STRUCTURAL DESIGN OF A FOUR-STOREY RESIDENTIAL BUILDING AT THE

TECHNICAL UNIVERSITY OF MOMBASA.

OMUCHUMA LLOYD TIMOTHY,

BTCE/208J/2019.

SUPERVISOR
ENG. JOHN CHEGE

TECHNICAL UNIVERSITY OF MOMBASA

DECEMBER 2023

Project submitted to the department of Building and Civil Engineering in the school of
Engineering and Technology in partial fulfillment for the award of a degree of
Bachelor of technology in Civil Engineering at the Technical University of
Mombasa.

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DECLARATION

This project proposal is entirely my own original work and has not been presented for a degree award at
any other university, in whole or in part. It is not to be replicated without my permission or that of the
Technical University of Mombasa.

OMUCHUMA LLOYD TIMOTHY BTCE/208J/2019

Sign………………………… Date………………….

This research proposal has been submitted with approval from our university supervisor;

Supervisor: ……………………………

Signature…………………. Date………………………

Department Building and Civil Engineering

Technical University of Mombasa.

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DEDICATION

I dedicate this to my loving parents Phanuel Omuchuma Webi and Jane Nekesa Nyongesa for funding my
project and whole heartedly helping me conduct and execute it through constant encouragement. I would
also like to thank my lecturer as well supervisor Mr. Chege and colleagues for time spent consultations
advice while compiling this proposal and more importantly skills gained in the general field of engineering
practice.

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ACKNOWLEDGEMENT

I wish to acknowledge my lecturers and colleagues for their great support during the preparation of this
research proposal. May God bless you abundantly for the support, interaction and making this project a
success.

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ABSTRACT

According to the 2009 - 2019 census statistics, there has been an annual population change of 2.5% in
Mombasa County. It is the second largest city in the country and considerably fast-growing county due to
rural to urban migration and search for job opportunities being a commercial hub and gateway to regional
market. These factors have contributed to an increase in population and a substantial number of those
population live within the suburbs of Mombasa city thus increasing demand for modern residential houses.
Most of the suburb's parts of Mombasa County are covered with traditional Swahili flat houses that has a
vast variety of limitations to meeting housing requirement i.e., security, space, accessibility, drainage and
aesthetics etc.

This project attempts to address this demand in residential housing design by proposing construction of a
four-storey building mainly targeting the newly employed people and those relocating from either rural area
for job opportunities or those from other urban centers. This is by optimizing the quality, functionality, and
affordability of residential housing while considering social, economic, and environmental factors. Through
an in-depth analysis of the construction, energy efficiency, space utilization, accessibility and urban
planning, the project aims to contribute to the advancement of residential housing practices resulting in
improved living standards and a sustainable future.

The proposed site is Technical University of Mombasa, Mombasa. The project report involves analysis of
the architectural drawing, structural analysis and design carried out in accordance with the standard codes
of practice specifically BS Codes. Calculations were done manually, structural drawing were done with the
aid of Auto Cad 2016 and Archi cad for architectural drawing.

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Table of contents
1.0 INTRODUCTION ..................................................................................................................... 7
1.1 Background .................................................................................................................................. 7
1.2 Problem Statement ...................................................................................................................... 7
1.3 Objectives ..................................................................................................................................... 8
1.4 Scope............................................................................................................................................. 8
2.0 Literature Review ........................................................................................................................... 8
2.1 Introduction................................................................................................................................ 8
2.2 Structural design ........................................................................................................................... 9
2.2.1 Limit State Design ................................................................................................................. 10
2.2.2 DESIGN LOADS...................................................................................................................... 11
2.3 Detailing of structural members ................................................................................................. 12
3.0 Research Methodology................................................................................................................. 13
3.1 Introduction.............................................................................................................................. 13
3.2 Reconnaissance study............................................................................................................... 13
3.3 Preparation of the architectural drawing ................................................................................. 14
3.4 Structural drawing .................................................................................................................... 14
3.5 Design loads ............................................................................................................................. 14
3.6 Analysis of the structure ........................................................................................................... 14
4.0 DESIGN AND ANALYSIS OF STRUCTURAL ELEMENTS ....................................................................... 14
4.1 SLAB .......................................................................................................................................... 16
4.2 Beams ......................................................................................................................................... 27
4.3 Column Designs ......................................................................................................................... 135
4.3.1 COLUMN SCHEDULE ........................................................................................................... 161
4.4 Foundations Bases .................................................................................................................... 161
4.5 Stair case ................................................................................................................................... 183
4.6 Waste water treatment plant ................................................................................................... 185
4.6.1 DESIGN OF THE SEPTIC TANK .............................................................................................. 186
4.6.2 Reinforced Concrete design for tank slab ........................................................................... 187
5.0 Lab work:..................................................................................................................................... 189
5.1 SOIL TEST .............................................................................................................................. 189
The Shear Box Test. .......................................................................................................................... 205
5.2 Design the concrete mix ............................................................................................................ 212

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6.0 unpriced BQ & specifications for substructure work ..................................................................... 227
6.1 Taking OFF ................................................................................................................................. 229
6.2 Abstracting ................................................................................................................................ 233
6.3 Billing ........................................................................................................................................ 237
7.0 Work plan and Budget .................................................................................................................. 242
7.1 Workplan .................................................................................................................................. 242
7.2 Budget ....................................................................................................................................... 242

1.0 INTRODUCTION

1.1 Background
The basic needs of human existence are food, clothing's and shelter. From times immemorial, man has
been making efforts in improving their standard of living. The point of his effort has been to provide an
economic and efficient shelter. The possession of shelter besides being a basic need, used, gives a feeling
of security, responsibility and shown the social status of man.

Every human being has an inherent liking for a peaceful environment needed for his pleasant living, this
objective is achieved by having a place of living situated at the safe and convenient location, such a place
for comfortable and pleasant living requires considerations and kept in view.

• A peaceful environment
• Safety from all-natural source and climate conditions
• General facilities for the community of his residential area

The engineer has to keep in mind the municipal conditions, building bye laws, environment, financial
capacity, water supply, sewerage arrangement, provision of future, aeration, ventilation etc. in suggestion
a particular type of plan to any client.

1.2 Problem Statement

Due to the rapid growth in population and urbanization trends in Mombasa city, there is a significant
shortage of affordable and suitable housing options to accommodate the increasing demand. This shortage
has led to various issues including overcrowded living conditions, increased housing costs and inadequate
access to quality housing for individuals and families. As a result, there is an urgent need to design and

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construct a residential structure that effectively addresses the housing demand, provides sustainable and
affordable housing solutions and contributes to the overall development and well-being of the community.

1.3 Objectives
General
The aim of this project is to plan and design the framed structure of a residential building.
Specific
• Identification of the architectural and structural arrangement of the plan
• Design of structural members
• Detailed structural analysis of the building structural components manually and with the help of Auto
Cad software
• Better acquaintance with the code provisions for reinforced concrete design of structures according to
(BS 8110 Part 1)

1.4 Scope
This project was done in wide view of the different activities required to facilitate the structural design of
structures according to the British design standard manual and in this case a residential apartment is
designed. It entailed the following activities:

• Survey – transfer of levels to site (Technical University of Mombasa)

• Generation of architectural drawings.

• Analysis of the structural members both manually with the aid of relevant software and producing
structural drawings.

• Lab tests - Soil lab tests as well as Concrete mix design.

2.0 Literature Review

2.1 Introduction
Civil engineering design process generally has two design stages; Initial and Final design involving
different design parameters that govern all the design stages. (ISE/ ICE)

a) Initial Design

In the initial stages of the design of building structures it is necessary, often at short notice to produce
alternative schemes that can be assessed for architectural and functional suitability and which can be
compared for cost.

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Usually based on vague and limited information on matter affecting the structure such as imposed loads
and nature of finishes, let alone firm dimensions but it is nevertheless expected that viable schemes be
produced on which reliable cost estimates can be based. It follows that initial design methods should be
simple, quick, conservative and reliable while avoiding analytical methods.

b) Final Design

It entails the actual design from the initial design carried out by Marshall all the information in it for use
in final design in the following sequence:

• Checking of all information

• Preparation of a list of design data
• Amendment of drawings as a basis for final calculations

It is carried in logical sequence according to BS8110 code except in situations when this sequence cannot
be adhered to. The following order is mostly used:

i. Slabs
ii. Structural frames
iii. Beams
iv. Columns
v. Walls
vi. Staircases
vii. Retaining wall, basements
viii. Foundations
ix. Robustness and Detailing

2.2 Structural design

Guidance on the design of building and civil engineering structures is given in various British Standards
and Codes of practice. These play an important role in the provision of structural designs which are both
safe and economic and which comply with the building regulations and other statutory requirements.
(Draycot)

The design is generally to limit state theory in accordance with BS8110:1985: Structural Use of Concrete
Part 1: Code of Practice for Design and Construction.

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The design of sections for strength is according to plastic theory based on behavior at ultimate loads.
Elastic analysis of sections is also covered because this is used in calculations for deflections and crack
width in accordance with BS8110:1985: Structural Use of Concrete Part 2: Code of Practice for Special
Circumstances. (Chod/Mac Ginley)

2.2.1 Limit State Design

Aims and methods of design

BS 8110 – 1:1997 Clause 2.1.1 of the code states that the aim of design is the achievement of an
acceptable probability that the structure will perform satisfactorily during its life. It must carry the loads
safely, not deform excessively and have adequate durability and resistance to effects of misuse and fire.
The clause recognizes that no structure can be made completely safe and that it is only possible to reduce
the probability of failure to an acceptably low level.
Clause 2.1.2 states that the method recommended in the code is limit state design where account is taken
of theory, experiment and experience. It adds that calculations alone are not sufficient to produce a safe,
serviceable and durable structure. Correct selection of materials, quality control and supervision of
construction are equally important.

The criterion for a safe design is that the structure should not become unfit for use, i.e., that it should not
reach a limit state during its design life. This is achieved, in particular, by designing the structure to ensure
that it does not reach: (Chod/Mac Ginley)

1) The ultimate limit state—the whole structure or its elements should not collapse,
overturn or buckle when subjected to the design loads. It focuses on strength, stability
and robustness of the structure.
2) serviceability limit states—the structure should not become unfit for use due to
excessive deflection, cracking or vibration.

The serviceability limit states are discussed in BS8110: Part 1, section 2.2.3. The code states that account
is to be taken of temperature, creep, shrinkage, sway and settlement. The main serviceability limit states
and code provisions are as follows.

a) Deflection
b) Cracking

The loading on structures conforms to BS6399:1984: Design Loading for Building

Part 1: Code of Practice for Dead and Imposed Loads CP3:1972:

Chapter V: Loading

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2.2.2 DESIGN LOADS
The characteristic or service loads are the actual loads that the structure is designed to carry. These are
normally thought of as the maximum loads which will not be exceeded during the life of the structure.

In statistical terms the characteristic loads have a 95% probability of not being exceeded. The
characteristic loads used in design and defined in BS8110: Part 1, clause 2.4.1, are as follows:
(Chod/Mac Ginley)

1) The characteristic dead load Gk is the self-weight of the structure and the weight of finishes,
ceilings, services and partitions;

2) The characteristic imposed load Qk is caused by people, furniture, equipment etc. on floors and
snow on roofs. Imposed loads for various types of buildings are given in BS6399: Part 1;

3) The wind load Wk depends on the location, shape and dimensions of the buildings. Wind loads
are estimated using CP3: Chapter V: Part 2. The code states that nominal earth loads are to be
obtained in accordance with normal practice. Reference should be made to BS8004:1986: Code
of Practice for Foundations and textbooks on geotechnics.

Load combination Load type

(including earth and water
loading where present)

Dead load Imposed load Earth and Wind

water
Adverse Beneficial Adverse Beneficial
1. Dead and imposed 1.4 1.0 1.6 0 1.4 —

2. Dead and wind 1.4 1.0 — — 1.4 1.4

3. Dead, wind and

1.2 1.2 1.2 1.2 1.2 1.2
imposed

Table 2.1 Load combination

The structure must also be able to resist the notional horizontal loads defined in clause 3.1.4.2 of the code

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Design load characteristic load x partial safety factor for loads =Fkyf The partial safety factor yf takes
account of
➢ possible increases in load
➢ inaccurate assessment of the effects of loads
➢ unforeseen stress distributions in members
➢ the importance of the limit state being considered

2.3 Detailing of structural members

Slabs

Concrete slabs are plate structures that are reinforced to span either in one way or both directions of a
structural bay. They are classified according to their method of spanning and the form in which they are
cast. Because of their no combustibility, concrete slabs can be used in all types of construction. (Ching)

Beams
Reinforced concrete beams are designed to act together with longitudinal and web reinforcement in
resisting applied forces. Cast-in-situ place concrete beams are almost always formed and placed along
with the slab they support. Because a portion of the slab acts as an integral part of the beam, the depth of
the beam is measured to the top of the slab. (Ching)

Columns

Column are rigid, relatively slender structural members designed primarily to support axial compressive
loads applied to the ends of the members. Relatively short, thick columns are subject to failure by
crushing rather than by buckling. Failure occurs when the direct stress from an axial load exceeds the
compressive strength of the material available in the cross section.

Long, slender columns are a subject to failure by buckling rather than by crushing. As opposed to
bending, buckling is the sudden lateral or torsional instability of a slender structural member induced by
the action of an axial load before the elastic limit of the material is reached.

Under a buckling load, a column begins to deflect laterally and cannot generate the internal forces
necessary to restore its original linear condition. Any additional loading would cause the column to
deflect further until collapse occurs in bending. The higher the slenderness ratio of a column, the lower is
the critical stress that will cause it to buckle. A primary objective in the design of a column is to reduce its
slenderness ratio by shortening its effective length or maximizing the radius of gyration of its cross
section.

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Intermediate columns have a mode of failure between that of a short column and a long column, often
partly inelastic by crushing and partly elastic by buckling. (Ching)

Walls

Walls are the vertical enclosures for building frames. They are not usually or necessarily made of
concrete but of any material that aesthetically fulfills the form and functional needs of the structural
system. Additionally structural concrete walls are often necessary as foundation walls, stairwell walls, and
shear walls that resist horizontal wind loads and earthquake induced loads. (G. Nawy)

Foundation

Foundation are the structural concrete elements that transmit the weight of the super structure to the
supporting soil. They could be in many forms simplest being isolated footing. It can be viewed as an
inverted slab transmitting a distributed load from the soil to the column. Other forms of foundations are
piles driven to rock, combined footing supporting more than one column.

Robustness and detailing

All members of the structure should be effectively tied together in the longitudinal, transverse and vertical
directions. A well-designed and well-detailed cast-in situ structure will normally satisfy the detailed tying
requirements.
Elements whose failure would cause collapse of more than a limited part of the structure adjacent to them
should be avoided. Where this is not possible, alternative load paths should be identified or the element in
question strengthened. (ISE/ICE)

3.0 Research Methodology

3.1 Introduction
This chapter involves the methods that were undertaken, tools and procedures used to come up with the
design of the four-storey residential structure at Technical University of Mombasa.

3.2 Reconnaissance study

Assessment study was carried out to know whether the proposed project would fit in the proposed area
(Technical University of Mombasa upper pitch). This was done by simple observation using the naked
eye. The proposed building dimensions were considered to ensure the proposed plan can fit in the given
space given that the prosed site is large enough to accommodate the construction of other similar
structures on the same piece of land.

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3.3 Preparation of the architectural drawing
Architectural drawings were prepared using Arch 24, a platform from which the plans, elevations and
pictorial views were prepared. Arch 24 was chosen due to the presence of latest features in it compared to
the previous versions. Rendering was done using same software also.

3.4 Structural drawing

The structural layout was prepared using Auto Cad where the structural elements were allocated in
different positions on the structural plan with grid lines used to aid easy identification. The reinforcement
bars and the cross sections of the structural elements were also produced

3.5 Design loads

Two types of design loads will be considered as follows;
i) Dead load
Dead loads for structural members were assessed based on the forces due to:
1. Weight of the member itself
2. Weight of all materials of construction incorporated into the building to be supported permanently
by the member.
3. Weight of permanent partitions and weight of fixed service equipment.

Dead loads were calculated from the unit weights given in BS6399 or from actual known weights of the
materials used. Dead loads of beams, column and bases were calculated depending on their sectional
areas.

ii) Live load

3.6 Analysis of the structure

The method of moment distribution using substitute frame analysis which involves distributing the fixed
end moments to the beams and column at the end was used to analyze the structure.

4.0 DESIGN AND ANALYSIS OF STRUCTURAL ELEMENTS

1 Multistorey Building in concrete:
(i) Location: TUM main campus
(ii) Materials: concrete 𝑓𝑐𝑢 = 40 N/mm2, steel 𝑓𝑦 = 460 N/mm2, cement: 42.5 N/mm2

(iii) Survey: Transfer levels to the site

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Coordinates

2 Architecture: (see architectural drawing)

(i) Ground floor plan, typical upper floor plan (scale: 1: 100)
(ii) Elevations

3 Engineers drawing: (see structural drawing)

(i) Ground floor plan, typical upper floor plan (scale: 1: 50)
(ii) design elements (beams, cols)
4 Structural Design:
(i) Slab,
(ii) Beams,
(iii) Columns,
(iv) Foundations,
(v) Stair case,
(vi) Waste water treatment system.

4.1 SLAB

Introduction

BS 648: 1964 weights of building materials

BS 6399 – 1: 1996 imposed and dead load

Tiling, floor Asphalt 0.125 in (3.2 mm) thick 1.3 lb/ft2 (6.3 kg/𝑚 2)

Clay 0.5 in (12.7 mm) thick 5.6 lb/ft2 (27.3 kg/𝑚 2 )

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0.875 in (22.2 mm) thick 9.2 lb/ft2 (44.9 kg/𝑚 2 )

Exposure conditions:
(i) Internal - mild,
(ii) External – moderate.
Fire resistance: 1 hr (Kenya Building Code pg. 37)
Dead loads:
Slab weight – 0.180 × 24 = 4.32 kN/m2
Finishes – 0.025 × 24 = 0.6 kN/m2
partitions - 1.0 kN/m2
Imposed load: 1.5 kN/m2,
Design load = 1.4gk + 1.6qk = 1.4(4.32+0.6+1.0) × (1.6×1.5) =10.688 kN/m2,
Characteristic strengths: 𝑓𝑐𝑢 = 40 N/m2, 𝑓𝑦 = 460 N/m2

Slab (A1, B1-A2, B2)

S/No Calculations Comments

Lx = 5150 mm

Ly= 8250 mm

fcu = 40N/mm2 , fy = 460 N/mm2,

n = 8.87 kN/m2, and initial trial thickness = 180 mm.

ly 8250
= = 1.60 < 2.0
lx 5150

Therefore two- way design.

Ultimate moments:
BS 8110 – Supports:
1:1997 Table From Table 3.14:
3.14 bending
moment sx = 0.087 and sy = 0.045
coefficients
Msx = sx nl2x = 0.087 × 10.688 × 5.152 = 24.66 kNm Msx = 24.66kNm

Msy = sy nl2x = 0.045 × 10.688 × 5.152 = 12.76 kNm

Msy = 12.76kNm

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 Mid-spans:
From Table 3.14:

sx = 0.047 and sy = 0.034

Msx = sx nl2x = 0.065 × 10.688 × 5.152 = 18.43 kNm Msx=18.43 kNm

Msy = 9.64kNm
Msy = sy nl2x = 0.034 × 10.688 × 5.152 = 9.64kNm

S/No Calculations Comments

Main reinforcement:
BS 8110 – Assume concrete cover = 20 mm (otherwise conc. cover =
1:1997 Clause obtained from Table 3.1)
3.3.7 control Therefore, effective depth Main, d = 155
cover 1 1
d = 180 − 20 − 2 (bar dia. )=180 − 20 − 2 (10) =
155 mm
Secondary reinforcement
secondary
1 d = 155
d = 180 − 20 − (bar dia. ) − bar dia=180 − 20 − 10 −
2
1
(10) = 145 mm
2

Supports:
Shorter span:
M 24.66 (106 )
K= = = 0.026
fcu bd2 (40)(1000)(155)2

From Table 4.6-1,

z
= 0.94
d

M 24.66 (106 )
Therefore As = =
0.87fy z (0.87)(460)(0.94)(155)
= 422.92 mm2 ⁄m

From Table 3.25:

0.13Ac (0.13)(1000)(180)
As(min) = = = 234 mm2 ⁄m Provide B1
100 100 As = 449 mm2⁄m

≤ 422.92 mm2 ⁄m
Provide Y10 −
Therefore, As = 422.92 mm2 ⁄m 175 − T1 (As =
449 mm2 ⁄m)
Therefore, Provide Y10 − 175 − T1 (As = 449 mm2 ⁄m)

S/No Calculations comments

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 Supports:
longer span:
M 12.76(106 ) K = 0.015
K= = = 0.015
fcu bd2 (40)(1000)(145)2
From Table 4.6-1:
z
= 0.94
d

M 12.76(106 )
Therefore As = =
0.87fy z (0.87)(460)(0.94)(145)
= 194.14 mm2 ⁄m

As(min) = 234 mm2 ⁄m ≤ 233.93 mm2 ⁄m

Provide Y8 −
Therefore, As = 234 mm2 ⁄m 200 − T1 (As =
252 mm2 ⁄m)
Therefore, Provide Y8 − 200 − T1 (As = 252 mm2 ⁄m)

Mid-span
Shorter span:
M 18.43(106 ) K = 0.019
K= = = 0.019
fcu bd2 (40)(1000)(155)2
From Table 4.6-1:
z
= 0.94
d

M 18.43(106 )
Therefore As = =
0.87fy z (0.87)(460)(0.94)(155)
= 316.07 mm2 ⁄m

As(min) = 234 mm2 ⁄m ≤ 316.07 mm2 ⁄m Provide Y10 −

225 − B1 (As =
Therefore, As = 316.07 mm2 ⁄m 349 mm2⁄m)

Therefore, Provide Y10 − 225 − B1 (As = 349 mm2 ⁄m)

S/No Calculations comments

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 Mid-span
longer span:
M 9.64 (106 ) K = 0.011
K= = = 0.011
fcu bd2 (40)(1000)(145)2
From Table 4.6-1:
z
= 0.94
d

M 9.64(106 )
Therefore As = =
0.87fy z (0.87)(460)(0.94)(145)
= 176.73 mm2 ⁄m

As(min) = 234 mm2 ⁄m ≤ 176.73 mm2 ⁄m

Provide Y8 − 200 −
Therefore, As = 234 mm2 ⁄m
B2 (As =
252 mm2 ⁄m)
Therefore, Provide Y8 − 200 − B2 (As = 252 mm2 ⁄m)

Distribution steel at supports:

ProvideY8 − 250 −
As(min) = 234 mm m, 2⁄ T2 (As = 252 mm2⁄m)
Therefore, ProvideY8 − 250 − T2 (As = 252 mm2 ⁄m)

S/NO. CALCULATIONS REMARK

BS 8110 – Shear:
1:1997 From Table 3.16 (BS 8110):
Table 3.15 vx = 0.57, vy = 0.40
shear force
coefficients : . Vsx = vx nlx = (0.57)(10.688)(5.15) = 31.37 kN
and Vsy = vy nlx = (0.40)(10.688)(5.15) = 22.02 kN

vx = 0.38, vy = 0.26

Vsx =20.92 kN
Vsx = vx nlx = (0.38)(10.688)(5.15) = 20.92 kN Vsy =14.31 kN
and Vsy = vy nlx = (0.26)(10.688)(5.15) = 14.31 kN

𝐒𝐡𝐨𝐫𝐭𝐞𝐫 𝐬𝐩𝐚𝐧 (𝐥𝐱 ):

V (31.37)(1000)
design shear v = = = 0.202 N⁄mm2 design shear v
dbv 155(1000) = 0.202 N⁄mm2
vmax = 0.8√fcu = (0.8)√40 = 5.1 N⁄mm2 or 5.0 N⁄mm2

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 whichever is less.
:. vmax = 5.0 N⁄mm2
Design shear v = 0.202 N⁄mm2 < 5.0 N⁄mm2
:. Section is adequate in shear.

100As 100(449)
bv d
= 155(1000) = 0.29%
From Table 6.4-1:
Design concrete shear stress vc Design concrete
0.73 − 0.57 shear stress vc
=( ) (0.29 − 0.25) + 0.6
0.5 − 0.25 = 0.6256 N⁄mm2
= 0.6256 N⁄mm2 > 0.202 N⁄mm2
:. Shear resistance O.K.

𝐥𝐨𝐧𝐠𝐞𝐫 𝐬𝐩𝐚𝐧 (𝐥𝐲 ):

V (22.02 )(1000) design shear v

design shear v = = = 0.15 N⁄mm2 = 0.15 N⁄mm2
dbv 145(1000)

100A s 100(349)
= = 0.24%
bv d 145(1000)

From Table 6.4-1:

Design concrete shear stress vc
0.60 − 0.50
=( ) (0.24 − 0.15) + 0.5
0.25 − 0.15
= 0.536 N⁄mm2 > 0.15 N⁄mm2
:. Shear resistance O.K.

S/NO CALCULATIONS REMARKS

BS 8110 – Deflection: Check
1:1997 From table 5.3-1: deflection in
Clause 3.5.7 Basic span/ depth ratio = 26 the shorter
deflection M shorter span moment 24.66(106 ) span only
= = = 1.03 N⁄mm2
bd2 bd2 1000 (155)2
From Table 5.3-2:
Tension steel modification factor
(1.38 − 1.21)
= (1.5 − 1.03) + 1.38 = 1.54
1.5 − 1.0
: . Allowable span⁄depth ratio = 26(1.54) = 40.03
The actual span⁄depth ratio = 5150 ⁄155 = 33.23 < 40.03
:. Deflection O.K.
BS 8110 – Cracking: Widest
1:1997 max. bar spacing = 3d = (3)(155) = 465 mm spacing
Clause 3.5.8 Clear space between bars = 275 – bar diameter between bars
crack = 275 − 10 = 265 mm < 465 mm = 265 mm
control :. Cracking O.K.

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Slab (A3, B3-A4, B4)

S/No Calculations Comments

Lx = 4200 mm

Ly= 8250 mm

fcu = 40N/mm2 , fy = 460 N/mm2,

n = 8.66 kN/m2, and initial trial thickness = 180 mm.
ly 8250
= = 1.96 < 2.0
lx 4200

Therefore two- way design.

Ultimate moments:
Supports:
BS 8110 – From Table 3.14:
1:1997
Table 3.14 sx = 0.067 and sy = 0.037
bending Msx
moment Msx = sx nl2x = 0.067 × 10.688 × 4.22 = 12.63 kNm = 12.63 kNm
coefficients
Msy = sy nl2x = 0.037 × 10.688 × 4.22 = 6.98 kNm Msy =6.98 kNm

Mid-spans:
From Table 3.14:

sx = 0.036 and sy = 0.028

Msx = sx nl2x = 0.050 × 10.688 × 4.22 = 9.43 kNm

Msx =9.43 kNm

Msy = sy nl2x = 0.028 × 10.688 × 4.22 = 5.28kNm Msy =5.28kNm

S/No Calculations Comments

Main reinforcement:

BS 8110 Assume concrete cover = 20 mm (otherwise conc. cover = obtained effective depth
– 1:1997 from Table 3.1)
Clause Therefore, effective depth Main d, =
3.3.7 1 1 155mm
control d = 180 − 20 − 2 (bar dia. )=180 − 20 − 2 (10) = 155 mm
cover

22 | P a g e
 Secondary reinforcement
1 1 Main d, =
d = 180 − 20 − 2 (bar dia. ) − bar dia=180 − 20 − 10 − 2 (10) = 145mm
145 mm

Supports:
Shorter span:
M 12.63 (106 ) K = 0.013
K= = = 0.013
fcu bd2 (40)(1000)(155)2

From Table 4.6-1,

z
= 0.94
d

M 12.63 (106 )
Therefore As = =
0.87fy z (0.87)(460)(0.94)(155)
= 216.60 mm2 ⁄m

From Table 3.25:

0.13Ac (0.13)(1000)(180)
As(min) = = = 234 mm2 ⁄m
100 100
≤ 216.60 mm2 ⁄m Provide 𝐘𝟖 −
𝟐𝟎𝟎 −
Therefore, As = 234 mm2 ⁄m 𝐓𝟏 (𝐀 𝐬 =
𝟐𝟓𝟐 𝐦𝐦𝟐 ⁄𝐦)
Therefore, Provide 𝐘𝟖 − 𝟐𝟎𝟎 − 𝐓𝟏 (𝐀𝐬 = 𝟐𝟓𝟐 𝐦𝐦𝟐⁄𝐦)

S/No Calculations comments

23 | P a g e
 Supports:
longer span:
M 6.98(106 )
K= = = 0.0083 K = 0.0083
fcu bd2 (40)(1000)(145)2
From Table 4.6-1:
z
= 0.94
d

M 6.98(106 )
Therefore As = =
0.87fy z (0.87)(460)(0.94)(145)
= 127.96 mm2 ⁄m

As(min) = 234 mm2 ⁄m ≤ 127.96 mm2 ⁄m

Provide Y8 −
Therefore, As = 234 mm2 ⁄m 200 −
T1 (As =
Therefore, Provide Y8 − 200 − T1 (As = 252 mm2 ⁄m) 252 mm2⁄m)

Mid-span
Shorter span:
M 9.43(106 ) K = 0.00981
K= = = 0.00981
fcu bd2 (40)(1000)(155)2
From Table 4.6-1:
z
= 0.94
d

M 9.43(106 )
Therefore As = =
0.87fy z (0.87)(460)(0.94)(155)
= 161.72 mm2 ⁄m

As(min) = 234 mm2 ⁄m ≤ 161.72 mm2 ⁄m Provide Y8 −

200 −
Therefore, As = 234 mm2 ⁄m B1 (As =
252 mm2⁄m)
Therefore, Provide Y8 − 200 − B1 (As = 252 mm2 ⁄m)

S/No Calculations comments

24 | P a g e
 Mid-span
longer span:
M 5.28 (106 ) K = 0.0063
K= = = 0.0063
fcu bd2 (40)(1000)(145)2
From Table 4.6-1:
z
= 0.94
d

M 5.28(106 )
Therefore As = = = 96.80 mm2 ⁄m
0.87fy z (0.87)(460)(0.94)(145)

As(min) = 234 mm2 ⁄m ≤ 96.80 mm2 ⁄m

Therefore, As = 234 mm2 ⁄m Provide Y8 − 200 −

B2 (As =
Therefore, Provide Y8 − 200 − B2 (A s = 252 mm2 ⁄m) 252 mm2⁄m)

Distribution steel at supports: ProvideY8 −

250 − T2 (As =
As(min) = 234 mm2 ⁄m, 252 mm2 ⁄m)
Therefore, ProvideY8 − 250 − T2 (A s = 252 mm2 ⁄m)

S/NO. CALCULATIONS REMARK

BS 8110 – Shear:
1:1997 Table From Table 3.15 (BS 8110):
3.15 shear vx = 0.52, vy = 0.36
force Vsx = 23.34 kN
coefficients : . Vsx = vx nlx = (0.52)(10.688)(4.2) = 23.34 kN Vsy = 16.16 kN
and Vsy = vy nlx = (0.36)(10.688)(4.2) = 16.16 kN
vx = − , vy = 0.24

and Vsy = vy nlx = (0.24)(10.688)(4.2) = 10.77 kN

𝐒𝐡𝐨𝐫𝐭𝐞𝐫 𝐬𝐩𝐚𝐧 (𝐥𝐱 ):

design shear v =
V (23.34)(1000)
BS 8110 – design shear v = = = 0.15 N⁄mm2 0.15 N⁄mm2
dbv 155(1000)
1:1997 Clause
3.6.4.6 vmax = 0.8√fcu = (0.8)√40 = 5.1 N⁄mm2 or 5.0 N⁄mm2
maximum
design shear whichever is less.
stress :. vmax = 5.0 N⁄mm2

25 | P a g e
 Design shear v = 0.15 N⁄mm2 < 5.0 N⁄mm2
:. Section is adequate in shear.

100As 100(252)
bv d
= 155(1000) = 0.163%
From Table 6.4-1: Design concrete
0.57−0.48
Design concrete shear stress vc = (0.25−0.15) (0.163 − shear stress vc
= 0.51 N⁄mm2
0.15) + 0.5
= 0.51 N⁄mm2 > 0.15 N⁄mm2
:. Shear resistance O.K.

