CON4332 REINFORCED CONCRETE DESIGN
│CHAPTER 5│
Design of R C Slabs
Learning Objectives
 Differentiate one-way and two-way slabs
 Design simply-supported and continuous one-way slabs by
integrating the processes of
o determining design loads
o determining design forces by force coefficients
o determining of reinforcement for bending and shear
o checking of deflection by span-to-depth ratio
 Extend the design method to design simple R C stairs
CONTENTS
5.1
5.2
5.2.1
5.2.2
5.3
5.3.1
5.3.2
5.3.3
5.3.4
5.3.5
5.3.6
5.4
5.4.1
5.4.2
5.4.3
5.4.4
Types of Slabs
Design Loads and Forces
Example – Design Loads and Forces for Simply-supported One-way Slab
Force Coefficients for Continuous One-way Slab
Design of Slabs
Design for Moment and Shear
Deflection Check by Span-to-depth Ratio
Distribution Bars
Examples – Simply-supported One-way Slab
Examples – Continuous One-way Slab
Example – Simply-supported Two-way Slab
Stairs
Types of Stairs
Design for Stair Slabs
Example – Self-weight of Stair Flight
Examples – Longitudinally-spanned Stairs
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5.1
Types of Slabs
When a slab is supported on two opposite edges as shown in the
left-hand-side of the following figure, it bends in one direction only. It is
called a one-way slab. The arrows marked on the slab indicate the
direction of span and are pointing to the supporting edges. The main
reinforcement bars for resisting bending are provided in the direction of
span.
Support
Support
Support
Support
Support
One-way Slab
Support
Two-way Slab
On the other hand, if the slab is supported on four edges as shown in the
right-hand-side of the figure, it bends in two directions as indicated by the
arrows marked on the slab. It is called a two-way slab. Reinforcements in
two directions have to be designed to resist the bending in their respective
direction.
In reality, most of the slabs are surrounded by beams on four edges. If the
length-to-width ratio, i.e. aspect ratio, of the slab is more than 2 as shown in
Figure 5.1 below, the supporting beams on shorter edges can be ignored,
and the slab is therefore treated as a one-way slab being supported on the
two opposite longer edges only.
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L
> 2L
Effective Support
Effective Support
Figure 5.1 – Aspect Ratio of One-way Slab
In addition to the slab supported on two or four edges as mentioned above,
slab can also be supported on a single edge, i.e. cantilever slab, or directly
on columns, i.e. flat slab. If the span of the slab is very large, it will become
so thick that it would be more economical to change the solid slab into ribbed,
waffle or hollow slab.
In this chapter, we will focus on the design of one-way simply-supported and
continuous slabs to illustrate the essences of R.C. slab design. The
relevant design formulae, rules and tables you have learnt in Chapters 1 to 3
are applicable. You can also refer to the “Annex – R C Design Formulae
and Data”.
5.2
Design Loads and Forces
When a slab is subjected to uniformly distributed load (udl), we can take a
unit width, say 1 m, of the slab to design as a beam. In other words, the
slab can be visualized as a series of beams of 1 m width placed side-by-side
as shown below. The design loads and forces are then presented in per
meter width.1
1
Slab subjected to uniformly distributed load (udl) is adopted in this chapter to illustrate the fundamental
procedures in slab design.
Slabs may be subjected to concentrated load in the forms of point load or line load.
An effective width of the slab has to be determined to design for the extra moment and shear induced by the
concentrated load. Details can be found in the design code.
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CON4332 REINFORCED CONCRETE DESIGN
1m
1m
Figure 5.2 – Unit Width of Slab for Design
5.2.1
Example – Design Loads and Forces for a Simply-supported
One-way slab
For simply-supported one-way slab under uniformly distributed load (udl),
the force coefficients, similar to those for beam design, are:
Mid-span Moment, M = 0.125 F L
Shear at Support, V = 0.5 F
or
0.125 w L2
or
0.5 w L
Question
Determine the design forces for the following simply-supported one-way slab.
Design parameters
Slab thickness, h :
c/c distance btw supports :
Width of the supporting beam, Sw :
Allowance for finishes :
Usage of the floor :
150 mm
3 000 mm
350 mm (same for both ends)
1.5 kPa
Department Store
Solution
Dead Load
1.50 kN/m2
Finishes :
Self-weight :
24.5 x 0.15 =
3.68 kN/m2
gk =
5.18 kN/m2
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CON4332 REINFORCED CONCRETE DESIGN
Imposed Load
Department Store:
Design Load, w =
=
As h < Sw
Effective Span, L =
=
Design Moment, M =
=
Design Shear, V =
!
The design moment
=
qk=
5.00 kN/m2
1.4 x 5.18 + 1.6 x 5.00
15.25 kN/m2
3000 – 350 + 150
2 800 mm
0.125 x 15.25 x 2.82
14.95 kN-m
0.5 x 15.25 x 2.8
21.35 kN
and shear are in per
Q.1
meter width of the
slab.
5.2.2
?
Force Coefficients for Continuous One-way Slab
For continuous slabs with approximately equal spans under udl, the
following force coefficients (extracted from Table 6.4 of HKCP-2013) can be
adopted:
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At outer
Near middle
At first
At middle of
At interior
support
of end span
interior
interior span
supports
(simply
support
supported)
Moment
0
0.086FL
-0.086FL
0.063FL
-0.063FL
Shear
0.4F
-
0.6F
-
0.5F
Notes:
1. Area of each bay exceeds 30 m2.
2. Characteristic imposed load does not exceed 5kPa.
3. The ratio of characteristic imposed load to the characteristic dead load does not
exceed 1.25.
4. An allowance of 20% redistribution of the moments at the supports has been made.
5. Load should be substantially uniformly distributed over three or more spans.
Table 5.1 – Force Coefficients for Continuous One-way Slabs
with Approximately Equal Span under udl
(Extracted from Table 6.4 of HKCP-2013)2
A bay mentioned in "Note 1" of the above table is defined as a strip across
the full width of a structure bounded on the other two sides by lines of
supports as illustrated below:
Figure 5.3 – Definitions of Panel and Bay
(Figure 6.5 of HKCP-2013)
The values in the above table can be presented in the form of moment and
2
The force coefficients for continuous end support is omitted from this table. If necessary, refer to the original
table in the design code for details.
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shear force diagrams as shown below.
