9. Final Project Report

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1. INTRODUCTION

The population explosion and advent of industrial revolution led to the exodus of

people from villages to urban areas. This urbanisation led to a new problem – less space for

housing, work and more people. Because of the demand of the land, the land costs got

skyrocketed. So, under the changed circumstances, the vertical growth of buildings i.e.

constructions of multi-storeyed buildings has become inevitable both for residential and as

well as office purposes.

For multi-storeyed buildings, the conventional load bearing structures become

uneconomical as they require larger sections to resist huge moments and loads. But in a

framed structure, the building frame consists of a network of beams and columns which are

built monolithically and rigidly with each other at their joints. Because of this rigidity at the

joints, there will be reduction in moments and also the structure tends to distribute the loads

more uniformly and eliminate the excessive effects of localised loads. Therefore in non-load

bearing framed structures, the moments and forces become less which in turn reduces the

sections of the members. As the walls don‟t take any load, they are also of thinner

dimensions. So, the lighter structural components and walls reduce the self weight of the

whole structure which necessitates a cheaper foundation. Also, the lighter walls which can

easily be shifted about provide flexibility in space utilisation. In addition to the above

mentioned advantages the framed structure is more effective in resisting wind loads and earth

quake loads.

Work done in this project:

A plot of 369.75 m2 has been selected for the construction of a multi-storeyed office

building. In the office building the functions will be different and it plays a major role

because of different loads acts on different slabs. The frame analysis requires the dimensions

of the members. For the analysis, 6 substitute frames taken in transverse direction and in

longitudinal direction the net moment acting is zero because of the same span and assumed as

a simply supported. For the maximum mid span moment obtained from the analysis, a T-

beam has been designed and for the support moment „doubly reinforced‟ section has been

provided. Stair case has also been designed. Isolated rectangular sloped footings have been

designed to transfer the load to the ground strata.

Analysis of structure:



2

Kani‟s method and substitute frame method is generally used to analyse a multi -

storeyed frame. The substitute frame method requires less computations and easier to carry

out the analysis. Therefore, here substitute frame method has been employed to carry out the

frame analysis and method is discussed in the following paragraphs.

Theoretically, a load applied at any point of the structure cause reactions at all

sections of the frame, but a close study of this aspect has shown that the moments in any

beam or column are mainly due to the loads on spans very close to it. The effect of loads on

distant panels is small. To facilitate the determination of moments in any member of a frame,

it is usual to analyse only a small portion of the frame consisting of adjacent members only.

Such a small portions are termed „SUBSTITUTE FRAMES‟. Their form and the end

conditions of their members are so adopted that the reaction in question works out to be the

greatest.

The reactions worked out with the help of substitute frame may sometimes appear to

be lower by about 10% compared with the value obtained from exact analysis. But it may be

mentioned that for analysis all the panels are located at the same time. Such a combination of

loading is highly improbable in practice. Thus the substitute frames give results are safe for

all practical purposes.

Design concept:

There are three design philosophies to design a reinforced concrete structures. They

are:

1. Working stress method,

2. Ultimate load method and

3. Limit state method.

In the „working stress‟ method it is seen 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. A linear variation stress form zero at the neutral axis to the

maximum stress at the extreme fibre is assumed.

Practically, the stress strain curve for concrete is not linear as it was assumed in

working stress method. So, in „ultimate load‟ design an idealised form of actual stress



3

strain diagram is used and the working loads are increased by multiplying them with the

load factors.

The basis for „limit state‟ method is a structure with appropriate degrees of reliability

should be able to withstand safely all loads that are liable to act on it throughout its life

and it should also satisfy the serviceability requirements such as limitations on deflection

and cracking.

Limit state method is the most rational method of the three methods. It considers the

actual behaviour of the materials at failure and also it takes serviceability also into

consideration. Therefore, limit state method has been employed in this work.



4

2. DESIGN OF SLABS

Typically we divided the slabs into two types:

i.

Roof Slab andii. Floor Slab

In case of roof slab the live load obtained is less compared to the floor slab. Therefore we

first design the roof slab and then floor slabs.

We have two types of supports. They are:

1. Ultimate support and

2. Penultimate support

Ultimate support is the end support and the penultimate supports are the intermediate

supports.

Ultimate support tends to have a bending moment of Wu x L

2

10 and the penultimate supports

haveWu x L

2

12

Design of roof slab:

It is a continuous slab on the top of the building which is also known as terrace.

Generally terrace has less live load and it is empty in most of the time except some occasions

in case of any residential building. In case of office buildings it will be empty and live load

act is very less.

According to the end conditions and the dimensions, the slabs are divided into 4 types. They

are Roof S1, Roof S2, Roof S3 and Roof S4.

Slab Dimensions (M x M)

Roof S1 8.62 X 3.05

Roof S2 8.62 X 3.05

Roof S3 5.78 X 3.05

Roof S4 5.78 X 3.05



5

We can observe the slab panels in the above figure and all the slabs are designed as one way

slab for the easy arrangement of the reinforcement and ease of work.

Roof S1 and Roof S2 are the slabs with same dimensions but different in end conditions.

Roof S3 and Roof S4 are also the slabs with the same conditions as mentioned above.

But the point to be noted is that all the Slabs have same shorter span and in design of one way

slab shorter span is of more importance. Therefore we design any two slabs with different end

conditions and the remaining two slabs also follow same design.

Design of Roof Slab S1:

Calculation of Depth (D) by using modification factor:

Assume the percentage of the tension reinforcement (P t) provided is 0.4%

From IS456-2000, P38 Fig4, we get the modification factor (α) = 1.4

Required Depth (D) =

L

ra + d

1



6

Where,L

r a=

Span

allowableL

dratio

d1 = Centre of the reinforcement to the end fibre (= 20mm for slab)

From IS456-2000, P39, Clause 24 & Clause 23.2 for continuous span we have

Span

Effective depth=

L

d= 26

r a = 26 x 1.4 = 36.4

Therefore, D =3.05 x 103

36.4

+ 20 = 103.79mm say 110 mm

Effective depth (d) = D – d1 = 110 – 20 = 90mm

Loads:

Dead loads (From IS875 – Part 1):

Terrace water proofing = 2.5 KN/m2

Self weight of the slab = 1 x 1 x D x 25 = 1101000

x 25

= 2.75 KN/m2

Live loads (From IS875 – Part 2):

Roof = 1.5 KN/m2

Total load (W) = 2.5 +2.75 + 1.5 = 6.75 KN/m2

Ultimate load or limit state load or design load (Wu) = 1.5 x W = 1.5 x 6.75

=10.125 KN/m2

Design moment: (for end panel)

Mu =Wu x L2

10=

10.125 x (3.05)2

10= 9.42 KN-m



7

Calculation of area of steel:

From IS456-2000, P96, Clause G-1.1 (b) we have

Mu = 0.87 x f y x Ast x d x (1 -f y x Ast

f ck x b x d )

Or

Ast = 0.5 xf ck

f y(1-(1-4.6

Mu

f ck x b x d)1/2

) b x d

9.42 x 106 = 0.87 x 415 x Ast x 90 x (1 -415 x Ast

20 x 1000 x 90)

Ast = 312.4 mm2

Spacing of 8mm φ bars =ast x 1000

Ast=

π4

x 82 x1000

312.4= 160.9mm

Therefore, Provide 8mm φ @ 150mm c/c.

Distribution steel:

Ast = 0.12% of Ag

=0.12

1000x 110 x 1000 = 132 mm2.

Spacing of 8mm φ bars =π4

x 82 x1000

132=380mm




8

Design of Roof slab S2:

Depth D = 110mm

Total load (W) = 6.75 KN/m2

Limit state load (Wu) = 1.5 x 6.75 = 10.125 KN/m2

Design moment: (for intermediate panel)

Mu =Wu x L2

12=

10.125 x (3.05)2

12= 7.85 KN-m



f ck x b x d)

7.85 x 106 = 0.87 x 415 x Ast x 90 x (1 -415 x Ast

20 x 1000 x 90)

Ast = 256.78 mm2

Spacing of 8mm φ bars =π

4x 82 x

1000

256.78= 195.75mm


Distribution steel:

Ast = 0.12% of Ag

=0.12

1000x 110 x 1000 = 132 mm2.


x 82 x1000

132=380mm




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Design of Roof Slab S3 is same as Roof Slab S1.

Design of Roof Slab S4 is same as Roof Slab S2.

Area of steel at support next to end support:

From IS 456-2000, moment = Wu x L2

10

Total Load acting on the support (Wu) = 10.125 KN/m

Therefore, Moment =10.125 x (3.05)2

10= 9.42 KN-m

Calculation of Ast:


f ck x b x d)

9.42 x 106 = 0.87 x 415 x Ast x 90 x (1 -415 x Ast

20 x 1000 x 90)

Ast = 312.4 mm2

Area of steel available by bending up the alternate bars of mid span steel:

From Slab S1 =1

2x 312.4 = 160.7 mm2

From Slab S2 =1

2x 256.78 = 128.39 mm2

Total Ast (available) = 160.7 + 128.39 = 284.59 mm2

Therefore, extra bars required for Ast = 312.4 – 284.59 = 27.81 mm2



10

Area of steel at any other interior support:

From IS 456-2000, moment =Wu x L2

12



12= 7.85 KN-m

Calculation of Ast:


f ck x b x d)

7.85 x 106 = 0.87 x 415 x Ast x 90 x (1 -415 x Ast

20 x 1000 x 90)

Ast = 256.78 mm2


From Slab S2 =1

2x 256.78 = 128.39 mm2

From Slab S2 =1

2x 256.78 = 128.39 mm2


Therefore Ast (available) = Ast (required)

No need of providing extra bars.



11

Design of Floor Slab:

It is the slab in which live load is more when compared to the roof slab. In this projectthe slab is divided into 9 types according to the end condition and function of slab.

S1 - Toilet and WC‟s

S2 - Office

S3 - Office sup dept.

S4 - Assembly hall

S5 (a), S5 (b) - Office chamber and waiting chamber

S6 (a), S6 (b) - Office

S7 - Library

S8 - Secretary Room

S9 (a), S9 (b) - Officers chamber



12

Floor Slab Dimensions

S1 to S5 (b) 8.62 x 3.05

S6 (a) to S9 (b) 5.78 x 3.05

DESIGN OF FLOOR SLAB (S1):

Calculation of Depth (D) by using modification factor:

Assume the percentage of the tension reinforcement (P t) provided is 0.4%

From IS456-2000, P38 Fig4, we get the modification factor (α) = 1.4

Required Depth (D) = Lra

+ d1

Where,L

r a=

Span

allowableL

dratio

d1 = Centre of the reinforcement to the end fibre (= 20mm for slab)

From IS456-2000, P39, Clause 24 & Clause 23.2 for continuous span we have

Span

Effective depth=

L

d= 26

r a = 26 x 1.4 = 36.4

Therefore, D =3.05 x 103

36.4+ 20 = 103.79mm say 120 mm


Loads:


Floor finish = 1 KN/m2

Sanitary Blocks including filling = 2.5 KN/m2



13

Self weight of the slab = 1 x 1 x D x 25 =120

1000x 25

= 3 KN/m2


Sanitary blocks public = 3 KN/m2

Corridor = 5 KN/m2

Maximum = 5 KN/m2

For Partition Wall = 1.5 KN/m2

Total load (W) = 1 + 2.5 +3 + 5 + 1.5 = 13 KN/m2

Ultimate load or limit state load or design load (Wu) = 1.5 x W = 1.5 x 13

=19.5 KN/m2


Mu =

Wu x L2

10 =

19.5 x (3.05)2

10 = 18.14 KN-m




f ck x b x d)

Or

Ast = 0.5 xf ck

f y(1-(1-4.6

Mu

f ck x b x d)1/2

) b x d

18.14 x 106 = 0.87 x 415 x Ast x 100 x (1 -415 x Ast

20 x 1000 x 100)

Ast = 569.79 mm2

Spacing of 10mm φ bars =a

stx 1000

Ast=

π4 x 102 x

1000

569.79 = 137.83mm



14


Distribution steel:

Ast = 0.12% of Ag

=0.12

1000x 120 x 1000 = 144 mm2.


x 82 x1000

144=349mm



D =3.05 x 103

36.4 + 20 = 103.79mm say 120 mm


Loads:




1000x 25

= 3 KN/m2


Office = 4 KN/m2

Corridor = 5KN/m2



15

Therefore, Maximum load = 5 KN/m2

For Partition wall = 1.5 KN/m2

Total load (W) = 5 +3 + 1 + 1.5 = 10.5 KN/m2


=15.75 KN/m2

Design moment:

Mu =Wu x L2

12=

15.75 x (3.05)2

12= 12.21 KN-m



f ck x b x d)

12.21 x 106 = 0.87 x 415 x Ast x 100 x (1 -415 x Ast

20 x 1000 x 100)

Ast = 365.97 mm2

Spacing of 10mm φ bars = π4

x 102 x1000

365.97= 214.61mm


Distribution steel:

Ast = 0.12% of Ag

= 0.121000 x 120 x 1000 = 144 mm2.


x 82 x1000

144=349mm




16

Area of steel at support next to end support (between S1 and S2):


10

Total Load acting on the support (Wu) =

13

2 +

15.75

2 = 14.375 KN/m


10= 13.372 KN-m

Calculation of Ast:


f ck x b x d)

13.372 x 106 = 0.87 x 415 x Ast x 100 x (1 -415 x Ast

20 x 1000 x 100)

Ast = 404.28 mm2


From Slab S1 =1

2x 569.79 = 284.895 mm2

From Slab S2 =1

2x 365.97 = 182.985 mm2




D = 3.05 x 103

36.4+ 20 = 103.79mm say 120 mm



17


Loads:




1000x 25

= 3 KN/m2


Private = 2 KN/m2

Corridor = 5 KN/m2

Maximum load = 5 KN/m2


Total load (W) = 5 +3 + 1 + 1.5 = 10.5 KN/m2


=15.75 KN/m2

Design moment:

Mu =Wu x L2

12=

15.75 x (3.05)2

12= 12.21 KN-m



f ck x b x d)

12.21 x 106 = 0.87 x 415 x Ast x 100 x (1 -415 x Ast

20 x 1000 x 100)

Ast = 365.97 mm2



18


x 102 x1000

365.97= 214.61mm


Distribution steel:

Ast = 0.12% of Ag

=0.12

1000x 120 x 1000 = 144 mm2.


x 82 x1000

144=349mm


Area of steel at any other interior support: (Between S2 and S3)


12

Total Load acting on the support (Wu) =15.75

2+

15.75

2= 15.75 KN/m

Therefore, Moment = 15.75 x (3.05)

2

12= 12.21 KN-m

Calculation of Ast:


f ck x b x d)

12.21 x 106 = 0.87 x 415 x Ast x 100 x (1 -415 x Ast

20 x 1000 x 100)

Ast = 365.97 mm2



19


From Slab S2 =1

2x 365.97 = 182.985 mm2

From Slab S3 =1

2x 365.97 = 182.985 mm2





D =3.05 x 103

36.4+ 20 = 103.79mm say 120 mm


Loads:




1000x 25

= 3 KN/m2


Assembly = 5 KN/m2

Corridor = 5 KN/m2

Maximum load = 5 KN/m2


Total load (W) = 5 +3 + 1 + 1.5 = 10.5 KN/m2



20


=15.75 KN/m2

Design moment:

Mu =Wu x L2

12=

15.75 x (3.05)2

12= 12.21 KN-m



f ck x b x d)

12.21 x 106 = 0.87 x 415 x Ast x 100 x (1 - 415 x Ast20 x 1000 x 100 )

Ast = 365.97 mm2


x 102 x1000

365.97= 214.61mm


Distribution steel:

Ast = 0.12% of Ag

=0.12

1000x 120 x 1000 = 144 mm2.


x 82 x1000

144=349mm




21

Area of steel at any other interior support: (Between S3 and S4)

Same as between S2 and S3

DESIGN OF FLOOR SLAB (S5 (a)):

D =3.05 x 103

36.4+ 20 = 103.79mm say 120 mm


Loads:




1000x 25

= 3 KN/m2


Office chamber = 4 KN/m2

Private = 2 KN/m2


Total load (W) = 2 + 4 +3 + 1 + 1.5 = 11.5 KN/m2


=17.25 KN/m2

Design moment:

Mu =Wu x L2

12=

17.25 x (3.05)2

12= 13.372 KN-m


Mu = 0.87 x f y x Ast x d x (1 - f y x Astf ck x b x d )



22

13.372 x 106 = 0.87 x 415 x Ast x 100 x (1 -415 x Ast

20 x 1000 x 100)

Ast = 404.27 mm2


x 102 x1000

347.05= 194.27mm


Distribution steel:

Ast = 0.12% of Ag

= 0.121000 x 120 x 1000 = 144 mm2.


x 82 x1000

144=349mm


Area of steel at any other interior support: (Between S4 and S5 (a))


12


2+

17.25

2= 16.5 KN/m


12= 12.79 KN-m

Calculation of Ast:

Mu = 0.87 x f y x Ast x d x (1 -

f y x Ast

f ck x b x d )



23

12.79 x 106 = 0.87 x 415 x Ast x 100 x (1 -415 x Ast

20 x 1000 x 100)

Ast = 385 mm2


From Slab S4 =1

2x 365.97 = 182.985 mm2

From Slab S5 (a) =1

2x 404.27 = 202.135 mm2




DESIGN OF FLOOR SLAB (S5 (b)):

D =3.05 x 103

36.4+ 20 = 103.79mm say 120 mm

Effective depth (d) = D – d1

= 120 – 20 = 100mm

Loads:




1000x 25

= 3 KN/m2



Private = 2 KN/m2




24

Total load (W) = 2 + 4 +3 + 1 + 1.5 = 11.5 KN/m2


=17.25 KN/m2


Mu =Wu x L2

10=

17.25 x (3.05)2

10= 16.047 KN-m



f ck x b x d

)

16.047 x 106 = 0.87 x 415 x Ast x 100 x (1 -415 x Ast

20 x 1000 x 100)

Ast = 495.37 mm2


x 102 x1000

495.37= 158.55mm


Distribution steel:

Ast = 0.12% of Ag

=0.12

1000x 120 x 1000 = 144 mm2.

