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DESIGN OF FOUNDATIONS
Dr. Izni Syahrizal bin IbrahimFaculty of Civil Engineering
Universiti Teknologi Malaysia
Email: [email protected]
Introduction
• Foundation – Part of structure which transmits load from the structure to the underlying soil or rock
• All soils compress noticeably when loadedcausing 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
PILE
PILE
PILE
Lower Density
Medium Density
High Density
Pile Cap
Design of Pad Footing
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 = 1.35Gk + 1.5Qk
vrd,max = 0.5vfcd = 0.5 0.6 1 −𝑓𝑐𝑘
250
𝑓𝑐𝑘
1.5
uo = Column perimeter
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
STR Combination 1:
𝑵 = 𝟏. 𝟑𝟓𝑮𝒌 + 𝟏. 𝟓𝑸𝒌
x
x
y
y
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
dh
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 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
• Shear stress, 𝒗𝑬𝒅 =𝑽𝑬𝒅
𝒖∙𝒅where u = Critical perimeter
• 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 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
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 + 𝑘𝑀𝐸𝑑
𝑉𝐸𝑑
𝑢1
𝑊1
k = coefficient depending on the ratio between column dimension c1 & c2
u1 = length of basic control perimeterW1 = 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(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)
• 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 kNQk = 400 kN
B
h
Column size: 300 300 mm
H
Durability & Bond Requirements
Min cover regards to bond, cmin,b = 12 mmMin cover regards to durability, cmin,dur = 25 mmAllowance 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)
Example 1: Pad Footing (Axial Load)
Size
Service load, N = 1000 kNAssumed selfweight 10% of service load , W = 100 kN
Area of footing required = 𝑁+𝑊
𝛾𝑠𝑜𝑖𝑙=1000+100
150= 7.33 𝑚2
Try footing size, B H h = 3 m 3 m 0.45 mArea, A = 9 m2
Selfweight, W = 9 0.45 25 = 101 kN
Check Service Soil Bearing Capacity = 𝑵+𝑾
𝑨=𝟏𝟎𝟎𝟎+𝟏𝟎𝟏
𝟗
= 122 kN/m2 150 kN/m2 OK
Example 1: Pad Footing (Axial Load)
Analysis
Ultimate axial force, NEd = 1.35Gk + 1.5Qk
= 1.35 (600) + 1.5 (400) = 1410 kN
Soil pressure at ultimate load, P = 𝑁𝐸𝑑
𝐴=1410
9= 157 kN/m2
Soil pressure per m length, w = 157 3 m = 470 kN/m
1.35 m1.35 m
w = 470 kN/m
0.3 m
Example 1: Pad Footing (Axial Load)
Analysis
Ultimate axial force, NEd = 1.35Gk + 1.5Qk
= 1.35 (600) + 1.5 (400) = 1410 kN
Soil pressure at ultimate load, P = 𝑁𝐸𝑑
𝐴=1410
9= 157 kN/m2
Soil pressure per m length, w = 157 3 m = 470 kN/m
1.35 m1.35 m
w = 470 kN/m
0.3 m
MEd
𝑴𝑬𝒅 = 𝟒𝟕𝟎 × 𝟏. 𝟑𝟓 ×𝟏. 𝟑𝟓
𝟐= 428 kNm
Example 1: Pad Footing (Axial Load)
Main Reinforcement
Effective depth, d = h – c – 1.5bar = 450 – 35 – (1.5 12) = 397 mm
𝐾 =𝑀𝐸𝑑
𝑓𝑐𝑘𝑏𝑑2 =
428×106
25×3000×3972= 0.036 Kbal = 0.167
Compression reinforcement is NOT required
𝑧 = 𝑑 0.25 −𝐾
1.134= 0.97𝑑 0.95d
𝐴𝑠,𝑟𝑒𝑞 =𝑀𝐸𝑑
0.87𝑓𝑦𝑘𝑧=
428×106
0.87×500×0.95×397= 𝟐𝟔𝟏𝟏mm2
Example 1: Pad Footing (Axial Load)
Minimum & Maximum Area of Reinforcement
𝐴𝑠,𝑚𝑖𝑛 = 0.26𝑓𝑐𝑡𝑚𝑓𝑦𝑘𝑏𝑑 = 0.26
2.56
5000.0013𝑏𝑑 ≥ 0.0013𝑏𝑑
As,min = 0.0013bd = 0.0013 3000 397 = 1589 mm2
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
