con4332 reinforced concrete designtycnw01.vtc.edu.hk/cbe2022/r5_design of r c slabs...

CON4332 REINFORCED CONCRETE DESIGN

Chapter 5 1

HD in Civil Engineering (Aug 2014)

│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 Types of Slabs

5.2 Design Loads and Forces 5.2.1 Example – Design Loads and Forces for Simply-supported One-way Slab

5.2.2 Force Coefficients for Continuous One-way Slab

5.3 Design of Slabs 5.3.1 Design for Moment and Shear

5.3.2 Deflection Check by Span-to-depth Ratio

5.3.3 Distribution Bars

5.3.4 Examples – Simply-supported One-way Slab

5.3.5 Examples – Continuous One-way Slab

5.3.6 Example – Simply-supported Two-way Slab

5.4 Stairs 5.4.1 Types of Stairs

5.4.2 Design for Stair Slabs

5.4.3 Example – Self-weight of Stair Flight

5.4.4 Examples – Longitudinally-spanned Stairs


Chapter 5 2


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.

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.

One-way Slab Two-way Slab

Support

Support

Support

Support Support

Support


Chapter 5 3


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.

L > 2L

Effective Support

Effective Support


Chapter 5 4


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 or 0.125 w L2

Shear at Support, V = 0.5 F or 0.5 w L

Question Determine the design forces for the following simply-supported one-way slab.

Design parameters

Slab thickness, h : 150 mm

c/c distance btw supports : 3 000 mm

Width of the supporting beam, Sw : 350 mm (same for both ends)

Allowance for finishes : 1.5 kPa

Usage of the floor : Department Store

Solution

Dead Load

Finishes : 1.50 kN/m2

Self-weight : 24.5 x 0.15 = 3.68 kN/m2

gk = 5.18 kN/m2

1 m

1 m


Chapter 5 5


Imposed Load

Department Store: qk= 5.00 kN/m2

Design Load, w = 1.4 x 5.18 + 1.6 x 5.00

= 15.25 kN/m2

As h < Sw Effective Span, L = 3000 – 350 + 150

= 2 800 mm

Design Moment, M = 0.125 x 15.25 x 2.82

= 14.95 kN-m

Design Shear, V = 0.5 x 15.25 x 2.8

= 21.35 kN

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:

The design moment

and shear are in per

meter width of the

slab.

! ?Q.1


Chapter 5 6


At outer

support

(simply

supported)

Near middle

of end span

At first

interior

support

At middle of

interior span

At interior

supports

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.


Chapter 5 7


shear force diagrams as shown below.

0.086FL

0.086FL

0.063FL

0.063FL

0.4F

0.6F

0.5F

0.5F

Bending Moment Diagram

Shear Force Diagram

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.)


Chapter 5 8


(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.

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:

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.

?Q.3

?Q.2


Chapter 5 9


T10-100 (As,pro = 78.5 / 0.100 = 785 mm2 /m)

The steel area can be read from the following table.

Bar

Size

Bar Spacing in mm

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.


Chapter 5 10


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

(b) The spacing shall not be more than 3h & 400mm.

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

Refer to Chapter

2 for bar spacing

requirement for

main bars.


Chapter 5 11


c/c distance btw supports = 4000 mm

Width of supporting beams, Sw = 250 mm

fcu = 30 MPa

fy = 500 MPa

Cover = 25 mm

Preferred bar size = 10

Characteristic imposed load = 3.0 kPa (Offices)

Allowance for finishes = 1.0 kPa

Partition load = 1.0 kPa (Lightweight undefined)

Solution Dead Load

Finishes : 1.00 kN/m2

Self-weight : 24.5 x 0.175 = 4.29 kN/m2

gk = 5.29 kN/m2

Imposed Load

Partition load : 1.00

Offices : 3.00 kN/m2

qk= 4.00 kN/m2

Design Load, w = 1.4 x 5.29 + 1.6 x 4.00

= 13.81 kN/m2

As h < Sw Effective Span, L = 4000 – 250 + 175

= 3925 mm


= 26.6 kN-m

Design Shear, V = 0.5 x 13.81 x 3.925

= 27.1 kN

Design for Bending Moment

Effective depth, d = 175 – 25 – 10/2

= 145 mm

K = M / (bd2fcu)

= 26.6 x 106 / (1000 x 1452 x 30)

= 0.042

βb = 1.0 < 0.156 (Singly reinforced)

b = 1000 !