Spacing of 8mm φ bars =

π4 x 8

2

x

1000

144 =349mm




25

Area of steel at support next to end support (between S5 (a) and S5 (b)):


10


2+

17.25

2= 17.25 KN/m


10= 16.05 KN-m

Calculation of Ast:


f ck x b x d

)

16.05 x 106 = 0.87 x 415 x Ast x 100 x (1 -415 x Ast

20 x 1000 x 100)

Ast = 495.47 mm2


From Slab S5 (a) =1

2x 404.27 = 202.135 mm2

From Slab S5 (b) =1

2x 495.47 = 247.735 mm2




D =3.05 x 103

36.4+ 20 = 103.79mm say 120 mm


Loads:





26


1000x 25

= 3 KN/m2


Office = 4 KN/m2

Total load (W) = 4 +3 + 1 = 8 KN/m2


=12 KN/m2


Mu =Wu x L2

10=

12 x (3.05)2

10= 11.163 KN-m



f ck x b x d)

11.163 x 106 = 0.87 x 415 x Ast x 100 x (1 -415 x Ast

20 x 1000 x 100)

Ast = 332.06 mm2


x 102 x1000

332.06= 236.53mm


Distribution steel:

Ast = 0.12% of Ag

=0.12

1000x 120 x 1000 = 144 mm2.


x 82 x1000

144=349mm



27



D =3.05 x 103

36.4+ 20 = 103.79mm say 120 mm


Loads:



Self weight of the slab = 1 x 1 x D x 25 = 1201000 x 25

= 3 KN/m2


Office = 4 KN/m2

Total load (W) = 4 +3 + 1 = 8 KN/m2


=12 KN/m2


Mu =Wu x L2

12=

12 x (3.05)2

12= 9.3025 KN-m



28



f ck x b x d)

9.3025 x 106 = 0.87 x 415 x Ast x 100 x (1 -415 x Ast

20 x 1000 x 100)

Ast = 273.13 mm2


x 102 x1000

273.13= 287.55mm


Distribution steel:

Ast = 0.12% of Ag

=0.12

1000x 120 x 1000 = 144 mm2.


x 82 x1000

144=349mm




10

Total Load acting on the support (Wu) = 12 KN/m

Therefore, Moment =12 x (3.05)2

10= 11.163 KN-m



29

Calculation of Ast:


f ck x b x d)

11.163 x 106 = 0.87 x 415 x Ast x 100 x (1 -415 x Ast

20 x 1000 x 100)

Ast = 332.06 mm2


From Slab S6 (a) =1

2x 332.06 = 166.03 mm2

From Slab S6 (b) =1

2x 273.23 = 136.565 mm2




D = 3.05 x 103

36.4+ 20 = 103.79mm say 120 mm


Loads:




1000x 25

= 3 KN/m2


Library = 10 KN/m2

Total load (W) = 10 +3 + 1 = 14 KN/m2



30


=21 KN/m2


Mu =Wu x L2

10=

12 x (3.05)2

10= 19.54 KN-m



f ck x b x d)

19.54 x 106 = 0.87 x 415 x Ast x 100 x (1 - 415 x Ast20 x 1000 x 100 )

Ast = 621.3 mm2


x 102 x1000

621.3= 126.41mm


Distribution steel:

Ast = 0.12% of Ag

=0.12

1000x 120 x 1000 = 144 mm2.


x 82 x1000

144=349mm




31


D =3.05 x 103

36.4+ 20 = 103.79mm say 120 mm


Loads:




1000

x 25

= 3 KN/m2


Private = 2 KN/m2

Total load (W) = 2 +3 + 1 = 6 KN/m2


= 9 KN/m2

Design moment:

Mu =Wu x L2

12=

9 x (3.05)2

12= 6.97 KN-m



f ck x b x d)

6.97 x 106 = 0.87 x 415 x Ast x 100 x (1 -415 x Ast

20 x 1000 x 100)

Ast = 201.47 mm2

Spacing of 10mm φ bars =π4 x 102 x

1000

201.47 = 389.8mm



32


Distribution steel:

Ast = 0.12% of Ag

=0.12

1000x 120 x 1000 = 144 mm2.


x 82 x1000

144=349mm


Area of steel at support next to end support (between S7and S8):


10

Total Load acting on the support (Wu) =21

2+

9

2= 15 KN/m

Therefore, Moment =15 x (3.05)2

10= 13.95 KN-m

Calculation of Ast:


f ck x b x d)

13.95 x 106 = 0.87 x 415 x Ast x 100 x (1 -415 x Ast

20 x 1000 x 100)

Ast = 423.61 mm2




33

From Slab S7 =1

2x 621.3 = 310.65 mm2

From Slab S8 =1

2x 201.47 = 100.735 mm2




D =3.05 x 103

36.4+ 20 = 103.79mm say 120 mm


Loads:




1000x 25

= 3 KN/m2



Total load (W) = 3 +3 + 1 = 7 KN/m2


= 10.5 KN/m2

Design moment:

Mu =Wu x L2

12=

10.5 x (3.05)2

12= 8.14 KN-m



34



f ck x b x d)

8.14 x 106 = 0.87 x 415 x Ast x 100 x (1 -415 x Ast

20 x 1000 x 100)

Ast = 237.12 mm2


x 102 x1000

237.12= 331.23mm


Distribution steel:

Ast = 0.12% of Ag

=0.12

1000x 120 x 1000 = 144 mm2.


x 82 x1000

144=349mm


Area of steel at any other interior support: (Between S8 and S9 (a))


12

Total Load acting on the support (Wu) =9

2+

10.5

2= 9.75 KN/m

Therefore, Moment = 9.75 x (3.05)2

12= 7.56 KN-m



35

Calculation of Ast:


f ck x b x d)

7.56 x 106 = 0.87 x 415 x Ast x 100 x (1 -415 x Ast

20 x 1000 x 100)

Ast = 219.38 mm2


From Slab S8 =1

2x 201.47 = 100.735 mm2

From Slab S9 (a) =1

2x 237.12 = 118.56 mm2





D =3.05 x 103

36.4+ 20 = 103.79mm say 120 mm


Loads:




1000x 25

= 3 KN/m2





36

Total load (W) = 3 +3 + 1 = 7 KN/m2


= 10.5 KN/m2


Mu =Wu x L2

10=

10.5 x (3.05)2

10= 9.77 KN-m



f ck x b x d

)

9.77 x 106 = 0.87 x 415 x Ast x 100 x (1 -415 x Ast

20 x 1000 x 100)

Ast = 287.79 mm2


x 102 x1000

287.79= 272.91mm


Distribution steel:

Ast = 0.12% of Ag

=0.12

1000x 120 x 1000 = 144 mm2.

Spacing of 8mm φ bars =

π4 x 8

2

x

1000

144 =349mm




37



10



10= 9.77 KN-m

Calculation of Ast:


f ck x b x d)

9.77 x 106 = 0.87 x 415 x Ast x 100 x (1 -415 x Ast

20 x 1000 x 100)

Ast = 287.79 mm2


From Slab S9 (a) =1

2x 237.12 = 118.56 mm2

From Slab S9 (b) =1

2x 287.79 = 143.895 mm2



Detail of reinforcement provided in slabs:

Slab Function Main steel Distribution steel End shears

Long

span

(Wu x L

2)

KN/m

Short

span

(Wu x L

6)

KN/m

Roof S1 and

Roof S3

Terrace 8mmφ bars @

150mm c/c

8mmφ bars @

300mm c/c 15.441 5.1467



38

Roof S2 and

Roof S4

Terrace 8mmφ bars @

190mm c/c

8mmφ bars @

300mm c/c 15.441 5.1467

Floor S1 Toilet 10mmφ bars @

130mm c/c

8mmφ bars @

300mm c/c

29.738 9.913

Floor S2 Office 10mmφ bars @

210mm c/c

8mmφ bars @

300mm c/c

24.019 8

Floor S3 Office SupDt. 10mmφ bars @

210mm c/c

8mmφ bars @

300mm c/c

24.019 8

Floor S4 Assembly Hall 10mmφ bars @

210mm c/c

8mmφ bars @

300mm c/c

24.019 8

Floor S5 (a) Office chamber

and Waiting space

10mmφ bars @

190mm c/c

8mmφ bars @

300mm c/c 26.306 8.77

Floor S5 (b) Office chamber

and Waiting space

10mmφ bars @

150mm c/c

8mmφ bars @

300mm c/c 26.306 8.77

Floor S6 (a) Office 10mmφ bars @

230mm c/c

8mmφ bars @

300mm c/c

18.3 6.1

Floor S6 (b) Office 10mmφ bars @

280mm c/c

8mmφ bars @

300mm c/c

18.3 6.1

Floor S7 Library 10mmφ bars @

120mm c/c

8mmφ bars @

300mm c/c

32.025 10.675

Floor S8 Secretary room 10mmφ bars @

300mm c/c

8mmφ bars @

300mm c/c

13.725 4.575

Floor S9 (a) Office chamber 10mmφ bars @

300mm c/c

8mmφ bars @

300mm c/c

16.01 5.3375

Floor S9 (b) Office chamber 10mmφ bars @

270mm c/c

8mmφ bars @

300mm c/c

16.01 5.3375



39

3. Analysis of frames

We have many number frames from the plan and the need to be analysed. We have two

different types of frames:

1. Longitudinal direction frame

2. Transverse direction frame

Transverse frame:

The frames are chosen in such a way that the loads vary from one frame to the other and we

have 6 transverse frames.

i. Frame 19-10-01

ii. Frame 20-11-02

iii. Frame 24-15-06

iv. Frame 25-16-07

v. Frame 26-17-08

vi. Frame 27-18-09

In every frame we need to analyse the three types of loading cases and each frame consists of

roof and floor analysis.

Here we assumed the cross sections of beams and columns in advance and with the help of

the assumed dimensions, we calculate the stiffness of the members and there by the

distribution factors for the members especially at the joints.

To analyse the frame we use the substitute frame method and there by applying the moment

distribution method to know the moments carried by the member at the joints.

We take each floor span and we assume the top and bottom storeys are fixed by the substitute

frame principle.



40

Typical building frame:

We have G+5frame with all the details shown in the above figure.



41

Type of

member

b(mm) x

D(mm)

Length

(mm)

M.O.I

(mm4) =

b x D3

12

Multiplying

factor for

flanged

section

Final

M.O.I

(mm4)

K =

final M.O.I

LENGTH

(mm3)

Beam 230 x

300

6000 517.5 x

106

2 1035 x 10 172.5 x 10

230 x

300

8410 517.5 x

106

2 1035 x 10 123.07 x 10

Beam 230 x

450

6000 1746.56 x

106

2 3493.125

x 106

582.18 x 10

230 x

450

8410 1746.56 x

106

2 3493.125

x 106

415.35 x 10

Beam 230 x

600

6000 4140 x

106

2 8280 x 10 1.38 x 10

230 x

600

8410 4140 x

106

2 8280 x 10 0.985 x 10

Column 230 x

600

3350 4140 x

106

- 4140 x 106 1235.82 x

103

Frame 19-10-01:

Calculation of loads:

Roof:

Slab: 110mm

Dead load = self weight + F.F

= (0.11 x 25) + 2.5 = 5.25 KN/m2

Live load = 1.5 KN/m2

Beam (19-10) (230mm X 300mm)

Dead load:

Self weight = 0.23 x 25 x (0.3-0.11) = 1.1 KN/m



42

Due to Slab =W x L

2=

5.25 x 3.05

2= 8 KN/m

250mm thick wall of 1m height = 0.25 x 1 x 20 = 5 KN/m

Live load:

Due to Slab =W x L

2=

1.5 x 3.05

2= 2.29KN/m

Therefore, Maximum load = 1.5 x (D.L + L.L) = 1.5 x (1.1 + 8 + 2.29 + 5)

= 24.6 KN/m

Minimum load = 0.9 x D.L = 0.9 x (1.1 + 8 + 5) = 13 KN/m

Beam (10-01) (230mm X 300mm)

Dead load:


Due to Slab =W x L

2=

5.25 x 3.05

2= 8 KN/m

250mm thick wall of 1m height = 0.25 x 1 x 20 = 5 KN/m

Live load:

Due to Slab =W x L

2=

1.5 x 3.05

2= 2.29KN/m

Therefore, Maximum load = 1.5 x (D.L + L.L) = 1.5 x (1.1 + 8 + 2.29 + 5)

= 24.6 KN/m

Minimum load = 0.9 x D.L = 0.9 x (1.1 + 8 + 5) = 13 KN/m

Case 1:

Maximum load on beam 19-10 and minimum load beam 10-01



43

Calculation of distribution factors:

Joint Member K ΣK D.F

1 Column 1235.82 x 10 1.41 x 10 0.88

Beam 172.5 x 10 0.12

2 Beam 172.5 x 10 1.53 x 10 0.11

Column 1235.82 x 10 0.81

Beam 123.07 x 103 0.08

3 Beam 123.07 x 103 1.35 x 106 0.09

column 1235.82 x 10 0.91

Calculation of moments:



44

Calculation of support reactions:

From beam 1-2, ΣM1=0

64.85 + V2 x 6 = 24.6 x 6 x6

2+ 78.52

V2 = 76.07 KN

V1 + V2 = 24.6 x 6 = 147.6 KN

V1 = 71.54 KN

Maximum Span moment,

X=71.54

24.6= 2.91m from the left support

Therefore, maximum moment =71.54 x 2.91

2- 64.85 = 39.24 KN-m


79.93 + V3 x 8.41 = 69.78 + 13 x 8.41 x8.41

2

V3 = 53.46 KN

V2 + V3 = 13 x 8.41 =109.33 KN

V2 = 55.87 KN



45


X=53.46

13= 4.12m from the right support


2- 69.78 = 40.35 KN-m

Case 2:

Minimum load on beam 19-10 and maximum load on 10-01

Distribution factors are same as calculate in case 1.




46



28.96 + V2 x 6 = 13 x 6 x

6

2 + 53.16

V2 = 42.87 KN

V1 + V2 = 13 x 6 = 78 KN

V1 = 35.13 KN


X=35.13

13= 2.7m from the left support


2- 28.16 = 19.27 KN-m


142.85 + V3 x 8.41 = 136.01 + 24.6 x 8.41 x8.41

2

V3 = 102.63 KN

V2 + V3 = 24.6 x 8.41 = 206.89 KN

V2 = 104.26 KN


X= 102.6324.6

= 4.17m from the right support



47


2- 136.01 = 77.97 KN-m

Case 3:

Max load on beam 19-10 and beam 10-01

Distribution factors are same as calculated in case 1.

Calculations of moments:



48



61.38 + V2 x 6 = 24.6 x 6 x6

2 + 86.09

V2 = 77.92 KN

V1 + V2 = 24.6 x 6 = 147.6 KN

V1 = 69.68 KN


X=69.68



2- 61.38 = 37.22 KN-m


145.76 + V3 x 8.41 = 134.65 + 24.6 x 8.41 x8.41

2

V3 = 102.05 KN

V2 + V3 = 24.6 x 8.41 = 206.89 KN

V2 = 104.84 KN


X=102.05

24.6= 4.15m from the right support


2- 134.65 = 77.1 KN-m





50

Due to Slab =W x L

2=

4 x 3.05

2= 6.1 KN/m

250mm thick wall = 0.25 x 20 x (3.35-0.3) = 15.25 KN/m

Live load:

Due to Slab =W x L

2=

4 x 3.05

2= 6.1KN/m

Therefore, Maximum load = 1.5 x (D.L + L.L) = 1.5 x (1.9+6.1+15.25+6.1)

= 44.025 KN/m

Minimum load = 0.9 x D.L = 0.9 x (1.9+6.1+15.25) = 21 KN/m

Beam (10-01) (230mm X 450mm)

Dead load:


Due to Slab =W x L

2=

4 x 3.05

2= 6.1 KN/m


Live load:

Due to Slab =W x L

2=

5 x 3.05

2= 7.625KN/m


= 46.5 KN/m

Minimum load = 0.9 x D.L = 0.9 x (1.9+6.1+15.25) = 21KN/m



51

Case 1:

Maximum load on beam 19-10 and minimum load on beam 10-01



4 Column 1235.82 x 10 3.05 x 10 0.4

Beam 582.18 x 10 0.2

Column 1235.82 x 10 0.4

5 Beam 582.18 x 10 3.47 x 10 0.17

Column 1235.82 x 103 0.36

Beam 415.35 x 103 0.11

Column 1235.82 x 10 0.36

6 Beam 415.35 x 10 2.89 x 10 0.14

Column 1235.82 x 10 0.43

Column 1235.82 x 10 0.43



52




106.54 + V5 x 6 = 44.025 x 6 x 62

+ 143.19

V5 = 138.18 KN

V4 + V5 = 44.025 x 6 = 264.15 KN

V4 = 125.97 KN



53


X=125.97



2- 106.54 = 73.58 KN-m


133.85 + V6 x 8.41 = 105.76 + 21 x 8.41 x8.41

2

V6 = 84.96 KN

V5 + V6 = 21 x 8.41 = 176.61 KN

V5 = 91.65 KN


X=84.96


Therefore, maximum moment = 84.96 x 4.052

- 105.76 = 66.28 KN-m

Case 2:

Minimum load on beam 19-10 and maximum load on beam 10-01

Distribution factors are same as calculated in case-1.



54




34.99 + V5 x 6 = 21 x 6 x 62 + 105.91

V5 = 74.81 KN

V4 + V5 = 21 x 6 = 126 KN

V4 = 51.19 KN



55


X=51.19



2- 34.99 = 27.72 KN-m


269.10 + V6 x 8.41 = 246.62 + 46.5 x 8.41 x8.41

2

V6 = 192.86 KN

V5 + V6 = 46.5 x 8.41 = 391.07 KN

V5 = 198.21 KN


X=192.86



- 246.62= 153.57 KN-m

Case 3:

Maximum load on both beams




56




96.02 + V5

x 6 = 44.025 x 6 x6

2+ 169.53

V5 = 144.33 KN

V4 + V5 = 44.025 x 6 = 264.15 KN

V4 = 119.82 KN


X=119.82




57


2- 96.02 = 66.94 KN-m


277.32 + V6 x 8.41 = 242.87 + 46.5 x 8.41 x8.41

2

V6 = 191.43 KN

V5 + V6 = 46.5 x 8.41 = 391.07 KN

V5 = 199.64 KN


X=191.43



2- 242.87= 151.48 KN-m

Beam and column bending moments (KN-m):

Joint 4 Joint 5 Joint 6

Case Column Beam Max

span

moment

Beam Column Beam Max

span

moment

Beam Column

1 53.27 106.54 73.58 143.19 4.67 133.84 65.86 105.75 53.13

2 17.5 34.99 27.2 105.91 81.6 269.1 153.57 246.62 122.6

3 48.01 96.02 66.94 169.53 53.9 277.32 151.48 242.87 121.11

Max 53.27 106.54 73.58 169.53 81.6 277.32 153.57 246.62 122.6

Shear or support reaction (KN):

Case no. Joint 4 Joint 5 Joint 6

1 125.6 138.18 91.65 84.96

2 51.19 74.81 198.21 192.85

3 119.82 144.33 199.64 191.43

Maximum shear 125.6 144.33 199.64 192.85

Maximum column load 125.6 343.97 192.85



58

Frame 20-11-02:

Calculation of loads:

Roof:

Slab: 110mm


= (0.11 x 25) + 2.5 = 5.25 KN/m2


Beam (20-11) (230mm X 450mm)

Dead load:


Due to Slab = W x L= 5.25 x 3.05= 16.0125 KN/m

Live load:


Therefore, Maximum load = 1.5 x (D.L + L.L) = 1.5 x (1.2+16.0125+4.6)

= 33 KN/m

Minimum load = 0.9 x D.L = 0.9 x (1.2+16.0125) = 15.49 KN/m

Beam (11-02) (230mm X 450mm)

Dead load:



Live load:




59


= 33 KN/m


Case 1:

Maximum load on beam 20-11 and minimum load beam 11-02



1 Column 1235.82 x 10 1.818 x 10 0.68

Beam 582.18 x 10 0.32

2 Beam 582.18 x 10 2.234 x 10 0.26

Column 1235.82 x 10 0.55

Beam 415.35 x 10 0.19

3 Beam 415.35 x 10 1.651 x 10 0.25

column 1235.82 x 103 0.75



60




68.42 + V2 x 6 = 33 x 6 x6

2+ 111.87

V2 = 106.24 KN

V1 + V2 = 33 x 6 = 198 KN

V1 = 91.76 KN



61


X=91.76



2- 68.42 = 59.13 KN-m


105.03 + V3 x 8.41 = 67.38 + 15.49 x 8.41 x8.41

2

V3 = 60.66 KN

V2 + V3 = 15.49 x 8.41 =130.27 KN

V2 = 69.61 KN


X=60.66



- 67.38 = 51.51 KN-m

Case 2:

Minimum load on beam 20-11 and maximum load on 11-02

Distribution factors are same as calculate in case 1.



62




16.56 + V2 x 6 = 15.49 x 6 x 62

+ 94.59

V2 = 59.48 KN

V1 + V2 = 15.49 x 6 = 92.94 KN

V1 = 33.46 KN



63


X=33.46



2- 16.56 = 19.58 KN-m


188.13 + V3 x 8.41 = 158.87 + 33 x 8.41 x8.41

2

V3 = 135.28 KN

V2 + V3 = 33 x 8.41 = 277.53 KN

V2 = 142.25 KN


X=135.28



- 158.81 = 118.52 KN-m

Case 3:

Max load on beam 20-11 and beam 11-02


Calculations of moments:



64



57.83 + V2 x 6 = 33 x 6 x6

2+ 140.51

V2 = 112.78 KN

V1 + V2 = 33 x 6 = 198 KN

V1 = 85.22 KN


X= 85.2233

= 2.58m from the left support



65


2- 57.83 = 52.1 KN-m


199.52 + V3 x 8.41 = 153.93 + 33 x 8.41 x8.41

2

V3 = 133.34 KN

V2 + V3 = 33 x 8.41 = 277.53 KN

V2 = 144.19 KN


X=133.34



2- 153.93 = 115.42 KN-m




span

moment


span

moment

Beam Column

1 68.42 68.42 59.13 111.87 6.84 105.0 51.51 67.38 68.22

2 16.58 16.58 19.58 94.59 93.54 188.13 118.52 158.81 155.58

3 57.85 57.83 52.1 140.51 59 199.51 115.42 153.93 152.3

Max 68.42 68.42 59.13 140.51 93.54 199.51 118.52 158.81 155.58



1 91.76 106.24 69.61 60.66

2 33.46 59.48 142.25 135.28

3 85.22 112.78 144.11 133.34

Maximum shear 91.76 112.78 144.11 135.28






67

Therefore, Maximum load = 1.5 x (D.L + L.L) = 1.5x(2.76+12.2+14.5+15.25)

= 67.1 KN/m

Minimum load = 0.9 x D.L = 0.9x(2.76+12.2+14.5) = 26.5 KN/m

Case 1:




4 Column 1235.82 x 10

3.85 x 106

0.32

Beam 1.38 x 10 0.36

Column 1235.82 x 10 0.32

5 Beam 1.38 x 10

4.84 x 106

0.29

Column 1235.82 x 10 0.26

Beam 0.985 x 106 0.19

Column 1235.82 x 103 0.26

6 Beam 0.985 x 10

3.46 x 106

0.28

Column 1235.82 x 10 0.36

Column 1235.82 x 10 0.36



68




75.53 + V5 x 6 = 41 x 6 x6

2+ 154.22

V5 = 136.12 KN

V4 + V5 = 41 x 6 = 246 KN

V4 = 109.88 KN



69


X=109.88



2- 75.53 = 71.71 KN-m


172.08 + V6 x 8.41 = 114.85 + 26.5 x 8.41 x8.41

2

V6 = 104.63 KN

V5 + V6 = 26.5 x 8.41 = 222.87 KN

V5 = 118.24 KN


X=104.63



- 114.85 = 91.8 KN-m

Case 2:





70




16.29 + V5 x 6 = 13.5 x 6 x6

2+ 158.17

V5 = 64.15 KN

V4 + V5 = 13.5 x 6 = 81 KN

V4 = 16.85 KN


X= 16.8513.5




71


2- 12.92 = -2.39 KN-m


375.67 + V6 x 8.41 = 315.93 + 67.1 x 8.41 x8.41

2

V6 = 275.05 KN

V5 + V6 = 67.1 x 8.41 = 564.31 KN

V5 = 289.26 KN


X=275.05



2- 315.93= 247.92 KN-m

Case 3:





72




49.25 + V5 x 6 = 41 x 6 x6

2 + 228.89

V5 = 152.94 KN

V4 + V5 = 41 x 6 = 246 KN

V4 = 93.06 KN


X= 93.0641




73


2- 49.25 = 56.37 KN-m


393.9 + V6 x 8.41 = 308.24 + 67.1 x 8.41 x8.41

2

V6 = 271.97 KN

V5 + V6 = 67.1 x 8.41 = 564.31 KN

V5 = 292.34 KN


X=271.97



2- 308.24= 242.5 KN-m




span

moment


span

moment

Beam Column

1 37.77 75.53 71.71 154.21 8.9 172.08 91.8 114.85 57.45

2 6.42 12.92 3.092 158.17 108.74 375.67 247.92 315.93 153.37

3 24.66 49.25 56.37 228.89 82.5 393.9 242.5 308.24 150.94

Max 37.77 75.53 71.71 228.89 108.74 393.9 247.92 315.93 153.37



1 109.8 136.12 118.24 104.63

2 16.85 64.15 289.26 275.05

3 93.06 152.94 292.34 271.97

Maximum shear 109.8 152.94 292.34 275.05




74

Frame 24-15-06

Roof:

Same as frame 20-11-02

Floor:

Slab = 120mm


= (0.12 x 25) + 1 = 4 KN/m2

Live load = 10 KN/m2

for Beam 24-15

= 2 KN/m2 for Beam 24-15


Beam (24-15) (230mm X 600mm)

Dead load:


Due to Slab = W x L= 4 x 3.05= 12.2 KN/m

150mm wall of 2.2m height = 0.15 x 20 x 2.2 = 6.6 KN/m

Live load:

Due to Slab =5 x 3.05

2+

2 x 3.05

2= 10.675 KN/m


= 48.5 KN/m

Minimum load = 0.9 x D.L = 0.9 x (2.76+12.2+6.6) = 19.5 KN/m



75

Beam (15-06) (230mm X 600mm)

Dead load:



Live load:

Due to Slab = W x L= 5 x 3.05= 15.25KN/m


= 45.5 KN/m


Case 1:




76



4 Column 1235.82 x 103

3.85 x 106

0.32

Beam 1.38 x 10 0.36

Column 1235.82 x 10 0.32

5 Beam 1.38 x 10

4.84 x 106

0.29

Column 1235.82 x 10 0.26

Beam 0.985 x 10 0.19

Column 1235.82 x 10 0.26

6 Beam 0.985 x 106

3.46 x 106

0.28

Column 1235.82 x 10 0.36

Column 1235.82 x 10 0.36




77



95.08 + V5 x 6 = 48.5 x 6 x62

+ 166.11

V5 = 157.34 KN

V4 + V5 = 48.5 x 6 = 291 KN

V4 = 133.66 KN


X=133.66



2- 95.08 = 89.37 KN-m


155.12 + V6 x 8.41 = 93.64 + 22.5 x 8.41 x 8.412

V6 = 87.3 KN

V5 + V6 = 22.5 x 8.41 = 189.23 KN

V5 = 101.93 KN



78


X=87.3



2- 93.64 = 75.72 KN-m

Case 2:






79



14.68 + V5 x 6 = 19.5 x 6 x6

2+ 133.86

V5 = 78.37 KN

V4 + V5 = 19.5 x 6 = 117 KN

V4 = 38.63 KN


X=38.63



2- 14.68 = 23.56 KN-m


261.6 + V6 x 8.41 = 211.34 + 45.5 x 8.41 x 8.412

V6 = 185.35 KN

V5 + V6 = 45.5 x 8.41 = 382.66 KN

V5 = 197.31 KN



80


X=185.35



2- 211.34= 165.85 KN-m

Case 3:






81



80.19 + V5 x 6 = 44.5 x 6 x 62

+ 208.43

V5 = 166.87 KN

V4 + V5 = 48.5 x 6 = 291 KN

V4 = 124.13 KN


X=124.13



2- 80.19 = 78.7 KN-m


280.82 + V6 x 8.41 = 203.22 + 45.5 x 8.41 x 8.412

V6 = 182.1 KN

V5 + V6 = 67.1 x 8.41 = 382.66 KN

V5 = 200.56 KN


X=182.145.5

= 4m from the right support



82

Therefore, maximum moment =182.1 x 4

2- 203.22= 160.98 KN-m




span

moment


span

moment

Beam Column

1 47.54 95.08 89.37 166.11 5.5 155.12 75.72 93.64 47.54

2 7.34 14.63 23.56 133.86 63.86 261.6 166.85 211.34 103.08

3 40.11 80.19 78.7 208.43 36.19 280.82 160.98 203.22 100.52

Max 47.54 95.08 89.37 208.43 63.86 280.82 166.85 211.34 103.08



1 133.66 157.34 101.93 87.3

2 38.63 78.37 197.31 185.35

3 124.13 166.87 200.56 182.1

Maximum shear 133.66 166.87 200.56 185.35


Frame 25-16-07

Roof:


Floor:

Slab = 120mm


= (0.12 x 25) + 1 = 4 KN/m2

Live load = 2 KN/m2 for Beam 25-16



83




Beam (25-16) (230mm X 600mm)

Dead load:




Live load:


2+

3 x 3.05

2= 7.625 KN/m


= 44 KN/m


Beam (16-07) (230mm X 600mm)

Dead load:




Live load:


2+

6 x 3.05

2= 16.775 KN/m


= 57.51 KN/m



84


Case 1:




4 Column 1235.82 x 10

3.85 x 106

0.32

Beam 1.38 x 10 0.36

Column 1235.82 x 10 0.32

5 Beam 1.38 x 10

4.84 x 106

0.29

Column 1235.82 x 10 0.26

Beam 0.985 x 10 0.19

Column 1235.82 x 103 0.26

6 Beam 0.985 x 106

3.46 x 106

0.28

Column 1235.82 x 10 0.36

Column 1235.82 x 10 0.36



85




85.69 + V5 x 6 = 44 x 6 x6

2+ 152.33

V5 = 143.11 KN

V4 + V5 = 44 x 6 = 264 KN

V4 = 120.89 KN



86


X=120.89



2- 85.69 = 80.45 KN-m


145.58 + V6 x 8.41 = 89.17 + 21.3 x 8.41 x8.41

2

V6 = 82.86 KN

V5 + V6 = 21.3 x 8.41 = 179.13 KN

V5 = 96.27 KN


X=82.86



- 89.17 = 72 KN-m

Case 2:




87





6.85 + V5 x 6 = 19.5 x 6 x6

2+ 155.96

V5 = 83.35 KN

V4 + V5 = 19.5 x 6 = 117 KN

V4 = 33.65 KN



88


X=33.65



2- 6.85 = 22.26 KN-m


327.23 + V6 x 8.41 = 268.6 + 57.51 x 8.41 x8.41

2

V6 = 234.86 KN

V5 + V6 = 57.51 x 8.41 = 483.66 KN

V5 = 248.8 KN


X=234.86



- 268.6= 210.5 KN-m

Case 3:





89




62.24 + V5 x 6 = 44 x 6 x6

2+ 218.95

V5 = 158.11 KN

V4 + V5 = 48.5 x 6 = 264 KN

V4 = 105.89 K



90


X=105.89



2- 62.24 = 65.36 KN-m


343.47 + V6 x 8.41 = 257.17 + 57.51 x 8.41 x8.41

2

V6 = 231.56 KN

V5 + V6 = 67.1 x 8.41 = 483.66 KN

V5 = 252.1 KN


X=231.56



- 257.17= 209.42 KN-m




span

moment


span

moment

Beam Column

1 42.84 85.69 80.45 152.33 3.38 145.56 72 89.17 45.17

2 3.46 6.85 22.26 155.96 85.63 327.23 210.5 268.55 130.75

3 31.15 62.24 65.36 218.95 62.25 343.47 209.42 261.7 128.6

Max 42.84 85.69 80.45 218.95 85.63 343.47 210.5 268.55 130.75



91



1 120.89 143.11 96.27 82.86

2 33.65 83.35 248.8 234.86

3 105.89 158.11 251.1 231.56

Maximum shear 120.89 158.11 251.1 234.86


Frame 26-17-08

Roof:


Floor:

Slab = 120mm


= (0.12 x 25) + 1 = 4 KN/m2



Beam (26-17) (230mm X 600mm)

Dead load:



Live load:

Due to Slab = 3 x 3.05= 9.15 KN/m

Therefore, Maximum load = 1.5 x (D.L + L.L) = 1.5x (2.76+12.2+9.15)

= 36.2 KN/m



92


Beam (17-08) (230mm X 600mm)

Dead load:



Live load:

Due to Slab = 6 x 3.05= 18.3 KN/m

Therefore, Maximum load = 1.5 x (D.L + L.L) = 1.5x (2.76+12.2+18.3)

= 50 KN/m


Case 1:




93



4 Column 1235.82 x 10

3.85 x 106

0.32

Beam 1.38 x 10 0.36

Column 1235.82 x 10 0.32

5 Beam 1.38 x 106

4.84 x 106

0.29

Column 1235.82 x 103 0.26

Beam 0.985 x 10 0.19

Column 1235.82 x 10 0.26

6 Beam 0.985 x 10

3.46 x 106

0.28

Column 1235.82 x 10 0.36

Column 1235.82 x 10 0.36




94



73.1+ V5 x 6 = 36.2 x 6 x6

2+ 117.93

V5 = 116.07 KN

V4 + V5 = 46.2 x 6 = 217.2 KN

V4 = 101.13 KN


X=101.13



2- 73.1 = 68.48 KN-m


97.78 + V6 x 8.41 = 54.2 + 13.5 x 8.41 x

8.41

2

V6 = 51.59KN

V5 + V6 = 13.5 x 8.41 = 113.54 KN

V5 = 61.95 KN


X= 51.5913.5




95


2- 54.2 = 44.34 KN-m

Case 2:






96



1.84 + V5 x 6 = 13.5 x 6 x62

+ 126.71

V5 = 61.31 KN

V4 + V5 = 13.5 x 6 = 81 KN

V4 = 19.69 KN


X=19.69



2- 1.84 = 12.54 KN-m


282.21 + V6

x 8.41 = 234.45 + 50 x 8.41 x8.41

2

V6 = 204.57 KN

V5 + V6 = 50 x 8.41 = 420.5 KN

V5 = 215.93 KN


X=204.57




97


2- 234.45= 183.9 KN-m

Case 3:






98



49.5 + V5 x 6 = 36.2 x 6 x6

2+ 185.08

V5 = 131.2 KN

V4 + V5 = 36.2 x 6 = 217.2 KN

V4 = 86 KN


X=86


Therefore, maximum moment =86 x 2.38

2- 49.5 = 52.84 KN-m


297.26 + V6 x 8.41 = 228.1 + 50 x 8.41 x

8.41

2

V6 = 202.03 KN

V5 + V6 = 50 x 8.41 = 420.5 KN

V5 = 218.47 KN


X= 202.0350




99


2- 228.1= 180 KN-m




span

moment


span

moment

Beam Column

1 36.55 73.1 68.48 117.93 10.07 97.78 44.34 54.2 27.9

2 0.9 1.85 12.54 126.71 77.74 282.21 183.9 234.45 113.98

3 24.76 49.5 52.84 185.08 56.08 297.26 180 228.1 111.97

Max 36.55 73.1 68.48 185.08 77.74 297.26 183.9 234.45 113.98



1 101.13 116.07 61.95 51.59

2 19.69 61.31 215.93 204.57

3 86 131.2 218.47 202.03

Maximum shear 101.13 131.2 218.47 204.57


Frame 27-18-09

Roof:

Same as of roof frame 19-10-01

Floor:

Slab: 120mm


= (0.12 x 25) + 1 = 4 KN/m2




100


Beam (27-18) (230mm X 450mm)

Dead load:


Due to Slab =W x L

2=

4 x 3.05

2= 6.1 KN/m


Live load:

Due to Slab =W x L

2=

3 x 3.05

2= 4.6KN/m


= 42 KN/m

Minimum load = 0.9 x D.L = 0.9 x (1.9+6.1+15.25) = 21 KN/m

Beam (18-09) (230mm X 450mm)

Dead load:


Due to Slab =W x L

2=

4 x 3.05

2= 6.1 KN/m


Live load:

Due to Slab =W x L

2=

6 x 3.05

2= 9.15KN/m


= 48.6 KN/m

Minimum load = 0.9 x D.L = 0.9 x (1.9+6.1+15.25) = 21KN/m



101

Case 1:




4 Column 1235.82 x 10 3.05 x 10 0.4

Beam 582.18 x 10 0.2

Column 1235.82 x 103 0.4

5 Beam 582.18 x 10 3.47 x 10 0.17

Column 1235.82 x 10 0.36

Beam 415.35 x 10 0.11

Column 1235.82 x 10 0.36

6 Beam 415.35 x 10 2.89 x 10 0.14

Column 1235.82 x 10 0.43

Column 1235.82 x 10 0.43



102




101.22 + V5 x 6 = 42 x 6 x6

2+ 137.6

V5 = 132.06 KN

V4 + V5 = 42 x 6 = 252 KN

V4 = 119.94 KN



103


X=119.94



2- 101.22 = 70.3 KN-m


133.13 + V6 x 8.41 = 106.52 + 21 x 8.41 x8.41

2

V6 = 85.14 KN

V5 + V6 = 21 x 8.41 = 176.61 KN

V5 = 91.47 KN


X=85.14



- 106.52 = 65.89 KN-m

Case 2:




104





34.07 + V5 x 6 = 21 x 6 x6

2+ 108.08

V5 = 75.34 KN

V4 + V5 = 21 x 6 = 126 KN

V4 = 50.66 KN



105


X=50.66



2- 34.07 = 26.98 KN-m


280.96 + V6 x 8.41 = 257.96 + 48.6 x 8.41 x8.41

2

V6 = 201.63 KN

V5 + V6 = 48.6 x 8.41 = 408.73 KN

V5 = 207.1 KN


X=201.63



- 257.96= 160.42 KN-m

Case 3:




106





89.24 + V5 x 6 = 42 x 6 x6

2+ 166.05

V5 = 138.8 KN

V4 + V5 = 42 x 6 = 252 KN

V4 = 113.2 KN



107


X=113.2



2- 89.24 = 63.58 KN-m


288.45 + V6 x 8.41 = 254.55 + 48.6 x 8.41 x8.41

2

V6 = 200.33 KN

V5 + V6 = 48.6 x 8.41 = 408.73 KN

V5 = 208.4 KN


X=200.33



- 254.55= 158.33 KN-m




span

moment


span

moment

Beam Column

1 50.61 101.22 70.3 137.6 2.24 133.12 65.89 106.09 53.26

2 17.04 34.07 26.98 108.08 86.44 280.96 160.42 257.96 128.22

3 44.62 89.24 63.58 166.06 61.2 288.45 158.33 254.55 126.86

Max 50.61 101.22 70.3 166.06 86.44 288.45 160.42 257.96 128.22



108



1 119.94 132.06 91.47 85.14

2 50.66 75.34 207.28 201.63

3 113.2 138.8 208.4 200.33

Maximum shear 119.94 138.8 208.4 200.33




109

4. DESIGN OF BEAMS

Bending moment diagram for entire building:



110

1. Longitudinal beams

a. Beam 23-24

Roof :

Section: 230mm x 300mm (b x D)

Cover: 25mm

Effective depth (d‟) = 300 – 25 = 275mm

Concrete: M20 Steel: Fe415

Loads:

Slab: 110mm thick

Dead load = self weight + floor finish

= 0.11 x 25 +2.5 = 5.25 KN/m2


Acting on beam:

Dead load:

Self weight = 0.23 x 25 x (0.3 – 0.11) = 1.1 KN/m

Slab =5.25 x 3.05

6= 2.67 KN/m

Parapet wall (250mm wall including plastering of 1m height)

= 20 x 0.25 x 1 = 5 KN/m

Live load:

Slab =1.5 x 3.05

6= 0.76 KN/m

Therefore, Total load (W) = 5 + 2.67 + 1.1 + 0.76 = 9.53 KN/m

Factored load (Wu) = 1.5 x 9.53 = 14.3 KN/m



111

Maximum load on column removing the triangular load = 1.5 (1.1 +5)

= 9.15 KN/m

Shear on column =column load x l

2

=9.15 x 3.05

2

= 13.96 KN (from each side)

Design moment:

Mu =14.3 x (3.05)2

12= 11.09 x 106 N-mm = 11.09 KN-m

Calculation of Ast:

From IS456-2000, P96a, clause G-1.1 (b), we have


f ck x b x d)

11.09 x 106 = 0.87 x 415 x Ast x 275 x (1-415 x Ast

20 x 230 x 275)

Ast = 116.12 mm2

No. of 10mm diameter bars =

4 x 116.12

π x 102 = 1.5 say 2 bars

Therefore, provided Ast = 2 xπ4

x 102 = 157.08 mm2

Design of shear reinforcement:

Vu =Wu x Lx

2=

14.3 x 3.05

2= 21.81KN

Nominal shear stress, v =Vu

b x d=

21.81 x 103

230 x 275= 0.35 N/mm2

Ast = 157.08 mm2

Consider Pt =100 xAst

b x d=

100 x 157.08

230 x 275= 0.25



112

From IS 456-2000, Table 19:

Pt c

0.25 0.36

Therefore, c = 0.36 N/mm2

v approximately equals to c

Therefore, provide minimum shear reinforcement.

Assuming 8mmφ- 2legged stirrups

Asv = 2 x π x 82

4= 100.53mm2

Min shear reinforcement (IS456-2000, clause 26.5.1.6, P48)

Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 300

0.87 x 415= 76.44mm2 <100.5mm2 (safe)

Provide 8mm dia stirrups @ 300mm c/c.

Floor:


Cover: 25mm





113

Loads:

Slab: 120mm thick


= 0.12 x 25 +1 = 4 KN/m2

Live load = 10 KN/m2

Acting on beam:

Dead load:


Slab =4 x 3.05

6= 2.03 KN/m

Parapet wall (250mm wall including plastering of 2.2m height)

= 20 x 0.25 x (3.35-0.3) = 15.25 KN/m

Live load:

Slab =10 x 3.05

6= 5.08 KN/m

Therefore, Total load (W) = 1.9 + 2.03 + 15.25 + 5.08 = 24.26 KN/m


Maximum load on column removing the triangular load = 1.5 (1.9 +5.08)

= 25.725 KN/m


2=

25.725 x 3.05

2


Design moment:

Mu =36.39 x (3.05)2

12= 28.21 x 106 N-mm = 28.21 KN-m



114

Calculation of Ast:


f ck x b x d)

28.21 x 106 = 0.87 x 415 x Ast x 425 x (1-415 x Ast

20 x 230 x 425)

Ast = 191.64 mm2

No. of 10mm diameter bars =4 x 191.64

π x 102 = 2.33 say 3 bars

Therefore, provided Ast = 3 xπ

4

x 102 = 235.62 mm2


Vu =Wu x Lx

2=

36.39 x 3.05

2= 55.5KN


b x d=

55.5 x 103

230 x 425= 0.57 N/mm2

Ast = 235.62 mm2


b x d=

100 x 235.62

230 x 425= 0.24


Pt c

0.15 0.28

0.24 ?

0.25 0.36

Therefore, c = 0.28 +0.36-0.28

0.25-0.15x (0.24-0.15) = 0.352 N/mm2

Therefore, v > c,



115

From IS456-2000(Clause 40.4, P72)

When v exceeds c, shear reinforcement shall be provided

Shear reinforcement shall be provided to carry a shear equal to Vu - c bd. The strength of

shear reinforcement Vus shall be calculated as follows:

Vus =0.87 x f y x Asv x d

sv

Stirrups are designed for shear Vus = Vu - c bd

= (55.5 x 103) – (0.32 x 230 x 425)

= 24220N


Asv =2 x π x 82

4= 100.53mm2

Therefore,

Spacing of stirrups, sv =

0.87 x f y x Asv x d

Vus =

0.87 x 415 x 100.53 x 425

24220

=636.9mm

From IS456-2000, (clause 26.5.1.5, P47)

The maximum spacing of shear reinforcement measured along the axis of the member shall

not be exceed 0.75d for vertical stirrups where d is the effective depth of the section under

consideration. In no case shall the spacing exceed 300mm.