Critical shear at 1.0d from face of column:
Design shear force, VEd = 470 0.953 = 448 kN
3 m
0.953 m
3 m953 mm1.35 m
w = 470 kN/m
d=
39
7 m
m
VEd
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 mm2
𝜌𝑙 =𝐴𝑠𝑙𝑏𝑑=2715
3000 × 397= 0.0023 ≤ 0.02
Example 1: Pad Footing (Axial Load)
(i) Vertical Shear
𝑉𝑅𝑑,𝑐 = 0.12𝑘 100𝜌𝑙𝑓𝑐𝑘1/3 𝑏𝑑
= 0.12 × 1.71 100 × 0.0023 × 25 1/3 3000 × 397 = 436463 N = 436 kN
𝑉𝑚𝑖𝑛 = 0.035𝑘3/2 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
Control perimeter, u = (4 300) + (2 806) = 6265 mm
Area within perimeter, A = (0.30 0.30) + (4 0.30 0.806) + ( 0.8062) = 3.10 m2
1350
2d = 806 300
300
2d = 806
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
Punching shear resistance:
𝑉𝑅𝑑,𝑐 = 𝑉𝑚𝑖𝑛 = 0.035𝑘3/2𝑓𝑐𝑘
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
Aperimeter = 3.10 m2
Aall = 9 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
Maximum shear resistance:
𝑉𝑅𝑑,𝑚𝑎𝑥 = 0.5𝑢𝑑 0.6 1 −𝑓𝑐𝑘250
𝑓𝑐𝑘1.5
= 0.5 4 × 300 × 403 0.6 1 −25
250
25
1.5
= 2176 kN VEd,max OK
Soil pressure = 157 kN/m2
Acolumn = 0.09 m2
Aall = 9 m2
Example 1: Pad Footing (Axial Load)
Cracking
h = 450 mm 200 mm
Steel stress, 𝑓𝑠 =𝐺𝑘+0.3𝑄𝑘
1.35𝐺𝑘+1.5𝑄𝑘
𝐴𝑠.𝑟𝑒𝑞
𝐴𝑠,𝑝𝑟𝑜𝑣
𝑓𝑦𝑘
1.15
=600+0.3×400
1.35×600+1.5×400
2611
2715
500
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= 127mm 200 mm
Cracking OK
Max bar spacing
Example 1: Pad Footing (Axial Load)
Detailing
3000
30
00
24
H1
2
24H123000
450
24H12
Plan View Section View
Example 2
PAD FOOTING(AXIAL LOAD & MOMENT)
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 kNMoment = 50 kNm
B
h
Column size: 250 350 mm
H
Durability & Bond Requirements
Min cover regards to bond, cmin,b = 12 mmMin cover regards to durability, cmin,dur = 25 mmAllowance in design for deviation, cdev = 10 mm
Nominal cover, cnom = cmin + cdev = 25 + 10 = 35 mm
cnom = 35 mm
Example 2: Pad Footing (Axial Load & Moment)
cmin = 25 mm
Size
Service axial, N = 1500 kN / 1.40 = 1071 kNService moment, M = 50 kNm / 1.40 = 36.1 kNmAssumed selfweight 10% of service load , W = 100 kN
Area of footing required = 𝑁+𝑊
𝛾𝑠𝑜𝑖𝑙=1071+107.1
150= 7.85 𝑚2
Try footing size, B H h = 2.80 m 3.50 m 0.65 mArea, A = 9.80 m2
Selfweight, W = 9.80 0.65 25 = 159 kN
Example 2: Pad Footing (Axial Load & Moment)
Size (Continued)
𝐼𝑥𝑥 =𝐵𝐻3
12=2.8×3.53
12= 10.0m4
𝑦 =𝐻
2=3.5
2= 1.75m
Maximum soil pressure, 𝑃 =𝑁+𝑊
𝐴+𝑀𝑦
𝐼=1071+159
9.80+50×1.75
10.0
= 132 kN/m2 150 kN/m2 OK
Example 2: Pad Footing (Axial Load & Moment)
B
H
x
x
Analysis
Ultimate soil pressure, 𝑃 =𝑁
𝐴±𝑀𝑦
𝐼=1500
9.80±
50×1.75
10.0= 153 ± 8.7 kN/m2
Pmin = 144 kN/m2 and Pmax = 162 kN/m2
1.575 m1.575 m
144
0.35 m
1.275
x
x
y y
162154
Example 2: Pad Footing (Axial Load & Moment)
144
162
Analysis (Continued)
𝑀𝑥𝑥
= 154 ×1.5752
2
+ 162 − 1541.575
2× 1.575 ×
2
3
= 197 kNm/m 2.80 m = 553 kNm
𝑀𝑦𝑦 = 𝟏𝟓𝟑 ×1.2752
2= 124 kNm/m 3.50 m =
435 kNm 1.575 m1.575 m
144
0.35 m
1.275
x
x
y y
162154
Example 2: Pad Footing (Axial Load & Moment)
=144 + 162
2
Effective Depth
dx = h – c – 0.5bar = 650 – 35 – (0.5 12) = 609 mmdy = h – c – 1.5bar = 650 – 35 – (1.5 12) = 597 mm
Main Reinforcement – Longitudinal Bar
𝐾 =𝑀𝑥𝑥
𝑓𝑐𝑘𝑏𝑑2 =
553×106
