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PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
REINFORCED CONCRETE STRUCTURE
COURSE STR 302A
CL0SED BOOK EXAM
THIRD YEAR CIVIL FIRST
TERM YEAR 2012 – 2013
COURSE CONTENTS : DESIGN OF DIFFERENT SLABS
TYPES OF SLABS
- Solid Slab - Hollow Block Slab
- Paneled Beam Slab - Stairs
- Flat Slab
GRADES
TUTORIAL & ASSIGNMENT
20%
MED TERMEXAM
10%
FINAL EXAM 70% TOTAL 100%
ناحتمالا ىف ةیرورض( میمصتلا تادعاسم – لامحالا دوك – ةناسرخلا دوك)
ظنام حصص الشرح كام ھو عمروف
SOLVED EXAMPLES
PREPARED BY
PROF. DR. MAGDY KASSEM
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
3rd Year Civil 1 st Term - 2012 - 2013
No Date Lecture Tutorial
1 15-9-2012 Solid Slabs Solid Slabs
2 22-9-2012 Solid Slabs Solid Slabs
3 29-9-2012 Solid Slabs Quiz + Grading S slabs
4 6-10-2012 Paneled Beams Paneled Beams
5 13-10-2012 H.B Slabs Quiz + Grading P.B
6 20-10-2012 H.B Slabs H.B Slabs
عید اضألحى المبارك 10-2012-(26-29) 7
إمتحانن صف الفصل ادلراسى 2012- 3-11 8
9 10-11-2012 H.B Slabs H.B Slabs
10 17-11 -2012 Flat Slabs Quiz + Grading H.B
11 24-11-2012 Flat Slabs Flat Slabs
12 1-12- 2012 Flat Slabs Flat Slabs
13 8-12-2012 Stairs Quiz + Grading F.S
14 15-12-2012 Stairs Stairs
15 22-12-2012 Quiz + Grading Stairs
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
One-Way Ribbed or Hollow-Block Slab
System
If the ribs are arranged in one direction, loads are transferred in one direction; the slab is called one-way ribbed (joist) slab, see Figures 3 and 4. The ribs are supported on girders that rest on columns.
Structural Plan
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Figure 3: One-Way hollow-block slab system
Figure 4: Isometric of one-way hollow-block slab system
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Two-Way Ribbed or Hollow-Block Slab
System
If ribs are arranged in two directions, loads are transferred in two directions; the slab is called two-way ribbed (joist) slab.
Different systems can be provided of two-way ribbed slab as
follows, see Figure 5.
i. Two-way rib system with permanent fillers between ribs that
produce flat slab is called two-way hollow-block (joist) slab system, see Figure 5a1 and 5a2.
Figure 5a-1: Structural plan of two-way hollow-block slab
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Figure 5a-2: Isometric of two-way hollow-block slab system
ii. Two-way rib system with voids between the ribs that obtained by using special removable forms (pans), normally square in shape, is called two-way ribbed (joist)slab system, Figure 5b.
Figure 5b: Structural plan of two-way ribbed slab
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
iii. Two-way rib system, with voids between the ribs, has the ribs continuing in both directions without supporting beams and
resting directly on columns through solid panels above the columns. This type is called a waffle slab system, Figure 5c.
Figure 5c: Interior panel of waffle slab system
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Reinforcement details and blocks
arrangement
Figure 10 shows the reinforcement details of one-way hollow-block slab systems.
Sec. (1-1)
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Sec. (2-2)
Sec. (3-3)
Figure 10: Reinforcement details of one-way hollow-block slab system on plan and in cross-sections
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Reinforcement details and blocks
arrangement The arrangement of hollow-block and shape of reinforcement details of two-way slab systems are shown in Figure 12.
Figure 12: Plan of reinforcement details of two-way simply
supported hollow-block slab system
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Paneled Beam Slab
Design of Edge Beams
The floor "mother" slab a × b is divided into small slabs s1, s2, s3
by a system of paneled beams. Each of the baby slabs s1, s2, s3 ….. is designed as a solid (or hollow blacks) slab.
