Final

Chapter 1

INTRODUCTION

1.1 Background

The sky rocketing population and haphazard land use has decreased the land availability

for construction of any structure requiring large plinth area. With due consideration to

this fact, a High-Rise building seemed to be one of the best options. Taking into account

this fact, we have come up with a project work on “Computer Aided Structural Analysis

and Design of High-Rise Hospital Building” located in Kathmandu Valley. It is intended

to design the hospital so as to provide health facilities to the general public. The hospital

will surely aid in the development of the society and hence country as a whole. The

people will also be benefitted by the quality service provided by the hospital.

A designer has to deal with various structures ranging from simple to more complex ones

like a multistoried frame building. All the structural members are subjected to various

loads like concentrated loads, uniformly distributed loads, live loads, earthquake load,

wind load, etc. The structure transfers the loads acting on it to the supports and ultimately

to the ground. While transferring the loads acting on the structure, the members of the

structure are subjected to the internal forces like axial forces, shear forces, bending and

torsion moments.

Structural Analysis deals with analyzing these internal forces in the structural member

developed as a result of various loading conditions or combinations.

Structural Design deals with sizing various members of the structures to resist the

internal forces to which they are subjected in the course of their life cycle. Unless the

proper structural detailing method is adopted the structural design will be no more

effective. A Standard Code of practice (Indian Standard code in our case) should be

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thoroughly followed and implemented for proper analysis, design and detailing with

respect to safety, economy, stability, strength.

Besides the scarcity of land, earthquake is one of the dominant constraints while

designing the multistory frame building in earthquake prone zone like Kathmandu.

According to IS1893: 2002, Kathmandu lies on V zone, the severest one, hence the effect

of earthquake is predominant to wind load. So, the building is analyzed for earthquake as

lateral load. The seismic coefficient design method as stipulated in IS 1893:2002 is

applied to analyze the building for earthquake. The 3- dimensional moment resistance

frame with shear wall is considered as the main structural system of the building.

This project work has been undertaken for the partial fulfillment of requirements for the

Bachelor’s Degree in Civil Engineering. This project work contains structural analysis,

design and detailing of a hospital building located in Kathmandu Valley. All the

theoretical knowledge on analysis and design acquired during the course works are

utilized with practical application. The main objective of the project work is to acquaint

us in the practical aspects of Civil Engineering.

1.2 Theme of Project work

This group under the project work has undertaken the Computer Aided Structural Analysis

and Design of High-Rise Hospital Building. The main aim of the project work under the

title is to acquire knowledge and skill with an emphasis on practical application. Besides

the utilization of analytical methods and design approaches, exposure and application of

various available codes of practices is another aim of the work.

1.3 Objectives

The specific objectives of the project work are:

i. Preparation of the plan of the building to meet the requirements for its intended

use.

ii. Identification of the structural arrangement of the plan.

iii. Modeling the building for structural analysis.

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iv. Analyzing the structure using structural analysis program.

v. Sectional design of the structural members.

vi. Preparation of detail structural drawing of the design.

1.4 Building Description

Building Type : High-Rise Hospital Building

Structural System : RCC Space Frame with Shear Wall

Plinth area covered : 1418 m2

Type of Foundation : Mat Foundation

No. of Storey : 16 + 1 (basement)

Floor Height : 3m

Type of Sub-Soil : Medium Soil

Seismic zone : V

Expansion Joints : Expansion joints are provided

According to IS 456-2000, Clause 27, structures in which changes in plan dimensions

take place abruptly shall be provided with expansion joints at the section where such

changes occur. Reinforcement shall not extend across an expansion joints and the break

between the sections shall be completed. Normally structure exceeding 45m in length is

designed with one or more expansion joints.

The design is intended to serve quality service to the general public round the clock.

1.5 Identification of Loads

i. Dead loads are calculated as per IS: 875( Part I) – 1987

ii. Seismic load is calculated according to IS :1893 (Part I) – 2002 considering

Kathmandu located at Zone V

iii. Imposed loads according to IS : 875(Part II) – 1987

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1.6 Code of Practices

Following codes of practices developed by Bureau of Indian Standards were followed in

the analysis and design of building:

i. IS 456:2000 (Code of practice for plain and reinforced concrete)

ii. IS 1893 (part 1):2002 (Criteria for earthquake resistant design of structures)

iii. IS 13920: 1993 (Code of practice for ductile detailing of reinforced concrete

structures subjected to seismic forces)

iv. IS 875 (part 1):1987 (to assess dead loads)

v. IS 875 (part 2):1987 (to assess live loads)

vi. IS 875 (part 5):1987 (for load combinations)

vii. SP 16 and SP 34 (design aids and hands book)

1.7 Idealization and Assumption in Analysis and Design

Various assumptions have been made in analysis and design of the structures, for consideration of simplicity and economy, viz.:

1. Tensile strength of concrete is ignored.

2. Shrinkage and temperature strength are negligible.

3. Adhesion between concrete and steel is adequate to develop full strength.

4. Seismic and wind load do not occur simultaneously.

5. Centerlines of beams, columns and shear walls are concurrent everywhere.

1.8 Method of Analysis

The building is modeled as a space frame. SAP2000 V12 is adopted as the basic tool for

the execution of analysis. SAP2000 program is based on Finite Element Method. Due to

possible actions in the building, the stresses, displacements and fundamental time periods

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are obtained using SAP2000 which are used for the design of the members. Mat

foundation, staircase and slabs are analyzed separately.

1.9 Design

The following materials are adopted for the design of the elements:

i. Concrete Grade :

M30 for the beams, columns, slabs and shear walls

M20 for the foundation

ii. Reinforcement Steel :

Fe500 for staircase

Fe415 for all other structural members

The design of RC elements is carried out using Limit State Method. The design is based

on various Indian Standards Code of Practice for Plain and Reinforced Concrete IS 456-

2002, Design Aids for Reinforced Concrete to IS 456:1987(SP-16), Criteria for

Earthquake Resistant Design Structures IS 1893-2002, Ductile Detailing of Reinforced

concrete Structures Subjected to Seismic Forces IS13920:1993, Handbook on Concrete

Reinforcement and Detailing SP-34. Various handbooks like Reinforced Concrete

Designer’s Handbook – Charles E. Reynolds and James C. Stedman, are extensively used

in the progress of design.

The design moments, shear forces, axial forces and torsions are taken as computed by

computer software program “SAP2000 V12” for the worst possible combination and a

number of hand calculations are done so as to verify the reliability of the design results

suggested by the software.

1.10 Detailing

The space frame is considered as a Special Moment Resisting Frame (SMRF) with a

special detailing to provide ductile behavior and comply with the requirements given in

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IS 13920:1993. Handbook on Concrete Reinforcement and Detailing SP-34 has been

extensively used for the detailing of beams and columns.

1.11 Scope

The project work is limited to the structural analysis and design only.

Design and Detailing of following structural elements is performed:

1. Slab

2. Beam

3. Column

4. Staircase

5. Mat plus Foundation

Design and layout of the building services like pipeline, electrical appliances, sanitary

and sewage system are not covered.

The required parking facilities are assumed to be provided in a separate parking structure,

however its design is not concerned with this project.

The project is not concerned with the existing soil condition of the locality.

The bearing capacity of the soil is assumed.

The environmental, social and economical condition of the locality is not taken into

consideration.

The project work is only related with the practical application of the studied courses in

the field.

Detail cost estimate of the project is not included in this report.

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Chapter 2

STRUCTURAL SYSTEM AND PRELIMINARY DESIGN

2.1 Structural System

Any structure is made up of structural elements (load carrying, such as beams and

columns and non-structural elements (such as partitions, false ceilings, doors). The

structural elements put together, constitute the structural systems. Its function is to resist

effectively the action of gravitational and environmental loads, and to transmit the

resulting forces to the supporting ground without significantly disturbing the geometry,

integrity and serviceability of the structure.

2.2 Structural Arrangement Plan

The planning of the building has been done by the group and is not an existing plan. The

final plan of the building was a result of review of various literatures like Timesavers

Standard, various codes of practice and some other reference books suggested by our

Supervisor. The positioning of the columns, staircases, toilets and bathrooms, lifts are

appropriately done and accordingly beam arrangement is carried out so that the whole

building will be aesthetically, functionally and economically feasible.

The aim of the design is the achievement of an acceptable probability that structures

being design will perform satisfactorily during their intended life with an appropriate

degree of safety, they should sustain all the loads and deformations of normal

construction and use and have adequate durability and resistance to the effect of misuse

and fire.

