User guide slab

CITY UNIVERSITY OF HONG KONG
Department of Building and Construction

Web-Based Reinforced Concrete Design (Part II): R.C. Slab Systems Design

Abstract

Reinforced concrete slabs are used in floors, roofs and walls of buildings and as the decks of

bridges. The floor system of a structure can take many forms such as in-situ solid slab, ribbed

slabs or precast units. Slabs may span in one direction or in two directions and they may be

supported on monolithic concrete beams, steel beams, walls or directly by the structure’s

columns. In this user guide, some common design methods, general assumptions and

considerations for one-way slabs and two-way slabs will be introduced. Typical demonstrations

will be shown. It will illustrate the programs with some concrete examples. The basic input

requirements and output characteristics of programs will be also introduced in this user guide.

Table of Content

2.1 Introduction to Slab Systems ....................................................................................... 4
2.1.1 Types of slab................................................................................................ ...... 4

2.1.2 Design Methods ................................ ................................ ................................ . 5

2.1.3 General design considerations and assumptions.................................................. 6

2.1.4 Resistance Moment of Solid Slabs................................................................ ...... 7

2.1.5 Resistance Moment of Solid Slabs................................................................ .... 10

2.1.6 Design algorithm.............................................................................................. 11

2.2 R.C. Slab Systems Design Examples ................................ ................................ .......... 13
2.2.1 Design for One-way Slab ................................ ................................ ................. 14

2.2.2 Design for Two-way Slab................................ ................................ ................. 19

CITY UNIVERSITY OF HONG KONG 2


Figures

Figure 1 - One-way slab on beams and girders ................................ ................................ .......... 7

Figure 2 - Design procedure for one-way slab design. ................................ ............................. 11

Figure 3 - Design procedures for two-way design................................................................ .... 12

Figure 4 - Loading input interface for slab systems design................................. ...................... 15

Figure 5 – Choose different location................................ ................................ ........................ 16

Figure 6 – Selection of distribution steel reinforcement ........................................................... 16

Figure 7 – Selection of tensile steel reinforcement................................ ................................... 17

Figure 8 - Results of moment resistance for one-way slab. ................................ ...................... 18

Figure 9 - Loading input interface for slab systems design................................. ...................... 20

Figure 10 – Choose different connection condition for two-way slab....................................... 21

Figure 11 - Selection of distribution steel reinforcement................................ .......................... 22

Figure 12 – Selection of tension steel reinforcement of resisting moment 1 ............................. 23

Figure 13 – Confirmation of reinforcements in different locations ........................................... 24

Figure 14 – View reinforcement for resisting moment in different locations ............................ 25

Figure 15 – Calculation of reinforcement for resisting moment 4................................ ............. 26

Tables

Table 1 – Comparison between one-way and two-way slab ....................................................... 4

Table 2 - Ultimate bending moment and shear forces in one-way spanning slab ........................ 7



2.1 Introduction to Slab Systems

2.1.1 Types of slab

Slabs are plate elements forming floors and roofs in buildings which normally carry uniformly
distributed loads. Slab may be simply supported or continuous over one or more supports and are
classified according to the method of support as follows:

1. Spanning on way between beams or walls
2. Spanning two ways between the support beams or walls
3. Flat slabs carried on columns and edge beams or walls with no interior beams

Slabs may be solid of uniform thickness or ribbed with ribs running in one or two directions.
Slabs with varying depth are generally not used. In this application, one-way and two-way solid
slabs are discussed.

Determination of slab type depends on the ratio of length of longer side to that of shorter side.

The comparison between one-way and two-way slab is shown in Table 1.

One-way slab Two-way slab

Symbol

ly/lx >2 ≤2

Distribution of reactions

on to supports

Table 1 – Comparison between one-way and two-way slab



2.1.2 Design Methods

Slabs may be analyzed using the following methods:-

1. Elastic methods

It covers three techniques.

a) Idealization into strips or beams spanning one way or a grid with the strips spanning two

ways.

b) Elastic plate analysis.

c) Finite element analysis. It is the best method for irregularly shaped slabs or slabs with

non-uniform loads.

2. Method of design coefficients

The moment and shear coefficients are selected from the code, which have been obtained from

yield line analysis

3. Yield line method

The yield line method is a powerful procedure for the design of slabs. It is an ultimate load

method of analysis that is based on plastic yielding of an under-reinforced concrete slab section.

