IRJET- A Technical Approach to Flat Slab Multistorey Building under Wind Speed of 39 M/S

International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 06 Issue: 05 | May 2019 www.irjet.net p-ISSN: 2395-0072
© 2019, IRJET | Impact Factor value: 7.211 | ISO 9001:2008 Certified Journal | Page 7629
A Technical Approach to Flat Slab Multistorey Building under Wind
Speed of 39 m/s
Mariyam1, Sagar jamle2
1 M. Tech. Scholar, Department of Civil Engineering, Oriental University, Indore (M.P.), India.
2Assistant Professor, Department of Civil Engineering, Oriental University, Indore (M.P.), India
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Abstract - In this scenario, for multistorey building and
skyscraper building, beam slab structure isnotused inIndia, it
is replaced by flat slab construction. The flat slab is a
reinforced concrete slab whichisdirectlysupportedoncolumn
so for aesthetic purpose, it is decent as well as it is efficient.
Flat slab is more flexible as compared to R.C. slab so it’s
advantages are more to design the flat slab. The flat slab has
many advantages but the main problem is that the flat slab is
weak against lateral loading such as wind and earthquake
loading and with the help of equivalent frame method;theflat
slab design is performed. In this work, taking the G+ 20 model
building plan, which is rectangular in shape 36m x 44m in
plan and this plan, is made with help of AutoCAD software.
After fixing the plan, it has divided into different panels and
each panels is designed by manual approach using equivalent
frame method. This manual data inserted in Staad pro and
analyze with providing shear wall at two different locations
i.e. lift area and maximum stress in plate area of the building
to minimize the same.
Key Words: Column Stresses, Displacement, Equivalent
frame method, Flat slab, Shear wall, Wind load.
1.INTRODUCTION
A concrete slabs are a common structural element which is
used generally in modern structural buildings. These slabs
are horizontal and it is generally made up of concrete or
steel typically between 100 and 500mm thick as per
requirement, are most often used floors and ceiling.Thetwo
types of slab are basically used in present time in structural
building-
 Common type slab
 Flat slab

The common type of slab is supported by beamsandbeamis
attached with the columns, this types of construction called
as simple beam slab construction. The slab which is directly
supported by columns known as flat slab. Flat slab is a
reinforced concrete slab supported by column, it may be
added or not added drops or the column may be added
column heads or without column heads. Drop is a local
thickening of the slab in the region of the column. In the
current scenario flat slab is used instead of beam column
construction because of its advantages over beam column
connections. In architectural point of view, flat slab are
better, also it permits flexibility in building construction. It
takes clear space, low height, easy framework and takes less
time therefore flat slab buildings are used now-a-days in
India. Flat slab structures are weak against lateral loading
such as seismic loading and wind loading so that the design
and analysis of flat slab is very important. Therefore
analysing the different types of flat slab, provided shearwall
at various points in different types of flat slab under wind
load condition using software Staad pro. In present time flat
slab buildings are used in high rise buildings because of its
advantages as it reduces time,costeffective,easyinstallation
and required the least storey height. To increase the
performance of buildings wind load behaviour of building
should be properly checked.
Fig -1: Flat Slab with Column Head
Mainly there are four types of flat slabs-
1. Simple flat slab
2. Flat slab added drop
3. Flat slab added column head
4. Flat slab added drop and column head
1. Simple flat slab - This type of flat slab having no dropand
no column head so that this type of flat slab is known as
simple flat slab. This type of flat slab is used in residential