The building consists of two blocks separated by an expansion joint.

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2.3 Need of Preliminary Design

It is necessary to know the preliminary section of the structure for the detail analysis. As

the section should be given initially while doing analysis in almost all software, the need

of preliminary design is vital. Only dead loads and live loads are considered while doing

preliminary design. Preliminary design is carried out to estimate approximate size of the

structural members before analysis of structure.

2.4 Preliminary Design

Slab

Each floor slabs are divided into different types as per end conditions defined in IS

456:2000.

Design of General use Slab type –S1

Size of slab = 6.7 X 6.1 m2

Ly/Lx = 6.7/6.1 = 1.098 <2

Therefore, the slab is two-way slab.

Calculating, depth of slab

d = span/αβγδλ

where,

α depends on support condition

β depends on span

γ depends on % tension steel

δ depends on compression steel

λ is considered for flanged beam

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or, d = 6100/(26*1*1.8*1*1)

d = 124.9 mm

So take D= 150 mm

Thus effective depth of 125 mm is adopted. So, take overall depth, D = 150 mm.

(Clear cover = 25mm)

Beam

The design is as per IS-456:2000. Same section of beam is provided throughout the

building.

Span of beam = 6.1m

Since Depth = span/(12~15)

= 6100/13

= 470 mm

Adopt 500 mm

Column

It is designed on the basis of IS 456:2000 & IS 456:1978.

Column size: grade M25

Pu = 0.4 *fck* Ac + 0.67*fy * Ast

= 0.4*25*(Ag-0.02Ag)+ 0.67*415*Ag*0.02 (Taking pt=2%)

= 15.61Ag

~ 16 Ag

9

Table 1: Calculation of preliminary loads on column

Load(KN) Size of column (mm2)

948 250*250

1887 360*360

3545 510*510

3785 510*510

7089 690*690

7075 690*690

2862 440*440

6285 650*650

2600 420*420

Thus, adopting the sizes as listed below:

450*450 mm2

500*500mm2

700*700mm2

Staircase design

15 cm X 30 cm

Width of flight=1.2m

Height of each flight =3/2=1.5m

10

No. of riser = 1.5/.15=10

Tread: 9

Space for tread: 9X30 = 270

Fig.1: Staircase Plan

Passage= 6-2.7-1.2=2.1

Room Size=6 m X 6.7m

Stair case Load

Live load = 5 KN/m2

R X T=15 cm X 30 cm

Height of floor: 3.0 m

Riser = 11

Tread = 10

10 X 30=300 cm = 3m

11

Passage = 6.7-3.2=1.7m

Effective span:

Leff = c/c distance between support

= 6.7 + .25/2 = 0.25/2 = 6.95 m

Assume slab thickness= L/26

= 6950/26*1.4

= 200 mm

Assume overall depth= 200 mm

Load on going:

√(T2 + R2) = √0.30 2 + 0.15 2

= 0.335 m

i. Self weight of slab = 25 * 0.2 * 0.335/0.3 = 5.6 KN/m2

ii. Step weight of steps = 25 * (1/2*0.15) = 1.9 KN/m2

iii. Finish (thickness = 25 mm) = 0.67 KN/m2

iv. Live load = 5 KN/m2

_______________

13.15 KN/m2

Load on landing:

i. Self weight of slab = 25 X 0.150 = 3.75 KN/m2

ii. Finish = 0.67 KN/m2

iii. Live load = 5 KN/m2

______________________

9.42 KN/m2

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Determination of Stiffness Centre (SC) and Mass Centre (MC)

Calculation of Centre of Stiffness ( block 1)

For First floor:

From IS:456-2000,

Modulus of Elasticity of Concrete (Ec) = 5000√ f ck =5000√25= 25000 N/mm2

=2.5E10 N/m2

Moment of Inertia (I1) =0.45*0.45**3/12 =3.417E-3 m4

Moment of Inertia (I2) = 0.5*0.5**3/12 =5.208E-3 m4

Moment of Inertia (I3) = 0.7*0.7**3/12=2.0E-2 m4

Height (h) =3m

Stiffness (k1) =12EI1/h3=3.797E7=k (say)

Stiffness (k2) =5.787E7=1.524k


_X = (k2+4k3+k1)*20+ (k1+k2+4k3)*40+ (k1+k2+5k3)*60+(k1+k2+5k3)*80

+ (k1+k2+5k3)*100+ (k1+6k2)*120

(7k1+16k2+23k3)

= 62.66f t= 19.105m

_Y= (2k2+5k3)*8+(2k2+5k3)*30+ (2k2+5k3)*40+(2k2+5k3)*62+ (4k2+3k3)*72+4k2*94

(7k1+16k2+23k3)

13

=40.963 ft

=12.49 m

Similarly, centre of stiffness of block2 is = (12.19, 8.52) m.

From excel sheet data centre of mass of block1 is = (20.38, 12.29) m.

And centre of mass of block2 is = (12.21, 9.22) m.

For block 1

Eccentricity ex= 1.275 m

ey= 0.2 m

For block 2

Eccentricity ex = 0.02 m

ey = 0.7 m

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CHAPTER 3

LOAD ASSESSMENT

3.1 Introduction

As described earlier, the building is a RCC framed structure, located in the

Kathmandu valley. Thus wind loads, snow loads, and other special types of loads

described by IS 875 (part 5):1987 can be taken as negligible as compared to the

dead, live and seismic loads.

3.1.1 Dead Loads:

According to the IS 875:1964:

The dead load in a building shall comprise the weights of all walls,

partitions, floors and roofs and shall include the weights of all other

permanent features in the building.

3.1.2 Live Loads:

It means the load assumed or known resulting from the occupancy or use

of a building and includes the load on balustrades and loads from movable

goods, machinery and plant that are not an integral part of the building.

3.1.3 Seismic Loads:

These are the load resulting from the vibration of the ground underneath

the superstructure during the earthquake. The earthquake is an

unpredictable natural phenomenon. Nobody knows the exact timing and

magnitude of such loads. Seismic loads are to be determined essentially to

produce an earthquake resistant design.

Seismic loads on the building may be incorporated by-

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1. Response Spectrum Method: In this method the design earthquake

forces are determined adopting IS 1893:2002. These design forces for the

buildings located along two perpendicular directions may be assumed to

act separately along each of these two horizontal directions.

2. Time History Analysis: In it the ground is subjected to a

predetermined acceleration and subsequent stress in the structural

elements are determined by appropriate methods.

3.2 Unit Weights for Dead Load Calculation

1. RCC: (IS 875 (part 1) :1987 table 1)

a) For slabs and shear walls:

γRCC = 25 KN/m3

b) For columns:

γRCC = 25 KN/m3

c) For Beams:

γRCC = 25 KN/m3

2. Plaster (20mm thickness):

γplaster = 20.40 KN/m3

3.3 Live Loads

1. On Floors: (IS875(part2):1987 table1)

2. On Partition walls: Live Load=1KN/m2

(Assuming a minimum live load as per IS 875(part2):1987, 3)

3. On roofs: Live Load=1.5 KN/m2

(Assuming access not provided except for the case of maintenance),

(IS875(part2):1987)

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3.4 Seismic Load

Seismic weight is the total dead load plus appropriate amount of specified

imposed load. While computing the seismic load weight of each floor, the weight

of columns and walls in any story shall be equally distributed to the floors above

and below the storey. The seismic weight of the whole building is the sum of the

seismic weights of all the floors. It has been calculated according to IS: 1893(Part

I) – 2002.

IS: 1893(Part I) – 2002 states that for the calculation of the design seismic forces

of the structure the imposed load on roof need not be considered

Base Shear Calculation

According to IS 1893 (Part I): 2002 Cl. No. 6.4.2 the design horizontal seismic

coefficient Ah for a structure shall be determined by the following expression:

Ah=Z I Sa

2 R g

Where,

Z = Zone factor given by IS 1893 (Part I):2002 Table 2, Here for Zone V,Z = 0.36

I = Importance Factor, I = 1.5 for high rise hospital building

R = Response reduction factor given by IS 1893 (Part I): 2002 Table 7, R = 5.0

Sa/g = Average response acceleration coefficient which depends on Fundamental

natural period of vibration (Ta).

According to IS 1893 (Part I): 2002 Cl. No. 7.6.1

The approximate fundamental natural period of vibration (Ta) in second of

moment resisting frame building without brick infill panel may be estimated by

empirical expression:

Ta = 0.075h0.75 for RC frame building

Where,

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h= height of building in metre. This excludes the basement storeys, where

basement walls are connected with the ground floor deck or fitted between the

building columns. But, it includes the basement storeys, when they are not so

connected.

