Reinforced Concrete Design
Lecture 13
Dr. Nader Okasha
1
One Way Slabs
2
Regula
(3
y
Introduction Plate/Shell (2D) z
x z x
A slab is a structural element whose thickness is small compared to
t<<(x,z)
its own length and width.
h
t L , S zS
t t
L
x
Slabs in buildings are usually used to transmit the loads on floors and
roofs to the supporting beams Loads
Dimensional Hierarchy of Structural
Beam Beam Column
Slab Beam
Column Beam Beam
Footing
Slab
Beam Beam
Soil
3
Introduction
Slabs are flexural members and as such, their flexure strength
requirement may be expressed by M M u n
Types of Slabs
Solid slabs :- which is divided into
- One way solid slabs
- Two way solid slabs One-way slab
Ribbed slabs :- which is divided into
- One way ribbed slabs
- Two way ribbed slabs
Two-way slab
4
One-way slabs
A one-way slab curves in one direction only under load. Accordingly
slabs supported on two opposite sides only and slabs supported on all
four sides, but L/S ≥ 2 are classified as one-way slabs.
shrinkage Reinft.
Main Reinft.
Main reinforcement is placed in the shorter direction, while the
longer direction is provided with shrinkage reinforcement to limit
cracking.
5
Two-way slabs
A two-way slab curves in two directions only under load.
Accordingly slabs supported on all four sides, and L/S < 2 are
classified as two-way slabs.
S
L
Bending will take place in the two directions in a dish-like form.
Accordingly, main reinforcement is required in the two directions.
6
One-way slabs
Solid slab
L
Two way slab L 2 One-way slab 2
S S
Ribbed Slab
8
Two way slab One-way slab
Ribbed Slab
One-way ribbed slab
10
Slab – Minimum thickness
11
Minimum Cover
12
One-way solid slabs
One-way solid slabs are designed as a number of independent 1 m
wide strips which span in the short direction and are supported on
crossing beams.
1m
L
S1 S2
S1 S2
13
One-way solid slabs
One-way solid slabs
Maximum Reinforcement Ratio
One-way solid slabs are designed as rectangular sections subjected to shear
and moment. Thus, the maximum reinforcement ratio is
3 f '
max (0.85 1 ) c
8 fy
Shrinkage Reinforcement Ratio
According to ACI Code and fy =4200 kg/cm2
shrinkage 0.0018 As , shrinkage 0.0018 b h
where, b = width of strip, and h = slab thickness
Minimum Reinforcement Ratio
According to ACI Code
15
As ,m in 0.0018 b h
One-way solid slabs
Spacing of Reinforcement Bars
a- Flexural Reinforcement Bars
Flexural reinforcement is to be spaced not farther than three times the slab
thickness (hs), nor farther apart than 45 cm, center-to-center.
3 hs
Sm ax smaller of 45cm
b- Shrinkage Reinforcement Bars
Shrinkage reinforcement is to be spaced not farther than five times the slab
thickness, nor farther apart than 45 cm, center-to-center.
5 hs
Sm ax smaller of 45cm
16
One-way solid slabs
Need to confirm that the thickness is adequate for shear.
Vu Vc
Vc 0.53 f c ' bw d fc '
b w , d in cm
Vc 0.17 f c ' bw d f c ' in MPa
17 b w , d in mm
One-way solid slabs
Loads Assigned to Slabs
wu=1.2 D.L + 1.6 L.L
a- Dead Load (D.L)
1- Own weight of slab (per unit area)= c h = 2.5 h t/m2
2- Weight of slab covering materials =0.2315 t/m2
tiles (2.5cm thick) =0.025×2.30
cement mortar (2.5cm thick) =0.025×2.10
sand (5.0cm thick) =0.05×1.80
plaster (1.5cm thick) =0.015×2.10
3-Equivalent Partition Weight
This load is usually taken as the weight of all walls carried by the slab
divided by the floor area and treated as a dead load rather than a live
load.
18
Minimum live Load values on slabs
Type of Use Uniform Live Load
kg/m2
Residential 200
One-way solid slabs Residential balconies 300
Computer use 500
b- Live Load (L.L) Offices 250
Warehouses
It depends on the purpose for Light storage 600
which the floor is constructed. Heavy Storage 1200
Schools
Classrooms 200
Libraries
rooms 300
Stack rooms 600
Hospitals 200
Assembly Halls
Fixed seating 250
Movable seating 500
Garages (cars) 250
Stores
Retail 400
wholesale 500
Exit facilities 500
Manufacturing
Light 400
19
Heavy 600
One-way solid slabs
Loads Assigned to Beams
The beams are usually designed to carry the following loads
- Their own weights.