30×2800×6092= 0.018 Kbal = 0.167
Compression reinforcement is NOT required
𝑧 = 𝑑 0.25 −𝐾
1.134= 0.98𝑑 0.95d
𝐴𝑠,𝑟𝑒𝑞 =𝑀𝑥𝑥
0.87𝑓𝑦𝑘𝑧=
553×106
0.87×500×0.95×609= 𝟐𝟏𝟗𝟕mm2
Example 2: Pad Footing (Axial Load & Moment)
Minimum & Maximum Area of Reinforcement
𝐴𝑠,𝑚𝑖𝑛 = 0.26𝑓𝑐𝑡𝑚𝑓𝑦𝑘𝑏𝑑 = 0.26
2.90
5000.0013𝑏𝑑 ≥ 0.0013𝑏𝑑
As,min = 0.0013bd = 0.0013 2800 609 = 2217 mm2
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
𝑓𝑐𝑘𝑏𝑑2 =
435×106
30×3500×5972= 0.018 Kbal = 0.167
Compression reinforcement is NOT required
𝑧 = 𝑑 0.25 −𝐾
1.134= 0.99𝑑 0.95d
𝐴𝑠,𝑟𝑒𝑞 =𝑀𝑦𝑦
0.87𝑓𝑦𝑘𝑧=
435×106
0.87×500×0.95×597= 𝟏𝟕𝟔𝟓mm2
Example 2: Pad Footing (Axial Load & Moment)
Minimum & Maximum Area of Reinforcement
𝐴𝑠,𝑚𝑖𝑛 = 0.26𝑓𝑐𝑡𝑚𝑓𝑦𝑘𝑏𝑑 = 0.26
2.90
5000.0013𝑏𝑑 ≥ 0.0013𝑏𝑑
As,min = 0.0013bd = 0.0013 3500 597 = 3147 mm2
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
Critical shear at 1.0d from face of column:
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
Example 2: Pad Footing (Axial Load & Moment)
144
2.891
162159
0.966d=
0.6
09
2.8
Note:Bar extend beyond critical section at = 966 – 35 = 931 mm 𝑙𝑏𝑑 + 𝑑 = 36∅ + 𝑑 = 36 × 12 + 609 = 1041 mm Asl = 0 mm2
(i) Vertical Shear
𝑘 = 1 +200
𝑑= 1 +
200
609= 1.57 2.0
𝜌𝑙 =𝐴𝑠𝑙𝑏𝑑= 0
𝑉𝑅𝑑,𝑐 = 0.12𝑘 100𝜌𝑙𝑓𝑐𝑘1/3 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
Example 2: Pad Footing (Axial Load & Moment)
(ii) Punching Shear
Critical shear at 2.0d from face of column:
Average 𝑑 =609+597
2= 603 mm
2d = 1206 mm
Control perimeter;u = 2(350 + 250) + (2 1206) = 8779 mm
Area within perimeter;A = (0.35 0.25) + (2 0.35 1.206)+ (2 0.25 1.206) + ( 1.2062) = 6.10 m2
𝑙𝑏𝑑 + 𝑑 = 36∅ + 𝑑 = 36 × 12 + 609 = 1041 mm Reinforcement NOT contributed to punching resistance
3500
2d = 1206 350
250
2d = 1206
369
69
28
00
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:
𝑣𝐸𝑑 =𝛽𝑉𝐸𝑑𝑢𝑑
Where 𝛽 = 1 + 𝑘𝑀𝐸𝑑
𝑉𝐸𝑑
𝑢1
𝑊1
k = 0.65 𝑐1
𝑐2=350
250= 1.4
W1 = 0.5𝑐12 + 𝑐1𝑐2 + 4𝑐2𝑑 + 16𝑑
2 + 2𝜋𝑑𝑐1= 0.5(3502) + 350 × 250 + 4 × 250 × 603 +
16 6032 + 2𝜋 × 603 × 350 = 7.9 106 mm2
𝛽 = 1 + 0.6550×106
566×1038779
7.9×106= 1.06
Therefore, 𝒗𝑬𝒅 =𝟏.𝟎𝟔×𝟓𝟔𝟔×𝟏𝟎𝟑
𝟖𝟕𝟕𝟗×𝟔𝟎𝟗= 𝟎. 𝟏𝟏 N/mm2
Example 2: Pad Footing (Axial Load & Moment)
Soil pressure = 153 kN/m2
Aall = 2.8 3.5 m
Aperimeter = 6.10 m2
(ii) Punching Shear
Punching shear resistance:
𝑘 = 1 +200
𝑑= 1 +
200
609= 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 (0.11 N/mm2) OK
Example 2: Pad Footing (Axial Load & Moment)
Soil pressure = 153 kN/m2
Aall = 2.8 3.5 m
Aperimeter = 6.10 m2
(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:
𝑣𝐸𝑑 =𝛽𝑉𝐸𝑑𝑢𝑜𝑑
Where 𝛽 = 1 + 𝑘𝑀𝐸𝑑
𝑉𝐸𝑑
𝑢𝑜
𝑊1
k = 0.65 𝑐1
𝑐2=350
250= 1.4
W1 = 0.5𝑐12 + 𝑐1𝑐2
= 0.5(3502) + 350 × 250 = 0.15 106 mm2
Soil pressure = 153 kN/m2
Aall = 2.8 3.5 m = 9.8 m2
Acolumn = 0.09 m2
Example 2: Pad Footing (Axial Load & Moment)
(iii) Maximum Punching Shear at Column Perimeter
𝛽 = 1 + 0.6550×106
1500×1031200
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 1 −𝑓𝑐𝑘250
𝑓𝑐𝑘1.5
= 0.5 0.6 1 −30
250
30
1.5
= 5.28 N/mm2 vEd OK
Soil pressure = 153 kN/m2
Aall = 2.8 3.5 m = 9.8 m2
Acolumn = 0.09 m2
Example 2: Pad Footing (Axial Load & Moment)
Cracking
h = 650 mm 200 mm
Assume steel stress is under quasi-permanent loading:
= 0.6𝑓𝑦𝑘
1.15
𝐴𝑠,𝑟𝑒𝑞
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
Actual bar spacing at x-x = 2800−2 35 −12
20= 136mm 150 mm
ctual bar spacing at y-y = 3500−2 35 −12
27= 126mm 150 mm
Cracking OK
Max bar spacing
Example 2: Pad Footing (Axial Load & Moment)
Detailing
Example 2: Pad Footing (Axial Load & Moment)
3500
28
00
28
H1
2
21H123500
650
28H12
Plan View Section View
21H12
Example 3
DESIGN OF COMBINED FOOTING
Example 3: Design of Combined Footing
NA = 1610 kN(Ultimate)
B
h
Column size: 300 300 mm
H
NB = 1950 kN(Ultimate)
3.4 m
Column size: 400 400 mm
• 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 = 1610 kN / 1.40 = 1150 kNNB = 1950 kN / 1.40 = 1393 kN
Total service axial, Ntotal = 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 mArea, A = 16.2 m2
Selfweight, W = 16.2 0.65 25 = 252.7 kN
Example 3: Design of Combined Footing
Size (Continued)
Check Service Soil Bearing Capacity = 𝑁+𝑊
𝐴=2543+252.7
16.2
= 173 kN/m2 200 kN/m2 OK
Arrange position of footing so that the distribution of soil pressure is uniform:
3.4 m
NA = 1150 kN NB = 1393 kN
Ntotal = 2543 kN
x
MA = 01393(3.4 ) + 2543x = 0x = 1.86 m
Example 3: Design of Combined Footing
Analysis
Soil pressure at ultimate load, P = 𝑁𝐸𝑑
𝐴=(1610+1950)
16.2= 219.8 kN/m2
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
0.6
5
0.3 1.71 1.34 0.40.99 1.26
1610 kN 1950 kN
3 m 3 m
x = 1.86
593.5 kN/m
384
353
634
289250
430 473
BMD (kNm)
587 676
934 846
1082964
868 749
1.43 1.62
SFD (kN)
Example 3: Design of Combined Footing
Main Reinforcement
Longitudinal Reinforcement: Bottom
Effective depth: dx = h – c – 0.5bar = 650 – 40 – (0.5 12) = 604 mm
𝐾 =𝑀𝐸𝑑
𝑓𝑐𝑘𝑏𝑑2 =
473×106
35×2700×6042= 0.014 Kbal = 0.167
Compression reinforcement is NOT required
𝑧 = 𝑑 0.25 −𝐾
1.134= 0.99𝑑 0.95d
𝐴𝑠,𝑟𝑒𝑞 =𝑀𝐸𝑑
0.87𝑓𝑦𝑘𝑧=
473×106
0.87×500×0.95×604= 𝟏𝟖𝟗𝟒mm2 As,min = 2722 mm2
Example 3: Design of Combined Footing
Main Reinforcement
Longitudinal Reinforcement: Top
𝐾 =𝑀𝐸𝑑
𝑓𝑐𝑘𝑏𝑑2 =
353×106
35×2700×6042= 0.010 Kbal = 0.167
Compression reinforcement is NOT required
𝑧 = 𝑑 0.25 −𝐾
1.134= 0.99𝑑 0.95d
𝐴𝑠,𝑟𝑒𝑞 =𝑀𝐸𝑑
0.87𝑓𝑦𝑘𝑧=
353×106
0.87×500×0.95×604= 𝟏𝟒𝟏𝟑mm2 As,min = 2722 mm2
Example 3: Design of Combined Footing
Minimum & Maximum Area of Reinforcement
𝐴𝑠,𝑚𝑖𝑛 = 0.26𝑓𝑐𝑡𝑚𝑓𝑦𝑘𝑏𝑑 = 0.26
3.2
5000.0017𝑏𝑑 ≥ 0.0013𝑏𝑑
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
1.2 m1.2 m
w = 219.8 kN/m
0.3 m
MEd
𝑴𝑬𝒅 = 𝟐𝟏𝟗. 𝟖 × 𝟏. 𝟐 ×𝟏. 𝟐
𝟐= 158 kNm/m
Consider b = 1000 mm: Soil pressure per 1 m width, w = 219.8 1.0 m = 219.8 kN/m
Example 3: Design of Combined Footing
Main Reinforcement
Transverse Reinforcement: Bottom
Effective depth: dy = h – c – 1.5bar = 650 – 40 – (1.5 12) = 592 mm
𝐾 =𝑀𝐸𝑑
𝑓𝑐𝑘𝑏𝑑2 =
158×106
35×𝟏𝟎𝟎𝟎×5922= 0.013 Kbal = 0.167
Compression reinforcement is NOT required
𝑧 = 𝑑 0.25 −𝐾
1.134= 0.99𝑑 0.95d
𝐴𝑠,𝑟𝑒𝑞 =𝑀𝐸𝑑
0.87𝑓𝑦𝑘𝑧=
158×106
0.87×500×0.95×592= 𝟔𝟒𝟕mm2/m As,min = 988 mm2/m
Example 3: Design of Combined Footing
Minimum & Maximum Area of Reinforcement
𝐴𝑠,𝑚𝑖𝑛 = 0.26𝑓𝑐𝑡𝑚𝑓𝑦𝑘𝑏𝑑 = 0.26
3.2
5000.0017𝑏𝑑 ≥ 0.0013𝑏𝑑
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/mProvide H12-100 (As,prov = 1131 mm2/m)
