design of proposed auditorium-project report
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DESIG� OF PROPOSED AUDITORIUM
ACK�OWLEDGEME�T
We are delighted to express our hearty thanks to our honorable Principal,
Dr.S.Joseph Sekhar, Ph.D for providing facilities to undertake this mini
project work.
We are grateful to Mrs.S.Judes Sujatha, B.E, M.Tech., Head of the
Department, for extending all possible help in the execution of this work.
We are extremely indebted to our internal guide,Mr.A.B.Danie Roy,
M.E., who has suggested this topic, provided all the expertise, facilities and
help in shaping this project into a successful one.
We acknowledge the help rendered by Mr.A.Jinu Antony, Engineer
associated with auditorium design
We also express our thanks to all the staff members of the Civil
Engineering department, for their support.
ABSTRACT
This project deals with the analysis and design of the Auditorium of
St.Xavier’s Catholic College of Engineering with special emphasis on Slabs,
Beams, Columns, Footing and Staircase.
Analysis is carried out using Substitute Frame Analysis and
preliminary analysis of Beams is carried out using Moment Distribution
method.
Concrete mix used for the RCC members is M20 and steel used is
high yield strength deformed bars of grade Fe415. Limit State Method is
adopted for the design of all structural members in the building.
Safe bearing capacity of soil is taken as 200kN/m2. Footing is
designed as Isolated type. Plan and detailing of reinforcement are enclosed
in this report.
TABLE OF CO�TE�TS
CHAPTER �O. TITLE PAGE �O.
ABSTRACT
LIST OF SYMBOLS
LIST OF FIGURES
1. I�TRODUCTIO� 1.1 GENERAL 1
1.2 OBJECTIVES 2
1.3 DESIGN OF RC STRUCTURES 2
1.3.1 LIMIT STATE DESIGN 2
1.4 SLABS 3
1.4.1 CLASSIFICATION OF SLABS 3
1.5 BEAMS 4
1.5.1 DESIGN OF BEAMS 4
1.6 COLUMN 5
1.6.1 SHORT COLUMN 5
1.6.2 SLENDER COLUMN 5
1.6.3 CLASSIFICATION OF COLUM 5
1.7 FOOTING 6
1.7.1 TYPES OF COLUMN FOOTING 6
1.8 STAIRCASE 7
1.8.1 CLASSIFICATION OF STAIRS 7
2. PLA��I�G OF PROPOSED AUDITORIUM
2.1 LAYOUT OF SITE
2.2 PLANS
2.3 SECTION
2.4 ELEVATION
3 METHODOLOGY
3.1 LIMIT STATE DESIGN
3.2 PARTIAL SAFETY FACTOR
4 A�ALYSIS
INTRODUCTION
METHOD OF SUBSTITTE FRAME ANALYSIS
ANALYSIS OF FRAMES
5 DESIG�
SLABS
5.1.1. DESIGN OF SLABS
BEAMS
5.2.1 TYPES OF BEAMS
5.2.2 DESIGN OF L-BEAMS
5.2.3 DESIGN OF T-BEAMS
5.3 STAIRCASE
5.3.1 TYPES OF STAIRS
5.3.2 DESIGN OF DOGLEGGED STAIRCASE
5.4 COLUMN
5.4.1 TYPES OF COLUMN
5.4.2 DESIGN OF COLUMN
5.4.2.1 DESIGN OF AXIALLY LOADED COLUMN
5.4.2.2DESIGN OF UNIAXIALLY LOADED COLUMN
5.4.2.3DESIGN OF BI AXIALLY LOADED COLUMN
5.5 FOOTING
5.5.1 TYPES OF FOOTING
5.5.2 DESIGN OF FOOTING
6 CO�CLUSIO�
REFERE�CE
LIST OF FIGURES
Serial �o. Title Page �o.
figure 1 analysis using substitute frame method - frame1 ................... 17 figure 2 analysis using substitute frame method-frame2 ..................... 31 figure 3 analysis using substitute frame method-frame3 ..................... 45 figure 4 analysis using substitute frame method-frame4 ..................... 59 figure 5 reinforcement details of two way slab-section ..................... 105 figure 6 reinforcement details of two way slab- plan ......................... 105 figure 7 reinforcement details of l-beams- longitudinal section ........ 111 figure 8 reinforcement details of l-beams- cross section.................... 111 figure 9 reinforcement details of t-beam-longitudinal design ............ 116 figure 10 cross section of t-beam ......................................................... 116 figure 11 reinforcement details of doglegged staircase ........................ 122 figure 12 reinforcement details of axially loaded column ................... 126 figure 13 reinforcement details of uniaxially loaded column .............. 129 figure 14 reinforcement details of biaxially loaded columns ............... 132 figure 15 reinforcement details of axially loaded column ................... 137
LIST OF SYMBOLS
Mx - Moment in shorter direction
My - Moment in shorter direction
d - Effective depth
D - Overall depth
Ast - Area of Steel
P - Load
Wu (or) Pu - Design load
Mu - Design moment
Asc - Area of concrete
fy - Characteristic strength of steel
fck - Characteristic strength of concrete
B.M - Bending Moment
b - Breadth of beam
D - Overall depth
Vus - Strength of shear reinforcement
L - Clear span
Le - Effective span
N.A - Neutral Axis
MF - Modification factor
Q - Angle of repose of soil
M - Modular of rupture
τc - Permissible shear stress in concrete
τv - Nominal shear stress
CHAPTER 1
1. I�TRODUCTIO�
1.1 GE�ERAL
Auditorium, Conference hall, Library and Indoor Games are
necessary for an Engineering college. In St.Xavier’s Catholic College of
Engineering, Library, Conference hall are located at different locations and
also there is no special building for Auditorium. This project reports on the
analysis and design of Auditorium, Library and Indoor Games hall in one
separate block.
All structural components for the building such as beams,
columns, slabs, staircase etc are analysed and designed. Isolated footing is
adopted for all columns. Safe bearing capacity is taken as 200kN/m2. The
structure is designed by using limit state method, adopting M20 concrete and
Fe415 HYSD bars.
Site plan, plan showing various floors, section of plan,
elevation of plan and detailing of reinforcements for Beam, Column, Slab,
Staircase and Footing are also enclosed.
OBJECTIVES
2. To analyse the frames in the building.
3. To design the structural components of the five storey building.
4. To prepare the detailed drawing for the design carried out.
1.3 DESIG� OF RC STRUCTURES
Reinforced cement concrete members can be designed by the
following methods:
1. Working stress method
2. Limit state method
1.3.1 Limit state design
• Limit state method of design is based on elastic theory.
• Partial safety factors are used in this method to determine the design
loads and design strength of materials from their characteristics
values.
• The design aids to IS:456, published by the bureau of Indian
standards. The design of limit state method is very simple and hence
widely used in practice.
• This method gives economical results when compared with the
conventional working stress method.
1.4 SLABS
• Slabs are primary members of a structure, which support the imposed
load directly on them and transfer the same safety to the supporting
elements such as beams, walls, columns etc.
• A slab is a thin flexural member used in floor and roof of a structure
to support the imposed loads.
1.4.1 Classification of slabs
1.4.1.1Solid slab
1.4.1.2Hollow slab
1.4.1.3Ribbed slab
1.5 BEAM
• A beam has to be generally designed for the actions such as bending
moments, shear forces and twisting moments developed by the lateral
loads.
• The size of the beam is designed considering the maximum moment
in it and generally kept uniform throughout its length.
• IS:456:2000 recommends that the minimum grade of concrete should
not be less than M20 in RC works.
1.5.1 Design of beams
• When there is a Reinforced concrete slab over a concrete beam, then
the beam and the slab can be constructed in such a way that they act
together.
• The combined beam and slab are called as flanged beams. It may be
‘T’ or ‘L’ beams. Here both T-beams and L-beams are designed.
1.6 COLUM�S
• Vertical members in compression are called as columns and struts.
• The term column is reserved for member which transfer load to the
ground. Classification of column, depending upon slenderness ratio is
1.6.1Short columns
1.6.2Slender columns
1.6.1 Short column
IS:456:2000 classifies rectangular column as short when the
ratio of effective length(Le) to the least dimension is less than 12.
This ratio is called slenderness ratio of the column.
1.6.2Slender columns
The ratio of Le to the least dimension is less than 12 are called
as slender column.
Classification of column
1. Axially loaded column
2. Eccentrically loaded column
3. Column subjected to axial load and moment
1.7 Footing
• Foundation is the most important component of a structure.
• It should be well planned and carefully designed to ensure the
safety and stability of the structure.
• Foundation provided for RCC columns are called as column base.
1.7.1Types of column base
1. Isolated footing
2. Combined footing
3. Strap footing
4. Solid raft foundation
5. Annular raft foundation
1.8 Staircase
A staircase is a flight of steps leading from one floor to another.
It is provided to afford the means of ascent and descent between various
floors of the building. It should be suitably located in a building. In a
domestic building the stair should be centrally located to provide easy
access to all rooms. In public buildings stairs should be located near the
entrance. In big building there can be more than one stairs. Fire
protection to stairs is important too. Stairs are constructed using timber,
bricks, stone, steel or reinforced cement concrete.
1.8.1Classification of stairs
1. Single flight stairs
2. Quarter turn stairs
3. Dog legged stairs
4. Open well type stairs
5. Biffurcated stairs
6. Circular stairs
7. Spiral stairs
CHAPTER 3
3.METHODOLOGY
Various methods are available for the design of a
structure. Limit state method is adopted in this project.
3.1 Limit state design
The acceptable limit for safety and serviceability requirement
before failure occur is called limit state. The aim of design is to achieve
acceptable probabilities that the structure will not become unfit for use.
All relevant limit state shall be considered in the design to ensure
adequate degree of safety and serviceability.
3.2Partial safety factor
The value of load which has a 95% probability of a structure of
structural member for the limit state of collapse the following values
of partial safety factor is applied for limit state of collapse.
Ym = 1.5 for concrete
Ym = 1.15 for steel
CHAPTER 4
4 A�ALYSIS
4.1 Introduction
A multistoried frame is a complicated statically
indeterminate structure. The analysis by moment distribution method is
very lengthy and difficult. Hence substitute frame analysis is adopted
for better and easier calculation.
4.2 Method of substitute frames
In this method only a part of the frame is
considered for the analysis. The part considered is called as substitute
frame. The moments for each floor are separately computed. It is
assumed that the moments transferred from one floor to another are
small. Each floor is taken as connected to columns above and below
with their far ends fixed. The frame taken this way is analysed for the
moments and shears in the beams and columns.
The column will carry the maximum bending moment when
any one series of alternate spans should be loaded. The moment
distribution for the substitute frame analysis is performed only for two
cycles and hence, the method is sometimes referred to as, the two cycle
method.
When it is required to find the maximum negative
moment at a joint, the spans meeting at the joint are loaded with dead
and live load. The other spans are loaded with dead load alone.