𝐥𝐨𝐧𝐠𝐞𝐫 𝐬𝐩𝐚𝐧 (𝐥𝐲 ):

V (16.16 )(1000)
design shear v = = = 0.11 N⁄mm2
dbv 145(1000)

100As 100(252)
= = 0.174%
bv d 145(1000)

From Table 6.4-1: concrete shear stress

Design concrete shear stress vc vc = 0.524 N⁄mm2
0.60 − 0.50
=( ) (0.174 − 0.15) + 0.5
0.25 − 0.15
= 0.524 N⁄mm2 > 0.11 N⁄mm2
:. Shear resistance O.K.

S/NO CALCULATIONS REMARKS

BS 8110 – Deflection: Check deflection in the
1:1997 From table 5.3-1: shorter span only
Clause Basic span/ depth ratio = 26
3.5.7 M shorter span moment 12.63(106 )
deflection = =
bd2 bd2 1000 (155)2
2
= 0.526 N⁄mm
From Table 5.3-2:
Tension steel modification factor
(1.68 − 1.5)
= (0.75 − 0.526) + 1.68 = 1.84
0.75 − 0.5
: . Allowable span⁄depth ratio = 26(1.84) = 47.87
The actual span⁄depth ratio = 4200 ⁄155 = 27.1
< 47.87
:. Deflection O.K.
BS 8110 – Cracking: Widest spacing between
1:1997 max. bar spacing = 3d = (3)(155) = 465 mm bars = 192 mm
Clause Clear space between bars = 200 – bar diameter
3.5.8 = 200 − 8 = 192 mm < 465 mm
crack :. Cracking O.K.
control

26 | P a g e
4.2 Beams

27 | P a g e
 BEAM A1–C1/BEAM A6-C6

Height of the beam = 600 mm

bw = 300 mm
hf = 180mm

Dimensions of the L – beam

BS 8110 –
1:1997
Clause
3.4.1.5
effective
width of
flanged
beam

b = bw + (0.1) (0.7) (l)

b =887.5m
= 0.3 + (0.1×0.7×8.25 = 877.5 mm

Calculation of the areas of influence on the L – beam

Beam A1-B1 = beam B1-C1 which is the same for Beam A6-B6 = B6-C6
Area ab
(a+b)h (3.1+8.25)2.575
2
= 2
= 14.61m2
Area ab = bc = ab9 = bc9 = 14.61m2

Calculation of dead load on the beam along the axis

28 | P a g e
 Static chart calculation
BS 6399 – A1 B1 C1
1:1996
(Dead and 8.25m 8.25m
imposed
loads)
Dead load slab AB = slab BC
• Self-weight of slab on span A1-B1 = 14.61×0.180×
BS 648: 24kn/m3 = 63.115 kN
1964
weights of • Finishes = 14.61×0.6kn/m2 = 8.766KN
building • Masonry wall = 8.25×2.5×0.2×22kn/m3 = 90.75KN
materials • Self-weight of beam on span A1-B1=B1-C1
= (0.6-0.18) ×8.25×0.3×24kn/m3= 24.948KN
Total dead load on span (A1-B1) = (B1-C1) = (A6-B6) = (B6-C6)
=63.115+8.766+90.75+24.948 = 187.579KN N
=187.57KN

Calculations of live load on the beam along the axis

Because of the purpose of the building (residential apartment), we assume live load is
taken as 1.50kn/m2

Calculation of dead load per meter on the beam (uniformly distributed load)

187.579
Span A1-B1 = B1-C1= A6-B6 = B6-C6 = 8.25
= 22.74kn/m
BS 8110 – Case (I) Case (i) = 34.23 Kn/
1:1997
Calculation of combination of load
Clause
3.2.1.2.2 Design load, N = 1.4gk + 1.6qk
choice of N = (1.4×22.74) + (1.6×1.5) = 34.23kn/m
critical
loading
arrangement Calculation of moments and shear forces using cross’s moment distribution method
Fixed end moment (FEM) calculations

29 | P a g e
 A1 B1
wL2 34.23×8.25×8.25
MA1-B1 = − =− = −194.16kNm
12 12
MB1-A1 = +194.16KNm

B1 C1
wL2 34.23×8.25×8.25
MB1-C1 = − 12
=− 12
= −194.16kNm
MC1-B1 = +194.16KNm

Case (II)
(a) Calculation of combination of load
Case (ii) =24.24kn/m
Design load, N = 1.4gk + 1.6qk
N = (1.4×22.74) + (1.6×1.5) = 34.23kn/m

(b) Calculation of combination of load

Design load, N = 1.0gk + 1.0qk
N = (1.0×22.74) + (1.0×1.5) = 24.24kn/m

Calculation of moments and shear forces using cross’s moment distribution method
Fixed end moment (FEM) calculations

A1 B1 FEM=-194.16
wL2 34.23×8.25×8.25
MA1-B1 = − 12
=− 12
= −194.16kNm
MB1-A1 = +194.16KNm

30 | P a g e
 B1 C1 FEM=-194.16
wL2 24.24×8.25×8.25
MB1-C1 = − =− = −137.486kNm
12 12
MC1-B1 = +137.486KNm

31 | P a g e
 Stiffness factors for the beams and columns

Beam trial - 300×600 mm

Column trial - 500×300 mm
Beam flexural rigidity is given by;
bd2 0.3 × 0.63
I= = = 5.4 × 10−3 m4 I=5.4 × 10−3 m4
12 12
flexural rigidity
Beam stiffness k = length
Span A1-B1 and span B1-C1
5.4 × 10−3 m4
k A1−B1 = k B1−C1 = = 6.545 × 10−4
8.25
Columns
bd2 0.3 × 0.53
I= = = 3.125 × 10−3 m4 k Col =1.25 × 10−3
12 12
3.125 × 10−3 m4
k Col = = 1.25 × 10−3
2.5

Calculation of distribution factors

Joint A1 = Joint C1
Total joint stiffness = beam stiffness + (column stiffness)2
(6.545 × 10−4 ) + (1.25 × 10−3 )2 = 3.1545 × 10−3
6.545 × 10−4
DFA1−B2 = DFC1−B2 = = 0.21
3.1545 × 10−3
2.5 × 10−3
DFCOL = = 0.79
3.1545 × 10−3

Joint B1 attaching span B1-A1 ad span B1-C1

Total joint stiffness = beam stiffness + (column stiffness)2
(6.545 × 10−4 ) + (6.545 × 10−4 ) + (1.25 × 10−3 )2 = 3.809 × 10−3
6.545 × 10−4
DFB1−A1 = DFB1−C1 = = 0.17
3.809 × 10−3
2.5 × 10−3
DFCOL = = 0.66
3.809 × 10−3

Moment distribution table using balanced method

32 | P a g e
 Case (I)

JOINT A1 B1 C1
MEMBER COL A1-B1 B1-A1 COL B1-C1 C1-B1 COL
DF 0.79 0.21 0.17 0.66 0.17 0.21 0.79
FEM 0 -194.16 194.16 0 -194.16 194.16 0
BAL 153.3864 40.7736 0 0 0 -40.7736 -153.386
CO 0 0 20.3868 0 -20.3868 0 0
BAL 0 0 0 0 0 0 0
TOTAL 153.3864 -153.386 214.5468 0 -214.547 153.3864 -153.386

Case (II)

JOINT A1 B1 C1
MEMBER COL A1-B1 B1-A1 COL B1-C1 C1-B1 COL
DF 0.79 0.21 0.17 0.66 0.17 0.21 0.79
FEM 0 -194.16 194.16 0 -137.486 137.486 0
BAL 153.3864 40.7736 -9.63458 -37.4048 -9.63458 -28.8721 -108.614
CO 0 -4.81729 20.3868 0 -14.436 -4.81729 0
BAL 3.805659 1.011631 -1.01163 -3.92751 -1.01163 1.011631 3.805659
CO -0.50582 0.505815 0.505815 -0.50582
BAL 0.399594 0.106221 -0.17198 -0.66768 -0.17198 0.106221 0.399594
TOTAL 157.5917 -157.592 204.2344 -42 -162.234 104.4087 -104.409

Final Beam moments

Joint A1 B1 C1
Member COL A1-B1 A1-B1 COL B1-C1 C1-B1 COL
Case (i) loading 153.3864 -153.386 214.5468 0 -214.547 153.3864 -153.386
Case (ii) loading 157.592 -157.592 204.234 -42 -162.234 104.4087 -104.409
Output Moments 157.592 -153.386 214.5468 -42 -214.547 153.3864 -153.386

Calculation of reactions in the beam

153.386 214.5468

A1 B1

Moment about A1, MA1;

8.25
= - RB1×8.25 + 214.5468 – 153.386 + (282.3975 × 2 )

33 | P a g e
 8.25RB1 = 1226.05
RB1 = 148.61KN RB1 = 148.61KN
Moment about B1, MB1;
= 8.25RA1 + 214.5468 – 153.386 – 1164.889
RA1 = 133.785KN
RA1 = 133.785KN

214.5468 153.386

B1 C1
Moment about B, MB; RC1 = 133.785KN
8.25
= - RC1×8.25 – 214.5468 + 153.386 + (282.3975 × 2 )
RC1 = 133.785KN
RB1 =148.61 KN
RB1 = 282.3975 – 133.785 = 148.61 KN

Shear forces and bending moment diagrams

a) Shear forces diagram

Span Design moment

Span A1-B1/A6-B6 Design

moment=261.432kN

34 | P a g e
 133.785
x= = 3.91 m from A1
34.23
Design moment = Reaction moment - Load moment
3.91
Span design moment = (133.785×3.91) – (34.23× 3.91 × ) = 261.432kNm
2
Design
Span B1-C1/B6-C6 moment=322.596kN
148.61
x= = 4.34 m from B1
34.23
Design moment = Reaction moment - Load moment
4.34
Span design moment = (148.61×4.34) – (34.23× 4.34 × 2 ) = 322.596kNm

35 | P a g e
 b) Bending moment diagram

Required steel reinforcement along the beam Axis

Span A1-B1/A6-B6
Maximum hogging moment = maximum moment at support B1 = M max =
214.5468KN
= maximum moment at support A1 = M max = 153.386 kN
Maximum sagging moment = maximum moment at midspan = 261.432kN
Maximum shear force = 148.61kN

A1 B1

➢ Cover = 20 mm
➢ Bar diameter = 20mm
20
➢ Effective depth = 600 – 20 – 2
= 570mm
➢ Hf = 180
➢ Beam type – L-beam

36 | P a g e
 ➢ Flange width b = 877.5mm
Moment of resistance MU
h
= 0.45fcu bhf (d − 2f) 10−6
180
0.45 × 40 × 877.5 × 180(570 − 2
) × 10−6 MU = 1364.688KN
=1364.688KN
1364.688kn > 261.432KN/m
Therefore, no compression reinforcement required
h
m = 0.87fy As (d − f)
2
Provide T3Y25 bottom
m 261.432×106 bars (As = 1473mm2 )
As = h = 0.87×460×480 = 1360.95mm2
0.87fy (d− f )
2
Provide T2Y12 hanger
Provide T3Y25 bottom bars (As = 1473mm2 ) bars (As = 226mm2 )
Provide T2Y8 hanger bars (As = 101mm2 )

Design for support steel at B1

M = 214.5468kNm
Section size = 600×300 MU=608.2128kn
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 214.5468kn
Compression steel not required at supports

m 214.5468 × 106
k= = = 0.055
fcu bd2 40 × 300 × 5702
z
= 0.93
d
Tension reinforcement;
Provide T3Y25 bottom
m 214.5468 × 106 bars (As = 1473mm2 )
As = = = 1011.32mm2
0.87fy z 0.87 × 460 × 0.93 × 570
Provide T3Y25 bottom bars (As = 1473mm2 )

Design for support steel at A1

M = 153.386kNm
Section size = 600×300 MU=608.2128kn
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 153.386kn
Compression steel not required at supports

m 153.386 × 106
k= = = 0.0393
fcu bd2 40 × 300 × 5702
z
= 0.94
d
Tension reinforcement;
m 153.386 × 106 Provide T3Y20 bottom
As = = = 715.329mm2 bars (As = 982mm2 )
0.87fy z 0.87 × 460 × 0.94 × 570

37 | P a g e
 Provide T2Y25 bottom bars (As = 982mm2 )

The design Shear stress Design shear stress V,

v
Design shear stress b d at support B1 =0.869N/mm2
BS 8110 – v

1:1997 V =148.61kN
Clause Bv = 300mm
3.4.5.2 D = 570mm
shear stress
148.61 × 1000
in beams = 0.869N/mm2
300 × 570

V =0.8√fcu = 40 = 5.1N/mm2

0.869 < 5.1N/mm2 therefore, shear stress ok.

Design concrete shear

Design concrete shear stress, VC. stress, VC=0.70
100AS 100×1473
bv d
= 300×570
= 0.86
BS 8110 –
0.74−0.67
1:1997 VC = (0.86 − 0.75) + 0.67 = 0.70
1.0−0.75
Clause
3.4.5.4 V < 0.5VC = 0.5 × 0.70 = 0.35N/mm2
concrete v = 0.869 > 0.35 = 0.5vc
shear stress (VC + 0.4) ˃ V
table 3.8
Therefore, provide links as follows;
Asv (0.4)bv
∴ (min. links) ≥
Sv 0.87f yv
0.4×300
0.87×460
= 0.2998 therefore, 0.33 ok Provide y8@300 c/c for
links spacing

Provide y8@300 c/c for links spacing

Minimum shear links will be used up to V = VC + 0.4 = 0.70 + 0.4 = 1.10 N/mm2
Shear load (minimum links) = (1.10) (300) (600)10−3 = 198 kN
BS 8110 –
1:1997 198 > 148.61 > 133.785 hence minimum links to be used throughout
Clause
3.4.4.4 table Anchorage of links
3.7 spacing
for links It should pass round another bar of at least its own size, through an angle of 180°,
and continue for a length of at least four times its own size

Deflection
b = 877.5 mm
bw =300
Span/effective depth ratio for the beam
BS 8110 – bw 300
bw
=0.342
1:1997 b
= 877.5 = 0.342 therefore, it is continuous -21.5 b

Clause 3.4.6 span

Basic ratio = 21.5
deflection depth
in beams Tension steel modification factor
table 3.9

38 | P a g e
 m 261.432 × 106
= = 0.92
bd2 877.5 × 5702
1.5 − 1.38
1.38 + ( (1.0 − 0.92)
1.0 − 0.75
1.38 + 0.0384 = 1.42
Therefore, allowable span/depth ratio; allowable span/depth
= 1.42×21.5 = 30.50 ratio=30.50
span 8.25
Actual depthratio = 0.57 = 14.474 < 30.50
Deflection ok.
Deflection ok.
Designed section output

Span B1-C1/B6-C6
Maximum hogging moment = maximum moment at support B1 = M max =
214.5468kN
= maximum moment at support C1 = M max
=153.386kN
Maximum sagging moment = maximum moment at midspan = 322.596kN
Maximum shear force = 148.61kN

39 | P a g e
 B1 C1

➢ Cover = 20 mm
➢ Bar diameter = 20mm
20
➢ Effective depth = 600 – 20 – 2
= 570mm
➢ hf = 180
➢ Beam type – L-beam
➢ Flange width b = 877.5mm
Moment of resistance MU
h
= 0.45fcu bhf (d − 2f) 10−6
180
0.45 × 40 × 877.5 × 180(570 − 2
) × 10−6
=1364.688KN
1364.688kn > 322.596KN/m
Therefore, no compression reinforcement required
h
m = 0.87fy As (d − f)
2

m 322.596×106
As = h = = 1679.35mm2
0.87fy (d− f ) 0.87×460×480
2
Provide T4Y25 bottom bars (As = 1963mm2 )
Provide T2Y8 hanger bars (As = 101mm2 )

40 | P a g e
 Design for support steel at B1
M = 214.5468kNm
Section size = 600×300 MU=608.2128kn
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 214.5468kn
Compression steel not required at supports

m 214.5468 × 106 k = 0.055

k= = = 0.055
fcu bd2 40 × 300 × 5702
z
= 0.93
d
Tension reinforcement;
m 214.5468 × 106
As = = = 1011.32mm2
0.87fy z 0.87 × 460 × 0.93 × 570
Provide T3Y25 bottom
Provide T3Y25 bottom bars (As = 1473mm2 ) bars (As = 1473mm2 )

Design for support steel at C1

M = 153.386kNm
Section size = 600×300
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 153.386kn
Compression steel not required at supports
k = 0.0393
m 153.386 × 106
k= = = 0.0393
fcu bd2 40 × 300 × 5702
z
= 0.94
d
Tension reinforcement;
m 153.386 × 106
As = = = 715.329mm2 Provide T3Y20 bottom
0.87fy z 0.87 × 460 × 0.94 × 570 bars (As = 942mm2 )
Provide T2Y25 bottom bars (As = 982mm2 )

BS 8110 –
The design Shear stress
v
1:1997 Design shear stress at support B1
bv d
Clause Design shear
V =148.61kN
3.4.5.2 stress=0.869N/mm2
shear stress Bv = 300mm
in beams D = 570mm
148.61 × 1000
= 0.869N/mm2
300 × 570

V =0.8√fcu = 40 = 5.1N/mm2

0.869 < 5.1N/mm2 therefore, shear stress ok.

Design concrete shear

stress, VC=0.77

41 | P a g e
 Design concrete shear stress, VC.
BS 8110 – 100AS 100 × 1963
1:1997 = = 1.148
Clause bv d 300 × 570
3.4.5.4 0.84 − 0.74
VC = (1.148 − 1.0) + 0.74 = 0.77
concrete 1.5 − 1.0
shear stress V < 0.5VC = 0.5 × 0.77 = 0.385N/mm2
table 3.8
v = 0.869 > 0.385 = 0.5vc

42 | P a g e
 (VC + 0.4) ˃ V
Therefore, provide links as follows;
Asv (0.4)bv
∴ (min. links) ≥
Sv 0.87fyv
0.4×300
0.87×460
= 0.2998 therefore, 0.33 ok

BS 8110 – Provide
Provide y8@300 c/c for links spacing y8@300 c/c for
1:1997 Clause
3.4.4.4 table 3.7 Minimum shear links will be used up to V = VC + 0.4 = 0.77 + links spacing
spacing for 0.4 = 1.17 N/mm2
links Shear load (minimum links) = (1.17) (300) (600)10−3 = 210.6
kN
210.6 > 148.61 > 133.785 hence minimum links to be used
throughout

Anchorage of links
It should pass round another bar of at least its own size, through
an angle of 180°, and continue for a length of at least four times
its own size

BS 8110 – Deflection
1:1997 Clause b = 877.5 mm
3.4.6 deflection bw =300
in beams table
3.9 Span/effective depth ratio for the beam
bw 300
= = 0.342 therefore, it is continuous -21.5
b 877.5
span bw
Basic depth ratio = 21.5 = 0.342
b
Tension steel modification factor
m 322.596 × 106
= = 1.13
bd2 877.5 × 5702
1.38 − 1.21
1.38 − ( (1.5 − 1.13)
1.5 − 1.0
1.39 – 0.1258 = 1.2542
Therefore, allowable span/depth ratio;
= 1.2542×21.5 = 26.9653
span 8.25
Actual depthratio = 0.57 = 14.474 < 26.9653
Deflection ok.
Deflection ok.
Designed section output

43 | P a g e
44 | P a g e
 BEAM A2 – C2/BEAM A5-C5

Height of the beam = 600 mm

bw = 300 mm
hf = 180mm

Dimensions of the T – beam

BS 8110 –
1:1997
Clause
3.4.1.5
effective
width of
flanged
beam

b = bw + (0.2) (0.7)(l)
b =1455 mm
= 0.3 + (0.2×0.7×8.25 = 1455 mm

Calculation of the areas of influence on the T – beam

Beam A2-C2 = beam A5-C5
Area ab1
(a+b)h (3.1+8.25)2.575
2
= 2
= 14.61m2
Area ab1 = bc1 = ab8 = bc8 = 14.61m2
Area 3
(a+b)h (4.05+8.25)2.1
2
= 2
= 12.915m2
Area ab2 = bc2 = ab7 = bc7 = 12.915m2
Total area on each beam span A2-B2=B2-C2 = A5-B5 = B5-C5 = 12.915+14.61 =
27.525m2

45 | P a g e
 BS 6399 – Calculation of dead load on the beam along the axis
1:1996
(Dead and Static chart calculation
imposed
A2 B2 C2
loads)
8.25m 8.250m
BS 648:
1964 Dead load slab AB = slab BC
weights of • Self-weight of slab on span A2-B2 = 27.525×0.180×
building
materials 24kn/m3 = 118.908 kN
• Finishes = 27.525×0.6kn/m2 = 16.515KN
• Masonry wall = 8.25×2.5×0.2×22kn/m3 = 90.75KN
• Self-weight of beam on span AB=BC
= (0.6-0.18) ×8.25×0.3×24kn/m3= 24.948KN
Total dead load on span A2-B2 = B2-C2 = A5-B5 = B5-C5
=118.908+16.515+90.75+24.948 = 251.121KN

Calculations of live load on the beam along the axis

Because of the purpose of the building (residential apartment), we assume live load is
taken as 1.50kn/m2

Calculation of dead load per meter on the beam (uniformly distributed load)

251.121
Span A2-B2 = 8.25
= 30.44kn/m

Case (I)
BS 8110 – Calculation of combination of load
1:1997 Design load, N = 1.4gk + 1.6qk
Clause
3.2.1.2.2 N = (1.4×30.44) + (1.6×1.5) = 45.01kn/m
choice of Case (I)=
critical Calculation of moments and shear forces using cross’s moment distribution 45.01kn/m
loading method
arrangement
Fixed end moment (FEM) calculations

46 | P a g e
 FEM
A2 B2 =255.32KNm
wL2 45.01×8.25×8.25
MA2-B2 = − =− = −255.32kNm
12 12
MB2-A2 = +255.32KNm

B2 C2 FEM
wL2 45.01×8.25×8.25 =255.32KNm
MB2-C2 = − 12
=− 12
= −255.32kNm
MC2-B2 = +255.32KNm

Case (II)
(a) Calculation of combination of load
Case (II)=
Design load, N = 1.4gk + 1.6qk
45.01kn/m
N = (1.4×30.44) + (1.6×1.5) = 45.01kn/m

(b) Calculation of combination of load

Case (II)=
Design load, N = 1.0gk + 1.0qk
31.94kn/m
N = (1.0×30.44) + (1.0×1.5) =31.94kn/m

Calculation of moments and shear forces using cross’s moment distribution

method
Fixed end moment (FEM) calculations

FEM
=255.32KNm
A2 B2
wL2 45.01×8.25×8.25
MA2-B2 = − 12
=− 12
= −255.32kNm
MB2-A2 = +255.32KNm

47 | P a g e
 FEM
=181.16KNm

B2 C2
wL2 31.94×8.25×8.25
MB2-C2 = − =− = 181.16kNm
12 12
MC2-B2 = +181.16KNm

48 | P a g e
 Stiffness factors for the beams and columns

Beam trial - 300×600 mm

Column trial - 500×300 mm
Beam flexural rigidity is given by; I=5.4 ×
bd2 0.3 × 0.63 10−3 m4
I= = = 5.4 × 10−3 m4
12 12
flexural rigidity
Beam stiffness k = length
Span A2-B2 and span B2-C2
5.4 × 10−3 m4
k A2−B2 = k B2−C2 = = 6.545 × 10−4
8.25
Columns
bd2 0.3 × 0.53
I= = = 3.125 × 10−3 m4
12 12
k Col
3.125 × 10−3 m4
k Col = = 1.25 × 10−3 = 1.25
2.5 × 10−3

Calculation of distribution factors

Joint A2/Joint A6
Total joint stiffness = beam stiffness + (column stiffness)2
(6.545 × 10−4 ) + (1.25 × 10−3 )2 = 3.1545 × 10−3
6.545 × 10−4
DFA2−B2 = DFC2−B2 = = 0.21
3.1545 × 10−3
2.5 × 10−3
DFCOL = = 0.79
3.1545 × 10−3

Joint B2/Joint B6
Total joint stiffness = beam stiffness + (column stiffness)2
(6.545 × 10−4 ) + (6.545 × 10−4 ) + (1.25 × 10−3 )2 = 3.809 × 10−3
6.545 × 10−4
DFB2−A2 = DFB2−C2 = = 0.17
3.809 × 10−3
2.5 × 10−3
DFCOL = = 0.66
3.809 × 10−3

Moment distribution table using balanced method

49 | P a g e
 Case (I)

JOINT A2 B2 C2
MEMBER COL A2-B2 B2-A2 COL B2-C2 C2-B2 COL
DF 0.79 0.21 0.17 0.66 0.17 0.21 0.79
FEM 0 -255.32 255.32 0 -255.32 255.32 0
BAL 201.7028 53.6172 0 0 0 -53.6172 -201.703
CO 0 0 26.8086 0 -26.8086 0 0
BAL 0 0 0 0 0 0 0
TOTAL 201.7028 -201.703 282.1286 0 -282.129 201.7028 -201.703

Case (II)

JOINT A2 B2 C2
MEMBER COL A2-B2 B2-A2 COL B2-C2 C2-B2 COL
DF 0.79 0.21 0.17 0.66 0.17 0.21 0.79
FEM 0 -255.32 255.32 0 -181.16 181.16 0
BAL 201.7028 53.6172 -12.6072 -48.9456 -12.6072 -38.0436 -143.116
CO 0 -6.3036 26.8086 0 -19.0218 -6.3036 0
BAL 4.979844 1.323756 -1.32376 -5.13929 -1.32376 1.323756 4.979844
CO -0.66188 0.661878 0.661878 -0.66188
BAL 0.522884 0.138994 -0.22504 -0.87368 -0.22504 0.138994 0.522884
TOTAL 207.2055 -207.206 268.6345 -54.9586 -213.676 137.6137 -137.614

Final Beam moments

Joint A2 B2 C2
Member COL A2-B2 B2-A2 COL B2-C2 C2-B2 COL
Case (I) 201.7028 -201.703 282.1286 0 -282.129 201.7028 -201.703
loading
Case (II) 207.2055 -207.206 268.6345 -54.9586 -213.676 137.6137 -137.614
loading
Output 207.2055 -201.703 282.1286 -54.9586 -282.129 201.7028 -201.703
Moments

Calculation of reactions in the beam

201.702 282.1286

A2 B2
Moment about A, MA;
8.25
= - RB2×8.25 + 282.1286 – 201.702 + (371.3325× 2 )
8.25RB2 = 1612.17

50 | P a g e
 RB2 = 195.41KN
Moment about B2, MB2; RB2 =
= 8.25RA2 + 282.1286 – 201.702 – 1531.75 195.41KN
RA2 = 175.92KN

282.1286 201.702 RA2 =

175.92KN

B2 C2
Moment about B, MB;
8.25
= - RC2×8.25 – 282.1286 + 201.702 + (371.3325× 2 )
RC2 = 175.92KN
RB2 = 371.3325 – 175.92 = 195.41 KN

Shear forces and bending moment diagrams

a) Shear forces diagram

Span Design moment

Span A2-B2/A5-B5

175.92
x= = 3.91 m from A2
45.01
Design moment = Reaction moment - Load moment Design
3.91 moment
Span design moment = (175.92×3.91) – (45.01× 3.91 × ) = 343.789kNm =343.789kNm
2

Span B2-C2/B5-C5

51 | P a g e
 195.41
x= = 4.34 m from B2
45.01
Design moment = Reaction moment - Load moment Design
4.34
Span design moment = (195.41×4.34) – (45.01× 4.34 × ) = 424.18kNm moment
2

52 | P a g e
 b) Bending moment diagram

Required steel reinforcement along the beam Axis

Span A2-B2/A5-B5
Maximum hogging moment = maximum moment at support B2= M max = 282.129KN
= maximum moment at support A2= M max = 201.702kN
Maximum sagging moment = maximum moment at midspan = 343.789kN
Maximum shear force = 195.41kN

A2 B2

➢ Cover = 20 mm
➢ Bar diameter = 20mm
20
➢ Effective depth = 600 – 20 – 2
= 570mm

53 | P a g e
 ➢ hf = 180
➢ Beam type – T-beam
➢ Flange width b = 1455mm

Moment of resistance MU
hf MU
= 0.45fcu bhf (d − ) 10−6 =2262.816KN
2
180
0.45 × 40 × 1455 × 180(570 − 2
) × 10−6
=2262.816KN
2262.816kn > 343.789KN/m
Therefore, no compression reinforcement required

h
m = 0.87fy As (d − 2f)

m 343.789×106
As = h = = 1789.67mm2
0.87fy (d− f ) 0.87×460×480
2
Provide
T4Y25 bottom
Provide T4Y25 bottom bars (As = 1963mm2 ) bars (As =
Provide T2Y8 hanger bars (A s = 101mm2 ) 1963mm2 )

Design for support steel at B2

M = 282.1286kNm M=
Section size = 600×300 282.1286kNm
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
MU
= 608.2128kn > 282.1286kn =608.2128kn
Compression steel not required at supports

m 282.1286 × 106
k= = = 0.072
fcu bd2 40 × 300 × 5702
z
= 0.90
d
Tension reinforcement;
m 282.1286 × 106 Provide
As = = = 1374.21mm2 T3Y25 bottom
0.87fy z 0.87 × 460 × 0.90 × 570 bars (As =
Provide T3Y25 bottom bars (As = 1473mm2 ) 1473mm2 )

Design for support steel at A2

M = 201.702kNm M=
Section size = 600×300 201.702kNm
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
MU
= 608.2128kn > 201.702kn =608.2128kn
Compression steel not required at supports

54 | P a g e
 m 201.702 × 106
k= = = 0.052 k = 0.052
fcu bd2 40 × 300 × 5702
z
= 0.93
d
Tension reinforcement;
m 201.702 × 106
As = = = 950.769mm2
0.87fy z 0.87 × 460 × 0.93 × 570 Provide
Provide T2Y25 bottom bars (As = 982mm2 ) T2Y25 bottom
bars (As =
982mm2 )
BS 8110 – The design Shear stress
1:1997 v
Design shear stress at support B2/B6
Clause bv d
3.4.5.2 V =195.41kN Design shear
shear stress bv = 300mm stress V, =
in beams
D = 570mm 1.143N/mm2
195.41 × 1000
= 1.143N/mm2
300 × 570

V =0.8√fcu = 40 = 5.1N/mm2

1.143 < 5.1N/mm2 therefore, shear stress ok.

BS 8110 – Design concrete shear stress, VC.

1:1997 100A S 100 × 1963 Design
Clause = = 1.148 concrete shear
bv d 300 × 570
3.4.5.4 0.84 − 0.74 stress,
concrete VC = (1.148 − 1.0) + 0.74 = 0.77 VC=0.77
shear stress 1.5 − 1.0
table 3.8 V < 0.5VC = 0.5 × 0.77 = 0.385N/mm2
v = 1.143 > 0.385 = 0.5vc

(VC + 0.4) > V

Therefore, provide links as follows;
Asv (0.4)bv
BS 8110 – ∴ (min. links) ≥
1:1997 Sv 0.87fyv
Clause 0.4×300
0.87×460
= 0.2998 therefore, 0.33 ok
3.4.4.4 table
3.7 spacing Provide y8@300 c/c for links spacing
for links Minimum shear links will be used up to V = VC + 0.4 = 0.77 + 0.4 = 1.227 N/mm2 Provide
Shear load (minimum links) = (1.17) (300) (600)10−3 = 210.6 kN y8@300 c/c
for links
210.6 > 195.41 > 175.92 hence minimum links to be used throughout spacing

Anchorage of links
It should pass round another bar of at least its own size, through an angle of 180°,
and continue for a length of at least four times its own size

55 | P a g e
 BS 8110 – Deflection
1:1997 b = 1455 mm
Clause 3.4.6
deflection bw =300
in beams Span/effective depth ratio for the beam
table 3.9 bw 300
b
= 1455 = 0.206 therefore, it is continuous -20.8
span bw
Basic depth ratio = 20.8 = 0.206
b
Tension steel modification factor
m 343.789 × 106
= = 0.727
bd2 1455 × 5702
1.50 − 1.38
1.38 + ( (1.0 − 0.727)
1.0 − 0.75
1.38+ 0.131 = 1.51
Therefore, allowable span/depth ratio;
= 1.51×20.8 = 31.43
span 8.25
Actual depthratio = 0.57 = 14.474 < 31.43 allowable
span/depth
Deflection ok. ratio= 31.43

Designed section output

Deflection ok.

