DESIGN OF FOUNDATIONS
Dr. Izni Syahrizal bin Ibrahim
Faculty of Civil Engineering
Universiti Teknologi Malaysia
Email: iznisyahrizal@utm.my
Introduction
• Foundation – Part of structure which
transmits load from the structure to the
underlying soil or rock
• All soils compress noticeably when loaded
causing structure to settle
Introduction
• Requirements in the design of foundations:
(i) Total settlement of the structure to be
limited to a tolerably small amount
(ii) Differential settlement of various parts
of structure shall be eliminated
Introduction
• To limit settlement, it is necessary to transmit the
structure load to a soil stratum of sufficient
strength
• Spread the structure load over a sufficiently large
area of stratum to minimize bearing pressure
• Satisfactory soil: Use footings
• Adequate soil: Use deep foundations i.e. piles
Introduction
• Pressure distribution under a footing
Uniform
distributed
Cohesive soil
Cohesionless soil
Types of Foundation
Pad Footings
• Transmit load from piers and
columns
• Simplest and cheapest type
• Use when soil is relatively strong or
when column loads are relatively
light
• Normally square or rectangular
shape in plan
• Has uniform thickness
Types of Foundation
Combine Footings
• Use when two columns are closed
together
• Combine the footing to form a
continuous base
• Base to be arranged so that its
centreline coincides with the centre
of gravity of the load – provide
uniform pressure on the soil
Types of Foundation
Strap Footings
• Use where the base for an exterior
column must not project beyond
the property line
• Strap beam is constructed between
exterior footing & adjacent interior
footing
• Purpose of strap – to restrain
overturning forces due to load
eccentricity on the exterior footing
Types of Foundation
Strap Footings (continued)
• Base area of the footings are
proportioned to the bearing
pressure
• Resultant of the loads on the two
footings should pass through the
centroid of the area of the two
bases
• Strap beam between the two
footings should NOT bear against
the soil
Types of Foundation
Strip Footings
• Use for foundations to load-bearing
wall
• Also use when pad footings for
number of columns are closely
spaced
• Also use on weak ground to
increase foundation bearing area
Types of Foundation
Raft Foundations
• Combine footing which covers the
whole building
• Support all walls & columns
• Useful where column loads are
heavy or bearing capacity is low –
need large base
• Also used where soil mass contains
compressible layers or soil is
variable – differential settlement
difficult to control
Types of Foundation
Pile Foundations
• More economic to be used when
solid bearing stratum i.e. rock is
deeper than about 3 m
• Pile loads can either be transmitted
to a stiff bearing layer (some
distance below surface) or by
friction along the length of pile
• Pile types – precast (driven into the
soil) or cast in-situ (bored)
• Soil survey is important to provide
guide on the length of pile and safe
load capacity of the pile
Types of Foundation
Pile Foundations
Load from
Structure
Pile Cap
Lower Density
Medium Density
High Density
P
I
L
E
P
I
L
E
P
I
L
E
P
I
L
E
Design of Pad Footing
Thickness and Size of Footing
Area of pad:
𝑮𝒌 + 𝑸𝒌 + 𝑾
𝑨=
𝑺𝒐𝒊𝒍 𝒃𝒆𝒂𝒓𝒊𝒏𝒈 𝒄𝒂𝒑𝒄𝒊𝒕𝒚
Minimum effective depth of pad:
𝒅=
𝑵𝑬𝒅
𝒗𝒓𝒅,𝒎𝒂𝒙 ∙ 𝒖𝒐
NEd = Ultimate vertical load = 1.35Gk + 1.5Qk
𝑓
vrd,max = 0.5vfcd = 0.5 0.6 1 − 𝑐𝑘
250
uo = Column perimeter
𝑓𝑐𝑘
1.5
Design of Pad Footing
Design for Flexure
• Critical section for bending – At the face of the column
• Moment is taken on a section passing completely across the
footing and due to ultimate load on one side of the section
• Moment & shear is assessed using STR (Structure) combination
x
y
y
x
STR Combination 1:
𝑵 = 𝟏. 𝟑𝟓𝑮𝒌 + 𝟏. 𝟓𝑸𝒌
Design of Pad Footing
Check for Shear
• May fail in shear as vertical shear or punching shear
Vertical shear
sections
Punching shear
perimeters
2d
d
h
Bends may be
required
d
Design of Pad Footing
Check for Shear
(i)
Vertical Shear
• Critical section at distance d from the face of column
• Vertical shear force =  Load acting outside the section
• If VEd  VRd,c = No shear reinforcement is required
Design of Pad Footing
Check for Shear
(ii) Punching Shear
Axial Force Only
•
•
•
•
•
Critical section at a perimeter 2d from the face of the column
Punching shear force =  Load outside the critical perimeter
𝑽𝑬𝒅
Shear stress, 𝒗𝑬𝒅 =
where u = Critical perimeter
𝒖∙𝒅
If vEd  vRd,c = No shear reinforcement is required
Also ensure that VEd  VRd,max
Design of Pad Footing
Check for Shear
(ii) Punching Shear (continued)
Axial Force & Bending Moment
• Punching shear resistance can be significantly reduced of a coexisting bending, MEd
• However, adverse effect of the moment will give rise to a nonuniform shear distribution around the control perimeter
• Refer to Cl. 6.4.3(3) of EC2
Design of Pad Footing
Check for Shear
(ii) Punching Shear (continued)
Shear stress, 𝒗𝑬𝒅 =
𝜷𝑽𝑬𝒅
𝒖𝟏 ∙𝒅
where;