4.3A�ALYSIS OF FRAMES
A B C D E F G
FRAME 1
Figure 1-Analysis using substitute frame method - frame1
4.3.1DISTRIBUTIO� FACTOR
4.3.2 LOAD CALCLATIO�
Joint Member Relative stiffness Total stiffness Stiffness for each member
A
AB
A1
A2
I/3.2
I/4
I/4
0.8125I
0.38
0.31
0.31
B
BA
BC
B1
B2
I/3.2
I/3.2
I/4
I/4
1.125I
0.28
0.28
0.22
0.22
C
CB
CD
C1
C2
I/3.2
I/6.4
I/4
I/4
0.9688I
0.32
0.16
0.26
0.26
D
DC
DE
D1
D2
I/6.4
I/6.4
I/4
I/4
0.8125I
0.32
0.32
0.31
0.31
E
ED
EF
E1
E2
I/6.4
I/6.4
I/4
I/4
0.8125I
0.32
0.32
0.31
0.31
F
FE
FG
F1
F2
I/6.4
I/6.4
I/4
I/4
0.8125 I
0.32
0.32
0.31
0.31
G
GF
G1
G2
I/6.4
I/4
I/4
0.656I
0.24
0.38
0.38
35.9635.9635.9635.9626.97526.975Total load
kN/m2
17.2317.2317.2317.2314.97514.975Total DL
kN/m2
4.54.54.54.53.3753.375Dead load
due to rib
kN/m2
18.7318.7318.7318.731212Total LL
kN /m
10.310.310.310.36.66.6Total DL
kN/ m
6.46.46.46.43.23.2Length m
7.997.997.997.992.562.56Area m2
151515151515LL kN/m2
8.258.258.258.258.258.25DL kN / m2
FGEFDECDBCABBeam name
4.3.3 MOME�T CALCULATIO� AT JOI�TS
JOINT-A
-15.16Net BM due
to A
-24.45
9.29
Distribution
-1.4336Distribution
+ Carryover
+23.02-23.02FEM due to
LL
-12.78FEM due to
DL
0.320.320.160.320.280.280.38DF
DEDCCDCBBCBAABMember
DCBAjoint
JOINT-B
-20.12124.563Net BM due
to A
-17.294
-2.827
27.39
-2.827
Distribution
5.7264.37Distribution
+ Carryover
+23.02-23.02+23.02-23.02FEM due to
LL
-58.81FEM due to
DL
0.320.320.160.320.280.280.38DF
DEDCCDCBBCBAABMember
DCBAjoint
JOINT-C
-133.8136.48Net BM
due to A
-132.97
-0.8256
138.13
-1.6512
Distributio
n
-10.2315.39Distributio
n +
Carryover
122.74-122.74+122.74-122.74FEM due
to LL
-58.8112.78FEM due
to DL
0.320.320.320.320.160.320.280.28DF
EFEDDEDCCDCBBCBAMember
EDCBjoint
JOINT-D
-132.51132Net BM
due to A
-132.97
0.46
131.54
0.46
Distributi
on
-10.238.8Distributi
on +
Carryove
r
122.74-122.74+122.74-122.74FEM due
to LL
-58.8112.78FEM due
to DL
0.320.320.320.320.160.320.280.28DF
EFEDDEDCCDCBBCBAMember
EDCBjoint
JOINT-E
FE G
-132.97132.97Net BM
due to A
-132.97
0
132.97
0
Distributi
on
-10.2310.23Distributi
on +
Carryove
r
122.74-122.74122.74-122.74FEM due
to LL
-58.81+58.81FEM due
to DL
0.240.320.320.320.320.320.32DF
GFFGFEEFEDDEDCMember
Djoint
JOINT-F
FE G
-136.03134.41Net BM
due to A
-132.97
1.44
132.9
1.44
Distributi
on
-14.7310.23Distributi
on +
Carryove
r
122.74-122.74122.74-122.74FEM due
to LL
58.81FEM due
to DL
0.240.320.320.320.320.320.32DF
GFFGFEEFEDDEDCMember
Djoint
JOINT-G
FE G
101.06Net BM
due to A
132.97
-31.91
Distributi
on
10.23Distributi
on +
Carryove
r
122.74-122.74FEM due
to LL
58.81FEM due
to DL
0.240.320.320.320.320.320.32DF
GFFGFEEFEDDEDCMember
Djoint
MID SPAN MOMENT AB
+18.375-15.16Net moment
-6.150.545distribtion
17.594.375-1.435carry over
moment
35.18-2.878.75distribution
moment
122.74-122.7423.02-23.02FEM due to
TL
12.78-12.78FEM due to
DL
0.320.320.160.320.280.280.38DF
DEDCCDCBBCBAABMember
DCBAjoint
Free BM at the centre of the span AB
= Wl2/8
=(26.975 * 3.22) / 8
=34.53 kNm
Net BM at centre of span AB
=34.53-[ (15.16+18.375) / 2]
=17.76 kNm
MID SPAN MOMENT OF BC
30.224-30.696Net moment
-2.8160.924distribtion
10.23-1.43-5.732.43carry over
moment
20.4611.45-2.874.86distribution
moment
-122.7423.02-23.02FEM due to
TL
58.81-58.8112.78-12.78FEM due to
DL
0.320.320.160.320.280.280.38DF
DEDCCDCBBCBAABMember
DCBAjoint
Free BM at the centre of the span BC
= Wl2/8
=(26.975 * 3.22) / 8
=34.53 kNm
Net BM at centre of span BC
=34.53-[ (30.696+30.224) / 2]
=4.07 kNm
MID SPAN MOMENT CD
104.99-95.923Net
moment
-6.0881.86distributio
n
10.238.795-10.23-1.43carry
over
moment
20.46-20.4617.59-2.86distributio
n
moment
-122.74122.74-122.7423.02FEM due
to TL
58.81-58.8112.78-12.78FEM due
to DL
0.320.320.320.320.160.320.280.28DF
EFEDDE DCCD CBBC BAmember
ED C BJoint
Free BM at the centre of the span CD
= Wl2/8
=(35.96 * 6.42) / 8
=184.12 kNm
Net BM at centre of span CD
=184.12-[ (95.92+104.99) / 2]
=83.66 kNm
MID SPAN MOMENT DE
105.97-105.97Net
moment
-6.546.54distribtion
10.2310.23-10.23-10.23carry
over
moment
20.46-20.4620.46-20.46-20.46distributio
n
moment
-122.74122.74-122.74122.74FEM due
to TL
58.81-58.8158.81-58.81FEM due
to DL
0.320.320.320.320.160.320.280.28DF
EFEDDE DCCD CBBC BAmember
ED C BJoint
Free BM at the centre of the span DE
= Wl2/8
=(35.96 * 6.42) / 8
=184.12 kNm
Net BM at centre of span DE
=184.12-[ (105.97+105.97) / 2]
=78.14 kNm
MID SPAN MOMENT EF
G
111.51-105.96Net
moment
-1.0016.55Distribtion
-7.110.23-10.23-10.23carry over
moment
-14.11-20.4620.46-20.46distributio
n moment
122.74-122.74122.74FEM due
to TL
58.81-58.8158.81-58.81FEM due
to DL
0.240.320.320.320.320.320.32DF
GFFGFEEFEDDEDCMember
EDCjoint
Free BM at the centre of the span EF
= Wl2/8
=(35.96 * 6.42) / 8
=184.12 kNm
Net BM at centre of span EF
=184.12-[ (105.97+111.51) / 2]
=75.4 kNm
MID SPAN MOMENT FG
G
101 05-109.03Net
moment
-2.4557.98distribtion
10.23-14.73-10.23carry over
moment
-29.4620.46-20.46distributio
n moment
122.74-122.74122.74FEM due
to TL
58.81-58.81FEM due
to DL
0.240.320.320.320.320.320.32DF
GFFGFEEFEDDEDCMember
FEDjoint
Free BM at the centre of the span FG
= Wl2/8
=(35.96 * 6.42) / 8
=184.12 kNm
Net BM at centre of span FG
=184.12-[ (109.03+101.05) / 2]
=79.08 kNm
NEGETIVE MOMENT AT CENTRE OF DE
75.59-75.12Net
moment
6.55-6.08distribtion
-10.23-10.2310.238.795carry
over
moment
-20.4620.4620.46-20.46-20.4617.59distributio
n
moment
122.74-122.74122.74-122.74FEM due
to TL
-58.8158.81-58.8112.78FEM due
to DL
0.320.320.320.320.320.320.160.32DF
FGFEEFEDDE DCCD CBmember
FE D CJoint
Free BM at the centre of the span DE
= Wl2/8
=(17.23 * 6.42) / 8
=88.21 kNm
Net BM at centre of span DE
=88.21-[ (75.12+75.59) / 2]
=12.86 kNm
4.3.4 BENDING MOMENT IN COLUMN LOADING
joint A B C D E F G
Column DF
Above floor
Below floor
0.31
0.31
0.22
0.22
0.26
0.26
0.3
0.3
0.3
0.3
0.3
0.3
0.38
0.38
Member
DF
AB
0.38
BA
0.28
BC
0.25
CB
0.32
CD
0.16
DC
0.19
DE
0.19
ED
0.19
EF
0.19
FE
0.19
FG
0.19
GF
0.24
FEM due to
DL
-12.8 12. 8 -58.8 58.8 -58.8 58.8
FEM due TL -23.02 23.02 -122.7 122.7 -122.7 122.7
Distribution+
carryover
-1.435 4.37 17.59 -1.28 -6.07 8.79 6.07 -6.07 -6.07 6.07 -7.06 6.07
Total -24.45 27.39 4.81 11.5 -128.8 131.53 -52.7 52.74 -128.8 128.81 -65.9 64.88
Distribution to column Above floor
7.58
7.084
30.5
-23.64
22.82
-18.882
24.65
Distribution to column below floor
7.58
7.084
30.5
-23.64
22.82
-18.882
24.65
Joint A B C D E F G
Column DF
Above floor
Below floor
0.31
0.31
0.22
0.22
0.26
0.26
0.3
0.3
0.3
0.3
0.3
0.3
0.38
0.38
Member
DF
AB
0.38
BA
0.28
BC
0.25
CB
0.32
CD
0.16
DC
0.19
DE
0.19
ED
0.19
EF
0.19
FE
0.19
FG
0.19
GF
0.24
FEM due to
DL
-12.8 12. 8 -58.8 58.8 -58.8 58.81
FEM due TL -
23.02
23.02 -122.7 122.7 -122.7 122.7
Distribution+
carryover
1.43 2.43 5.73 1.28 -6.07 2.86 -6.07 6.07 6.07 -6.07 -14.73 6.07
Total -11.4 15.21 -17.3 24.3 -64.9 61.7 -128.8 128.8 -52.7 52.7 -137.5 128.8
Distribution to column Above floor
3.52
0.46
10.55
20.14
22.82
25.42
-48.9
Distribution to column below floor
3.52
0.46
10.55
20.14
22.82
25.42
-48.9
H
FRAME 2
I J K L M N
Figure 2-Analysis using substitute frame method-frame2
4.3.5 DISTRIBUTIO� FACTOR
Joint Member Relative stiffness Total stiffness Stiffness for each member
H
HI
H1
H2
I/3.72
I/4
I/4
0.769I
0.35
0.33
0.33
I
IH
IJ
I1
I2
I/3.72
I/3.55
I/4
I/4
1.05I
0.26
0.27
0.24
0.24
J
JI
JK
J1
J2
I/3.55
I/6.33
I/4
I/4
0.94I
0.3
0.17
0.27
0.27
K
KJ
KL
K1
K2
I/6.33
I/6.43
I/4
I/4
0.81I
0.2
0.19
0.31
0.31
L
LK
LM
L1
L2
I/6.43
I/3.55
I/4
I/4
0.937I
0.17
0.3
0.27
0.27
M
ML
MN
M1
M2
I/3.55
I/3.72
I/4
I/4
1.05 I
0.27
0.26
0.24
0.24
N
NM
N1
N2
I/3.72
I/4
I/4
0.769I
0.35
0.33
0.33
4.3. 6 LOAD CALCLATIO�
45.7744.24579.2478.1144.24545.77Total load
kN/m2
18.4217.87531.0230.6217.87518.42Total DL
kN/m2
3.3753.3754.54.53.3753.375Dead load
due to rib
kN/m2
27.3526.3748.2247.4926.3727.35Total LL
kN /m
15.04514.526.5226.1214.515.045Total DL
kN/ m
3.723.556.436.333.553.72Length m
6.7846.2420.67220.046.246.784Area m2
151515151515LL kN/m2
8.258.258.258.258.258.25DL kN / m2
MNLMKLJKIJHIBeam name
4.3.7 MOME�T CALCULATIO� AT JOI�TS
JOINT-H
-37.18Net BM due
to H
-57.2
20.02
Distribution
-4.42Distribution
+ Carryover
+52.78-52.78FEM due to
LL
-18.77FEM due to
DL
0.190.20.170.30.270.260.35DF
KLKJJKJIIJIHHIMember
KJIHjoint
JOINT-I
-44.56455.8Net BM due
to I
-38.104
-6.46
62.02
-6.22
Distribution
8.3669.237Distribution
+ Carryover
46.47-46.47+52.78-52.78FEM due to
LL
-102.24FEM due to
DL
0.190.20.170.30.270.260.35DF
KLKJJKJIIJIHHIMember
KJIHjoint
JOINT-J
-235.87110.12Net BM
due to J
-276.2
40.33
38.94
71.18
Distributio
n
-15.39-7.53Distributio
n +
Carryover
260.81-260.8146.47-46.47FEM due
to LL
-102.24102.24FEM due
to DL
0.30.170.190.20.170.30.270.26DF
LMLKKL KJJKJIIJIHMember
LKJIjoint
JOINT-K
-292.10284.03Net BM
due to K
-294.62
2.516
281.38
2.648
Distributi
on
-21.6120.57Distributi
on +
Carryove
r
273.01-273.01260.81-260.81FEM due
to LL
-58.8118.77FEM due
to DL
0.30.170.190.20.170.30.270.26DF
LMLKKL KJJKJIIJIHMember
LKJIjoint
JOINT-L
ML N
