FOOTING,COLUMN,BEAM,SLAB,STAIR DESIGN FOR ANALYSIS BY MANUAL AND ETABS,SAFE.

1. 1
Presidency University
Azimur Rahman School of Engineering
DEPARTMENT OF CIVIL ENGINEERING
TITLE OF THE CAPSTONE DESIGN
DESIGN OF SIX STORED RESIDENTIAL BUILDING
Capstone Report (CE-492&CE-493)
DEPARTMENT OF CIVIL ENGINEERING
Presidency University, Dhaka, Bangladesh
January-2017

2. 2
TITLE OF THE CAPSTONE DESIGN
DESIGN OF SIX STORIED RESIDENTIAL BUILDING
A Capstone Design Submitted to the
Department of Civil Engineering
Presidency University
Azimur Rahman School of Engineering
Submitted By
Name Student ID
Sharmin Aktar 132219046
Md. Rajib Hossain 141054046
Md. Sumon Miah 141057046
Md.Mehedi Amin 141073046
Md. Kawser Ahmed Rony 141081046
Mostakim Alam 141112046
Submitted to the Department of Civil Engineering of Presidency University, Dhaka,
Bangladesh in partial fulfillment of the requirements for the degree of
BSC IN CIVIL ENGINEERING

3. 3
DECLARATION
It is stated that the work “DESIGN OF SIX STORIED RESIDENTIAL BUILDING “reported in this
capstone has been performed under supervision of Md. Rokon Hasan, Lecturer, Department
of Civil Engineering, Presidency University, Dhaka. To the best of our knowledge and belief,
the capstone contains no materials previously published or written by another person
except where due reference is made in the capstone itself.
SharminAktar Md. Rajib Hossain
Md. Sumon Miah Md. Mehedi Amin
Kawser Ahmed Rony Mostakim Alam

4. 4
ACKNOWLEDGEMENT
It gives us a great pleasure to express our greatest appreciation to all of those people who
had helped us along the training until writing this report.
First and foremost, we would like to express our great full to Allah SWT because of bring us
here and give us good health. Secondly, we would like to thanks our family for all their
support, love and prayer.
It would not have been possible to write that thesis without the help and support our
supervisor Md. Rokon Hasan, Lecturer, Department of civil Engineering, PRESSIDENCY
UNIVERSITY, Dhaka, Bangladesh for his valuable continuous guidance, helpful suggestions
and constant encouragement for this going of thesis.
We would like to acknowledge the academic and technical support of Presidency University
and its staff. Thanks, because guide and give us to perform here. With their Accompany this
report would a reality. Once again, we are great full for all their helping us to complete our
report successfully.
Thanks, you.

5. 5
DEDICATED
TO MY
BELOVED PARENTS

6. 6
TABLE OF CONTENTS
Sl. No. Title
CHAPTER 1
INTRODUCTION
1.1 GENERAL
1.2 OBJECTIVES
1.3 PRELIMINNARY DESIGN AND OPTIMIZATION
1.4 INITIAL SELECTION OF STRUCTURAL SYSTEM
1.5 ADVANTAGES AND LIMITATION STRUCTURAL OPTIMIZATION
CHAPTER 2
SLAB DESIGN
2.1 DESIGN DATA
2.2 LOAD CALCULATION
2.3 CHECK ONE WAY OR TWO -WAY SLAB
2.4 All Slab Maximum Moment Ageist Thickness Check
2.5 All Slab Maximum Moment Ageist Main Reinforcement Calculation
2.6 SLAB DESIGN HEAD CALCULATION
CHAPTER 3
BEAM DESIGN
3.1 STIRRUP DESIGN
3.2 MOMENT AND SHEAR CHART
3.3.0 BEAM DESIGN DATA
3.3.1 DESIGN OF BEAM FOR SHEAR
CHAPTER 4
4.1 GENERAL
4.2.2 COLUMN DESIGN CANCULATION
4.3 COLUMN DESIGN CANCULATION SUMMARY
4.4 AXIAL FORCE, MOMENT,2-2 AND MOMENT 3-3 CHART
4.5 INTERACTION DIAGRAM
4.6 STRIUP DESIGN
CHAPTER 5
STAIR DESIGN
5.1 STAIR DESIGNDATA
CHARTER 6
FOOTING DESIGN
6.1 FOORTING DESIGN DATA
CHARTER 7
UNDERGROUND WATER TANK DESIGN
7.1 DESIGN DATA
7.2 BASE SLAB DESIGN
7.3 COVER SLAB DESIGN

7. 7
7.4 DESIGN OF SIDE WALL
OVERHEAD WATER TANK DESIGN
7.5 DESIGN DATA
7.6 BASE SLAB DESIGN
7.7 COVER SLAB DESIGN
7.8 DESIGN OF SIDE WALL
CHAPTER 8
REFERENCES
CHAPTER 9
APPENDEX
9.1 EARTHQUAKE LOAD CALCULATION
9.2 WIND LOAD CALCULATION
9.3 BUILDING DESIGN USING ETABS
LIST OF FOGURE
Sl.No Title
Figure-1.1 3D View of Project
Figure-1.2 Typical Floor Lay out Plane
Figure-1.3 Typical Floor Working Lay out Plane
Figure-1.4 Ground Floor Lay out Plan
Figure-1.5 Stair Case Section View
Figure-1.6 Front Elevation
Figure-3.1 Slab Reinforcement in Short Direction
Figure-3.2 Slab Reinforcement in Long Direction
Figure-3.3 Slab Reinforcement Details (Both Long and Short Direction)

8. 8
Figure-4.1 Floor Beam Reinforcement Section and Details
Figure-4.2 Floor Beam Lay out Plan
Figure-5.1 Column lay out Plan
Figure-5.2 Column Reinforcement Details
Figure 6.1 Stair Reinforcement Details and Layout Plan
Figure-7.1 Footing and Column Layout Plane
Figure-7.2 Footing Layout Plane Cad View
Figure-7.3 Footing Details
Figure-7.4 Footing Reinforcement Details Section A-A
Figure-8.1 Under Ground Water Tank Design Detailing
Figure-8.2 Overhead Tank Design Detailing
Figure-12.1 ETABS Floor Plan
Figure-12.2 ETABS Floor Plan 3D
Figure-12.3 ETABS Elevation View
Figure-12.4 ETABS Elevation View 3D
Figure-12.5 ETABS Column and Beam 3D View
Figure-12.6 ETABS Axial Force Diagram with Value X-X
Figure-12.7 ETABS Axial Force Diagram with Value Y-Y
Figure-12.8 ETABS Shear Force 2-2 Diagram
Figure-12.9 ETABS Moment 2-2 Diagram
Figure-12.10 ETABS Moment 3-3 Diagram
Figure-12.11 Other Appendix Diagram

9. 9
LIST OF ABBREVIATION
BNBC = Bangladesh National Building Code.
ACI = American Concrete Institute.
ASTM = American Society for Testing of Materials.
AASHTO = American Associate State Highway Transport Officials.
DCSS = Design Criteria for Structural System.
BSS = Building Shapes and Setbacks
RCC=Reinforcement Cement Concrete
CC=Cement Concrete

10. 10
Symbols and Notation
The following symbols and notation apply only to the provision of Section 2.4
a = depth of equivalent rectangular stress block as defined in Sec 6.2.3.7
A = effective tension area of concrete surrounding the flexural tension
reinforcement and having the same centroid as that of the reinforcement,
divided by the number of bars. When the flexural reinforcement consists of
different bar sizes the number of bars or wires shall be computed as the
total area of reinforcement divided by the area of the largest bar used
Ag = gross area of section
= total area of longitudinal reinforcement to resist torsion
As = area of tension reinforcement
= area of compression reinforcement
As1 = area of tension reinforcement corresponding to moment of resistance M1
As2 = area of additional tension steel
Asf = area of reinforcement required to balance the longitudinal compressive
force in the overhanging portion of the flange of a T-beam
Ask = area of skin reinforcement per unit height in a side face
At = area of one leg of a closed stirrup resisting torsion within a distance s
Av = area of shear reinforcement within a distance
b = width of compression face of member
bt = width of that part of cross-section containing the closed stirrups resisting
torsion
bw = web width, or diameter of circular section
c = distance from extreme compression fiber to neutral axis
Ct = factor relating shear and torsional stress properties =
d = distance from extreme compression fiber to centroid of tension
reinforcement
dc = thickness of concrete cover measured from extreme tension fiber to Centre
of bar or wire located closest thereto
Es = modulus of elasticity of reinforcement
fs = calculated stress in reinforcement at service loads
fr = modulus of rupture of concrete
fy = specified yield strength of reinforcement
h = overall thickness of member
Icr = moment of inertia of cracked section transformed to concrete
Ie = effective moment of inertia for computation of deflection
Ig = moment of inertia of gross concrete section about centroidal axis,
neglecting reinforcement
M1 = moment of resistance of a section without compression steel
M2 = additional moment of resistance due to added compression steel and
additional tension steel As2

11. 11
Mn1 = moment of resistance developed by compression in the overhanging
portion of the T-flange
Mn2 = moment of resistance developed by the web of a T-beam
Ma = maximum moment in member at stage deflection is computed
Mcr = cracking moment
Mm = modified moment
Mmax = maximum moment at section due to externally applied loads
Mn = nominal flexural strength
Mu = factored moment at section
Nu = axial load normal to cross-section occurring simultaneously with Vu; to be
taken as positive for compression, negative for tension and to include
effects of tension due to creep and shrinkage
s = spacing of shear or torsion reinforcement in direction parallel to
longitudinal reinforcement
Tc = torsional moment strength provided by concrete
Tn = torsional moment strength
Ts = torsional moment strength provided by torsion reinforcement
Tu = torsional moment at section
Vu = shear at section
Vc = shear strength provided by concrete
Vn = shear strength
Vs = nominal shear strength provided by shear reinforcement
wu = factored load per unit length of beam or per unit area of slab
x = shorter overall dimension of rectangular part of cross-section
x1 = shorter Centre to Centre dimension of closed rectangular stirrup
y = longer overall dimension of rectangular part of cross-section
y1 = longer Centre to Centre dimension of closed rectangular stirrup
z = quantity limiting distribution of flexural reinforcement, see Eq (6.2.35)
a = depth of equivalent rectangular stress block as defined in Sec 6.2.3.7
Ab = area of an individual bar, mm2
As = area of tension reinforcement, mm2
Atr = total cross-sectional area of transverse reinforcement (stirrup or tie) within
a spacing s and perpendicular to plane of bars being spliced or developed,
mm2
Av = area of shear reinforcement within a distance s, mm2
bw = web width, or diameter of circular section, mm
d = distance from extreme compression fiber to centroid of tension
reinforcement, mm
db = nominal diameter of bar, mm
h = overall thickness of member, mm
a = additional embedment length at support or at point of inflection, mm
d = development length, mm

