1. 1
SELECTION OF FOUNDATION FOR A BUILDING
FRAME
DEPARTMENT OF CIVIL ENGINEERING
NED UNIVERSITY OF ENGINEERING AND
TECHNOLOGY
KARACHI, PAKISTAN

2. 2
SELECTION OF FOUNDATION FOR A BUILDING
FRAME
BATCH 2011-2012
By
NAME SEAT NO
1. ABBAS KHAN CE-11036
2. HASSNAIN CE-11047
3. OSAID MAJEED CE-11054
4. MUHAMMAD YOUSUF IQBAL CE-11057
5. ZEHRA SHAWOO CE-11199
6. JAVERIA ALI CE-11308
DEPARTMENT OF CIVIL ENGINEERING
NED UNIVERSITY OF ENGINEERING AND
TECHNOLOGY
KARACHI, PAKISTAN

3. 3
CERTIFICATE
It is certified that the following students of batch 2011-2012 have successfully completed the
final year project in partial fulfilment of the requirements for four years degree of Bachelor of
Civil Engineering from NED University of Engineering and Technology, Karachi, Pakistan.
ABBAS KHAN CE-11036
HASSNAIN CE-11047
OSAID MAJEED CE-11054
MUHAMMAD YOUSUF IQBAL CE-11057
ZEHRA SHAWOO CE-11199
JAVERIA ALI CE-11308
PROJECT ADVISOR
_______________________
Aftab Ahmad Farooqi
Associate Professor
Department of Civil Engineering
NED University of Engineering &
Technology, Karachi.
__________________________
Prof. Dr. Asad-ur-Rehman Khan
Chairman
Department of Civil Engineering
NED University of Engineering &
Technology, Karachi.

4. 4
TABLE OF CONTENTS
CHAPTER NO. 1____________________________________________________________ 10
PRELIMINARY INFORMATION__________________________________________________ 10
1.1 INTRODUCTION ____________________________________________________________________ 10
1.2 OBJECTIVES _______________________________________________________________________ 11
1.3 LIMITATIONS ______________________________________________________________________ 11
1.4 APPROACH _________________________________________________________________________ 11
1.5 STRUCTURAL LOADS ______________________________________________________________ 11
1.5.1 DEAD LOAD_______________________________________________________________________ 12
1.5.3 LIVE LOADS ______________________________________________________________________ 12
1.5.4 LATERAL LOADS__________________________________________________________________ 13
1.5.5 EARTHQUAKE LOADS_____________________________________________________________ 13
1.6 STOREY FORCES ___________________________________________________________________ 13
1.7 CENTRE OF MASS __________________________________________________________________ 14
1.8 CENTRE OF RIGIDITY ______________________________________________________________ 14
1.9 WEIGHT OF THE BUILDING FRAME _________________________________________________ 14
1.9.1 BUILDING DESCRIPTION __________________________________________________________ 14
1.9.2 MATERIAL PROPERTIES __________________________________________________________ 15
1.9.3 UNIT LOAD _______________________________________________________________________ 15
1.9.3.1 DEAD LOAD FOR UNIT SLAB AREA_______________________________________________ 15
1.9.3.2 BEAM LOAD PER FOOT LENGTH _________________________________________________ 16
1.9.3.3 COLUMN LOAD PER UNIT HEIGHT_______________________________________________ 17
1.10 TRIBUTARY AREA METHOD _______________________________________________________ 18
1.10.2 EARTHQUAKE FORCE____________________________________________________________ 19
1.10.6 CALCULATIONS FOR CENTER OF MASS AND CENTER OF RIGIDITY _______________ 22
1.10.7 SUPERPOSITIONS ________________________________________________________________ 23
1.11 LATERAL EARTHQUAKE ANALYSIS________________________________________________ 23
1.12 GRAVITY ANALYSIS _______________________________________________________________ 23
1.13 COLUMN DESIGN __________________________________________________________________ 24
1.14 STRUT BEAM DESIGN FOR RESISTING TORSION ___________________________________ 25
CHAPTER NO. 2____________________________________________________________ 27
2.1 DESIGN OF FOUNDATIONS __________________________________________________ 27
2.1.1 ISOLATED FOOTING DESIGN ______________________________________________________ 28
2.1.2 PILE FOUNDATION DESIGN _______________________________________________________ 35

5. 5
LIST OF TABLES
Table 1. 1 Typical uniformly distributed design loads 12
Table 1. 2 Dimensions of Structural Elements 15
Table 1. 3 Slab and Finishes Self Weight 16
Table 1. 4 Summary of loads 18
Table 1. 5 Loads on Column 18
Table 1. 6 Earthquake Factors 19
Table 1. 7 Distribution of Base Shear into Storey forces 21
Table 1. 8 Distribution of storey shear into frame forces 22