-121.65248.56Net BM
due to L
-49.88
-71.77
298.23
-40.67
Distributi
on
-3.4116.22Distributi
on +
Carryove
r
46.47-46.47273.01-273.01FEM due
to LL
-21.24+102.2
4
FEM due
to DL
0.350.260.270.30.170.190.2DF
NMMNMLLMLKKLKJMember
Kjoint
JOINT-M
ML N
-60.545.834Net BM
due to M
-68.61
8.112
37.41
8.424
Distributi
on
-15.83-9.06Distributi
on +
Carryove
r
52.78-52.7846.47-46.47FEM due
to LL
106.88FEM due
to DL
0.350.260.270.30.170.190.2DF
NMMNMLLMLKKLKJMember
Kjoint
JOINT-N
ML N
40.04Net BM due
to N
57.2
-17.16
Distribution
4.42Distribution
+ Carryover
52.78-52.78FEM due to
LL
18.77FEM due to
DL
0.350.260.270.30.170.190.2DF
NMMNMLLMLKKLKJMember
Kjoint
MID SPAN MOMENT HI
41.18-37.23Net moment
-11.841.573distribution
36.39.235-4.495carry over
moment
72.6-8.9918.47distribution
moment
260.81-260.8152.78-52.78FEM due to
TL
18.77-18.77FEM due to
DL
0.190.20.170.30.270.260.35DF
KLKJJKJIIJIHHIMember
KJIHjoint
Free BM at the centre of the span HI
= Wl2/8
=(45.77 * 3.722) / 8
=79.2 kNm
Net BM at centre of span HI
=79.2-[ (37.23+41.18) / 2]
=39.995 kNm
MID SPAN MOMENT OF IJ
52.8-60.26Net moment
-2.87-7.09distribution
17.1-7.538.3617.89carry over
moment
34.1516.73-15.0635.78distribution
moment
-273.0146.47-46.47FEM due to
TL
102.24-102.24102.24-102.24FEM due to
DL
0.190.20.170.30.270.260.35DF
KLKJJKJIIJIHHIMember
KJIHjoint
Free BM at the centre of the span IJ
= Wl2/8
=(44.245 * 3.552) / 8
=69.699 kNm
Net BM at centre of span IJ
=69.7-[ (60.26+52.8) / 2]
=13.2 kNm
MID SPAN MOMENT JK
247.52-231.66Net
moment
-3.13.4distribution
-5.13520.6-15.4-4.59carry over
moment
-10.27-30.7941.15-9.18distribution
moment
-46.47260.81-260.8152.78FEM due
to TL
106.9-106.918.77-18.77FEM due
to DL
0.30.170.190.20.170.30.270.26DF
LMLKKL KJJKJIIJIHmember
LKJIJoint
Free BM at the centre of the span JK
= Wl2/8
=(78.11 * 6.332) / 8
=391.22 kNm
Net BM at centre of span JK
=391.22-[ (231.7+247.52) / 2]
=151.63 kNm
MID SPAN MOMENT KL
243.18-257.26Net
moment
-3.683.21distribution
4.5917.07-21.614.74carry over
moment
9.18-43.2234.159.48distribution
moment
-52.78273.01-273.0146.47FEM due
to TL
18.77-18.77102.24-102.24FEM due
to DL
0.30.170.190.20.170.30.270.26DF
LMLKKL KJJKJIIJIHmember
LKJIJoint
Free BM at the centre of the span KL
= Wl2/8
=(79.24 * 6.432) / 8
=409.52 kNm
Net BM at centre of span KL
=409.52-[ (257.26+243.18) / 2]
=159.3 kNm
MID SPAN MOMENT LM
N
34.05-62.591Net
moment
3.455.412Distribution
-3.717-9.06-3.41-14.63carry over
moment
-7.434-6.812-18.123-29.25distribution moment
46.47-46.47260.81FEM due
to TL
21.24-21.24106.88-106.88FEM due
to DL
0.350.260.270.30.170.190.2DF
NMMNMLLMLKKLKJMember
MLKjoint
Free BM at the centre of the span LM
= Wl2/8
=(44.25* 3.552) / 8
=69.71 kNm
Net BM at centre of span LM
=69.71-[ (62.59+34.05) / 2]
=21.39 kNm
MID SPAN MOMENT MN
N
37.15-40.87Net
moment
-1.5512.32distribution
4.42-9.25-38.15carry over
moment
-18.58.843-76.3distribution
moment
52.78-52.78273.01FEM due to
TL
18.77-18.77FEM due to
DL
0.350.260.270.30.170.190.2DF
NMMNMLLMLKKLKJMember
MLKjoint
Free BM at the centre of the span MN
= Wl2/8
=(45.77 * 3.722) / 8
=79.2 kNm
Net BM at centre of span MN
=79.2-[ (40.87+37.15) / 2]
=40.19kNm
NEGETIVE MOMENT AT CENTRE OF KL
84.41-147.4Net
moment
3.2-2.19distribtion
-3.4-15.4-9.0620.6carry
over
moment
-6.81-18.12-10.27-29.25-30.7941.15distributio
n
moment
41.47-41.47260.81-260.81FEM due
to TL
-21.24106.9-106.918.77FEM due
to DL
0.260.270.30.170.190.20.170.3DF
MNMLLMLKKLKJJKJImember
MLKJJoint
Free BM at the centre of the span KL
= Wl2/8
=(31.02 * 6.432) / 8
=160.31 kNm
Net BM at centre of span KL
=160.31-[ (147.4+84.41) / 2]
=44.61 kNm
4.3.8 BENDING MOMENT IN COLUMN LOADING
joint H I J K L M N
Column DF
Above floor
Below floor
0.33
0.33
0.24
0.24
0.27
0.27
0.31
0.31
0.27
0.27
0.24
0.24
0.33
0.33
Member
DF
HI
0.35
IH
0.26
IJ
0.27
JI
0.3
JK
0.17
KJ
0.19
KL
0.19
LK
0.17
LM
0.3
ML
0.27
MN
0.26
NM
0.35
FEM due to
DL
-18.8 18.8 -106.9 106.9 -21.24 21.24
FEM due TL -52.78 52.78 -260.8 260.8 -46.5 46.5
Distribution+
carryover
-4.421 9.24 36.31 -4.56 -14.62 20.57 -5.13 -14.6 -3.41 -9.06 -3.72 -3.28
Total -57.2 62.02 17.51 14.18 -275.4 281.4 -112.0 92.26 -49.9 37.41 -24.96 17.96
Distribution to column Above floor
18.88
-19.09
70.54
-52.50
-11.44
-2.988
-5.93
Distribution to column below floor
18.88
-19.09
70.54
-52.50
-11.44
-2.988
-5.93
joint H I J K L M N
Column DF
Above floor
Below floor
0.33
0.33
0.24
0.24
0.27
0.27
0.31
0.31
0.27
0.27
0.24
0.24
0.33
0.33
Member
DF
HI
0.35
IH
0.26
IJ
0.27
JI
0.3
JK
0.17
KJ
0.19
KL
0.19
LK
0.17
LM
0.3
ML
0.27
MN
0.26
NM
0.35
FEM due to
DL
-21.24 21.24 -
102.24
102.2
4
-18.77 18.77
FEM due TL -46.5 46.5 -273 273 -52.78 52.78
Distribution+
carryover
3.28 3.72 8.37 3.41 16.22 4.74 -21.61 16.22 4.59 -38.2 -9.24 4.42
Total -17.96 24.96 -38.1 49.88 -86.02 106.9 -294.6 289.2 -14.18 -19.4 -62.02 57.2
Distribution to column Above floor
5.93
3.15
9.76
58.17
-74.26
19.53
-18.9
Distribution to column below floor
5.93
3.15
9.76
58.17
-74.26
19.53
-18.9
O
FRAME 3
P Q R S T U
Figure 3-Analysis using substitute frame method-frame3
4.3.9 DISTRIBUTIO� FACTOR
Joint Member Relative stiffness Total stiffness Stiffness for each member
O
OP
O1
O2
I/3.72
I/4
I/4
0.769I
0.35
0.33
0.33
P
PO
PQ
P1
P2
I/3.72
I/3.55
I/4
I/4
1.05I
0.26
0.27
0.24
0.24
Q
QP
QR
Q1
Q2
I/3.55
I/6.33
I/4
I/4
0.94I
0.3
0.17
0.27
0.27
R
RQ
RS
R1
R2
I/6.33
I/6.43
I/4
I/4
0.81I
0.19
0.19
0.31
0.31
S
SR
ST
S1
S2
I/6.43
I/3.55
I/4
I/4
0.937I
0.17
0.3
0.27
0.27
T
TS
TU
T1
T2
I/3.55
I/3.72
I/4
I/4
1.05 I
0.27
0.26
0.24
0.24
U
UT
U1
U2
I/3.72
I/4
I/4
0.769I
0.35
0.33
0.33
4.3.10 LOAD CALCLATIO�
54.72547.79564.0563.4647.79554.725Total load
KN/m2
48.08522.04527.192722.04548.085Total DL
KN/m2
3.3753.3754.54.53.3753.375Dead load
due to rib
KN/m2
6.6425.7536.8636.4725.756.64Total LL
KN /m
40.4114.1620.2720.0614.1640.41Total DL
KN/ m
3.723.556.436.333.553.72Length m
8.2286.09515.815.396.0958.228Area m2
3151515153LL KN/m2
18.278.258.258.258.2518.27DL KN / m2
TUSTRSQRPQOPBeam name
4
.3.11 MOME�T CALCULATIO� AT JOI�TS
JOINT-O
-44.41Net BM due
to O
-68.32
23.912
Distribution
-5.2Distribution
+ Carryover
+63.12-63.12FEM due to
LL
-23.15FEM due to
DL
0.190.190.170.30.270.260.35DF
RSRQQRQPPQPOOPMember
RQPOjoint
JOINT-P
-52.28466.376Net BM due
to P
-44.194
-8.09
74.166
-7.79
Distribution
5.99611.046Distribution
+ Carryover
50.19-50.19+63.12-63.12FEM due to
LL
-90.16FEM due to
DL
0.190.190.170.30.270.260.35DF
RSRQQRQPPQPOOPMember
RQPOjoint
JOINT-Q
-193.61101.575Net BM due
to Q
-223.13
29.521
49.48
52.095
Distribution
-11.23-0.71Distribution
+ Carryover
211.9-211.950.19-50.19FEM due to
LL
-93.6855.45FEM due to
DL
0.190.190.170.30.270.260.35DF
RSRQQRQPPQPOOPMember
RQPOjoint
JOINT-R
-235.66229.754Net BM
due to R
-237.47
1.81
227.944
1.81
Distributio
n
-16.7916.04Distributio
n +
Carryover
+220.68-220.68211.9-211.9FEM due
to LL
-23.1523.15FEM due
to DL
0.260.270.30.170.190.190.170.3DF
TUTSSTSRRSRQQRQPMember
TSRQjoint
JOINT-S
-104.56201.87Net BM
due to S
-49.48
-55.08
233.08
-31.212
Distributio
n
0.7112.4Distributio
n +
Carryover
50.19-50.19+220.68-220.68FEM due
to LL
-55.4590.16FEM due
to DL
0.260.270.30.170.190.190.170.3DF
TUTSSTSRRSRQQRQPMember
TSRQjoint
JOINT-T
TS U
-64.5447.13Net BM
due to T
-74.17
9.63
37.14
9.99
Distributi
on
-11.05-13.05Distributi
on +
Carryove
r
63.12-63.1250.19-50.19FEM due
to LL
93.68FEM due
to DL
0.350.260.270.30.170.190.19DF
UTTUTSSTSRRSRQMember
Rjoint
JOINT U
TS U
44.41Net BM
due to U
68.32
-23.912
Distributi
on
5.2Distributi
on +
Carryove
r
63.12-63.12FEM due
to LL
23.15FEM due
to DL
0.350.260.270.30.170.190.19DF
UTTUTSSTSRRSRQMember
Rjoint
MID SPAN MOMENT OP
53.55-44.4Net moment
-10.231.818distribution
28.311.05-5.195carry over
moment
56.6-10.3922.1distribution
moment
211.9-211.963.12-63.12FEM due to
TL
23.15-23.15FEM due to
DL
0.190.190.170.30.270.260.35DF
RSRQQRQPPQPOOPMember
RQPOjoint
Free BM at the centre of the span OP
= Wl2/8
=(54.725 * 3.722) / 8
=94.66 kNm
Net BM at centre of span OP
=94.66-[ (44.4+53.55) / 2]
=45.68 kNm
MID SPAN MOMENT OF PQ
57.97-49.86Net moment
-3.5-4.24distribution
12.35-0.715.999.71carry over
moment
20.4611.99-1.4219.41distribution
moment
-220.6850.19-50.19FEM due to
TL
90.16-90.1655.45-55.45FEM due to
DL
0.190.190.170.30.270.260.35DF
RSRQQRQPPQPOOPMember
RQPOjoint
Free BM at the centre of the span PQ
= Wl2/8
=(47.79 * 3.552) / 8
=75.3 kNm
Net BM at centre of span PQ
=75.3-[ (49.86+57.97) / 2]
=21.4 kNm
MID SPAN MOMENT QR
203.14-188.2Net
moment
-2.352.83distributio
n
-3.716.05-11.23-5.4carry
over
moment
-7.4-22.4632.1-10.79distributio
n
moment
-50.19211.9-211.963.12FEM due
to TL
93.68-93.6823.15-23.15FEM due
to DL
0.30.170.190.190.170.30.270.26DF
STSRRSRQQRQPPQPOmember
SRQPJoint
Free BM at the centre of the span QR
= Wl2/8
=(63.46 * 6.332) / 8
=317.85 KNm
Net BM at centre of span QR
=317.85-[ (188.2+203.14) / 2]
=122.18 KNm
MID SPAN MOMENT RS
196.25-210.13Net
moment
-3.032.54distributio
n
5.412.4-16.793.4carry over
moment
