DESIG OF PROPOSED AUDITORIUM ACKOWLEDGEMET
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 COTETS CHAPTER O.
TITLE
PAGE O.
ABSTRACT LIST OF SYMBOLS LIST OF FIGURES
1.
ITRODUCTIO 1.1 GENERAL
1
1.2
OBJECTIVES
2
1.3
DESIGN OF RC STRUCTURES
2
1.3.1 1.4
SLABS 1.4.1
1.5
1.7
2 3 3 4 4 5
1.6.1
SHORT COLUMN
5
1.6.2
SLENDER COLUMN
5
1.6.3
CLASSIFICATION OF COLUM
5
FOOTING TYPES OF COLUMN FOOTING
STAIRCASE 1.8.1
2.
DESIGN OF BEAMS
COLUMN
1.7.1 1.8
CLASSIFICATION OF SLABS
BEAMS 1.5.1
1.6
LIMIT STATE DESIGN
CLASSIFICATION OF STAIRS
PLAIG OF PROPOSED AUDITORIUM 2.1
LAYOUT OF SITE
2.2
PLANS
2.3
SECTION
2.4
ELEVATION
6 6 7 7
3
METHODOLOGY
4
3.1
LIMIT STATE DESIGN
3.2
PARTIAL SAFETY FACTOR
AALYSIS 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
6
FOOTING 5.5.1
TYPES OF FOOTING
5.5.2
DESIGN OF FOOTING
COCLUSIO REFERECE
LIST OF FIGURES
Serial o. figure 1 figure 2 figure 3 figure 4 figure 5 figure 6 figure 7 figure 8 figure 9 figure 10 figure 11 figure 12 figure 13 figure 14 figure 15
Title
Page o.
analysis using substitute frame method - frame1 ................... 17 analysis using substitute frame method-frame2 ..................... 31 analysis using substitute frame method-frame3 ..................... 45 analysis using substitute frame method-frame4 ..................... 59 reinforcement details of two way slab-section ..................... 105 reinforcement details of two way slab- plan ......................... 105 reinforcement details of l-beams- longitudinal section ........ 111 reinforcement details of l-beams- cross section.................... 111 reinforcement details of t-beam-longitudinal design ............ 116 cross section of t-beam ......................................................... 116 reinforcement details of doglegged staircase........................ 122 reinforcement details of axially loaded column ................... 126 reinforcement details of uniaxially loaded column .............. 129 reinforcement details of biaxially loaded columns ............... 132 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. ITRODUCTIO
1.1 GEERAL 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 COLUMS • 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 AALYSIS 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.
4.3AALYSIS OF FRAMES
A
B
C
D
E
F
FRAME 1 Figure 1-Analysis using substitute frame method - frame1
G
4.3.1DISTRIBUTIO FACTOR
Joint A
B
C
D
E
F
G
Member
Relative stiffness
AB
I/3.2
A1
I/4
A2
I/4
0.31
BA
I/3.2
0.28
BC
I/3.2
B1
I/4
0.22
B2
I/4
0.22
CB
I/3.2
0.32
CD
I/6.4
C1
I/4
0.26
C2
I/4
0.26
DC
I/6.4
0.32
DE
I/6.4
D1
I/4
0.31
D2
I/4
0.31
ED
I/6.4
0.32
EF
I/6.4
E1
I/4
0.31
E2
I/4
0.31
FE
I/6.4
0.32
FG
I/6.4
F1
I/4
0.31
F2
I/4
0.31
GF
I/6.4
0.24
G1
I/4
G2
I/4
4.3.2 LOAD CALCLATIO
Total stiffness
Stiffness for each member 0.38
0.8125I
1.125I
0.9688I
0.8125I
0.8125I
0.8125 I
0.656I
0.31
0.28
0.16
0.32
0.32
0.32
0.38 0.38
Beam name
AB
BC
CD
DE
EF
FG
8.25
8.25
8.25
8.25
8.25
8.25
15
15
15
15
15
15
2.56
2.56
7.99
7.99
7.99
7.99
Length m
3.2
3.2
6.4
6.4
6.4
6.4
Total DL kN/ m
6.6
6.6
10.3
10.3
10.3
10.3
Total LL kN /m
12
12
18.73
18.73
18.73
18.73
Dead load due to rib kN/m2
3.375
3.375
4.5
4.5
4.5
4.5
Total DL kN/m2
14.975
14.975
17.23
17.23
17.23
17.23
Total load kN/m2
26.975
26.975
35.96
35.96
35.96
35.96
DL kN / LL
m2
kN/m2
Area
m2
4.3.3 MOMET CALCULATIO AT JOITS JOINT-A joint Member
A AB
B BA
DF
0.38
0.28
FEM due to DL
BC 0.28 -12.78
FEM due to LL
-23.02
Distribution + Carryover
-1.4336
Distribution
-24.45 9.29
Net BM due to A
-15.16
+23.02
C CB CD 0.32
0.16
D DC 0.32
DE 0.32
JOINT-B joint Member
A AB
B BA
DF
0.38
0.28
C CB CD
BC 0.28
0.32
DC
0.16
FEM due to DL FEM due to LL
D DE
0.32
0.32
-58.81 -23.02
+23.02
-23.02
+23.02
Distribution + Carryover
4.37
5.726
Distribution
27.39 -2.827
-17.294 -2.827
Net BM due to A
24.563
-20.121
JOINT-C joint Member
B BA
DF
0.28
FEM due to DL
12.78
FEM due to LL
BC 0.28
C CB 0.32
D CD 0.16
DC 0.32
DE 0.32 -58.81
-122.74 +122.74 -122.74
Distributio n+ Carryover
15.39
-10.23
Distributio n
138.13 -1.6512
-132.97 -0.8256
Net BM due to A
136.48
-133.8
122.74
E ED EF 0.32
0.32
JOINT-D joint Member DF
B
C
BA BC CB CD 0.28
0.28
FEM due to DL
0.32
0.16
D DC 0.32
E DE
ED
0.32
0.32
12.78
FEM due to LL
EF 0.32 -58.81
-122.74 +122.74 -122.74 122.74
Distributi on + Carryove r
8.8
-10.23
Distributi on
131.54 0.46
-132.97 0.46
Net BM due to A
132
-132.51
JOINT-E joint Member
D DC
DF
0.32
FEM due to DL
+58.81
FEM due to LL
DE 0.32
ED 0.32
E EF
F FE
0.32
0.32
FG 0.32 -58.81
-122.74
122.74
-122.74 122.74
Distributi on + Carryove r
10.23
-10.23
Distributi on
132.97 0
-132.97 0
Net BM due to A
132.97
-132.97
G GF 0.24
JOINT-F joint Member DF
D DC 0.32
DE 0.32
FEM due to DL
ED 0.32
E EF
F FE
0.32
0.32
FG
G GF
0.32
0.24
-122.74
122.74
58.81
FEM due to LL
-122.74 122.74
Distributi on + Carryove r
10.23
-14.73
Distributi on
132.9 1.44
-132.97 1.44
Net BM due to A
134.41
-136.03
JOINT-G joint Member DF FEM due to DL FEM due to LL
D DC 0.32
DE 0.32
ED 0.32
E EF
F FE
0.32
0.32
FG
G GF
0.32
0.24
-122.74
122.74
58.81
Distributi on + Carryove r
10.23
Distributi on
132.97 -31.91
Net BM due to A
101.06
MID SPAN MOMENT AB joint Member DF
A AB
B BA
0.38
0.28
FEM due to DL FEM due to TL
-23.02
23.02
distribution moment
8.75
-2.87
carry over moment
-1.435
4.375
distribtion
0.545
-6.15
Net moment -15.16
C BC
CB
0.28
0.32
-12.78
12.78
D CD
0.16
0.32
-122.74
122.74
35.18 17.59
+18.375
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
DC
DE 0.32
MID SPAN MOMENT OF BC joint
A AB
BA
DF
0.38
0.28
FEM due to DL
-12.78
12.78
Member
B
FEM due to TL
C BC 0.28
CB 0.32
-23.02
23.02
-2.87
11.45
-5.73
-1.43
distribtion
0.924
-2.816
Net moment
-30.696 30.224
distribution moment
4.86
carry over moment
2.43