Span B2-C2/B5-C5
Maximum hogging moment = maximum moment at support B2 = M max = 282.129kN
= maximum moment at support C2 = M max = 201.702kN
Maximum sagging moment = maximum moment at midspan = 424.18kN
Maximum shear force = 195.41kN

B2 C2

➢ Cover = 20 mm

56 | P a g e
 ➢ Bar diameter = 20mm
20
➢ Effective depth = 600 – 20 – 2
= 570mm
➢ hf = 180
➢ Beam type – T-beam
➢ Flange width b = 1455mm

Moment of resistance MU
hf
= 0.45fcu bhf (d − ) 10−6
2
180
0.45 × 40 × 1455 × 180(570 − 2
) × 10−6
=2262.816KN
2262.816kn > 424.18KN/m
Therefore, no compression reinforcement required

h
m = 0.87fy As (d − 2f)

m 424.18×106
As = h = 0.87×460×480 = 2208.167mm2
0.87fy (d− f )
2

Provide
Provide T5Y25 bottom bars (As = 2454mm2 ) T5Y25 bottom
Provide T2Y8 hanger bars (A s = 101mm2 ) bars (As =
2454mm2 )

57 | P a g e
 Design for support steel at C2
M = 201.702kNm M=
201.702kNm
Section size = 600×300
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6 MU
= 608.2128kn > 201.702kn =608.2128kn
Compression steel not required at supports

m 201.702 × 106
k= = = 0.052
fcu bd2 40 × 300 × 5702 k = 0.052
z
= 0.93
d
Tension reinforcement;
m 201.702 × 106
As = = = 950.769mm2
0.87fy z 0.87 × 460 × 0.93 × 570
Provide
Provide T2Y25 bottom bars (As = 982mm2 ) T2Y25 bottom
bars (As =
Design for support steel at B2 982mm2 )
M = 282.1286kNm
Section size = 600×300 M=
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6 282.1286kNm
= 608.2128kn > 282.1286kn
Compression steel not required at supports

m 282.1286 × 106
k= = = 0.072
fcu bd2 40 × 300 × 5702
z
= 0.90
d
Tension reinforcement;
m 282.1286 × 106
As = = = 1374.21mm2
0.87fy z 0.87 × 460 × 0.90 × 570
Provide T3Y25 bottom bars (As = 1473mm2 ) Provide
T3Y25 bottom
The design Shear stress bars (As =
BS 8110 – v 1473mm2 )
1:1997 Design shear stress at support B2/B5
bv d
Clause
3.4.5.2 V =195.41kN
shear stress Bv = 300mm
in beams D = 570mm Design shear
195.41 × 1000 stress
= 1.143N/mm2 v=1.143N/
300 × 570 mm2

V =0.8√fcu = 40 = 5.1N/mm2

1.143 < 5.1N/mm2 therefore, shear stress ok.

58 | P a g e
 BS 8110 – Design concrete shear stress, VC.
1:1997 100A S 100 × 2454
Clause = = 1.435 Design
3.4.5.4 bv d 300 × 570 concrete shear
concrete 0.84 − 0.74 stress, VC.=
VC = (1.435 − 1.0) + 0.74 = 0.827
shear stress 1.5 − 1.0 0.827
table 3.8 V < 0.5VC = 0.5 × 0.827 = 0.4135N/mm2
v = 1.143 > 0.4135 = 0.5vc

59 | P a g e
 (VC + 0.4) > V
BS 8110 – Therefore, provide links as follows;
1:1997 Asv (0.4)bv
Clause ∴ (min. links) ≥
Sv 0.87fyv
3.4.4.4
0.4×300
table 3.7
0.87×460
= 0.2998 therefore, 0.33 ok
spacing for Provide
Provide y8@300 c/c for links spacing
links y8@300 c/c for
Minimum shear links will be used up to V = VC + 0.4 = 0.827 + links spacing
0.4 = 1.227 N/mm2
Shear load (minimum links) = (1.227) (300) (600)10−3 = 220.86
kN
220.86 > 195.41 > 175.92 hence minimum links to be used
throughout

Anchorage of links
It should pass round another bar of at least its own size, through an
BS 8110 – angle of 180°, and continue for a length of at least four times its
1:1997 own size
Clause
3.4.6 Deflection
deflection bw
in beams b = 1455 mm = 0.206
b
table 3.9 bw =300
Span/effective depth ratio for the beam
bw 300
b
= 1455 = 0.206 therefore, it is continuous -20.8
span
Basic depth ratio = 20.8
Tension steel modification factor
m 424.18 × 106
= = 0.897
bd2 1455 × 5702
1.50 − 1.38 allowable
1.38 + ( (1.0 − 0.897)
1.0 − 0.75 span/depth
1.38+ 0.04944 = 1.4294 ratio=29.73
Therefore, allowable span/depth ratio;
= 1.42944×20.8 = 29.73
span 8.25 Deflection ok.
Actual depthratio = 0.57 = 14.474 < 29.73
Deflection ok.

Designed section output

60 | P a g e
61 | P a g e
 BEAM A3-C3/A4-C4

Height of the beam = 600 mm

bw = 300 mm
hf = 180mm

Dimensions of the T – beam

BS 8110 –
1:1997
Clause b = bw + (0.2) (0.7) (l) b =1455 mm
3.4.1.5 = 0.3 + (0.2×0.7×8.25 = 1455 mm
effective
width of Calculation of the areas of influence on the T – beam
flanged Beam A3-C3 = beam A4-C4
beam Area ab3
(a+b)h (4.05+8.25)2.1
2
= 2
= 12.915m2
Area ab3 = bc3 = ab4 = bc4 = ab5 = bc5 = 12.915m2

Total area on each beam span = 12.915+12.915 = 25.83m2

62 | P a g e
 Calculation of dead load on the beam along the axis

BS 6399 – Static chart calculation

1:1996 A3 B3 C3
(Dead and
imposed
loads) 8.25m 8.250m

Dead load slab AB = slab BC

BS 648: • Self-weight of slab on each span = 25.83×0.180×
1964
weights of 24kn/m3 = 111.5856 kN
building • Finishes = 25.83×0.6kn/m2 = 15.498KN
materials • Masonry wall = 8.25×2.5×0.2×22kn/m3 = 90.75KN
• Self-weight of beam on span AB=BC
= (0.6-0.18) ×8.25×0.3×24kn/m3= 24.948KN
Total dead load on each of the above beam spans
=111.5856+15.498+90.75+24.948 = 242.7816KN

Calculations of live load on the beam along the axis

Because of the purpose of the building (residential apartment), we assume live load
is taken as 1.50kn/m2

Calculation of dead load per meter on the beam (uniformly distributed load)

242.7816
Span UDL = 8.25
= 29.428kn/m

Case (I)
Calculation of combination of load
Design load, N = 1.4gk + 1.6qk
N = (1.4×29.428) + (1.6×1.5) = 43.60kN/m
Case (I)
BS 8110 – =43.60kN/m
1:1997 Calculation of moments and shear forces using cross’s moment distribution
Clause method
3.2.1.2.2
choice of
Fixed end moment (FEM) calculations
critical
loading
arrangement

A3 B3
wL2 43.60×8.25×8.25
MA3-B3 = − =− = −247.28kNm
12 12

63 | P a g e
 MB3-A3 = +247.28KNm

B3 C3 FEM
wL2 43.60×8.25×8.25 =247.28KNm
MB3-C3 = − =− = −247.28kNm
12 12
MC3-B3 = +247.28KNm

Case (II)
(a) Calculation of combination of load Case (II)=
Design load, N = 1.4gk + 1.6qk 43.60kN/m
N = (1.4×29.428) + (1.6×1.5) = 43.60kN/m

(b) Calculation of combination of load

Design load, N = 1.0gk + 1.0qk
Case (II)=
N = (1.0×29.428) + (1.0×1.5) = 30.928kN/m 30.928kN/m

Calculation of moments and shear forces using cross’s moment distribution

method
Fixed end moment (FEM) calculations

A3 B3 FEM
=247.28KNm

wL2 43.60×8.25×8.25
MA3-B3 = − 12
=− 12
= −247.28kNm
FEM
MB3-A3 = +247.28KNm =175.42kNm

B3 C3
wL2 30.928×8.25×8.25
MB3-C3 = − 12
=− 12
= −175.42kNm
MC3-B3 =
+175.42KNm

64 | P a g e
 Stiffness factors for the beams and columns

Beam trial - 300×600 mm

Column trial - 500×300 mm
Beam flexural rigidity is given by;
bd2 0.3 × 0.63 I= 5.4 ×
I= = = 5.4 × 10−3 m4
12 12 10−3 m4
flexural rigidity
Beam stiffness k = length
Span A3-B3 and span B3-C3
5.4 × 10−3 m4
k A3−B3 = k B3−C3 = = 6.545 × 10−4
8.25
Columns
bd2 0.3 × 0.53
I= = = 3.125 × 10−3 m4 k Col =1.25 ×
12 12
10−3
3.125 × 10−3 m4
k Col = = 1.25 × 10−3
2.5

Calculation of distribution factors

Joint A3/Joint A4
Total joint stiffness = beam stiffness + (column stiffness)2
(6.545 × 10−4 ) + (1.25 × 10−3 )2 = 3.1545 × 10−3
6.545 × 10−4
DFA3−B3 = DFC3−B3= = 0.21
3.1545 × 10−3
2.5 × 10−3
DFCOL = = 0.79
3.1545 × 10−3

Joint B3/B4
Total joint stiffness = beam stiffness + (column stiffness)2
(6.545 × 10−4 ) + (6.545 × 10−4 ) + (1.25 × 10−3 )2 = 3.809 × 10−3
6.545 × 10−4
DFB3−A3 = DFB3−C3 = = 0.17
3.809 × 10−3
2.5 × 10−3
DFCOL = = 0.66
3.809 × 10−3

Moment distribution table using balanced method

65 | P a g e
 Case (I)

JOINT A3 B3 C3
MEMB COL A3- B3-A3 CO B3-C3 C3-B3 COL
ER B3 L
DF 0.79 0.21 0.17 0.6 0.17 0.21 0.79
6
FEM 0 - 247.28 0 - 247.28 0
247.2 247.28
8
BAL 195.35 51.92 0 0 0 - -
12 88 51.928 195.351
8
CO 0 0 25.964 0 - 0 0
4 25.964
4
BAL 0 0 0 0 0 0 0
TOTAL 195.35 - 273.24 0 - 195.35 -
12 195.3 44 273.24 12 195.351
51 4

Case (II)

JOINT A3 B3 C3
MEMB COL A3-B3 B3-A3 COL B3-C3 C3-B3 COL
ER
DF 0.79 0.21 0.17 0.66 0.17 0.21 0.79
FEM 0 - 247.28 0 - 175.42 0
247.28 175.42
BAL 195.35 51.928 - - - - -
12 8 12.216 47.42 12.216 36.838 138.58
2 76 2 2 2
CO 0 - 25.964 0 - - 0
6.1081 4 18.419 6.1081
1
BAL 4.8253 1.2827 - - - 1.2827 4.8253
99 01 1.2827 4.979 1.2827 01 99
9
CO - 0.6413 0.6413 -
0.6413 51 51 0.6413
5 5
BAL 0.5066 0.1346 - - - 0.1346 0.5066
67 84 0.2180 0.846 0.2180 84 67
6 58 6
TOTA 200.68 - 260.16 - - 133.24 -
L 33 200.68 88 53.25 206.91 97 133.25
3 41 5

Final Beam moments

Joint A3 B3 C3
Member COL A3-B3 B3-A3 COL B3-C3 C3-B3 COL

66 | P a g e
 Case (I) 195.3512 - 273.2444 0 - 195.3512 -195.351
loading 195.351 273.244
Case (II) 200.6833 - 260.1688 - - 133.2497 -133.25
loading 200.683 53.2541 206.915
Output 200.6833 - 273.2444 - - 195.3512 -195.351
Moments 195.351 53.2541 273.244
RB3 = 189.29KN

Calculation of reactions in the beam

RA3 = 170.41KN

195.351 273.244

A3 B3
Moment about A3, MA3;
8.25 RC3 = 170.41KN
= - RB3×8.25 + 273.244 – 195.351 + (359.7× )
2
8.25RB3 = 1561.655
RB3 = 189.29KN RB3 =189.29 KN
Moment about B, MB;
8.25
= 8.25RA + 273.244 – 195.351 – (359.7 × 2 )
RA3 = 170.41KN

273.244 195.351

B3 C3
Moment about B, MB;
8.25
= - RC3×8.25 – 273.244 + 195.351 + (359.7× )
2
RC3 = 170.41KN
RB3 = 359.7 – 170.41 = 189.29 KN

Shear forces and bending moment diagrams

a) Shear forces diagram

67 | P a g e
 Design moment
=333.02kNm

Design moment
=410.90kNm

Span Design moment

Span A3-B3/A4-B4

170.41
x= = 3.91 m from A3
43.60
Design moment = Reaction moment - Load moment
3.91
Span design moment = (170.41×3.91) – (43.60× 3.91 × ) = 333.02kNm
2

Span B3-C3/B4-C4

189.29
x= = 4.34 m from B3
43.60
Design moment = Reaction moment - Load moment
4.34
Span design moment = (189.29×4.34) – (43.60× 4.34 × 2 ) = 410.90kNm

68 | P a g e
 b) Bending moment diagram

Required steel reinforcement along the beam Axis

Span A3-B3/A4-B4
Maximum hogging moment= maximum moment at support B3= M max =
273.244KN
= maximum moment at support A3= M max =
195.351kN
Maximum sagging moment = maximum moment at midspan = 333.02kN
Maximum shear force = 189.29KN
A3 B3

➢ Cover = 20 mm
➢ Bar diameter = 20mm
20
➢ Effective depth = 600 – 20 – = 570mm
2
➢ hf = 180 MU
➢ Beam type – T-beam =2262.816KN
➢ Flange width b = 1455mm

69 | P a g e
 Moment of resistance MU
hf
= 0.45fcu bhf (d − ) 10−6
2
180
0.45 × 40 × 1455 × 180(570 − ) × 10−6
2
=2262.816KN
2262.81kn > 333.02KN/m
Therefore, no compression reinforcement required

h
m = 0.87fy As (d − 2f)

m 333.02×106
As = h = 0.87×460×480 = 1733.61mm2
0.87fy (d− f )
2

Provide T4Y25 bottom bars (As = 1963mm2 ) Provide T4Y25

Provide T2Y8 hanger bars (A s = 101mm2 ) bottom bars
(As =
1963mm2 )
Design for support steel at B3
M = 273.244kNm
Section size = 600×300
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 273.244kN
Compression steel not required at supports

m 273.244×106 k = 0.070
k=f bd2
= 40×300×5702 = 0.070
cu
z
d
= 0.91
Provide T3Y25
Tension reinforcement;
m 273.244×106
bottom bars
As = 0.87f = 0.87×460×0.91×570 = 1316.31mm2 (As =
yz
1473mm2 )
Provide T3Y25 bottom bars (As = 1473mm2 )

Design for support steel at A3

M = 195.351kNm
Section size = 600×300
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 195.351kN
Compression steel not required at supports k = 0.050

m 195.351×106
k=f bd2
= 40×300×5702 = 0.050
cu
z
d
= 0.94
Tension reinforcement;

70 | P a g e
 m 195.351×106
As = 0.87f = 0.87×460×0.94×570 = 911.036mm2
y z Provide T2Y25
Provide T2Y25 bottom bars (As = 982mm2 ) bottom bars
(As = 982mm2 )

BS 8110 – The design Shear stress

v
1:1997 Design shear stress b d at support B3
Clause v

3.4.5.2 V =189.29kN
shear stress bv = 300mm
in beams d = 570mm
The design Shear
189.29 × 1000
= 1.107N/mm2 stress v
300 × 570 =1.107N/mm2

V =0.8√fcu = 40 = 5.1N/mm2

1.107 < 5.1N/mm2 therefore, shear stress ok.

Design concrete shear stress, VC.

100AS 100×1963
bv d
= 300×570
= 1.148 Design concrete
BS 8110 – shear stress, VC
0.84−0.74
1:1997 VC = (1.148 − 1.0) + 0.74 = 0.77 =0.77
1.5−1.0
Clause
3.4.5.4 V < 0.5VC = 0.5 × 0.77 = 0.385N/mm2
concrete v = 1.107 > 0.385 = 0.5vc
shear stress
table 3.8
(VC + 0.4) > V
Therefore, provide links as follows;
A sv (0.4)bv
∴ (min. links) ≥
Sv 0.87fyv
0.4×300
= 0.2998 therefore, 0.33 ok
0.87×460

BS 8110 –
Provide y8@300 c/c for links spacing
1:1997 Minimum shear links will be used up to V = VC + 0.4 = 0.77 + 0.4 = 1.17 N/mm2
Clause Shear load (minimum links) = (1.17) (300) (600)10−3 = 210.6 kN
3.4.4.4 table 210.6 > 189.29 > 170.41 hence minimum links to be used throughout
3.7 spacing
for links
Anchorage of links Provide y8@300
It should pass round another bar of at least its own size, through an angle of 180°, c/c for links
and continue for a length of at least four times its own size spacing

Deflection
b = 1455 mm
bw =300
Span/effective depth ratio for the beam
bw 300
b
= 1455 = 0.206 therefore, it is continuous -20.8

71 | P a g e
 BS 8110 – span
Basic depth ratio = 20.8
1:1997
Clause 3.4.6 Tension steel modification factor
deflection m 333.02×106
in beams bd2
= 1455×5702
= 0.70
table 3.9 1.50−1.38
1.38 + ( 1.0−0.75 )(1.0 − 0.70)
1.38+ 0.144 = 1.524 bw
= 0.206
Therefore, allowable span/depth ratio; b
= 1.524×20.8 = 31.70
span 8.25
Actual ratio = = 14.474 < 30.70
depth 0.57
Deflection ok.

Designed section output allowable

span/depth
ratio=31.70

Deflection ok.

Span B3-C3/B4-C4
Maximum hogging moment = maximum moment at support B3=M max =
273.244KN
= maximum moment at support C3= M max =
195.351kN
Maximum sagging moment = maximum moment at midspan = 410.90kN
Maximum shear force = 189.29KN

B3 C4

➢ Cover = 20 mm
➢ Bar diameter = 20mm
20
➢ Effective depth = 600 – 20 – = 570mm
2
➢ hf = 180

72 | P a g e
 ➢ Beam type – T-beam
➢ Flange width b = 1455mm

Moment of resistance MU
h
= 0.45fcu bhf (d − 2f) 10−6
180
0.45 × 40 × 1455 × 180(570 − 2
) × 10−6
=2262.816KN
2262.81kn > 410.90KN/m
Therefore, no compression reinforcement required
MU=2262.816KN
hf
m = 0.87fy As (d − 2 )

m 410.90×106
As = h = 0.87×460×480 = 2139.03mm2
0.87fy (d− f )
2

Provide T5Y25 bottom bars (As = 2454mm2 )

Provide T2Y8 hanger bars (A s = 101mm2 )
Provide T5Y25
bottom bars
(As =
2454mm2 )

73 | P a g e
 Design for support steel at B3
M = 273.244kNm
Section size = 600×300 MU
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6 =608.2128kn
= 608.2128kn > 273.244kN
Compression steel not required at supports

m 273.244×106
k=f bd2
= 40×300×5702 = 0.070
cu
z
= 0.91
d
Tension reinforcement;
m 273.244×106 Provide T3Y25
As = 0.87f z
= 0.87×460×0.91×570 = 1316.31mm2
y bottom bars
Provide T3Y25 bottom bars (As = 1473mm2 ) (As =
1473mm2 )
Design for support steel at C3
M = 195.351kNm
Section size = 600×300
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 195.351kN
Compression steel not required at supports

m 195.351×106 k = 0.050
k=f bd2
= 40×300×5702 = 0.050
cu
z
d
= 0.91
Tension reinforcement;
m 195.351×106
As = = = 911.036mm2 Provide T2Y25
0.87fyz 0.87×460×0.91×570
bottom bars
Provide T2Y25 bottom bars (As = 982mm2 ) (As = 982mm2 )

The design Shear stress

v
Design shear stress b d at support B3 The design Shear
v
stress v
V =189.29kN =1.107N/mm2
Bv = 300mm
D = 570mm
189.29×1000
300×570
= 1.107N/mm2

V =0.8√fcu = 40 = 5.1N/mm2
BS 8110 –
1:1997 1.107 < 5.1N/mm2 therefore, shear stress ok.
Clause
3.4.5.4 Design concrete shear stress, VC.
concrete 100AS 100×2454
shear bv d
= 300×570
= 1.435

74 | P a g e
 stress table 0.84−0.74 Design concrete
VC = (1.435 − 1.0) + 0.74 = 0.827
1.5−1.0 shear stress, VC.=
3.8
V < 0.5VC = 0.5 × 0.827 = 0.4135N/mm2 0.827

v = 1.107 > 0.4135 = 0.5vc

75 | P a g e
 (VC + 0.4) > V
BS 8110 –
Therefore, provide links as follows;
1:1997
Clause Asv (0.4)bv
3.4.4.4 table ∴ (min. links) ≥
Sv 0.87fyv
3.7 spacing
0.4×300
for links = 0.2998 therefore, 0.33 ok
0.87×460 Provide
Provide y8@300 c/c for links spacing y8@300 c/c
Minimum shear links will be used up to V = VC + 0.4 = 0.827 + 0.4 = for links
1.227 N/mm2 spacing
Shear load (minimum links) = (1.227) (300) (600)10−3 = 220.86 kN
220.86 > 189.29 > 170.41 hence minimum links to be used
throughout

Anchorage of links
It should pass round another bar of at least its own size, through an
angle of 180°, and continue for a length of at least four times its own
size

Deflection
b = 1455 mm
bw =300
Span/effective depth ratio for the beam
bw 300
= 1455 = 0.206 therefore, it is continuous -20.8 bw
b
b
= 0.206
span
Basic ratio = 20.8
depth
Tension steel modification factor
m 410.90 × 106
= = 0.869
bd2 1455 × 5702
1.50 − 1.38
1.38 + ( )(1.0 − 0.869)
1.0 − 0.75
1.38+ 0.06288 = 1.44288 allowable
Therefore, allowable span/depth ratio; span/depth
= 1.44288×20.8 = 30.01 ratio = 30.01
span 8.25
Actual depthratio = 0.57 = 14.474 < 30.01
Deflection ok.
Deflection
ok.
Designed section output

76 | P a g e
77 | P a g e
 BEAM A1-A6/C1-C6

Height of the beam = 600 mm

bw = 300 mm
hf = 180mm

Dimensions of the L – beam

b for span A1-A2 = A5-A6 = C1-C2 = C5-C6 = bw + (0.1) (0.7) (l)

= 0.3 + (0.1×0.7×5.15 = 660.5 mm

b for span A2-A3, A3-A4, A4-A5 = C2-C3 = C3-C4 =C4-C5 = bw + (0.1) (0.7) (l)
= 0.3 + (0.1×0.7×4.2 = 594 mm

Calculation of the areas of influence on the L – beam

Beam A1-A6 = beam C1-C6
Area a1
1 1
2
bh = 2 × 5.15 × 2.575 = 6.63m2

78 | P a g e
 Area a2
1 1
2
bh = 2 × 4.2 × 2.1 = 4.41m2
Area a3
1 1
2
bh = 2 × 4.2 × 2.1 = 4.41m2
Area a4
1 1
2
bh = 2 × 4.2 × 2.1 = 4.41m2
Area a5
1 1
2
bh = 2 × 5.15 × 2.575 = 6.63m2

Calculation of dead load on the beam along the axis

Dead load on A1-A2

• Self-weight of slab on span A1-A2 = 6.63×0.180×
24kn/m3 = 28.6416 kN
• Finishes = 6.63×0.6kn/m2 = 3.978KN
• Masonry wall = 5.15×2.5×0.2×22kn/m3 = 56.65KN
• Self-weight of beam on span A1-A1=A5-A6
= (0.6-0.18) ×5.15×0.3×24kn/m3= 15.5736KN
Total dead load on span A1-A2 = A5-A6
=28.6416+3.978+56.65+15.5736 = 104.8432KN

Calculations of live load on the beam along the axis

Because of the purpose of the building (residential apartment), we assume live load is taken as
1.50kn/m2

Calculation of dead load per meter on the beam (uniformly distributed load)

104.8432
Span A1-A2=A5-A6 = 5.15
= 20.358kn/m
Case (I)
Calculation of combination of load
Design load, N = 1.4gk + 1.6qk
N = (1.4×20.358) + (1.6×1.5) = 30.90kn/m

Case (II)
Calculation of combination of load
Design load, N = 1.0gk + 1.0qk
N = (1.0×20.358) + (1.0×1.5) = 21.858kn/m

Dead load ON A2-A3

• Self-weight of slab on span A2-A3 = 4.41×0.180×
24kn/m3 = 19.0512 kN
• Finishes = 4.41×0.6kn/m2 = 2.646KN
• Masonry wall = 4.20×2.5×0.2×22kn/m3 = 46.2KN

79 | P a g e
 • Self-weight of beam on span A2-A3/A3-A4/A4-A5
= (0.6-0.18) ×4.20×0.3×24kn/m3= 12.70KN
Total dead load on spans between A2-A5
=19.0512+2.646+46.2+12.70 = 80.598KN

Calculations of live load on the beam along the axis

Because of the purpose of the building (residential apartment), we assume live load is taken as
1.50kn/m2

Calculation of dead load per meter on the beam (uniformly distributed load)
80.598
Span UDL = 4.2
= 19.19kn/m

Case (I)
Calculation of combination of load
Design load, N = 1.4gk + 1.6qk
N = (1.4×19.19) + (1.6×1.5) = 29.266kn/m

Case (II)
Calculation of combination of load
Design load, N = 1.0gk + 1.0qk
N = (1.0×19.19) + (1.0×1.5) = 20.69kn/m

Calculation of moments and shear forces using cross’s moment distribution method
Fixed end moment (FEM) calculations

Case (I) – FEM

wL2 30.90×5.15×5.15
MA1-A2 = − =− = −68.30kNm
12 12
MA2-A1 = +68.30KNm

wL2 29.266×4.2×4.2
MA2-A3 = − 12
=− 12
= −43.02kNm
MA3-A4 = +43.02KNm

80 | P a g e
 Case (II) – FEM

wL2 30.90×5.15×5.15
MA1-A2 = − 12
=− 12
= −68.30kNm
MA2-A1 = +68.30KNm

wL2 20.69×4.2×4.2
MA3-A2 = − 12
=− 12
= −30.41kNm
MA2-A3 = +30.41KNm

wL2 29.266×4.2×4.2
MA4-A3 = − =− = −43.02kNm
12 12
MA3-A4 = +43.02KNm

wL2 20.69×4.2×4.2
MA5-A4 = − 12
=− 12
= −30.41kNm
MA4-A5 = +30.41KNm

81 | P a g e
 Stiffness factors for the beams and columns

Beam trial - 300×600 mm

Column trial - 500×300 mm
Beam flexural rigidity is given by;
bd2 0.3 × 0.63
I= = = 5.4 × 10−3 m4
12 12
flexural rigidity
Beam stiffness k = length
Span A1-A2 and span A5-A6
5.4 × 10−3 m4
k A1−A2 = k A5−A6 = = 1.0485 × 10−3
5.15
Span 2-3, 3-4 and span 4-5
5.4 × 10−3 m4
k A2−A1 = k A2−A3 = = 1.2857 × 10−3
4.2
Columns
bd2 0.3 × 0.53
I= = = 3.125 × 10−3 m4
12 12
3.125 × 10−3 m4
k Col = = 1.25 × 10−3
2.5

Calculation of distribution factors

Joint A1/A6/C1/C6.
Total joint stiffness = beam stiffness + (column stiffness)2
(1.0485× 10−3 ) + (1.25 × 10−3 )2 = 3.5485 × 10−3
1.0485 × 10−3
DFA1−A2 = DFA6−A5= = 0.3
3.5485 × 10−3
2.5 × 10−3
DFCOL = = 0.70
3.5485 × 10−3

Joint A2/C2
Total joint stiffness = beam stiffness + (column stiffness)2
(6.545 × 10−4 ) + (1.0485 × 10−3 ) + (1.25 × 10−3 )2 = 4.203 × 10−3
6.545 × 10−4
DFA2−A1 = = 0.16
4.203 × 10−3
1.0485 × 10−3
DFA2−A3 = = 0.25
4.203 × 10−3
2.5 × 10−3
DFCOL = = 0.59
4.203 × 10−3

Joint A3/C3
Total joint stiffness = beam stiffness + (column stiffness)2
(1.0485× 10−3 ) + (1.0485 × 10−3 ) + (1.25 × 10−3 )2 = 4.597 × 10−3
1.0485 × 10−3
DFA3−A2 = = 0.23
4.597 × 10−3

82 | P a g e
 1.0485 × 10−3
DFA2−A3 = = 0.23
4.597 × 10−3
2.5 × 10−3
DFCOL = = 0.54
4.597 × 10−3
Joint A4/C4
Total joint stiffness = beam stiffness + (column stiffness)2
(1.0485× 10−3 ) + (1.0485 × 10−3 ) + (1.25 × 10−3 )2 = 4.597 × 10−3
1.0485 × 10−3
DFA4−A3 = = 0.23
4.597 × 10−3
1.0485 × 10−3
DFA4−A5 = = 0.23
4.597 × 10−3
2.5 × 10−3
DFCOL = = 0.54
4.597 × 10−3

Joint A5/C5
Total joint stiffness = beam stiffness + (column stiffness)2
(1.0485× 10−3 ) + (6.545 × 10−4 ) + (1.25 × 10−3 )2 = 4.203 × 10−3
1.0485 × 10−3
DFA5−A4 = = 0.25
4.203 × 10−3
6.545 × 10−4
DFA5−A6 = = 0.16
4.203 × 10−3
2.5 × 10−3
DFCOL = = 0.59
4.203 × 10−3

Moment distribution table using balanced method

Case (I)

JOIN A1 A2 A3 A4 A5 A6
T
MEM COL A1- A2- COL A2- A3- COL A3- A4- COL A4- A5- COL A5- A6- COL
BER A2 A1 A3 A2 A4 A3 A5 A4 A6 A5
DF 0.7 0.3 0.16 0.59 0.25 0.23 0.54 0.23 0.23 0.54 0.23 0.25 0.59 0.16 0.3 0.7

FEM 0 -68.3 68.3 0 - 43.02 0 - 43.02 0 - 43.02 0 -68.3 68.3 0

43.0 43.02 43.02
2
BAL 47.81 20.4 - - -6.32 0 0 0 0 0 0 6.32 14.91 4.044 - -
9 4.044 14.9 52 8 20.49 47.8
8 152 1
CO 0 - 10.24 0 0 -3.16 0 0 0 0 3.16 0 0 - 2.022 0
2.02 5 10.24 4
24 5
BAL 1.415 0.60 - - - 0.726 1.706 0.726 - - - 2.561 6.044 1.639 - -
68 672 1.639 6.04 2.56 8 4 8 0.726 1.70 0.726 25 55 2 0.606 1.41
2 455 125 8 64 8 72 568
CO 0 - 0.303 0 0.36 - 0 - 0.363 0 1.280 - 0 - 0.819 0
0.81 36 34 1.280 0.363 4 625 0.363 0.303 6
96 63 4 4 36
BAL 0.573 0.24 - - - 0.378 0.887 0.378 - - - 0.166 0.393 0.106 - -
72 588 0.106 0.39 0.16 126 774 126 0.378 0.88 0.378 69 388 682 0.245 0.57
68 339 669 13 777 13 88 372
TOTA 49.61 - 73.09 - - 39.64 2.801 - 42.42 - - 51.56 21.56 - 49.85 -
L 611 49.6 122 21.5 51.5 298 393 42.42 559 2.78 39.63 569 88 73.13 719 49.8
161 379 533 76 78 78 45 572

Case (II)