10.8-33.5824.86.8distributio
n moment
-63.12220.68-220.6850.19FEM due
to TL
23.15-23.1590.16-90.16FEM due
to DL
0.260.270.30.170.190.190.170.3DF
TUTSSTSRRSRQQRQPmember
TSRQJoint
Free BM at the centre of the span RS
= Wl2/8
=(64.05 * 6.432) / 8
=331.02 KNm
Net BM at centre of span RS
=331.02-[ (210.13+196.25) / 2]
=127.8 KNm
MID SPAN MOMENT ST
U
59-33.27Net
moment
0.863.16Distribution
-9.716.530.71-11.23carry over
moment
-19.411.4213.05-22.46distribution
moment
50.19-50.19211.9FEM due to
TL
55.45-55.4593.68-93.68FEM due to
DL
0.350.260.270.30.170.190.19DF
UTTUTSSTSRRSRQMember
TSRjoint
Free BM at the centre of the span ST
= Wl2/8
=(47.79 * 3.552) / 8
=75.3 kNm
Net BM at centre of span ST
=75.3-[ (33.27+59) / 2]
=29.17 kNm
MID SPAN MOMENT TU
U
44.4-53.18Net
moment
-1.8210.6distribution
5.2-11.05-29.65carry over
moment
-22.110.39-59.3distribution
moment
63.12-63.12220.68FEM due
to TL
23.15-23.15FEM due
to DL
0.350.260.270.30.170.190.19DF
UTTUTSSTSRRSRQMember
TSRjoint
Free BM at the centre of the span TU
= Wl2/8
=(54.725 * 3.72) / 8
=94.66 kNm
Net BM at centre of span TU
=94.66-[ (53.18+44.4) / 2]
=45.87 kNm
NEGETIVE MOMENT AT CENTRE OF RS
76.83-122.29Net
moment
1.79-2.45Distributi
on
0.71-11.24-3.7-16.05carry
over
moment
1.42-13.05-7.422.46-22.4632.1distributio
n
moment
50.19-50.19211.9-211.9FEM due
to TL
-55.4593.68-93.6823.15FEM due
to DL
0.260.270.30.170.190.190.170.3DF
TUTSSTSRRSRQQRQPmember
TSRQJoint
Free BM at the centre of the span RS
= Wl2/8
=(27.19 * 6.432) / 8
=140.521 kNm
Net BM at centre of span RS
=140.521-[ (122.29+76.83) / 2]
=40.96 kNm
4.3.12 BENDING MOMENT IN COLUMN LOADING
joint O P Q R S T U
Column DF
Above floor
Below floor
0.33
0.33
0.24
0.24
0.27
0.27
0.31
0.31
0.27
0.27
0.24
0.24
0.33
0.33
Member
DF
OP
0.35
PO
0.26
PQ
0.27
QP
0.3
QR
0.17
RQ
0.19
RS
0.19
SR
0.17
ST
0.3
TS
0.27
TU
0.26
UT
0.35
FEM due to
DL
-23.2 23.2 -93.68 93.68 -55.45 55.45
FEM due TL -63.12 63.12 -211.9 211.9 -50.19 50.19
Distribution+
carryover
-5.2 11.05 28.31 -5.4 -11.23 16.04 -3.7 -11.2 0.71 -6.52 -9.7 0.68
Total -68.32 74.17 -5.16 17.75 -223.1 227.9 -97.38 82.45 -49.48 43.67 -65.15 56.13
Distribution to column Above floor
22.55
-16.56
55.45
-40.47
-8.9
5.15
-18.5
Distribution to column below floor
22.55
-16.56
55.45
-40.47
-8.9
5.15
-18.5
joint O P Q R S T U
Column DF
Above floor
Below floor
0.33
0.33
0.24
0.24
0.27
0.27
0.31
0.31
0.27
0.27
0.24
0.24
0.33
0.33
Member
DF
OP
0.35
PO
0.26
PQ
0.27
QP
0.3
QR
0.17
RQ
0.19
RS
0.19
SR
0.17
ST
0.3
TS
0.27
TU
0.26
UT
0.35
FEM due to
DL
-55.45 55.45 -90.16 90.16 -23.2 23.2
FEM due TL -50.2 50.2 -220.7 220.7 -63.12 63.12
Distribution+
carryover
-0.68 9.7 5.6 -0.71 12.4 3.4 -16.79 12.4 5.4 -29.6 -11.05 5.2
Total -56.13 65.15 -44.2 49.48 -77.76 93.56 -237.5 233.1 -17.75 -6.48 -74.17 68.32
Distribution to column Above floor
18.52
-5.03
7.63
44.61
-58.14
19.36
-22.6
Distribution to column below floor
18.52
-5.03
7.63
44.61
-58.14
19.36
-22.6
FRAME 4
A1 F1 G1 J1 K1 N1H1 O1 S1R1P1M1L1I1E1C1 D1B1 Q1 T1 U1
Figure 4-Analysis using substitute frame method-frame4
4.3.13 DISTRIBUTIO� FACTOR
Joint Member Relative stiffness Total stiffness Stiffness for each member A1
A1B1 A11 A12
I/3.2 I/4 I/4
0.8125I
0.38 0.31 0.31
B1
B1A1 B1C1 B11 B12
I/3.2 I/3.2 I/4 I/4
1.125I
0.28 0.28 0.22 0.22
C1
C1B1 C1D1 C11 C12
I/3.2 I/3.2 I/4 I/4
1.125I
0.28 0.28 0.22 0.22
D1
D1C1 D1E1 D11 D12
I/3.2 I/3.2 I/4 I/4
1.125I
0.28 0.28 0.22 0.22
E1
E1D1 E1F1 E11 E12
I/3.2 I/3.2 I/4 I/4
1.125I
0.28 0.28 0.22 0.22
F1
F1E1 F1G1 F11 F12
I/3.2 I/3.2 I/4 I/4
1.125I
0.28 0.28 0.22 0.22
G1
G1F1 G1H1 G11 G12
I/3.2 I/3.2 I/4 I/4
1.125I
0.28 0.28 0.22 0.22
H1
H1G1 H1I1 H11 H12
I/3.2 I/3.2 I/4 I/4
1.125I
0.28 0.28 0.22 0.22
I1
I1H1 I1J1 I11 I12
I/3.2 I/3.2 I/4 I/4
1.125I
0.28 0.28 0.22 0.22
J1
J1I1 J1K1 J11 J12
I/3.2 I/3.2 I/4 I/4
1.125I
0.28 0.28 0.22 0.22
K1J1 I/3.2 0.28
K1 K1L1 K11 K12
I/3.2 I/4 I/4
1.125I 0.28 0.22 0.22
L1
L1K1 L1M1 L11 L12
I/3.2 I/3.2 I/4 I/4
1.125I
0.28 0.28 0.22 0.22
M1
M1L1 M1N1 M11 M12
I/3.2 I/3.2 I/4 I/4
1.125I
0.28 0.28 0.22 0.22
N1
N1M1 N1O1 N11 N12
I/3.2 I/3.2 I/4 I/4
1.125I
0.28 0.28 0.22 0.22
O1
O1N1 O1P1 O11 O12
I/3.2 I/3.2 I/4 I/4
1.125I
0.28 0.28 0.22 0.22
P1
P1O1 P1Q1 P11 P12
I/3.2 I/3.2 I/4 I/4
1.125I
0.28 0.28 0.22 0.22
Q1
Q1P1 Q1R1 Q11 Q12
I/3.2 I/3.2 I/4 I/4
1.125I
0.28 0.28 0.22 0.22
R1
R1Q1 R1S1 R11 R12
I/3.2 I/3.2 I/4 I/4
1.125I
0.28 0.28 0.22 0.22
S1
S1R1 S1T1 S11 S12
I/3.2 I/6.4 I/4 I/4
0.9688I
0.32 0.16 0.26 0.26
T1
T1S1 T1U1 T11 T12
I/6.4 I/8.16 I/4 I/4
0.7788I
0.2 0.16 0.32 0.32
U1
U1T1 U11 U12
I/8.16 I/4 I/4
0.6225I
0.2 0.4 0.4
4.3.14 LOAD CALCLATIO�
No Beam name
Dead load kN/m2
Live load kN/m2
Area m2
Length m
slab DL kN/m
Total LL kN/m
Rib DL kN/m
Total DL kN/m
Total load kN/m
1 A1B1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575
2 B1C1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575
3 C1D1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575
4 D1E1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575
5 E1F1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575
6 F1G1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575
7 G1H1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575
8 H1I1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575
9 I1J1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575
10 J1K1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575
11 K1L1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575
12 L1M1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575
13 M1N1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575
14 N1O1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575
15 O1P1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575
16 P1Q1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575
17 Q1R1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575
18 R1S1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575
19 S1T1 8.25 15 16.2 6.4 20.88 37.97 4.5 25.38 63.35
20 T1U1 8.25 15 13.74 8.16 13.89 25.26 4.5 18.39 43.65
JOINT-A1
-23.245Net BM
due to A1
-37.492
14.247
Distributio
n
-2.868Distributio
n +
Carryover
34.624-34.624FEM due
to LL
-12.78FEM due
to DL
0.280.280.280.280.280.280.38DF
D1E1D1C1C1D1C1B1B1C1B1A1A1B1Member
D1C1B1A1joint
JOINT-B1
-38.53340.165Net BM due
to B1
-37.494
-1.0388
41.204
-1.0388
Distribution
-2.876.58Distribution
+ Carryover
34.624-34.62434.624-34.624FEM due to
LL
-14.14FEM due to
DL
0.280.280.280.280.280.280.38DF
D1E1D1C1C1D1C1B1B1C1B1A1A1B1Member
D1C1B1A1joint
JOINT-C1
-37.4937.49Net BM
due to C1
-37.49
0
37.49
0
Distributio
n
-2.8682.868Distributio
n +
Carryover
+34.62-34.62+34.62-34.62FEM due
to LL
-14.1414.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
E1F1E1D1D1E1D1C1C1D1C1B1B1C1B1A1Member
E1D1C1B1
joint
JOINT-D1
-37.49237.492Net BM
due to D1
-37.492
0
37.492
0
Distributi
on
-2.9682.868Distributi
on +
Carryove
r
+34.62
4
-34.624+34.62
4
-34.624FEM due
to LL
-58.8114.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
E1F1E1D1D1E1D1C1C1D1C1B1B1C1B1A1Member
E1D1C1B1
joint
JOINT-E1
-37.49237.492Net BM
due to E1
-37.492
0
37.492
0
Distributio
n
-2.9682.868Distributio
n +
Carryover
+34.624-34.624+34.62
4
-
34.624
FEM due
to LL
-14.1414.1
4
FEM due
to DL
0.280.280.280.280.280.280.280.28DF
G1H1G1F1F1G1F1E1E1F1E1D1D1E1D1C1Member
G1F1E1
D1joint
JOINT-F1
-37.49237.492Net BM
due to F1
-37.492
0
37.492
0
Distributi
on
-2.9682.868Distributi
on +
Carryove
r
+34.62
4
-34.624+34.62
4
-34.624FEM due
to LL
-14.1414.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
G1H1G1F1F1G1F1E1E1F1E1D1D1E1D1C1Member
G1F1E1
D1joint
JOINT-G1
-37.49237.492Net BM
due toG1
-37.492
0
37.492
0
Distributi
on
-2.9682.868Distributi
on +
Carryove
r
+34.624-34.624+34.624-
34.624
FEM due
to LL
-14.1414.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
I1J1I1H1H1I1H1G1G1H1G1F1F1G1F1E1Member
I1H1G1F1
joint
JOINT-H1
-37.49237.492Net BM
due to H1
-37.492
0
37.492
0
Distributi
on
-2.9682.868Distributi
on +
Carryove
r
+34.62
4
-34.624+34.62
4
-34.624FEM due
to LL
-14.1414.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
I1J1I1H1H1I1H1G1G1H1G1F1F1G1F1E1Member
I1H1G1F1
joint
JOINT-I1
-37.49237.492Net BM
due to I1
-37.492
0
37.492
0
Distributio
n
-2.9682.868Distributio
n+
Carryover
34.624-34.62434.624-34.624FEM due
to LL
-14.1414.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
K1L1K1J1J1K1J1I1I1J1I1H1H1I1H1G1Member
K1J1I1H1
joint
JOINT-J1
-37.49237.492Net BM
due to J1
-37.492
0
37.492
0
Distributi
on
-2.9682.868Distributi
on +
Carryove
r
+34.62
4
-34.624+34.62
4
-34.624FEM due
to LL
-14.1414.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
K1L1K1J1J1K1J1I1I1J1I1H1H1I1H1G1Member
K1J1I1H1
joint
JOINT-K1
-37.49237.492Net BM
due to K1
-37.492
0
37.492
0
Distributio
n
-2.9682.868Distributio
n +
Carryover
34.624-34.62434.624-34.624FEM due
to LL
-14.1414.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
M1N1
M1L1L1M1L1K1K1L1K1J1J1K1J1I1Member
M1L1K1
J1joint
JOINT-L1
-37.49237.492Net BM
due to L1
-37.492
0
37.492
0
