Geotechnical Pad Foundation (11-1-2021).pptx

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

• Tolimit 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

• Pressure distribution under a footing
Uniform
distributed
Cohesivesoil Cohesionlesssoil

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

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

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

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

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

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

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

Pile Foundations
Load from
Structure
P
I
L
E
P
I
L
E
P
I
L
E
P
I
L
E
Lower Density
Medium Density
High Density
Pile Cap

Thickness and Size of Footing
Area of pad:
𝑨 =
𝑮𝒌 + 𝑸𝒌 +𝑾
𝑺𝒐𝒊𝒍 𝒃𝒆𝒂𝒓𝒊𝒏𝒈 𝒄𝒂𝒑𝒄𝒊𝒕𝒚
Minimum effective depth of pad:
𝒅 =
𝑵𝑬𝒅
𝒗𝒓𝒅,𝒎𝒂𝒙 ∙ 𝒖𝒐
NEd = Ultimate vertical load = 1.35 Gk +1.5 Qk
1 − 𝑓𝑐𝑘
250
𝑓𝑐𝑘
1.5
vrd,max = 0.5vfcd = 0.5 0.6
uo = Columnperimeter

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
STR Combination 1:
𝑵 = 𝟏. 𝟑𝟓𝑮𝒌 + 𝟏.𝟓𝑸𝒌
x
x
y
y

Check for Shear
• May fail in shear as vertical shear or punching shear
Vertical shear
sections
Punching shear
perimeters
2d
d
d
h
Bends may be
required

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 isrequired

Check for Shear
(ii) Punching Shear
Axial Force Only
𝑬𝒅 = 𝑽𝑬𝒅
𝒖∙𝒅
where u = Critical perimeter
• Critical section at a perimeter 2d from the face of the column
• Punching shear force =  Load outside the critical perimeter
• Shear stress, 𝒗
• If vEd  vRd,c = No shear reinforcement is required
• Also ensure that VEd  VRd,max

Check for Shear
(ii) Punching Shear (continued)
Axial Force & Bending Moment
• Punching shear resistance can be significantly reduced of a co-
existing bending, MEd
• However, adverse effect of the moment will give rise to a non-
uniform shear distribution around the control perimeter
• Refer to Cl. 6.4.3(3) of EC2

Check for Shear
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
u1 = length of basic control perimeter
W1 = function of basic control perimeter corresponds to the distribution of
shear = 0.5𝑐1
2 + 𝑐1𝑐2 + 4𝑐2𝑑 + 16𝑑2 + 2𝜋𝑑𝑐1
c1/c2  0.5 1.0 2.0  3.0
k 0.45 0.60 0.70 0.80

Design of Pad Footing
Check for Shear

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)

• 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
Axial Force,N:
Gk = 600 kN
Qk = 400kN
B
h
Column size:
300  300 mm
H

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 = 10mm
Nominal cover, cnom = cmin + cdev = 25 + 10 = 35mm
 cnom = 35mm
cmin = 25mm

Size
Service load, N
Assumed selfweight 10% of service load , W
= 1000 kN
= 100 kN
𝛾𝑠𝑜𝑖𝑙
𝑁+𝑊 1000+100
150
Area of footing required = = = 7.33 𝑚2
 Try footing size, B  H  h = 3 m  3 m  0.45 m
Area, A = 9m2
Selfweight, W = 9 0.45  25 = 101 kN
𝑵+𝑾
𝑨
=
𝟏𝟎𝟎𝟎+𝟏𝟎𝟏
𝟗
Check Service Soil Bearing Capacity =
= 122 kN/m2  150 kN/m2  OK

