analysis and design of a multi-storeyed residential building

CHAPTER - 1

INTRODUCTION

The basic need for a person is to provide with proper food, clothing and shelter.

The first two aspects are very essential to each and every person and people who are

sufficient with the food and clothing they will have thought of a shelter. In shelter aspect

it has been observed that lot of money is involved which a single common cannot afford

because so of many reasons. One them is land cost i t-self being rising in exponential way,

which a common man cannot afford with the available financial resources to him. Because

of the foresaid reasons people have started purchasing a common land and same being

used to construct a common building in which individual family activities are limited to a

part of the building which is popularly known as Apartments. This aspect being mostly

prevails in modern urbanization of cities.

On emphasizing requirements of modern urbanization this project has been

chosen to design of multi-storied Residential apartment provides shelter to twelve

families in three floors and a cellar which is provided for common usage such as parking

etc.

Keeping in view the future requirements the structure is analyzed and designed

with all care so that the structure gives safety, comfortable utili ty and good shelter to the

occupants as per the relevant Indian Standard norms. The analysis of all the frames

carried out by using rotation contribution method with substitute frames. This project

covers all the aspects regarding Analysis and Design of multi-storied residential building

which can use as reference guide in my project.

1

REVIEW OF LITERATURE

The concept of the multistory building has changed throughout history, depending on the heights

of urban structures, which, in turn, have been determined by social, economic, and urban-construction

requirements.

Residential and public multistory buildings became widespread in ancient Greek and Roman

cities as a result of the need for accelerated construction of inexpensive dwellings to house the low-income

population (for example, ancient Roman insula). They were widespread in medieval cities because only a

limited amount of territory could be protected by town walls. Examples of the medieval multistory building

were the residences of wealthy European city-dwellers, which had living quarters, workshops, and

commercial establishments on the first two floors and warehouses on the upper floors.

During the capitalist period the rapid growth of cities and the considerable rise in costs of urban

real estate resulted in great expansion of the construction of multistory buildings. Engineering advances,

particularly the invention of the elevator, made it possible to increase heights significantly (16-story

Monadnock Building in Chicago, 1891; architects D. H. Burnham and J. W. Root).

In the late 19th and early 20th centuries buildings several dozen stories high (skyscrapers) were

constructed in the United States. They were used for offices, banks, hotels, and residences. The Empire State

Building (architectural firm of Shreve, Lamb, and Harmon), which was built in New York City in 1930–31,

has 102 stories; its height (without the television antenna, which was erected in 1951) is approximately 380

m. Beginning in the late 1940’s, in conjunction with intensive urbanization and, at times, as a result of

shortages of space, multistory buildings became widespread in many countries. The typical multistory

building has from nine to 17 stories. Taller buildings, which are known as high-rises, frequently serve

multiple purposes. For example, the 100-story John Hancock Building in Chicago (1971, architects L.

Skidmore, N. A. Owings, and J. O. Merrill) contains stores, a bank, a garage, offices, and apartments.

Under conditions of capitalist urban construction, the spontaneous concentration of multistory

buildings in a limited amount of territory and the accumulation of considerable masses of people and means

of transport result in the destruction of the functional, hygienic, and aesthetic qualities of the urban

environment (transport congestion; noisy, narrow streets devoid of fresh air; a sensation of chaos created by

viewing the dense construction of multistory buildings that are of various heights and are frequently

architecturally unexpressive).

In the USSR and other socialist countries, multistory buildings are usually erected in accordance

with urban construction requirements and in harmony with the general plan of the city. They are built

specifically to save space in the center of the city, which is particularly valuable because of the

concentration there of expensive communications and engineering equipment. In the late 1940’s and early 2

1950’s seven high-rise buildings with 26 to 32 stories were built in Moscow in accordance with a unified

urban construction plan (architects V. G. Gel’freikh, A. N. Dushkin, B. S. Mezentsev, M. A. Minkus, A. G.

Mordvinov, L. M. Poliakov, L. V. Rudnev, D. N. Chechulin). The erection of these buildings promoted

innovations in the construction industry. Placed in key parts of the capital and crowned with spires, the high-

rises imparted to the city a new silhouette and scale. The buildings are characterized by complex

compositions of masses of varying heights, abundance of decoration on facades and in the interiors, and a

low percentage of living space. The construction of multistory buildings by mass-production methods

sharply increased in the USSR during the late 1960’s (in 1973, 20 percent of the total construction of

residential buildings). In addition to buildings with the typical nine to 17 stories, structures of 25 or more

floors are being erected. Sometimes, multistory buildings form entire complexes (for example, Kalinin

Prospect in Moscow, 1964—69; architects M. V. Posokhin and A. A. Mndoiants).

There is no unified classification of multistory buildings. It is customary to consider qualitative

changes (as a result of great heights) in planning, design, and technical equipment as criteria for putting a

building in the category of multistory buildings. Such buildings require special provisions for fire safety

(structures with increased fire resistance, smoke-free staircases, systems of water supply for fire-fighting,

and smoke removal equipment) and structural stability under the impact of wind loads (including dynamic

loads). Elevator facilities and mechanical equipment are particularly complex in multistory buildings.

Structural stability of residential multistory buildings is achieved, for the most part, through load-

bearing cross walls or a braced frame (in the USSR mainly of sectional reinforced concrete). In public

buildings structural stability generally is combined with what is referred to as the rigid core (a reinforced-

concrete casing enclosing the elevator shafts and mechanical systems, which are assembled together). In

high-rise buildings outside of the USSR, core-casing structures are widespread, in which the casing a load-

bearing facade enclosure of the lattice type, made of steel or pre - stressed reinforced concrete elements is

linked by the floors to a core situated in the center. These elements form a single system of great rigidity (for

example, the two 110-story towers of the World Trade Center in New York City; architects M. Yamasaki

and others, 1971–73).

It is very difficult to find expressive architectural solutions for multistory buildings because their

enormous size and the repetition of thousands of identical facade elements have a great and, at times,

negative influence on the traditional appearance of old cities. Striving to overcome superhuman scales and

monotony, architects employ contrasting masses of various heights and, at times, curvilinear contours. They

seek expressive proportion and silhouette and have recourse to rhythmic organization of facade elements

(for example, the grouping of balconies and their enclosures or windows in an ornamental composition) and

effective decoration of facades with stainless steel, aluminum, bronze, or glass (for example, the 38-story

Seagram Building in New York City, 1958; architect L. Mies Van der Rohe).

3

CHAPTER - 2

DESIGN PHILOSOPHIES

Introduction:

For a given structural system the design problem consists of following steps.

1. Idealization of structure for analysis.

2. Estimation of loads.

3. Analysis of idealized structural mode to determine actual thrust, shears, bending

moments and deflections,

4. Design of structural elements and

5. Detailed structural drawings and scheduling of reinforcing bars.

There are three philosophies for design of reinforced concrete, pre stressed

concrete as well as steel structures.

i. The working stress method

ii. Ultimate load method

iii . Limit state method

i. The working stress method:

This has been the traditional method used for reinforced concrete design where it is

assumed that concrete is elastic, steel and concrete act together elastically and

relationship between loads and stresses is linear right up to the collapse of the structure.

The basis of the method is that the permissible stresses for concrete and steel are not

exceeded anywhere in the structure when it is subjected to the worst combination of

working loads. The sections are designed in accordance with elastic theory of bending

assuming that both materials obey Hook’s law.

The permissible stresses are prescribed by a building code to provide suitable

factors of uncertainties in the estimation of working loads and variation in properties of

materials. IS 456-1978 used a factor of safety equal to 3 on the 28 days cubes strength to

obtain the permissible tensile stress in reinforcement. Thus for properly designed

structural elements stresses computed under fraction of working loads will be within

elastic range.

4

ii . Ultimate Load Method:

In the ult imate load method the working loads are increased by suitable factors to

obtain ultimate loads. These factors are called load factors. The structure is than designed

to resist the desired ultimate loads. This method takes into account the nonlinear stress

strain behavior of concrete.

The term safety factors has been used in the working stress method to denote the

ratio between yield stresses and permissible stresses .It had li ttle meaning as far the ratio

between collapse load and working load was concerned. The term load factor has been

rationally used to denote the ratio between the collapse or ultimate load to working load.

The knowledge of load factor is more important than the knowledge of factor of safety. In

this method the designer can able to predict the excess load, which a given structure can

carry beyond the working load without collapse. The level of stresses is immaterial.

Whitney’s theory has been the most popular ult imate load theory dose we to its

simplicity. The theory is based on the assumption that ult imate strain in concrete is 0.3%

and compressive stress at the extreme edge of the section corresponds to this strain. He

replaced actual parabolic stress diagram by a rectangle stress diagram such that the center

of gravity of both diagrams lies at the same point and their areas are also equal. The

ultimate load design of reinforced concrete sections is based on the assumptions in

accordance with IS 456-1964.

iii . Limit state method:

In the method of design based on limit state concept, the structure shall be

designed to withstand safely all loads liable to act on its throughout i ts l ife. It shall also

satisfy the serviceabili ty requirements such as l imits on deflection and cracking. The

acceptable limit for the safety and requirements before failure occurs is called a ’LIMIT

STATE.’