Distributio
n
-2.9682.868Distributio
n +
Carryover
34.624-34.62434.624-34.624FEM due
to LL
-14.1414.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
M1N1M1L1L1M1L1K1K1L1K1J1J1K1J1I1Member
M1L1K1
J1joint
JOINT-M1
-37.49237.492Net BM
due to M1
-37.492
0
37.492
0
Distributi
on
-2.9682.868Distributi
on +
Carryove
r
34.624-34.62434.624-34.624FEM due
to LL
-14.1414.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
O1P1O1N1N1O1N1M1M1N1M1L1L1M1L1K1Member
O1N1M1
L1joint
JOINT-N1
-37.49237.492Net BM
due to N1
-37.492
0
37.492
0
Distributi
on
-2.9682.868Distributi
on +
Carryove
r
34.624-34.62434.624-34.624FEM due
to LL
-14.1414.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
O1P1O1N1N1O1N1M1M1N1M1L1L1M1L1K1Member
O1N1M1
L1joint
JOINT-O1
-37.49237.492Net BM
due to O1
-37.492
0
37.492
0
Distributi
on
-2.9682.868Distributi
on +
Carryove
r
34.624-34.62434.624-34.624FEM due
to LL
-14.1414.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
Q1R1Q1P1P1Q1P1O1O1P1O1N1N1O1N1M1Member
Q1P1O1
N1joint
JOINT-P1
-37.49237.492Net BM
due to P1
-37.492
0
37.492
0
Distributi
on
-2.9682.868Distributi
on +
Carryove
r
34.624-34.62434.624-34.624FEM due
to LL
-14.1414.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
Q1R1Q1P1P1Q1P1O1O1P1O1N1N1O1N1M1Member
Q1P1O1
N1joint
JOINT-Q1
-37.49237.492Net BM
due to Q1
-37.492
0
37.492
0
Distributi
on
-2.9682.868Distributi
on +
Carryove
r
34.624-34.62434.624-34.624FEM due
to LL
-14.1414.14FEM due
to DL
0.160.320.280.280.280.280.280.28DF
S1T1S1R1R1S1R1Q1Q1R1Q1P1P1Q1P1O1Member
S1R1Q1
P1joint
JOINT-R1
-29.43434.362Net BM
due to R1
-26.304
-3.13
37.492
-3.13
Distributi
on
8.322.868Distributi
on +
Carryove
r
34.624-34.62434.624-34.624FEM due
to LL
-86.6314.14FEM due
to DL
0.160.320.280.280.280.280.280.28DF
S1T1S1R1R1S1R1Q1Q1R1Q1P1P1Q1P1O1Member
S1R1Q1
P1joint
JOINT-S1
T1S1 U1
-197.2298.342Net BM
due to S1
-227.65
30.43
37.492
60.85
Distributi
on
-11.422.868Distributi
on +
Carryove
r
216.23-216.2334.624-34.624FEM due
to LL
-102.0414.14FEM due
to DL
0.20.160.20.160.320.280.28DF
U1T1T1U1T1S1S1T1S1T1R1S1R1Q1Member
R1joint
JOINT-T1
T1S1 U1
-260.98239.207Net BM
due to T1
-266.42
5.44
232.397
6.81
Distributio
n
-24.2216.167Distributio
n +
Carryover
242.2-242.2216.23-216.23FEM due
to LL
14.14FEM due
to DL
0.20.160.20.160.320.280.28DF
U1T1T1U1T1S1S1T1S1T1R1S1R1Q1Member
R1joint
JOINT-U1
T1S1 U1
203.72Net BM
due to U1
254.65
-50.93
Distributio
n
12.45Distributio
n +
Carryover
242.2-242.2FEM due
to LL
86.63FEM due
to DL
0.20.160.20.160.320.280.28DF
U1T1T1U1T1S1S1T1S1T1R1S1R1Q1Member
R1joint
MID SPAN MOMENT A1B1
+32.814-23.24Net moment
-2.651.09distribution
2.876.58-2.87carry over
moment
5.74-5.7413.16distribution
moment
34.624-34.62434.624-34.624FEM due to
TL
14.14-14.14FEM due to
DL
0.280.280.280.280.280.280.38DF
D1E1D1C1C1D1C1B1B1C1B1A1A1B1Member
D1C1B1A1joint
F
ree BM at the centre of the span A1B1
= Wl2/8
=(40.575 * 3.22) / 8
=51.936 kNm
Net BM at centre of span A1B1
=51.936-[ (23.24+32.814) / 2]
=23.909 kNm
MID SPAN MOMENT OF B1C1
30.14-31.702Net moment
-1.610.052distribution
2.87-2.872.872.685carry over
moment
5.74-5.745.745.37distribution
moment
-34.62434.624-34.624FEM due to
TL
14.14-14.1414.14-14.14FEM due to
DL
0.280.280.280.280.280.280.38DF
D1E1D1C1C1D1C1B1B1C1B1A1A1B1Member
D1C1B1A1joint
Free BM at the centre of the span B1C1
= Wl2/8
=(40.575 * 3.22) / 8
=51.936 kNm
Net BM at centre of span B1C1
=51.936-[ (31.702+30.14) / 2]
=21.015 kNm
MID SPAN MOMENT C1D1
33.35-33.35Net
moment
1.6-1.6distributio
n
2.872.87-2.87-2.87carry
over
moment
5.74-5.745.74-5.74distributio
n
moment
-34.62434.624-34.62434.62
4
FEM due
to TL
14.14-14.1414.14-14.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
E1F1E1D1D1E1D1C1C1D1C1B1B1C1B1A1member
E1D1C1B1
Joint
Free BM at the centre of the span C1D1
= Wl2/8
=(40.575 * 3.22) / 8
=51.936 kNm
Net BM at centre of span C1D1
=51.936-[ (33.35+33.35) / 2]
=18.586 kNm
MID SPAN MOMENT D1E1
33.35-33.35Net
moment
1.6-1.6distribution
2.872.87-2.87-2.87carry over
moment
5.74-5.745.74-5.74distribution
moment
-34.62434.624-34.62434.624FEM due
to TL
14.14-14.1414.14-14.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
F1G1F1E1E1F1E1D1D1E1D1C1C1D1C1B1member
F1E1D1C1
Joint
Free BM at the centre of the span D1E1
= Wl2/8
=(40.575 * 3.22) / 8
=51.936 kNm
Net BM at centre of span D1E1
=51.936-[ (33.35+33.35) / 2]
=18.586 kNm
MID SPAN MOMENT E1F1
33.35-33.35Net
moment
1.6-1.6distribution
2.872.87-2.87-2.87carry over
moment
5.74-5.745.74-5.74distribution
moment
-34.62434.624-34.62434.624FEM due
to TL
14.14-14.1414.14-14.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
G1H1G1F1F1G1F1E1E1F1E1D1D1E1D1C1member
G1F1E1
D1Joint
Free BM at the centre of the span E1F1
= Wl2/8
=(40.575 * 3.22) / 8
=51.936 kNm
Net BM at centre of span E1F1
=51.936-[ (33.35+33.35) / 2]
=18.586 kNm
MID SPAN MOMENT F1G1
33.35-33.35Net
moment
1.6-1.6distribution
2.872.87-2.87-2.87carry over
moment
5.74-5.745.74-5.74distribution
moment
-34.62434.624-34.62434.624FEM due
to TL
14.14-14.1414.14-14.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
H1I1H1G1G1H1G1F1F1G1F1E1E1F1E1D1member
H1G1F1E1
Joint
F
ree BM at the centre of the span F1G1
= Wl2/8
=(40.575 * 3.22) / 8
=51.936 kNm
Net BM at centre of span F1G1
=51.936-[ (33.35+33.35) / 2]
=18.586 kNm
MID SPAN MOMENT G1H1
33.35-33.35Net
moment
1.6-1.6distribution
2.872.87-2.87-2.87carry over
moment
5.74-5.745.74-5.74distribution
moment
-34.62434.624-34.62434.624FEM due
to TL
14.14-14.1414.14-14.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
I1J1I1H1H1I1H1G1G1H1G1F1F1G1F1E1member
I1H1G1F1
Joint
Free BM at the centre of the span G1H1
= Wl2/8
=(40.575 * 3.22) / 8
=51.936 kNm
Net BM at centre of span G1H1
=51.936-[ (33.35+33.35) / 2]
=18.586 kNm
MID SPAN MOMENT H1I1
33.35-33.35Net
moment
1.6-1.6distribution
2.872.87-2.87-2.87carry over
moment
5.74-5.745.74-5.74distribution
moment
-34.62434.624-34.62434.624FEM due
to TL
14.14-14.1414.14-14.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
J1K1J1I1I1J1I1H1H1I1H1G1G1H1G1F1member
J1I1H1G1Joint
F
ree BM at the centre of the span H1I1
= Wl2/8
=(40.575 * 3.22) / 8
=51.936 kNm
Net BM at centre of span H1I1
=51.936-[ (33.35+33.35) / 2]
=18.586 kNm
MID SPAN MOMENT I1J1
33.35-33.35Net
moment
1.6-1.6distribution
2.872.87-2.87-2.87carry over
moment
5.74-5.745.74-5.74distribution
moment
-34.62434.624-34.62434.624FEM due
to TL
14.14-14.1414.14-14.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
K1L1K1J1J1K1J1I1I1J1I1H1H1I1H1G1member
K1J1I1H1
Joint
Free BM at the centre of the span I1J1
= Wl2/8
=(40.575 * 3.22) / 8
=51.936 kNm
Net BM at centre of span I1J1
=51.936-[ (33.35+33.35) / 2]
=18.586 kNm
MID SPAN MOMENT J1K1
33.35-33.35Net
moment
1.6-1.6distribution
2.872.87-2.87-2.87carry over
moment
5.74-5.745.74-5.74distribution
moment
-34.62434.624-34.62434.624FEM due
to TL
14.14-14.1414.14-14.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
L1M1L1K1K1L1K1J1J1K1J1I1I1J1I1H1member
L1K1J1I1
Joint
Free BM at the centre of the span J1K1
= Wl2/8
=(40.575 * 3.22) / 8
=51.936 kNm
Net BM at centre of span J1K1
=51.936-[ (33.35+33.35) / 2]
=18.586 kNm
MID SPAN MOMENT K1L1
33.35-33.35Net
moment
1.6-1.6distribution
2.872.87-2.87-2.87carry over
moment
5.74-5.745.74-5.74distribution
moment
-34.62434.624-34.62434.624FEM due
to TL
14.14-14.1414.14-14.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
M1N1M1L1L1M1L1K1K1L1K1J1J1K1J1I1member
M1L1K1
J1Joint
F
ree BM at the centre of the span K1L1
= Wl2/8
=(40.575 * 3.22) / 8
=51.936 kNm
Net BM at centre of span K1L1
=51.936-[ (33.35+33.35) / 2]
=18.586 kNm
MID SPAN MOMENT L1M1
33.35-33.35Net
moment
1.6-1.6distribution
2.872.87-2.87-2.87carry over
moment
5.74-5.745.74-5.74distribution
moment
-34.62434.624-34.62434.624FEM due
to TL
14.14-14.1414.14-14.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
N1O1N1M1M1N1M1L1L1M1L1K1K1L1K1J1member
N1M1L1K1
Joint
F
ree BM at the centre of the span L1M1
= Wl2/8
=(40.575 * 3.22) / 8
=51.936 kNm
Net BM at centre of span L1M1
=51.936-[ (33.35+33.35) / 2]
=18.586 kNm
MID SPAN MOMENT M1N1
33.35-33.35Net
moment
1.6-1.6distribution
2.872.87-2.87-2.87carry over
moment
5.74-5.745.74-5.74distribution
moment
-34.62434.624-34.62434.624FEM due
to TL
14.14-14.1414.14-14.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
O1P1O1N1N1O1N1M1M1N1M1L1L1M1L1K1member
O1N1M1
L1Joint
F
ree BM at the centre of the span M1N1
= Wl2/8
=(40.575 * 3.22) / 8
=51.936 kNm
Net BM at centre of span M1N1
=51.936-[ (33.35+33.35) / 2]
=18.586 kNm
MID SPAN MOMENT N1O1
33.35-33.35Net
moment
1.6-1.6distribution
2.872.87-2.87-2.87carry over
moment
5.74-5.745.74-5.74distribution
moment
-34.62434.624-34.62434.624FEM due
to TL
14.14-14.1414.14-14.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
P1Q1P1O1O1P1O1N1N1O1N1M1M1N1M1L1member
P1O1N1M1
Joint
F
ree BM at the centre of the span N1O1
= Wl2/8
=(40.575 * 3.22) / 8
=51.936 kNm
Net BM at centre of span N1O1
=51.936-[ (33.35+33.35) / 2]
=18.586 kNm
