CHAPTER-I
INTRODUCTION
1.1 General
The safe and accurate design of a 10 storied building for a frame structure, largely depends on
the size/shape of column for the effect of the lateral load (caused by the wind load and
earthquake load), maximum or minimum bending moment for the shape and size of the column,
which may help the design criteria of the structure because lateral load does not affect on slab
and significantly beam design.
1.2 Objectives of study
The objectives of the study are as follows:
a) Comparative study of building design with lateral loads and without lateral loads.
b) To get introduced with different way used for analyzing superstructure.
c) To compare the results (reinforcement) of without considering lateral load and considering
lateral loads.
1.3 Scope of works
Without lateral load condition column has been design only gravity loads. Beams only for singly
design method are used in this study. In this study slab design not required because lateral load
does not affect on slab.
1.4 Methodology
Keeping a close look to the followings, the project has been accomplished:
1) Analysis of the superstructure.
2) Design of the superstructure without considering lateral load.
3) Analysis of the superstructure by using ETABS.
1
4) Design of the Superstructure considering lateral load.
5) Comparison of the results for both conditions.
1.5 Limitations
In this study no dynamic analysis is done for lateral loads. Dynamic analysis for lateral loads
would give more accurate and reliable results. If dynamic analysis was performed then this study
would have been more safe, economic and aesthetic.
2
CHAPTER-II
LITERATURE REVIEW
2.1 General
Reinforced Cement Concrete (RCC) has become the best choice to the structural engineers as a
design option over the decades around the world. It gives better performance, easy to construct
can be given any shape we like. Concrete is weak in tension but good in compression. On the
other hand steel is both good in tension and compression. Concrete is good against corrosion but
steel is vulnerable against corrosion. Again, concrete has good fire resistance properties but steel
is very poor in this case. So, in can be inferred that concrete has some good qualities which steel
does not possess and steel has some weakness which concrete can cover. And luckily and
interestingly when they are used together in combination they offer us all the properties and
qualities we need.
In all sorts of structural members there are
always tensions and compressions, either on the
top or at the bottom, depending on the support
conditions and loading conditions. When the
members are heavily loaded, on the compression
Zone concrete can take a substantial amount of
compression up to a certain limit (when limit
crosses steel is introduced which is known as
doubly reinforced as steel is good in
compression too)
But on the tension zone concrete can not take tension. Tensile strength of normal weight
concrete ranges from 3 to 5√f 'c (10% of √f 'c as a thumb rule). That means a 4000 psi concrete
can take only 400 psi in tension. But the tensile stress comes is as high as 60000 psi or more
which is well beyond concrete’s ability. This is why steel is introduced in tension zone as a
reinforcement to take tension. Steel can take up to 75000 psi or even more in tension. This is
why the name RCC comes from.
Again, Concrete is alkaline in character. It has a typical PH value ranging 12 to 13. When steel is
used in RCC, steel does not get rusted until its PH value goes below 9.0. So steel is better
protected in concrete.
3
2.2 Flexural analysis and design of beams
Flexural analysis got its theoretical ground when Euler–Bernoulli proposed their beam theory.
Apart from this stress- strain analysis, some assumptions when stress becomes non-linear and
using of equation of static equilibrium paved the way of today’s Flexural Analysis and Design
procedure.
The basic assumptions of Flexural Analysis of Beam can be stated as –
1) Euler–Bernoulli beam theory (also known as engineer's beam theory, classical beam theory or
just beam theory) is a simplification of the linear theory of elasticity. In the Euler-Bernoulli
theory of slender beams, a major assumption is that 'plane sections remain plane before and after
bending' where any deformation due to shear across the section is not accounted for. Also, this
linear distribution is only applicable if the maximum stress is less than the yield stress of the
material.
2) Beam behaves elastically up to proportional limit .Neutral axis passes through the center of
gravity of the cross section. Stress at any point in the section can be represented by the equation
–
f = My / I
In the figure below the beam illustrates how plane section remains plane in a beam
ρ
θ
Distorted section of bent beam
Bent element from which relation for elastic curve
is obtained
4
2.2.1 Conditions of designing tension reinforced rectangular beam
In order to ensure that a tension reinforced rectangular beam can resist hypothetical loads.
Required strength must be less than or equal to design strength. Design strength can be found by
multiplying nominal strength with a strength reduction factor Φ.
Therefore –
Mu ≤ ΦMn
Pu ≤ ΦPn
Vu ≤ ΦVn
2.2.2 USD (Ultimate Strength Design)
It has been found from experiments and analysis that distribution of internal stress and strain of
a beam becomes non –linear near ultimate load. For the first mood of failure, that means steel
stress equal yield point, concrete will either fail by crushing of concrete or yielding of steel in
the compression zone. The exact behavior of concrete is yet to be known. But it has been
possible to measure concrete strain from 0.0030 to 0.004 before failure for a rectangular beam.
In order to ensure yielding of steel before crushing of concrete, it is not really necessary to know
the exact shape of concrete. The most important thing to know is for a given distance c of the
neutral axis 1) total resultant compression force C in the concrete and 2) its vertical location
(Location of Resultant Force).
In a rectangular beam compressive strength on area bc can be expressed as –
C = favbc.
In order to express the equation in terms of f'c, a ratio α has been introduced which can be
expressed as α = fav / f'c , therefore –
C = α f'c bc
c
γ = αa
γ f'c
C = α f'c cb
C = γ f'c ab
When concrete reaches its ultimate load it takes a parabolic stress distribution. This nonlinear
stress distribution can be converted to an equivalent rectangular stress distribution of depth a and
constant stress γf'c (γ= stress intensity factor).
5
Where, γ = α c/a
=> αc = γa
C = γ f'c ab
The value of stress intensity factor γ is 0.85 regardless of f’c . The concrete compression force at
failure in a rectangular beam of width b is –
C = 0.85 f'c ab
From the condition of equilibrium
C=T
Again, tensile force can be written as –
T = Asfs
If, failure is initiated by yielding, then fs=fy
So,
T = Asfy
Therefore,
0.85 f'c ab
= Asfy
=> a = Asfy / 0.85 f'c b
Substituting, As = ρbd
=> a = ρbdfy / 0.85 f'c b = ρdfy / 0.85 f'c
Nominal moment of an under reinforced beam considering yielding of steel fs = fy at failure
Mn = Asfy (d – a/2)
Substituting, a = ρdfy / 0.85 f'c
Mn = ρbd fy (d – 1/2(ρdfy / 0.85 f'c ))
= ρbd2 fy (1 – 0.59ρfy / f'c )
From experimentally obtained values β / α = 0.59, which gives –
Mn = ρbd2fy (1 - 0.59 ρfy / f'c )
Mn = Rbd2
R = ρfy (1 - 0.59 ρfy / f'c )
This R is termed as flexural resistant factor.
6
2.3 Moment coefficients
Discontinuous end
unrestrained:
Spandrel:
Column:
Discontinuous end
unrestrained:
Spandrel:
Column:
0
1
1
11
1
1
1
1
1
1
24
1
14
1
10
11
16
11
11
16
14
1
1
1
11
1
1
1
11
1
24
1
14
1
9
9
14
1
24
1
16
14
14
16
0
0
1
2.4 Column
A primarily compression member, which may or may not be designed to carry simultaneous
flexural forces, shall be termed a column.
2.4.1 Design of short columns
For concentrically axially loaded columns design strength of a column is –
Pu =ΦPn
= αΦ Ag [0.85f`c + (1 - ρg) + ρg fy]
ρg = ratio of area of distributed longitudinal reinforcement to gross concrete area perpendicular
to that reinforcement
ACI code specifies - 0.01 ≤ ρg ≤ 0.08
But for practice ρg should no go beyond 0.04 for bar splicing and steel congestion.
Where:
Φ
α
Tied Columns
0.65
0.80
Spiral
Columns
0.70
0.85
7
2.5 ETABS
ETABS means Extended 3D Analysis of Building Systems. ETABS is a sophisticated, yet easy
to use, special purpose analysis and design program developed specifically for building systems.
ETABS Version 9 features an intuitive and powerful graphical interface coupled with
unmatched modeling, analytical, and design procedures, all integrated using a common database.
Although quick and easy for simple structures, ETABS can also handle the largest and most
complex building models, including a wide range of nonlinear behaviors, making it the tool of
choice for structural engineers in the building industry.
2.5.1 Overview of ETABS
1. Object-oriented intuitive 2D/3D graphical model generation.
2. Pull down menus, floating tool bars, and tool tip help.
3. Quick data input through property sheets and spreadsheets.
4. Customizable structural templates for creating a model.
5. Full range of analysis including static, concrete.
6. Supports truss and beam members, plates, solids, linear and non-linear cables and
curvilinear beams.
7. Flexible zoom, pan and multiple views.
8. Isometric and perspective views 3D shapes.
9. Toggle display of loads, supports, properties, joints, members, etc.
10. Built-in command file editor.
11. Copy &paste through clipboard.
12. Simple command language.
13. State-of-the-art graphical pre and post processors.
14. Rectangular/cylindrical coordinate systems with mix and match capabilities.
15. Joint, member/element, mesh generation with flexible user- controlled numbering
scheme.
16. Presentation quality printer plots of geometry and results as part of run output.
17. Performs multiple analysis in the same run - perfect for phase or stage construction.
8
2.5.2 Design Codes
2.5.2.1 Concrete design
1. Design of Concrete Beam, Column per ACI 318. Optional codes include BS8007,
BS8110, Canadian, French, German, Spanish, Scandinavian, Japanese, Chinese,
Australian, Singaporean and Indian Codes.
2. Numerical and Graphical Design Outputs with reinforcement details.
3. Interactive concrete design and detailing with bar scheduling and interactive rebar
layout.
4. Design "physical members" by combining separate beam entities as one with full
step-by-step calculation sheets for verification.