6. 6
LIST OF FIGURES
Figure 1. 1 Typical section of slab 15
Figure 1. 2 Typical Beam Section 16
Figure 1. 3 Typical Section of Column 17
Figure 1. 4 Typical Slab Beam Masonry and column 17
Figure 1. 5 Superposition of earthquake and gravity moments 23
Figure 1. 6 Typical Section of a Column Error! Bookmark not defined.
Figure 1. 7 Cross section of Strut Beam 26
Figure 2. 1 Preliminary Footing Plan 27
Figure 2. 2 Isolated Footing 28
Figure 2. 2 Isolated Footing 28
Figure 2. 3 One way shear 30
Figure 2. 3 One way shear 30
Figure 2. 4 Two Way Shear Check 31
Figure 2. 4 Two Way Shear Check 31
Figure 2. 5 Dimensions for flexure Design 32
Figure 2. 5 Dimensions for flexure Design 32
Figure 2. 6 Reinforcement for Isolated Footing Error! Bookmark not defined.
Figure 2. 7 Plan with Final dimensions of Footing 34
Figure 2. 8 Cross section with Water Table 37
Figure 2. 8 Cross section with Water Table 37
Figure 2. 9 Pile Cap 40

7. 7
NOTATIONS
ACI American Concrete Institute
As Area of Steel
ASCE American Society of Civil Engineers
d Effective depth
Ca Seismic Coefficient
Cv Seismic Coefficient
D Dead Loads
E Earthquake Loads
EC Modulus of Elasticity of concrete
f’c Standard Cylinder Strength of Concrete
fy Yield Strength of Steel reinforcement
I Importance Factor
IMRF Intermediate Moment Resisting Frame
Kips Kilo pounds
L Live load
ld Development length
Psf Pounds per square foot
Psi Pounds per square inch
Pu Factored load
q Soil bearing capacity
qu Ultimate bearing capacity of soil using factored load
Tu Torque provided by factored load
V Base shear

8. 8
ABSTRACT
Foundation play a vital role in the stability of a building frame subjected to gravity as well as
seismic loads. For a particular building frame, choices/ selection of foundation and its design is a
very important task of the structure engineer.
A Model reinforced concrete building frame subjected to gravity and earthquake loads was
selected for design of its footing. In the first phase, reinforced concrete isolated footings were
chosen and designed as per ACI Code of practice. As a second choice, reinforced concrete pile
foundations were designed for the same building frame. The building frame selected was under
the action of torsion with respect to vertical axis. This torsional effect is resisted within the frame
through mechanism of beams and columns. Building frame was first analysed subjected to gravity
plus seismic loads to find loads and moment to be transferred on the foundations. Uniform bearing
capacity of the soil beneath the building foundation is assumed.
Reader of this report can acquire knowledge about procedure of analysis of building frame and
design of isolated footings as well as Pile foundation.

9. 9
DEDICATION
To our beloved Teachers of
NED University of Engineering and Technology, Karachi

10. 10
CHAPTER NO. 1
PRELIMINARY INFORMATION
1.1 INTRODUCTION
Foundation is the part of the engineered system that interfaces the load-carrying components to
the ground. It is the part that transmits to, and into the underlying soil or rock, the loads supported
by foundation and it’s self-weight.
Foundation elements must be proportioned both to interface with the soil at a safe stress level
and to limit settlements to an acceptable amount.
We have designed two types of footings for the given building frame as described below:-
 Isolated footing/Spread footing
 Pile footing
The super structure of the given building frame was analysed to find the moments and loads
transferred to the foundation level. For that purpose, we used Portal Frame Method for
calculating seismic loads and Matrix Displacement Method for gravity loads; UBC-1997 is used
for earthquake analysis whereas ACI 318-05 is used for design of Reinforced Concrete Footings
and piled foundation.
.
Methods used for the design of above mentioned footings are given below:-
 Isolated footings are designed by Ultimate Strength Design (USD) Method.
 Pile foundations are designed by Ultimate Limit State Design Method.
All calculations of analysis and design are performed manually. Microsoft Excel has been used for
solution of Matrices.