- Weights of partitions applied directly on them.
- Floor loads.
The floor loads on beams
supporting the slab in the shorter
L
direction may be assumed
uniformly distributed throughout
their spans.
S1 S2
20
Loads Assigned to Slabs
Example
Assume depth = 25 cm (20cm block +5cm toping slab)
Weight of concrete per unit volume = 2500 kg/m3
Weight of one hollow block= 20 kg
Hollow blocks are 40 cm × 25 cm × 20 cm in
dimension
Assume ribs have 10 cm width of web
Assume equivalent partition load =75 kg/m2
Consider live load = 200 kg/m2.
Loads Assigned to Slabs
Dead Load (Slab own weight )
• Total volume (hatched) = 0.5 × 0.25 × 0.25 =
0.03125 m3
• Volume of one hollow block = 0.4 × 0.20 × 0.25
= 0.02 m3
• Net concrete volume = 0.03125 - 0.02 =
0.01125 m3
• Weight of concrete = 0.01125 × 2.5= 0.028125 t
• Weight of concrete /m2 = 0.028125 /(0.5)(0.25)
= 0.225 t/ m2
• Weight of hollow blocks /m2 =
[20/(1000)]/[(0.5)(0.25)] = 0.16 t/ m2
• Total slab own weight= 0.225 + 0.16= 0385 t/m2
Loads Assigned to Slabs
Load per rib
Total slab own weight= 0.385 + 0.2315 + 0.075 = 0.6915 t/m2
Ultimate load = 1.2(0.6915) + 1.6(0.2) = 1.15 t/m2
Ultimate load per rib = 1.15 × 0.5 = 0.575 t/m
Approximate Structural Analysis
24
Approximate Structural Analysis
Bending Moment
More than two slabs
25
Approximate Structural Analysis
Bending Moment
Two slabs
26
Approximate Structural Analysis
Shear
More than two slabs
27
Approximate Structural Analysis
Bending Moment
Two slabs
28
Typical reinforcement in a one-way slab
Cutoffs
If slab satisfies ACI 8.3.3
30
Summary of One-way Solid Slab Design Procedure
1- Select representative 1m wide design strip/strips to span in the
short direction.
2- Choose a slab thickness to satisfy deflection control requirements.
When several numbers of slab panels exist, select the largest
calculated thickness.
3- Calculate the factored load wu by magnifying service dead and
live loads according to this equation wu=1.20wD +1.60wL .
4- Draw the shear force and bending moment diagrams for each of
the strips.
5- Check adequacy of slab thickness in terms of resisting shear by
satisfying the following equation: V 0.53 f ' b d
u c
If the previous equation is not satisfied, go ahead and enlarge the
thickness to do so
31
Summary of One-way Solid Slab Design Procedure
6- Design flexural and shrinkage reinforcement:
Flexural reinforcement ratio is calculated from the following
equation
0.85 f c 2 10 M u
5
0.85 f b d 2
1 1
fy
c w
Make sure that the reinforcement ratio is not larger than ρmax
Compute the area of shrinkage reinforcement, where
Ashrinkage=0.0018bh
7- Draw a plan of the slab and representative cross sections showing
the dimensions and the selected reinforcement
32
Example 1
Using the ACI-Code approximate structural analysis, design for a
warehouse, a continuous one-way solid slab supported on beams 4.0 m
apart as shown in Figure. Assume that the beam webs are 30 cm wide.
The dead load is 0.3t/m2 in addition to the own weight of the slab, and the
live load is 0.3t/m2.
Use fc’=280 kg/cm2, fy=4200kg/cm2
8.0 m
4.0 m 4.0 m 4.0 m
33
Solution:
1- Select a representative 1 m wide slab strip:
The selected representative strip is shown in Figure
2- Select slab thickness:
1.0 m
The clear span length, ln
8.0 m
ln = 4.0 – 0.30 = 3.70 m
17cm
For one-end continuous spans,
hmin = l/24 =4.0/24=0.167m 4.0 m 4.0 m 4.0 m
Slab thickness is taken as 17 cm Wu
34
Solution:
3- Calculate the factored load wu per unit length of the selected strip:
Own weight of slab = 0.17× 2.50 = 0.425 ton/m2
wu= 1.20 (0.30+0.425) +1.60 (0.30)= 1.35 ton/m2
For a strip 1 m wide, wu=1.35 ton/m
4- Evaluate the maximum factored shear forces and bending moments
in the strip:
wu=1.35 ton/m & ln = 3.7m
35
Solution:
1.85 1.85
1.68 1.68
0.77 0.77
1.68
1.68
36
Solution:
2.5 2.87
2.5
2.5 2.5
2.87
37
Solution:
5- Check slab thickness for beam shear:
Effective depth d = 17 – 2 – 0.60 = 14.40 cm, assuming Φ12 mm bars.