0.086FL
0.063FL
Bending
Moment
Diagram
0.063FL
0.086FL
0.5F
0.5F
0.4F
Shear Force
Diagram
0.6F
0.5F
Figure 5.4 - Moment and Shear Force Diagrams
for Continuous One-way Slabs
with Approximately Equal Span under udl
In the above table, the design moment at the outer support is zero, i.e.
simply supported.
However, reinforced concrete slabs are usually
constructed monolithically with the supporting beam. In order to avoid
unsightly cracks due to the bending arising from partial fixity at the support, a
minimum design moment of at least 50% of the mid-span moment is
recommended by Cl.9.3.1.3 of HKCP-2013.
Take note of the following differences when compared with that for beam in
Chapter 4 (i.e. Table 4.1 and Figure 4.1):
(a)
The force coefficients for slab are in general smaller than those for
beam. The support moments are about 22% lesser, and the
mid-span moments are 5% to 10% lesser.
(b)
A redistribution of 20% is allowed for the support moments of slab,
i.e. βb = 0.8, and therefore K' is reduced to 0.132 for designing the
section at supports. (There is no redistribution for mid-span
moments, and the value of K’ for mid-span moment remains
0.156.)
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(c)
The nominal design moment at the outer (or end) support for slab
(simply-supported) is 50% instead of 15% (for beam) of the
mid-span moment.
If a slab does not fulfill the conditions to use the force coefficients in the
above table, structural analysis has to be performed to determine the design
forces. The most unfavorable arrangement of design loads as described in
Section 1.5 of Chapter 1 has to be designed for. However, if the slab fulfills
conditions 1 to 3 stated in the notes of Table 5.1 above, a single-load case of
maximum design load on all spans can be adopted for design.
Q.2
5.3
?
Design of Slab
The design method for slabs is quite similar to that for beam with the
following differences:
(a)
The breadth of the section, b = 1 000 mm, i.e. taking one meter
width for design.
(d)
No compression bar is usually designed for unless the slab is very
thick, i.e. h > 200mm, and heavily loaded. Simply check if K < K';
otherwise, increase the slab thickness and re-design the slab.
(e)
For slab supported on beams, the design shear stress is usually
very small and not critical.3 Simply check if v < vc; otherwise,
increase the slab thickness and re-design the slab.
(f)
The steel area, As, obtained is per meter width of the slab, i.e. in
mm2 per m. The reinforcement is provided in the terms of bar
spacing instead of number of bars.
Example
If As,req = 723 mm2, we can provide:
Q.3
T12-150 (As,pro = 113 / 0.150 = 754 mm2 /m); or
3
?
Design for shear is critical for slabs supported directly on columns, i.e. flat slab, or slabs subjected to high
magnitude concentrated load.
The thickness of this types of slab is always controlled by punching shear stress
at the perimeter of the column or concentrated load.
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T10-100 (As,pro = 78.5 / 0.100 = 785 mm2 /m)
The steel area can be read from the following table.
Bar Spacing in mm
Bar
Size
100
125
150
175
200
225
250
275
300
350
8
503
402
335
287
251
223
201
183
168
144
10
785
628
524
449
393
349
314
286
262
224
12
1131
905
754
646
565
503
452
411
377
323
16
2011
1608
1340
1149
1005
894
804
731
670
574
Table 5.2 – Steel Area in mm2 per m Width
5.3.1
Design for Moment and Shear
In general, the procedures to design for moment, M, are:
1. Identify the effective sectional dimensions and design parameters.
2. Calculate the K value and check if K < K'.
3. Calculate the lever arm z and check its limits.
4. Calculate the amount of steel required, As.
5. Determine the bar size and spacing.
6. Check if the limits to steel area are complied with.
In general, the procedures to design for shear, V, are:
1. Identify the effective sectional dimensions and design parameters.
2. Calculate vc.
3. Check if the shear stress v exceeds vc.
If v < vc, no shear reinforcement is required.
If v > vc, increase the thickness of the slab and re-design it.
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5.3.2
Deflection Check by Span-to-depth Ratio
In general, the procedures to check deflection by span-to-depth ratio are:
1. Determine the basic L/d ratio.
2. Determine the modification factors mt.
3. Determine the allowable L/d ratio.
4. Check if the actual L/d ratio exceeds the allowable or not.
5.3.3
Distribution Bars
For one-way slab, the reinforcement bars designed to resist the bending
moment are placed in one direction only, i.e. in the direction of span. In
addition to these main reinforcement bars, secondary reinforcement bars
have to be provided in the direction at right angle, i.e. transverse, to the main
bars to tie the slab together and to distribute uneven loading or any
accidental concentrated load that may arise during its life of usage. These
secondary reinforcement bars are called distribution bars, Asd, which has to
fulfill the following requirements:
(a)
The steel area shall not be less than:
0.13%bh, &
20% of the main steel
Refer to Chapter
(b)
The spacing shall not be more than 3h & 400mm.
2 for bar spacing
requirement for
main bars.
5.3.4
Examples – Simply-supported One-way Slab
Question A
Design the reinforcement and check if the deflection is acceptable for the following
simply-supported one-way slab.
Design parameters
Slab thickness, h =
175 mm
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c/c distance btw supports =
4000 mm
Width of supporting beams, Sw =
250 mm
fcu =
30 MPa
fy =
500 MPa
Cover =
Preferred bar size =
Characteristic imposed load =
Allowance for finishes =
Partition load =
25 mm
10
3.0 kPa (Offices)
1.0 kPa
1.0 kPa (Lightweight undefined)
Solution
Dead Load
1.00 kN/m2
Finishes :
Self-weight :
24.5 x 0.175 =
4.29 kN/m2
gk =
5.29 kN/m2
Imposed Load
Partition load :
1.00
3.00 kN/m2
Offices :
qk=
Design Load, w =
=
As h < Sw
Effective Span, L =
=
Design Moment, M =
=
Design Shear, V =
=
4.00 kN/m2
1.4 x 5.29 + 1.6 x 4.00
13.81 kN/m2
4000 – 250 + 175
3925 mm
0.125 x 13.81 x 3.9252
26.6 kN-m
0.5 x 13.81 x 3.925
27.1 kN
Design for Bending Moment
Effective depth, d =
=
!
b = 1000
βb = 1.0
K=
175 – 25 – 10/2
145 mm
M / (bd2fcu)
=
26.6 x 106 / (1000 x 1452 x 30)
=
0.042
<
0.156
(Singly reinforced)
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K < 0.0428
z=
=
Tension steel req'd, As,req =
0.95d = 0.95 x 145
137.8 mm
M / (0.87 fy z)
=
26.6 x 106 / (0.87 x 500 x 137.8)
=
444 mm2
(Provide T10-175 Bottom)
As,pro =
Alternatively, As,pro
can be read from
=
table.