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
CD
D DC DE
0.16
0.32
-58.81
58.81
0.32
-122.74 20.46 10.23
MID SPAN MOMENT CD Joint member DF
B BA 0.28
FEM due to DL FEM due to TL distributio n moment carry over moment
BC 0.28
C CB CD 0.32
0.16
D DC 0.32
-12.78 12.78 23.02
E DE 0.32
ED 0.32
0.32
-58.81 58.81 -122.74 122.74
-2.86
EF
17.59
-1.43 -10.23
-122.74
-20.46
8.795
distributio n
1.86
Net moment
-95.923 104.99
20.46
10.23
-6.088
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 Joint member DF
B BA 0.28
FEM due to DL FEM due to TL
C BC 0.28
CB 0.32
D CD
0.16
DC 0.32
-58.81 58.81 122.74
distributio -20.46 n moment
E DE 0.32
ED EF 0.32
-58.81 58.81 -122.74 122.74
-20.46
carry over moment
20.46
-10.23 -10.23
-122.74
-20.46
10.23
distribtion
6.54
Net moment
-105.97 105.97
-6.54
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
0.32
20.46
10.23
MID SPAN MOMENT EF joint
C
Member
D
DC 0.32
DF FEM due to DL
FEM due 122.74 to TL distributio n moment
DE
ED
0.32
0.32
-58.81
58.81
E EF
FE
0.32
0.32
G FG
GF
0.32
0.24
-58.81
58.81
-122.74 122.74 -20.46
20.46
-20.46
-10.23
10.23
Distribtion
6.55
-1.001
Net moment
-105.96 111.51
carry over moment
-10.23
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
-14.11 -7.1
MID SPAN MOMENT FG joint
D
Member
DC 0.32
DF
E DE 0.32
ED 0.32
FEM due to DL FEM due to TL distributio n moment
F EF
FE
0.32
0.32
-58.81
58.81
122.74
G FG
0.32
GF 0.24
-122.74 122.74 -20.46
20.46
-29.46
-14.73
10.23
distribtion
7.98
-2.455
Net moment
-109.03 101 05
carry over moment
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
-10.23
NEGETIVE MOMENT AT CENTRE OF DE Joint
C
member
CB
DF
0.32
FEM due to DL
12.78
D CD
0.16
DC 0.32
FEM due to TL
-122.74 122.74
distributio n moment
17.59
DE
E ED EF
0.32
0.32
-58.81
58.81
0.32
0.32
-122.74 122.74 -20.46
20.46
8.795
10.23
-10.23 -10.23
distribtion
-6.08
6.55
Net moment
-75.12
75.59
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
FE
FG 0.32 -58.81
-20.46
carry over moment
F
20.46
-20.46
4.3.4 BENDING MOMENT IN COLUMN LOADING joint
A
B
C
D
E
F
G
Column DF Above floor
0.31
0.22
0.26
0.3
0.3
0.3
0.38
Below floor
0.31
0.22
0.26
0.3
0.3
0.3
0.38
Member
AB
BA
BC
CB
CD
DC
DE
ED
EF
FE
FG
GF
DF
0.38
0.28
0.25
0.32
0.16
0.19
0.19
0.19
0.19
0.19
0.19
0.24
-12.8
12. 8
-58.8
58.8
-58.8
58.8
FEM due to DL FEM due TL
-23.02
23.02
Distribution+
-1.435
4.37
17.59
-24.45
27.39
4.81
-122.7
122.7
-1.28
-6.07
8.79
6.07
11.5
-128.8
131.53
-52.7
-122.7
122.7
-6.07
-6.07
6.07
-7.06
6.07
52.74
-128.8
128.81
-65.9
64.88
carryover Total Distribution to column Above floor Distribution to column below floor Joint
7.58
7.084
30.5
-23.64
22.82
-18.882
24.65
7.58
7.084
30.5
-23.64
22.82
-18.882
24.65
A
B
C
D
E
F
G
Column DF Above floor
0.31
0.22
0.26
0.3
0.3
0.3
0.38
Below floor
0.31
0.22
0.26
0.3
0.3
0.3
0.38
Member
AB
BA
BC
CB
CD
DC
DE
ED
EF
FE
FG
GF
DF
0.38
0.28
0.25
0.32
0.16
0.19
0.19
0.19
0.19
0.19
0.19
0.24
FEM due to
-12.8
12. 8
-58.8
58.8
-58.8
58.81 -122.7
122.7
DL FEM due TL
-
23.02
-122.7
122.7
23.02 Distribution+
1.43
2.43
5.73
1.28
-6.07
2.86
-6.07
6.07
6.07
-6.07
-14.73
6.07
-11.4
15.21
-17.3
24.3
-64.9
61.7
-128.8
128.8
-52.7
52.7
-137.5
128.8
carryover Total Distribution to column Above floor Distribution to column below floor
3.52
0.46
10.55
20.14
22.82
25.42
-48.9
3.52
0.46
10.55
20.14
22.82
25.42
-48.9
H
I
J
K
L
M
FRAME 2
Figure 2-Analysis using substitute frame method-frame2
N
4.3.5 DISTRIBUTIO FACTOR Joint H
I
J
K
L
M
N
Member
Relative stiffness
Total stiffness
Stiffness for each member
HI
I/3.72
H1
I/4
H2
I/4
0.33
IH
I/3.72
0.26
IJ
I/3.55
I1
I/4
0.24
I2
I/4
0.24
JI
I/3.55
0.3
JK
I/6.33
J1
I/4
0.27
J2
I/4
0.27
KJ
I/6.33
0.2
KL
I/6.43
K1
I/4
0.31
K2
I/4
0.31
LK
I/6.43
0.17
LM
I/3.55
L1
I/4
0.27
L2
I/4
0.27
ML
I/3.55
0.27
MN
I/3.72
M1
I/4
0.24
M2
I/4
0.24
NM
I/3.72
0.35
N1
I/4
N2
I/4
0.35 0.769I
1.05I
0.94I
0.81I
0.937I
1.05 I
0.769I
0.33
0.27
0.17
0.19
0.3
0.26
0.33 0.33
4.3. 6 LOAD CALCLATIO Beam name
HI
IJ
8.25
8.25
8.25
8.25
8.25
8.25
15
15
15
15
15
15
6.784
6.24
20.04
20.672
6.24
6.784
Length m
3.72
3.55
6.33
6.43
3.55
3.72
Total DL kN/ m
15.045
14.5
26.12
26.52
14.5
15.045
Total LL kN /m
27.35
26.37
47.49
48.22
26.37
27.35
Dead load due to rib kN/m2
3.375
3.375
4.5
4.5
3.375
3.375
Total DL kN/m2
18.42
17.875
30.62
31.02
17.875
18.42
Total load kN/m2
45.77
44.245
78.11
79.24
44.245
45.77
DL kN / LL
m2
kN/m2
Area
m2
JK
KL
LM
MN
4.3.7 MOMET CALCULATIO AT JOITS JOINT-H joint Member DF
H HI
I IH
IJ
0.35
0.26
0.27
FEM due to DL
J
-18.77
FEM due to LL
-52.78
Distribution + Carryover
-4.42
Distribution
-57.2 20.02
Net BM due to H
-37.18
+52.78
JI 0.3
K JK
KJ
0.17
0.2
KL 0.19
JOINT-I joint Member DF
H HI
I
J
IH
IJ
JI
JK
0.35
0.26
0.27
0.3
0.17
FEM due to DL FEM due to LL
K KJ 0.2
KL 0.19
-102.24 -52.78
+52.78
-46.47
46.47
Distribution + Carryover
9.237
8.366
Distribution
62.02 -6.22
-38.104 -6.46
Net BM due to I
55.8
-44.564
JOINT-J joint Member
I
J
K
IH
IJ
JI
JK
KJ
DF
0.26
0.27
0.3
0.17
0.2
FEM due to DL
102.24
FEM due to LL
L KL
0.19 -102.24
-46.47 46.47
-260.81
Distributio n+ Carryover
-7.53
-15.39
Distributio n
38.94 71.18
-276.2 40.33
Net BM due to J
110.12
-235.87
260.81
LK LM 0.17 0.3
JOINT-K joint Member DF
I
J
K
IH
IJ
JI
JK
KJ
0.26
0.27
0.3
0.17
0.2
FEM due to DL
L KL 0.19
LK
LM
0.17
0.3
18.77
FEM due to LL
-58.81 -260.81 260.81
-273.01 273.01
Distributi on + Carryove r
20.57
-21.61
Distributi on
281.38 2.648
-294.62 2.516
Net BM due to K
284.03
-292.10
JOINT-L joint Member
K KJ
DF
0.2
FEM due to DL
+102.2 4
FEM due to LL
KL
LK
L LM
0.19
0.17
0.3
M ML
MN
N NM
0.27
0.26
0.35
-21.24 -273.01
273.01
-46.47
Distributi on + Carryove r
16.22
-3.41
Distributi on
298.23 -40.67
-49.88 -71.77
Net BM due to L
248.56
-121.65
46.47
JOINT-M joint
K KL
LK
L LM