MID SPAN MOMENT O1P1
33.35-33.35Net
moment
1.6-1.6distribution
2.872.87-2.87-2.87carry over
moment
5.74-5.745.74-5.74distribution
moment
-34.62434.624-34.62434.624FEM due
to TL
14.14-14.1414.14-14.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
Q1R1Q1P1P1Q1P1O1O1P1O1N1N1O1N1M1member
Q1P1O1
N1Joint
Free BM at the centre of the span O1P1
= Wl2/8
=(40.575 * 3.22) / 8
=51.936 kNm
Net BM at centre of span O1P1
=51.936-[ (33.35+33.35) / 2]
=18.586 kNm
MID SPAN MOMENT P1Q1
33.35-33.35Net
moment
1.6-1.6distribution
2.872.87-2.87-2.87carry over
moment
5.74-5.745.74-5.74distribution
moment
-34.62434.624-34.62434.624FEM due
to TL
14.14-14.1414.14-14.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
R1S1R1Q1Q1R1Q1P1P1Q1P1O1O1P1O1N1member
R1Q1P1O1
Joint
F
ree BM at the centre of the span P1Q1
= Wl2/8
=(40.575 * 3.22) / 8
=51.936 kNm
Net BM at centre of span P1Q1
=51.936-[ (33.35+33.35) / 2]
=18.586 kNm
MID SPAN MOMENT Q1R1
37.084-33.35Net
moment
5.33-1.6distribution
16.162.87-2.87-2.87carry over
moment
32.33-5.745.74-5.74distribution
moment
-216.2334.624-34.62434.624FEM due
to TL
14.14-14.1414.14-14.14FEM due
to DL
0.160.320.280.280.280.280.280.28DF
S1T1S1T1R1S1R1Q1Q1R1Q1P1P1Q1P1O1member
S1R1Q1
P1Joint
F
ree BM at the centre of the span Q1R1
= Wl2/8
=(40.575 * 3.22) / 8
=51.936 kNm
Net BM at centre of span Q1R1
=51.936-[ (33.35+37.084) / 2]
=16.719 kNm
MID SPAN MOMENT R1S1
60.03-19.038Net
moment
5.8961.526distribution
15.552.878.32-2.87carry over
moment
31.1116.645.74-5.74distribution
moment
-242.234.624-34.62434.624FEM due
to TL
86.63-86.6314.14-14.14FEM due
to DL
0.160.20.160.320.280.280.280.28DF
T1U1T1S1S1T1S1T1R1S1R1Q1Q1R1Q1P1member
T1S1R1Q1
Joint
Free BM at the centre of the span R1S1
= Wl2/8
=(40.575 * 3.22) / 8
=51.936 kNm
Net BM at centre of span R1S1
=51.936-[ (19.038+60.03) / 2]
=12.402 kNm
MID SPAN MOMENT S1T1
U1
210.75-197.61Net moment
1.19-2.29distribution
16.165-11.42-2.87carry over
moment
-22.83832.33-5.74distribution
moment
216.23-216.2334.624FEM due to
TL
102.04-102.0414.14-14.14FEM due to
DL
0.20.160.20.160.320.280.28DF
U1T1T1U1T1S1S1T1S1R1R1S1R1Q1Member
T1S1R1joint
F
ree BM at the centre of the span S1T1
= Wl2/8
=(216.23 * 3.22) / 8
=276.77 kNm
Net BM at centre of span S1T1
=276.77-[ (197.61+210.75) / 2]
=72.594kNm
MID SPAN MOMENT T1U1
U1
30.792-54.42Net moment
-0.6880.48Distribution
-3.44-4.377.37carry over
moment
-8.73-68814.74distribution
moment
43.65-43.6540.575FEM due to
TL
86.63-86.63FEM due to
DL
0.20.160.20.160.320.280.28DF
U1T1T1U1T1S1S1T1S1R1R1S1R1Q1Member
T1S1R1joint
F
ree BM at the centre of the span T1U1
= Wl2/8
=(43.65 * 8.162) / 8
=363.3kNm
Net BM at centre of span T1U1
=363.3-[ (54.42+30.79) / 2]
=320.69kNm
MID SPAN MOMENT K1L1
18.62-18.62Net
moment
1.67-1.61distribution
-2.87-2.872.872.87carry over
moment
-5.745.745.74-5.74-5.745.74distribution
moment
34.62
4
-34.62434.624-34.624FEM due
to TL
-14.1414.14-14.1414.14FEM due
to DL
0.280.280.280.280.280.280.280.28DF
M1N1M1L1L1M1L1K1K1L1K1J1J1K1J1I1member
M1L1K1
J1Joint
Free BM at the centre of the span K1L1
= Wl2/8
=(40.575 * 3.22) / 8
=51.936 kNm
Net BM at centre of span K1L1
=51.936-[ (18.62+18.62) / 2]
=33.316 kNm
4.3.15 BENDING MOMENT IN COLUMN LOADING
joint A1 B1 C1 D1 E1 F1 G1
Column DF
Above floor
Below floor
0.31
0.31
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
Member
DF
A1 B1
0.39
B1A1
0.28
B1 C1
0.28
C1 B1
0.28
C1 D1
0.28
D1 C1
0.28
D1 E1
0.28
E1D1
0.28
E1 F1
0.28
F1E1
0.28
F1 G1
0.28
G1F1
0.28
G1H1
0.28
FEM due to
DL
-14.2 14.2 -14.2 14.2 -14.2 14.2
FEM due TL -34.62 34.62 -34.62 34.62 -34.62 34.62 -34.62
Distribution+
carryover
-2.87 6.75 2.87 -2.87 -2.87 2.87 2.87 -2.87 -2.87 2.87 2.87 -2.87 -2.87
Total -37.49 41.37 -11.3 11.3 -37.49 37.49 -11.3 11.3 -37.49 37.49 -11.3 11.3 -37.49
Distribution to column Above floor
11.62
-6.62
5.77
-5.77
5.77
-5.77
5.77
Distribution to column below floor
11.62
-6.62
5.77
-5.77
5.77
-5.77
5.77
joint H1 I1 J1 K1 L1 M1 N1
Column DF
Above floor
Below floor
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
Member
DF
H1 G1
0.28
H1 I1
0.28
I1 H1
0.28
I1 J1
0.28
J1 I1
0.28
J1 K1
0.28
K1 J1
0.28
K1 E1
0.28
L1 K1
0.28
L1 M1
0.28
M1 L1
0.28
M1 N1
0.28
N1M1
0.28
N1O1
0.28
FEM due to
DL
-14.2 14.2 -14.2 14.2 -14.2 14.2 -14.2
FEM due
TL
34.62 -
34.62
34.62 -
34.62
34.62 -
34.62
34.62
Distribution
+ carryover
2.87 2.87 -2.87 -2.87 2.87 2.87 -2.87 -2.87 2.87 2.87 -2.87 -2.87 2.87 2.87
Total 37.5 -11.3 11.3 -37.5 37.5 -11.3 11.3 -37.5 37.5 -11.3 11.3 -37.5 37.5 -11.3
Distribution to column Above floor
-5.77
5.77
-5.77
5.77
-5.77
5.77
-5.77
Distribution to column below floor
-5.77
5.77
-5.77
5.77
-5.77
5.77
-5.77
joint O1 P1 Q1 R1 S1 T1 U1
Column DF
Above floor
Below floor
0.33
0.33
0.24
0.24
0.27
0.27
0.31
0.31
0.27
0.27
0.24
0.24
0.33
0.33
Member
DF
O1N1
0.28
O1P1
0.28
P1O1
0.28
P1Q1
0.28
Q1P1
0.28
Q1R1
0.28
R1Q1
0.28
R1S1
0.28
S1R1
0.32
S1T1
0.16
T1S1
0.2
T1U1
0.16
U1T1
0.2
FEM due to
DL
14.2 -14.2 14.2 -14.2 14.2 -102.1 102.1
FEM due TL -34.62 34.62 -34.62 34.62 -216.2 216.2
Distribution+
carryover
-2.87 -2.87 2.87 2.87 -2.87 -2.87 2.87 32.33 -2.87 -11.42 16.17 -10.2 -9.14
Total 11.3 -37.5 37.5 -11.3 11.3 -37.5 37.5 -11.3 11.3 227.65 232.4 -112.2 92.9
Distribution to column Above floor
5.77
-5.77
5.77
-12.25
-62.12
-38.45
-37.2
Distribution to column below floor
5.77
-5.77
5.77
-12.25
-62.12
-38.45
-37.2
joint A1 B1 C1 D1 E1 F1 G1
Column DF
Above floor
Below floor
0.31
0.31
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
Member
DF
A1 B1
0.39
B1A1
0.28
B1 C1
0.28
C1 B1
0.28
C1 D1
0.28
D1 C1
0.28
D1 E1
0.28
E1D1
0.28
E1 F1
0.28
F1E1
0.28
F1 G1
0.28
G1F1
0.28
G1H1
0.28
FEM due to
DL
-14.2 14.2 -14.2 14.2 -14.2 14.2 -14.2
FEM due TL -34.6 34.62 -34.62 34.62 -34.62 34.62 -34.62
Distribution+
carryover
2.87 2.76 -2.87 2.87 2.87 -2.87 -2.87 2.87 2.87 -2.87 -2.87 2.87 2.87
Total 11.3 16.9 -37.5 37.49 -11.3 11.3 -37.49 37.49 -11.3 11.3 -37.49 37.49 -11.3
Distribution to column Above floor
-3.49
4.53
-5.77
5.77
-5.77
5.77
-5.77
Distribution to column below floor
-3.49
4.53
-5.77
5.77
-5.77
5.77
-5.77
joint H1 I1 J1 K1 L1 M1 N1
Column DF
Above floor
Below floor
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
Member
DF
H1 G1
0.28
H1 I1
0.28
I1 H1
0.28
I1 J1
0.28
J1 I1
0.28
J1 K1
0.28
K1 J1
0.28
K1 E1
0.28
L1 K1
0.28
L1 M1
0.28
M1 L1
0.28
M1 N1
0.28
N1M1
0.28
N1O1
0.28
FEM due to
DL
14.2 -14.2 14.2 -14.2 14.2 -14.2 14.2
FEM due
TL
-
34.62
34.62 -
34.62
34.62 -
34.62
34.62 -34.62
Distribution
+ carryover
-2.87 -2.87 2.87 2.87 -2.87 -2.87 2.87 2.87 -2.87 -2.87 2.87 2.87 -2.87 -2.87
Total 11.3 -37.5 37.5 -11.3 11.3 -37.5 37.5 -11.3 11.3 -37.5 37.5 -11.3 11.3 -37.5
Distribution to column Above floor
5.77
-5.77
5.77
-5.77
5.77
-5.77
5.77
Distribution to column below floor
5.77
-5.77
5.77
-5.77
5.77
-5.77
5.77
joint O1 P1 Q1 R1 S1 T1 U1
Column DF
Above floor
Below floor
0.33
0.33
0.24
0.24
0.27
0.27
0.31
0.31
0.27
0.27
0.24
0.24
0.33
0.33
Member
DF
O1N1
0.28
O1P1
0.28
P1O1
0.28
P1Q1
0.28
Q1P1
0.28
Q1R1
0.28
R1Q1
0.28
R1S1
0.28
S1R1
0.32
S1T1
0.16
T1S1
0.2
T1U1
0.16
U1T1
0.2
FEM due to
DL
-14.2 14.2 -14.2 14.2 -86.63 86.63
FEM due TL 34.62 -34.6 34.62 -34.62 34.62 -242.2 242.2
Distribution+
carryover
2.87 2.87 -2.87 -2.87 2.87 2.87 -2.87 8.32 2.87 15.56 4.16 -24.22 12.45
Total 37.5 -11.3 11.3 -37.5 37.5 -11.3 11.3 -26.31 37.5 -71.1 90.79 -266.4 254.7
Distribution to column Above floor
-5.77
5.77
-5.77
3.31
8.73
56.2
-102
Distribution to column below floor
-5.77
5.77
-5.77
3.31
8.73
56.2
-102
CHAPTER 5
5.DESIG�
5.1SLABS
The most common type of structural element used to cover floors and
roofs of building are reinforced concrete slabs of different types. One
way slabs are those supported on the two opposite sides so that the
loads are carried along one direction only. Two way slabs are
supported on all four sides with such dimensions such that the loads
are carried to the supports along both directions.
If Ly/Lx < 2, then the slab is designed as two way slab
If Ly/Lx >2, then the slab is designed as one way slab.
Where, Ly = longer span dimension of the slab.
Lx = shorter span dimension of slab.
Restrained slabs are referred to as slabs whose corners are prevented from
lifting. They may be supported on continuous or discontinuous edges.
5.1.1 DESIG� OF SLAB
Dimensions
Lx =3.2
Ly =5.5
Span ratio =5.5 /3.2
=1.1<2
ie, Two way slab
Assume ,
Overall depth =150-20
D =130mm
Load calculation
Live load =10kN/m2
Light patrician =1kN/m2
D.L =(0.15*25)
=3.75kN/m2
Floor finish =0.75kN/m2
Total load =15.5kN/m2
Factored load =1.5*15.5
= 23.25kN/m2
1) Lx =3.2+(0.23/2)+(0.23/2)
=3.43
2) Ly =3.55+(0.23/2)+(0.23/2)
=3.78
Lx =3.33
Ly =3.78
Bending moment
Mx =αx wlx2
My =αy wlx2
Lx / Ly =1.1
Lx =0.074
Ly =0.061
Mx =0.074*23.25*3.432
=20.24kNm
My =0.061*23.25*3.432
=16.69kNm
Check for depth
Mulim =0.138fckbd2
20.24*103 =0.138*20*1000*d2
D =85.63
D < 130mm
Hence the effective depth selected is sufficient to resist the design
ultimate moment.