For h= 51 m

Ta= 1.43 sec.

And Sa/g = 0.95 (From IS Code 1893:2000) Fig 2

Now,

A h= z2

×IR

×Sag

=0.362

×15

× 0.95=.0513

According to IS 1893 (Part I) : 2002 Cl. No. 7.5.3 the total design lateral force or

design seismic base shear (VB) along any principle direction is given by

VB = Ah x W

Where, W = Seismic weight of the building

Wblock 1= 149259.12KN

Wblock 2 = 81603.644 KN

Then,(Vb)block 1=0.513*149259.12KN=7656.99KN

(Vb)block 2=0.513*81603.644KN=4186.27KN

The total base shear is firstly distributed horizontally in basement in proportion to

the stiffness. Then according to IS 1893 (part I):2002 C.L. No. 7.7.1 the design

base shear (VB) computed above shall be distributed along the height of the

building as per the following expression:

Qi=Vb∗(Wi∗h i2)

∑j=1

n

Wi∗hi2

Where, Qi=Design lateral force at floor i

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W1

W2

W3

W4

W5

W1

W2

W3

W4

W5

Wi=Seismic weight of floor i

hi=Height of floor I measured from base

n=no. of stories in the building

Additional Shear Calculation Due to Torsion in Building

Center of Rigidity (CR) - A point through which a horizontal force is applied

resulting in translation of the floor without any rotation

Determination of Stiffness Centre (SC) and Mass Centre (MC)

Calculation of Centre of Stiffness ( block1)

For First floor:

From IS:456-2000,

Modulus of Elasticity of Concrete (Ec) = 5000√ f ck =5000√25= 25000 N/mm2

=2.5E10 N/m2

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Fig.2 Storey Shear in X direction Fig.3 Storey Shear in Y direction

Moment of Inertia (I1) =0.45*0.45**3/12 =3.417E-3 m4

Moment of Inertia (I2) = 0.5*0.5**3/12 =5.208E-3 m4

Moment of Inertia (I3) = 0.7*0.7**3/12=2.0E-2 m4

Height (h) =3m

Stiffness (k1) =12EI1/h3=3.797E7=k (say)



_

X = (k2+4k3+k1)*20+ (k1+k2+4k3)*40+ (k1+k2+5k3)*60+(k1+k2+5k3)*80

+ (k1+k2+5k3)*100+ (k1+6k2)*120

(7k1+16k2+23k3)

= 62.66f t= 19.105m

_

Y=(2k2+5k3)*8+(2k2+5k3)*30+(2k2+5k3)*40+(2k2+5k3)*62+ (4k2+3k3)*72+

4k2*94

(7k1+16k2+23k3)

=40.963 ft

=12.49 m

Similarly, centre of stiffness of block2 is = (12.19, 8.52) m.

From excel sheet data centre of mass of block1 is = (20.38, 12.29) m.

20

And centre of mass of block 2 is = (12.21, 9.22) m.

For block 1

Eccentricity ex= 1.275 m

ey= 0.2 m

For block 2

Eccentricity ex = 0.02 m

ey = 0.7 m

Response spectrum Analysis:

The response history analysis provides structural response r(t) as a function of

time, but the structural design is usually based on the peak values of forces and

deformations over the duration of earthquake induced response. The peak

response can be determined directly from the response spectrum for the ground

motion in case of single degree of freedom. The peak response of multi-degree

freedom systems can be calculated from the response spectrum.

The exact peak value of Nth mode response rn(t)=-rnstAn

Where, An is the ordinate of the pseudo acceleration spectrum corresponding to

the natural period Tn and Damping ratio ε.

The peak value ro of the total response can be estimated by combining the modal

peaks rno according to one of the modal combination rules. Because the natural

frequencies of transverse vibration of a beam are well separated, the SRSS

combination is satisfactory. Thus,

Ro=¿)

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3.5 Load Combination

In the course of analysis, different load cases and combinations are considered to

obtain the most critical stresses in the element of the structure. The load cases

considered for the structural analysis are:

i. Dead Load (DL)

ii. Live Load (LL)

iii. Earthquake load in X (EQx) static

iv. Earthquake load in Y (EQy) static

v. Eathquake load in X (Rx) response spectrum method

vi. Eathquake load in y (Ry) response spectrum method

Following load combination as per IS 1893(Part I):2002 are adopted:

i. 1.5(DL + LL)

ii. 1.2(DL + LL + EQx)

iii. 1.2(DL + LL - EQx)

iv. 1.2(DL + LL + EQy)

v. 1.2(DL + LL - EQy)

vi. 1.5(DL + EQx)

vii. 1.5(DL - EQx)

viii. 1.5(DL + EQy)

ix. 1.5(DL - EQy)

x. 0.9DL+1.5EQx

xi. 0.9DL-1.5EQx

xii. 0.9DL+1.5EQy

xiii. 0.9DL-1.5EQy

xiv. 1.5(DL + Rx)

xv. 1.5(DL - Rx)

xvi. 1.5(DL + Ry)

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xvii. 1.5(DL - Ry)

xviii. 1.2(DL + LL + Rx)

xix. 1.2(DL + LL - Rx)

xx. 1.2(DL + LL + Ry)

xxi. 1.2(DL + LL - Ry)

After checking the results, it was found that the stress developed are most critical

for the following load combinations:

i. 1.5 (DL + LL)

ii. 1.2 (DL + LL + EQx)

iii. 1.2 (DL + LL - EQx)

iv. 1.2 (DL + LL + EQy)

v. 1.2 (DL + LL - EQy)

vi. 1.2(DL + LL + Rx)

vii. 1.2(DL + LL - Rx)

viii. 1.2(DL + LL + Ry)

ix. 1.2(DL + LL - Ry)

The characteristic loads considered in the design of foundation are:

i. Dead Load plus Live Load

ii. Dead Load plus Earthquake Load

iii. Dead Load minus Earthquake Load

To find the stress at the various points of the foundation, depth of footing and

reinforcements most critical factored loads are taken into account.

23

Chapter 4

MODELING AND STRUCTURAL ANALYSIS

4.1 Salient Feature of Sap2000

SAP2000 represents the most sophisticated and user-friendly release of SAP series of

computer programs. Creation and modification of the model, execution of the analysis,

and checking and optimization of the design are all done through this single interface.

Graphical displays of the results, including real-time display of time-history

displacements are easily produced.

The finite element library consists of different elements out of which the three

dimensional FRAME element was used in this analysis. The Frame element uses a

general, three dimensional, beam-column formulation which includes the effects of

biaxial bending, torsion, axial deformation, and biaxial shear deformations. Structures

that can be modeled with this element include:

i. Three-dimensional frames

ii. Three-dimensional trusses

iii. Planar frames

iv. Planar grillages

v. Planar trusses

A Frame element is modeled as a straight line connecting two joints. Each element has its

own local coordinate system for defining section properties and loads, and for

interpreting output.

Each Frame element may be loaded by self-weight, multiple concentrated loads, and

multiple distributed loads. End offsets are available to account for the finite size of beam

and column intersections. End releases are also available to model different fixity

conditions at the ends of the element. Element internal forces are produced at the ends of

each element and at a user specified number of equally-spaced output stations along the

length of the element.

24

Loading options allow for gravity, thermal and pre-stress conditions in addition to the

usual nodal loading with specified forces and or displacements. Dynamic loading can be

in the form of a base acceleration response spectrum, or varying loads and base

accelerations.

4.2 Structural Analysis

The structural analysis is done using the SAP 2000(V12). Since the design is aimed for

the adequate resistance to the earthquake, the analysis is done by both methods of

analysis namely Static and Dynamic analysis.

For the static analysis, the design lateral force is distributed to the various floor levels

which is then distributed to individual lateral load resisting elements depending on floor

diaphragm action. In SAP 2000, the modeling for rigid floor diaphragm is done. A floor

diaphragm is modeled as rigid horizontal plane parallel to each other in X – Y plane.

Each floor diaphragm is established by a joint in the plane of the diaphragm called

Master Joint of the diaphragm. The location of the master joint on each floor diaphragm

is arbitrary and is selected by the user. For the convenience, the master joint is located at

the center of mass of each floor. All the joints that exists on the diaphragm are connected

to the master joint by rigid links and their displacement are dependent on the

displacement of the master joint. Therefore these points are called dependent joint or

slave joint. The lateral forces calculated earlier are applied to each master joint located

on each floor of the building.

