CH1 Limit State Design (2014-08-01)atycnw01.vtc.edu.hk/cbe2022/R1_Limit State Design...

CON4332 REINFORCED CONCRETE DESIGN

Chapter 1 1

HD in Civil Engineering (Aug 2014)

│CHAPTER 1│

Limit State Design of Reinforced Concrete Structures

Learning Objectives

Appreciate the concept of limit state design and the application of partial factors of safety

Determine the design forces for simple structural elements

Appreciate the concept of load combination and envelope of design forces

CONTENTS

1.0 Design Code and References 1.0.1 The Design Code and Scope

1.0.2 Other References

1.1 Reinforced Concrete Structure

1.2 Limit State Design 1.2.1 Limit States

1.2.2 Design Approach

1.3 Material Strength 1.3.1 Concrete

1.3.2 Reinforcing Steel

1.3.3 Partial Safety Factor for Material Strength

1.3.4 Example – Tension Capacity of a Reinforcement Bar

1.4 Design Loads 1.4.1 Characteristic Dead Load

1.4.2 Characteristic Imposed Load

1.4.3 Design Load

1.4.4 Examples – Design Load for a Slab

1.4.5 Examples – Design Forces for a Simply-supported Beam

1.5 Load Pattern 1.5.1 Example – Loading for a 2-span Continuous Beam

1.5.2 Loading Arrangement for Design of Continuous Beam

1.5.3 Example – Design Moment and Shear Envelope

1.6 Effective Span 1.6.1 Example – Clear Span and Effective Span

1.7 Moment Redistribution


Chapter 1 2


1.0 Design Code and References

1.0.1 The Design Code and Scope

The Hong Kong Code of Practice for Structural Use of Concrete – 2013

(hereinafter called HKCP-2013) published by the Buildings Department of

HKSAR is adopted in this course.

Although HKCP-2013 covers the strength of concrete up to grade 100, for

simplicity, concrete not higher than grade 45 is adopted in this course to

illustrate the basic principles of design.

Essential design data, formulae and tables, which are useful for studying this

course, are extracted and summarized in the “Annex – R C Design

Formulae and Data”.

1.0.2 Other References

Other design codes commonly used in Hong Kong are:

(a) BS8110: 1985 and BS8110: 1997, Structural Use of Concrete – Part 1:

Code of Practice for Design and Construction.

The design rules of HKCP-2013 are quite similar to that of

BS8110: 1985.

The major change in the 1997 code is that the partial safety

factor for steel reinforcement is changed from 1.15 to 1.05.

However, HKCP-2013 retains the 1.15 partial factor of safety for

steel.

Both versions of BS8110 have been superseded by Eurocode 2

in UK and European countries.

(b) Eurocode 2 (BSEN1992-1-1:2004): Design of Concrete Structures:

General Rules and Rules for Buildings.

It is the RC design code for UK and European countries.

(c) 混凝土結構設計規範(GB50010-2010)/中華人民共和國國家標準.


Chapter 1 3


It is the code of practice for design of RC structures in Mainland

China.

(d) Structures Design Manual for Highways and Railways published by

the Highways Department of HKSAR.

It provides specific requirements on the design of bridges and

associated structures. BS5400: Part 4: Code of Practice for

Design of Concrete Bridges is adopted in this manual.

(e) Code of Practice for Dead and Imposed Loads – 2011 published by

the Buildings Department of HKSAR.

It specifies the dead loads and imposed loads for design of

buildings and street works in Hong Kong.

(f) Code of Practice for Fire Safety in Building – 2011 Part C – Fire

Resisting Construction published by the Buildings Department of

HKSAR.

It specifies the minimum size of structural elements and the

minimum concrete cover to reinforcement bars for specified fire

resisting construction of building.


Chapter 1 4


1.1 Reinforced Concrete Structure

Reinforced concrete (R.C.) is commonly referred to concrete embedded with

steel bars1. The beauty of R.C. is the perfect complementary of these two

materials. The relatively poor tensile strength and ductility of concrete are

improved by the inclusion of steel bars while the relatively vulnerable to

corrosion and fire damage of steel are protected by concrete cover,

rendering the composite one of the most versatile construction materials.

The following figure shows a typical R. C. beam and how it behaves under

loads.

Elevation of a Concrete Beam under Load

Elevation of the Reinforcement

Section X-X

X

X

Figure 1.1 – Cracking Pattern and Reinforcement Details

of an R.C. Beam

1 Reinforcement can also be provided in other forms, e.g. welded fabric wire mesh, plates, etc. This course

focuses on the design using steel bars.