For secondary bar reinforcement:Provide H12-100 (As,prov = 1131 mm2/m)
(i) Vertical Shear
Critical shear at 1.0d from face of column:
𝑉𝐸𝑑 =1.02
1.62× 964 = 𝟔𝟎𝟓 kN
𝑘 = 1 +200
𝑑= 1 +
200
604= 1.6 2.0
𝜌𝑙 =𝐴𝑠𝑙𝑏𝑑=2828
2700 × 604= 0.0017 ≤ 0.02
𝑉𝑅𝑑,𝑐 = 0.12𝑘 100𝜌𝑙𝑓𝑐𝑘1/3 𝑏𝑑
= 0.12 × 1.6 100 × 0.0017 × 35 1/3 2700 × 604 = 562369 N = 562.4 kN
𝑉𝑚𝑖𝑛 = 0.035𝑘3/2 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
587 676
934846
964
868749
1.43
1.62
1082VEd
d = 0.604
1.02
(ii) Punching Shear
Critical shear at 2.0d from face of column:
Average 𝑑 =604+592
2= 598 mm
2d = 1196 mm
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
2d = 1196 400
400
2d = 1196
104
27
00
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
Punching shear resistance:
𝑘 = 1 +200
𝑑= 1 +
200
609= 1.57 2.0
𝑣𝑅𝑑,𝑐 = 𝑣𝑚𝑖𝑛 = 0.035𝑘3/2𝑓𝑐𝑘
1/2
= 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
Aperimeter = 6.57 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
Maximum shear resistance:
𝑉𝑅𝑑,𝑚𝑎𝑥 = 0.5𝑢𝑑 0.6 1 −𝑓𝑐𝑘250
𝑓𝑐𝑘1.5
= 0.5 × 1600 × 598 0.6 1 −35
250
35
1.5
= 2880 kN VEd,max OK
Soil pressure = 219.8 kN/m2
Acolumn = 0.16 m2
Example 3: Design of Combined Footing
Cracking
h = 650 mm 200 mm
Assume steel stress is under quasi-permanent loading:
= 0.6𝑓𝑦𝑘
1.15
𝐴𝑠,𝑟𝑒𝑞
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= 109mm 250 mm
Actual bar spacing 2 = 2700−2 40 −12
24= 109mm 250 mm
Actual bar spacing 3 = 100 mm 250 mm
Cracking OK
Max bar spacing
Example 3: Design of Combined Footing
Example 4
DESIGN OF STRAP FOOTING
Example 4: Design of Strap Footing
• 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
NA
Gk = 800 kN, Qk = 275 kN
700
Column size: 300 300 mm
S
6000
Beam size: 300 900 mm
Column size: 300 300 mm
2000
H
S
RA RB
NB
Gk = 1000 kN, Qk = 525 kN
Loading
Example 4: Design of Strap Footing
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
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
Size
Example 4: Design of Strap Footing
NA = 1075 kN
0.7 m
S
6 m
2 m
H
S
RA RB
NB = 1525 kN
0.3 m0.3 m
𝑀@𝐵 = 0
𝑅𝐴 −𝑊𝐴 6.15 − 1.0 − 𝑁𝐴 6.0 −𝑊𝑅 3.0 = 0𝑅𝐴 − 108 6.15 − 1.0 − 1075 6.0 − 38 3.0= 0𝑅𝐴 − 108 = 1275 RA = 1382 kN
𝑅𝐴2.0𝐻= 200
1382
2.0𝐻= 200 H = 3.46 mm
Footing A size: 2.00 × 3.50 m
WA WB
WR
𝑅𝐴 + 𝑅𝐵 = 𝑁𝐴 + 𝑁𝐵 +𝑊𝐴 +𝑊𝐵 +𝑊𝑅𝑅𝐴 + 𝑅𝐵 = 1075 + 1525 + 108 + 153 + 38𝑅𝐵 = 2898 − 𝑅𝐴𝑅𝐵 = 2898 − 1382 RA = 1516 kN
𝑅𝐵𝑆2= 200
1516
𝑆2= 200 S = 2.75 mm
Footing B size: 2.75 × 2.75 m
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
Footing Self Weight
Example 4: Design of Strap Footing
NA = 1075 kN6 m
2 m
NB = 1525 kN
X’
0.3 m
WA = 123 kN WB = 132 kNWR = 38 kN
2.75 m
0.3 m
R = 2893 kN
WA = 25 (2.0 3.5 0.7) = 123 kNWB = 25 (2.75 2.75 0.7) = 132 kN
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
Analysis
Design Load:
NA = (1.35 800) + (1.50 275) = 1493 kNNB = (1.35 1000) + (1.50 525) = 2138 kNWA = 1.35 123 = 165 kNWB = 1.35 132 = 179 kNWR = 1.35 38 = 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
Example 4: Design of Strap Footing
1493 kN
-1227
2138 kN
0.3 0.3
276.5 −165
2.0×3.5= 253 × 3.5 = 885 kN/m 276.5 −
179
2.75×2.75= 253 × 2.75 = 695 kN/m
52
5.7= 9.1 kN/m
1.2251.2252.7751.7
2621.40
237
1077
-852
.