Analysis
Ultimate axial force, NEd = 1.35Gk + 1.5Qk
= 1.35 (600) + 1.5 (400) = 1410 kN
𝐴 9
𝑁𝐸𝑑 1410
Soil pressure at ultimate load, P = = = 157 kN/m2
Soil pressure per m length, w = 157  3 m = 470 kN/m
1.35 m
1.35 m
w = 470 kN/m
MEd
0.3 m
𝟏.𝟑𝟓
𝟐
𝑴𝑬𝒅 = 𝟒𝟕𝟎 × 𝟏. 𝟑𝟓 ×
= 428 kNm

Main Reinforcement
Effective depth, d = h – c – 1.5bar = 450 – 35 – (1.5  12) = 397mm
𝐾 =
𝑀𝐸𝑑 428×106
𝑓𝑐𝑘𝑏𝑑2 25×3000×3972 bal
= = 0.036  K = 0.167
 Compression reinforcement is NOT required
𝑧 = 𝑑 0.25 −
𝐾
1.134
= 0.97𝑑  0.95d
𝐴𝑠,𝑟𝑒
𝑞
=
𝑀𝐸𝑑
0.87𝑓𝑦𝑘𝑧
=
428×106
0.87×500×0.95×397
= 𝟐𝟔𝟏𝟏 mm2

Minimum & Maximum Area of Reinforcement
𝐴𝑠,𝑚𝑖
𝑛
= 0.26
𝑓𝑐𝑡𝑚
𝑓𝑦𝑘
2.56
500
𝑏𝑑 = 0.26 0.0013𝑏𝑑 ≥0.0013𝑏𝑑
 As,min = 0.0013bd = 0.0013  3000  397 = 1589mm2
As,max = 0.04Ac = 0.04bh = 0.04  3000  450 = 54000mm2
Provide 24H12 (As,prov = 2715mm2)

(i) Vertical Shear
Critical shear at 1.0d from face of column:
 Design shear force, VEd = 470  0.953 = 448kN
3 m
0.953 m
3 m
953 mm
1.35 m
w = 470 kN/m
d
=
397
mm
VEd

(i) Vertical Shear
𝑘 = 1+
200
= 1+
𝑑 397
200
= 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 = 2715mm2
𝑙
𝜌 =
𝑠
𝑙
𝑏𝑑
𝐴 2715
3000 × 397
= = 0.0023 ≤ 0.02

(i) Vertical Shear
𝑉𝑅𝑑,𝑐 = 0.12𝑘 100𝜌𝑙𝑓𝑐𝑘
1/3 𝑏𝑑
= 0.12 × 1.71 100 × 0.0023 × 25 1/3 3000 × 397 = 436463 N = 436 kN
𝑉𝑚𝑖𝑛 = 𝑓𝑐𝑘
0.035𝑘3/2 𝑏𝑑
= 0.035 × 1.713/2 25 3000 × 397 = 465970 𝑁 = 466 kN
VEd (448 kN)  Vmin (466 kN)  OK

(ii) Punching Shear
Average d = 450 – 35 – 12 = 403 mm
 2d = 806 mm
Control perimeter, u = (4  300) + (2  806) =
6265 mm
Area within perimeter, A = (0.30  0.30) + (4 
0.30  0.806) + (  0.8062) = 3.10m2
1350
2d = 806
300
300 2d = 806
544
 𝑙𝑏𝑑 + 𝑑 = 40∅ + 𝑑 = 40 × 12 + 397 = 877
mm
 Reinforcement NOT contributed to punching
resistance

(ii) Punching Shear
Punching shear force:
VEd = 157 (32 – 3.10) = 925kN
Punching shear resistance:
𝑉𝑅𝑑,𝑐 = 𝑉𝑚𝑖𝑛 = 0.035𝑘3/2𝑓𝑐𝑘
1/2
𝑢𝑑
25 1/2
= 0.035 1.71 3/2 6265 × 403
= 983199 N = 983 kN  VEd (925 kN)  OK
Soil pressure = 157 kN/m2
Aperimeter = 3.10 m2
Aall = 9m2