 = factor used to include effect of eccentric load & bending moment =
1+𝑘
𝑀𝐸𝑑
𝑉𝐸𝑑
𝑢1
𝑊1
k = coefficient depending on the ratio between column dimension c1 & c2
c1/c2
 0.5
1.0
2.0
 3.0
k
0.45
0.60
0.70
0.80
u1 = length of basic control perimeter
W1 = function of basic control perimeter corresponds to the distribution of
shear = 0.5𝑐1 2 + 𝑐1 𝑐2 + 4𝑐2 𝑑 + 16𝑑 2 + 2𝜋𝑑𝑐1
Design of Pad Footing
Check for Shear
(ii) Punching Shear (continued)
Design of Pad Footing
Cracking & Detailing Requirements
• All reinforcements should extend the full length of the footing
• If 𝐿𝑥 > 1.5 𝑐𝑥 + 3𝑑 , at least two-thirds of the reinforcement parallel
to Ly should be concentrated in a band width 𝑐𝑥 + 3𝑑 centred at
column where Lx & Ly and cx & cy are the footing and column dimension
in x and y directions
• Reinforcements should be anchored each side of all critical sections for
bending. Usually possible to achieve using straight bar
• Spacing between centre of reinforcements  20 mm for fyk = 500
N/mm2
• Reinforcements normally not provided in the side face nor in the top
face (except for balanced & combined foundation)
• Starter bar should terminate in a 90 bend tied to the bottom
reinforcement, or in the case of unreinforced footing spaced 75 mm
off the building
Example 1
PAD FOOTING
(AXIAL LOAD ONLY)
Example 1: Pad Footing (Axial Load)
Axial Force, N:
Gk = 600 kN
Qk = 400 kN
Column size:
300  300 mm
h
B
H
•
•
•
•
•
•
•
fck = 25 N/mm2
fyk = 500 N/mm2
soil = 150 N/mm2
Unit weight of concrete = 25 kN/m3
Design life = 50 years
Exposure Class = XC2
Assumed bar = 12 mm
Example 1: Pad Footing (Axial Load)
Durability & Bond Requirements
Min cover regards to bond, cmin,b = 12 mm
Min cover regards to durability, cmin,dur = 25 mm
Allowance in design for deviation, cdev = 10 mm
Nominal cover, cnom = cmin + cdev = 25 + 10 = 35 mm
 cnom = 35 mm
cmin = 25 mm
Example 1: Pad Footing (Axial Load)
Size
Service load, N
Assumed selfweight 10% of service load , W
Area of footing required =
𝑁+𝑊
𝛾𝑠𝑜𝑖𝑙
=
= 1000 kN
= 100 kN
1000+100
150
= 7.33 𝑚2
 Try footing size, B  H  h = 3 m  3 m  0.45 m
Area, A = 9 m2
Selfweight, W = 9 0.45  25 = 101 kN
Check Service Soil Bearing Capacity =
= 122 kN/m2  150 kN/m2
 OK
𝑵+𝑾
𝑨
=
𝟏𝟎𝟎𝟎+𝟏𝟎𝟏
𝟗
Example 1: Pad Footing (Axial Load)
Analysis
Ultimate axial force, NEd
= 1.35Gk + 1.5Qk
= 1.35 (600) + 1.5 (400) = 1410 kN
𝑁
1410
Soil pressure at ultimate load, P = 𝐸𝑑 =
= 157 kN/m2
𝐴
9
Soil pressure per m length, w = 157  3 m = 470 kN/m
0.3 m
1.35 m
w = 470 kN/m
1.35 m
Example 1: Pad Footing (Axial Load)
Analysis
Ultimate axial force, NEd
= 1.35Gk + 1.5Qk
= 1.35 (600) + 1.5 (400) = 1410 kN
𝑁
1410
Soil pressure at ultimate load, P = 𝐸𝑑 =
= 157 kN/m2
𝐴
9
Soil pressure per m length, w = 157  3 m = 470 kN/m
MEd
0.3 m
1.35 m
w = 470 kN/m
1.35 m
𝟏. 𝟑𝟓
𝑴𝑬𝒅 = 𝟒𝟕𝟎 × 𝟏. 𝟑𝟓 ×
𝟐
= 428 kNm
Example 1: Pad Footing (Axial Load)
Main Reinforcement
Effective depth, d = h – c – 1.5bar = 450 – 35 – (1.5  12) = 397 mm
𝐾=
𝑀𝐸𝑑
𝑓𝑐𝑘 𝑏𝑑 2
=
428×106
25×3000×3972
= 0.036  Kbal = 0.167
 Compression reinforcement is NOT required
𝑧 = 𝑑 0.25 −
𝐴𝑠,𝑟𝑒𝑞 =
𝐾
1.134
𝑀𝐸𝑑
0.87𝑓𝑦𝑘 𝑧
=
= 0.97𝑑  0.95d
428×106
0.87×500×0.95×397
= 𝟐𝟔𝟏𝟏 mm2
Example 1: Pad Footing (Axial Load)
Minimum & Maximum Area of Reinforcement
𝑓𝑐𝑡𝑚
2.56
𝑏𝑑 = 0.26
0.0013𝑏𝑑 ≥ 0.0013𝑏𝑑
𝑓𝑦𝑘
500
= 0.0013bd = 0.0013  3000  397 = 1589 mm2
𝐴𝑠,𝑚𝑖𝑛 = 0.26
 As,min
As,max = 0.04Ac = 0.04bh = 0.04  3000  397 = 54000 mm2
Provide 24H12 (As,prov = 2715 mm2)
Example 1: Pad Footing (Axial Load)
(i) Vertical Shear
3m
1.35 m
d = 397 mm
VEd
953 mm
3m
w = 470 kN/m
0.953 m
Critical shear at 1.0d from face of column:
 Design shear force, VEd = 470  0.953 = 448 kN
Example 1: Pad Footing (Axial Load)
(i) Vertical Shear
𝑘 =1+
200
𝑑
=1+
200
397
= 1.71  2.0
Note:
Bar extend beyond critical section at = 953 – 35 = 918 mm
 𝑙𝑏𝑑 + 𝑑 = 40∅ + 𝑑 = 40 × 12 + 397 = 877 mm
𝐴𝑠𝑙
2715
𝜌𝑙 =
=
= 0.0023 ≤ 0.02
𝑏𝑑 3000 × 397
 Asl = 2715 mm2
Example 1: Pad Footing (Axial Load)
(i) Vertical Shear
𝑉𝑅𝑑,𝑐 = 0.12𝑘 100𝜌𝑙 𝑓𝑐𝑘 1/3 𝑏𝑑
= 0.12 × 1.71 100 × 0.0023 × 25
1/3
3000 × 397 = 436463 N = 436 kN
𝑉𝑚𝑖𝑛 = 0.035𝑘 3/2 𝑓𝑐𝑘 𝑏𝑑
= 0.035 × 1.713/2 25 3000 × 397 = 465970 𝑁 = 466 kN
VEd (448 kN)  Vmin (466 kN)
 OK
Example 1: Pad Footing (Axial Load)
(ii) Punching Shear
Critical shear at 2.0d from face of column:
Average d = 450 – 35 – 12 = 403 mm
 2d = 806 mm
1350
2d = 806
Control perimeter, u = (4  300) + (2  806) =
6265 mm
300
2d = 806
300
Area within perimeter, A = (0.30  0.30) + (4 
0.30  0.806) + (  0.8062) = 3.10 m2
544
 𝑙𝑏𝑑 + 𝑑 = 40∅ + 𝑑 = 40 × 12 + 397 = 877
mm
 Reinforcement NOT contributed to punching
resistance
Example 1: Pad Footing (Axial Load)
(ii) Punching Shear
Punching shear force:
VEd = 157 (32 – 3.10) = 925 kN
Aall = 9 m2
Aperimeter = 3.10 m2
Punching shear resistance:
𝑉𝑅𝑑,𝑐 = 𝑉𝑚𝑖𝑛 = 0.035𝑘 3/2 𝑓𝑐𝑘 1/2 𝑢𝑑
= 0.035 1.71 3/2 25 1/2 6265 × 403
= 983199 N = 983 kN  VEd (925 kN)
 OK
Soil pressure = 157 kN/m2
Example 1: Pad Footing (Axial Load)
(iii) Maximum Punching Shear at Column Perimeter
Maximum punching shear force:
VEd,max = 157 (32 – 0.09) = 1400 kN
Aall = 9 m2
Acolumn = 0.09 m2
