2014B4A2813P-Bhabha Atomic Research Centre (Visakhapatnam)

1
A REPORT
ON
ANALYSIS OF RCC FRAMED STRUCTURES USING FE SOFT-
WARE (ETABS) AND INTRODUCTION TO SEISMIC AND WIND
LOADS.
BY
Name(s) of the Student(s) ID.No(s)
Anand Morlawar 2014B4A2813P
UNDER THE SUPERVISION OF
Shri P. Ramesh
AT
Bhabha Atomic Research Centre, Visakhapatnam
A Practice School – I station of
BIRLA INSTITUE OF TECHNOLOGY & SCIENCE, PILANI
23rd
May 2016 - 16th
July 2016

2
A REPORT
ON
ANALYSIS OF RCC FRAMED STRUCTURES USING FE SOFT-
WARE(ETABS) AND INTRODUCTION TO SEISMIC AND WIND
LOADS.
BY
Name of the ID.No Discipline
Student
Anand Morlawar 2014B4A2813P M.Sc.(Hons.) in Mathematics +
B.E.(Hons) in Civil
Prepared in partial fulfillment of the Practice
School-I Course No. F221
AT
Bhabha Atomic Research Centre, Visakhapatnam
A Practice School – I station of
BIRLA INSTITUE OF TECHNOLOGY & SCIENCE, PILANI
23rd
May 2016 - 16th
July 2016

3
BIRLA INSTITUTE OF TECHNOLOGY AND SCIENCE
PILANI (RAJASTHAN)
Practice School Division
Station: Bhabha Atomic Research Centre Place: Visakhapatnam
Duration: From: 23rd
May 2016 To: 16th
July 2016
Date of Submission: 15th
July 2016
Title of the Project: ANALYSIS OF RCC FRAMED STRUCTURES USING FE SOFT-
WARE (ETABS) AND INTRODUCTION TO SEISMIC AND WIND LOADS.
ID No. Name of Student(s) Discipline
Anand Morlawar 2014B4A2813P M.Sc.(Hons.) in Mathematics +
B.E.(Hons) in Civil
Name of the expert: Shri. P. Ramesh
Designation: Scientific Officer 'F'
Name of the PS Faculty: Dr. Vinay Budhraja
Key Words: RCC, Bending moment, Shear Force, ETABS, wind, seismic, Response spectrum
and Time history.
Abstract : This report contains manual analysis of basic beams and frames for different types of
loading and same beams and frames were modelled using ETABS software and results were com-
pared.
A simple one storey building –Security Building was modelled using ETABS software and basic
gravity loads were applied, later the study of seismic loads as per IS 1893(part 1) was done and
the seismic load was also applied in the FE model. The Time period of the vibration motion of the
building due to the seismic load on the building was calculated in ETABS along the to extend the
study on seismic loads .The base shear of a simple one storey building was calculated manually,
same was modelled in ETABS and found that results were in order.
Industrial truss of length 15m was modelled using STAAD pro software and basic gravity loads
were applied, later the study of wind loads as per IS 875-(part3) was done and the same wind load
was applied in the FE model. The analysis of the truss for the combined effect of gravity and wind
loads using STAAD.pro software and same analysis was carried out manually by method of sec-
tions and method of joints. Member forces from STAAD pro and manual calculations were com-
pared and found that the results were matching.
Signature of Student Signature of PS Faculty
Date: Date:

4
Acknowledgments
First, I would like to express my gratitude to my mentor, Shri. P.Ramesh. His vast knowledge
helped us to cross many difficulties in our project. His constant guidance and insights have been
extremely valuable. He provided me with a wonderful research experience and the knowledge
gained from him has taught me how to approach a problem in an easier way.
I would also like to thank my other mentors, Shri PGK Chaitanya, Shri Abhishek Kumar and Shri
Shiva for their guidance and their experience helped me develop and learn a lot of things.I would
also like to thank Shri T. V. Murthy sir for making this amazing opportunity available for us and
helping us at every step.
I would also like to thank Dr Vinay Budhraja (BITS PS instructor) for his constant support and
care in the times of need, his constant motivation to improve and his guidance during the course of
seven weeks. I would like to express my appreciation and regard to Bhabha Atomic Research Cen-
tre (B.A.R.C.), Visakhapatnam for providing us with the facilities to do our project and a pleasant
atmosphere for research. Finally, I would thank our university Birla Institute of Technology and
Science, Pilani (BITS, Pilani) for providing me opportunity to work in research institution and or-
ganizing this whole Practice School program.

5
Contents
1. Introduction .................................................................................................................................................8
1.1 Type of beams........................................................................................................................................8
1.2 Type of loadings ................................................................................................................................. 14
1.3 Type of supports ............................................................................................................................. 16
1.4 Moment distribution and Double integration Method ........................................................................ 18
1.4.1Moment distribution...................................................................................................................... 18
1.4.2 Double Integration Method.......................................................................................................... 19
1.5 Indian Standard Codes........................................................................................................................ 19
1.6 Finite element method......................................................................................................................... 19
1.7 Finite element software....................................................................................................................... 19
1.7.1 STAAD.Pro.................................................................................................................................. 19
1.7.2 ETABS......................................................................................................................................... 20
1.8. Truss Analysis........................................................................................................................................ 20
1.8.1 Introduction...................................................................................................................................... 20
2.1 Seismic Analysis..................................................................................................................................... 21
2.1.1 Introduction...................................................................................................................................... 21
2.1.2 Design Spectrum Formula ............................................................................................................... 22
2.2 Calculation of Average Response Acceleration Coefficient............................................................... 24
2.3 Time period......................................................................................................................................... 26
2.3.1 Static Method............................................................................................................................... 26
2.3.2 Dynamic Method ......................................................................................................................... 26
2.3.2.1 Time History Method................................................................................................................ 26
2.3.2.2 Response Spectrum Method...................................................................................................... 27
2.4 Design Lateral Force........................................................................................................................... 27
2.4.1 Design Seismic Base Shear.......................................................................................................... 27
2.4.2 Distribution of Design Force........................................................................................................ 28
2.5 Wind Analysis......................................................................................................................................... 28
2.5.1 Introduction...................................................................................................................................... 28
2.5.2 Basic Wind Speed............................................................................................................................ 29
2.5.3 Design Wind Speed.......................................................................................................................... 30
2.5.4 Risk Coefficient Factor................................................................................................................ 31
2.5.5 Terrain, Height & Structure size factor........................................................................................ 32
2.5.6 Topography factor........................................................................................................................ 32
2.6 Design Wind Pressure......................................................................................................................... 32
2.7 Pressure Coefficients .......................................................................................................................... 32

6
2.7.1 External Pressure Coefficient....................................................................................................... 33
2.7.2 Internal Pressure Coefficient........................................................................................................ 35
2.7.3 Design Wind Force ......................................................................................................................... 35
2.8 Introduction......................................................................................................................................... 35
2.8.1 Different loads on the building ....................................................................................................... 36
2.8.2 Materials used................................................................................................................................. 37
2.8.3 Preliminary sizes of beams & columns considered......................................................................... 37
2.9 3-D View, Plan, Elevation .................................................................................................................. 38
3.1 Analysis and Validation of some frames & beams................................................................................. 40
3.2. Analysis of RCC Structure (Service Building-City Plot) ...................................................................... 66
3.2.1Shear Force Diagrams....................................................................................................................... 66
3.2.2 Bending Moment Diagrams............................................................................................................ 66
3.2.3 Seismic Analysis of Building........................................................................................................... 67
2.13. Wind Analysis of Building .............................................................................................................. 67
3.3 Truss Analysis..................................................................................................................................... 68
3.3.1 Method of Section/Joint Analysis.................................................................................................... 70
3.3.2 STAAD.Pro Results......................................................................................................................... 72
3.3.2. (a) Live Load Results.................................................................................................................. 72
3.3.2 (b) Dead Load Results ................................................................................................................. 74
3.3.2 (c) Wind Load Results ................................................................................................................. 76
3.3.4 Comparison of manual and STAAD results .................................................................................... 78
3.4 Fundamental Time Period....................................................................................................................... 80
3. 5 Calculation of Base & Storey Shears..................................................................................................... 82
3.5.1 Lump mass at storey levels.............................................................................................................. 84
3.5.1.1 Plinth Level............................................................................................................................... 86
3.5.1.2 Storey-I ..................................................................................................................................... 88
3.5.2 Other addition to lump weight ......................................................................................................... 89
3.5.2.1 Plinth Level............................................................................................................................... 89
3.5.2.2 Storey-I ..................................................................................................................................... 91
3.5.3 Total Seismic Weight of the Building.............................................................................................. 92
3.5.3.1 Plinth Level............................................................................................................................... 92
3.5.3.2 Storey-I ..................................................................................................................................... 92
3.5.4 Calculation of horizontal acceleration coefficient ........................................................................... 92
3.5.5 Base Shear........................................................................................................................................ 93
3.5.6 Distribution of Base Shears (Storey Shears).................................................................................... 93
3.5.7 Results (ETABS) ............................................................................................................................. 94

7
3.5.8 Comparison of manual and ETABS Result ..................................................................................... 94
4. Conclusion................................................................................................................................................. 95
References .................................................................................................................................................... 96

8
1. Introduction
Structural analysis means determination of the general shape and all the specific dimensions of a
particular structure so that it will perform the function for which it is created and will safely with-
stand the influences which will act on it throughout its useful life. The input, output and numerical
solution techniques of ETABS are specifically designed to take advantage of the unique physical
and numerical characteristics associated with building type structures.
Buildings are structurally designed to support anticipated loads adequately and safely in addition
to fulfill clients’ needs which include functional and aesthetic requirements. Progressive collapse
can be defined by a chain failure of structural members triggered by local failure or damage and
causing partial or entire collapse of the structure).
The local failure or damage in well-engineered structures and buildings usually results from unan-
ticipated abnormal loads. The abnormal loads arise from extraordinary events which are character-
ized by low probability of occurrence, short time effect and high intensity. Abnormal loads may
include pressure loads (gas explosions and bomb blasts), impact loads (aircraft and vehicular colli-
sion and failing debris) and deformation loads (softening members resulting from fire and founda-
tion subsidence). In many cases, extraordinary events are indirectly avoided by nonstructural
measures. However, the increase in potential extraordinary events and the difficulty of applying
nonstructural measures increase the risk of the progressive collapse. Also, recent facilities and ar-
chitectural requirements to construct buildings with large panels and needs for high rise building
increase the hazards of the extraordinary events that may lead to the progressive collapse.
The structural design of high-rise buildings is often governed by dynamic performance in winds,
and in regions of high seismicity, by seismic performance. Conventional practice is to stiffen a
building in order to reduce the dynamic response under wind loading. However, this has the effect
of increasing the seismic forces the building must be designed for. By adding a robust supplemen-
tary damping system to the structure instead of stiffening, both wind responses and seismic forces
are reduced – which leads to construction cost savings. The behavior of a building during earth-
quakes depends critically on its overall shape, size and geometry, in addition to how the earth-
quake forces are carried to the ground. The earthquake forces developed at different floor levels in
a building need to be brought down along the height to the ground by the shortest path. Any devia-
tion or discontinuity in this load transfer path results in poor performance of the building. Loads
considered are taken in accordance with the IS-875(Part-1, Part-2, Part-3)(1987), IS-1893(Part-
1)(2002),code and combinations are acc. to IS-875(Part-5).Post analysis of the structure, maxi-
mum shear forces, bending moments, and maximum storey displacement are computed and then
compared for all the analyzed cases. Also, the truss analysis is also done on STAAD.Pro consider-
ing a commercial building and assuming dead, live and Wind load on the truss.
1.1 Type of beams
There are different types of beams that are used in the building.
Classification of beams is basically based on:-
1.1.1 Classification of beams based on supports
1.1.2 Classification of beams based on geometry
1.1.3 Classification of beams based on the shape of cross section
1.1.4 Classification of beams based on equilibrium conditions
1.1.5 Classification of beams based on material

9
1.1.1 Based on supports
 Simply Supported beam: - a beam with one end hinged and other end roller supported.
The ends are free to rotate at the supports and have no moment of resistance
(MOR).Figure 1.1.1 depicts a simply supported beam.
Fig.1.1.1 Simply supported beam.
 Fixed beam: - A beam which is supported at both the ends and all the possible degrees of
freedom are restricted at the fixed ends. Figure 1.1.2 depicts a fixed beam.
Fig.1.1.2 Fixed beam
 Overhanging beam: - a simple beam extending beyond its support on one end. Figure
1.1.3 depicts an overhanging beam.
Fig. 1.1.3 Overhanging beam
 Double overhanging beam: - a simple beam with both ends extending beyond its sup-
port on both ends. Figure 1.1.4 depicts a double overhanging beam.