For ensuring the above objective, the design should be based on characteristics

values for material strengths and applied loads, which take into the account the variation

in material strengths and in the loads to be supported. The design values are derived from

the characteristic values through the use of partial safety factors, one for material

strengths and the other for loads.

5

Limit state of collapse:

Limit state of collapse or the part of the structure could be accessed from the

rupture of one or more crit ical sections and from buckling due to elastic or plastic

instability or overturning.

LOADS: The design load ‘F d’ is given by F d = F ( γf ).

Where

F = Characteristic load.

γf = partial safety factor appropriate to the nature of loading and limit being considered.

Partial Safety factors:

(i) γm=1.5 for concrete

(ii) γm=1/5 for steel.

Stress block parameters:

Area of stress block = 0.36fckXu

Depth of Centre of compressive force from the extreme fiber in compression =

(0.42)Xu

Xu = depth of neutral axis.

f ck = characteristic compressive strength of concrete.

The tensile strength of concrete is ignored. The limiting values of the neutral

axis for different grades of steel area given by

f y (Xu/f:Max)

250 0.53

415 0.48

500 0.46

6

Limit state of serviceability:

Control of deflection:

The deflection of a structure or part there of shall not adversely affect appearance

or efficiency of the structure. The final deflection due to all measured from the as-cost

level of the supports should not normally exceed.

SPAN/250 the final deflection after erection of structure should not normally exceed

SPAN/350 or 20mm whichever is less.

Cracking:

Cracking of concrete not adversely affect the appearance or durability of the

structure. The surface width of the cracks should not, in general, exceed 0.3mm.

Limit state of collapse:

Flexure: Assumptions

Plane sections normal to the axis remain plane after bending.

(i) The maximum strain in concrete at the outermost compression fiber is taken

as 0.0035 in bending.

(ii) For design purposes, the compressive strength of concrete in the structures

shall be assumed to be 0.67 t ime the characteristic strength.

The partial safety factor γm shall be applied in addition to this. Main reinforcement

should not exceed 0.004 times the nominal cover to main reinforcement.

Characteristic strength of materials:

7

Load combination Limit state of collapse Limit state of

serviceability

(1) DL WL LL DL WL LL

DL+LL 1.5 1.5 --- 1.0 1.0 ----

DL+WL 1.5 ---- 1.5 1.0 ---- 1.0

DL+LL+WL 1.2 1.2 1.2 1.0 0.8 0.8

The term “characteristic strength” means that value of the strength of material

below which not more than 5percent of the test results are expected to fall.

Characteristic loads:

The term “Characteristic load” means that value of load which has a 95%

probability of not being exceeded during the l ife of structure.

Design values:

Material: The design strength of the material fd is given by fd=f/m .

Where f=Characteristic strength of material.

m =Partial safety factor appropriate to the material and the l imit state being considered.

LIMIT STATE OF COLLAPSE: COMPRESSION:

Assumptions:

(a) The maximum compressive strain in concrete in axial compression is taken as

0.002.

(b) The maximum compressive strain at the highly compressed extreme fibre in

concrete subjected to axial compression and bending and when there is no tension

on the section shall be (0.0035-0.75) times the strain at the least compressed

extreme fiber.

CHAPTER - 3

ANALYSIS METHOD

8

The various methods available for the analysis of building frames are:

(i) Slope-deflection method,

(ii) Moment-distribution method,

(iii) Kani’s method,

(iv) Substitute frame method (Approximate method).

In structural frame analysis code provides a simplifying assumption for

arrangement of l ive loads referring clauses 21.4 IS 456-1978 has given below.

(a) Consideration may be limited to combination of

(i) Design dead load on all spans with full design live load on two adjacent spans.

(ii) Design dead load on all spans with full design live load on two alternative spans.

(b) When design live load does not exceed ¾ of the design dead load the load

arrangements may be design dead load and design live load on all spans.

The code IS 875-1984 provides, for wind analysis The ratio of the height of the building

to least lateral dimension if less than 2 no wind analysis is to be carried out.

The Height of the proposed building (H) = 15.00m

Least plan dimension (B) = 10.46m

(i) Slope-deflection method:

This method can be used to analysis all types of statically in determined rigid

frames .In this method all joints are considered rigid and the rotation of the joints are

treated as unknowns. The end moments of the any member bounded by two joints can be

expressed in terms of end rotations. By applying the equilibrium equations at a juice the

unknown rotations of the joint are found out. The end moments can thus be found from the

slope deflection equations. This method becomes very cumbersome for analysis multistory

frames since number of unknown rotations is to be solved.

(ii) Moment-distribution method:

The moment distribution method, also known as the Hardy cross method

provides a convenient means of analysis statically in terminate beams and frames by

simple hand calculations. This is basically an iterative process. The procedure in general 9

involves retraining all the joints temporarily against rotation and writing down the fixed

end moments for all the members. The joints are than released one by one in succession at

each released joint. The unbalanced moments are distribute to all the ends of the members

meeting at that joint. A certain fractions of these disturbed moments are carried over to

the far end of the members. The released joints are again restrained temporarily before

proceeding to the next joint. The same sets of operations are carried out at each joint of

the frame. This completed one cycle of operations. The process is repeated for a number

of cycles til l the values obtained are within the desired accuracy.

The moment distribution method is also a displacement method of analysis. However this

method does not involve solving any equations. This method is highly popular among

engineers as the calculations involved are minimum and free from solving simultaneous

equations even if the frame.

(iii) Rotation contribution method:

Gasper Kani’s of Germany developed this method in 1947.This is an iterative

method to solve the slope deflection equation Kani’s method iterates the member end

moments themselves rather iterates their increments. It consists of a single, simple

numerical operation repeated at the joints of a structure in a chosen sequence. In the

present project Kani’s method is used for the analysis of various frames.

The moments are determined by going through the following stages:

1. The both ends of member are first regarded as fixed Corresponding to this

fixed end moments M ab at B determined.

2. Now maintaining fixity of the end B, the end A is rotated through an angle θ a

by the application of the moment 2M ab at A.For this condition a moment of

M ab is induced at a end B.The moment M ab is called the rotation contribution

of the end A.

3. In this stage the end A is considered a fixed and b is rotated an angle θb by

application of a moment 2M ba at B for this condition a moment of M ba is

induced at the end A the moment is called the rotation contribution of the end

B.

Final moments:

M ab = M ab+2M ab+M ba

10

M ba = M ba+2M ba+M ab

Final moments are = 0.5× (K a b /∑K a b) × (∑M b a+∑M a b)

Where = 0.5× (K a b /∑K a b) is called the rotation factor.

Once the rotation displacement contributions are known the final moments may

be determined by the above equations.

(iv) Substitute frame method:

A multi-stored frames a complicated statically indeterminate structure. The

extent of analysis by the moment distribution method is very lengthy and difficult . To

overcome the difficulties stated above by adopting the SUBTITUTE FRAME METHOD

can be used.

In this method only a part of frame is considered for the analysis. The moments

for each floor to another floor are separately computed. It will be assumed that the

moments transferred from one floor to another floor are small . Each floor will be taken as

connected to column above and below with their far ends fixed. The frame taken this way

is analyzed for the moments and shears in the beams and columns.

Generally to get maximum positive bending moment at middle of the span a

loading pattern can be adopted in such a way loading over the span and no loading is

taken over adjacent spans on either side. For getting the maximum negative bending

moment by adopting loading is considered on that particular span and loading will be

taken on adjacent spans.

To get maximum negative moment at support a loading pattern can be

considered in such a way loading on both sides of the support will be taken.

The moment distribution for the substi tute frame is performed only for two

cycles and hence the method is sometimes referred as the two-cycle method.

11

CHAPTER - 4

DESIGN OF SLABS

12

DESIGN OF SLABS

Classification of slabs:

Lx=2.97m

Ly=3.89m

Ly /Lx=3.89 /2.97=1.31<2

The slab is to be designed as two way slab.

Edge condition: Two adjacent edges discontinuous.

Calculation of Depth of Slab:

Short span Lx=2.97m

Long Span Ly=3.89m

Live Load = 2 KN/m². < 3.0 KN/m²

Steel grade = fe415

D = 2.97/40 = 0.074 m

= 74.30 mm. (For continuous slab Fe415)

Assume overall depth (D) = 100 mm

Assume effective cover 20 mm

Effective depth (d) = 100-20 = 80 mm.

16

Loads:

Dead Load: Self weight of Slab = 0.1×1×25

= 2.5 KN/m²

Wt of floor finish = 1.0 KN/m²

Total dead load = 3.5 KN/m²

Live load for residential buildings = 2.0 KN/m².

∴ Total working load = 5.5 KN/m².

Factored load = 1.5×5.5 = 8.25 KN/m².∴

Bending Moment Coefficients:

For Ly /Lx=3.89 /2.97=1.31 [From table 26, IS 456-2000]

i. Along short span (α x ):

-ve moment at continuous edge

α x ¿

M x ¿ = 0.065×8.25 × (2.97)²

= 4.73 KN-m (at supports)

+ve moment at mid span

α x ¿

M x ¿ = 0.049×8.25 × (2.97)²

= 3.56 KN-m (at mid span).