(iii) Maximum Punching Shear at Column Perimeter
Maximum punching shear force:
VEd,max = 157 (32 – 0.09) = 1400kN
Maximum shear resistance:
𝑉𝑅𝑑,𝑚𝑎𝑥 = 0.5𝑢𝑑 0.6
𝑓𝑐𝑘
1 −
𝑓𝑐𝑘
250 1.5
= 0.5 4 × 300 × 403 0.6
25
25
1 −
250 1.5
= 2176 kN  VEd,max  OK
Acolumn = 0.09m2
Aall = 9m2

Cracking
h = 450 mm  200 mm
𝑠
Steel stress, 𝑓 =
𝐺𝑘+0.3𝑄𝑘 𝐴𝑠.𝑟𝑒𝑞
𝐴𝑠,𝑝𝑟𝑜𝑣
𝑓𝑦𝑘
1.15
=
600+0.3×400
1.35𝐺𝑘+1.5𝑄𝑘
2611 500
1.35×600+1.5×400 2715 1.15
= 213 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
Max bar spacing

Detailing
3000
3000
24H12
24H12
3000
450
24H12
Plan View Section View

Example 2
PAD FOOTING
(AXIAL LOAD & MOMENT)

• 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
Axial Force, N = 1500 kN
Moment = 50 kNm
B
h
Column size:
250  350 mm
H

Size
Service axial, N
Service moment, M
Assumed selfweight 10% of service load , W
= 1500 kN / 1.40 = 1071 kN
= 50 kNm / 1.40 = 36.1 kNm
= 100 kN
𝛾𝑠𝑜𝑖𝑙
𝑁+𝑊 1071+107.1
150
Area of footing required = = = 7.85 𝑚2
 Try footing size, B  H  h = 2.80 m  3.50 m  0.65 m
Area, A = 9.80m2
Selfweight, W = 9.80 0.65  25 = 159 kN

Size (Continued)
𝐼𝑥
𝑥
3 3
= 𝐵𝐻
= 2.8×3.5
= 10.0 m4
12 12
𝑁+𝑊
𝐴
+
𝑀𝑦
𝐼
=
1071+159
9.80
+
50×1.75
10.0
𝑦 = 𝐻
= 3.5
= 1.75m
2 2
Maximum soil pressure, 𝑃 =
= 132 kN/m2  150 kN/m2  OK
B
H
x
x

Analysis
Ultimate soil pressure, 𝑃 = 𝑁
± 𝑀𝑦
= 1500
±
𝐴 𝐼 9.80
10.0
50×1.75
= 153 ± 8.7 kN/m2
 Pmin = 144 kN/m2 and Pmax = 162 kN/m2
1.575 m
1.575 m
144
1.275
x
0.35 m
x
y y
162
154
144
162

Analysis (Continued)
𝑀𝑥𝑥
= 154 ×
1.5752
2
+ 162 − 154
1.575
2
2
× 1.575 ×
3
= 197 kNm/m  2.80 m = 553kNm
2
2
𝑀 = 𝟏𝟓𝟑× 1.275
= 124 kNm/m  3.50 m=
𝑦𝑦
435 kNm 1.575 m
1.575 m
144
0.35 m
1.275
x
x
y y
162
154
144 + 162
=
2

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 553×106
𝑓𝑐𝑘𝑏𝑑2 30×2800×6092 bal
= = 0.018  K = 0.167
𝑧 = 𝑑 0.25 −
𝐾
1.134
= 0.98𝑑  0.95d
𝐴𝑠,𝑟𝑒
𝑞
=
𝑀𝑥𝑥
=
553×106
0.87×500×0.95×609
= 𝟐𝟏𝟗𝟕 mm2