10
.
Fig. 1.1.4 Double overhanging beam
 Continuous beam: - a beam extending over more than 2 supports. Figure 1.1.5 depicts a
continuous beam.
Fig. 1.1.5 Continuous beam.
 Cantilever beam: A beam whose all the degrees of freedom are restricted at one end and
and other end is left free.
Fig. 1.1.6 Cantilever beam.
1.1.2. Classification of beams based on geometry
 Straight beam – Beam with straight profile.

11
Fig. 1.1.7 Straight beam.
 Curved beam – Beam with curved profile
Fig.1.1.8 Curved beam.
 Tapered beam – Beam with tapered cross section.
Fig. 1.1.9 Tapered beam.
1.1.3. Classification of beams based on the shape of cross section
 I-beam – Beam with ‘I’ cross section
Fig. 1.1.10 I-beam.

12
 T-beam – Beam with ‘T’ cross section.
Fig. 1.1.11 T-beam.
 C-beam – Beam with ‘C’ cross section.
Fig. 1.1.12 C-beam.
1.1.4. Classification of beams based on equilibrium conditions
 Statically determinate beam– For a statically determinate beam, equilibrium conditions
alone can be used to solve reactions.
Fig. 1.1.13 Statically determinate beam.

13
 Statically indeterminate beam – For a statically indeterminate beam, equilibrium condi-
tions are not enough to solve reactions. Additional deflections are needed to solve reac-
tions. Fig. 1.2.14 depicts a statically indeterminate beam.
Fig. 1.1.14 Statically indeterminate beam.
1.1.5. Classification of beams based on material
 Timber beam: - Timber beam is made completely of timber and has a long life span.
Fig. 1.1.15 Timber beam.
 Steel beam:-This beam is made of steel and is used for purposes where more stress is de-
veloped.
Fig.1.1.16 Steel beam

14
 Concrete Beam: - Concrete beams are the most commonly used beams in all the build-
ings.
Fig. 1.1.17 Concrete beam.
1.2 Type of loadings
The loads are broadly classified as vertical loads, horizontal loads and longitudinal loads. The ver-
tical loads consist of dead load, live load and impact load. The horizontal loads comprises of wind
load and earthquake load. The longitudinal loads i.e. tractive and braking forces are considered in
special case of design of bridges, gantry girders etc.
1.2.1. Dead load:
Dead loads are permanent or stationary loads which are transferred to structure throughout the life
span. Dead load is primarily due to self weight of structural members, permanent partition walls,
fixed permanent equipments and weight of different materials.
1.2.2. Imposed loads or live loads:
Live loads are either movable or moving loads without any acceleration or impact. There are as-
sumed to be produced by the intended use or occupancy of the building including weights of mov-
able partitions or furniture etc. The floor slabs have to be designed to carry either uniformly dis-
tributed loads or concentrated loads whichever produce greater stresses in the part under consider-
ation. Since it is unlikely that any one particular time all floors will not be simultaneously carrying
maximum loading, the code permits some reduction in imposed loads in designing columns, load
bearing walls, piers supports and foundations.
1.2.3. Impact loads:
Impact load is caused by vibration or impact or acceleration. Thus, impact load is equal to im-
posed load incremented by some percentage called impact factor or impact allowance depending
upon the intensity of impact.
1.2.4. Wind loads:
Wind load is primarily horizontal load caused by the movement of air relative to earth. Wind load
is required to be considered in design especially when the heath of the building exceeds two times
the dimensions transverse to the exposed wind surface.
For low rise building say up to four to five storeys, the wind load is not critical because the mo-
ment of resistance provided by the continuity of floor system to column connection and walls pro-
vided between columns are sufficient to accommodate the effect of these forces. Further in limit
state method the factor for design load is reduced to 1.2 (DL+LL+WL) when wind is considered
as against the factor of 1.5(DL+LL) when wind is not considered.

15
1.2.5. Earthquake load:
Earthquakes are caused by sudden breaks in the earth’s crust due to continuous strain movement
between one section of the earth’s crust and the adjacent one. This disaster leads to an extra verti-
cal load on the building which it has to whit stand to prevent accidents and safeguard the life of
the people living in the building.
Two common measures of earthquakes are:
(a) Magnitude (Richter): This is a measure of total energy released during the earthquake. This is
defined as the common logarithm (base 10) of the trace amplitude (in microns) of a standard seis-
mograph located on firm ground at a distance of 100 km from the epicenter.
(b) Intensity: A subjective measure of the ground motion on the basis of the effect of earthquake
on human beings and structures. This is measured on the XII point Modified Mercalli scale
(MM).
We are generally concerned about the intensity and the exact location of the earthquake on the
building and not primarily on the magnitude of the earthquake obtained through Richter scale.
(a) (b)
Fig. 1.2.1 Effect of inertia in a building when shaken at its base due to earthquake.
The following codes were referred for respective loads and were studied thoroughly.
IS- 875 (1987) part 1- Dead Load
IS- 875 (1987) part 2- Live Load (Imposed Load)
IS- 875 (1987) part 3- Wind Load
IS- 1893 (2002) part 1- Earthquake Load

16
1.3 Type of supports
The three common types of connections which join a built structure to its foundation
are; roller, pinned and fixed. This is often idealized as a frictionless surface. All of these supports
can be located anywhere along a structural element. They are found at the ends, at midpoints, or at
any other intermediate points. The type of support connection determines the type of load that the
support can resist. The support type also has a great effect on the load bearing capacity of each
element, and therefore the system.
1.3.1 Roller Supports
Roller supports are free to rotate and translate along the surface upon which the roller rests. The
surface can be horizontal, vertical, or sloped at any angle. The resulting reaction force is always a
single force that is perpendicular to, and away from, the surface. The most common use of a roller
support is in a bridge. In civil engineering, a bridge will typically contain a roller support at one
end to account for vertical displacement and expansion from changes in temperature. This is re-
quired to prevent the expansion causing damage to a pinned support.
Fig.1.3.1 Roller support
Fig.1.3.2 reactions at the roller support
1.3.2 Pinned Supports
A pinned support can resist both vertical and horizontal forces but not a moment. They will allow
the structural member to rotate, but not to translate in any direction. The knee can be idealized as a
connection which allows rotation in only one direction and provides resistance to lateral move-
ment. The design of a pinned connection is a good example of the idealization of the reality. A
single pinned connection is usually not sufficient to make a structure stable. Another support must
be provided at some point to prevent rotation of the structure. The representation of a pinned sup-
port includes both horizontal and vertical forces. Pinned supports can be used in trusses. By link-
ing multiple members joined by hinge connections, the members will push against each other; in-
ducing an axial force within the member. The benefit of this is that the members contain no inter-
nal moment forces, and can be designed according to their axial force only.

17
Fig. 1.3.3 Pinned support.
Fig 1.3.4 reactions at pinned support
1.3.3 Fixed Supports
Fixed supports can resist vertical and horizontal forces as well as a moment. Since they restrain
both rotation and translation, they are also known as rigid supports. This means that a structure
only needs one fixed support in order to be stable. All three equations of equilibrium can be satis-
fied. A flagpole set into a concrete base is a good example of this kind of support. The representa-
tion of fixed supports always includes two forces (horizontal and vertical) and a moment. : Fixed
supports are extremely beneficial when you can only use a single support. The fixed support pro-
vides all the constraints necessary to ensure the structure is static. It is most widely used as the on-
ly support for a cantilever.
Fig 1.3.5Fixed support
Fig. 1.3.6 reactions at fixed support

18
1.4 Moment distribution and Double integration Method
1.4.1Moment distribution
In this method, first we calculate the distribution factor that is a function of stiffness and length of
members. We also calculate the fixed end moments and list them down in the table. Then, we dis-
tribute the fixed end moments in different members and carry over (50%) of the moment and con-
tinue doing so. Then, we add all the columns to get moments in different members.
Fig. 1.4.1 Simple Portal Frame
JOINT A B C D
MEMBER AB BA BC CB CD DC
Distribution Factor
FEM
Distribution
CO
Distribution
CO
Distribution
MOMENT
Table 1.4.1 Moment distribution table.

19
1.4.2 Double Integration Method
In this method, first we draw the bending moment diagram of the members as a function of length.
Then, we use this formula to get the slope and deflection of the member.
𝐸𝐼
𝑑2 𝑦
𝑑𝑥2
= 𝑀
 E = Modulus of Elasticity
 I = Area Moment of Inertia
 M = Moment
 y = Deflection
1.5 Indian Standard Codes
There are a lot of IS codes that are written and updated by Bureau of Indian Standards (BIS).These
codes are basically the guidelines that should be followed while designing and construction of the
building.
IS 875 (1987) - Part 1-This is the code that is referred to for the dead load analysis of a building.
IS 875 (1987) - Part 2 - This is the code that is referred to for the live load analysis of a building.
IS 1893 (2002) - Part 1 - This is the code that is referred to for the seismic load analysis of a build-
ing.
IS 875 (1987) - Part 3 - This is the code that is referred to for the wind load analysis of a building.
1.6 Finite element method
The finite element method (FEM) is a numerical technique for finding approximate solutions
to boundary value problems for partial differential equations. It is also referred to as finite ele-
ment analysis (FEA). FEM subdivides a large problem into smaller, simpler, parts, called finite
elements. The simple equations that model these finite elements are then assembled into a larger
system of equations that models the entire problem. FEM then uses variation methods from
the calculus of variations to approximate a solution by minimizing an associated error function.
Write Steps followed in finite element method
1.7 Finite element software
A software that follows finite element method and does the analysis of a building and gives you
the result. There are many software; two of them are listed below:
1.7.1 STAAD.Pro
STAAD or (STAAD.Pro) is a structural analysis and design computer program. The commercial
version STAAD.Pro is one of the most widely used structural analysis and design software. It
supports several steel, concrete and timber design codes.
It can make use of various forms of analysis from the traditional 1st order static analysis, 2nd or-
der p-delta analysis, geometric non linear analysis or a buckling analysis. It can also make use of
….Eq. 1.4.1

20
various forms of dynamic analysis from modal extraction to time history and response spectrum
analysis.
1.7.2 ETABS
ETABS is an engineering software product that caters to multi-story building analysis and design.
Modeling tools and templates, code-based load prescriptions, analysis methods and solution tech-
niques, all coordinate with the grid-like geometry unique to this class of structure. Basic or ad-
vanced systems under static or dynamic conditions may be evaluated using ETABS. For a sophis-
ticated assessment of seismic performance, modal and direct-integration time-history analyses
may couple with P-Delta and Large Displacement effects. Nonlinear links and concentrated PMM
or fiber hinges may capture material nonlinearity under monotonic or hysteretic behavior. Intuitive
and integrated features make applications of any complexity practical to implement. Interoperabil-
ity with a series of design and documentation platforms makes ETABS a coordinated and produc-
tive tool for designs which range from simple 2D frames to elaborate modern high-rises.
1.8. Truss Analysis
1.8.1 Introduction
A truss is a structure composed of slender members joined together at their end points. The mem-
bers commonly used in construction consist of wooden struts, metal bars, angles, or channels.
The joint connections are usually formed by bolting or welding the ends of the members to a
common plate, called a gusset plate, as shown in Fig. A, or by simply passing a large bolt or pin
through each of the members. Planar trusses lie in a single plane and are often used to support
roofs and bridges.
Fig.6.1.1 sample truss.
I applied simultaneous forces of dead load, live load, and wind load on a industrial roof structure
called truss. Truss is generally provided where a large slab is needed without columns.
I calculated the member forces in each member by method of joints and method of sections.

21
2. Methodology and Modelling
2.1 Seismic Analysis
2.1.1 Introduction
IS 1893(Part-1) is followed for the seismic analysis of a building.
Fig. 2.1.1 Good Performance during EQ
Fig. 2.1.2 Poor performance during EQ, if it is not properly engineered.
{Ref. EQ tips by CVR Murthy}
Importance Factor (I): It is a factor used to obtain the design seismic force Depending on the func-
tional use of the structure, Characterized by hazardous consequences of its failure, its post-
earthquake functional need, historic value, or economic importance.
Design Horizontal Acceleration Coefficient (Ah): It is a horizontal acceleration coefficient that
shall be used for design of structures.
Structural Response Factors (Sa/g): It is a factor denoting the acceleration response spectrum of
the structure subjected to earthquake ground vibrations, and depends on natural period of vibration
and damping of the structure.
Zone Factor (Z): It is a factor to obtain the design spectrum depending on the perceived maximum
seismic risk characterized by Maximum Considered Earthquake (MCE) in the zone in which the
structure is located. The basic zone factors included in this standard are reasonable estimate of ef-
fective peak ground acceleration.
Response Reduction Factor (R): It is the factor by which the actual base shears force that would be
generated if the structure were to remain elastic during its response to the Design Basis Earth-
quake (DBE) shaking, shall be reduced to obtain the design lateral force.