Chapter 1 5


The main features in the design of reinforced concrete are:

a. Reinforcement bars (rebars) are designed to take up the tensile stress

in the structural element. Tensile stress is induced by structural

action, like direction tension, bending, shear, torsion and also by

shrinkage, temperature effect, etc. Rebars are designed and

positioned in the tension zone of the structural elements to resist the

tensile stress. Concrete is assumed not to take up any tensile

stress.

b. Rebars are used to prevent brittle failure of concrete, or, in other

words, it provides ductility to the concrete structure. It is undesirable

for a structure to collapse suddenly without excessive deformation.

The ability of a structure to undergo "plastic deformation", i.e. large

deformation without actual breakage, (i) allows the structure to

re-distribute its internal forces, (ii) dissipates the energy of the

external force and (iii) gives warning for the occupants to escape

before failure. Ductility is an important requirement in structural

design.

c. Rebars may be used to improve the compressive strength of concrete

element provided that the compression bars are adequately

restrained from buckling.

d. Rebars can be properly detailed to disperse cracks in concrete so as

to render them unnoticeable. It is usually accomplished by limiting

the minimum steel ratio and the maximum clear spacing of rebars

near the surface of concrete.

e. Rebars can improve the stiffness of concrete element, i.e. reduce

deflection. The elastic modulus of steel is much higher than that of

concrete and therefore its inclusion increases the sectional modulus

of the concrete element.

f. Rebars are used to tie concrete structural elements together to form a

robust structure so that it will not fall apart with its elements still

hanged together by rebars, when part of the structure is damaged by

accidental load.


Chapter 1 6


g. Rebars have to be adequately embedded in concrete for protection

from corrosion and fire; the alkaline concrete passivizes the corrosion

activities, and the low thermal conductivity of concrete insulates steel

from fire. Provide adequate concrete cover to rebars.

h. Rebars have to be provided with adequate length of interface with

concrete for effective transfer of stresses between steel and concrete

so that they can work together to take up the loads. In other words,

provide adequate bond length to rebars.

Key Words

Tensile stress

Ductility

Buckling

Cracks

Min steel ratio

Max clear bar spacing

Stiffness and deflection

Robustness

Concrete cover

Fire resistance

Corrosion

Bond length

(Identify the key words, which are

printed in italics, when you read

through the text of the teaching

notes.)


Chapter 1 7


1.2 Limit State Design

1.2.1 Limit States

Before the advent of limit state design, structural design is based on the

concept of permissible stress design, by which the structure is designed

such that the stresses in any parts of the structure would not exceed the

elastic limit of the materials. In other words, it aims to ensure all the

materials in the structure remain linear elastic.

However, it is found that a structure may not collapse or even can still

perform satisfactorily if certain parts of the materials in a structure have

stressed beyond the elastic limit. Hence, a more rational and realistic

assessment of the uncertainties in structural design, the Limit state design

(LSD), is advocated. It aims to ensure an acceptable probability that a

structure will perform satisfactorily during its design life. In other words, it

ensures the structure would not exceed its limit states, which are broadly

classified into two: (i) ultimate limit state (ULS) and (ii) serviceability limit

state (SLS).

Ultimate limit state (ULS) is the state when the structure collapses. It

concerns with the strength and stability of the structure.

Serviceability limit state (SLS) is the state when the structure fails to serve its

purposes. It concerns with deflection, cracking, durability, vibration, etc. of

the structure.

In design, both limit states have to be checked. For commonly

encountered building structures, the usual approach is to design for the

strength under ULS first, and then check if other limit states under SLS, e.g.

deflection and cracking, will not be exceeded.2

2 For some special structures, the most critical limit state may not be the strength under ULS. For examples,

control of crack width dominates the design of water retaining structure; deflection dominates the design of

long-span prestressed concrete girder; settlement dominates the design of footing, etc.


Chapter 1 8


1.2.2 Design Approach

Two important parameters for structural strength design are loads and

material strengths. Loads induce forces in the structure while material

strengths provide capacity for the structure to resist the forces. However,

the values of loads and strengths cannot be ascertained with definite values.

They are statistical values, and therefore their "characteristic values"

together with "safety factors" come in play in structural design.

The characteristic load (Fk) is a magnitude of load that is sufficiently larger

than the average load so that only a very low probability it will be exceeded

during the design life of the building, as illustrated in Figure 1.2 below. The

characteristic load is further multiplied by partial safety factor (f) to obtain

the design load for calculating the design forces of the structural elements.

Design Load = Characteristic Load x f

Figure 1.2 – Distribution Curve of Imposed Load

The characteristic strength of the material (fk) is a value of the strength of the

material that is sufficiently lower than the mean value so that only a small

portion of the materials in the structure is expected to fall below it, as

illustrated in Figure 1.3 below. The characteristic strength is further

reduced by partial safety factor (m) to arrive at the design strength for

Mean load,

Fm

Characteristic

load, Fk

Frequency of

occurrences

Load

Not more than 5%

of the occurrences

exceed the

characteristic load

Design Load,

f F

k


Chapter 1 9


calculating the design capacity of the member.