Provide spacing = 300mm

Therefore, Vus =0.87 x f y x Asv x d

sv=

0.87 x 415 x 100.53 x 425

300= 51419.83N


Asv

b x Sv

≥0.4

0.87 x f y



116

Asv =0.4 x 230 x 300

0.87 x 415= 76.44mm2 <100.5mm2 (safe)


b. Beam 14-15

Roof :


Cover: 25mm



Loads:

Slab: 110mm thick


= 0.11 x 25 +2.5 = 5.25 KN/m2


Acting on beam:

Dead load:


Slab = 2 x

5.25 x 3.05

6 = 5.34 KN/m



117

Live load:

Slab = 2 x1.5 x 3.05

6= 1.53 KN/m

Therefore, Total load (W) = 1.1 + 5.34 + 1.53 = 7.97 KN/m


Maximum load on column removing the triangular load = 1.5 (1.1)

= 1.65 KN/m


2=

1.65 x 3.05

2= 2.52 KN (from each side)

Design moment:

Mu =11.96 x (3.05)2

12= 9.27 x 106 N-mm = 9.27 KN-m

Calculation of Ast:


f ck x b x d)

9.27 x 106 = 0.87 x 415 x Ast x 275 x (1-415 x Ast

20 x 230 x 275)

Ast = 96.41 mm2


π x 82 = 1.91 say 2 bars

Therefore, provided Ast = 2 xπ4 x 82 = 100.53 mm2


Vu =Wu x Lx

2=

11.96 x 3.05

2= 18.239KN


b x d=

18.239 x 103

230 x 275= 0.29 N/mm2

Ast = 100.53 mm2



118


b x d=

100 x 100.53

230 x 275= 0.15


Pt c

0.15 0.28





Asv =2 x π x 82

4= 100.53mm2



sv=

0.87 x 415 x 100.53 x 275

300= 33371.66KN


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 300

0.87 x 415= 76.44mm2 <100.5mm2 (safe)




119

Floor:


Cover: 25mm



Loads:

Slab: 120mm thick

Dead load = self weight + floor finish + wall

= 0.12 x 25 +1 + 1.5 = 5.5 KN/m2

Live load = 10 KN/m2 + 5 KN/m2

Acting on beam:

Dead load:


Slab =5.5 x 3.05

6+

4 x 3.05

6= 4.83 KN/m

Internal wall (150mm wall including plastering of 2.2m height)

= 20 x 0.15 x 2.2 = 6.6 KN/m

Live load:

Slab =15 x 3.05

6= 7.625 KN/m

Therefore, Total load (W) = 1.9 + 4.83 + 6.6 + 7.625 = 20.955 KN/m



= 12.75 KN/m



120


2=

12.75 x 3.05

2


Design moment:

Mu =31.4325 x (3.05)2

12= 24.37 x 106 N-mm = 24.37 KN-m

Calculation of Ast:


f ck x b x d)

24.37 x 106 = 0.87 x 415 x Ast x 425 x (1-415 x Ast

20 x 230 x 425)

Ast = 164.5 mm2


π x 102 = 2.1 say 3 bars


4x 102 = 235.62 mm2


Vu =Wu x Lx

2=

20.955 x 3.05

2= 31.96KN


b x d=

31.96 x 103

230 x 425= 0.33 N/mm2

Ast = 235.62 mm2


b x d=

100 x 235.62

230 x 425= 0.24


Pt c

0.15 0.28



121

0.24 ?

0.25 0.36

Therefore, c = 0.28 +0.36-0.28

0.25-0.15 x (0.24-0.15) = 0.352 N/mm2

Therefore, v approximately equals to c,


Asv =2 x π x 82

4= 100.53mm2


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 300

0.87 x 415= 76.44mm2 <100.5mm2 (safe)


c. Beam 7-8

Roof :


Cover: 25mm





122

Loads:

Slab: 110mm thick


= 0.11 x 25 +2.5 = 5.25 KN/m2


Acting on beam:

Dead load:


Slab =5.25 x 3.05

6= 2.67 KN/m

Parapet wall (250mm wall including plastering of 1m height)

= 20 x 0.25 x 1 = 5 KN/m

Live load:

Slab =1.5 x 3.05

6= 0.76 KN/m

Therefore, Total load (W) = 5 + 2.67 + 1.1 + 0.76 = 9.53 KN/m


Maximum load on column removing the triangular load = 1.5 (1.1 +5)

= 9.15 KN/m


2=

9.15 x 3.05

2= 13.96 KN (from each side)

Design moment:

Mu =14.3 x (3.05)2

12= 11.09 x 106 N-mm = 11.09 KN-m



123

Calculation of Ast:

From IS456-2000, P96a, clause G-1.1 (b), we have


f ck x b x d )

11.09 x 106 = 0.87 x 415 x Ast x 275 x (1-415 x Ast

20 x 230 x 275)

Ast = 116.12 mm2


π x 102 = 1.5 say 2 bars

Therefore, provided Ast = 2 xπ4

x 102 = 157.08 mm2


Vu =Wu x Lx

2=

14.3 x 3.05

2= 21.81KN


b x d=

21.81 x 103

230 x 275= 0.35 N/mm2

Ast = 157.08 mm2


b x d=

100 x 157.08

230 x 275= 0.25


Pt c

0.25 0.36







124

Asv =2 x π x 82

4= 100.53mm2


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 300

0.87 x 415= 76.44mm2 <100.5mm2 (safe)


Floor:


Cover: 25mm



Loads:

Slab: 120mm thick

Dead load = self weight + floor finish + wall

= 0.12 x 25 +1 + 1.5 = 5.5 KN/m2

Live load = 6 KN/m2

Acting on beam:

Dead load:




125

Slab =5.5 x 3.05

6= 2.8 KN/m

External wall (250mm wall including plastering)

= 20 x 0.25 x (3.35 – 0.3) = 15.25 KN/m

Live load:

Slab =6 x 3.05

6= 3.05 KN/m

Therefore, Total load (W) = 1.9 + 2.8 + 15.25 + 3.05 = 23 KN/m

Factored load (Wu) = 1.5 x 23 = 34.5 KN/m


= 25.73 KN/m


2=

25.73 x 3.05

2


Design moment:

Mu =34.5 x (3.05)2

12= 26.75 x 106 N-mm = 26.75 KN-m

Calculation of Ast:


f ck x b x d)

26.75 x 106 = 0.87 x 415 x Ast x 425 x (1-415 x Ast

20 x 230 x 425)

Ast = 181.3 mm2


π x 102 = 2.31 say 3 bars


4

x 102 = 235.62 mm2



126


Vu =Wu x Lx

2=

34.5 x 3.05

2= 52.61KN


b x d=

52.61 x 103

230 x 425= 0.54 N/mm2

Ast = 235.62 mm2


b x d=

100 x 235.62

230 x 425= 0.24


Pt c

0.15 0.28

0.24 ?

0.25 0.36

Therefore, c = 0.28 +0.36-0.28

0.25-0.15x (0.24-0.15) = 0.352 N/mm2

Therefore, v > c,


= (52.61 x 103) – (0.35 x 230 x 425)

= 18397.5N


Asv =2 x π x 82

4= 100.53mm2

Therefore,

Spacing of stirrups, sv =0.87 x f y x Asv x d

Vus=

0.87 x 415 x 100.53 x 425

18397.5

= 838.48mm



127



sv=

0.87 x 415 x 100.53 x 425

300= 51419.83N


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 300

0.87 x 415= 76.44mm2 <100.5mm2 (safe)






129

ROOF: (section 230mm x 300mm, d' = 25mm)

From analysis of frames the moments acting on the supports are:

At left support (Ast1):

Moment due to external load (Mu) = 64.65 KN-m

From IS456-2000, P96, clause G-1.1 (c), moment of resistance of the section is given by

Mu‟Limit = 0.36 xxumax

dx (1- 0.42 x

xumax

d) x b x d2 x f ck

For Fe 415,xumax

d= 0.48

Mu‟limit = 0.138 x f ck x b x d2

= 0.138 x 20 x 230 x (275)2 = 48 x 106 N-mm = 48 KN-m

Mu > Mu‟limit , therefore design as doubly reinforced beam.

Consider d'

d=

25

275= 0.1

Therefore, f sc = 353 N/mm2

Area of steel in compression (Asc):

Asc =Mu - Mu‟limit

f sc x (d - d')=

(64.65 - 48) x 106

353 x (275 - 25)=188.67 mm2

To find Ast1:

Mu‟limit = 0.87 x f y x Ast1 x d x (1 - f y x Ast1

f ck x b x d)



130

48 x 106 = 0.87 x 415 x Ast1 x 275 x (1 -415 x Ast1

20 x 230 x 275)

Ast1 = 602.55 mm2

To find Ast2:

Ast2 =f sc x Asc

0.87 x f y=

353 x 188.67

0.87 x 415=184.46 mm2

Ast = Ast1 + Ast2 = 602.55 + 184.46 = 787.01 mm2

Between the support:

Span moment = 39.24 KN-m

For T-beam, from IS456-2000, P37, clause 23.1.2 (a)

Effective width of flange (bf ) =l0

6+ (6 x Df ) + bw

l0 = 0.7 x l = 0.7 x 6000 = 4200 mm

Therefore, bf =4200

6+ (6 x 110) + 230

= 1590 mm

Therefore Mu‟limit = 0.36 x f ck x bf x Df x(d – 0.42Df )

= 0.36 x 20 x 1590 x 110 x (275 – 0.42 x 110)

= 288.12 KN-m

Mu < < Mu‟limit

Ast:

39.24 x 106 = 0.87 x 415 x Ast2 x 275 x (1-415 x Ast2

20 x 230 x 275)

Ast2 = 466.65 mm2



131

At right support (Ast3):



= 0.138 x 20 x 230 x (275)2 = 48 x 106 N-mm = 48 KN-m


Consider d'

d=

25

275= 0.1




f sc x (d - d')=

(86.09 - 48) x 106

353 x (275 - 25)=431.62 mm2

To find Ast1:

Mu‟limit = 0.87 x f y x Ast1 x d x (1 -f y x Ast1

f ck x b x d)

48 x 106 = 0.87 x 415 x Ast1 x 275 x (1 -415 x Ast1

20 x 230 x 275)

Ast1 = 602.55 mm2

To find Ast2:

Ast2 =f sc x Asc

0.87 x f y=

353 x 431.62

0.87 x 415=422 mm2

Ast3 = Ast1 + Ast2 = 602.55 + 422 = 1024.55 mm2

At support (Ast4):



= 0.138 x 20 x 230 x (275)2 = 48 x 106 N-mm = 48 KN-m



132


Consider d'

d=

25

275= 0.1




f sc x (d - d')=

(145.76 - 48) x 106

353 x (275 - 25)=1107.76 mm2

To find Ast1:

Mu‟limit = 0.87 x f y x Ast1 x d x (1 - f y x Ast1f ck x b x d )

48 x 106 = 0.87 x 415 x Ast1 x 275 x (1 -415 x Ast1

20 x 230 x 275)

Ast1 = 602.55 mm2

To find Ast2:

Ast2 = f sc x Asc

0.87 x f y= 353 x 1107.76

0.87 x 415=1083.06 mm2

Ast4 = Ast1 + Ast2 = 602.55 + 1083 = 1685.61 mm2

Between the support (Ast5):


Effective width of flange (bf ) =

l0

6 + (6 x Df ) + bw

l0 = 0.7 x l = 0.7 x 8410 = 5887 mm

Therefore, bf =5887

6+ (6 x 110) + 230

= 1871.16 mm



133


= 0.36 x 20 x 1871.16 x 110 x (275 – 0.42 x 110)

= 339.07 KN-m

Mu < Mu‟limit

Ast:

77.97 x 106 = 0.87 x 415 x Ast5 x 275 x (1-415 x Ast5

20 x 230 x 275)

Ast5 = 1411.66 mm2

At right end (Ast6):



= 0.138 x 20 x 230 x (275)2 = 48 x 106 N-mm = 48 KN-m


Consider d'

d=

25

275= 0.1




f sc x (d - d')=

(136.01 - 48) x 106

353 x (275 - 25)=997.28 mm2

To find Ast1:


f ck x b x d)

48 x 106 = 0.87 x 415 x Ast1 x 275 x (1 -415 x Ast1

20 x 230 x 275)

Ast1

= 602.55 mm2



134

To find Ast2:

Ast2 =f sc x Asc

0.87 x f y=

353 x 997.28

0.87 x 415=975.05 mm2

Ast6 = Ast1 + Ast2 = 602.55 + 975.05 = 1577.6 mm2


At left support:

End shear = (Vu) = 71.54 KN


b x d=

71.54 x 103

230 x 275= 1.13 N/mm2

Ast = 787.01 mm2


b x d=

100 x 787.01

230 x 275= 1.25


Pt c

1.25 0.67


Therefore, v > c,


= (71.54 x 103) – (0.67 x 230 x 275)

= 29162.5N



135


Asv =2 x π x 82

4= 100.53mm2

Therefore,


Vus=

0.87 x 415 x 100.53 x 275

29162.5

= 342.27mm


Therefore, Vus = 0.87 x f y x Asv x dsv= 0.87 x 415 x 100.53 x 275300 = 33261.73N


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 300

0.87 x 415= 76.44mm2 <100.5mm2 (safe)

Therefore, provide 8mm dia stirrups @ 300mm c/c.

At middle support:



b x d=

77.92 x 103

230 x 275= 1.23 N/mm2

Ast = 1024.55 mm2


b x d=

100 x 1024.55

230 x 275= 1.62


Pt c

1.5 0.72



136

1.62 ?

1.75 0.75

Therefore, c = 0.72 +0.75-0.72

1.75-1.5 x (1.62-1.5) = 0.734 N/mm2

Therefore, v > c,


= (77.92 x 103) – (0.734 x 230 x 275)

= 31494.5N


Asv =2 x π x 82

4= 100.53mm2

Therefore,


Vus=

0.87 x 415 x 100.53 x 275

31494.5

= 316.93mm



sv=

0.87 x 415 x 100.53 x 275

300= 33261.73N


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 300

0.87 x 415= 76.44mm2 <100.5mm2 (safe)




137

At middle support:



b x d =104.84 x 103

230 x 275 = 1.66 N/mm2

Ast = 1685.61 mm2


b x d=

100 x 1685.61

230 x 275= 2.67


Pt c

2.67 0.82


Therefore, v > c,


= (104.84 x 103) – (0.82 x 230 x 275)

= 52975N


Asv =2 x π x 82

4= 100.53mm2

Therefore,


Vus=

0.87 x 415 x 100.53 x 275

52975

= 188.42mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x f y



138

Asv =0.4 x 230 x 180

0.87 x 415= 45.87mm2 <100.5mm2 (safe)


At right support:



b x d=

102.63 x 103

230 x 275= 1.63 N/mm2

Ast = 1577.6 mm2

Consider Pt = 100 xAst b x d = 100 x 1577.6230 x 275 = 2.5


Pt c

2.5 0.82


Therefore, v > c,


= (102.63 x 103) – (0.82 x 230 x 275)

= 50765N


Asv =2 x π x 82

4= 100.53mm2

Therefore,


Vus=

0.87 x 415 x 100.53 x 275

50765

= 196.62mm < 300mm



139


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 190

0.87 x 415= 48.41mm2 <100.5mm2 (safe)


Reinforcement for roof 19-10-01:

FLOOR : (section: 230mm x 450mm, d‟= 25mm)

The moments acting on the beam are:




= 0.138 x 20 x 230 x (425)2 = 114.66 x 106 N-mm = 114.66 KN-m

Mu < Mu‟limit , therefore design as singly reinforced beam.



140

To find Ast1:


f ck x b x d)

106.54 x 106 = 0.87 x 415 x Ast1 x 425 x (1 -415 x Ast1

20 x 230 x 425)

Ast1 = 546 mm2

Between the supports (Ast2):


Effective width of flange (bf ) = l06 + (6 x Df ) + bw

l0 = 0.7 x l = 0.7 x 6000 = 4200 mm

Therefore, bf =4200

6+ (6 x 120) + 230

= 1650 mm

Therefore Mu‟limit = 0.36 x f ck x bf x Df x (d – 0.42Df )

= 0.36 x 20 x 1650 x 120 x (425 – 0.42 x 120)

= 534.03 KN-m

Mu < Mu‟limit

Ast:

73.58 x 106 = 0.87 x 415 x Ast2 x 425 x (1-415 x Ast2

20 x 230 x 425)

Ast2 = 852.07 mm2





141


= 0.138 x 20 x 230 x (425)2 = 114.66 x 106 N-mm = 114.66 KN-m


Consider d'

d=

25

425= 0.05




f sc x (d - d')

=(169.53 - 114.66) x 106

353 x (425 - 25)

=386.84 mm2

To find Ast1:


f ck x b x d)

114.66 x 106 = 0.87 x 415 x Ast1 x 425 x (1 -415 x Ast1

20 x 230 x 425)

Ast1 = 932 mm2

To find Ast2:

Ast2 =f sc x Asc

0.87 x f y=

353 x 386.84

0.87 x 415= 380.36 mm2

Ast3 = Ast1 + Ast2 = 932 + 380.6 = 1312.36 mm2

At support (Ast4):



= 0.138 x 20 x 230 x (425)2 = 114.66 x 106 N-mm = 114.66 KN-m


Consider d'

d =25

425 = 0.05



142




f sc x (d - d') =(277.32 -11 4.66) x 106

353 x (425 - 25) =1145.5 mm2

To find Ast1:


f ck x b x d)

114.66 x 106 = 0.87 x 415 x Ast1 x 425 x (1 -415 x Ast1

20 x 230 x 425)

Ast1 = 932 mm2

To find Ast2:

Ast2 =f sc x Asc

0.87 x f y=

355 x 1145.5

0.87 x 415=1126.31 mm2

Ast4 = Ast1 + Ast2 = 923 + 1126.31 = 2058.31 mm2




6+ (6 x Df ) + bw

l0 = 0.7 x l = 0.7 x 8410 = 5887 mm

Therefore, bf =

5887

6 + (6 x 120) + 230

= 1931.67 mm


= 0.36 x 20 x 1871.16 x 120 x (425 – 0.42 x 120)

= 625.2 KN-m

Mu < Mu‟limit



143

Ast:

153.57 x 106 = 0.87 x 415 x Ast5 x 425 x (1-415 x Ast5

20 x 230 x 425)

Ast5 = 1442.53 mm2




= 0.138 x 20 x 230 x (425)2 = 114.66 x 106 N-mm = 114.66 KN-m


Consider d'

d=

25

425= 0.05



Asc = Mu - Mu‟limitf sc x (d - d')

= (246.62 - 114.66) x 106

355 x (425 - 25)=929.29 mm2

To find Ast1:


f ck x b x d)

114.66 x 106 = 0.87 x 415 x Ast1 x 425 x (1 -415 x Ast1

20 x 230 x 425)

Ast1 = 932 mm2

To find Ast2:

Ast2 =f sc x Asc

0.87 x f y=

355 x 929.29

0.87 x 415=913.72 mm2

Ast6 = Ast1 + Ast2 = 932 + 913.72 = 1845.72 mm2



144


At left support:



b x d=

125.6 x 103

230 x 425= 1.29 N/mm2

Ast = 546 mm2


b x d=

100 x 546

230 x 425= 0.56


Pt c

0.5 0.48

0.56 ?

0.75 0.56

Therefore, c = 0.48 +0.56-0.48

0.75-0.5x (0.56-0.5) = 0.5 N/mm2

Therefore, v > c,


= (125.6 x 103) – (0.5 x 230 x 425)

= 76725N



145


Asv =2 x π x 82

4= 100.53mm2

Therefore,


Vus=

0.87 x 415 x 100.53 x 425

76725

= 184.87mm <300mm


Asvb x Sv

≥ 0.40.87 x f y

Asv =0.4 x 230 x 180

0.87 x 415= 45.86mm2 <100.5mm2 (safe)


At middle support:



b x d=

144.35 x 103

230 x 425= 1.48 N/mm2

Ast = 1312.16 mm2


b x d=

100 x 1312.16

230 x 425= 1.34


Pt c

1.25 0.67

1.34 ?

1.5 0.72



146

Therefore, c = 0.67 +0.72-0.67

1.5-1.25x (1.34-1.25) = 0.69 N/mm2

Therefore, v > c,


= (144.32 x 103) – (0.69 x 230 x 425)

= 78827.5N


Asv =2 x π x 82

4

= 100.53mm2

Therefore,


Vus=

0.87 x 415 x 100.53 x 425

78827.5

= 195.7mm


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 190

0.87 x 415= 48.41mm2 <100.5mm2 (safe)


At middle support:



b x d=

199.64 x 103

230 x 425= 2.04 N/mm2

Ast = 2058.31 mm2


b x d=

100 x 2058.31

230 x 425= 2



147


Pt c

2.00 0.79


Therefore, v > c,


= (199.64 x 103) – (0.79 x 230 x 425)

= 122417.5N


Asv =2 x π x 102

4= 157.07mm2

Therefore,



Vus =

0.87 x 415 x 157.07 x 425

122417.5

= 196.9mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =

0.4 x 230 x 190

0.87 x 415 = 48.41mm2

<157.07mm2

(safe)


At right support:



b x d=

192.85 x 103

230 x 425= 1.97 N/mm2



148

Ast = 1845.72 mm2


b x d=

100 x1845.72

230 x 425= 1.89


Pt c

1.75 0.75

1.89 ?