83 | P a g e
 JOIN A1 A2 A3 A4 A5 A6
T
MEM COL A1- A2- COL A2- A3- COL A3- A4- COL A4- A5- COL A5- A6- COL
BER A2 A1 A3 A2 A4 A3 A5 A4 A6 A5
DF 0.7 0.3 0.16 0.59 0.25 0.23 0.54 0.23 0.23 0.54 0.23 0.25 0.59 0.16 0.3 0.7

FEM 0 -68.3 68.3 0 - 30.41 0 - 43.02 0 - 31.41 0 -68.3 68.3 0

30.41 43.02 30.41
BAL 47.81 20.49 - - - 2.900 6.809 2.900 - - - 9.222 21.76 5.902 - -
6.062 22.3 9.472 3 4 3 2.900 6.80 2.900 5 51 4 20.49 47.8
4 551 5 3 94 3 1
CO 0 - 10.24 0 1.450 - 0 - 1.450 0 4.611 - 0 - 2.951 0
3.031 5 15 4.736 1.450 15 25 1.450 10.24 2
2 25 15 15 5
BAL 2.121 0.909 - - - 1.422 3.340 1.422 - - - 2.923 6.900 1.871 - -
84 36 1.871 6.90 2.923 872 656 872 1.394 3.27 1.394 788 139 224 0.885 2.06
22 014 79 12 316 12 36 584
CO 0 - 0.454 0 0.711 - 0 - 0.711 0 1.461 - 0 - 0.935 0
0.935 68 436 1.461 0.697 436 894 0.697 0.442 612
61 89 06 06 68
BAL 0.654 0.280 - - - 0.496 1.165 0.496 - - - 0.284 0.672 0.182 - -
928 684 0.186 0.68 0.291 56 836 56 0.499 1.17 0.499 935 447 359 0.280 0.65
58 801 53 87 36 87 68 493
CO 0 0.327 0.140 0 0.248 - 0 - 0.248 0 0.142 - 0 - 0.091 0
464 342 28 0.145 0.249 28 468 0.249 0.140 179
76 93 93 34
BAL - - - - - 0.091 0.213 0.091 - - - 0.097 0.230 0.062 - -
0.229 0.098 0.062 0.22 0.097 01 677 01 0.089 0.21 0.089 569 262 444 0.027 0.06
22 24 18 929 16 87 1 87 35 383
TOTA 50.38 - 70.91 - - 28.95 11.61 - 40.57 - - 41.50 29.61 - 50.62 -
L 003 50.38 115 30.1 40.73 157 218 40.53 687 11.5 29.04 28 365 71.11 076 50.6
806 06 97 3 69 65 208

Final Beam moments

JOINT A1 A2 A3 A4 A5 A6

MEMBE COL A1- A2- COL A2- A3- COL A3- A4- CO A4- A5- COL A5- A6- COL
R A2 A1 A3 A2 A4 A3 L A5 A4 A6 A5
Case (I) 49.61 - 73.09 - - 39.64 2.801 - 42.42 - - 51.56 21.56 - 49.85 -
Loading 611 49.6 122 21.5 51.5 298 393 42.4 559 2.78 39.6 569 88 73.1 719 49.8
161 379 533 276 78 378 345 572
Case (II) 50.38 - 70.91 - - 28.95 11.61 - 40.57 - - 41.50 29.61 - 50.62 -
loading 0 50.3 115 30.1 40.7 157 218 40.5 687 11.5 29.0 28 365 71.1 076 50.6
8 806 306 397 3 469 165 208
Output 50.38 - 73.09 - - 39.64 11.61 - 42.42 - - 51.56 29.61 - 49.85 50.6
moments 0 49.6 12 30.1 51.5 298 218 42.4 559 11.5 39.6 569 365 73.1 719 208
161 806 533 276 3 378 345

84 | P a g e
 Calculation of reactions in the beam

49.62 73.09

A1 A2
Moment about A1, MA1;
5.15
= - RA2×5.15 + 73.09 – 49.62 + (159.135× )
2
5.15RA2 = 433.24
RA2 = 84.12KN
Moment about A2, MA2;
5.15
= 5.15RA1 + 73.496 – 49.62– (159.135 × )
2
RA1 = 74.93KN

51.55 39.64

A2 A3
Moment about A2, MA2;
4.2
= - RA3×4.2 – 51.55 + 39.64 +(122.917× )
2
RA3 = 58.63KN
RA2 = 122.917 – 58.63 = 64.289 KN

42.43 42.43

A3 A4
Moment about A3, MA3;
4.2
= - RA4×4.2 – 42.43 + 42.43+(122.917× 2 )
RA4 = 61.4585KN
RA3 = 122.917 – 61.4585 = 61.4585 KN

39.64 51.566

A4 A5
Moment about 4, M4;
4.2
= - RA5×4.2 – 39.64 + 51.566 +(122.917× 2 )
RA5 = 64.298KN
RA4 = 122.917 – 68.448 = 58.62 KN
73.13 49.86

A5 A6

85 | P a g e
 Moment about A5, MA5;
5.15
= - RA6×5.15 – 73.13 + 49.86 + (159.135× )
2
RA6 = 75.049KN
RA5 = 159.135 – 66.488 = 84.10 KN

Shear forces and bending moment diagrams

a) Shear forces diagram

Span Design moment

Span A1-A2
74.93
x = 30.90 = 2.42 m from A1
Design moment = Reaction moment - Load moment
2.42
Span design moment = (74.93×2.42) – (30.90× 2.42 × ) =90.849kNm
2

Span A2-A3
64.289
x = 29.266 = 2.20 m from A2
Design moment = Reaction moment - Load moment
2.20
Span design moment = (64.289×2.20) – (29.266× 2.20 × 2 ) = 70.61kNm

Span A3-A4

61.4585
x = 29.266 = 2.10 m from A3
Design moment = Reaction moment - Load moment
2.10
Span design moment = (61.46×2.10) – (29.266× 2.10 × 2 ) = 64.53kNm

Span A4-A5
58.62
x = 29.266 = 2.00 m from A4
Design moment = Reaction moment - Load moment
2.00
Span design moment = (58.62×2.00) – (29.266× 2.00 × ) = 58.708kNm
2

86 | P a g e
 Span A5-A6

84.10
x = 30.90 = 2.72 m from A5
Design moment = Reaction moment - Load moment
2.72
Span design moment = (84.10×2.72) – (30.90× 2.72 × ) = 114.447kNm
2

b) Bending moment diagram

87 | P a g e
 Required steel reinforcement in the beam

Span A5-A6
Maximum hogging moment = maximum moment at support A5= M max = 73.13KN
= maximum moment at support A6= M max = 49.857KN
Maximum sagging moment = maximum moment at midspan = 114.447kNm
Maximum shear force = 84.10
A5 A6

➢ Cover = 20 mm
➢ Bar diameter = 20mm
20
➢ Effective depth = 600 – 20 – = 570mm
2
➢ hf = 180
➢ Beam type – L-beam
➢ Flange width b = 660.5mm

Moment of resistance MU
h
= 0.45fcu bhf (d − f) 10−6
2
180
0.45 × 40 × 660.5 × 180(570 − 2
) × 10−6
=1027.2096KN
1027.2096kn > 114.447KN/m
Therefore, no compression reinforcement required

h
m = 0.87fy A s (d − 2f)

m 114.447×106
As = h = 0.87×460×480 = 595.78mm2
0.87fy (d− f )
2

Provide T2Y20 bottom bars (As = 628mm2 )

Provide T2Y8 hanger bars (As = 101mm2 )

88 | P a g e
 Design for support steel at A5
M = 73.10knm
Section size = 600×300
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 73.10kN
Compression steel not required at supports

m 73.10 × 106
k= = = 0.0187
fcu bd2 40 × 300 × 5702
z
= 0.94
d
Tension reinforcement;
m 73.10 × 106
As = = = 340.90mm2
0.87fy z 0.87 × 460 × 0.94 × 570
Provide T2Y16 bottom bars (As = 402mm2 )

Design for support steel at A6

M = 49.857knm
Section size = 600×300
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 49.857kN
Compression steel not required at supports

m 49.857 × 106
k= = = 0.0127
fcu bd2 40 × 300 × 5702
z
= 0.94
d
Tension reinforcement;
m 49.857 × 106
As = = = 232.51mm2
0.87fy z 0.87 × 460 × 0.94 × 570
Provide T2Y16 bottom bars (As = 402mm2 )

The design Shear stress

v
Design shear stress at support A5
bv d
V =84.10kN
Bv = 300mm
D = 570mm
84.10 × 1000
= 0.492N/mm2
300 × 570

V =0.8√fcu = 40 = 5.1N/mm2

0.492 < 5.1N/mm2 therefore, shear stress ok.

Design concrete shear stress, VC.

89 | P a g e
 100AS 100 × 628
= = 0.367
bv d 300 × 570
0.59 − 0.47
VC = (0.367 − 0.25) + 0.47 = 0.526
0.5 − 0.25
V < 0.5VC = 0.5 × 0.526 = 0.263N/mm2
v = 0.492 > 0.263 = vc

90 | P a g e
 (VC + 0.4) < V
Therefore, provide links as follows;
Asv (0.4)bv
∴ (min. links) ≥
Sv 0.87fyv
0.4×300
0.87×460
= 0.2998 therefore, 0.33 ok
Provide y8@300 c/c for links spacing
Minimum shear links will be used up to V = VC + 0.4 = 0.526 + 0.4 = 0.926 N/mm2
Shear load (minimum links) = (0.926) (300) (600)10−3 = 166.68 kN
166.68 > 84.10 > 75.049 hence minimum links to be used throughout
Anchorage of links
It should pass round another bar of at least its own size, through an angle of 180,
and continue for a length of at least four times its own size

Deflection
b = 660.5 mm
bw =300
Span/effective depth ratio for the beam
bw 300
= = 0.454 therefore, it is continuous -22.3
b 660.5
span
Basic ratio = 22.3
depth
Tension steel modification factor
m 114.447 × 106
= = 0.533
bd2 660.5 × 5702
From table
0.533< 0.5
Therefore, allowable span/depth ratio;
1.68 − 1.5
1.68 − ( )(0.75 − 0.533)
0.75 − 0.5
= 1.68 – 0.15624 = 1.523
= 1.523×22.3 = 33.98
span 5.15
Actual depthratio = 0.57 = 9.035 < 33.98
Deflection ok.

Designed section output

91 | P a g e
 Span A4-A5
Maximum hogging moment = maximum moment at support A5= M max = 51.57KN
= maximum moment at support A4=M max = 39.638KN
Maximum sagging moment = maximum moment at midspan = 58.708kNm
Maximum shear force = 64.298
A4 A5

➢ Cover = 20 mm
➢ Bar diameter = 20mm
20
➢ Effective depth = 600 – 20 – 2
= 570mm
➢ hf = 180
➢ Beam type – L-beam
➢ Flange width b = 594mm

Moment of resistance MU
hf
= 0.45fcu bhf (d − ) 10−6
2
180
0.45 × 40 × 594 × 180(570 − 2
) × 10−6
=923.78KN
923.78kn > 58.708KN/m
Therefore, no compression reinforcement required

h
m = 0.87fy As (d − f)
2

m 58.708×106
As = h = = 305.62mm2
0.87fy (d− f ) 0.87×460×480
2

92 | P a g e
 Provide T2Y16 bottom bars (As = 402mm2 )
Provide T2Y8 hanger bars (As = 101mm2 )

Design for support steel at A5

M = 51.5656knm
Section size = 600×300
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 51.5656kn
Compression steel not required at supports

m 51.5656 × 106
k= = = 0.0132
fcu bd2 40 × 300 × 5702
z
= 0.94
d
Tension reinforcement;
m 51.5656 × 106
As = = = 240.48mm2
0.87fy z 0.87 × 460 × 0.94 × 570
Provide T2Y16 bottom bars (As = 402mm2 )

Design for support steel at A4

M = 39.638knm
Section size = 600×300
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 39.638kn
Compression steel not required at supports

m 39.638 × 106
k= = = 0.010
fcu bd2 40 × 300 × 5702
z
= 0.94
d
Tension reinforcement;
m 39.638 × 106
As = = = 184.86mm2
0.87fy z 0.87 × 460 × 0.94 × 570
Provide T2Y12 bottom bars (As = 226mm2 )

The design Shear stress

v
Design shear stress b d at support A5
v
V =64.298kN
Bv = 300mm
D = 570mm
64.298 × 1000
= 0.376N/mm2
300 × 570

V =0.8√fcu = 40 = 5.1N/mm2

93 | P a g e
 0.376 < 5.1N/mm2 therefore, shear stress ok.

Design concrete shear stress, VC.

100AS 100 × 402
= = 0.235
bv d 300 × 570
0.47 − 0.40
VC = (0.235 − 0.15) + 0.40 = 0.4595
0.25 − 0.15
V < 0.5VC = 0.5 × 0.4595 = 0.229N/mm2
v = 0.376 > 0.229 = vc

(VC + 0.4) < V

Therefore, provide links as follows;
Asv (0.4)bv
∴ (min. links) ≥
Sv 0.87fyv
0.4×300
0.87×460
= 0.2998 therefore, 0.33 ok
Provide y8@300 c/c for links spacing
Minimum shear links will be used up to V = VC + 0.4 = 0.4595 + 0.4 = 0.8595
N/mm2
Shear load (minimum links) = (0.8595) (300) (600)10−3 = 154.71 kN
154.71 > 64.298 > 58.62 hence minimum links to be used throughout
Anchorage of links
It should pass round another bar of at least its own size, through an angle of 180°,
and continue for a length of at least four times its own size

Deflection
b= 594 mm
bw =300
Span/effective depth ratio for the beam
bw 300
b
= 594 = 0.505 therefore, it is continuous -22.3
span
Basic depth ratio = 22.3
Tension steel modification factor
m 58.708 × 106
= = 0.304
bd2 594 × 5702
From table
0.304< 0.5
Therefore, allowable span/depth ratio;
= 1.68 ok
= 1.68×22.3 = 37.464
span 5.15
Actual depthratio = 0.57 = 9.035 < 37.464
Deflection ok.

Designed section output

94 | P a g e
 Span A3-A4
Maximum hogging moment = maximum moment at support A3= M max = 42.43KN
= maximum moment at support A4= M max = 42.43KN
Maximum sagging moment = maximum moment at midspan = 64.53kNm
Maximum shear force = 61.4585
A3 A4

➢ Cover = 20 mm
➢ Bar diameter = 20mm
20
➢ Effective depth = 600 – 20 – = 570mm
2
➢ hf = 180
➢ Beam type – L-beam
➢ Flange width b = 594mm

Moment of resistance MU
hf
= 0.45fcu bhf (d − ) 10−6
2
180
0.45 × 40 × 594 × 180(570 − 2
) × 10−6
=923.78KN
923.78kn > 64.53KN/m
Therefore, no compression reinforcement required

h
m = 0.87fy As (d − 2f)

m 64.53×106
As = h = = 335.93mm2
0.87fy (d− f ) 0.87×460×480
2

95 | P a g e
 Provide T2Y16 bottom bars (As = 402mm2 )
Provide T2Y8 hanger bars (As = 101mm2 )

Design for support steel at A3

M = 42.43knm
Section size = 600×300
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 42.43kn
Compression steel not required at supports

m 42.43 × 106
k= = = 0.01
fcu bd2 40 × 300 × 5702
z
= 0.94
d
Tension reinforcement;
m 42.43 × 106
As = = = 197.88mm2
0.87fy z 0.87 × 460 × 0.94 × 570
Provide T2Y12 bottom bars (As = 226mm2 )

Design for support steel at A4

M = 42.43knm
Section size = 600×300
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 42.43kn
Compression steel not required at supports

m 42.43 × 106
k= = = 0.01
fcu bd2 40 × 300 × 5702
z
= 0.94
d
Tension reinforcement;
m 42.43 × 106
As = = = 197.88mm2
0.87fy z 0.87 × 460 × 0.94 × 570
Provide T2Y12 bottom bars (As = 226mm2 )

The design Shear stress

v
Design shear stress at support A4
bv d
V =61.4585kN
Bv = 300mm
D = 570mm
61.4585 × 1000
= 0.359N/mm2
300 × 570

V =0.8√fcu = 40 = 5.1N/mm2

96 | P a g e
 0.359 < 5.1N/mm2 therefore, shear stress ok.

Design concrete shear stress, VC.

100AS 100 × 402
= = 0.235
bv d 300 × 570
0.47 − 0.40
VC = (0.235 − 0.15) + 0.40 = 0.4595
0.25 − 0.15
V < 0.5VC = 0.5 × 0.4595 = 0.229N/mm2
v = 0.359 > 0.229 = vc

(VC + 0.4) < V

Therefore, provide links as follows;
Asv (0.4)bv
∴ (min. links) ≥
Sv 0.87fyv
0.4×300
0.87×460
= 0.2998 therefore, 0.33 ok
Provide y8@300 c/c for links spacing
Minimum shear links will be used up to V = VC + 0.4 = 0.4595 + 0.4 = 0.8595
N/mm2
Shear load (minimum links) = (0.8595) (300) (600)10−3 = 154.71 kN
154.71 > 61.4585 hence minimum links to be used throughout
Anchorage of links
It should pass round another bar of at least its own size, through an angle of 180°,
and continue for a length of at least four times its own size

Deflection
b = 594 mm
bw =300
Span/effective depth ratio for the beam
bw 300
b
= 594 = 0.505 therefore, it is continuous -22.3
span
Basic ratio = 22.3
depth
Tension steel modification factor
m 64.53 × 106
= = 0.334
bd2 594 × 5702
From table
0.334< 0.5
Therefore, allowable span/depth ratio;
= 1.68 ok
= 1.68×22.3 = 37.464
span 5.15
Actual depthratio = 0.57 = 9.035 < 37.464
Deflection ok.

97 | P a g e
 Designed section output

Span A2-A3
Maximum hogging moment = maximum moment at support A2= M max = 51.55KN
= maximum moment at support A3= M max = 39.64KN
Maximum sagging moment = maximum moment at midspan = 70.61kNm
Maximum shear force = 64.289
A2 A3

➢ Cover = 20 mm
➢ Bar diameter = 20mm
20
➢ Effective depth = 600 – 20 – = 570mm
2
➢ hf = 180
➢ Beam type – L-beam
➢ Flange width b = 594mm

Moment of resistance MU
hf
= 0.45fcu bhf (d − ) 10−6
2
180
0.45 × 40 × 594 × 180(570 − ) × 10−6
2
=923.78KN
923.78kn > 70.61KN/m
Therefore, no compression reinforcement required

h
m = 0.87fy As (d − 2f)

98 | P a g e
 m 70.61×106
As = h = 0.87×460×480 = 367.58mm2
0.87fy (d− f )
2

Provide T2Y16 bottom bars (As = 402mm2 )

Provide T2Y8 hanger bars (As = 101mm2 )

Design for support steel at A2

M = 51.55knm
Section size = 600×300
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 51.55kn
Compression steel not required at supports

m 51.55 × 106
k= = = 0.0132
fcu bd2 40 × 300 × 5702
z
= 0.94
d
Tension reinforcement;
m 51.55 × 106
As = = = 240.41mm2
0.87fy z 0.87 × 460 × 0.94 × 570
Provide T2Y16 bottom bars (As = 402mm2 )

Design for support steel at A3

M = 39.64knm
Section size = 600×300
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 39.64kn
Compression steel not required at supports

m 39.64 × 106
k= = = 0.010
fcu bd2 40 × 300 × 5702
z
= 0.94
d
Tension reinforcement;
m 39.64 × 106
As = = = 184.86mm2
0.87fy z 0.87 × 460 × 0.94 × 570
Provide T2Y12 bottom bars (As = 226mm2 )

The design Shear stress

v
Design shear stress b d at support A2
v
V =64.289kN
Bv = 300mm
D = 570mm
64.289 × 1000
= 0.376N/mm2
300 × 570

99 | P a g e
 V =0.8√fcu = 40 = 5.1N/mm2

0.376 < 5.1N/mm2 therefore, shear stress ok.

Design concrete shear stress, VC.

100AS 100 × 402
= = 0.235
bv d 300 × 570
0.47 − 0.40
VC = (0.235 − 0.15) + 0.40 = 0.4595
0.25 − 0.15
V < 0.5VC = 0.5 × 0.4595 = 0.229N/mm2
v = 0.376 > 0.229 = vc

(VC + 0.4) < V

Therefore, provide links as follows;
Asv (0.4)bv
∴ (min. links) ≥
Sv 0.87fyv
0.4×300
0.87×460
= 0.2998 therefore, 0.33 ok
Provide y8@300 c/c for links spacing
Minimum shear links will be used up to V = VC + 0.4 = 0.4595 + 0.4 = 0.8595
N/mm2
Shear load (minimum links) = (0.8595) (300) (600)10−3 = 154.71 kN
154.71 > 64.289 > 58.63 hence minimum links to be used throughout
Anchorage of links
It should pass round another bar of at least its own size, through an angle of 180°,
and continue for a length of at least four times its own size

Deflection
b = 594 mm
bw =300
Span/effective depth ratio for the beam
bw 300
b
= 594 = 0.505 therefore, it is continuous -22.3
span
Basic depth ratio = 22.3
Tension steel modification factor
m 70.61 × 106
= = 0.366
bd2 594 × 5702
From table
0.366< 0.5
Therefore, allowable span/depth ratio;
= 1.68 ok
= 1.68×22.3 = 37.464

100 | P a g e
 span 5.15
Actual depthratio = 0.57 = 9.035 < 37.464
Deflection ok.

Designed section output

Span A1-A2
Maximum hogging moment = maximum moment at support A2=M max = 73.09KN
= maximum moment at support A1= M max = 49.616KN
Maximum sagging moment = maximum moment at midspan = 90.849kNm
Maximum shear force = 84.12
A1 A2

➢ Cover = 20 mm
➢ Bar diameter = 20mm
20
➢ Effective depth = 600 – 20 – 2
= 570mm
➢ hf = 180
➢ Beam type – L-beam
➢ Flange width b = 660.5mm

Moment of resistance MU
hf
= 0.45fcu bhf (d − ) 10−6
2
180
0.45 × 40 × 660.5 × 180(570 − ) × 10−6
2
=1027.2096KN
1027.2096kn > 90.849KN/m
Therefore, no compression reinforcement required

h
m = 0.87fy As (d − f)
2

101 | P a g e
 m 90.849×106
As = h = = 472.94mm2
0.87fy (d− f ) 0.87×460×480
2

Provide T2Y20 bottom bars (As = 628mm2 )

Provide T2Y8 hanger bars (As = 101mm2 )

Design for support steel at A2

M = 73.09knm
Section size = 600×300
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 73.09kn
Compression steel not required at supports

m 73.09 × 106
k= = = 0.0187
fcu bd2 40 × 300 × 5702
z
= 0.94
d
Tension reinforcement;
m 73.09 × 106
As = = = 340.86mm2
0.87fy z 0.87 × 460 × 0.94 × 570
Provide T2Y16 bottom bars (As = 402mm2 )

Design for support steel at A1

M = 49.616knm
Section size = 600×300
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 49.616kn
Compression steel not required at supports

m 49.616 × 106
k= = = 0.0127
fcu bd2 40 × 300 × 5702
z
= 0.94
d
Tension reinforcement;
m 49.616 × 106
As = = = 231.389mm2
0.87fy z 0.87 × 460 × 0.94 × 570
Provide T2Y16 bottom bars (As = 402mm2 )

The design Shear stress

v
Design shear stress b d at support A2
v
V =84.12kN
Bv = 300mm
D = 570mm

102 | P a g e
 84.12 × 1000
= 0.492N/mm2
300 × 570

V =0.8√fcu = 40 = 5.1N/mm2

0.492 < 5.1N/mm2 therefore, shear stress ok.

Design concrete shear stress, VC.

100AS 100 × 628
= = 0.367
bv d 300 × 570
0.59 − 0.47
VC = (0.367 − 0.25) + 0.47 = 0.526
0.5 − 0.25
V < 0.5VC = 0.5 × 0.526 = 0.263N/mm2
v = 0.492 > 0.263 = vc

(VC + 0.4) < V

Therefore, provide links as follows;
Asv (0.4)bv
∴ (min. links) ≥
Sv 0.87fyv
0.4×300
= 0.2998 therefore, 0.33 ok
0.87×460
Provide y8@300 c/c for links spacing
Minimum shear links will be used up to V = VC + 0.4 = 0.526 + 0.4 = 0.926 N/mm2
Shear load (minimum links) = (0.926) (300) (600)10−3 = 166.68 kN
166.68 > 84.12 > 74.93 hence minimum links to be used throughout
Anchorage of links
It should pass round another bar of at least its own size, through an angle of 180°,
and continue for a length of at least four times its own size

Deflection
b = 660.5 mm
bw =300
Span/effective depth ratio for the beam
bw 300
b
= 660.5 = 0.454 therefore, it is continuous -22.3
span
Basic depth ratio = 22.3
Tension steel modification factor
m 90.849 × 106
= = 0.423
bd2 660.5 × 5702
From table
0.423< 0.5
Therefore, allowable span/depth ratio;
= 1.68 ok
= 1.68×22.3 = 37.464

103 | P a g e
 span 5.15
Actual depthratio = 0.57 = 9.035 < 37.464
Deflection ok.

Designed section output

104 | P a g e
 BEAM E - E

Height of the beam = 600 mm

bw = 300 mm
hf = 180mm

Dimensions of the L – beam

b for span B1-B2 and B5-B6 = bw + (0.2) (0.7) (l)

105 | P a g e
 = 0.3 + (0.2×0.7×5.15) = 1021 mm

b for span B2-B3, B3-B4 and B4-B5 = bw + (0.2) (0.7) (l)

= 0.3 + (0.2×0.7×4.2) = 888 mm

Calculation of the areas of influence on the L – beam

Total area B1 =b1+b1
1 1
2
bh = 2 × 5.15 × 2.575 = 6.63m2 × 2 = 13.26m2

Total area B2 = b2+b2

1 1
bh = × 4.2 × 2.1 = 4.41m2 × 2 = 8.82m2
2 2

Total area B3 = b3+b3

1 1
2
bh = 2 × 4.2 × 2.1 = 4.41m2 × 2 = 8.82m2

Total area B4 = b4+b4

1 1
2
bh = 2 × 4.2 × 2.1 = 4.41m2 × 2 = 8.82m2
Total area B5 = b5+b5
1 1
2
bh = 2 × 5.15 × 2.575 = 6.63m2 × 2 = 13.26m2

Calculation of dead load on the beam along the axis

Static chart calculation

Dead load beam B1-B2 and beam B5-B6
• Self-weight of slab on span B1-B2 = 13.26×0.180×
24kn/m3 = 57.2832 kN
• Finishes = 13.26×0.6kn/m2 = 7.956KN
• Masonry wall = 5.15×2.5×0.2×22kn/m3 = 56.65KN
• Self-weight of beam on span B1-B2=B5-B6
= (0.6-0.18) ×5.15×0.3×24kn/m3= 15.5736KN
Total dead load on span B1-B2 = B5-B6
=57.2832+7.956+56.65+15.5736 = 137.4628KN

Calculations of live load on the beam along the axis

Because of the purpose of the building (residential apartment), we assume live load is taken as
1.50kn/m2

106 | P a g e
 Calculation of dead load per meter on the beam (uniformly distributed load)
137.4628
Span B1-B2=B5-B6 = 5.15
= 26.691kn/m

Case (I)
Calculation of combination of load
Design load, N = 1.4gk + 1.6qk
N = (1.4×26.691) + (1.6×1.5) = 39.769kn/m

Case (II)
Calculation of combination of load
Design load, N = 1.0gk + 1.0qk
N = (1.0×26.691) + (1.0×1.5) = 28.191kn/m
Dead load on beam B2-B3, B3-B4, and B4-B5
• Self-weight of slab on span B2-B3 = 8.82×0.180×
24kn/m3 = 38.1024 kN
• Finishes = 8.82×0.6kn/m2 = 5.292KN
• Masonry wall = 4.20×2.5×0.2×22kn/m3 = 46.2KN
• Self-weight of beam on each span between B2 and B5
= (0.6-0.18) ×4.20×0.3×24kn/m3= 12.70KN
Total dead load on each span
=38.1024+5.292+46.2+12.70 = 102.2944KN
Calculations of live load on the beam along the axis
Because of the purpose of the building (residential apartment), we assume live load is taken as
1.50kn/m2
Calculation of dead load per meter on the beam (uniformly distributed load)

102.2944
Span UDL = = 24.356kn/m
4.2

Case (I)
Calculation of combination of load
Design load, N = 1.4gk + 1.6qk
N = (1.4×24.356) + (1.6×1.5) = 36.498kn/m

Case (II)

107 | P a g e
 Calculation of combination of load
Design load, N = 1.0gk + 1.0qk
N = (1.0×24.356) + (1.0×1.5) = 25.856kn/m

Calculation of moments and shear forces using cross’s moment distribution method
Fixed end moment (FEM) calculations
Case (I) - FEM

B1 B2
wL2 39.769×5.15×5.15
MB1-B2 = − =− = −87.90kNm
12 12
MB2-B1 = +87.90KNm

B3 C4
wL2 36.498×4.2×4.2
MB3-B4 = − =− = −53.65kNm
12 12
MB4-B3 = +53.65KNm

Case (II) - FEM

108 | P a g e
 B1 B2
wL2 39.769×5.15×5.15
MB1-B2 = − 12
=− 12
= −87.90kNm
MB2-B1 = +87.90KNm

B3 C4
wL2 25.856×4.2×4.2
MB3-B4 = − =− = −38.00kNm
12 12
MB4-B3 = +38.00KNm

wL2 36.498×4.2×4.2
MB3-B4 = − =− = −53.65kNm
12 12
MB4-B3 = +53.65KNm

wL2 25.856×4.2×4.2
MB3-B4 = − 12
=− 12
= −38.00kNm
MB4-B3 = +38.00KNm

109 | P a g e
 Stiffness factors for the beams and columns

Beam trial - 300×600 mm

Column trial - 500×300 mm
Beam flexural rigidity is given by;
bd2 0.3 × 0.63
I= = = 5.4 × 10−3 m4
12 12
flexural rigidity
Beam stiffness k =
length

Span B1-B2 and span B5-B6

5.4 × 10−3 m4
k B1−B2 = k B5−B6 = = 1.0485 × 10−3
5.15
Span 2-3, 3-4 and span 4-5
5.4 × 10−3 m4
k B2−B3 = k B3−B4 = k B4−B5 = = 1.2857 × 10−3
4.2
Columns
bd2 0.3 × 0.53
I= = = 3.125 × 10−3 m4
12 12
3.125 × 10−3 m4
k Col = = 1.25 × 10−3
2.5

Calculation of distribution factors

Joint B1 and Joint B6.
Total joint stiffness = beam stiffness + (column stiffness)2
(1.0485× 10−3 ) + (1.25 × 10−3 )2 = 3.5485 × 10−3
1.0485 × 10−3
DFB1−B2 = DFB6−B5= = 0.3
3.5485 × 10−3
2.5 × 10−3
DFCOL = = 0.70
3.5485 × 10−3
Joint B2
Total joint stiffness = beam stiffness + (column stiffness)2
(6.545 × 10−4 ) + (1.0485 × 10−3 ) + (1.25 × 10−3 )2 = 4.203 × 10−3
6.545 × 10−4
DFB2−B1 = = 0.16
4.203 × 10−3
1.0485 × 10−3
DFB2−B3 = = 0.25
4.203 × 10−3

110 | P a g e
 2.5 × 10−3
DFCOL = = 0.59
4.203 × 10−3

Joint B3
Total joint stiffness = beam stiffness + (column stiffness)2
(1.0485× 10−3 ) + (1.0485 × 10−3 ) + (1.25 × 10−3 )2 = 4.597 × 10−3
1.0485 × 10−3
DFB3−B2 = = 0.23
4.597 × 10−3
−3
1.0485 × 10
DFB2−B3 = = 0.23
4.597 × 10−3
2.5 × 10−3
DFCOL = = 0.54
4.597 × 10−3
Joint B4
Total joint stiffness = beam stiffness + (column stiffness)2
(1.0485× 10−3 ) + (1.0485 × 10−3 ) + (1.25 × 10−3 )2 = 4.597 × 10−3
1.0485 × 10−3
DFB4−B3 = = 0.23
4.597 × 10−3
1.0485 × 10−3
DFB4−B5 = = 0.23
4.597 × 10−3
2.5 × 10−3
DFCOL = = 0.54
4.597 × 10−3