Design Strength = Characteristic Strength / m

Figure 1.3 – Distribution Curve of Test Results of Material Strength

Structural element is designed such that its design capacity or resistance,

which is calculated from the reduced characteristic strength of the materials,

is larger than the design forces, which is calculated from the increased

characteristic loads:

Design Capacity (fk / m) > Design Force (f Fk)

Different partial factors of safety are adopted for different types of load and

different material stresses to account for their variability and their effect on

the probability and consequence of structural failure.

The above approach is for ULS checking. On the other hands, SLS

checking in principle uses mean values instead of characteristic values and

almost always does not apply partial factor of safety (i.e. partial factor of

safety for SLS = 1.0.)

Mean

strength, fm

Characteristic

strength, fk

Frequency

of test

results

Strength

Not more than 5%

of the test results

fall below the

characteristic

strength

Design

strength, fk / m


Chapter 1 10


1.3 Material Strength

1.3.1 Concrete

Characteristic strength of concrete is:

28-day cube strength

Not more than 5% of test results will fall below it

Denoted by fcu in N/mm2 (or MPa)

For example, Grade C40 concrete, fcu = 40 N/mm2 (or 40 MPa)

For simplicity in illustrating the basic principles of design, grade of concrete

not higher than Grade C45 is adopted in this teaching material. Higher

grade concrete requires stricter design on ductility, of which the design

formulae and detailing requirements are slightly modified, that can be found

in the design code.

As the testing condition and the shape and size of test specimen for

compressive cube test is quite different from the actual effect on the concrete

in the structure, in order to accommodate the differences, the cube strength

fcu, is modified by a coefficient for deriving the design formulae in the design

code. In the UK codes and HKCP-2013, a coefficient of 0.67 is adopted

and therefore,

Compressive strength of concrete in the structural element

= 0.67fcu

1.3.2 Reinforcing Steel

Characteristic strength of steel is:

Yield strength3

Not more than 5% of test results will fall below it

Denoted by fy in N/mm2 (or MPa)

Two specified grades of steel are used in Hong Kong as given the following

3 For certain type of steel, where a yield is not present, 0.2% proof strength is adopted.


Chapter 1 11


table4:

Mild Steel High Tensile Steel

Grade 250 500B or 500C5

Specified characteristic strength, fy

(N/mm2)

250 500

Appearance Plain Ribbed

Notation6 R T

Table 1.1 – Properties of Reinforcement Bars

Preferred nominal size (mm) Nominal cross-sectional area (mm2)

(8) 50.3

10 78.5

12 113.1

16 201.1

20 314.2

25 490.9

32 804.2

40 1256.6

(50) 1963.5

Notes: Sizes in brackets are not commonly used.

For design purpose, the values of area for T12 or above are rounded to nearest 1 mm2.

Table 1.2 – Sizes of Reinforcement Bars

(Extracted from Table 2 of CS2:2012)

4 In the previous version of HKCP, there are two grades of steel, 250 and 460. HKCP-2013 retains grade 250 but

replaces 460 with 500B and 500C. Grade 250 steel is seldom used nowadays. 5 BS4449:2005 replaces grades 250 and 460 with three grades of steel, namely 500A, 500B & 500C, with the

same characteristic proof or yield strength of 500 MPa but of different levels of ductility. Grade 500B is

commonly used. Grade 500A is cold form steel with low ductility while 500C is hot rolled steel with very high

ductility. 6 The notation is according to BS4466:1989. Some engineers in Hong Kong prefer to use "Y" to denote high

yield bars. However, BS4466 has been superseded by BS8666:2005, which uses H to denote Grade 500 steel

and is further subdivided to HA, HB & HC to denote Grade 500A, 500B & 500C. It is expected the current

notation, using T and R, will be used in parallel with the new system for certain period of time.


Chapter 1 12


1.3.3 Partial Safety Factor for Material Strength

In estimating the capacity or resistance of a structural element, the

characteristic strengths of the material are reduced by the following partial

factors of safety (m):

Material/design consideration Values of m for ULS

Reinforcement 1.15

Concrete in flexure or axial load 1.50

Concrete in shear strength without shear

reinforcement

1.25

Bond strength 1.40

Others (e.g. bearing stress) >= 1.50

Table 1.3 – Values of Partial Safety Factors for

Material Strength (m) for ULS

(Extracted from Table 2.2 of HKCP-2013)

The partial factors of safety for material strength (m) are usually incorporated

in the design formulae or design table provided in the design code.