Maximum shear resistance:
𝑓𝑐𝑘
= 0.5𝑢𝑑 0.6 1 −
250
𝑓𝑐𝑘
𝑉𝑅𝑑,𝑚𝑎𝑥
1.5
25
= 0.5 4 × 300 × 403 0.6 1 −
250
= 2176 kN  VEd,max
 OK
25
1.5
Soil pressure = 157 kN/m2
Example 1: Pad Footing (Axial Load)
Cracking
h = 450 mm  200 mm
Max bar spacing
𝐴𝑠.𝑟𝑒𝑞
𝐺𝑘 +0.3𝑄𝑘
Steel stress, 𝑓𝑠 =
1.35𝐺𝑘 +1.5𝑄𝑘
𝐴𝑠,𝑝𝑟𝑜𝑣
600+0.3×400
2611
500
=
= 213
1.35×600+1.5×400
2715
1.15
𝑓𝑦𝑘
1.15
N/mm2
For design crack width 0.3 mm:
Maximum allowable bar spacing = 200 mm
Actual bar spacing =
3000−2 35 −12
23
= 127 mm  200 mm
Cracking OK
Example 1: Pad Footing (Axial Load)
Detailing
24H12
3000
24H12
3000
450
3000
24H12
Plan View
Section View
Example 2
PAD FOOTING
(AXIAL LOAD & MOMENT)
Example 2: Pad Footing (Axial Load &
Moment)
Axial Force, N = 1500 kN
Moment = 50 kNm
Column size:
250  350 mm
h
B
H
• Design Life = 50 years (Table 2.1: EN
1990)
• Exposure Class = XC3
• fck = 30 N/mm2
• fyk = 500 N/mm2
• soil = 150 N/mm2
• Unit weight of concrete = 25 kN/m3
• Assumed bar = 12 mm
Example 2: Pad Footing (Axial Load &
Moment)
Durability & Bond Requirements
Min cover regards to bond, cmin,b = 12 mm
Min cover regards to durability, cmin,dur = 25 mm
Allowance in design for deviation, cdev = 10 mm
Nominal cover, cnom = cmin + cdev = 25 + 10 = 35 mm
 cnom = 35 mm
cmin = 25 mm
Example 2: Pad Footing (Axial Load &
Moment)
Size
Service axial, N
Service moment, M
Assumed selfweight 10% of service load , W
Area of footing required =
𝑁+𝑊
𝛾𝑠𝑜𝑖𝑙
=
1071+107.1
150
= 1500 kN / 1.40 = 1071 kN
= 50 kNm / 1.40 = 36.1 kNm
= 100 kN
= 7.85 𝑚2
 Try footing size, B  H  h = 2.80 m  3.50 m  0.65 m
Area, A = 9.80 m2
Selfweight, W = 9.80 0.65  25 = 159 kN
Example 2: Pad Footing (Axial Load &
Moment)
Size (Continued)
𝐵𝐻 3
2.8×3.53
𝐼𝑥𝑥 =
=
=
12
12
𝐻
3.5
𝑦= =
= 1.75 m
2
2
10.0 m4
Maximum soil pressure, 𝑃 =
= 132 kN/m2  150 kN/m2
𝑁+𝑊
𝐴
𝑀𝑦
+
=
𝐼
 OK
1071+159
9.80
+
50×1.75
10.0
B
x
H
x
Example 2: Pad Footing (Axial Load &
Moment)
Analysis
Ultimate soil pressure, 𝑃 =
50×1.75
10.0
𝑁
𝐴
±
𝑀𝑦
𝐼
=
1500
9.80
x
±
= 153 ± 8.7 kN/m2
1.275
 Pmin = 144 kN/m2 and Pmax = 162 kN/m2
y
y
x
0.35 m
1.575 m
1.575 m
144
162
144
154
162
Example 2: Pad Footing (Axial Load &
Moment)
Analysis (Continued)
𝑀𝑥𝑥
x
1.5752
= 154 ×
2
1.275
1.575
2
× 1.575 ×
2
3
= 197 kNm/m  2.80 m = 553 kNm
+
162 − 154
𝑀𝑦𝑦 = 𝟏𝟓𝟑
1.2752
×
2
y
y
x
= 124 kNm/m  3.50 m =
0.35 m
1.575 m
435 kNm
=
144 + 162
2
144
1.575 m
154
162
Example 2: Pad Footing (Axial Load &
Moment)
Effective Depth
dx = h – c – 0.5bar = 650 – 35 – (0.5  12) = 609 mm
dy = h – c – 1.5bar = 650 – 35 – (1.5  12) = 597 mm
Main Reinforcement – Longitudinal Bar
𝐾=
𝑀𝑥𝑥
𝑓𝑐𝑘 𝑏𝑑 2
=
553×106
30×2800×6092
= 0.018  Kbal = 0.167
 Compression reinforcement is NOT required
𝑧 = 𝑑 0.25 −
𝐴𝑠,𝑟𝑒𝑞 =
𝐾
1.134
𝑀𝑥𝑥
0.87𝑓𝑦𝑘 𝑧
=
= 0.98𝑑  0.95d
553×106
0.87×500×0.95×609
= 𝟐𝟏𝟗𝟕 mm2
Example 2: Pad Footing (Axial Load &
Moment)
Minimum & Maximum Area of Reinforcement
𝑓𝑐𝑡𝑚
2.90
𝑏𝑑 = 0.26
0.0013𝑏𝑑 ≥ 0.0013𝑏𝑑
𝑓𝑦𝑘
500
= 0.0013bd = 0.0013  2800  609 = 2217 mm2
𝐴𝑠,𝑚𝑖𝑛 = 0.26
 As,min
As,max = 0.04Ac = 0.04bh = 0.04  2800  609 = 72800 mm2
Since As  As,min, Use As,min = 2217 mm2
Provide 21H12 (As = 2375 mm2)
Example 2: Pad Footing (Axial Load &
Moment)
Main Reinforcement – Transverse Bar
𝐾=
𝑀𝑦𝑦
𝑓𝑐𝑘 𝑏𝑑 2
=
435×106
30×3500×5972
= 0.018  Kbal = 0.167
 Compression reinforcement is NOT required
𝑧 = 𝑑 0.25 −
𝐴𝑠,𝑟𝑒𝑞 =
𝐾
1.134
𝑀𝑦𝑦
0.87𝑓𝑦𝑘 𝑧
=
= 0.99𝑑  0.95d
435×106
0.87×500×0.95×597
= 𝟏𝟕𝟔𝟓mm2
Example 2: Pad Footing (Axial Load &
Moment)
Minimum & Maximum Area of Reinforcement
𝑓𝑐𝑡𝑚
2.90
𝑏𝑑 = 0.26
0.0013𝑏𝑑 ≥ 0.0013𝑏𝑑
𝑓𝑦𝑘
500
= 0.0013bd = 0.0013  3500  597 = 3147 mm2
𝐴𝑠,𝑚𝑖𝑛 = 0.26
 As,min
As,max = 0.04Ac = 0.04bh = 0.04  3500  597 = 91000 mm2
Since As  As,min, Use As,min = 3147 mm2
Provide 28H12 (As = 3167 mm2)
Example 2: Pad Footing (Axial Load &
Moment)
(i) Vertical Shear
2.891
Critical shear at 1.0d from face of column:
2.8
Average pressure at critical section:
2.891
3.50
× 18 = 159 kN/m2
d = 0.609
= 144 +
 Design shear force, VEd = 159  0.966 
2.80 = 431 kN
0.966
144
159
Note:
Bar extend beyond critical section at = 966 – 35 = 931 mm
 𝑙𝑏𝑑 + 𝑑 = 36∅ + 𝑑 = 36 × 12 + 609 = 1041 mm
162
 Asl = 0 mm2
Example 2: Pad Footing (Axial Load &
Moment)
(i) Vertical Shear
𝑘 = 1+
𝜌𝑙 =
200
𝑑
=1+
200
609
= 1.57  2.0
𝐴𝑠𝑙
=0
𝑏𝑑
 𝑉𝑅𝑑,𝑐 = 0.12𝑘 100𝜌𝑙 𝑓𝑐𝑘 1/3 𝑏𝑑
= 0.12 × 1.57 100 × 0 × 30 1/3 2800 × 609 = 0 N = 0 kN
 𝑉𝑚𝑖𝑛 = 0.035𝑘 3/2 𝑓𝑐𝑘 𝑏𝑑
= 0.035 × 1.573/2 30 2800 × 609 = 644949 𝑁 = 645 kN
VEd (430 kN)  Vmin (645 kN)
 OK
Example 2: Pad Footing (Axial Load &
Moment)
(ii) Punching Shear
Critical shear at 2.0d from face of column:
609+597
= 603 mm
3500
Control perimeter;
u = 2(350 + 250) + (2  1206) = 8779 mm
2d = 1206
350
2d = 1206
250
Area within perimeter;
A = (0.35  0.25) + (2  0.35  1.206)
+ (2  0.25  1.206) + (  1.2062) = 6.10 m2
2800
Average 𝑑 =
2
 2d = 1206 mm
69
369
 𝑙𝑏𝑑 + 𝑑 = 36∅ + 𝑑 = 36 × 12 + 609 = 1041 mm
 Reinforcement NOT contributed to punching resistance
Example 2: Pad Footing (Axial Load &
Moment)
(ii) Punching Shear
Average punching shear force at control perimeter:
VEd = 153 [(2.80  3.50) – 6.10] = 566 kN