100As / bh =
>
Distribution Bar
Asd =
=
78.5 / 0.175
449 mm2
100 x 449 / (1000 x 175) = 0.256
0.13 and < 4.0
(Steel ratio ok)
Max(0.13x1000x175/100 or 444x20%)
227 mm2
(Provide T10-300 DB)
Check Shear at Support
v=
=
27.1 x 103 / (1000 x 145)
0.187 MPa
Calculate the design concrete shear stress, vc :
100As/(bvd) =
!
In this case, v < 0.34, the
smallest value in Table 6.3 of
HKCP-2013. The calculation of
vc can be omitted.
(400/d)1/4 =
vc =
(Table 6.3)
100 x 449 / (1000 x 145) = 0.31 < 3
(400 / 145)1/4 = 1.289 (> 0.67)
0.79 x (0.31)1/3 x 1.289 / 1.25 x (30/25)1/3
=
0.551 x 1.06
=
0.584 MPa
>
0.187 MPa
(No shear reinforcement is req'd)
Check Deflection by Span-to-depth Ratio
Basic L /d =
M/(bd2) =
=
As,req / As,pro =
fs =
mt =
20 (Simply Supported Slab)
(Table 7.3)
26.6 x 106 / (1000 x 1452)
1.264 N/mm2
444 / 449 = 0.99
2/3 x 500 x 0.99 = 330 MPa
0.55 + (477-330)/[120(0.9+1.26)]
=
0.55 + 0.567
=
1.12
(Table 7.4)
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Allowable L / d =
Actual L / d =
=
1.12 x 20 = 22.3
3925 / 145
27.1 > 22.3
(Deflection not ok)
(Increase the tension steel to T10-125 Bottom)
As,pro =
As,req / As,pro =
fs =
mt =
78.5 / 0.125 = 628 mm2
444 / 628 = 0.707
2/3 x 500 x 0.707 = 236 MPa
0.55 + (477-236)/[120(0.9+1.26)]
=
0.55 + 0.93
=
1.48
Allowable L / d =
1.48 x 20 = 29.6 > 27.1
Comment: For this slab, deflection controls the amount the steel required.
(Deflection ok)
Q.4
?
Question B
Present the reinforcement detail of the slab in Question A in proper engineering drawing.
Solution
Notes on the detailing:4
(a) Top bars are provided at the supports for anti-cracking purposes. The nominal
4
The rules of reinforcement detailing is beyond the scope of this chapter. Refer to the design code for details.
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CON4332 REINFORCED CONCRETE DESIGN
requirement is 50% of the steel required at mid-span. They are provided with full
anchorage length into the supports and extend 0.15L or 45 into the span.
(b) The bar spacing of 125mm for main bars and 250mm for top bars deem appropriate.
Refer to chapter 2 for details.
(c) Pay attention that the slab may be designed to act as the top flange of the
supporting beams to take up flexural compressive stress. If it is the case, the
amount of top bars has to be increased to 015% and extends into the slab over the
whole effective flange width of the flanged section.
(d) Theoretically, 50% of the bottom bars can be curtailed at about 0.1L from the
support. However, for simplicity, all the bottom bars are extended into the support
in this case.
(e) The bottom bars have to extend 12 beyond the centerline of the support.
5.3.5
Example – Continuous One-way Slab
Question A
Design the end span of the continuous slab. 5S1, shown on the framing plan in DWG-01 of
Chapter 1. The following are the design parameters for the slab.
Design parameters
Slab thickness, h =
c/c distance btw supports =
Width of support, Sw =
160 mm
3 300 mm
300 mm (similar at both ends)
fcu =
35 MPa
fy =
500 MPa
Cover =
Preferred bar size =
Density of concrete =
25 mm
10
24.5 kN/m3
Allowance for finishes =
2.0 kPa
Characteristic imposed load =
5.0 kPa
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CON4332 REINFORCED CONCRETE DESIGN
Solution
Effective Span
As h < Sw
L=
=
3300 – 300 + 160
3 160 mm
Loading
Dead Load
Finishes:
2.00 kN/m
Slab S/W:
24.5 x 0.16 =
3.92 kN/m
gk =
5.92 kN/m
qk =
5.00 kN/m
Imposed Load
Design load, F =
=
(1.4 x 5.92 + 1.6 x 5.00) x 3.16
51.5 kN per m width
Design Forces
Bay size 
=
9 x 9.9
98.1 m2 > 30 m2
Imposed load is not greater than 1.25 dead load.
Imposed load is not greater than 5kPa.
The force coefficients in Table 6.4 of HKCP-2013 can be used.
Design Moment, M =
0.086 x 51.5 x 3.16
=
14.0 kN-m
Design Shear, V =
0.6 x 51.5
=
Effective Depth
d=
=
30.9 kN
The moment
coefficients for both
span and support
are the same.
160 – 25 – 10/2
130 mm
Design for Bending Moment
K=
Support moment is
adopted for rebar
design, which has a
redistribution of 20%.
βb = 0.8
K < 0.0428
M / (bd2fcu)
=
14.0 x 106 / (1000 x 1302 x 35)
=
0.024
<
0.13
z=
=
Tension steel req'd, As,req =
=
(Singly reinforced)
0.95d = 0.95 x 130
123.5 mm
M / (0.87 fy z)
14.0 x 106 / (0.87 x 500 x 123.5)
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CON4332 REINFORCED CONCRETE DESIGN
=
260 mm2
(Provide T10-250 Top at supports
and T10-250 Bottom at mid-span)
As,pro =
100As / bh =
>
Distribution Bar
Ads =
=
2
314 mm per meter
100 x 314 / (1000 x 160) = 0.196
0.13 and < 4.0
(Steel ratio ok)
Max(0.13x1000x160/100 or 260x20%)
208 mm2
(Provide T10-300 DB)
Check Shear at Support
v=
30.9 x 103 / (1000 x 130)
=
0.23 MPa
<
0.34, the smallest value of vc in Table 6.3 of HKCP-2013
(No shear reinforcement required)
Check Deflection by Span-to-depth Ratio
Basic L /d =
M/(bd2) =
=
As,req / As,pro =
fs =
!