11. 11
1.2 OBJECTIVES
 To Analyse and design the in situ isolated footings receiving gravity and earthquake
loads from superstructure of a given model building frame.
 To Analyse and design the in situ piled foundation receiving gravity and earthquake
loads from superstructure of the given model building frame.
1.3 LIMITATIONS
 Isolated and pile foundations are made of Reinforced concrete.
1.4 APPROACH
Work of this project proceeded in the steps given under:
 Gravity analysis of the frame by using Matrix Displacement Method.
 Earthquake lateral analysis by using Portal Frame Method.
 Design of Isolated footings
 Design of Pile Foundation
1.5 STRUCTURAL LOADS
Structural loads could be of two type like i) Live Loads and ii) Dead Loads; specifications and
intensity of design live load defined by UBC are reproduced under.

12. 12
1.5.1 DEAD LOAD
Dead load is primarily due to self-weight of structural members, permanent partition walls, fixed
permanent equipment and weight of materials, floor surfacing materials and other finishes. It can
be worked out precisely from the known weights of the materials and the dimensions on the
working drawings.
1.5.3 LIVE LOADS
All the movable objects in a building such as people, desks, cupboards and filing cabinets produce
an imposed load on the structure. This loading may come and go with the result that its intensity
will vary considerably.
Table 1. 1 Typical uniformly distributed design live loads

13. 13
1.5.4 LATERAL LOADS
The Lateral loads on the super structure of a building frame generally considered are due to
blowing wind, action of earthquake, lateral earth pressure etc.
Frame considered in this report is subjected to earth quake forces.
1.5.5 EARTHQUAKE LOADS
The wave effect of an earthquake on the base of a building frame is converted into a static lateral
force called base shear. UBC 1997 is used for determination of base shear as well as further
onwards distribution of lateral Force. As per UBC 1997, the base shear can be calculated as,
𝑽 =
𝑪𝒗 ∗ 𝑰
𝑹𝑻
∗ 𝑾
1.6 STOREY FORCES
Base shear acting at the lowest level of frame is further distributed at each storey level as defined
by UBC- 1997.
It can be calculated as:
𝑭 𝒙 =
(𝑽 − 𝑭𝒕)(𝑾 𝒙 ∗ 𝒉 𝒙)
∑( 𝑾 𝒙 ∗ 𝒉 𝒙)
x= storey level
V = Base Shear
Ft = Extra Force at the top
Wx = Portion of total weight acting on each floor
hx = Height of each floor

14. 14
1.7 CENTRE OF MASS
In a continua, the total mass of the body can be lumped on the centre of mass which can be
calculated as under,
COM = ∑ Area * Moment
∑Area
1.8 CENTRE OF RIGIDITY
Point of total resistance against lateral forces is termed as Centre of Rigidity and is calculated as
given under,
COR = ∑ (Moment of the I with respect to give reference)
∑I
1.9 WEIGHT OF THE BUILDING FRAME
1.9.1 BUILDING DESCRIPTION
The main features of building are stated below:
a) Covered area = 4056ft2
b) Number of stories = Ground+4
c) No. of levels below ground = Zero
d) Storey height for ground and 1st
floor is 16ft and for 2nd
, 3rd
and 4th
floor is 14ft.

15. 15
1.9.2 MATERIAL PROPERTIES
Material = Reinforced Concrete
fc’ = 4,000 psi Ec= (57000) √𝑓𝑐′
2
γRC = 150pcf
fy = 60,000 psi Es= 29 x 103
ksi γCC = 144psf
1.9.3 UNIT LOAD
1.9.3.1 DEAD LOAD FOR UNIT SLAB AREA
i. Self-weight =
𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠
12
× 150 𝑝𝑠𝑓
=
6"
12
× 150 𝑝𝑠𝑓
= 75 𝑝𝑠𝑓
Finishes =
𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠
12
× 144 𝑝𝑠𝑓
=
3"
12
× 144 𝑝𝑠𝑓
= 36 𝑝𝑠𝑓
Where
t = thickness of slab = 6”
Load of Partition Wall = 30 𝑝𝑠𝑓
Slab Beam Column Masonry
6” thick 8” x 24” 18” x 18”
24” x 24”
30” x 30”
13’-6" height
11’-6” height
Table 1. 2 Dimensions of Structural Elements
Figure 1. 1 Typical section of slab

16. 16
Figure 1. 2 Typical Beam
Section
1.9.3.2 BEAM LOAD PER FOOT LENGTH
Beam Dimension = 8” x 24”
Self –weight =
𝐴𝑟𝑒𝑎
144
× 𝑢𝑛𝑖𝑡 𝑙𝑒𝑛𝑔𝑡ℎ × 150 𝑝𝑠𝑓
=
8" × 24"
144
× 1′ × 150 𝑝𝑠𝑓
= 200
𝑙𝑏
𝑓𝑡
Beam height is taken as 24” rather than 30” because the slab thickness
which is 6” is excluded from the total height of beam.
Table 1. 3 Slab and Finishes Self Weight