For the designer of edge beams, consider the whole slab a × b to be carried by the four beams at the four edges: For analysis of edge beams, there are two approaches:
1. distribute load on edge beams from slab as if there is no
paneled beams inside, i.e. assuming fracture lines at 45o
from corners.
• this approach becomes more appropriate as the number of paneled beams not supported directly on columns increases.
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
2. Transfer reactions of panelled beams to edge beams and solve the edge beam subjected to:
1. reaction of panelled beams
2. slab loads of slabs adjacent to the edge beam.
• this approach becomes more appropriate than the former
one as the number of panelled beams supported directly on
column increases.
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Flat Slabs
Introduction
Flat slabs are beamless slabs resting directly on columns. They may have drop panels and/ or columns may have column capitals
as shown in Figure (1). They may be solid or ribbed in two
directions with or without hollow blocks. Generally speaking, flat slabs behave like two continuous
orthogonal frame system. Accordingly, the bigger moments occur in the direction of bigger spans. This behavior is different than two- way solid slabs supported on all four sides.
Flat slabs Roofs
Solid Slabs Flat Slabs
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Flat Slabs without drop Panel ( L.L 500 kg/cm2)
B
A A
B
ts
Section A - A
Flat slabs with drop Panel ( L.L 1000 kg/cm2)
A A
ts (drop panel) ts
Section A - A
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Flat Slabs
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
o Stairs Slab and Beam Type
Slab Type Cantilever Type
Cantilever Type
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Solid Slabs
O ne Way Slabs
Effective Span
Loads
Minimum Thickness
Bending Moments
Reinforcement
Supports
Two Way Slabs
General
Effective Spans
Minimum Thickness
Simplified Method for Calculating
B.M. in Two Way Slabs
Reinforcement
Calculations
Practical Cases
Concluding Remarks
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Waffled slabs (ribbed slabs without hollow
blocks , spacing about 1 m each direction)
• Solid Slabs
• Hollow Block Slabs
• Flat Slabs
• Waffle Slabs
• Paneled Beams Slabs
•Loads
•Spans
•Design Requirements
Structural system must be chosen
Ly to satisfy economic requirements
according to these factors
Lx
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
One-way slabs
Thickness t = b / 30
= b / 35
= b / 40
= b / 10
Two-way slabs
Thickness t = b / 35
= b / 40
= b / 45
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
One Way Slabs
* Slabs supported on two opposite sides only
* Slabs supported on four sides with effective length (be) greater than twice the effective
length (ae)
* One way slabs are to be calculated based on strips of unit length in the short
direction
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Effective Span
i) Le for simple or continues slab is the larger of :
c s
or 1.05 Lc
or Le < L
ii) Le for cantilever slabs is the smaller of :
L
or Lc + ts
iii) Continuous or simply supported slabs supported on beams with b > 0.2 Lc can
be
designed as fixed at support, and every span can be calculated separately
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Loads
* Total load on slab: W = WD + WL
* Dead loads:
WD = slab self weight + flooring = ts (mm) × 2.5 + (1.5 → 3.5)
kN/m2
* Live loads:
WL: depends on the function of the building.
* If designing using Limit State Design Method,
use Wu = 1.4 WD + 1.6 WL
L.L IS TO BE TAKEN FROM FROM CODE ACCORDING TO THE
FUNCTION OF THE BUILDING
Supports
The width of the support for slab is the larger of :
ts or 150 mm …..for brick walls
100 mm …………for R.C. beams
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Minimum Thickness
1. Slab thickness shall not be less than the absolute minimum of:
tmin = Le/30 simply supported at both ends tmin = Le/35 continuous at one end
tmin = Le/40 continuous at both ends Where Le is the effective span as defined above
2. For ordinary buildings and cast-in-situ slabs:
ts > 80 mm under static loads ts > 120 mm under dynamic loads
3. Smaller thicknesses can be used in pre-cast slabs.
4. Minimum thickness to satisfy the deflection limits in ordinary buildings (i.e. no need to
check deflection if the slab thickness is not less than the values indicated) is as follows:
Table (4-10) – ECCS 203-2001
Span / depth ratio; L/d
* Table (4-10) applies only if the following conditions are satisfied:
1. ordinary buildings
2. spans ≤ 10 m
3. uniform, but not heavy, live load, (live load ≤ twice dead load)
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Bending Moments
N.B.