building that reduces the available net clear ceiling height.
Hence in warehouses, offices and public halls sometimes
beams are avoided and slabs are directly supported by
column are called flat slab.
2. Flat slab added drop - Drops are providedtoincreasethe
shear strength of slab. In flat slab bending moments are
generated more near to the column, so that provided
thickness to the slab near to the column by providing the
drops. Sometimes the drops are known as the capital of the
column.
3. Flat slab added column capital - The column capital is
provided sometimes widened because to reduce the
punching shear in the slab. The column head is provided in
any angle for architectural purpose but for designpurposeit
is provided on 45 degree from vertical. Therefore in
multistorey buildings, to reduce the punching shear column
head is provided in the slab.
4. Flat slab added drop and column capital - Both are the
combinations are the best for the design of flat slab because
of the advantages of drops and column heads. This type of
flat slab has high strength in shear. It is provided stiffening
to the slab so that it reduces deflection.
2. OBJECTIVES
The main purpose is to find the economical model case to
counteract wind forces and analysis is done using software
Staad pro. So for this, different loads applied and parametric
values obtained are considered and point of comparison on
different models is as follows:
1. To find maximum Nodal Displacement in X Direction
and Z Direction.
2. To show the maximum Axial ForceinColumnatGround
Level.
3. To compare maximum Shear ForceinColumnSyandSz
for all model cases.
4. To relate maximum Compressive Stress in Column.
5. To find and observe maximum Tensile Stress in
Column.
6. To show and relate maximum Torsional Moment in
Column for all model cases.
7. To obtain economical model among all model cases by
observing and comparing their parametric values.
3. STRUCTURE CONFIGURATION AND
METHODOLOGY
In this paper, taking G+20 model buildingwithoverall height
of 80.01m with plan area (36mx44m) for four model cases.
For this, the foundation depth is 3m and total height of each
storey is 3.81m. Four different model cases are selected and
modelled in Staad pro under basic windspeedof39m/swith
reference to Indian Standard code IS 875 Part 3. The main
aim is to design the flat slab so for this, firstly the whole plan
is differentiated into different panels and each panels are
design by manuallyusingEquivalentFrameMethodanddata
obtained is provided to Staad pro for the detailed analysis of
the structure. All panels are designed on the basis of:
 Roof
 Exterior wall
 Interior wall
The data selected such as Grade of concrete M35, Grade of
steel Fe 415 is selected. The bar diameter selected as 12 mm
with a Clear cover of 25 mm throughout the structure. Unit
wt. of brick taken as 20 KN/m3, height of floor selected as
3.81m for all the subsequent levels. Thickness of external
wall and internal wall are 0.228m and 0.15m respectively
with plaster thickness of 0.24m with 20KN/m3 unit weight.
Also, parapet height of 0.75 m is used. 10 mm mortar unit
weight 0.42 KN/m3 for ceiling and 10 mm thick terrazzo
flooring with weight of 0.24 KN/m2 is selected. Column size
selected as 500 mm x 400 mm by hit and trial method. For
load consideration, live load for floor and roof are 3.5KN/m2
and 1.6KN/m2.
DESIGN OF FLAT SLAB FOR PANEL SIZE 6X8
Step1- Thickness of Flat Slab-
Equivalent Frame M/D = Modification Factor (M.F) = 33.8
Overall depth (D) = Span/Ratio = 8000/33.8 = 237 mm
D Approx. = 294 mm
Let Effective Depth (d) = D - (Dia. of Bar / 2) - Clear
Cover = 294 - (12/2) - 25
In Longer Direction (dl) = 263mm or .263m
In Shorter direction (ds) = Dl – Dia. of Bar = 263 - 12
ds = 251mm or .251m
Step 2 - Load Calculation
1 - Dead Load
A - Self load of slab = D x unit weight of concrete =.294 x 25
= 7.4 KN/m2
B - Plate area load
1) Parapet wall load
PWL = (thickness x height x unit weight of brick) /plate area
PWL = [(.228 x 20 + .024 x 20) x .75] / (6 x 8) = .078
KN/m2
C- for 10 mm mortar both side of roof and floor = .42
KN/m2
D- Terrazzo floor tiles load 10 mm thick = 0.24 KN/m2
Total dead load
For roof level dead load = 7.4 + .078 + .42 + 0.24 = 8.1
KN/m2
2 - Live load-
For roof = 1.6 KN/m2
Total load-