Punching shear stress:
𝛽𝑉𝐸𝑑
𝑣𝐸𝑑 =
𝑢𝑑
𝑀
Where 𝛽 = 1 + 𝑘 𝑉 𝐸𝑑
k = 0.65 
𝐸𝑑
𝑐1
𝑐2
Aall = 2.8  3.5 m
Aperimeter = 6.10 m2
𝑢1
𝑊1
350
= 250
= 1.4
W1 = 0.5𝑐1 2 + 𝑐1 𝑐2 + 4𝑐2 𝑑 + 16𝑑 2 + 2𝜋𝑑𝑐1
= 0.5(3502 ) + 350 × 250 + 4 × 250 × 603 +
16 6032 + 2𝜋 × 603 × 350 = 7.9  106 mm2
 𝛽 = 1 + 0.65
50×106
566×103
Therefore, 𝒗𝑬𝒅 =
𝟏.𝟎𝟔×𝟓𝟔𝟔×𝟏𝟎𝟑
𝟖𝟕𝟕𝟗×𝟔𝟎𝟗
8779
7.9×106
= 1.06
= 𝟎. 𝟏𝟏 N/mm2
Soil pressure = 153 kN/m2
Example 2: Pad Footing (Axial Load &
Moment)
(ii) Punching Shear
Punching shear resistance:
Aall = 2.8  3.5 m
𝑘 = 1+
200
𝑑
=1+
200
609
= 1.57  2.0
𝑣𝑅𝑑,𝑐 = 𝑣𝑚𝑖𝑛 = 0.035𝑘 3/2 𝑓𝑐𝑘 1/2
= 0.035 1.57 3/2 30 1/2
= 0.38 N/mm2  vEd (0.11 N/mm2)
Aperimeter = 6.10 m2
 OK
Soil pressure = 153 kN/m2
Example 2: Pad Footing (Axial Load &
Moment)
(iii) Maximum Punching Shear at Column Perimeter
Maximum punching shear force:
VEd,max = 1500 kN
Column perimeter, uo = 2(350 + 250) = 1200 mm
Punching shear stress:
𝛽𝑉𝐸𝑑
𝑣𝐸𝑑 =
𝑢𝑜 𝑑
𝑀
Where 𝛽 = 1 + 𝑘 𝐸𝑑
k = 0.65 
𝑉𝐸𝑑
𝑐1
=
𝑐2
𝑢𝑜
𝑊1
350
250
Aall = 2.8  3.5 m = 9.8 m2
Acolumn = 0.09 m2
Soil pressure = 153 kN/m2
= 1.4
W1 = 0.5𝑐1 2 + 𝑐1 𝑐2
= 0.5(3502 ) + 350 × 250 = 0.15  106 mm2
Example 2: Pad Footing (Axial Load &
Moment)
(iii) Maximum Punching Shear at Column Perimeter
 𝛽 = 1 + 0.65
50×106
1500×103
Therefore, 𝒗𝑬𝒅 =
𝟏.𝟏𝟕×𝟏𝟓𝟎𝟎×𝟏𝟎𝟑
𝟏𝟐𝟎𝟎×𝟔𝟎𝟑
1200
0.15×106
Maximum shear resistance:
𝑓𝑐𝑘
𝑣𝑅𝑑,𝑚𝑎𝑥 = 0.5 0.6 1 −
250
30
30
= 0.5 0.6 1 −
250
1.5
= 5.28 N/mm2  vEd
= 1.17
= 𝟐. 𝟒𝟒
N/mm2
Aall = 2.8  3.5 m = 9.8 m2
Acolumn = 0.09 m2
𝑓𝑐𝑘
1.5
Soil pressure = 153 kN/m2
 OK
Example 2: Pad Footing (Axial Load &
Moment)
Cracking
h = 650 mm  200 mm
Max bar spacing
Assume steel stress is under quasi-permanent loading:
= 0.6
𝑓𝑦𝑘
𝐴𝑠,𝑟𝑒𝑞
1.15
𝐴𝑠,𝑝𝑟𝑜𝑣
= 0.6
500
1.15
2197
2375
= 241 N/mm2
For design crack width 0.3 mm:
Maximum allowable bar spacing = 150 mm
2800−2 35 −12
= 136 mm  150 mm
20
3500−2 35 −12
= 126 mm  150 mm
27
Actual bar spacing at x-x =
ctual bar spacing at y-y =
Cracking OK
Example 2: Pad Footing (Axial Load &
Moment)
Detailing
21H12
2800
28H12
3500
650
3500
21H12
28H12
Plan View
Section View
Example 3
DESIGN OF COMBINED
FOOTING
Example 3: Design of Combined Footing
NA = 1610 kN
(Ultimate)
NB = 1950 kN
(Ultimate)
3.4 m
Column size:
300  300 mm
Column size:
400  400 mm
h
B
H
•
•
•
•
•
•
fck = 35 N/mm2
fyk = 500 N/mm2
soil = 200 N/mm2
Unit weight of concrete = 25 kN/m3
Cover = 40 mm
Assumed bar = 12 mm
Example 3: Design of Combined Footing
Size
Service axial:
NA
NB
Total service axial, Ntotal
= 1610 kN / 1.40 = 1150 kN
= 1950 kN / 1.40 = 1393 kN
= 1150 + 1393 = 2543 kN
Assumed selfweight 10% of service load , W = 254.3 kN
Area of footing required =
𝑁+𝑊
𝛾𝑠𝑜𝑖𝑙
=
2543+254.3
200
= 14.0 𝑚2
 Try footing size, B  H  h = 2.70 m  6.00 m  0.65 m
Area, A = 16.2 m2
Selfweight, W = 16.2 0.65  25 = 252.7 kN
Example 3: Design of Combined Footing
Size (Continued)
𝑁+𝑊
𝐴
Check Service Soil Bearing Capacity =
= 173 kN/m2  200 kN/m2
 OK
=
2543+252.7
16.2
Arrange position of footing so that the distribution of soil pressure is uniform:
Ntotal = 2543 kN
NA = 1150 kN
NB = 1393 kN
x
3.4 m
MA = 0
1393(3.4 ) + 2543x = 0
x = 1.86 m
Example 3: Design of Combined Footing
Analysis
𝑁
(1610+1950)
Soil pressure at ultimate load, P = 𝐸𝑑 =
= 219.8 kN/m2
𝐴
16.2
Soil pressure per m width, w
= 219.8  2.7 m = 593.5 kN/m
Example 3: Design of Combined Footing
Shear Force & Bending Moment Diagram
3m
3m
1610 kN
0.3
0.99
1950 kN
1.34
1.71
1.26
0.4
0.65
x = 1.86
593.5 kN/m
1082
964
587
676
1.62
1.43
SFD (kN)
868
934
749
846
353
289
250
384
BMD (kNm)
473
430
634
Example 3: Design of Combined Footing
Main Reinforcement
Longitudinal Reinforcement: Bottom
Effective depth: dx = h – c – 0.5bar = 650 – 40 – (0.5  12) = 604 mm
𝐾=
𝑀𝐸𝑑
𝑓𝑐𝑘 𝑏𝑑 2
=
473×106
35×2700×6042
= 0.014  Kbal = 0.167
 Compression reinforcement is NOT required
𝑧 = 𝑑 0.25 −
𝐴𝑠,𝑟𝑒𝑞 =
𝐾
1.134
𝑀𝐸𝑑
0.87𝑓𝑦𝑘 𝑧
=
= 0.99𝑑  0.95d
473×106
0.87×500×0.95×604
= 𝟏𝟖𝟗𝟒 mm2  As,min = 2722 mm2
Example 3: Design of Combined Footing
Main Reinforcement
Longitudinal Reinforcement: Top
𝐾=
𝑀𝐸𝑑
𝑓𝑐𝑘 𝑏𝑑 2
=
353×106
35×2700×6042
= 0.010  Kbal = 0.167
 Compression reinforcement is NOT required
𝑧 = 𝑑 0.25 −
𝐴𝑠,𝑟𝑒𝑞 =
𝐾
1.134
𝑀𝐸𝑑
0.87𝑓𝑦𝑘 𝑧
=
= 0.99𝑑  0.95d
353×106
0.87×500×0.95×604
= 𝟏𝟒𝟏𝟑 mm2  As,min = 2722 mm2
Example 3: Design of Combined Footing
Minimum & Maximum Area of Reinforcement
𝑓𝑐𝑡𝑚
3.2
𝐴𝑠,𝑚𝑖𝑛 = 0.26
𝑏𝑑 = 0.26
0.0017𝑏𝑑 ≥ 0.0013𝑏𝑑
𝑓𝑦𝑘
500
 As,min = 0.0017bd = 0.0017  2700  604 = 2722 mm2
As,max = 0.04Ac = 0.04bh = 0.04  2700  650 = 70200 mm2
Provide 25H12 at both top and bottom (As,prov = 2828 mm2)
Example 3: Design of Combined Footing
Main Reinforcement
Transverse Reinforcement: Bottom
Consider b = 1000 mm:
 Soil pressure per 1 m width, w = 219.8  1.0 m = 219.8 kN/m
MEd
0.3 m
1.2 m
w = 219.8 kN/m
1.2 m
𝟏. 𝟐
𝑴𝑬𝒅 = 𝟐𝟏𝟗. 𝟖 × 𝟏. 𝟐 ×
𝟐
= 158 kNm/m
Example 3: Design of Combined Footing
Main Reinforcement
Transverse Reinforcement: Bottom
Effective depth: dy = h – c – 1.5bar = 650 – 40 – (1.5  12) = 592 mm
𝐾=
𝑀𝐸𝑑
𝑓𝑐𝑘 𝑏𝑑 2
=
158×106
35×𝟏𝟎𝟎𝟎×5922
= 0.013  Kbal = 0.167
 Compression reinforcement is NOT required
𝑧 = 𝑑 0.25 −