For the details of the theory and yield line analysis, please refer to “MacGinley, T.J., and Choo,

B.S., Reinforced Concrete: Design Theory and Examples, 2nd edition, E & F N Spon, London,

1990.”



2.1.3 General design considerations and assumptions

1) Uniformly loaded slabs

Slabs carrying predominantly uniform load are designed on the assumption that they consist of a

series of rectangular beams 1 m wide spanning between supporting beams or walls.

2) Arrangement of loads

Although the code states that in principle the slab should be designed to resist the most

unfavorable arrangement of loads, usually it is only necessary to design for the single-load case

of maximum design load on all spans or panels. Design load = 1.4G k + 1.6Q k This is permitted

subject to the following conditions:

• The area of each bay exceeds 30 m2.

• The ratio of characteristic imposed load to characteristic dead load does not exceed 1.25.

• The characteristic imposed load does not exceed 5kN/m2 excluding partitions.

3) Shear

Shear stresses are usually low, except where are heavy concentrated loads. But in my FYP, only

uniform distributed loads (including dead load and live load) are considered so that there is not

any shear reinforcement will be considered.

4) Distribution reinforcement

The functions of distribution reinforcement are typing the slab together, distributing non-uniform

loads through slabs and taking the possible bending moments in the long span.



2.1.4 Resistance Moment of Solid Slabs

2.1.4.1 One-way Solid Slabs

The Figure 1 shows the typical one-way slab.

Figure 1 - One-way slab on beams and girders

Slabs behave primarily as flexural members and design of the cross-section is similar to beams.

Breadth is fixed since a unit value of one meter is normally used in calculations. The design

ultimate moment and shear force are given in Table 2 here.

One important note should be mentioned here is 20% redistribution is allowed when using the

table.

End support/slab connection
Simple Continuous At first Middle Interior
interior interior supports
At outer Near middle At outer Near middle
support spans
support of end span support of end
support
Moment 0 0.086FL -0.04FL 0.075FL -0.086FL 0.063FL -0.063FL

Shear 0.4F -- 0.46F -- 0.6F -- 0.5F

Note: F is the total design ultimate load ( 1.4G k + 1.6Qk )
L is the effective span
Table 2 - Ultimate bending moment and shear forces in one-way spanning slab



2.1.4.2 Two-way Solid Slabs

The design of two-way slab presents varying degrees of difficulty depending on the boundary

conditions. General, there are two types of two-way slab:

• Simply supported slabs

• Restrained slabs

When a slab is supported on all four of its sides it effectively spans in both directions, and it is

sometimes more economical to design the slab on this basis. The amount of bending in each

direction will depend on the ratio of the two spans and the conditions of restraint at each support.

Moment in each direction of span are generally calculated using coefficients which are tabulated

in the codes of practice. Areas of reinforcement to resist the moments are determined

independently for each direction of span.

2.1.4.2.1 Simply supported slabs

A slab simply supported on its four sides will deflect about both axes under load and the corners

will tend to lift and curl up from the supports, causing torsional moments. When no provision

has been made to prevent this lifting or to resist the torsion then the moment coefficients

( α sx , α sy ) may be used and the maximum moments are given by:

m sx = α sx nl x
2
in direction of span l x

m sy = α sy nl y in direction of span l y
2

Where msx and msy are the moments at mid-span on strips of unit width with spans lx (the length

of longer side) and ly (the length of shorter side) respectively. And n is the total ultimate load per

unit area: n = (1.4G k + 1.6Qk )



The value of the moment coefficients are derived from the following equations:

ly ly
( )4 ( )2
lx lx
α sx = α sy =
 ly   ly 
81 + ( ) 4  81 + ( ) 4 
 lx   lx 

The area of reinforcement in direction l x and l y respectively are

m sx m sy
Asx = Asy = (per meter width)
0.95 f y z 0.95 f y z

2.1.4.2.2 Restrained slab spanning in two direction

When the slabs have fixity at the supports and reinforcement is added to resist the maximum

moments per unit width are given by

m sx = β sx nl x in direction of span l x
2

m sy = β sy nl y in direction of span l y
2

Where β sx and β sy are the moment coefficients and n is the total ultimate load per unit area:

n = (1.4G k + 1.6Qk ) .