𝐴𝑠,𝑚𝑖
𝑛
= 0.26
𝑓𝑐𝑡𝑚
𝑓𝑦𝑘
2.90
500
𝑏𝑑 = 0.26 0.0013𝑏𝑑 ≥0.0013𝑏𝑑
 As,min = 0.0013bd = 0.0013  2800  609 = 2217mm2
As,max = 0.04Ac = 0.04bh = 0.04  2800  609 = 72800mm2
Since As  As,min, Use As,min = 2217 mm2
Provide 21H12 (As = 2375mm2)

Main Reinforcement – Transverse Bar
𝐾 =
𝑀𝑦𝑦 435×106
𝑓𝑐𝑘𝑏𝑑2 30×3500×5972 bal
= = 0.018  K = 0.167
𝑧 = 𝑑 0.25 −
𝐾
1.134
= 0.99𝑑  0.95d
𝐴𝑠,𝑟𝑒
𝑞
=
𝑀𝑦𝑦
=
435×106
0.87×500×0.95×597
= 𝟏𝟕𝟔𝟓mm2

𝐴𝑠,𝑚𝑖
𝑛
= 0.26
𝑓𝑐𝑡𝑚
𝑓𝑦𝑘
2.90
500
𝑏𝑑 = 0.26 0.0013𝑏𝑑 ≥0.0013𝑏𝑑
 As,min = 0.0013bd = 0.0013  3500  597 = 3147mm2
As,max = 0.04Ac = 0.04bh = 0.04  3500  597 = 91000mm2
Since As  As,min, Use As,min = 3147 mm2
Provide 28H12 (As = 3167mm2)

(i) Vertical Shear
Average pressure at critical section:
= 144 +
2.891
3.50
× 18 = 159 kN/m2
 Design shear force, VEd = 159  0.966
2.80 = 431 kN
144
2.891
162
159
0.966
d
=
0.609
2.8
Note:
Bar extend beyond critical section at = 966 – 35 = 931 mm
 𝑙𝑏𝑑 + 𝑑 = 36∅ + 𝑑 = 36 × 12 + 609 = 1041 mm  Asl = 0mm2

(i) Vertical Shear
𝑘 = 1 +
𝑑
200
= 1 +
609
200
= 1.57  2.0
𝑙
𝜌 =
𝐴𝑠
𝑙
𝑏𝑑
= 0
 𝑉𝑅𝑑,𝑐 = 0.12𝑘 100𝜌𝑙𝑓𝑐𝑘
1/3 𝑏𝑑
= 0.12 × 1.57 100 × 0 × 30 1/3 2800 × 609 = 0 N = 0 kN
 𝑉𝑚𝑖𝑛 = 0.035𝑘3/2 𝑓𝑐𝑘 𝑏𝑑
= 0.035 × 1.573/2 30 2800 × 609 = 644949 𝑁 = 645 kN
VEd (430 kN)  Vmin (645 kN)  OK

(ii) Punching Shear
Average 𝑑 =
609+597
2
= 603 mm
 2d = 1206mm
Control perimeter;
u = 2(350 + 250) + (2  1206) = 8779mm
Area within perimeter;
A = (0.35  0.25) + (2  0.35 1.206)
+ (2  0.25  1.206) + ( 1.2062) = 6.10 m2
3500
2d = 1206 350
250
2d = 1206
69
369
 𝑙𝑏𝑑 + 𝑑 = 36∅ + 𝑑 = 36 × 12 + 609 = 1041 mm
 Reinforcement NOT contributed to punching resistance
2800

