M3c  278.567kN m
M3c  0.278MN m
b  3.85m


fcd  12MPa
d  0.27m
M3c

2
 0.0249

 0.083
b d fcd
Ast 
  b d fcd 
MN
2
Ast
2
100cm
Ast  31.06cm
3.85
2
 8.068cm
Among the columned strip M 4c:
M4c  185.711kN m
M4c  0.185MN m
fcd  12MPa
d  0.27m


M4c

2
b  3.85m
 0.055

b d fcd
Ast 
  b d fcd 
2
100cm
MN
Investigation replacement frame in y Frame 2
Calculation Model
2
Ast  19.809cm
 0.01588
Sabah Shawkat Cabinet of Structural Engineering Figure: 3.2.1-6
Load calculation
q2d  qd   l1  l2
1
q2d  97.886
2
kN
m
Calculation internal forces
Support part:
1
2
Ma    q2d ly
 12 
Ma  483.64kN m
Among the supports:
Mc' 
1
16
q2d ly
2
Mc'  362.73kN m
a magnification between support:
Mc  Mc' 1.25
Mc  453.412kN m
Transformation moments for the part columned strip and among columned
support t of Ma2
p
 0.75
Ma1   p Ma
Ma2   1   p  Ma
Ma1  362.73kN m
Ma2  120.91kN m
Between the support of M c
m
 0.6
Mc1   mMc
Mc1  272.047kN m
2017 Sabah Shawkat Cabinet of Structural Engineering Mc2   1   m Mc
2017 Mc2  181.365kN m
Dimensioning of reinforcement
Upper reinforcement of moment:
Effective depth:
d  hd  3cm
Column strip M 1a:
The width on which acting the moment:
b 
l1
4

l2
4
b  2.825m
Column strip M 1a:
M1a  0.518MN m


d  0.27m
M1a

2
 0.06687
fcd  12MPa
b  2.825m
 0.21

b d fcd
Ast 
  b d fcd 
MN
2
2
100cm
Ast  61.206cm
Between the column strip M 2a:
M2a  0.172MN m


d  0.27m
M2a

2
 0.02037
fcd  12MPa

b  2.825m
 0.07
b d fcd
Ast 
  b d fcd 
MN
2
100cm
2
Ast  18.645cm
Sabah Shawkat Cabinet of Structural Engineering 2017 Column strip M 1a:
Mc1  0.272MN m


Mc1
2
d  0.27m

 0.03277
fcd  12MPa
b  2.825m
 0.11

b d fcd
Ast 
  b d fcd 
MN
2
2
100cm
Ast  29.994cm
Between column strip M 2a:
Mc2  0.181MN m


Mc2
2
d  0.27m

 0.0216
fcd  12MPa

b  2.825m
 0.073
b d fcd
Ast 
  b d fcd 
MN
2
100cm
Investigation extreme frame replacement
Calculation Model:
2
Ast  19.77cm
Sabah Shawkat Cabinet of Structural Engineering Figure: 3.2.1-7
Calculation of load:
From the slab:


q3do  qd  lk 
l1 

2
q3do  106.549
kN
m
Peripheral masonry thickness of 400 mm YTONG:
F1  10
kN
m
3
kv 400mm1.4
F1  15.96
kN
m
Total load replacement frame:
qkd  q3do  F1
Calculation of internal forces
Moment of the end strip:
q kd  122.509 
kN
m
2017 Sabah Shawkat Cabinet of Structural Engineering 2017 Support bending moment:
Mka  
1
12
qkd ly
2
Mka  605.295kN m
Between the column bending moment:
Mkc 
1
16
qkd ly
2
Mkc  453.971kN m
Transformation moments for the part columned bands and among columned
columned strip width:
l1
bp3  lk 
2
bp3  6.15m
Moments over support:
Mexta 
Mka 
lk 
 1  2 