17

ii. Along long span (α y ):

-ve moment at continuous edge

α y ¿

M y ¿ = 0.047×8.25× (2.97)²

= 3.42 KN-m (at supports)

+ve moment at mid span

α y ¿

M y ¿ = 0.035×8.25× (2.97)²

= 2.55 KN-m (at mid span)

iii. Effective Depth Calculation:

Taking maximum moment M x=4.73×10 ⁶ N-mm

B.M = 0.138 f ck bd ²

d=√M /0.138 f ck b=√4.73×10⁶/2.07×1000

= 47.80mm < 80mm (safe)

Provide effective depth is sufficient.

Width of Strips:

i. Along short span:

The slab is divided into three strips

Width of middle strip = 3/4 l y = 3/4 × 3.89

= 2.92 m

Width of edge strip = l y/8 = 3.89/8

= 0.48 m

18

ii. Along long span:

Width of middle strip = 3/4 lx = 3/4 × 2.97

= 22.23 m

Width of edge strip = lx/8 = 2.97/8

= 0.37 m

Design of Reinforcement:

Short span:

Steel for negative B.M. at continuous edge

M x (−ve )=0.87 f y A st d ⌊1−f y A st

ƒCK×bd⌋

4.73×106=0.87×415× A st 80 ⌊1−415 A st

15×1000×80⌋

A st=174.46mm ²

Using 8mm ϕ bars,A st=π d ²/4=π ×8 ² /4=50.26mm ²

Spacing ( s )=1000×50.26/174.46=287.60mm>75mm¿

Maximum spacing = 3d = 3×80 =240mm.

Hence provide 8mm ϕ bars of spacing 240mm c/c

Steel for positive B.M.at mid span

M x (+ve )=0.87 f y A st d ⌊1−f y A st

ƒCK×bd⌋

3.56×106=0.87×415× Ast 80 ⌊1−415 A st

15×1000×80⌋

19

A st=129.11mm ²


Spacing ( s )=1000×50.26/129.11=389.28mm>75mm¿



Long span:

Steel for negative B.M. at continuous edge

M x (−ve )=0.87 f y A st d ⌊1−f y A st

ƒCK×bd⌋

3.42×106=0.87×415× A st 80 ⌊1−415 Ast

15×1000×80⌋

A st=123.75mm ²


Spacing ( s )=1000×50.26/123.75=406.14mm>75mm¿



Steel for positive B.M.at mid span

M x (+ve )=0.87 f y A st d ⌊1−f y A st

ƒCK×bd⌋

2.55×106=0.87×415× A st 80 ⌊1−415 A st

15×1000×80⌋

20

A st=91.07mm ²


Spacing ( s )=1000×50.26/91.072=551.88>75mm¿



Provide A st=¿ 50.26240

×= 209.42 ≈210mm²

Main Reinforcement:

A stmin=0.12 % of the gross sectional area of slab

= 0.12/100 × 1000 × 100

= 120 mm²

Reinforcement in Edge Strip:

Reinforcement in edge strip A stmin=180 mm²

Using 8mm ϕ bars of spacing 240mm c/c is provided

Torsion Reinforcement:

At each corner at top and bottom mesh of reinforcement shall be provided covering of

square area

lx5×lx5=2.97

5× 2.97

5=¿ 0.353mm²

For layers should be provided, 2 layers at top and 2 layers at bottom

21

∴ Steel required for meter width = 3/4 × steel required for the maximum mid span

= 3/4 × 174.46 = 130.85 mm²


Spacing ( s )=1000×50.26/120=418.83>75mm


Hence provide 8mm ϕ bars, @ 320 mm c/cat edge strips.

22

AL

ON

G L

ON

G S

PAN

spac

ing 240

240

240

240

75

A st¿

12 3 15 6 15 6 15 6 62 8

spac ing

240

240

240

240

75

A st¿

210

156

156

156

713

M y¿

2.25

2.04

2.12

2.47

14.2

M y¿

3.42

2.69

2.82

3.26

15.5

α y¿

0.03

5

0.02

8

0.02

4

0.02

8

α y¿0.

047

0.03

7

0.03

2

0.03

7

AL

ON

G S

HO

RT

SPA

N

spac

ing 240

240

240

240

112

A st¿

129

209

156

156

445

spa

cin g 240

240

240

240

85

A st¿

174.

4

209

156

156

581

M x¿

3.56

3.20

2.82

3.17

10.8

M x¿

4.73

4.15

3.79

4.23

13.4

α x¿

0.04

9

0.04

4

0.03

2

0.03

6

α x¿

0.06

5

0.05

7

0.43

0.04

8

FAC

TO

R

8.25

8.25

8.25

8.25

9.75

SPA

N

Ly

3.89

3.89

3.89

3.89

3.89

Lx

2.97

2.97

3.27

3.27

1.33

S.N

O

S1 S2 S3 S4 S5

23

CHAPTER - 5

ANALYSIS OF FRAMES

26

METHOD OF CACULATION OF LOADS ON EACH BEAM OF

SUB STITUTE FRAME:

Load from slab:

If the beam is supporting long edge of the one way slab = (W dl x)/2

If the beam is supporting short edge of the one way slab = (W dl x)/6

If the beam is supporting long edge of the two way slab = W d lx

2×(1− 1

3 {Ly /Lx¿2 ¿)

If the beam is supporting short edge of the two way slab = (W dl x)/3

Load from walls (230mm thick) = 0.23×height×20

Fixed end moments:

MFAB=−¿

MFBA=+¿

Frame analysis of internal frame (A)-(A) is transverse direction (fig 2.)

29

LOADING ON BEAMS (FRAME NO.1)

For intermediate beams:-

i. Load on beam B11−12:

From Slabs = 2×(W d lx )/3 ¿2×5.5×2.97 /3 ¿10.89KN /m

From walls = 0.23×H ×20 = 0.23×2.75×20 = 12.65 KN/m

Self wt of beam = 0.23×0.3×25 = 1.725 KN/m

Total load 26.265 KN/m

Ultimate load ¿1.5×26.265=37.90KN /m

ii. Load on beam B12−13:


From walls = 0.23×H ×20 = 0.23×2.75×20 = 12.65 KN/m




iii. Load on beam B13−14:


From walls = 0.23×H ×20 = 0.23×2.75×20 = 12.65 KN/m



30


iv. Load on beam B14−15:

From Slabs = 2×(W d lx )/2 ¿2×6.5×1.33/2 ¿8.645KN /m

From walls = 0.23×H ×20 = 0.23×2.75×20 = 12.65 KN/m




For roof beams:

i. Load on beam B1−2:




Ultimate load ¿1.5×12.615=18.92 KN /m

ii. Load on beam B2−3:





31

iii. Load on beam B3−4:





iv. Load on beam B4−5:

From Slabs = 2×(W d lx )/2 ¿2×6.5×1.33/2 ¿8.645KN /m




Fixed end moments:

Intermediate frame:

Mf 11−12=−Wl ²/12=−¿

Mf 12−11=+Wl ²/12=+¿

Mf 12−13=−Wl ² /12=−¿

Mf 13−12=+Wl ² /12=+¿

Mf 13−14=−Wl ²/12=−¿

Mf 14−13=+Wl ² /12=+¿

Mf 14−15=−Wl ²/12=−¿

Mf 15−14=0

32

Roof frame:

Mf 1−2=−Wl ²/12=−¿

Mf 2−1=+Wl ² /12=+¿

Mf 2−3=−Wl ²/12=−¿

Mf 3−2=+Wl ² /12=+¿

Mf 3− 4=−Wl ² /12=−¿

Mf 4−3=+Wl ² /12=+¿

Mf 4−5=−Wl ² /12=−¿

Mf 5−4=0

33

Distribution Factor

Joint Member LengthRelativeStiffness

K

TotalRelativeStiffness

RotationalFactor

K∑ K

⌊−12⌋

Intermediate Frame

1111-0611-1211-16

305029703050

3.28 × 10−4

3.37 × 10−4

3.28 × 10−4

9.93 × 10−40.33×− 0.5 = − 0.170.34×− 0.5 = − 0.170.33×− 0.5 = − 0.17

1212-1112-0712-1312-17

2970305032703050

3.37 × 10−4

3.28 × 10−4

3.06 × 10−4

3.28 × 10−4

12.99 × 10−40.26×− 0.5 = − 0.130.25×− 0.5 = − 0.130.24×− 0.5 = − 0.120.25×− 0.5 = − 0.13

1313-1213-0813-1413-18

3270305029703050

3.06 × 10−4

3.28 × 10−4

3.37 × 10−4

3.28 × 10−4

12.99 × 10−40.24×− 0.5 = − 0.120.25×− 0.5 = − 0.130.26×− 0.5 = − 0.130.25×− 0.5 = − 0.13

1414-1314-0914-1514-19

2970305013303050

3.37 × 10−4

3.28 × 10−4

7.52 × 10−4

3.28 × 10−4

17.45 × 10−40.19×− 0.5 = − 0.100.19×− 0.5 = − 0.100.43×− 0.5 = − 0.220.19×− 0.5 = − 0.10