IS 1893:2002 specifies that for the zone IV & V, the dynamic analysis shall be done for

the regular buildings greater than 40 m in height and for the irregular buildings greater

than 12 m. So the dynamic analysis is also done for the building by the use of response

spectrum method in SAP 2000. The design base shear (Vb) is compared with the base

shear (Vb’) calculated using fundamental period (Ta). Where Vb is less than Vb’,all the

response quantities like member forces, store shears and base reactions shall be

multiplied by Vb’/Vb

The natural period of the building calculated is 1.43 seconds whereas the period obtained

from the analysis result of the SAP 2000 is 1.14 seconds.

25

4.2.1 Storey Drift

It is the displacement of one level relative to the other level above or below.

According to IS 1893:2002 Clause 7.11.1, the storey drift due to the minimum

specified design lateral force with partial load factor of 1.0 shall not exceed 0.004

times the storey height which comes to be 12 mm. However, the storey drift value

obtained from SAP analysis of the building is found out to be 11.3 mm which is

lower than the limited value as specified by code. Hence, our building is safe

against the storey drift.

4.2.2 Stress Envelope

Stress envelope is the diagrammatic representation of maximum and minimum

stress values in any given structural member resulting from different load

combinations assigned for the analysis. The stresses in the structural elements are

interpreted with the stress envelope which represents the maximum values of the

stresses in the member.

4.3 Inputs and Outputs

The design of earthquake resistant structure should aim at providing appropriate dynamic

and structural characteristics so that acceptable response level results under the design

earthquake. The aim of design is the achievement of an acceptable probability that

structures being designed will perform satisfactorily during their intended life. With an

appropriate degree of safety, they should sustain all the loads and deformations of normal

construction and use and have adequate durability and adequate resistance to the effects

of misuse and fire.

For the purpose of seismic analysis of our building we used the structural analysis

program SAP2000. SAP2000 has a special option for modeling horizontal rigid floor

diaphragm system. A floor diaphragm is modeled as a rigid horizontal plane parallel to

global X-Y plane, so that all points on any floor diaphragm cannot displace relative to

each other in X-Y plane. This type of modeling is very useful in the lateral dynamic

26

analysis of building. The base shear and earthquake lateral force are calculated as per

code IS 1893(part1)2002 and are applied at each master joint located on every storey of

the building.

Table 2: Sample output of SAP2000

FrameDesignTyp

e LocationFTopComb

o FTopAreaFBotComb

o FBotArea VCombo VRebarText Text m Text m2 Text m2 Text m2/m

4668 Beam 0.0000 envelope 0.0013 envelope 0.0009 envelope 0.0000

4668 Beam 0.4877 envelope 0.0008 envelope 0.00061.5DL+1.5LL 0.0003


4668 Beam 1.4630 envelope 0.0006 envelope 0.0006 envelope 0.00004668 Beam 1.9507 envelope 0.0006 envelope 0.0006 envelope 0.00004668 Beam 2.4384 envelope 0.0007 envelope 0.0009 envelope 0.0000



4.4 Joint Displacement at Expansion Joint

After the analysis of structure using SAP2000 the maximum displacement of nodes at the

expansion joint was found out. It is clear from table below that the available gap for

expansion joint is much greater relative displacement of the nodes at joint. In order to

reduce the pounding effect between the two units, the adequate spacing is provided. The

separation between the adjacent units of the same buildings in between shall be separated

by a distance equal to the amount R times the sum of the calculated storey displacements

to avoid the damaging contact when the two units deflect towards each other. Since the

elevation levels of both units are same in our case the factor R is replaced by R/2. Hence

the building will not collide at the expansion joint during earthquake condition.

27

Drift Calculation:

Table 3: Drift calculation of Block 1

storey drift Drift ratiofloor U1 U2 D1 D2 DR1 DR2roof 0.1165 0.1548 15th 0.1127 0.1498 0.0038 0.005 0.001267 0.00166714th 0.108 0.1429 0.0047 0.0069 0.001567 0.002313th 0.1026 0.1355 0.0054 0.0074 0.0018 0.00246712th 0.0964 0.1271 0.0062 0.0084 0.002067 0.002811th 0.0896 0.1178 0.0068 0.0093 0.002267 0.003110th 0.082 0.1076 0.0076 0.0102 0.002533 0.00349th 0.074 0.0968 0.008 0.0108 0.002667 0.00368th 0.0655 0.0855 0.0085 0.0113 0.002833 0.0037677th 0.0576 0.0737 0.0079 0.0118 0.002633 0.0039336th 0.0478 0.0619 0.0098 0.0118 0.003267 0.0039335th 0.0388 0.05 0.009 0.0119 0.003 0.0039674th 0.0299 0.0383 0.0089 0.0117 0.002967 0.00393th 0.0214 0.0271 0.0085 0.0112 0.002833 0.0037332th 0.0134 0.0116 0.008 0.0113 0.002667 0.0051671th 0.0066 0.0077 0.0068 0.0039 0.002267 0.0013ground 0.0015 0.0012 0.0051 0.0065 0.0017 0.002167basement0 0 0 0.0015 0.0012 0.0005 0.0004

Table 4: Drift calculation of Block 2

storey drift Drift ratiofloor U1 U2 D1 D2 DR1 DR2roof -0.1201 -0.1417 15th -0.1147 -0.1404 -0.0054 -0.0013 -0.0018 -0.0004314th -0.1083 -0.1331 -0.0064 -0.0073 -0.00213 -0.0024313th -0.1021 -0.125 -0.0062 -0.0081 -0.00207 -0.002712th -0.095 -0.1163 -0.0071 -0.0087 -0.00237 -0.002911th -0.0874 -0.1069 -0.0076 -0.0094 -0.00253 -0.0031310th -0.0793 -0.0968 -0.0081 -0.0101 -0.0027 -0.003379th -0.0708 -0.0863 -0.0085 -0.0105 -0.00283 -0.00358th -0.062 -0.0754 -0.0088 -0.0109 -0.00293 -0.00363

28

7th -0.0531 -0.0643 -0.0089 -0.0111 -0.00297 -0.00376th -0.0441 -0.0532 -0.009 -0.0111 -0.003 -0.00375th -0.0353 -0.0432 -0.0088 -0.01 -0.00293 -0.003334th -0.0319 -0.0319 -0.0034 -0.0113 -0.00113 -0.003773th -0.0189 -0.022 -0.013 -0.0099 -0.00433 -0.00332th -0.0118 -0.0134 -0.0071 -0.0086 -0.00237 -0.002871th -0.0058 -0.0062 -0.006 -0.0072 -0.002 -0.0024ground -0.0015 -0.0011 -0.0043 -0.0051 -0.00143 -0.0017basement0 0 0 -0.0015 -0.0011 -0.0005 -0.00037

Minimum spacing of expansion joint = Maximum deflection of Block 1 in

positive X-direction + Maximum deflection of Block 2 in negative X- direction

= 0.0113 + 0.0113

= 0.0226 m

So we provided expansion joint of 30 cm.

29

CHAPTER 5

SECTION DESIGN

5.1 Limit State Method

` In the method if design based on limit state concept, the structure shall be designed to

withstand safely all loads liable to act on it throughout its life; it shall also satisfy the

serviceability requirements, such as limitations on deflection and cracking. The

acceptable limit for the safety and serviceability requirements before failure occurs is

called a ‘limit state’. The aim of design is to achieve acceptable probabilistic that the

structure will not become unfit for the use for which it is intended, that is, that it will not

reach a limit state.

Assumptions for flexural member:

i) Plane sections normal to the axis of the member remain plane after bending.

ii) The maximum strain in concrete at the outermost compression fiber is 0.0035.

iii) The relationship between the compressive stress distribution in concrete and the

strain in concrete may be assumed to be rectangle, trapezoidal, parabola or any

other shape which results in prediction of strength in substantial agreement with

the result of test. For design purposes, the compressive strength of concrete in the

structure shall be assumed to be 0.67 times the characteristic strength. The partial

safety factor γm = 1.5 shall be applied in addition to this.

iv) The tensile strength of concrete is ignored.

v) The design stresses in reinforcement are derived from representative stress-strain

curve for the type of steel used. For the design purposes the partial safety factor γm

=1.15 shall be applied.

vi) The maximum strain in the tension reinforcement in the section at failure shall not

be less than: fy/1.15Es + 0.002

30

Where, fy = characteristic strength of steel

Es = modulus of elasticity of steel

Limit state of collapse for compression:

Assumption:

In addition to the assumptions given above from i) to v), the following shall be assumed:

i.) The maximum compressive strain in concrete in axial compression is taken

as 0.002.

ii.) The maximum compressive strain at highly compressed extreme fiber in concrete

subjected to axial compressive and bending and when there is no tension on the

section shall be 0.0035 minus 0.75 times the strain at the least compressed

extreme fiber.