0.19
0.17
0.3
0.27
0.26
0.35
-46.47
46.47
-52.78
52.78
Distributi on + Carryove r
-9.06
-15.83
Distributi on
37.41 8.424
-68.61 8.112
Net BM due to M
45.834
-60.5
Member DF
KJ 0.2
FEM due to DL
M ML MN
N NM
106.88
FEM due to LL
JOINT-N joint Member DF FEM due to DL FEM due to LL
KJ 0.2
K KL 0.19
LK
L LM
0.17
0.3
M ML MN
N NM
0.27
0.26
0.35
-52.78
52.78
18.77
Distribution + Carryover
4.42
Distribution
57.2 -17.16
Net BM due to N
40.04
MID SPAN MOMENT HI joint Member DF
H HI
I IH
IJ
0.35
0.26
0.27
0.3
-18.77
18.77
FEM due to DL FEM due to TL
-52.78
52.78
distribution moment
18.47
-8.99
carry over moment
-4.495
9.235
distribution
1.573
-11.84
Net moment -37.23
J JI
JK
KJ
0.17
0.2
-260.81 72.6 36.3
41.18
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
K
260.81
KL 0.19
MID SPAN MOMENT OF IJ joint
H HI
IH
IJ
DF
0.35
0.26
0.27
FEM due to DL
-102.24
102.24
Member
I
FEM due to TL
J JI 0.3
46.47
-15.06
16.73
8.36
-7.53
distribution
-7.09
-2.87
Net moment
-60.26
52.8
35.78
carry over moment
17.89
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
JK
KJ
0.17
0.2
KL 0.19
-102.24 102.24 -46.47
distribution moment
K
-273.01 34.15 17.1
MID SPAN MOMENT JK Joint member DF
I
distribution moment
K
IH
IJ
JI
JK
KJ
0.26
0.27
0.3
0.17
0.2
FEM due to DL FEM due to TL
J
-18.77 18.77 52.78
KL LK 0.19
0.17
carry over moment
41.15
-4.59 -15.4
-46.47
-30.79
20.6
distribution
3.4
Net moment
-231.66 247.52
-3.1
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
0.3
-106.9 106.9 -260.81 260.81
-9.18
L LM
-10.27
-5.135
MID SPAN MOMENT KL Joint
I
member DF
J
K
IH
IJ
JI
JK
KJ
0.26
0.27
0.3
0.17
0.2
FEM due to DL
-102.24 102.24
FEM due to TL
46.47
distribution moment
L KL LK
0.19
0.17
34.15
-43.22
-21.61
17.07
distribution
3.21
-3.68
Net moment
-257.26 243.18
carry over moment
4.74
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
0.3
-18.77 18.77 -273.01 273.01
9.48
LM
-52.78 9.18
4.59
MID SPAN MOMENT LM joint
K
Member
KJ
DF FEM due to DL
0.2
FEM due to TL
260.81
distribution moment
L
M
N
KL
LK
LM
ML
MN
NM
0.19
0.17
0.3
0.27
0.26
0.35
-106.88
106.88
-21.24
21.24
-46.47 -29.25
46.47
-18.123 -6.812 -3.41
-9.06
Distribution
5.412
3.45
Net moment
-62.591 34.05
carry over moment
-14.63
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
-7.434 -3.717
MID SPAN MOMENT MN joint
K
Member
KJ
L
M
N
KL
LK
LM
ML
MN
NM
0.19
0.17
0.3
0.27
0.26
0.35
-18.77
18.77 -52.78
52.78
8.843
-18.5
-9.25
4.42
distribution
12.32
-1.55
Net moment
-40.87
37.15
0.2
DF FEM due to DL FEM due to TL distribution moment
273.01 -76.3
carry over moment
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
-38.15
NEGETIVE MOMENT AT CENTRE OF KL Joint
J
member
JI
JK
DF
0.3
0.17
FEM due to DL
KJ
K KL
LK
L LM
M ML MN
0.19
0.17
0.3
0.27
-106.9
106.9
0.2
18.77
FEM due to TL
-260.81 260.81
distributio n moment
41.15
-21.24 -41.47
41.47 -6.81
-30.79
-29.25
-10.27 -18.12
20.6
-9.06
-15.4
distribtion
-2.19
3.2
Net moment
-147.4
84.41
carry over moment
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
0.26
-3.4
4.3.8 BENDING MOMENT IN COLUMN LOADING joint
H
I
J
K
L
Above floor
0.33
Below floor
0.33
Member
HI
IH
IJ
JI
JK
KJ
KL
LK
DF
0.35
0.26
0.27
0.3
0.17
0.19
0.19
0.17
-18.8
18.8
-106.9
106.9
M
N
0.24
0.27
0.31
0.27
0.24
0.33
0.24
0.27
0.31
0.27
0.24
0.33
Column DF
FEM due to
LM 0.3
ML
MN
0.27
NM
0.26
0.35
-21.24
21.24
DL FEM due TL
-52.78
52.78
Distribution+
-4.421
9.24
36.31
-57.2
62.02
17.51
-260.8
260.8
-4.56
-14.62
20.57
-5.13
14.18
-275.4
281.4
-112.0
-46.5
46.5
-14.6
-3.41
-9.06
-3.72
-3.28
92.26
-49.9
37.41
-24.96
17.96
carryover Total Distribution to column Above floor Distribution to column below floor
18.88
-19.09
70.54
-52.50
-11.44
-2.988
-5.93
18.88
-19.09
70.54
-52.50
-11.44
-2.988
-5.93
H
I
J
K
L
Above floor
0.33
0.24
0.27
0.31
0.27
0.24
0.33
Below floor
0.33
0.24
0.27
0.31
0.27
0.24
0.33
Member
HI
IH
IJ
JI
JK
KJ
KL
LK
DF
0.35
0.26
0.27
0.3
0.17
0.19
0.19
0.17
FEM due to
-21.24
21.24
-
102.2
102.24
4
joint
M
N
Column DF
DL FEM due TL Distribution+
-46.5
46.5
-273
273
LM
ML
0.3
0.27
-18.77
18.77
MN
NM
0.26
0.35
-52.78
52.78
3.28
3.72
8.37
3.41
16.22
4.74
-21.61
16.22
4.59
-38.2
-9.24
4.42
-17.96
24.96
-38.1
49.88
-86.02
106.9
-294.6
289.2
-14.18
-19.4
-62.02
57.2
carryover Total Distribution to column Above floor Distribution to column below floor
5.93
3.15
9.76
58.17
-74.26
19.53
-18.9
5.93
3.15
9.76
58.17
-74.26
19.53
-18.9
O
P
Q
R
S
T
FRAME 3
Figure 3-Analysis using substitute frame method-frame3
4.3.9 DISTRIBUTIO FACTOR
U
Joint O
P
Q
R
S
T
U
Member
Relative stiffness
OP
I/3.72
O1
I/4
O2
I/4
0.33
PO
I/3.72
0.26
PQ
I/3.55
P1
I/4
0.24
P2
I/4
0.24
QP
I/3.55
0.3
QR
I/6.33
Q1
I/4
0.27
Q2
I/4
0.27
RQ
I/6.33
0.19
RS
I/6.43
R1
I/4
0.31
R2
I/4
0.31
SR
I/6.43
0.17
ST
I/3.55
S1
I/4
0.27
S2
I/4
0.27
TS
I/3.55
0.27
TU
I/3.72
T1
I/4
0.24
T2
I/4
0.24
UT
I/3.72
0.35
U1
I/4
U2
I/4
4.3.10 LOAD CALCLATIO
Total stiffness
Stiffness for each member 0.35
0.769I
1.05I
0.94I
0.81I
0.937I
1.05 I
0.769I
0.33
0.27
0.17
0.19
0.3
0.26
0.33 0.33
Beam name DL KN /
m2
OP
PQ
QR
RS
ST
TU
18.27
8.25
8.25
8.25
8.25
18.27
3
15
15
15
15
3
8.228
6.095
15.39
15.8
6.095
8.228
Length m
3.72
3.55
6.33
6.43
3.55
3.72
Total DL KN/ m
40.41
14.16
20.06
20.27
14.16
40.41
Total LL KN /m
6.64
25.75
36.47
36.86
25.75
6.64
Dead load due to rib KN/m2
3.375
3.375
4.5
4.5
3.375
3.375
Total DL KN/m2
48.085
22.045
27
27.19
22.045
48.085
Total load KN/m2
54.725
47.795
63.46
64.05
47.795
54.725
LL
KN/m2
Area
m2
4 .3.11 MOMET CALCULATIO AT JOITS
JOINT-O joint
O
Member
OP
PO
PQ
QP
QR
RQ
RS
DF
0.35
0.26
0.27
0.3
0.17
0.19
0.19
P
FEM due to DL
Q
-23.15
FEM due to LL
-63.12
Distribution + Carryover
-5.2
Distribution
-68.32 23.912
Net BM due to O
-44.41
+63.12
R
JOINT-P joint
O
Member
OP
PO
PQ
QP
QR
RQ
RS
DF
0.35
0.26