Joint B5
Total joint stiffness = beam stiffness + (column stiffness)2
(1.0485× 10−3 ) + (6.545 × 10−4 ) + (1.25 × 10−3 )2 = 4.203 × 10−3
1.0485 × 10−3
DFB5−B4 = = 0.25
4.203 × 10−3
6.545 × 10−4
DFB5−B6 = = 0.16
4.203 × 10−3
2.5 × 10−3
DFCOL = = 0.59
4.203 × 10−3

Moment distribution table using balanced method

Case (I)

JOIN B1 B2 B3 B4 B5 B6
T
MEM COL B1- B2- COL B2- B3- COL B3- B4- COL B4- B5- COL B5- B6- COL
BER B2 B1 B3 B2 B4 B3 B5 B4 B6 B5
DF 0.7 0.3 0.16 0.59 0.25 0.23 0.54 0.23 0.23 0.54 0.23 0.25 0.59 0.16 0.3 0.7

FEM 0 - 87.9 0 - 53.65 0 - 53.65 0 - 53.65 0 -87.9 87.9 0

87.9 53.65 53.65 53.65
BAL 61.5 26.3 -5.48 - - 0 0 0 0 0 0 8.562 20.20 5.48 - -
3 7 20.2 8.562 5 75 26.37 61.5
075 5 3
CO 0 - 13.18 0 0 - 0 0 0 0 4.281 0 0 - 2.74 0
2.74 5 4.281 25 13.18
25 5

111 | P a g e
 BAL 1.91 0.82 - - - 0.984 2.311 0.984 - - - 3.296 7.779 2.109 - -
8 2 2.109 7.77 3.296 688 875 688 0.984 2.31 0.984 25 15 6 0.822 1.91
6 915 25 69 188 69 8
CO 0 - 0.411 0 0.492 - 0 - 0.492 0 1.648 - 0 - 1.054 0
1.05 344 1.648 0.492 344 125 0.492 0.411 8
48 13 34 34
BAL 0.73 0.31 - - - 0.492 1.155 0.492 - - - 0.225 0.532 0.144 - -
836 644 0.144 0.53 0.225 308 853 308 0.492 1.15 0.492 836 973 535 0.316 0.73
54 297 84 31 585 31 44 836
TOTA 63.9 - 93.80 - - 49.14 3.740 - 52.85 - - 65.06 28.79 - 64.26 -
L 507 63.9 443 28.7 65.04 061 429 52.85 633 3.72 49.13 069 989 93.86 352 64.2
507 599 46 89 245 39 06 635

Case (II)

JOIN B1 B2 B3 B4 B5 B6
T
MEM COL B1- B2- COL B2- B3- COL B3- B4- COL B4- B5- COL B5- B6- COL
BER B2 B1 B3 B2 B4 B3 B5 B4 B6 B5
DF 0.7 0.3 0.16 0.59 0.25 0.23 0.54 0.23 0.23 0.54 0.23 0.25 0.59 0.16 0.3 0.7

FEM 0 -87.9 87.9 0 -38 38 0 - 53.65 0 -38 38 0 -87.9 87.9 0

53.65
BAL 61.53 26.37 - - - 3.599 8.451 3.599 - - - 12.47 29.44 7.984 - -
7.984 29.4 12.47 5 5 3.599 8.45 3.599 5 1 26.37 61.5
41 5 5 1 5 3
CO 0 - 13.18 0 1.799 - 0 - 1.799 0 6.237 - 0 - 3.992 0
3.992 5 75 6.237 1.799 75 5 1.799 13.18
5 75 75 5
BAL 2.794 1.197 - - - 1.848 4.340 1.848 - - - 3.746 8.841 2.397 - -
4 6 2.397 8.84 3.746 568 115 568 1.848 4.34 1.848 188 003 56 1.197 2.79
56 1 19 57 012 57 6 44
CO 0 - 0.598 0 0.924 - 0 - 0.924 0 1.873 - 0 - 1.198 0
1.198 8 284 1.873 0.924 284 094 0.924 0.598 78
78 09 28 28 8
BAL 0.839 0.359 - - - 0.643 1.510 0.643 - - - 0.380 0.898 0.243 - -
146 634 0.243 0.89 0.380 397 584 397 0.643 1.51 0.643 771 619 693 0.359 0.83
69 862 77 4 058 4 63 915
CO 0 0.419 0.179 0 0.321 - 0 - 0.321 0 0.190 - 0 - 0.121 0
573 817 698 0.190 0.321 698 385 0.321 0.179 847
39 7 7 82
BAL - - - - - 0.117 0.276 0.117 - - - 0.125 0.295 0.080 - -
0.293 0.125 0.080 0.29 0.125 779 525 779 0.117 0.27 0.117 379 894 242 0.036 0.08
7 87 24 589 38 78 653 78 55 529
CO 0 - - 0 0.058 - 0 - 0.058 0 0.062 - 0 - 0.040 0
0.040 0.062 89 0.062 0.058 89 689 0.058 0.018 121
12 94 69 89 89 28
TOTA 64.89 - 91.09 - - 35.87 14.68 - 50.52 - - 51.63 39.53 - 65.28 -
L 891 64.89 847 39.4 51.61 581 587 50.53 694 14.6 35.86 093 641 91.16 257 65.2
89 873 12 01 594 75 73 826

Final Beam moments

JOINT B1 B2 B3 B4 B5 B6

MEMBE COL B1- B2- COL B2- B3- COL B3- B4- COL B4- B5- COL B5- B6- COL
R B2 B1 B3 B2 B4 B3 B5 B4 B6 B5
Case (I) 63.95 - 93.80 - - 49.14 3.740 - 52.85 - - 65.06 28.79 - 64.26 -
Loading 07 63.9 443 28.7 65.0 061 429 52.8 633 3.72 49.1 069 989 93.8 352 64.2
507 599 446 589 245 339 606 635
Case (II) 64.89 - 91.09 - - 35.87 14.68 - 50.52 - - 51.63 39.53 - 65.28 -
loading 891 64.8 847 39.4 51.6 581 587 50.5 694 14.6 35.8 093 641 91.1 257 65.2
989 873 112 301 594 675 673 826
Output 64.89 - 93.80 - - 49.14 14.68 - 52.85 - - 65.06 39.53 - 64.26 -
moments 891 63.9 443 39.4 65.0 061 587 52.8 633 14.6 49.1 069 641 93.8 352 65.2
507 873 446 589 594 339 606 826

112 | P a g e
 Calculation of reactions in the beam

63.95 93.80

B1 B2
Moment about B1, MB1;
5.15
= - RB2×5.15 + 93.80 – 63.95 + (204.81× 2
)

5.15RB2 = 557.23575
RB2 = 108.20KN
Moment about B2, MB2;
5.15
= 5.15RB1 + 93.80 – 63.95– (204.81 × 2
)

RB1 = 96.61KN

65.04 49.14

B2 B3
Moment about B2, MB2;
4.2
= - RB3×4.2 – 65.04 + 49.14+(153.29× 2
)
RB3 = 72.859KN
RB2 = 153.29 – 72.859 = 80.43 KN

52.86 52.86

B3 B4
Moment about B3, MB3;
4.2
= - RB4×4.2 – 52.86 + 52.86+(153.29× 2
)
RB4 = 76.645KN
RB3 = 153.29 – 76.645 = 76.645 KN

113 | P a g e
 49.13 65.06

B4 B5
Moment about B4, MB4;
4.2
= - RB5×4.2 – 49.13 + 65.06+(153.29× )
2

RB5 = 85.845KN
RB4 = 153.29 – 85.845 = 67.445 KN

93.86 64.26

B5 B6
Moment about B5, MB5;
5.15
= - RB6×5.15 – 93.86 + 64.26 + (204.81× )
2

RB6 = 96.657KN
RB5 = 204.81 – 96.657 = 108.15 KN

Shear forces and bending moment diagrams

a) Shear forces diagram

114 | P a g e
 Span Design moment

Span B1-B2

96.61
x = 39.769 = 2.429 m from B1

Design moment = Reaction moment - Load moment

2.43
Span design moment = (96.61×2.43) – (39.769× 2.43 × ) = 117.35kNm
2

Span B2-B3

80.43
x= = 2.20 m from B2
36.498

Design moment = Reaction moment - Load moment

2.20
Span design moment = (80.43×2.20) – (36.498× 2.20 × 2
) = 88.62kNm

Span B3-B4

76.645
x= = 2.10 m from B3
36.498

Design moment = Reaction moment - Load moment

2.10
Span design moment = (76.645×2.10) – (36.498× 2.10 × 2
) = 80.48kNm

Span B4-B5

67.445
x= = 1.85 m from B4
36.498

Design moment = Reaction moment - Load moment

1.85
Span design moment = (67.445×1.85) – (36.498× 1.85 × ) = 62.32kNm
2

Span B5-B6

108.15
x = 39.769 = 2.72 m from B5

Design moment = Reaction moment - Load moment

115 | P a g e
 2.72
Span design moment = (108.15×2.72) – (39.769× 2.72 × 2
) = 147.05kNm

b) Bending moment diagram

116 | P a g e
 Required steel reinforcement in the beam
Span B5-B6
Maximum hogging moment = maximum moment at support B5= M max = 93.86KN
= maximum moment at support B6= M max = 64.263KN
Maximum sagging moment = maximum moment at midspan = 147.05kn
Maximum shear force = 108.15
B5 B6

➢ Cover = 20 mm
➢ Bar diameter = 20mm
20
➢ Effective depth = 600 – 20 – = 570mm
2

➢ hf = 180
➢ Beam type – L-beam
➢ Flange width b = 1021mm
Moment of resistance MU
hf
= 0.45fcu bhf (d − ) 10−6
2
180
0.45 × 40 × 1021 × 180(570 − 2
) × 10−6

=1587.8592KN
1587.8592kn > 147.05KN/m
Therefore, no compression reinforcement required
h
m = 0.87fy A s (d − 2f)
m 147.05×106
As = h = 0.87×460×480 = 765.53mm2
0.87fy (d− f )
2

Provide T3Y20 bottom bars (As = 942mm2 )

Provide T2Y8 hanger bars (As = 101mm2 )

117 | P a g e
 Design for support steel at B5
M = 93.86KN
Section size = 600×300
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 93.86kn
Compression steel not required at supports
m 93.86 × 106
k= = = 0.024
fcu bd2 40 × 300 × 5702
z
= 0.94
d
Tension reinforcement;
m 93.86 × 106
As = = = 437.72mm2
0.87fy z 0.87 × 460 × 0.94 × 570
Provide T2Y20 bottom bars (As = 628mm2 )
Design for support steel at B6
M = 64.263KN
Section size = 600×300
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 64.263kn
Compression steel not required at supports
m 64.263 × 106
k= = = 0.0165
fcu bd2 40 × 300 × 5702
z
= 0.94
d
Tension reinforcement;
m 64.263 × 106
As = = = 299.696mm2
0.87fy z 0.87 × 460 × 0.94 × 570

Provide T2Y16 bottom bars (As = 402mm2 )

The design Shear stress
v
Design shear stress b at support B5
vd

V =108.15kN
Bv = 300mm
D = 570mm

118 | P a g e
 108.15 × 1000
= 0.632N/mm2
300 × 570

V =0.8√fcu = 40 = 5.1N/mm2

0.632 < 5.1N/mm2 therefore, shear stress ok.

Design concrete shear stress, VC.

100AS 100 × 942
= = 0.55
bv d 300 × 570
0.67 − 0.59
VC = (0.55 − 0.50) + 0.59 = 0.606
0.75 − 0.50
V < 0.5VC = 0.5 × 0.606 = 0.303N/mm2
v = 0.632 > 0.303 = vc

119 | P a g e
 (VC + 0.4) < V
Therefore, provide links as follows;
Asv (0.4)bv
∴ (min. links) ≥
Sv 0.87fyv
0.4×300
0.87×460
= 0.2998 therefore, 0.33 ok

Provide y8@300 c/c for links spacing

Minimum shear links will be used up to V = VC + 0.4 = 0.606 + 0.4 = 1.00
N/mm2
Shear load (minimum links) = (1.00) (300) (600)10−3 = 180 kN
180 > 108.15 > 96.657 hence minimum links to be used throughout
Anchorage of links
It should pass round another bar of at least its own size, through an angle of
180°, and continue for a length of at least four times its own size
Deflection
b = 1021 mm
bw =300
Span/effective depth ratio for the beam
bw 300
b
= 1021 = 0.2938 therefore, it is continuous -20.8
span
Basic depth ratio = 20.8

Tension steel modification factor

m 147.05 × 106
= = 0.443
bd2 1021 × 5702
From table
0.443< 0.5= 1.68
Therefore, allowable span/depth ratio;
= 1.68×20.8 = 34.944
span 5.15
Actual depthratio = 0.57 = 9.035 < 34.944

Deflection ok.
Designed section output

120 | P a g e
121 | P a g e
 Span B4-B5
Maximum hogging moment = maximum moment at support B5= M max = 65.06KN
= maximum moment at support B4= M max = 49.134KN
Maximum sagging moment = maximum moment at midspan = 62.32kn
Maximum shear force = 85.845
B4 B5

➢ Cover = 20 mm
➢ Bar diameter = 20mm
20
➢ Effective depth = 600 – 20 – 2
= 570mm
➢ hf = 180
➢ Beam type – L-beam
➢ Flange width b = 888mm

Moment of resistance MU
hf
= 0.45fcu bhf (d − ) 10−6
2
180
0.45 × 40 × 888 × 180(570 − 2
) × 10−6
=1381.02KN
1381.02kn > 62.32KN/m
Therefore, no compression reinforcement required

h
m = 0.87fy As (d − f )
2

m 62.32×106
As = h = 0.87×460×480 = 324.42mm2
0.87fy (d− f )
2

Provide T2Y16 bottom bars (As = 402mm2 )

Provide T2Y8 hanger bars (As = 101mm2 )

Design for support steel at B5

M = 65.06KN
Section size = 600×300
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 65.06kn
Compression steel not required at supports

m 65.06 × 106
k= = = 0.016
fcu bd2 40 × 300 × 5702
z
= 0.94
d
Tension reinforcement;

122 | P a g e
 m 65.06 × 106
As = = = 303.41mm2
0.87fy z 0.87 × 460 × 0.94 × 570

Provide T2Y16 bottom bars (As = 402mm2 )

Design for support steel at B4

M = 49.134KN
Section size = 600×300
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 49.134kn
Compression steel not required at supports

m 49.134 × 106
k= = = 0.0126
fcu bd2 40 × 300 × 5702
z
= 0.94
d
Tension reinforcement;
m 49.134 × 106
As = = = 229.14mm2
0.87fy z 0.87 × 460 × 0.94 × 570

Provide T2Y16 bottom bars (As = 402mm2 )

The design Shear stress

v
Design shear stress b d at support B5
v
V =85.845kN
Bv = 300mm
D = 570mm
85.845 × 1000
= 0.50N/mm2
300 × 570

V =0.8√fcu = 40 = 5.1N/mm2

0.50 < 5.1N/mm2 therefore, shear stress ok.

Design concrete shear stress, VC.

100AS 100 × 402
= = 0.235
bv d 300 × 570
0.47 − 0.40
VC = (0.235 − 0.15) + 0.40 = 0.4595
0.25 − 0.15
V < 0.5VC = 0.5 × 0.4595 = 0.229N/mm2
v = 0.50 > 0.229 = vc
(VC + 0.4) < V
Therefore, provide links as follows;

123 | P a g e
 Asv (0.4)bv
∴ (min. links) ≥
Sv 0.87fyv
0.4×300
0.87×460
= 0.2998 therefore, 0.33 ok
Provide y8@300 c/c for links spacing
Minimum shear links will be used up to V = VC + 0.4 = 0.4595 + 0.4 = 0.8595 N/mm2
Shear load (minimum links) = (0.8595) (300) (600)10−3 = 154.71 kN
154.71 > 85.845 > 67.445 hence minimum links to be used throughout

Anchorage of links
It should pass round another bar of at least its own size, through an angle of 180°, and
continue for a length of at least four times its own size

Deflection
b = 888 mm
bw =300
Span/effective depth ratio for the beam
bw 300
b
= 888 = 0.338 therefore, it is continuous -21.5
span
Basic depth ratio = 21.5
Tension steel modification factor
m 65.06 × 106
= = 0.226
bd2 888 × 5702
From table
0.226< 0.5= 1.68
Therefore, allowable span/depth ratio;
= 1.68×21.5 = 36.12
span 5.15
Actual depthratio = 0.57 = 9.035 < 36.12
Deflection ok.

Designed section output

124 | P a g e
 Span B3-B4
Maximum hogging moment = maximum moment at support B3= M max = 52.86KN
= maximum moment at support B4= M max = 52.86KN
Maximum sagging moment = maximum moment at midspan = 80.48kn
Maximum shear force = 76.645

B3 B4

➢ Cover = 20 mm
➢ Bar diameter = 20mm
20
➢ Effective depth = 600 – 20 – = 570mm
2
➢ hf = 180
➢ Beam type – L-beam
➢ Flange width b = 888mm

Moment of resistance MU
hf
= 0.45fcu bhf (d − ) 10−6
2
180
0.45 × 40 × 888 × 180(570 − 2
) × 10−6
=1381.02KN
1381.02kn > 80.48KN/m
Therefore, no compression reinforcement required

h
m = 0.87fy As (d − 2f )

m 80.48×106
As = h = 0.87×460×480 = 418.95mm2
0.87fy (d− f )
2

Provide T2Y20 bottom bars (As = 628mm2 )

Provide T2Y8 hanger bars (As = 101mm2 )

Design for support steel at B4

M = 52.86KN
Section size = 600×300
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 52.86kn
Compression steel not required at supports

m 52.86 × 106
k= = = 0.0135
fcu bd2 40 × 300 × 5702
z
= 0.94
d
Tension reinforcement;

125 | P a g e
 m 52.86 × 106
As = = = 246.52mm2
0.87fy z 0.87 × 460 × 0.94 × 570

Provide T2Y16 bottom bars (As = 402mm2 )

Design for support steel at B3

M = 52.86KN
Section size = 600×300
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 52.86kn
Compression steel not required at supports

m 52.86 × 106
k= = = 0.0135
fcu bd2 40 × 300 × 5702
z
= 0.94
d
Tension reinforcement;
m 52.86 × 106
As = = = 246.52mm2
0.87fy z 0.87 × 460 × 0.94 × 570

Provide T2Y16 bottom bars (As = 402mm2 )

The design Shear stress

v
Design shear stress b d at support B4
v
V =76.645kN
bv = 300mm
d = 570mm
76.645 × 1000
= 0.448N/mm2
300 × 570

V =0.8√fcu = 40 = 5.1N/mm2

0.448 < 5.1N/mm2 therefore, shear stress ok.

Design concrete shear stress, VC.

100AS 100 × 628
= = 0.367
bv d 300 × 570
0.59 − 0.47
VC = (0.367 − 0.25) + 0.47 = 0.526
0.5 − 0.25
V < 0.5VC = 0.5 × 0.526 = 0.263N/mm2
v = 0.632 > 0.263 = vc
(VC + 0.4) < V
Therefore, provide links as follows;

126 | P a g e
 Asv (0.4)bv
∴ (min. links) ≥
Sv 0.87fyv
0.4×300
0.87×460
= 0.2998 therefore, 0.33 ok
Provide y8@300 c/c for links spacing
Minimum shear links will be used up to V = VC + 0.4 = 0.526 + 0.4 = 0.926 N/mm2
Shear load (minimum links) = (0.926) (300) (600)10−3 = 166.68 kN
166.68 > 76.645 hence minimum links to be used throughout

Anchorage of links
It should pass round another bar of at least its own size, through an angle of 180°, and
continue for a length of at least four times its own size

Deflection
b = 888 mm
bw =300
Span/effective depth ratio for the beam
bw 300
b
= 888 = 0.338 therefore, it is continuous -21.5
span
Basic depth ratio = 21.5
Tension steel modification factor
m 80.48 × 106
= = 0.279
bd2 888 × 5702
From table
0.279< 0.5= 1.68
Therefore, allowable span/depth ratio;
= 1.68×21.5 = 36.12
span 5.15
Actual depthratio = 0.57 = 9.035 < 36.12
Deflection ok.

Designed section output

127 | P a g e
 Span B2-B3
Maximum hogging moment = maximum moment at support B2= M max = 65.04KN
= maximum moment at support B3= M max = 49.14KN
Maximum sagging moment = maximum moment at midspan = 88.62kN
Maximum shear force = 80.43
B2 B3

➢ Cover = 20 mm
➢ Bar diameter = 20mm
20
➢ Effective depth = 600 – 20 – = 570mm
2
➢ hf = 180
➢ Beam type – L-beam
➢ Flange width b = 888mm

Moment of resistance MU
hf
= 0.45fcu bhf (d − ) 10−6
2
180
0.45 × 40 × 888 × 180(570 − ) × 10−6
2
=1381.02KN
1381.02kn > 88.62KN/m
Therefore, no compression reinforcement required

h
m = 0.87fy As (d − f )
2

m 88.62×106
As = h = 0.87×460×480 = 461.33mm2
0.87fy (d− f )
2

Provide T2Y20 bottom bars (As = 628mm2 )

Provide T2Y8 hanger bars (As = 101mm2 )

Design for support steel at B2

M = 65.04KN
Section size = 600×300
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 65.04kn
Compression steel not required at supports

m 65.04 × 106
k= = = 0.0167
fcu bd2 40 × 300 × 5702
z
= 0.94
d

128 | P a g e
 Tension reinforcement;
m 65.04 × 106
As = = = 303.32mm2
0.87fy z 0.87 × 460 × 0.94 × 570

Provide T2Y16 bottom bars (As = 402mm2 )

Design for support steel at B3

M = 49.14KN
Section size = 600×300
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 49.14kn
Compression steel not required at supports

m 49.14 × 106
k= = = 0.0126
fcu bd2 40 × 300 × 5702
z
= 0.94
d
Tension reinforcement;
m 49.14×106
As = 0.87f z
= 0.87×460×0.94×570 = 229.169mm2
y

Provide T2Y16 bottom bars (As = 402mm2 )

The design Shear stress

v
Design shear stress at support B2
bv d
V =80.43kN
Bv = 300mm
D = 570mm
80.43 × 1000
= 0.47N/mm2
300 × 570

V =0.8√fcu = 40 = 5.1N/mm2

0.47 < 5.1N/mm2 therefore, shear stress ok.

Design concrete shear stress, VC.

100AS 100 × 628
= = 0.367
bv d 300 × 570
0.59 − 0.47
VC = (0.367 − 0.25) + 0.47 = 0.526
0.5 − 0.25
V < 0.5VC = 0.5 × 0.526 = 0.263N/mm2
v = 0.47 > 0.263 = vc
(VC + 0.4) < V
Therefore, provide links as follows;

129 | P a g e
 Asv (0.4)bv
∴ (min. links) ≥
Sv 0.87fyv
0.4×300
0.87×460
= 0.2998 therefore, 0.33 ok
Provide y8@300 c/c for links spacing
Minimum shear links will be used up to V = VC + 0.4 = 0.526 + 0.4 = 0.926 N/mm2
Shear load (minimum links) = (0.926) (300) (600)10−3 = 166.68 kN
166.68 > 80.43 > 72.859 hence minimum links to be used throughout

Anchorage of links
It should pass round another bar of at least its own size, through an angle of 180°, and
continue for a length of at least four times its own size

Deflection
b = 888 mm
bw =300
Span/effective depth ratio for the beam
bw 300
b
= 888 = 0.338 therefore, it is continuous -21.5
span
Basic depth ratio = 21.5
Tension steel modification factor
m 88.62 × 106
= = 0.307
bd2 888 × 5702
From table
0.307< 0.5= 1.68
Therefore, allowable span/depth ratio;
= 1.68×21.5 = 36.12
span 5.15
Actual depthratio = 0.57 = 9.035 < 36.12
Deflection ok.

Designed section output

130 | P a g e
 Span B1-B2
Maximum hogging moment = maximum moment at support B2= M max = 93.80KN
= maximum moment at support B1= M max = 63.95KN
Maximum sagging moment = maximum moment at midspan = 117.35kn
Maximum shear force = 108.20

B1 B2

➢ Cover = 20 mm
➢ Bar diameter = 20mm
20
➢ Effective depth = 600 – 20 – = 570mm
2
➢ hf = 180
➢ Beam type – L-beam
➢ Flange width b = 888mm

Moment of resistance MU
hf
= 0.45fcu bhf (d − ) 10−6
2
180
0.45 × 40 × 888 × 180(570 − 2
) × 10−6
=1381.02KN
1381.02kn > 117.35KN/m
Therefore, no compression reinforcement required

h
m = 0.87fy As (d − 2f )

m 117.35×106
As = h = 0.87×460×480 = 610.89mm2
0.87fy (d− f )
2

Provide T2Y20 bottom bars (As = 628mm2 )

Provide T2Y8 hanger bars (As = 101mm2 )

Design for support steel at B2

M = 93.80KN
Section size = 600×300
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 93.80kn
Compression steel not required at supports
m 93.80 × 106
k= = = 0.024
fcu bd2 40 × 300 × 5702

131 | P a g e
 z
= 0.94
d
Tension reinforcement;
m 93.80 × 106
As = = = 437.44mm2
0.87fy z 0.87 × 460 × 0.94 × 570

Provide T2Y20 bottom bars (As = 628mm2 )

Design for support steel at B1

M = 63.95KN
Section size = 600×300
MU = 0.156fcu bd2 = (0.156×40×300×5702 )10−6
= 608.2128kn > 63.95kn
Compression steel not required at supports

m 63.95 × 106
k= = = 0.029
fcu bd2 40 × 300 × 5702
z
= 0.94
d
Tension reinforcement;
m 63.95 × 106
As = = = 298.236mm2
0.87fy z 0.87 × 460 × 0.94 × 570

Provide T2Y16 bottom bars (As = 402mm2 )

The design Shear stress

v
Design shear stress b d at support B2
v
V =108.20kN
bv = 300mm
d = 570mm
108.20 × 1000
= 0.633N/mm2
300 × 570

V =0.8√fcu = 40 = 5.1N/mm2

0.633 < 5.1N/mm2 therefore, shear stress ok.

Design concrete shear stress, VC.

100AS 100 × 628
= = 0.367
bv d 300 × 570
0.59 − 0.47
VC = (0.367 − 0.25) + 0.47 = 0.526
0.5 − 0.25
V < 0.5VC = 0.5 × 0.526 = 0.263N/mm2
v = 0.633 > 0.263 = vc

132 | P a g e
 (VC + 0.4) < V
Therefore, provide links as follows;
Asv (0.4)bv
∴ (min. links) ≥
Sv 0.87fyv
0.4×300
0.87×460
= 0.2998 therefore, 0.33 ok
Provide y8@300 c/c for links spacing
Minimum shear links will be used up to V = VC + 0.4 = 0.526 + 0.4 = 0.926 N/mm2
Shear load (minimum links) = (0.926) (300) (600)10−3 = 166.68 kN
166.68 > 108.20 > 96.61 hence minimum links to be used throughout

Anchorage of links
It should pass round another bar of at least its own size, through an angle of 180°, and
continue for a length of at least four times its own size

Deflection
b = 888 mm
bw =300
Span/effective depth ratio for the beam
bw 300
= = 0.338 therefore, it is continuous -21.5
b 888
span
Basic depth ratio = 21.5
Tension steel modification factor
m 117.35 × 106
= = 0.407
bd2 888 × 5702
From table
0.407< 0.5= 1.68
Therefore, allowable span/depth ratio;
= 1.68×21.5 = 36.12
span 5.15
Actual depthratio = 0.57 = 9.035 < 36.12
Deflection ok.

Designed section output

133 | P a g e
134 | P a g e
 4.3 Column Designs (See on structural plan)

Clear height of ground floor column = 2400 mm (see architectural plan)

End conditions
Condition at top
End of column is connected monolithically to beams on either side and are at least as deep
at the overall
Condition at bottom
End of column is connected monolithically to beams or to footing on either side and are
at least as deep as the overall
Dimension of the column (minimum cross section of column = 500 * 300 mm)
h = 500 mm b = 300 mm
H = Total height of column
BS 8110: cl. 3.8.1.3 defines a braced short column as a braced column for which both
the ratios lex/h and ley/b are less than 15.
Where lex = effective height in respect for the major axis,
ley= effective height in respect of the minor axis’
h = depth in respect to major axis,
b = width of the column.
3000×0.75 3000×0.75
Therefore 500
= 4.5 < 15 𝑎𝑛𝑑 300
= 7.5 < 15
Hence column is to be designed as short braced axially loaded column (short column)
Limits for main reinforcement (BS 8110: clauses 3.12.5 and 3.12.6)
(a) The total area of longitudinal bars Asc ≥ 0.4 % cross sectional area of the
column,
(b) The total area of longitudinal bars Asc should not exceed 6 % of the cross-
sectional area of the vertically cast column and 8 % of that of a horizontally cast
column except at laps where the limit is 10 % for both cases
(c) A minimum of 4 no. longitudinal bars to be used for a rectangular column,

(d) Longitudinal bars should not be smaller than size 12 mm

(e) Longitudinal bars spacing should not exceed 250 mm.

135 | P a g e
 BS 3.8.4.6 Shear in columns The design shear strength of columns may be
checked in accordance with 3.4.5.12. For rectangular sections in compression no
check is required provided that M/N does not exceed 0.6h and v does not exceed the
maximum value given in 3.4.5.12.

BS3.8.5 Deflection of columns No check is necessary under the following conditions.

a) Braced columns. Within the recommended limits of slenderness, no specific check is
necessary.

BS 3.8.6 Crack control in columns Cracks due to bending in a column designed for
design ultimate axial load greater than 0.2fcuAc are unlikely to occur and therefore no
check is required. A more lightly-loaded column subject to bending should be considered
as a beam for the purpose of crack control.

Design of ground floor columns

136 | P a g e
 Design analysis of column A1
Mx=50.38kNm
My=157.59kNm
Beam A1-B1 (reaction)=133.785kN
Beam A1-A2 (reaction)=74.93kN
Column self-weight =0.3(0.5) (2.5)24=9.0kN
FLOOR LOAD (kN) MX MY

4th Floor Beams=208.715 25.19 78.796

Colum self weight=9.0
Total 217.715 25.19 78.796

3rd Floor Beams=208.715 25.19 78.796

Colum self weight=9.0
Total 435.43 25.19 78.796

2nd Floor Beams=208.715 25.19 78.796

Colum self weight=9.0
Total 653.145 25.19 78.796

1st Floor Beams=208.715 25.19 78.796

Colum self-weight=9.0
Total 870.86 25.19 78.796
Ground Floor Beams=208.715 25.19 78.796
Column self weight=9.0
Total 1088.575 25.19 78.796

Column section size – 500mm × 300mm

Concrete cover to center of bars = 30 mm
hI = 500 − 30 = 470 mm
bI = 500 − 30 = 270 mm
My 78.796×106
hI
= 470
= 167651.0638N

Mx 25.19×106
= = 93296.2963N
bI 270

137 | P a g e
 My Mx
hI
> bI
therefore, design N, My
N (1088.575)(103 )
fcubh
= (40)(300)(500)
= 0.18

Values of β

N 0 0.1 0.2 0.3 0.4 0.5 ≥ 0.6

fcu bh
β 1.0 0.88 0.77 0.65 0.53 0.42 0.30

From the table, β = 0.77 by interpolation

bI 270
Enhanced moment = MyI = My + β Mx = 78.796+ (0.77 × × 25.19)
hI 470

= 89.94kNm
Design for reinforcement
N = 1088.575 kN
MyI = 89.94 kNm
N 1088.575×103
bh
= 300×500
= 7.26N/mm2

MIy 89.94×106
bh2
= 300×5002 = 1.2N/mm2

138 | P a g e
 From the column design chart, NO: 40
(100)(Asc )
(b)(h)
= 0.21% < 0.4% (minimum)

Hence 0.4% ok
(0.4)(b)(h) (0.4)(300)(500)
therefore, Asc = = = 600 mm2
100 100

Hence provide reinforcement as shown below:

provide 6Y12 (ASC = 679 mm2)
Link size: ≥ (¼) (12), therefore 8mm bar, ok.