Roof Frame

1 1-21-6

29703050

3.37 × 10−4

3.28 × 10−46.65 × 10−4 0.51×− 0.5 = − 0.26

0.49×− 0.5 = − 0.25

22-12-32-7

297032703050

3.37 × 10−4

3.06 × 10−4

3.28 × 10−4

9.71 × 10−40.35×− 0.5 = − 0.180.32×− 0.5 = − 0.160.34×− 0.5 = − 0.17

33-23-43-8

327029703050

3.06 × 10−4

3.37 × 10−4

3.28 × 10−4

9.71 × 10−40.32×− 0.5 = − 0.160.35×− 0.5 = − 0.180.34×− 0.5 = − 0.17

44-34-54-9

297013303050

3.37 × 10−4

7.52 × 10−4

3.28 × 10−4

14.17× 10−40.24×− 0.5 = − 0.120.53×− 0.5 = − 0.270.23×− 0.5 = − 0.12

34

LOADING ON BEAMS (FRAME NO: 2)

For Intermediate Beams:

Load on beam F f−g:


From walls = 0.23×H ×20 = 0.23×2.75×20 = 12.65 KN/m

From Slabs = (W dl x)/2 ⌊1− 13 (l y /l x) ²

⌋

⌊ 5.5×3.892

⌋ ⌊1− 13(1.31)²

⌋+⌊ 5.5×2.972

⌋ ⌊1− 13(1.31) ²

⌋ = 15.20 KN /m

Total load = 29.58 KN/m


Loading on beams Fg−h ,Fh−i ,F i− j are same as F f−g

For Roof Beams:

Load on beam Fa−b:



⌋

⌊ 5.5×3.892

⌋ ⌊1− 13(1.31) ²

⌋+⌊ 5.5×2.972

⌋ ⌊1− 13(1.31) ²

⌋ = 15.20 KN /m



Loading on beams Fb−cF c−dFd−c are same as frame Fa−b:

38

Fixed end moments:

Intermediate Frames:

Mf f−g=−Wl ²/12=−¿

Mf g−f=+Wl ²/12=+¿

Mf g−h=−Wl ² /12=−¿

Mf h−g=+Wl ² /12=+¿

Mf h−i=−Wl ² /12=−¿

Mf i−h=+Wl ² /12=+¿

Mf i− j=−Wl ² /12=−¿

Mf j−i=+Wl ² /12=+¿

Roof Frame:

Mf a−b=−Wl ² /12=−¿

Mf b−a=+Wl ²/12=+¿

Mf b−c=−Wl ²/12=−¿

Mf c−b=+Wl ² /12=+¿

Mf c−d=−Wl ² /12=−¿

Mf d−c=+Wl ² /12=+¿

Mf d−e=−Wl ² /12=−¿

Mf e−d=+Wl ² /12=+¿

39

Distribution Factor


K


RotationalFactor

K∑ K

⌊−12⌋

Intermediate Frame

F f-gf-af-k

389030503050

2.57 × 10−4

3.28 × 10−4

3.28 × 10−4

9.13 × 10−40.28×− 0.5 = − 0.140.36×− 0.5 = − 0.180.36×− 0.5 = − 0.18

G g-fg-bg-bg-i

3890305038903050

2.57 × 10−4

3.28 × 10−4

2.57 × 10−4

3.28 × 10−4

11.70 × 10−40.22×− 0.5 = − 0.110.28×− 0.5 = − 0.140.22×− 0.5 = − 0.110.28×− 0.5 = − 0.14

H h-gh-ch-ih-m

3890305038903050

2.57 × 10−4

3.28 × 10−4

2.57 × 10−4

3.28 × 10−4

11.70 × 10−40.22×− 0.5 = − 0.110.28×− 0.5 = − 0.140.22×− 0.5 = − 0.110.28×− 0.5 = − 0.14

I i-hi-di-ji-n

3890305038903050

2.57 × 10−4

3.28 × 10−4

2.57 × 10−4

3.28 × 10−4

11.70 × 10−40.22×− 0.5 = − 0.110.28×− 0.5 = − 0.140.22×− 0.5 = − 0.110.28×− 0.5 = − 0.14

J j-ij-ej-o

389030503050

2.57 × 10−4

3.28 × 10−4

3.28 × 10−4

9.13 × 10−40.28×− 0.5 = − 0.140.36×− 0.5 = − 0.180.36×− 0.5 = − 0.18

Roof Frame

A a-b 38903050

2.57 × 10−4

3.28 × 10−45.85 × 10−4 0.44×− 0.5 = − 0.22

0.56×− 0.5 = − 0.28

B b-ab-gb-c

389030503890

2.57 × 10−4

3.28 × 10−4

2.57 × 10−4

8.42 × 10−40.31×− 0.5 = − 0.160.39×− 0.5 = − 0.200.31×− 0.5 = − 0.16

C c-bc-hc-d

389030503890

2.57 × 10−4

3.28 × 10−4

2.57 × 10−4

8.42 × 10−40.31×− 0.5 = − 0.160.39×− 0.5 = − 0.200.31×− 0.5 = − 0.16

D d-cd-id-e

389030503890

2.57 × 10−4

3.28 × 10−4

2.57 × 10−4

8.42 × 10−40.31×− 0.5 = − 0.160.39×− 0.5 = − 0.200.31×− 0.5 = − 0.16

E e-de-j

38903050

2.57 × 10−4

3.28 × 10−45.85 × 10−4 0.44×− 0.5 = − 0.22

0.56×− 0.5 = − 0.28

40

FRAME ANALYSIS OF EXTERNAL FRAME (A)-(A) IN TRANSVERSE

DIRECTION (FRAME: 3)

44

Loading on Beams:

i. Beam,A11−12:

From Slabs = (W dl x)/3 ¿5.5×2.97 /3 ¿5.45KN /m

From walls = 0.23×H ×20 = 0.23×2.75×20 = 12.65 KN/m



Factored load ¿1.5×19.83=29.74KN /m

ii. Beam,A12−13:


From walls = 0.23×H ×20 = 0.23×2.75×20 = 12.65 KN/m




iii. Beam,A13−14:


From walls = 0.23×H ×20 = 0.23×2.75×20 = 12.65 KN/m




45

iv. Load on beam A14−15:

From Slabs = (W dl x)/2 ¿2×6.5×1.33/2 ¿4.32KN /m

From walls = 0.23×H ×20 = 0.23×2.75×20 = 12.65 KN/m




For Roof Frame:

i. Beam,A1−2:

From Slabs = (W lx )/3 ¿5.5×2.97 /3 ¿5.45KN /m




ii. Beam,A2−3:

From Slabs = (W lx )/3 ¿5.5×3.27 /3 ¿5.99KN /m




iii. Beam,A3−4:



Total load = 7.175KN/m

46

Factored load ¿1.5×=7.175=10.76KN /m

iv. Beam,A4−5:

From Slabs = ¿(W d lx )/2 ¿6.5×1.33/2 ¿4.32KN /m




47

Fixed end moments:

Intermediate frame:

Mf 11−12=−Wl ²/12=−¿

Mf 12−11=+Wl ²/12=+¿

Mf 12−13=−Wl ² /12=−¿

Mf 13−12=+Wl ² /12=+¿

Mf 13−14=−Wl ²/12=−¿

Mf 14−13=+Wl ² /12=+¿

Mf 14−15=−Wl ²/12=−¿

Mf 15−14=0

Roof frame:

Mf 1−2=−Wl ²/12=−¿

Mf 2−1=+Wl ² /12=+¿

Mf 2−3=−Wl ²/12=−¿

Mf 3−2=+Wl ² /12=+¿

Mf 3−4=−Wl ² /12=−¿

Mf 4−3=+Wl ² /12=+¿

Mf 4−5=−Wl ² /8=−¿

Mf 5−4=0

48

Distribution Factor


K


RotationalFactor

K∑ K

⌊−12⌋

Intermediate Frame

1111-0611-1211-16

305029703050

3.28 × 10−4

3.37 × 10−4

3.28 × 10−4

9.93 × 10−40.33×− 0.5 = − 0.170.34×− 0.5 = − 0.170.33×− 0.5 = − 0.17

1212-1112-0712-1312-17

2970305032703050

3.37 × 10−4

3.28 × 10−4

3.06 × 10−4

3.28 × 10−4

12.99 × 10−40.26×− 0.5 = − 0.130.25×− 0.5 = − 0.130.24×− 0.5 = − 0.120.25×− 0.5 = − 0.13

1313-1213-0813-1413-18

3270305029703050

3.06 × 10−4

3.28 × 10−4

3.37 × 10−4

3.28 × 10−4

12.99 × 10−40.24×− 0.5 = − 0.120.25×− 0.5 = − 0.130.26×− 0.5 = − 0.130.25×− 0.5 = − 0.13

1414-1314-0914-1514-19

2970305013303050

3.37 × 10−4

3.28 × 10−4

7.52 × 10−4

3.28 × 10−4

17.45 × 10−40.19×− 0.5 = − 0.100.19×− 0.5 = − 0.100.43×− 0.5 = − 0.220.19×− 0.5 = − 0.10