Vu,max = 2.87 ton.
0.75 0.53 280 100 14 .4
Vc 0.53 f c ' bw d 3
9.58 ton
10
i.e. , slab thickness is adequate in terms of resisting beam shear.
6- Design flexural and shrinkage reinforcement:
Assume that Φ=0.9
0.85 f c 2 105 M u
0.85 f b d 2
1 1
fy
c w
Where b = 100cm & d = 14.4cm
38
Solution:
For max. negative moment, Mu = 1.85 t.m
0.85 280 2 105 1.85
ρ 0.85 0.9 280 100 14.42 0.00241 ρ max
1 1
4200
3 f ' 3 280
ρ max (0.851 ) c (0.85 0.85) 0.01806 ρ 0.90
8 fy 8 4200
[if ρ max ρ , 0.90 and if not 0.90]
A s , ve 0.00241100 14.4 3.47 cm2 A s ,min
A s ,min 0.0018100 17 3.06 cm2
3.47 0.79φ10
S 22.75cm
100cmstrip S
Smax min(45 or 3 17) 45 use 10@20cm
39
Solution:
For max. positive moment, Mu = 1.68 t.m
0.85 280 2 105 1.68
ρ 0.85 0.9 280 100 14.42 0.00219 ρ m ax
1 1
4200
3 f ' 3 280
ρ m ax (0.851 ) c (0.85 0.85) 0.01806 ρ 0.90
8 fy 8 4200
[if ρ m ax ρ , 0.90 and if not 0.90]
A s , ve 0.00219100 14.4 3.15cm2 A s ,m in
A s ,m in 0.0018100 17 3.06 cm2
3.15 0.79φ10
S 25.1cm
100cmstrip S
Sm ax min(45 or 3 17) 45 use 10@25cm
40
Solution:
Calculate the area of shrinkage reinforcement:
Area of shrinkage reinforcement = 0.0018 (100) (17) = 3.06 cm2/m
For shrinkage reinforcement use Φ 10 mm @ 25 cm
Shrinkage reinft.
Φ10@25 Φ10@25 Φ10@20 Φ10@20 Φ10@25
17cm
Φ10@25 Φ10@25 Φ10@25
41
Solution:
Φ10@25 Φ10@20 Φ10@20 Φ10@25
8.0 m
Φ10@25 Φ10@25 Φ10@25
4.0 m 4.0 m 4.0 m
42
One-way ribbed slabs
Ribbed slab consists of regularly spaced ribs monolithically built
with a top floor slab. The voids between the ribs may be either light
material such as hollow blocks [figure 1] or it may left unfilled
[figure 2].
Topping slab
Rib Hollow block Temporary form
Figure [1] Hollow block floor Figure [2] Moulded floor
The use of these blocks makes it possible to have smooth ceiling
which is often required for architectural or hygienic considerations
and have good sound and temperature insulating properties besides
reducing the dead load of the slab greatly.
One-way ribbed slabs
If the ribs are provided in one direction only, the slab is classified as
being one-way, regardless of the ratio of longer to shorter panel
dimensions.
44
Key components of one-way ribbed slabs
a. Topping slab:
According to ACI Code, topping slab thickness (t) is not to be less
than 1/12 the clear distance (Lc) between ribs, nor less than 5.0
cm. Slab thickness (t)
Lc
The topping slab is designed as a continuous beam supported by the
ribs and the maximum bending moment = wuLc/12.
Shrinkage reinforcement is provided in the topping slab in both
directions in a mesh form.
45
Key components of one-way ribbed slabs
b. Regularly spaced ribs:
Minimum dimensions:
According to ACI Code, ribs are not to be less than 10 cm in width,
and a depth of not more than 3.5 times the minimum web width and
Clear spacing between ribs is not to exceed 75.0 cm.
Lc ≤ 75 cm
h ≤ 3.5 bw
bw ≥ 10
Loads:
The dead load includes own weight of the slab, weight of the surface
finish, and equivalent partition load.
The live load is dependant on the intended use of the building.
46
Key components of one-way ribbed slabs
Shear strength:
According to ACI-Code, shear strength provided by rib concrete Vc
may be taken 10 % greater than those for beams. It is permitted to
increase shear strength using shear reinforcement or by widening the
ends of ribs.
Flexural strength:
Ribs are designed as rectangular beams in the regions of negative
moment at the supports and as T-shaped beams in the regions of
positive moments between the supports.