12. 12
db = basic development length, mm
dh = development length of standard hook in tension, measured from the
critical section to the farthest point on the bar, parallel to the straight part
of the bar, mm
hb = basic development length of standard hook in tension, mm
Mn = nominal moment strength at section, N-mm
N = number of bars, in a layer, being spliced or developed at a critical section
s = spacing of stirrups or ties, mm
Vu = factored shear force at section, N
b = ratio of area of reinforcement cut off to total area of tension
reinforcement at section.
cc = creep coefficient
Ec = modulus of elasticity of concrete
Es = modulus of elasticity of reinforcement
Et = modulus of elasticity of concrete at the age of loading t
ƒy = specified yield strength of reinforcement
K = coefficient of shrinkage
s = standard deviation
wc = unit weight of concrete
cc = creep strain in concrete
sh = shrinkage of plain concrete
a = depth of equivalent rectangular stress block for strength design
Ab = cross-sectional area of anchor bolt
Ae = effective area of masonry
Ag = gross area of wall
Amv= net area of masonry section bounded by wall thickness and length of section
in the direction of shear force considered
Ap = area of tension (pullout) cone of an embedded anchor bolt projected into
the surface of masonry
As = effective cross-sectional area of reinforcement in a flexural member
Av = area of steel required for shear reinforcement perpendicular to the
longitudinal reinforcement
b = effective width of rectangular member or width of flange for T and I section
bt = computed tension force on anchor bolt
bv = allowable shear force on anchor bolt
bw = width of web in T and I member
Bt = allowable tension force on anchor bolt
Bv = computed shear force on anchor bolt
c = distance from the neutral axis to extreme fiber
Cd = masonry shear strength coefficient
d = distance from the compression face of a flexural member to the centroid of
longitudinal tensile reinforcement
db = diameter of the reinforcing bar, diameter of bolt

13. 13
e = eccentricity of Pu
emu= maximum usable compressive strain of masonry
Em = modulus of elasticity of masonry
Es = modulus of elasticity of steel
fa = computed axial compressive stress due to design axial load
fb = computed flexural stress in the extreme fiber due to design bending load
only
fmd = computed compressive stress in masonry due to dead load only
fr = modulus of rupture
fs = computed stress in reinforcement due to design load
fy = tensile yield stress of reinforcement
fv = computed shear stress due to design load
F = loads due to weight and pressure of fluids or related moments and forces
Fa = allowable average axial compressive stress for centroid ally applied axial
load only
Fb = allowable flexural compressive stress if members were carrying bending load
only
Fbr = allowable bearing stress
Fs = allowable stress in reinforcement
Fsc = allowable compressive stress in column reinforcement
Ft = allowable flexural tensile stress in masonry
Fv = allowable shear stress in masonry
G = shear modulus of masonry
h = height of wall between points of support
h = effective height of a wall or column
H = actual height between lateral supports
H' = height of opening
I = moment of inertia about the neutral axis of the cross-sectional area
Ig,Icr= gross, cracked moment of inertia of the wall cross-section
j = ratio or distance between centroid of flexural compressive force and
centroid of tensile forces to depth, d
k = ratio of depth of the compression zone in flexural member to depth, d;
stiffening coefficient
 = length of a wall or segment
b = embedment depth of anchor bolt
be = anchor bolt edge distance, the least length measured from the edge of
masonry to the surface of the anchor bolt
d = required development length of reinforcement
L = actual length of wall
M = design moment
Mc = moment capacity of the compression steel in a flexural member about the
centroid of the tensile force
Mcr = cracking moment strength of the masonry wall

14. 14
Mm = the moment of the compressive force in the masonry about the centroid of
the tensile force in the reinforcement
Mn = nominal moment strength of the masonry wall
Ms = the moment of the tensile force in the reinforcement about the centroid of
the compressive force in the masonry
Mser = service moment at the mid-height of the panel, including P-Delta effects
Mu = factored moment
n = modular ratio = Es/Em
P = design axial load
Pa = allowable centroidal axial load for reinforced masonry columns
Pb = nominal balanced design axial strength
Pf = load from tributary floor or roof area
Po = nominal axial load strength with bending
Pu = factored axial load
Puf = factored load from tributary floor or roof loads
Puw= factored weight of the wall tributary to the section under consideration
Pw = weight of the wall tributary to the section under consideration
rb = ratio of the area of bars cut off to the total area of bars at the section
s = spacing of stirrups or bent bars in a direction parallel to that of the main
reinforcement
S = section modulus
t = effective thickness of a Wythe, wall or column
u = bond stress per unit of surface area of bar
V = total design shear force
Vn = nominal shear strength
Vm = nominal shear strength provided by masonry
Vs = nominal shear strength provided by shear reinforcement
∆u = horizontal deflection at mid-height under factored load; P-Delta effects shall
be included in deflection calculation
 = steel ratio = As/bd
n = ratio of distributed shear reinforcement on a plane perpendicular to the
plane of Amv
∑o = sum of the perimeters of all the longitudinal reinforcement
A = tributary area, square meters.
a = width of pressure coefficient zone used in Fig 6.2.7 and 6.2.8, meters
B = horizontal dimension of buildings and structures measured normal to
wind direction, meters.
c = average horizontal dimension of the building or structure in a direction
normal to the wind, meters.
Cc = velocity-to-pressure conversion coefficient =47.2x10-6
CG = gust coefficient
CI = structure importance coefficient

15. 15
Cp = pressure coefficient to be used for determination of wind loads on
buildings and structures.
Ct = local topographic coefficient given in Sec 2.4.6.8.
Cz = combined height and exposure coefficient for a building at height z above
ground
D = diameter of a circular structure or member, meters
d = diameter of a circular structure or member, meters
Do = surface drag coefficient given in Table 6.2.12.
f = fundamental frequency of buildings or structures in a direction parallel to
the wind, Hz
F1, F2 = design wind forces on primary framing system, KN
Gh = gust response factor for primary framing systems evaluated at height z =
h
Gz = gust response factor for components and cladding evaluated at height
zabove ground
h = mean roof height or height to top of parapet whichever is greater of a
building or structure, except that eaves height may be used for roof slope
of less than 10 degrees, meters.
J = pressure profile factor as a function of ratio
L = horizontal dimension of a building or structure measured parallel to wind
direction, meters
M = larger dimension of a sign, meters
N = smaller dimension of asign, meters
p = design pressure to be used in determination of wind loads for
buildings,kN/m2
ph = design pressure evaluated at height z =h, kN/m2
pi = internal pressure, kN/m2
pz = design wind pressure evaluated at height z above ground, kN/m2
q = sustained wind pressure, kN/m2
qh = sustained wind pressure evaluated at height z=h,kN/m2
qz = sustained wind pressure evaluated at height z above ground, in kN/m2
r = rise-to-span ratio for arched roofs
s = surface friction factor given in Table 6.2.12
S = structure size factor given in Fig 6.2.4
TI = turbulence intensity factor evaluated at two-thirds of the mean roof
height or parapet height of the structure (see Eq 2.4.11)
V = basic wind speed, km/h
X = distance to Centre of pressure from windward edge, meters
Y = response factor as a function of the ratio and the ratio c/h given in Fig
6.2.3
z = height above ground level, meters
zg = gradient height given in Table 6.2.12, meters

16. 16
 = power-law coefficient given in Table 6.2.12
 = structural damping coefficient (fraction of critical damping)
 = ratio obtained from Table 6.2.12
 = ratio of solid area to gross area for open sign face of a trussed tower, or
lattice structure
 = angle of the plane of roof from horizontal, degrees
 = height-to-width ratio for sign or hoarding
 = angle between wind direction and chord of tower guy, degrees.
C = numerical coefficient specified in Sec 2.5.6.1
D = dead load on a member including self-weight and weight of components,
materials and permanent equipment’s supported by the member
E = earthquake load
E' = amplified earthquake load equal to (0.375R)E
Fi = lateral force applied at level-iof a building
Ft = a portion of the seismic base shear, V, considered concentrated at the top
of the building in addition to the force Fn
h = height of a building or a structure above ground level inmeters
hi,hn,hx = height in meters above ground level to level-i, -n or -x respectively
level-i = i-th level of a structure above the base; i=1 designates the first level
above the base
level-n = upper most level of a structure
level-x = x-th level of a structure above the base; x=1 designates the first level
above the base.
L = live load due to intended use or occupancy
Mx = overturning moment at level-x
V = the total design lateral force or shear at the base
Vx = the story shear at story level-x
R = response modification coefficient for structural system given in Table
6.2.26 for seismic design.
T = fundamental period of vibration in seconds
W = the weight of an element or component
Z = seismic zone coefficient given in Table 6.2.22
∆ = story lateral drift.

17. 17
CHAPTER 1

18. 18
Figure 1.1: 3D View Project

19. 19
Figure-1.2 Typical Floor Lay out Plane

20. 20
Figure-1.3Typical Floor Working Lay out Plane

21. 21
Figure- 1.4 Ground Floor Lay out Plan

22. 22
Figure-1.5 Stair Case Section View

23. 23
Figure-1.6 Front Elevation

24. 24
CHAPTER 2
SLAB DESIGN

25. 25
2.1 DESIGN DATA
Concrete Design Strength 𝑓′ 𝑐 = 3.0 𝐾𝑠𝑖
Yield Strength of Steel 𝑓𝑦 = 60 𝐾𝑠𝑖
𝜀 𝑢 = 0.003
𝜀𝑡 = 0.005
𝛽1 = 0.85
∅ = 0.90
𝑏 = 12"
2.2 LOAD CALCULATION
Sl.No Load Psf
1 Live Load LL 40
2 Partition Wall Load PW 50
3 Floor Finishing FF 25
4 Total Factored Uniform Live Load Wull=LLX1.7 68
5 Self-Weight= (thicknessX150)/12 62.5
6 Total Dead Load TDL= DL+PW+FF 137.5
7 Total Factored Uniform Deal Load Wull=TDLX1.4 192.5
8 Total Uniform Load, Wu= (WuLL+WuDL) 260.5
2.3 CHECK ONE WAY OR TWO-WAY SLAB
CHECH ONE & TWO-WAY SLAB:
Panels
No.
Clear Span
in Long
Direction(lb)
Clear Span
in short
Direction(la)
Ration of the length
Bright(lb/la)
Remark
Unit ft ft
Slab-1 11.75 10.42 1.12763916 < 2 So this panel is two way slab design.
Slab-2 12.92 11.75 1.09957447 < 2 So this panel is two way slab design.
Slab-3 11.75 11.17 1.0519248 < 2 So this panel is two way slab design.
Slab-4 14.42 11.5 1.25391304 < 2 So this panel is two way slab design.
Slab-5 15.83 9.67 1.63702172 < 2 So this panel is two way slab design.