2 0.79

Therefore, c = 0.75 +0.79-0.75

2-1.75x (1.89-1.75) = 0.77 N/mm2

Therefore, v > c,


= (192.85 x 103) – (0.77 x 230 x 425)

= 117582.5N


Asv =2 x π x 102

4= 157.07mm2

Therefore,



Vus =

0.87 x 415 x 157.07 x 425

117582.5

= 204.98mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 200

0.87 x 415 = 50.96mm2 <157.07mm2 (safe)



149


Reinforcement for floor 19-10-01:



150

b. Beam 20-11 and Beam 11-02



151

ROOF (section: 230mm x 450mm, d‟ = 25mm)





= 0.138 x 20 x 230 x (425)2 = 114.66 x 106 N-mm = 114.66 KN-m


To find Ast1:


f ck x b x d)

68.42 x 106 = 0.87 x 415 x Ast1 x 425 x (1 -415 x Ast1

20 x 230 x 425)

Ast1 = 498.67 mm2




6+ (6 x Df ) + bw

l0 = 0.7 x l = 0.7 x 6000 = 4200 mm

Therefore, bf =4200

6+ (6 x 110) + 230 = 1590 mm



152


= 0.36 x 20 x 1590 x 110 x (425 – 0.42 x 110)

= 477.02 KN-m

Mu < Mu‟limit

Ast:

59.13 x 106 = 0.87 x 415 x Ast2 x 425 x (1-415 x Ast2

20 x 230 x 425)

Ast2 = 423.4 mm2




= 0.138 x 20 x 230 x (425)2 = 114.66 x 106 N-mm = 114.66 KN-m


Consider d'

d=

25

425= 0.05




f sc x (d - d')=

(140.51 - 114.66) x 106

353 x (425 - 25)=181.97 mm2

To find Ast1:


f ck x b x d)

114.66 x 106 = 0.87 x 415 x Ast1 x 425 x (1 -415 x Ast1

20 x 230 x 425)

Ast1 = 932 mm2



153

To find Ast2:

Ast2 =f sc x Asc

0.87 x f y=

355 x 181.97

0.87 x 415= 178.92 mm2

Ast3 = Ast1 + Ast2 = 932 + 178.92 = 1110.4 mm2

At support (Ast4):



= 0.138 x 20 x 230 x (425)2 = 114.66 x 106 N-mm = 114.66 KN-m


Consider d'

d=

25

425= 0.05



Asc = Mu - Mu‟limitf sc x (d - d')

= (199.51 -11 4.66) x 106

353 x (425 - 25)=597.46 mm2

To find Ast1:


f ck x b x d)

114.66 x 106 = 0.87 x 415 x Ast1 x 425 x (1 -415 x Ast1

20 x 230 x 425)

Ast1 = 932 mm2

To find Ast2:

Ast2 =f sc x Asc

0.87 x f y=

355 x 597.46

0.87 x 415=587.45 mm2

Ast4 = Ast1 + Ast2 = 923 + 587.45 = 1518.93 mm2



154




6 + (6 x Df ) + bw

l0 = 0.7 x l = 0.7 x 8410 = 5887 mm

Therefore, bf =5887

6+ (6 x 110) + 230

= 1871.17 mm


= 0.36 x 20 x 1871.16 x 110 x (425 – 0.42 x 110)

= 561.37 KN-m

Mu < Mu‟limit

Ast:

118.5 x 106 = 0.87 x 415 x Ast5 x 425 x (1- 415 x Ast5

20 x 230 x 425)

Ast5 = 973.38 mm2




= 0.138 x 20 x 230 x (425)2 = 114.66 x 106 N-mm = 114.66 KN-m


Consider d'

d=

25

425= 0.05




155



f sc x (d - d')=

(158.81 - 114.66) x 106

355 x (425 - 25)=310.85 mm2

To find Ast1:


f ck x b x d)

114.66 x 106 = 0.87 x 415 x Ast1 x 425 x (1 -415 x Ast1

20 x 230 x 425)

Ast1 = 932 mm2

To find Ast2:

Ast2 =f sc x Asc

0.87 x f y=

355 x 310.85

0.87 x 415=305.64 mm2

Ast6 = Ast1 + Ast2 = 932 + 305.64 = 1237.12 mm2


At left support:



b x d=

91.76 x 103

230 x 425= 0.94 N/mm2

Ast = 498.67 mm2


b x d=

100 x 498.67

230 x 425= 0.51



156


Pt c

0.5 0.48

0.51 ?

0.75 0.56

Therefore, c = 0.48 +0.56-0.48

0.75-0.5x (0.51-0.5) = 0.483 N/mm2

Therefore, v > c,


= (91.76 x 103) – (0.483 x 230 x 425)

= 44546.75N


Asv =

2 x π x 82

4 = 100.53mm

2

Therefore,


Vus=

0.87 x 415 x 100.53 x 425

44546.75

= 346.29mm >300mm



sv=

0.87 x 415 x 100.53 x 425

300= 51419.84N


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 300

0.87 x 415 = 76.44mm2 <100.5mm2 (safe)



157


At middle support:



b x d=

112.78 x 103

230 x 425= 1.15 N/mm2

Ast = 1110.4 mm2


b x d=

100 x 1110.4

230 x 425= 1.14


Pt c

1.00 0.62

1.15 ?

1.25 0.67

Therefore, c = 0.62 + 0.67-0.621.25-1

x (1.15-1.00) = 0.65 N/mm2

Therefore, v > c,


= (112.78 x 103) – (0.65 x 230 x 425)

= 49242.5N


Asv =2 x π x 82

4= 100.53mm2



158

Therefore,


Vus=

0.87 x 415 x 100.53 x 425

49242.5

= 313.26mm < 300mm



sv=

0.87 x 415 x 100.53 x 425

300= 51419.84N


Asvb x Sv

≥ 0.40.87 x f y

Asv =0.4 x 230 x 300

0.87 x 415= 76.44mm2 <100.5mm2 (safe)


At middle support:



b x d=

144.11 x 103

230 x 425= 1.47 N/mm2

Ast = 1598.13 mm2


b x d=

100 x 1598.13

230 x 425= 1.64


Pt c

1.50 0.72

1.64 ?

1.75 0.75



159

Therefore, c = 0.72 +0.75-0.72

1.75-1.5x (1.64 – 1.5) = 0.74 N/mm2

Therefore, v > c,


= (144.11 x 103) – (0.74 x 230 x 425)

= 71775N


Asv =2 x π x 82

4

= 100.53mm2

Therefore,


Vus=

0.87 x 415 x 100.53 x 425

71775

= 214.92mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 210

0.87 x 415= 53.51mm2 <100.5mm2 (safe)


At right support:



b x d=

135.28 x 103

230 x 425= 1.38 N/mm2

Ast = 1237.12 mm2


b x d=

100 x1237.12

230 x 425= 1.27



160


Pt c

1.25 0.67

1.27 ?

1.5 0.72

Therefore, c = 0.67 +0.72-0.67

1.5-1.25x (1.27-1.25) = 0.674 N/mm2

Therefore, v > c,


= (135.28 x 103) – (0.674 x 230 x 425)

= 69396.5N


Asv =

2 x π x 82

4 = 100.53mm

2

Therefore,


Vus=

0.87 x 415 x 100.53 x 425

69396.5

= 222.28mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 220

0.87 x 415= 56.06mm2 <100.5mm2 (safe)




161


FLOOR (Section: 230mm x 600mm, d‟ = 50mm)





= 0.138 x 20 x 230 x (550)2 = 192.027 x 106 N-mm

= 192.027 KN-m


To find Ast1:


f ck x b x d)

75.53 x 106 = 0.87 x 415 x Ast1 x 550 x (1 -415 x Ast1

20 x 230 x 550)

Ast1 = 407.61 mm2



162




6 + (6 x Df ) + bw

l0 = 0.7 x l = 0.7 x 6000 = 4200 mm

Therefore, bf =4200

6+ (6 x 120) + 230

= 1650 mm


= 0.36 x 20 x 1590 x 120 x (550 – 0.42 x 120)

= 712.23 KN-m

Mu < Mu‟limit

Ast:

71.71 x 106 = 0.87 x 415 x Ast2 x 550 x (1- 415 x Ast2

20 x 230 x 550)

Ast2 = 385.5 mm2




= 0.138 x 20 x 230 x (550)2 = 192.027 x 106 N-mm

= 192.027 KN-m


Consider d'

d=

50

550= 0.1




163



f sc x (d - d')=

(228.89 - 192.027) x 106

353 x (550 - 50)=208.86 mm2

To find Ast1:


f ck x b x d)

192.027 x 106 = 0.87 x 415 x Ast1 x 550 x (1 -415 x Ast1

20 x 230 x 550)

Ast1 = 1205.32 mm2

To find Ast2:

Ast2 =f sc x Asc

0.87 x f y=

353 x 208.86

0.87 x 415= 201.21 mm2

Ast3 = Ast1 + Ast2 = 1205.32 + 204.21 = 1409.53 mm2

At support (Ast4):



= 0.138 x 20 x 230 x (550)2 = 192.027 x 106 N-mm

= 192.027 KN-m


Consider d'd = 50550 = 0.1




f sc x (d - d')=

(393.9 - 192.027) x 106

353 x (550 - 50)=1143.76 mm2





165




= 0.138 x 20 x 230 x (550)2 = 192.027 x 106 N-mm

= 192.027 KN-m


Consider d'

d=

50

550= 0.1




f sc x (d - d')=

(315.93 - 192.027) x 106

353 x (550 - 50)=701.76 mm2

To find Ast1:

Mu‟limit = 0.87 x f y x Ast1 x d x (1 -

f y x Ast1

f ck x b x d )

192.027 x 106 = 0.87 x 415 x Ast1 x 550 x (1 -415 x Ast1

20 x 230 x 550)

Ast1 = 1205.32 mm2

To find Ast2:

Ast2 = f sc x Asc0.87 x f y = 353 x 701.760.87 x 415 = 686.11 mm2

Ast6 = Ast1 + Ast2 = 1205.32 + 686.11 = 1891.43 mm2



166


At left support:



b x d=

109.8 x 103

230 x 550= 0.87 N/mm2

Ast = 407.61 mm2


b x d=

100 x 407.61

230 x 550= 0.32


Pt c

0.25 0.36

0.32 ?

0.5 0.48

Therefore, c = 0.36 +0.48-0.360.5-0.25

x (0.32-0.25) = 0.4 N/mm2

Therefore, v > c,


= (109.8 x 103) – (0.4 x 230 x 550)

= 59200N



167


Asv =2 x π x 82

4= 100.53mm2

Therefore,


Vus=

0.87 x 415 x 100.53 x 550

59200

= 337.21mm >300mm


Vus = 0.87 x f y x Asv x dsv= 0.87 x 415 x 100.53 x 550300 = 66543.32N


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 300

0.87 x 415= 76.44mm2 <100.5mm2 (safe)


At middle support:



b x d=

152.94 x 103

230 x 550= 1.21 N/mm2

Ast = 1409.33 mm2


b x d=

100 x 1409.33

230 x 550= 1.12



168


Pt c

1.00 0.62

1.12 ?

1.25 0.67

Therefore, c = 0.62 +0.67-0.62

1.25-1x (1.12-1) = 0.65 N/mm2

Therefore, v > c,


= (152.94 x 103) – (0.65 x 230 x 550)

= 70715N


Asv =

2 x π x 82

4 = 100.53mm

2

Therefore,


Vus=

0.87 x 415 x 100.53 x 550

70715

= 282.3mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 280

0.87 x 415= 71.35mm2 <100.5mm2 (safe)




169

At middle support:



b x d =292.34 x 103

230 x 550 = 2.31 N/mm2

Ast = 2323.58 mm2


b x d=

100 x 2323.58

230 x 550= 1.84


Pt c

1.75 0.75

1.84 ?

2.00 0.79

Therefore, c = 0.75 +0.79-0.75

2-1.75x (1.84-1.75) = 0.77 N/mm2

Therefore, v > c,


= (292.34 x 103) – (0.77 x 230 x 550)

= 194935N


Asv =2 x π x 102

4= 157.07mm2

Therefore,


Vus=

0.87 x 415 x 157.07 x 550

194935

= 160mm < 300mm



170


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 160

0.87 x 415= 40.77mm2 <157.07mm2 (safe)


At right support:


Nominal shear stress, v = Vub x d = 275.05 x 10

3

230 x 550 = 2.18 N/mm2

Ast = 1891.43 mm2


b x d=

100 x 1891.43

230 x 550= 1.5


Pt c

1.5 0.72


Therefore, v > c,


= (275.05 x 103) – (0.72 x 230 x 550)

= 183970N


Asv =2 x π x 102

4= 157.07mm2



171

Therefore,


Vus=

0.87 x 415 x 157.07 x 550

183970

= 169.54mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 160

0.87 x 415= 40.77mm2 <157.07mm2 (safe)





172

c. Beam 24-15 and Beam 15-06



173

ROOF – Same as in case of roof 20-11-12

FLOOR :





= 0.138 x 20 x 230 x (550)2 = 192.027 x 106 N-mm

= 192.027 KN-m


To find Ast1:


f ck x b x d)

95.08 x 106

= 0.87 x 415 x Ast1 x 550 x (1 -

415 x Ast1

20 x 230 x 550 )

Ast1 = 523.81 mm2




6+ (6 x Df ) + bw



174

l0 = 0.7 x l = 0.7 x 6000 = 4200 mm

Therefore, bf =4200

6+ (6 x 120) + 230

= 1650 mm


= 0.36 x 20 x 1590 x 120 x (550 – 0.42 x 120)

= 712.23 KN-m

Mu < Mu‟limit

Ast:

89.37 x 106 = 0.87 x 415 x Ast2 x 550 x (1-415 x Ast2

20 x 230 x 550)

Ast2 = 489.33 mm2




= 0.138 x 20 x 230 x (550)2 = 192.027 x 106 N-mm

= 192.027 KN-m


Consider d'd = 50550 = 0.1




f sc x (d - d')=

(208.43 - 192.027) x 106

353 x (550 - 50)=92.94 mm2



175

To find Ast1:


f ck x b x d)

192.027 x 106 = 0.87 x 415 x Ast1 x 550 x (1 -415 x Ast1

20 x 230 x 550)

Ast1 = 1205.32 mm2

To find Ast2:

Ast2 =f sc x Asc

0.87 x f y=

353 x 92.94

0.87 x 415= 90.87 mm2

Ast3 = Ast1 + Ast2 = 1205.32 + 90.87 = 1296.19 mm2

At support (Ast4):



= 0.138 x 20 x 230 x (550)2 = 192.027 x 106 N-mm

= 192.027 KN-m


Consider d'

d=

50

550= 0.1




f sc x (d - d')=

(280.82 - 192.027) x 106

353 x (550 - 50)=503.08 mm2

To find Ast1:


f ck x b x d)

192.027 x 106 = 0.87 x 415 x Ast1 x 550 x (1 - 415 x Ast1

20 x 230 x 550)



176

Ast1 = 1205.32 mm2

To find Ast2:

Ast2 =f sc x Asc

0.87 x f y =353 x 503.08

0.87 x 415 = 491.86 mm2

Ast4 = Ast1 + Ast2 = 1205.32 + 491.86 = 1697.18 mm2




6

+ (6 x Df ) + bw

l0 = 0.7 x l = 0.7 x 8410 = 5887 mm

Therefore, bf =5887

6+ (6 x 120) + 230

= 1931.17 mm


= 0.36 x 20 x 1931.7 x 120 x (550 – 0.42 x 120)

= 833.83 KN-m

Mu < Mu‟limit

Ast:

166.85 x 106

= 0.87 x 415 x Ast2 x 550 x (1-

415 x Ast2

20 x 230 x 550 )

Ast5 = 1006.35 mm2






177

= 0.138 x 20 x 230 x (550)2 = 192.027 x 106 N-mm

= 192.027 KN-m


Consider d'

d=

50

550= 0.1




f sc x (d - d')=

(211.34 - 192.027) x 106

353 x (550 - 50)=109.42 mm

2

To find Ast1:


f ck x b x d)

192.027 x 106 = 0.87 x 415 x Ast1 x 550 x (1 -415 x Ast1

20 x 230 x 550)

Ast1 = 1205.32 mm2

To find Ast2:

Ast2 =f sc x Asc

0.87 x f y=

353 x 109.42

0.87 x 415= 106.98 mm2

Ast6 = Ast1 + Ast2 = 1205.32 + 106.98 = 1312.3 mm2





179


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 250

0.87 x 415= 63.7mm2 <100.5mm2 (safe)


At middle support:



3

230 x 550 = 1.32 N/mm2

Ast = 1296.19 mm2


b x d=

100 x 1296.19

230 x 550= 1.03


Pt c

1.00 0.62

1.03 ?

1.25 0.67

Therefore, c = 0.62 +0.67-0.62

1.25-1x (1.03-1) = 0.63 N/mm2

Therefore, v > c,


= (166.87 x 103) – (0.63 x 230 x 550)

= 87175N




180

Asv =2 x π x 82

4= 100.53mm2

Therefore,


Vus=

0.87 x 415 x 100.53 x 550

87175

= 229mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 220

0.87 x 415= 56.05mm2 <100.5mm2 (safe)


At middle support:


Nominal shear stress, v = Vu

b x d= 200.56 x 10

3

230 x 550= 1.59 N/mm2

Ast = 1697.18 mm2


b x d=

100 x 1697.18

230 x 550= 1.34


Pt c

1.25 0.67

1.34 ?

1.50 0.72

Therefore, c = 0.67 +0.72-0.67

1.5-1.25x (1.34-1.25) = 0.69 N/mm2



181

Therefore, v > c,


= (200.56 x 103) – (0.69 x 230 x 550)

= 113275N


Asv =2 x π x 102

4= 157.07mm2

Therefore,


Vus=

0.87 x 415 x 157.07 x 550

113275

= 275.35mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 270

0.87 x 415= 68.8mm2 <157.07mm2 (safe)


At right support:



b x d=

185.35 x 103

230 x 550= 1.47 N/mm2

Ast = 1312.3 mm2


b x d=

100 x 1312.3

230 x 550= 1



182


Pt c

1 0.62


Therefore, v > c,


= (185.35 x 103) – (0.62 x 230 x 550)

= 106920N


Asv =2 x π x 102

4= 157.07mm2

Therefore,



Vus =

0.87 x 415 x 157.07 x 550

106920

= 291mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =

0.4 x 230 x 290

0.87 x 415 = 73.9mm2

<157.07mm2

(safe)




183




184

d. Beam 25-16 and Beam 16-07



185

ROOF – Same as 20-11-02

FLOOR :





= 0.138 x 20 x 230 x (550)2 = 192.027 x 106 N-mm

= 192.027 KN-m


To find Ast1:


f ck x b x d)

85.69 x 106 = 0.87 x 415 x Ast1 x 550 x (1 -415 x Ast1

20 x 230 x 550)

Ast1 = 467.34 mm2




6+ (6 x Df ) + bw

l0 = 0.7 x l = 0.7 x 6000 = 4200 mm



186

Therefore, bf =4200

6+ (6 x 120) + 230

= 1650 mm


= 0.36 x 20 x 1590 x 120 x (550 – 0.42 x 120)

= 712.23 KN-m

Mu < Mu‟limit

Ast:

80.45 x 106 = 0.87 x 415 x Ast2 x 550 x (1-415 x Ast2

20 x 230 x 550)

Ast2 = 436.37 mm2




= 0.138 x 20 x 230 x (550)2 = 192.027 x 106 N-mm

= 192.027 KN-m


Consider d'

d=

50

550= 0.1




f sc x (d - d')=

(218.95 - 192.027) x 106

353 x (550 - 50)=152.54 mm2

To find Ast1:


f ck x b x d)



187

192.027 x 106 = 0.87 x 415 x Ast1 x 550 x (1 -415 x Ast1

20 x 230 x 550)

Ast1 = 1205.32 mm2

To find Ast2:

Ast2 =f sc x Asc

0.87 x f y=

353 x 152.54

0.87 x 415= 149.14 mm2

Ast3 = Ast1 + Ast2 = 1205.32 + 149.14 = 1354.46 mm2

At support (Ast4):



= 0.138 x 20 x 230 x (550)2 = 192.027 x 106 N-mm

= 192.027 KN-m


Consider

d'

d =

50

550 = 0.1




f sc x (d - d')=

(343.47 - 192.027) x 106

353 x (550 - 50)=858.03 mm2

To find Ast1:


f ck x b x d)

192.027 x 106 = 0.87 x 415 x Ast1 x 550 x (1 -415 x Ast1

20 x 230 x 550)

Ast1 = 1205.32 mm2



188

To find Ast2:

Ast2 =f sc x Asc

0.87 x f y=

353 x 858.03

0.87 x 415= 838.9 mm2

Ast4 = Ast1 + Ast2 = 1205.32 + 838.9 = 2044.22 mm2




6+ (6 x Df ) + bw

l0 = 0.7 x l = 0.7 x 8410 = 5887 mm

Therefore, bf =5887

6+ (6 x 120) + 230

= 1931.17 mm


= 0.36 x 20 x 1931.7 x 120 x (550 – 0.42 x 120)

= 833.83 KN-m

Mu < Mu‟limit

Ast:

210.5 x 106 = 0.87 x 415 x Ast2 x 550 x (1-415 x Ast2

20 x 230 x 550)

Ast5 = 1366.21 mm2




= 0.138 x 20 x 230 x (550)2 = 192.027 x 106 N-mm

= 192.027 KN-m



189


Consider d'

d=

50

550= 0.1




f sc x (d - d')=

(268.55 - 192.027) x 106

353 x (550 - 50)=433.56 mm2

To find Ast1:


192.027 x 106 = 0.87 x 415 x Ast1 x 550 x (1 -415 x Ast1

20 x 230 x 550)

Ast1 = 1205.32 mm2

To find Ast2:

Ast2 = f sc x Asc

0.87 x f y= 353 x 433.56

0.87 x 415= 423.9 mm2

Ast6 = Ast1 + Ast2 = 1205.32 + 423.9 = 1629.22 mm2


At left support:




190


b x d=

120.89 x 103

230 x 550= 0.96 N/mm2

Ast = 467.34 mm2


b x d=

100 x 467.34

230 x 550= 0.37


Pt c

0.25 0.36

0.37 ?