0.27
0.3
0.17
0.19
0.19
P
Q
FEM due to DL FEM due to LL
R
-90.16 -63.12
+63.12
-50.19
50.19
Distribution + Carryover
11.046
5.996
Distribution
74.166 -7.79
-44.194 -8.09
Net BM due to P
66.376
-52.284
JOINT-Q joint
O
Member
OP
PO
PQ
QP
QR
RQ
RS
DF
0.35
0.26
0.27
0.3
0.17
0.19
0.19
FEM due to DL FEM due to LL
P
Q
R
55.45
-93.68 -50.19
50.19
-211.9
Distribution + Carryover
-0.71
-11.23
Distribution
49.48 52.095
-223.13 29.521
Net BM due to Q
101.575
-193.61
211.9
JOINT-R joint
Q
R
S
T
Member
QP
QR
RQ
RS
SR
ST
TS
TU
DF
0.3
0.17
0.19
0.19
0.17
0.3
0.27
0.26
FEM due to DL
23.15
FEM due to LL
-23.15 -211.9
211.9
-220.68
Distributio n+ Carryover
16.04
-16.79
Distributio n
227.944 -237.47 1.81 1.81
Net BM due to R
229.754 -235.66
+220.68
JOINT-S joint
Q
R
S
T
Member
QP
QR
RQ
RS
SR
ST
TS
TU
DF
0.3
0.17
0.19
0.19
0.17
0.3
0.27
0.26
FEM due to DL FEM due to LL
90.16
-55.45 -220.68 +220.68 -50.19
Distributio n+ Carryover
12.4
0.71
Distributio n
233.08 -31.212
-49.48 -55.08
Net BM due to S
201.87
-104.56
50.19
JOINT-T joint
R
S
T
U
Member
RQ
RS
SR
ST
TS
TU
UT
DF
0.19
0.19
0.17
0.3
0.27
0.26
0.35
-50.19
50.19
-63.12
63.12
Distributi on + Carryove r
-13.05
-11.05
Distributi on
37.14 9.99
-74.17 9.63
Net BM due to T
47.13
-64.54
FEM due to DL
93.68
FEM due to LL
JOINT U joint
R
S
T
U
Member
RQ
RS
SR
ST
TS
TU
UT
DF
0.19
0.19
0.17
0.3
0.27
0.26
0.35
-63.12
63.12
FEM due to DL FEM due to LL
23.15
Distributi on + Carryove r
5.2
Distributi on
68.32 -23.912
Net BM due to U
44.41
MID SPAN MOMENT OP joint
O
Member
OP
PO
PQ
QP
QR
RQ
RS
DF
0.35
0.26
0.27
0.3
0.17
0.19
0.19
-23.15
23.15 -211.9
211.9
P
FEM due to DL FEM due to TL
-63.12
63.12
distribution moment
22.1
-10.39
carry over moment
-5.195
11.05
distribution
1.818
-10.23
Net moment -44.4
Q
R
56.6 28.3
53.55
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 joint
O
Member
OP
PO
PQ
QP
QR
RQ
RS
DF
0.35
0.26
0.27
0.3
0.17
0.19
0.19
FEM due to DL
-55.45
55.45
-90.16
90.16
P
FEM due to TL
Q
-50.19
50.19
-1.42
11.99
5.99
-0.71
distribution
-4.24
-3.5
Net moment
-49.86
57.97
distribution moment carry over moment
19.41 9.71
R
-220.68 20.46 12.35
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 Joint
P
Q
R
S
member
PO
PQ
QP
QR
RQ
RS
SR
ST
DF
0.26
0.27
0.3
0.17
0.19
0.19
0.17
0.3
FEM due to DL FEM due to TL
-23.15 23.15 -211.9
211.9
32.1
-22.46
-11.23
16.05
distributio n
2.83
-2.35
Net moment
-188.2
203.14
distributio n moment carry over moment
63.12
-93.68 93.68
-10.79
-5.4
-50.19 -7.4
-3.7
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 Joint
Q
R
S
T
member
QP
QR
RQ
RS
SR
ST
TS
TU
DF
0.3
0.17
0.19
0.19
0.17
0.3
0.27
0.26
FEM due to DL FEM due to TL distributio n moment
-90.16 90.16 50.19
-23.15 23.15 -220.68 220.68 24.8
-33.58
-16.79
12.4
distributio n
2.54
-3.03
Net moment
-210.13 196.25
carry over moment
6.8 3.4
-63.12 10.8 5.4
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 joint
R
S
T
U
Member
RQ
RS
SR
ST
TS
TU
UT
DF
0.19
0.19
0.17
0.3
0.27
0.26
0.35
-93.68
93.68
-55.45
55.45
FEM due to DL
FEM due to 211.9 TL -22.46 distribution moment
-50.19
50.19
13.05
1.42
0.71
6.53
Distribution
3.16
0.86
Net moment
-33.27
59
carry over moment
-11.23
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
-19.41 -9.71
MID SPAN MOMENT TU joint
R
S
T
U
Member
RQ
RS
SR
ST
TS
TU
UT
DF
0.19
0.19
0.17
0.3
0.27
0.26
0.35
-23.15
23.15 -63.12
63.12
10.39
-22.1
-11.05
5.2
distribution
10.6
-1.82
Net moment
-53.18
44.4
FEM due to DL FEM due to TL distribution moment
220.68 -59.3 -29.65
carry over moment
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 Joint
Q
R
S
T
member
QP
QR
RQ
RS
SR
ST
TS
TU
DF
0.3
0.17
0.19
0.19
0.17
0.3
0.27
0.26
FEM due to DL
23.15
-93.68
93.68
FEM due to TL
-211.9 211.9
distributio n moment
32.1
-55.45 -50.19
50.19
-13.05
1.42
-22.46
22.46
-7.4
-16.05
-3.7
-11.24 0.71
Distributi on
-2.45
1.79
Net moment
-122.29
76.83
carry over moment
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
0.33
0.24
0.27
0.31
0.27
0.24
0.33
Below floor
0.33
0.24
0.27
0.31
0.27
0.24
0.33
Member
OP
PO
PQ
QP
QR
RQ
RS
SR
DF
0.35
0.26
0.27
0.3
0.17
0.19
0.19
0.17
-23.2
23.2
-93.68
93.68
FEM due to
ST 0.3
TS
TU
UT
0.27
0.26
0.35
-55.45
55.45
DL FEM due TL
-63.12
63.12
Distribution+
-5.2
11.05
28.31
-68.32
74.17
-5.16
-211.9
211.9
-5.4
-11.23
16.04
-3.7
17.75
-223.1
227.9
-97.38
-50.19
50.19
-11.2
0.71
-6.52
-9.7
0.68
82.45
-49.48
43.67
-65.15
56.13
carryover Total Distribution to column Above floor Distribution to column below floor
joint
22.55
-16.56
55.45
-40.47
-8.9
5.15
-18.5
22.55
-16.56
55.45
-40.47
-8.9
5.15
-18.5
P
Q
R
O
S
T
U
Column DF Above floor
0.33
0.24
0.27
0.31
0.27
0.24
0.33
Below floor
0.33
0.24
0.27
0.31
0.27
0.24
0.33
Member
OP
PO
PQ
DF
0.35
0.26
0.27
FEM due to
-55.45
55.45
QP 0.3
QR
RQ
RS
SR
0.17
0.19
0.19
0.17
-90.16
90.16
ST
TS
TU
UT
0.3
0.27
0.26
0.35
-23.2
23.2 -63.12
63.12
DL FEM due TL Distribution+
-50.2
50.2
-220.7
220.7
-0.68
9.7
5.6
-0.71
12.4
3.4
-16.79
12.4
5.4
-29.6
-11.05
5.2
-56.13
65.15
-44.2
49.48
-77.76
93.56
-237.5
233.1
-17.75
-6.48
-74.17
68.32
carryover Total Distribution to column Above floor Distribution to column below floor
18.52
-5.03
7.63
44.61
-58.14
19.36
-22.6
18.52
-5.03
7.63
44.61
-58.14
19.36
-22.6
A1
B1
C1
D1
E1
F1
G1
H1
I1
J1
K1
L1
M1
N1
O1
P1
Q1
FRAME 4
Figure 4-Analysis using substitute frame method-frame4
R1
S1
T1
U1
4.3.13 DISTRIBUTIO FACTOR
Joint A1 B1
C1
D1
E1
F1
G1
H1
I1
J1
Member A1B1 A11 A12 B1A1 B1C1 B11 B12 C1B1 C1D1 C1 1 C1 2 D1C1 D1E1 D11 D12 E1D1 E1F1 E1 1 E1 2 F1E1 F1G1 F11 F12 G1F1 G1H1 G11 G12 H1G1 H1I1 H11 H12 I1H1 I1J1 I11 I12 J1I1 J1K1 J1 1 J1 2 K1J1
Relative stiffness I/3.2 I/4 I/4 I/3.2 I/3.2 I/4 I/4 I/3.2 I/3.2 I/4 I/4 I/3.2 I/3.2 I/4 I/4 I/3.2 I/3.2 I/4 I/4 I/3.2 I/3.2 I/4 I/4 I/3.2 I/3.2 I/4 I/4 I/3.2 I/3.2 I/4 I/4 I/3.2 I/3.2 I/4 I/4 I/3.2 I/3.2 I/4 I/4 I/3.2