* In designing with Ultimate Limit States Design Method: w = w ultimate
* In designing with Working Stress Design Method: w = wworking
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Bending moment of slab
According to ECCS 203-2001, the bending moment of equal span ribs or
with maximum difference not greater than 20% of the larger span is as follows:
M = wL2/k, where k is as follows:
- Simple span
- Two spans
- More than two spans
Reinforcement
The main reinforcement of a slab shall satisfy the requirements of equilibrium:
C = T Mu = C . yct = T. yct
cover = 15 – 25 mm
The reinforcement shall also be arranged along the direction of short span.Normal
to the main reinforcement, shrinkage and temperature (distributes) reinforcement
are provided.
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
1. Minimum main reinforcement:
* For mild steel :
* For high strength steel :
The main reinforcement should be arranged to cover all the tension zones and
extend beyond the end of these zones enough to develop the necessary
anchorage.
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
One way slab reinforcement details.
3. Maximum spacing between bars of main steel is 2 ts and < 200 mm.
For ts < 100 mm, spacing of 200 mm may be used. 4. A minimum of 1/3 of the main steel must be extended to the supports.
5. Distributors with minimum of 4 bars/m. 6. Minimum diameter for straight bars is 6 mm.
Minimum diameter for bent bars is 8 mm.
Bars of smaller diameters may be used for welded wire mesh and pre-cast concrete.
7. When using steel mesh, all conditions above must be satisfied.
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
8. For slabs with thicknesses ≥ 160 mm, top mesh must be provided. The top mesh must
be taken ≥ 20% of the bottom reinforcement, but not less than 5Ø8 /m' for mild steel
or 5Ø6 /m' for high strength steel.
Choice of Slab Reinforcement
1. Chosen bars diameters:
- mild steel (24/35) 8, 10, 13mm
- high grade steel(36/52) 8, 10, 12mm
2. Half of the steel is straight bars & the other half is bent bars
3. Min. number of bars (in 1.0 m) is 5 and max. number is 10, the
common number is 6 – 8 ( 6 10/m )
4. We can use bars of the same diameters or two successive diameters
( 310+ 38/m ) and the big diam is the bent bar
5. The number of bars should be the same in the same direction
6. Steel in long direction (secondary) 0.25 steel in short direction
(main)
7. Rules of bent bars in beams ( L / 7, L / 5, L / 4) are applied for slabs
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Practical Cases
1- In one way slab, difference between adjacent spans is greater than
20%: - Solve continuous strip (1.0 m width)
2- One way slab adjacent to two way slab:
- Design for the larger – ve moment at the common support
3- Washrooms: - Drop slab of the washrooms by (100-150) mm.
Hence slabs become not continuous with adjacent slab of the floor.
4- Slab with openings:
- For small openings less than spacing between bars; do nothing.
- For small openings greater than spacing between bars; use additional reinforcement.
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
* For large openings, use more accurate analysis.
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Two Way Slabs
Two-way slabs
Thickness t = b / 35
= b / 40
= b / 45
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Two Way Slabs
- A small element of medium – thick plate is shown in figure:
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
The following equation gives the basic equilibrium requirement for a medium thick plate:
Unlike one way slabs which deform under load into nearly a cylindrical surface, two way slabs bend into a dished surface. This means that at any point, the slab is curved in both directions. The applied distributed load “W” on the element is carried partly by bending as a beam strip in the x- direction producing “Mx”, partly by bending of the strip in the y-direction
producing “My”, and the remainder by interaction between the x and y
strips, i.e. by torsion producing Mxy .