22
2.1.2 Design Spectrum Formula
For the purpose of determining seismic forces, the country is divided into four zones as shown in
the figure below
Fig. 2.1.3 zones divided as per IS 1893(Part-1)
{Ref: IS 1893 Part 1(2002)}
A formula is given in IS 1893 Part 1 to determine the design horizontal seismic coefficient Ah.
𝐴ℎ =
𝑍𝐼𝑆 𝑎
2𝑅𝑔
…(3.2.1)
 The value of Zone Factor (Z) is known through the location.
 The value of Importance Factor (I) depends on the importance of building.(See Fig.3)

23
 The value of Response Reduction Factor(R) depends on the ductile detailing of frames.
 The value of Structural Response Factor (Sa/g) depends on the time period of the building and
the type of soil
Table 2.1.1 Value of Zone Factor for different zones in India
{Ref: IS 1893 Part 1(2002)}
Seismic Zone II III IV V
Seismic Intensity Low Moderate Severe Very Severe
Z 0.10 0.16 0.24 0.36
Table 2.1.2 Values of Importance Factor for different types of buildings
{Ref: IS 1893 Part 1(2002)}
Sl. No. Structure Importance Factor
1) Important service and com-
munity buildings, such as
hospitals, schools, monumen-
tal structures, emergency
buildings like telephone ex-
change, television stations,
radio stations, railway sta-
tions, fire station buildings,
large community halls like
cinemas, assembly halls and
subway stations, power sta-
tions
1.5
2) All other buildings 1.0

24
Table 2.1.3 Values of Response Reduction Factor for different types of frames.
{Ref: IS 1893 Part 1(2002)}
2.2 Calculation of Average Response Acceleration Coefficient
The Value of Sa/g is computed using the value of T using the graph and equations below.
Formulation for Calculating Sa /g Value for 5% damping

25
Fig. 2.3.1 Graph between Sa /g v/s T.
{Ref. IS 1893 Part-1(2002)}
Table 2.3.1 Multiplying Factor for 5% spectrum to obtain for other damping values
….(3.3.1)
…(3.3.2)
…(3.3.3)

26
{Ref: IS 1893 Part 1(2002)}
2.3 Time period
The value of time period is determined using two methods out of which one of them is a conserva-
tive method and the other involves a thorough analysis of the building.
Both the analyses are performed and the value which is higher is taken as the value of T and Sa/g
is computed accordingly.
2.3.1 Static Method
This is a conservative method to find the T using different formulas given in the IS Code.
The approximate fundamental natural period of vibration (T), in seconds, of a moment-resisting
frame building without brick infill panels may be estimated by the empirical expression:
𝑻 = 𝟎. 𝟎𝟕𝟓 × 𝒉.𝟕𝟓
for RC frame building ...(2.3.1)
=0. 𝟎𝟖𝟓 × 𝒉.𝟕𝟓
for steel frame building ...(2.3.2)
where,
h = Height of building, in m. This excludes the basement storeys, where basement walls are con-
nected with the ground floor deck or fitted between the building columns. But it includes the
basement storeys, when they are not so connected.
The approximate fundamental natural period of vibration (T), in seconds, of all other buildings,
including moment-resisting frame buildings with brick infill panels, may be estimated by the em-
pirical expression:
T =
.09×h
√d
…(2.3.3)
Where,
h=Height of building, defined above;
d=Base dimension of the building at the plinth level, in m, along the considered direction of the
lateral force.
2.3.2 Dynamic Method
Dynamic analysis of building is carried out including the strength and stiffness effects of infill and
inelastic deformations in the members.
Dynamic analysis may be performed either by the Time History Method or by the Response Spec-
trum Method. However, in either method, the design base shear (VB) shall be compared with a
base shear (VB) calculated using a fundamental period T
2.3.2.1 Time History Method
Time history method of analysis, when used, shall be based on an appropriate ground motion and
shall be performed using accepted principles of dynamics.
This involves the study of past earthquakes in the area and the future predictions and using the
time history graphs doing the seismic analysis and design of the building.

27
2.3.2.2 Response Spectrum Method
Response spectrum method of analysis shall be performed using the design spectrum specified in
3.2, or by a site-specific design spectrum.
We calculate the value using the time period formula.
𝑇 = 2 × 𝜋√
𝑚
𝑘
…(2.4.4)
Where,
m=Mass of the building
k=Stiffness of different components.
We calculate the stiffness of each column and add all of those in parallel and use moment area to
find the stiffness of the building. Using that we get the value of k.
The mass of the building is basically the sum of its self weight which includes all super dead load
as well and live load. The dead and super dead loads have a factor of 1 whereas the factor of live
load is taken according to the table below.
Table 2.4.1 %age of Imposed Load to be taken into account.
{Ref: IS 1893 Part 1(2002)}
Imposed Uniformity Distributed Floor
Loads (kN / m2
)
Percentage of Imposed
load
Up to and including 3.0 25
Above 3.0 30
After getting the value of T, we get the value of Sa/g as discussed in Sec.3.2
2.4 Design Lateral Force
Buildings and portions there of shall be designed and constructed, to resist the effects of design
lateral force. The design lateral force shall first be computed for the building as a whole. This de-
sign lateral force shall then be distributed to the various floor levels. The overall design seismic
force thus obtained at each floor level, shall then be distributed to individual later a load resisting
elements depending on the floor diaphragm action.
2.4.1 Design Seismic Base Shear
The total design lateral force or design seismic base shear (VB) along any principal direction shall
be determined by the following expression:
𝑉𝐵 = 𝐴ℎ × 𝑊 ...(2.5.1)
Where,
Ah=Design horizontal acceleration spectrum
Value
W=Seismic weight of the building.

28
2.4.2 Distribution of Design Force
The design base shear (VB) computed in 3.5.1 shall be distributed along the height of the building
as per the following expression:
𝑄𝑖 = 𝑉𝐵 ×
𝑊𝑖×ℎ 𝑖
2
∑ 𝑊 𝑗×ℎ 𝑗
2𝑛
𝑗=1
…(2.5.2)
Where,
Qi = Design lateral force at floor i,
Wi = Seismic weight of floor i,
hi = Height of floor i measured from base, and
n =Number of storeys in the building is the number of levels at which the masses are located.
Fig 2.4.1 elevation of a building
Fig 2.4.2 lump mass distribution of masses
2.5 Wind Analysis
2.5.1 Introduction
A large majority of structures met with in practice do not however, suffer wind induced oscilla-
tions and generally do not require to be examined for the dynamic effects of wind, including use
of gust factor method, Nevertheless, there are various types of structures or their components such
as some tall buildings, chimneys, latticed towers, cooling towers, transmission towers, guyed
masts, communication towers, long span bridges, partially or completely solid faced antenna dish,
etc, which require investigation of wind induced oscillations.
Distribution of forces

29
2.5.2 Basic Wind Speed
Figure below gives basic wind speed map of India, as applicable to 10 m height above mean
ground level for different zones of the country. Basic wind speed is based on peak gust velocity
averaged over a short time interval of about 3 seconds and corresponds to mean heights above
ground level in an open terrain (Category 2). Basic wind speeds presented in Fig. 1 have been
worked out for a 50 year return period. Basic wind speed for some important cities/towns is given
below.
Fig. 2.5.1 Wind Speed Zones in India
{Ref. IS 875 Part 3(2002)}

30
Fig. 2.5.2 basic Wind Speed of some major cities
{Ref: IS 875 Part 3(2002)}
2.5.3 Design Wind Speed
The basic wind speed (Vb) for any site shall be obtained from IS code and shall be modified to
include the following effects to get design wind velocity at any height for the chosen structure.
(a) Risk level
(b) Terrain roughness, height and size of structure
(c) Local topography

31
𝑉𝑧 = 𝑉𝑏 × 𝑘1 × 𝑘2 × 𝑘3 …(2.5.1)
Where,
Vz is design wind speed
Vb is basic wind speed
k1 is probability factor (risk coefficient)
k2 is terrain, height and structure size factor
k3 is topography factor
2.5.4 Risk Coefficient Factor
All general structures are designed for 50 years. k1 depends on the importance of the structure,
mean probable design life of structure (in years) and Vb.
For exceptionally important structures:
Return periods ranging from 100 to 1000 years (implying lower risk level) such as nuclear power
reactors and satellite communication.
k1=
𝑋N ,P
X50,0.63
=
A−B[ln{
−1
𝒩
ln(1−PN)} ]
𝐴+4𝐵
…(2.5.2)
where k1 is risk factor
𝒩 is mean probable design life of structure in years.
PNis risk level in 𝒩 consecutive years, nominal value 0.63
𝑋N ,Pis extreme wind speed for given values of 𝒩 and PN
X50,0.63is extreme wind speed for 𝒩 = 50 years and PN
A, B are coefficients having the following values for different Vb zones are given in Table 1
ZONE A B
33 m/sec 83.2 9.2
39 m/sec 84.0 14.0
44 m/sec 88.0 18.0
47 m/sec 88.0 20.5
50 m/sec 88.8 22.8
55 m/sec 90.8 27.3
Table 2.5.1 Values of coefficients for Eq 2.5.2.
{Ref: IS 875 Part 3(2002)}

32
2.5.5 Terrain, Height & Structure size factor
It depends on the height of the building, surrounding buildings and obstacles and terrain. Terrain
in which a specific structure stands shall be assessed as being one of the following terrain catego-
ries.
Category 1:
Exposed open terrain with few or no obstructions, average height of object surrounding is less
than 1.5 m. It includes open sea coasts and flat treeless plains.
Category 2:
Open terrain with well scattered obstructions having heights generally between 1.5 m to 10 m.
Category 3:
Terrain with numerous closely spaced obstructions having the size of building structures upto 10
m in height with or without a few isolated tall structures.
Category 4:
Terrain with numerous large high closely spaced obstructions.
2.5.6 Topography factor
The basic wind speed Vb given in the above fig. takes account of the general level of site above
sea level. This does not allow for local topographic features such as hills, valleys, cliffs, escarp-
ments, or ridges which can significantly affect wind speed in their vicinity. The effect of topogra-
phy is to accelerate wind near the summits of hills or crests or cliffs, escarpments or ridges and
decelerate the wind in valleys or near the foot of cliff, steep escarpments, or ridges.
The effect of topography will be significant at a site when the upwind slope (6) is greater than
about 3”, and below that, the value of k3 may be taken to be equal to 1 O. The value of k3 is con-
fined in the range of 1-O to 1.36 for slopes greater than 3”. A method of evaluating the value of k3
for values greater than 1.0 is given in Appendix C. It may be noted that the value of k3 varies with
height above ground level, at a maximum near the ground, and reducing to 1.0 at higher levels.
2.6 Design Wind Pressure
The design wind pressure at any height above mean ground level shall be obtained by the follow-
ing relationship between wind pressure and wind velocity:
𝑝𝑧 = 0.6 × 𝑉𝑧
2
…(2.6.1)
Where,
pz = design wind pressure in N/ms at height z,
vz = design wind velocity in m/s at height z.
2.7 Pressure Coefficients
The pressure coefficients are always given for a particular surface or part of the surface of a build-
ing. The wind load acting normal to a surface is obtained by multiplying the area of that surface or

33
its appropriate portion by the pressure coefficient (Cp) and the design wind pressure at the height
of the surface from the ground.
2.7.1 External Pressure Coefficient
Fig. 2.7.1 External pressure coefficient for walls of rectangular clad buildings.
{Ref: IS 875 Part 3(2002)}

34
Fig. 2.7.2 External pressure coefficient for walls of rectangular clad buildings.
(Continued)
{Ref: IS 875 Part 3(2002)}
Fig. 2.7.3 External pressure coefficient of pitched roof of rectangular clad building.
{Ref: IS 875 Part 3(2002)}

35
2.7.2 Internal Pressure Coefficient
The 𝐶 𝑝 𝑖
is calculated through different tables given in IS-875(Part-3)
2.7.3 Design Wind Force
When calculating the wind load on individual structural elements such as roofs and walls, and in-
dividual cladding units and their fittings, it is essential to take account of the pressure difference
between opposite faces of such elements or units. For clad structures, it is, therefore, necessary to
know the internal pressure as well as the external pressure. Then the wind load, F, acting in a di-
rection normal to the individual structural element or cladding unit is:
𝐹 = (𝐶 𝑝 𝑒
− 𝐶 𝑝𝑖
)𝐴𝑝 𝑎 …(2.7.1)
Where,
F=Design Wind Force
𝐶 𝑝 𝑒
=External Pressure Coefficient
𝐶 𝑝 𝑖
=Internal Pressure Coefficient
A= Surface Area of structural element
Pa= Design Wind Pressure
2.8 Introduction
A 11X6 1 storey structure is considered for the study. Modeling, analysis and design of the struc-
ture is done in ETABS software.