Link spacing: ≤ (12)(12) = 144mm, therefore 125mm ok

Provide same reinforcement for all columns to forth storey

139 | P a g e
140 | P a g e
 Design analysis of column A2
Mx=30.18kNm
My=207.204kNm
Beam A1-A2 (reaction)=84.12kN
Beam A2-A3 (reaction)=64.289kN
Beam A2-B2 (reaction)=175.92kN
Column self-weight =0.3(0.5) (2.5)24=9.0kN
FLOOR LOAD (kN) MX MY

4th Floor Beams=324.329 15.09 103.602

Colum self weight=9.0
Total 333.329 15.09 103.602

3rd Floor Beams=324.329 15.09 103.602

Colum self weight=9.0
Total 666.658 15.09 103.602

2nd Floor Beams=324.329 15.09 103.602

Colum self weight=9.0
Total 999.987 15.09 103.602

1st Floor Beams=324.329 15.09 103.602

Colum self-weight=9.0
Total 1333.316 15.09 103.602
Ground Floor Beams=324.329 15.09 103.602
Column self weight=9.0
Total 1666.645 15.09 103.602

Column section size – 500mm × 300mm

Concrete cover to center of bars = 30 mm
hI = 500 − 30 = 470 mm
bI = 500 − 30 = 270 mm
My 103.602×106
hI
= 470
=220429.7872N

141 | P a g e
 Mx 15.09×106
bI
= 270
= 55888.889N

My Mx
hI
> bI
therefore, design N, My
N (1666.645)(103 )
fcu bh
= (40)(300)(500)
= 0.28

Values of β

N 0 0.1 0.2 0.3 0.4 0.5 ≥ 0.6

fcu bh
β 1.0 0.88 0.77 0.65 0.53 0.42 0.30

From the table, β = 0.65 by interpolation

bI 270
Enhanced moment = MyI = My + β hI Mx = 103.602+ (0.65 × 470 × 15.09)

= 109.2367kNm
Design for reinforcement
N = 1666.645 kN
MyI = 109.2367 kNm
N 1666.645×103
bh
= 300×500
= 11.11N/mm2

MIy 109.2367×106
bh2
= 300×5002
= 1.456N/mm2

142 | P a g e
 From the column design chart, NO: 40
(100)(Asc )
(b)(h)
= 0.25% < 0.4% (minimum)

Hence 0.4% ok
(0.4)(b)(h) (0.4)(300)(500)
therefore, Asc = = = 600 mm2
100 100

Hence provide reinforcement as shown below:

provide 6Y12 (ASC = 679 mm2)
Link size: ≥ (¼) (12), therefore 8mm bar, ok.

Link spacing: ≤ (12)(12) = 144mm, therefore 125mm ok

Provide same reinforcement for all columns to forth storey

143 | P a g e
144 | P a g e
 Design analysis of column B1
Mx=64.898kNm
My=42kNm
Beam A1-B1 (reaction)=148.61kN
Beam B1-C1 (reaction)=148.61kN
Beam B1-B2 (reaction)=96.61kN
Column self-weight =0.3(0.5) (2.5)24=9.0kN
FLOOR LOAD (kN) MX MY

4th Floor Beams=393.83 32.449 21

Colum self weight=9.0
Total 402.83 32.449 21

3rd Floor Beams=393.83 32.449 21

Colum self weight=9.0
Total 805.66 32.449 21

2nd Floor Beams=393.83 32.449 21

Colum self weight=9.0
Total 1208.49 32.449 21

1st Floor Beams=393.83 32.449 21

Colum self-weight=9.0
Total 1611.32 32.449 21
Ground Floor Beams=393.83 32.449 21
Column self weight=9.0
Total 2014.15 32.449 21

Column section size – 500mm × 300mm

Concrete cover to center of bars = 30 mm
hI = 500 − 30 = 470 mm
bI = 500 − 30 = 270 mm
My 21×106
hI
= 470
= 44680.85N

145 | P a g e
 Mx 32.449×106
bI
= 270
= 120181.4815N

Mx My
bI
> hI
therefore, design N, Mx
N (2014.15)(103 )
fcu bh
= (40)(300)(500) = 0.34

Values of β

N 0 0.1 0.2 0.3 0.4 0.5 ≥ 0.6

fcu bh
β 1.0 0.88 0.77 0.65 0.53 0.42 0.30

From the table, β = 0.53 by interpolation

bI 270
Enhanced moment = MxI = Mx + β hI My = 32.449+ (0.53 × 470 × 21)

= 38.84kNm
Design for reinforcement
N = 2014.15 kN
MxI = 38.84 kNm
N 2014.15×103
bh
= 300×500
= 13.43N/mm2

MIx 38.84×106
bh2
= 300×5002 = 0.52N/mm2

146 | P a g e
 From the column design chart, NO: 40
(100)(Asc )
(b)(h)
= 0.18% < 0.4% (minimum)

Hence 0.4% ok
(0.4)(b)(h) (0.4)(300)(500)
therefore, Asc = = = 600 mm2
100 100

Hence provide reinforcement as shown below:

provide 6Y12 (ASC = 679 mm2)
Link size: ≥ (¼) (12), therefore 8mm bar, ok.

Link spacing: ≤ (12)(12) = 144mm, therefore 125mm ok

Provide same reinforcement for all columns to forth storey

147 | P a g e
148 | P a g e
 Design analysis of column B2
Mx=39.488kNm
My=54.9586kNm
Beam B1-B2 (reaction)=108.20kN
Beam B2-B3 (reaction)=80.43kN
Beam A2-B2 (reaction)=195.41kN
Beam B2-C2 (reaction)=195.41kN
Column self-weight =0.3(0.5) (2.5)24=9.0kN
FLOOR LOAD (kN) MX MY

4th Floor Beams=579.45 19.744 27.4793

Colum self weight=9.0
Total 588.45 19.744 27.4793

3rd Floor Beams=579.45 19.744 27.4793

Colum self weight=9.0
Total 1176.90 19.744 27.4793

2nd Floor Beams=579.45 19.744 27.4793

Colum self weight=9.0
Total 1765.35 19.744 27.4793

1st Floor Beams=579.45 19.744 27.4793

Colum self-weight=9.0
Total 2353.8 19.744 27.4793
Ground Floor Beams=579.45 19.744 27.4793
Column self weight=9.0
Total 2942.25 19.744 27.4793

Column section size – 500mm × 300mm

Concrete cover to center of bars = 30 mm
hI = 500 − 30 = 470 mm
bI = 500 − 30 = 270 mm
My 27.4793×106
= = 58466.596N
hI 470

149 | P a g e
 Mx 19.744×106
bI
= 270
= 73125.926N

Mx My
> therefore, design N, Mx
bI hI
N (2942.25)(103 )
= (40)(300)(500) = 0.49
fcubh

Values of β

N 0 0.1 0.2 0.3 0.4 0.5 ≥ 0.6

fcu bh
β 1.0 0.88 0.77 0.65 0.53 0.42 0.30

From the table, β = 0.42 by interpolation

bI 270
Enhanced moment = MxI = Mx + β My = 19.744 + (0.42 × × 27.47932)
hI 470

= 26.374kNm
Design for reinforcement
N = 2942.25 kN
MxI = 26.374 kNm
N 2942.25×103
bh
= 300×500
= 19.615N/mm2

MIx 26.374×106
bh2
= 300×5002
= 0.352N/mm2

150 | P a g e
 From the column design chart, NO: 40
(100)(Asc )
(b)(h)
= 0.41% < 0.4% (minimum)

Hence 0.41% ok
(0.41)(b)(h) (0.41)(300)(500)
therefore, Asc = = = 615 mm2
100 100

Hence provide reinforcement as shown below:

provide 6Y12 (ASC = 679 mm2)
Link size: ≥ (¼) (12), therefore 8mm bar, ok.

Link spacing: ≤ (12)(12) = 144mm, therefore 125mm ok

Provide same reinforcement for all columns to forth storey

151 | P a g e
152 | P a g e
 Design analysis of column B3
Mx=14.68kNm
My=53.254kNm
Beam B2-B3 (reaction)=72.859kN
Beam B3-B4 (reaction)=76.645kN
Beam A3-B3 (reaction)=189.29kN
Beam B3-C3 (reaction)=189.29kN
Column self-weight =0.3(0.5) (2.5)24=9.0kN
FLOOR LOAD (kN) MX MY

4th Floor Beams=528.794 7.34 26.627

Colum self weight=9.0
Total 537.794 7.34 26.627

3rd Floor Beams=528.794 7.34 26.627

Colum self weight=9.0
Total 1075.588 7.34 26.627

2nd Floor Beams=528.794 7.34 26.627

Colum self weight=9.0
Total 1613.382 7.34 26.627

1st Floor Beams=528.794 7.34 26.627

Colum self-weight=9.0
Total 2151.176 7.34 26.627
Ground Floor Beams=528.794 7.34 26.627
Column self weight=9.0
Total 2688.97 7.34 26.627

Column section size – 500mm × 300mm

Concrete cover to center of bars = 30 mm
hI = 500 − 30 = 470 mm
bI = 500 − 30 = 270 mm
My 26.627×106
hI
= 470
= 56653.191N

153 | P a g e
 Mx 7.34×106
bI
= 270
= 27185.185N

My Mx
> therefore, design N, My
hI bI
N (2688.97)(103 )
= (40)(300)(500) = 0.448
fcubh

Values of β

N 0 0.1 0.2 0.3 0.4 0.5 ≥ 0.6

fcu bh
β 1.0 0.88 0.77 0.65 0.53 0.42 0.30

From the table, β = 0.42 by interpolation

bI 270
Enhanced moment = MyI = My + β Mx = 26.627+ (0.42 × × 7.34)
hI 470

= 28.3979kNm
Design for reinforcement
N = 2688.97 kN
MyI = 28.3979 kNm
N 28.3979×103
bh
= 300×500
= 17.93N/mm2

MIy 28.3979×106
bh2
= 300×5002
= 0.379N/mm2

154 | P a g e
 From the column design chart, NO: 40
(100)(Asc )
(b)(h)
= 0.35% < 0.4% (minimum)

Hence 0.4% ok
(0.4)(b)(h) (0.4)(300)(500)
therefore, Asc = = = 600 mm2
100 100

Hence provide reinforcement as shown below:

provide 6Y12 (ASC = 679 mm2)
Link size: ≥ (¼) (12), therefore 8mm bar, ok.

Link spacing: ≤ (12)(12) = 144mm, therefore 125mm ok

Provide same reinforcement for all columns to forth storey

155 | P a g e
156 | P a g e
 Design analysis of column A3
Mx=11.612kNm
My=200.6832kNm
Beam A2-A3 (reaction)=58.63kN
Beam A3-A4 (reaction)=61.4585kN
Beam A3-B3 (reaction)=170.41kN
Column self-weight =0.3(0.5) (2.5)24=9.0kN
FLOOR LOAD (kN) MX MY

4th Floor Beams=290.50 5.806 100.3416

Colum self weight=9.0
Total 299.50 5.806 100.3416

3rd Floor Beams=290.50 5.806 100.3416

Colum self weight=9.0
Total 599.0 5.806 100.3416

2nd Floor Beams=290.50 5.806 100.3416

Colum self weight=9.0
Total 898.50 5.806 100.3416

1st Floor Beams=290.50 5.806 100.3416

Colum self-weight=9.0
Total 1198.0 5.806 100.3416
Ground Floor Beams=290.50 5.806 100.3416
Column self weight=9.0
Total 1497.5 5.806 100.3416

Column section size – 500mm × 300mm

Concrete cover to center of bars = 30 mm
hI = 500 − 30 = 470 mm
bI = 500 − 30 = 270 mm
My 100.3416×106
hI
= 470
= 213492.766N

157 | P a g e
 Mx 5.806×106
bI
= 270
= 21503.7037N

My Mx
hI
> bI
therefore, design N, My
N (1497.5)(103 )
fcu bh
= (40)(300)(500) = 0.25

Values of β

N 0 0.1 0.2 0.3 0.4 0.5 ≥ 0.6

fcu bh
β 1.0 0.88 0.77 0.65 0.53 0.42 0.30

From the table, β = 0.65 by interpolation

bI 270
Enhanced moment = MyI = My + β hI Mx = 100.3416+ (0.65 × 470 × 5.806)

= 102.5096kNm
Design for reinforcement
N = 1497.5 kN
MyI = 102.5096 kNm
N 1497.5×103
bh
= 300×500
= 9.98N/mm2

MIy 102.5096×106
bh2
= 300×5002
= 1.367N/mm2

158 | P a g e
 From the column design chart, NO: 40
(100)(Asc )
(b)(h)
= 0.2% < 0.4% (minimum)

Hence 0.4% ok
(0.4)(b)(h) (0.4)(300)(500)
therefore, Asc = = = 600 mm2
100 100

Hence provide reinforcement as shown below:

provide 6Y12 (ASC = 679 mm2)
Link size: ≥ (¼) (12), therefore 8mm bar, ok.

Link spacing: ≤ (12)(12) = 144mm, therefore 125mm ok

Provide same reinforcement for all columns to forth storey

159 | P a g e
160 | P a g e
4.3.1 COLUMN SCHEDULE
Column mark Column load Column
(KN) moments (KN)
Beam mark Beam load Total load (KN) M xx M yy
(KN)
A1 Beam (A1-B1) 133.785
Beam (A1-A2) 74.93 1088.575 25.19 78.796
Self-weight 9.0
A2 Beam (A1-A2) 84.12
Beam (A2-A3) 64.289 1666.645 15.09 103.602
Beam (A2-B2) 175.95
Self-weight 9.0
B1 Beam (A1-B1) 148.61
Beam (B1-C1) 148.61 32.449 21
Beam (B1-B2) 96.61 2014.15
Self-weight 9.0
B2 Beam (B1B2) 108.20
Beam (B2-B3) 80.43
Beam (A2-B2) 195.41 2942.25 19.744 27.4793
Beam (B2-C2) 195.41
Self-weight 9.0
B3 Beam (B2-B3) 72.859
Beam (B3-B4) 76.645 2688.97 7.34 26.627
Beam (A3-B3) 189.29
Beam (B3-C3) 189.29
Self-weight 9.0
A3 Beam (A2-A3) 58.63
Beam (A3-A4) 61.4585 1497.5 5.806 100.3416
Beam (A3-B3) 170.41
Self-weight 9.0

4.4 Foundations Bases

161 | P a g e
 Column A1/A6 base
Step 1:
Footing self weight = 150 kN (assumed)
.: Total load = 1088.575 + 150 = 1238.575 kN
Total load 1238.575
Required base area = = = 5.734 m2
safe bearing capacity 216
Base lateral dimensions = (√5.734) = 2.39 m
Provide a base 2.450 m square, area = 6.0 m2
Base design load = (1.4)(150) = 210 kN
Total design load = 1088.575 + 210 = 1298.575 kN
1298.575
Earth pressure = = 216.34 kN/m2
2.452
Assume a 600 mm thick footing.
Net upward pressure = 216.34 − (h)(24)(γF )
= 216.34 − (0.6)(24)(1.4) = 196.18 kN/m2
Step 2: Anchorage length for the dowels:
Compression anchorage length = (26)(bar size)
= (26)(16) = 416 mm
∴ A base thickness of 600 mm would adequately allow for concrete cover and two
layers of reinforcement below the dowels.
Step 3: Punching shear
Assume the footing is to be constructed on a blinding layer of concrete so that the
minimum cover is 40 mm.
Effective depth d = 600 − cover − bar size
Effective depth d = 600 − 40 − 20 = 540 mm
Critical perimeter = column perimeter + 3πh
(500 × 2) + (300 × 2) + (3 ∗ π ∗ 600)
= 7254.87 mm
Area within perimeter = (400 + 3h)2 − (4 − π)(1.5h)2
= (400 + 1800)2 − (4 − π)(900)2

= (4.1)106 mm2
Punching shear force V = (196.18)(2.452 − 4.1)
= 373.23 kN

162 | P a g e
 (373.23)1000
Punching shear stress = = 0.095 N/mm2
(7254.87)(540)
From Table 6.4-1 (design II), this shear stress is not excessive, therefore h =
600 mm will be suitable.

Step 4: Bending reinforcement

Critical section is at the column face:

1.075
Design moment M = (196.18 )(2.45)(1.075)
2
= 277.72 kNm
Ultimate moment of the section M = 0.156fcu bd2
= (0.156)(40)(2450)(540)2 (10−6 )
= 4457.98 kNm > 277.72 kNm
. : Compression reinforcement not required

M
K=
fcu bd2
(277.72)106
= = 0.0097
(40)(2450)(5402 )
From Table 4.6 -1 (design II)
z
= 0.94
d
M
Tension steel area A s =
0.87fy z

163 | P a g e
 (277.72)106
= = 1367.126 mm2
(0.87)(460)(0.94)(540)
Minimum tension steel area ∶
100As
= 0.13
bh
0.13(2450)(600)
As = = 1911 mm2
100
Provide 10 Y16 – 240 (both ways) − (As = 2011 mm2 )
Step 5: Local Bond
At the critical section for bending
shear V = (196.18)(2.450)(1.075) = 516.69 kN
V (516.69 )103
Local bond stress fbs = = = 1.906 N/mm2
Ʃus d 10(50.2)(540)
𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑏𝑜𝑛𝑑 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝛽√𝑓𝑐𝑢
𝐹𝑟𝑜𝑚 𝑇𝑎𝑏𝑙𝑒 6.6 − 1: 𝐵𝑜𝑛𝑑 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝛽 (𝐵𝑆 8110: 𝐶𝑙𝑎𝑢𝑠𝑒 3.12.8.4), 𝛽 = 0.5
∴ 𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑏𝑜𝑛𝑑 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝛽√𝑓𝑐𝑢 = 0.5√40 = 3.2 𝑁mm2 / > 1.906 𝑁/mm2
∴ 𝐵𝑜𝑛𝑑 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝑂k
Step 6: Final check of punching shear
From Table 6.4-1: (design II)
100As (100)2011
= = 0.152
bd (2450)540
Ultimate shear stress vc = 0.4 N/mm2
Punching shear stress was 0.095 N/mm2
Therefore, a 600 mm thick pad is adequate.

Step7: Shear stress

164 | P a g e
 1.5d = (1.5)540 = 810 mm
1075 − 810 = 265 mm
At the critical section for shear, 1.5d from the column face
V = (196.18)(2.45)(0.265) = 127.37 kN
V (127.37)1000
𝐯= =
bd (2450)540
= 0.096 N/mm2 < 0.4 N/mm2
∴ The section is adequate in shear

165 | P a g e
 Column B1/B6 base
Step 1:
Footing self weight = 150 kN (assumed)
.: Total load = 2014.15 + 150 = 2164.15 kN
Total load 2164.15
Required base area = = = 10.02 m2
safe bearing capacity 216
Base lateral dimensions = (√10.02) = 3.165 m
Provide a base 3.25 m square, area = 10.56 m2
Base design load = (1.4)(150) = 210 kN
Total design load = 2014.15 + 210 = 2224.15 kN
2224.15
Earth pressure = = 210.57 kN/m2
3.252
Assume a 600 mm thick footing.
Net upward pressure = 210.57 − (h)(24)(γF )
= 210.57 − (0.6)(24)(1.4) = 190.41 kN/m2
Step 2: Anchorage length for the dowels:
Compression anchorage length = (26)(bar size)
= (26)(16) = 416 mm
∴ A base thickness of 600 mm would adequately allow for concrete cover and two
layers of reinforcement below the dowels.
Step 3: Punching shear
Assume the footing is to be constructed on a blinding layer of concrete so that the
minimum cover is 40 mm.
Effective depth d = 600 − cover − bar size
Effective depth d = 600 − 40 − 20 = 540 mm
Critical perimeter = column perimeter + 3πh
(500 × 2) + (300 × 2) + (3 ∗ π ∗ 600)
= 7254.87 mm
Area within perimeter = (400 + 3h)2 − (4 − π)(1.5h)2
= (400 + 1800)2 − (4 − π)(900)2

= (4.1)106 mm2
Punching shear force V = (190.41)(3.252 − 4.1)

166 | P a g e
 = 1230.53 kN
(1230.53)1000
Punching shear stress = = 0.314 N/mm2
(7254.87)(540)
From Table 6.4-1 (design II), this shear stress is not excessive, therefore h =
600 mm will be suitable.
Step 4: Bending reinforcement

Critical section is at the column face:

1.475
Design moment M = (190.41 )(3.25)(1.475)
2
= 673.17 kNm
Ultimate moment of the section M = 0.156fcu bd2
= (0.156)(40)(3250)(540) 2 (10−6 )
= 5913.648 kNm > 673.17 kNm
. : Compression reinforcement not required

M
K=
fcu bd2
(673.17)106
= = 0.0177
(40)(3250)(5402 )
From Table 4.6 -1 (design II)
z
= 0.94
d

167 | P a g e
 M
Tension steel area A s =
0.87fy z
(673.17)106
= = 3313.80 mm2
(0.87)(460)(0.94)(540)
Minimum tension steel area ∶
100A s
= 0.13
bh
0.13(3250)(600)
As = = 2535 mm2
100
Provide 17 Y16 – 185 (both ways) − (As = 3417 mm2 )
Step 5: Local Bond
At the critical section for bending
shear V = (190.41)(3.25)(1.475) = 912.78 kN
V (912.78 )103
Local bond stress fbs = = = 1.98 N/mm2
Ʃus d 17(50.2)(540)
𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑏𝑜𝑛𝑑 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝛽√𝑓𝑐𝑢
𝐹𝑟𝑜𝑚 𝑇𝑎𝑏𝑙𝑒 6.6 − 1: 𝐵𝑜𝑛𝑑 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝛽 (𝐵𝑆 8110: 𝐶𝑙𝑎𝑢𝑠𝑒 3.12.8.4), 𝛽 = 0.5
∴ 𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑏𝑜𝑛𝑑 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝛽√𝑓𝑐𝑢 = 0.5√40 = 3.2 𝑁mm2 / > 1.98 𝑁/mm2
∴ 𝐵𝑜𝑛𝑑 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝑂k
Step 6: Final check of punching shear
From Table 6.4-1: (design II)
100A s (100)3417
= = 0.195
bd (3250)540
Ultimate shear stress vc = 0.4 N/mm2
Punching shear stress was 0.314 N/mm2
Therefore, a 600 mm thick pad is adequate.

168 | P a g e
 Step7: Shear stre5d = (1.5)540 = 810 mm
1475 − 810 = 665 mm
At the critical section for shear, 1.5d from the column face
V = (190.41)(3.25)(0.665) = 411.52 kN
V (411.52)1000
𝐯= =
bd (3250)540
= 0.234 N/mm2 < 0.4 N/mm2
∴ The section is adequate in shear

Column A2/A5 base

Step 1:
Footing self weight = 150 kN (assumed)
.: Total load = 1666.65 + 150 = 1816.65 kN
Total load 1816.65
Required base area = = = 8.41 m2
safe bearing capacity 216
Base lateral dimensions = (√8.41) = 2.90 m
Provide a base 3.0 m square, area = 9.0 m2
Base design load = (1.4)(150) = 210 kN
Total design load = 1666.65 + 210 = 1876.65 kN
1876.65
Earth pressure = = 208.52 kN/m2
3.02

169 | P a g e
 Assume a 600 mm thick footing.
Net upward pressure = 208.52 − (h)(24)(γF )
= 208.52 − (0.6)(24)(1.4) = 188.36 kN/m2
Step 2: Anchorage length for the dowels:
Compression anchorage length = (26)(bar size)
= (26)(16) = 416 mm
∴ A base thickness of 600 mm would adequately allow for concrete cover and two
layers of reinforcement below the dowels.
Step 3: Punching shear
Assume the footing is to be constructed on a blinding layer of concrete so that the
minimum cover is 40 mm.
Effective depth d = 600 − cover − bar size
Effective depth d = 600 − 40 − 20 = 540 mm
Critical perimeter = column perimeter + 3πh
(500 × 2) + (300 × 2) + (3 ∗ π ∗ 600)
= 7254.87 mm
Area within perimeter = (400 + 3h)2 − (4 − π)(1.5h)2
= (400 + 1800)2 − (4 − π)(900)2

= (4.1)106 mm2
Punching shear force V = (188.36)(3.02 − 4.1)
= 922.95 kN
(922.95)1000
Punching shear stress = = 0.236 N/mm2
(7254.87)(540)
From Table 6.4-1 (design II), this shear stress is not excessive, therefore h =
600 mm will be suitable.
Step 4: Bending reinforcement

170 | P a g e
 Critical section is at the column face:
1.35
Design moment M = (188.36 )(3.0)(1.35)
2
= 514.93 kNm
Ultimate moment of the section M = 0.156fcu bd2
= (0.156)(40)(3000)(540) 2 (10−6 )
= 5458.752 kNm > 514.93 kNm
. : Compression reinforcement not required

M
K=
fcu bd2
(514.93)106
= = 0.0147
(40)(3000)(5402 )
From Table 4.6 -1 (design II)
z
= 0.94
d
M
Tension steel area A s =
0.87fy z
(514.93)106
= = 2534.834 mm2
(0.87)(460)(0.94)(540)
Minimum tension steel area ∶
100A s
= 0.13
bh
0.13(3000)(600)
As = = 2340 mm2
100
Provide 13 Y16 – 225 (both ways) − (As = 2613 mm2 )

171 | P a g e
 Step 5: Local Bond
At the critical section for bending
shear V = (188.36)(3.0)(1.35) = 762.858 kN
V (762.858 )103
Local bond stress fbs = = = 2.16 N/mm2
Ʃus d 13 (50.2)(540)
𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑏𝑜𝑛𝑑 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝛽√𝑓𝑐𝑢
𝐹𝑟𝑜𝑚 𝑇𝑎𝑏𝑙𝑒 6.6 − 1: 𝐵𝑜𝑛𝑑 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝛽 (𝐵𝑆 8110: 𝐶𝑙𝑎𝑢𝑠𝑒 3.12.8.4), 𝛽 = 0.5
∴ 𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑏𝑜𝑛𝑑 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝛽√𝑓𝑐𝑢 = 0.5√40 = 3.2 𝑁mm2 / > 2.16 𝑁/mm2
∴ 𝐵𝑜𝑛𝑑 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝑂k
Step 6: Final check of punching shear
From Table 6.4-1: (design II)
100A s (100)2613
= = 0.161
bd (3000)540
Ultimate shear stress vc = 0.4 N/mm2
Punching shear stress was 0.236 N/mm2
Therefore, a 600 mm thick pad is adequate.
Step7: Shear stress

1.5d = (1.5)540 = 810 mm

1350 − 810 = 540 mm
At the critical section for shear, 1.5d from the column face
V = (188.36)(3.0)(0.540) = 305.1432 kN
V (305.1432)1000
𝐯= =
bd (3000)540
= 0.188 N/mm2 < 0.4 N/mm2

172 | P a g e
 ∴ The section is adequate in shear

Column A3/A4 base

Step 1:
Footing self weight = 150 kN (assumed)
.: Total load = 1497.5 + 150 = 1238.575 kN
Total load 1647.5
Required base area = = = 7.63 m2
safe bearing capacity 216
Base lateral dimensions = (√7.63) = 2.76 m
Provide a base 2.850 m square, area = 8.1 m2
Base design load = (1.4)(150) = 210 kN
Total design load = 1497.5 + 210 = 1707.5 kN
1707.5
Earth pressure = = 210.22 kN/m2
2.852
Assume a 600 mm thick footing.
Net upward pressure = 210.22 − (h)(24)(γF )
= 210.22 − (0.6)(24)(1.4) = 190.06 kN/m2
Step 2: Anchorage length for the dowels:
Compression anchorage length = (26)(bar size)
= (26)(16) = 416 mm
∴ A base thickness of 600 mm would adequately allow for concrete cover and two
layers of reinforcement below the dowels.
Step 3: Punching shear
Assume the footing is to be constructed on a blinding layer of concrete so that the
minimum cover is 40 mm.
Effective depth d = 600 − cover − bar size
Effective depth d = 600 − 40 − 20 = 540 mm
Critical perimeter = column perimeter + 3πh
(500 × 2) + (300 × 2) + (3 ∗ π ∗ 600)

173 | P a g e
 = 7254.87 mm
Area within perimeter = (400 + 3h)2 − (4 − π)(1.5h)2
= (400 + 1800)2 − (4 − π)(900)2
= (4.1)106 mm2
Punching shear force V = (190.06)(2.852 − 4.1)
= 764.51 kN
(764.51)1000
Punching shear stress = = 0.195 N/mm2
(7254.87)(540)
From Table 6.4-1 (design II), this shear stress is not excessive, therefore h =
600 mm will be suitable.
Step 4: Bending reinforcement

Critical section is at the column face:

1.275
Design moment M = (190.06 )(2.85)(1.275)
2
= 440.28 kNm
Ultimate moment of the section M = 0.156fcu bd2
= (0.156)(40)(2850)(540)2 (10−6 )
= 5185.81 kNm > 440.28 kNm
. : Compression reinforcement not required
M
K=
fcu bd2
(440.28)106
= = 0.0132
(40)(2850)(5402 )
From Table 4.6 -1 (design II)

174 | P a g e
 z
= 0.94
d
M
Tension steel area As =
0.87fy z
(440.28)106
= = 2167.36 mm2
(0.87)(460)(0.94)(540)
Minimum tension steel area ∶
100A s
= 0.13
bh
0.13(2850)(600)
As = = 2223 mm2
100
Provide 12 Y16 – 245 (both ways) − (As = 2412 mm2 )
Step 5: Local Bond
At the critical section for bending
shear V = (190.06)(2.850)(1.275) = 690.63 kN
V (690.63 )103
Local bond stress fbs = = = 2.12 N/mm2
Ʃus d 12(50.2)(540)
𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑏𝑜𝑛𝑑 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝛽√𝑓𝑐𝑢
𝐹𝑟𝑜𝑚 𝑇𝑎𝑏𝑙𝑒 6.6 − 1: 𝐵𝑜𝑛𝑑 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝛽 (𝐵𝑆 8110: 𝐶𝑙𝑎𝑢𝑠𝑒 3.12.8.4), 𝛽 = 0.5
∴ 𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑏𝑜𝑛𝑑 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝛽√𝑓𝑐𝑢 = 0.5√40 = 3.2 𝑁mm2 / > 2.12 𝑁/mm2
∴ 𝐵𝑜𝑛𝑑 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝑂k
Step 6: Final check of punching shear
From Table 6.4-1: (design II)
100A s (100)2412
= = 0.157
bd (2850)540
Ultimate shear stress vc = 0.4 N/mm2
Punching shear stress was 0.195 N/mm2
Therefore, a 600 mm thick pad is adequate.