βb = 0.8 is for support
moment.
For L/d checking,
mid-span moment is used.
mt =
23 (End span of continuous slab)
14.0 x 106 / (1000 x 1302)
0.828 N/mm2
260 / 314 = 0.828
2/3 x 500 x 0.828 = 276 MPa
0.55 + (477-276)/[120(0.9+0.828)]
=
0.55 + 0.969
=
1.52
Allowable L / d =
(Table 7.3)
(Table 7.4)
1.52 x 23 = 34.9
Actual L / d =
3160 / 130
=
24.3 ≤ 34.9
(Deflection ok)
Question B
Design the interior span of the continuous slab. 5S1, shown on the framing plan in DWG-01
of Chapter 1. The design parameters in Question A are still applicable.
Solution
The effective span, effective depth and the design loads of the interior span of this beam are
the same as that for the end-span in Question A. The only differences are the design
moment, design shear and span-to-depth ratio. Although they are not critical for this case,
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CON4332 REINFORCED CONCRETE DESIGN
the calculation is presented below as an illustration of the complete process of design.
Design Forces
Design Moment, M =
0.063 x 51.5 x 3.16
=
10.3 kN-m
Design Shear, V =
0.5 x 51.5
=
25.8 kN
Design for Bending Moment
K=
βb = 0.8
K < 0.0428
M / (bd2fcu)
=
10.3 x 106 / (1000 x 1302 x 35)
=
0.017
<
0.132
z=
=
Tension steel req'd, As,req =
(Singly reinforced)
0.95d = 0.95 x 130
123.5 mm
M / (0.87 fy z)
=
10.3 x 106 / (0.87 x 500 x 123.5)
=
192 mm2
(Provide T10-300 Top at the supports
and T10-300 Bottom at mid-span)
As,pro =
100As / bh =
>
Distribution Bar
Ads =
=
2
262 mm per meter
100 x 262 / (1000 x 160) = 0.164
0.13 and < 4.0
(Steel ratio ok)
Max(0.13x1000x160/100 or 192x20%)
208 mm2
(Provide T10-300 DB)
Check Deflection by Span-to-depth Ratio
Basic L /d =
M/(bd2) =
=
As,req / As,pro =
fs =
mt =
=
26 (Interior span of continuous slab )
(Table 7.3)
10.3 x 106 / (1000 x 1302)
0.609 N/mm2
192 / 262 = 0.733
2/3 x 500 x 0.733 = 244 MPa
0.55 + (477-244)/[120(0.9+0.609)]
(Table 7.4)
0.55 + 1.287
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=
Allowable L / d =
5.3.6
1.84
1.84 x 26 = 47.8
Actual L / d =
3160 / 130
=
24.3 ≤ 47.8
(Deflection ok)
Example – Simply-supported Two-way Slab
When a slab is supported on four edges and the aspect ratio of the slab is
smaller than 2, the four edges are considered effective in supporting the slab,
which is then considered to be spanned in two directions. Reinforcement
bars have to be provided in two directions to resist the bending moment in
their respectively direction as shown below.
msx
Ly
Long span
msy
Lx
Short Span
If the four edges are simply supported and the four corners are not
prevented from uplifting and there is no provision for torsion, the maximum
moments per unit width are given by the following equations of HKCP-2013.5
5
Moment in the shorter span
msx = αsxnLx2
Moment in the longer span
msy = αsynLx2
Refer to the design code for the moment coefficients for slab with continuous edges and the requirements on the
details to restrain corners from uplifting and torsion.
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The bending moment coefficients αsx & αsy are given in the following table.
Ly / L x
1.0
1.1
1.2
1.3
1.4
1.5
1.75
2.0
αsx
0.062
0.074
0.084
0.093
0.099
0.104
0.113
0.118
αsy
0.062
0.061
0.059
0.055
0.051
0.046
0.037
0.029
Table 5.3 – Bending Moment Coefficients
for Simply-Supported Two-way slabs without Restrain at the Corners
(Table 6.5 of HKCP-2013)
where
n=
Design ultimate load per unit area
Lx =
Effective span of shorter span
Ly =
Effective span of longer span
When Ly/Lx > 2.0, the slab can be treated as one-way slab and the force
coefficients described in the previous paragraphs can then be adopted. For
simply supported one-way slab the moment coefficient is 0.125.
Question
Design the reinforcement and check the deflection of the following simply-supported
two-way slab.
Design parameters
Slab thickness, h =
200 mm
Effective spans, Lx =
4 200 mm
Ly =
5 460 mm
fcu =
35 MPa
fy =
500 MPa
Cover =
Preferred bar size =
Density of concrete =
Allowance for finishes =
Characteristic imposed load =
25 mm
12
24.5 kN/m3
2.0 kPa
10.0 kPa
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CON4332 REINFORCED CONCRETE DESIGN
Solution
Loading
Dead Load
2.00 kN/m2
Finishes:
Slab S/W:
24.5 x 0.20 =
4.90 kN/m2
gk =
6.90 kN/m2
qk =
10.00 kN/m2
Imposed Load
Design load, n =
=
(1.4 x 6.90 + 1.6 x 10.00)
25.7 kN/m (per meter width)
Design for Bending Moment (Short Span)
Effective depth, d =
Ly/Lx =
αsx =
msx =
=
K=
βb = 1.0
K < 0.0428
200 – 25 – 12/2 = 169mm
5460 / 4200 = 1.3
0.093
0.093 x 25.7 x 4.2
Table 6.5
2
42.2 kN-m (per meter width)
M / (bd2fcu)
=
42.2 x 106 / (1000 x 1692 x 35)
=
0.0422
<
0.156
z=
=
Tension steel req'd, As,req =
(Singly reinforced)
0.95d = 0.95 x 169
161 mm
M / (0.87 fy z)
=
42.2 x 106 / (0.87 x 500 x 161)
=
603 mm2
(Provide T12-150 Bottom)
As,pro =
100As / bh =
>
753 mm2
100 x 753 / (1000 x 200) = 0.377
0.13 and < 4.0
(Steel ratio ok)
Design for Bending Moment (Long Span)
d=
!