Total stiffness 0.8125I 1.125I
1.125I
1.125I
1.125I
1.125I
1.125I
1.125I
1.125I
1.125I
Stiffness for each member 0.38 0.31 0.31 0.28 0.28 0.22 0.22 0.28 0.28 0.22 0.22 0.28 0.28 0.22 0.22 0.28 0.28 0.22 0.22 0.28 0.28 0.22 0.22 0.28 0.28 0.22 0.22 0.28 0.28 0.22 0.22 0.28 0.28 0.22 0.22 0.28 0.28 0.22 0.22 0.28
K1
L1
M1
N1
O1
P1
Q1
R1
S1
T1
U1
K1L1 K11 K12 L1K1 L1M1 L11 L12 M1L1 M1N1 M1 1 M1 2 N1M1 N1O1 N11 N12 O1N1 O1P1 O11 O12 P1O1 P1Q1 P1 1 P1 2 Q1P1 Q1R1 Q11 Q12 R1Q1 R1S1 R1 1 R1 2 S1R1 S1T1 S1 1 S1 2 T1S1 T1U1 T1 1 T1 2 U1T1 U11 U12
I/3.2 I/4 I/4 I/3.2 I/3.2 I/4 I/4 I/3.2 I/3.2 I/4 I/4 I/3.2 I/3.2 I/4 I/4 I/3.2 I/3.2 I/4 I/4 I/3.2 I/3.2 I/4 I/4 I/3.2 I/3.2 I/4 I/4 I/3.2 I/3.2 I/4 I/4 I/3.2 I/6.4 I/4 I/4 I/6.4 I/8.16 I/4 I/4 I/8.16 I/4 I/4
4.3.14 LOAD CALCLATIO
1.125I
1.125I
1.125I
1.125I
1.125I
1.125I
1.125I
1.125I
0.9688I
0.7788I
0.6225I
0.28 0.22 0.22 0.28 0.28 0.22 0.22 0.28 0.28 0.22 0.22 0.28 0.28 0.22 0.22 0.28 0.28 0.22 0.22 0.28 0.28 0.22 0.22 0.28 0.28 0.22 0.22 0.28 0.28 0.22 0.22 0.32 0.16 0.26 0.26 0.2 0.16 0.32 0.32 0.2 0.4 0.4
Beam name
Live load kN/m2 15
Area m2 5.12
Length slab DL m kN/m 3.2 13.2
Total LL kN/m 24
Rib DL kN/m 3.375
Total DL kN/m 16.575
Total load kN/m 40.575
1
A1B1
Dead load kN/m2 8.25
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
No
JOINT-A1 joint
A1
B1
C1
D1
Member
A 1B 1
B 1A 1
B1C1
C1B1
C1D1
D1C1
D1E1
DF
0.38
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL
-12.78
FEM due to LL
-34.624 34.624
Distributio n+ Carryover
-2.868
Distributio n
-37.492 14.247
Net BM due to A1
-23.245
JOINT-B1 joint
A1
B1
C1
D1
Member
A 1B 1
B 1A 1
B1C1
C1B1
C1D1
D1C1
D1E1
DF
0.38
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL FEM due to LL
-14.14 -34.624 34.624
-34.624 34.624
Distribution + Carryover
6.58
-2.87
Distribution
41.204 -37.494 -1.0388 -1.0388
Net BM due to B1
40.165
-38.533
joint
JOINT-C1 C1
B1
D1
E1
Member
B 1A 1
B1C1
C1B1
C1D1
D1C1
D1E1
E1D1
E 1F 1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL
14.14
FEM due to LL
-14.14 -34.62
+34.62 -34.62
Distributio n+ Carryover
2.868
-2.868
Distributio n
37.49 0
-37.49 0
Net BM due to C1
37.49
-37.49
+34.62
JOINT-D1 joint
B1
C1
D1
E1
Member
B 1A 1
B1C1
C1B1
C1D1
D1C1
D1E1
E1D1
E 1F 1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL FEM due to LL
14.14
-58.81 -34.624 +34.62 -34.624 +34.62 4 4
Distributi on + Carryove r
2.868
-2.968
Distributi on
37.492 -37.492 0 0
Net BM due to D1
37.492 -37.492
JOINT-E1 joint
D1
F1
E1
G1
Member
D1C1 D1E1
E1D1
E 1F 1
F 1E 1
F 1G 1
G 1F 1
G1H1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL
14.1 4
FEM due to LL
0.28
-14.14 +34.62 34.624 4
-34.624 +34.624
Distributio n+ Carryover
2.868
-2.968
Distributio n
37.492 0
-37.492 0
Net BM due to E1
37.492
-37.492
JOINT-F1 joint
D1
F1
E1
G1
Member
D1C1
D1E1
E1D1
E 1F 1
F 1E 1
F 1G 1
G 1F 1
G1H1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL FEM due to LL
14.14
-14.14 -34.624 +34.62 -34.624 +34.62 4 4
Distributi on + Carryove r
2.868
-2.968
Distributi on
37.492 -37.492 0 0
Net BM due to F1
37.492 -37.492
JOINT-G1 joint
F1
H1
G1
I1
Member
F 1E 1
F 1G 1
G 1F 1
G1H1
H1G1
H1I1
I1H1
I 1J 1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL
14.14
FEM due to LL
-14.14 +34.624 -34.624 +34.624 34.624
Distributi on + Carryove r
2.868
-2.968
Distributi on
37.492 0
-37.492 0
Net BM due toG1
37.492
-37.492
JOINT-H1 joint
F1
H1
G1
I1
Member
F 1E 1
F 1G 1
G 1F 1
G 1H 1
H1G1
H1I1
I1H1
I 1J 1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL FEM due to LL
14.14
-14.14 -34.624 +34.62 -34.624 +34.62 4 4
Distributi on + Carryove r
2.868
-2.968
Distributi on
37.492 -37.492 0 0
Net BM due to H1
37.492 -37.492
JOINT-I1 joint
H1
J1
I1
K1
Member
H1G1
H1I1
I1H1
I 1J 1
J 1I 1
J 1K 1
K 1J 1
K1L1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL
14.14
FEM due to LL
-14.14 -34.624 34.624
-34.624 34.624
Distributio n+ Carryover
2.868
-2.968
Distributio n
37.492 0
-37.492 0
Net BM due to I1
37.492
-37.492
JOINT-J1 joint
H1
J1
I1
K1
Member
H1G1
H1I1
I1H1
I 1J 1
J 1I 1
J 1K 1
K 1J 1
K1L1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL FEM due to LL
14.14
-14.14 -34.624 +34.62 -34.624 +34.62 4 4
Distributi on + Carryove r
2.868
-2.968
Distributi on
37.492 -37.492 0 0
Net BM due to J1
37.492 -37.492
JOINT-K1 joint Member
J1 J 1I 1
J 1K 1
L1
K1 K 1J 1
K1L1
L1K1
M1 L1M1
M1L1
M 1N 1
DF
0.28
FEM due to DL
14.14
FEM due to LL
0.28
0.28
0.28
0.28
0.28
0.28
0.28
-14.14 -34.624 34.624 -34.624 34.624
Distributio n+ Carryover
2.868
-2.968
Distributio n
37.492 -37.492 0 0
Net BM due to K1
37.492 -37.492
JOINT-L1 joint
J1
L1
K1
M1
Member
J 1I 1
J 1K 1
K 1J 1
K 1L 1
L1K1
L1M1
M1L1
M1N1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL FEM due to LL
14.14
-14.14 -34.624 34.624 -34.624 34.624
Distributio n+ Carryover
2.868
-2.968
Distributio n
37.492 -37.492 0 0
Net BM due to L1
37.492 -37.492
JOINT-M1 joint
L1
N1
M1
O1
Member
L1K1
L1M1
M1L1
M1N1
N1M1
N1O1
O1N1
O 1P 1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL
14.14
FEM due to LL
-14.14 -34.624 34.624 -34.624 34.624
Distributi on + Carryove r
2.868
-2.968
Distributi on
37.492 -37.492 0 0
Net BM due to M1
37.492 -37.492
JOINT-N1 joint
L1
N1
M1
O1
Member
L1K1
L1M1
M1L1
M1N1
N1M1
N1O1
O1N1
O 1P 1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL FEM due to LL
14.14
-14.14 -34.624
34.624 -34.624
Distributi on + Carryove r
2.868
-2.968
Distributi on
37.492 -37.492 0 0
Net BM due to N1
37.492 -37.492
34.624
JOINT-O1 joint
N1
P1
O1
Q1
Member
N1M1
N1O1
O1N1
O 1P 1
P 1O 1
P 1Q 1
Q 1P 1
Q1R1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL
14.14
FEM due to LL
-14.14 -34.624 34.624 -34.624 34.624
Distributi on + Carryove r
2.868
-2.968
Distributi on