2.5.2.2 Analysis and design
1. Plate elements consider inclined supports.
2. Full and partial moment releases (excellent for steel frames where releases defined by
springs are hard to determine).
3. Member and spring specification.
4. Fixed, pinned and spring supports with releases. Also Inclined Supports.
5. Automatic spring support generator for mat foundations.
2.5.2.3 Load types and load generation
1. Loading for Joints, Members/Elements including Concentrated, Uniform Linear,
Trapezoidal, Temperature, Strain, Support displacement, Pre-stress and Fixed-end Loads.
2. Global, Local and Projected Loading Directions.
3. Uniform or varying Element Pressure Loading on entire or selected portion of plate
elements.
4. Floor/Area Load converts load-per-area to member loads based on one-way or two-way
actions.
9
2.5.2.4 Model verification
1. 2D/3D drawings on screen as well as on plotter/printer.
2. Graphical representation of the unstable area of a structure due to improper modeling.
3. Full 3D shapes for Beams, Plates and Solids with shading and lighting.
4. Isometric or any rotations for full 3D viewing.
Display of Loads, Supports, Orientations, Properties, Hidden line removed, Joint/Member
numbering, Dimensions etc.
2.6 Loads for different types of building
In designing a floor the following loads are considered:
1. Dead load.
2. Live load.
3. Wind load.
4. Earthquake load
2.6.2 Dead load
Dead load is the vertical load due to the weight of permanent structural and non-structural
components of building such as walls, floors, ceilings, permanent partitions and fixed service
equipment etc. In designing a floor slab, the actual weight of the materials of construction must
be added to the live load to obtain the design load.
2.6.2 Live load
The live loads are governed by the type of occupancy of the building. Minimum live loads for
floors and roofs are given the building regulations of the various cities and the designer is
required to make his computations in accordance with the loads specified. In general, live loads
are due to human occupancy, furniture, equipment, stored materials, and occasionally movable
partitions.
10
2.6.3 Wind load and earthquake load
For Buildings having greater height, lateral loads should have to be considered for the design of
the building component. Mainly wind loads and Earthquake loads are the major forms of lateral
loads. To make the design structurally safe these lateral loads should be considered very
cautiously.
2.6.3.1 Wind load
Design wind pressure, Pz = CG Cp qz
Where, Pz=Design wind pressure at height z, kN/m2
CG=Gust coefficient which shall be Gz or Gh
Cp=Pressure coefficient for structures or components
qz=Sustained wind pressure
=CcCtCzVb2
Ct=Structure importance coefficient
Cc=Velocity to pressure conversion coefficient
Cz=Combined height and exposure coefficient
Vb=Basic wind speed
Design wind load for all framing system,
F1=ΣPAz
Where, F1=Wind force acting normal to a surface
P=Design wind pressure on building surfaces
Az=Area of the building surface at height z upon which design pressure P operates
Design wind load for any building or structure as a whole,
F2=ΣPzAz
Where, F2=Total wind force on the framed system of the building in a specified direction
Pz=Design wind pressure for the cross-sectional shapes of the building
Az=Projected frontal area normal to the wind at height z
2.6.3.2 Earthquake load
Total design base shear in a given direction,
V=
ZIC
R
W
Where, Z=Seismic zone coefficient
I=Structure importance coefficient
R=Response modification coefficient for the structural systems
C=Numerical coefficient
1.25𝑆
= 𝑇 2/3 ≤<2.75
S= Site coefficient for soil characteristics
T=fundamental period of vibration of the structure
11
=Ct (hn) ¾
Ct=0.083 (for steel moment resisting frames)
=0.073 (for reinforced concrete resisting frames and eccentric braced steel frames)
=0.049 (for all other structural systems)
hn=Height above the base to level n
The vertical distribution of the total lateral force, which is the base shear v, can be given along
the height of the structure as:
V=Ft+∑𝑛𝑖=1 Fi
Where, Fi=Lateral load applied at story level i
Ft=Concentrated lateral force considered at the top of the building in addition to the force Fn
=0.07TV ≤ 0.25V (for T > 0.7 second)
=0
(for T ≤ 0.7 second)
The remaining portion of the base shear (V-Ft) shall be distributed over the height of the
building including level n as follows:
𝐹𝑥 =
(V−Ft )Wx hx
∑n
i=1 wi hi
At each storey level x, the force Fx shall be applied over the area of the building in proportion to
the mass distribution at that level.
12
CHAPTER-III
BEAM LOAD CALCULATION AND DESIGN
WITHOUT LATERAL LOAD
Fig. 3.1 Plan
13
Fig. 3.2 Beam area details
14
Fig. 3.3 Grid and beam, column layout
15
3.1 Concrete and steel properties
Strength of Concrete f ’c = 4000 psi
Yield strength of Steel fy = 60000 psi
Unit weight of Concrete = 150 pcf
3.2 Slab thickness calculation
Slab thickness "h"= Perimeter/180
= [{2 ∗ (16.75 + 16.50) ∗ 12}/180]
= 4.43 in
𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 = 5 in
3.3 Load calculation
3.3.1 Dead load calculation:Self weight for slab = 62.5 psf
Floor finish = 25 psf
False celing = 30 psf
Partition wall load = 66.50 psf
Total Dead Load = 184 psf
𝑇𝑜𝑡𝑎𝑙 𝐷𝑒𝑎𝑑 𝐿𝑜𝑎𝑑 𝐹𝑎𝑐𝑡𝑜𝑟𝑒𝑑 "𝑊𝐷𝐿 " = 1.2 ∗ D. L psf
= 1.2 ∗ 184 psf
= 221 psf
3.3.2 Live load calculation:Assume Live Load = 60 psf
𝑇𝑜𝑡𝑎𝑙 𝐿𝑖𝑣𝑒 𝐿𝑜𝑎𝑑 𝐹𝑎𝑐𝑡𝑜𝑟𝑒𝑑 "WLL " = 1.6 ∗ L. L psf
= 1.6 ∗ 60 psf
= 96 psf
Total Factored Load = (1.2 ∗ D. L + 1.6 ∗ L. L) psf
Total Factored Load = (1.2 ∗ 184 + 1.6 ∗ 60) psf
𝑇𝑜𝑡𝑎𝑙 𝐹𝑎𝑐𝑡𝑜𝑟𝑒𝑑 𝐿𝑜𝑎𝑑 "𝑊" = 317 psf
16
3.4 Loading area calculation
A1 = (7.875)2 = 62 ft 2
2
A2 = (8.75) = 76.56 ft 2
Similarly,