Lx instead of Ly
Ly/Lx =
200 – 25 – 12 – 12/2 = 157mm
5460 / 4200 = 1.3
is used to
αsy =
0.055
calculate msy
msy =
0.055 x 25.7 x 4.22
=
Table 6.5
24.9 kN-m (per meter width)
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CON4332 REINFORCED CONCRETE DESIGN
K=
βb = 1.0
K < 0.0428
M / (bd2fcu)
=
24.9 x 106 / (1000 x 1572 x 35)
=
0.029
<
0.156
z=
=
Tension steel req'd, As,req =
(Singly reinforced)
0.95d = 0.95 x 157
149 mm
M / (0.87 fy z)
=
24.9 x 106 / (0.87 x 500 x 149)
=
384 mm2
As,pro =
2
(Provide T12-275 Bottom)
100As / bh =
>
411 mm
100 x 411 / (1000 x 200) = 0.21
0.13 and < 4.0
(Steel ratio ok)
Check Shear at Support
Design Shear, V =
=
v=
0.5 x 25.7 x 4.2
54.0 kN
54.0 x 103 / (1000 x 169)
=
0.32 MPa
<
the smallest value of vc in Table 6.3 of HKCP-2013
(No shear reinforcement required)
Check Deflection by Span-to-depth Ratio
Basic L / d =
M/(bd2) =
=
As,req / As,pro =
fs =
mt =
=
Allowable L / d =
20 (Simply-supported Slab)
(Table 7.3)
42.2 x 106 / (1000 x 1692)
1.478 N/mm2
603 / 753 = 0.801
2/3 x 500 x 0.801 = 267MPa
0.55 + (477-267)/[120(0.9+1.478)]
(Table 7.4)
0.55 + 0.736 = 1.286
1.286 x 20 = 25.7
Actual L / d =
4200 / 169
=
24.9 ≤ 25.7
(Deflection ok)
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CON4332 REINFORCED CONCRETE DESIGN
5.4
Stairs
A stair is composed of the following elements:
Flight
The inclined slab supporting steps
Waist
The thickness of flight slab measured perpendicular to the
soffit
Steps
Divisions of the total vertical rise of a flight, composed of
treads and risers
Tread
The horizontal length or depth of a step
Riser
The vertical dimension of a step
Landing
The horizontal slab connecting flights
Handrails
On both sides of the flight to facilitate climbing of the stairs
and in some cases in the form of parapet to prevent falling
out of the stairs
Cross Section of a Stair
Steps and handrail/parapet are usually not regarded as parts of the structure
of a stair unless they are specifically designed as structural elements.
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5.4.1
Structural Forms of Stairs
A stair can be transversely-spanned
and supported on both sides of the
stairs as illustrated in Figure 5.5.
The direction of span is perpendicular
to, or transverse to, the direction of the
flight. In the design of this type of
stairs, the main bars are placed in the
direction of the steps.
Figure 5.5 –
Transversely-spanned Stair Slab
A stair can be longitudinally-spanned
in the direction of the flight and is
supported at the ends of the flight and
landing as illustrated in Figure 5.6.
The main bars of this type of stair are
placed along the flight direction and
extend to the end supports.
Figure 5.6 –
Longitudinally-spanned flight can be
supported by the landing slabs, which,
in turn, are transversely-spanned and
supported on the walls at both sides of
the landing as illustrated in Figure 5.7.
There are other possible structural
schemes for stair design, e.g.
cantilever steps, flight supported by
stringer beams, cantilever stairs, etc.
In this Chapter, longitudinally-spanned
stairs are used to highlight the salient
points in stair design.
Longitudinally-spanned Stair Slab
Supported by End Walls
Figure 5.7 –
Longitudinally-spanned Stair Slab
Supported by Landing Slabs
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CON4332 REINFORCED CONCRETE DESIGN
5.4.2
Design of Stair Slabs
The design of stair slabs is quite similar to that for solid slab with the
following points highlighted:
(a) The following dead load have to be taken into account:
i.
The weight of the steps (see 5.4.3)
ii.
Adjustment to the self-weight of the stair slab for the increased
length of the inclined flight slabs (see 5.4.3)
iii.
The weight of handrail/parapet, if any
(b) Although the slab is inclined, as
the loads are acting vertically
downward under gravitational
action, the projected horizontal
distance is used in determining
the effective span as illustrated in
the following examples.
(c)
The depth of the section, h, used
for design is the minimum
thickness perpendicular to the
soffit of the inclined stair slab, i.e. the waist. The effective depth, d,
is then determined by using this value of h. The steps are usually
ignored in the calculation of the structural capacity of stair slab.
(d) The allowable span-to-depth ratio can be increased by 15% if the
stair flight occupies at least 60% of the span (Cl.6.6.2.1 of
HKCP-2013).
As a consequence of (a) mentioned above, the dead weight of flight slab is
usually larger than that of landing slab.
For simply-supported
longitudinally-spanned stairs, the following formulae are useful in
determining the mid-span moment and support shear.
For a simply-supported beam subjected to partial udl loads, w1 and w2,
symmetrically loaded as shown in the figure below,
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CON4332 REINFORCED CONCRETE DESIGN
W2
W1
W1
L2
L1
L1
the reactions or shears at the supports are:
V = w1L1 + 0.5 w2L2
[5.1]
The mid-span moment is:
M = w1L1(L1 + 0.5L2) – w1L1(0.5L1 + 0.5L2)
+ 0.5w2L2(L1 + 0.5L2) – 0.5w2L2(0.25L2)
= 0.5w1L12 + 0.5w2L2(L1 + 0.25L2)
5.4.3
[5.2]
Example – Self-weight of Stair Flight
Question
Determine the self-weight of the following stair flight:
Tread =
250 mm
Riser =
150 mm
Waist thickness, h =
175 mm
Solution
For a meter width of the stair flight
For a single step
horizontal length =
Inclined length =
=
Weight of a step
Weight of the waist
250 mm
(2502 + 1502)1/2
292 mm
24.5 x 0.150 x 0.250 /2 =
0.459 kN
24.5 x 0.292 x 0.175 =
1.252 kN
Total =
1.711 kN
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CON4332 REINFORCED CONCRETE DESIGN
Therefore, udl per meter horizontal length of the flight =
=
1.711 / 0.250
6.84 kN/m
Alternatively,
For a meter width of the stair flight
Ratio of inclined length to horizontal length =
=
Weight of a step
1.166
24.5 x 0.150 /2 =
1.84 kN/m
24.5 x 0.175 x 1.166 =
5.00 kN/m
Total udl per meter horizontal length of the flight =
6.84 kN/m
Weight of the waist
5.4.4
(2502 + 1502)1/2 / 250
Q.5
?