Roof Frame

1 1-2 29703050

3.37 × 10−4

3.28 × 10−46.65 × 10−4 0.51×− 0.5 = − 0.26

0.49×− 0.5 = − 0.25

22-12-32-7

297032703050

3.37 × 10−4

3.06 × 10−4

3.28 × 10−4

9.71× 10−40.35×− 0.5 = − 0.180.32×− 0.5 = − 0.160.34×− 0.5 = − 0.17

33-23-43-8

327029703050

3.06 × 10−4

3.37 × 10−4

3.28 × 10−4

9.71× 10−40.32×− 0.5 = − 0.160.35×− 0.5 = − 0.180.34×− 0.5 = − 0.17

44-34-54-9

297013303050

3.37 × 10−4

7.52 × 10−4

3.28 × 10−4

14.17× 10−40.24×− 0.5 = − 0.120.53×− 0.5 = − 0.270.23×− 0.5 = − 0.12

49

FRAME ANALYSIS OF EXTERNAL FRAME (E)-(A) IN

LONGITUDINAL DIRECTION (FRAME: 4)

Loading on Beams:

For intermediate beams:

BeamEA f−g:


From walls = 0.23×2.75×20 = 12.65KN/m


⌋ ¿6.58 KN /m



Loading on beams EA g−h ,EA h−i , EAi− j are same as frame EA f−g

For roof frame beams:

BeamEA a−b:



⌋ ¿6.58 KN /m

Total load = 8.31KN/m


Load on beams EAb−c EA c−d EA d−c are same as frame EA a−b

53

Fixed end moments:

Intermediate frame:

Mf f−g=−Wl ²/12=−(31.43×3.89 ²)/12=−39.63KN−m

Mf g−f=+Wl ² /12=+(31.43×3.89 ²)/12=+39.63KN−m

Mf g−h=−Wl ² /12=−(31.43×3.89 ²) /12=−39.63KN−m

Mf h−g=+Wl ² /12=+(31.43×3.89 ²)/12=+39.63KN−m

Mf h−i=−Wl ² /12=−(31.43×3.89 ²)/12=−39.63KN−m

Mf i−h=+Wl ² /12=+(31.43×3.89 ²)/12=+39.63 KN−m

Mf i− j=−Wl ² /12=−(31.43×3.89 ²)/12=−39.63KN−m

Mf j−i=0

Roof frame:

Mf a−b=−Wl ² /12=−¿

Mf b−a=+Wl ²/12=+¿

Mf b−c=−Wl ²/12=−¿

Mf c−b=+Wl ² /12=+¿

Mf c−d=−Wl ² /12=−¿

Mf d−c=+Wl ² /12=+¿

Mf d−e=−Wl ² /12=−¿

Mf 5−4=0

54

Distribution Factor


K


RotationalFactor

K∑ K

⌊−12⌋

Intermediate Frame

F f-gf-af-k

389030503050

2.57 × 10−4

3.28 × 10−4

3.28 × 10−4

9.13 × 10−40.28×− 0.5 = − 0.140.36×− 0.5 = − 0.180.36×− 0.5 = − 0.18

G g-fg-bg-bg-i

3890305038903050

2.57 × 10−4

3.28 × 10−4

2.57 × 10−4

3.28 × 10−4

11.70 × 10−40.22×− 0.5 = − 0.110.28×− 0.5 = − 0.140.22×− 0.5 = − 0.110.28×− 0.5 = − 0.14

H h-gh-ch-ih-m

3890305038903050

2.57 × 10−4

3.28 × 10−4

2.57 × 10−4

3.28 × 10−4

11.70 × 10−40.22×− 0.5 = − 0.110.28×− 0.5 = − 0.140.22×− 0.5 = − 0.110.28×− 0.5 = − 0.14

I i-hi-di-ji-n

3890305038903050

2.57 × 10−4

3.28 × 10−4

2.57 × 10−4

3.28 × 10−4

11.70 × 10−40.22×− 0.5 = − 0.110.28×− 0.5 = − 0.140.22×− 0.5 = − 0.110.28×− 0.5 = − 0.14

J j-ij-ej-o

389030503050

2.57 × 10−4

3.28 × 10−4

3.28 × 10−4

9.13 × 10−40.28×− 0.5 = − 0.140.36×− 0.5 = − 0.180.36×− 0.5 = − 0.18

Roof Frame

A a-b 38903050

2.57 × 10−4

3.28 × 10−45.85 × 10−4 0.44×− 0.5 = − 0.22

0.56×− 0.5 = − 0.28

B b-ab-gb-c

389030503890

2.57 × 10−4

3.28 × 10−4

2.57 × 10−4

8.42 × 10−40.31×− 0.5 = − 0.160.39×− 0.5 = − 0.200.31×− 0.5 = − 0.16

C c-bc-hc-d

389030503890

2.57 × 10−4

3.28 × 10−4

2.57 × 10−4

8.42 × 10−40.31×− 0.5 = − 0.160.39×− 0.5 = − 0.200.31×− 0.5 = − 0.16

D d-cd-id-e

389030503890

2.57 × 10−4

3.28 × 10−4

2.57 × 10−4

8.42 × 10−40.31×− 0.5 = − 0.160.39×− 0.5 = − 0.200.31×− 0.5 = − 0.16

E e-de-j

38903050

2.57 × 10−4

3.28 × 10−45.85 × 10−4 0.44×− 0.5 = − 0.22

0.56×− 0.5 = − 0.28

55

CHAPTER - 6

DESIGN OF BEAMS

60

DESIGN OF BEAMS

Model calculation:

BeamD7−8: (Intermediate floor – Frame No.1) (External)

Total load on beam = 39.55 KN/m

Beam size is assumed as 230×300 mm

Assuming an effective cover of 30 mm

We have effective depth of beam = 260 mm

Length of beam = 3.27 m

61

At B12(Hogging Moment ):

M u/bd2=¿ 1.5×35.44×106

230×2602 ¿3.42>2.06→ Double reinforced beam

230×2602 Referring table 49 of SP 16-1980 for Fe415,M 15 we have

d ' /d=0.153

Pt=1.156

Pc=0.472

Area of tension reinforcement, A st = (1.156×230×260)/100 = 691.29 mm²

Provide 3-12 ϕ +2.16 ϕ bars

Area of compression reinforcement, A sc=¿(0.472×230×260)/100 = 282.26 mm²

Provide 3-12 ϕAt B13(Hogging Moment ):

M u/bd2=¿ 1.5×33.26×106

230×2602 ¿3.26>2.06→Double reinforced beam

230×2602 Referring table 49 of SP 16-1980 for Fe415,M 15 we have

d ' /d=0.15

Pt=1.104

Pc=0.416


Provide 3-12 ϕ +2.16 ϕ bars

62

Area of compression reinforcement, A sc=¿(0.416×230×260)/100 = 248.77 mm²

Provide 3-12 ϕAt center of beam B12−13:

M u/bd2=¿ 1.5×17.62×106

230×2602 ¿1.70<2.06→Single reinforced beam

d ' /d=0.15

Pt=0.558


Provide 2.16 ϕ bars

63

Design of Beams For Flexure

Name Beam Span Loading on beam

KN/m

Simply supported beamKN-m

Design momentfor beam

KN/m

M u/bd2

×1.5

% of reinforcement

Area of steel No. of bars

Pt Pc A st A sc A st A sc

Intermediate

B11−122.97 37.90 41.79

L:18.81 1.81 0.602 - 359.99 - 2-16

M:13.93 1.34 0.420 - 25.16 - 2-16

R:32.98 3.18 1.078 0.388 644.64 232.02 3-121-16

3-12

B12−133.27 39.55 52.86

L:35.44 3.42 1.156 0.472 691.29 282.26 3-123-12

2-16

M:17.62 1.70 0.558 - 333.68 - 2-16

R:33.75 3.26 1.104 0.416 660.19 248.77 3-123-12

2-16

B13−142.97 37.90 41.79

L:31.80 3.07 1.042 0.349 623.12 208.70 3-123-12

2-16

M:13.93 1.34 0.420 - 251.16 - 3-12 2-16

R:23.01 2.22 0.765 0.052 457.47 31.096 2-122-16

2-12

B14−151.33 34.53 16.09 1.34 0.420 - 251.16 - 2-12

2-162-12

Roof beams

B1−22.97 18.92 20.86

L:8.58 0.83 0.247 - 147.71 - 1-16

M:6.95 0.67 0.196 - 117.21 - 2-122-16

2-12

R:17.09 1.65 0.538 - 321.72 - 2-124-16

3-12

B2−33.27 20.57 27.49

L:18.46 1.78 0.590 - 352.82 - 1-122-121-16

4-16

M:9.16 0.88 0.632 - 377.94 - 4-161-10

2-12

R:17.12 1.65 0.538 - 321.72 - 2-122-12

2-16

B3−42.97 18.92 13.75

L:16.76 1.62 0.526 - 314.55 -

M:6.95 0.67 0.196 - 117.21 - 4-164-17

2-121-10

R:10.75 1.04 0.316 - 188.96 - 2-12

B4−51.33 15.55 8.65 0.83 0.247 - 147.70 14-16

2-12

65

Name Beam Span Loading on

beamKN/m

Simply supported

beamKN-m


KN/m

M v /bd2

×1.5

% of reinforcement


Pt Pc A st A sc A st A scIntermediate

F f−g3.89 44.37 83.93

L:42.76 4.13 1.384 0.718 827.63 429.36 3-12 2-122-16 1-10

M:27.98 2.699 0.918 0.216 548.96 129.17 3-12 3-122-16 1-10

R:61.39 5.92 1.971 1.348 102.49 806.10 5-16 2-124-16 -

Fg−h3.89 44.37 83.93

L:57.49 5.55 1.851 1.045 1106.9 806.10 2-12 1-122-16 -

M:27.98 2.699 0.918 0.216 548.96 129.17 5-16 3-121-10 2-16

R:55.18 5.32 1.775 1.138 1061.5 680.25 5-16 3-121-10 2-16

Fh−i3.89 44.37 83.93

L:55.18 5.32 1.775 1.138 1061.5 680.25 2-12 2-161-12

M:27.98 2.699 0.918 0.216 548.96 129.17 5-16 4-162-12 -

R:57.49 5.55 1.851 1.045 1106.9 806.10 5-16 1-122-12 4-16

F i− j3.89 44.37 83.93

L:61.39 5.92 1.971 1.348 102.49 806.10 3-12 2-122-16 1-10

M:27.98 2.699 0.918 0.216 548.96 129.17 3-12 2-122-16 1-10

R:42.76 4.13 1.384 0.718 827.63 429.36 3-12 2-122-16 1-10

Roof beams

Fa−b3.89 25.39 48.03

L:21.94 2.12 0.732 0.017 432.35 10.16 3-12 2-122-16 1-10

M:16.01 1.54 0.495 - 296.01 - 2-12 -1-10 -

R:34.02 3.28 1.11 0.423 663.78 252.92 2-16 -- -

Fb−c3.89 25.39 48.03

L:32.29 3.11 1.05 0.363 630.89 217.17 3-12 2-122-16 1-10

M:16.01 1.54 0.495 - 296.01 - 3-12 2-122-16 1-10

R:29.99 2.89 0.983 0.286 587.83 171.03 3-12 -- -

F c−d3.89 25.39 48.03

L:29.99 2.89 0.983 0.286 587.83 171.03 2-12 1-122-16 1-10

M:16.01 1.54 0.495 - 296.01 - 3-12 -- -

R:32.29 3.11 1.05 0.363 630.89 217.17 3-12 3-122-16 -

Fd−e3.89 25.39 48.03

L:34.02 3.28 1.11 0.423 663.78 252.92 3-12 2-12M:16.01 1.54 0.495 - 296.01 - 2-12 1-10R:21.94 2.12 0.732 0.017 432.35 10.16

66


KN/m

Simply supported beamKN-m


KN/m

M u/bd2

×1.5

% of reinforcement


Pt Pc A st A sc A st A scInter

mediateA11−12

2.97 29.73 32.78

L:14.56 1.40 0.442 - 264.32 - 2-16

M:10.93 1.05 0.319 - 190.76 - 2-16

R:25.87 2.50 0.857 0.150 512.49 89.70 3-121-16

3-12

A12−133.27 30.55 43.51

L:27.45 2.65 0.905 0.202 541.19 120.79 3-123-12

2-16

M:11.23 1.08 0.329 - 196.74 - 2-16

R:26.10 2.52 0.863 0.157 516.07 93.89 3-123-12

2-16

A13−142.97 29.73 32.78

L:24.80 2.40 0.824 0.115 492.75 68.77 3-123-12

2-16

M:10.93 1.05 0.319 - 190.76 - 3-12 2-16

R:18.15 1.75 0.578 - 345.64 - 2-122-16

2-12

A14−151.33 28.05 30.93 12.77 1.23 0.381 - 227.84 - 2-12

2-162-12

Roof beams

A1−22.97 10.76 11.86

L:5.41 0.52 0.15 - 89.7 - 1-16

M:3.95 0.38 0.108 - 64.58 - 2-122-16

2-12

R:9.50 0.92 0.276 - 165.04 - 2-124-16

3-12

A2−33.27 11.57 15.46

L:10.32 1.00 0.303 - 181.19 - 1-122-121-16

4-16

M:5.15 0.50 0.144 - 86.11 - 4-161-10

2-12

R:9.68 0.93 0.279 - 166.84 - 2-122-12

2-16

A3−42.97 10.76 11.86

L:9.50 0.92 0.276 - 165.04 -

M:3.95 0.38 0.108 - 64.58 - 4-164-17

2-121-10

R:6.14 0.59 0.171 - 102.21 - 2-12

A4−51.33 9.07 2.01 4.94 0.48 0.138 - 82.52 14-16

2-12

67


KN/m

Simply supported

beamKN-m


KN/m

M v /bd2

×1.5

% of reinforcement


Pt Pc A st A sc A st A sc

Intermediate

EA f−g3.89 31.43 59.45

L:29.83 2.87 0.976 0.279 583.65 166.84 3-12 2-122-16 1-10

M:19.81 1.91 0.645 - 385.71 - 3-12 3-122-16 1-10

R:43.68 4.21 1.41 0.75 843.18 447.90 5-16 2-124-16 -

EA g−h3.89 31.43 59.45

L:46.77 4.51 1.52 0.854 904.17 508.30 2-12 1-122-16 -

M:19.81 1.91 0.645 - 385.71 - 5-16 3-121-10 2-16

R:39.06 3.77 1.270 0.595 759.46 355.81 5-16 3-121-10 2-16

EAh−i3.89 31.43 59.45

L:39.06 3.77 1.270 0.595 759.46 355.81 2-12 2-161-12

M:19.81 1.91 0.645 - 385.71 - 5-16 4-162-12 -

R:46.77 4.51 1.52 0.854 904.17 508.30 5-16 1-122-12 4-16

EAi− j3.89 31.43 59.45

L:43.68 4.21 1.41 0.75 843.18 447.90 3-12 2-122-16 1-10

M:19.81 1.91 0.645 - 385.71 - 3-12 2-122-16 1-10

R:29.83 2.87 0.976 0.279 583.65 166.84 3-12 2-122-16 1-10

Roof beams

EA a−b3.89 12.46 23.57

L:12.05 1.162 0.357 - 213.48 - 3-12 2-122-16 1-10

M:7.85 0.75 0.221 - 132.15 - 2-12 -1-10 -

R:17.06 1.64 0.534 - 319.93 - 2-16 -- -

EAb−c3.89 12.46 23.57

L:16.34 1.57 0.506 - 302.58 - 3-12 2-122-16 1-10

M:7.85 0.75 0.221 - 132.15 - 3-12 2-122-16 1-10

R:15.39 1.48 0.472 - 282.25 - 3-12 -- -

EA c−d3.89 12.46 23.57

L:15.39 1.48 0.472 - 282.25 - 2-12 1-122-16 1-10

M:7.85 0.75 0.221 - 132.15 - 3-12 -- -

R:16.34 1.57 0.506 - 302.58 - 3-12 3-122-16 -

EA d−e3.89 12.46 23.57

L:17.06 1.64 0.534 - 319.93 - 3-12 2-12M:7.85 0.75 0.221 - 132.15 - 2-12 1-10R:12.05 1.162 0.357 - 213.48 -

MODEL SHEAR CALCULATION

68

BeamB12−13:

Maximum shear force (V u ¿=Wl ⁄ 2=39.55×3.27 /2=64.66 KN

Shear stress( τ v )=¿ V u/bd

= 64.66×103 /230×260=1.081N /mm ²

τ v<¿ τ c Maximum

From table SP 16-1980

Pt=1.156 for left support

= 1.104 for right support

From SP 16-1980

Design shear strength τ c=0.626 left support

= 0.620 right support

τ cbd=0.626×230×260=37.435KN (left support)

= 0.620×230×260=37.08 KN (right support)

V us=V u−τ c bd

= 64.66−¿ 37.435 = 27.23 (left)

= 64.66−¿ 37.08 = 27.58 (right)

V us /d=27.23 /26=1.05 (left)

¿27.58/26=1.06 (right)

As per SP-16 reinforcement details = 8 mm ϕ @ 195 c/c

69

SHEAR FORCE DEVELOPED ON BEAMSS.No Name of Beam Span Total Load

(w)Maximum Shear Force on

Beam = WL/2Frame No = 1 (Internal) Intermediate beams

1 B11−12 2.97 37.90 56.282 B12−13 3.27 39.55 64.663 B13−14 2.97 37.90 56.284 B14−15 1.33 34.53 45.92

Roof Frame

5 B1−2 2.97 18.92 28.106 B2−3 3.27 20.57 33.637 B3−4 2.27 18.92 28.108 B4−5 1.33 15.55 20.68

Frame No = 2 (External) Intermediate beams

9 F f−g 3.89 44.37 82.2910 Fg−h 3.89 44.37 82.2911 Fh−i 3.89 44.37 82.2912 F i− j 3.89 44.37 82.29

Roof Frame

13 Fa−b 3.89 25.39 49.3814 Fb−c 3.89 25.39 49.3815 F c−d 3.89 25.39 49.3816 Fd−e 3.89 25.39 49.38


17 A11−12 2.97 29.73 44.1518 A12−13 3.27 30.55 49.9519 A13−14 2.97 29.73 44.1520 A14−15 1.33 28.05 37.31

Roof Frame

21 A1−2 2.97 10.76 15.9822 A2−3 3.27 11.57 18.9223 A3−4 2.97 10.76 15.9824 A4−5 1.33 9.07 12.06


25 EA f−g 3.89 31.43 61.1326 EAg−h 3.89 31.43 61.1327 EAh−i 3.89 31.43 61.1328 EAi− j 3.89 31.43 61.13

Roof Frame70

29 EA a−b 3.89 12.46 24.2330 EAb−c 3.89 12.46 24.2331 EA c−d 3.89 12.46 24.2332 EA d−e 3.89 12.46 24.23

71

CHAPTER - 7

DESIGN OF COLUMNS

72

DESIGN OF COLUMNS

The column in a structure are usually carrying axial compressive load i.e,, relation from

beams and some moments in both directions. Rectangular cross section is selected considering

the biaxial and some moments in both directions. Rectangular cross section is selected

considering the biaxial bending for columns. All the columns have been divided in six groups

according to the loads falling in a certain range. Each group of columns is designed taking the

maximum axial compressive load and moments.