For roof level = 8.1 + 1.6 = 9.7 KN/m2
Total Factored Load -
For roof level = 1.5 x 9.7 = 14.6 KN/m2
Step 3 - Calculation of stiffness and alpha c (αc)
Along longer direction
For slab
Ks = (4 x E x I)/LL = (4 x E x 12665474635) /8000 =
6332737 x E
∑ks = 2 x 6332737 = 12665475
For column
Kc = (4 x E x I)/CH = (4 x E x 4166666667) / 3810 =
437474453 x E
∑kc = 2 x 437474453 = 8748906
Then, αC = ∑kc / ∑ks = (8748906 / 12665475 x E) = .7
Along shorter direction
A. For slab
Ks = (4 x E x I) / LL = ( 4 x E x 16887299514) / 6000 =
11258200 E
∑ks = 2 x 11258200 E = 22516399
B. For column
Kc = (4 x E x I) / CH = (4 x E x 2666666667) / 3810 =
2799650 E
∑kc = 2 x 2799650 = 5599300
Then, αC = ∑kc / ∑ks = 5599300 / 22516399 = .25
Step-4 Check for correction due to pattern loading
If ratio of Live Load and Dead Load is greater than 0.5, then
pattern loading required
Live Load / Dead Load < = .5
At roof level = live load / dead load = 1.6 / 8.1 = .2 (not
Required)
Step-4 Check for correction due to pattern loading
If ratio of Live Load and Dead Load is greater than 0.5, then
pattern loading required
Live Load / Dead Load < = .5
At roof level = live load / dead load = 1.6 / 8.1 = .2 (not
required)
Step- 5 Total moment calculation
In longer direction
Ln = 7.5 M L2 = 6 M Ln2 = 56.25 m
Mo = (W x Ln x L2) / 8 or (w x L2 x Ln2) / 8 = (14.6 x 6 x
56.25) / 8 = 613
In shorter direction
Ln = 5.6 m L1 = 8 m Ln2 = 31.36 m
Mo = (W x Ln x L1) / 8 or (w x L1 x Ln2) / 8 = (14.6 x 6 x
31.36) / 8 = 456
Step-6 Column strip and middle strips
In longer direction
Column strips
A- 2(.25 x L2) = 2(.25 x 6000) = 3000 mm
B- 2(.25 x L1) = 2(.25 x 8000) = 4000 mm
Lesser value will be taken (a or b) column strip = 3000 mm
Middle strips = L2 - column strips = 6000 - 3000 = 3000 mm
In shorter direction
Column strips
A- 2 (.25 x L1) = 2(.25 x 8000) = 4000 mm
B- 2 (.25 x L2) = 2(.25 x 6000) = 3000 mm
Lesser value will be taken (a or b) column strip = 3000 mm
Middle strips = L1 - column strip = 8000 - 3000 = 5000 mm
Step- 7 Reinforcement along longer direction
Moment in longer direction
Pt % =
Table 1: Moment in Longer Direction
Mu Mucn
= .65 x
.75 x Mo
= .65 x
.75 x
613
= 300
Mucp
= .35 x .6 x
Mo
= .35 x .6 x
613
= 130
Mumn
= .65 x Mo -
Mucn
= .65 x 613 –
300
= 100
Mump
= .35 x
Mo - Mucp
= .35 x
613 -130
= 86
Pt .42 % .17 % .13 % .12 %
Total
Ast
(Pt x b x
d) /100
= (.42 x
263 x
3000) /
100
= 3310
(Pt x b x d)
/100
= (.17 x
263 x
3000) /
100
= 1340
(Pt x b x d)
/100
= (.13 x 3000 x
263) / 100
= 1025
(Pt x b x
d) /100
= (.12 x
3000 x
263) /
100
= 946
Ast/m 1105 447 342 316
Step-8 Reinforcement along shorter direction
Table 2 : Moment in Longer Direction: For roof
Mu Mucn = .65
x .75 x Mo
= .65 x .75
x 456
= 223
Mucp = .35 x
.6 x Mo
= .35 x .6 x
456
= 96
Mumn = .65
x Mo -Mucn
= .65 x 456
– 22
= 75
Mump =
.35 x Mo -
Mucp
= .35 x
456 – 96
= 64
Pt .34 % .15 % .12 % .06 % but
take .12
%
Total
Ast
(Pt x b x d)
/100
= (.34 x 251
x 3000) /
100
= 2560
(Pt x b x d)
/100
= (.15 x 251
x 3000) /
100
= 1130
(Pt x b x d)
/100
= (.12 x
5000 x251)
/ 100
= 1505
(Pt x b x
d) /100
= (.12 x
5000 x
251) /
100
= 1505
Ast/m 855 380 300 300
Step- 9 Check for two way shear or punching shear
Shear force calculation
Vu = (L1 x L2 - critical section area) x factored load
= (6 x 8 - .750 x .650) 14.6 = 690 KN
Bo= 2 x critical section area = ( 650 + 750) x 2 = 2803
Bo x d = 2803 x 251 = 702610
Tau c = Vu / Bo x d = (690 / 702610) x 1000 = .98 N/mm2