Chapter 5 12


K < 0.0428 z = 0.95d = 0.95 x 145

= 137.8 mm

Tension steel req'd, As,req = M / (0.87 fy z)

= 26.6 x 106 / (0.87 x 500 x 137.8)

= 444 mm2

(Provide T10-175 Bottom)

As,pro = 78.5 / 0.175

= 449 mm2

100As / bh = 100 x 449 / (1000 x 175) = 0.256

> 0.13 and < 4.0 (Steel ratio ok)

Distribution Bar Asd = 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 : (Table 6.3)

100As/(bvd) = 100 x 449 / (1000 x 145) = 0.31 < 3

(400/d)1/4 = (400 / 145)1/4 = 1.289 (> 0.67)

vc = 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 = 20 (Simply Supported Slab) (Table 7.3)

M/(bd2) = 26.6 x 106 / (1000 x 1452)

= 1.264 N/mm2

As,req / As,pro = 444 / 449 = 0.99

fs = 2/3 x 500 x 0.99 = 330 MPa

mt = 0.55 + (477-330)/[120(0.9+1.26)] (Table 7.4)

= 0.55 + 0.567

= 1.12

Alternatively, As,pro

can be read from

table.

In this case, v < 0.34, the

smallest value in Table 6.3 of

HKCP-2013. The calculation of

vc can be omitted.

!


Chapter 5 13


Allowable L / d = 1.12 x 20 = 22.3

Actual L / d = 3925 / 145

= 27.1 > 22.3 (Deflection not ok)

(Increase the tension steel to T10-125 Bottom)

As,pro = 78.5 / 0.125 = 628 mm2

As,req / As,pro = 444 / 628 = 0.707

fs = 2/3 x 500 x 0.707 = 236 MPa

mt = 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 (Deflection ok)

Comment: For this slab, deflection controls the amount the steel required.

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.

?Q.4


Chapter 5 14


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


c/c distance btw supports = 3 300 mm

Width of support, Sw = 300 mm (similar at both ends)

fcu = 35 MPa

fy = 500 MPa

Cover = 25 mm


Density of concrete = 24.5 kN/m3


Characteristic imposed load = 5.0 kPa


Chapter 5 15


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

Imposed Load qk = 5.00 kN/m

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.


= 14.0 kN-m

Design Shear, V = 0.6 x 51.5

= 30.9 kN

Effective Depth d = 160 – 25 – 10/2

= 130 mm


K = M / (bd2fcu)

= 14.0 x 106 / (1000 x 1302 x 35)

= 0.024


K < 0.0428 z = 0.95d = 0.95 x 130

= 123.5 mm


= 14.0 x 106 / (0.87 x 500 x 123.5)

The moment

coefficients for both

span and support

are the same.

Support moment is

adopted for rebar

design, which has a

redistribution of 20%.


Chapter 5 16


= 260 mm2

(Provide T10-250 Top at supports

and T10-250 Bottom at mid-span)

As,pro = 314 mm2 per meter

100As / bh = 100 x 314 / (1000 x 160) = 0.196


Distribution Bar Ads = Max(0.13x1000x160/100 or 260x20%)

= 208 mm2



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)


Basic L /d = 23 (End span of continuous slab) (Table 7.3)

M/(bd2) = 14.0 x 106 / (1000 x 1302)

= 0.828 N/mm2

As,req / As,pro = 260 / 314 = 0.828

fs = 2/3 x 500 x 0.828 = 276 MPa

mt = 0.55 + (477-276)/[120(0.9+0.828)] (Table 7.4)

= 0.55 + 0.969

= 1.52

Allowable L / d = 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,

βb = 0.8 is for support

moment. For L/d checking,

mid-span moment is used. !