26. 26
Slab-6 15.83 10.42 1.51919386 < 2 So this panel is two way slab design.
Slab-7 15.83 10.42 1.51919386 < 2 So this panel is two way slab design.
Slab-8 15.83 9.67 1.63702172 < 2 So this panel is two way slab design.
Slab-9 14.42 11.5 1.25391304 < 2 So this panel is two way slab design.
Slab-10 11.75 11.17 1.0519248 < 2 So this panel is two way slab design.
Slab-11 12.92 11.75 1.09957447 < 2 So this panel is two way slab design.
Slab-12 11.75 10.42 1.12763916 < 2 So this panel is two way slab design.
2.4 SLAB THICKNESS
CHECK FOR SLAB THICKNESS
Penel
e No
Clear Span
in Long
Direction
(lb)
Clear
Span in
short
Direction
(la)
Perimeter
of the
Panel.
P=(la+lb)X2
Thickness
of Slab.
t=(P/180)
X12
Minimum
Thickness
of Slab
Private
Slab
Thickness
t
Effectiv
e
Depth
(d)
Panel
Ratio
(la/lb)
Case
No.
Unit ft ft ft inch inch inch inch ratio Case
Slab-1 11.75 10.42 44.34 2.95 3 5 4 0.89 Case-4
Slab-2 12.92 11.75 49.34 3.29 3.5 5 4 0.91 Case-8
Slab-3 11.75 11.17 45.84 3.05 3.5 5 4 0.96 Case-4
Slab-4 14.42 11.5 51.84 3.45 3 5 4 0.80 Case-4
Slab-5 15.83 9.67 51 3.4 4 5 4 0.62 Case-2
Slab-6 15.83 10.42 52.5 3.5 4 5 4 0.66 Case-8
Slab-7 15.83 10.42 52.5 3.5 3.5 5 4 0.66 Case-8
Slab-8 15.83 9.67 51 3.4 4 5 4 0.62 Case-2
Slab-9 14.42 11.5 51.84 3.45 3 5 4 0.80 Case-4
Slab-
10
11.75 11.17 45.84 3.05 3.5 5 4 0.96 Case-4
Slab-
11
12.92 11.75 49.34 3.29 3.5 5 4 0.91 case-8
Slab-
12
11.75 10.42 44.34 2.95 3 5 4 0.89 Case-4
2.5 All Slab Maximum Moment Ageist Thickness Check:
Plan No Mu(-)a Mu(-)b Effective
Depth (d)La
Effective
Depth (d)Lb
Thickness
La(t)
Thickness
Lb(t)
Slab-1 1741.822 1002.0107 2.759 1.587 3.759 2.587

27. 27
Slab-2 1512.546 970.86490 2.396 1.538 3.396 2.538
Slab-3 1785.551 1016.2456 2.829 1.610 3.829 2.610
Slab-4 2454.630 1431.4115 3.889 2.268 4.889 3.268
Slab-5 2146.783 966.08262 3.401 1.530 4.401 2.530
Slab-6 2065.046 1335.3797 3.272 2.115 4.272 3.115
Slab-7 2065.046 1335.3797 3.272 2.115 4.272 3.115
Slab-8 2146.783 966.08262 3.401 1.530 4.401 2.530
Slab-9 2454.630 1431.4115 3.889 2.268 4.889 3.268
Slab-10 1785.551 1016.2456 2.829 1.610 3.829 2.610
Slab-11 1512.546 970.86490 2.396 1.538 3.396 2.538
Slab-12 1741.822 1002.0107 2.759 1.587 3.759 2.587
2.6 All Slab Maximum Moment Ageist Main Reinforcement
Calculation:
Plan No Maximum
Moment
(Mu)
Total
Reinforcement
Area (As)
Minimum
Reinforcement
Area (As’)
Bar Number Bar Spacing
C/C
Slab-1 1741.822 0.0991 0.108 #3-10mm 12”C/C
Slab-2 1512.546 0.0858 0.108 #3-10mm 12”C/C
Slab-3 1785.551 0.1017 0.108 #3-10mm 12”C/C
Slab-4 2454.630 0.1412 0.108 #3-10mm 9”C/C
Slab-5 2146.783 0.1229 0.108 #3-10mm 10.5”C/C
Slab-6 2065.046 0.1181 0.108 #3-10mm 11”C/C
Slab-7 2065.046 0.1181 0.108 #3-10mm 11”C/C
Slab-8 2146.783 0.1229 0.108 #3-10mm 10.5”C/C
Slab-9 2454.630 0.1412 0.108 #3-10mm 9”C/C
Slab-10 1785.551 0.1017 0.108 #3-10mm 12”C/C
Slab-11 1512.546 0.0858 0.108 #3-10mm 12”C/C
Slab-12 1741.822 0.0991 0.108 #3-10mm 12”C/C
Maximum Spacing, 1) 12”C/C
2) 2t= 2*5=10” C/C
2.7 SLAB DESIGN HEAD CALCULATION
1. Load Calculation:
Wu=1.4DL+1.7LL
Assume, Slab Thickness t=5 inch

28. 28
Self-Weight=
𝑇𝑕𝑖𝑐𝑘𝑛𝑒𝑠𝑠
12
∗ 150 (𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔𝑕𝑡)
=
5
12
∗ 150 = 62.5 𝑃𝑠𝑓
Partition Wall (PW)=50 Psf
Floor Finish (FF)= 25 Psf
Total Dead Load, DL = Self weight+ partition wall +Floor Finish
= 62.5+50+25 = 137.5 Psf
Live load (LL)=40 Psf
Total Factor Uniform Live Load (LL)= 1.7*40 = 68 Psf
Total Factor Uniform Dead Load (DL)= 1.4*137.5 = 192.5 Psf
Total Factor Uniform Load,Wu= 1.4*137.5+1.7*40 = 260.5 Psf
2. Moment Calculation:
Negative Moment Calculation:
Mu − la = CaWula
2
=0.072 ∗ 260.50 ∗ 11.5 2
=2480.48 Ib-ft
Mu − lb = CbWulb
2
=0.029 ∗ 260.50 ∗ 14.42 2
=1570.85 Ib-ft
Positive Moment Calculation:
Mu + la = 1.4 Ca. DL ∗ WDL ∗ la
2
+ 1.7(Ca. LL ∗ WLL ∗ la
2
)
=1.4 0.0392 ∗ 137.5 ∗ 11.5 2
+ 1.7(0.0482 ∗ 40 ∗ 11.5)2
=1431.42 Ib-ft
Mu + lb = 1.4 Cb. DL ∗ WDL ∗ lb
2
+ 1.7(Cb. LL ∗ WLL ∗ lb
2
)
=1.4(0.0158 ∗ 137.5 ∗ (14.42)2
) + 1.7(0.0198 ∗ 40 ∗ (14.42)2
)
=915.40 Ib-ft
3. Check for Slab Thickness:

29. 29
𝜌 𝑚𝑎𝑥 = 0.85 ∗ 𝛽1 ∗
𝑓′
𝑐
𝑓𝑦
∗
𝜀 𝑢
𝜀 𝑢 + 𝜀𝑡
= 0.85 ∗ 0.85 ∗
3000
60000
∗
0.003
0.003+0.005
= 0.014
𝑀𝑢 = ∅𝑀𝑛 = ∅𝜌𝑓𝑦 𝑏𝑑2
1 − 0.59 ∗
𝑓𝑦
𝑓′
𝑐
∗ 𝜌
𝑑 =
𝑀𝑢
∅𝜌𝑓𝑦 𝑏 1 − 0.59 ∗
𝑓𝑦
𝑓′
𝑐
∗ 𝜌
𝑑 =
2480.48 ∗ 12
0.90 ∗ 0.014 ∗ 60000 ∗ 12 ∗ (1 − 0.59 ∗
60000
3000
∗ 0.014)
d=2” ; then, Cover=1”
𝑡 𝑟𝑒𝑞 = 2"+1" = 3"<5" ; 𝑂𝑘
4. Total Reinforcement Area (for Bending Moment):
𝐴 𝑠 = 0.0018 𝑏𝑡
=0.0018*12*5
=0.108 𝑖𝑛𝑐𝑕2
𝐴 𝑠 =
𝑀𝑢
∅𝑓𝑦 (𝑑 −
𝑎
2
)
∶ 𝑎 =
𝐴 𝑠 𝑓𝑦
0.85𝑓′ 𝑐 𝑏
𝐴 𝑠 =
2480.48 ∗ 12
0.90 ∗ 60000(4 −
𝑎
2
)
∶ 𝑎 =
𝐴 𝑠 ∗ 60000
0.85 ∗ 3000 ∗ 12
𝑆𝑜, 𝐴 𝑠 = 0.143 𝑖𝑛𝑐𝑕2
; 𝑎 = 0.28 𝑖𝑛𝑐𝑕
5. Spacing of Main Reinforcement:
Use of #-4 bar 𝑎 𝑠 = 0.11 𝑖𝑛𝑐𝑕2
Spacing, 𝑆 =
𝑎 𝑠
𝐴 𝑠
∗ 12
𝑆 =
0.11
0.143
∗ 12
S= 9.23 inch
Maximum Spacing 1) 12”C/C
2) 2t= 2*5=10” C/C
Use of #-3 bar @9” C/C

30. 30
6. Detailing of Slab:
Shown below

31. 31
Figure-3.1 Slab Reinforcement in Short Direction

32. 32
Figure-3.2 Slab Reinforcement in Long Direction

33. 33
Figure-3.3 Slab Reinforcement Details (Both Long and Short Direction)