4
bp3 

Mexta  264.509kN m
Minta  Mka  Mexta
Mk4a   p Minta
Mk3a   1   p  Minta
Mk3a  85.197kN m
Mk4a  255.59kN m
Between the column moments:
Mextc 
Mkc 
lk 
 1  2

4
bp3 

Mextc  198.382kN m
Mintc  Mkc  Mextc
Mk4c   mMintc
Mk3c   1   m Mintc
Mk3c  102.236kN m
Design the reinforcement to the reinforced concrete slab
The top reinforcement for moments:
Mk4c  153.354kN m
Sabah Shawkat Cabinet of Structural Engineering 2017 effective height:
d  d  3cm
Column extreme strip Mexta:
width which act moment
b  lk
b  2.3 m
Column extreme strip Mexta: see diagram B3-B3.3
Mexta  0.265MN m


d  0.24m
Mexta

2
 0.0509
fcd  12MPa
b  2.3 m
 0.166

b d fcd
Ast 
  b d fcd 
MN
2
2
100cm
Ast  33.716cm
Column strip inside M k4a:
width, which acts moment, see diagram B3-B3.3
b 
l1
b  1.925m
4
Mk4a  0.256MN m


d  0.24m
Mk4a

2
 0.05965
fcd  12MPa

b  1.925m
 0.192
b d fcd
Ast 
  b d fcd 
MN
2
100cm
Among the columned strip M k3a:
width, which acts moment
2
Ast  33.07cm
Ast
1.925
2
 17.179cm
Sabah Shawkat b 
l1
Cabinet of Structural Engineering b  1.925m
4
Mk3a  0.085MN m


2017 fcd  12MPa
d  0.24m
Mk3a

2
 0.01739

b  1.925m
 0.064
b d fcd
Ast 
  b d fcd 
MN
2
2
100cm
Ast  9.641cm
Ast
1.925
2
 5.008cm
The lower reinforcement for moments:
Column extreme strip Mextc:
width, which acts moment, see diagram B3-B3.3
b  lk
b  2.3 m
Mextc  0.198MN m


d  0.24m
Mextc

2
 0.03758
fcd  12MPa

b  2.3 m
 0.125
b d fcd
Ast 
  b d fcd 
MN
2
100cm
2
Ast  24.893cm
Ast
2.3
Column strip inside M k4c:
width, which acts moment
b 
l1
4
Mk4c  0.153MN m
b  1.925m
d  0.24m
fcd  12MPa
2
 10.823cm
b  1.925m
Sabah Shawkat 

Cabinet of Structural Engineering Mk4c

2
 0.03436

2017  0.115
b d fcd
Ast 
  b d fcd 
MN
2
Ast
2
100cm
Ast  19.049cm
1.925
2
 9.896cm
Among the columned strip M k3c:
width, which acts moment
b 
l1
b  1.925m
4
See diagram B3-B3.3
Mk3c  0.102MN m


Mk3c

2
 0.02189
fcd  12MPa

b  1.925m
 0.077
b d fcd
Ast 
d  0.24m
  b d fcd 
MN
2
100cm
2
Ast  12.136cm
Ast
1.925
2
 6.304cm
Sabah Shawkat Cabinet of Structural Engineering 2017 Example 3.2-2: In the example we are considering reinforced concrete slab flat, floor slab
thickness is hd = 0.3m, Column diameter (round column) d =0.50 m, the maximum force applied
one column at Nd= 1800 kN.
d 
hd 
0.5  m
0.3  m
b 
Nd 
1 m
1800  kN
Material characteristics:
fcd 
17  MPa
fctm 
1.2  MPa
fyd 
375  MPa
Figure: 2.3.2‐1 Coefficient of shear strength
 1 
18  mm
 stw 
Ast
As1   
1
2
4
n 
Ast  n  As1
5
Ast  0.00127 m
2
 stw  0.004241 m
b  hd
in both directions
On 1m plate
 stmin 
1 fctm