General
1. Rectangular slabs supported on all four sides can be considered as two way slabs if the
rectangularity ratio “r” is smaller than 2, or large than 0.5.
2. Slabs can be solved according to the elastic theories on condition that
the reinforcement resisting –ve moments is placed in the right position during casting..
3. The following method of design is valid only for ordinary buildings with
small uniformly
distributed live load up to 5 kN/m2.
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Loads
* Total load on slab: W = WD + WL
* Dead loads:
WD = slab self weight + flooring = ts (mm) × 2.5 + (1.5 → 3.5)
kN/m2
* Live loads:
WL: depends on the function of the building.
* If designing using Limit State Design Method,
use Wu = 1.4 WD + 1.6 WL
L.L IS TO BE TAKEN FROM FROM CODE
ACCORDING TO THE FUNCTION OF THE BUILDING
Supports
The width of the support for slab is the larger of :
ts or 150 mm …..for brick walls
100 mm …………for R.C. beams
Effective Spans
- Refer to one way slabs.
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Minimum Thickness
1. Slab thickness shall not be less than the absolute minimum of:
tmin = Le/35 for freely supported slab
tmin = Le/40 for slab continuous at one end
tmin = Le/45 for continuous or fixed slab
Where Le = the smaller effective span of the slab as indicated in one way
slabs.
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Simplified Method for Calculating B.M. in Two Way
Slabs
Rectangular slabs, monolithically cast with beams and supported on all four sides, with rectangularity ratio “r” less than 2 and greater than 0.5 subjected to uniform loads can be calculated according to the following simplified method:
where: mb = ratio of length between points of inflection for a loaded strip in
direction
of span “b” to the effective span “b”. ma = same but for span “a”
a = the effective span in direction a.
b = the effective span in direction b.
The values of ma & mb are to be determined from theory of elasticity. For slabs, the following values may be adopted.
Knowing the value of “r”, the load carried in each of the two perpendicular directions can be calculated using the values of α and β of Table (6-1) (ECCS
203-2001)
Table (6-1) EC 2001
For slabs monolithically cast with beams
α = 0.5r – 0.15 β = 0.35/r2
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
The bending moment in continuous slabs can be calculated as follows:
where: k = 10 for slabs continuous at one end. = 12 for slabs continuous at both ends.
N.B.: negative moment of at exterior support must be considered. For slabs resting on masonry walls, the load distribution factors, α and β are determined according to Table (6-2) ECCS 203-2001.
Table (6-2) (ECCS 203-2001) For slabs resting on masonry walls
* If live load ≥ 500 kg/m2, use Table (6-3) instead of Tables (6-1) and (6-2).
Table (6-3) (ECCS 203-2001)
For solid slabs resting with L.L. ≥ 500 kg/m2
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Two-way slabs
Thickness t = b / 35
= b / 40
= b / 45
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
1. Loadings :
* Dead Load (wD) - concrete weight : 0.12 x 2500 kg/m2
- flooring 150 – 200 kg/m2
* Live Load (wL) = according to building type
* Total slab load :
ws = wD + wL
2. Load distribution according to Code :
* distribution factor “ 1 r 2 ”
* continuity factor “ m ”
m1 * L1 m = 1.00 r =
m2 * L2 m = 0.87
m = 0.76
The loads in each direction are as:
For two way slabs
w = w s ---- short direction
w = w s ---- long direction
Using “ r “ ---- Get , from
Code table (6-1)…………..page(5)
in the curves
For one way slabs
w = w s
w = 0.0
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
3. bending moments in slabs
w L2
M = 8
w L2
M = 10
w L2
M = 12
w L2
M = 2
4. Design of sections
dshort = ts – 1.5 cm
dlong = ts – 2.5 cm
dlong
dshort t
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Reinforcement
The main reinforcement of a slab shall satisfy the requirements of equilibrium
C = T
Mu = C . yct = T. yct
cover = 15 – 25 mm
The reinforcement shall also be arranged along the direction of short span.Normal
to the main reinforcement, shrinkage and temperature (distributes) reinforcement
are provided.