37.492 -37.492 0 0
Net BM due to O1
37.492 -37.492
JOINT-P1 joint
N1
P1
O1
Q1
Member
N1M1
N1O1
O1N1
O 1P 1
P 1O 1
P 1Q 1
Q 1P 1
Q1R1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL FEM due to LL
14.14
-14.14 -34.624
34.624 -34.624
Distributi on + Carryove r
2.868
-2.968
Distributi on
37.492 -37.492 0 0
Net BM due to P1
37.492 -37.492
34.624
JOINT-Q1 joint
P1
R1
Q1
S1
Member
P 1O 1
P 1Q 1
Q 1P 1
Q1R1
R1Q1
R1S1
S1R1
S 1T 1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.32
0.16
FEM due to DL
14.14
FEM due to LL
-14.14 -34.624 34.624 -34.624 34.624
Distributi on + Carryove r
2.868
-2.968
Distributi on
37.492 -37.492 0 0
Net BM due to Q1
37.492 -37.492
JOINT-R1 joint
P1
R1
Q1
S1
Member
P 1 O 1 P 1 Q 1 Q 1P 1
Q1R1
R1Q1
R1S1
S1R1
S 1T 1
DF
0.28
0.28
0.28
0.28
0.32
0.16
FEM due to DL FEM due to LL
0.28
0.28 14.14
-86.63 -34.624 34.624
-34.624
Distributi on + Carryove r
2.868
8.32
Distributi on
37.492 -3.13
-26.304 -3.13
Net BM due to R1
34.362
-29.434
34.624
JOINT-S1 joint
R1
S1
Member
R1Q1
R1S1
S 1T 1
S 1T 1
T 1S 1
DF
0.28
0.28
0.32
0.16
0.2
FEM due to DL
14.14
FEM due to LL
U1
T1 T1U1 0.16
U1T1 0.2
-102.04 -34.624
34.624
-216.23 216.23
Distributi on + Carryove r
2.868
-11.42
Distributi on
37.492 60.85
-227.65 30.43
Net BM due to S1
98.342
-197.22
JOINT-T1 joint
R1
S1
T1
U1
R1Q1
R1S1
S 1T 1
S 1T 1
T 1S 1
DF
0.28
0.28
0.32
0.16
0.2
0.16
0.2
-216.23
216.23
-242.2
242.2
Distributio n+ Carryover
16.167
-24.22
Distributio n
232.397 6.81
-266.42 5.44
Net BM due to T1
239.207
-260.98
FEM due to DL FEM due to LL
T1U1
U1T1
Member
14.14
JOINT-U1 joint
R1
S1
T1
Member
R1Q1
R1S1
S 1T 1
S 1T 1
T 1S 1
DF
0.28
0.28
0.32
0.16
0.2
FEM due to DL FEM due to LL
U1 T1U1
U1T1
0.16
0.2
-242.2
242.2
86.63
Distributio n+ Carryover
12.45
Distributio n
254.65 -50.93
Net BM due to U1
203.72
MID SPAN MOMENT A1B1 B1 C1
joint
A1
Member
A 1B 1
B 1A 1
B 1C 1
C1B1
C1D1
D1C1
D1E1
DF
0.38
0.28
0.28
0.28
0.28
0.28
0.28
-14.14
14.14 -34.624
34.624
FEM due to DL FEM due to TL
-34.624
34.624
distribution moment
13.16
-5.74
carry over moment
-2.87
6.58
distribution
1.09
-2.65
Net moment -23.24
D1
5.74 2.87
+32.814
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 joint
A1
Member
A 1B 1
B 1A 1
B 1C 1
C1B1
C1D1
D1C1
D1E1
DF
0.38
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL
-14.14
14.14
-14.14
14.14
B1
FEM due to TL distribution moment
C1
-34.624 34.624 5.74
-5.74
2.87
-2.87
distribution
0.052
-1.61
Net moment
-31.702 30.14
carry over moment
D1
5.37 2.685
-34.624 5.74 2.87
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 Joint
B1
C1
D1
E1
member
B 1A 1
B1C1
C1B1
C1D1
D1C1
D1E1
E1D1
E 1F 1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL FEM due to TL distributio n moment carry over moment
-14.14 14.14 34.62 4
-14.14 14.14 -34.624 34.624
-5.74
5.74
-2.87 -2.87
-34.624
-5.74
2.87
distributio n
-1.6
1.6
Net moment
-33.35
33.35
5.74
2.87
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 Joint
C1
D1
F1
E1
member
C1B1
C1D1
D1C1
D 1E 1
E1D1
E 1F 1
F 1E 1
F 1G 1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL FEM due to TL distribution moment
-14.14 14.14 34.624
-14.14 14.14 -34.624 34.624 5.74
-5.74
-2.87
2.87
distribution
-1.6
1.6
Net moment
-33.35
33.35
carry over moment
-5.74
-2.87
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
-34.624 5.74
2.87
MID SPAN MOMENT E1F1 Joint
D1
F1
E1
G1
member
D1C1
D1E1
E1D1
E 1F 1
F 1E 1
F 1G 1
G 1F 1
G1H1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL FEM due to TL distribution moment
-14.14 14.14 34.624
-14.14 14.14 -34.624 34.624 5.74
-5.74
-2.87
2.87
distribution
-1.6
1.6
Net moment
-33.35
33.35
carry over moment
-5.74 -2.87
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
-34.624 5.74 2.87
MID SPAN MOMENT F1G1 Joint
F1
E1
H1
G1
member
E1D1
E 1F 1
F 1E 1
F 1G 1
G 1F 1
G1H1
H1G1
H1I1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL FEM due to TL distribution moment
-14.14 14.14 34.624
-14.14 14.14 -34.624 34.624 5.74
-5.74
-2.87
2.87
distribution
-1.6
1.6
Net moment
-33.35
33.35
carry over moment
-5.74 -2.87
-34.624 5.74 2.87
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 Joint
F1
H1
G1
I1
member
F 1E 1
F 1G 1
G 1F 1
G1H1
H1G1
H1I1
I1H1
I 1J 1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL FEM due to TL distribution moment
-14.14 14.14 34.624
-14.14 14.14 -34.624 34.624 5.74
-5.74
-2.87
2.87
distribution
-1.6
1.6
Net moment
-33.35
33.35
carry over moment
-5.74
-2.87
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
-34.624 5.74
2.87
MID SPAN MOMENT H1I1 Joint
H1
G1
J1
I1
member
G 1F 1
G1H1
H1G1
H 1I 1
I1H1
I 1J 1
J 1I 1
J 1K 1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL FEM due to TL distribution moment
-14.14 14.14 34.624
-14.14 14.14 -34.624 34.624 5.74
-5.74
-2.87
2.87
distribution
-1.6
1.6
Net moment
-33.35
33.35
carry over moment
-5.74 -2.87
-34.624 5.74 2.87
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 Joint
H1
J1
I1
K1
member
H1G1
H1I1
I1H1
I 1 J1
J 1I 1
J 1K 1
K 1J 1
K1L1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL FEM due to TL distribution moment
-14.14 14.14 34.624
-14.14 14.14 -34.624 34.624 5.74
-5.74
-2.87
2.87
distribution
-1.6
1.6
Net moment
-33.35
33.35
carry over moment
-5.74 -2.87
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
-34.624 5.74 2.87
MID SPAN MOMENT J1K1 Joint
J1
I1
L1
K1
member
I1H1
I 1J 1
J 1I 1
J 1K 1
K 1J 1
K1L1
L1K1
L1M1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL FEM due to TL distribution moment
-14.14 14.14 34.624
-14.14 14.14 -34.624 34.624 5.74
-5.74
-2.87
2.87
distribution
-1.6
1.6
Net moment
-33.35
33.35
carry over moment
-5.74 -2.87
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
-34.624 5.74 2.87
MID SPAN MOMENT K1L1 Joint
J1
L1
K1
M1
member
J 1I 1