A3 = 62 ft 2
A4 = 124 ft 2
A5 = 152.17 ft 2
A6 = 123.97 ft 2
A7 = 75.61 ft 2
A8 = 61.95 ft 2
A9 = 77.77 ft 2
A10 = 58.14 ft 2
A11 = 158.88 ft 2
A12 = 116.28 ft 2
A13 = 156.88 ft 2
A14 = 116.28 ft 2
A15 =
1
15
∗ (17.75 + 2) ∗ ( ) = 77.77 ft 2
2
2
3.5 Uniformly distributed load calculation for beam
Assume Beam size =12"X24"
Slab load (factored) = 317 psf
Self Weight for Beam Stem (factored) =
12 ∗ 19
∗ 150 ∗ 1.2 = 285 lb/ft
144
17
Load/ft of Beam
WB1 =
62 ∗ 317
+ 285 = 1533 lb/ft
15.75
WB2 =
76.56 ∗ 317
+ 285 = 1672 lb/ft
17.5
WB3 =
62 ∗ 317
+ 285 = 1533 lb/ft
15.75
WB4 =
124 ∗ 317
+ 285 = 2781 lb/ft
15.75
WB5 =
152.56 ∗ 317
+ 285 = 3041 lb/ft
17.5
WB6 =
123.97 ∗ 317
+ 285 = 2780 lb/ft
15.75
WB7 =
75.61 ∗ 317
+ 285 = 1655 lb/ft
17.5
WB8 =
61.95 ∗ 317
+ 285 = 1532 lb/ft
15.75
WB9 =
77.77 ∗ 317
+ 285 = 1674 lb/ft
17.75
WB10 =
58.14 ∗ 317
+ 285 = 1494 lb/ft
15.75
WB11 =
156.88 ∗ 317
+ 285 = 3087 lb/ft
17.75
WB12 =
116.28 ∗ 317
+ 285 = 2702 lb/ft
15.25
WB13 =
156.88 ∗ 317
+ 285 = 3087 lb/ft
17.75
WB14 =
116.28 ∗ 317
+ 285 = 2702 lb/ft
15.25
WB15 =
77.77 ∗ 317
+ 285 = 1674 lb/ft
17.75
18
3.6 Moment calculation
3.6.1 Grid 3
B1
−M =
+M =
−M =
WB1 l2
16
WB1 l2
14
WB1 l2
10
=
=
=
1533∗15.752
16
1533∗15.752
14
1533∗15.752
10
= 23768 lb-ft = 285 kip-in
= 27163 lb-ft = 326 kip-in
= 38028 lb-ft = 456 kip-in
B2
−M =
+M =
−M =
WB2 l2
11
WB2 l2
16
WB2 l2
11
=
=
=
1672∗17.52
11
1672∗17.52
16
1672∗17.52
11
= 46550 lb-ft = 559 kip-in
= 32003 lb-ft = 384 kip-in
= 46550 lb-ft = 559 kip-in
B3
−M =
+M =
−M =
WB3 l2
10
WB3 l2
14
WB3 l2
16
=
=
=
1533∗15.752
10
1533∗15.752
14
1533∗15.752
16
= 38028 lb-ft = 456 kip-in
= 27163 lb-ft = 326 kip-in
= 23768 lb-ft = 285 kip-in
3.6.2 Grid 2
B4
−M =
+M =
−M =
WB4 l2
16
WB4 l2
14
WB4 l2
10
=
=
=
2781∗15.752
16
2781∗15.752
14
2781∗15.752
10
= 43116 lb-ft = 517 kip-in
= 49276 lb-ft = 591 kip-in
= 68986 lb-ft = 828 kip-in
B5
−M =
+M =
−M =
WB5 l2
11
WB5 l2
16
WB5 l2
11
=
=
=
3041∗17.52
11
3041∗17.52
16
3041∗17.52
11
= 84664 lb-ft = 1016 kip-in
= 58207 lb-ft = 699 kip-in
= 84664 lb-ft = 1016 kip-in
19
B6
−M =
+M =
−M =
WB6 l2
10
WB6 l2
14
WB6 l2
16
=
=
=
2780∗15.752
10
2780∗15.752
14
2780∗15.752
16
= 76624 lb-ft = 920 kip-in
= 49258 lb-ft = 591 kip-in
= 43101 lb-ft = 517 kip-in
3.6.3 Grid 1
B7
−M =
+M =
−M =
WB7 l2
16
WB7 l2
14
WB7 l2
9
=
=
=
1655∗17.52
16
1655∗17.52
14
1655∗17.52
9
= 31678 lb-ft = 380 kip-in
= 36203 lb-ft = 434 kip-in
= 56316 lb-ft = 676 kip-in
B8
−M =
+M =
−M =
WB8 l2
9
WB8 l2
14
WB8 l2
16
=
=
=
1532∗15.752
9
1532∗15.752
14
1532∗15.752
16
= 42226 lb-ft = 507 kip-in
= 27145 lb-ft = 326 kip-in
= 23752 lb-ft = 285 kip-in
3.6.4 Grid D
B9
−M =
+M =
−M =
WB9 l2
16
WB9 l2
14
WB9 l2
9
=
=
=
1674∗17.752
16
1674∗17.752
14
1674∗17.752
9
= 32963 lb-ft = 396 kip-in
= 37672 lb-ft = 452 kip-in
= 58602 lb-ft = 703 kip-in
B10
−M =
+M =
WB10 l2
9
WB10 l2
14
=
=
1494∗15.252
9
1494∗15.252
14
= 38605 lb-ft = 463 kip-in
= 24818 lb-ft = 298 kip-in
20
−M =
WB10 l2
16
=
1494∗15.252
16
= 21716 lb-ft = 261 kip-in
3.6.5 Grid C
B11
−M =
+M =
−M =
WB11 l2
16
WB11 l2
14
WB11 l2
9
=
=
=
3087∗17.752
16
3087∗17.752
14
3087∗17.752
9
= 60787 lb-ft = 730 kip-in
= 69471 lb-ft = 834 kip-in
= 108066 lb-ft = 1297 kip-in
B12
−M =
+M =
−M =
WB12 l2
9
WB12 l2
14
WB12 l2
16
=
=
=
2702∗15.252
9
2702∗15.252
14
2702∗15.252
16
= 69820 lb-ft = 838 kip-in
= 44885 lb-ft = 539 kip-in
= 39274 lb-ft = 471 kip-in
3.6.6 Grid B
B13
−M =
+M =
−M =
WB13 l2
16
WB13 l2
14
WB13 l2
9
=
=
=
3087∗17.752
16
3087∗17.752
14
3087∗17.752
9
= 60787 lb-ft = 730 kip-in
= 69471 lb-ft = 834 kip-in
= 108066 lb-ft = 1297 kip-in
B14
−M =
+M =
−M =
WB14 l2
9
WB14 l2
14
WB14 l2
16
=
=
=
2702∗15.252
9
2702∗15.252
14
2702∗15.252
16
= 69820 lb-ft = 838 kip-in
= 44885 lb-ft = 539 kip-in
= 39274 lb-ft = 471 kip-in
21
3.6.7 Grid A
B15
−M =
+M =
−M =
WB15 l2
16
WB15 l2
14
WB15 l2
9
=
=
=
1674∗17.752
16
1674∗17.752
14
1674∗17.752
9
= 32964 lb-ft = 396 kip-in
= 37673 lb-ft = 452 kip-in
= 58602 lb-ft = 703 kip-in
3.7 Size determination and reinforcement calculation
3.7.1 Grid C
Beam ID: B11 & B12
-Mmax = 1297 kip-in
ø=0.9
Mu
d = √(
)
øRb
ρ = 0.85 ∗ β ∗
d = √(
1297
)
0.9 ∗ 0.912 ∗ 12
f′c
€c
∗
fy €c + €t
ρ = 0.85 ∗ 0.85 ∗
4
0.003
∗
60 0.003 + 0.005
ρ= 0.0181
d= 11.48"
ρ ∗ fy
)
f ′c
Total depth “h” = d+ clear cover
R = ρ ∗ fy ∗ (1 − 0.588 ∗
h= 11.48"+2.5"
R = 0.0181 ∗ 60 ∗ (1 − 0.588 ∗
h=13.98 ≈ 14" (So ok)
0.0181 ∗ 60
)
4
R= 0.912 ksi
Effective Depth d = 14"-2.5"
d= 11.50"
Check for Singly or Doubly Beam:
M= *R*b*d2
M=0 .9*0.912*12*11.502
M= 1303 kip-in > Mu
So it is singly reinforcement beam.
Let, a= 3.65"
As =
Mu
a
ø ∗ fy ∗ (d − 2)
22
1297
As =
0.9 ∗ 60 ∗ (11.5 –
3.65
2 )
As= 2.48 in2
Check a:
As ∗ fy
a=
0.85 ∗ f`c ∗ b
a=
2.48 ∗ 60
0.85 ∗ 4 ∗ 12
a= 3.65"
C=
a
3.65
=
= 4.30 in
β1 0.85
∈t =∈u ×
d−C
C
∈t = 0.003 ×
12.5 − 4.30
4.30
∈t = 0.00502
So ∈t = 0.005 (ok)
So As= 2.48 in2
∴ Beam size = 12" X 14"
Design of T beam
+Mmax = 834 kip-in
Calculation effective width of flange
a) ¼ * Span = ¼ *17.75*12 = 53.24 ≈ 54 in
b) bw+16*t = 12+16*5 = 92 in
c) Spacing of beam = 15*12 =180 in
∴ bf = 54 in
Let, a = 0.45
834
As =
0.9 ∗ 60 ∗ (11.5 –
0.45
)
2
As = 1.37 in2
23
Check a:
As ∗ fy
a=
0.85 ∗ f`c ∗ bf
a=
1.37 ∗ 60
0.85 ∗ 4 ∗ 54
a= 0.45"
∴ As = 1.37 in
3.7.2 Grid A
Beam ID: B15
-Mmax = 703 kip-in
ø=0.9
Mu
d = √(
)
øRb
d = √(
ρ = 0.85 ∗ β ∗
703
)
0.9 ∗ 0.912 ∗ 12
f′c
€c
∗
fy €c + €t
ρ = 0.85 ∗ 0.85 ∗
4
0.003
∗
60 0.003 + 0.005
d= 8.45"
ρ= 0.0181
Total depth “h” = d+ clear cover
R = ρ ∗ fy ∗ (1 − 0.588 ∗
ρ ∗ fy
)
f ′c
h= 8.45"+2.5"
h=10.95 ≈ 12" (So ok)
Effective Depth d = 12"-2.5"
R = 0.0181 ∗ 60 ∗ (1 − 0.588 ∗
0.0181 ∗ 60
)
4
R= 0.912 ksi
d= 9.5"
Check for Singly or Doubly Beam:
M= *R*b*d2
M=0 .9*0.912*12*9.52
M= 889 kip-in > Mu
So it is singly reinforcement beam.