-1044
492 522
-1004
-313 BMD (kNm)
SFD (kN)
Example 4: Design of Strap Footing
DESIGN OF FOOTING A
Main Reinforcement
1.6 m1.6 m
w = 506 kN/m
0.3 m
MEd
𝑴𝑬𝒅 = 𝟓𝟎𝟔 × 𝟏. 𝟔 ×𝟏. 𝟔
𝟐= 647 kNm
Soil pressure, w = 253 2.0 m = 506 kN/m
Effective depth: d = h – c – 0.5bar = 700 – 40 – (0.5 16) = 652 mm
Example 4: Design of Strap Footing
𝐾 =𝑀𝐸𝑑
𝑓𝑐𝑘𝑏𝑑2 =
647×106
30×2000×6522= 0.025 Kbal = 0.167
Compression reinforcement is NOT required
𝑧 = 𝑑 0.25 −𝐾
1.134= 0.98𝑑 0.95d
𝐴𝑠,𝑟𝑒𝑞 =𝑀𝐸𝑑
0.87𝑓𝑦𝑘𝑧=
647×106
0.87×500×0.95×652= 𝟐𝟒𝟎𝟐mm2/m As,min
Minimum & Maximum Area of Reinforcement
𝐴𝑠,𝑚𝑖𝑛 = 0.26𝑓𝑐𝑡𝑚𝑓𝑦𝑘𝑏𝑑 = 0.26
2.90
5000.0015𝑏𝑑 ≥ 0.0013𝑏𝑑
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
Critical shear at 1.0d from face of column:
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
2.552
0.948d=
0.6
52
2.0
3.5
Example 4: Design of Strap Footing
(i) Vertical Shear
𝑘 = 1 +200
𝑑= 1 +
200
652= 1.55 2.0
𝜌𝑙 =𝐴𝑠𝑙𝑏𝑑= 0
𝑉𝑅𝑑,𝑐 = 0.12𝑘 100𝜌𝑙𝑓𝑐𝑘1/3 𝑏𝑑
= 0.12 × 1.55 100 × 0 × 30 1/3 3500 × 652 = 0 N = 0 kN
𝑉𝑚𝑖𝑛 = 0.035𝑘3/2 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
Assume steel stress is under quasi-permanent loading:
= 0.59𝑓𝑦𝑘
1.15
𝐴𝑠,𝑟𝑒𝑞
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= 146mm 200 mm
Cracking OK
Max bar spacing
Example 4: Design of Strap Footing
DESIGN OF FOOTING B
Main Reinforcement
1.225 m1.225 m
w = 695.75 kN/m
0.3 m
MEd
𝑴𝑬𝒅 = 𝟔𝟗𝟓. 𝟕𝟓 × 𝟏. 𝟐𝟐𝟓 ×𝟏. 𝟐𝟐𝟓
𝟐= 522 kNm
Soil pressure, w = 253 2.75 m = 695.75 kN/m
Effective depth: d = h – c – 1.5bar = 700 – 40 – (1.5 16) = 636 mm
Example 4: Design of Strap Footing
𝐾 =𝑀𝐸𝑑
𝑓𝑐𝑘𝑏𝑑2 =
522×106
30×2750×6362= 0.016 Kbal = 0.167
Compression reinforcement is NOT required
𝑧 = 𝑑 0.25 −𝐾
1.134= 0.99𝑑 0.95d
𝐴𝑠,𝑟𝑒𝑞 =𝑀𝐸𝑑
0.87𝑓𝑦𝑘𝑧=
522×106
0.87×500×0.95×636= 𝟏𝟗𝟖𝟓mm2/m As,min = 2623 mm2
Minimum & Maximum Area of Reinforcement
𝐴𝑠,𝑚𝑖𝑛 = 0.26𝑓𝑐𝑡𝑚𝑓𝑦𝑘𝑏𝑑 = 0.26
2.90
5000.0015𝑏𝑑 ≥ 0.0013𝑏𝑑
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
Critical shear at 1.0d from face of column:
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
2.161
0.589d=
0.6
36
2.7
5
2.75
Example 4: Design of Strap Footing
(i) Vertical Shear
𝑘 = 1 +200
𝑑= 1 +
200
636= 1.56 2.0
𝜌𝑙 =𝐴𝑠𝑙𝑏𝑑= 0
𝑉𝑅𝑑,𝑐 = 0.12𝑘 100𝜌𝑙𝑓𝑐𝑘1/3 𝑏𝑑
= 0.12 × 1.56 100 × 0 × 30 1/3 2750 × 636 = 0 N = 0 kN
𝑉𝑚𝑖𝑛 = 0.035𝑘3/2 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
Assume steel stress is under quasi-permanent loading:
= 0.54𝑓𝑦𝑘
1.15
𝐴𝑠,𝑟𝑒𝑞
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= 189mm 200 mm
Cracking OK
Max bar spacing
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
PILE
PILE
PILE
Lower Density
Medium Density
High Density
Pile Cap
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 pilesb) Cast in-situ reinforced concrete pilesc) Timber pilesd) Steel pilese) Bakau piles
Design of Pile Foundation
Determination of Pile Capacity
• Safe load:a) Determined from test loading of a pile or using a
pile formulab) Ultimate load divided by safety factor between 2 &
3c) 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 & SpacingFoundation Subjected to Axial Load Only
Load applied at centroid of the group of piles:
𝑭𝒂 =𝑵+𝑾
𝒏
N = Axial load from columnW = Self weight of pile capn = Number of piles
Design of Pile Foundation
Determination of Piles Number & SpacingFoundation 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
𝒏±𝑴𝒙𝒊𝑰𝒚
Fai = Load per pile iM = Momentxi = Distance from pile i to centroid of pile capIy = Moment of inertial of pile group = 𝑥1