17. 17
1.9.3.3 COLUMN LOAD PER UNIT HEIGHT
a. For Column dimension = 18” x 18”
Self-weight =
𝐴𝑟𝑒𝑎
144
× 𝑢𝑛𝑖𝑡 𝑙𝑒𝑛𝑔𝑡ℎ × 150 𝑝𝑠𝑓
=
18" × 18"
144
× 1′ × 150 𝑝𝑠𝑓
= 337.5
𝑙𝑏
𝑓𝑡
b. For Column dimension = 24” x 24”
Self-weight =
𝐴𝑟𝑒𝑎
144
× 𝑢𝑛𝑖𝑡 ℎ𝑒𝑖𝑔ℎ𝑡 × 150 𝑝𝑠𝑓
=
24" × 24"
144
× 1′ × 150 𝑝𝑠𝑓
= 600
𝑙𝑏
𝑓𝑡
1.9.3.4 MASONRY LOAD PER UNIT HEIGHT
Self-weight =
𝐴𝑟𝑒𝑎
144
× 𝑢𝑛𝑖𝑡 ℎ𝑒𝑖𝑔ℎ𝑡 × 150 𝑝𝑠𝑓
=
"11.5 ∗ 0.5
1
× 1′ × 150 𝑝𝑠𝑓
= 840
𝑙𝑏
𝑓𝑡
=
"13.5 ∗ 0.5
1
× 1′ × 150 𝑝𝑠𝑓
= 1012
𝑙𝑏
𝑓𝑡
Figure 1. 3 Typical Section of
Column
Figure 1. 4 Typical Slab Beam Masonry
and Column

18. 18
1.10 TRIBUTARY AREA METHOD
Loads form floors supported by beams are transformed to the lower column through Tributary
Area Technique. The summary of loads is shown under:
1.10.1 LOADS ON COLUMN
Table 1. 4 Summary of loads
Table 1. 5 Loads on Column

19. 19
1.10.2 EARTHQUAKE FORCE
Zone, Z 2B
Soil profile type
Sc
0.32 Very Dense Soil and
Soft Rock
Seismic zone factor Z 0.2 Ref: Appendix C
Seismic importance
factor
I
1.0 Ref: Appendix C
Structural system
(moment resisting
concrete frame)
R
5.5 Ref: Appendix C
Seismic coefficient Cv
Ct
0.32
0.030
Ref: Appendix C
Time period 𝑇 = 𝐶𝑡 ( ℎ𝑠𝑡)3/4 0.757 Sec
Dead Weight W 9310 Kips
Base Shear V 792.4 kips
Weight of Building
The weight of the building is 9310 Kips. (Detailed calculations are attached in Appendix D).
Table 1. 6 Earthquake Factors

20. 20
1.10.3 Base Shear
Base shear is calculated using Eq. 2.3
𝑉 =
Cv x I
RT
x W
Where
T = Fundamental period of structure
T = Ct * h3/4
(Using Eq. 2.4)
Cv = Seismic Co-efficient
Cv = 0.32 (Table 6, Appendix C)
Ct = 0.030 (for concrete moment frame)
I = Seismic Importance Factor
I = 1.0 (Table 6, Appendix C)
R = Ductility and Over strength factor for Intermediate Moment Resisting Frame
R = 5.5 (Table 5 , Appendix C)
W = Cumulative service D.L
W = 9310 Kips
h = 74’ (Height of building)
Z = Seismic zone co-efficient
Z = 0.2 (Table 6, Appendix C)
T = Time Period T = 0.757 Sec
V = Base shear V = 792.5 Kips

21. 21
1.10.4 Storey Forces
Storey forces are calculated by using Eq. 2.5
Fx =
(V−Ft)(Wx ∗ hx)
∑(Wx ∗ hx)
Where,
V = Base Shear =792.5 Kips
Ft = 0.07 TV ≤ 0.25 V (if T>0.7 sec)
T = Time period of Vibration = 0.757 sec
Since T > 0.7 sec, so Ft = 0.07 TV ≤ 0.25 V
Ft = 0.07 TV = 0.07(0.757*794.5) ≤ 0.25V = 0.25*794.5
Ft = 42.1 < 198.6 (ok).
Table 1. 7 Variation of Storey forces