𝐴𝑠,𝑟𝑒𝑞 =
𝐾
1.134
𝑀𝐸𝑑
0.87𝑓𝑦𝑘 𝑧
=
= 0.99𝑑  0.95d
158×106
0.87×500×0.95×592
= 𝟔𝟒𝟕 mm2/m  As,min = 988 mm2/m
Example 3: Design of Combined Footing
Minimum & Maximum Area of Reinforcement
𝑓𝑐𝑡𝑚
3.2
𝐴𝑠,𝑚𝑖𝑛 = 0.26
𝑏𝑑 = 0.26
0.0017𝑏𝑑 ≥ 0.0013𝑏𝑑
𝑓𝑦𝑘
500
 As,min = 0.0017bd = 0.0017  1000  592 = 988 mm2/m
As,max = 0.04Ac = 0.04bh = 0.04  1000  650 = 26000 mm2/m
Since As  As,min, then use As,min = 988 mm2/m
Provide H12-100 (As,prov = 1131 mm2/m)
For secondary bar reinforcement:
Provide H12-100 (As,prov = 1131 mm2/m)
Example 3: Design of Combined Footing
(i) Vertical Shear
Critical shear at 1.0d from face of column:
VEd
964
𝑉𝐸𝑑 =
1.02
1.62
× 964 = 𝟔𝟎𝟓 kN
587
1082
676
1.43
1.02
𝑘 = 1+
𝜌𝑙 =
200
𝑑
=1+
200
604
= 1.6  2.0
934
846
d = 0.604
868
749
1.62
𝐴𝑠𝑙
2828
=
= 0.0017 ≤ 0.02
𝑏𝑑 2700 × 604
𝑉𝑅𝑑,𝑐 = 0.12𝑘 100𝜌𝑙 𝑓𝑐𝑘 1/3 𝑏𝑑
= 0.12 × 1.6 100 × 0.0017 × 35 1/3 2700 × 604 = 562369 N = 562.4 kN
 𝑉𝑚𝑖𝑛 = 0.035𝑘 3/2 𝑓𝑐𝑘 𝑏𝑑
= 0.035 × 1.63/2 35 2700 × 604 = 6667734 𝑁 = 667 kN
VEd (605 kN)  Vmin (667 kN)
 OK
Example 3: Design of Combined Footing
(ii) Punching Shear
Critical shear at 2.0d from face of column:
604+592
= 598 mm
2d = 1196
400
2d = 1196
400
Control perimeter;
u = 2(400 + 400) + (2  1196) = 9116 mm
Area within perimeter;
A = (0.40  0.40) + (2  0.40  1.196)
+ (2  0.40  1.196) + (  1.1962) = 6.57 m2
104
2700
Average 𝑑 =
2
 2d = 1196 mm
Example 3: Design of Combined Footing
(ii) Punching Shear
Average punching shear force at control perimeter:
VEd = 1950 – (219.8  6.57) = 506 kN
Punching shear stress:
𝑣𝐸𝑑 =
𝑉𝐸𝑑
𝑢𝑑
=
506×103
9116×598
= 0.09 N/mm2
Aperimeter = 6.57 m2
Punching shear resistance:
𝑘 =1+
200
𝑑
=1+
200
609
3/2
= 1.57  2.0
1/2
𝑣𝑅𝑑,𝑐 = 𝑣𝑚𝑖𝑛 = 0.035𝑘 𝑓𝑐𝑘
= 0.035 1.6 3/2 35 1/2
= 0.41 N/mm2  vEd (0.09 N/mm2)  OK
Soil pressure = 219.8 kN/m2
Example 3: Design of Combined Footing
(iii) Maximum Punching Shear at Column Perimeter
Maximum punching shear force:
VEd,max = 1950 kN
Column perimeter, uo = 4  400 = 1600 mm
Acolumn = 0.16 m2
Maximum shear resistance:
𝑉𝑅𝑑,𝑚𝑎𝑥
𝑓𝑐𝑘
= 0.5𝑢𝑑 0.6 1 −
250
35
250
 OK
= 0.5 × 1600 × 598 0.6 1 −
= 2880 kN  VEd,max
𝑓𝑐𝑘
1.5
35
1.5
Soil pressure = 219.8 kN/m2
Example 3: Design of Combined Footing
Cracking
h = 650 mm  200 mm
Max bar spacing
Assume steel stress is under quasi-permanent loading:
= 0.6
𝑓𝑦𝑘
𝐴𝑠,𝑟𝑒𝑞
1.15
𝐴𝑠,𝑝𝑟𝑜𝑣
= 0.6
500
1.15
1894
2828
= 175 N/mm2
For design crack width 0.3 mm:
Maximum allowable bar spacing = 250 mm
Actual bar spacing 1 =
2700−2 40 −12
24
2700−2 40 −12
24
= 109 mm  250 mm
Actual bar spacing 2 =
= 109 mm  250 mm
Actual bar spacing 3 = 100 mm  250 mm
Cracking OK
Example 4
DESIGN OF STRAP FOOTING
Example 4: Design of Strap Footing
NB
Gk = 1000 kN, Qk = 525 kN
NA
Gk = 800 kN, Qk = 275 kN
6000
•
•
•
•
•
•
Beam size:
300  900 mm
Column size:
300  300 mm
Column size:
300  300 mm
700
RA
RB
H
S
2000
S
fck = 30 N/mm2
fyk = 500 N/mm2
soil = 200 N/mm2
Unit weight of concrete = 25
kN/m3
Cover = 40 mm
Assumed bar = 12 mm
Example 4: Design of Strap Footing
Loading
Column
Size (mm × mm)
Load (kN)
Gk
Qk
Total (1.0Gk + 1.0Qk)
Column A
300 × 300
800
275
1075
Column B
300 × 300
1000
525
1525
Example 4: Design of Strap Footing
Size
Assumed self weight of footing is 10% of service load:
 WA = 108 kN and WB = 153 kN
Beam self weight, WR = 25 (0.3  0.9  5.7) = 38 kN
Example 4: Design of Strap Footing
𝑀@𝐵 = 0
Size
𝑅𝐴 − 𝑊𝐴 6.15 − 1.0 − 𝑁𝐴 6.0 − 𝑊𝑅 3.0 = 0
𝑅𝐴 − 108 6.15 − 1.0 − 1075 6.0 − 38 3.0
=0
𝑅𝐴 − 108 = 1275
 RA = 1382 kN
NB = 1525 kN
NA = 1075 kN
6m
𝑅𝐴
= 200
2.0𝐻
1382
= 200
2.0𝐻
WR
0.3 m
WA
RA
Footing A size: 2.00 × 3.50 m
0.3 m
WB
0.7 m
𝑅𝐴 + 𝑅𝐵 = 𝑁𝐴 + 𝑁𝐵 + 𝑊𝐴 + 𝑊𝐵 + 𝑊𝑅
𝑅𝐴 + 𝑅𝐵 = 1075 + 1525 + 108 + 153 + 38
𝑅𝐵 = 2898 − 𝑅𝐴
𝑅𝐵 = 2898 − 1382
 RA = 1516 kN
RB
H
S
2m
S
 H = 3.46 mm
𝑅𝐵
= 200
𝑆2
1516
= 200
𝑆2
 S = 2.75 mm
Footing B size: 2.75 × 2.75 m
Example 4: Design of Strap Footing
Footing Size Checks
Total service load = 1075 + 1525 + 123 + 132 + 38 = 2893 kN
Area of footing = (2.0  3.5) + (2.75  2.75) = 14.56 m2
Soil pressure =
2893
14.56
= 199 kN/m2  200 kN/m2
 OK
Example 4: Design of Strap Footing
NA = 1075 kN
NB = 1525 kN
6m
R = 2893 kN
Footing Self Weight
X’
0.3 m
0.3 m
WA = 123 kN
WA = 25 (2.0  3.5  0.7) = 123 kN
WB = 25 (2.75  2.75  0.7) = 132 kN
WR = 38 kN
2m
Distance from resultant force R to centroid of column B:
1075 × 6.0 + 123 × 5.15 + 38 × 3.0 = 2893𝑋 ′
X’ = 2.48 m
Distance from centroid of footings to centroid of column B:
2.0 × 3.50 5.15 = 2.0 × 3.5 + 2.75 × 2.75 𝑋′
X’ = 2.48 m
WB = 132 kN
2.75 m
Example 4: Design of Strap Footing
Analysis
Design Load:
NA = (1.35  800) + (1.50  275)
NB = (1.35  1000) + (1.50  525)
WA = 1.35  123
WB = 1.35  132
WR = 1.35  38
= 1493 kN
= 2138 kN
= 165 kN
= 179 kN
= 52 kN
Total Design Load, Ptotal
= 4026 kN
Ultimate soil pressure =
𝑃𝑡𝑜𝑡𝑎𝑙
𝐴𝑟𝑒𝑎 𝑜𝑓 𝐹𝑜𝑜𝑡𝑖𝑛𝑔
=
4026
14.56
= 276 kN/m2
Example 4: Design of Strap Footing
Shear Force & Bending Moment Diagram
1493 kN
2138 kN
2.775
1.7
1.225
0.3
0.3
52
5.7
276.5 −
1.225
165
2.0×3.5
= 9.1 kN/m
= 253 × 3.5 = 885 kN/m
1.40
276.5 −
179
2.75×2.75
= 253 × 2.75 = 695 kN/m
1077
262
SFD (kN)
237
-852
-1227
-1044
.