Let, a= 2.30"
As =
As =
Mu
a
ø ∗ fy ∗ (d − 2)
703
0.9 ∗ 60 ∗ (9.5 –
2.30
2 )
24
As= 1.56 in2
Check a:
As ∗ fy
a=
0.85 ∗ f`c ∗ b
a=
1.65 ∗ 60
0.85 ∗ 4 ∗ 12
a= 2.30"
C=
a
2.30
=
= 2.71 in
β1 0.85
∈t =∈u ×
d−C
C
∈t = 0.003 ×
9.5 − 2.71
2.71
∈t = 0.0075
So ∈t = 0.005 (ok)
So As= 1.56 in2
∴ Beam size = 12" X 12"
(Note: The minimum width of a beam should be 12 inches. Bibliography No. 11)
+Mmax = 452 kip-in
As= 0.89 in2
Similarly
3.7.3 Grid B
Beam ID: B13 & B14
-Mmax = 1297 kip-in
Mu
d = √(
)
øRb
d = √(
ø=0.9
R= 0.912 ksi
1297
)
0.9 ∗ 0.912 ∗ 12
d= 11.48"
Total depth “h” = d+ clear cover
25
h= 11.48"+2.5"
h=13.98 ≈ 14" (So ok)
Effective Depth d = 14"-2.5"
d= 11.50"
As =
Mu
a
ø ∗ fy ∗ (d − 2)
1297
As =
0.9 ∗ 60 ∗ (11.5 –
3.65
2 )
As= 2.48 in2
∴ Beam size = 12" X 14"
+Mmax = 834 kip-in
As= 1.37 in2
3.7.4 Grid D
Beam ID: B9 & B10
-Mmax = 703 kip-in
d = √(
Mu
)
øRb
d = √(
703
)
0.9 ∗ 0.912 ∗ 12
d= 8.45"
Total depth “h” = d+ clear cover
h= 8.45"+2.5"
h=10.95 ≈ 12" (So ok)
Effective Depth d = 12"-2.5"
d= 9.5"
Let, a= 2.30"
As =
Mu
a
ø ∗ fy ∗ (d − 2)
26
703
As =
0.9 ∗ 60 ∗ (9.5 –
2.30
2 )
As= 1.56 in2
∴ Beam size = 12" X 12"
+Mmax = 452 kip-in
As= 0.89 in2
3.7.5 Grid 1
Beam ID: B7 & B8
-Mmax = 676 kip-in
d = √(
Mu
)
øRb
d = √(
676
)
0.9 ∗ 0.912 ∗ 12
d= 8.30"
Total depth “h” = d+ clear cover
h= 8.30"+2.5"
h=10.8 ≈ 12" (So ok)
Effective Depth d = 12"-2.5"
d= 9.5"
676
As =
0.9 ∗ 60 ∗ (9.5 –
2.20
2 )
As= 1.50 in2
∴ Beam size = 12" X 12"
+Mmax = 434 kip-in
As= 0.86 in2
27
3.7.6 Grid 2
Beam ID: B4, B5 & B6
-Mmax = 1016 kip-in
d = √(
Mu
)
øRb
d = √(
1016
)
0.9 ∗ 0.912 ∗ 12
d= 10.16"
Total depth “h” = d+ clear cover
h= 10.16"+2.5"
h=12.66 ≈ 14" (So ok)
Effective Depth d = 14"-2.5"
d= 11.5"
1016
As =
0.9 ∗ 60 ∗ (11.5 –
2.73
2 )
As= 1.86 in2
∴ Beam size = 12" X 14"
+Mmax = 699 kip-in
As= 1.14 in2
3.7.7 Grid 3
Beam ID: B1, B2 & B3
-Mmax = 559 kip-in
d = √(
Mu
)
øRb
d = √(
559
)
0.9 ∗ 0.912 ∗ 12
d= 7.53"
Total depth “h” = d+ clear cover
h= 7.53"+2.5"
28
h=10.03 ≈ 12" (So ok)
Effective Depth d = 12"-2.5"
d= 9.5"
559
As =
0.9 ∗ 60 ∗ (9.5 –
1.76
2 )
As= 1.20 in2
∴ Beam size = 12" X 12"
+Mmax = 384 kip-in
As= 0.76 in2
Table 3.1
Summary of the beam design for without lateral load condition
Beam ID
Beam Size
B1
B2
B3
B4
B5
B6
B7
B8
B9
B10
B11
B12
B13
B14
B15
12" X 12"
12" X 12"
12" X 12"
12" X 14"
12" X 14"
12" X 14"
12" X 12"
12" X 12"
12" X 12"
12" X 12"
12" X 14"
12" X 14"
12" X 14"
12" X 14"
12" X 12"
Positive Steel Area
in in2
0.64
0.76
0.64
0.97
1.14
0.97
0.86
0.64
0.89
0.60
1.37
0.89
1.37
0.89
0.89
Negative Steel Area
in in2
0.98
1.20
0.98
1.51
1.86
1.70
1.50
1.12
1.56
1.03
2.48
1.61
2.48
1.61
1.56
29
CHAPTER-IV
COLUMN DESIGN WITHOUT LATERAL LOAD
4.1 Loading area calculation of columns
CA1 = (
17.75
15.75
)∗(
) = 69.89 ft 2
2
2
CA2 = (
15.75 + 17.5
17.75
)∗(
) = 147.55 ft 2
2
2
CA3 = (
15.75 + 17.5
17.75
)∗(
) = 147.55 ft 2
2
2
CA4 = (
17.75
15.75
)∗(
) = 69.89 ft 2
2
2
CA5 = (
17.75 + 15.25
15.75
)∗(
) = 129.94 ft 2
2
2
CA6 = (
17.75 + 15.25
15.75 + 17.5
)∗(
) = 274.31 ft 2
2
2
CA7 = (
17.75 + 15.25
15.75 + 17.5
)∗(
) = 274.31 ft 2
2
2
CA8 = (
17.75 + 15.25
15.75
)∗(
) = 129.94 ft 2
2
2
CA9 = (
15.25
15.75
)∗(
) = 60 ft 2
2
2
17.5 + 15.75
15.25
CA10 = (
)∗(
) = 126.77 ft 2
2
2
17.5 + 15.75
15.25
CA11 = (
)∗(
) = 126.77 ft 2
2
2
15.25
15.75
CA12 = (
)∗(
) = 60 ft 2
2
2
30
4.1.1 Beam load
Beam size
Beam Stem Load Per ft
12"X12"
12 ∗ 7
∗ 150 = 87.5 lb/ft
144
12"X14"
12 ∗ 9
∗ 150 = 112.5 lb/ft
144
12"X15"
12 ∗ 10
∗ 150 = 125 lb/ft
144
4.1.2 Self wt of Column
Column size
Column load/floor
12"X12"
12 ∗ 12
∗ 9.5 ∗ 150 = 1425 lb
144
15"X15"
15 ∗ 15
∗ 9.5 ∗ 150 = 2227 lb
144
20"X20"
20 ∗ 20
∗ 9.5 ∗ 150 = 3958 lb
144
24"X24"
24 ∗ 24
∗ 9.5 ∗ 150 = 5700 lb
144
4.2 Load calculation on column for 1st storey
C1 = (69.89 ∗ 317 + 87.5 ∗
15.75
2
15.75
C2 = {147.55 ∗ 317 + 87.5 ∗ (
17.75
+ 87.5 ∗
2
+
2
+ 1425) ∗ 10 = 250457 lb = 250 kips
17.5
17.75
2
2
) + 87.5 ∗
+ 2227} ∗ 10 = 512316 lb
= 512 kips
15.75
C3 = {147.55 ∗ 317 + 87.5 ∗ (
2
+
17.5
17.75
2
2
) + 87.5 ∗
+ 2227} ∗ 10 = 512316 lb
= 512 kips
C4 = (69.89 ∗ 317 + 87.5 ∗
15.75
2
17.75
C5 = {129.94 ∗ 317 + 87.5 ∗ (
17.75
+ 87.5 ∗
2
+
2
+ 1425) ∗ 10 = 250457 lb = 250 kips
15.25
15.75
2
2
) + 112.5 ∗
+ 2227} ∗ 10 = 457476 lb
= 458 kips
C6 = {274.31 ∗ 317 + 125 ∗ (
17.75
2
+
15.25
15.75
2
2
) + 112.5 ∗ (
+
17.5
2
) + 5700} ∗ 10 = 965891 lb
= 966 kips
31
15.75
C7 = {274.31 ∗ 317 + 112.5 ∗ (
2
∗
17.5
17.75
2
2
) + 125 ∗ (
+
15.25
2
) + 5700} ∗ 10 = 965891 lb
= 966 kips
17.75
C8 = {129.94 ∗ 317 + 87.5 ∗ (
2
+
15.25
15.75
2
2
) + 112.5 ∗
+ 2227} ∗ 10 = 457476 lb
= 458 kips
C9 = (60 ∗ 317 + 87.5 ∗
15.25
2
+ 87.5 ∗ 9 + 1425) ∗ 10 = 218997 lb = 219 kips
C10 = {126.77 ∗ 317 + 87.5 ∗
C11 = {126.77 ∗ 317 + 87.5 ∗
15.25
2
15.25
2
+ 87.5 ∗ (
+ 87.5 ∗ (
17.5
2
) + 2227} ∗ 10 = 438460 lb = 439 kips
15.75
2
+
17.5
2
) + 2227} ∗ 10 = 445350 lb
= 445 kips
C12 = (60 ∗ 317 + 87.5 ∗
15.25
2
+ 87.5 ∗
15.75
2
+ 1425) ∗ 10 = 218013 lb = 218 kips
4.3 Column design for 1st storey
4.3.1 Column ID: C6 & C7
Load Pn = 966 kips
Assume Column size=20"X20"
ACI code specifies 0.01 ≤ ρg ≤ 0.08
Using ρg=0.03
Pu = φPn(max) = 0.80φAg [0.85fc′ (1 − ρg ) + ρfy ]
Pu = 0.80 ∗ 0.65 ∗ 400 ∗ {0.85 ∗ 4 ∗ (1 − 0.03) + 0.03 ∗ 60} = 1060.5 kips > 966 kips
so ok
AS = ρ ∗ Ag = 0.03 ∗ 400 = 12 in2
4.3.2 Column ID: C1, C4, C9 & C12
Load Pn = 250 kips
Assume Column size=12"X12"
Pu = φPn(max) = 0.80φAg [0.85fc′ (1 − ρg ) + ρfy ]
Pu = 0.80 ∗ 0.65 ∗ 144 ∗ {0.85 ∗ 4 ∗ (1 − 0.03) + 0.03 ∗ 60} = 381 kips > 250 kips
so ok
AS = ρ ∗ Ag = 0.03 ∗ 144 = 4.32 in2
(Note: The minimum column dimension to 12 inches. Bibliography No. 2)
32
4.3.3 Column ID: C2 & C3
Load Pn = 512 kips
Assume Column size=12"X15"
Pu = φPn(max) = 0.80φAg [0.85fc′ (1 − ρg ) + ρfy ]
Pu = 0.80 ∗ 0.65 ∗ 180 ∗ {0.85 ∗ 4 ∗ (1 − 0.03) + 0.03 ∗ 60} = 477 kips < 512 kips
not ok
Assume Column size=12"X18"