0.5 0.48

Therefore, c = 0.36 +0.48-0.36

0.5-0.25x (0.37-0.25) = 0.42 N/mm2

Therefore, v > c,


= (120.89 x 103) – (0.42 x 230 x 550)

= 67760N


Asv =2 x π x 82

4= 100.53mm2

Therefore,


Vus=

0.87 x 415 x 100.53 x 550

67760

= 294mm <300mm


Asvb x Sv

≥0.4

0.87 x f y



191

Asv =0.4 x 230 x 290

0.87 x 415= 73.9mm2 <100.5mm2 (safe)


At middle support:



b x d=

158.11 x 103

230 x 550= 1.25 N/mm2

Ast = 1354.46 mm2

Consider Pt = 100 xAst b x d = 100 x 1354.46230 x 550 = 1.07


Pt c

1.00 0.62

1.07 ?

1.25 0.67

Therefore, c = 0.62 +0.67-0.62

1.25-1x (1.07-1) = 0.634 N/mm2

Therefore, v > c,


= (158.11x 103) – (0.634 x 230 x 550)

= 77909N


Asv =2 x π x 82

4= 100.53mm2



192

Therefore,


Vus=

0.87 x 415 x 100.53 x 550

77909

= 256.24mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 250

0.87 x 415= 63.7mm2 <100.5mm2 (safe)


At middle support:



b x d=

251.1 x 103

230 x 550= 2 N/mm2

Ast = 2044.2 mm2


b x d=

100 x 2044.2

230 x 550= 1.62


Pt c

1.5 0.72

1.62 ?

1.75 0.75

Therefore, c = 0.72 +0.75-0.72

1.75-1.5x (1.62-1.5) = 0.73 N/mm2

Therefore, v > c,



193


= (251.1 x 103) – (0.73 x 230 x 550)

= 158755N


Asv =2 x π x 102

4= 157.07mm2

Therefore,


Vus

=0.87 x 415 x 157.07 x 550

158755

= 196.5mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 190

0.87 x 415= 48.41mm2 <157.07mm2 (safe)


At right support:



b x d=

234.86 x 103

230 x 550= 1.86 N/mm2

Ast = 1629.22 mm2


b x d=

100 x 1629.22

230 x 550= 1.29


Pt c

1.25 0.67



194

1.29 ?

1.5 0.72

Therefore, c = = 0.67 +0.72-0.67

1.5-1.25 x (1.29-1.25) = 0.68 N/mm2

Therefore, v > c,


= (234.86 x 103) – (0.68 x 230 x 550)

= 148840N


Asv =2 x π x 102

4= 157.07mm2

Therefore,


Vus=

0.87 x 415 x 157.07 x 550

148840

= 209.56mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 200

0.87 x 415= 50.96mm2 <157.07mm2 (safe)




195

Reinforcement for roof 25-16-07



196

e. Beam 26-17 and Beam 17-08



197


FLOOR:





= 0.138 x 20 x 230 x (550)2 = 192.027 x 106 N-mm

= 192.027 KN-m


To find Ast1:


f ck x b x d)

73.1 x 106 = 0.87 x 415 x Ast1 x 550 x (1 -415 x Ast1

20 x 230 x 550)

Ast1 = 393.52 mm2




6+ (6 x Df ) + bw

l0 = 0.7 x l = 0.7 x 6000 = 4200 mm



198

Therefore, bf =4200

6+ (6 x 120) + 230

= 1650 mm


= 0.36 x 20 x 1590 x 120 x (550 – 0.42 x 120)

= 712.23 KN-m

Mu < Mu‟limit

Ast:

68.48 x 106 = 0.87 x 415 x Ast2 x 550 x (1-415 x Ast2

20 x 230 x 550)

Ast2 = 366.94 mm2




= 0.138 x 20 x 230 x (550)2 = 192.027 x 106 N-mm

= 192.027 KN-m


To find Ast3:


185.08 x 106 = 0.87 x 415 x Ast3 x 550 x (1 -415 x Ast3

20 x 230 x 550)

Ast3 = 1148.33 mm2

At support (Ast4):




199


= 0.138 x 20 x 230 x (550)2 = 192.027 x 106 N-mm

= 192.027 KN-m


Consider d'

d=

50

550= 0.1



Asc = Mu - Mu‟limit

f sc x (d - d')= (297.26 - 192.027) x 106

353 x (550 - 50)=596.22 mm2

To find Ast1:


f ck x b x d)

192.027 x 106 = 0.87 x 415 x Ast1 x 550 x (1 -415 x Ast1

20 x 230 x 550)

Ast1 = 1205.32 mm2

To find Ast2:

Ast2 =f sc x Asc

0.87 x f y=

353 x 596.22

0.87 x 415= 582.93 mm2

Ast4 = Ast1 + Ast2 = 1205.32 + 582.93 = 1788.25 mm2




6+ (6 x Df ) + bw

l0 = 0.7 x l = 0.7 x 8410 = 5887 mm



200

Therefore, bf =5887

6+ (6 x 120) + 230

= 1931.17 mm


= 0.36 x 20 x 1931.7 x 120 x (550 – 0.42 x 120)

= 833.83 KN-m

Mu < Mu‟limit

Ast:

183.9 x 106 = 0.87 x 415 x Ast2 x 550 x (1-415 x Ast2

20 x 230 x 550)

Ast5 = 1138.82 mm2




= 0.138 x 20 x 230 x (550)2 = 192.027 x 106 N-mm

= 192.027 KN-m


Consider d'

d=

50

550= 0.1




f sc x (d - d')=

(234.45 - 192.027) x 106

353 x (550 - 50)=240.36 mm2

To find Ast1:


f ck x b x d)



201

192.027 x 106 = 0.87 x 415 x Ast1 x 550 x (1 -415 x Ast1

20 x 230 x 550)

Ast1 = 1205.32 mm2

To find Ast2:

Ast2 =f sc x Asc

0.87 x f y=

353 x 240.36

0.87 x 415= 235 mm2

Ast6 = Ast1 + Ast2 = 1205.32 + 235 = 1440.32 mm2


At left support:



b x d=

101.13 x 103

230 x 550= 0.8 N/mm2

Ast = 393.52 mm2

Consider Pt = 100 xAst b x d

= 100 x 393.52230 x 550

= 0.31


Pt c

0.25 0.36

0.31 ?

0.5 0.48



202

Therefore, c = 0.36 +0.48-0.36

0.5-0.25x (0.31-0.25) = 0.39 N/mm2

Therefore, v > c,


= (101.13 x 103) – (0.39 x 230 x 550)

= 51795N


Asv =2 x π x 82

4

= 100.53mm2

Therefore,


Vus=

0.87 x 415 x 100.53 x 550

51795

= 385.42mm >300mm



sv=

0.87 x 415 x 100.53 x 550

300= 66543.32N


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =

0.4 x 230 x 300

0.87 x 415 = 76.44 mm2

<100.5mm2

(safe)


At middle support:



b x d=

131.2 x 103

230 x 550= 1.04 N/mm2



203

Ast = 1148.33 mm2


b x d=

100 x 1148.33

230 x 550= 0.91


Pt c

0.75 0.56

0.91 ?

1.00 0.62

Therefore, c = 0.56 +0.62-0.56

1-0.75x (0.91-0.75) = 0.6 N/mm2

Therefore, v > c,


= (131.2x 103) – (0.6 x 230 x 550)

= 55300N


Asv =2 x π x 82

4= 100.53mm2

Therefore,



Vus =

0.87 x 415 x 100.53 x 550

55300

= 361mm > 300mm



sv=

0.87 x 415 x 100.53 x 550

300= 66543.32N






205

Therefore,


Vus=

0.87 x 415 x 157.07 x 550

128655

= 242.5mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 240

0.87 x 415= 61.16mm2 <157.07mm2 (safe)


At right support:



b x d=

204.57 x 103

230 x 550= 1.62 N/mm2

Ast = 1400.32 mm2


b x d=

100 x 1400.32

230 x 550= 1.11


Pt c

1.00 0.62

1.11 ?

1.25 0.67

Therefore, c = = 0.62 +0.67-0.62

1.25-1.00x (1.11-1.00) = 0.64 N/mm2

Therefore, v > c,



206


= (204.57 x 103) – (0.64 x 230 x 550)

= 123610N


Asv =2 x π x 102

4= 157.07mm2

Therefore,


Vus

=0.87 x 415 x 157.07 x 550

123610

= 252.33mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 250

0.87 x 415= 63.7mm2 <157.07mm2 (safe)





207

f. Beam 27-18 and Beam 18-09



208


FLOOR :





= 0.138 x 20 x 230 x (550)2 = 192.027 x 106 N-mm

= 192.027 KN-m


To find Ast1:


f ck x b x d)

101.22 x 106 = 0.87 x 415 x Ast1 x 550 x (1 -415 x Ast1

20 x 230 x 550)

Ast1 = 561.43 mm2




6+ (6 x Df ) + bw

l0 = 0.7 x l = 0.7 x 6000 = 4200 mm



209

Therefore, bf =4200

6+ (6 x 120) + 230

= 1650 mm


= 0.36 x 20 x 1590 x 120 x (550 – 0.42 x 120)

= 712.23 KN-m

Mu < Mu‟limit

Ast:

70.3 x 106 = 0.87 x 415 x Ast2 x 550 x (1-415 x Ast2

20 x 230 x 550)

Ast2 = 377.38 mm2




= 0.138 x 20 x 230 x (550)2 = 192.027 x 106 N-mm

= 192.027 KN-m


To find Ast3:


166.06 x 106 = 0.87 x 415 x Ast3 x 550 x (1 -415 x Ast3

20 x 230 x 550)

Ast3 = 1000.42 mm2

At support (Ast4):




210


= 0.138 x 20 x 230 x (550)2 = 192.027 x 106 N-mm

= 192.027 KN-m


Consider d'

d=

50

550= 0.1



Asc = Mu - Mu‟limit

f sc x (d - d')= (288.45 - 192.027) x 106

353 x (550 - 50)=546.31 mm2

To find Ast1:


f ck x b x d)

192.027 x 106 = 0.87 x 415 x Ast1 x 550 x (1 -415 x Ast1

20 x 230 x 550)

Ast1 = 1205.32 mm2

To find Ast2:

Ast2 =f sc x Asc

0.87 x f y=

353 x 546.31

0.87 x 415= 534.13 mm2

Ast4 = Ast1 + Ast2 = 1205.32 + 534.13 = 1739.45 mm2




6+ (6 x Df ) + bw

l0 = 0.7 x l = 0.7 x 8410 = 5887 mm



211

Therefore, bf =5887

6+ (6 x 120) + 230

= 1931.17 mm


= 0.36 x 20 x 1931.7 x 120 x (550 – 0.42 x 120)

= 833.83 KN-m

Mu < Mu‟limit

Ast:

160.42 x 106 = 0.87 x 415 x Ast2 x 550 x (1-415 x Ast2

20 x 230 x 550)

Ast5 = 958.56 mm2




= 0.138 x 20 x 230 x (550)2 = 192.027 x 106 N-mm

= 192.027 KN-m


Consider d'

d=

50

550= 0.1




f sc x (d - d')=

(257.96 - 192.027) x 106

353 x (550 - 50)=373.56 mm2

To find Ast1:


f ck x b x d)



212

192.027 x 106 = 0.87 x 415 x Ast1 x 550 x (1 -415 x Ast1

20 x 230 x 550)

Ast1 = 1205.32 mm2

To find Ast2:

Ast2 =f sc x Asc

0.87 x f y=

353 x 373.56

0.87 x 415= 365.23 mm2

Ast6 = Ast1 + Ast2 = 1205.32 + 365.23 = 1570.55 mm2


At left support:



b x d=

119.94 x 103

230 x 550= 0.95 N/mm2

Ast = 561.43 mm2


b x d=

100 x 561.43

230 x 550= 0.44


Pt c

0.25 0.36

0.44 ?

0.5 0.48

Therefore, c = 0.36 +0.48-0.36

0.5-0.25x (0.44-0.25) = 0.45 N/mm2

Therefore, v > c,



213


= (119.94 x 103) – (0.45 x 230 x 550)

= 63015N


Asv =2 x π x 82

4= 100.53mm2

Therefore,


Vus

=0.87 x 415 x 100.53 x 550

63015

= 316.8mm >300mm



sv=

0.87 x 415 x 100.53 x 550

300= 66543.32N


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 300

0.87 x 415= 76.44 mm2 <100.5mm2 (safe)


At middle support:



b x d=

138.8 x 103

230 x 550= 1.1 N/mm2

Ast = 1000.42 mm2


b x d=

100 x 1000.42

230 x 550= 0.79



214


Pt c

0.75 0.56

0.79 ?

1.00 0.62

Therefore, c = 0.56 +0.62-0.56

1-0.75x (0.79-0.75) = 0.57 N/mm2

Therefore, v > c,


= (138.8x 103) – (0.57 x 230 x 550)

= 66695N


Asv =

2 x π x 82

4 = 100.53mm

2

Therefore,


Vus=

0.87 x 415 x 100.53 x 550

66695

= 300mm


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 300

0.87 x 415= 76.44 mm2 <100.5mm2 (safe)




215

At middle support:



b x d =208.4 x 103

230 x 550 = 1.65 N/mm2

Ast = 1739.5mm2


b x d=

100 x 1739.5

230 x 550= 1.38


Pt c

1.25 0.67

1.38 ?

1.5 0.72

Therefore, c = 0.67 +0.72-0.67

1.5-1.25x (1.38-1.25) = 0.7 N/mm2

Therefore, v > c,


= (208.4 x 103) – (0.7 x 230 x 550)

= 119850N


Asv =2 x π x 102

4= 157.07mm2

Therefore,


Vus=

0.87 x 415 x 157.07 x 550

119850

= 260.25mm < 300mm



216


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 260

0.87 x 415= 66.25mm2 <157.07mm2 (safe)


At right support:



3

230 x 550 = 1.58 N/mm2

Ast = 1570.5 mm2


b x d=

100 x 1570.5

230 x 550= 1.24


Pt c

1.00 0.62

1.24 ?

1.25 0.67

Therefore, c = = 0.62 +0.67-0.62

1.25-1.00x (1.24-1.00) = 0.67 N/mm2

Therefore, v > c,


= (200.33 x 103) – (0.67 x 230 x 550)

= 115575N




217

Asv =2 x π x 102

4= 157.07mm2

Therefore,


Vus=

0.87 x 415 x 157.07 x 550

115575

= 269.87mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x f y

Asv =0.4 x 230 x 260

0.87 x 415= 66.25mm2 <157.07mm2 (safe)





218

5. DESIGN OF COLUMNS

Since the loads and moments in the three columns in a frame are different. Each of the

Column is required to be designed separately. However, when entire building is to be

designed, there will be a number of other columns along with each of the above columns to

form a group.

Since exact values of Pu and Mu are known for all storeys for all columns, the column

section will be designed using exact method using charts and tables. Charts are useful for any

which of column. It is advisable to have curves plotted of Pu-Mu for standard sections

normally used in building design to avoid calculations.

All the columns are subjected to axial loads and uni-axial bending. They will be

designed to resist Pu and Mu for bending @ x-axis which is the major axis.

For frame 19-10-01:

Columns are C19 at left end, C10 at middle and C01 at right end of the frame.

The moments and axial forces are calculated in analysis of frames and design of

beams. We have transverse frames and in these frames the plinth level transverse beams are

absent and longitudinal beams are present at the plinth level.

In each level the types of loads are:

1. Max shear from transverse beams

2. Shear from longitudinal beams

3. Self weight of columns.

But in plinth level shear from transverse beams is absent. We have only two values at the plinth level

Section: 230mm x 600mm

Cover (d‟) = 50mm

Minimum eccentricity (ex,min) =unsupported length

500+

lateral dimension

30or

= 20mm



219

Pu – Load acting on the column

Mux,min = ex,min x Pu

MuD = Maximum (Mu, Mux,min)

f ck x b x D = 20 x 230 x 600 = 2760000N = 2760 KN

f ck x b x D2 = 20 x 230 x (600)2 = 1676 x 106 N-mm = 1676 KN-m



220

For Column 19:

To find MuD:

Storey Length

(mm)

ex,min (mm) Pu (KN) Mux (KN-

m)

Mux,min

(KN-m)

MuD

Roof – 5 3050 26.1 102.83 64.85 2.68 64.85

5-4 2900 25.8 271.485 53.27 7 53.27

4-3 2900 25.8 440.14 53.27 11.36 53.27

3-2 2900 25.8 608.795 53.27 15.71 53.27

2-1 2900 25.8 777.45 53.27 20.06 53.27

1- Plinth 4900 29.8 946.11 53.27 28.19 53.27

Plinth –

Footing

- 29.8 982.185 0 29.27 29.27

By the help of SP16, chart no:32,

FromPu

f ck x b x Dand

MuD

f ck x b x D2 Plot theP

f ck

Where p is the percentage of steel reinforcement

Storey Pu

f ck x b x D

MuD

f ck x b x D2

P

f ck From chart

P (%)

Roof – 5 0.04 0.04 0.015 0.3

5-4 0.1 0.03 0.01 0.2

4-3 0.16 0.03 0.01 0.2

3-2 0.22 0.03

0.01

0.22-1 0.28 0.03 0.01 0.2

1- Plinth 0.34 0.03 0.01 0.2

Plinth – Footing 0.36 0.02 0.01 0.2

But from Is 456-2000, P48, Clause 26.5.3.1 (a)

The cross-sectional area of longitudinal reinforcement, shall be not less than

0.8% nor more than 6% of the gross cross-sectional area (b x D) of the column.



221

Therefore, provide P = 0.8%

Ast =0.8

100x 230 x 600 = 1104 mm2

No. of bars of 16mmφ bars =4 x 1104

π x (16)2 = 5.49 say 6 bars

Therefore provide 6 bars of 16mmφ bars.

Design of lateral ties:

1. Should not be less than 6mm

2. 1

4x diameter of max main steel bar =

1

4x 16 = 4mm

Maximum of two values =6mm.

But in the recent construction the usage of 6mm bars is outdated. Therefore consider

minimum 8mm dia bars.

Design of pitch:

1.

Least lateral dimension = 230mm2. 16 x diameter of main steel bar = 16 x 16 = 256

3. 300mm

Minimum of the above values is considered.

Therefore, provide 8mm φ bars @230mm c/c.

No. of bars:

Storey Ast (mm ) No.of bars:

Roof-5 1104 6 bars of 16mm φ

5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1104 6 bars of 16mm φ

2-1 1104 6 bars of 16mm φ

1-Plinth 1104 6 bars of 16mm φ

Plinth – footing 1104 6 bars of 16mm φ



222

For Column 10:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 3050 26.1 202.61 89.7 5.29 89.70

5-4 2900 25.8 583.35 81.6 15.05 81.60

4-3 2900 25.8 964.09 81.6 24.87 81.60

3-2 2900 25.8 1344.83 81.6 34.70 81.60

2-1 2900 25.8 1725.57 81.6 44.52 81.60

1- Plinth 4900 29.8 2106.31 81.6 62.77 81.60

Plinth –

Footing - 29.8 2136.1 0 63.66 63.66


FromPu

f ck x b x Dand

MuD


f ck


Storey Pu

f ck x b x D

MuD

f ck x b x D2

P

f ck From chart

P (%)

Roof – 5 0.07 0.05 0.02 0.4

5-4 0.21 0.05 0.02 0.4

4-3 0.35 0.05 0.02 0.4

3-2 0.49 0.05 0.04 0.82-1 0.63 0.05 0.08 1.6

1- Plinth 0.76 0.05 0.12 2.4

Plinth – Footing 0.77 0.04 0.12 2.4

But from Is 456-2000, P48, Clause 26.5.3.1 (a)

The cross-sectional area of longitudinal reinforcement, shall be not less than

0.8% nor more than 6% of the gross cross-sectional area (b x D) of the column.