The limiting values of the depth of neutral axis for different grades of steel based

on the assumptions are as follows:

Table 5: Limiting values of depth of neutral axis

Fy Xu, max

250 0.53

415 0.48

500 0.46

Materials adopted in our design:

M20 (1:1.5:3)

M30 (proportion according to Mix Design)

Fe415

Use of SP16, IS456-2000, IS1893-2002, IS13920-1993, SP34:

After analyzing the given structure using the software SAP2000 the structural

elements are designed by Limit state Method. Account should be taken of

accepted theories, experiment, experience as well as durability.

The code we use for the design is IS456-2000; IS1893-2002, IS13920-1993 and

Design aids are SP16 and SP34. Suitable material, quality control, adequate

31

detailing and good supervision are equally important during implementation of

the project.

Use of different handbook for the design:

The structural elements (special staircases, lift wall, basement wall) which are not

described by the above mentioned codes and design aids were handled with the

help of the handbooks viz. Reinforced concrete Designer’s Handbook – Charles

E. Reynolds.

5.2 Design Of Structural Elements

The aim of design is the achievement of an acceptable probability that structures being

designed will perform satisfactorily during their intended life. With an appropriate degree

of safety, they should sustain all the loads and deformations of normal construction and

use and have adequate durability and adequate resistance to the effects of misuse and fire.

After analyzing the structure using SAP 2000, the structural elements are designed by

Limit State Method. In the limit state design concept, the structure shall be designed to

withstand safely all loads liable to act on it throughout its life; it shall also satisfy the

serviceability requirements such as limitation on deflection and cracking. The acceptable

limit for the safety and serviceability requirements before failure occurs is called a limit

state. The aim of limit state design is to achieve acceptable probabilities that the structure

will not become unfit for the use for which it is intended, i.e. it will not reach a limit

state. The design is in compliance with clearly defined standards for materials,

production, workmanship, and maintenance and use of the structure in service.

Following are the sample calculation of the design process of single element as

i. Slab

ii. Beam

iii. Column

iv. Staircase

v. Shear Wall

vi. Basement Wall

vii. Mat Foundation

32

5.2.1 Design of slab

Grade of concrete = M30 Grade of steel = Fe415

Ref Step Calculations Output

Slab ID S1

IS456:2000

Clause

23.2.1

1

Interior Panel

Thickness of slab and durability consideration

Clean spans

Lx=6.1 m

Ly=6.7 m

Depth (d)= Shorter span/(α*β*γ*δ*λ)

= 6100/(26 *1*1.6*1*1)

= 146.6 mm

Provide D= 150 mm

And clear cover = 15mm

Provide 10 mm Φ

d = 150 – 15 - 5 = 130 mm

Since,

Ly/Lx = 6700/6100

= 1.1 < 2

(Design as two way slab)

D=

150

mm

d =

130

mm

2 Design load

Self weight of slab= 0.15* 25 = 3.75 KN/m2

Plaster = 1 KN/m2

Others = 1 KN/m2

__________________ 5.75 KN/m2

Live load = 3.5 KN/m2

Design load = 1.5 *(DL + LL)

33

= 1.5 *(5.75 + 3.5)

= 13.875 KN/m2

Considering unit length of slab

w = 13. 875 KN/m

w =

13.

875

KN/m

IS 456,

Table 26

IS 456,

Annex

Clause

D.1.0

3. Moment calculation

For Ly/Lx= 1.1

-ve BM coefficient at continuous edge

αx=- 0.037

αy =- 0.032

+ve moment at mid span

αx = 0.028

αy= 0.024

For shorter span

Support Moment Ms = - αx wLx2

= -.037 * 13.875* 6.12

= -19.1 KNm

Mid span moment Mm = αx w Lx2

= 0.028 * 13.875* 6.12

= 14.456 KNm

For larger span

Support Moment Ms=- αy *w*Ly 2

= -0.032 * 13.875 * 6.72

= - 19.93 KNm

Mid span moment Mm = αy *w*Ly 2

= 0.024 * 13.875 * 6.72

= 14.948 KNm

34

IS

456 :2000

Annex

G.1.1

4 Check for depth from moment consideration

Mu= 0.36 * xu,max /d*(1 – 0.42 xu,max /d)* bd2 *fck

19.93 X 10^6 = 0.36 * 0.48 *(1-0.42 * 0.48) * 1000 * d2*

30

d = 69.4 mm < 130 mm OK

IS 456

Clause

26.5.2.1

IS

456 :2000

Annex

G.1.1

5 Calculation of Area of steel

Minimum area of steel (Ast,min)= 0.12 % of bD

= 180 mm2

For shorter span,

Area of steel at support ((Top bars)

Mu= 0.87 * fy* Ast * d(1- Ast*fy/bd*fck)

19.1 X 106 = 0.87 * 415 * Ast * 130 *(1-

Ast*415/(1000*130*30))

Ast= 426.3 mm2 > Ast

Provide 10 mm Φ bars

Ab= 78.5 mm2

Spacing Sv= 184.14

Providing 10 mm Φ @ 180 mm c/c

Then Area of steel Ast = 436 mm2

Pt= 0.29 %

Again,

Area of steel at mid span (bottom bar)

14.456X10^6 = 0.87*415*Ast * 130 *(1- Ast*415/

(1000*130*30))

Ast = 318. 8 mm2 > Ast min

10

mm Ǿ

@

180

mm

c/c

Area

of

steel

Ast =

436

mm2

Pt=

35

IS

456 :2000

Annex

G.1.1

Providing 10 mm Φ bars

Ab= 78.5 mm2

Spacing of bars Sv= 246.2 mm

Providing 10 mm Φ @ 240 c/c

Then Actual Area Ast= 327 mm2

Pt = 0.22 %

For longer span

Area of steel at support ( Top bars)


19.93X 10^6 = 0.87 * 415*Ast * 120 *(1- Ast *

415/(1000*120*30))

Ast=446 mm2 >Ast min


Ab = 78.5 mm2

Spacing of bar, Sv= Ab/Ast * 100

= 78.5 *1000/446

= 176 mm


Actual Ast = Ab/ Sv * 1000

= 446 mm2

Pt= 0.3 %

Again,

Area of steel at mid span ( Bottom bars)


14.948X 10^6 = 0.87* 415*Ast *120*(1- Ast *

415/(1000*120*30))

Ast=330 mm2 >Ast min


Ab = 78.5 mm2

0.29 %

36

Spacing of bar, = Sv = Ab/Ast * 100

= 78.5 *1000/330

=238 mm



= 341.3 mm2

Pt= 0.23 %

IS 456,

Table 19

Clause

40.2.1.1

6 Check for shear

For short span,

Shear force at the face of support

V = w Ly

= 13.875 * 6.7

= 46.48 KN

Shear at critical section

46.48/3350 = x/3220

X = 44.67 KN

Hence tension reinforcement of slab contribute in shear

For Pt = 0.29 %

τc = 0.39 N/mm2

And K = 1.3

K τc *bd = 1.3 * 0.39 * 1000* 130/1000

37

= 65.91 KN > Vu OK .

7 Check for deflection

Along the shorter span

Since, both ends are continuous the basic value may be

taken as 26.

fs= 0.58 fy*Ast,req / Ast,prov

= 0.58 * 415 * 426/436

= 235.2 N/mm2

Pt = 0.26 %

Modification factor (MF) = 1.85

dper= 6100/1.85*26 = 126.82 < 130 mm OK.

IS 456,

Clause

26.2.1

IS 456,

Clause

26.2.3.2

8 Check for development length

Ld= Φ σs/1.6*4 *τbd

= 10 * 0.87 * 415/1.6*4*1.5

= 350 mm

Ld< M1/V + Lo

For short span

Ld< M1/V + Lo

= (19.93/2)/44.67 + 0.15

= 373 mm > Ld OK.