Pu = 0.80 ∗ 0.65 ∗ 216 ∗ {0.85 ∗ 4 ∗ (1 − 0.03) + 0.03 ∗ 60} = 573 kips > 512 kips
So ok
AS = ρ ∗ Ag = 0.03 ∗ 216 = 6.48 in2
4.3.4 Column ID: C5, C8, C10 & C11
Load Pn = 458 kips
Assume Column size=12"X16"
Pu = φPn(max) = 0.80φAg [0.85fc′ (1 − ρg ) + ρfy ]
Pu = 0.80 ∗ 0.65 ∗ 192 ∗ {0.85 ∗ 4 ∗ (1 − 0.03) + 0.03 ∗ 60} = 509 kips > 458 kips
So ok
AS = ρ ∗ Ag = 0.03 ∗ 192 = 5.76 in2
4.4 Load calculation on column for 3rd storey
C1 = (69.89 ∗ 317 + 87.5 ∗
15.75
2
+ 87.5 ∗
15.75
C2 = {147.55 ∗ 317 + 87.5 ∗ (
2
+
17.75
2
+ 1425) ∗ 7 = 175320 lb = 175 kips
17.5
17.75
2
2
) + 87.5 ∗
+ 2227} ∗ 7 = 358621 lb
= 359 kips
15.75
C3 = {147.55 ∗ 317 + 87.5 ∗ (
2
+
17.5
17.75
2
2
) + 87.5 ∗
+ 2227} ∗ 7 = 358621 lb
= 359 kips
C4 = (69.89 ∗ 317 + 87.5 ∗
15.75
2
+ 87.5 ∗
17.75
2
+ 1425) ∗ 7 = 175320 lb = 175 kips
33
17.75
C5 = {129.94 ∗ 317 + 87.5 ∗ (
+
2
15.25
15.75
2
2
) + 112.5 ∗
+ 2227} ∗ 7 = 320234 lb
= 320 kips
C6 = {274.31 ∗ 317 + 125 ∗ (
17.75
2
+
15.25
15.75
2
2
) + 112.5 ∗ (
+
17.5
) + 5700} ∗ 7 = 676124 lb
2
= 676 kips
15.75
C7 = {274.31 ∗ 317 + 112.5 ∗ (
2
∗
17.5
17.75
2
2
) + 125 ∗ (
+
15.25
2
) + 5700} ∗ 7 = 676124 lb
= 676 kips
17.75
C8 = {129.94 ∗ 317 + 87.5 ∗ (
2
+
15.25
15.75
2
2
) + 112.5 ∗
+ 2227} ∗ 7 = 320234 lb
= 320 kips
C9 = (60 ∗ 317 + 87.5 ∗
15.25
2
+ 87.5 ∗ 9 + 1425) ∗ 7 = 153298 lb = 153 kips
C10 = {126.77 ∗ 317 + 87.5 ∗
C11 = {126.77 ∗ 317 + 87.5 ∗
C12 = (60 ∗ 317 + 87.5 ∗
15.25
2
15.25
2
15.25
2
+ 87.5 ∗ (
+ 87.5 ∗ (
+ 87.5 ∗
15.75
2
17.5
2
) + 2227} ∗ 10 = 214845 lb = 215 kips
15.75
2
+
17.5
2
) + 2227} ∗ 7 = 218222 lb
= 218 kips
+ 1425) ∗ 7 = 106826 lb = 107 kips
4.5 Column design for 3rd storey
4.5.1 Column ID: C6 & C7
Load Pn = 676 kips
Assume Column size=16"X16"
Pu = φPn(max) = 0.80φAg [0.85fc′ (1 − ρg ) + ρfy ]
Pu = 0.80 ∗ 0.65 ∗ 256 ∗ {0.85 ∗ 4 ∗ (1 − 0.03) + 0.03 ∗ 60} = 679 kips > 676 𝑘𝑖𝑝𝑠
so ok
AS = ρ ∗ Ag = 0.03 ∗ 256 = 7.68 in2
34
4.5.2 Column ID: C1, C4, C9 & C12
Load Pn = 175 kips
Assume Column size=12"X12"
Pu = φPn(max) = 0.80φAg [0.85fc′ (1 − ρg ) + ρfy ]
Pu = 0.80 ∗ 0.65 ∗ 144 ∗ {0.85 ∗ 4 ∗ (1 − 0.03) + 0.03 ∗ 60} = 382 kips > 175 kips
so ok
AS = ρ ∗ Ag = 0.03 ∗ 144 = 4.32 in2
4.5.3 Column ID: C2 & C3
Load Pn = 359 kips
Assume Column size=12"X12"
Pu = φPn(max) = 0.80φAg [0.85fc′ (1 − ρg ) + ρfy ]
Pu = 0.80 ∗ 0.65 ∗ 144 ∗ {0.85 ∗ 4 ∗ (1 − 0.03) + 0.03 ∗ 60} = 382 kips > 359 kips
So ok
AS = ρ ∗ Ag = 0.03 ∗ 144 = 4.32 in2
4.5.4 Column ID: C5, C8, C10 & C11
Load Pn = 320 kips
Assume Column size=12"X12"
Pu = φPn(max) = 0.80φAg [0.85fc′ (1 − ρg ) + ρfy ]
Pu = 0.80 ∗ 0.65 ∗ 144 ∗ {0.85 ∗ 4 ∗ (1 − 0.03) + 0.03 ∗ 60} = 382 kips > 320 kips
So ok
AS = ρ ∗ Ag = 0.03 ∗ 144 = 4.32 in2
(Note: The minimum column dimension to 12 inches. Bibliography No. 2)
35
4.6 Load calculation on column for 5th storey
C1 = (69.89 ∗ 317 + 87.5 ∗
15.75
2
15.75
C2 = {147.55 ∗ 317 + 87.5 ∗ (
17.75
+ 87.5 ∗
+
2
2
+ 1425) ∗ 5 = 125229 lb = 125 kips
17.5
17.75
2
2
) + 87.5 ∗
+ 2227} ∗ 5 = 256158 lb
= 256 kips
15.75
C3 = {147.55 ∗ 317 + 87.5 ∗ (
+
2
17.5
17.75
2
2
) + 87.5 ∗
+ 2227} ∗ 5 = 256158 lb
= 256 kips
C4 = (69.89 ∗ 317 + 87.5 ∗
15.75
2
17.75
C5 = {129.94 ∗ 317 + 87.5 ∗ (
17.75
+ 87.5 ∗
+
2
2
+ 1425) ∗ 5 = 125229 lb = 125 kips
15.25
15.75
2
2
) + 112.5 ∗
+ 2227} ∗ 5 = 228738 lb
= 229 kips
C6 = {274.31 ∗ 317 + 125 ∗ (
17.75
2
+
15.25
15.75
2
2
) + 112.5 ∗ (
+
17.5
) + 5700} ∗ 5
2
= 482945.4 lb = 483 kips
15.75
C7 = {274.31 ∗ 317 + 112.5 ∗ (
2
∗
17.5
17.75
2
2
) + 125 ∗ (
+
15.25
2
) + 5700} ∗ 5
= 482945.4 lb = 483 kips
17.75
C8 = {129.94 ∗ 317 + 87.5 ∗ (
2
+
15.25
15.75
2
2
) + 112.5 ∗
+ 2227} ∗ 5 = 228738 lb
= 229 kips
C9 = (60 ∗ 317 + 87.5 ∗
15.25
2
+ 87.5 ∗ 9 + 1425) ∗ 5 = 109499 lb = 110 kips
C10 = {126.77 ∗ 317 + 87.5 ∗
C11 = {126.77 ∗ 317 + 87.5 ∗
15.25
2
15.25
2
+ 87.5 ∗ (
+ 87.5 ∗ (
17.5
2
) + 2227} ∗ 5 = 219230 lb = 219 kips
15.75
2
+
17.5
2
) + 2227} ∗ 5 = 222675 lb
= 223 kips
C12 = (60 ∗ 317 + 87.5 ∗
15.25
2
+ 87.5 ∗
15.75
2
+ 1425) ∗ 5 = 109007 lb = 110 kips
36
4.7 Column design for 5th storey
4.7.1 Column ID: C6 & C7
Load Pn = 483 kips
Assume Column size=14"X14"
Pu = φPn(max) = 0.80φAg [0.85fc′ (1 − ρg ) + ρfy ]
Pu = 0.80 ∗ 0.65 ∗ 196 ∗ {0.85 ∗ 4 ∗ (1 − 0.03) + 0.03 ∗ 60} = 520 kips > 483 kips
so ok
AS = ρ ∗ Ag = 0.03 ∗ 196 = 5.88 in2
4.7.2 All columns are minimum size = 12" X 12"
Steel area is,
AS = ρ ∗ Ag = 0.03 ∗ 144 = 4.32 in2
(Note: The minimum column dimension to 12 inches. Bibliography No. 2)
Table 4.1
Column design summary without lateral load condition
Column
ID
C1
C2
C3
C4
C5
C6
C7
C8
C9
C10
C11
C12
Steel Area in in2
Column Size
1st Storey
12" X 12"
12" X 18"
12" X 18"
12" X 12"
12" X 16"
20" X 20"
20" X 20"
12" X 16"
12" X 12"
12" X 16"
12" X 16"
12" X 12"
3rd Storey
12" X 12"
12" X 12"
12" X 12"
12" X 12"
12" X 12"
16" X 16"
16" X 16"
12" X 12"
12" X 12"
12" X 12"
12" X 12"
12" X 12"
5th Storey
12" X 12"
12" X 12"
12" X 12"
12" X 12"
12" X 12"
14" X 14"
14" X 14"
12" X 12"
12" X 12"
12" X 12"
12" X 12"
12" X 12"
1st Storey
4.32
6.48
6.48
4.32
5.76
12
12
5.76
4.32
5.76
5.76
4.32
3rd Storey
4.32
4.32
4.32
4.32
4.32
7.68
7.68
4.32
4.32
4.32
4.32
4.32
5th Storey
4.32
4.32
4.32
4.32
4.32
5.88
5.88
4.32
4.32
4.32
4.32
4.32
37
CHAPTER-V
ETABS ANALYSIS
5.1 Introduction of ETABS
ETABS will continue to be the professional engineer's choice for an end-to-end, multi-material
structural software solution. For static, dynamic, P-delta, non-linear, buckling or cable analysis,
ETABS is the industry standard. Complex models can be quickly and easily generated through
powerful graphics, text and spreadsheet interfaces that provide true interactive model generation,
editing, and analysis. ETABS easily generates comprehensive custom reports for management,
architects, owners, etc. Reports contain only the information you want, where you want it. Add
your own logo as well as graphical input and output results. Export all data to Microsoft Word.