From IS code 456 -2000 page no 58 (cl. 31.6.3.1)
Ks =1.3
Tau c = .35 x (fck).5 = 1.47
Tau c' = 1.47
For roof Tau c = .98 N/mm2
4. LOADING DETAILS
With the help of IS 456-2000 and IS 875 Part-3, the load
selected along with their combinations with appropriate
partial factor of safety. Load taken in this work are as
follows:
1. Wind in + X direction
2. Wind in - X direction
3. Wind in + Z direction
4. Wind in - Z direction
5. D.L.
6. L.L.
7. 1.5 (D.L + L.L)
8. 1.2 (D.L. + L.L ± Wind X)
9. 1.2 (D.L. + L.L ± Wind Z)
10. 1.5 (D.L. ± WindX)
11. 1.5 (D.L ± WindZ)
12. 0.9 (D.L. ± 1.5 Wind X)
13. 0.9 (D.L. ± 1.5 Wind Z)
5. STRUCTURE MODELING
In this work, the G+20 Model building plan selected and
designed simple flat slab and added drop flat slab which is
further extended into two other cases on the basis of stress
location in flat slab. Different types of model are shown in
Table 3.
Table 3 : Different Building Model Cases
Model No. Name of models
Model M1
G+20 storey building with simple flat slab
providing shear wall around the lift
Model M2
G+20 storey building with simple flat slab
providing shear wall around the lift and the
core
Model M3
G+20 storey building with added drop flat
slab providing shear wall around the lift
Model M4
G+20 storey building with added drop flat
slab providing shear wall around the lift and
the core
Fig -2: Plan of Building Model Case
Fig -3: 3D view of Building Model Case

Fig -4: Maximum Stress Occurring in Model Case M1
Fig -5: Shear Wall Location in G + 20 Storey Building
6. RESULTS AND DISCUSSION
When building analyzed under the influence of Wind load,
the four different model case’s result parameters are
compared to find the most economical model therefore as
per the objective of this work, the results obtained are
shown in graphical form as well as in tabular form for
different parameters which are as follows:
Table 4: Nodal Displacement in X and Z Direction
Maximum
Displacement
X-Direction (in
mm)
Cases
Model
case M1
Model
case M2
Model
case M3
Model
case M4
92.600 59.889 100.661 67.331
Maximum
Displacement
Z-Direction (in
mm)
Model
case M1
Model
case M2
Model
case M3
Model
case M4
113.159 104.184 118.624 109.113
Chart -1: Nodal Displacement in X and Z Direction
In model case M2, the nodal displacement in X and Z
Direction is least among all of four Model Cases M1, M3 and
M4 in both directions.
Table 5: Axial Force in Column at Ground Level
Cases
Axial Force In Column At Ground Level (KN)
Model case
M1
Model case
M2
Model case
M3
Model case
M4
12210.335 7874.994 12682.515 8213.113