IS 456,

ANNEX D,

clause

D.1.10

9 Torsion reinforcement:

Since this is interior panel, no need of torsion

reinforcement

38

Ref Ste

p

Calculations Outpu

t

Slab ID S2

IS456:200

0

Clause

23.2.1

1

Exterior Panel (three edges continuous and one long edge

discontinuous)

Thickness of slab and durability consideration

Clean spans

Lx=6.1 m

Ly=6.7 m

Depth (d)= Shorter span/(α*β*γ*δ*λ)

= 6100/(26 *1*1.6*1*1)

= 146.6 mm

Provide D= 150 mm

And clear cover = 15mm

Provide 10 mm Φ

d = 150 – 15 - 5 = 130 mm

Effective length= Ly/Lx

= 6700/6100

= 1.1 < 2

(Design as two way slab)

2 Design load

Self weight of slab= 0.15* 25 = 3.75 KN/m2

Plaster = 1 KN/m2

Others = 1 KN/m2

____________ 5.75 KN/m2

Live load = 3.5 KN/m2

Design load = 1.5 *(DL + LL)

= 1.5 *(5.75 + 3.5)

39

= 13.875 KN/m2

Considering unit length of lab

w = 13. 875 KN/m

IS 456,

Table 26

IS 456,

Annex

Clause

D.1.0

3. Moment calculation

For Ly/Lx= 1.1

-ve BM coefficient at continuous edge

αx= -0.037

αy = -0.037

+ve moment at mid span

αx = 0.028

αy= 0.028

For shorter span

Support Moment Ms = - αx wLx2

= -.037 * 13.875* 6.12

= -19.1 KNm

Mid span moment Mm = αx w Lx2

= 0.028 * 13.875* 6.12

= 14.456 KNm

For larger span

Support Moment Ms=- αy *w*Ly 2

= -0.037 * 13.875 * 6.72

= - 23.045 KNm

Mid span moment Mm = αy *w*Ly 2

= 0.028 * 13.875 * 6.72

= 17.44 KNm

4 Check for depth from moment consideration

Mu= 0.36 * xu,max /d*(1 – 0.42 xu,max /d)* bd2 *fck

40

IS

456 :2000

Annex

G.1.1

Or, 23.045 X 106 = 0.36 * 0.48 *(1-0.42 * 0.48) * 1000 *

d2* 30

Or, d = 74.62 mm < 130 mm. OK.

IS 456,

Clause

26.5.2.1

IS

456 :2000

Annex

G.1.1

5 Calculation of Area of steel

Minimum area of steel ( Ast,min) = 0.12 % of bD

= 180 mm2

For shorter span

Area of steel at support ((Top bars)


19.1 X 106 = 0.87 * 415 * Ast * 130 *(1-

Ast*415/(1000*130*30))

Ast= 426.3 mm2 > Ast

Provide 10 mm Φ bars

Ab= 78.5 mm2

Spacing Sv= 184.14


Then Area of steel Ast = 436 mm2

Pt= 0.29 %

Again,

Area of steel at mid span (bottom bar)

14.456X106 = 0.87*415*Ast * 130 *(1- Ast*415/

(1000*130*30))

Ast = 318. 8 mm2 > Ast min


Ab= 78.5 mm2

Spacing of bars Sv= 246.2 mm

Providing 10 mm Φ @ 240 c/c

Then Actual Area Ast= 327 mm2

Pt = 0.22 %

41

IS

456 :2000

Annex

G.1.1

For longer span

Area of steel at support ( Top bars)


Or, 23.045X 10^6 = 0.87 * 415*Ast * 120 *(1- Ast *

415/(1000*120*30))

Or, Ast=519 mm2 >Ast min


Ab = 78.5 mm2

Spacing of bar, = Sv = Ab/Ast * 100

= 78.5 *1000/446

= 151 mm



= 523.33 mm2

Pt= 0.35 %

Again,

Area of steel at mid span ( Bottom bars)


Or, 17.44 X 106 = 0.87* 415*Ast *130*(1- Ast *

415/(1000*130*30))

or, Ast=387.6 mm2 >Ast min


Ab = 78.5 mm2

Spacing of bar, Sv= Ab/Ast * 1000

= 78.5 *1000/387.6

=202.58 mm



42

= 392.5 mm2

Pt= 0.26 %

IS 456,

Table 19

Clause

40.2.1.1

6 Check for shear

For short span,

Shear force at the face of support

V = w Ly

= 13.875 * 6.7/2

= 46.48 KN

Shear at critical section

46.48/3350 = x/3220

X = 44.67 KN

Hence tension reinforcement of slab contribute in shear

For Pt = 0.29 %

τc = 0.39 N/mm2

And K = 1.3

K τc *bd = 1.3 * 0.39 * 1000* 130/1000

= 65.91 KN > Vu OK.

43


Along the shorter span

Since, both ends are continuous the basic value may be

taken as 26

fs= 0.58 fy*Ast,req / Ast,prov

= 0.58 * 415 * 519/523.33

= 238 N/mm2

Pt = 0.35 %

Modification factor (MF) = 1.85

dper= 6100/1.85*26 = 126.82 < 130 mm OK.

IS 4565,

Clause

26.2.1

IS 456,

Clause

26.2.3.2

8 Check for development length

Ld= Φ σs/1.6*4 *τbd

= 10 * 0.87 * 415/1.6*4*1.5

= 350 mm

Ld< M1/V + Lo

For short span

Ld< M1/V + Lo

= (23.045/2)/44.67 + 0.12

= 378 mm > Ld OK.

IS

456,Annex

D, D.1.8

9 Torsion reinforcement

Area of torsion reinforcement

= 0.75 /2 *area required for maximum mid span

moment

= 0.75/2* 387.6

=145.35 mm2

So, provide 10 mm Φ @ 300mm c/c at a distance 0.2lx

from the edge.

44

Fig.4 Design of Slab

Design of Beams

Grade of concrete = M30 Grade of steel = Fe415

46

47

Ref. Calculation

Beam (4687)

ID

IS456:2000

Annex 6.4.

IS456,

clause38.1,

2000

IS456: 2000

Annex G-1.1

At section L=0

D=600 mm

d = 560 mm

b= 300 mm

L=0

M3 V2 T P

292.115 50.013 4.9247 5.393

-416.214 -190.135 -6.448 -3.54

Mt = T(1+D/b)/1.7

= ( 4.9247(1 + 600/300))/1.7

= 8.690 KNm (+ve)

Mt = 6.448 (1 + 600/300)/1.7

= 11.38 KN-m (-ve)

Maximum moment= 292.115 + 8.690 = 300.805 KNm (+ve)

Minimum moment = -416.214 – 11.38 = -427.59 KNm (-ve)

Moment of Resistance of the beam

Mu,lim = 0.36 x u,lim /d*(1-0.42*xu,max/d)*bd2 fck

= 389.38 X 106 Nmm

For +ve moment,

Since Md< Mu, beam is singly reinforced.

For singly reinforced beam

Md= 0.87 * 415*Ast * 560 * (1- Ast * 415/300*560*30)

300.805 X 106 = 0.87* 415*Ast * 560 * (1- Ast * 415/300*560*30)

Ast = 1735.8 mm2

Assuming 12 mm Φ bar

Asb= π * 122 / 4 = 113.04 mm2

No. of bar = 1735.8 mm2 / 113.04 mm2

= 15.35 Say 16

Again, Take 25 mm Φ bar

Asb= π * 252 / 4 = 490.87 mm2

The above design also fulfill the standards of IS 13920 : 1993, clause 6.1.1, clause

6.1.2,clause 6.1.3, clause 6.1.4, clause 6.2.1, clause 6.2.2, clause 6.2.3 and detailing is

done also considering the standards of code IS 13920 : 1993, clause 6.2 and 6.3.

Design of column

Grade of concrete M30 Grade of steel Fe415

48

49

Reference Calculation Rema

rks

IS

456:2000

(cl.25.4)

Design of Longitudinal Reinforcement:

Column of size 450 X 450

Sample Calculation: For Column ID -5374

Ultimate Load, Pu = 180.739KN

Mux = 178.2433KNm

Muy = 193.2118KNm

Column Section =450mm × 450mm

Grade of concrete =M30

Grade of Steel = Fe 415

Design Steps:

Assume following data

Clear cover = 40mm

Diameter of longitudinal Reinforcement Φ = 20mm

So, effective cover =40+20/2 =50mm

Effective length =.65*3 =1.95m

Effective length/least lateral dimension =1.95*1000/450 =4.33<12(short

column)

Minimum eccentricity =L/500+Least dimension/30≥20mm

eminx =3000/500+450/30 =21mm>20mm

eminy=3000/500+450/30 =21mm>20mm

Moment due to minimum eccentricity

Mux=Pu*eminx =180.739*.021 =3.795 KNm<178.243KNm

Muy= Pu*eminy =180.739*.021 =3.795 KNm<193.212KNm

Therefore,

Mux = 178.2433 KN-m

(FRO

M

SAP)

The above design also fulfill the standards of IS 13920 : 1993, clause 7.1.1, clause

7.1.2,clause 7.1.3, and detailing is done also considering the standards of code IS 13920 :

1993, clause 7.2, clause 7.3 and clause 7.4.