22. 22
1.10.5 Transformation of Story forces into 2 D FRAMES
The distribution of storey forces into frame forces is determined by using the Portal frame method.
As a sample, calculations of Shear Forces and Bending Moments of Frame 1 are reproduced under. The
other all values are attached in the Appendix.
Frame 1 Shear forces Bending moments
Forces X 2x x 2x
Storey 5 31.7 7.925 15.85 55.475 110.95
Storey 4 30.9 15.65 31.3 109.55 219.1
Storey 3 23.1 21.425 42.85 149.975 299.95
Storey 2 16.3 25.5 51 204 408
Storey 1 8.3 27.575 55.15 220.6 441.2
1.10.6 CALCULATIONS FOR CENTER OF MASS AND CENTER OF RIGIDITY
To check whether building frame is under the action of any torsion due to lateral forces,
determination of centre of mass and centre of rigidity is required. Calculate under
a) Centre of Mass
The coordinates for Centre of Mass are:
Cmx = 42.8’
Cmy = 39.3’
b) Centre of Rigidity
Coordinate for centre of rigidity are calculated by using Eq. 2.2
𝑪𝑶𝑹 =
∑(𝐈 𝐗 𝐌𝐨𝐦𝐞𝐧𝐭 𝐚𝐫𝐦)
∑𝐈
Table 1. 8 Distribution of storey shear into frame forces

23. 23
The Coordinate for center of rigidity are:
Crx = 39.3’
Cry = 42.8’
Tabulated determination of Centre of Rigidity is attached in (Table 10, Appendix C)
1.10.7 SUPERPOSITIONS
Earthquake moments and Gravity moments are superimposed on each other.
(Refer Appendix D for detailed calculation)
1.11 LATERAL EARTHQUAKE ANALYSIS
 Each foundation is subjected to the vertical load coming through column and the moment
due to effect of lateral forces on the frame.
 Once lateral seismic force on each storey are calculated, they are further distributed in
each 2D frame with respect to the relativity of the frame to that of the storey.
 For analysis of frame Portal Frame method is used.
1.12 GRAVITY ANALYSIS
Matrix displacement method is adopted for gravity analysis of building frame.
Figure 1. 5 Superposition of earthquake and gravity moments

24. 24
1.13 Check for adequacy of columns for Torsion
Applied Torque
Applied Torque, Ta = 792.4 x 3.4 = 2695 K-ft
Resisting Torque
The effect of applied torque is transferred to each column. The calculation of equilibrium is
given under:
Ta = 2695 K-ft < Tr = 86402.78 K-ft
Therefore, building frame is safe in torsion.
Note: Resisting torque created in each individual column is being transferred to Plinth level, i.e.
bottom of column.

25. 25
1.14 STRUT BEAM DESIGN FOR RESISTING TORSION (At Plinth Level)
Shear Stress = 480 psi
Pu = 432 Kip
Using the theory of strut design,
Pu = φ Pn
Pu = φ K Ag (0.85fc’+ρ (fy – 0.85 fc’))
432 = 0.65*0.8*Ag {0.85*4+0.05*(60-0.85*4)}
Ag= in2
Let b =
Ag= b*h
= in2
h=
Dimension is ” x ”
For Ast,
Pu = φ*K* [Ag*0.85*fc’+Ast*(fy – 0.85 fc’)]
432 = 0.65*0.8{*0.85*4+Ast*(60-0.85*4)}
Ast = in2
Asmin = 0.01Ag
Asmin = 0.01* = 0.72 in2
No. of bars = Ast/Ab (using # 4 bar, Ab = 0.196 in2
)

26. 26
No. of bars = bars
USE bars
SECTIONS
bmin = 2(1.5”) + 1(1”) + 2(0.5”) + 2(3/8)”
bmin = 4.75” < 6” OK
hmin = 2(1.5”) + 4(1”) + 5(4/8)” + 2(3/8)”
hmin = 10.25” < h= 12” .OK.
Figure 1. 6 Cross section of Strut Beam

27. 27
CHAPTER NO. 2
2.1 DESIGN OF FOUNDATIONS
Figure 2. 1 Preliminary Footing Plan

28. 28
2.1.1 ISOLATED FOOTING DESIGN
FOUNDATION 2
COLUMN 3C
NET UPWARD PRESSURE
Allowable Bearing Capacity of Soil = 𝑞 𝑎𝑙𝑙
𝑞 𝑎𝑙𝑙=3 tsf Refer Appendix B
=3 ∗
2240
1000
=6.72 ksf
Assume Area of Footing be 10 ft * 16 ft
Net effective pressure beneath the footing = 𝑞 𝑢
For Biaxial Moments, following conditions are applied:
Condition 1: 𝑞 𝑢=
𝑃
𝐴
+
𝑀 𝑥 𝑌
𝐼 𝑥
+
𝑀 𝑌 𝑋
𝐼 𝑦
=
1260
10∗16
+
834∗(
16
2
)
16∗
163
12
+
205.2∗(
10
2
)
10∗
103
12
=10.33 Ksf
Condition 2: 𝑞 𝑢=
𝑃
𝐴
+
𝑀 𝑥 𝑌
𝐼 𝑥
-
𝑀 𝑌 𝑋
𝐼 𝑦
=
1260
10∗16
+
834∗(
16
2
)
16∗
163
12
-
205.2∗(
10
2
)
10∗
103
12
=7.86 Ksf
Condition 3: 𝑞 𝑢=
𝑃
𝐴
-
𝑀 𝑥 𝑌
𝐼 𝑥
+
𝑀 𝑌 𝑋
𝐼 𝑦
Figure 2. 2 Isolated Footing
Figure 2. 3 Isolated Footing