2 + 𝑥22 +⋯+ 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
𝒂𝒗where 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 0.5𝑣1𝑓𝑐𝑑𝑢𝑑 =
0.5𝑣1𝑓𝑐𝑘
1.5𝑢𝑑 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
/5 Vertical shear @ 1.0d from column face
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
Links for starter bars
Horizontal links around up-standing end of main bars (12 mm @ 250 mm, say)
Vertical links between bars extending from piles
Starter bars for columns
Bars projecting into cap from piles
Main bars design to resist tensile forces
Example 5
DESIGN OF PILE CAP – TRUSS THEORY
Example 5: Design of Pile Cap – Truss Theory
• 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
N
Example 5: Design of Pile Cap – Truss Theory
Size of Pile Cap
Service load = 3500 + 2500 = 6000 kNAssumed self weight of pile cap, say W = 200 kNPile 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 = 1500mm
Width, B = 𝑘 + 1 ∅𝑝 + 300 = 3 + 1 × 500 + 300 = 2300mm
Length, H = 𝑘 + 1 ∅𝑝 + 300 = 3 + 1 × 500 + 300 = 2300mm
Depth, h = 2∅𝑝 + 100 = 2 500 + 100 = 1100mm
Example 5: Design of Pile Cap – Truss Theory
Size of Pile Cap (continued)
Try size: B H h = 2.3 2.3 1.1 mSelf weight = 25(2.3 2.3 1.1) = 145 kN 200 kN OK
av = 750 – 250 – 250 + 500/5 = 350
p/5 = 100
400
40
04
00
15
00
400 1500
Critical shear section
h=
11
00
d
Example 5: Design of Pile Cap – Truss Theory
Main Reinforcement
Effective depth, d = h – c – 1.5bar = 1100 – 75 – 1.5(25) = 988 mmUltimate load, N = 1.35Gk + 1.5Qk = 1.35(3500) + 1.5(2500) = 8475 kN
From truss analogy:
Tension force, 𝑇 =𝑁𝑙
8𝑑=8475×1.5
8×0.988= 𝟏𝟔𝟎𝟗 kN
Area of reinforcement, 𝐴𝑠 =𝑇
0.87𝑓𝑦𝑘=1609×103
0.87×500= 𝟑𝟔𝟗𝟗mm2
For the whole width of pile cap, 𝐴𝑠 = 2 × 3699 = 𝟕𝟑𝟗𝟗mm2
Example 5: Design of Pile Cap – Truss Theory
Main Reinforcement (continued)
Minimum & Maximum Area of Reinforcement
𝐴𝑠,𝑚𝑖𝑛 = 0.26𝑓𝑐𝑡𝑚𝑓𝑦𝑘𝑏𝑑 = 0.26
2.90
5000.0015𝑏𝑑 ≥ 0.0013𝑏𝑑
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)
Shear
(i) Vertical Shear
Critical shear at p/5 section inside pile:
Load per pile = 8475
4= 2119 kN
2 pile outside critical section Shear force, VEd = 2 2119 = 4238 kN
Pile spacing 3p 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
Example 5: Design of Pile Cap – Truss Theory
av = 750 – 250 – 250 + 500/5 = 350
100
400
40
04
00
15
00
400 1500
Critical section
475
h=
11
00
1388
Critical section
475
Shear
𝑘 = 1 +200
𝑑= 1 +
200
988= 1.45 2.0
𝑉𝑅𝑑,𝑐 = 𝑉𝑚𝑖𝑛 = 0.035𝑘3/2 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 3p
No punching shear check in necessary
Example 5: Design of Pile Cap – Truss Theory
Shear
(iii) Maximum Punching Shear at Column Perimeter
𝑉𝑅𝑑,𝑚𝑎𝑥 = 0.5 0.6 1 −𝑓𝑐𝑘250
𝑓𝑐𝑘1.5𝑢𝑑
= 0.5 0.6 1 −30
250
30
1.54 × 500 × 988 = 10428 kN VEd,max = 8475 kN
OK
Example 5: Design of Pile Cap – Truss Theory
Cracking
h = 1100 mm 200 mm
Assume steel stress is under quasi-permanent loading:
= 0.50𝑓𝑦𝑘
1.15
𝐴𝑠,𝑟𝑒𝑞
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= 142mm 200 mm
Cracking OK
Max bar spacing
Example 5: Design of Pile Cap – Truss Theory
Detailing
Example 5: Design of Pile Cap – Truss Theory
h=
11
00
5H16 Binders
500 mm spun pile
400
40
04
00
15
00
400 1500
16H25 @ 140 mm
16H25 @ 140 mm
Example 5
DESIGN OF PILE CAP – BEAM THEORY
Example 5: Design of Pile Cap – Beam Theory
• 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 mmo Working load = 600 kN
• Safety factor = 1.40