The analysis of the building has done by computer software ETABS for 3D analysis of the
structure. A manual design has been done by USD methods for design beam & column.
5.2 Modeling a building by ETABS
Step-1: Geometry
New page-File- New modal
Building plan Grid system and spacing-Grid only-Edit Grid
Unit – (kip-ft)
X Grid
Display Grid as Spacing
Bubble size-(60)
Y grid- Same as
Simple story data
No of story-11
Typical story height-10 ft
Bottom story height-8 ft
Ok
38
Step-2: Assign property
Column
Define
Frame section
Add rectangular
Section name (CC)
Material Properties (Column)
Dimension-Depth (12)-Width (12)
Reinforcement
Design Type (column)
Check/Design-(Reinforcement to be designed),
Ok
(Define Beam, Slab, and Share Wall as same)
Step-3: Insert support
Support
Selects below all column
Column symbol select
Assign Restrain
First restrain
Fixed
Ok
Step-4: Assign load
Define
Static load Cases
Type
Dead load select
Load (DL)
Add new load- (Ok)
39
Modify load- (OK)
(Define LL and WL same as)
Step-5: Load combination
Define
Load combinations
Add New Combination Data
Case name (like as DEAD static load)
Scale Factor (like as 1.2)
Add
Ok
5.3 Analysis of a ten storied frame structure residential building
Analyze
Check model (all select)
Ok
Analyze
Run analysis
Ok
Beam moment and Shear force out put
Display
Show member forces/stress diagram
Frame/Pier/Spandrel forces
Load (Com-1)
Beam moment (3-3)
Show value
OK
40
Fig. 5.1.1 Three dimensional view on ETABS
41
Fig. 5.1.2 Axial force diagram
42
Fig. 5.1.3 Bending moment diagram
43
5.4 Summary after analysis by ETABS
Table 5.1 Maximum moment on beams
Beam ID
B1
B2
B3
B4
B5
B6
B7
B8
B9
B10
B11
B12
B13
B14
B15
Positive Moment
(Max) in kip-in
629
392
566
660
868
762
721
317
620
818
1059
1031
969
680
356
Negative Moment
(Max) in kip-in
1258
783
1133
1319
1736
1525
1442
633
610
1636
2119
2062
1938
1359
712
Table 5.2 Data for column
1st Story
Column
ID
3rd Story
5th Story
C1
Vertical
Mx
Load in
in
kips
kip-in
340
326
My
in
kip-in
689
Vertical
Mx
Load in
in
kips
kip-in
285
548
My
Vertical
Mx
in
Load in
in
kip-in
kips
kip-in
471
209
614
My
in
kip-in
386
C2
686
776
782
543
1119
619
404
1135
460
C3
702
674
800
565
1120
644
406
1250
463
C4
378
939
363
309
982
296
232
920
223
C5
359
672
562
320
153
95
193
160
70
C6
1385
1828
916
1229
1229
339
490
827
273
C7
1473
1945
85
1179
1556
345
484
248
499
C8
609
953
875
482
574
905
356
404
861
C9
220
167
216
88
129
286
71
75
290
C10
440
191
565
225
224
350
98
29
38
C11
538
963
613
410
917
467
291
775
331
C12
337
822
324
262
794
252
188
180
242
44
CHAPTER-VI
BEAM DESIGN WITH LATERAL LOAD
6.1 Beams design with lateral load conditions
6.1.1 Grid C
Beam ID: B11 & B12
-Mmax = 2119 kip-in
ø=0.9
Mu
d = √(
)
øRb
ρ = 0.85 ∗ β ∗
2119
)
0.9 ∗ 0.912 ∗ 12
d= 14.67"
d = √(
Total depth “h” = d+ clear cover
f′c
€c
∗
fy €c + €t
ρ = 0.85 ∗ 0.85 ∗
4
0.003
∗
60 0.003 + 0.005
ρ= 0.0181
R = ρ ∗ fy ∗ (1 − 0.588 ∗
h= 14.67"+2.5"
ρ ∗ fy
)
f ′c
h=17.17 ≈ 18" (So ok)
R = 0.0181 ∗ 60 ∗ (1 − 0.588 ∗
Effective Depth d = 18"-2.5"
R= 0.912 ksi
0.0181 ∗ 60
)
4
d= 15.50"
Check for Singly or Doubly Beam:
M= *R*b*d2
M=0 .9*0.912*12*15.502
M= 2366 kip-in > Mu
So it is singly reinforcement beam.
Let, a= 4.32"
As =
Mu
a
ø ∗ fy ∗ (d − 2)
2119
As =
0.9 ∗ 60 ∗ (15.5 –
4.32
2 )
As= 2.94 in2
45
Check a:
As ∗ fy
a=
0.85 ∗ f`c ∗ b
a=
2.94 ∗ 60
0.85 ∗ 4 ∗ 12
a= 4.32"
C=
a
4.32
=
= 5.10 in
β1 0.85
∈t =∈u ×
d−C
C
∈t = 0.003 ×
15.5 − 5.10
5.10
∈t = 0.0061
So ∈t = 0.005 (ok)
So As= 2.94 in2
∴ Beam size = 12" X 18"
Design of T beam
+Mmax = 1059 kip-in
Calculation effective width of flange
a) ¼ * Span = ¼ *17.75*12 = 53.24 ≈ 54 in
b) bw+16*t = 12+16*5 = 92 in
c) Spacing of beam = 15*12 =180 in
∴ bf = 54 in
a = 0.42
1059
As =
0.9 ∗ 60 ∗ (15.5 –
0.42
2 )
As= 1.28 in2
Check a:
As ∗ fy
a=
0.85 ∗ f`c ∗ bf
a=
1.28 ∗ 60
0.85 ∗ 4 ∗ 54
46
a= 0.42"
∴ As = 1.28 in
6.1.2 Grid A
Beam ID: B15
-Mmax = 712 kip-in
ø=0.9
Mu
d = √(
)
øRb
ρ = 0.85 ∗ β ∗
f′c
€c
∗
fy €c + €t
712
)
0.9 ∗ 0.912 ∗ 12
d= 8.51"
ρ = 0.85 ∗ 0.85 ∗
Total depth “h” = d+ clear cover
ρ= 0.0181
d = √(
4
0.003
∗
60 0.003 + 0.005
R = ρ ∗ fy ∗ (1 − 0.588 ∗
h= 8.51"+2.5"
h=11.01 ≈ 12" (So ok)
ρ ∗ fy
)
f ′c
R = 0.0181 ∗ 60 ∗ (1 − 0.588 ∗
Effective Depth d = 12"-2.5"
0.0181 ∗ 60
)
4
R= 0.912 ksi
d= 9.5"
Check for Singly or Doubly Beam:
M= *R*b*d2
M=0 .9*0.912*12*9.52
M= 889 kip-in > Mu
So it is singly reinforcement beam.
Let, a= 2.35"
As =
Mu
a
ø ∗ fy ∗ (d − 2)
712
As =
0.9 ∗ 60 ∗ (9.5 –
2.50
2 )
As= 1.60 in2
Check a:
As ∗ fy
a=
0.85 ∗ f`c ∗ b
a=
1.6 ∗ 60
0.85 ∗ 4 ∗ 12
47
a= 2.35"
C=
a
2.35
=
= 2.76 in
β1 0.85
∈t =∈u ×
d−C
C
∈t = 0.003 ×
9.5 − 2.76
2.76
∈t = 0.007
So ∈t = 0.005 (ok)
So As= 1.60 in2
∴ Beam size = 12" X 12"
(Note: The minimum width of a beam should be 12 inches. Bibliography No. 11)
+Mmax = 356 kip-in
As= 0.80 in2
Similarly
6.1.3 Grid B
Beam ID: B13 & B14
-Mmax = 1938 kip-in
Mu
d = √(
)
øRb
ø=0.9
R= 0.912 ksi
1938
)
0.9 ∗ 0.912 ∗ 12
d= 11.61"
d = √(
Total depth “h” = d+ clear cover
h= 14.03"+2.5"
h=16.53 ≈ 18" (So ok)
Effective Depth d = 18"-2.5"
d= 15.5"
48
As =
Mu
a
ø ∗ fy ∗ (d − 2)
1938
As =
0.9 ∗ 60 ∗ (15.5 –
4.20
2 )
As= 2.70 in2
∴ Beam size = 12" X 18"
+Mmax = 969 kip-in
As= 1.35 in2
6.1.4 Grid D
Beam ID: B9 & B10
-Mmax = 1636 kip-in
d = √(
Mu
)
øRb
1636
)
0.9 ∗ 0.912 ∗ 12
d= 12.89"
d = √(
Total depth “h” = d+ clear cover
h= 12.89"+2.5"
h=15.39 ≈ 16" (So ok)
Effective Depth d = 16"-2.5"
d= 13.5"
1636
As =
0.9 ∗ 60 ∗ (13.5 –
3.85
2 )
As= 2.62 in2
∴ Beam size = 12" X 16"
+Mmax = 818 kip-in
As= 1.14 in2
6.1.5 Grid 1
Beam ID: B7 & B8
-Mmax = 1442 kip-in
49
d = √(
Mu
)
øRb
1442
)
0.9 ∗ 0.912 ∗ 12
d= 12.1"
d = √(
Total depth “h” = d+ clear cover
h= 12.1"+2.5"
h=14.6 ≈ 15" (So ok)
Effective Depth d = 15"-2.5"
d= 12.5"
1442
As =
0.9 ∗ 60 ∗ (12.5 –
3.70
2 )
As= 2.51 in2
∴ Beam size = 12" X 15"
+Mmax = 721 kip-in
As= 1.08 in2
6.1.6 Grid 2
Beam ID: B4, B5 & B6
-Mmax = 1736 kip-in
d = √(
Mu
)
øRb
1736
)
0.9 ∗ 0.912 ∗ 12
d= 13.28"
d = √(
Total depth “h” = d+ clear cover
h= 13.28"+2.5"
h=15.78 ≈ 16" (So ok)
Effective Depth d = 16"-2.5"
d= 13.5"
50
As =
1736
4.1
0.9 ∗ 60 ∗ (13.5 – 2 )
As= 2.81 in2
∴ Beam size = 12" X 16"
+Mmax = 868 kip-in
As= 1.21 in2
6.1.7 Grid 3
Beam ID: B1, B2&B3
-Mmax = 1258 kip-in
d = √(
Mu
)
øRb
1258
)
0.9 ∗ .912 ∗ 12
d= 11.30"
d = √(
Total depth “h” = d+ clear cover