223


Ast =0.8

100x 230 x 600 = 1104 mm2


π x (16)2 = 5.49 say 6 bars


For P = 1.6%

Ast =1.6

100x 230 x 600 = 2208 mm2


π x (20)2 = 7 bars

Therefore provide 8 bars of 20mmφ bars. (Each side 4bars)

For P = 2.4%

Ast =2.4

100x 230 x 600 = 3312 mm2


π x (25)2 = 6.75 bars say 8 bars




2. 1

4 x diameter of max main steel bar =1

4 x 25 = 6.25mm say 8mm


Design of pitch:

1. Least lateral dimension = 230mm

2. 16 x diameter of main steel bar = 16 x 16 = 256

3. 300mm



224



No. of bars:

Storey Ast (mm2) No.of bars:


5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1104 6 bars of 16mm φ

2-1 2208 8 bars of 20mm φ



For Column 01:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 3050 26.1 133.97 136.67 3.50 136.67

5-4 2900 25.8 383.39 122.6 9.89 122.60

4-3 2900 25.8 632.81 122.6 16.33 122.60

3-2 2900 25.8 882.23 122.6 22.76 122.60

2-1 2900 25.8 1131.65 122.6 29.20 122.60

1- Plinth 4900 29.8 1381.07 122.6 41.16 122.60

Plinth –

Footing

-

29.8 1430.66 0 42.63 42.63


FromPu

f ck x b x Dand

MuD


f ck




225

Storey Pu

f ck x b x D

MuD

f ck x b x D2

P

f ck From chart

P (%)

Roof – 5 0.05 0.08 0.02 0.4

5-4 0.14 0.07 0.02 0.4

4-3 0.23 0.07 0.02 0.4

3-2 0.32 0.07 0.04 0.8

2-1 0.41 0.07 0.04 0.8

1- Plinth 0.50 0.07 0.06 1.2

Plinth – Footing 0.52 0.03 0.06 1.2

Provide min P = 0.8%

Ast = 0.8100

x 230 x 600 = 1104 mm2


π x (16)2 = 5.49 say 6 bars

For P=1.2%

Ast =1.2

100x 230 x 600 = 1656 mm2


π x (20)2 = 5.27 say 6 bars




2. 14 x diameter of max main steel bar = 14 x 20 = 5mm




Design of pitch:




226


3. 300mm


Therefore, provide 8mm φ bars @230mm c/c

No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1104 6 bars of 16mm φ

2-1 1104 6 bars of 16mm φ





227

For frame 20-11-02:



Cover (d‟) = 50mm


500+

lateral dimension

30or

= 20mm



228

f ck x b x D = 20 x 230 x 600 = 2760000N = 2760 KN


For Column 20:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 2900 25.8 137.01 68.42 3.53 68.42

5-4 2750 25.5 315.59 37.77 8.05 37.77

4-3 2750 25.5 494.17 37.77 12.60 37.77

3-2 2750 25.5 672.75 37.77 17.16 37.77

2-1 2750 25.5 851.33 37.77 21.71 37.77

1- Plinth 4750 29.5 1029.91 37.77 30.38 37.77

Plinth –

Footing

-

29.5 1091.71 0 32.21 32.21

By the help of SP16, chart no: 32,

FromPu

f ck x b x Dand

MuD


f ck


Storey Pu

f ck x b x D

MuD

f ck x b x D2

P

f ck From chart

P (%)

Roof – 5 0.05 0.04 0.01 0.2

5-4 0.11 0.02 0.01 0.2

4-3 0.18 0.02 0.01 0.2

3-2 0.24 0.02 0.01 0.2

2-1 0.31 0.02 0.01 0.2

1- Plinth 0.37 0.02 0.01 0.2

Plinth – Footing 0.40 0.02 0.01 0.2



229


Ast =0.8

100x 230 x 600 = 1104 mm2


π x (16)2 = 5.49 say 6 bars




2. 1


1

4x 16 = 4mm




Design of pitch:

1.


3. 300mm



No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1104 6 bars of 16mm φ

2-1 1104 6 bars of 16mm φ





230

For Column 11:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 2900 25.8 279.26 93.54 7.20 93.54

5-4 2750 25.5 806.49 108.74 20.57 108.74

4-3 2750 25.5 1307.98 108.74 33.35 108.74

3-2 2750 25.5 1809.47 108.74 46.14 108.74

2-1 2750 25.5 2310.96 108.74 58.93 108.74

1- Plinth 4750 29.5 2812.45 108.74 82.97 108.74

Plinth –

Footing - 29.5 2861.68 0 84.42 84.42


FromPu

f ck x b x Dand

MuD


f ck


Storey Pu

f ck x b x D

MuD

f ck x b x D2

P

f ck From chart

P (%)

Roof – 5 0.10 0.06 0.02 0.4

5-4 0.29 0.07 0.02 0.4

4-3 0.47 0.07 0.06 1.2

3-2 0.66 0.07 0.1 22-1 0.84 0.07 0.16 3.2

1- Plinth 1.02 0.07 0.2 4

Plinth – Footing 1.04 0.05 0.2 4

For P less than 0.8, Provide P = 0.8%

Ast =0.8

100 x 230 x 600 = 1104 mm2



231


π x (16)2 = 5.49 say 6 bars


For P = 1.2%

Ast =1.2

100x 230 x 600 = 1656 mm2


π x (20)2 = 5.27, say 6 bars


For P = 2%

Ast =2

100x 230 x 600 = 2760 mm2




For P = 3.2%

Ast =3.2

100x 230 x 600 = 4416 mm2


π x (25)2 = 9 bars


For P = 4%

Ast =4

100x 230 x 600 = 5520 mm2


π x (25)2 = 11.24 say 12 bars




232



2. 1

4

x diameter of max main steel bar =1

4

x 25 = 6.25mm say 8mm


Design of pitch:


2. 16 x diameter of main steel bar = 16 x 25 = 400 mm

3. 300mm



No. of bars:



5-4 1104 6 bars of 16mm φ 4-3 1656 6 bars of 20mm φ

3-2 2760 6 bars of 25mm φ

2-1 4416 10 bars of 25mm φ





233

For Column 02:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 2900 25.8 180.53 155.8 4.66 155.80

5-4 2750 25.5 551.39 153.37 14.06 153.37

4-3 2750 25.5 922.25 153.37 23.52 153.37

3-2 2750 25.5 1293.11 153.37 32.97 153.37

2-1 2750 25.5 1663.97 153.37 42.43 153.37

1- Plinth 4750 29.5 2034.83 153.37 60.03 153.37

Plinth –

Footing

-

29.5 2123.66 0 62.65 62.65


FromPu

f ck x b x Dand

MuD


f ck


Storey Pu

f ck x b x D

MuD

f ck x b x D2

P

f ck From chart

P (%)

Roof – 5 0.07 0.09 0.04 0.8

5-4 0.20 0.09 0.04 0.8

4-3 0.33 0.09 0.06 1.2

3-2 0.47 0.09 0.08 1.6

2-1 0.60 0.09 0.1 2

1- Plinth 0.74 0.09 0.14 2.8

Plinth – Footing 0.77 0.04 0.16 3.2

For P = 0.8%

Ast =0.8

100x 230 x 600 = 1104 mm2



234


π x (16)2 = 5.49 say 6 bars


For P=1.2%

Ast =1.2

100x 230 x 600 = 1656 mm2


π x (20)2 = 5.27 say 6 bars


For P=1.6%

Ast =1.6

100x 230 x 600 = 2208 mm2


π x (25)2 = 4.5 say 6 bars


For P = 2%

Ast =2

100x 230 x 600 = 2760 mm2




For P = 2.8%

Ast =2.8

100x 230 x 600 = 3864 mm2






235

For P = 3.2%

Ast =3.2

100x 230 x 600 = 4416 mm2


π x (25)2 = 9 bars




2. 1


1

4x 25 = 6.25mm say 8mm


Design of pitch:


2. 16 x diameter of main steel bar = 16 x 25 = 400mm

3. 300mm



No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1656 6 bars of 20mm φ

3-2 2208 6 bars of 25mm φ

2-1 2760 6 bars of 25mm φ





236

For frame 24-15-06:



Cover (d‟) = 50mm


500+

lateral dimension

30or

= 20mm



237

f ck x b x D = 20 x 230 x 600 = 2760000N = 2760 KN


For Column 24:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 2900 25.8 137.01 68.42 3.53 68.42

5-4 2750 25.5 339.45 47.54 8.66 47.54

4-3 2750 25.5 541.89 47.54 13.82 47.54

3-2 2750 25.5 744.33 47.54 18.98 47.54

2-1 2750 25.5 946.77 47.54 24.14 47.54

1- Plinth 4750 29.5 1149.21 47.54 33.90 47.54

Plinth –

Footing

-

29.5 1211.01 0 35.72 35.72


FromPu

f ck x b x Dand

MuD


f ck


Storey Pu

f ck x b x D

MuD

f ck x b x D2

P

f ck From chart

P (%)

Roof – 5 0.05 0.04 0.01 0.2

5-4 0.12 0.03 0.01 0.2

4-3 0.20 0.03 0.01 0.2

3-2 0.27 0.03 0.01 0.2

2-1 0.34 0.03 0.01 0.2

1- Plinth 0.42 0.03 0.01 0.2

Plinth – Footing 0.44 0.02 0.01 0.2



238


Ast =0.8

100x 230 x 600 = 1104 mm2


π x (16)2 = 5.49 say 6 bars




2. 1


1

4x 16 = 4mm




Design of pitch:

1.


3. 300mm



No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1104 6 bars of 16mm φ

2-1 1104 6 bars of 16mm φ





239

For Column 15:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 2900 25.8 279.26 93.54 7.20 93.54

5-4 2750 25.5 702.9 63.86 17.92 63.86

4-3 2750 25.5 1126.54 63.86 28.73 63.86

3-2 2750 25.5 1550.18 63.86 39.53 63.86

2-1 2750 25.5 1973.82 63.86 50.33 63.86

1- Plinth 4750 29.5 2397.46 63.86 70.73 70.73

Plinth –

Footing - 29.5 2446.69 0 72.18 72.18


FromPu

f ck x b x Dand

MuD


f ck


Storey Pu

f ck x b x D

MuD

f ck x b x D2 P

f ck From chart

P (%)

Roof – 5 0.10 0.06 0.01 0.2

5-4 0.25 0.04 0.01 0.2

4-3 0.41 0.04 0.02 0.4

3-2 0.56 0.04 0.06 1.22-1 0.72 0.04 0.1 2

1- Plinth 0.87 0.04 0.16 3.2

Plinth – Footing 0.89 0.04 0.16 3.2


Ast =0.8

100 x 230 x 600 = 1104 mm2



240


π x (16)2 = 5.49 say 6 bars


For P = 1.2%

Ast =1.2

100x 230 x 600 = 1656 mm2


π x (20)2 = 5.27, say 6 bars


For P = 2%

Ast =2

100x 230 x 600 = 2760 mm2




For P = 3.2%

Ast =3.2

100x 230 x 600 = 4416 mm2


π x (25)2 = 9 bars




2. 1


1

4x 25 = 6.25mm say 8mm




241

Design of pitch:



3. 300mm



No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1656 6 bars of 20mm φ

2-1 2760 6 bars of 25mm φ



For Column 06:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 2900 25.8 180.53 155.8 4.66 155.80

5-4 2750 25.5 461.69 103.08 11.77 103.08

4-3 2750 25.5 742.85 103.08 18.94 103.08

3-2 2750 25.5 1024.01 103.08 26.11 103.08

2-1 2750 25.5 1305.17 103.08 33.28 103.08

1- Plinth 4750 29.5 1586.33 103.08 46.80 103.08

Plinth –

Footing

-

29.5 1675.16 0 49.42 49.42



242


FromPu

f ck x b x Dand

MuD


f ck


Storey Pu

f ck x b x D

MuD

f ck x b x D2

P

f ck From chart

P (%)

Roof – 5 0.07 0.09 0.04 0.8

5-4 0.17 0.06 0.02 0.4

4-3 0.27 0.06 0.02 0.4

3-2 0.37 0.06 0.02 0.4

2-1 0.47 0.06 0.04 0.8

1- Plinth 0.57 0.06 0.08 1.6

Plinth – Footing 0.61 0.03 0.08 1.6

For P = 0.8% and P <0.8%

Ast =0.8

100x 230 x 600 = 1104 mm2

No. of bars of 16mmφ bars = 4 x 1104 π x (16)2 = 5.49 say 6 bars


For P=1.6%

Ast =1.6

100x 230 x 600 = 2208 mm2


π x (25)2 = 4.5 say 6 bars




2. 1


1

4x 25 = 6.25mm say 8mm



243


Design of pitch:



3. 300mm



No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1104 6 bars of 16mm φ

2-1 1104 6 bars of 16mm φ





244

For frame 25-16-07:



Cover (d‟) = 50mm


500+

lateral dimension

30or

= 20mm





246

For P < 0.8%, Provide P = 0.8%

Ast =0.8

100x 230 x 600 = 1104 mm2


π x (16)2 = 5.49 say 6 bars




2. 1


1

4x 16 = 4mm




Design of pitch:

1.


3. 300mm



No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1104 6 bars of 16mm φ

2-1 1104 6 bars of 16mm φ





247

For Column 16:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 2900 25.8 279.26 68.42 7.20 68.42

5-4 2750 25.5 744.68 42.84 18.99 42.84

4-3 2750 25.5 1210.1 42.84 30.86 42.84

3-2 2750 25.5 1675.52 42.84 42.73 42.84

2-1 2750 25.5 2140.94 42.84 54.59 54.59

1- Plinth 4750 29.5 2606.36 42.84 76.89 76.89

Plinth –

Footing - 29.5 2655.59 0 78.34 78.34


FromPu

f ck x b x Dand

MuD


f ck


Storey Pu

f ck x b x D

MuD

f ck x b x D2

P

f ck From chart

P (%)

Roof – 5 0.10 0.04 0.01 0.2

5-4 0.27 0.03 0.01 0.2

4-3 0.44 0.03 0.02 0.4

3-2 0.61 0.03 0.08 1.62-1 0.78 0.03 0.12 2.4

1- Plinth 0.94 0.05 0.18 3.6

Plinth – Footing 0.96 0.05 0.18 3.6


Ast =0.8

100 x 230 x 600 = 1104 mm2



248


π x (16)2 = 5.49 say 6 bars


For P = 1.6%

Ast =1.6

100x 230 x 600 = 2208 mm2


π x (20)2 = 4.5 say 6 bars


For P = 2.4%

Ast =2.4

100x 230 x 600 = 3312 mm2




For P = 3.6%

Ast =3.6

100x 230 x 600 = 4968 mm2


π x (25)2 = 10.12, say 12 bars




2. 1


1

4x 25 = 6.25mm say 8mm




249

Design of pitch:



3. 300mm



No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 2208 6 bars of 25mm φ

2-1 3312 8 bars of 25mm φ



For Column 07:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 2900 25.8 180.53 155.8 4.66 155.80

5-4 2750 25.5 511.2 130.75 13.04 130.75

4-3 2750 25.5 841.87 130.75 21.47 130.75

3-2 2750 25.5 1172.54 130.75 29.90 130.75

2-1 2750 25.5 1503.21 130.75 38.33 130.75

1- Plinth 4750 29.5 1833.88 130.75 54.10 130.75

Plinth –

Footing

-

29.5 1922.71 0 56.72 56.72



250


FromPu

f ck x b x Dand

MuD


f ck


Storey Pu

f ck x b x D

MuD

f ck x b x D2

P

f ck From chart

P (%)

Roof – 5 0.07 0.09 0.04 0.8

5-4 0.19 0.08 0.02 0.4

4-3 0.31 0.08 0.04 0.8

3-2 0.42 0.08 0.06 1.2

2-1 0.54 0.08 0.08 1.6

1- Plinth 0.66 0.08 0.12 2.4

Plinth – Footing 0.70 0.03 0.12 2.4

For P < 0.8%, Provide P = 0.8%

Ast =0.8

100x 230 x 600 = 1104 mm2

No. of bars of 16mmφ bars = 4 x 1104 π x (16)2 = 5.49 say 6 bars


For P=1.2%

Ast =1.2

100x 230 x 600 = 1656 mm2


π x (20)2 = 5.27 say 6 bars


For P=1.6%

Ast =1.6

100x 230 x 600 = 2208 mm2



251


π x (25)2 = 4.5 say 6 bars


For P = 2.4%

Ast =2.4

100x 230 x 600 = 3312 mm2






2. 1


1

4x 25 = 6.25mm say 8mm


Design of pitch:



3. 300mm





252

No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1656 6 bars of 20mm φ

2-1 2208 6 bars of 25mm φ





253

For frame 26-17-08:



Cover (d‟) = 50mm


500+

lateral dimension

30or

= 20mm



254

f ck x b x D = 20 x 230 x 600 = 2760000N = 2760 KN


For Column 26:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 2900 25.8 137.01 68.42 3.53 68.42

5-4 2750 25.5 306.92 36.55 7.83 36.55

4-3 2750 25.5 476.83 36.55 12.16 36.55

3-2 2750 25.5 646.74 36.55 16.49 36.55

2-1 2750 25.5 816.65 36.55 20.82 36.55

1- Plinth 4750 29.5 986.56 36.55 29.10 36.55

Plinth –

Footing

-

29.5 1048.36 0 30.93 30.93


FromPu

f ck x b x Dand

MuD


f ck


Storey Pu

f ck x b x D

MuD

f ck x b x D2 P

f ck From chart

P (%)

Roof – 5 0.05 0.04 0.01 0.2

5-4 0.11 0.02 0.01 0.2

4-3 0.17 0.02 0.01 0.2

3-2 0.23 0.02 0.01 0.2

2-1 0.30 0.02 0.01 0.2

1- Plinth 0.36 0.02 0.01 0.2

Plinth – Footing 0.38 0.02 0.01 0.2



255

For P < 0.8%, provide P = 0.8%

Ast =0.8

100x 230 x 600 = 1104 mm2


π x (16)2 = 5.49 say 6 bars




2. 1


1

4x 16 = 4mm




Design of pitch:

1.


3. 300mm



No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1104 6 bars of 16mm φ

2-1 1104 6 bars of 16mm φ





256

For Column 17:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 2900 25.8 279.26 68.42 7.20 68.42

5-4 2750 25.5 685.14 36.55 17.47 36.55

4-3 2750 25.5 1091.02 36.55 27.82 36.55

3-2 2750 25.5 1496.9 36.55 38.17 38.17

2-1 2750 25.5 1902.78 36.55 48.52 48.52

1- Plinth 4750 29.5 2308.66 36.55 68.11 68.11

Plinth –

Footing - 29.5 2357.89 0 69.56 69.56


FromPu

f ck x b x Dand

MuD


f ck


Storey Pu

f ck x b x D

MuD

f ck x b x D2 P

f ck From chart

P (%)

Roof – 5 0.10 0.04 0.01 0.2

5-4 0.25 0.02 0.01 0.2

4-3 0.40 0.02 0.01 0.2

3-2 0.54 0.02 0.04 0.82-1 0.69 0.03 0.1 2

1- Plinth 0.84 0.04 0.14 2.8

Plinth – Footing 0.85 0.04 0.14 2.8


Ast =0.8

100 x 230 x 600 = 1104 mm2



257


π x (16)2 = 5.49 say 6 bars


For P = 2%

Ast =2

100x 230 x 600 = 2760 mm2




For P = 2.8%

Ast =2.8

100x 230 x 600 = 3864 mm2


π x (25)2 = 7.87, say 8 bars




2. 1


1

4x 25 = 6.25mm say 8mm


Design of pitch:



3. 300mm





258

No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1104 6 bars of 16mm φ

2-1 2760 6 bars of 25mm φ



For Column 08:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 2900 25.8 180.53 155.8 4.66 155.80

5-4 2750 25.5 480.91 113.98 12.26 113.98

4-3 2750 25.5 781.29 113.98 19.92 113.98

3-2 2750 25.5 1081.67 113.98 27.58 113.98

2-1 2750 25.5 1382.05 113.98 35.24 113.98

1- Plinth 4750 29.5 1682.43 113.98 49.63 113.98

Plinth –

Footing

-

29.5 1771.26 0 52.25 52.25


FromPu

f ck x b x Dand

MuD


f ck




259

Storey Pu

f ck x b x D

MuD

f ck x b x D2

P

f ck From chart

P (%)

Roof – 5 0.07 0.09 0.04 0.8

5-4 0.17 0.07 0.02 0.4

4-3 0.28 0.07 0.02 0.4

3-2 0.39 0.07 0.04 0.8

2-1 0.50 0.07 0.06 1.2

1- Plinth 0.61 0.07 0.1 2

Plinth – Footing 0.64 0.03 0.1 2

For P = 0.8%

Ast = 0.8100

x 230 x 600 = 1104 mm2


π x (16)2 = 5.49 say 6 bars


For P=1.2%

Ast = 1.2100

x 230 x 600 = 1656 mm2


π x (20)2 = 5.27 say 6 bars


For P = 2%

Ast =2

100x 230 x 600 = 2760 mm2








260

2. 1


1

4x 25 = 6.25mm say 8mm


Design of pitch:



3. 300mm



No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1104 6 bars of 16mm φ

2-1 1656 6 bars of 20mm φ





261

For frame 27-18-09:



Cover (d‟) = 50mm


500+

lateral dimension

30or

= 20mm



262

f ck x b x D = 20 x 230 x 600 = 2760000N = 2760 KN


For Column 27:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 3050 26.1 102.83 64.85 2.68 64.85

5-4 2900 25.8 265.825 50.61 6.86 50.61

4-3 2900 25.8 428.82 50.61 11.06 50.61

3-2 2900 25.8 591.815 50.61 15.27 50.61

2-1 2900 25.8 754.81 50.61 19.47 50.61

1- Plinth 4900 29.8 917.805 50.61 27.35 50.61

Plinth –

Footing

-

29.8 953.88 0 28.43 28.43


FromPu

f ck x b x Dand

MuD


f ck


Storey Pu

f ck x b x D

MuD

f ck x b x D2

P

f ck From chart

P (%)

Roof – 5 0.04 0.04 0.01 0.2

5-4 0.10 0.03 0.01 0.2

4-3 0.16 0.03 0.01 0.2

3-2 0.21 0.03 0.01 0.2

2-1 0.27 0.03 0.01 0.2

1- Plinth 0.33 0.03 0.01 0.2

Plinth – Footing 0.35 0.02 0.01 0.2



263

For P < 0.8, provide P = 0.8%

Ast =0.8

100x 230 x 600 = 1104 mm2


π x (16)2 = 5.49 say 6 bars




2. 1


1

4x 16 = 4mm




Design of pitch:

1.