Design of Shear wall

Sample detailing of 2.6 m structural wall with boundary element at lower

basement according to IS13920:1993

Materials: Grade of concrete = M30

Grade of steel = Fe415

Ref. Step Calculations Output

Shell

element :1079(

1 Direction Dimension no. Mu(kNm) Pu(kN)

y 2.6 2 89.44 8271.05

50

57) Description: yz plane.

IS:13920:1993

Cl.9.4.1

2 Check whether boundary elements are required:

Extreme fibre compressive stress in the wall should

not exceed 0.2 fck

L = 3050 mm

tw = 200 mm

Ag = 610000 mm2

Ix = 4.73*E11 mm4

(200*30503/12)

y = 1525 mm

0.2 fck = 6 N/mm2

fc = Pu /Ag + Mu.y/Iy

= 13.55mm2>0.2 fck

Hence,

boundar

y

elements

required.

3 Check whether two curtains are required:

IS:13920:1993

Cl.9.1.5

Two curtains of bars is required if thickness of wall is

more than 200mm or if the factored shear stress is

more than 0.25√ fck.

1.25 fck = 37.50 > Vu

Shear stress = Vu = 16.05kN

Providing two curtains.

Provide

two

curtains.

IS 13920:1993

Cl. 9.1.4

4 Minimum reinforcement required in longitudinal

and transverse direction:

0.25 % of gross area

5 Shear strength requirement:

51

IS 456:2000

Cl. 40.1

Tb. 19

Tb.20

IS 13920:1993

Cl.9.1.7

Vu = 16.05kN

τ v = Vu / twdw

=0.0328 N/mm2

τc = 0.37 N/mm2 for 0.25% of steel

τc,max = 4 N/mm2

τ v < τc

Hence, providing minimum reinforcement in each

direction.

Maximum spacing of reinforcement:

Description lw/5 3tw 450

max. spacing for

3.05wall 488 600 450

max. spacing for 6.1m

wall 976 600 450

max. dia. Of bar 20 provide 12mm

Gross area per meter of wall = tw * 1000 mm2 (per

meter of wall)

= 200 * 1000 mm2

= 200000 mm2

The required area of reinforcement area of

reinforcement in each direction per meter of wall is,

0.0025 * 200000 = 500 mm2 per meter.

Required spacing of 12mm dia. Bar (in two curtains,

As = 2 * 113.04 mm2) :

S (required) = (2 * 113.04) * 1000/ 500

= 452.16 mm

dw=

effective

depth of

wall

section

taken as

0.8 * lw

lw=.8L

52

Hence,

Providing spacing of 400 mm for 3.05m shear

wall.

S > max.

spacing

as

specified

by code.

IS 13920:1993

Annex A

6 Flexure strength requirement :

Mu = 89.44 kNm

Pu = 8271.059kN

d' = 26

d'/D = 0.1

λ = Pu/fck.tw.lw = 0.636375

p/ fck = 0.0000625

φ = (0.87*fy*p ) / fck

= 0.002256

xu/lw = (φ +λ)/(2 φ + 0.36)

= 1.626

xu*/lw = 0.69

Moment of resistance = Muv/fck tw lw2

β = 0.5157

α1 = 0.3548

α2 = 0.4166

α3 = -0.01739

therefore,

Mux/fck tw lw2 = 0.1116

Muv = 6036.08985 kNm

Mux >

Mu

Safe in

flexure

53

7 Checking adequacy of boundary element:

a. Calculations for the longitudinal bars:

Let us assume that the size of column is adequate for

boundary element.

Size of boundary element :

Length = 800 mm

Breathe = 800 mm

Take clear cover = 40 mm

Dia. of bar = 25 mm

Farzad Naiem

IS 13920:1993

Cl. 9.4.4

SP 16

Axial load on boundary element is calculated as,

Axial load on wall/2 + moment / width of wall

Ultimate axial load on boundary wall = 6652 kN

Percentage of vertical reinforcement in the boundary

elements shall not be less than 0.8%, nor greater than

6%. In order to avoid the congestion, practical limit

would be 4%.

Taking 2.5% of gross area as reinforcement bar.

Ultimate load capacity of boundary element:

Pu, = 0.4fck Ag + (0.67 fy -0.4fck) As

where,

Ag = 800 * 800 mm2

= 640000 mm2

As = 2.5% of Ag

= 0.0025 * 640000

= 16000 mm2

Therefore,

Pu,max = 11936.8 kN ( > 6652 kN )

Pu,capacity

> Pu

54

Hence, provided size of boundary element is adequate

for the ultimate load.

Then,

Dia. of bar = 25mm

Ast = 491.07 mm2

no. of bars = As / Ast

= 16000 / 491.07

= 32

Actual area of steel = 16198.84 mm2

Actual % of steel = 2.53%

Hence,

Provide 32 bars of 25 mm diameter distributed on the

four sides of section .

Min. spacing = dia. of bar = 25mm

= 5 mm more than the nominal

maximum size of aggregate.

Spacing in 800mm length= (800 – 2*40 – 25 )/(8– 1)

= 99 mm

Hence, provide:

Dimensio

n 800 mm 800 mm

no. of

bars 8 8

clear

cover 40 40

spacing 99 mm 99 mm

No. of

bars

provided

= 32

S j 100

mm

55

IS 13920:1993

Cl. 9.4.5

Cl.7.4.6

Cl. 7.4.8

b. Calculations for the confinement bars:

Boundary elements, where required, as per Cl.9.4.1

(already defined above), shall be provided throughout

their height, with special confining reinforcement.

Spacing of hoops shall not exceed 1/4 th of min.

member dimension but need not be less than 75mm

nor more than 100 mm

¼ of min. dimension = ¼ *800

= 200 mm

Hence, Providing 100mm spacing.

The area of cross section , Ash of the bar forming

rectangular hoop, to be used as special confining

reinforcement shall not be less than,

Ash = 0.18 S h fck / fy [ Ag / Ak – 1 ]

Where,

S = 100 mm

h = 272 mm

Providing 12mm Φ rectangular hoops

Ak = area of concrete core

= 744 * 744 mm2

=553536 mm2

Ag = 800 * 800

= 640000 mm2

Therefore,

Ash = 0.18 S h fck / fy [ Ag / Ak – 1 ]

Spacing

of

confine

ment

bars =

100 mm

Ash =

113.143

mm2

800 –

2*40

+2*12 =

56

= 55.28mm2 < 113.143mm2

Hence, providing 12mmφ confining bars at 100mm

spacing as confining reinforcement.

744 mm2

O.K

Design of Basement wall

Ref. Step Calculation Output

1 Design constants

Concrete grade M20

Steel Fe415

Clear height of the basement wall , H =3

Specific weight of soil, γsoil = 16.5 kN/m3

Angle of internal friction, φ = 30.

Surcharge load , = 19 kN/m2

Safe bearing capacity, qs = 150 kN/m2

2 Approximate design of section

57

IS

456:2000

Cl. 32.3.4

Let effective depth of wall, d = H/15 = 3000/15 =

200 mm

Take clear 25 mm, diameter of bar = 16 mm

Provide overall depth, D = 233 mm

Effective depth, d = 200 mm

Slenderness ratio = H/d = 15 < 30

D = 240 mm,

d = 207 mm

O.K

3 Moment Calculations

Coeff. of earth pressure Ka = (1 – sin φ )/(1 +sin φ

) = 0.3333

Lateral load due to soil pressure pa = 1/2kaγH2 =

24.50 kN/m

Lateral load due to surcharge load ps = kawH =

18.81 kN/m

Characteristic moment at the base of the wall, Mc

= paH/3 + psH/2 = 52.715 kNm

Design moment Md = 1.5 Mc = 79.07 kNm

pa = 24.50 kN/m

ps = 18.81 kN/m

Md = 79.07 kNm

4 Check for the depth

Considering unit length of wall,

Mu,lim = 0.138fckbd2 = 118.26 kNm > Md

Depth of wall from moment consideration, d =

(Md/0.138fckb)0.5 = 169.26 mm < d provided

O.K

O.K

58

IS

456:2000

Cl.32.5

Cl.26.5.2.2

5 Calculation of main steel reinforcement

Ast = 0.5*fck/fy*b*d*(1-(1-4.6Md/fck/b*d2)0.5) =

1147.22mm2

Min. area of tensile reinforcement = 0.12% of bd

= 248.4 mm2

Max. spacing = 3d = 621 mm or 450 mm

Max. dia of reinforcement = D/8 = 30 mm

Providing 16 mm Φ bar, Ab = 201.143 mm2

Spacing of bar Sv = 175.25

Actual Ast = 1183.19mm2

Actual pt = 0.4605%

For front face provide nominal vertical

reinforcement Φ 12mm @ 300mm c/c

Provide Φ

16mm@ 170 mm

c/c

Front face Φ

12mm @ 300mm

c/c

IS

456:2000

Cl.31.6.2.1

table 19

6 Check for shear

Shear force V = 0.5kaγH2 = 23.98 kN

Design shear force Vu = 1.5V = 35.96 kN

Nominal shear stress τv =Vu/bd = 0.174 N/mm2

Permissible shear stress τc = 0.469 N/mm2 >τv No shear

reinforcement

IS

456:2000

cl.23.2


Leff = H+d = 3.207 m

Allowable deflection = Leff/250 = 12.83mm

Actual deflection = Psleff4/8EI+Paleff

4/30EI =

11.5mm < 12.83

O.K

IS 8 Calculation of horizontal steel bars

59

456:2000

cl 32.5

i.

ii.