34. 34
CHAPTER-3
BEAM DESIGN

35. 35
3.1 Stirrup Design
Maximum shear=
Vu at critical distance=
φVc = 2φf′
cbwd = 2X0.75X 3500X12X15.5 = 16.50 kip
φVc
2
=
16.50
2
= 8.25 kip
φVc
2
< 𝑉𝑢 𝑠𝑜 𝑠𝑕𝑒𝑎𝑟 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑚𝑒𝑛𝑡 𝑖𝑠 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑
φVs = Vu − φVc = 19.64 − 16.50 = 4.14 Kip
so S, req =
φAvfyd
φVs
=
0.11X60000X15.50
4.14
= 24.71
So S, max. =
Avfy
50bw
=
0.11X60
50X12
22" C/C
S, max =
d
2
=
15.50
2
= 7.5 C/C ≈7"C/C
S, max=22” C/C
Use #3 no bar@7 in C/C
3.2 MOMENT AND SHEAR CHART

36. 36

37. 37

38. 38
1. Check Effective Depth (d) of Beam:
𝜌 𝑚𝑎𝑥 = 0.85 ∗ 𝛽1 ∗
𝑓′
𝑐
𝑓𝑦
∗
𝜀 𝑢
𝜀 𝑢 + 𝜀𝑡
= 0.85 ∗ 0.85 ∗
3500
60000
∗
0.003
0.003+0.005
= 0.016
𝑀𝑢 = ∅𝑀𝑛 = ∅𝜌𝑓𝑦 𝑏𝑑2
1 − 0.59 ∗
𝑓𝑦
𝑓′
𝑐
∗ 𝜌
𝑑 =
𝑀𝑢
∅𝜌𝑓𝑦 𝑏 1 − 0.59 ∗
𝑓𝑦
𝑓′
𝑐
∗ 𝜌
𝑑 =
163.86 ∗ 12
0.9 ∗ 0.016 ∗ 60 ∗ 12 1 − 0.59 ∗
60
3
∗ 0.014
d= 15.30”=15.5” ; Cover = 2”
t=15.5”+2=17.5” > 16” not ok (Doubly Reinforcement Beam Design)
2. Total Reinforcement Area (for Bending Moment):
𝐴 𝑠 =
𝑀𝑢
∅𝑓𝑦 (𝑑 −
𝑎
2
)
∶ 𝑎 =
𝐴 𝑠 𝑓𝑦
0.85𝑓′ 𝑐 𝑏
𝐴 𝑠 =
163.86 ∗ 12
0.90 ∗ 60(15 −
𝑎
2
)
∶ 𝑎 =
𝐴 𝑠 ∗ 60
0.85 ∗ 3.5 ∗ 12
𝑆𝑜, 𝐴 𝑠 = 2.897 𝑖𝑛𝑐𝑕2
; 𝑎 = 4.870 𝑖𝑛𝑐𝑕
3. Minimum Reinforcement Area (for ACI Code):
𝐴𝑠 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 =
3 𝑓′ 𝑐
𝑓𝑦
∗ 𝑏𝑑
As=0.532 inch^2
𝐴𝑠 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 =
200
𝑓𝑦
∗ 𝑏𝑑
As=0.600 inch^2
4. Number of bar :
Use #-5 bar -16 mm
𝑎 𝑠 = 0.31 𝑖𝑛𝑐𝑕2
𝑁 =
𝐴 𝑠
𝑎 𝑠
= 𝑛𝑜𝑠
5. Beam Detailing:

39. 39
Figure-4.1 Floor Beam Reinforcement Section and Details

40. 40
3.3.1 DESIGN OF BEAM FOR SHEAR

41. 41
Figure-4.2 Floor Beam Lay out Plan

42. 42
CHAPTER 4
COLUMN DESIGN

43. 43
4.1 COLUMN DESIGN CALCULATION
Design Data
Dimension of column= 12X20 in
E= Modulus of Elasticity= 29000 psi
Concrete strength, f’c= 3500 psi
Yield strength, fy =60000 psi
P=3%
Column Load Calculation
P Mx My
Dead Load.DL -53.01 1.22 5.34
Live Load.LL -18.22 0.48 1.95
Wind Load (x-dir,+ve)Wx -5.34 91.94 0
Wind Load (y-dir,+ve)Wy -11.12 0 120.78
Earthquake Load (x-dir,+ve),Ex -5.94 101.17 0
Earthquake Load (y-dir,+ve),Ey -13.68 0 95.67
Load Combination
1.4*DL -74.214 1.71 7.48
1.4*DL+1.7*LL -105.18 2.52 10.79
1.05*DL+1.275*LL+1.275*WX -85.69 119.11 8.09
1.05*DL+1.275*LL-1.275*WX -72.08 -115.33 8.09
1.05*DL+1.275*LL+1.275*WY -93.06 1.89 162.08
1.05*DL+1.275*LL-1.275*WY -64.71 1.89 -145.9
1.05*DL+1.275*WX -62.46 -115.94 5.61
1.05*DL-1.275*WX -48.85 -115.94 5.61
1.05*DL+1.275*WY -69.84 1.28 159.6
1.05*DL-1.275*WY -41.48 1.28 -148.38
.9*DL+1.3*WX -54.65 120.62 4.8
.9*DL-1.3*WX -40.76 -118.42 4.8
.9*DL+1.3*WY -62.17 1.09 161.82
.9*DL-1.3*WY -33.25 1.09 -152.2
1.05*DL+1.275*LL+1.4025*EX -87.22 1.89 149.88

44. 44
1.05*DL+1.275*LL-1.4025*EX -70.56 1.89 -133.79
1.05*DL+1.275*LL+1.4025*EY -98.08 136.07 8.09
1.05*DL+1.275*LL-1.4025*EY -59.7 -132.28 8.09
1.05*DL+1.4025*EX -63.14 1.28 147.49
1.05*DL-1.4025*EX -47.32 1.28 -136.28
1.05*DL+1.4025*EY -74.85 135.46 5.61
1.05*DL-1.4025*EY -36.47 -132.89 5.61
.9*DL+1.3*EX -55.43 1.09 132.61
.9*DL-1.3*EX -39.98 1.09 -130.42
.9*DL+1.3*EY -65.49 125.46 4.81
.9*DL-1.3*EY -29.93 -123.27 4.81
Maximum -105.18 136.07 161.82
From Ground Floor
(C2)
Steel ratio =0.03
Load
Com
Mx
(k-f)
My
(k-f)
Pu
(k)
hx
(in)
hy
(in)
Ag
(in²)
ey=My/Pu ey/hy Ƴy=cy/hy ex=Mx/Pu
1 1.71 7.48 -
74.21
12 18 216 -0.1008 -0.0056 0.83 -0.02304
2 2.52 10.79 -
105.2
12 18 216 -0.10262 -0.0057 0.83 -0.02397
3 119.1 8.09 -
85.69
12 18 216 -0.09441 -0.00525 0.83 -1.39001
4 -115 8.09 -
72.08
12 18 216 -0.11224 -0.00624 0.83 1.600028
5 1.89 162.1 -
93.06
12 18 216 -1.74167 -0.09676 0.83 -0.02031
6 1.89 -
145.9
-
64.71
12 18 216 2.254675 0.12526 0.83 -0.02921
7 -116 5.61 -
62.46
12 18 216 -0.08982 -0.00499 0.83 1.856228
8 -116 5.61 -
48.85
12 18 216 -0.11484 -0.00638 0.83 2.373388
9 1.28 159.6 -
69.84
12 18 216 -2.28522 -0.12696 0.83 -0.01833
10 1.28 -
148.4
-
41.48
12 18 216 3.577146 0.19873 0.83 -0.03086
11 120.6 4.8 -
54.65
12 18 216 -0.08783 -0.00488 0.83 -2.20714
12 -118 4.8 - 12 18 216 -0.11776 -0.00654 0.83 2.905299

45. 45
40.76
13 1.09 161.8 -
62.17
12 18 216 -2.60286 -0.1446 0.83 -0.01753
14 1.09 -
152.2
-
33.25
12 18 216 4.577444 0.254302 0.83 -0.03278
15 1.89 150 -
87.22
12 18 216 -1.71956 -0.09553 0.83 -0.02167
16 1.89 -
133.8
-
70.56
12 18 216 1.896117 0.10534 0.83 -0.02679
17 136.1 8.09 -
98.08
12 18 216 -0.08248 -0.00458 0.83 -1.38734
18 -132 8.09 -59.7 12 18 216 -0.13551 -0.00753 0.83 2.215745
19 1.28 147.5
-
63.14 12 18 216 -2.33592 -0.12977 0.83 -0.02027
20 1.28
-
136.3
-
47.32 12 18 216 2.879966 0.159998 0.83 -0.02705
21 -136 5.61
-
74.85 12 18 216 -0.07495 -0.00416 0.83 1.815498
22 -133 5.61
-
36.47 12 18 216 -0.15383 -0.00855 0.83 3.643817
23 1.09 132.6
-
55.43 12 18 216 -2.39239 -0.13291 0.83 -0.01966
24 1.09
-
130.4
-
39.98 12 18 216 3.262131 0.18123 0.83 -0.02726
25 125.5 4.81
-
65.49 12 18 216 -0.07345 -0.00408 0.83 -1.91571
26 -123 4.81
-
29.93 12 18 216 -0.16071 -0.00893 0.83 4.11861
27 136.1
-
152.2
-
150.2 12 18 216 1.013451 0.056303 0.83 -0.90605
ex/hx
Ƴx=cx
/hx
Kn=Pn
/f'c*Ag
Kn=Pn/
f'c*Agx
Φpnox Φpnoy Kn ΦPo
kip
ΦPn kip
Remark
s
-0.00192 0.75 1.22 1.22 790.56 790.56 1.27 822.96 760.61 OK
-0.002 0.75 1.22 1.22 790.56 790.56 1.27 822.96 760.61 OK
-0.11583 0.75 1.22 0.68 440.64 790.56 1.27 822.96 431.18 OK
0.133336 0.75 1.22 0.68 440.64 790.56 1.27 822.96 431.18 OK
-0.00169 0.75 0.48 1.22 790.56 310.65 1.27 822.96 305.92 OK