3 fyd
 b 
1
 h 
1.4 
 stmin  0.001067
 s 
2 hd

3 m
 g   s  n   h   f
1.159
 h  1.2
 g  1.74
 n 
1.0
 f 
1.25
 b 
1
Sabah Shawkat Cabinet of Structural Engineering 2017 Carrying capacity of the concrete section
qbu 
0.42  hd   g   b  fctm
qbu 
262.86 m
1
 kN
Assess the resistance of the concrete section
Maximum force per columns
Vcd  Nd
Vcd  1800 kN
Basic critical perimeter
ucr  2.51 m
Shear force on the critical perimeter
qd 
Vcd
qd  716.2 m
ucr
1
 kN
Shear resistance of concrete
qbu  262.86 m
1
 kN
qd  716.2 m
1
 kN
qd  qbu
We suggest shear reinforcement
qd  2  qbu
2  qbu  525.72 m
1
 kN
Incorrect design, head to be designed so that they apply condition:
qd  2  qbu
correct proposal
Proposal of hidden head
Maximum critical perimeter with hidden head
Ucrmax 
1.9  ucr
Ucrmax  4.78 m
qda 
Vcd
Ucrmax
qda  qbu
qda  376.95 m
1
 kN
2  qbu  525.72 m 1  kN
qbu  262.86 m
1
 kN
Sabah Shawkat Cabinet of Structural Engineering 2017 If we want to make a proposal without head, subject to the following parameters:
d 
fctm40 
1.0  m

ucr40     d 
1.40  MPa

2
hd 

2
ucr40  4.08 m
Carrying capacity of the concrete section
qbu40 
0.42  hd   g   b  fctm40  ucr40
qbu40  1252.47 kN
2  qbu40  2504.94 kN
Vcd 
1800 kN
2qbu40  Vcd
Proposal visible head
Geometry head
 
45  deg
hh 
0.6  m
qd2 
Vcd
sin ( )  0.71
cos ( )  0.71
dh 
Ucr2    dh
2.0  m
qd2  286.48 m
Ucr2
1
Ucr2  6.28 m
 kN
qbu  262.86 m
1
 kN
qd2  qbu
Figure: 2.3.2‐2 We suggest shear reinforcement
qd2  2  qbu
Proposal shear reinforcement - reinforced by bins
 qsu
q
q
fyd 
190  MPa
d2
bu
qsu
n  A ss   ss   s  fyd
 ss 
1
 s 
1
q su  q d2  q bu
A ss 
1  m2
n 
q su 
23.62 m
1
 kN
1
n is the number of bins reinforced, Ass area of reinforcement to a bin
Given
qsu
n  Ass  ss  s fyd  m
1
 
Ass  Find Ass
Ass  0.000124 m
2
Sabah Shawkat Cabinet of Structural Engineering n1 
5 number of bars in one bin / m 'ss = 0.25m
 1 
8  mm
diameter of one profile
2017 1
Asssku  n1   
2
Asssku  0.00025133 m
2
Asssku  Ass
4
Assessment of the punching according to EC 2
design value of shear resistance of plate without shear reinforcement (per unit length of critical
perimeter)
 Rd  k 
v Rd1
1.2  40  1  d
shear resistance
 Rd 
0.3 
MN
m
bt 
1 m
fyk 
410 
hd 
2
0.3  m
k 
1.6 
hd
m
k  1.3
average width tension section
MN
m
min2 
min1 
2
0.0015  bt 
hd
m
min 
 min1 


 min2 
2

min  max min

0.6  bt 
min2 
0.00045
min 
0.00045
concrete area
Ac  hd  bt
Ac 
0.3 m2
The maximum degree of reinforcement
max 
0.04 
Ac
m
2
max 
0.01
The average degree of reinforcement
1 
min  max
2
hd kN

fyk
4
m
1 
0.01
min1  0.00000044
Sabah Shawkat vRd1   Rd  k 
Cabinet of Structural Engineering 1.2  40  1  hd
vRd1 
2017 169.53 m 1  kN
The maximum design value of shear resistance of plate with shear reinforcement (per unit
length of critical perimeter)
vRd2 
1.6  vRd1
vRd2 
271.25 m 1  kN
Design value of shear resistance of plate with shear reinforcement (per unit length of critical
perimeter)
 A f
s yd  sin ( )
vRd1 
vRd3

i
u
Column diameter
Ps 
0.5  m
Diameter of critical perimeter
Pu 
2  1.5  hd  Ps
Pu 
1.4 m
Critical perimeter
u    Pu
Acw   
u
Pu
2
4
4.4 m
2
 