36
Length x Width 20mX 18m
No. of storeys 2
Storey heights 4.5m
Foundation 1.5m
Support conditions Fixed
Wall thickness 230mm
2.8.1 Different loads on the building
Loads acting on the structure are dead load (DL), Live Load (LL), Seismic Load (EL) & Wind
Load (WL).
 Floor Finishes= 3 KN/Sq. m is considered
 LL=1.5 KN/Sq. m is considered
 A super dead load of brickwork is also considered which has two different values:-
 18.63kN
 17.94kN
 A seismic force for the zone Visakhapatnam (Zone-II) and Importance factor of 1.5 along with a
response reduction factor of 5 is taken into account and modal analysis is done by ETABS and the
time period was found out.

37
 A wind force for the basic wind speed of 50m/s is found and combining with different factors such
as topography, terrain is taken into account and wind load is found out.
2.8.2 Materials used
Beam RCC(M-30+FE400)
Column RCC(M-30+FE400)
M-30(Concrete)
Weight per unit volume=25kN/𝑚3
Modulus of Elasticity= 27386.13 MPa
Poisson’s Ratio, U=0.2
FE400 (Steel)
Modulus of Elasticity= 200000.03 MPa
Brick (Brickwork)
2.8.3 Preliminary sizes of beams & columns considered
Beams 300X600mm, 300X450mm
Columns 300X600mm, 300X450mm
Slab thickness 125mm

38
2.9 3-D View, Plan, Elevation
Fig. 2.9.1 Plan of Service Building

39
Fig. 2.9.2 3-D View of Service Building
.

40
3. Results and Discussions
3.1 Analysis and Validation of some frames & beams.
Problem 3.1.1
Cantilever beam of length 5m and a force of 10kN is acting at the free end (as in Fig 3.1.1.a). Cal-
culate the shear force, moment, slope and deflection in the beam. Also show the Shear force dia-
gram and bending moment diagram.Assume flexural rigidity (EI) constant.
Fig 3.1.1.a
Solution :
First the free body diagram (FBD) of the given beam is drawn and reactions are calculated.
Fig.3.1.1.bFree body diagram of the beam.
To calculate the reactions at the fixed end,
∑ 𝑭 𝑿 = 𝟎
RAX = 0 kN
∑ 𝐹𝑌 = 0
RAY = 10 kN
To calculate Moment at A,
∑ 𝑀 = 0 at A
MA – 10(5) = 0

41
MA= 50kNm
To calculate shear force and bending moment at any point on the beam, we cut a section and cal-
culate shear force and bending moment at that point (generally at a distance x from one end)
Fig. 3.1.1.cFreebody diagram at a distance ‘x’ m.
Shear force (V) at a distance x from fixed end
0 < x < 5m
∑ 𝐹𝑌 = 0
V= 10 kN
Fig 3.1.1.dShear Force Diagram for this beam
Moment (M) at a distance x from fixed end
0 < x < 5m
M+ 50 -10x = 0
M = 10x – 50

42
Fig 3.1.1.eBending moment diagram for this beam
To calculate the deflection and slope, we are using Double Integration method,
𝑀 = 𝐸𝐼
𝑑2
𝑦
𝑑2 𝑥
where‘M’ is moment ( a function of x )
‘E’ is Young’s modulus of elasticity
‘I’ is moment of inertia
10𝑥 − 50 = 𝐸𝐼
𝑑2
𝑦
𝑑2 𝑥
Integrating w.r.t x
5𝑥2
− 50𝑥 + 𝐶1 = 𝐸𝐼
𝑑𝑦
𝑑𝑥
Integrating w.r.t x
5𝑥3
3
− 25𝑥2
+ 𝐶1 𝑥 + 𝐶2 = 𝐸𝐼𝑦
Boundary conditions
1) When x=0, y=0
 C2 =0
2) When x =5, y=0
C1 =
−250
3
Slope =
𝑑𝑦
𝑑𝑥
=
1
𝐸𝐼
(5𝑥2
− 50𝑥 −
250
3
)
Deflection = 𝑦 =
1
𝐸𝐼
(
5
3
𝑥3
+ 25𝑥2
−
250
3
𝑥)

43
Validation(ETABS)
Fig. 3.1.1.f ETABS MODEL
Fig 3.1.1.g ETABS Shear Force Diagram.
Fig.3.1.1.h ETABS Bending Moment Diagram
------------------------------------------------------------------------------------------------------------------
Problem 3.1.2
A UDL of 4kN/m is acting on a cantilever beam of length 5m as shown in Fig 3.1.2.a. Compute
the shear force, moment, slope and deflection in the beam. Also draw the Shear force and bending
moment diagram. Assume flexural rigidity (EI) constant.
Fig 3.1.2.a
Solution :
First the free body diagram (FBD) of the given beam is drawn and reactions are calculated.

44
Fig.3.1.2.bFree body diagram of the beam showing support reactions
∑ 𝐹𝑋 = 0
RAX = 0 kN
∑ 𝐹𝑌 = 0
RAY = 20 kN
∑ 𝑀 = 0 Moment about ‘A’
MA – 20(2.5) = 0
MA= 50kNm
To calculate shear force and bending moment at any section in the beam, we cut a section and cal-
culate shear force and bending moment at that point (generally at a distance ‘x’ from either end)
Fig.3.1.2.cFree body diagram of the beam at a distance ‘x’ m.
0 < x < 5m

45
∑ 𝐹𝑌 = 0
V -20 + 4x =0
V= 20 - 4x
Fig 3.1.2.dShear Force Diagram
0 < x < 5m
M+ 50 -20x + 2𝑥2
= 0
M = −2𝑥2
+ 20x – 50
Fig 3.1.2.eBending moment diagram for this beam
To calculate the deflection and slope we are using Double Integration method,
𝑀 = 𝐸𝐼
𝑑2
𝑦
𝑑2 𝑥
where‘M’ is moment at a section (expressed as a function of x )

46
−2𝑥2
+ 20x – 50 = 𝐸𝐼
𝑑2
𝑦
𝑑2 𝑥
Integrating w.r.t x
−2𝑥3
3
+ 10𝑥2
− 50𝑥 + 𝐶1 = 𝐸𝐼
𝑑𝑦
𝑑𝑥
Integrating w.r.t x
−𝑥4
6
+
10𝑥3
3
− 25𝑥2
+ 𝐶1 𝑥 + 𝐶2 = 𝐸𝐼𝑦
Boundary Conditions
1) When x=0 , y=0
 C2 =0
2) When x=5, y=0
C1 = 62.5
Slope =
𝑑𝑦
𝑑𝑥
=
1
𝐸𝐼
(
−2𝑥3
3
+ 10𝑥2
− 50𝑥 + 62.5)
Deflection = 𝑦 =
1
𝐸𝐼
(
−𝑥4
6
+
10𝑥3
3
− 25𝑥2
+ 62.5𝑥)
Validation(ETABS)
Fig. 3.1.2.f ETABS Model
Fig.3.1.2.g ETABS Shear Force Diagram

47
Fig.3.1.2.h ETABS Bending Moment Diagram
------------------------------------------------------------------------------------------------------------------
Problem 3.1.3
Given a Cantilever beam of length 5m and a UDL of 4kN is acting along its length, the other end
of the beam is pinned (as in Fig 3.1.3.a). Calculate the shear force, moment, slope and deflection
in the beam. Also show the Shear force diagram and bending moment diagram.Assume flexural
rigidity (EI) constant.
Fig 3.1.3.a
Solution :
Since it is an indeterminate structure, the beam is tranformed as in Fig.2.3.b and reactions are
calculated
Fig.3.1.3.b
Applying the condition of geometry, the deflection at support ‘B’ shall be zero
𝑤𝑙4
8𝐸𝐼
=
𝑅𝑙3
3𝐸𝐼
𝑅 =
3𝑤𝑙
8
Here, R = RBY =
3𝑤𝑙
8
=
3×4×5
8
= 7.5
∑ 𝐹𝑌 = 0
RAY+ RBY= 20 kN
RAY = 20-7.5=12.5 kN

48
Fig.3.1.3.c Free body diagram
∑ 𝑀 = 0about point ‘A’
MA+7.5(5) - 20(2.5) = 0
MA=12.5 kNm
culate shear force and bending moment at that point (at a distance x from one end)
Fig.3.1.3.dFree body diagram at a distance ‘x’ m.
0 < x < 5m
∑ 𝐹𝑌 = 0

49
V -12.5 + 4x =0
V= 12.5 - 4x
Fig 3.1.3.eShear Force Diagram for this beam
0 < x < 5m
∑ 𝑀 = 0
M+ 12.5- 12.5x + 2𝑥2
= 0
M = −2𝑥2
+ 12.5x – 12.5
Fig 3.1.3.fBending moment diagram for this beam
To calculate the deflection and slope we are using Double Integrationmethod,
𝑀 = 𝐸𝐼
𝑑2
𝑦
𝑑2 𝑥

50
−2𝑥2
+ 12.5x – 12.5 = 𝐸𝐼
𝑑2
𝑦
𝑑2 𝑥
Integrating w.r.t x
−2𝑥3
3
+
12.5
2
𝑥2
− 12.5𝑥 + 𝐶1 = 𝐸𝐼
𝑑𝑦
𝑑𝑥
Integrating w.r.t x
−𝑥4
6
+
12.5𝑥3
6
−
12.5
2
𝑥2
+ 𝐶1 𝑥 + 𝐶2 = 𝐸𝐼𝑦
Boundary conditions:
1) When x=0, y=0
 C2 =0
2) When x=5, y=0
C1 = 0
Slope =
𝑑𝑦
𝑑𝑥
=
1
𝐸𝐼
(
−2𝑥3
3
+
12.5
2
𝑥2
− 12.5𝑥)
Deflection = 𝑦 =
1
𝐸𝐼
(
−𝑥4
6
+
12.5𝑥3
6
−
12.5
2
𝑥2
)
Validation(ETABS)
Fig.3.1.3.g ETABS Model
Fig 3.1.3.h ETABS Shear Force Diagram

51
Fig 3.1.3.i ETABS Bending Moment Diagram
------------------------------------------------------------------------------------------------------------------
Problem 3.1.4
Fixed beam of length 5m and a force of 8kN is acting at the midpoint (as in Fig 3.1.4.a). Calculate
the shear force, moment, slope and deflection in the beam. Also draw the Shear force diagram and
bending moment diagram.Assume flexural rigidity (EI) constant.
Fig 3.1.4.a
Solution :
Fig.3.1.4.b Free body diagram
∑ 𝐹𝑋 = 0
RAX =RBX= 0 kN

52
∑ 𝐹𝑌 = 0
RAY=RBY = 4 kN
MA =
𝑃𝐿
8
= 5kN
MB =
𝑃𝐿
8
= 5kN
0 < x < 2.5m
Fig3.1.4.c Free body diagram of beam for 0<x<2.5m
∑ 𝐹𝑌 = 0
V= 4 kN
∑ 𝑀 = 0
M+5 –4x=0
M=4x-5
𝑀 = 𝐸𝐼
𝑑2
𝑦
𝑑2 𝑥