29. 29
=
1260
10∗16
-
834∗(
16
2
)
16∗
163
12
+
205.2∗(
10
2
)
10∗
103
12
= 7.88 Ksf
Condition 4: 𝑞 𝑢=
𝑃
𝐴
-
𝑀 𝑥 𝑌
𝐼 𝑥
-
𝑀 𝑌 𝑋
𝐼 𝑦
=
1260
10∗16
-
834∗(
16
2
)
16∗
163
12
-
205.2∗(
10
2
)
10∗
103
12
= 5.4 Ksf
Since 𝑞 𝑢 >𝑞 𝑎𝑙𝑙, the condition is not satisfied.
Therefore Area of footing may be increased.
Assume Area of footing =16’ *16’
For Biaxial Moments, following conditions are applied:
Condition 1: 𝑞 𝑢=
𝑃
𝐴
+
𝑀 𝑥 𝑌
𝐼 𝑥
+
𝑀 𝑌 𝑋
𝐼 𝑦
=
1260
16∗16
+
834∗(
16
2
)
16∗
163
12
+
205.2∗(
12
2
)
16∗
163
12
= 6.44 Ksf
Condition 2: 𝑞 𝑢=
𝑃
𝐴
+
𝑀 𝑥 𝑌
𝐼 𝑥
-
𝑀 𝑌 𝑋
𝐼 𝑦
=
1260
16∗16
+
834∗(
16
2
)
16∗
163
12
-
205.2∗(
12
2
)
16∗
163
12
= 5.41 Ksf
Condition 3: 𝑞 𝑢=
𝑃
𝐴
-
𝑀 𝑥 𝑌
𝐼 𝑥
+
𝑀 𝑌 𝑋
𝐼 𝑦

30. 30
=
1260
16∗16
+
834∗(
16
2
)
16∗
163
12
-
205.2∗(
12
2
)
16∗
163
12
= 3.92 Ksf
Condition 4: 𝑞 𝑢=
𝑃
𝐴
-
𝑀 𝑥 𝑌
𝐼 𝑥
-
𝑀 𝑌 𝑋
𝐼 𝑦
=
1260
16∗16
-
834∗(
16
2
)
16∗
163
12
-
205.2∗(
12
2
)
16∗
163
12
= 3.47 Ksf
𝑞 𝑢 < 𝑞 𝑎𝑙𝑙
Therefore, selected size of Area of Footing is 16’ *16’
CHECK FOR ONE WAY SHEAR
Assume depth of footing =1’-6”
𝑑= h- cover- db (1.5)
=1.5*12 -3- 1(1.5)
= 13.5”
According to ACI 318-05.
 Shear capacity = фVc
ф𝑉𝑐= ф2√𝑓𝑐′ bd
=0.75 * 2*1* √4000 *(16*12) *
13.5
1000
=246 k
 Applied shear = Vu
𝑉𝑢= 𝑞 𝑢 ∗ 𝐵 ∗ (
𝐿
2
-
𝑐
2
- d)
=6.444 *16* (
16
2
-
30
2∗12
-
13.5
12
)
=579.96 K
Assuming #8 bars (db=1)
Figure 2. 4 One way shear
Figure 2. 5 One way shear