Area of horizontal steel reinforcement =

0.0012DH = 864mm2

As the temperature changes occurs at the front

face of the basement wall,2/3 of the horizontal

reinforcement Is provided at the front face and

remaining at inner face.

So front face horizontal reinforcement = 2/3 of

area of ho. Reinforcement = 576mm2

Providing 8mm dia bar

Ab = 50.286 mm2

No of bars N = 576/50.286 = 12

Spacing = (H – clear cover – Φ )/(N-1) = 269

mm

Max spacing 3d = 621or 450 mm

Inner face reinforcement= 1/3*1176 = 288 mm2

Providing 8mm bars

No of bars N = 6

Spacing = 166.7 mm

Max spacing 3d = 621 or 450

Provide Φ 8mm @

260mmc/c

O.K

Provide Φ 8mm

@160c/c

O.K

9 Curtailment of Reinforcement

No bars can be curtailed in less than Ld distance

from the bottom of the stem,

Ld = 564mm

The curtailment of bars can be done in two layers

1/3 and 2/3 height of steam above the base.

Let us curtail bars at 1/3 distance i.e. 1000mm

from base,

Lateral load due to soil pressure,

60

Pa = 10.65KN/m

Lateral load due to surcharge load,

Ps = 12.65KN/m

Characteristics BM at base of the wall is,

Mc = paH/3 + psH/2 = 29.625 kNm

Design moment ,M=1.5 Mc=44.44KNm

Ast = 620mm2

No. of bars = 8

Spacing = 142.8

Provide Φ

10mm@140mm c/c

Design of Mat foundation

Required data:

Case considered = 1.5(DL + LL)

Total design load (P) = 403065.4 KN (From SAP)

From soil investigation report,

Safe soil bearing capacity = 150 KN/m2

61

Grade of concrete = M20

Grade of steel = Fe415

Calculation of centre of gravity of plan area

Taking moment of column load about outer face of the mat,

X = = 95.31 ft

Y = = 36.96 ft

Geometric centre of mat foundation,

X = = 100 ft

Y = = 37 ft

ex = X –X = 100 - 95.31 = 4.69 ft =1.43 m

ey = Y – Y = 37 – 36.96 = 0.04 ft ≈ 0 m

Calculation of MOI

Ixx = 82190.29 m4

Iyy = 505873.19 m4

Area coverage of mat = 1532.17 m2

Mxx = ∑P × ey = 403065.4 × 0 = 0

Myy = ∑P × ex = 403065.4 × 1.43 = 576383.52 KNm

= 263.07 KN/m2

Soil pressure calculation at different points:

At corner A-2

A-2 = 263.07 + × (18.79 - 30.48) = 249.18 KN/m2

At corner A-3

A-3 = 263.07 + × (37.08 - 30.48) = 270.6 KN/m2

At corner B-1

62

B-1 = 263.07 + × (0 - 30.48) = 228.33 KN/m2

At corner B-4

B-4 = 263.07 + × (66.57 - 30.48) = 304.21 KN/m2

At corner C-1

C-1 = 263.07 + × (0 - 30.48) = 228.33 KN/m2

At corner C-4

C-4 = 263.07 + × (66.57 - 30.48) = 304.21 KN/m2

In x-direction the raft is divided into three strips, that is three equivalent beam

(i) Beam A-A with 5.35m width and soil pressure of 270.6 KN/m2

(ii) Beam B-B with (5.35+19.9/2) = 15.3 m width and soil pressure of 0.5(270.6+304.21) = 287.41 KN/m2

(iii) Beam C-C with 9.95 m width and soil pressure of 0.5(304.21+304.21) = 304.21 KN/m2

The bending moment is obtained by using a coefficient of 1/10 as a centre of column,

+M = -M =wl2/2

For strip A-A,

Maximum moment = 270.6*6.12/10= 1006.25 KNm per meter

For strip B-B,

Maximum moment = 287.41*6.12/10 = 1069.45 KNm per meter

For strip C-C,

Maximum moment = 304.21*6.12/10 = 1131.96 KNm per meter

For any strip in Y-direction, taking coefficient of 1/10 with maximum centre to centre length of 6.7m

For strip 1-1,

Maximum moment = 228.33*6.72/10 = 1024.97KNm per meter

For strip 2-2,

63

Maximum moment = 0.5*(228.33+249.13)*6.72/10 = 1071.61KNm per meter

Similarly, for strip 3-3, maximum moment = 1166.53 KNm per meter

For strip 4-4, maximum moment = 1290.16 KNm per meter

(i) Calculation of depth of foundation

Depending upon moment criteria, taking maximum moment in either direction,

Maximum moment is occurring for strip

M = 1920.16 KNm per meter

Mu,lim /(fck* b*d2) = 0.138

Mu,lim = 0.138*30*1000*d2

1290.16*10^6 = 0.138*30*1000*d2

d = 558.24 m

(ii) Calculation of depth from two way shear,

The depth of the raft will be governed by two-way shear at one of the exterior column, in case location of critical shear Is not obvious it may be necessary to check all possible location. When shear reinforcement is not provided, the calculated shear stress at critical section shall not excess Ksτc

i.e, τv = Ksτc (clause 31.6.3, IS 456, 2006)

Ks = 0.5 + β3 but not greater than 1

Then, β3 = (19.90+10.75)/66.57 = 0.46

Then, Ks =0.5+0.46 = 0.96

Shear strength of concrete,

τc = 0.25 *√fck (clause 31.6.3, IS 456, 2000)

= 0.25 *√30

= 1.37 N/mm2

64

= 1369.30 KN/m2

For corner, column say A-1

Column load = 4918KN (from SAP)

Perimeter = Po = 2(d/2 + 750) = d + 1500

τv = column load/(Po*d)

1.37 = 4918*1000/(d + 1500)*d

d = 1287.71mm

For side column say A-2

Column load = 6954.92 KN (from sap)

Perimeter = 2(0.5d + 750) + (d+560)

= 2d + 2000

τv = column load/(Po*d)

Solving, d = 1191.32mm

Adopt, effective depth of 1300mm

Total depth= 1350mm

Calculation of reinforcement in foundation:-

Alone x- direction

Maximum negative moment = maximum positive moment=1131.96KNm/m

Then, area of steel, Ast = 2476.97 mm2

Minimum reinforcement = 0.12 * 1000 * 1300 = 1560 mm2

65

` ok

Provide 20mm bar, area of single bar = 314.16 mm2

Providing 20 mm Ǿ @ 140mm c/c in X-direction

Alone Y- direction

Moment = 1290.16 KNm/m

Then, area of steel, Ast =2834.21 mm2

So, providing 20 mm Ǿ @ 110mm c/c in Y-direction

Table 6: Summary of Calculation of reinforcement in foundation

SN Description Value Remarks1 Mux (moment in X-direction) 1131.96 KNm Max –ve moment=

Max +ve moment2 Muy (moment in Y-direction) 1290.96 KNm Max –ve moment=

Max +ve moment3 Area of Steel in X- direction(Astx) 20mm Φ @ 140

mm c/c Provided in per metre width

4 Area of Steel in Y-direction(Asty) 20 mm Φ @ 110 mm c/c

Provided in per metre width

5 Number of bar in X-direction 8 Provided in per metre width

6 Number of bar in Y-direction 10 Provided in per metre width

66

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