β y = (24 + 2 N d + 1.5 N d )1000
2
Nd is the number of discontinuous edges

2 l
γ = {3 − 18 x [ β y + β 1 + β y + β 2 ]}
9 ly

γ = β x + β3 + β x + β4

Note: β 1 and β 2 take values of 4 / 3β y for continuous edges or zero for discontinuous edges.

β 3 and β 4 take values of 4 / 3β x for continuous edges or zero for discontinuous edges.



The area of reinforcement in direction l x and l y respectively are

m sx m sy
Asx = Asy = (per meter width)
0.95 f y z 0.95 f y z

2.1.5 Resistance Moment of Solid Slabs

The theories and procedures regarding shear reinforcement design of the cross-section are

similar to beams. It would not repeat here. The maximum shear force per unit width are given by

v sx = β sx nl x in direction of span l x

v sy = β sy nl y in direction of span l y

Shear reinforcement ratio:

Asv bv (v − v c )
=
Sv 0.95 f yv

And the maximum spacing of stirrups in the direction of span is less that 0.75 times the depth of

the beam. It makes sure that at least one link intercepts a diagonal crack. The area of shear

reinforcement in slabs depends on the value of applied shear stress. For details, please refer to

BS8110: Part 1, Table 3.16 (Form and area of shear reinforcements in solid slabs).



2.1.6 Design algorithm

This section will focus on the discussion of design algorithm for slab systems design including
design for one-way and two-way slab.

2.1.6.1 One-way Slab System Design

In one-way slab design, calculation of steel reinforcement for resisting bending moment is very

similar to the beam design. In the beam design, shear links arrangement was also considered. But

in one-way slab, we assume that there is no shear link in the slab system. Actually, we still check

the shear resistance. When the shear stress is larger than the concrete shear resistance, the slab

will fail in shear.

Figure 2 - Design procedure for one-way slab design.



2.1.6.2 Two-way Slab System Design

In two-way slab analysis, the support condition will affect loading and bending factors. In order

to find the ratio, a database should be set up which contain those values. Computer will find out

the ratio from the database. After the ratio is determined, the calculation is very much similar to

the beam design. So the design procedure will follow the beam design.

Figure 3 - Design procedures for two-way design



2.2 R.C. Slab Systems Design Examples

The user guide provides a few typical examples for slab systems design. More concrete examples
with different slab types, such as one-way and two-way slab, and assumptions are available at:
http://bccw.cityu.edu.hk/rc.design/example.asp. The completed list of examples is listed as
follows.

R.C. Slab Systems Design Examples

Example Assumptions/Situations

• One-way spanning solid slab

• Continuous slab
1
• Equal spans

• Two-way spanning solid slab

• Simply supported
2
• No provision to resist torsion at the corners

• Two-way spanning solid slab

• Restrained edge
3
• Corner portion



2.2.1 Design for One-way Slab

2.2.1.1 Input

In this section, a typical example for one-way slab design is shown. Reader may follow the
detailed procedures.

Example 1
A continuous one-way slab has three equal
spans of 3.5 meter each. The slab depth is
assumed to be 140mm. The loading is as
3.5m
followings:

Dead load (including self-weight, screed, finish,
10m
partitions, ceiling) = 5.2 kN/m2
Imposed load = 3.0 kN/m2

The construction materials are Grade 30
concrete and Grade 460 reinforcement. The
conditions of exposure are mild and the cover
required is 25mm. Design the reinforcement for
the positions of near middle point end span
and middle interior span.



Detail Procedures

1. Input basic parameters, such as include loads, sections properties, etc.
2. Chick “Submit” when you finish inputting basic parameters. (See Figure 4)
3. Select different location of slab connection. (See Figure 5)
4. Select an appropriate value of distribution reinforcement. In this example, 223mm2 is
chosen. (See Figure 6)
5. Select an appropriate value of tension/compression reinforcement. In this example,
335mm2 of tension reinforcement is chosen.
6. Chick “Submit” when you finish choosing areas of reinforcement. (See Figure 7)

In this example, since the dead load includes self-weight, screed, finish, partitions and ceiling,
therefore, zero value should be inputted for the density of slab. After entering the design
parameters, click the “Submit” button to proceed to the next step – selection of slab connection.

Figure 4 - Loading input interface for slab systems design.