Step7: Shear stress

175 | P a g e
 1.5d = (1.5)540 = 810 mm
1275 − 810 = 465 mm
At the critical section for shear, 1.5d from the column face
V = (190.06)(2.85)(0.465) = 251.88 kN
V (251.88)1000
𝐯= =
bd (2850)540
= 0.164 N/mm2 < 0.4 N/mm2
∴ The section is adequate in shear

Column B2, B5 base

Step 1:
Footing self weight = 150 kN (assumed)
.: Total load = 2942.25 + 150 = 3092.25 kN
Total load 3092.25
Required base area = = = 14.32 m2
safe bearing capacity 216
Base lateral dimensions = (√14.32) = 4.0 m
Provide a base 4.0 m square, area = 16.0 m2

176 | P a g e
 Base design load = (1.4)(150) = 210 kN
Total design load = 2942.25 + 210 = 3152.25 kN
3152.25
Earth pressure = = 197.0 kN/m2
4.02
Assume a 800 mm thick footing.
Net upward pressure = 197.0 − (h)(24)(γF )
= 197.0 − (0.8)(24)(1.4) = 170.12 kN/m2
Step 2: Anchorage length for the dowels:
Compression anchorage length = (26)(bar size)
= (26)(16) = 416 mm
∴ A base thickness of 800 mm would adequately allow for concrete cover and two
layers of reinforcement below the dowels.
Step 3: Punching shear
Assume the footing is to be constructed on a blinding layer of concrete so that the
minimum cover is 40 mm.
Effective depth d = 800 − cover − bar size
Effective depth d = 800 − 40 − 20 = 740 mm
Critical perimeter = column perimeter + 3πh
(500 × 2) + (300 × 2) + (3 ∗ π ∗ 800)
= 9139.82 mm
Area within perimeter = (400 + 3h)2 − (4 − π)(1.5h)2
= (400 + 1800)2 − (4 − π)(900)2
= (4.1)106 mm2
Punching shear force V = (163.73)(4.02 − 4.1)
= 1948.387 kN
(1948.387)1000
Punching shear stress = = 0.288 N/mm2
(9139.82)(740)
From Table 6.4-1 (design II), this shear stress is not excessive, therefore h =
800 mm will be suitable.
Step 4: Bending reinforcement

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 Critical section is at the column face:
1.85
Design moment M = (170.12 )(4.0)(1.85)
2
= 1164.47 kNm
Ultimate moment of the section M = 0.156fcu bd2
= (0.156)(40)(4000)(740)2 (10−6 )
= 9139.82 kNm > 1164.47 kNm
. : Compression reinforcement not required

M
K=
fcu bd2
(1164.47)106
= = 0.013
(40)(4000)(7402 )
From Table 4.6 -1 (design II)
z
= 0.94
d
M
Tension steel area As =
0.87fy z
(1164.47)106
= = 4183.03 mm2
(0.87)(460)(0.94)(740)
Minimum tension steel area ∶
100A s
= 0.13
bh
0.13(4000)(600)
As = = 3120 mm2
100
Provide 21 Y16 – 140 (both ways) − (As = 4221 mm2 )

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 Step 5: Local Bond
At the critical section for bending
shear V = (163.73)(4.0)(1.85) = 1211.602 kN
V (1211.602 )103
Local bond stress fbs = = = 1.6 N/mm2
Ʃus d 28 (50.2)(540)
𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑏𝑜𝑛𝑑 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝛽√𝑓𝑐𝑢
𝐹𝑟𝑜𝑚 𝑇𝑎𝑏𝑙𝑒 6.6 − 1: 𝐵𝑜𝑛𝑑 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝛽 (𝐵𝑆 8110: 𝐶𝑙𝑎𝑢𝑠𝑒 3.12.8.4), 𝛽 = 0.5
∴ 𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑏𝑜𝑛𝑑 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝛽√𝑓𝑐𝑢 = 0.5√40 = 3.2 𝑁mm2 / > 1.6 𝑁/mm2
∴ 𝐵𝑜𝑛𝑑 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝑂k
Step 6: Final check of punching shear
From Table 6.4-1: (design II)
100A s (100)5628
= = 0.26
bd (4000)540
Ultimate shear stress vc = 0.4 N/mm2
Punching shear stress was 0.497 N/mm2
Therefore, a 600 mm thick pad is adequate.
Step7: Shear stress

1.5d = (1.5)540 = 810 mm

1850 − 810 = 1040 mm
At the critical section for shear, 1.5d from the column face
V = (163.73)(4.0)(1.040) = 681.11 kN
V (681.11)1000
𝐯= =
bd (4000)540

179 | P a g e
 = 0.315 N/mm2 < 0.4 N/mm2
∴ The section is adequate in shear

Column B3, B4 base

Step 1:
Footing self weight = 150 kN (assumed)
.: Total load = 2688.97 + 150 = 2838.97 kN
Total load 2838.97
Required base area = = = 13.14 m2
safe bearing capacity 216
Base lateral dimensions = (√13.14) = 3.625 m
Provide a base 3.7 m square, area = 13.7 m2
Base design load = (1.4)(150) = 210 kN
Total design load = 2688.97 + 210 = 2898.97 kN
2898.97
Earth pressure = = 211.76 kN/m2
3.72
Assume a 600 mm thick footing.
Net upward pressure = 211.76 − (h)(24)(γF )
= 211.76 − (0.6)(24)(1.4) = 191.60 kN/m2
Step 2: Anchorage length for the dowels:
Compression anchorage length = (26)(bar size)
= (26)(16) = 416 mm
∴ A base thickness of 600 mm would adequately allow for concrete cover and two
layers of reinforcement below the dowels.
Step 3: Punching shear
Assume the footing is to be constructed on a blinding layer of concrete so that the
minimum cover is 40 mm.
Effective depth d = 600 − cover − bar size
Effective depth d = 600 − 40 − 20 = 540 mm
Critical perimeter = column perimeter + 3πh

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 (500 × 2) + (300 × 2) + (3 ∗ π ∗ 600)
= 7254.87 mm
Area within perimeter = (400 + 3h)2 − (4 − π)(1.5h)2
= (400 + 1800)2 − (4 − π)(900)2
= (4.1)106 mm2
Punching shear force V = (191.60)(3.72 − 4.1)
= 1837.43 kN
(1837.43)1000
Punching shear stress = = 0.469 N/mm2
(7254.87)(540)
From Table 6.4-1 (design II), this shear stress is not excessive, therefore h =
600 mm will be suitable.
Step 4: Bending reinforcement

Critical section is at the column face:

1.7
Design moment M = (191.60 )(3.7)(1.7)
2
= 1024.389 kNm
Ultimate moment of the section M = 0.156fcu bd2
= (0.156)(40)(3700)(540)2 (10−6 )
= 6732.46 kNm > 1024.389 kNm
. : Compression reinforcement not required

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 M
K=
fcu bd2
(1024.389)106
= = 0.024
(40)(3700)(5402 )
From Table 4.6 -1 (design II)
z
= 0.94
d
M
Tension steel area As =
0.87fy z
(1024.389)106
= = 5042.74 mm2
(0.87)(460)(0.94)(540)
Minimum tension steel area ∶
100A s
= 0.13
bh
0.13(3700)(600)
As = = 2886 mm2
100
Provide 26 Y16 – 140 (both ways) − (As = 5226 mm2 )
Step 5: Local Bond
At the critical section for bending
shear V = (191.60)(3.70)(1.70) = 1205.164 kN
V (1205.164 )103
Local bond stress fbs = = = 1.71 N/mm2
Ʃus d 26(50.2)(540)
𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑏𝑜𝑛𝑑 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝛽√𝑓𝑐𝑢
𝐹𝑟𝑜𝑚 𝑇𝑎𝑏𝑙𝑒 6.6 − 1: 𝐵𝑜𝑛𝑑 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝛽 (𝐵𝑆 8110: 𝐶𝑙𝑎𝑢𝑠𝑒 3.12.8.4), 𝛽 = 0.5
∴ 𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑏𝑜𝑛𝑑 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝛽√𝑓𝑐𝑢 = 0.5√40 = 3.2 𝑁mm2 / > 1.71 𝑁/mm2
∴ 𝐵𝑜𝑛𝑑 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝑂k
Step 6: Final check of punching shear
From Table 6.4-1: (design II)
100A s (100)5226
= = 0.262
bd (3700)540
Ultimate shear stress vc = 0.4 N/mm2
Punching shear stress was 0.469 N/mm2
Therefore, a 600 mm thick pad is adequate.

Step7: Shear stress

182 | P a g e
 1.5d = (1.5)540 = 810 mm
1700 − 810 = 890 mm
At the critical section for shear, 1.5d from the column face
V = (191.60)(3.7)(0.890) = 630.94 kN
V (630.94)1000
𝐯= =
bd (3700)540
= 0.316 N/mm2 < 0.4 N/mm2
∴ The section is adequate in shear

4.5 Stair case

183 | P a g e
 Flight 1 = flight 2
Data:
(a) Height of flight = 1500 mm
(b) Spine trial thickness = 175 mm
(c) Concrete:
(i) Unit weight =24 kN/m3
(ii) fcu = 40 N/mm2
(iii) Concrete cover = 20 mm
(d) 12.7 mm thick cement/sand (1:3) Screed = 0.293 kN/m2
(e) Steel: fy = 460 N/mm2
(f) Imposed load: qk = 1.5 kN/m2

Length of the spine = √(2.5)2 + (1.50)2 = 2.94 m

𝐋𝐨𝐚𝐝𝐬:
Steps = (0.150)(0.25)(1.2)(0.5)(10)(24) = 5.58 kN
spine = (1.2)(0.175)(2.94)(24) = 14.82 kN
Landing = (1.5)(1.2)(24)(0.175) = 7.56 kN
Screed = (2.50 + 1.4)(1.2)(0.293) = 1.37 kN
Gk = (4.95) + (11.56) + 4.158 + (1.133) = 28.83 kN
Qk = (2.50 + 1.5)(1.2)(1.5) = 7.2 kN
Design load F = (1.4)(Gk ) + (1.6)(Qk )
= (1.4)(28.83 ) + (1.6)(7.2) = 40.362 + 11.232 = 51.594 kN/m

184 | P a g e
 span l = (0.25)(10) + (1.4) + 0.10 = 4.0 m
Fl (51.594 )(4.0)
Design moment = = = 25.797kNm
8 8
12
Effective depth = 175 − 20 − = 149 mm
2
Mu = 0.156fcu bd2 = 0.156(40)(1000)(149)2 × 10−6 = 138.53 kNm > 25.797 kNm
. : Compression reiforcement not reqd.
Area of steel As :
M (25.797)(106 )
K= 2 = = 0.029
bd fcu (1000)(1492 )(40)
From Table 4.6 -1: Lever arm factors (Design II)
z
= 0.94
d
M = 0.87fy As z
.: (25.797)(106 ) = 0.87(460)As (0.94)(149)
. : As = 460.23 mm2 /m width of the stair
. : Provide Y12 − 225 − B1 (A s = 503 mm2 /m)
Distribution steel:
From Table 1.2:
100As
= 0.13%
Ac
(1000)(175)
A s = (0.13) = 227.5 mm2 /m
100
provide Y8 − 200 − B2 (As = 252 mm2 /m)

4.6 Waste water treatment plant

185 | P a g e
 4.6.1 DESIGN OF THE SEPTIC TANK
Design Details
Number of populations to be considered
• Two-bedroom house assumed number = 5 people per house
Therefore = 2×5 = 10 people.
• One-bedroom house assumed number = 4 people per house
Therefore =4×2 =8 people.
Total per floor = 8 + 10 = 18
Number of people in the apartment = 18 × 5 floors = 90
Total population =90+ allowance of 20 people=110 people
Average Water consumption =150 l/h/d (Barnes, 2005)
Sewage generation = assume 80% total water consumption
Detention period=24 hours
Desludging interval=2 years
Sludge deposit =30 l/h/year
Minimum Free board=0.3m
Capacity due to retention of sewage
Total sewage generated=0.8× 110 × 150=13200 l/d= 13.2𝑚 3 /𝑑
Since detention period is 24hrs, Volume = 13.2𝑚 3
Capacity due to retention of sludge
=110× 30 × 2=6600 l/d= 6.6𝑚 3 /𝑑
Total capacity =13.2+6.6=19.8𝑚 3
Depth of tank as 2.3m(chosen)
19.8𝑚 3
Area of tank = 2.3
= 8.6𝑚 2

Take ratio of breadth to length as 1:1.5

L=1.5w,
1.5w× 𝑤 = 8.6𝑚 2
W = 2.39 = 2.4m
Therefore; breadth=2.4m, Length=3.6m
∴ Provide 3.60m by 2.4m by 2.3m septic tank.

Connecting sewer
Q =AV= 13.2𝑚 3 /𝑑

186 | P a g e
 13.2
In 𝑚 2/s 24×60×60 = 1.5278 × 10−4 𝑚 3 /𝑠

Use a self-cleansing velocity as 0.75 m/s.

𝜋
1.5278 × 10−4 𝑚3 /𝑠 = 4 × 𝑑 2 × 0.75

d = 0.01610 =16mm hence use 100mm

Soak Pit Calculation
Waste water coming out from septic tank = 13.2𝑚 3
Percolation rate=1500 𝑙 ⁄𝑚3 /𝑑𝑎𝑦
13.2
Volume of filter media = 1.5
= 8.8𝑚 3

Assume depth of pit=3m

8.8
Area of soak pit = 3
= 2.933𝑚 2

Diameter of soak well required =2m

4.6.2 Reinforced Concrete design for tank slab

Dead load:
Concrete: (0.15) (24) = 3.6𝑘𝑁/𝑚 2
Live load:
Live load Qk = 1.0 kN/m2
Design load n = 1.4Gk + 1.6Qk = (1.4) (3.6) + (1.6) (1) = 6.64
n = 6.64kN/m2

𝑙𝑦 3600
= = 1.5 therefore, it a two-way slab.
𝑙𝑥 2400

Moments
Support moments:
Table 3.13 (BS 8110
𝑎𝑠𝑥 = 0.104 and 𝑎𝑠𝑦 = 0.046

187 | P a g e
 Shorter span moment 𝑚𝑠𝑥 = 𝑎𝑠𝑥 𝑛𝑙𝑥 2= (0.104) (6.64) ×2.32 = 3.653 kNm
Longer span moment 𝑚𝑠𝑦 = 𝑎𝑠𝑦 𝑛𝑙𝑥 2 = (0.046) (6.64) × 2.32 = 1.6158
kNm
Main reinforcement:
Concrete cover = 25 mm
1 1
Effective depth: d = 150 − cover − 2
(bar dia.) =150 − 25 − 2
(12) = 119

mm
𝑚 3.653×106
K=𝑓 𝑏𝑑2
= 40×1000×1192 = 0.00645
𝑐𝑢
𝑧
𝑑
= 0.94.
𝑀 3.653×106
Therefore As = = = 81.60𝑚𝑚2
0.87𝑓𝑦 𝑍 0.87×460×0.94×119

From Table 3.25 (BS 8110)

0.13𝐴𝐶 0.13×1000×150
As(min) = = = 195𝑚𝑚 2 /𝑚 > 81.60𝑚𝑚 2 /𝑚
100 100

Therefore, As = 195𝑚𝑚 2 /𝑚
Therefore,
Provide Y8 @ 250 mm C/C − T1 (As = 201 𝑚𝑚 2 /𝑚)
Provide Y8 @ 250 mm C/C− T2 (As = 201𝑚𝑚 2 /𝑚 )
Provide Y8 @ 250 mm C/C− 𝐵1 (As = 201𝑚𝑚 2 /𝑚 )
Provide Y8 @ 250 mm C/C− 𝐵2 (As = 201 𝑚𝑚 2 /𝑚)

Side walls is to be made of masonry wall

Use A-142 mesh for bottom slab reinforcement
More details are indicated in the structural drawing of the septic tank.

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5.0 Lab work:
5.1 SOIL TEST
Soil Sampling

Obtaining high-quality undisturbed samples is one of the biggest challenges faced in the field of
geotechnology. Poor quality samples may result in non-representative results being obtained in lab tests,
the use of which may give rise to faulty designs.

Types of samplers adopted in each part of the world depend on the state of development of the area, its
sampling tradition, economics, and its principal soil types. For instance, in the heavily developed South
East and Midlands of England, soil types are typically stiff or very stiff clays interstratified in and weak
rocks. This calls for organized and detailed geotechnical investigations to be carried out if any sound
economic and safe design is to be realized.

All samples should be labeled clearly to show project name, date, location, borehole number, depth and
method of sampling used. In addition, each sample should be allocated serial number for easy
identification, transportation and storage.

12.6.1 Undisturbed soil samples

Undisturbed soil samples refer to soil samples that preserve in situ structure and pore water content. Such
samples are best needed for determination of soil shear strength and soil consolidation parameters. Soils
such as Soft clays are extremely sensitive to sampling disturbance. The effect is more pronounce in clays
of low plasticity than in those of high plasticity. Release of in situ total stresses occurs immediately after
sampling. This results into dissipation pore water pressure causing the outer zone to be more relatively
disturbed than the central zone. Water is then drawn from the inner zone leading to swelling effect.
Redistribution of pore water takes place accompanied by reduction in effective stress resulting into less
resistance to shear.

Undisturbed samples can be obtained from bore holes when boring tools are withdrawn. Tube samplers
are then used to extract the samples. Molten wax about 25mm thick is applied around each sample at end
of the sampling tube. The tube ends are then covered by protective caps so as to sustain the undisturbed
state. It is recommended that undisturbed samples be placed in airtight non-corrosive containers for
effective preservation of natural moisture content.

12.6.2: Disturbed soil samples

189 | P a g e
A disturbed soil sample has same particle size distribution as the one for in situ but has its soil structure
significantly damaged and water content altered. Such samples are used mainly for visual classification
and compaction. They can be excavated from trial pits or obtained from tools that are used to advance
bore holes such as augers.

12.6.3 Typical types of tube samplers

There are currently several types of tube samplers that can be used to excavate soil samples from in situ
under different conditions. Typical samplers include the following:

12.6.3.1 Open drive sampler

An open dive sampler consists of a steel tube with a screw thread at each end. It has an internal diameter
of 100mm and a length of 450mm. A cutting shoe is attached to one end of the tube. The other end of the
tube is screwed onto a sampler head which is intern screwed on to boring rod(s). Non-return valve is fifed
onto the sampler so as to allow air and pore ware to escape in the sampling process (fig.12.5 (a)). This
sampler is suitable for all clay soils. When used to sample sands a core catcher is fitted. The catcher
consists of a short tube with spring-loaded flaps which prevent loss of the soil excavated.

Prior to sampling, all loose material at bottom of in a bore hole should be removed. The sampler can then
be driven statically by hydraulic or mechanical jacking using the boring rig. Care should be taken not to
drive the sampler beyond its capacity to avoid compression on sampler head. After withdrawal, cutting
shoe and sampler head are detached and ends of sampler are sealed.

12.6.3.2 Split-barrel sampler

A split–barrel sampler consists of a tube which is split longitudinally into two halves. It has an internal
diameter of 35mm and external diameter of 50mm respectively. A sampler head, also provided, has air
lease holes (fig.12.5 (b)). In addition, the sampler has a cutting shoe which is used to beak the soil in situ.
The two halves can be separated when the shoe and the head have been detaches so as to easily remove
the sample from the sampler. The split barrel sampler is used mainly in sand soils. It is the specified
standard tool employed in the standard penetration test (SPT).

190 | P a g e
 Fig.12.5: Typical types of sampling tubes for soil samples

12.6.3.3 Compresses Air Sampler

A compressed air sampler consists of a sample tube 60mm in diameter attached to a sampler head that has
a release valve which can be closed by a rubber diaphragm. It is enclosed in an outer tube (‘bell’) and
attached to a weight which slides in a hollow guide rod (fig.12.5(c)). The guide rod is attached to the
sampler head and connected to the boring rod. When in use compressed air, produced by a foot pump, is
forced down the hollow guide rod to the bell.

The sampler is used to obtain undisturbed samples of sand below water table. It is lowered to the bottom
of the bore hole and pushed into the soil by means of boring rods. Compressed air is then used to expel
water from the bell and to close valve in the sampler head. After the tube and the bell have been
withdrawn from the hole, a plug is placed in the bottom of the tube before sanction is released and
sampling tube removed from sampler head. Usually, volume of soil displaced by the sampler as
proportion of the sample volume is represented by area ratio (Ar) of the sampler as shown in expression
12.1

Ar = …………………………..……………….12.1

191 | P a g e
The lower the ratio, the lower the degree of disturbance caused to a sample. In the open drive sampler, the
ratio is lower than for the compressed air sampler.

The particle size distribution

Particle size analysis of a soil sample involves determining the percentage by mass of particles within the
different size ranges. The particle size distribution of a course soil can be determined by the point of
sieving. If fine particles are on the soil the sample should be treated with a deflocculating agent and
washed through the sieves.

The particle size distribution of a fine soil or fine fraction of a course soil can be determined by the
method of sedimentation. This method is based on Stokes’ law which governs the velocity at which
spherical particles settle in a suspension.

The particle size distribution is presented as a curve on a semilogarithmic plot, the ordinates being the
percentage by mass of particles smaller than the size given by abscicca. A coarse soil is described as well
graded if there is no excess of the particles in any size range and if no intermediate sizes are lacking. in
general, a well graded soil is represented by a smooth, concave distribution curve. A course soil is
described as poorly graded if:

A high proportion of the particles have sizes within narrow limits (a uniform soil).

Particles of both large and small sizes are present but with a relatively low proportion of particles of
intermediate size (a gap-graded or step-graded soil).

Particle size distribution is represented on a logarithmic scale so that two soils having the same degree of
uniformity are represented by curves of the same shape regardless of their positions on the particle size
distribution plot.

The particle size corresponding to any specified value on the ‘percentage smaller’ scale can be read from
the particle size distribution curve. The size such that 10% of the particles are smaller than that size is
denoted by D10. Other sizes such as D30 and D60 can be defined in a similar way.

The size D10 is defined as the effective size. The general slope and shape of the distribution curve can be
described by means of the coefficient of uniformity (CU) and the coefficient of curvature (CZ), defined as
follows:

Cu=D60/D10

CZ=D²30/D60D10

192 | P a g e
The higher the value of the coefficient of uniformity the larger the range of particle sizes in the soil. A
well graded soil has a coefficient of curvature of between 1 and 3.

SAMPLE SIZE – 450 g

Sieve size (mm) Mass retained (g) Mass passing (g) % Passing
2.00 0 450 100
1.18 10.15 439.85 97.7
0.600 25.10 414.75 94
0.425 72.00 342.75 82.64
0.250 85.25 257.50 75.13
0.075 111.50 146.0 56.69
Pan 146.00 0 0

PARTICLE SIZE DISTRIBUTION CHART.

193 | P a g e
INDEX TESTS

LIQUID LIMIT TEST

The liquid limit is the empirically established moisture content at which the soil passes from the liquid
state to the plastic. It provides a means of classifying a soil especially when the plastic limit is also
known. Two main types of tests are specified. the first is the cone penetrometer method, which is

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fundamentally more satisfactory than the alternative because it is essentially a static test depending on soil
shear strength. It is also easier to perform and gives more reproducible results.

The second is much earlier Casagrande type of test which has been used for years as a basis for soil
classification and correlation of engineering properties. This test introduces dynamic effects and is more
susceptible to discrepancies between operators.

Sample preparation

Wherever possible the test shall be carried out on the soil in its natural state. With many clay soils it is
practicable and shall be permissible to remove by hand any course particles present, i.e., particles retained
on a 425um.

Apparatus

➢ Test sieves of sizes 425um and 2mm, with receiver.

➢ Apparatus for determination of moisture content
➢ A sharp knife
➢ An implement for shredding cohesive soil
➢ Two palette knives
➢ A corrosion resistant airtight container, large enough to take 200g to 300g of wet soil.
➢ A flat, glass plate, a convenient size being 10mm thick and about 500mm square.
➢ A wash bottle containing distilled water.
➢ Cone penetrometer
➢ Stop clockk

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CONE PENETROMETER METHOD

This method covers the determination of the liquid limit of a sample of soil in its natural state, or of
sample of soil from which material retained on 425um test sieve has been removed.

Procedure

i. Take a sample of about 300g from the soil paste prepared and place it on the glass plate.
ii. Mix the paste for at least 10min using the two palette knives. If necessary, add more distilled
water so that the first cone penetration reading is about 15mm.
iii. Push a portion of the mixed soil into the cup with a palette knife taking care not to trap air. Strike
off excess soil with the straightedge to give a smooth level surface.
iv. With the penetration cone locked in the raised position lower the supporting assembly so that the
tip of the cone just touches the surface of the soil. When the cone is in the correct position a slight
movement of the cup will just mark the soil surface. Lower the stem of the dial gauge to contact
the cone shaft and record the reading of the dial gauge to the nearest 0.1 mm.
v. Release the cone for a period of 5s. After locking the cone in position lower the stem of the dial
gauge to contact the cone shaft and record the reading of the dial gauge to the nearest 0.1mm.
vi. Lift out the cone and clean it carefully to avoid scratching.
vii. Add a little more wet soil into the cup, and repeat the process.
viii. If the difference between the first and second readings is not more than 0.5mm, record the
average of the two penetrations
ix. Take a moisture content sample of about 10g from the area penetrated by the cone and determine
the moisture content.

RESULTS.

1st

Reading on dial gauge before release – 393

Reading after 5 sec – 588

2nd

Reading on dial gauge before release – 393

Reading after 5 sec – 592

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3rd

Reading on dial gauge before release – 393

Reading after 5 sec – 594

Average

Reading on dial gauge before release – 393

Reading after 5 sec – 591

197 | P a g e
Data Analysis

Tin + Wet Soil = 120g

Tin + Dry Soil= 96g

Tin weight =25g

Wet mass (Mw) = (120 －25) = 95g

Dry Mass (Md) = (96－25) = 71g

Moisture weight= (Mw－Md) = 95 －71= 24g

Moisture Content =(24/71)100= 33.8

Plastic limit test

The plastic limit is the empirically established moisture content at which the soil becomes too dry be
plastic.

Procedure

Take a sample of about 20g from the soil paste prepared as in the case of liquid limit and place it on the
glass mixing plate.

Allow the soil to dry partially on the plate until it becomes plastic enough to be shaped into a ball.

Mould the ball of the soil between the fingers and roll it between the palms of the hands until the heat of
the hands has dried the soil sufficiently for slight cracks to appear on the surface.

Mould the soil in the fingers to equalize the distribution of moisture then form the soil into a thread about
6mm diameter between the first finger and thumb of each hand.

Roll the thread between the fingers and on the surface of glass rolling plate. Use enough pressure to
reduce the diameter of the thread to about 3mm in 5 to 10 complete forward and back movements of the
hand until it cracks.

Collect broken pieces of the thread into two containers, and determine the moisture content.

Report the average moisture content (in whole number) as the plastic limit.

Data Analysis

198 | P a g e
Tin + Thread =39 g

Tin + Dry Thread= 35g

Tin weight =20g

Wet mass (Mw) = (39 －20) =19g

Dry Mass (Md) = (35－20) = 15g

Moisture weight= (Mw－Md) = 19 －15= 4g

Moisture Content =(4/15)100= 26.60 %

Plasticity Index (PI)

Plasticity is the ability of a soil to undergo unrecoverable deformation without cracking or crumbling.

Plasticity Index is the range of moisture content at which soils remain plastic (mouldable condition)

PI = LL－PL

33.8% － 26.6% = 7.2

199 | P a g e
200 | P a g e
SOIL CLASSIFICATION

It is necessary for the foundation Engineer to classify the site soils for use as a foundation for the
following reasons:

To be able to use database of others in predicting foundation performance

To build one’s local database of successes and failures.

To maintain the permanent record that can be understood by others should problems later develop and
outside parties be required to investigate the original design.

To be able to contribute to the general body of knowledge in common terminology via journal papers or
conference presentations.

Soil description

The principal material characteristics are particle size distribution and plasticity, from which the soil
name can be deduced. secondary soil characteristics are color of the soil and shape, texture and
composition of the particles. Mass characteristics should ideally be determined in the field but in many
cases, they can be detected in undisturbed samples.

In Soil classification a soil is allocated to one of a limited number of groups on the basis of material
characteristics only. Soil classification is thus independent of the in-situ condition of the soil mass.

Soil description details

According to BS 5930, the basic soil types are boulders, cobbles, gravel, sand, silt and clay, defined in
terms of the particle size ranges; added to these are organic clay, silt, silt or sand, and peat. The names
are always written in capital letters in soil description mixtures of the basic soil types are referred to as
composite types.

A soil is of basic type sand or gravel if, after removal of any cobbles or boulders, over 65% of the
material is of sand and gravel sizes. A soil is of basic type silt or clay (fine soils) if, after the removal of
any cobbles or boulders, over 35% of the material is of silt and clay sizes. Sand and gravel may be each
be sub divided into coarse, medium and fine fractions.

The state of sand can be described as well graded, poorly graded, uniform or gap-graded. Fine soils
containing 35-65% coarse material are described as sandy and/or gravelly SILT (or CLAY). Deposits
containing over 50% of boulders and cobbles are referred to as very coarse and normally can be
described in excavations and exposures.

201 | P a g e
Soil classification systems

There are two elaborate classification systems that use grain size distribution and plasticity of the soils.
They are;

American Association of State Highway and Transport Official System. (AASHTO).

Unified Soil classification System. -this is the system preferred by Geotechnical Engineers.

The unified soil classification system

In this system the main soil types are designated by capital letters

• G-gravel
• S-sand
• M-silt
• C-Clay

The liquid and plastic limits are used to classify the soils, employing the plasticity chart. The axes of
the plasticity chart are plasticity index and liquid limit; therefore, the plasticity characteristics of a
particular soil can be represented by a point on the chart. Classification letters are allotted to the soil

202 | P a g e
according to the zone within which the point lies. The chart is divided into five ranges of liquid limit. The
four ranges I, H, V and E can be combined as an upper range(U) if closer designation is not required or if
the rapid assessment procedure has been used to assess plasticity.

The diagonal line on the chart known as the A-line, should not be regarded as a rigid boundary between
clay and silt for purposes of soil classification, as opposed to classification. A group symbol may consist
of two or more letters eg.

• SW-well graded SAND

• SCL-very clayey SAND
• CIS- sandy clay of intermediate plasticity.

MHSO-organic sandy SILT of high plasticity

203 | P a g e
The name of the soil group should always be given as above in addition to the symbol, the extent of
subdivision depending on the particular situation.

The term FINE SOIL or FINES(F) is used when it is not required, or not possible to differentiate between
SILT(M) and CLAY(C). SILT(M) plots below the A-line on the plasticity chart. SILT or CLAY is
qualified as gravelly if more than 50% of the coarse fraction is gravel size and as sandy if more than 50%
of the coarse fraction is of sand size. The alternative term M-SOIL is introduced to describe material
which, regardless of its particle size distribution, plots below the A-line of the plasticity chart. fine soils
containing significant amounts of organic matter usually have high to extremely high liquid limits and
plot below the A-line as organic silt. Peats usually have very high or extremely high liquid limits.

Any cobbles or boulders (particles retained on a 63-mm BS sieve) are removed from the soil before the
classification tests are carried out but their percentages in the total sample should be determined or
estimated.

From the data obtained in sieve analysis;

Sample Size = 450 g

Sieve size (mm) 2.0 0.425 0.075

Percentage passing 100 82.64 56.69

Liquid Limit = 33.8 %

Plasticity Index (P.I) = 7.2%

Since materials passing sieve number 200 (0.075mm) =56.69% which is greater than 50% therefore the
soil is fine grained soil.

Referring to the plasticity chart of LL = 33.8% and PI = 7.2 % the soil plots below the A-line thus can be
classified as ML.

Group Name.

Percentage passing 0.075 mm sieve =56.69% therefore some 43.31% sand exists

Gravel content = 0%

The soil is therefore Silt with Low Plasticity (ML)

204 | P a g e
The Shear Box Test.

From the building code table 1 page 23

1 ton/ft2=95760.52Pascals (N/M 2 )

3 × 95760.52 × 10−3 = 216KN/M 2

1. INTRODUCTION

Experiment: Laboratory testing for shear strengths by the direct shear box.

Aim of the practical: To determine the shear strength of soil materials. It is defined as the maximum
resistance that a soil material can withstand when subjected to shearing.

2. THEORY BEHIND THE PRACTICAL

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Shear box test is a testing method used to determine the shear strength of a soil sample. Since granular
soils cannot be retrieved undisturbed, the soil is re-compacted inside a shear box.

The soil friction angle is then determined by direct shear and the tests are carried out under saturated
conditions. Thus, is performed as per ASTM D4080 and is performed on three or four specimen of
undisturbed soil samples.

3. METHODOLOGY

3.1. Apparatus

➢ Shear box
➢ Shear box container
➢ Base plate with cross groves on its top
➢ Load pad with steel ball
➢ Weighing machine
➢ Proving ring
➢ Dial gauge
➢ Weights
➢ Sampler
➢ The apparatus consists of a brass box, split horizontally at the Centre of soil specimen. The usual
plan size of the sample is 60 × 60 𝑚𝑚 2 . A vertical load is applied to the top of the sample by
means of weights. As the shear plane is predetermined in the horizontal direction, the vertical
load is also the normal load on the plane of failure. Having applied the required vertical load a
shearing force is gradually exerted on the box until failure occurs.

Apparatus continuation

206 | P a g e
 3.2. Procedure
1. The shear box is placed in a large container and is tightly held in position at the bottom of
container. The container is supported over rollers to facilitate lateral movement of lower-half of
the shear box when shear force is applied to the lower shear box through a geared jack.
2. Placing of the soil specimen in the direct shear box apparatus
3. A normal load is applied through a loading yoke, placed over the pressure ball on the pressure
pad. The required normal stress of 0.5kgf/cm2 is applied. The shear deformation dial gauge is
placed in position. For drained tests, the soil specimen is allowed to consolidate under normal
load.
4. The locking pins are removed and the upper box is slightly raised using spacing screws and then
the shear load is applied to the lower-half of the box through a geared jack such that the lower-
half moves at a constant rate of strain. The proving dial gauge readings are taken at regular
intervals of deformation dial gauge readings.
5. The test is continued till the shear load reaches a maximum value and then decreases.
6. The shear load is then released, and the proving ring and deformation dial gauges as well as the
shear box are dismantled.
7. Repeat the previous experiment by adding a 5 kg normal stress and continue the experiment till
failure.
8. Record carefully all the readings. Set the dial gauges zero, before starting the experiment.