h= 11.30"+2.5"
h=13.80 ≈ 14" (So ok)
Effective Depth d = 14"-2.5"
d= 11.50"
As =
Mu
a
ø ∗ fy ∗ (d − )
2
1258
As =
0.9 ∗ 60 ∗ (11.50 –
3.50
2 )
So As= 2.40 in2
∴ Beam size = 12" X 14"
+Mmax = 629 kip-in
As= 1.03 in2
51
Table 6.1
Summary of the beam design for with lateral load condition
Beam ID
Beam Size
B1
B2
B3
B4
B5
B6
B7
B8
B9
B10
B11
B12
B13
B14
B15
12" X 14"
12" X 14"
12" X 14"
12" X 16"
12" X 16"
12" X 16"
12" X 15"
12" X 15"
12" X 16"
12" X 16"
12" X 18"
12" X 18"
12" X 18"
12" X 18"
12" X 12"
Positive Steel Area
in in2
1.03
0.75
0.93
1.07
1.21
1.05
1.08
0.55
1.00
1.14
1.28
1.44
1.35
0.94
0.80
Negative Steel Area
in in2
2.40
1.49
2.16
2.13
2.81
2.47
2.51
1.10
1.00
2.62
2.94
2.86
2.70
1.88
1.60
52
CHAPTER-VII
COLUMN DESIGN WITH LATERAL LOAD
7.1 For 1st storey column design
Column ID: C1
Assume Column size=14"X14"
Ag = 14 X 14 = 196 in2
ρ = 0.03
Pu = 340 kips
MX = 326 kip-in
My = 689 kip-in
ex =
MX 326
=
= 0.959 in
Pu
340
ey =
My 689
=
= 2.026 in
Pu
340
ex = 0
γ=
14 − 5
= 0.64
14
ey 2.026
=
= 0.1447
h
14
Pnx0 = 0.8629 ∗ 4 ∗ 196 = 676.48 kips
ey = 0
γ=
14 − 5
= 0.64
14
ex 0.959
=
= 0.064
h
14
Pny0 = 1.071 ∗ 4 ∗ 196 = 840 kips
P0 = 1.27 ∗ 4 ∗ 196 = 995.69 kips
53
1
1
1
1
1
1
1
=
+
− =
+
−
Pn Pnx0 Pny0 P0 676.48 840 995.69
Pn = 600.82 kips
Pu = 0.65 X 600.84 = 390 kips > 340 𝑘𝑖𝑝𝑠 (𝑆𝑜 𝑜𝑘)
∴ As = ρAg = 0.03 X 196 = 5.88 in2
Column ID: C2
Assume Column size=14"X20"
Ag = 14 X 20 = 280 in2
ρ = 0.03
Pu = 686 kips
MX = 776 kip-in
My = 782 kip-in
ex =
MX 776
=
= 1.13 in
Pu
686
ey =
My 782
=
= 1.14 in
Pu
686
ex = 0
γ=
20 − 5
= 0.75
20
ey 1.14
=
= 0.06
h
20
Pnx0 = 1.1 ∗ 4 ∗ 280 = 1232 kips
ey = 0
γ=
14 − 5
= 0.64
14
ex 1.13
=
= 0.08
h
14
Pny0 = 1.069 ∗ 4 ∗ 280 = 1197 kips
54
P0 = 1.27 ∗ 4 ∗ 280 = 1422 kips
1
1
1
1
1
1
1
=
+
− =
+
−
Pn Pnx0 Pny0 P0 1232 1197 1422
Pn = 1059.08 kips
Pu = 0.65 X 1059.08 = 688.4 kips > 686 𝑘𝑖𝑝𝑠 (𝑆𝑜 𝑜𝑘)
∴ As = ρAg = 0.03 X 280 = 8.4 in2
Similarly
Column ID: C3
Column size=15"X20"
∴ As = 9.00 in2
Column ID: C4
Column size=15"X15"
∴ As = 6.75 in2
Column ID: C5
Column size=14"X16"
∴ As = 6.72 in2
Column ID: C6
Column size=24"X24"
∴ As = 17.28 in2
Column ID: C7
Column size=24"X24"
∴ As = 17.28 in2
Column ID: C8
Column size=16"X20"
∴ As = 9.60 in2
55
Column ID: C9
Column size=12"X12"
∴ As = 4.32 in2
(Note: The minimum column dimension to 12 inches. Bibliography No. 2)
Column ID: C10
Column size=14"X18"
∴ As = 7.56 in2
Column ID: C11
Column size=14"X20"
∴ As = 8.4 in2
Column ID: C12
Column size=14"X14"
∴ As = 5.88 in2
56
7.2 For 3rd storey column design
Column ID: C1
Assume Column size=14"X14"
Ag = 14 X 14 = 196 in2
ρ = 0.03
Pu = 285 kips
MX = 548 kip-in
My = 471 kip-in
ex =
MX 548
=
= 1.923 in
Pu
285
ey =
My 471
=
= 1.653 in
Pu
285
ex = 0
γ=
14 − 5
= 0.64
14
ey 1.653
=
= 0.118
h
14
Pnx0 = 0.918 ∗ 4 ∗ 196 = 720.16 kips
ey = 0
γ=
14 − 5
= 0.64
14
ex 1.923
=
= 0.14
h
14
Pny0 = 0.889 ∗ 4 ∗ 196 = 687 kips
P0 = 1.27 ∗ 4 ∗ 196 = 995.69 kips
1
1
1
1
1
1
1
=
+
− =
+
−
Pn Pnx0 Pny0 P0 720.16 687 995.69
Pn = 549.54 kips
Pu = 0.65 X 549.54 = 357 kips > 285 𝑘𝑖𝑝𝑠 (𝑆𝑜 𝑜𝑘)
57
∴ As = ρAg = 0.03 X 196 = 5.88 in2
Column ID: C2
Assume Column size=14"X20"
Ag = 14 X 20 = 280 in2
ρ = 0.03
Pu = 543 kips
MX = 1119 kip-in
My = 619 kip-in
ex =
MX 1119
=
= 2.06 in
Pu
543
ey =
My 619
=
= 1.14 in
Pu
543
ex = 0
γ=
20 − 5
= 0.75
20
ey 1.14
=
= 0.06
h
20
Pnx0 = 1.12 ∗ 4 ∗ 280 = 1254 kips
ey = 0
γ=
14 − 5
= 0.64
14
ex 2.06
=
= 0.15
h
14
Pny0 = 0.864 ∗ 4 ∗ 280 = 968 kips
P0 = 1.27 ∗ 4 ∗ 280 = 1422 kips
1
1
1
1
1
1
1
=
+
− =
+
−
Pn Pnx0 Pny0 P0 1254 968 1422
Pn = 887 kips
58
Pu = 0.65 X 887 = 576.6 kips > 543 𝑘𝑖𝑝𝑠 (𝑆𝑜 𝑜𝑘)
∴ As = ρAg = 0.03 X 280 = 8.4 in2
Similarly
Column ID: C3
Column size=15"X20"
∴ As = 9.00 in2
Column ID: C4
Column size=14"X14"
∴ As = 5.88 in2
Column ID: C5
Column size=12"X14"
∴ As = 5.04 in2
Column ID: C6
Column size=20"X20"
∴ As = 12 in2
Column ID: C7
Column size=22"X22"
∴ As = 14.52 in2
Column ID: C8
Column size=14"X18"
∴ As = 7.56 in2
59
Column ID: C9
Column size=12"X12"
∴ As = 4.32 in2
(Note: The minimum column dimension to 12 inches. Bibliography No. 2)
Column ID: C10
Column size=12"X14"
∴ As = 5.05 in2
Column ID: C11
Column size=14"X18"
∴ As = 7.56 in2
Column ID: C12
Column size=14"X14"
∴ As = 5.88 in2
60
7.3 For 5th storey column design
Column ID: C1
Assume Column size=14"X14"
Ag = 14 X 14 = 196 in2
ρ = 0.03
Pu = 209 kips
MX = 614 kip-in
My = 386 kip-in
ex =
MX 614
=
= 2.938 in
Pu
209
ey =
My 386
=
= 1.847 in
Pu
209
ex = 0
γ=
14 − 5
= 0.64
14
ey 1.847
=
= 0.132
h
14
Pnx0 = 0.89 ∗ 4 ∗ 196 = 697.76 kips
ey = 0
γ=
14 − 5
= 0.64
14
ex 2.938
=
= 0.21
h
14
Pny0 = 0.7 ∗ 4 ∗ 196 = 549 kips
P0 = 1.27 ∗ 4 ∗ 196 = 995.69 kips
1
1
1
1
1
1
1
=
+
− =
+
−
Pn Pnx0 Pny0 P0 697.76 549 995.69
Pn = 444 kips
Pu = 0.65 X 444 = 288 kips > 285 𝑘𝑖𝑝𝑠 (𝑆𝑜 𝑜𝑘)
61
∴ As = ρAg = 0.03 X 196 = 5.88 in2
Column ID: C2
Assume Column size=14"X18"
Ag = 14 X 18 = 252 in2
ρ = 0.03
Pu = 404 kips
MX = 1135 kip-in
My = 460 kip-in
ex =
Mx 1135
=
= 2.81 in
Pu
404
ey =
My 460
=
= 1.14 in
Pu
404
ex = 0
γ=
18 − 5
= 0.722
18
ey 1.14
=
= 0.063
h
20
Pnx0 = 1.06 ∗ 4 ∗ 252 = 1073 kips
ey = 0
γ=
14 − 5
= 0.64
14
ex 2.81
=
= 0.2
h
14
Pny0 = 0.75 ∗ 4 ∗ 252 = 755 kips
P0 = 1.27 ∗ 4 ∗ 252 = 1280 kips
1
1
1
1
1
1
1
=
+
− =
+
−
Pn Pnx0 Pny0 P0 1073 755 1280
62
Pn = 677 kips
Pu = 0.65 X 677 = 440 kips > 404 𝑘𝑖𝑝𝑠 (𝑆𝑜 𝑜𝑘)
∴ As = ρAg = 0.03 X 252 = 7.56 in2
Column ID: C3
Column size=14"X18"
∴ As = 7.56 in2
Column ID: C4
Column size=14"X14"
∴ As = 5.88 in2
Column ID: C5
Column size=12"X12"
∴ As = 4.32 in2
Column ID: C6
Column size=15"X15"
∴ As = 6.75 in2
Column ID: C7
Column size=15"X15"
∴ As = 6.75 in2
Column ID: C8
Column size=14"X16"
∴ As = 6.75 in2
63
Column ID: C9
Column size=12"X12"
∴ As = 4.32 in2
Column ID: C10
Column size=12"X12"
∴ As = 4.32 in2
Column ID: C11
Column size=12"X16"
∴ As = 5.75 in2
Column ID: C12
Column size=12"X12"
∴ As = 4.32 in2
(Note: The minimum column dimension to 12 inches. Bibliography No. 2)
64
Table 7.1
Column design summary with lateral load condition
Column
ID
C1
C2
C3
C4
C5
C6
C7
C8
C9
C10
C11
C12
Steel Area in in2
Column Size
1st Storey
14" X 14"
14" X 20"
15" X 20"
15" X 15"
14" X 16"
24" X 24"
24" X 24"
16" X 20"
12" X 12"
14" X 18"
14" X 20"
14" X 14"
3rd Storey
14" X 14"
14" X 20"
15" X 20"
14" X 14"
12" X 14"
20" X 20"
22" X 22"
14" X 18"
12" X 12"
12" X 14"