Chapter 5 17


the calculation is presented below as an illustration of the complete process of design.

Design Forces


= 10.3 kN-m

Design Shear, V = 0.5 x 51.5

= 25.8 kN


K = M / (bd2fcu)

= 10.3 x 106 / (1000 x 1302 x 35)

= 0.017


K < 0.0428 z = 0.95d = 0.95 x 130

= 123.5 mm


= 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 = 262 mm2 per meter

100As / bh = 100 x 262 / (1000 x 160) = 0.164



= 208 mm2



Basic L /d = 26 (Interior span of continuous slab ) (Table 7.3)

M/(bd2) = 10.3 x 106 / (1000 x 1302)

= 0.609 N/mm2

As,req / As,pro = 192 / 262 = 0.733

fs = 2/3 x 500 x 0.733 = 244 MPa

mt = 0.55 + (477-244)/[120(0.9+0.609)] (Table 7.4)

= 0.55 + 1.287


Chapter 5 18


= 1.84

Allowable L / d = 1.84 x 26 = 47.8

Actual L / d = 3160 / 130


5.3.6 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.

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

Moment in the shorter span msx = αsxnLx2

Moment in the longer span msy = αsynLx2

5 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.

Lx Ly

msy msx

Long span Short Span


Chapter 5 19


The bending moment coefficients αsx & αsy are given in the following table.

Ly / Lx 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


Effective spans, Lx = 4 200 mm

Ly = 5 460 mm

fcu = 35 MPa

fy = 500 MPa

Cover = 25 mm






Chapter 5 20


Solution Loading

Dead Load

Finishes: 2.00 kN/m2

Slab S/W: 24.5 x 0.20 = 4.90 kN/m2

gk = 6.90 kN/m2

Imposed Load qk = 10.00 kN/m2

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 = 200 – 25 – 12/2 = 169mm

Ly/Lx = 5460 / 4200 = 1.3

αsx = 0.093 Table 6.5

msx = 0.093 x 25.7 x 4.22

= 42.2 kN-m (per meter width)

K = M / (bd2fcu)

= 42.2 x 106 / (1000 x 1692 x 35)

= 0.0422


K < 0.0428 z = 0.95d = 0.95 x 169

= 161 mm


= 42.2 x 106 / (0.87 x 500 x 161)

= 603 mm2


As,pro = 753 mm2

100As / bh = 100 x 753 / (1000 x 200) = 0.377


Design for Bending Moment (Long Span)

d = 200 – 25 – 12 – 12/2 = 157mm

Ly/Lx = 5460 / 4200 = 1.3

αsy = 0.055 Table 6.5

msy = 0.055 x 25.7 x 4.22


Lx instead of Ly

is used to

calculate msy !


Chapter 5 21


K = M / (bd2fcu)

= 24.9 x 106 / (1000 x 1572 x 35)

= 0.029


K < 0.0428 z = 0.95d = 0.95 x 157

= 149 mm


= 24.9 x 106 / (0.87 x 500 x 149)

= 384 mm2


As,pro = 411 mm2

100As / bh = 100 x 411 / (1000 x 200) = 0.21




= 54.0 kN

v = 54.0 x 103 / (1000 x 169)

= 0.32 MPa

< the smallest value of vc in Table 6.3 of HKCP-2013



Basic L / d = 20 (Simply-supported Slab) (Table 7.3)

M/(bd2) = 42.2 x 106 / (1000 x 1692)

= 1.478 N/mm2

As,req / As,pro = 603 / 753 = 0.801

fs = 2/3 x 500 x 0.801 = 267MPa

mt = 0.55 + (477-267)/[120(0.9+1.478)] (Table 7.4)

= 0.55 + 0.736 = 1.286

Allowable L / d = 1.286 x 20 = 25.7

Actual L / d = 4200 / 169



Chapter 5 22


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

Steps and handrail/parapet are usually not regarded as parts of the structure

of a stair unless they are specifically designed as structural elements.