Ps
Acw  1.34 m
4
2
concrete shear area
Assumption degree of shear reinforcement
´w 
0.0013  0.6
As  ´w Acw
vRd3  vRd1 
As  0 m
2
As  fyd  sin ( ) 
u
 

2
vRd3 
fyd 
360 
MN
m
255.28 m 1  kN
2
carrying capacity
Sabah Shawkat Cabinet of Structural Engineering 2017 The load effects
Vsd 
1800  kN
 
Computing shear force
Figure: 2.3.2‐3 1.15
internal columns
vsd 
Vsd  
vsd 
u
470.64 m 1  kN
as being applicable condition
vsd  vRd3
incorrect design, design head
vsd  vRd3
Geometry head
lh 
0.9  m
hh 
0.6  m
d2crit 
3.11  m
Critical perimeter with head
ucrit    d2crit
ucrit 
9.77 m
Concrete shear area
Acwh   
d2crit
2
2
 
Ps
Acwh 
7.4 m2
4
4
The expected level of reinforcement by shear reinforcement
Ash   ´w Acwh
vRd3  vRd1 
Ash  0.01 m
Ash  fyd  sin ( ) 
ucrit
2
vRd3 
vsd 
Vsd  
ucrit
382.21 m 1  kN
Slab with shear reinforcement to a void punching.
vsd 
211.87 m 1  kN
vsd  vRd3
Sabah Shawkat  A
Cabinet of Structural Engineering sw  sin ( )
2017 
i
w
Ax
Space inside the critical perimeter less the contact surface
Ax   
d2crit
2
4
2
 
Ps
Ax  7.4 m
4
2
For dimensioning elements requiring shear reinforcement
0.5    fcd   bw  0.9  d  1  cotg ()
VRd2
 
0.7 
fck
200  MPa
  0.58
fck 
25  MPa
  if    0.5 0.5  
fcd 
13.3  MPa
  0.58
  0.5
Smallest section width in the range of effective height
bw 
1.0  m
Height of the floor slab
hd 
0.3  m
cot ( )  0
VRd2 
0.5    fcd   bw  0.9 hd  1  cot ()
VRd2  1032.41 kN
Maximum distance of stirrups
s max 
0.3  hd
s max  0.09 m

s max  if s max  0.2  m s max 0.2  m
2
 VRd2  688.27 kN
3
Maximum diameter of reinforcement stirrups with a smooth surface
Vsd  1800 kN
s 
0.012  m
Sectional area of shear reinforcement in the length range

s max  0.09 m
Sabah Shawkat 2
Asw   
w 
s
4
Asw  sin ( )
Ax
Cabinet of Structural Engineering Asw  0.00011 m
2
 
wtab 
w  0.00001528
2017 
2
wmin  0.6 wtab
0.0013
wmin  0.00078
Necessary degree of reinforcement EC2:


w  if w  wmin wmin w
w  0.00078
Minimum design values of moments on columns in contact with the plate at the eccentric load
x  0.125
y  0.125
force msdx  x  Vsd
Internal Column, top moment
Internal Column, top moment
msdx  225 kN
msdy  y  Vsd
Vsd  1800 kN
acting shear
msdy  225 kN
Figure: 2.3.2‐4 Sabah Shawkat Cabinet of Structural Engineering 2017 Example 3.2-3: In the middle columns of dimensions as x bs from adjacent reinforced flat slab
of thickness hs at a critical cross-section carries a full load slab, shear force Vcd = 400 kN,
shear force from accidental load Vcd = 325 kN and the bending moment Mcd = 20 kNm (moment
transmitted from slab to reinforced column).
Figure: 2.3.3-1
Material characteristics:
fckcub  20  MPa
fckcyl  0.8  fckcub
fckcyl
fcd  0.85 
fckcyl  16  MPa
fcd  9.067  MPa
1.5
2
3
 fckcyl 
  MPa
 10  MPa 
fctm  1.4  
fctm  1.915 MPa
fyk  345  MPa
fyd 
fyk
1.15
fyd  300  MPa
where fctk is the characteristic tensile strength of concrete (5-percent fractile), fctm is the mean
tensile strength and fck is the characteristic compressive strength of concrete measured on
cylinders.
The depth of reinforced concrete slabs
hs  0.2  m
Dimension columns:
as  0.40  m
bs  0.40  m
Sabah Shawkat Cabinet of Structural Engineering 2017 Bending moment and shearing forces:
Vcd1  400  kN
hs
Uc1  as 
2
Vcd  325  kN
Mcd  20  kN m
Uc1  0.6 m
2
Uc2  bs 
Ucr  2   Uc1  Uc2
Vcd1
qdmax 
Ucr
Ucr  2.4 m
Mkontr  0.2  Vcd  hs
Mkontr  13 m  kN
1
qdmax  166.667 m
 kN
If Mkontr less than the Mcd, should be respected Mcd
Icr 
U c1
3

6
U c2  U c1
2
2
Icr  0.144 m
3
Figure: 2.3.3-2
n 
1
 U c2 
 
1 

2
 U c1 
n  0.4
3
 dmax

V cd
U cr

M cd  n  0.5  U c1
Calculation of Qbu
Icr
 dmax
1
 152.08m
 kN
hs
2
2
Uc2  0.6 m
Sabah Shawkat Cabinet of Structural Engineering 2017 2
d
 18  mm
As1   

 stmin
 sty
s
4
n  6
Astd  0.00152681 m
c2  bs

 stx
1 fctm

3 fyd
 stmin
Astd

 c2  4  hs  hs

  stx   sty 
 stm
2
As1  0.00025447 m
2
Astd  n As1
c1  as \
d
 stm
qbu  0.42   s   h   n  fctm 
1
qbu  246.91096132 m
hs
m
 kN
 stx
 c2  4  hs  hs
 0.00636173
 0.00212797
 0.00636173
 sty
 1  50   b    stm   stmin 
Astd
s
 0.00636173
 1.212
h
 1.4  m  2 
qbur  0.42  fctm 
hs
3
h
 1.267 m
hs
m
2
qbur  160.87453706 m
 kN
The reliability condition
1
qdmax  166.667 m
1
 kN
2
qbur  160.87453706 m
qbu  246.911 m
 kN
 kN
2
qbueur  160.87453706 m
 kN
The cross-section without shear reinforcement does not comply
I suggest shear reinforcement in the form of welded of mesh
 ss
 1
ss 
s
 8  mm
 ss  Ass1   s  fyd
qdmax  0.5  qbu
Ass1   
s
2
4
ss  0.34897549 m
2
Ass1  0.00005027 m
Sabah Shawkat Cabinet of Structural Engineering 2017 Perimeter displaced the critical cross section
Ucrp  2   c1  hs  2  ss   c2  hs  2  ss
qdmaxp 
Vcd1
Ucrp  5.19180391 m
1
qdmaxp  77.04451232 m
Ucrp
 kN
It is less than qbu, that is, the cross section satisfies without shear reinforcement.
Alternative we suggest shear reinforcement consisting of a flexible conduit at an angle
 =60.deg.
1
qbu  246.911 m
Asb 
1
 kN
qdmax  166.667 m
 qdmax  0.5  qbu  Ucr
 kN