1. Minimum main reinforcement:
* For mild steel :
* For high strength steel :
2. The main reinforcement should be arranged to cover all the tension zones a nd
extend beyond the end of these zones enough to develop the necessary
anchorage.
:
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
EACH DIRECTIN OF X AND Y
One way slab reinforcement details. 3. Maximum spacing between bars of main steel is 2 ts and < 200 mm.
For ts < 100 mm, spacing of 200 mm may be used. 4. A minimum of 1/3 of the main steel must be extended to the supports.
5. Minimum diameter for straight bars is 6 mm.
Minimum diameter for bent bars is 8 mm.
Bars of smaller diameters may be used for welded wire mesh and pre-cast concrete.
6. When using steel mesh, all conditions above must be satisfied.
7. For slabs with thicknesses ≥ 160 mm, top mesh must be provided. The top mesh must be taken ≥ 20% of the bottom reinforcement, but not less than 5Ø8
/m' for mild steel or 5Ø6 /m' for high strength steel.
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Reinforcement
1. Maximum spacing of main reinforcement in the middle zone of the span
is 2ts but not more than 200 mm. However, For ts < 100 mm , 5 bars/m may be
used.
2. Reinforcement in the secondary direction should not be less than 0.25 of
the main reinforcement with minimum of 5 bars/m'.
3. Positive reinforcement adjacent and parallel to a continuous edge may be reduced
by 25% for a width of not more than ¼ the shorter span of the panel.
4. For other details, refer to one way slabs.
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Choice of Slab Reinforcement
1. Chosen bars diameters:
- mild steel (24/35) 8, 10, 13mm
- high grade steel(36/52) 8, 10, 12mm
2. Half of the steel is straight bars & the other half is bent bars
3. Min. number of bars (in 1.0 m) is 5 and max. number is 10, the
common number is 6 – 8 ( 6 10/m )
4. We can use bars of the same diameters or two successive diameters
( 310+ 38/m ) and the big diam is the bent bar
5. The number of bars should be the same in the same direction
6. Steel in long direction (secondary) 0.25 steel in short direction
(main)
7. Rules of bent bars in beams ( L / 7, L / 5, L / 4) are applied for slabs
ANALYSIS AND DESIGN APPROACHE:
- TAKING STRIPS IN BOTH DIRECTION X & Y
- USING DESIGN TABLES
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Calculations
Calculations may be carried out in a table form as follows:
1. Load distribution
2. Bending moment and reinforcement For each direction (i.e. main and secondary directions) a table in the
following form may be used.
i) For short direction
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
10-
300
(31
0/m
) 1
0-30
0 (31
0/m
)
Slab Reinforcement
L / 4
10-400 (2.5 10/m)
10-400 (2.5 10/m)
L / 7 L / 5
L
Slab Reinforcement
1.5 L
10 – 100 ( 10 10 / m )
• • •
• • •
• • • • • •
• • • • • • • • •
• •• • •
Main reinforcement
(10 – 100 )
Secondary reinforcement
( 8 – 200 T & B )
Cantilever Slab
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Practical Cases
1- In one way slab, difference between adjacent spans is greater than
20%: - Solve continuous strip (1.0 m width)
2- One way slab adjacent to two way slab:
- Design for the larger – ve moment at the common support
3- Washrooms:
- Drop slab of the washrooms by (100-150) mm. Hence slabs become not continuous with adjacent slab of the floor.
4- Slab with openings:
- For small openings less than spacing between bars; do nothing.
- For small openings greater than spacing between bars; use additional reinforcement.
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
* For large openings, use more accurate analysis.