Cross Section of a Stair


Chapter 5 23


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.

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.

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.

Figure 5.5 –

Transversely-spanned Stair Slab

Figure 5.6 –

Longitudinally-spanned Stair Slab

Supported by End Walls

Figure 5.7 –

Longitudinally-spanned Stair Slab

Supported by Landing Slabs


Chapter 5 24


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,


Chapter 5 25


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) [5.2]

5.4.3 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 = 250 mm

Inclined length = (2502 + 1502)1/2

= 292 mm

Weight of a step 24.5 x 0.150 x 0.250 /2 = 0.459 kN

Weight of the waist 24.5 x 0.292 x 0.175 = 1.252 kN

Total = 1.711 kN

W1 W1

W2

L1 L1 L

2


Chapter 5 26


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 = (2502 + 1502)1/2 / 250

= 1.166

Weight of a step 24.5 x 0.150 /2 = 1.84 kN/m

Weight of the waist 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

5.4.4 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 = 14

Flight horizontal length, L2 = 250 x 14 = 3500 mm

Flight width, W = 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 = 25 mm




Allowance for handrail/parapet = (assume negligible)


?Q.5


Chapter 5 27


Solution Effective Span

As h > Sw, Effective L1 = 1200 + 100 = 1300mm

Overall L = 3500 + 2 x 1300

= 6100 mm

Loading (landing, w1)

Dead Load

Self-weight 24.5 x 0.275 = 6.74 kN/m2

Finishes 1.50 kN/m2

gk = 8.24 kN/m2


The design load, w1 = 1.4 x 8.24 + 1.6 x 5.00


Loading (flight, w2)

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 0.275 x 1.166 = 7.86 kN/m2

Finishes 1.50 kN/m2

gk = 11.19 kN/m2



Chapter 5 28


The design load, w2 = 1.4 x 11.19 + 1.6 x 5.00


Design Forces

Design Mid-span Mt, M = 0.5 x 19.53 x 1.32

+ 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

K = M / (bd2fcu)

= 107 x 106 / (1000 x 2422 x 35)

= 0.052


Lever arm, z = [0.5 + (0.25 – K/0.9)0.5] d

= [0.5 + (0.25 – 0.052/0.9)0.5] x 242

= 0.938 x 242

= 227 mm


= 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



= 358 mm2



Chapter 5 29



v = 67 x 103 / (1000 x 242)

= 0.28 MPa

< 0.8 √ 35 = 4.73 MPa (Concrete does not crush)

and < the smallest value of vc in Table 6.3 of HKCP-2013



Basic L /d = 20 (simply-supported slab) (Table 7.3)

M/(bd2) = 107 x 106 / (1000 x 2422)

= 1.83 N/mm2

As,req / As,pro = 1084 / 1608 = 0.67

fs = 2/3 x 500 x 0.67 = 224 MPa

mt = 0.55 + (477-224)/[120(0.9+1.83)] (Table 7.4)

= 0.55 + 0.772

= 1.32

L2 / L = 3500 / 6100 = 0.57 < 60% (no increase in L/d Ratio)

Allowable L / d = 1.32 x 20 = 26.4

Actual L / d = 6100 / 242


Comments:

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.

?Q.6


Chapter 5 30


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 = 14

Horizontal length of the flight, L2 = 250 x 14 = 3500 mm

Width of the flight = 1200 mm

Thickness of the landing slab = 200 mm (same as the waist)

Width of landing, Lb = 1200 mm (same at both ends)

Clear span of the landing, Ln = 2600 mm


fcu = 35 MPa

fy = 500 MPa

Cover = 25 mm







Chapter 5 31


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 = min(1200 / 2 or 1800) = 600

Overall L = 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

Finishes 1.50 kN/m2

gk = 9.05 kN/m2


The design load, w = 1.4 x 9.05 + 1.6 x 5.00


Design Forces

Design Mid-span Mt, M = 0.5 x 20.7 x 3.5 x (0.6 + 0.25 x 3.5)