For SLS, m is generally taken as 1.0.

1.3.4 Example – Tension Capacity of a Reinforcement Bar

The design tension capacity of steel is:

Ts = (fy / m) As

= (fy / 1.15) As

= 0.87 fy As

where As = Cross-sectional area of reinforcement bars

or simply called steel area

Question A Determine the design tension capacity of 2T32 rebar.


Chapter 1 13


Solution fy = 500 MPa

Steel area, As = 2 x 804 mm2

= 1608 mm2

Design Tension Capacity = 0.87 fy As

= 0.87 x 500 x 1608 / 103

= 700 kN

Question B Determine the design tension capacity of a T16-150 rebars.

Solution (T16-150 means T16 bars at 150mm center-to-center spacing)

fy = 500 MPa

Steel area, As = 201 / 0.15

= 1340 mm2 per meter width

Design Tension Capacity = 0.87 fy As

= 0.87 x 500 x 1340 / 103

= 583 kN/m

1.4 Design Loads

In general, there are three types of load:

Dead load

Imposed load

Wind load.

There are other types of load, like water pressure, earth pressure,

construction load, etc. Details can be found in the relevant design codes.

For the purpose of this course, only dead and imposed loads are

considered.

?Q.1 – Q.3


Chapter 1 14


1.4.1 Characteristic Dead Load (Gk , gk)

Dead loads are the self-weight of permanent items, e.g. structural element

itself (or called self-weight, s/w), partitions, finishes, etc.

It is usually calculated by multiplying the nominal dimensions of the element

with the density of the materials, which is usually specified in the design

code. Examples of material density are given in the table below:

Materials Density (kN/m3)

Reinforced concrete 24.5

Cement mortar 23

Natural stone (granite) 29

Soil 20

Table 1.4 – Examples of Density of Material

(Extracted from Appendix A of the

Code of Practice for Dead and Imposed Loads – 2011)

1.4.2 Characteristic Imposed Load (Qk , qk)

Imposed load7 arises from the usage of the building. It is highly variable

and depends on the type of occupancy. It is usually specified by the

building regulations or design code. Examples are given in the table below:

7 In the previous design codes, it is called live load. So, the abbreviation, LL, is still frequently used.


Chapter 1 15


Usage qk (kPa) Qk (kN)

Domestic 2.0 2.0

Offices for general use 3.0 4.5

Department stores, shops, etc. 5.0 4.5

Table 1.5 – Examples of Imposed Load

(Extracted from Table 3.2 of the

Code of Practice for Dead and Imposed Loads – 2011)

Either the uniformly distributed load qk (kPa) or concentrated load Qk (kN)

whichever produces the most adverse effect shall be used for design.

1.4.3 Design Load (F, w)

The design load is obtained by summation of the characteristic loads

multiplied by their corresponding partial safety factors (f):

F = (f Fk)

If the structure is designed for dead and imposed loads only, the partial

safety factors for ULS are:8

Dead Load Imposed Load

Adverse effect 1.4 1.6

Beneficial effect 1.0 0

Table 1.6 – Values of Partial Safety Factors

for Load (f) for ULS

(Extracted from Table 2.1 of HKCP – 2013)

8 For simplicity, only partial safety factors for dead load and imposed load are considered in the course. There are

other partial safety factors for other loads, like wind load, earth load, fire load, etc. Details can be found in the

relevant design code.


Chapter 1 16


Therefore, the maximum design load for ULS can be expressed as:

F = 1.4Gk + 1.6Qk (in kN)

w = 1.4gk + 1.6gk (in kN/m or kN/m2)

For SLS, f is generally taken as 1.0.

1.4.4 Examples – Design Load for a Slab

Question A Determine the design load for the following slab:

Overall slab thickness, h : 175 mm

Weight of finishes : 1.5 kPa

Usage : Offices for general use

Solution

Dead Load

Finishes : 1.50 kN/m2

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

gk = 5.79 kN/m2

Imposed Load

Office for general usage: qk= 3.00 kN/m2

Design Load, w = 1.4 x 5.79 + 1.6 x 3.00

= 12.91 kN/m2

Question B Determine the design load for the following slab:

Overall slab thickness, h : 225 mm

Finishes : 20 mm granite + 25 mm cement mortar

Usage : Department stores

Other loads : Allow 1.5 kPa for movable light-weight partitions


Chapter 1 17


Solution

Dead Load

Finishes : 29 x 0.02 = 0.58 kN/m2

23 x 0.025 = 0.58 kN/m2

Self-weight : 24.5 x 0.225 = 5.51 kN/m2

gk = 6.67 kN/m2

Imposed Load

Department stores : 5.00 kN/m2

Partitions9 : 1.50 kN/m2

qk = 6.50 kN/m2

Design Load, w = 1.4 x 6.67 + 1.6 x 6.50

= 19.74 kN/m2

1.4.5 Examples – Design Forces for a Simply-Supported Beam

The design forces for a simply-supported beam are mainly mid-span

moment, M, and support shear, V. For simply-supported beam subjected to

uniformly distributed load (udl), their formulae are as follows:

Mid-span Moment, M = 0.125 F L or 0.125 w L2 or w L2

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

where

L = Effective span (in m)

F = 1.4Gk + 1.6Qk (in kN)

w = 1.4gk + 1.6gk (in kN/m or kN/m2)

9 Partitions which are permanent in nature with its construction and position indicated on the building plan are

considered as dead load. Partition, the location of which is not defined in the building plan and subject to

change during the usage of the building, shall be regarded as imposed load.

?Q.4 – Q.5


Chapter 1 18


Question A Determine the design forces for the following simply-supported beam under uniformly

distributed load (udl):

Effective Span, L = 8 000 mm

Characteristic Dead Load, gk = 29.2 kN/m

Characteristic Imposed Load, qk = 23.5 kN/m

Solution

Gk = 29.2 x 8 = 233.6 kN

Qk = 23.5 x 8 = 188.0 kN

The design load, F = 1.4 x 233.6 + 1.6 x 188.0

= 627.8 kN

Design Mid-span Moment, M = 0.125 F L

= 0.125 x 627.8 x 8

= 627.8 kN-m

Design Shear at Support, V = 0.5 F

= 0.5 x 628

= 313.9 kN

Question B Determine the design forces for the following simply-supported beam under uniformly


Effective Span, L = 9 000 mm

Overall depth of the beam, h = 750 mm

Breadth of the beam, b = 300 mm

Details of the slab supported by the beam:

Slab thickness = 160 mm

Finishes = 2.0 kPa

Imposed load = 5.0 kPa

Width of slab supported by the beam = 3 300 mm

[Refer to the beam 5B2 on the framing plan in DWG-01 attached at the end of this Chapter

for the details. It is adapted from the 2012/13 examination paper.]


Chapter 1 19


Solution

Dead Load

Finishes: 2.0 x 3.3 = 6.6 kN/m

Slab S/W: 24.5 x 0.16 x 3.3 = 12.9 kN/m

Beam S/W: 24.5 x 0.3 x (0.75-0.16) = 4.3 kN/m

gk = 23.8 kN/m

Imposed Load

5.0 x 3.3 = 16.5 kN/m

qk = 16.5 kN/m

The design load, w = 1.4 x 23.8 + 1.6 x 16.5

= 59.7 kN/m

Design Mid-span Moment, M = 0.125 w L2

= 0.125 x 59.7 x 92

= 604.5 kN-m

Design Shear at Support, V = 0.5 w L

= 0.5 x 59.7 x 9

= 268.7 kN

?Q.6 – Q.11


Chapter 1 20


1.5 Load Pattern

For a continuous beam, i.e. beam with more than one span, all spans being

loaded with maximum design ultimate load, i.e. [1.4Gk + 1.6Qk], may not

necessarily produce the most adverse bending moment for design. As

illustrated in the following example for a 2-span continuous beam, the most

critical mid-span moment occurs when only one span is maximum loaded,

i.e. [1.4Gk + 1.6Qk] and another span is minimum loaded, i.e. [1.0Gk].

1.5.1 Example – Loading for a 2-span Continuous Beam

Question Determine the mid-span design moment for the following 2-span continuous beam.

Solution Maximum design load = 1.4gk + 1.6qk

= 1.4 x 11.5 + 1.6 x 6.0

= 25.7 kN/m

Minimum design load = 1.0gk

= 1.0 x 11.5

= 11.5 kN/m

gk = 11.5 kN/m

qk= 6.0 kN/m

6 000 6 000


Chapter 1 21


Case I – All spans are maximum loaded

Case II – Only one span is maximum loaded while the other span is minimum loaded

Therefore, the design mid-span moment is 78.4 kN-m.

25.7 kN/m 25.7 kN/m

115.7 kN-m

65.0 kN-m

25.7 kN/m 11.5 kN/m

83.7 kN-m

78.4 kN-m

Moment Diagram of Case I


Chapter 1 22


1.5.2 Loading Arrangement for Design of Continuous Beam

The design code requires the following three load arrangements have to be

considered to determine the design forces, i.e. moment and shear, for

continuous beam (Cl. 5.2.5.2 of HKCP-2013):

Case I: all spans maximum loaded to obtain the maximum support reactions

Case 2: alternate spans loaded with maximum and minimum load to obtain

the maximum sagging mid-span moments

Case 3: any two adjacent spans maximum loaded and all the other spans

minimum loaded to obtain the maximum hogging support moment

Figure 1.4 – Loading Arrangement

for Design of Continuous Beam

1.5.3 Example – Design Moment and Shear Envelopes

The concept of bending moment and shear force envelopes is illustrated by

the following example.