Effective flange width be is taken as half the distance between ribs,
center-to-center. b
e
47
Key components of one-way ribbed slabs
c. Hollow blocks:
Hollow blocks are made of lightweight concrete or other lightweight
materials. The most common concrete hollow block sizes are 40 × 25
cm in plan and heights of 14, 17, 20, and 24 cm.
48
Summary of one-way ribbed slab design procedure
1. The direction of ribs is chosen.
2. The overall slab thickness h is determined based on deflection
control requirement.
3. The factored load on each of the ribs is computed.
4. The shear force and bending moment diagrams are drawn.
5. The strength of web in shear is checked.
6. Design positive and negative moment reinforcement.
49
Example
Design a one-way ribbed slab to cover a 3.8 m x 10 m panel, shown in the
figure below. The covering materials weigh 225kg/m2, equivalent partition
load is equal to 75 kg/m2, and the live load is 200 kg/m2.
Use fc’=280 kg/cm2, fy=4200kg/cm2
3.8 m
5.0 m 5.0 m
50
Solution
1. The direction of ribs is chosen:
Ribs are arranged in the short direction as shown in Figure
3.8 m
5.0 m 5.0 m
2. The overall slab thickness h is determined:
According to ACI Table 9.5(a), hmin =380/16 = 23.75cm, use an overall
slab thickness of 24 cm.
Using concrete hollow blocks 40 cm × 25 cm × 17 cm (weight=17kg)
Topping slab thickness = 24 – 17 = 7 cm
Let width of web be equal to 10 cm
51
Solution
Design of topping slab:
Area of shrinkage reinforcement, As=0.0018(100)7=1.26 cm2
Use 5 Φ 6 mm/m in both directions.
3. The factored load on each of the ribs is to be computed:
Total volume 1.0 m
= 1.0 × 1.0 × 0.24 = 0.24 m3
Volume of hollow block
= 8 × 0.4 × 0.25 × 0.17 = 0.136 m3 0.05 m
Net concrete volume
= 0.24- 0.136 = 0.104 m3
1.0 m
0.25 m
Weight of concrete
= 0.104 × 2.5 = 0.26 ton
Weight of hollow blocks
= 8 × 17/103 = 0.136 ton/m
7 cm
0.4 m 0.1 m 0.4 m
Total dead load
= 0.225 + 0.075 + 0.26 + 0.136
0.24 m
= 0.70 t/m2
52
Solution
wu=1.2(0.7)+1.6(0.2)=1.16 t/m2
wu/rib=1.16x0.5= 0.58 t/m
4. Critical shear forces and bending moments are determined:
Maximum factored shear force = 0.58 (3.8/2) = 1.1 ton
Maximum factored bending moment = 0.58 (3.8)2/8 = 1.05 ton. m
5. Check rib strength for beam shear:
Effective depth d = 24–2–0.60–0.6 =20.8 cm, assuming 12mm
reinforcing bars and Φ 6 mm stirrups.
1.1 0.75 0.53 250 10 20.8
1.1ΦVc 3
1.44 ton Vu,m ax 1.1 ton
10
Though shear reinforcement is not required, 4 6 mm U-stirrups per meter
run are to be used to carry the bottom flexural reinforcement.
53
Solution
6. Design flexural reinforcement:
Since the maximum factored moment creates compression in the flange,
the section of maximum positive moment is to be designed as a T-section
Assuming that a<7cm and Φ=0.90 →Rectangular section with b=beff =50 cm
0.85 250 50
1 1 2 105 1.05
ρ 7
4200
1.05 t.m
0.9 0.85 250 10 20.82
24
As
0.00689
10
As ρ beff d 0.0068910 20.8 1.44 cm 2
Use 210mm (As,used= 1.57 cm2) one is straight and the other is bent-up in
each rib at its bottom side
As f y 1.57 4200
a 3.10cm 7cm
0.85f c ' beff 0.85 250 10
The assumption is right
54
Solution
Check As,min
fc ' 1.4
A s,min bw d bw d
4f y fy
A s,min 0.7 cm2 A s,used 1.57 cm2 OK
Check Φ=0.9
a 3.10
c 3.65 cm
β1 0.85
dc 20.8 3.65
εt 0.003 0.003
c 3.65
0.0141 0.005 0.9 OK
55
Solution
7. Neat sketches showing arrangement of ribs and details of the reinforcement are to be
prepared
1Φ10 m
1Φ10 m
1Φ10 m
1Φ10 m
3.8 m
A A
5.0 m 5.0 m
Φ6mm stirrups Φ6mm mesh
@25 cm @20 cm
7cm
24cm
17cm
2Φ10mm 10 40 cm 10 2Φ10mm
Section A-A
56