= 36.2 kN (per meter width)



= 169 mm


Chapter 5 32


K = M / (bd2fcu)

= 53.4 x 106 / (1000 x 1692 x 35)

= 0.053


Lever arm, z = [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


= 53.4 x 106 / (0.87 x 500 x 158)

= 775 mm2


As,pro = 113 / 0.100

= 1130 mm2

100As / bh = 100 x 1130 / (1000 x 200) = 0.565


Distribution Bar, Asd = 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






M/(bd2) = 53.4 x 106 / (1000 x 1692)

= 1.87 N/mm2

As,req / As,pro = 775 / 1130 = 0.686


Chapter 5 33


fs = 2/3 x 500 x 0.686 = 228 MPa

mt = 0.55 + (477-228)/[120(0.9+1.87)] (Table 7.4)

= 0.55 + 0.75

= 1.30

L2 / L = 3500 / 4700 = 0.74 > 60%

Allowable L / d = 1.30 x 1.15 x 20 = 29.9

Actual L / d = 4700 / 169


(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

Loading (landing)

Width of the landing = 1200 mm

Dead Load


Chapter 5 34


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

Imposed Load

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

Design load, w = (1.4 x 23.52 + 1.6 x 14.75) / 1.2


Design Forces

Design Mid-span Mt, M = 0.125 x 47.11 x 2.82

= 46.2 kN-m


= 66 kN



= 169 mm

K = M / (bd2fcu)

= 46.2 x 106 / (1000 x 1692 x 35)

= 0.046


Lever arm, z = [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


= 46.2 x 106 / (0.87 x 500 x 160)

= 664 mm2


As,pro = 113 / 0.150


Chapter 5 35


= 754 mm2

100As / bh = 100 x 754 / (1000 x 200) = 0.377


Check Shear

Max shear at the face of support

v = 66 x 103 / (1000 x 169)

= 0.39 MPa






M/(bd2) = 46.2 x 106 / (1000 x 1692)

= 1.62 N/mm2

As,req / As,pro = 664 / 754 = 0.881

fs = 2/3 x 500 x 0.881 = 294 MPa

mt = 0.55 + (477-294)/[120(0.9+1.62)] (Table 7.4)

= 0.55 + 0.605

= 1.16

Allowable L / d = 1.16 x 20 = 23.2

Actual L / d = 2800 / 169


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.


Chapter 5 36



Chapter 5 37



Chapter 5 38


│Self-Assessment Questions│

Q.1 Given the following design parameters of a one-way simply-supported slab:

Slab thickness, h : 175 mm

c/c distance btw supports : 3 300 mm

Width of the supporting beam, Sw : 400 mm (same for both ends)

Allowance for finishes : 1.5 kPa

Partition: 1.5 kPa (light weigh, undefined)

Usage of the floor : 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.

Q.2 (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:

(a) (b) (c)

Slab thickness 150mm 175mm 200mm

Steel area required (mm2/m) 338 634 1265

Bar size

Spacing (mm)

As,pro

100As/bh


Chapter 5 39


Q.4 Determine the self-weight of the following stair flight:

Tread = 225 mm

Riser = 175 mm


Q.5 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:


Cover = 35 mm


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

Q2a: (i) one-way slab with bay size > 30m2, (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

Q6: As,req = 1279mm2/m, provide T16-150, Asd = 364mm2/m, provide T10-200, comment: pay attention to L/d ratio.


Chapter 5 40


│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) Identify the essential design parameters from the drawing and

determine the design load for one span in kN per m width.

(b) 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

(c) Check if shear reinforcement is required.

(d) Check if the deflection of the slab is acceptable by span-to-depth

ratio.


Chapter 5 41


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 = 14

Flight horizontal length, L2 = 260 x 14 = 3640 mm

Flight width, W = 1250 mm

Landing slab thickness = 275 mm (same as waist)

Landing clear width, Ln = 1250 mm (same at both ends)


fcu = 40 MPa

fy = 500 MPa

Cover = 25 mm






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