1.4Gk+1.6Qk 1.4Gk+1.6Qk 1.4Gk+1.6Qk 1.4Gk+1.6Qk 1.4Gk+1.6Qk 1.4Gk+1.6Qk

1.4Gk+1.6Qk 1.0Gk

1.4Gk+1.6Qk 1.0Gk

1.4Gk+1.6Qk 1.0Gk

1.0Gk 1.0Gk 1.4Gk+1.6Qk 1.4Gk+1.6Qk

1.0Gk 1.0Gk


Chapter 1 23


Question Find the shear force and bending moment envelopes of the beam 1-2-3-4 simply supported

at 2 and 3 with overhang over each support as shown in the following figure.

Solution Maximum design load = 1.4gk + 1.6qk

= 1.4 x 15 + 1.6 x 23

= 57.8 kN/m (say 58 kN/m)

Minimum design load = 1.0gk

= 1.0 x 15.0

= 15.0 kN/m

The following table lists 5 possible load patterns and Figures A, B, C, D & E in the following

pages show their load patterns and their shear force and bending moment diagrams.

Load

Pattern

Span with

max. load Effects

A 2-3 Max sagging moment of span 2-3

Max shear of span 2-3

B 1-2 & 3-4 Max hogging moment of span 2-3

Max shear and moment of cantilevers 1-2 & 3-4

C All Max support reactions

D 1-2 & 2-3 Max support moment at 2 (but not controlling)

E 2-3 & 3-4 Max support moment at 3 (but not controlling)

By plotting all the shear force diagrams for different load patterns on a single drawing, we

can obtain the shear force envelope which is the outer boundary of all the shear force

diagrams, as shown in Figure F. In similar manner, we can obtain the bending moment

envelope as shown in Figure G. These envelops will then be used for reinforcement

design and detailing.

1 2 3 4

2 000 7 000 2 000

gk = 15.0 kN/m

qk = 23.0 kN/m


Chapter 1 24


Figure A - Load Pattern A

Figure B - Load Pattern B


Chapter 1 25


Figure C - Load Pattern C

Figure D - Load Pattern D


Chapter 1 26


Figure E - Load Pattern E

Figure F – Shear Envelope

Figure G – Bending Moment Envelope


Chapter 1 27


1.6 Effective Span

In the analysis of beams and slabs, the supports are idealized as a point

without width. However, in reality supports have width, so we have to

identify the effective span for analysis.10 The design code defines the

effective span as follows (Cl.5.2.1.2(b) of HKCP-2013):

Effective span, L = Ln + a1 +a2

where Ln = Clear span, i.e. distance between faces of

support

a1, a2 = lesser of h/2 or Sw/2 at each support

h = overall depth of the beam

Sw = width of the support

Except for beam seating on bearing, where the center of bearing should be

used to assess the effective span.

Elevation

h

Sw Sw

Clear Span, Ln

Effective Span, La1 = Sw/2 (if Sw < h)

a2 = h/2 (if h < Sw)

BEAM

SUPPORT 1 SUPPORT 2

Figure 1.5 – Effective Span

10 The design code allows moment reduction over supports to account for the width of the support. They are not

taken into consideration in this course. Details refer to Cl. 5.2.1 of HKCP-2013.


Chapter 1 28


1.6.1 Example – Clear Span and Effective Span

Question Determine the effective spans of the beam 5B2 and the slab 5S1 as shown in drawing

DWG-01 attached at the end of this Chapter.

Solution Beam 5B2

Clear Span Ln = 9000 – 250 – 250 = 8 500 mm

a1 = a2 = Min(750/2 or 500/2) = 250 mm

Effective Span L = Ln + a1 + a2

= 8500 + 250 + 250

= 9 000 mm

Slab 5S1

Clear Span Ln = 3300 – 150 – 150 = 3 000 mm

a1 = a2 = Min(300/2 or 160/2) = 80 mm

Effective Span L = Ln + a1 + a2

= 3 000 + 80 + 80

= 3 160 mm

1.7 Moment Redistribution

The concept of moment redistribution is illustrated with the 2-span

continuous beam in Example 1.5.1. The design moment envelope of the

beam is as follows:

?Q.12 – Q.17


Chapter 1 29


It has to note that this moment envelope is generated by linear elastic

analysis with the assumption that no part of the beam has reached the

ultimate moment of resistance, or, in other words, the materials are linear

elastic.