3. 300mm



No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1104 6 bars of 16mm φ

2-1 1104 6 bars of 16mm φ





264

For Column 18:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 3050 26.1 202.61 89.7 5.29 89.70

5-4 2900 25.8 586.58 86.44 15.13 86.44

4-3 2900 25.8 970.55 86.44 25.04 86.44

3-2 2900 25.8 1354.52 86.44 34.95 86.44

2-1 2900 25.8 1738.49 86.44 44.85 86.44

1- Plinth 4900 29.8 2122.46 86.44 63.25 86.44

Plinth –

Footing - 29.8 2152.25 0 64.14 64.14


FromPu

f ck x b x Dand

MuD


f ck


Storey Pu

f ck x b x D

MuD

f ck x b x D2

P

f ck From chart

P (%)

Roof – 5 0.07 0.05 0.01 0.2

5-4 0.21 0.05 0.01 0.2

4-3 0.35 0.05 0.02 0.4

3-2 0.49 0.05 0.04 0.82-1 0.63 0.05 0.08 1.6

1- Plinth 0.77 0.05 0.14 2.8

Plinth – Footing 0.78 0.04 0.14 2.8


Ast =0.8

100 x 230 x 600 = 1104 mm2



265


π x (16)2 = 5.49 say 6 bars


For P = 1.6%

Ast =1.6

100x 230 x 600 = 2208 mm2


π x (20)2 = 8 bars


For P = 2.8%

Ast =2.8

100x 230 x 600 = 3864 mm2






2. 1


1

4x 25 = 6.25mm say 8mm


Design of pitch:



3. 300mm





266

No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1104 6 bars of 16mm φ

2-1 2208 8 bars of 20mm φ



For Column 09:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 3050 26.1 133.97 136.67 3.50 136.67

5-4 2900 25.8 390.87 128.22 10.08 128.22

4-3 2900 25.8 647.77 128.22 16.71 128.22

3-2 2900 25.8 904.67 128.22 23.34 128.22

2-1 2900 25.8 1161.57 128.22 29.97 128.22

1- Plinth 4900 29.8 1418.47 128.22 42.27 128.22

Plinth –

Footing

-

29.8 1468.06 0 43.75 43.75


FromPu

f ck x b x Dand

MuD


f ck




267

Storey Pu

f ck x b x D

MuD

f ck x b x D2

P

f ck From chart

P (%)

Roof – 5 0.05 0.08 0.04 0.8

5-4 0.14 0.08 0.02 0.4

4-3 0.23 0.08 0.02 0.4

3-2 0.33 0.08 0.04 0.8

2-1 0.42 0.08 0.04 0.8

1- Plinth 0.51 0.08 0.06 1.2

Plinth – Footing 0.53 0.03 0.06 1.2

Provide min P = 0.8%

Ast = 0.8100

x 230 x 600 = 1104 mm2


π x (16)2 = 5.49 say 6 bars

For P=1.2%

Ast =1.2

100x 230 x 600 = 1656 mm2


π x (20)2 = 5.27 say 6 bars




2. 14 x diameter of max main steel bar = 14 x 20 = 5mm




Design of pitch:




268


3. 300mm



No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1104 6 bars of 16mm φ

2-1 1104 6 bars of 16mm φ





269

6. DESIGN OF FOOTINGS

In this design of footing, the loads are known from the column analysis. The working load is

used for the footing design than the ultimate load. Footing is a member through which the

load of the superstructure is transferred to the sub soil. Therefore the safe bearing capacity is

the main factor in design of footings. From the information provided and conducting test on

sub soil, it is decided that the Safe Bearing Capacity (SBC) of the soil is 250 KN/m2.

Therefore the footing is isolated rectangular sloped footing. The slope is provided to decrease

the amount of concrete in the construction which results into an economic construction.

From the analysis of frames and columns, we considered 6 frames and each frame

consists of 3 footings so totally 18 footings are to be designed. But in this project we designthe most critical any 3 footings based on the load carried are selected and the design of

remaining footings follow the same.

Footing F19:

Steel : Fe415

Concrete : M20

Column : 230mm x 600mm (bmm x Dmm)

Load from column (Pu) : 982.185 KN

Working load (P) :982.185

1.5= 654.79 KN

Self weight of footing : 10% of Pu

SBC of soil : 250 KN/m2

Area of footing (Af ) :1.1 x P

SBC of soil=

1.1 x 654.79

250= 2.88 m2

Consider,D-b

2=

600 - 230

2= 185 mm

Length of footing (Lf ) = 185mm + (√(185)2 + 2.88 x 106 ) = 1892.11

say 1900mm



270

Breadth of footing (Bf ) =Af

Lf =

2.88 x 106

1900= 1515.79 say 1550 mm

Area of footing provided = Lf x Bf = 2000 x 1520 = 2.95 x 106 = 2.95 m2

Wu =Pu

Af =

982.185

2.95= 332.95 KN/m2

X1 =

Lf - D

2 =

1900 - 600

2 = 650 mm

Y1 =Bf - b

2=

1550 - 230

2= 660 mm

Depth of footing for 2-way bending:

Mux =Wu x Bf x X1

2

2=

332.94 x 1.550 x (0.65)2

2= 109.02 KN-m

Muy =Wu x Lf x Y1

2

2=

332.94 x 1.9 x (0.66)2

2= 137.78 KN-m

Therefore, max bending moment = Mu = 137.78 KN-m



271

b' = b + 2 x e = 230 + 2 x 50 = 330 mm

D' = D + 2 x e = 600 + 2 x 50 = 700 mm

Clear cover = 50mm

Assume 10mm bars

d'x = 50 +10

2= 55mm

d'y = 50 + (3 x102

)= 65mm



272

Depth of foundation:

(Df )x =

+ d'x = √109.02 x 106

0.138 x 20 x 330+ 55 = 400.97 mm

(Df )y =

+ d'y = √ 137.78 x 10

6

0.138 x 20 x 700 + 65 = 332.05 mm

Therefore, maximum (Df ) = 400.97 mm say 450mm

Therefore dx = 450 – 55 = 395mm

dy = 450 – 65 = 385mm

Provide Df,min = 150mm

df,min ( x-direction) = 150 - 55 = 95mm

df,min ( y-direction) = 150 - 65 = 85mm



273



274

Check for two way shear:

Consider a section „dy

2 ‟ from the face of the column.

L2 = D +dy

2+

dy

2= 600 + 385 = 985 mm

B2 = b +dy

2+

dy

2= 230 + 385 = 615 mm

d2 = Df – (Df – Df,min) x

(dy

2- 50)

(X1 - 50) – d'y

= 450 – (450 – 150) x

(385

2- 50)

(650 - 50) – 65 = 313.75 mm

Area of resisting shear (A2) = 2 x(L2 + B2) x d2

= 2 x (985 + 615) x 313.75

= 1004000 mm2

Shear strength of concrete for two way:

τuc2 = k s x τuc

where, τuc = 0.25 x √f ck = 0.25 x √20 = 1.12 N/mm2

k s = (0.5 + βc) < 1 else =1

βc =

b

D =

230

600 = 0.385

therefore, k s = (0.5 + βc) = 0.5 + 0.385 = 0.885 <1

k s = 0.885

τuc2 = k s x τuc = 0.885 x 1.12 = 0.99 N/mm2

Internal load carrying capacity (Vuc2) = 1004000 x 0.99 = 993960 N

= 993.96 KN



275

External load (VuD2) = Wu x (Lf x Bf – L2 x B2)

= 332.94 x (1.900 x 1.550 – 0.985 x 0.615)

= 778.82 KN

Therefore, Vuc2 > VuD2 (SAFE)

Design of Steel mesh:

Mux = 0.87 x f y x Astx x dx x (1-f y x Astx

f ck x b' x dx)

109.02 x 106 = 0.87 x 415 x Astx x 395 x (1-415 x Astx

20 x 330 x 395

)

Astx = 890.72 mm2

No. of 10mmφ bars =4 x 890.74

π x 102 = 11.34 say 12 bars

Muy = 0.87 x f y x Asty x dy x (1-f y x Astx

f ck x D' x dy)

137.78 x 106 = 0.87 x 415 x Astx x 385 x (1- 415 x Astx

20 x 700 x 385)

Asty = 1081.2 mm2


π x 102 = 13.77 say 14 bars

Check for one way shear:

Consider a section at distance„d‟ from the face of column

d1 = Df – (Df – Df,min) x(dy- 50)

(Y1 - 50) – d'y

= 450 – (450 – 150) x(385- 50)

(660 - 50) – 65 = 220.24 mm

L1 = D + 2 x dy = 600 + 2 x 385 = 1370 mm



276

Area resisting shear (A1) = (Lf x Dy,min) + (d1 – Dy,min) x (Lf + L1

2)

= (1900 x 150) + (220.24 – 150) x (1900 + 1370

2)

= 399842.4 mm2

Asty = 14 bars of 10mm dia = 1099.56 mm2

Pt =100 x Asty

A1=

100 x 1099.56

399842.4= 0.28

From IS456-2000, P73, Table 19

P τc (N/mm2)

0.25 0.36

0.28 ?

0.50 0.48

τc = 0.36 +(0.48 - 0.36)

(0.5 - 0.25)x (0.28 – 0.25) = 0.38 N/mm2

Internal shear resisting capacity =0.38 x 399842.4

1000= 151.94 KN

External shear = Wu x Lf x (Y1 – dy)

= 332.94 x 1.900 x (0.660 – 385)

= 173.96 KN

Internal shear resisting capacity < External Shear (NOT SAFE)

Increase the depth.

Shear =load

area

0.38 =173.96 x 103

A1



277

A1 =457789.47 mm2

To calculate d1:

A1 = (Lf x Dy,min) + (d1 – Dy,min) x (Lf + L1

2 )

457789.47 = (1900 x 150) + (d1 – 150) x (1900 + 1370

2)

d1 = 255.68 mm

Footing F11:



1.5= 1907.79 KN




SBC of soil=

1.1 x 1907.79

250= 8.4 m2

Consider,D-b

2=

600 - 230

2= 185 mm


say 3100mm


Lf =

8.4 x 106

3100= 2709.67 say 2725 mm


Wu =Pu

Af =

2861.68

8.45= 338.66 KN/m2

X1 =Lf - D

2=

3100 - 600

2= 1250 mm

Y1 = Bf - b2

= 2725 - 2302

= 1247.5 mm



278


Mux =Wu x Bf x X1

2

2=

338.66 x 2.725 x (1.25)2

2= 720.98 KN-m

Muy =Wu x Lf x Y1

2

2=

338.66 x 3.1 x (1.2475)2

2= 816.92 KN-m


b' = b + 2 x e = 230 + 2 x 50 = 330 mm

D' = D + 2 x e = 600 + 2 x 50 = 700 mm

Clear cover = 50mm

Assume 16mm bars

d'x = 50 +16

2= 58 mm

d'y = 50 + (3 x16

2) = 74 mm


(Df )x =

+ d'x = √720.98 x 106

0.138 x 20 x 330+ 58 = 947.71 mm

(Df )y =

+ d'y = √816.92 x 106

0.138 x 20 x 700+ 74 = 724.26 mm


Therefore dx = 1000 – 58 = 942 mm

dy = 1000 – 74 = 926 mm


df,min ( x-direction) = 150 - 58 = 92 mm

df,min ( y-direction) = 150 - 74 = 76 mm



279




L2 = D +dy

2+

dy

2= 600 + 926 = 1526 mm

B2 = b +dy

2+

dy

2= 230 + 926 = 1156 mm


(dy

2- 50)

(X1 - 50) – d'y

= 1000 – (1000 – 150) x

(926

2- 50)

(1250 - 50) – 74 = 633.45 mm


= 2 x (1526 + 1156) x 633.45

= 3.4 x 106 mm2


τuc2 = k s x τuc


k s = (0.5 + βc) < 1 else =1

βc =

b

D =

230

600 = 0.385

Therefore, k s = (0.5 + βc) = 0.5 + 0.385 = 0.885 <1

k s = 0.885

τuc2 = k s x τuc = 0.885 x 1.12 = 0.99 N/mm2

Internal load carrying capacity (Vuc2) = 3.4 x 106 x 0.99 = 3.37 x 106 N

= 3370 KN



280


= 338.66 x (3.1 x 2.725 – 1.526 x 1.156)

= 2263.41 KN




f ck x b' x dx)

720.98 x 106 = 0.87 x 415 x Astx x 942 x (1-415 x Astx

20 x 330 x 942

)

Astx = 2555.91 mm2


π x 162 = 12.71 say 13 bars


f ck x D' x dy)

816.92 x 106 = 0.87 x 415 x Astx x 926 x (1- 415 x Astx

20 x 700 x 926)

Asty = 2671.98 mm2


π x 162 = 13.29 say 14 bars



d1 = Df – (Df – Df,min) x(dy- 50)

(Y1 - 50) – d'y

= 1000 – (1000 – 150) x(926- 50)

(1247.5 - 50) – 74 = 304.21 mm

L1 = D + 2 x dy = 600 + 2 x 926 = 2452 mm



281


2)

= (3100 x 150) + (304.21 – 150) x (3100 + 2452

2)

= 893086 mm2


Pt =100 x Asty

A1=

100 x 2814.87

893086= 0.32


P τc (N/mm2)

0.25 0.36

0.32 ?

0.5 0.48

τc = 0.36 +(0.48 - 0.36)

(0.5 - 0.25)x (0.32 – 0.25) = 0.4 N/mm2

Internal shear resisting capacity =0.4 x 893086

1000= 357.23 KN


= 338.66 x 3.1 x (1.2475 – 0.926)

= 337.53 KN

Internal shear resisting capacity > External Shear (SAFE)

Footing F07:



1.5= 1281.81 KN




282



SBC of soil=

1.1 x 1281.81

250= 5.64 m2

Consider,D-b

2=

600 - 230

2= 185 mm


say 2575mm


Lf =

5.64 x 106

2575= 2190.29 say 2200 mm


Wu =Pu

Af =

1922.71

5.665= 339.4 KN/m2

X1 =Lf - D

2=

2575 - 600

2= 987.5 mm

Y1 =Bf - b

2=

2200 - 230

2= 985 mm


Mux =Wu x Bf x X1

2

2=

339.4 x 2.2 x (0.9875)2

2= 364.06 KN-m

Muy =Wu x Lf x Y1

2

2=

339.4 x 2.575 x (0.985)2

2= 423.97 KN-m


b' = b + 2 x e = 230 + 2 x 50 = 330 mm

D' = D + 2 x e = 600 + 2 x 50 = 700 mm

Clear cover = 50mm

Assume 12mm bars

d'x = 50 + 122

= 56 mm



283

d'y = 50 + (3 x12

2) = 68 mm


(Df )x =

+ d'x = √364.06 x 106

0.138 x 20 x 330+ 56 = 688.23 mm

(Df )y =

+ d'y = √423.97 x 106

0.138 x 20 x 700+ 68 = 536.45 mm


Therefore dx = 700 – 56 = 644 mm

dy = 700 – 68 = 632 mm


df,min ( x-direction) = 150 - 56 = 94 mm

df,min ( y-direction) = 150 - 68 = 82 mm




L2 = D +dy

2+

dy

2= 600 + 632 = 1232 mm

B2 = b +dy

2+

dy

2= 230 + 632 = 862 mm


(dy

2- 50)

(X1 - 50) – d'y

= 700 – (700 – 150) x

(632

2- 50)

(987.5 - 50) – 68 = 475.95 mm



284


= 2 x (1232 + 862) x 475.95

= 1993278.6 mm2


τuc2 = k s x τuc


k s = (0.5 + βc) < 1 else =1

βc = bD

= 230600

= 0.385

therefore, k s = (0.5 + βc) = 0.5 + 0.385 = 0.885 <1

k s = 0.885

τuc2 = k s x τuc = 0.885 x 1.12 = 0.99 N/mm2

Internal load carrying capacity (Vuc2) = 1993278.6x 0.99 = 1973345.814N

= 1973.35 KN


= 339.4 x (2.575 x 2.2 – 1.232 x 0.862)

= 1562.27 KN




f ck x b' x dx)

364.06 x 106 = 0.87 x 415 x Astx x 644 x (1-415 x Astx

20 x 330 x 644)

Astx = 1929.08 mm2



285


π x 122 = 17.1 say 18 bars


f ck x D' x dy)

423.06 x 106 = 0.87 x 415 x Astx x 632 x (1-415 x Astx

20 x 700 x 632)

Asty = 2051.42 mm2


π x 122 = 18.13 say 19 bars



d1 = Df – (Df – Df,min) x(dy- 50)

(Y1 - 50) – d'y

= 700 – (700 – 150) x(632- 50)

(985 - 50) – 68 = 289.65 mm

L1 = D + 2 x dy = 600 + 2 x 632 = 1864 mm


2)

= (2575 x 150) + (289.65 – 150) x (2575 + 1864

2)

= 696203.175 mm2


Pt =100 x Asty

A1=

100 x 2148.84

696203.175= 0.31



286


P τc (N/mm2)

0.25 0.36

0.31 ?

0.5 0.48

τc = 0.36 +(0.48 - 0.36)

(0.5 - 0.25)x (0.31 – 0.25) = 0.39 N/mm2

Internal shear resisting capacity =0.39 x 696203.175

1000

= 271.52 KN


= 339.4 x 2.575 x (0.985 – 0.632)

= 308.51 KN

Internal shear resisting capacity < External Shear (NOT SAFE)

Increase the depth.

Shear =load

area

0.39 =308.51 x 103

A1

A1 =791051.28 mm2

To calculate d1:

A1 = (Lf x Dy,min) + (d1 – Dy,min) x (Lf + L1

2)

791051.28 = (2575 x 150) + (d1 – 150) x (2575 +1864

2)

d1 = 332.38 mm



287

7. DESIGN OF STAIR CASE

Stairs consist of steps arranged in a series for purpose of giving access to different floors of a

building. Since a stair is often the only means of communication between the various floors

of a building, the location of the stair requires good and careful consideration. In a residential

house, the staircase may be provided near the main entrance. In a public building, the stairs

must be from the main entrance itself and located centrally, to provide quick accessibility to

the principal apartments. All staircases should be adequately lighted and properly ventilated.

Various types of Staircases

Straight stairs

Dog-legged stairsOpen newel stair

Geometrical stair

RCC design of a Dog-legged staircase

In this type of staircase, the succeeding flights rise in opposite directions. The two

flights in plan are not separated by a well. A landing is provided corresponding to the level at

which the direction of the flight changes.

Dimensions:

B x L x H = 2820mm x 6000mm x 3350mm

Assume, Rise = 150mm

Tread = 300mm

sec θ =√(150)2+(300)2

300= 1.12

No. of risers =3350

150 = 22.33 say 22



288

Provide 11 + 11.

For flight-1: 11 and for Flight-2: 11

Going = 11 x treads = 11 x 300 = 3300mm

Total width of landings = 6000 – 3300 = 2700 mm

Therefore, width of landing at each end = 1350 mm



289

Design of Flight -1:

Type: One way single span simply supported inclined slab

Span (L): 3300 + 1350 = 4650mm

Trail Depth:

From IS456-2000, P39, Clause 24 & Clause 23.2 for Simply supported span we have

Span

Effective depth=

L

d= 20

r a = 20 x 1.4 = 28

Therefore, D =4750

28+ 20 = 189.65mm say 200 mm

Effective depth (d) = D – d1

= 200 – 20 = 180mm



290

Loads:

Self weight of slab (inclined) = 25 x D x secθ

= 25 x 0.2 x 1.12

= 5.6 KN/m2

Weight of steps =25 x rise x secθ

2=

25 x 0.15 x 1.1

2

= 2.09 KN/m2

Live load = 5 KN/m2

Floor Finish = 1 KN/m2

Total Load (W) = 13.69 KN/m2

Ultimate load (Wu) = 1.5 x 13.69 = 20.54 KN/m2

Design moment (Mu) =Wu x L2

10=

20.54 x (4.65)2

10= 44.41 KN-m

Mu,limit = 0.138 x f ck x b x d2 = 0.138 x 20 x 1000 x (180)2

= 89.42 x 106 N-mm = 89.42 KN-m

Mu < Mu,limit

Main steel:



f ck x b x d)

44.7 x 106 = 0.87 x 415 x Ast x 180 x (1 -415 x Ast

20 x 1000 x 180)

Ast = 753.21 mm2

Spacing of 12mm φ bars =ast x 1000

Ast=

π4

x 122 x1000

753.21= 150.15mm



291


Distribution steel:

Ast = 0.12% of Ag

=0.12

1000x 200 x 1000 = 240 mm2.


x 82 x1000

240= 209.44mm


Design of flight-2:

Same as design of flight-1.



292

8. CONCLUSION

The complete details and the necessity of the Multi-Storeyed building have been

explained in the introduction part. Live loads and dead loads are taken into consideration and

analysis is done manually. Manual analysis comprises of load distribution of slabs on to

beams and calculation of bending moment and shear force by any approximate method.

Analysis of frames is done by the substitute frame method and is furnished in the

introduction part. With the help of moment distribution method the moments carried at each

joints are known.

The design method adopted is limit state method. The bending moments obtained from

chapter 4 are used in calculating the area of steel in each frame section. The percentage of

steel obtained by limit state method is minimum as mentioned in limit state method of IS

456-2000.

Design of columns and footings are done depending on the load acting on them.

This project explains the basic concept behind the software. With the help of any softwarewe can do this analysis within no time but the basic thing is to know the concept behind it.

The same result can be achieved using the software like “Staad-Pro”. The wind load and

seismic analysis can be included for the future extension of the project.



Documents

9. Final Project Report