As all columns are subjected to moment in both directions, columns are designed for

biaxial bending in addition to the axial compressive load. Columns are designed using interaction

diagrams, which are readily available in SP:16-1980 for different combinations of grade of steel,

grade of concrete percentage of reinforcement and size of the column to avoid cumbersome

calculations. Longitudinal and transverse reinforcement providing in according with norms laid

down clause 25.5.3 of IS 456 1978. While designing the columns it is assumed that the reinforce

is distributed equally on all sides.

Columns re so oriented that the maximum depth is available perpendicular to the axis

about which the maximum moment is acting. M 20 grade and Fe415 grade of steel is used for all

columns.

74

DESIGN OF COLUMNS

Materials:

M 20 grade concrete and Fe415 grade steel

Column:C1 : (Between first floor and plinth) Axial load and biaxial bending

Eccentricity: In x-direction emin=L/500+D /30

¿3050/500+300 /30

¿ 16.10 < 20 mm

In Y-direction E¿ h=L/500+D /30

¿3050/500+230 /30

¿ 13.76 < 20 mm

End conditions: Fixed @both ends

Effective length = 0.65×L = 2.275 m

Pu=388.05 KN

B=230mm

D=300mm

d1=46mm

Pu×emin=388.05×20/1000=7.761 KN-m

F ck=20

L/D = 2.275/0.23 = 9.89 < 12

Hence column should be design as short column

75

Assuming % of reinforcement equal to two times the minimum reinforcement,

i.e. P=0.8×2=1.6 %

P/ f ck=1.6 /20=0.08

Pu

f ck bd= 3.8805×103

20×230×300=0.281

In X- direction:

d ' /D=46 /300=0.153

M ux/ f ck bd2=0.13 (From chart)

M ux1=0.13×20×230×300 ²=53.82 KN-m

In Y- direction:

d ' /D=46 /230=0.2

M uy / f ck bd2=0.16

M uy1=0.16×20×300×230 ² /100=50.78 KN-m

( Mux

M ux1)∝n

+( Muy

Muy 1)∝n

≤ 1.0

∝n=Pu/Puz

Referring to chart corresponding to P = 1.6, f y=415∧f ck=20

We get Puz /A g=14 N /mm ²

Puz=14×230×300=966KN

Pu

Puz=388.05

966=0.402

From code IS.456 : 2000, clause 39.6

76

∝n=1.32

¿( 22.3253.82 )

1.32

+(37.5950.78 )

1.32

¿0.312+0.672

¿0.984 ≤ 1.0

Hence safe

Provide A s=pbD /100=1.6×230×300/100

¿1104mm ²

Provide 12-12 ϕ bars distributed equally on four sides.

Lateral ties:

Provide 6mm ϕ lateral ties,

Pitch = Least of

i. Least lateral dimensions = 230

ii. 16×dia of main bars = 16×12 = 192mm

iii. 48×dia of ties = 48×6 =288mm

Provide a pitch of 6mm ϕ @ 190 mm c/c.

77

AXIAL LOADS ON COLUMNS

DESCRIPTION C1 C2 C3 C4

Column between Roof & Third floor

Factored self weight of column 8.00 8.00 8.00 8.00

Shear from longitudinal beams 24.23 48.46 49.38 98.28

Shear from transverse beams 15.98 28.10 34.90 61.73

Load from above - - - -

Total load 48.21 84.56 92.28 168.01

Column between Third floor & Second floor




Load from above 48.21 84.56 92.28 168.01

Total load 161.49 271.1 276.67 454.01

Column between Second floor & First floor




Load from above 161.49 271.1 276.67 454.01

Total load 274.77 457.64 461.06 747.53

Column between First floor & Ground floor




Load from above 274.77 457.64 461.06 747.53

Total load 388.05 644.18 645.45 1041.05

78

Name For eccentricity

M uxe=M uye

Pu×emin

Calculated Total

M uxe M uye M ux M uy

C1

Between roof and 3rd floor 0.96 0.96 5.41 12.05 6.37 13.01

Between 3rd and 2nd floor 3.23 3.23 14.56 29.83 17.79 33.06

Between 2nd and 1st floor 5.50 5.50 14.56 29.83 20.06 35.33

Between 1st and Ground 7.76 7.76 14.56 29.83 22.32 37.59

C2





C3





C4





79

BIAXIAL BENDING

( Mux

M ux1)∝n

+( Muy

Muy 1)∝n

≤

1.0

0.4

91

≤

1.0

0.98

2

≤

1.0

0.98

6

≤

1.0

0.98

4

≤

1.0

0.40

5

≤

1.0

0.55

7

≤

1.0

0.56

1

≤

1.0

0.70

3

≤

1.0

0.91

4

≤

1.0

0.99

1

≤

1.0

0.98

2

≤

1.0

0.99

4

≤

1.0

0.31

2

≤

1.0

0.38

8

≤

1.0

0.60

4

≤

1.0

0.27

4

≤

1.0

∝n

1.0

1.0

1.14

1.32 1.0

1.12

1.35

1.50 1.0

1.14

1.40

1.52 1.0

1.38

1.64

1.84Pu

Puz

0.06

0.18

0.28

0.40

0.10

0.27

0.43

0.52

0.11

0.28

0.45

0.53

0.18

0.44

0.60

0.72

Puz

Ag N

mm

²

12 13.2

0

14 14 12 14.5

15.5 18 12 14.5 15 17.6

13.0

2

15 18 21

Una

xial

mom

ent c

apac

ity o

f the

se

ctio

n y-

y ax

is

M uy1

22.2

2

50.7

8

50.7

8

50.7

8

25.3

9

38.0

9

39.0

9

39.0

9

28.5

7

53.9

6

57.1

3

60.3

1

28.5

7

34.9

1

31.7

4

60.3

1

Mu

f ck bd ²

0.07

0.16

0.16

0.16

0.08

0.12

0.12

0.12

0.09

0.17

0.18

0.19

0.09

0.11

0.10

0.19

Pu

f ck bd

0.04

0.18

0.20

0.28

0.06

0.20

0.34

0.47

0.07

0.20

0.33

0.48

0.12

0.33

0.54

0.75

d '

D

0.20

0.20

0.20

0.20

0.20

0.20

0.20

0.20

0.20

0.20

0.20

0.20

0.20

0.20

0.20

0.20

Una

xial

mom

ent c

apac

ity o

f the

se

ctio

n x-

x ax

is

M ux1

31.0

1

53.8

2

53.8

2

53.8

2

33.1

2

53.8

2

53.8

2

55.4

8

33.1

2

49.6

8

53.8

2

62.1

0

35.5

4

49.6

8

49.6

8

91.0

8

Mu

f ck bd ²

0.08

0.13

0.13

0.13

0.08

0.13

0.13

0.13

0.08

0.12

0.13

0.15

0.11

0.12

0.12

0.22

Pu

f ck bd0.04

0.18

0.20

0.28

0.06

0.20

0.34

0.47

0.07

0.20

0.33

0.48

0.12

0.33

0.54

0.75d '

D

0.15

0.15

0.15

0.15

0.15

0.15

0.15

0.15

0.15

0.15

0.15

0.15

0.15

0.15

0.15

0.15

Pf ck

0.05

0.08 6 0.08

0.08

0.05

0.09

0.11

0.15

0.05

0.09

0.10

0.14

0.07 5 0.10

0.15

0.20

% o

f ste

el

1.0

1.5

1.6

1.6

1.0

1.8

2.2

3.0

1.0

1.8

2.0

2.8

1.5

1.0

2.0

4.0

Col

umn

C1 1 2 3 4 C2 1 2 3 4 C3 1 2 3 4 C4 1 2 3 4

80

DETAILS OF STEEL REINFORCEMENT IN COLUMNS

Name Area of longitudinal

reinforcement mm²

No. Of bars Lateral ties

C1

1 690 12mm ϕ -8no’s 6mm ϕ -@190mm c/c no’s2 1035 12mm ϕ -12no’s 6mm ϕ -@190mm c/c no’s3 1104 12mm ϕ -12no’s 6mm ϕ -@190mm c/c no’s4 1104 12mm ϕ -12no’s 6mm ϕ -@190mm c/c no’s