Figure 5 – Choose different location

Users may assign appropriate area of reinforcement by selecting from the table in the “Areas of
groups of bars” section or defining at the “User Define” section. The required and maximum
areas of tension reinforcement are shown at the top of the window as shown in Figure 6. In this
example, 182mm2 is required and 223mm2 is chosen. After selecting distribution reinforcements,
then go to selection of tension / compression reinforcements.

Figure 6 – Selection of distribution steel reinforcement



The required and maximum areas of tension reinforcement are shown at the top of the window as
shown in Figure 7. In this example, 243mm2 is required. After selecting reinforcements, then
chick the “Submit” bottom.

Figure 7 – Selection of tensile steel reinforcement



2.2.1.1 Output

The application will show the results quickly. Numerical result and graphical output can be
shown in the output part. The section properties and design loadings are displayed at the top of
reinforcement calculation. The detailed calculations, including K value, tension and compression
reinforcement and checking of shear resistance are also displayed. A typical output is shown in
Figure 8.

Figure 8 - Results of moment resistance for one-way slab.



2.2.2 Design for Two-way Slab

2.2.2.1 Input

In this section, a typical example for one-way slab design is shown. Reader may follow the
detailed procedures.

Example 2
A part floor plan for an office building measuring 6m
x 6m. (As shown in the right hand side) It consists of
restrained slabs poured monolithically with the edge
beams. The slab is 175mm thick and the loading is as
6m
follows:

Total dead load = 6.2 kN/m2
Imposed load = 2.5 kN/m2 6m

Design the corner slab using Grade 35 concrete and
Grade 460 steel reinforcement.



Detail Procedures

1. Input basic parameters, such as include loads, sections properties, etc.
2. Chick “Submit” when you finish inputting basic parameters. (See Figure 9)
3. Select different location of slab connection. (See Figure 10)
4. Select an appropriate value of distribution reinforcement. In this example, 223mm2 is
chosen. (See Figure 11)
5. Chick “Submit” when you finish choosing areas of reinforcement.
6. Select an appropriate value of tension/compression reinforcement for resisting moment
1 (m1), moment 2 (m2), moment 3 (m3), moment 4 (m4), moment 5 (m5) and moment
6 (m6). (See Figure 12)
7. Chick “Confirm” when you finish choosing areas of reinforcement for different location.

In this example, since the dead load includes self-weight, screed, finish, partitions and ceiling,
therefore, zero value should be inputted for the density of slab. After entering the design
parameters, click the “Submit” button to proceed to the next step – selection of slab connection.

Figure 9 - Loading input interface for slab systems design.



There are two types of two-way slab, which includes: simply supported slab and restrained slabs.
When considering restrained slabs, user may need to determine the continuity condition of the
four edges.

Figure 10 – Choose different connection condition for two-way slab

There are nine type of panel are considered:

• Interior panels

• One short edge discontinuous

• One long edge discontinuous

• Two adjacent edges discontinuous

• Two short edges discontinuous

• Two long edges discontinuous

• Three edges discontinuous (one long edge continuous)

• Three edges discontinuous (one short edge continuous)

• Four edges discontinuous



Selection of distribution steel reinforcement is shown in Figure 11. In this example, 251mm2 is
chosen., and then chick “Submit”.

Figure 11 - Selection of distribution steel reinforcement



Users may assign appropriate area of reinforcement by selecting from the table or defining by
themselves. The required and maximum areas of tension reinforcement are shown at the top of
the table.

Figure 12 – Selection of tension steel reinforcement of resisting moment 1



Summary will be displayed after selection of steel reinforcement, please chick “Confirm” if there
is no any mistake.

Figure 13 – Confirmation of reinforcements in different locations



2.2.2.2 Output

After selecting reinforcements in different location, designer can view the detailed calculation of
reinforcement in each location. In this case, there are six locations, they are from m1 to m6
according to Figure 14.

For example, suppose that the user want to view the calculation for resisting moment 4, than
select “Moment 4” and then chick “View Detailed Calculation”. The detail calculation is
displayed the page. See Figure 14 and Figure 15.

Figure 14 – View reinforcement for resisting moment in different locations



Figure 15 – Calculation of reinforcement for resisting moment 4


User guide slab

Recommandé

Recommandé

Contenu connexe

Tendances

Tendances (20)

En vedette

En vedette (20)

Similaire à User guide slab

Similaire à User guide slab (20)

User guide slab