(ii) Punching Shear
Average punching shear force at control perimeter:
VEd = 153 [(2.80  3.50) – 6.10] = 566kN
𝑣𝐸𝑑
Punching shear stress:
𝛽𝑉
= 𝐸𝑑
𝑢𝑑
Where 𝛽 = 1 +𝑘
𝑀𝐸𝑑
𝑉𝐸𝑑
𝑢1
𝑊1
𝑐2 250
k = 0.65  𝑐1
= 350
= 1.4
W1 = 0.5𝑐1
2 + 𝑐1𝑐2 + 4𝑐2𝑑 + 16𝑑2 + 2𝜋𝑑𝑐1
= 0.5(3502) + 350 × 250 + 4 × 250 × 603
16 6032 + 2𝜋 × 603 × 350 = 7.9  106 mm2
 𝛽 = 1 +0.65
50×106 8779
566×103 7.9×106
= 1.06
Therefore,𝒗𝑬𝒅 𝟖𝟕𝟕𝟗×𝟔𝟎𝟗
𝟑
= 𝟏.𝟎𝟔×𝟓𝟔𝟔×𝟏𝟎
= 𝟎. 𝟏𝟏 N/mm2
+
Aall = 2.8  3.5m
Aperimeter =6.1 0 m2

(ii) Punching Shear
Punching shear resistance:
𝑘 = 1 +
𝑑
200
= 1 +
609
200
= 1.57  2.0
𝑣𝑅𝑑,𝑐 = 𝑣𝑚𝑖𝑛 = 0.035𝑘3/2𝑓𝑐𝑘
1/2
= 0.035 1.57 3/2 30 1/2
= 0.38 N/mm2  vEd (0.11N/mm2)  OK
Aall = 2.8  3.5m
Aperimeter = 6.10m2

Maximum punching shear force:
VEd,max = 1500kN
Column perimeter, uo = 2(350 + 250) = 1200mm
Punching shear stress:
𝐸𝑑
𝛽𝑉𝐸𝑑
𝑣 =
Where
𝑀𝐸𝑑
𝑉𝐸𝑑
𝑢𝑜
𝑊1
𝑢𝑜𝑑
𝛽 = 1 +𝑘
k = 0.65 
𝑐2 250
𝑐1
= 350
= 1.4
W1 = 0.5𝑐1
2 +𝑐1𝑐2
= 0.5(3502) + 350 × 250 = 0.15  106mm2
all
A = 2.8  3.5 m = 9.8 m2
Acolumn = 0.09 m2

 𝛽 = 1 +0.65
50×106 1200
1500×103 0.15×106
= 1.17
Therefore, 𝒗𝑬𝒅 =
𝟏.𝟏𝟕×𝟏𝟓𝟎𝟎×𝟏𝟎𝟑
𝟏𝟐𝟎𝟎×𝟔𝟎𝟑
= 𝟐. 𝟒𝟒 N/mm2
Maximum shear resistance:
𝑣𝑅𝑑,𝑚𝑎𝑥 = 0.5 0.6
𝑓𝑐𝑘
1 −
𝑓𝑐𝑘
250 1.5
30
30
= 0.5 0.6 1 −
250 1.5
= 5.28 N/mm2 vEd  OK
all
A = 2.8  3.5 m = 9.8 m2
Acolumn = 0.09m2

Cracking
h = 650 mm  200mm
= 0.6
1.15
𝑓𝑦𝑘 𝐴𝑠,𝑟𝑒𝑞
𝐴𝑠,𝑝𝑟𝑜𝑣
500 2197
1.15 2375
= 0.6 = 241 N/mm2
For design crack width 0.3 mm:
Maximum allowable bar spacing = 150 mm
Actual bar spacing at x-x =
2800−2 35 −12
20
= 136 mm  150 mm
ctual bar spacing at y-y =
3500−2 35 −12
27
= 126 mm  150 mm
Cracking OK
Max bar spacing
Assume steel stress is under quasi-permanent loading:

Detailing
3500
2800
28H12
21H12 3500
650
28H12
Plan View Section View
21H12

Geotechnical Pad Foundation (11-1-2021).pptx

Recommended

Recommended

More Related Content

Similar to Geotechnical Pad Foundation (11-1-2021).pptx

Similar to Geotechnical Pad Foundation (11-1-2021).pptx (20)

Recently uploaded

Recently uploaded (20)

Geotechnical Pad Foundation (11-1-2021).pptx