Examples – Longitudinally-spanned Stairs
Question A – Longitudinally-spanned Stair Supported by End Walls
Design the reinforcement and check the deflection of the stair slab as shown in DWG-04
with the following design parameters.
Design parameters
Waist, h =
275 mm
Tread =
250 mm
Riser =
150 mm
Number of risers, N =
Flight horizontal length, L2 =
Flight width, W =
14
250 x 14 = 3500 mm
1200 mm
Landing slab thickness =
275 mm (same as waist)
Landing clear width, Ln =
1200 mm (same at both ends)
Width of support, Sw =
200 mm (same at both ends)
fcu =
35 MPa
fy =
500 MPa
Cover =
Preferred bar size =
Density of concrete =
Allowance for finishes =
Allowance for handrail/parapet =
Characteristic imposed load =
25 mm
16
24.5 kN/m3
1.5 kPa
(assume negligible)
5.0 kPa
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CON4332 REINFORCED CONCRETE DESIGN
Solution
Effective Span
As h > Sw,
Effective L1 =
Overall L =
=
1200 + 100 = 1300mm
3500 + 2 x 1300
6100 mm
Loading (landing, w1)
Dead Load
Self-weight
24.5 x 0.275 =
1.50 kN/m2
Finishes
Imposed Load
The design load, w1 =
6.74 kN/m2
gk =
8.24 kN/m2
qk =
5.00 kN/m2
1.4 x 8.24 + 1.6 x 5.00
=
19.53 kN/m (per meter width)
Inclined length ratio =
(2502 + 1502)0.5 / 250 = 1.166
Loading (flight, w2)
Dead Load
Steps
24.5 x 0.150 / 2 =
1.84 kN/m2
Self-weight
24.5 x 0.275 x 1.166 =
7.86 kN/m2
1.50 kN/m2
Finishes
Imposed Load
gk =
11.19 kN/m2
qk =
5.00 kN/m2
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CON4332 REINFORCED CONCRETE DESIGN
The design load, w2 =
1.4 x 11.19 + 1.6 x 5.00
=
23.67 kN/m (per meter width)
Design Forces
0.5 x 19.53 x 1.32
Design Mid-span Mt, M =
+ 0.5 x 23.67 x 3.5 x (1.3 + 0.25 x 3.5)
=
16.5 + 90.1
=
107 kN-m (per meter width)
Design Shear, V =
19.53 x 1.3 + 0.5 x 23.67 x 3.5
=
67 kN (per meter width)
Design for Mid-span Bending Moment
Effective Depth, d =
275 – 25 – 16/2
=
242 mm
M / (bd2fcu)
K=
βb = 1.0
=
107 x 106 / (1000 x 2422 x 35)
=
0.052
<
0.156
(Singly reinforced)
[0.5 + (0.25 – K/0.9)0.5] d
Lever arm, z =
=
[0.5 + (0.25 – 0.052/0.9)0.5] x 242
=
0.938 x 242
=
227 mm
Tension steel req'd, As,req =
M / (0.87 fy z)
=
107 x 106 / (0.87 x 500 x 227)
=
1084 mm2
(Provide T16-125 bottom)
As,pro =
201 / 0.125
1608 mm2
=
100As / bh =
100 x 1608 / (1000 x 275) = 0.585
>
Distribution Bar
0.13 and < 4.0
Ads =
=
(Steel ratio ok)
Max(0.13x1000x275/100 or 1084x20%)
358 mm2
(Provide T12-275 DB)
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CON4332 REINFORCED CONCRETE DESIGN
Check Shear at Support
v=
67 x 103 / (1000 x 242)
=
0.28 MPa
<
0.8 √ 35 = 4.73 MPa
and <
(Concrete does not crush)
the smallest value of vc in Table 6.3 of HKCP-2013
(No shear reinforcement required)
Check Deflection by Span-to-depth Ratio
Basic L /d =
M/(bd2) =
=
As,req / As,pro =
fs =
mt =
20 (simply-supported slab)
107 x 106 / (1000 x 2422)
1.83 N/mm2
1084 / 1608 = 0.67
2/3 x 500 x 0.67 = 224 MPa
0.55 + (477-224)/[120(0.9+1.83)]
=
0.55 + 0.772
=
1.32
L2 / L =
Allowable L / d =
3500 / 6100 = 0.57
(Table 7.4)
< 60% (no increase in L/d Ratio)
1.32 x 20 = 26.4
Actual L / d =
6100 / 242
=
25.2 ≤ 26.4
Comments:

(Table 7.3)
(Deflection ok)
Q.6
?
In this example, it is the deflection that controls the amount of steel required. The
amount of steel provided is about 48% more than that required for resisting the
design moment so as to reduce the service stress in the bars and therefore
increase the modification factor to the L/d ratio.

The width of the flight is not large in this example; it may be more convenient to use
the actual width of 1200mm instead of a unit width of 1000mm for design. If the
actual width is used in the design calculations, i.e. b = 1200mm, the As obtained is
then the total area for the whole section instead of per meter width and therefore
rebars to be provided will then be in terms of number of bars instead of spacing.
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CON4332 REINFORCED CONCRETE DESIGN
Question B – Longitudinally-spanned Stair Supported by Landing
Slabs
Design the reinforcement and check the deflection of the stair slab as shown in DWG05 with
the following design parameters.
Design parameters
Waist, h =
200 mm
Tread =
250 mm
Riser =
150 mm
No. of Risers =
Horizontal length of the flight, L2 =
Width of the flight =
Thickness of the landing slab =
Width of landing, Lb =
Clear span of the landing, Ln =
Width of support, Sw =
14
250 x 14 = 3500 mm
1200 mm
200 mm (same as the waist)
1200 mm (same at both ends)
2600 mm
200 mm (same at both ends)
fcu =
35 MPa
fy =
500 MPa
Cover =
Preferred bar size =
Density of concrete =
Allowance for finishes =
Allowance for handrail/parapet =
Characteristic imposed load =
25 mm
12
24.5 kN/m3
1.5 kPa
(assume negligible)
5.0 kPa
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CON4332 REINFORCED CONCRETE DESIGN
Solution
In this example, the flight slab is supported by landing slabs. Hence, two slabs have to be
designed: (I) flight slab and then (II) landing slab. The widths of the landing slabs, Lb, will
be regarded as the widths of the supports to the flight slab and will be used for determining
the effective span, L2 + 2L1, of the flight slab, where L1 is the lesser of Lb/2 or 1800mm, (Cl.