Chart -2: Axial Force in Column at Ground Level
The value of the Axial Force in column at ground level in
Model Case M2 is 7874.994 KN, this value is lesser amongall
the model cases such as Model Case M1, M3 and M4.
Table 6: Shear Force in Column Sy and Sz
Shear Force
In Column
Sy (KN)
Cases
Model
case M1
Model
case M2
Model
case M3
Model
case M4
346.813 242.059 360.085 251.007
Shear Force
In Column
Sz (KN)
Model
case M1
Model
case M2
Model
case M3
Model
case M4
191.817 109.343 199.098 115.351
Chart -3: Shear Force in Column Sy and Sz
Comparing all Model Cases, Model Case M2 shows least
values among all for Shear Forces Sy and Sz. Hence the
optimum case will be Model Case M2.
Table 7: Maximum Compressive Stress in column
Cases
Maximum Compressive Stress In Column (N/mm2)
Model case
M1
Model case
M2
Model case
M3
Model
case M4
61.598 40.82 63.985 42.221
Chart -4: Maximum Compressive Stress in Column
The Maximum Compressive Stress in column seems to be
minimum in Model Case M2 with a value of 40.82 N/mm2 as
compared to other models cases such as Model Case M1, M3
and M4.
Table 8: Maximum Tensile Stress In Column
Cases
Maximum Tensile Stress In Column (N/mm2)
Model case
M1
Model case
M2
Model case
M3
Model
case M4
49.24 36.974 51.098 38.317
Chart -5: Maximum Tensile Stress in Column

The Maximum Tensile Stress in column observed maximum
in Model Case M3 which is 51.098 N/mm2 and lesser in
Model Case M2, the value is 36.974 N/mm2 which is
minimum among all the Model Cases.
Table 9: Torsional Moment In Column
Cases
Torsional Moment in Column (KNm)
Model case
M1
Model case
M2
Model case
M3
Model
case M4
2.500 2.143 2.808 2.582
Chart -6: Maximum Torsional Moment in Column
The Torsional Moment in column is maximuminModel Case
M3 which is 2.808 KNm but lesser in model case M2 and this
value is smaller in all the Model Cases.
7. CONCLUSIONS
The some conclusions are written below according to some
results parameters for four different cases:
1. In Model Case M2, the value of Nodal Displacement in X
direction is least among all the Model Cases and the
maximum value of Nodal Displacement in Model Case
M3.
2. The Nodal Displacement in Z direction is minimum in
Model Case M2 and Model Case M4 but maximum in
model case M3.
3. The Axial Force in column at ground level is maximum
in Model Case M3 but minimum in model case M2 and
M4.
4. The Shear Force in column in Y direction is minimum in
Model Case M2 which is lesser among all the model
cases. Shear Force value in Z direction is maximum in
Model Case M3 but lesser in Model Case M2 and M4.
5. The Maximum Compressive Stress in column is least in
Model Case M2 but maximum in model case M3.
Maximum Tensile in Column is least in model case M2
and model case M4.
6. The torsional moment in column is maximum in model
case M3 but least in model case M2 and M1.
7. Observing all the result parameters Model Case M2
seems to be efficient among all four cases. Hence in
G+20 storey building with simple flat slab providing
shear wall around the lift and the core should be
preferred.
ACKNOWLEDGEMENT
I would like to thank my guide Mr. Sagar Jamle, Assistant
Professor, Department of Civil Engineering, Oriental
University, Indore (M.P.). He always gave me unremitting
and enthusiastic support in this work. He gave me excessive
support and self-determination as an M.Tech scholar.
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IRJET- A Technical Approach to Flat Slab Multistorey Building under Wind Speed of 39 M/S

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IRJET- A Technical Approach to Flat Slab Multistorey Building under Wind Speed of 39 M/S