2
Asb  0.00039917 m
sin (  )   s  fyd
The proposal
2
 oh
 14  mm
Asb
Asoh
 2.59304223
Asoh   
 oh
4
2
Asoh  0.00015394 m
 60  deg
Sabah Shawkat Cabinet of Structural Engineering 2017 Figure: 2.3.3-3
Internal column of 500 x 500
hd  25  cm
fctm  1.2  MPa
fyd  375  MPa
P1  856  kN
bs  50  cm
hs  50  cm
step 1:
ucr1   bs  hd   hs  hd  2
Qbu1  0.42  hd  fctm  ucr1
P  P1  0.5  Qbu1
P  667 kN
Asb 
As1 
P
0.86  fyd

2
Asb  0.00206822 m
2

4
V1  0.42  hd  fctm  ucr1
Qbu1  378 kN
 25  mm

2
Asb
As1  0.00049087 m
n 
V1  378 kN
P  667 kN
As1
Step 2
ucr2   bs  3  hd   hs  3  hd  2
ucr2  5 m
Qbu2  0.42  hd  fctm  ucr2
Qbu2  630 kN
n  4.21333718
Sabah Shawkat Cabinet of Structural Engineering P  P1  0.5  Qbu2
n 
Asb
n  3.42
As1

P
0.86  fyd
Asb 
P  541 kN
 25  mm
As1 

2017 2
Asb  0.00167752 m
2

4
2
As1  0.00049087 m
V2  0.42  hd  fctm  ucr2 V2  630 kN
ucr 
P
0.42  hd  fctm
ucr  4.2937 m
ucr
4
 1.0734 m
Column of 400x 500 extreme
hd  25  cm
fctm  1.2  MPa
bs  40  cm
hs  50  cm
fyd  375  MPa
Qbu1  0.42  hd  fctm  ucr1
Qbu1  226.8 kN
Step 1:
ucr1   bs  0.5  hd  2   hs  hd
0.5  Qbu1  113.4 kN
When applied to the plate even bending moment, then we take 0.5 qbu
P1  577  kN
P  P1  0.5  Qbu1
P  463.6 kN
For P we calculate the required shear reinforcement.
Asb 
As1 
P
0.86 fyd

2
Asb  0.00143752m

 20  mm
2

4
V1  0.42  hd  fctm  ucr1
2
As1  0.00031416m
V1  226.8 kN
n 
Asb
n  4.575766
As1
P  463.6 kN V1  P does not comply
Sabah Shawkat Cabinet of Structural Engineering 2017 Step 2
ucr2   hs  3  hd   bs  1.5  hd  2
ucr2  2.8 m
Qbu2  0.42  hd  fctm  ucr2
Qbu2  352.8 kN
We expand the circumference in order to prevent the creation of a new shear crack
P  P1  0.5  Qbu2

 20  mm
n 
Asb
As1
P  400.6 kN
As1 
P
0.86  fyd
2

4
V2  P
2
Asb  0.00124217 m
2
As1  0.00031416 m
V2  0.42  hd  fctm  ucr2
n  3.95395164
P  400.6 kN

Asb 
V2  352.8 kN
does not comply
Figure 2.3.3-4: Shear reinforcement at slab-column connection
Step 3:
ucr3   hs  5  hd   bs  2.5  hd  2
Qbu3  0.42  hd  fctm  ucr3
Qbu3  478.8 kN
ucr3  3.8 m
P  P1  0.5  Qbu3
P  337.6 kN
Sabah Shawkat Asb 

Cabinet of Structural Engineering P
0.86  fyd
 25  mm
2017 2
Asb  0.00104682 m
As1 
V3  0.42  hd  fctm  ucr3

2

4
2
As1  0.00049087 m
V3  478.8 kN
n 
Asb
As1
n  2.13
P  337.6 kN V3  P OK V3 is greater than P, thus the determination of the reinforcement to avoid the punching in
reinforced concrete slab flat over the column is o