53
4x − 5 = 𝐸𝐼
𝑑2
𝑦
𝑑2 𝑥
Integrating w.r.t x
2𝑥2
− 5𝑥 + 𝐶1 = 𝐸𝐼
𝑑𝑦
𝑑𝑥
Integrating w.r.t x
2𝑥3
3
−
5
2
𝑥2
+ 𝐶1 𝑥 + 𝐶2 = 𝐸𝐼𝑦
Boundary Conditions
1) When x=0, y=0
 C2 =0
2) When x=2.5,
𝑑𝑦
𝑑𝑥
= 0
C1 = 0
Slope =
𝑑𝑦
𝑑𝑥
=
1
𝐸𝐼
(2𝑥2
− 5𝑥)
Deflection = 𝑦 =
1
𝐸𝐼
(
2𝑥3
3
−
5
2
𝑥2
)
2.5m< x < 5m
Fig.3.1.4.d Freebody diagram of beam for 2.5m< x<5m
∑ 𝐹𝑌 = 0
V= -4 kN

54
∑ 𝑀 = 0
M+5 –4(5-x)=0
M= -4x + 15
Fig 3.1.4.eShear force diagram for this beam
Fig 3.1.4.fBending moment diagram
𝑀 = 𝐸𝐼
𝑑2
𝑦
𝑑2 𝑥
Where ‘M’ is moment (a function of x)
‘E’ is elasticity
−4x + 15 = 𝐸𝐼
𝑑2
𝑦
𝑑2 𝑥
Integrating w.r.t x
−2𝑥2
+ 15𝑥 + 𝐶1 = 𝐸𝐼
𝑑𝑦
𝑑𝑥
Integrating w.r.t x

55
−2𝑥3
6
+
15
2
𝑥2
+ 𝐶1 𝑥 + 𝐶2 = 𝐸𝐼𝑦
Boundary Conditions
1) When x=2.5,
𝑑𝑦
𝑑𝑥
= 0
C1 = −25
2) When x=5, y=0
 C2 =20.833
Slope =
𝑑𝑦
𝑑𝑥
=
1
𝐸𝐼
(−2𝑥2
+ 15𝑥 − 25)
Deflection = 𝑦 =
1
𝐸𝐼
(
𝑥3
3
+
15
2
𝑥2
− 25x + 20.833)
Validation(ETABS)
Fig. 3.1.4.g ETABS Model
Fig. 3.1.4.h ETABS Shear Force Diagram
Fig.3.1.4.i ETABS Bending Moment Diagram
------------------------------------------------------------------------------------------------------------------
Problem 3.1.5
A beam of length 3.5m is considered that is fixed at both ends and a UDL of 10kN/m is acting
along its length. Find the slope and deflection also draw the SFD and BMD. Assume flexural ri-
gidity (EI) constant.

56
Fig 3.1.5.a
Solution :
Fig.3.1.5.b Free body diagram
∑ 𝐹𝑋 = 0
RAX = RBX = 0 kN
∑ 𝐹𝑌 = 0
RAY = RBY =
𝑤𝑙
2
= 35/2 kN
To calculate Moment at A, B
MA = MB =
𝑤𝑙2
12
= 10.21 kNm
0 < x < 3.5m
∑ 𝐹𝑌 = 0
V +10x = 35/2

57
𝑉 =
35
2
− 10𝑥
Fig 3.1.5.c Shear Force Diagram for this beam
0 < x < 3.5m
M+ 10.21+ 5𝑥2
−
35
2
𝑥 = 0
M = −5𝑥2
+
35
2
x – 10.21
Fig 3.1.5.d Bending moment diagram for this beam
𝑀 = 𝐸𝐼
𝑑2
𝑦
𝑑2 𝑥
where M is moment ( a function of x )
E is elasticity

58
−5𝑥2
+
35
2
x – 10.21 = 𝐸𝐼
𝑑2
𝑦
𝑑2 𝑥
Integrating w.r.t x
−5𝑥3
3
+ 35/4𝑥2
− 10.21𝑥 + 𝐶1 = 𝐸𝐼
𝑑𝑦
𝑑𝑥
Integrating w.r.t x
−5𝑥4
12
+
35𝑥3
8
− 5.205𝑥2
+ 𝐶1 𝑥 + 𝐶2 = 𝐸𝐼𝑦
Boundary Conditions
1) When x=0, y=0
 C2 =0
2) When x=3.5, y=0
C1 = -17.8617
Slope =
𝑑𝑦
𝑑𝑥
=
1
𝐸𝐼
(
−5𝑥3
3
+
35𝑥2
4
− 10.21𝑥 − 17.8617)
Deflection = 𝑦 =
1
𝐸𝐼
(
−5𝑥4
12
+
35𝑥3
8
− 5.205 𝑥2
− 17.8617𝑥)
Validation(ETABS)
Fig 3.1.5.e ETABS Model
Fig 3.1.5.f ETABS Shear Force Diagram

59
Fig 3.1.5.g ETABS Bending Moemnt Diagram
-----------------------------------------------------------------------------------------------------------------
Problem 3.1.6
A frame is considered with the dimensions as mentioned in Fig 3.1.6.a. Calculate the moment in
the beams and columns. Assume flexural rigidity (EI) constant.
Fig 3.1.6.a
DF AB = 0; DFDC = 0 (Fixed ends)
KBA =
4𝐸𝐼
𝐿
=
4𝐸𝐼
3
KBC =
3𝐸𝐼
𝐿
=
3𝐸𝐼
4
KCB =
3𝐸𝐼
𝐿
=
3𝐸𝐼
4
KCD =
4𝐸𝐼
𝐿
=
4𝐸𝐼
3
DFBA =
𝐾 𝐵𝐴
𝐾 𝐵𝐴 + 𝐾 𝐵𝐶
=
4
3
𝐸𝐼
4
3
𝐸𝐼 + 3
4
𝐸𝐼
=
16
25

60
DFBC =
𝐾 𝐵𝐶
𝐾 𝐵𝐴 + 𝐾 𝐵𝐶
=
3
4
𝐸𝐼
4
3
𝐸𝐼 + 3
4
𝐸𝐼
=
9
25
DFCB =
𝐾 𝐶𝐵
𝐾 𝐶𝐵 + 𝐾 𝐶𝐷
=
3
4
𝐸𝐼
4
3
𝐸𝐼 + 3
4
𝐸𝐼
=
9
25
DFCD =
𝐾 𝐶𝐵
=
4
3
𝐸𝐼
4
3
𝐸𝐼 + 3
4
𝐸𝐼
=
16
25
JOINT A B C D
MEMBER AB BA BC CB CD DC
Distribution Factor 0 0.64 0.36 0.36 0.64 0
FEM 0 0 -16 16 0 0
Distribution 0 10.24 5.76 -5.76 -10.24 0
CO 5.12 0 -2.88 2.88 0 -5.12
Distribution 0 1.8432 1.037 -1.037 -1.843 0
CO 0.9216 0 -0.518 0.518 0 -0.922
Distribution 0 0.3315 0.187 -0.187 -0.3315 0
MOMENT 6.3365 12.441 -12.414 12.414 -12.441 -6.337
Validation(ETABS)

61
Fig 3.1.6.b ETABS Model
Fig 3.1.6.c ETABS Bending Moment Diagram
------------------------------------------------------------------------------------------------------------------
Problem 3.1.7
A frame is considered with the dimensions as mentioned in Fig 3.1.6.a. Calculate the moment in
the beams and columns. Assume flexural rigidity (EI) constant.

62
Fig 3.1.7.a
𝐷𝐹𝐴𝐵 = 0(Fixed end)
𝐷𝐹𝐹𝐸 = 0(Fixed end)
𝐾 𝐵𝐶 =
3𝐸𝐼
𝑙
=
3𝐸𝐼
4
𝐾 𝐵𝐴 =
4𝐸𝐼
𝑙
=
4𝐸𝐼
4
𝐾 𝐵𝐸 =
3𝐸𝐼
𝑙
=
3𝐸𝐼
4
𝐾𝐶𝐷 =
3𝐸𝐼
𝑙
=
3𝐸𝐼
4
𝐾𝐶𝐵 =
3𝐸𝐼
𝑙
=
3𝐸𝐼
4
𝐾 𝐷𝐶 =
3𝐸𝐼
𝑙
=
3𝐸𝐼
4
𝐾 𝐸𝐵 =
3𝐸𝐼
𝑙
=
3𝐸𝐼
4
𝐾 𝐸𝐷 =
3𝐸𝐼
𝑙
=
3𝐸𝐼
4
𝐾 𝐷𝐸 =
3𝐸𝐼
𝑙
=
3𝐸𝐼
4

63
𝐾 𝐸𝐹 =
3𝐸𝐼
𝑙
=
4𝐸𝐼
4
DFBA =
𝐾 𝐵𝐴
𝐾 𝐵𝐴 + 𝐾 𝐵𝐶 + 𝐾 𝐵𝐸
=
1𝐸𝐼
1𝐸𝐼 + 3𝐸𝐼
4
+ 3𝐸𝐼
4
= 0.4
DFBE =
𝐾 𝐵𝐸
=
3
4
𝐸𝐼
1𝐸𝐼 + 3
4
𝐸𝐼 + 3
4
𝐸𝐼
= 0.3
DFBC =
𝐾 𝐵𝐶
=
3
4
𝐸𝐼
1𝐸𝐼 + 3
4
𝐸𝐼 + 3
4
𝐸𝐼
= 0.3
DFCB =
𝐾 𝐶𝐵
=
3
4
𝐸𝐼
3
4
𝐸𝐼 + 3
4
𝐸𝐼
= 0.5
DFCD =
𝐾 𝐶𝐷
=
3
4
𝐸𝐼
3
4
𝐸𝐼 + 3
4
𝐸𝐼
= 0.5
DFDC =
𝐾 𝐷𝐶
𝐾 𝐷𝐶 + 𝐾 𝐷𝐸
=
3
4
𝐸𝐼
3
4
𝐸𝐼 + 3
4
𝐸𝐼
= 0.5
DFDE =
𝐾 𝐷𝐸
𝐾 𝐷𝐶 + 𝐾 𝐷𝐸
=
3
4
𝐸𝐼
3
4
𝐸𝐼 + 3
4
𝐸𝐼
= 0.5
DFED =
𝐾 𝐸𝐷
𝐾 𝐸𝐷 + 𝐾 𝐸𝐵 + 𝐾 𝐸𝐹
=
3
4
𝐸𝐼
3
4
𝐸𝐼 + 3
4
𝐸𝐼 + 1𝐸𝐼
= 0.3
DFEB =
𝐾 𝐸𝐷
=
3
4
𝐸𝐼
3
4
𝐸𝐼 + 3
4
= 0.3

64
DFEF =
𝐾 𝐸𝐷
=
𝐸𝐼
3
4
𝐸𝐼 + 3
4
= 0.4
JOINT A B C D E F
MEMBER AB BA BE BC CB CD DC DE ED EB EF FE
Distribution
Factor 0 0.4 0.3 0.3 0.5 0.5 0.5 0.5 0.3 0.3 0.4 0
FEM 0 0 -4 0 0 -4 4 0 0 4 0 0
Distribution 0 1.6 1.2 1.2 2 2 -2 -2 -1.2 -1.2 -1.6 0
CO 0.8 0 -0.6 1 0.6 -1 1 -0.6 -1 0.6 0 -0.8
Distribution 0 -0.16 -0.12 -0.12 0.2 0.2 -0.2 -0.2 0.12 0.12 0.16 0
CO -0.08 0 0.06 0.1 -0.06 -0.1 0.1 0.06 -0.1 -0.06 0 0.08
Distribution 0 -0.064 -0.048 -0.048 0.08 0.08 -0.08 -0.08 0.048 0.048 0.064 0
CO
-
0.032 0 0.024 0.04
-
0.024 -0.04 0.04 0.024 -0.04
-
0.024 0 0.032
Distribution 0
-
0.0256
-
0.0192
-
0.0192 0.032 0.032
-
0.032
-
0.032 0.0192 0.019 0.0256 0
MOMENT 0.688 1.3504
-
3.5032 2.1528 2.828
-
2.828 2.828
-
2.828
-
2.1528 3.503
-
1.3504 -0.688
Validation(ETABS)

65
Fig 3.1.7.b ETABS Model
Fig 3.1.7.c ETABS Bending Moment Diagram

66
3.2. Analysis of RCC Structure (Service Building-City Plot)
3.2.1Shear Force Diagrams
Fig. 3.2.1 theshear force of different members.
3.2.2 Bending Moment Diagrams
Fig. 3.2.2 shear force of different members.