-1004
BMD (kNm)
-313
492
522
Example 4: Design of Strap Footing
DESIGN OF FOOTING A
Main Reinforcement
Soil pressure, w = 253  2.0 m = 506 kN/m
MEd
0.3 m
1.6 m
1.6 m
𝟏. 𝟔
𝑴𝑬𝒅 = 𝟓𝟎𝟔 × 𝟏. 𝟔 ×
𝟐
= 647 kNm
w = 506 kN/m
Effective depth: d = h – c – 0.5bar = 700 – 40 – (0.5  16) = 652 mm
Example 4: Design of Strap Footing
𝐾=
𝑀𝐸𝑑
𝑓𝑐𝑘 𝑏𝑑2
=
647×106
30×2000×6522
= 0.025  Kbal = 0.167
 Compression reinforcement is NOT required
𝑧 = 𝑑 0.25 −
𝐴𝑠,𝑟𝑒𝑞 =
𝐾
1.134
𝑀𝐸𝑑
0.87𝑓𝑦𝑘 𝑧
=
= 0.98𝑑  0.95d
647×106
0.87×500×0.95×652
= 𝟐𝟒𝟎𝟐 mm2/m  As,min
Minimum & Maximum Area of Reinforcement
𝑓𝑐𝑡𝑚
2.90
𝐴𝑠,𝑚𝑖𝑛 = 0.26
𝑏𝑑 = 0.26
0.0015𝑏𝑑 ≥ 0.0013𝑏𝑑
𝑓𝑦𝑘
500
 As,min = 0.0015bd = 0.0015  2000  652 = 1964 mm2 or 982 mm2/m
As,max = 0.04Ac = 0.04bh = 0.04  2000  700 = 56000 mm2
Main Reinforcement: Provide 14H16 (As,prov = 2815 mm2)
Secondary Reinforcement: H16-200 (As,prov = 1005 mm2/m)
Example 4: Design of Strap Footing
(i) Vertical Shear
3.5
2.552
Critical shear at 1.0d from face of column:
d = 0.652
2.0
0.948
 Design shear force, VEd = 253  0.948  2.0 = 479.4 kN
Note:
Bar extend beyond critical section at = 948 – 40 = 908 mm
 𝑙𝑏𝑑 + 𝑑 = 36∅ + 𝑑 = 36 × 16 + 652 = 1228 mm
 Asl = 0 mm2
Example 4: Design of Strap Footing
(i) Vertical Shear
𝑘 = 1+
𝜌𝑙 =
200
𝑑
=1+
200
652
= 1.55  2.0
𝐴𝑠𝑙
=0
𝑏𝑑
 𝑉𝑅𝑑,𝑐 = 0.12𝑘 100𝜌𝑙 𝑓𝑐𝑘 1/3 𝑏𝑑
= 0.12 × 1.55 100 × 0 × 30 1/3 3500 × 652 = 0 N = 0 kN
 𝑉𝑚𝑖𝑛 = 0.035𝑘 3/2 𝑓𝑐𝑘 𝑏𝑑
= 0.035 × 1.553/2 30 3500 × 652 = 484194 𝑁 = 484.2 kN
VEd (479.4 kN)  Vmin (484.2 kN)
 OK
Example 4: Design of Strap Footing
Cracking
h = 700 mm  200 mm
Max bar spacing
Assume steel stress is under quasi-permanent loading:
= 0.59
𝑓𝑦𝑘
𝐴𝑠,𝑟𝑒𝑞
1.15
𝐴𝑠,𝑝𝑟𝑜𝑣
= 0.59
500
1.15
2402
2815
= 219 N/mm2
For design crack width 0.3 mm:
Maximum allowable bar spacing = 200 mm
Actual bar spacing =
2000−2 40 −16
13
= 146 mm  200 mm
Cracking OK
Example 4: Design of Strap Footing
DESIGN OF FOOTING B
Main Reinforcement
Soil pressure, w = 253  2.75 m = 695.75 kN/m
MEd
0.3 m
1.225 m
1.225 m
𝟏. 𝟐𝟐𝟓
𝑴𝑬𝒅 = 𝟔𝟗𝟓. 𝟕𝟓 × 𝟏. 𝟐𝟐𝟓 ×
𝟐
= 522 kNm
w = 695.75 kN/m
Effective depth: d = h – c – 1.5bar = 700 – 40 – (1.5  16) = 636 mm
Example 4: Design of Strap Footing
𝐾=
𝑀𝐸𝑑
𝑓𝑐𝑘 𝑏𝑑2
522×106
30×2750×6362
=
= 0.016  Kbal = 0.167
 Compression reinforcement is NOT required
𝑧 = 𝑑 0.25 −
𝐴𝑠,𝑟𝑒𝑞 =
𝐾
1.134
𝑀𝐸𝑑
0.87𝑓𝑦𝑘 𝑧
=
= 0.99𝑑  0.95d
522×106
0.87×500×0.95×636
= 𝟏𝟗𝟖𝟓 mm2/m  As,min = 2623 mm2
Minimum & Maximum Area of Reinforcement
𝑓𝑐𝑡𝑚
2.90
𝐴𝑠,𝑚𝑖𝑛 = 0.26
𝑏𝑑 = 0.26
0.0015𝑏𝑑 ≥ 0.0013𝑏𝑑
𝑓𝑦𝑘
500
 As,min = 0.0015bd = 0.0015  2750  636 = 2623 mm2
As,max = 0.04Ac = 0.04bh = 0.04  2750  700 = 77000 mm2
Main Reinforcement: Provide 15H16 (As,prov = 3016 mm2)
Example 4: Design of Strap Footing
(i) Vertical Shear
2.75
d = 0.636
2.75
2.161
Critical shear at 1.0d from face of column:
0.589
 Design shear force, VEd = 253  0.589  2.75 = 409.5 kN
Note:
Bar extend beyond critical section at = 589 – 40 = 549 mm
 𝑙𝑏𝑑 + 𝑑 = 36∅ + 𝑑 = 36 × 16 + 636 = 1212 mm
 Asl = 0 mm2
Example 4: Design of Strap Footing
(i) Vertical Shear
𝑘 = 1+
𝜌𝑙 =
200
𝑑
=1+
200
636
= 1.56  2.0
𝐴𝑠𝑙
=0
𝑏𝑑
 𝑉𝑅𝑑,𝑐 = 0.12𝑘 100𝜌𝑙 𝑓𝑐𝑘 1/3 𝑏𝑑
= 0.12 × 1.56 100 × 0 × 30 1/3 2750 × 636 = 0 N = 0 kN
 𝑉𝑚𝑖𝑛 = 0.035𝑘 3/2 𝑓𝑐𝑘 𝑏𝑑
= 0.035 × 1.563/2 30 2750 × 636 = 653774 𝑁 = 653.8 kN
VEd (409.5 kN)  Vmin (653.8 kN)
 OK
Example 4: Design of Strap Footing
Cracking
h = 700 mm  200 mm
Max bar spacing
Assume steel stress is under quasi-permanent loading:
= 0.54
𝑓𝑦𝑘
𝐴𝑠,𝑟𝑒𝑞
1.15
𝐴𝑠,𝑝𝑟𝑜𝑣
= 0.54
500
1.15
2623
3016
= 204 N/mm2
For design crack width 0.3 mm:
Maximum allowable bar spacing = 200 mm
Actual bar spacing =
2750−2 40 −16
14
= 189 mm  200 mm
Cracking OK
Example 4: Design of Strap Footing
DESIGN OF TIE BEAM
Design similar to beam design. Refer to RC1.
Moment Design
for Flexural
Reinforcement
5H32
Shear Design for
Shear Links
H6-150
Cracking Checks
DESIGN OF PILE
FOUNDATIONS
Design of Pile Foundation
• To be used when the soil conditions are poor and
uneconomical, or not possible to provide adequate
spread foundations
• The piles must extend down to firm soil
• Load carried by either end bearing, friction or
combination of both
Design of Pile Foundation
Load from
Structure
Pile Cap
Lower Density
Medium Density
High Density
P
I
L
E
P
I
L
E
P
I
L
E
P
I
L
E
Design of Pile Foundation
Selection of Piles Type
• Depends on loading, type of structure, soil strata & site
conditions
• Main types:
a) Precast reinforced or pre-stressed concrete piles
b) Cast in-situ reinforced concrete piles
c) Timber piles
d) Steel piles
e) Bakau piles
Design of Pile Foundation
Determination of Pile Capacity
• Safe load:
a) Determined from test loading of a pile or using a
pile formula
b) Ultimate load divided by safety factor between 2 &
3
c) Depends on size & depth, & whether the pile is of
the end bearing or friction type
• Pile formula – gives resistance from the energy of the
driving force & the final set or penetration of the pile
per blow
Design of Pile Foundation
Determination of Piles Number & Spacing
• Pile load  Single pile capacity
• Piles are usually arranged symmetrically with respect to
the column axis
Design of Pile Foundation
Determination of Piles Number & Spacing
Foundation Subjected to Axial Load Only
Load applied at centroid of the group of piles:
𝑵+𝑾
𝑭𝒂 =
𝒏
N = Axial load from column
W = Self weight of pile cap
n = Number of piles
Design of Pile Foundation
Determination of Piles Number & Spacing
Foundation Subjected to Axial Load & Moment
•
•
Pile cap assumed to rotate about the centroid of the pile group
Piles load resisting moment vary uniformly from zero at the
centroidal axis to a maximum for the piles farthest away
𝑭𝒂𝒊
𝑵+𝑾
𝑴𝒙𝒊
=
±
𝒏
𝑰𝒚
Fai = Load per pile i
M = Moment
xi = Distance from pile i to centroid of pile cap
Iy = Moment of inertial of pile group = 𝑥1 2 + 𝑥2 2 + ⋯ + 𝑥𝑛 2
Design of Pile Cap
Design of Pile Cap
pile cap
Design of Pile Cap
Size & Thickness
• Depends of number of piles used, arrangement of
piles & shape of piles
• Thickness of the cap should be sufficient to provide
adequate bond length for bars projecting from the
piles as well for the dowel bars of the column
Design of Pile Cap
Main Reinforcement
• Using bending theory or truss theory
• Truss Theory: Force from the supported column is
assumed to be transmitted by a triangular truss action
– concrete providing the compressive members of the
truss & steel reinforcement providing tensile force
Design of Pile Cap
Truss Theory – Axial Force
Design of Pile Cap
Truss Theory – Axial Force
Design of Pile Cap
Pile Cap Size
Design of Pile Cap
Pile Cap Size
Design of Pile Cap
Design for Shear
• Critical section taken to be 20%  pile diameter (or
p/5) inside the face of the column
• Shear enhancement may be considered such that the
shear resistance of the concrete maybe increased to
𝑽𝑹𝒅,𝒄 ×
𝟐𝒅
𝒂𝒗
where av = distance from the face of column
to the critical section
• For spacing of piles  3  pile diameter, this
enhancement may be applied across the whole critical
section
Design of Pile Cap
Design for Shear
• For spacing  3  pile diameter, then the pile cap
should be checked for punching shear on the
perimeter
• Check shear force at column face  0.5𝑣1 𝑓𝑐𝑑 𝑢𝑑 =
𝑓𝑐𝑘
0.5𝑣1
1.5
𝑢𝑑 where u is the perimeter of the column
and strength reduction factor, 𝑣1 = 0.6 1 −
𝑓𝑐𝑘
250
Design of Pile Cap
Design for Shear
/5
Vertical shear @ 1.0d from
column face
/5
Punching shear @ 2.0d from
column face
Design of Pile Cap
Detailing of Reinforcement
• Main tension reinforcement should continue pass each
pile and bent-up vertically to provide full anchorage
length beyond the centre line of each pile
• Normal to provide fully lapped horizontal link of size 
12 mm @ spacing  250 mm
Design of Pile Cap
Detailing of Reinforcement
Starter bars for columns
Links for starter bars
Bars projecting into cap
from piles
Vertical links between bars
extending from piles
Horizontal links around upstanding end of main bars (
12 mm @ 250 mm, say)
Main bars design to
resist tensile forces
Example 5
DESIGN OF PILE CAP – TRUSS
THEORY
Example 5: Design of Pile Cap – Truss
Theory
N
• Axial Load, N:
o Gk = 3500 kN & Qk = 2500 kN
• fck = 30 N/mm2
• fyk = 500 N/mm2
• soil = 200 N/mm2
• Unit weight of concrete = 25 kN/m3
• Column size = 500 mm  500 mm
• Cover = 75 mm
• Assumed bar = 25 mm
• Pile:
o Pre-stressed spun pile: 500 mm dia.