31. 31
𝑉𝑢 579.96k > ф 𝑉𝑐= 246k
Therefore, the section is not adequate
Now, take depth of footing =3’-0”
𝑑= h-cover- db (1.5)
=3*12 -3- 1(1.5)
= 31.5”
A/c to ACI 318-05
 Shear capacity = фVc
=0.75 * 2*1* √4000 *(16*12) *
31.5
1000
=573.76 k
 Applied shear = Vu
𝑉𝑢= 𝑞 𝑢 ∗ 𝐵 ∗ (
𝐿
2
-
𝑐
2
- d)
=6.444 *16* (
16
2
-
30
2∗12
-
31.5
12
)
=425 K
𝑆𝑖𝑛𝑐𝑒, 𝑉𝑢 = 435k < ф 𝑉𝑐 =573.76k
Section is adequate in one way shear for h=3’-0”
CHECK FOR TWO WAY SHEAR:
𝑏0 = 2(𝑐1+d) + 2(𝑐2+d)
= 2 (30+31.5) + 2(30+31.5)
= 246 in
ф 𝑉𝑐= ф 4λ√𝑓𝑐′ 𝑏0d
=0.75 * 4*1* √4000 *(246) *
31.5
1000
=1470.3 k
𝑉𝑢= 𝑃𝑢 - 𝑞 𝑢(𝑐1+d) (𝑐2+d)
Assuming #8 bars (db=1)
Assuming #8 bars (db=1)
Figure 2. 6 Two Way Shear Check

32. 32
= 1260 – 6.44
(30+31.5) (30+31.5)
144
= 1090.75 K
𝑉𝑢= 1090.75k < ф𝑉𝑐= 1470.3k
Section is Adequate in Two Way Shear
FLEXURAL DESIGN:
𝑀 𝑢= 𝑞 𝑢 ∗ (
𝐿
2
-
𝑐
2
)*
𝐵
2
=6.44 * (
16
2
-
30
2∗12
) *
16
2
=348 k-ft
X and Y -direction:
𝑅 𝑢=
𝑀 𝑢
𝐵𝑑2
=
348∗1000∗12
16∗12∗31.52
= 21.92 psi
ρ =
0.85 𝑓𝑐′
𝑓𝑦
(1-√1 −
2𝑅 𝑢
∅0.85𝑓𝑐′ )
=
0.85∗4
60
(1-√1 −
2∗21.92
0.9∗0.85∗4000
)
=0.000433
𝐴 𝑠 = ρbd
= (0.000433) (16*12) (32.5)
= 2.7 𝑖𝑛2
𝐴 𝑠 𝑚𝑖𝑛=
200
𝑓𝑦
Bd
=
200
60,000
* 16*12 *32.5
= 20.8 𝑖𝑛2
𝑈𝑠𝑒𝑑 𝐴 𝑠 = 20.8 𝑖𝑛2
Using #9 Bars, 𝐴 𝑏 = 0.994 𝑖𝑛2
Figure 2. 8 Dimensions for flexure Design
Figure 2. 9 Dimensions for flexure Design

33. 33
No of Bars =
𝐴 𝑠
𝐴 𝑏
=
20.8
0.984
=21 bars
Spacing =
16∗12−2∗3
20
= 9.3” ≈ 9”
Again,
No of Bars (revised) =
16∗12−2∗3
9
= 21 bars
Provide 21#9 bars @9” c/c in both X and Y direction.
DEVELOPMENT LENGTH OF MAIN REINFORCEMENT:
𝑙 𝑑=
𝑑 𝑏 𝑓𝑦
20√ 𝑓𝑐
′
=
(1)(60,000)
20√4000
= 47.43” ≈ 48”
Figure 2. 10 Dimensions for flexure Design

34. 34
Figure 2. 11 Plan with Final dimensions of Footing

35. 35
2.1.2 PILE FOUNDATION DESIGN
Procedure used for pile design is described in Appendix D.
2.1.2.1 Basic Data for Plie Design
Values given under are taken for the book “Principles of Foundation Engineering “written by
Braja.M.Das, Appendix D.
Soil Type: Clayey Sand (φ=30⁰)
1. Cohesion, c 200
𝑙𝑏
𝑓𝑡2
2. Angle of
Friction
300
3. Nc 30.14 4. Nq 18.40
5. 𝑁𝑟 22.40 6. ɣ 110
𝑙𝑏
𝑓𝑡3
7. tan φ 0.58 8. 𝑓
′
𝑐 4 ksi
9. 𝑓𝑦 60 ksi 10. Dia of Pile 18 in
11. Earth pressure
coefficient(K=1- sin φ)
0.5 12. ɣw 62.4
𝑙𝑏
𝑓𝑡3
Pile Design
Dia of Pile=18 in
𝑓
′
𝑐= 4 ksi, 𝑓𝑦=60 ksi
𝑸 𝒖= 𝑸 𝒑 + 𝑸 𝒔
𝑄 𝑝=𝑞 𝑝 𝐴 𝑝
𝑞 𝑝=C𝑁𝑐 + 𝑞𝑁𝑐 + 0.5ɣ𝐷𝑁𝑐
𝑞 𝑝= (200) (30.14) + (110) (20) (18.40) + (0.5) (110) (18 12)⁄ (22.4)
𝑞 𝑝=50384 𝑙𝑏 𝑓𝑡2⁄