67
3.2.3 Seismic Analysis of Building
ETABS has a feature of automatic seismic loading by just inputting the different values in the
formula:-
𝐴ℎ =
𝑍𝐼𝑆 𝑎
2𝑅𝑔
…Eq 3.2.1
The values are being inputted as per IS 1893(Part-1).
Zone is taken as Zone-II. (Zone of Vizag)
Zone Factor (Z) = 0.10(As per table)
Importance Factor(I)=1.0(as per table)
Response Reduction Factor(R) = 3(As per above table put table no.)
The soil type is taken as to be medium soil. (Soil Type-II)
Fig. 3.2.3 auto-lateral seismic load in ETABS.44
2.13. Wind Analysis of Building
ETABS can also do automated wind analysis; it just requires some values such as basic wind
speed, pressure coefficients, terrain category and structure class etc.
Wind Speed (VB) =50m/s(Wind speed of Vizag)
External Pressure Coefficient (Ce) =0.8
Internal Pressure Coefficient (Ci) =0.5
Risk Coefficient Factor (k1) =1
Topography Factor (k2) =1
Terrain Category = II (As per IS 875 Part-3)
Structure Class = B(As per IS 875 Part-3)

68
Fig. 3.2.4 auto-lateral wind load in ETABS.
3.3 Truss Analysis
There is a building that has some trusses with a span of 6m and total length of 12m with 1.2m
horizontal length of members.
Fig. 3.3.1 Truss system.
I have assumed a dead load of 0.25kN/m2
, which includes weight of sheets, self weights of mem-
bers, nuts & bolts.
The dead load at one node= (Dead load) x (Area)
= (0.25) x (6X1.2)
=1.8kN
T1
T1
T1
T1
T1
T1
6m
6m
6m
6m
6m

69
1.2m 1m
Fig3.3.2 Dead load at nodes of the truss.
I have assumed a live load of 0.75kN/m2
(for inaccessible roofs) from IS 875 Part-2.
Live Load at nodes= (Live Load) x (Area)
= (0.75) x (6x1.2)
= 5.4kN
1.2m 1m
Fig. 3.3.3 Live load at nodes of the truss.
Wind Load was calculated for a basic wind speed of 50m/s with different factors k1, k2, k3 as-
sumed as 1.
𝑉𝑍 = 𝑘1 𝑘2 𝑘3 × 𝑉𝑏 ….Eq 3.3.1
VZ=Design wind speed
Therefore, VZ=Vb=50m/s
Using IS 875(Part-3),
For these type of truss roofs,
External pressure coefficient,𝐶 𝑝 𝑒
=0.7
Internal pressure coefficient,𝐶 𝑝 𝑖
=-0.5
Wind Force, F=(𝐶 𝑝 𝑒
− 𝐶 𝑝 𝑖
)𝐴𝑝 𝑎 …Eq 3.3.2
300mm
300mm

70
A=Area= (6x1.2) =7.2m2
𝑝 𝑎 = 0.6𝑉𝑍
2
=Design Wind Pressure
pa =1.5 kN/m
F= (0.7+0.5) x 7.2 x 1.5=12.96kN
Which is taken as 12kN and the analysis is performed.
Fig. 3.3.4 Wind load at nodes of the truss.
3.3.1 Method of Section/Joint Analysis
’
Fig. 3.3.5 Truss with the load combination.
300mm
1.2m
1m

71
∑ 𝐹𝑥 = 0
R3=0
∑ Fy = 0
11(7.2) – 9(12) – R1 – R2 = 0
R1 + R2 = -28.8
∑ 𝑀 = 0 at F22
7.2(1.8+3.6+5.4+7.2+9) -12(1.8+3.6+5.4+7.2) –R1 (9) -7.2(1.8+3.6+5.4+7.2+9)
+12(1.8+3.6+5.4+7.2) + R2 (9) = 0
 R1 = R2
R1 = R2 = -14.4 kN
At Joint 22
Fig. 3.3.6 FBD of joint 22(See fig.3.3.5)
F22x21=0
R1=-14.4kN
F22x11=14.4kN
Taking a section that cuts 10x11, 22x21 & 11x21
Fig. 3.3.7 FBD of joint 11 and 22
R1
F11x10
F11x21
F22x21
22
11 ɸ1
θ
F22x11
F22x21
R1
22

72
Angle of 10x11 with horizontal is θ
Tanθ= 7/60
θ=6.654
Angle 10x11x21 is ɸ1
tanɸ1= ¼
ɸ1=14.036
Resolving each into horizontal and vertical directions and using the equilibrium equations we get
the three forces at this section.
Similarly, the same results are calculated for each member sections.
3.3.2 STAAD.Pro Results
3.3.2. (a) Live Load Results
Sign Convention:
 Negative (-ve) sign indicates Tensile Force.
 Positive (+ve) sign indicates Compressive Force.
Table 3.3.1 shows the member forces in truss only due to live load.
Beam L/C Node Fx kN Fy kN Fz kN
Mx
kNm
My
kNm
Mz
kNm
1 LIVE 1 29.7 0 0 0 0 0
12 -29.7 0 0 0 0 0
2 LIVE 12 0 0 0 0 0 0
13 0 0 0 0 0 0
3 LIVE 13 -66.273 0 0 0 0 0
14 66.273 0 0 0 0 0
4 LIVE 14 -89.379 0 0 0 0 0
15 89.379 0 0 0 0 0
5 LIVE 15 -94.5 0 0 0 0 0
16 94.5 0 0 0 0 0
6 LIVE 16 -90.419 0 0 0 0 0
17 90.419 0 0 0 0 0
7 LIVE 17 -90.419 0 0 0 0 0
18 90.419 0 0 0 0 0
8 LIVE 18 -94.5 0 0 0 0 0
19 94.5 0 0 0 0 0
9 LIVE 19 -89.379 0 0 0 0 0
20 89.379 0 0 0 0 0
10 LIVE 20 -66.273 0 0 0 0 0
21 66.273 0 0 0 0 0
11 LIVE 21 0 0 0 0 0 0

73
22 0 0 0 0 0 0
12 LIVE 11 66.722 0 0 0 0 0
10 -66.722 0 0 0 0 0
13 LIVE 11 29.7 0 0 0 0 0
22 -29.7 0 0 0 0 0
14 LIVE 10 89.986 0 0 0 0 0
9 -89.986 0 0 0 0 0
15 LIVE 9 95.141 0 0 0 0 0
8 -95.141 0 0 0 0 0
16 LIVE 8 91.032 0 0 0 0 0
7 -91.032 0 0 0 0 0
17 LIVE 7 81.549 0 0 0 0 0
6 -81.549 0 0 0 0 0
18 LIVE 6 81.549 0 0 0 0 0
5 -81.549 0 0 0 0 0
19 LIVE 5 91.032 0 0 0 0 0
4 -91.032 0 0 0 0 0
20 LIVE 4 95.141 0 0 0 0 0
3 -95.141 0 0 0 0 0
21 LIVE 3 89.986 0 0 0 0 0
2 -89.986 0 0 0 0 0
22 LIVE 2 66.722 0 0 0 0 0
1 -66.722 0 0 0 0 0
23 LIVE 2 16.568 0 0 0 0 0
13 -16.568 0 0 0 0 0
24 LIVE 14 8.472 0 0 0 0 0
3 -8.472 0 0 0 0 0
25 LIVE 4 2.475 0 0 0 0 0
- 15 -2.475 0 0 0 0 0
26 LIVE 16 -2.449 0 0 0 0 0
5 2.449 0 0 0 0 0
27 LIVE 6 -13.5 0 0 0 0 0
17 13.5 0 0 0 0 0
28 LIVE 18 -2.449 0 0 0 0 0
7 2.449 0 0 0 0 0
29 LIVE 19 2.475 0 0 0 0 0
8 -2.475 0 0 0 0 0
30 LIVE 20 8.472 0 0 0 0 0
9 -8.472 0 0 0 0 0
31 LIVE 21 16.568 0 0 0 0 0
10 -16.568 0 0 0 0 0
32 LIVE 13 -68.312 0 0 0 0 0
1 68.312 0 0 0 0 0
33 LIVE 14 -24.611 0 0 0 0 0
2 24.611 0 0 0 0 0
34 LIVE 15 -5.687 0 0 0 0 0
3 5.687 0 0 0 0 0

74
35 LIVE 16 4.76 0 0 0 0 0
4 -4.76 0 0 0 0 0
36 LIVE 17 11.588 0 0 0 0 0
5 -11.588 0 0 0 0 0
37 LIVE 17 11.588 0 0 0 0 0
7 -11.588 0 0 0 0 0
38 LIVE 18 4.76 0 0 0 0 0
8 -4.76 0 0 0 0 0
39 LIVE 19 -5.687 0 0 0 0 0
9 5.687 0 0 0 0 0
40 LIVE 20 -24.611 0 0 0 0 0
10 24.611 0 0 0 0 0
41 LIVE 21 -68.312 0 0 0 0 0
11 68.312 0 0 0 0 0
3.3.2 (b) Dead Load Results
Sign Convention:
Table 3.3.2 shows the member forces in truss only due to dead load
Beam L/C Node Fx kN Fy kN Fz kN Mx kNm
My
kNm
Mz
kNm
1 DEAD 1 9.9 0 0 0 0 0
12 -9.9 0 0 0 0 0
2 DEAD 12 0 0 0 0 0 0
13 0 0 0 0 0 0
3 DEAD 13 -22.091 0 0 0 0 0
14 22.091 0 0 0 0 0
4 DEAD 14 -29.793 0 0 0 0 0
15 29.793 0 0 0 0 0
5 DEAD 15 -31.5 0 0 0 0 0
16 31.5 0 0 0 0 0
6 DEAD 16 -30.14 0 0 0 0 0
17 30.14 0 0 0 0 0
7 DEAD 17 -30.14 0 0 0 0 0
18 30.14 0 0 0 0 0
8 DEAD 18 -31.5 0 0 0 0 0
19 31.5 0 0 0 0 0
9 DEAD 19 -29.793 0 0 0 0 0
20 29.793 0 0 0 0 0
10 DEAD 20 -22.091 0 0 0 0 0
21 22.091 0 0 0 0 0
11 DEAD 21 0 0 0 0 0 0

75
22 0 0 0 0 0 0
12 DEAD 11 22.241 0 0 0 0 0
10 -22.241 0 0 0 0 0
13 DEAD 11 9.9 0 0 0 0 0
22 -9.9 0 0 0 0 0
14 DEAD 10 29.995 0 0 0 0 0
9 -29.995 0 0 0 0 0
15 DEAD 9 31.714 0 0 0 0 0
8 -31.714 0 0 0 0 0
16 DEAD 8 30.344 0 0 0 0 0
7 -30.344 0 0 0 0 0
17 DEAD 7 27.183 0 0 0 0 0
6 -27.183 0 0 0 0 0
18 DEAD 6 27.183 0 0 0 0 0
5 -27.183 0 0 0 0 0
19 DEAD 5 30.344 0 0 0 0 0
4 -30.344 0 0 0 0 0
20 DEAD 4 31.714 0 0 0 0 0
3 -31.714 0 0 0 0 0
21 DEAD 3 29.995 0 0 0 0 0
2 -29.995 0 0 0 0 0
22 DEAD 2 22.241 0 0 0 0 0
1 -22.241 0 0 0 0 0
23 DEAD 2 5.523 0 0 0 0 0
13 -5.523 0 0 0 0 0
24 DEAD 14 2.824 0 0 0 0 0
3 -2.824 0 0 0 0 0
25 DEAD 4 0.825 0 0 0 0 0
15 -0.825 0 0 0 0 0
26 DEAD 16 -0.816 0 0 0 0 0
5 0.816 0 0 0 0 0
27 DEAD 6 -4.5 0 0 0 0 0
17 4.5 0 0 0 0 0
28 DEAD 18 -0.816 0 0 0 0 0
7 0.816 0 0 0 0 0
29 DEAD 19 0.825 0 0 0 0 0
8 -0.825 0 0 0 0 0
30 DEAD 20 2.824 0 0 0 0 0
9 -2.824 0 0 0 0 0
31 DEAD 21 5.523 0 0 0 0 0
10 -5.523 0 0 0 0 0
32 DEAD 13 -22.771 0 0 0 0 0
1 22.771 0 0 0 0 0
33 DEAD 14 -8.204 0 0 0 0 0
2 8.204 0 0 0 0 0
34 DEAD 15 -1.896 0 0 0 0 0
3 1.896 0 0 0 0 0