C21 690 12mm ϕ-8no’s 6mm ϕ -@190mm c/c no’s2 1242 12mm ϕ-12no’s 6mm ϕ -@190mm c/c no’s3 1518 12mm ϕ-16no’s 6mm ϕ -@190mm c/c no’s4 2070 12mm ϕ-12no’s 6mm ϕ -@190mm c/c no’s

C31 690 12mm ϕ-8no’s 6mm ϕ -@190mm c/c no’s2 1242 12mm ϕ-12no’s 6mm ϕ -@190mm c/c no’s3 1380 12mm ϕ-16no’s 6mm ϕ -@190mm c/c no’s4 1932 12mm ϕ -12no’s 6mm ϕ -@190mm c/c no’s

C41 1035 12mm ϕ -12no’s 6mm ϕ -@190mm c/c no’s2 1580 12mm ϕ -16no’s 6mm ϕ -@190mm c/c no’s3 2070 16mm ϕ -12no’s 6mm ϕ -@190mm c/c no’s4 2760 16mm ϕ -16no’s 6mm ϕ -@190mm c/c no’s

81

CHAPTER - 8

DESIGN OF FOOTINGS

83

DESIGN OF FOUNDATIONS

The foundation is a structure or the part of the structure, which transverse the load to the

soil on which it rests. If forms a very important part of the structure. A foundation should be

designed such that there is no possibility of tilting of structure. At the – edge close to the center

of gravity of the loads, the pressure intensity will be higher resulting in greater settlement of the

soil there. This will result in greater tilting of the structure in the direction. Hence it is better to

design the foundation are such that the center of gravity of the foundation area so that the soil

reaction will be of uniform intensity.

As there are several types foundations available isolated sloped footings is selected in this

project considering safe bearing capacity of soil and other factors such as distance between

columns, space available and loads to which it is subjecting,

Footing shall be designed to sustain the applied loads, moments, forces, induced reactions

and ensure that the any settlement, which may occur, shell be, as nearly uniform as possible and

the safe bearing capacity of the soil is not exceeding. In slopped footings the effective cross

section in compression shall be limited by the area above the neutral plane and the angle of slope

or depth shall be such that the design requirements are satisfied at every section. Sloped footings

are designed, as a unit shall be constructed to assure action as a unit.

84

DESIGN OF FOOTING

MODEL CALCULATION:

Size of Column ¿230×300mm.

Ultimate load from column ¿388.05 KN .

Working load from column ¿388.05/1.5=258.70KN

Self-weight of the column ¿10%of working load

¿10× 258.70

100 ¿25.87KN

Total load ¿258.70+25.87=284.57KN

Assuming the soil is cohesive moist clay and sand clay mixture.

Safe baring capacity (SBC) P0=180 KN/m²

Area of footing required = (258.70+25.87)/180 = 1.58 m² ≈ 1.60 m²

Size of footing = 1.4×1.6 m

Net upward pressure = q0=¿ (388.05/1.4×1.6) = 173.23 KN/m²

Determination of Depth of footing:

M x=¿

¿¿ = 41.50 KN-m

M y=¿

¿¿ = 58.55 KN-m

86

Effective Depth dx=√M x

RB=√ 41.50×106

2.07×1400 ¿119.66mm say 200mm

Effective Depth dy=√M y

RB=√ 58.55×106

2.07×1600 ¿132.96mmsay 200mm

Assuming bar diameter = 12mm

Effective cover d '=50+ 122

=56mm

Total depth = 200+56 = 256 ≈ 270mm

Calculation of area of steel:

A stx=0.5 f ckf y [1−√1− 4.6MD

f ck×bd2 ]bd

¿ 0.5×20415 [1−√1−4.6×41.50×106

20×1400×2002 ]1400×200=901.37mm ²

A stx=0.5×20

415 [1−√1−4.6×58.55×106

20×1600×2002 ]1600×200=855.66mm ²

A stx : Use 12mm ϕ bars,

No. of bars used = 901.37/113 = 8.80 ≈ 12

A sty : Use 12mm ϕ bars,

No. of bars used = 855.66/113 = 7.57 ≈ 10

Check for one way shear:

87

Along X-X axis:

V u=q0B (L/2−a /2−d ) = 173.23×1.4(1.6/2-0.23/2-0.2)

= 117.62 KN

Nominal shear stress τ v=V u /(B dx) = 117.62 ×103/(1400×200) = 0.42 N/mm²

ForM 20 ¿table∧%of steel Pt=0.44

τ c=0.48N /mm ²

D ≥ 300

K = 1

τ v<K τc

Hence safe.

Check for two way shear: (Punching shear)

Area = (a + d) (b + d)

= (0.3+0.2) (0.23+0.2) = 0.215 m²

Punching shear = P0(B×L) = 173.23(1.4×1.6) = 388.05 KN

Perimeter = 2[(a + d) (b + d)] = 2[(0.3+0.2) (0.23+0.2)] = 1.86 m

τ v=388.05/ (1.86×200) = 1.04 N/mm²

88

DESIGN OF FOOTINGS

Safe bearing capacity of soil = 180 KN/m

Use 10 mm bars

Column Axial load

Area of

footing

Size (B×L)

Upward

pressure P0

KN/m²

M xKN/m

D xmm

M yKN/m

D ymm

A stxmm²

No. of

bars

A stymm²

No. of

bars

C1 388.05 1.60 1.4×1.6 173.23 41.45 200 58.55 200 901.37 12 855.66 10

C2 644.18 2.62 1.6×2.0 201.31 75.57 200 145.43 250 2373.8 21 1730.3 14

C3 645.45 2.64 1.6×2.0 201.62 75.68 200 145.67 250 2373.8 21 1730.3 14

C4 1041.05 4.24 2.0×2.4 216.88 169.86 250 286.93 300 3755.3 29 2879.7 21

90

CHAPTER - 9

DESIGN OF STAIRCASE

STAIRCASE

91

For Fe415 steel, f y=415 N/mm²

For M 20 concrete f ck=20 N/mm²

Height of floor = 3.05 m

Assume – rise 150mm & tread 280 mm

Width of stair = 1.2 m

Height of each flight = 3.05/2 = 1.52 m

No. of rises required = 1.5/0.15 = 10

No. of rises in each flight = 10

No. of treads in each flight = 10-1 =9

Let the thickness of waist slab = 150 mm

Space occupied by treads = 9×0.28 = 2.52 m

Calculation of loads:

Bearing of the landing slab in the wall = 0.16 m

Effective span = 2.52+1.25+0.16/2 = 3.85 ≈ 39 m

Let the thickness of waist slab = 150 mm

Weight of slab on slope = 0.15×1.2×1×25 = 4.5 KN/m²

Dead weight on horizontal area = 4.5×√ 150²+2802

280² = 5105 N/m²

Dead weight of steps = 1/2 ×(0.15×25) = 1875 N/m

Total dead weight for horizontal meter run =1.105+5.105 = 7000 N

92

Weight of finishing = 100 N (assume)

Live load =2000 N

Total load = 9100 N/m

Factored load (W u=1.5×9100=13650 N/m

Mu = Wu l²/8 = 13650×(3.9)²/8 = 25.95×106 N-mm

Design of waist slab:

d = √ M u

Rub

¿√25.95×106/(2.76×1000)=97mm

Provide over all depth as 150 mm

Adopt 150 mm overall depth

Using 20 mm normal cover and 10 mm ϕ bars

Effective depth (d) =150−¿20−¿10/2 = 125 mm

So effective depth is less than provided d for required B.M

So we have an under reinforced section

Area of reinforcement:

A st=0.5×20

415 [1−√1−4.6×25.95×106

20×1000×1252 ]×1000×125

= 644.20 mm²

93

Using 10 mm ϕ bars

Number of bars required in 1.2 m width = 1.2×644.2/78.54 = 9.84 ≈ 10

Spacing of bars = 1200/10 = 120 mm

Distribution requirement

Asd = 0.12×150×1000/100

= 180 mm²

Using 6 mm ϕ bars = 1000×28.3/180

= 157.22 ≈ 160 mm

Hence provide 6 mm ϕ bars @ 160 mm c/c.

96

CHAPTER - 10CONCLUSION

This project concluded that basic need of a human being is having a home. So there is

a need of plan a building to meet all the requirements and comfort. So we design a multi-storied

building that which meets all the requirements.

The plan of the muli-storied building consists of similar features hence we get

symmetrical frames in the structures. By taking the advantage of symmetry we calculate the

loads falling on the structure and observing the safe bearing capacity (SBR) of the soil, we

designed the structure manually that which consists of beam, columns, slabs, staircase and

footings, to carry loads safely by using INDIAN STANDARDS and SPECIFICATIONS.

97

CHAPTER - 11REFERENCES

1. THEORY OF STRUCTURES:

S.RAMAMRUTHAM & R.NARAYANA

2. CONCRETE STRUCTURE:

P.DAYARATNAM

3. LIMIT STATE OF DESIGN:

DR.RAMCHANDRA A.K.JAIN

CODES USED:

IS: 456-2000, CODE OF PRACTICE FOR PLAN AND RCC

SP: 16-1980, DESIGN AIDS OF RCC TO IS: 456-2000

98

analysis and design of a multi-storeyed residential building

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