Find the spacing
Use 10mm dia bars
Spacing =(1000*π*102/4)/465.86
=168.59mm
Provide 150 mm c/c spacing
Find Ast provided
Ast provided =1000ast/spacing
=(1000*π*102/4)/150*4
=523.59mm2
For longer span
16069*106 =0.87*415* Ast *130[1-( Ast *415/20*1000*130)]
Ast =378mm2
Using 10mm dia bars
Spacing =(1000*π/4*102)/378
=207mm.
Adopt the spacing 200mm
Checks,
Ast min =0.12% of c/s of Fe 415
=0.12/100[1000*150]
=180mm2
Ast min< Ast for longer and shorter span,
Hence, provide 10mm dia bars @150mmc/c
Ast =465.86mm2[shorter span]
Hence, provide 10mm dia bars @ 200mm c/c
Ast =378mm2[longer span]
Check for shear
Shear stress
τv =Vu/bd
Vu =wl/2
=23.25*3.43/2=39.87Kn
τv =39.87*103/1000*100
τv =0.398N/mm2
Pt =100 Ast pro / bd
=100*523.59/1000*130
To find τc
From IS456:2000,
τc = 0.432
kτc > τv
k*τc=1.3*0.43
=0.5616
0.398 > 0.5616N/mm2
Hence it is safe.
Check for deflection
(L/d) basic =20
Pt =100 Ast pro/bd
=100*523.59/1000*130
=0.4
Fs =0.58*415*523.59/465.86
=270
kc =1
kf =1
kt =1.2
(L/d)max =(L/d)basic*kt*kc*kt
=20*1.3*1*1
=26
(L/d)act =3200/130=24.16
(L/d)act < (L/d)max
Hence safe against deflection.
Check for control
Reinforcement provided is more than, the minimum % of c/s area
Ast =(0.12/100)*1000*150
=180mm2
Spacing of main reinforcement should not be greater than 3d
ie, 3*130 =390mm
Diameter of reinforcement should be less than D/8
150/8 =18.75
Hence cracks will be with in safe permissible limits
Torsion reinforcement at corner
Area of torsion steel at each of the corners in 4 layers is computed as
=0.75* Ast along shorter span
=0.75*523.59
=393mm2
Length cover which torsion steel is provided
=1/5*shorter span
=1/5*3200
=640mm
Using 6mm dia bars
Spacing =1000ast/ Ast
=(1000*π*62/4)393
=71.9mm
Provide 6mm bars at 100mm c/c for length and 640mmat all corners in
4 layers
Reinforcement in end strips
Ast =0.12% of c/s
=180mm2
Assume 10mm dia bars
Spacing =(1000*π/4*102)/180
=436 > 300
As per code spacing should not exceed 300mm
Provide 10mm dia bars at 300mm c/c
Ast =(1000*π/4*102)/300
Ast =262mm2
10mm dia bars @ 150mm c/c
6mm dia bars @ 100mm c/c
150
Figure 5-Reinforcement details of two way slab-section
REINFORCEMENT DETAILS OF TWO WAY SLAB
6mm dia @ 100mm c/c
10mm dia bars @ 250mm c/c
10mm dia bars @ 150mm c/c
400
3500
2400
2625
437.5
437.5
400
ly/8
ly/8
lx/8 lx/8
3ly/4
3lx/4
3200
Figure 6-Reinforcement details of two way slab- plan
5.2 BEAMS
Beams are defined as structural members subjected to transverse load
that caused bending moment and shear force along the length.
The plane of transverse loads is parallel to the plane of symmetry of the
cross section of the beam and it passes through the shear centre so that the
simple bending of beams occurs. The bending moments and shear forces
produced by the transverse loads are called as internal forces.
5.2.1Types of beams
Depending upon the supports and end condition, beams are classified as
below.
5.2.1.1simply supported beams
5.2.1.2over hanging beams
5.2.1.3cantilever beam
5.2.1.4fixed beam
The reinforced concrete beams, in which the steel reinforced is placed
only on tension side, are known as singly reinforced beams, the tension
developed due to bending moment is mainly resisted by steel reinforcement
and compression by concrete.
When a singly reinforced beam needs considerable depth to exist large
bending moment, then the beam is also reinforced in the compression zone.
The beams having reinforcement in compression and tension zone is called
as doubly reinforced beam.
5.2.2 DESIG� OF L-BEAMS
Dimensions
c/c of support = 3.2+(0.3/2)+(0.3/2) = 3.5m
Thickness of slabs = 150 mm
fy = 20 N/mm2
fck = 415 N/mm2
Width of beam = 300 mm
Overall depth = 300 mm
Effective cover = 25 mm
Effective depth = 300-25-10=265mm
Effective span
a) c/c of supports = 3.2 +(0.3/2) +(0.32/2) = 3.5 m
b) Clear span + d = 3.2 +0.265 = 3.465m
Hence, l = 3.465 m
Load calculation
Dead load of slab = (3.465/2)*0.15*25 = 6.5 kN/m
Floor finish = 0.75*(3.465/2) = 1.3 kN/m
Self weight of rib = 0.3 *0.15 *25 = 1.125 kN/m
Live load = 4*(3.465/2) = 6.93 kN/m
Total load = 16.855 kN/m
Effective flange width
a)bf = (Lo/12)+bw+3Df = 952.125 mm
b) bf = bw +0.5 times spacing b/w ribs = 1900 mm
Ultimate BM and SF
At support,
Mu = 1.5 * wl2/12 = 25.3 kNm
Vu = 1.5 * wl / 2 = 43.8 kN
At centre of span section,
Mu = 1.5 * wl2 / 24 = 12.65kNm
Vu = 1.5 * wl / 2 = 43.8 kN
Torsion moment produced due to dead load of span and live load on it
= working load parameter-rib wt = 16.855-1.125=15.73 kN/m
Ultimate load on slab = 15.73 *3.465 * 1.5 = 81.8 kN
Total ultimate load = 82/2 =41 kN
Distance of centroid of SF from the centre line of the Beam=(952.125/2)-
150
= 326.06mm
Ultimate tortional moment = 4 * 103 *326.06= 13.37 kNm
Equivalent BM and SF
According to IS456 2000 clause 41.4.2
Mel = Mu +Mt
Mt = Tu*(1+D/b)/1.7 = 15.73 kNm
Mel = 13.37 + 15.73 = 29.09 kNm
Equivalent SF
Ve = Vu + 1.6(Tu/b)
=115.1 kN
Main reinforcement
Mu (lim) = 0.138*fck*bd2 = 58.15 kNm
Mel < Mu (lim)
Hence the section is under reinforced
To find Ast
Mu = 0.87*fy* Ast *d[1-( Ast *fy/bd*fck)]
Ast = 332.83mm2
20mm dia rods are used
Ast pro = 628.32 mm2
Ast min = 0.85*bw*d/ fy = 162.83 mm2
Assume 20mm dia bars, Ast pro = 628 mm2
Provide 2 nos of 20mm dia bars @ side face
Reinforcement
Shear reinforcement
τve =Ve/ bw *d = 1.45 N/ mm2
Pt= (100* Ast)/( bw *d) = 0.79
Ref table 19 of IS456 2000
τc=0.56N/ mm2
Hence shear reinforcement are required using 10mm dia 2 legged
stirrups with side cover 25mm top+ bottom cover of 25mm
b1= 300-25-25 = 250mm
d1= 300-25-25 = 250mm
Asv= 157 mm2
σc = Asv *0.87*fy/ (τv – τc)*b = 214.6
Provide 10mm dia 2 legged stirrups @200mm spacing
Check for deflection
(L/d)max = (L/d)basic *kt * kc* kf
(L/d)basic = 20 [for simply supported]
(L/d)max = 20*1*1*1.04 = 20.8
(L/d)actual = 3200/300 = 10.66
(L/d)max > (L/d)actual
Hence the design is safe
LONGITUDINAL SECTION OF L-BEAM
2 nos of 20mm dia bars
3 nos of 20mm dia bars10mm dia 2 legged stirrups @ 200mm c/c
3500
Figure 7-Reinforcement details of L-beams- longitudinal section
CROSS SECTION OF L-BEAM
SUPPORT SECTION
2 bars of 20mm dia
3 bars of 20mm dia
10mm dia 2 legged stirrups @ 200mm c/c
150
MID SECTION
2 bars of 20mm dia
3 bars of 20mm dia
10mm dia 2 legged stirrups @ 200mm c/c
150
Figure 8- Reinforcement details of L-beams- cross section
5.2.3 DESIG� OF T- BEAM
Dimensions
Slab thickness =150mm
c/c of support =3.2+(0.3/2)+(0.3/2)
=3.5m
fy =20N/mm2
fck =415N/mm2
Cross sectional dimension
Width of beam =300mm
Overall depth =300mm
Effective cover =25mm
Effective depth =300-2-10
=265mm
Effective span
1. c/c of support =3.2+(0.3/2)+(0.3/2)
=3.5m
2. clear span+depth =3.2+0.265
=3.465m
Load calculation
Dead load of slab =(3.465/2)*0.15*25
=6.5kN/m
Floor finish =0.75*(3.465/2)
=1.3kN/m
Self weight of rib =0.3*0.15*25
=1.125kN/m
Live load =4*(3.465/2)
=6.93kN/m
Light partition =1kN/m
Total load =16.855kN/m
Ultimate moment and shear
Mu =1.5wl2/8
=(1.5*16.855*3.4652)/8
=37.935kN/m
Vu =wl/2
=(16.855*3.465)/2
=43.8kN/m
Effective width of flange
Refer page no 36 clause 23,
1. bf =L0/b+bw+6Df
=(3.465*0.7)/6+300+(6*150)
=1604.25mm
2. c/c of rib =3000-(300/2)-(300/2)
=2700mm
Ie, bf =1604.25mm
Moment capacity of flange
Assume N.A lies with in the flange
Xu(max)=Df , b=bf
Mu(limit) =(0.36*Xu(max))/d *(1-(0.42Xu(max)/d))*(bd2fck)
=0.36*(150/265) *[1-(0.42*150)/26]*(1604.25*2652*20)
=349.98kNm
Mu < Mu(limit)
Hence the section is under reinforced.
Since the section should design as a singly reinforced beam.
Find Ast
Mu =0.87fy Ast d [1-( Ast fy /bf d fck)]
37.935*106=0.87*415* Ast *265*[1-( Ast *415/1604*265*20)]
Ast =404.46mm2
Check for Ast min
Ast min/bw d =0.85/fy
Ast =(0.85*300*265) /415
=162.83mm2
Ast > Ast min
Ast =404.46mm2
N*πd2/4 =404.46
N =2 nos.
Ast provide,
Provide 2 nos of 20mm dia bars
=2*π202/4
=628mm2
And two longer bars of 12mm dia on the compression face .
Shear reinforcement
τv =Vu/bw d
=43.8*103/300*265
=0.55
Pt =100 Ast /bw d
=(100*π/4*202*2)/300*265
=0.79
τc =0.56+((0.62-0.56)/(1-0.75))*(0.79-0.75)
=0.57
τv<τc
Minimum shear reinforcement in the form of stirrups shall be provided.
Design of shear reinforcement
Asv/b Sv =0.4/0.87*fy
Sv =302.47mm
The spacing should not exceed 300mm
Sv =300mm
Provide 8mm dia bars 2 legged stirrups at 300mmc/c.
Check for deflection control
(L/d)max =(L/d)basic*kt*kf*kc
(L/d)basic =20*0.8
=16
fsc =0.58*fy* Ast req/ Ast pro
=0.58*415*404.46/628
fsc =155
kt =1.5,pt=0.78,kf=1,kc=0.8
(L/d)act =3200/265
=12.075
(L/d)max =16*1.5*1*0.8
=19.2
(L/d)act < (L/d)max .
Hence safe.
LONGITUDINAL SECTION OF T-BEAM
2 nos of 12mm dia bars
3 nos of 20mm dia bars
8mm dia 2 legged stirrups @300mm c/c
150
3500
Figure 9-Reinforcement details of T-beam-longitudinal design
CROSS SECTION OF T-BEAM
SUPPORT SECTION
2 bars of 12mm dia
3 bars of 20mm dia
8mm dia 2 legged stirrups @ 300mm c/c
150
MID SECTION
2 bars of 12mm dia
3 bars of 20mm dia
8mm dia 2 legged stirrups @ 300mm c/c
150
Figure 10- Cross section of T-beam
5.3 STAIRCASE
Stairs are needed for ascending and descending from floor to floor.
The stairs in a structure consists of a series of steps with or without landing
and give access from floor to floor. A flight between the landings is also
called as stair. There may be 3 to 12 steps in one flight. Each step has one
tread and riser. A landing is usually provided after 12 to 15 steps. The width
of landing should not be less than the width of stair.