46. 46
-0.00257 0.75 0.41 1.22 790.56 265.68 1.27 822.96 262.21 OK
-0.18393 0.75 1.22 0.43 278.64 790.56 1.27 822.96 274.82 OK
0.242108 0.75 1.22 0.4 259.2 790.56 1.27 822.96 255.89 OK
-0.00146 0.75 0.64 1.22 790.56 414.2 1.27 822.96 431.71 OK
-0.00273 0.75 0.4 1.22 790.56 259.2 1.27 822.96 255.89 OK
-0.00181 0.75 0.44 1.22 790.56 285.12 1.27 822.96 281.13 OK
-0.00223 0.75 0.7 1.22 790.56 453.6 1.27 822.96 443.58 OK
-0.11561 0.75 1.22 0.71 460.08 790.56 1.27 822.96 449.78 OK
0.184645 0.75 1.22 0.4 259.2 790.56 1.27 822.96 255.89 OK
-0.00169 0.75 0.68 1.22 790.56 440.64 1.27 822.96 431.18 OK
-0.00225 0.75 0.64 1.22 790.56 414.72 1.27 822.96 406.33 OK
0.151291 0.75 1.22 0.65 421.2 790.56 1.27 822.96 412.55 OK
0.303651 0.75 1.22 0.35 226.8 790.56 1.27 822.96 224.27 OK
-0.00164 0.75 0.7 1.22 790.56 453.6 1.27 822.96 453.58 OK
-0.00227 0.75 0.4 1.22 790.56 259.2 1.27 822.96 255.89 OK
-0.15964 0.75 1.22 0.64 414.72 790.56 1.27 822.96 432.21 OK
0.343218 0.75 1.22 0.38 216.24 790.56 1.27 822.96 213.94 OK
-0.0755 0.75 0.57 0.7 453.6 369.36 1.27 822.96 270.5 OK
Design Colum C2 At Ground Floor:
Here,we use reciprocal load equation.Bresler,s reciprocal equation are given below
1/Pn=1/Pnxo+1/Pnyo-1/po
Where,
Pn=approximate value of nominal load in biaxial bending with eccentricity ex
and ey. Pnxo=nominal load when only eccentricity ey is present (cx=0)
Pnyo=nominal load when only eccentricity ex is present (cy=0)
Po=nominal load for concentrically loaded column.
1.5” covering all side
Cx=12-1.5*2=9Cy=18-1.5*2=15
hx=12” hy=18” Ƴx=9/12=.75, Ƴy=15/18=.833
Ag=12*18=216in²use,pg=.03
Main reinforcement,Ast=pg*Ag
-0.00243 0.75 0.7 1.22 790.56 453.6 1.27 822.96 443.58 OK
0.154686 0.75 1.22 0.66 427.68 790.56 1.27 822.96 418.76 OK
0.197782 0.75 1.22 0.41 302.58 790.56 1.27 822.96 298.09 OK
-0.00153 0.75 0.7 1.22 790.56 453.6 1.27 822.69 443.58 OK
Column Size: 12” x 18”

47. 47
=.03*216
=6.48in²
So we use 8#9 bar
For lateral ties we use #3 bar
Spacing is the minimum of-
 16db=16*9/8=18”
 48tb=48*3/8=18”
 12” (Governs)

48. 48
Figure-5.1 Column Lay out Plan

49. 49
Figure-5.2 Column Reinforcement Details

50. 50
CHAPTER-5
STAIR DESIGN

51. 51
5.1 STAIR DESIGN DATA
Riser,(R)=6”
Tread,(T)=10”
Assume, Waist Slab Thickness, t=6”
Floor to Floor Height, H=10’
Total Span (Run)=16’-5”
Effective Span of Stair C/C Distance of Supporting Beam,
L= 4’-4”+7’-6”+3’-7”=15’-5”
𝑓′ 𝑐 = 3.0 𝐾𝑠𝑖
𝑓𝑦 = 60 𝐾𝑠𝑖
𝜀 𝑢 = 0.003
𝜀𝑡 = 0.005
𝛽1 = 0.85
∅ = 0.90
𝑏 = 12"
1. Load Calculation:
Load from Waist Slab= 𝑡 ∗
𝑅2+𝑇2
𝑇
∗ 𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔𝑕𝑡
=
6
12
∗
62+102
10
∗ 150
= 87.46 Psf
Load from Step =
1
2
∗ 𝑅 ∗ 𝑇 ∗ 𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔𝑕𝑡
=
1
2
∗
6
12
∗
10
12
∗ 150
=31.125 Psf
=31.125 ∗
12
10
= 37.35 Psf
Live Load,(LL) = 100 Psf
Floor Finish,(FF) =25 Psf
So, Total Dead Load= Load from Waist Slab+Load from Step+Floor Finish
DL= 87.46+37.35+25 = 149.81 Psf
Design Load 𝑊𝑢 = 1.4𝐷𝐿 + 1.7𝐿𝐿
Wu= 1.4*149.81+1.7*100
Wu=379.73 Psf
2. Moment Calculation:

52. 52
Mu =
WuL2
8
Ib − ft
=
379.72∗(15.42)2
8
= 11286.35 Ib − ft
3. Check Effective Depth (d) of Waist slab :
𝜌 𝑚𝑎𝑥 = 0.85 ∗ 𝛽1 ∗
𝑓′
𝑐
𝑓𝑦
∗
𝜀 𝑢
𝜀 𝑢 + 𝜀𝑡
= 0.85 ∗ 0.85 ∗
3000
60000
∗
0.003
0.003+0.005
= 0.014
𝑀𝑢 = ∅𝑀𝑛 = ∅𝜌𝑓𝑦 𝑏𝑑2
1 − 0.59 ∗
𝑓𝑦
𝑓′
𝑐
∗ 𝜌
𝑑 =
𝑀𝑢
∅𝜌𝑓𝑦 𝑏 1 − 0.59 ∗
𝑓𝑦
𝑓′
𝑐
∗ 𝜌
𝑑 =
11286.35 ∗ 12
0.90 ∗ 0.014 ∗ 60000 ∗ 12 ∗ (1 − 0.59 ∗
60000
3000
∗ 0.014)
d=4.22” ;then,Cover=1”
𝑡 𝑟𝑒𝑞 = 4.22"+1" = 5.22"<6" ; 𝑂𝑘
4. Total Reinforcement Area (for Bending Moment):
𝐴 𝑠 =
𝑀𝑢
∅𝑓𝑦 (𝑑 −
𝑎
2
)
∶ 𝑎 =
𝐴 𝑠 𝑓𝑦
0.85𝑓′ 𝑐 𝑏
𝐴 𝑠 =
11286.35 ∗ 12
0.90 ∗ 60000(5 −
𝑎
2
)
∶ 𝑎 =
𝐴 𝑠 ∗ 60000
0.85 ∗ 3000 ∗ 12
𝑆𝑜, 𝐴 𝑠 = 0.564 𝑖𝑛𝑐𝑕2
; 𝑎 = 1.106 𝑖𝑛𝑐𝑕
5. Spacing of Main Reinforcement:
Use of #-4 bar 𝑎 𝑠 = 0.20 𝑖𝑛𝑐𝑕2
Spacing, 𝑆 =
𝑎 𝑠
𝐴 𝑠
∗ 12
S= 4.25 inch
Use of #-4 bar @4” C/C
6. Reinforcement Area of Temperature bar :
𝐴 𝑠 = 0.0018 𝑏𝑡
=0.0018*12*6

53. 53
= 0.1296 𝑖𝑛𝑐𝑕2
Use of #-4 bar 𝑎 𝑠 = 0.20 𝑖𝑛𝑐𝑕2
Spacing, 𝑆 =
𝑎 𝑠
𝐴 𝑠
∗ 12
S= 18 inch
Maximum Spacing, 1) 12” C/C
2) 3t C/C = 3*6 =18” C/C
Use of #-4 bar @12” C/C
7. Detailing of Stair:

54. 54
Figure-6.1 Stair Reinforcement Details and Layout Plan

55. 55
CHAPTER-6
FOOTING DESIGN

56. 56
6.1 FOOTING DESIGN DATA
R.C.C Single Footing Design:
Given,
DL=137.32 kip
LL=50 Kip
F’c=3.5 Ksi
Fy=60 Ksi
Allowable soil pressure qa=5 Ksf
Assume Footing Thickness= 24 inch
Step-1:
Effective Soil Pressure qe= pa-5X110-2X150=4.15 Ksf
Step-2:
Area Required=
Unfactored Load
Effective Soil Pressure
=
𝐷𝐿+𝐿𝐿
𝑞𝑒
=
137.32+50
4.15
= 45.13 𝑠𝑓𝑡
Let,
One Side Length, L=8’
So other Side Length = 45.13/8 = 5.64 =6’
New Area = 8X6= 48 sft
Step-3:
Ultimate Soil Pressure qu=
Factored Load
Area
=
1.2X137.32+1.6X50
48
= 5.09Ksf
Step-4:
Punching Shear Check
Here’d(effective)=Footing Thickness -Clear Cover.
d=24-3=21”
Vu=qu X Area reman
Vu=5.09X[(7x7)-(43/12X31/12)]=197.20 Kip
Concrete resistance φVc = 4φf′
cbd = 0.75x4x 3500x 43 + 31 x2x21 = 551.62 Kip
Step-5:
One way on beam shear
3’-5/12-21/12 = 0.83

57. 57
Vu=qu X A
Vu=5.09X0.83X8=33.80
Concrete resistance φVc = 2φf′cbd = 0.75X2X 3500 X8X21X12 = 178.90 Kip
Step-6:
Moment in long direction
Here, A=22”=1.83’ ,B=8’
W = quxL = 5.09X6 = 30.54
ML =
1
8
X30.54(8 − 1.83)2
= 145.33 Kip − ft
Moment in short direction
Here, A=10”=83’ ,B=6’
W = quxL = 5.09X8 = 4074
ML =
1
8
X30.54(6 − .83)2
= 136.12 Kip − ft
Step-7:
Reinforcement
Assume,a=1”
As =
ML
φfy(d −
a
2
)
= 1.58 in2
a =
Asfy
0.85ficb
= 5.31 in2
AginAs=1.79 in2 Forlongdirection
Asforshortdirection=
1.79
145.33
𝑋136.12 = 1.68 𝑖𝑛2
CheckForminimum, As
𝐴𝑠 𝑚𝑖𝑛 =
3 𝑓′𝑐
𝑓𝑦
𝑏𝑑 ≥
200
𝑓𝑦
𝑏𝑑 = 4.47 𝑖𝑛2 ≥ 5.04 𝑖𝑛2
SouseminimumReinforcement 5.04 in2 inbothDirections
Use#5 Nobarinbothlongandshortdirections
SpacingofBar (LongDirection),S=
𝟎.𝟑𝟏
𝟓.𝟎𝟒
𝑿𝟔𝑿𝟏𝟐 = 𝟒" 𝑪/𝑪
SpacingofBar (ShortDirection),S=
𝟎.𝟑𝟏
𝟓.𝟎𝟒
𝑿𝟖𝑿𝟏𝟐 = 𝟔" 𝑪/𝑪