36. 36
𝐴 𝑝=
⊼
4
𝐷2
𝐴 𝑝=
⊼
4
(1.5)2
𝐴 𝑝=1.77 𝑓𝑡2
𝑄 𝑝=𝑞 𝑝 𝐴 𝑝
𝑄 𝑝= (50384) (1.77)
𝑄 𝑝=89.2 kips = 40.5 tons
𝑸 𝒔:
𝑓𝑠= K 𝜃 𝑣tan δ
𝑓𝑠=(0.5) (110*20) (0.58)
𝑓𝑠=638 𝑙𝑏 𝑓𝑡2⁄
𝐴 𝑝=𝐶𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙 𝐴𝑟𝑒𝑎
𝐴 𝑝= ⊼ DL
𝐴 𝑝= ⊼ (18) (20)/12
𝐴 𝑝= 94.2 𝑓𝑡2
𝑄𝑠=𝑓𝑠 𝐴 𝑝
= (638) (94.2)
=60.1 kips = 27.32 tons
𝑸 𝒖= 𝑸 𝒑 + 𝑸 𝒔
= 27.32+40.5
= 67.8 tons ≈68 tons

37. 37
With Water Table at h=15’
ɣ′
= ɣ-ɣ 𝑤
Where,
ɣ′
= ɣ of submerge
ɣ 𝑤= ɣ of water
ɣ′
= 110 − 62.4
ɣ′
=47.6
𝑙𝑏
𝑓𝑡3
Now,
𝑞 𝑤=ɣ′
ℎ 𝑤
= 47.6 (5)
=238 𝑙𝑏 𝑓𝑡2⁄ = 0.238 ksf
𝑄 𝑤=𝑞 𝑤 𝐴 𝑝
= (0.238) (1.77)
= 0.421 kips =0.191 tons
𝑸 𝒖= 𝑸 𝒑 + 𝑸 𝒔 + 𝑸 𝒘
= 27.32 + 40.5 +0.191
= 68.191 tons
Figure 2. 12 Cross section with Water Table
Figure 2. 13 Cross section with Water Table

38. 38
DESIGN OF PILE CAP:
Column 2C
Load=1845 kips Refer Appendix B
=
1845
2.204
tons
= 837.11 tons ≈ 840 tons
Total load = 840 + 18= 858 tons
Taking 60 tons pile capacity per pile
No of Piles =
858
60
= 16 piles
Pile diameter =18”
ℎ 𝑐=30”
Use #9 bars (𝐴 𝑏 = 1.0)
No of bars=
𝐴 𝑠
𝐴 𝑏
=
13.15
1
= 14 bars
Spacing =
11.5∗12−2(3)
13
= 10.15” ≈ 10”
Therefore, Provide 14 #9 @10” c/c
ℎ𝑓 = d+
𝑑 𝑏
2
+ 𝑑 𝑏+cover +pile cap embedded

39. 39
= (3.875*12) + (1.125
2⁄ ) + 1.125 + 2 + 4
= 36”
APPLY CHECK FOR REINFORCEMENT
ρ=
𝐴 𝑠
𝑏ℎ 𝑓
ρ =
13.14
138∗36
ρ = 0.0017
Min required ratio =0.0018 >0.0017 Not Satisfied!
Use ρ = 0.0018
MANUAL CHECK FOR SHEAR PARAMETER:
𝑑 𝑃=18” (FOR 60 ton pile)
𝑃𝑢= 60*2204*1.55= 212 kips
CHECK FOR EFFECTIVE DEPTH:
d=
𝑃 𝑢
∅4√𝑓𝑐′((𝑑 𝑃
+𝑑)
⊼
4
+2(𝑒𝑑𝑔𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒))
d=
246848
4∗0.75∗√4000⌊(18+46)
⊼
4
+2(15)⌋
d= 22.16” < 46” OK!

40. 40
Check for Beam Shear:
d=
𝑃 𝑢
∅2√𝑓𝑐′((𝑑 𝑃
+2𝑑)+2 √2(𝑒𝑑𝑔𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒))
d=
212000
0.75∗2∗√4000⌊(18+2(46)+2√2(15)⌋
d= 15” < 46” OK!
Figure 2. 14 Pile Cap

41. 41
REFERENCES
 Hassoun, Nadim M., Al Manaseer,. Akhtem., “Structural Concrete: Theory and Design”,
4th
Edition.
 Das, Braja M., “Principles of Foundation Engineering”, 3rd
Edition.
 ACI 318-05
 Uniform Building code (UBC-1997)