However, if the design moment of resistance of the beam section at the

support is only 92.6 kN-m, under load case I, the support moment will not be

able to reach 115.7 kN-m. If it can maintain at 92.6 kN-m and continue to

deform without rupture, the corresponding moment at midspan will then be

increased to 74.1 kN-m, which is still within the lower boundary of the

original envelope. The design moment envelope is then changed to as

follows:

Hence, the beam is safe to design for a reduced support moment, i.e.

115.7 kN-m

78.4 kN-m

92.6 kN-m

78.4 kN-m


Chapter 1 30


92.6kN-m instead of 115.7kN-m, as shown in the above envelope, and

therefore, the amount of reinforcement at the support is reduced. However,

it has to take note of the following condition when using redistribution11 to

design:

a. The beam section has adequate ductility

b. The crack widths are properly controlled

c. The deflection is within the limit

Cl.5.2.9 of HKCP-2013 allows moment redistribution for design and the limit

is 30%. The percentage of redistribution for the above beam is (1 -

92.5/115.7) = 20%. It is within this limit.

In the design code, the effect of redistribution is taken into account by the

following parameter:

βb = moment at the section after redistribution

[1.1]moment at the section before redistribution

For the above example, βb = 92.6/115.7 = 0.80.

11 In addition to plastic redistribution as illustrated in the example, bending moment in beam, or indeterminate

structure, will also undergo redistribution when its sections cracks, or even under service load. The section

modulus reduces when the section cracks and therefore the stiffness of the beam is no more uniform, and the

moment will then be redistributed according to the changes in stiffness along the beam. This effect is ignored

and outside the scope of this course.


Chapter 1 31



Chapter 1 32


│Key Concepts/Terms│

Ultimate Limit State and Serviceability Limit State ULS & SLS

Characteristic Dead Load and Imposed Load Gk & Qk

gk & qk

Concrete Strength and Steel Strength fcu & fy

Partial Safety Factors for Load and Materials f & m

Clear Span and Effective Span Ln & L

Design Load F & w

Design Moment and Shear M & V

Moment Redistribution βb


Chapter 1 33


│Self-Assessment Questions│

Q.1 What is the design tension capacity of a T25?

A. 123 kN

B. 213 kN

C. 245 kN

D. 107 kN

Q.2 What is the design tension capacity of 5R12?

A. 141 kN

B. 246 kN

C. 123 kN

D. 57 kN

Q.3 What is the steel area, As, of T12-200?

A. 565 mm2 /m

B. 113 mm2

C. 22.6 mm2

D. 5656 mm2

Q.4 Determine the characteristic dead load, gk, of a 150 mm thick R C slab with 1.0 kPa finishes.

A. 4.68 kN/m

B. 6.75 kN/m2

C. 6.55 kN/m2

D. 4.68 kN/m2


Chapter 1 34


Q.5 Determine the characteristic dead load, gk, of a 175mm thick R C slab with 25 mm thick

cement mortar and 450 mm thick soil on it.

A. 13.9 kPa

B. 19.4 kPa

C. 4.86 kN/m2

D. 13.9 kN/m

Q.6 Determine the design moment, M, for the following simply-supported beam under a uniformly


Effective Span = 7 000 mm

gk = 21.5 kN/m; qk = 15.2 kN/m

A. 381 kN

B. 225 kN-m

C. 3.33 x 105 kN-m

D. 333 kN-m

Q.7 Determine the design moment, M, for the following simply-supported beam under udl:

Effective Span = 7.0 m

Gk = 151 kN; Qk = 106 kN

A. 381 kN

B. 225 kN-m

C. 3.33 x 105 kN-m

D. 333 kN-m


Chapter 1 35


Q.8 Determine the design load, F, for the following simply-supported beam under udl:


gk = 21.5 kN/m; qk = 15.2 kN/m

A. 381 kN

B. 257 kN

C. 3.33 x 105 kN-m

D. 333 kN-m

Q.9 Determine the design moment, M, for the following cantilever beam under udl:


gk = 21.5 kN/m; qk = 15.2 kN/m

A. 381 kN-m

B. 257 kN-m

C. 3.33 x 105 kN-m

D. 333 kN-m


Effective cantilever span = 4 000 mm

Characteristic dead load = 15.2 kN/m

Characteristic imposed load = 10.3 kN/m

A. 302 kN-m

B. 151 kN

C. 3.03 x 105 kN-m

D. 204 kN-m


Chapter 1 36



Effective cantilever span = 3.5 m

Characteristic dead load = 53 kN

Characteristic imposed load = 35 kN

A. 456 kN-m

B. 154 kN-m

C. 228 kN-m

D. 797 kN-m

Q.12 Determine the effective span, L, of the following simply-supported beam:

Center-to-center distance between supports = 5 400 mm

Width of the supports at both ends = 300 mm

Effective depth of the beam = 395 mm


A. 5400 mm

B. 5100 mm

C. 5495 mm

D. 5550 mm

Q.13 Determine the effective span, L, of the following simply-supported beam:





A. 5400 mm

B. 4900 mm

C. 5295 mm

D. 5350 mm


Chapter 1 37


Q.14 Determine the clear span, Ln, of the following simply-supported beam:





A. 5400 mm

B. 4900 mm

C. 5235 mm

D. 5550 mm

Q.15 Determine the design moment, M, for the following simply-supported beam under udl:

Clear Span = 6 700 mm



gk = 21.5 kN/m; qk = 15.2 kN/m

A. 381 kN-m

B. 257 kN-m

C. 3.33 x 105 kN-m

D. 333 kN-m

Q.16 Determine the design shear, Vs, at the face of the support for the following simply-supported

beam under udl:

Center-to-center span = 7 000 mm

Width of supports at both ends = 300 mm


gk = 21.5 kN/m; qk = 15.2 kN/m

A. 381 kN

B. 182 kN

C. 190 kN

D. 333 kN-m


Chapter 1 38


Q.17 Determine the design shear, Vs, at the face of the support for the following simply-supported

beam under udl:

Center-to-center span = 9 500 mm

Width of supports at both ends = 400 mm


Design load, w = 73.2 kN/m

A. 381 kN

B. 311 kN

C. 348 kN

D. 333 kN

Answers:

Q1 B T: Grade 500 steel, fy = 500MPa; area of a 25 bar = 491mm2; design tension = 500 x 491 / 1.15 = 213 kN

Q2 C T: Grade 250 steel, fy = 250MPa; area of a 12 bar = 113mm2; design tension = 5 x 500 x 113 / 1.15 = 123 kN

Q3 A Area of a 12 bar = 113mm2; total area for 12 bar at 200 spacing = 113 / 0.200 = 565 mm2 per meter

Q4 D gk = 24.5 x 0.150 + 1.0 = 4.68 kN/m2

Q5 A gk = 24.5 x 0.175 + 23 x 0.025 + 20 x 0.450 = 13.9 kN/m2

Q6 D M = 0.125 x (1.4 x 21.5 + 1.6 x 15.2) x 7.02 = 333 kN-m

Q7 D M = 0.125 x (1.4 x 151 + 1.6 x 106) x 7 = 333 kN-m

Q8 A F = (1.4 x 21.5 + 1.6 x 15.2) x 7 = 382 kN

Q9 A; M = 0.5 x (1.4 x 21.5 + 1.6 x 15.2) x 3.7402 = 381 kN-m

Q10 A M = 0.5 x (1.4 x 15.2 + 1.6 x 10.3) x 4.02 = 302 kN-m

Q11 C M = 0.5 x (1.4 x 53 + 1.6 x 35) x 3.5 = 228 kN-m

Q12 A As Sw = 300 < h = 450, Sw controls, and therefore, L = c/c distance btw supports = 5400 mm

Q13 D As h = 450 < Sw = 500, h controls, and therefore, L = 5400 – 500 + 450 = 5350 mm

Q14 B Ln = 5400 – 500 = 4900 mm

Q15 D L = 6700 + 300 = 7000mm; M = 0.125 x (1.4 x 21.5 + 1.6 x 15.2) x 72 = 333 kN-m

Q16 B Ln = 7000 – 300 = 6700mm, Vs = 0.5 x (1.4 x 21.5 + 1.6 x 15.2) x 6.7 = 182 kN

Q17 D Ln = 9500 – 400 = 9100mm, Vs = 0.5 x 73.2 x 9.1 = 333 kN


Chapter 1 39


│Tutorial Questions│

[Present your answers with detailed working steps in a neat, tidy and logical manner.]

AQ1 For the slab 5S1 shown on DWG-01:

(a) Determine the effective span for the slab.

(b) Determine the design udl load in kPa.

(c) Determine the total design load in kN per m width of the slab.

AQ2 Determine the design forces for the beam 5B2 shown in DWG-01 with the

following changes:

i. The center-to-center distance between adjacent 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.

The other design parameters remain unchanged.

(Reference: Question B of Example 1.4.5)

AQ3 Figure AQ3 shows a three-span continuous beam subjected to uniformly

distributed characteristic dead load (Gk) and characteristic imposed load (Qk).

Sketch the load patterns for obtaining:

(a) Maximum span moment of BC.

(b) Maximum support moment at B.

(c) Maximum span moments of span AB & CD.

A B C D

Figure AQ3

CH1 Limit State Design (2014-08-01)atycnw01.vtc.edu.hk/cbe2022/R1_Limit State Design...

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