58. 58
Figure-7.1 Footing and Column Layout Plane

59. 59
Figure-7.2 Footing Layout Plane Cad View

60. 60
Figure-7.3 Footing Details

61. 61
Figure-7.4 Footing Reinforcement Details Section A-A

62. 62
CHAPTER-7
UNDER GROND WATER TANK DESIGN

63. 63
7.1 DESIGN DATA
Water Consumption = 235 lpcd
Volume of water to be stored = 235 lpcd * 12 persons/floor *6 floor*2 day
= 235*12*6*2 =33840 Litre
= (33840/1000)*35.315 = 1195.05 cft/day
Assume, Length, L=15’
Width, B=10’
Height, H= 8’
fy = 60 Ksi
f′c = 3 Ksi
β = 0.85
∅ = 0.90
∈u= 0.003
∈t= 0.005
b=12”
Cove slab Thickness, t=5”
Bottom slab thickness t=15” *then ACI code minimum thickness=9”+
Downward force =Wt of cover slab+sidewall+Wt of soil+Wt of base slab
Upward force= Wt of Water*base slab area
Factor of Safety against uplifting = downward force / upward force >1.25 OK
7.2 BASE SLAB DESIGN
ACI Code Minimum Thickness (t) mim=9”
Let’s Assuming Thickness, t=15”
1. Load Calculation
Self weigth,DL=
15
12
∗ 150 = 187.5 psf
H=8’+6”+5”+1’= 9.92 -ft

64. 64
Live Load, LL= γwh = 62.4 ∗ 9.92 = 619 psf
Floor Finish, FF= 25 psf
Total Dead Load, DL= 187.5+25 = 212.5 psf
Factor Load,Wu= 1.4DL+1.7LL
Wu=1.4*212.5+1.7*619 = 1243.55 psf = 1.244 ksf
2. Moment calculation
Mmax(+)=
1
6
𝑃𝑣𝑕2
∗ 1.7 =
1
6
∗ 0.809 ∗ (3.28)2
∗ 1.7
= 2.47 kip -ft
Mmax(-)=
1
6
𝑃𝑣𝑕2
∗ 1.7 −
𝑊𝑢 𝐿2
8
=
1
6
∗ 0.809 ∗ (3.28)2
∗ 1.7 −
1.244∗152
8
= -32.52 kip-ft
3. Total Reinforcement Calculation
𝜌 𝑚𝑎𝑥 = 0.85𝛽
𝑓′𝑐
𝑓𝑦
.
𝜖 𝑢
𝜖 𝑢 +𝜖 𝑡
= 0.0014
𝑀𝑢 = ∅𝜌𝑓𝑦𝑏𝑑2
(1 − 0.59
𝑓𝑦
𝑓′ 𝑐
𝜌)
d=7.17 in < 12 in ; ok
As mim =0.0018 bt
As mim =0.0018∗12∗15=𝟎.𝟑𝟐𝟒 𝐢𝐧 𝟐
𝐴 𝑠 =
𝑀𝑢
∅𝑓𝑦(𝑑 −
𝑎
2
)
= 0.653 𝑖𝑛2
𝑎 =
𝐴 𝑠 𝑓𝑦
0.85𝑓′𝑐𝑏
= 0.1.877 𝑖𝑛𝑐𝑕
Use #5 bar 𝑎 𝑠 = 0.31 𝑖𝑛2
Spacing, 𝑆 =
𝑜.31
0.653
∗ 12 = 5.70" 𝐶/𝐶
Use of #-5 bar @ 5.5” C/C.
7.3 COVER SLAB DESIGN
La=B=10’

65. 65
Lb=L=15’
La/Lb=10/15=0.67 >0.5 ; so two way slab
1. Slab Thickness Calculation.
t =
2(la + lb)
180
t =
2(10′+15′)
180
X12 = 3.33”
t=5”
2. Load Calculation.
Slab Self Weigth, DL=
5
12
X150 = 62.5 psf
Live Load, LL= 20 Psf
Floor Finish,FF= 25 psf
Total Dead Load, DL=62.5+25 = 87.5 psf
Factor Load, Wu=1.4DL+1.7LL
Wu=1.4*87.5+1.7*40 = 190.5 Psf
3. Moment Calculation.
𝑀𝑎 + =
𝑊𝑢 𝑙𝑎2
8
= 2381.25 Ib-ft
𝑀𝑏 + =
𝑊𝑢 𝑙𝑏2
8
= 5357.82 Ib-ft
4. Thickness Check.
𝜌 𝑚𝑎𝑥 = 0.85𝛽
𝑓′𝑐
𝑓𝑦
.
𝜖 𝑢
𝜖 𝑢 +𝜖 𝑡
= 0.0014
𝑀𝑢 = ∅𝜌𝑓𝑦𝑏𝑑2
(1 − 0.59
𝑓𝑦
𝑓′ 𝑐
𝜌)
𝑑 = 3 𝑖𝑛𝑐𝑕 *Cover=1”+
t,req= 3+1=4 < 5 inch ; Ok
5. Total Reinforcement Calculation.
As mim =0.0018 bt
As mim =0.0018∗12∗5=𝟎.𝟏𝟎𝟖 𝐢𝐧 𝟐
d = 5 − 1 = 4 inch
Short Directions.

66. 66
𝐴 𝑠 =
𝑀𝑢
∅𝑓𝑦(𝑑 −
𝑎
2
)
= 𝟎. 𝟏𝟑𝟕 𝒊𝒏 𝟐
𝑎 =
𝐴 𝑠 𝑓𝑦
0.85𝑓′𝑐𝑏
= 0.268 𝑖𝑛𝑐𝑕
Long direction.
𝐴 𝑠 =
𝑀𝑢
∅𝑓𝑦(𝑑 −
𝑎
2
)
= 𝟎. 𝟑𝟐𝟒 𝒊𝒏 𝟐
𝑎 =
𝐴 𝑠 𝑓𝑦
0.85𝑓′𝑐𝑏
= 0.633 𝑖𝑛𝑐𝑕
6. Spacing Calculation.
Use of #3 bar 𝑎 𝑠 = 0.11 𝑖𝑛2
Short Direction Spacing.
𝑆 =
𝑎 𝑠
𝐴 𝑠
∗ 12 ; 9.63”@3# bar
Use of #-3 bar @ 9” C/C.
Long Direction Spacing.
𝑆 =
𝑎 𝑠
𝐴 𝑠
∗ 12 ; 4”@3# bar
Use of #-3 bar @ 4” C/C.
Maximum Spacing.
1. 12” C/C
2. 2t= 2*5 = 10” C/C
7.4 DESIGN OF SIDE WALL
The practical case would be to consider load acting on wall height only
Water, 𝛾𝑤𝐻′
= 62.4 ∗ 9.92 = 619 𝑝𝑠𝑓 = 𝟎. 𝟔𝟏𝟗 𝒌𝒔𝒇
𝐾𝑎 =
1 − sin 30
1 + sin 30
=
1
3
Soil, 𝐾𝑎 𝛾𝑠𝑡𝐻′
=
1
3
120 − 62.4 ∗ 9.92 = 190.46 𝑝𝑠𝑓 = 𝟎. 𝟏𝟗𝟎 𝒌𝒔𝒇
Pv=Water+Soil

67. 67
Pv= 0.619+0.190 = 0.809 ksf
𝑃𝑕 = 𝑃𝑣 ∗
𝐻′
𝐻
= 0.809 ∗
6.64
9.92
= 𝟎. 𝟓𝟒𝟐 𝒌𝒔𝒇
Reinforcement Calculation for Vertical Banding
Force =
1
2
𝑃𝑣𝑕
Moment, 𝑀 =
1
2
𝑃𝑣𝑕(
𝑕
3
)
M= 1.45 kip-ft/ft
Mu=1.7*1.45 =2.465 kip-ft
From ACI Code t mim= 9” ; Assume, t=10”
Here, d=10”-3” = 7”
𝜌 𝑚𝑎𝑥 = 0.85𝛽
𝑓′𝑐
𝑓𝑦
.
𝜖 𝑢
𝜖 𝑢 +𝜖 𝑡
= 0.0014
𝑀𝑢 = ∅𝜌𝑓𝑦𝑏𝑑2
(1 − 0.59
𝑓𝑦
𝑓′ 𝑐
𝜌)
d=1.97,=2”<8.5” ; ok
As mim =0.0018 bt
As mim =0.0018∗12∗10=𝟎.𝟐𝟏𝟔 𝐢𝐧 𝟐
𝐴 𝑠 =
𝑀𝑢
∅𝑓𝑦(𝑑 −
𝑎
2
)
= 0.079 𝑖𝑛2
𝑎 =
𝐴 𝑠 𝑓𝑦
0.85𝑓′𝑐𝑏
= 0.155 𝑖𝑛𝑐𝑕
Spacing calculation.
Use of #4 bar 𝑎 𝑠 = 0.2 𝑖𝑛2
𝑆 =
𝑎 𝑠
𝐴 𝑠
∗ 12 =
0.2
0.216
∗ 12 = 11.11"
Use of #-4 bar @ 11” C/C.
Reinforcement Calculation for Horizontal Banding

68. 68
Since the critical condition is the empty tank with outside soil pressure so here will be no
direct tension outside soil will give compression and its effect can be neglected in
reinforcement during since correct itself can take compression.
Ph=0.542
1.
PhL2
12
=
0.542∗152
12
= 10.16 kip/ft
2.
PhB2
12
=
0.542∗102
12
= 4.52 kip/ft
Slab action.
Now use Moment Distribution method.
𝐾𝐿 =
4𝐸𝐼
𝐿
x
1
2
=
4
15∗2
= 0.13
𝐾 𝐵 =
4𝐸𝐼
𝐵
x
1
2
=
4
10∗2
= 0.20
Distribution factor,DF(L)=
0.13
0.13+0.20
= 0.394
M(+)=+ 7.92 kip-ft
M(-)=-7.92 kip-ft
So moment at each corner (Slab action)
Mu= 1.7*7.92 = 13.464 kip-ft
Here, d=10”-3”-0.5” = 6.5”
𝐴 𝑠 =
𝑀𝑢
∅𝑓𝑦(𝑑 −
𝑎
2
)
= 0.497 𝑖𝑛2
𝑎 =
𝐴 𝑠 𝑓𝑦
0.85𝑓′𝑐𝑏
= 0.975 𝑖𝑛𝑐𝑕
As mim =0.0018 bt
As mim =0.0018∗12∗10=𝟎.𝟐𝟏𝟔 𝐢𝐧 𝟐
Use #4 bar 𝑎 𝑠 = 0.20 𝑖𝑛2
Spacing, 𝑆 =
𝑜.20
0.497
∗ 12 = 4.82" 𝐶/𝐶
Use of #-4 bar @ 4.5” C/C.