14" X 18"
14" X 14"
5th Storey
14" X 14"
14" X 18"
14" X 18"
14" X 14"
12" X 12"
15" X 15"
15" X 15"
14" X 16"
12" X 12"
12" X 12"
12" X 16"
12" X 12"
1st Storey
5.88
8.4
9
6.75
6.72
17.28
17.28
9.6
4.32
7.56
8.4
5.88
3rd Storey
5.88
8.4
9
5.88
5.04
12
14.52
7.56
4.32
5.04
7.56
5.88
5th Storey
5.88
7.56
7.56
5.88
4.32
6.75
6.75
6.72
4.32
4.32
5.76
4.32
65
CHAPTER VIII
RESULTS & DISCUSSIONS
8.1 Results
For beam:
Without lateral load
Positive
Negative Steel
Steel Total
Total
in in2
in in2
13.52
23.18
With lateral load
Positive Steel
Negative
Total
Steel Total
in in2
in in2
15.62
32.67
For column:
No of Storey
1st Storey
3rd Storey
5th Storey
Without lateral load
Total reinforcement of column in
in2
77.28
58.56
54.96
With lateral load
Total reinforcement of column in
in2
107.07
91.08
70.14
8.2 Discussions
From the results we can see that the amount of reinforcement varies in a significant amount while
considering lateral loads and not considering lateral loads in the design. The concrete section also varies
for both the condition. The inner beams carry more load than the perimeter beams. The inner columns
carry more load than the other columns. The perimeter columns carry medium load than the inner
columns. The corner columns carry smaller loads than other columns.
 For Beam design it is found that the amount of steel required in design while
considering lateral loads are about 28% higher of the amount of steel required in design
without considering lateral loads.
 In 1st storey columns design it is found that the amount of steel required in design while
considering lateral loads are about 38% higher of the amount of steel required in design
without considering lateral loads.
 In 3rd storey columns design it is found that the amount of steel required in design while
considering lateral loads are about 55% higher of the amount of steel required in design
66
 In 5th storey columns design it is found that the amount of steel required in design while
considering lateral loads are about 27% higher of the amount of steel required in design
without considering lateral loads.
 Average considering lateral loads are about 37% higher of the amount of steel required
in design without considering lateral loads.
8.3 Comparison chart & data
8.3.1 Positive steel area (in2) comparison chart on beam
Positive Steel Area (in2) Comparison on Beam
1.6
Without lateral load
1.4
With lateral load
Steel Area
1.2
1
0.8
0.6
0.4
0.2
0
Beam ID
B1
B2
B3
B4
B5
B6
B7
B8
B10
B11
Without lateral load 0.64 0.76 0.64 0.97 1.14 0.97 0.86 0.64 0.89
0.6
1.37 0.89 1.37 0.89 0.89
With lateral load
1.14 1.28 1.44 1.35 0.94
1.03 0.75 0.93 1.07 1.21 1.05 1.08 0.55
B9
1
B12
B13
B14
B15
0.8
67
8.3.2 Negative steel area (in2) comparison chart on beam
Negative Steel Area (in2) Comparison Chart on Beam
3.5
Without lateral load
3
With lateral load
Steel Area
2.5
2
1.5
1
0.5
Beam ID0
B1
B2
B3
B4
B5
B6
B7
1.7
1.5
Without lateral load 0.98
1.2
With lateral load
1.49 2.16 2.13 2.81 2.47 2.51
2.4
0.98 1.51 1.86
B8
B9
B10 B11 B12 B13 B14 B15
1.12 1.56 1.03 2.48 1.61 2.48 1.61 1.56
1.1
1
2.62 2.94 2.86
2.7
1.88
1.6
8.3.3 1st story comparison chart & data
Steel area comparison between two case (1st Storey)
20
18
Without lateral load
Steel Area
16
14
With lateral load
12
10
8
6
4
2
Column0ID C1
C2
C3
C4
C5
C6
C7
C8
C9
C10
C11
C12
Without lateral load 4.32
6.48
6.48
4.32
5.76
12
12
5.76
4.32
5.76
5.76
4.32
With lateral load
8.4
9
6.75
6.72
9.6
4.32
7.56
8.4
5.88
5.88
17.28 17.28
68
8.3.4 3rd storey comparison chart & data
Steel area comparison between two case (3rd Storey)
16
14
Without lateral load
With lateral load
Steel Area
12
10
8
6
4
2
0
Column ID
C1
C2
C3
C4
C5
C6
C7
C8
C9
C10
C11
C12
Without lateral load 4.32
4.32
4.32
4.32
4.32
7.68
7.68
4.32
4.32
4.32
4.32
4.32
With lateral load
8.4
9
5.88
5.04
12
14.52
7.56
4.32
5.04
7.56
5.88
5.88
8.3.5 5th storey comparison chart & data
Steel area comparison between two case (5th Storey)
8
Without lateral load
7
With lateral load
Steel Area
6
5
4
3
2
1
Column 0ID
C1
C2
C3
C4
C5
C6
C7
C8
C9
C10
C11
C12
Without lateral load
4.32
4.32
4.32
4.32
4.32
5.88
5.88
4.32
4.32
4.32
4.32
4.32
With lateral load
5.88
7.56
7.56
5.88
4.32
6.75
6.75
6.72
4.32
4.32
5.76
4.32
69
CHAPTER IX
CONCLUSION & RECOMMENDATION
9.1 Conclusion
The column section varies for different stories. Amount of reinforcement in column varies for
different stories. The beam section does not vary significantly for different stories. Amount of
reinforcement in beam varies for different stories. The cost is higher when lateral load is
considered for design than that is not considered for design. Though the overall cost is higher
but the design of building considering lateral load is more safe and reliable.
9.2 Recommendation
The analysis can be compared for flat plate design. Dynamic analysis for earthquake loading
was not considered in this study and hence should be considered in future study. The similar
investigation can be made by providing shear wall or other type of lateral bracing system.
70
BIBLIOGRAPHY
1. ACI Code, 2002, 1995.
2. ACI 318 - 21.6.1.1
3. Arthur H. Nilson, (2005) Design of Concrete Structures (12th Edition), TATA
McGraw-Hill Company Ltd, New Delhi.
4. Arthur H. Nilson, (2006) Design of Concrete Structures (13th Edition), TATA
McGraw-Hill Company Ltd, New Delhi.
5. Bangladesh National Building Code BNBC-2006. Housing and building Research
Institute. Darussalam, Mirpur, Dhaka 1218.and Bangladesh Standard and Testing
Institution. 16/A Tejgaon Industrial Area, Dhaka1208.
6. Evangelos Petroutsos, (1998) Mastering Visual Basic 6, 1st Indian edition.
7. Everard Noel J. and Tanner John L. “Reinforced Concrete Design” McGraw-Hill, New
York, USA, 1966.
8. Khan, A.F. Advanced Concrete Structures. Dhaka-1: Rizia Khatun 61, Hossani Dalan.
2002.
9. Leet Kenneth M. and Bernal Dionisio “Reinforced Concrete Design” 3 rd edition
McGraw-Hill, Singapore, 1997.
10. Park and Pauley, Reinforced Concrete Structure, The McGraw-Hill Companies,
Inc.International edition.
11. Refer to the table on page 6 from section 9 of the ACI 318.
12. S.N. Sinha, (1995) Handbook of Concrete Design, Tata McGraw-Hill Publishing
Companies.
13. Vazirani, V.N. Concrete Structures. Delhi: Khanna Publishers 2-B, Nath Market, Nai
Shrak, 1980.
71