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SabahShawkat CabinetofStructuralEngineering 20173.2 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. One-
way 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.
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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 cross-
beams 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
SabahShawkat CabinetofStructuralEngineering 20173.2.1 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).
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Figure 3.2.1-2: Combined punching shear and transfer of moments
Figure 3.2.1-3
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Figure 3.2.1-4
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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
SabahShawkat CabinetofStructuralEngineering 2017Since 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
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Figure: 3.2.1.1-3
Example: 3.2-1 Design and calculation of Flat Plate
Geometric Shapes
Slab thickness
The geometry of the building floor plans:
Construction height of object:
Dimensions columns:
The peripheral dimensions of the beam:
Figure: 3.2.1-1
Load calculation
Load per area
Reinforced concrete slab thickness of 300 mm
hd 300 mm
l1 7.7 m lk 2.3 m l2 3.6 m ly 7.7 m
kv 2.850 m
bs 400 mm hs bs
ho 0.5 m bo 0.30 m
qdo hd 25kN
m3 1.35 qdo 10.125
kN
m2
SabahShawkat CabinetofStructuralEngineering 2017floor layer:
Live load (apartments):
Total load on 1 m 2 of slab:
Force load Peripheral masonry thickness of 400 mm YTONG:
Total load acting on the console:
Investigation replacement frame in the X-axis Frame 1:
Calculation model
Figure: 3.2.1-2
load calculation
q1d 3kN
m2 1.4 q 1d 4.2
kN
m 2
vd 2.0kN
m2 1.5 vd 3
kN
m2
qd qdo q1d vd qd 17.325kN
m2
F1 10kN
m3 kv ly 400 mm 1.35 F1 118.503kN
F1d F1 F1d 118.503kN
SabahShawkat CabinetofStructuralEngineering 2017Load width in a direction perpendicular to the x:
Load in the x-direction:
Calculation of internal forces
Moment on a console:
Moment of inertia:
Transverse replacement frame:
Central girders replacement of frame:
column:
Bending stiffness:
Transverse replacement frame:
Central girders replacement frame:
Column
zsx ly
qdx qd zsx qdx 133.403kNm
Mk F1d lk qdxlk
2
2
Mk 625.407kN m
Iply hd
312
Ip 0.017m4
Ist Ip
Is112
bs hs3 Is 5.208 10 3 m
4
KpIpl1
1000kN
rad m2 Kp 2.25 kN
mrad
KstIstl2
1000kN
rad m2 Kst 4.813 kN
mrad
KsIskv
1000kN
rad m2 Ks 1.827 kN
mrad
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Figure: 3.2.1-3
Primary moments in node 9:
Primary moments in node 10:
Equilibrium conditions:
Node 9
Node 10:
9 1 10 1 M910 1 M108 1 M109 1 M1012 1M1010' 1 M10'10 1 M911 1 M97 1
M97o 0 kN m M910o112
qdx l12 M911o 0 kN m
M109o112
qdx l12 M108o 0 kN m M1012o 0 kN m
M1010112
qdx l22 M10'10o M1010
GivenM97 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 M1010 Kst 10 rad M10'10 kN m M10'10o Kst 10 rad
Mk M97 kN m M910 kN m M911 kN m 0 kN m
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The calculated moments of individual members of equilibrium conditions:
The computation of shear forces in the individual members:
Maximum moment between 9-10 Mmax
Maximum moment between 10-10 Mstr
M109 kN m M1012 kN m M1010' kN m M108 kN m 0 kN mv Find M97 M910 M911 M109 M1012 M1010' M108 9 10 M10'10
M910 v 1.0( ) kN m M910 691.408 kN mM10'10 v 9 0( ) kN m M10'10 282.659kN mM1010' v 5.0( ) kN m M1010' 282.659 kN mM911 v 2.0( ) kN m M911 26.401 kN mM109 v 3.0( ) kN m M109 545.787kN m
V910o qdxl12
V109o V910o
V910 V910oM910 M109
l1
V910 532.511kN
V109 V109oM910 M109
l1
V109 494.688 kN
V1010 qdxl14
V1010' V1010V1010' 256.8 kN
v
0
01
2
3
4
5
6
7
8
9
39.601-691.408
26.401
545.787
-105.251
-282.659
-157.877
7.223
-28.797
282.659
a V910l1
V910 V109 a 3.992m
Mmax V910 a M910 qdx a2
2 Mmax 371.423kN m
Mstr V1010'l14 M1010' qdx
l22