o Working load = 1800 kN
Example 5: Design of Pile Cap – Truss
Theory
Size of Pile Cap
Service load = 3500 + 2500 = 6000 kN
Assumed self weight of pile cap, say W = 200 kN
Pile capacity, Fa = 1800 kN
No. of pile required, 𝑛 =
𝑁+𝑊
𝐹𝑎
=
(6000 + 200)
1800
= 3.4  Use n = 4
Pile spacing, 𝑙 = 𝑘∅𝑝 = 3∅𝑝 = 3 × 500 = 1500 mm
Width, B = 𝑘 + 1 ∅𝑝 + 300 = 3 + 1 × 500 + 300 = 2300 mm
Length, H = 𝑘 + 1 ∅𝑝 + 300 = 3 + 1 × 500 + 300 = 2300 mm
Depth, h = 2∅𝑝 + 100 = 2 500 + 100 = 1100 mm
Example 5: Design of Pile Cap – Truss
Theory
Size of Pile Cap (continued)
Try size: B  H  h = 2.3  2.3  1.1 m
Self weight = 25(2.3  2.3  1.1) = 145 kN  200 kN
OK
av = 750 – 250 – 250 + 500/5 = 350
400
p/5 = 100
400
d
400
1500
400
h = 1100
1500
Critical shear
section
Example 5: Design of Pile Cap – Truss
Theory
Main Reinforcement
Effective depth, d = h – c – 1.5bar = 1100 – 75 – 1.5(25) = 988 mm
Ultimate load, N = 1.35Gk + 1.5Qk = 1.35(3500) + 1.5(2500) = 8475 kN
From truss analogy:
𝑁𝑙
8475×1.5
Tension force, 𝑇 = =
= 𝟏𝟔𝟎𝟗 kN
8𝑑
8×0.988
Area of reinforcement, 𝐴𝑠 =
𝑇
0.87𝑓𝑦𝑘
=
1609×103
0.87×500
= 𝟑𝟔𝟗𝟗 mm2
For the whole width of pile cap, 𝐴𝑠 = 2 × 3699 = 𝟕𝟑𝟗𝟗 mm2
Example 5: Design of Pile Cap – Truss
Theory
Main Reinforcement (continued)
Minimum & Maximum Area of Reinforcement
𝑓𝑐𝑡𝑚
2.90
𝐴𝑠,𝑚𝑖𝑛 = 0.26
𝑏𝑑 = 0.26
0.0015𝑏𝑑 ≥ 0.0013𝑏𝑑
𝑓𝑦𝑘
500
 As,min
= 0.0015bd = 0.0015  2300  988 = 3421 mm2
As,max = 0.04Ac = 0.04bh = 0.04  2300  1100 = 101200 mm2
Provide 16H25 (As,prov = 7855 mm2)
Example 5: Design of Pile Cap – Truss
Theory
Shear
(i) Vertical Shear
Critical shear at p/5 section inside pile:
8475
Load per pile =
= 2119 kN
4
2 pile outside critical section  Shear force, VEd = 2  2119 = 4238 kN
Pile spacing  3p  Consider shear enhancement on the whole width of the
section
 Reduced shear force, 𝑉𝐸𝑑 = 4238 ×
𝑎𝑣
2𝑑
= 4238 ×
350
2×988
= 𝟕𝟓𝟏 kN
Example 5: Design of Pile Cap – Truss
Theory
Shear
Note:
Bar extend beyond critical section at = 475 + 988 – 75 = 1388 mm
 𝑙𝑏𝑑 + 𝑑 = 36∅ + 𝑑 = 36 × 25 + 988 = 1888 mm  Asl = 0 mm2
av = 750 – 250 – 250 + 500/5 = 350
400
100
1500
Critical
section
400
1500
400
475
1388
h = 1100
400
475
Critical
section
Example 5: Design of Pile Cap – Truss
Theory
Shear
𝑘 =1+
200
𝑑
=1+
200
988
= 1.45  2.0
 𝑉𝑅𝑑,𝑐 = 𝑉𝑚𝑖𝑛 = 0.035𝑘 3/2 𝑓𝑐𝑘 𝑏𝑑
= 0.035 × 1.453/2 30 2300 × 988 × 10−3 = 760 kN
VEd (751 kN)  Vmin (760 kN)
 OK
Example 5: Design of Pile Cap – Truss
Theory
Shear
(ii)
Punching Shear at Perimeter 2.0d from Column Face
Since pile spacing  3p
 No punching shear check in necessary
Example 5: Design of Pile Cap – Truss
Theory
Shear
(iii)
Maximum Punching Shear at Column Perimeter
𝑉𝑅𝑑,𝑚𝑎𝑥
𝑓𝑐𝑘
= 0.5 0.6 1 −
250
= 0.5 0.6 1 −
30
250
30
1.5
𝑓𝑐𝑘
𝑢𝑑
1.5
4 × 500 × 988 = 10428 kN  VEd,max = 8475 kN
OK
Example 5: Design of Pile Cap – Truss
Theory
Cracking
h = 1100 mm  200 mm
Max bar spacing
Assume steel stress is under quasi-permanent loading:
= 0.50
𝑓𝑦𝑘
𝐴𝑠,𝑟𝑒𝑞
1.15
𝐴𝑠,𝑝𝑟𝑜𝑣
= 0.50
500
1.15
7399
7855
= 205 N/mm2
For design crack width 0.3 mm:
Maximum allowable bar spacing = 200 mm
Actual bar spacing =
2300−2 75 −25
15
= 142 mm  200 mm
Cracking OK
Example 5: Design of Pile Cap – Truss
Theory
Detailing
5H16
Binders
1500
h = 1100
400
16H25 @ 140 mm
400
16H25 @ 140 mm
400
1500
400
500 mm spun
pile
Example 5
DESIGN OF PILE CAP – BEAM
THEORY
Example 5: Design of Pile Cap – Beam
Theory
N
M
•
•
•
•
•
•
•
•
•
•
Axial Load, Nultimate = 4200 kN
Moment, Multimate = 75 kNm
fck = 30 N/mm2
fyk = 500 N/mm2
soil = 200 N/mm2
Unit weight of concrete = 25 kN/m3
Column size = 400 mm  400 mm
Cover = 75 mm
Assumed bar = 16 mm
Pile:
o Precast RC pile: 300 mm  300 mm
o Working load = 600 kN
• Safety factor = 1.40
Example 5: Design of Pile Cap – Beam
Theory
Size of Pile Cap
4200
Service load =
= 3000 kN
1.40
Assumed self weight of pile cap, say W = 200 kN
Pile capacity, Fa = 600 kN
No. of pile required, 𝑛 =
𝑁+𝑊
𝐹𝑎
=
(3000 + 200)
600
= 5.3  Use n = 6
Example 5: Design of Pile Cap – Beam
Theory
Size of Pile Cap
4200
Service load =
= 3000 kN
1.40
Assumed self weight of pile cap, say W = 200 kN
Pile capacity, Fa = 600 kN
No. of pile required, 𝑛 =
𝑁+𝑊
𝐹𝑎
=
(3000 + 200)
600
= 5.3  Use n = 6
Example 5: Design of Pile Cap – Beam