76
35 DEAD 16 1.587 0 0 0 0 0
4 -1.587 0 0 0 0 0
36 DEAD 17 3.863 0 0 0 0 0
5 -3.863 0 0 0 0 0
37 DEAD 17 3.863 0 0 0 0 0
7 -3.863 0 0 0 0 0
38 DEAD 18 1.587 0 0 0 0 0
8 -1.587 0 0 0 0 0
39 DEAD 19 -1.896 0 0 0 0 0
9 1.896 0 0 0 0 0
40 DEAD 20 -8.204 0 0 0 0 0
10 8.204 0 0 0 0 0
41 DEAD 21 -22.771 0 0 0 0 0
11 22.771 0 0 0 0 0
3.3.2 (c) Wind Load Results
Sign Convention:
Table 3.3.3 member forces in truss only due to wind load.
Beam L/C Node Fx kN Fy kN Fz kN
Mx
kNm
My
kNm
Mz
kNm
1 WIND 1 -54 0 0 0 0 0
12 54 0 0 0 0 0
2 WIND 12 0 0 0 0 0 0
13 0 0 0 0 0 0
3 WIND 13 147.273 0 0 0 0 0
14 -147.273 0 0 0 0 0
4 WIND 14 198.621 0 0 0 0 0
15 -198.621 0 0 0 0 0
5 WIND 15 210 0 0 0 0 0
16 -210 0 0 0 0 0
6 WIND 16 200.93 0 0 0 0 0
17 -200.93 0 0 0 0 0
7 WIND 17 200.93 0 0 0 0 0
18 -200.93 0 0 0 0 0
8 WIND 18 210 0 0 0 0 0
19 -210 0 0 0 0 0
9 WIND 19 198.621 0 0 0 0 0
20 -198.621 0 0 0 0 0
10 WIND 20 147.273 0 0 0 0 0
21 -147.273 0 0 0 0 0
11 WIND 21 0 0 0 0 0 0

77
22 0 0 0 0 0 0
12 WIND 11 -148.272 0 0 0 0 0
10 148.272 0 0 0 0 0
13 WIND 11 -54 0 0 0 0 0
22 54 0 0 0 0 0
14 WIND 10 -199.968 0 0 0 0 0
9 199.968 0 0 0 0 0
15 WIND 9 -211.424 0 0 0 0 0
8 211.424 0 0 0 0 0
16 WIND 8 -202.293 0 0 0 0 0
7 202.293 0 0 0 0 0
17 WIND 7 -181.221 0 0 0 0 0
6 181.221 0 0 0 0 0
18 WIND 6 -181.221 0 0 0 0 0
5 181.221 0 0 0 0 0
19 WIND 5 -202.293 0 0 0 0 0
4 202.293 0 0 0 0 0
20 WIND 4 -211.424 0 0 0 0 0
3 211.424 0 0 0 0 0
21 WIND 3 -199.968 0 0 0 0 0
2 199.968 0 0 0 0 0
22 WIND 2 -148.272 0 0 0 0 0
1 148.272 0 0 0 0 0
23 WIND 2 -24.818 0 0 0 0 0
13 24.818 0 0 0 0 0
24 WIND 14 -6.828 0 0 0 0 0
3 6.828 0 0 0 0 0
25 WIND 4 6.5 0 0 0 0 0
15 -6.5 0 0 0 0 0
26 WIND 16 17.442 0 0 0 0 0
5 -17.442 0 0 0 0 0
27 WIND 6 42 0 0 0 0 0
17 -42 0 0 0 0 0
28 WIND 18 17.442 0 0 0 0 0
7 -17.442 0 0 0 0 0
29 WIND 19 6.5 0 0 0 0 0
8 -6.5 0 0 0 0 0
30 WIND 20 -6.828 0 0 0 0 0
9 6.828 0 0 0 0 0
31 WIND 21 -24.818 0 0 0 0 0
10 24.818 0 0 0 0 0
32 WIND 13 151.805 0 0 0 0 0
1 -151.805 0 0 0 0 0
33 WIND 14 54.691 0 0 0 0 0
2 -54.691 0 0 0 0 0
34 WIND 15 12.639 0 0 0 0 0
3 -12.639 0 0 0 0 0

78
35 WIND 16 -10.577 0 0 0 0 0
4 10.577 0 0 0 0 0
36 WIND 17 -25.75 0 0 0 0 0
5 25.75 0 0 0 0 0
37 WIND 17 -25.75 0 0 0 0 0
7 25.75 0 0 0 0 0
38 WIND 18 -10.577 0 0 0 0 0
8 10.577 0 0 0 0 0
39 WIND 19 12.639 0 0 0 0 0
9 -12.639 0 0 0 0 0
40 WIND 20 54.691 0 0 0 0 0
10 -54.691 0 0 0 0 0
41 WIND 21 151.805 0 0 0 0 0
11 -151.805 0 0 0 0 0
3.3.4 Comparison of manual and STAAD results
Sign conventions:
Method of Joint/Section
 Negative (-ve) sign indicates Compressive Force
 Positive (+ve) sign indicates Tensile Force
STAAD Results
 Negative (-ve) sign indicates Tensile Force
 Positive (+ve) sign indicates Compressive Force
Table 3.3.4 compares the result obtained by method of section/joint with the STAAD
result.
Member ID Method of
Joint/Section
STAAD RESULTS Percentage
difference in
results
M11x10 59.42kN -59.3kN 0.202%
M11x22 14.4kN -14.4kN 0%
M11x21 -60.82kN 60.7kN 0.197%
M22x21 0kN 0kN 0%
M22X11 14.4kN -14.4kN 0%
M10x11 59.42kN -59.3kN 0.202%
M10x21 2.72kN -2.73kN 0.366%
M10x20 -21.9kN 21.9kN 0%
M10x9 80.027kN -80kN 0.0375%
M21x22 0kN 0kN 0%
M21x11 -60.82kN 60.7kN 0.197%
M21x10 2.72kN -2.73kN 0.366%

79
M21x20 -58.99kN 58.9kN 0.152%
M9x10 80.027kN -80kN 0.0375%
M9x20 -4.46kN 4.47kN 0.223%
M9x19 -5.066kN 5.06kN 0.118%
M9x8 84.59kN -84.6kN 0.011%
M20x21 -58.99kN 58.9kN 0.152%
M20x10 -21.9kN 21.9kN 0%
M20x9 -4.46kN 4.47kN 0.223%
M20x19 -79.448kN 79.4kN 0.060%
M8x9 84.59kN -84.6kN 0.011%
M8x7 80.95kN -80.9kN 0.061%
M8x19 -9.795kN 9.8kN 0.051%
M8x18 4.22kN -4.23kN 0.236%
M19x20 -79.448kN 79.4kN 0.060%
M19x9 -5.066kN 5.06kN 0.118%
M19x8 -9.795kN 9.8kN 0.051%
M19x18 -84kN 84kN 0%
M7x8 80.95kN -80.9kN 0.061%
M7x18 -14.2kN 14.2kN 0%
M7x6 72.51kN -72.5kN 0.013%
M7x17 10.315kN -10.3kN 0.14%
M18x19 -84kN 84kN 0%
M18x8 4.22kN -4.23kN 0.236%
M18x7 -14.2kN 14.2kN 0%
M18x17 -80.372kN 80.4kN 0.034%
M6x7 72.51kN -72.5kN 0.013%
M6x17 -23.99kN 24kN 0.041%
M6x5 72.51kN -72.5kN 0.013%
M17x18 -80.372kN 80.4kN 0.034%
M17x16 -80.372kN 80.4kN 0.04%
M17x5 10.315kN -10.3kN 0.14%
M17x6 -23.99kN 24kN 0.041%
M17x7 10.315kN -10.3kN 0.14%
M5x6 72.51kN -72.5kN 0.013%
M5x17 10.315kN -10.3kN 0.14%
M5x16 -14.2kN 14.2kN 0%
M5x4 80.95kN -80.9kN 0.061%
M16x17 -80.372kN 80.4kN 0.034%
M16x5 -14.2kN 14.2kN 0%
M16x4 4.22kN -4.23kN 0.236%
M16x15 -84kN 84kN 0%
M4x5 80.95kN -80.9kN 0.061%
M4x3 84.59kN -84.6kN 0.011%
M4x15 -9.795kN 9.8kN 0.051%
M4x16 4.22kN -4.23kN 0.236%
M15x16 -84kN 84kN 0%
M15x4 -9.795kN 9.8kN 0.051%
M15x3 -5.066kN 5.06kN 0.118%
M15x14 -79.448kN 79.4kN 0.060%
M3x4 84.59kN -84.6kN 0.011%
M3x2 80.027kN -80kN 0.0375%

80
M3x14 -4.46kN 4.47kN 0.223%
M3x15 -5.066kN 5.06kN 0.118%
M14x15 -79.448kN 79.4kN 0.060%
M14x3 -4.46kN 4.47kN 0.223%
M14x2 -21.9kN 21.9kN 0%
M14x13 -58.99kN 58.9kN 0.152%
M2x14 -21.9kN 21.9kN 0%
M2x13 2.72kN -2.73kN 0.366%
M2x3 80.027kN -80kN 0.0375%
M2x1 59.42kN -59.3kN 0.202%
M13x2 2.72kN -2.73kN 0.366%
M13x1 -60.82kN 60.7kN 0.197%
M13x14 -58.99kN 58.9kN 0.152%
M13x12 0kN 0kN 0%
M1x13 -60.82kN 60.7kN 0.197%
M1x12 14.4kN -14.4kN 0%
M1x2 59.42kN -59.3kN 0.202%
M12x1 14.4kN -14.4kN 0%
M12x13 0kN 0kN 0%
3.4 Fundamental Time Period
It is calculated manually as follows:
Fig. 3.4.1 Plan

81
Fig. 3.4.2 3-D view of the building.
Table 3.4.1 Dimensions of the building.
No. of storeys 1
Storey heights 2.5m
Beams 300mm*600mm
Columns 300mm*600mm
Slab thickness 150mm
Support conditions Fixed
Fig.3.4.3 Cross section of a column
b
d
x
y

82
• 𝐼𝑥𝑥 =
𝑑×𝑏3
12
…(3.4.1)
• 𝐼 𝑦𝑦 =
𝑏×𝑑3
12
…(3.4.2)
• 𝐸 = 5000√𝑓𝑐𝑘 …(3.4.3)
• For Parallel system of columns, k= k1 + k2 +..... + k6 = Total Stiffness ...(3.4.4)
• 𝑇 = 2𝜋
√ 𝑚
√𝑘
…(3.4.5)
• 𝑚𝑎𝑠𝑠 = 𝑐𝑜𝑙. 𝑤𝑖𝑑𝑡ℎ × 𝑐𝑜𝑙. 𝑑𝑒𝑝𝑡ℎ × 𝑐𝑜𝑙. ℎ𝑒𝑖𝑔ℎ𝑡 × 𝑤𝑒𝑖𝑔ℎ𝑡 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 …(3.4.6)
• 𝑘 =
12𝐸𝐼
𝐿3
…(3.4.7)
COLUMN b(mm) d(mm) L(mm) Ixx(𝑚𝑚4
) Iyy(𝑚𝑚4
) E(MPa) Kx (N/m) Ky (N/m)
1 300 600 2500 5400000000 1350000000 27386.12788 28393937.38 113575749.5
2 300 600 2500 5400000000 1350000000 27386.12788 28393937.38 113575749.5
3 300 600 2500 5400000000 1350000000 27386.12788 28393937.38 113575749.5
4 300 600 2500 5400000000 1350000000 27386.12788 28393937.38 113575749.5
5 300 600 2500 5400000000 1350000000 27386.12788 28393937.38 113575749.5
6 300 600 2500 5400000000 1350000000 27386.12788 28393937.38 113575749.5
Total stiff-
ness(Kx) =
∑Kx1,2,….,6
Total stiff-
ness(Ky) =
∑Ky1,2,….6
170363624.3 681454497.1
mass(kg) 184953
Time period X (sec) Y (sec)
0.207024681 0.10351234
Table 3.4.2 Manual calculation.
ETABS RESULT:
Tx=0.212sec Ty=0.109 sec
3. 5 Calculation of Base & Storey Shears
Plan:-

83
Fig. 3.5.1 Plan of the building.
Elevation:-
Fig. 3.5.2 Elevation of the building.
2.5m
4.5m

84
Table 3.5.1 shows the specifications of the building.
Plan Dimension 11mX6m
No. of storeys 1
No. of columns 8
No. of beams 10
No. of slabs 2
Sizes of beams 300mmx600mm(B1) & 300mmx450mm(B2)
Sizes of columns 300mmx600mm(C1) & 300mmx450mm(C2)
Slab Thickness 150mm
Brick Wall Thickness 230mm
Type of Soil Medium Soil
Zone II
Importance Factor(I) 1.5
Response Reduction Factor(R) 3
Table 3.5.2 shows the material of different members with their unit weights.
Material of Beams M30 (Unit weight=25kN/m3
)
Material of Columns M30 (Unit weight=25kN/m3
)
Material of walls Brick masonry(Unit Weight=20kN/m3
)
Table 3.5.3 shows the external load imposed on the building.
Live Load 2kN/m2
Floor Finishers Load 1kN/m2
The self weights and other loads will act as masses and for the calculation of base and storey
shears; we consider lump masses one at the Plinth level and the other at Storey-I.
3.5.1 Lump mass at storey levels.
We distribute the whole weight of all members into two seismic lumps at height of 2.5m and 7m
from ground. Column weights, beam weights, slab weights etc are divided accordingly. The figure
below shows the division.