5.3.1Types of stairs
The different types of reinforced cement concrete stairs are:
1. Straight stairs
2. Dog-legged stairs
3. Open well stairs
4. Quarter turn stairs
5. Geometrical stairs
6. Circular stairs
7. Spiral stair
Dog-legged type of stair is selected and designed for the proposed
auditorium.
5.3.2 DESIG� OF DOG-LEGGED STAIRCASE
Dimensions
Room size =6.4*3.6m
Height of room =4m
Live load =5kN/m2
fy =415N/mm2
fck =20N/mm2
Assume
Tread =300mm
Riser =125mm
It is proposed to provide 2 flights for the stairway, Hence,
The height of each flight =vertical distance/2
=4/2
=2m
No of steps required =2/0.125
=16nos
No of treads in each flight = No of riser-1
=16-1
=15
Space occupied by threads =15*0.3
=4.5m
Assume width of landing =1.25m
Hence space left for passage =6.4-4.5-1.25
=0.65m
Let as assume bearing of landing as 150mm
ie, Effective span of flight =4.5+1.25+0.15/2
=5.8m
Let the thickness of waist slab be 200mm,
This can be assumed as 40mm to 50mm/m span
Load calculation
Dead load of waist slab =w’*√(R2+T2)/T
=0.2*√(0.1252+0.32)/0.3
=5.5kN/m2
Dead load of slab =125/2
=62.5(average)
ie, dead load =0.0625*25
=1.56kN/m2
Main reinforcement
Mu=0.87fyAst*d[1-fy Ast/bdfck]
76.699*106=0.87*415*Ast*180*[1-(415*Ast/1000*20*180)]
Ast=1408.9mm2
Spacing with 10mm dia bars
Sv=(1000*¶/4*102)/1408.9
Sv=55.75mm
Provide 10mm dia bars 100mm c/c
Distributor Reinforcement
Provide Astmin =0.12% of cs area and assume 8mm dia bars
Astmin =0.12/100*1000 *200
=240mm2
Spacing =1000(π/4)*82/240
=209.44mm
Provide 8mm dia bars@200mm c/c
Assume top finish =0.1kN/m2
Live load =25kN/m2
ie, Total load =5.5+1.56+0.1+0.5
=12.16kN/m2
Factored load =1.5*12.16
=18.24kN/m2
Factored moment =wl2/8
=18.24*5.82/8
Distributor Reinforcement
Provide =0.12% and assume 8mm Dia bars
Astmin =0.12/100*1000 *200
=240mm2
Spacing =1000(π/4)*82/240
=209.44mm
Provide 8mm dia bars@200c/c
Assume top finish =0.1kN/m2
Live load =25kN/m2
ie, Total load =5.5+1.56+0.1+0.5
=12.16kN/m2
Factored load =1.5*12.16
=18.24kN/m2
Factored moment =wl2/8
=18.24*5.82/8
=76.699kNm
Check for depth for waist slab
D =√Mu/(0.138fckb)
=√76.699*106/(0.138*20*1000)
=166.7
But, D =200mm
D =200-20
=180>166.7
Hence ok.
Check for depth for waist slab
D =√Mu/(0.138fckb)
=√76.699*106/(0.138*20*1000)
=166.7mm
But D =200mm
D =200-20
=180>166.7
Hence ok.
DETAILING OF DOG LEGGED STAIRCASE
8mm dia bars @ 200mm c/c
Tread = 300mm
Riser =125mm
8mm dia bars @ 200mm c/c
Thickness of flight = 200mm12mm dia bars @ 100mm c/c
12mm dia bars @ 100mm c/c
Figure 11-Reinforcement details of doglegged staircase
5.4 COLUM�S
A column is defined as a structural member subjected to compressive
force in a direction parallel to its longitudinal axis. The columns are used
primarily to support compressive load. When the compression members
are over loaded then their failure may take place in direct compression
(crushing), excessive bending combined with twisting. Failure of column
depends upon slenderness ratio.
5.4.1 Types of columns
1) Short column
2) Long column
When slenderness ratio (lex/b) is less than 12, the compression member
(lex/b) is said to be short column and if the slenderness ratio is greater than
12, it is called as long column.
5.4.2DESIG� OF COLUM�
5.4.2.1 DESIG� OF AXIALLY LOADED COLUM�
Dimensions
Factored load =1200kN
Concrete grade =M20
Characteristic strength of reinforcement =415N/mm2
Unsupported length of column =3.55m
Cross sectional area of column =400*300
Leff =k*L
k =0.65(effectively held in
position at both ends)
Leff =0.65*3550
=2307mm
Slenderness ratio
Leff /D =2307/400
=5.8<12
Leff /d =2307/300=7.7<12
ie, column is designed as short column
Minimum eccentricity
emin1 =2307/500+400/30=17.9mm
emin2 =2307/500+300/30=14.6mm
From clause 39.3 Is 456-2000
400*0.05=20>17.9
300*0.05=15>14.6
ie, codal formula for axially compressed column can be used.
Longitudinal reinforcement
Pu =[0.4fck Ac+0.67fy-0.4fck)Ast]
1200*103 =[(0.4*20*400*300)+[(.67*415)-(0.4*20)]Asc
Asc =888.7mm2
Minimum reinforcement provided
=0.008*400*300=960mm2
ie, Provide 6 nos of 20 mm dia bars of longitudinal reinforcement
Lateral ties
Tie diameter >6mm
<16mm
Provide 8mm diameter ties
Tie spacing > 16*20=320mm
ie, provide 8mm dia ties @ 300mm c/c
AXIALLY LOADED COLUMN
6 nos of 20mm dia bars
Lateral ties 8mm dia bars @ 300mm c/c
Figure 12-Reinforcement details of axially loaded column
5.4.2.2 DESIG� OF U�IAXIALLY LOADED COLUM�
Dimensions
Size of column = 400mm×300mm
Load, Pu = 650 kN
Factored moment= 127 kNm
Eccentricity = 127/650 = 0.19
fck = 20 N/mm2
fy = 415 N/mm2
D =400mm
b =300mm
Assuming cover, d`=50mm
d`/D = 50/400 =0.125
Pu/ fck bD = (650*103)/(20*300*400)
= 0.27
Mu/ fck bD2 = (127*106)/ (20*300*4002)
= 0.132
From graph 45 of SP16,
p/ fck = 0.07
percentage of steel = 1.4
As = pbD/100
= (1.4*400*300)/100
=1800mm2
nπd2/4 = 1800mm2
n = 6 nos
Provide 8 nos. of 20mm dia bars and they are equally arranged on all four
sides
Spacing = 400-(50+50+20)
= 280mm < 300mm
Hence safe
Design of lateral ties
Dia of lateral ties not less than 6mm and not greater than 16mm
Take 8mm dia ties
Spacing should not be greater than 300mm or 16φ=16*20= 320mm
Hence provide 8mm φ bars @ 250mm c/c
UNIAXIALLY LOADED COLUMN
8 nos of 20mm dia bars
Lateral ties 8mm dia bars @ 250mm c/c
Figure 13-Reinforcement details of uniaxially loaded column
5.4.2.3 DESIG� OF BI AXIALLY LOADED COLUM�
Dimension
b = 450mm
D = 600mm
fck =20 N/mm2
fy = 414 N/mm2
Pu = 590.6 kN
Mux= 150 kN
Muy= 106 kN
Reinforcements
Reinforcements are distributed equally on four sides
As a trial adopt percentage of reinforcement in the CS as p =1%
As = pbD/100 = 1*450*600/100
= 2700mm2
Provide 10 bars of 20mm dia on each face
As = 10*π*202/4
= 3141.6mm2
P = (100*3141.6)/(450*600)
= 1.16
p/ fck = 1.16/20 =0.058
d`= 40+10 = 50mm
Pu/ fck bD = (590.6*103)/(20*450*300)
= 0.12
d`/D = 50/600 =0.08
from chart 44 of SP16
Mu/ fck bD2 = 0.09
For moments about minor axis yy
b= 450mm
d`=40+10=50mm
d`/D = 50/450 =0.111
Pu/ fck bD = 0.12
From chart 44 for d`/D=0.15
Mu/ fck bD2 = 0.09
Muy1=0.09*20*600*4502
= 218.7 kNm
Puz= 0.45 fck Ac+0.75 fy As
=0.45*20[(600*450)-314.6]+[0.75*415*3141.6*10-3]
= 3379.55kN
Pu/ Puz = (590.6/3379.55)=0.175
According to IS456 clause 39.6
αn =1.04
(Mux/ Mux1) αn +( Muy/ Muy1)
αn = 0.97<1
Hence the section is safe
Design of lateral ties
According to IS456:2000,
Dia of lateral ties not less than 6mm and not greater than 16mm
Take 8mm dia ties
Spacing should not be greater than 300mm or 16φ=16*20= 320mm
Hence provide 8mm φ bars @ 300mm c/c
BIAXIALLY LOADED COLUMN
10 nos of 20mm dia bars
Lateral ties 8mm dia bars @300mm c/c
Figure 14-Reinforcement details of biaxially loaded columns
5.5 FOOTI�G
The portion of the structure above the plinth is called superstructure
and the portion of the structure below the plinth is called as substructure.
The footing or foundation is the part of substructure which remains in
contact with the soil or rock. The footing or foundation transmits safely the
load to soil stratum and distributes the load over large area of the stratum so
that the bearing pressure developed in the soil remains less than the safe
bearing capacity of the soil and is ensured that the any settlement which may
occur shall be nearly uniform as possible and the differential settlement of
the various parts of the structure shall be eliminated as nearly as possible
5.5.1Types of footing
1) Isolated footing
2) Combined footing
3) Spread footing
4) Raft footing
5.5.2 DESIG� OF FOOTI�G
Dimension
Factored load, Pu = 1200 kN
Size of column = 400 × 300 mm
SBC of soil = 200 kN/m2
fck = 20 N/mm2
fy = 415 N/mm2
Size footing
Load on column = 1200 kN
Weight of footing and backfill at 10% = 120 kN
Area of footing =(1200+120)/(1.5*200)
= 4.4 m2
Adopt 2.5m×2m rectangular footing
Net soil pressure at ultimate load is given by,
qu = 1320/(2.5*2)
= 264 kN/ mm2
One way shear
Critical section is at a distance ‘d’ from the column face
Factored shear force, Vu1 = (0.264*2500)(1050-d)
= 660(1050-d)
Assuming percentage of reinforced cement, Pt = 0.25%
For M20 grade concrete, from table 19 IS456:2000
τc = 0.36 N/mm2
one way shear resistance, Vc1 = 0.36*2500*d
= 900d
Equating both,
660(1050-d) = 900d
Hence d = 442.3 mm
Two way shear
Assuming effective depth = 443 mm
Two way shear resistance at a critical section (d/2) from face of
column,
Then, Vu2 = 0.264{(2500*2000)-[(400+d)(300+d)]}
= 0.264{(2500*2000)-[(400+443)(300+443)]}
= 1154.6 kN
Two way shear resistance, Vc2 is computed on,
Vc2 = Ks τc [2(400+d)+2(300+d)]d
Ks = 1
τc = 0.25√k
= 1.118 N/mm2
Hence, Vc2 = 1565.2d+4.47d2
Equating both,
1154644=1565.2d+4.47 d2
d= 362 mm
Therefore one way shear is critical
Adopt effective depth = 450mm
Overall depth = 550mm
Ultimate moment at column face = 0.264*2000*10502/2
= 291.06 kNm
291.06*106 = 0.87fyAstd[1-(Astfy/bdfck)]
291.06*106 = 162472.5 Ast – 7.49 Ast2
Ast = 1970.4mm2
Pt = 100 Ast/bd
=(100*1970.4)/(1000*450)
= 0.43>0.25
Assuming 60mm dia bars,
Spacing = 1000ast/ Ast
=1000*π*162/(4*1970.4)
=105mm
Hence, provide 16mm dia bars @100mm c/c in both directions.
FOOTING FOR AXIALLY LOADED COLUMN
Column Reinforcement
16mm dia bars @ 100mm c/c (both ways)
GL
Figure 15-Reinforcement details of axially loaded column
CHAPTER 6
6. CO�CLUSIO�
The analysis and design of the structural components of the
college auditorium envisaged planning for each floor of the building with
detailed analyses of Beams, Columns, Slabs and Stairs. Isolated footings for
Columns were considered. This work throws an insight into the structural
components of the proposed college auditorium which will be constructed
soon.
REFERE�CES
1. “Advanced Reinforced Concrete Design”, by N.Krishna Raju.
2. “Strength of Materials”, by Ramamirtham and Narayanan.
3. “Reinforced Concrete Design”, by P.P.Vargheese
4. IS:875 part , “Code of Practice for design loads for buildings and
structures – Dead Loads”.
5. IS:875 part , “Code of Practice for design loads for buildings and
structures – Live Loads”.
6. IS:875 part , “Code of Practice for design loads for buildings and
structures – Wind Loads”.
7. IS:456: 2000, “Plain and Reinforced Concrete - Code of Practice”.
8. “Design of Concrete Structures”, by Shah
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