6.6.1.2 of HKCP-2013).
As the width of support is comparatively large, and there is no
loading at the support width, partial udl is adopted in the design. The support reaction, R,
from the flight slab will be transmitted to the landing slab for design.
(I) Design of Flight Slab
Effective Span
As h > Sw,
Effective L1 =
Overall L =
=
min(1200 / 2 or 1800) = 600
3500 + 2 x 600
4700 mm
Loading (flight, w)
Inclined length ratio =
(2502 + 1502)0.5 / 250 = 1.166
Dead Load
Steps
24.5 x 0.150 / 2 =
1.84 kN/m2
Self-weight
24.5 x 0.200 x 1.166 =
5.71 kN/m2
1.50 kN/m2
Finishes
Imposed Load
The design load, w =
=
gk =
9.05 kN/m2
qk =
5.00 kN/m2
1.4 x 9.05 + 1.6 x 5.00
20.7 kN/m (per meter width)
Design Forces
Design Mid-span Mt, M =
=
Design Shear, V =
=
0.5 x 20.7 x 3.5 x (0.6 + 0.25 x 3.5)
53.4 kN-m (per meter width)
0.5 x 20.7 x 3.5
36.2 kN (per meter width)
Design for Mid-span Bending Moment
Effective Depth, d =
=
200 – 25 – 12/2
169 mm
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K=
βb = 1.0
M / (bd2fcu)
=
53.4 x 106 / (1000 x 1692 x 35)
=
0.053
<
0.156
Lever arm, z =
(Singly reinforced)
[0.5 + (0.25 – K/0.9)0.5] d
=
[0.5 + (0.25 – 0.053/0.9)0.5] x 169
=
0.937 x 169
=
158 mm
Tension steel req'd, As,req =
M / (0.87 fy z)
=
53.4 x 106 / (0.87 x 500 x 158)
=
775 mm2
(Provide T12-100 bottom)
As,pro =
=
100As / bh =
>
Distribution Bar, Asd =
=
113 / 0.100
1130 mm2
100 x 1130 / (1000 x 200) = 0.565
0.13 and < 4.0
(Steel ratio ok)
Max (0.13bh or 0.2As,req)
Max (260 or 155) = 260 mm2
(Provide T10 -250 DB)
Asd,pro =
78.5 / 0.25 = 314 mm2
Check Shear
Max shear at the face of support
v=
36.2 x 103 / (1000 x 169)
=
0.21 MPa
<
0.8 √ 35 = 4.73 MPa
and <
(Concrete does not crush)
the smallest value of vc in Table 6.3 of HKCP-2013
(No shear reinforcement required)
Check Deflection by Span-to-depth Ratio
Basic L /d =
M/(bd2) =
=
As,req / As,pro =
20 (simply-supported slab)
(Table 7.3)
53.4 x 106 / (1000 x 1692)
1.87 N/mm2
775 / 1130 = 0.686
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CON4332 REINFORCED CONCRETE DESIGN
fs =
mt =
2/3 x 500 x 0.686 = 228 MPa
0.55 + (477-228)/[120(0.9+1.87)]
=
0.55 + 0.75
=
1.30
L2 / L =
Allowable L / d =
3500 / 4700 = 0.74
(Table 7.4)
> 60%
1.30 x 1.15 x 20 = 29.9
Actual L / d =
4700 / 169
=
27.8 ≤ 29.9
(Deflection ok)
(II) Design of Landing Slab
The landing slab supports two types of loads:
i.
Reactions from the flight slabs, R, which
are two partial udl separated by the gap of
the flights; and,
ii.
Self-weight, finishes, imposed load, etc.
that are acting directly on the landing slab,
w, which, are, to be precise, also a partial
udl over the clear span of the landing only.
Although the above two loads, to be precise, are
partial udl, as the width of support and the gap
between flights are usually comparatively very
small and can be ignored for simplicity, the loads
can therefore be assumed to be distributed
uniformly over the whole effective span as
illustrated in the following calculations.
Effective Span
As h = Sw,
Effective L=
2600 + 200
=
2800 mm
Width of the landing =
1200 mm
Loading (landing)
Dead Load
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CON4332 REINFORCED CONCRETE DESIGN
From flight slab
9.05 x 3.5 / 2 =
15.84 kN/m
Self-weight
24.5 x 0.20 x 1.2 =
5.88 kN/m
Finishes
1.5 x 1.2 =
1.80 kN/m
gk =
23.52 kN/m
From flight slab
5.00 x 3.5 / 2 =
8.75 kN/m
Landing slab
5.00 x 1.2 =
6.00 kN/m
qk =
14.75 kN/m
Imposed Load
Design load, w =
=
(1.4 x 23.52 + 1.6 x 14.75) / 1.2
47.11 kN/m (per meter width)
Design Forces
Design Mid-span Mt, M =
=
Design Shear, V =
=
0.125 x 47.11 x 2.82
46.2 kN-m
0.5 x 47.11 x 2.8
66 kN
Design for Mid-span Bending Moment
Effective Depth, d =
=
K=
βb = 1.0
200 – 25 – 12/2
169 mm
M / (bd2fcu)
=
46.2 x 106 / (1000 x 1692 x 35)
=
0.046
<
0.156
Lever arm, z =
(Singly reinforced)
[0.5 + (0.25 – K/0.9)0.5] d
=
[0.5 + (0.25 – 0.046/0.9)0.5] x 169
=
0.946 x 169
=
160 mm
Tension steel req'd, As,req =
M / (0.87 fy z)
=
46.2 x 106 / (0.87 x 500 x 160)
=
664 mm2
(Provide T12-150 bottom)
As,pro =
113 / 0.150
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CON4332 REINFORCED CONCRETE DESIGN
=
100As / bh =
>
754 mm2
100 x 754 / (1000 x 200) = 0.377
0.13 and < 4.0
(Steel ratio ok)
Check Shear
Max shear at the face of support
v=
66 x 103 / (1000 x 169)
=
0.39 MPa
<
0.8 √ 35 = 4.73 MPa
and <
(Concrete does not crush)
the smallest value of vc in Table 6.3 of HKCP-2013
(No shear reinforcement required)
Check Deflection by Span-to-depth Ratio
Basic L /d =
M/(bd2) =
=
As,req / As,pro =
fs =
mt =
20 (simply-supported slab)
46.2 x 106 / (1000 x 1692)
1.62 N/mm2
664 / 754 = 0.881
2/3 x 500 x 0.881 = 294 MPa
0.55 + (477-294)/[120(0.9+1.62)]
=
0.55 + 0.605
=
1.16
Allowable L / d =
(Table 7.3)
(Table 7.4)
1.16 x 20 = 23.2
Actual L / d =
2800 / 169
=
16.6 ≤ 23.2
(Deflection ok)
Comment:
Although the configuration of the stairs in Questions A and B are the same, different
structural arrangements can lead to substantial saving in materials. The thickness of the
stair slabs and the amount of steel required for the stair in Question B are reduced by about
27% and 30% respectively.