207 | P a g e
 4. OBSERVATIONS AND DATA ANALYSIS
Time Dial (ND) Strain
(seconds ) Gauge Shear Stress( ᵟ )
readings
5 kg 10 kg 15 kg 5 kg 10 kg 15 kg

0 0 0 0 0 0 0 0

30 88 85 150 14.784 14.28 25.2 0.525

60 170 220 255 28.56 36.96 42.84 1.05

90 220 300 346 36.96 50.4 58.128 1.575

120 254 350 415 42.672 58.8 69.72 2.1

150 277 387 465 46.536 65.016 78.12 2.625

180 295 413 501 49.56 69.384 84.168 3.15

210 306 430 526 51.504 72.24 88.368 3.625

240 314 441 542 52.752 74.088 91.056 4.2

270 316 449 555 53.088 75.432 93.24 4.725

300 312 454 561 52.416 76.272 94.248 5.25

330 317 456 565 53.256 76.608 94.92 5.775

360 317 456 569 53.256 76.608 95.592 6.3

390 316 455 571 53.088 76.44 95.928 6.825

420 312 448 575 52.416 75.264 96.6 7.35

Time Dial Strain

(seconds) Gauge (ND) Shear
readings Stress

208 | P a g e
 5 kg 10 kg 15 kg 5 kg 10 kg 15 kg

450 306 440 582 51.408 73.92 97.776 7.875

480 294 430 586 49.392 72.24 98.448 8.4

510 592 99.456 8.925

540 592 99.456 9.45

570 593 99.624 9.975

600 597 100.296 10.5

630 596 100.128 11.025

660 596 100.128 11.25

690 594 99.792 12.075

720 587 98.616 12.6

750 574 96.432 13.125

780 563 94.584 13.65

RESULTS AND DISCUSSIONS

I. Strain (ƹ%)
𝛿𝑙
= × 100%
10
(𝑓𝑒𝑒𝑑×𝑡𝑖𝑚𝑒)
= 10
× 100%
𝑡
1.05𝑚𝑚/𝑚𝑖𝑛×
60
= × 100%
100

Where t=time in seconds.

ii. Normal stress
[9(m+1)+(19.5+7.43)]×9.81×10−3
σn =
100×100×10−6

where : m= 5,10, and 15kg

209 | P a g e
 σ5 =79.40
σ10=123.54

σ15=167.68
iii. Shear stress
𝑁𝑑 ×𝑘𝑟
𝜏 = 𝐴
𝑘𝑁/𝑚2 𝜏

Where: Nd = dial gauge reading

Kr = ring constant (1.68 N/div) (10-3)
A= Area of shear box (100×100)mm2 (10-6)
𝑁𝑑 ×1.68×10−3
𝜏=
100×100×10−6

Iv. Graph of shear stress (𝛕) Vs strain (ƹ%)

Graph of Shear stress (𝜏) against Strain (ε%)

120

100
Shear stress (kN/m2)

80
(x10-³kN/m²)

60

40

20

0
0 2 4 6 8 10 12 14
Normal stress (kN/m2)

5 KG 10 KG 15 KG

210 | P a g e
v. Ultimate shear stress (𝛕f) Vs Normal stress (𝛔n)

120

100
Shear stress (kN/m2)

80

60

40

20

0
0 20 40 60 80 100 120 140 160 180
Normal stress (kN/m2)

vi. From ᶲ determine the bearing capacity factor (special graph)

From the graph:

Cohesion =

Angle 𝜑 = 280

6. From φ determine the bearing capacity factor (special graph/table)

Bearing capacity

From the chart of bearing capacity factors

Nr = 16

Nq = 18

Nc = 29

Ultimate bearing capacity = Qf = 1.2CNc + σ2Nq + 0.4ɤBNr

Where:

C = Cohesion

211 | P a g e
 ɤ = unit weight of soil (14kN/m2)

B = unit size of footing (1x1) m2

Z = depth of excavation (= 2m)

Qf = 0 + (14x2x18) + (0.4x14x1x16)

= 593.6kN/m2 (Ultimate value)

Allowable bearing capacity

Q = (Qf- ɤ2)/3 + ɤz

Where ɤz is the overburden pressure due to backfill.

Q = (593.6 -(14x2))/3 + 14x2

= 188.5 + 27.5

= 216kN/m2

5.2 Design the concrete mix: (grade 40 N/mm2

Characteristic Strength 30N/mm2 at 28 days

Standard Deviation

212 | P a g e
As shown in figure 3 above, we use a standard deviation of 8N/mm2 for s less than 20 results

Margin:

The constant k is derived from the mathematics of the normal distribution and increases as the proportion
of defectives is decreased, thus: The characteristic strength may be defined to have any proportion of
defectives; BS 5328 and BS 8110 adopt the 5% defective level in line with the CEB/FIP international
recommendations for the design and construction of concrete structures

K for 10% defectives = 1.28

k for 5% defectives = 1.64

k for 2.5% defectives = 1.96

k for 1% defectives = 2.33

Our k is 1.64 pg 7 BRE

Margin (C1) =K*S= 1.64*8=13.12 N/mm2

Target mean strength (C2) = fcu+C1

=30 +13.12= 43.12 N/mm2

Cement strength class =42.5

Aggregate type: coarse = crushed

Aggregate type: fine = uncrushed

Free –water/cement ratio

213 | P a g e
Next, a value is obtained from Table 2 for the strength of a mix made with a free-water/cement ratio of
0.5 according to the specified age, the strength class of the cement and the aggregate to be used. This
strength value is then plotted on Figure 4 and a curve is drawn from this point and parallel to the printed
curves until it intercepts a horizontal line passing through the ordinate representing the target mean
strength. We obtain 0.47

Slump test

It is done to determine the workability or consistency of concrete mix prepared at the laboratory or the
site during the progress of the work. Concrete slump test is carried out from batch to batch to check the
uniform quality of concrete during construction.

Result =30-60mm

Maximum aggregate size: 20mm

214 | P a g e
Free-water content:

2 1
Therefore; water content= × 180 + × 210 =190kg/m3
3 3

Cement content C3) = Free water content / maximum free water

=190/0.5

=380kg/m3 >275kg/m3 therefore cement content is OK

Minimum cement content

The following requirements are specified and thus entered under the relevant item on the mix design
form, as shown in Table 4

Minimum cement content, 275 kg/m3 (BS 8110-1997 Cl 3.3.3 Table 3.3.

Modified free-water/cement ratio is given by free-water content/minimum cement content

That is; 190/275

=0.68, use the max; 0.65

Relative density of aggregate (SSD) is assumed to be2.7

215 | P a g e
RDA= the relative density of the aggregate calculated on a saturated surface-dry basis. (When unknown
an approximation can be made by assuming a value of 2.6 or 2.7 as appropriate)

Concrete density: from fig5. Above =2430kg/m3

Total aggregate content (C4) = concrete density - cement content – free-water content

=2430-380-190=1860 kg/m3

Maximum cement content is not specified and thus there is no entry under Item 3.2. There are no previous
control data and thus a standard deviation of 8 N/mm2 obtained from Figure 3 is used in Item 1.2. The
fine and coarse aggregates to be used are uncrushed, the relative density is unknown and is assumed to be
2.7 as stated in 5.4, and the fine aggregate has 70% passing a 600 µm sieve

Grading of fine aggregate (percentage passing 600 μm sieve) =70%

Proportion of fine aggregates: =30% as shown below

216 | P a g e
Fine aggregate content C5 1860*0.3= 560kg/m3

Coarse aggregate content 1860-560=1300kg/m3

100mm cubes table

Quantities Cement(kg) Water(kg/l) Fine Coarse aggregates(kg)

aggregate(kg)
10mm 20mm 40mm
Per m3(to nearest 5kg) 380 190 560 435 865 N.A
3
Per trial mix of 0.012m 4.686 2.28 6.60 5.16 10.32 N.A

NOTE: volume of cubes per trial mix= 0.13*12 cubes=0.012m3

Design ratio: 380:560:1300

1:1.47:3.42

1 m3 of dry concrete = (1.54-1.57) m3 of wet concrete

Volume of cement = 1/5.89*1.57=0.27

Density of cement=1440 kg/m³

217 | P a g e
 Hence, Weight of cement = 1440*0.27=388.8kg

Volume of F. A=1.4/5.6*1.57=0.39m3

Density o f F. A=1600-1800 kg/m³: use 1700kg/m³

Hence, Weight of F. A=0.39*1700= 663 kg

Volume of C.A =3.2/5.6*1.57= 0.90

Density of C. A=1500-1800 kg/m³; use 1600 kg/m³

Hence, Weight of C.A =1600*0.9=1440kg

Hence, Weight of water= 190kg=190litres

Volumes

12 cubes =12(0.1³) =0.012 m³

Material Requirements

Cement=403.2*0.012m³=4.84kg -

F. A=663*0.012m³=8.0kg

C.A=1440kg*0.012m³=17.28kg

Water =190*0.012m³=2.28litres

ABSORPTION OF FINE AGGREGATES

Absorption: The increase in mass due to water in the pores of the material.

SSD -Saturated Surface Dry. The condition in which the aggregate has been soaked in water and has
absorbed water into its pore spaces. The excess, free surface moisture has been removed so that the
particles are still saturated, but the surface of the particle is essentially dry.

Apparatus

218 | P a g e
 i. A sample of sand
ii. Weighing balance
iii. Measuring cylinder
iv. Container for soaking sand

Procedure

1. Thoroughly mix the sample and reduce the sample to the required size in accordance with
AASHTO T 248 (Reducing Field samples of Aggregates to Test Size). The sample size for this
procedure is approximately 1000g of material passing the No. 4 (4.75 mm) sieve.
2. Dry test samples to constant weight in an oven set at 230 ± 9°F (110 ± 5°C). Cool the sample at
room temperature for 1 to 3 hours. After the cooling period, immerse the sand in water at room
temperature for a period of 15 to 19 hours.
Instead of completely immersing the sand in water, AASHTO considers sand to be "soaked" if
the sand is maintained at a moisture content of at least 6% for the prescribed period. This is the
recommended procedure to eliminate the need to decant excess water from the sand prior to
testing. The decantation process is time consuming and difficult, since great care must be taken to
avoid decanting some of the sample along with the water. Additionally, the sand will be much
closer to the SSD condition when soaked at 6% moisture, which expedites the dry procedure.
3. Decant water from sample, avoiding loss of fines. Spread the sample on a flat, non-absorbent
surface. Stir the sample occasionally to assist in homogeneous drying. A current of warm air may
be used to assist drying procedures (Figure 2); however, fine particles may be lost with this
procedure if not careful.
4. Determine the SSD condition of the sand using the Cone Test.
Note: Throughout the process of drying in Step 3, test the sand for SSD condition sing the cone
method. Place the cone with the large diameter down on a glass plate. Fill cone to overflowing
with drying sand. Lightly tamp the fine aggregate into the mold with 25 light drops of the tamper
(Figure 3). Each drop should start about 1/5 in. above the top surface of the fine aggregate.
Remove loose sand from base and carefully lift the mold vertically. If surface moisture is still
present, the fine aggregate will retain the molded shape. When the sand achieves an SSD
condition, the sand will slump.
If on the first trial the sand slumps, moisture must be re-added and the drying process repeated.
Record the weight of the sand as SSD mass when the sand slumps to the nearest 0.1 g.

Results;

219 | P a g e
 Sample ID Weight of dry sample Weight of saturated Absorption (%)
(g)
1 1000 1020 2.0
2 1000 1015 1.5
3 1000 1025 2.5

ABSORPTION OF COARSE AGGREGATES(10mm and 20mm)

Apparatus
1) 10mm and 20mm aggregate samples
2) Dry cloth
3) Measuring cylinder
4) Weighing balance
5) Soaking container
6) Water
7) Pans

Procedure

i. Weigh the dry aggregate. Take 1kg of dry 20mm and 10mm aggregates on different pans.
ii. Place the dry aggregate in a container of water, ensuring that the aggregate is completely
submerged.
iii. Let the aggregate soak in water for 24 hours
iv. After soaking remove the aggregates from the water, and wipe the aggregates using the dry
cloth.
v. Using the weighing balance, record the weight of the aggregates.

Absorption (%) = [wet weight- dry weight/dry weight] *100

Coarse Absorption Rate for 20mm aggregates.

Sample ID Weight of dry sample W of SSD (g) 20 mm absorption (%)

(g)
1 1000 1012 1.2
2 1000 1009 0.9
3 1000 1010 1.0

220 | P a g e
Coarse Absorption Rate for 10mm aggregates.

Sample ID Weight of dry sample W of SSD (g) 20 mm absorption (%)

(g)
1 1000 1011 1.1
2 1000 1010 1.0
3 1000 1010 1.0

Note:

➢ According to BS8007, aggregate should comply with either BS 882 OR BS 1047 and have
absorption, as measured in accordance with BS 812-2, generally not greater than 3%.

Average absorption rate: fine aggregates and coarse aggregates (10& 20mm)

Fine

2.0+1.5+2.5
Avg. = = 2% (which is within the range provided in the note above)
3

100+𝐴
100
∗ 𝑌 = 𝑋(𝐵𝑎𝑡𝑐ℎ 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦)

Where;

X = batch quantity

Y = known mass of the aggregate from the table

A = percentage by mass of water needed to bring the dry aggregates to a saturated surface-dry
condition

Coarse Aggregate (10mm)

1.1+1.0+1.0
Avg. = 3
= 1.03% (which is within the range provided in the note above)

Coarse Aggregate (20mm)

1.2+0.9+1.0
Avg. = 3
= 1.03% (which is within the range provided in the note above)

Final Table adjusted to SSD

221 | P a g e
 Quantities Cement(kg) Water(kg/l) Fine Coarse aggregates(kg)
aggregate(kg)
10mm 20mm 40mm
Per m3(to nearest 5kg) 380 190 560 435 865 N. A
3
Per trial mix of 0.012m 4.56 2.28 6.72 5.22 10.38 N. A
Adjustment per m3 380 213.33 549.02 429.42 8653.90 N. A
Per trial mix 0.012 4.56 2.61 6.59 5.15 10.25 N. A

Final Results Adjusted To 15%

First Trial Mix

Material Requirements

Cement=4.56kg * 1.15 = 5.244kg

F. A=6.59kg * 1.15 = 7.579 kg

C.A

10mm=5.15*1.15kg = 5.923kg

20mm = 10.25*1.15= 11.79kg

Water =2.61* 1.15 = 3.0 L

2ND TRIAL MIX

Add 50kg of cement to that of the first batch

380+50= 430kg

Fine aggregates =549.02kg

Coarse aggregates= 1283.32kg

Ratio 430:549:1283.32

=1:1.28:2.98
1 m3 of dry concrete = (1.54-1.57) m3 of wet concrete

Volume of cement = 1/5.26*1.57=0.3m3

222 | P a g e
 Density of cement=1440 kg/m³

Hence, Weight of cement = 1440*0.3= 432kg

Volume of F. A=1.28/5.26*1.57=0.382m3

Density of F. A=1600-1800 kg/m³: use 1700kg/m³

Hence, Weight of F. A=0.3*12700=649.4kg

Volume of C.A =2.98/5.26*1.57= 0.889m3

Density of C. A=1500-1800 kg/m³; use 1600 kg/m³

Hence, Weight of C.A =1600*0.889=1422.4kg

Hence, Weight of water= 190kg=190litres

Volumes

12 cubes =12(0.1³) =0.012 m³

Material Requirements

Cement=432*0.012m³=5.184kg

F. A=649.4*0.012m³= 7.79kg

C.A=1422.4kg *0.012m³=17.07kg

Water =190*0.012m³=2.56kg

Final Results Adjusted To 15%

Second Trial Mix

Material Requirements

Cement=5.18kg * 1.15 = 5.962kg

F. A=7.79kg * 1.15 = 8.959 kg

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C.A=17.07kg*1.15= 19.63 kg ( ratio of C.A is 1:2- that is 10mm:20mm)

Water =2.56* 1.15 = 2.94 L

3RD TRIAL MIX

430+50= 480kg

Fine aggregates =549.02kg

Coarse aggregates= 1283.32kg

Ratio 480:549.02:1283.32

=1:1.14:2.67
1 m3 of dry concrete = (1.54-1.57) m3 of wet concrete

Volume of cement = 1/4.81*1.57=0.326m3

Density of cement=1440 kg/m³

Hence, Weight of cement = 1440*0.326= 469.44kg

Volume of F. A=1.14/4.81*1.57=0.372m3

Density of F. A=1600-1800 kg/m³: use 1700kg/m³

Hence, Weight of F. A=0.372*1700=623.4 kg

Volume of C.A =2.67/4.81*1.57= 0.871m3

Density of C. A=1500-1800 kg/m³; use 1600 kg/m³

Hence, Weight of C.A =1600*0.871=1393.6kg

Hence, Weight of water= 190kg=190litres

Volumes

12 cubes =12(0.1³) =0.012 m³

224 | P a g e
 Material Requirements
Cement=469.44*0.012m³=5.633kg

F. A=623.4*0.012m³=7.48kg

C.A=1393.6kg *0.012m³=16.72kg

Water =190*0.012m³=2.56kg=2.56litre

Final Results Adjusted To 15%

Third Trial Mix

Material Requirements

Cement=5.63kg * 1.15 = 6.47kg

F. A=7.589kg * 1.15 = 8.727 kg

C.A=16.72kg*1.15= 19.23kg

Water =2.56* 1.15 = 2.94 L

APPENDICES
DESIGN OF CLASS 40 CONCRETE(C40)
Table 1. Concrete Mix Design Form (BRE method)

Concrete mix design form Job title .............................................

Reference
Stage Item or calculation Values

1 1.1 Characteristic strength Specified …30………N/mm2 at …28……Days

Proportion defective ……5……%
1.2 Standard deviation Fig 3 …8… N/mm2 or no data …N/A… N/mm2
1.3 Margin C1 (k=…1.64…) …1.64…×…8…= …13.12…N/mm2
or
Specified = …………N/mm2

225 | P a g e
 1.4 Target mean strength C2 …30… + …13… = …43…N/mm2
1.5 Cement strength class Specified 42.5/52.5
1.6 Aggregate type: coarse Crushed/uncrushed
Aggregate type: fine Crushed/uncrushed
1.7 Free-water/cement ratio Table 2, Fig 4 ………0.55…… Use the lower value 0.5
1.8 Maximum free-water/ Specified ………0.5………
cement ratio
2 2.1 Slump or Vebe time Specified Slump…30.60…mm or Vebe time…N/A…s
2.2 Maximum aggregate size Specified ………………mm
2 1
2.3 Free-water content Table 3 …… × 180 + × 210…190 kg/m3
3 3

3 3.1 Cement content C3 …190… ÷…0.5……= …380…kg/m3

3.2 Maximum cement content Specified ……………N/A……kg/m3
3.3 Minimum cement content Specified …………275………kg/m3
use 3.1 if ≤ 3.2
use 3.3 if > 3.1 380kg/m3
3.4 Modified free-water/cement ratio ………190/275………………0.65
4 4.1 Relative density of ………………2.7………known/assumed
aggregate (SSD)
4.2 Concrete density Fig 5 …2430…kg/m3
4.3 Total aggregate content C4 2430… – …380… – 190=…1835…Kg/m3
5 5.1 Grading of fine aggregate Percentage passing 600 µm sieve ………70………………%
5.2 Proportion of fine aggregate Fig 6 …………………30…………………%
5.3 Fine aggregate content …1835…… ×……0.3……… = kg/m3
C5
5.4 Coarse aggregate content …1835…… –………550……… = kg/m3

Cement Water Fine Aggregate Coarse aggregate (kg)

Quantities (kg) (kg or litres) (kg) 10 mm 20 mm 40 mm

per m3 (to nearest 5 kg) …480… …190… …550… …430…… …855…… ………

226 | P a g e
per trial mix of ..0.012…m3 …5.633… …2.56… …7.48… …5.57…… …11.15……
………

Adjustment to 20% 6.756 3.0 9.11 6.69 13.37

6.0 unpriced BQ & specifications for substructure work (up to and including

ground floor slab) (BTCE only).

SUBSTRUCTURE WORKS

227 | P a g e
228 | P a g e
6.1 Taking OFF

229 | P a g e
 SUBSTRUCTURE 18 3.50 Exca. Column pits commecg fr. Stripped
WORKS 3.50 330.7 level and n.exc 1.5 m deep
1.50 5

‘All provisional’ &

Backfill
Items to measure

-Clear site /veg soil exctn Column dept

-Excavations (column) 1.600
-Excavation fdtn (wall) Add 600+50 .650
-P&S and dewatering 2.250
-Concrete work (blinding) Less veg. soil
-Concrete and Cover(150) .150
reinforcement 2.100
Less
-Subs-column-concrete
1st stage 1.500
formwork/reinforcement 0.600
-Subs -walls
-Hardcore/hd core blinding
-Anti-termite insecticide 18 3.50
-D.P.M 3.50 Exctn column pits exc. 1.5m and n.exc
-Oversite concrete 0.60 132.3 3.0m
slab/fabric reinforcement
-Edge formwork &
-D.P.C
Backfill

L W
23.20 17.0
Add Subs-Wall-Fdtn-Exctn
2/1.6 3.2 3.2
26.40 20.2 Mean Girth L – 23.20
W – 17.00
2/ 40.20
Clear site of all bushes, 80.40
26.40 533. overgrowth, shrubs and uproot Less corners
20.20 28 roots & burn all the arising 4/ 1⁄2 /2/200 0.80
79.60

Ht. 1.20
.30
1.500
Less veg soil .150
Exaction to rem veg. soil av. 1.350
26.40 533. 150mm th. and S&L 50m fr.
20.20 28 Site

79.60 Exc fdtn commcg-fr- stripped level and

0.80 n.exc 1.5m deep
1.35 85.97

&
Columns
6 × 3 = 18 backfill

230 | P a g e
 Adjustment for over Add 3.440
measurement exctn hooks 2/6/16 0.192
0.8 3.632
18
4 3.50 Ddt 2 2484. 16mm dia – hi-yield st- to K.B.S/BS - - - -
0.80 Fdtn exctn as. Bef 19 3.632 29 including tying wire, spacing blocks and
1.35 15.12 described bends - - - × kg/m - - - kg

Item Provide for rem. Of gen . surf Column ht.

water fr. Exctn Oversite 1.600
Slab (150) .150
1.750
(500×300)
Item Provide for nec. P&S to sides
of column & foundation exctns
18 0.50 V.R.C – (M40) 20mm agg. In subs –
0.30 column
18 3.50 1.75 4.725
3.50 E.O column pit – exctn in Backfill adjustment
0.65 143.3 excavating in rock 1.75
2 Less – 0.15+0.15 .30
1.45
18 3.50 Plain conc (M40) 20mm agg
3.50 220.5 in 50mm th. blinding 18 0.50 Ddt
0.30 Backfill
& 1.45 3.915
Backfill - - - -× 0.05 - -m3 &
& Add
Load & rem . fr site - - - Load & rem fr. Site
0.05m3

Formwork
18 3.50 V.R.C (M40) – 20mm agg in Ht - 1.75
3.50 column bases L - 500 2/0.50 1.00
0.60 132.3 W- 300 2/0.30 0.60
& 1.600
18 1.60
Ddt 1.75 50.4 SSW formwork to vert sides of columns
Backfill
&
Add
Load & rem fr. Site
Subs – Column Reinf.
Ht - 1.75
Column base point Add bars in base 0.60
Y 16 – both ways@ 200 c/c 2.35
L 3.50 Add
Allow for cover Hooks 6/12 0.072
2/30 0.06 2.422
3.44
Number 18
3.44⁄ + 1 =18.2 6 2.422 261.5 12mm dia – hi yield st to K.B.S/BS/ - - - -/
0.2
= 19 7 Incldg tying wire, hooks, spacing blocks
and bends - - - × kg/m - - -kg

231 | P a g e
 Stirrups – 8mm R/bars L W
L W 26.40 20.20
500 300
Less
Cover 60 60 26.40 533.2 150mm th. hardcore in crushed concrete
440 240 20.20 8 blocks compacted in 3 layers each 50 mm
240 th.
2/680
1360
Hooks
2/6/8 96 26.40 50 mm th. hdcore blinding in generals
1456 20.20 533.2 rejects
8
Ht 2.35
No – 2.35⁄0.2 +1
= 11.75 + 1 = 12.75
=13 26.40 Apply anti-termite insecticide Aid – 48 - - -
18 20.20 533.2 - -strictly to manufacturers specifications
340.
13 1.456 7 8 mm dia mild st to K.B.S/BS 8 and instructions
- - - - including tying wire,
bends and spacer blocks - - - - &
kg/m - - -kg 500 g polythene d.p.m

&

79.60 Plain conc. (M40) 20mm agg

0.800 19.1 in fdtns Plain concrete (M40) in 150 mm thick
0.300 & oversite slab

Ddt
Backfill
&
Add
Load and rem fr. Site Fabric reinforcement
L W
26.40 20.20
Subst – wall
M.G 79.60 Less cover
Ht 1.35 2/30 0.060 0.060
26.34 20.14

79.60 200mm th. S.C blockwall

107.
1.350 46 bedded in C.S mortar (1:3)

26.34 Fabric reinf A142 / - - -/ - - - -/ including

Ht - 1.20 20.14 530.4 bends, tying wire and dist blocks
Less hardcore - .150 9
1.05

79.60 Ddt
0.20 16.7 Backfill
1.05 2
&
Add
Load and rem fr. site

232 | P a g e
 Edge formwork
S.M.M F 21
26.40
20.20
2/46.60
93.20

93.20 93.2 S.S.W formwork to edges of

beds, exc. 75 and n.exc 150
mm wide

DPC
M.G - 79.60

79.60 79.6 200 mm wide hessian based

bit -felt class A1 D.P.C
bedded in c:s mortar (1:3)

End of substructure work

6.2 Abstracting

Substructure Works
Residential Apartment TUM – Drawing Ref No. RB/05/07/C

233 | P a g e
EARTHWORKS/EXCAVATION Clear site of all bushes, Provide for removal of
AND FILLING overgrowth, shrubs and general surface water from
uproot roots and burn all the excavation
arisings
Excavate column commencing
from stripped level and not 533.28 (pg.)
exceeding 1.5m deep
330.75 (pg.)
Provide for necessary
planking
Excavate to remove And strutting to sides of
vegetation soil average excavation
Excavation of column pits 150mm thick and spread and
exceeding 1.5m and not exceeding level 50m from site
3.0m deep
132.3 (pg.) 533.28 (pg.)
200mm wide hessian-
based bit felt class A1
dampproof coarse in
Excavation of foundation cement sand mortar
commencing from stripped level 79.60 (pg.)
and not exceeding 1.5m deep
85.97 (pg.)

150mm thick hardcore in Application of anti-termite

crushed concrete blocks insecticide Aide-48 strictly
compacted in 3 layers 50mm to manufacturers
Extra over column pit in thick specifications and
excavation rock instructions
143.32 (pg.) 533.28 (pg.) 533.28 (pg.)

Backfilling
99.36 (pg.)
26.50 (pg.) 39.74 (pg.) 150mm thick hardcore 500g polythene dampproof
43.20 (pg.) 194.4 (pg.) blinding in gravel rejects membrane
3.312 (pg.) 533.28 (pg.)
6.07 (pg.) 533.28 (pg.)
37.80 (pg.)

234 | P a g e
 In Situ Concrete Works

Vibrated reinforced concrete

(M40) 20mm aggregate in
column bases
132.3 (pg.)
Plain concrete (M40) 20mm
aggregates in 50mm thick
blinding
220.5 (pg.)
Vibrated reinforced concrete
(M40) 20mm aggregates in
substructure columns
4.725 (pg.)
Plain concrete (M40) in 150mm
thick oversite slab

Plain concrete (M40) 20mm 533.28 (pg.)

aggregates in foundations
19.1 (pg.)

Reinforcement

16mm diameter high yield

steel to BS/K.B. S including
tying wire, hooks, spacing
blocks and bends - - - kg/m - -
kg
3922.69 (pg.)
Fabric reinforcement A142
including bends, tying wire and
spacer blocks

530.49 (pg.)
8mm diameter high yield steel
to K.B.S/BS including tying
wire, bends and spacer blocks
- - - kg/m - - - kg
134.576 (pg.)

235 | P a g e
 12mm diameter high yield
steel to K.B.S/BS including
tying wire, bends and spacer
blocks - - - kg/m - - - kg
232.274 (pg.)

Formworks
Soft sawn wood
formwork edges of beds
exceeding 75mm and not
Exceeding 150mm wide
93.20 (pg.)

Soft sawn wood formwork to

vertical sides of columns
50.4 (pg.)

Masonry Walling

200mm thick solid concrete

block wall bedded in cement
sand mortar

107.46 (pg.)

236 | P a g e
6.3 Billing

ITEM RESIDENTIAL APARTMENT TUM Quantity Unit Rate sh Cts

DRAWNG REF N0: RB/05/07/C

SUBSTRUCTURE WORKS

“ALL PROVISIONAL”

Excavation and filling (Earthworks)

A Excavation of column pits commencing from 331 m3

stripped level and not exceeding 1.5m deep

Excavation of column pits exceeding 1.5m

B 133 m3
and not exceeding 3.0m deep

C Foundation excavation commencing from

stripped level and not exceeding 1.5m deep 86 m3

Site clearance of all bushes, overgrowth,

D shrubs and uproot roots and burn all the 534 m2
arisings

E Excavation to remove vegetation soil average

150mm thick and spread and level 50m from 534 m2
site

F Provide for removal of general surface water Item - --

from excavation

237 | P a g e
 Quantity Unit Rate sh Cts

A Provide for necessary planking and strutting to Item -

sides of excavation

B 200mm wide hessian-based bit felt class A1 80 m

dampproof coarse bedded in cement sand
mortar

C Extra over column pit in excavation work 144 m3

D Backfilling m3

E Load and remove from site m3

F 150mm thick hardcore in crushed concrete 534 m2

blocks compacted in 3 layers 50mm thick

238 | P a g e
 Quantity Unit Rate sh Cts

A 150mm thick hardcore blinding in gravel 534 m2

rejects

B Application of anti-termite insecticide Aide- 534 M

48 strictly to manufacturers specification and
instructions

C 500g polythene dampproof membrane 534 M

Concrete works

A Vibrated reinforced concrete (M40) 20mm 133 m3

aggregates in column bases

B Vibrated reinforced concrete (M40) 20mm 5 m3

aggregates in substructure columns

C Plain concrete (M40) 20mm aggregate in 20 m3

foundation

239 | P a g e
 Quantity Unit Rate sh Cts

A Plain concrete (M40) 20mm aggregate in 221 m2

50mm thick blinding

B Plain concrete (M40) 20mm aggregate in 534 m2

150mm thick oversite slab

Reinforcement

A 16mm diameter high yield steel to BS/K.B. S 3923 Kg

including tying wire, hooks, bends and
spacer blocks

B 12mm diameter high yield steel to BS/K.B. S 233 Kg

including tying wire, hooks, bends and
spacer blocks

C 8mm diameter high yield steel to BS/K.B. S 135 Kg

including tying wire, bends and spacer blocks

D Fabric reinforcement A142 including bends, 531 m2

tying wire and spacer blocks

240 | P a g e
 Formwork Quantity Unit Rate sh Cts

A Soft sawn wood formwork to vertical sides of 51 m3

columns

B Soft sawn wood formwork to edges of beds 94 M

exceeding 75mm and not exceeding 150mm
wide

Masonry Walling

A 200mm thick solid concrete block wall bedded 108 m3

in cement sand mortar (1:3)

241 | P a g e
7.0 Work plan and Budget
7.1 Workplan

month July August September October November December

Introduction(project
briefing)
Architectural
drawing and
elevations
Analysis of the
project(calculations)
Structural drawings
preparation(output)
Soil lab test
Mix design tests
(M30)
Compilation of the
report

7.2 Budget

Printing costs:

➢ Architectural and structural drawings

5 A1 blue print papers @ sh - - -
➢ Project printing (243 pages)
243 pages @ sh - - - per page
➢ Binding cost
Hard cover binder @ sh

Transportation costs:

General transportation costs experienced

Other related costs:

Total cost experienced

242 | P a g e