N
M
Example 5: Design of Pile Cap – Beam Theory
Size of Pile Cap
Service load = 4200
1.40= 3000 kN
Assumed self weight of pile cap, say W = 200 kNPile 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
Service load = 4200
1.40= 3000 kN
Assumed self weight of pile cap, say W = 200 kNPile 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
Service load = 4200
1.40= 3000 kN
Assumed self weight of pile cap, say W = 200 kNPile 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 = 900mm
Width, B = 𝑘 + 1 ∅𝑝 + 300 = 3 + 1 × 300 + 300 = 1500mm
Length, H = 2𝑘 + 1 ∅𝑝 + 300 = 2 3 + 1 300 + 300 = 2400mm
Depth, h = 2∅𝑝 + 500 = 2 300 + 500 = 1100mm
Example 5: Design of Pile Cap – Beam Theory
Size of Pile Cap (continued)
Try size: B H h = 1.5 2.4 1.1 mSelf weight = 25(1.5 2.4 1.1) = 99 kN 200 kN OK
I = 4(0.9)2 = 3.24 m2
av = 1200 – 200 – 450 + 300/5 = 610
p/5 = 60
300
30
03
00
90
0
900
Critical shear section
900300
x
x
yy
Example 5: Design of Pile Cap – Beam Theory
Analysis
Service moment, Mservice = 75
1.40= 53.6 kNm
Maximum service load per pile, 𝐹 =𝑁+𝑊
𝑛+𝑀𝑥
𝐼
=3000+99
6+53.6×0.90
3.24= 531 kN 600 kN OK
Ultimate load per pile:
𝐹𝑥𝑥 =𝑁𝑢𝑙𝑡
𝑛+𝑀𝑢𝑙𝑡𝑥
𝐼=4200
6+75×0.90
3.24= 𝟕𝟐𝟏 kN
𝐹𝑦𝑦 =𝑁𝑢𝑙𝑡
𝑛=4200
6= 𝟕𝟎𝟎 kN
Maximum moment at column face:𝑀𝑥𝑥 = 2 721 × 0.90 − 0.20 = 𝟏𝟎𝟎𝟗 kNm𝑀𝑦𝑦 = 3 700 × 0.45 − 0.20 = 𝟓𝟐𝟓 kNm
Example 5: Design of Pile Cap – Beam Theory
Main Reinforcement
Effective depth:dx = h – c – 0.5bar = 1100 – 75 – 0.5(16) = 1017 mmdy = h – c – 1.5bar = 1100 – 75 – 1.5(16) = 1001 mm
Longitudinal Bar
𝐾 =𝑀𝐸𝑑
𝑓𝑐𝑘𝑏𝑑2=
1009×106
30×1500×10172= 0.022 Kbal = 0.167
Compression reinforcement is NOT required
𝑧 = 𝑑 0.25 −𝐾
1.134= 0.98𝑑 0.95d
𝐴𝑠,𝑟𝑒𝑞 =𝑀𝐸𝑑
0.87𝑓𝑦𝑘𝑧=
1009×106
0.87×500×0.95×1017= 𝟐𝟒𝟎𝟏mm2
Example 5: Design of Pile Cap – Beam Theory
Minimum & Maximum Area of Reinforcement
𝐴𝑠,𝑚𝑖𝑛 = 0.26𝑓𝑐𝑡𝑚𝑓𝑦𝑘𝑏𝑑 = 0.26
2.90
5000.0015𝑏𝑑 ≥ 0.0013𝑏𝑑
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
𝑓𝑐𝑘𝑏𝑑2 =
525×106
30×2400×10012= 0.007 Kbal = 0.167
Compression reinforcement is NOT required
𝑧 = 𝑑 0.25 −𝐾
1.134= 0.9𝑑 0.95d
𝐴𝑠,𝑟𝑒𝑞 =𝑀𝐸𝑑
0.87𝑓𝑦𝑘𝑧=
525×106
0.87×500×0.95×1001= 𝟏𝟐𝟔𝟗mm2
Example 5: Design of Pile Cap – Beam Theory
Minimum & Maximum Area of Reinforcement
𝐴𝑠,𝑚𝑖𝑛 = 0.26𝑓𝑐𝑡𝑚𝑓𝑦𝑘𝑏𝑑 = 0.26
2.90
5000.0015𝑏𝑑 ≥ 0.0013𝑏𝑑
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)
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 3p 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
Example 5: Design of Pile Cap – Beam Theory
1316
Critical section
315
av = 610
60
300
30
03
00
90
0
300 900
Critical section
315
900
h=
11
00
Shear
𝑘 = 1 +200
𝑑= 1 +
200
1001= 1.45 2.0
𝑉𝑅𝑑,𝑐 = 𝑉𝑚𝑖𝑛 = 0.035𝑘3/2 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 3p
No punching shear check in necessary
Example 5: Design of Pile Cap – Beam Theory
Shear
(iii) Maximum Punching Shear at Column Perimeter
𝑉𝑅𝑑,𝑚𝑎𝑥 = 0.5 0.6 1 −𝑓𝑐𝑘250
𝑓𝑐𝑘1.5𝑢𝑑
= 0.5 0.6 1 −30
250
30
1.54 × 400 × 1017 = 8456 kN VEd,max = 4200 kN
OK
Example 5: Design of Pile Cap – Beam Theory
Cracking
h = 1100 mm 200 mm
Assume steel stress is under quasi-permanent loading:
= 0.55𝑓𝑦𝑘
1.15
𝐴𝑠,𝑟𝑒𝑞
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 = 2300−2 75 −16
17= 131mm 200 mm
Actual bar spacing 2 = 1500−2 75 −16
11= 121mm 200 mm
Cracking OK
Max bar spacing
Example 5: Design of Pile Cap – Beam Theory
Detailing
Example 5: Design of Pile Cap – Beam Theory
300
30
03
00
90
0
900 900300
12H16
18H16
h=
11
00
4H12 Binders