85
Fig 3.5.1.1 Elevation of the building.
The weights above AA’ are lumped m2 and the weights between AA’ and BB’ are taken into
m1.The weight below BB’ is taken care by the ground and is not taken in the seismic weight of the
building.
Self weight
A
B
A’
B’

86
3.5.1.1 Plinth Level
Fig.3.5.1.1 Plan of the building. {Ground Floor}
Fig 3.5.1.2 Lump masses at different stories for the building.
Beams
W=Weight of building.
W= (Beam width) x (Beam Depth) x (Beam Length) x (Unit Weight of Concrete) …(3.5.1.1)
Ground
m1
2.5m
4.5m
m2
A
B
C D
E
F
G

87
WAB= (0.3x0.6x6x25) =27kN
WBC= (0.3x0.45x5x25) =16.875kN
WBH= (0.3x0.45x4x25) =13.5kN
WCD= (0.3x0.45x4x25) =13.5kN
WDH= (0.3x0.45x5x25) =16.875kN
WDE= (0.3x0.45x2x25) =6.75kN
WEF= (0.3x0.45x5x25) =16.875kN
WFH= (0.3x0.45x2x25) =6.75kN
WFG= (0.3x0.6x6x25) =27kN
WAG= (0.3x0.6x6x25) =27kN
Total Self Weight of Beams=172.125kN
Columns
W= (Column width) x Column Depth)x(Column Length)x(Unit Weight of Concrete) …(3.5.1.2)
Height=3.5m (See Fig.)
WC1= (0.3x0.6x3.5x25) kN =15.75kN
Total Self weight of C1=Wc1x4= (15.75x4) kN =63kN
Wc2= (0.3x0.45x3.5x25) kN =11.8125kN
Total Self weight of C2=Wc2x4= (11.8125x4) kN =47.25kN
Total Self weight of columns=63kN+47.25kN=110.25kN
Slab
W= (Slab Length) x (Column Width) x (Slab Thickness) x (Unit Weight of Concrete) …(3.5.1.3)
WSlab= (11x6x0.15x25) kN =247.5kN

88
3.5.1.2 Storey-I
Fig.3.5.1.1 Plan of the building.{Storey-I}
Fig 3.5.1.2 Lump masses at different stories for the building.
Beams
W= (Beam width) x (Beam Depth) x (Beam Length) x (Unit Weight of Concrete)
Ground
m1
2.5m
4.5m
m2

89
WAB= (0.3x0.6x6x25) =27kN
WBC= (0.3x0.45x5x25) =16.875kN
WBH= (0.3x0.45x4x25) =13.5kN
WCD= (0.3x0.45x4x25) =13.5kN
WDH= (0.3x0.45x5x25) =16.875kN
WDE= (0.3x0.45x2x25) =6.75kN
WEF= (0.3x0.45x5x25) =16.875kN
WFH= (0.3x0.45x2x25) =6.75kN
WFG= (0.3x0.6x6x25) =27kN
WAG= (0.3x0.6x6x25) =27kN
Total Self Weight of Beams=172.125kN
Columns
W= (Column width)x(Column Depth)x(Column Length)x(Unit Weight of Concrete)
Height=2.25m (See Fig.)
WC1= (0.3x0.6x2.25x25) kN=10.125kN
Total Self weight of C1=Wc1x4= (10.125x4)kN=40.5kN
Wc2= (0.3x0.45x2.25x25) kN=7.60kN
Total Self weight of C2=Wc2x4= (7.60x4)kN=30.375kN
Total Self weight of columns=40.5kN+30.375kN=70.875kN
Slab
W= (Slab Length)x(Column Width)x(Slab Thickness)x(Unit Weight of Concrete)
WSlab= (11x6x0.15x25) kN=247.5kN
3.5.2 Other addition to lump weight
Brickwork (Super dead weight)

90
Along x-direction
W=Brick Thickness x Wall Length x Height of wall x Unit Weight of Brick
W= (0.23x6x2.25x20) kN=62.1kN
No. of Walls=3
Total brickwork along x-direction= (62.1x3) kN=186.3kN
Along y-direction
W1= {(0.23x11x2.25x20) kN=113.85
No. of walls=2
W2= (0.23x5x2.25x20) kN=51.75kN
Total brickwork along y-direction= (W1x2)+W2=(113.85x2)kN+(51.75)kN
=279.45kN
Total Weight of brickwork= (Wx+Wy)
= (186.3+279.45)kN
=465.75kN
Floor Finishers Load (Super dead Weight)
Weight due to floor finishers= (Floor finishers load x Slab Area)
WFF= (1x6x11) kN =66kN
Live Load Weight
Weight due to live load= (Live load x Slab Area)
WLL= (2x6x11)kN=132kN
According to IS 1893 Part 1,(Table 3.4.1)
If LL<3kN/m2
, consider only 25% of the total weight.
Therefore, WLL=132x25%=33kN

91
3.5.2.2 Storey-I
Brickwork (Super deasd weight)
Along x-direction
W= (0.23x6x2.25x20) kN =62.1kN
No. of Walls=3
Total brickwork along x-direction=(62.1x3)kN=186.3kN
Along y-direction
W1={(0.23x11x2.25x20)kN=113.85
No. of walls=2
W2= (0.23x5x2.25x20)kN=51.75kN
Total brickwork along y-direction= (W1x2) +W2= (113.85x2) kN+ (51.75) kN
=279.45kN
Total Weight of brickwork= (Wx+Wy)
= (186.3+279.45) kN
=465.75kN
Floor Finishers Load (Super dead Weight)
Weight due to floor finishers= (Floor finishers load x Slab Area)
WFF= (1x6x11) kN =66kN
Live Load Weight
Weight due to live load= (Live load x Slab Area)
WLL= (2x6x11) kN =132kN
According to IS 1893 Part 1,
If LL<3kN/m2
, consider only 25% of the total weight.

92
Therefore, WLL=132x25%=33kN
3.5.3 Total Seismic Weight of the Building
WPLINTH = (Self weight of beams) + (Self weights of columns) + (Self weight of slab) + (Super
dead weight) + (Live load Weight)
WPLINTH= (172.125kN) + (110.25kN) + (247.5kN) + (465.75kN+66kN) + (33kN)
=1094.625kN
3.5.3.2 Storey-I
WStorey-I= (Self weight of beams) + (Self weights of columns) + (Self weight of slab) + (Super
dead weight) + (Live load Weight)
WPLINTH= (172.125kN) + (70.875kN) + (247.5kN) + (465.75kN+66kN) + (33kN)
=1055.25kN
3.5.4 Calculation of horizontal acceleration coefficient
𝑇 = 0.075 × ℎ0.75
(According to IS 1893 Part-I) …(3.5.4.1)
h=7m
T=0.3227s
Using Figure in seismic analysis, we get the value of Sa/g
Sa/g=2.5
The building lies in Zone-II.
Therefore, Z=0.10(Table 3.2.1)
I=1.5(Table 3.2.2)
R=3(Table 3.2.3)
𝐴ℎ =
𝑍𝐼𝑆 𝑎
2𝑅𝑔
…(7.4.2)
Putting values, we get Ah=0.0625

93
3.5.5 Base Shear
𝑉𝐵 = 𝐴ℎ × 𝑊 …(3.5.5.1)
Ah=0.0625
WPLINTH=1094.625Kn
WSTOREY-I=1055.25kN
W = WPLINTH+WSTOREY-i
= (1094.625kN) + (1055.25kN)
= 2149.875kN
VB = (0.0625 x 2149.875)
= 134.36kN
3.5.6 Distribution of Base Shears (Storey Shears)
Q1 = VB
W1h1
2
W1h1
2
+W2h2
2 …(3.5.6.1)
Q2 = VB
W2h2
2
W1h1
2
+W2h2
2 …(3.5.6.2)
h=height from the base
h1 = 2.5m
h2 = 7m
Putting values in eq (5.6.1), we get the value of Q1
Q1 = 15.70kN
Putting values in eq (5.6.2), we get the value of Q2
Q2 = 118.66kN

94
Fig 3.5.6.1 Storey shears.
3.5.7 Results (ETABS)
Fig 3.5.7.1 Results of storey shears in ETABS.
VB = 128.85kN
Q2 = 96.71kN
Q1 = 32.14
3.5.8 Comparison of manual and ETABS Result
Table 3.5.8.1 shows the comparison of manual and ETABS result.
Manual Result ETABS Result
VB(kN) 134.36 128.85
Q1(kN) 15.70 32.14
Q2(kN) 118.66 96.71
Height(m)
Storey Shear (kN)
2.5
7
15.70 134.36

95
4. Conclusion
 A cantilever beam of length of 5m and a point load of 10 kN action on the free end was analyzed
with and a point load of 10kN with one end fixed and other free .The hand calculation result Of
BMD,SFD, and deflection are exactly matching the ETABS result.
 A beam was analyzed with length of 5m and a UDL of 4kN/m was applied with one end fixed and
other free. The hand calculation result Of BMD,SFD, and deflection are exactly matching the
ETABS result
 A beam was analyzed with length of 5m and a UDL of 4kN/m was applied with one hand hinged
and other fixed. The hand calculation result Of BMD,SFD, and deflection are exactly matching the
ETABS result
 A beam was analyzed with length of 5m and a point load of 8kN with both ends fixed. The hand
calculation result Of BMD,SFD, and deflection are exactly matching the ETABS result
 A beam was analyzed with length of 3.5m and a UDL of 10kN/m was applied with both ends fixed.
The hand calculation result Of BMD,SFD, and deflection are exactly matching the ETABS result
 A portal frame was analyzed with one beam and two columns and a span of 4m and height 3m
with a UDL of 12kN/m. The hand calculation result Of BMD,SFD, and deflection are approximately
matching the ETABS result
 A portal frame was analyzed with two beam and two columns and a span of 4m and height 4m
with a UDL of 3kN/m. The hand calculation result Of BMD,SFD, and deflection are approximately
matching the ETABS result.
 IS 1893 Part-1(2002) and IS 875 Part-3(2002) were studied for the introduction to seismic and
wind loads respectively.
 Service building in city plot area was modelled in ETABS and different static loads were applied
both vertical(super dead, live) and horizontal(seismic and wind).
 A typical truss was analyzed using method of sections/joints for gravity and wind
loads,individually as well as combined and also the same analysis was done in STAAD.pro and
a %age variation of less than 1% were found.
 A simple building with 6 columns was analyzed for time period using dynamic method and ETABS
and a %age variation of 4.5% was found.
 Service building was also analyzed for storey shears using IS 1893 Part-1(2002) and a %age varia-
tion of 5% was found.
 two site visits to actupuram and city plot area was done. Observing the different plans being exe-
cuted and seeing various difficulties.

96
References
[1] Bureau of Indian standards: IS-875, part 1 (1987), Dead Loads on Buildings and Structures,
New Delhi, India.
[2] Bureau of Indian standards: IS-875, part 2 (1987), Live Loads on Buildings and Structures,
New Delhi, India.
[3] Bureau of Indian standards: IS-875, part 3 (1987), Wind Loads on Buildings and Structures,
New Delhi, India.
[4] Bureau of Indian standards: IS-1893, part 1 (2002), Seismic Loads on Buildings and Structures,
New Delhi, India.
[5] www.google.com/images
[6] www.wikipedia.org

2014B4A2813P-Bhabha Atomic Research Centre (Visakhapatnam)

Recommended

Recommended

More Related Content

Similar to 2014B4A2813P-Bhabha Atomic Research Centre (Visakhapatnam)

Similar to 2014B4A2813P-Bhabha Atomic Research Centre (Visakhapatnam) (20)

2014B4A2813P-Bhabha Atomic Research Centre (Visakhapatnam)