Theory
Size of Pile Cap
4200
Service load =
= 3000 kN
1.40
Assumed self weight of pile cap, say W = 200 kN
Pile capacity, Fa = 600 kN
No. of pile required, 𝑛 =
𝑁+𝑊
𝐹𝑎
=
(3000 + 200)
600
= 5.3  Use n = 6
Pile spacing, 𝑙 = 𝑘∅𝑝 = 3∅𝑝 = 3 × 300 = 900 mm
Width, B = 𝑘 + 1 ∅𝑝 + 300 = 3 + 1 × 300 + 300 = 1500 mm
Length, H = 2𝑘 + 1 ∅𝑝 + 300 = 2 3 + 1 300 + 300 = 2400 mm
Depth, h = 2∅𝑝 + 500 = 2 300 + 500 = 1100 mm
Example 5: Design of Pile Cap – Beam
Theory
Size of Pile Cap (continued)
Try size: B  H  h = 1.5  2.4  1.1 m
Self weight = 25(1.5  2.4  1.1) = 99 kN  200 kN OK
av = 1200 – 200 – 450 + 300/5 = 610
p/5 = 60
300
x
y
300
y
900
Critical shear
section
x
300
900
900
300
I = 4(0.9)2 = 3.24 m2
Example 5: Design of Pile Cap – Beam
Theory
Analysis
Service moment, Mservice =
75
1.40
= 53.6 kNm
Maximum service load per pile, 𝐹 =
=
3000+99
6
+
53.6×0.90
3.24
𝑁+𝑊
𝑛
+
𝑀𝑥
𝐼
= 531 kN  600 kN
Ultimate load per pile:
𝑁
𝑀 𝑥
4200
75×0.90
𝐹𝑥𝑥 = 𝑢𝑙𝑡 + 𝑢𝑙𝑡 =
+
= 𝟕𝟐𝟏 kN
𝐹𝑦𝑦 =
𝑛
𝑁𝑢𝑙𝑡
𝑛
=
𝐼
4200
6
6
3.24
= 𝟕𝟎𝟎 kN
Maximum moment at column face:
𝑀𝑥𝑥 = 2 721 × 0.90 − 0.20 = 𝟏𝟎𝟎𝟗 kNm
𝑀𝑦𝑦 = 3 700 × 0.45 − 0.20 = 𝟓𝟐𝟓 kNm
OK
Example 5: Design of Pile Cap – Beam
Theory
Main Reinforcement
Effective depth:
dx = h – c – 0.5bar = 1100 – 75 – 0.5(16) = 1017 mm
dy = h – c – 1.5bar = 1100 – 75 – 1.5(16) = 1001 mm
Longitudinal Bar
𝐾=
𝑀𝐸𝑑
𝑓𝑐𝑘 𝑏𝑑 2
=
1009×106
30×1500×10172
= 0.022  Kbal = 0.167
 Compression reinforcement is NOT required
𝑧 = 𝑑 0.25 −
𝐴𝑠,𝑟𝑒𝑞 =
𝐾
1.134
𝑀𝐸𝑑
0.87𝑓𝑦𝑘 𝑧
=
= 0.98𝑑  0.95d
1009×106
0.87×500×0.95×1017
= 𝟐𝟒𝟎𝟏 mm2
Example 5: Design of Pile Cap – Beam
Theory
Minimum & Maximum Area of Reinforcement
𝑓𝑐𝑡𝑚
2.90
𝐴𝑠,𝑚𝑖𝑛 = 0.26
𝑏𝑑 = 0.26
0.0015𝑏𝑑 ≥ 0.0013𝑏𝑑
𝑓𝑦𝑘
500
 As,min = 0.0015bd = 0.0015  1500  1017 = 2298 mm2
As,max = 0.04Ac = 0.04bh = 0.04  1500  1100 = 66000 mm2
Provide 12H16 (As,prov = 2413 mm2)
Example 5: Design of Pile Cap – Beam
Theory
Transverse Bar
𝐾=
𝑀𝐸𝑑
𝑓𝑐𝑘 𝑏𝑑 2
=
525×106
30×2400×10012
= 0.007  Kbal = 0.167
 Compression reinforcement is NOT required
𝑧 = 𝑑 0.25 −
𝐴𝑠,𝑟𝑒𝑞 =
𝐾
1.134
𝑀𝐸𝑑
0.87𝑓𝑦𝑘 𝑧
=
= 0.9𝑑  0.95d
525×106
0.87×500×0.95×1001
= 𝟏𝟐𝟔𝟗 mm2
Example 5: Design of Pile Cap – Beam
Theory
Minimum & Maximum Area of Reinforcement
𝑓𝑐𝑡𝑚
2.90
𝐴𝑠,𝑚𝑖𝑛 = 0.26
𝑏𝑑 = 0.26
0.0015𝑏𝑑 ≥ 0.0013𝑏𝑑
𝑓𝑦𝑘
500
 As,min = 0.0015bd = 0.0015  2400  1001 = 3618 mm2
As,max = 0.04Ac = 0.04bh = 0.04  2400  1100 = 105600 mm2
Since As,req = 1269 mm2  As, min = 3618 mm2
Provide 18H16 (As,prov = 3620 mm2)
Example 5: Design of Pile Cap – Beam
Theory
Shear
(i) Vertical Shear
Critical shear at p/5 section inside pile:
2 pile outside critical section  Shear force, VEd = 2  721 = 1442 kN
Pile spacing  3p  Consider shear enhancement on the whole width of the
section
 Reduced shear force, 𝑉𝐸𝑑 = 1442 ×
𝑎𝑣
2𝑑
= 1442 ×
610
2×1017
= 𝟒𝟑𝟐 kN
Example 5: Design of Pile Cap – Beam
Theory
Shear
Note:
Bar extend beyond critical section at = 315 + 1001 = 1316 mm
 𝑙𝑏𝑑 + 𝑑 = 36∅ + 𝑑 = 36 × 16 + 1001 = 1577 mm  Asl = 0 mm2
av = 610
300
60
Critical
section
300
315
300
900
900
300
315
1316
h = 1100
900
Critical
section
Example 5: Design of Pile Cap – Beam
Theory
Shear
𝑘 =1+
200
𝑑
=1+
200
1001
= 1.45  2.0
 𝑉𝑅𝑑,𝑐 = 𝑉𝑚𝑖𝑛 = 0.035𝑘 3/2 𝑓𝑐𝑘 𝑏𝑑
= 0.035 × 1.453/2 30 1500 × 1001 × 10−3 = 501 kN
VEd (432 kN)  Vmin (501 kN)
 OK
Example 5: Design of Pile Cap – Beam
Theory
Shear
(ii)
Punching Shear at Perimeter 2.0d from Column Face
Since pile spacing  3p
 No punching shear check in necessary
Example 5: Design of Pile Cap – Beam
Theory
Shear
(iii)
Maximum Punching Shear at Column Perimeter
𝑉𝑅𝑑,𝑚𝑎𝑥
𝑓𝑐𝑘
= 0.5 0.6 1 −
250
= 0.5 0.6 1 −
30
250
30
1.5
𝑓𝑐𝑘
𝑢𝑑
1.5
4 × 400 × 1017 = 8456 kN  VEd,max = 4200 kN
OK
Example 5: Design of Pile Cap – Beam
Theory
Cracking
h = 1100 mm  200 mm
Max bar spacing
Assume steel stress is under quasi-permanent loading:
= 0.55
𝑓𝑦𝑘
𝐴𝑠,𝑟𝑒𝑞
1.15
𝐴𝑠,𝑝𝑟𝑜𝑣
= 0.55
500
1.15
2401
2413
= 238 N/mm2
For design crack width 0.3 mm:
Maximum allowable bar spacing = 200 mm
Actual bar spacing 1 =
Actual bar spacing 2 =
2300−2 75 −16
17
1500−2 75 −16
11
= 131 mm  200 mm
= 121 mm  200 mm
Cracking OK
Example 5: Design of Pile Cap – Beam
Theory
Detailing
300
12H16
300
900
18H16
900
900
300
h = 1100
300
4H12
Binders