69. 69
Figure-8.1 Under Ground Water Tank design

70. 70
OVERHEAD WATER TANK DESIGN
7.5 DESIGN DATA
Water Consumption = 235 lpcd
Volume of water to be stored = 235 lpcd * 12 persons/floor *6 floor
= 235*12*6 =16920 Litre
= (16020/1000)*35.315 = 692.88cft/day
So that Assuming two time in a day if we pump so that the Volume of water per pump can
be Reserved = 692.88/2 = 346.5cft/day
Assume, Length, L=12’
Width, B=7’
Height, H= 4.25’
fy = 60 Ksi
f′c = 3 Ksi
β = 0.85
∅ = 0.90
∈u= 0.003
∈t= 0.005
b=12”
7.6 BASE SLAB DESIGN
ACI Code Minimum Thickness (t) mim=9”
Let’s Assuming Thickness, t=10”
4. Load Calculation
Self weigth,DL=
10
12
∗ 150 = 125 psf
Live Load, LL=20 psf + γwh = 20 + 62.4 ∗ 4.25 = 285.2 psf
Factor Load,Wu= 1.4DL+1.7LL
Wu=1.4*125+1.7*185.2 = 659.84 psf
5. Moment calculation

71. 71
Mmax(+)=
𝑊𝑢 𝐿2
8
−
1
6
𝛾𝑤𝐻𝑕2
=
659.84∗(12)2
8
−
1
6
∗ 62.4 ∗ 4.25 ∗ (3.28)2
= 11401.59 Ib-ft
Mmax(-)= −
1
6
𝛾𝑤𝐻𝑕2
= −
1
2
∗ 62.4 ∗ 4.25 ∗ (3.28)2
= -808.38 Ib-ft
6. Total Reinforcement Calculation
𝜌 𝑚𝑎𝑥 = 0.85𝛽
𝑓′𝑐
𝑓𝑦
.
𝜖 𝑢
𝜖 𝑢 +𝜖 𝑡
= 0.0014
𝑀𝑢 = ∅𝜌𝑓𝑦𝑏𝑑2
(1 − 0.59
𝑓𝑦
𝑓′ 𝑐
𝜌)
d=4.25 in < 8.5 in ; ok
As mim =0.0018 bt
As mim =0.0018∗12∗10=𝟎.𝟐𝟏𝟔 𝐢𝐧 𝟐
𝐴 𝑠1 =
𝑀𝑢
∅𝑓𝑦(𝑑 −
𝑎
2
)
= 0.309 𝑖𝑛2
𝑎 =
𝐴 𝑠 𝑓𝑦
0.85𝑓′𝑐𝑏
= 0.606 𝑖𝑛𝑐𝑕
For design T > 1.7 TB =
1
2
∗ 𝛾𝑤 ∗ 𝐻2
𝐵 ∗ 1.7 =
1
2
∗ 62.4 ∗ 4.252
∗ 7 1.7
= 6.71 kip-ft
T > 1.7 TL =
1
2
∗ 𝛾𝑤 ∗ 𝐻2
𝐿 ∗ 1.7 =
1
2
∗ 62.4 ∗ 4.252
∗ 12 1.7
= 11.50 kip-ft
So, T max= 11.50 kip-ft
𝐴 𝑠2 =
𝑇 𝑚𝑎𝑥
∅𝑓𝑦
=
11.50∗12
0.90∗60∗12
= 0.212𝑖𝑛2
𝑁𝑜𝑤, 𝐴 𝑠1 +
𝐴 𝑠2
2
= 0.309 +
0.212
2
= 0.415 𝑖𝑛2
if we want to cut off bar, then.
As2
2
Bar =
𝐴 𝑠1
2
+
𝐴 𝑠2
2
=
0.309
2
+
0.212
2
= 0.260 𝑖𝑛2
> 0.216 𝑖𝑛2
So bar cut off can’t be done.

72. 72
Use #4 bar 𝑎 𝑠 = 0.20 𝑖𝑛2
Spacing, 𝑆 =
𝑜.20
0.415
∗ 12 = 5.5" 𝐶/𝐶
Use of #-4 bar @ 5.5” C/C.
7.7 COVER SLAB DESIGN
La=B=7’
Lb=L=12’
La/Lb=7/12=0.58>0.5 ; so two way slab
7. Slab Thickness Calculation.
t =
2(la + lb)
180
t =
2(7′+12′)
180
X12 = 2.53”
t=3.5”
8. Load Calculation.
Slab Self Weigth, DL=
3.5
12
X150 = 43.75 psf
Live Load, LL= 20 Psf
Factor Load, Wu=1.4DL+1.7LL
Wu=1.4*43.75+1.7*20 = 95.25 Psf
9. Moment Calculation.
𝑀𝑎 + =
𝑊𝑢 𝑙𝑎2
8
= 583.41 Ib-ft
𝑀𝑏 + =
𝑊𝑢 𝑙𝑏2
8
= 1714.5 Ib-ft
10. Thickness Check.
𝜌 𝑚𝑎𝑥 = 0.85𝛽
𝑓′𝑐
𝑓𝑦
.
𝜖 𝑢
𝜖 𝑢 +𝜖 𝑡
= 0.0014
𝑀𝑢 = ∅𝜌𝑓𝑦𝑏𝑑2
(1 − 0.59
𝑓𝑦
𝑓′ 𝑐
𝜌)
𝑑 = 1.64 𝑖𝑛𝑐𝑕 *Cover=1”+
t,req= 1.64+1=2.75 <3.5 inch ; Ok
11. Total Reinforcement Calculation.

73. 73
As mim =0.0018 bt
As mim =0.0018∗12∗3.5=𝟎.𝟎𝟕𝟓𝟔 𝐢𝐧 𝟐
d = 3.5 − 1 = 2.5 inch
Short Directions.
𝐴 𝑠 =
𝑀𝑢
∅𝑓𝑦(𝑑 −
𝑎
2
)
= 0.0529 𝑖𝑛2
𝑎 =
𝐴 𝑠 𝑓𝑦
0.85𝑓′𝑐𝑏
= 0.103 𝑖𝑛𝑐𝑕
Long direction.
𝐴 𝑠 =
𝑀𝑢
∅𝑓𝑦(𝑑 −
𝑎
2
)
= 𝟎. 𝟏𝟔𝟐 𝒊𝒏 𝟐
𝑎 =
𝐴 𝑠 𝑓𝑦
0.85𝑓′𝑐𝑏
= 0.319 𝑖𝑛𝑐𝑕
12. Spacing Calculation.
Use of #3 bar𝑎 𝑠 = 0.11 𝑖𝑛2
Short Direction Spacing.
𝑆 =
𝑎 𝑠
𝐴 𝑠
∗ 12 ; 17”@3# bar
Long Direction Spacing.
𝑆 =
𝑎 𝑠
𝐴 𝑠
∗ 12 ; 8”@3# bar
Maximum Spacing.
3. 12” C/C
4. 2t= 2*3.5 = 7” C/C
Use of #-3 bar @ 7” C/C.
7.8 DESIGN OF SIDE WALL
Reinforcement Calculation for Vertical Banding
Vertical bending (cantilever action); 𝑕 =
𝐻
4
𝑜𝑟 1 𝑚
H=4.25 ft
h=3.28 ft

74. 74
H-h=4.25-3.28= 0.97 ft
𝑃𝑕 = 𝑓𝑜𝑟 𝑠𝑙𝑎𝑏 𝑎𝑐𝑡𝑖𝑜𝑛
𝑃𝑕 = 𝛾𝑤 𝐻 − 𝑕 = 62.4 4.25 − 3.28 = 60.53 𝑝𝑠𝑓
Pv=for Cantilever action
𝑃𝑣 = 𝛾𝐻 = 62.4 ∗ 4.25 = 265.2 𝑝𝑠𝑓
Force =
1
2
𝛾𝑤𝐻𝑕
Moment, 𝑀 =
1
2
𝛾𝑤𝐻𝑕(
𝑕
3
)
M= 475.52 Ib-ft/ft
Mu=1.7*475.52 = 808.4 Ib-ft
From ACI Code t mim= 9” ; Assume, t=10”
Here, d=10”-1.5” = 8.5”
𝜌 𝑚𝑎𝑥 = 0.85𝛽
𝑓′𝑐
𝑓𝑦
.
𝜖 𝑢
𝜖 𝑢 +𝜖 𝑡
= 0.0014
𝑀𝑢 = ∅𝜌𝑓𝑦𝑏𝑑2
(1 − 0.59
𝑓𝑦
𝑓′ 𝑐
𝜌)
d=1.13”<8.5” ; ok
As mim =0.0018 bt
As mim =0.0018∗12∗10=𝟎.𝟐𝟏𝟔 𝐢𝐧 𝟐
𝐴 𝑠 =
𝑀𝑢
∅𝑓𝑦(𝑑 −
𝑎
2
)
= 0.021 𝑖𝑛2
𝑎 =
𝐴 𝑠 𝑓𝑦
0.85𝑓′𝑐𝑏
= 0.041 𝑖𝑛𝑐𝑕
Spacing calculation.
Use of #3 bar𝑎 𝑠 = 0.11 𝑖𝑛2
𝑆 =
𝑎 𝑠
𝐴 𝑠
∗ 12 ; 6”@3# bar
Use of #-3 bar @ 6” C/C.

75. 75
Reinforcement Calculation for Horizontal Bending
3.
PhL2
12
=
60.53∗122
12
= 0.726 kip/ft
4.
PhB2
12
=
60.53∗72
12
= 0.247 kip/ft
Slab action.
Now use Moment Distribution method.
𝐾𝐿 =
4𝐸𝐼
𝐿
x
1
2
=
4
12∗2
= 0.17
𝐾 𝐵 =
4𝐸𝐼
𝐵
x
1
2
=
4
7∗2
= 0.29
Distribution factor,DF(L)=
0.17
0.17+0.29
= 0.369
M(+)= 0.549 kip-ft
M(-)=0.549 kip-ft
So moment at each corner (Slab action)
Mu= 1.7*0.549 = 0.933 kip-ft
Here, d=10”-1.5”-0.375” = 8.125”
𝐴 𝑠 =
𝑀𝑢
∅𝑓𝑦(𝑑 −
𝑎
2
)
= 0.00214 𝑖𝑛2
𝑎 =
𝐴 𝑠 𝑓𝑦
0.85𝑓′𝑐𝑏
= 0.00417 𝑖𝑛𝑐𝑕
Direct Tension
𝑇 𝑚𝑎𝑥 =
𝑃𝑕𝐿
2
=
60.53 ∗ 12
2
= 0.364 𝑘𝑖𝑝 − 𝑓𝑡
𝐴 𝑠2 =
𝑇 𝑚𝑎𝑥
∅𝑓𝑦
=
0.364∗1.7∗12
0.90∗60
= 0.137 𝑖𝑛2
𝑁𝑜𝑤, 𝐴 𝑠1 +
𝐴 𝑠2
2
= 0.00214 +
0.137
2
= 0.0706 𝑖𝑛2
As mim =0.0018 bt
As mim =0.0018∗12∗10=𝟎.𝟐𝟏𝟔 𝐢𝐧 𝟐
Use of #-3 bar @ 6” C/C.