J 1K 1
K 1J 1
K1L1
L1K1
L1M1
M1L1
M1N1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL FEM due to TL distribution moment
-14.14 14.14 34.624
-14.14 14.14 -34.624 34.624 5.74
-5.74
-2.87
2.87
distribution
-1.6
1.6
Net moment
-33.35
33.35
carry over moment
-5.74 -2.87
-34.624 5.74 2.87
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 Joint
L1
K1
N1
M1
member
K 1J 1
K1L1
L1K1
L 1M 1
M1L1
M1N1
N1M1
N1O1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL FEM due to TL distribution moment
-14.14 14.14 34.624
-14.14 14.14 -34.624 34.624 5.74
-5.74
-2.87
2.87
distribution
-1.6
1.6
Net moment
-33.35
33.35
carry over moment
-5.74 -2.87
-34.624 5.74 2.87
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 Joint
L1
N1
M1
O1
member
L1K1
L1M1
M1L1
M1N1
N1M1
N1O1
O1N1
O 1P 1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL FEM due to TL distribution moment
-14.14 14.14 34.624
-14.14 14.14 -34.624 34.624 5.74
-5.74
-2.87
2.87
distribution
-1.6
1.6
Net moment
-33.35
33.35
carry over moment
-5.74 -2.87
-34.624 5.74 2.87
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 Joint
N1
M1
P1
O1
member
M1L1
M1N1
N1M1
N 1O 1
O1N1
O 1P 1
P 1O 1
P 1Q 1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL FEM due to TL distribution moment
-14.14 14.14 34.624
-14.14 14.14 -34.624 34.624 5.74
-5.74
-2.87
2.87
distribution
-1.6
1.6
Net moment
-33.35
33.35
carry over moment
-5.74 -2.87
-34.624 5.74 2.87
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 Joint
N1
P1
O1
Q1
member
N1M1
N1O1
O1N1
O 1P 1
P 1O 1
P 1Q 1
Q 1P 1
Q1R1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL FEM due to TL distribution moment
-14.14 14.14 34.624
-14.14 14.14 -34.624 34.624 5.74
-5.74
-2.87
2.87
distribution
-1.6
1.6
Net moment
-33.35
33.35
carry over moment
-5.74 -2.87
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
-34.624 5.74 2.87
MID SPAN MOMENT P1Q1 Joint
P1
O1
R1
Q1
member
O1N1
O 1P 1
P 1O 1
P 1Q 1
Q 1P 1
Q1R1
R1Q1
R1S1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL FEM due to TL distribution moment
-14.14 14.14 34.624
-14.14 14.14 -34.624 34.624 5.74
-5.74
-2.87
2.87
distribution
-1.6
1.6
Net moment
-33.35
33.35
carry over moment
-5.74 -2.87
-34.624 5.74 2.87
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 Joint
P1
R1
Q1
S1
member
P 1O 1
P 1Q 1
Q 1P 1
Q1R1
R1Q1
R1S1
S 1T 1
S 1T 1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.32
0.16
FEM due to DL FEM due to TL distribution moment
-14.14 14.14 34.624
-14.14 14.14 -34.624 34.624 5.74
-5.74
-2.87
2.87
distribution
-1.6
5.33
Net moment
-33.35
37.084
carry over moment
-5.74 -2.87
-216.23 32.33 16.16
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 Joint
R1
Q1
S1
T1
member
Q 1P 1
Q1R1
R1Q1
R 1S 1
S 1T 1
S 1T 1
T 1S 1
DF
0.28
0.28
0.28
0.28
0.32
0.16
0.2
FEM due to DL FEM due to TL distribution moment
-14.14 14.14 34.624
5.74
16.64
8.32
2.87
distribution
1.526
5.896
Net moment
-19.038 60.03
carry over moment
-2.87
-242.2 31.11 15.55
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
0.16
-86.63 86.63 -34.624 34.624
-5.74
T1U1
MID SPAN MOMENT S1T1 joint Member DF
R1 R1Q1 0.28
FEM due to DL FEM due to TL distribution moment
R1S1
S1
T1
U1
S1R1
S1T1
T1S1
T1U1
U1T1
0.28
0.32
0.16
0.2
0.16
0.2
-14.14
14.14
34.624
-102.04 102.04 -216.23 216.23
-5.74
32.33
-22.838
-11.42
16.165
distribution
-2.29
1.19
Net moment
-197.61 210.75
carry over moment
-2.87
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 joint Member
R1
U1
S1T1
T1S1
T1U1
U1T1
0.32
0.16
0.2
0.16
0.2
-86.63
86.63 -43.65
43.65
-688
-8.73
-4.37
-3.44
Distribution
0.48
-0.688
Net moment
-54.42
30.792
0.28
R1S1
T1
S1R1
DF
R1Q1
S1
0.28
FEM due to DL FEM due to TL
40.575
distribution moment
14.74
carry over moment
7.37
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 Joint
J1
L1
K1
M1
member
J 1I 1
J 1K 1
K 1J 1
K1L1
L1K1
L1M1
M1L1
M1N1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to DL
14.14
-14.14 14.14
FEM due to TL
-34.624 34.624
distribution moment
5.74
-14.14 -34.624 34.62 4
-5.74
-5.74
5.74
5.74
2.87
2.87
-2.87
-2.87
distribution
-1.61
1.67
Net moment
-18.62 18.62
carry over moment
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
-5.74
4.3.15 BENDING MOMENT IN COLUMN LOADING joint
A1
B1
C1
D1
E1
F1
G1
Above floor
0.31
0.22
0.22
0.22
0.22
0.22
0.22
Below floor
0.31
0.22
0.22
0.22
0.22
0.22
0.22
Member
A1 B1
B1A1
B1 C1
C1 B1
C1 D1
D1 C1
D1 E 1
E 1D 1
E1 F1
F1E1
F1 G1
G1F1
G 1H 1
DF
0.39
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
-14.2
14.2
-14.2
14.2
-14.2
14.2
Column DF
FEM due to DL FEM due TL
-34.62
34.62
Distribution+
-2.87
6.75
2.87
-37.49
41.37
-11.3
-34.62
34.62
-2.87
-2.87
2.87
2.87
11.3
-37.49
37.49
-11.3
-34.62
34.62
-34.62
-2.87
-2.87
2.87
2.87
-2.87
-2.87
11.3
-37.49
37.49
-11.3
11.3
-37.49
carryover Total Distribution to column Above floor Distribution to column below floor joint
11.62
-6.62
5.77
-5.77
5.77
-5.77
5.77
11.62
-6.62
5.77
-5.77
5.77
-5.77
5.77
H1
I1
J1
K1
L1
M1
N1
Column DF Above floor
0.22
0.22
0.22
0.22
0.22
0.22
0.22
Below floor
0.22
0.22
0.22
0.22
0.22
0.22
0.22
Member
H1 G1
H1 I 1
I 1 H1
I1 J 1
J 1 I1
J 1 K1
K1 J 1
K1 E 1
L1 K1
L1 M1
M1 L1
M 1 N1
N 1M 1
N 1O 1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
-14.2
14.2
-14.2
14.2
-14.2
14.2
FEM due to
-14.2
DL FEM
due
34.62
-
TL Distribution
34.62
-
34.62
34.62
-
34.62
34.62
34.62
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
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
+ carryover Total Distribution to column Above floor Distribution to column below floor
-5.77
5.77
-5.77
5.77
-5.77
5.77
-5.77
-5.77
5.77
-5.77
5.77
-5.77
5.77
-5.77
joint
O1
P1
Q1
R1
S1
T1
U1
Above floor
0.33
0.24
0.27
0.31
0.27
0.24
0.33
Below floor
0.33
0.24
0.27
0.31
0.27
0.24
0.33
Column DF