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CON4332 REINFORCED CONCRETE DESIGN
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CON4332 REINFORCED CONCRETE DESIGN
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CON4332 REINFORCED CONCRETE DESIGN
│Self-Assessment Questions│
Q.1
Given the following design parameters of a one-way simply-supported slab:
Slab thickness, h :
c/c distance btw supports :
Width of the supporting beam, Sw :
Allowance for finishes :
Partition:
Usage of the floor :
Q.2
175 mm
3 300 mm
400 mm (same for both ends)
1.5 kPa
1.5 kPa (light weigh, undefined)
Offices
(a)
Determine the characteristic loads in kPa.
(b)
Determine the design load in kN per meter width of the slab.
(c)
Determine the design forces.
(a)
Identify the conditions under which a single-load case of maximum design load on all
spans can be used for slab design.
(b)
Identify the additional conditions to those you have identified in (a) for the usage of the
force coefficients in Table 5.1 (i.e. Table 6.4 of HKCP-2013).
Q.3
Determine the rebars for the following slabs:
Slab thickness
2
Steel area required (mm /m)
(a)
(b)
(c)
150mm
175mm
200mm
338
634
1265
Bar size
Spacing (mm)
As,pro
100As/bh
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CON4332 REINFORCED CONCRETE DESIGN
Q.4
Q.5
Determine the self-weight of the following stair flight:
Tread =
225 mm
Riser =
175 mm
Waist thickness, h =
150 mm
For the stair in Question A of 5.4.4, if the clear width of the landing of the staircase, Ln is
proposed to be increased to 1300mm.
(a)
Determine the new design bending moment.
(b)
Check if the original bar provided is adequate or not.
(c)
Determine the allowable L/d ratio and check if deflection is acceptable or not and give
advice.
Q.6
Determine the reinforcement for the flight slab of a stair with the following given:
Waist thickness, h =
Cover =
Preferred bar size =
280 mm
35 mm
16
fcu =
35 MPa
fy =
500 MPa
Design Moment, M =
122 kN-m per m width
Answers:
Q1a: gk=5.79kPa, qk = 4.50kPa; Q1b: 47.1kN; Q1c:M=18.1kN-m, V=23.6kN
2
Q2a: (i) one-way slab with bay size > 30m , (ii) Qk/Gk ≤ 1.25, (iii) qk ≤ 5kPa;
Q2b: (i) the load is substantially uniformly distributed, (ii) 3 or more spans, (iii) approximately equal span.
Q3a: T10-225, 0.23; Q2b: T12-175, 0.37; Q2c: T16-175, 0.57
Q4: 6.81kN/m per m width
Q5a: M=113.4kN-m; Q5b: As,req=1153, ok; Q5c: Allowable L/d = 25.0, unacceptable, increase h or As
2
2
Q6: As,req = 1279mm /m, provide T16-150, Asd = 364mm /m, provide T10-200, comment: pay attention to L/d ratio.
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CON4332 REINFORCED CONCRETE DESIGN
│Tutorial Questions│
(Present your calculations with detailed working steps in a logical, neat and tidy
manner.)
AQ1
Re-design the reinforcement and check the deflection of the end span of the
the continuous one-way slab, 5S1 as shown in DWG-01 of Chapter 1 with
the following changes (make reference to Question A of 5.3.5 for the original
design):
i. The center-to-center distance between beams is changed from 3300
mm to 3500 mm, i.e. the distance between gridlines 6 and 7 is changed
to 10 500 mm.
ii. An additional allowance for 300 mm thick soil is required.
iii. The width of the beam is increased to 400mm.
AQ2
If a very heavy equipment is to be placed on the slab, 5S1 of AQ1, at the
area marked "Area A" on the DWG-01 of Chapter 1. Give advice on the
possible implications to the design of the slab, without doing any detail
calculations. (Adapted from 2012/13 Sem 3 examination paper.)
AQ3
For the slab RB1 shown in DWG-03 in Chapter 3,
(a)
(b)
(c)
(d)
Identify the essential design parameters from the drawing and
determine the design load for one span in kN per m width.
Given the following force coefficients, check the adequacy of
providing T10-125 as top and bottom bars for the slab.
At outer support
Mid-span
Support (βb = 0.8)
Moment
0
0.080FL
-0.100FL
Shear
0.4F
-
0.6F
Check if shear reinforcement is required.
Check if the deflection of the slab is acceptable by span-to-depth
ratio.
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CON4332 REINFORCED CONCRETE DESIGN
AQ4
Design the reinforcement and check the deflection by span-to-depth ratio of
the stairs as shown in DWG-04 with the following design information:
Design parameters
Waist, h =
275 mm
Tread =
260 mm
Riser =
160 mm
Number of Risers, N =
Flight horizontal length, L2 =
Flight width, W =
14
260 x 14 = 3640 mm
1250 mm
Landing slab thickness =
275 mm (same as waist)
Landing clear width, Ln =
1250 mm (same at both ends)
Width of support, Sw =
200 mm (same at both ends)
fcu =
40 MPa
fy =
500 MPa
Cover =
Preferred bar size =
Density of concrete =
Allowance for finishes =
Allowance for handrail/parapet =
Characteristic imposed load =
25 mm
16
24.5 kN/m3
1.5 kPa
(assume negligible)
5.0 kPa
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