76. 76
Figure-8.2 Overhead Tank Design Detailing

77. 77
CHAPTER 8
REFFERENCES

78. 78
Agbayani, N,, Jayachandran, P, and Sriram ,D,, (1989), ``Novel Method for the evaluation of
K-Factor for effective length and column design`` ,Journal of the structural division, Proc.
ASCE., October.
‘’Bangladesh National Building Code ’’Final Draft, Prepared for Housing & Research Institute,
Consultants: Development Design Consultants Limited,1993. Chan CM ,(2007),’’
Computer Aided Design Optimization of Tall Buildings’’. CIEM 532 Course Notes,
Colacho, J.P.,(1978), ‘’Elastic Analysis’’, Tall Building Monographs, Tall Concrete Buildings,
Vol, CB., Chapter 9.
Heide Brecht, A.C and smith, B.S., (1973),``Approximate Methods of Analysis of tall wall-
frame Buildings, Journal of the structural Division, Proc .ASCE.
Lyenger, H.S Conference on Tall Buildings, Lehigh University., Vol.2.
Jayachandran, P.(2009), ``Design of Tall Buildings Preliminary Design and Optimization.’’
Khan , F.R. and S BAROUNIS (1964) ,``Interaction of Shear Walls and Frames’’, Journal of the
Structural Division , Proc., ASCE.
Nilson A.H. ``Design of concrete Structures’’13th
Edition, McGraw-Hill Companies, Inc.1997.
Sirigiri, Mamatha, (2014),``Preliminary Structural Design Optimization Of Tall Buildings.
Using DS-USAD © Frame 3D’’
Winter George, Urquhart L.C.,O’ Rourke C.E.,Nilson A.H. “Design of Concrete Structure. ‘’ 7th
Edition, Tata McGraw-Hill Publishing Company Ltd. New Delhi.1979.
Books:
 Bangladesh National Building code (BNBC)-2006
 American Concrete Institute (ACI)-2008
Software:
 ETABS
 STAAD. Pro
 Auto CAD
 Microsoft Office
 Microsoft Excel
 Adobe Acrobat 8 Professional
Website:

79. 79
 www.e-book.com
 www.google.com
 www.wikipedia.com
 www.academia.edu
 www.concrete.org
 www.yourspreadsheets.co.uk
 www.engineeringcivil.com/theory/design
 Previous Capstone report of undergraduate studies, Dept. of Civil Engineering,
Presidency University, Bangladesh.

80. 80
CHAPTER-9
APPENDEX

81. 81
9.1 EARTHQUAKE LOAD CALCULATION
Earthquake Load Calculation---BNBC-1993
Design Base Shear, V =
ZIC
R
∗ w − − − − − − − − − − − −1
Here, Z= Seismic Zone Coefficient = 0.15
I= Structural Importance Coefficient =1.0
R= Response Modification Coefficient=8.0 (IMRF)
Here, S=1.5
C =
1.25 ∗ S
T
2
3
− − − − − − − 2
T = Ct hn
3
4 − − − − − −3
T = 0.073(
67
3.28
)
3
4
T=0.701
C =
1.25 ∗ 1.5
(0.701)
2
3
C=2.37
Total Seismic Dead Load
1. ∈ Floor Self Weight.
2. ∈ Beam Self Weight.
3. ∈ Column Self Weight.
4. ∈ Overhead Water Tank.
1. ∈ Floor Self Weight = (60’*42’)*(150+40*0.25)*6= 2419 Kips.
2. ∈ Beam Self Weight = {(60*4+42*5)*
12"∗16"
144
*150 }*7 = 630 Kips
3. ∈ Column Self Weight =
12∗14
144
*150*22*10*6= 231 Kips
4. ∈ Overhead Water Tank = 35 Kips
So, W= (2419+630+231+35) = 3315 Kips
Design Base Shear, V =
0.15∗1.0∗2.37
8
∗ 3315 − − − − − − − − − 1
V=147.31 Kips
Vertical Distribution of Lateral Forces
Here, T= 0.701

82. 82
So, concentrated force Ft acting at the top of the Building
Ft=0.0
Now, Fx=
𝑉−𝐹𝑡 𝑤𝑥 ∗𝑕𝑥
∈𝑤𝑖∗𝑕𝑖
--------- 4
Fx=
147.31−0 𝑕𝑥
∈10+20+30+40+50+60+7
Fx= 0.701hx
Vertical Distribution of Lateral Forces Table.
Story Height Force (Kips)
GL 7 4.76
1ST 17 11.54
2ND 27 18.33
3RD 37 25.12
4TH 47 31.91
5TH 57 38.70
6th
67 45.47
9.2 WIND LOAD CALCULATION
Wind Load Calculation-------- BNBC-1993

83. 83
Basic Wind Pressure, qb =
Pair ∗Vb
2
2
---------------- 1
Where, P(air)=
0.0765
32.2
= 23.76X10−4
slug/ft3
Vb= Basic Wind Speed, ft/see= 1.467 X Basic Wind Speed.
qb =
23.76X10−4
X(1.467Vb
2
)
2
= 0.00254 Vb
2
− − − − − 2
Where (qb) is in psf (Ib/ft^2) and (Vb) is in mph (mile/hr).
Sustained Wind Pressure.𝑞 𝑧 = 0.00254𝐶1 𝐶𝑧 𝑉𝑏
2
− − − − − −3
Where, C1=Structural Importance Coefficient.=
Cz=Heght and exposure Coefficient.=?
Design Wind Pressure. Pz = CGCtCpqz − − − − − 4
Where, CG=Wind gust Coefficient,
Ct= Local Topography Coefficient.=
Cp= Pressure Coefficient.
Design Data
X-Direction
Width (B)ft Length (L)ft Hight (H)ft H/B L/B Cp
42 60 67 1.595238095 1.428571429 1.6
Cc=2.54*10^-3 Ct=1.0
C1=1.0 Cz=? Fz=B*Hff*Pz (kip)
Vb=130mph CG=?
𝑞 𝑧 = 0.00254𝐶1 𝐶𝑧 𝑉𝑏
2
(psf) Pz = CGCtCpqz(psf)
Y-Direction
Width (B)ft Length (L)ft Hight (H)ft H/B L/B Cp
60 42 67 1.116667 0.70 2.062
Cc=2.54*10^-3 Ct=1.0
C1=1.0 Cz=? Fz=B*Hff*Pz (kip)
Vb=130mph CG=?
𝑞 𝑧 = 0.00254𝐶1 𝐶𝑧 𝑉𝑏
2
(psf) Pz = CGCtCpqz(psf)
Sustained Wind Pressure
Story Z(ft) Cc C1 Cz Vb(mph) qZ(psf)

84. 84
Ground Floor 7 0.00256 1 0.368 130 15.92115
1st Floor 17 0.00256 1 0.382 130 16.52685
2nd Floor 27 0.00256 1 0.455 130 19.68512
3rd Floor 37 0.00256 1 0.528 130 22.84339
4th Floor 47 0.00256 1 0.601 130 26.00166
5th Floor 57 0.00256 1 0.675 130 29.2032
6th Floor 67 0.00256 1 0.748 130 32.36147
Roof
Design Wind Pressure
Story Z(ft) CG Ct Cp(X) Cp(Y) qz(psf) Pz in (X)Psf Pz in (Y)Psf
Ground Floor 7 1.654 1 1.6 2.062 15.92115 42.1337367 54.2998531
1st Floor 17 1.64 1 1.6 2.062 16.52685 43.3664492 55.8885113
2nd Floor 27 1.573 1 1.6 2.062 19.68512 49.54351 63.8491985
3rd Floor 37 1.505 1 1.6 2.062 22.84339 55.0068879 70.8901268
4th Floor 47 1.438 1 1.6 2.062 26.00166 59.8246285 77.09899
5th Floor 57 1.37 1 1.6 2.062 29.2032 64.0134144 82.4972878
6th Floor 67 1.303 1 1.6 2.062 32.36147 67.4671968 86.9483499
Roof
Design Wind Force
Story Z(ft) Hff(ft)
Width
(B-X)ft
Width
(B-Y)ft Pz in (X)Psf Pz in (Y)Psf Fz in (X) Kip Fz in (Y) Kip
Ground
Floor 7 8.5 42 60 42.13373665 54.29985311 15.041744 27.6929251
1st Floor 17 10 42 60 43.36644915 55.88851134 18.2139086 33.5331068
2nd Floor 27 10 42 60 49.54351002 63.84919853 20.8082742 38.3095191
3rd Floor 38 10 42 60 55.00688794 70.89012683 23.1028929 42.5340761
4th Floor 47 10 42 60 59.82462853 77.09899002 25.126344 46.259394
5th Floor 57 10 42 60 64.0134144 82.49728781 26.885634 49.4983727
6th Floor 67 5 42 60 67.46719683 86.94834991 14.1681113 26.084505
Roof
9.3 BUILDING DESIGN USING ETABS

85. 85
Figure-12.1 Floor Plan

86. 86
Figure-12.2 Floor Plan 3D

87. 87
Figure-12.3 Elevation View

88. 88
Figure-12.4 Elevation View 3D

89. 89
Figure-12.5 Column and Beam 3D View

90. 90
Figure-12.6 Axial Force Diagram with Value X-X

91. 91
Figure-12.7 Axial Force Diagram with Value Y-Y

92. 92
Figure-12.8 Shear Force 2-2 Diagram

93. 93
Figure-12.9 Moment 2-2 Diagram

94. 94
Figure-12.10 Moment 3-3 Diagram

95. 95

96. 96

97. 97

98. 98

99. 99

100. 100

101. 101

102. 102

103. 103

104. 104

105. 105

106. 106

107. 107

108. 108

109. 109

110. 110

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112. 112

113. 113

114. 114