Member
O 1N 1
O 1P 1
P 1O 1
P 1Q 1
Q 1P 1
Q1R1
R1Q1
R1S1
S1R1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.32
FEM due to
14.2
-14.2
14.2
-14.2
14.2
S1T1 0.16
T1S1
T 1U 1
U 1T 1
0.2
0.16
0.2
-102.1
102.1
DL FEM due TL Distribution+
-34.62
34.62
-2.87
-2.87
2.87
2.87
11.3
-37.5
37.5
-11.3
-34.62
34.62
-2.87
-2.87
2.87
32.33
11.3
-37.5
37.5
-11.3
-216.2
216.2
-2.87
-11.42
16.17
-10.2
-9.14
11.3
227.65
232.4
-112.2
92.9
carryover Total Distribution to column Above floor Distribution to column below floor
joint
5.77
-5.77
5.77
-12.25
-62.12
-38.45
-37.2
5.77
-5.77
5.77
-12.25
-62.12
-38.45
-37.2
A1
B1
C1
Above floor
0.31
0.22
Below floor
0.31
0.22
Member
A1 B1
B1A1
B1 C1
C1 B1
C1 D1
D1 C1
D1 E 1
E 1D 1
E1 F1
F1E1
F1 G1
G1F1
G 1H 1
DF
0.39
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to
-14.2
14.2
-14.2
14.2
-14.2
14.2
D1
E1
F1
G1
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
0.22
Column DF
-14.2
DL FEM due TL Distribution+
-34.6
34.62
-34.62
34.62
-34.62
34.62
-34.62
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
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
carryover Total Distribution to column Above floor Distribution to column below floor
-3.49
4.53
-5.77
5.77
-5.77
5.77
-5.77
-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
0.22
0.22
0.22
0.22
0.22
0.22
0.22
Below floor
0.22
0.22
0.22
0.22
0.22
0.22
0.22
Member
H1 G1
H1 I 1
I 1 H1
I1 J 1
J 1 I1
J 1 K1
K1 J 1
K1 E 1
L1 K1
L1 M1
M1 L1
M 1 N1
N 1M 1
N 1O 1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
FEM due to
14.2
-14.2
14.2
-14.2
14.2
-14.2
14.2
DL FEM
due
-
TL
34.62
-
34.62
Distribution
34.62
-
34.62
34.62
-34.62
34.62
-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
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
+ carryover Total Distribution to column Above floor Distribution to column below floor
5.77
-5.77
5.77
-5.77
5.77
-5.77
5.77
5.77
-5.77
5.77
-5.77
5.77
-5.77
5.77
joint
O1
P1
Q1
R1
Above floor
0.33
0.24
0.27
0.31
0.27
0.24
0.33
Below floor
0.33
0.24
0.27
0.31
0.27
0.24
0.33
S1
T1
U1
Column DF
Member
O 1N 1
O 1P 1
P 1O 1
P 1Q 1
Q 1P 1
Q1R1
R1Q1
R1S1
S1R1
DF
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.28
0.32
-14.2
14.2
-14.2
14.2
FEM due to
S1T1
T1S1
T 1U 1
U 1T 1
0.16
0.2
0.16
0.2
-86.63
86.63 -242.2
242.2
DL FEM due TL
34.62
Distribution+
2.87
2.87
37.5
-11.3
-34.6
34.62
-2.87
-2.87
2.87
2.87
11.3
-37.5
37.5
-11.3
-34.62
34.62
-2.87
8.32
2.87
15.56
4.16
-24.22
12.45
11.3
-26.31
37.5
-71.1
90.79
-266.4
254.7
carryover Total Distribution to column Above floor Distribution to column below floor
-5.77
5.77
-5.77
3.31
8.73
56.2
-102
-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 =0.061*23.25*3.432
My
=16.69kNm
Check for depth =0.138fckbd2
Mulim
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, =0.12% of c/s of Fe 415
Ast min
=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 Pt
=20 =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
ly/8 437.5
3500
6mm dia @ 100mm c/c 10mm dia bars @ 150mm c/c
3ly/4 2625
10mm dia bars @ 250mm c/c ly/8 437.5 lx/8 400
3lx/4 2400
lx/8 400
3200
REINFORCEMENT DETAILS OF TWO WAY SLAB
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
2 nos of 20mm dia bars 3 nos of 20mm dia bars 10mm dia 2 legged stirrups @ 200mm c/c
3500 LONGITUDINAL SECTION OF L-BEAM
Figure 7-Reinforcement details of L-beams- longitudinal section
2 bars of 20mm dia
2 bars of 20mm dia
150
150
10mm dia 2 legged stirrups @ 200mm c/c
10mm dia 2 legged stirrups @ 200mm c/c
3 bars of 20mm dia
3 bars of 20mm dia
SUPPORT SECTION
MID SECTION
CROSS SECTION OF L-BEAM 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 =1.5wl2/8
Mu
=(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.
8mm dia 2 legged stirrups @300mm c/c 2 nos of 12mm dia bars 150
3 nos of 20mm dia bars
3500
LONGITUDINAL SECTION OF T-BEAM
Figure 9-Reinforcement details of T-beam-longitudinal design
2 bars of 12mm dia
2 bars of 12mm dia
150
150
8mm dia 2 legged stirrups @ 300mm c/c
8mm dia 2 legged stirrups @ 300mm c/c
3 bars of 20mm dia
3 bars of 20mm dia
SUPPORT SECTION
MID SECTION
CROSS SECTION OF T-BEAM
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 =w’*√(R2+T2)/T
Dead load of waist slab
=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
Astmin =0.12% of cs area and assume 8mm dia bars =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.
12mm dia bars @ 100mm c/c 8mm dia bars @ 200mm c/c Tread = 300mm Riser =125mm Thickness of flight = 200mm
12mm dia bars @ 100mm c/c 8mm dia bars @ 200mm c/c
DETAILING OF DOG LEGGED STAIRCASE Figure 11-Reinforcement details of doglegged staircase
5.4 COLUMS 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
Lateral ties 8mm dia bars @ 300mm c/c 6 nos of 20mm dia bars
AXIALLY LOADED COLUMN
Figure 12-Reinforcement details of axially loaded column
5.4.2.2 DESIG OF UIAXIALLY LOADED COLUM
Dimensions Size of column = 400mm×300mm = 650 kN
Load, Pu
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
Lateral ties 8mm dia bars @ 250mm c/c 8 nos of 20mm dia bars
UNIAXIALLY LOADED COLUMN
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
Lateral ties 8mm dia bars @300mm c/c 10 nos of 20mm dia bars
BIAXIALLY LOADED COLUMN
Figure 14-Reinforcement details of biaxially loaded columns
5.5 FOOTIG 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 FOOTIG 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) Hence
= 900d 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.
GL Column Reinforcement
16mm dia bars @ 100mm c/c (both ways)
FOOTING FOR AXIALLY LOADED COLUMN Figure 15-Reinforcement details of axially loaded column
CHAPTER 6 6. COCLUSIO 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.
REFERECES 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