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
• Solid Slabs
• Hollow Block Slabs
• Flat Slabs
• Waffle Slabs
• Paneled Beams Slabs
•Loads
•Spans
•Design Requirements
Structural system must be chosen
Ly to satisfy economic requirements
according to these factors
Lx
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
One-way slabs
Thickness t = b / 30
= b / 35
= b / 40
= b / 10
Two-way slabs
Thickness t = b / 35
= b / 40
= b / 45
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
1. Loadings :
* Dead Load (wD) - concrete weight : 0.12 x 2500 kg/m2
- flooring 150 – 200 kg/m2
* Live Load (wL) = according to building type
* Total slab load :
ws = wD + wL
2. Load distribution according to Code :
* distribution factor “ 1 r 2 ”
* continuity factor “ m ”
m1 * L1 m = 1.00 r =
m2 * L2 m = 0.87
m = 0.76
The loads in each direction are as:
For two way slabs
w = w s ---- short direction
w = w s ---- long direction
Using “ r “ ---- Get , from
Code table (6-1)…………..page(5)
in the curves
For one way slabs
w = w s
w = 0.0
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
3. bending moments in slabs
w L2
M = 8
w L2
M = 10
w L2
M = 12
w L2
M = 2
4. Design of sections
dshort = ts – 1.5 cm
dlong = ts – 2.5 cm
dlong
dshort t
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
S1
9.00
0.00
0.00
--
--
--
cm2/m
--
5 8/m
S2 4.50 0.39 0.745 10.50 0.027 0.032 3.60 3 8/m+3 10/m
S3
2.25
0.975
0.365
10.50
0.0132
0.024
2.63 6 8/m
S4
2.25
0.975
0.438
10.50
0.0159
0.024
2.63 6 8/m
S5 3.50 0.712 1.011 10.50 0.0367 0.041 4.6 6 10/m
S6
5.50
0.312
0.94
09.50
0.04
0.05
4.95 3 10/m+3 13/m
S7
2.00
0.975
2.25
10.50
0.082
0.105
11.48 6 13/m
S8
5.50
0.224
0.81
09.50
0.036
0.04
4.10 6 10/m
Slab leffx wx t/m Mx d cm R as As chosen/m
x- direc.
Slab leffx
wx
t/m My
d (cm) R As
cm2/m
As chosen
/m
S1 1.50 0.975 1.322 10.5 0.048 0.058 6.35 6 10/m
S2 5.00 0.283 0.642 9.5 0.029 0.032 3.42 3 8/m++3 10/m
y- direc. S3 5.00 0.00 -- 9.5 -- -- -- 5 8/m
S4 5.00 0.00 -- 9.5 -- -- -- 5 8/m
S5 6.00 0.117 0.504 9.5 0.023 0.024 2.40 6 8/m
S6 6.00 0.366 1.051 10.5 0.0381 0.042 4.65 6 10/m
S7 6.00 0.00 -- 9.5 -- -- -- 5 8/m
S8 5.00 0.463 1.10 10.5 0.04 0.05 4.95 3 10/m+3 13/m
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Choice of Slab Reinforcement
1. Chosen bars diameters:
- mild steel (24/35) 8, 10, 13mm
- high grade steel(36/52) 8, 10, 12mm
2. Half of the steel is straight bars & the other half is bent bars
3. Min. number of bars (in 1.0 m) is 5 and max. number is 10, the
common number is 6 – 8 ( 6 10/m )
4. We can use bars of the same diameters or two successive diameters
( 310+ 38/m ) and the big diam is the bent bar
5. The number of bars should be the same in the same direction
6. Steel in long direction (secondary) 0.25 steel in short direction
(main)
7. Rules of bent bars in beams ( L / 7, L / 5, L / 4) are applied for slabs
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
1
0-3
00 (3
10/
m)
1
0-3
00 (
3
10/m
)
Slab Reinforcement
L / 4
10-400 (2.5 10/m)
10-400 (2.5 10/m)
L / 7 L / 5
L
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Slab Reinforcement
1.5 L
10 – 100 ( 10 10 / m )
• • •
• • •
• • • • • •
• • • • • •
• • •• •
• • •
Main reinforcement
(10 – 100 )
Secondary reinforcement
( 8 – 200 T & B )
Cantilever Slab
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013
Final Reinforcement
of Slab
PROF. Dr. MAGDY KASSEM - LECTURES OF 3RD YEAR CIVIL – 1ST TERM - 2012/2013