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DESIGN OF FOUNDATIONS

Dr. Izni Syahrizal bin IbrahimFaculty 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 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

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

𝑓𝑐𝑘

uo = Column perimeter

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

Check for Shear

• May fail in shear as vertical shear or punching shear

Vertical shear sections

Punching shear perimeters

Bends may be required

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

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

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

Shear stress, 𝒗𝑬𝒅 =𝜷𝑽𝑬𝒅

𝒖𝟏∙𝒅

where; = factor used to include effect of eccentric load & bending moment =

1 + 𝑘𝑀𝐸𝑑

𝑉𝐸𝑑

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

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

Column size: 300 300 mm

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

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

= 122 kN/m2 150 kN/m2 OK

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

Analysis

Ultimate axial force, NEd = 1.35Gk + 1.5Qk

= 1.35 (600) + 1.5 (400) = 1410 kN

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

𝑴𝑬𝒅 = 𝟒𝟕𝟎 × 𝟏. 𝟑𝟓 ×𝟏. 𝟑𝟓

𝟐= 428 kNm

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

Minimum & Maximum Area of Reinforcement

𝐴𝑠,𝑚𝑖𝑛 = 0.26𝑓𝑐𝑡𝑚𝑓𝑦𝑘𝑏𝑑 = 0.26

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)

(i) Vertical Shear

Critical shear at 1.0d from face of column:

Design shear force, VEd = 470 0.953 = 448 kN

0.953 m

3 m953 mm1.35 m

w = 470 kN/m

(i) Vertical Shear

𝑘 = 1 +200

𝑑= 1 +

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

(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

(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) = 3.10 m2

2d = 806 300

2d = 806

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) = 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

(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

= 2176 kN VEd,max OK

Acolumn = 0.09 m2

Aall = 9 m2

Cracking

h = 450 mm 200 mm

Steel stress, 𝑓𝑠 =𝐺𝑘+0.3𝑄𝑘

1.35𝐺𝑘+1.5𝑄𝑘

𝐴𝑠.𝑟𝑒𝑞

𝐴𝑠,𝑝𝑟𝑜𝑣

𝑓𝑦𝑘

=600+0.3×400

1.35×600+1.5×400

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

Detailing

24H123000

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

• Assumed bar = 12 mm

Axial Force, N = 1500 kNMoment = 50 kNm

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

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

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

Size (Continued)

𝐼𝑥𝑥 =𝐵𝐻3

12=2.8×3.53

12= 10.0m4

𝑦 =𝐻

2= 1.75m

Maximum soil pressure, 𝑃 =𝑁+𝑊

𝐴+𝑀𝑦

𝐼=1071+159

9.80+50×1.75

= 132 kN/m2 150 kN/m2 OK

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

0.35 m

162154

Analysis (Continued)

𝑀𝑥𝑥

= 154 ×1.5752

+ 162 − 1541.575

2× 1.575 ×

= 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

0.35 m

162154

=144 + 162

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

553×106

30×2800×6092= 0.018 Kbal = 0.167

𝑧 = 𝑑 0.25 −𝐾

1.134= 0.98𝑑 0.95d

𝐴𝑠,𝑟𝑒𝑞 =𝑀𝑥𝑥

553×106

0.87×500×0.95×609= 𝟐𝟏𝟗𝟕mm2

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)

Main Reinforcement – Transverse Bar

𝐾 =𝑀𝑦𝑦

435×106

30×3500×5972= 0.018 Kbal = 0.167

𝑧 = 𝑑 0.25 −𝐾

1.134= 0.99𝑑 0.95d

𝐴𝑠,𝑟𝑒𝑞 =𝑀𝑦𝑦

435×106

0.87×500×0.95×597= 𝟏𝟕𝟔𝟓mm2

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)

(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

162159

0.966d=

(i) Vertical Shear

𝑘 = 1 +200

𝑑= 1 +

609= 1.57 2.0

𝜌𝑙 =𝐴𝑠𝑙𝑏𝑑= 0

= 0.12 × 1.57 100 × 0 × 30 1/3 2800 × 609 = 0 N = 0 kN

= 0.035 × 1.573/2 30 2800 × 609 = 644949 𝑁 = 645 kN

(ii) Punching Shear

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

2d = 1206 350

2d = 1206

(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

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

Aall = 2.8 3.5 m

(ii) Punching Shear

𝑘 = 1 +200

𝑑= 1 +

609= 1.57 2.0

𝑣𝑅𝑑,𝑐 = 𝑣𝑚𝑖𝑛 = 0.035𝑘3/2𝑓𝑐𝑘

= 0.035 1.57 3/2 30 1/2

= 0.38 N/mm2 vEd (0.11 N/mm2) OK

Aall = 2.8 3.5 m

Maximum punching shear force:VEd,max = 1500 kN

Column perimeter, uo = 2(350 + 250) = 1200 mm

𝑣𝐸𝑑 =𝛽𝑉𝐸𝑑𝑢𝑜𝑑

Where 𝛽 = 1 + 𝑘𝑀𝐸𝑑

𝑉𝐸𝑑

𝑢𝑜

k = 0.65 𝑐1

𝑐2=350

250= 1.4

W1 = 0.5𝑐12 + 𝑐1𝑐2

= 0.5(3502) + 350 × 250 = 0.15 106 mm2

Aall = 2.8 3.5 m = 9.8 m2

Acolumn = 0.09 m2

𝛽 = 1 + 0.6550×106

1500×1031200

0.15×106= 1.17

Therefore, 𝒗𝑬𝒅 =𝟏.𝟏𝟕×𝟏𝟓𝟎𝟎×𝟏𝟎𝟑

𝟏𝟐𝟎𝟎×𝟔𝟎𝟑= 𝟐. 𝟒𝟒 N/mm2

𝑣𝑅𝑑,𝑚𝑎𝑥 = 0.5 0.6 1 −𝑓𝑐𝑘250

𝑓𝑐𝑘1.5

= 0.5 0.6 1 −30

= 5.28 N/mm2 vEd OK

Aall = 2.8 3.5 m = 9.8 m2

Acolumn = 0.09 m2

Cracking

h = 650 mm 200 mm

Assume steel stress is under quasi-permanent loading:

= 0.6𝑓𝑦𝑘

𝐴𝑠,𝑟𝑒𝑞

𝐴𝑠,𝑝𝑟𝑜𝑣= 0.6

2375= 241 N/mm2

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

Detailing

21H123500

Plan View Section View

Example 3

DESIGN OF COMBINED FOOTING

Example 3: Design of Combined Footing

NA = 1610 kN(Ultimate)

NB = 1950 kN(Ultimate)

• fck = 35 N/mm2

• fyk = 500 N/mm2

• soil = 200 N/mm2

• Cover = 40 mm• Assumed bar = 12 mm

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

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

Size (Continued)

Check Service Soil Bearing Capacity = 𝑁+𝑊

𝐴=2543+252.7

= 173 kN/m2 200 kN/m2 OK

Arrange position of footing so that the distribution of soil pressure is uniform:

NA = 1150 kN NB = 1393 kN

Ntotal = 2543 kN

MA = 01393(3.4 ) + 2543x = 0x = 1.86 m

Analysis

𝐴=(1610+1950)

16.2= 219.8 kN/m2

Soil pressure per m width, w = 219.8 2.7 m = 593.5 kN/m

Shear Force & Bending Moment Diagram

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

289250

430 473

BMD (kNm)

587 676

934 846

1082964

868 749

1.43 1.62

SFD (kN)

Main Reinforcement

Longitudinal Reinforcement: Bottom

Effective depth: dx = h – c – 0.5bar = 650 – 40 – (0.5 12) = 604 mm

𝐾 =𝑀𝐸𝑑

473×106

35×2700×6042= 0.014 Kbal = 0.167

𝑧 = 𝑑 0.25 −𝐾

1.134= 0.99𝑑 0.95d

473×106

0.87×500×0.95×604= 𝟏𝟖𝟗𝟒mm2 As,min = 2722 mm2

Main Reinforcement

Longitudinal Reinforcement: Top

𝐾 =𝑀𝐸𝑑

353×106

35×2700×6042= 0.010 Kbal = 0.167

𝑧 = 𝑑 0.25 −𝐾

1.134= 0.99𝑑 0.95d

353×106

0.87×500×0.95×604= 𝟏𝟒𝟏𝟑mm2 As,min = 2722 mm2

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)

Main Reinforcement

Transverse Reinforcement: Bottom

1.2 m1.2 m

w = 219.8 kN/m

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

Main Reinforcement

Transverse Reinforcement: Bottom

Effective depth: dy = h – c – 1.5bar = 650 – 40 – (1.5 12) = 592 mm

𝐾 =𝑀𝐸𝑑

158×106

35×𝟏𝟎𝟎𝟎×5922= 0.013 Kbal = 0.167

𝑧 = 𝑑 0.25 −𝐾

1.134= 0.99𝑑 0.95d

158×106

0.87×500×0.95×592= 𝟔𝟒𝟕mm2/m As,min = 988 mm2/m

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

𝑉𝐸𝑑 =1.02

1.62× 964 = 𝟔𝟎𝟓 kN

𝑘 = 1 +200

𝑑= 1 +

604= 1.6 2.0

𝜌𝑙 =𝐴𝑠𝑙𝑏𝑑=2828

2700 × 604= 0.0017 ≤ 0.02

= 0.12 × 1.6 100 × 0.0017 × 35 1/3 2700 × 604 = 562369 N = 562.4 kN

= 0.035 × 1.63/2 35 2700 × 604 = 6667734 𝑁 = 667 kN

587 676

934846

868749

1082VEd

d = 0.604

(ii) Punching Shear

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

2d = 1196

(ii) Punching Shear

Average punching shear force at control perimeter:VEd = 1950 – (219.8 6.57) = 506 kN

𝑣𝐸𝑑 =𝑉𝐸𝑑

𝑢𝑑=506×103

9116×598= 0.09 N/mm2

𝑘 = 1 +200

𝑑= 1 +

609= 1.57 2.0

𝑣𝑅𝑑,𝑐 = 𝑣𝑚𝑖𝑛 = 0.035𝑘3/2𝑓𝑐𝑘

= 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

Maximum punching shear force:VEd,max = 1950 kN

Column perimeter, uo = 4 400 = 1600 mm

𝑉𝑅𝑑,𝑚𝑎𝑥 = 0.5𝑢𝑑 0.6 1 −𝑓𝑐𝑘250

𝑓𝑐𝑘1.5

= 0.5 × 1600 × 598 0.6 1 −35

= 2880 kN VEd,max OK

Soil pressure = 219.8 kN/m2

Acolumn = 0.16 m2

Cracking

h = 650 mm 200 mm

= 0.6𝑓𝑦𝑘

2828= 175 N/mm2

Actual bar spacing 1 = 2700−2 40 −12

24= 109mm 250 mm

Actual bar spacing 3 = 100 mm 250 mm

Cracking OK

Max bar spacing

Example 4

DESIGN OF STRAP FOOTING

Example 4: Design of Strap Footing

• fck = 30 N/mm2

• fyk = 500 N/mm2

• soil = 200 N/mm2

• Cover = 40 mm• Assumed bar = 12 mm

Gk = 800 kN, Qk = 275 kN

Beam size: 300 900 mm

Gk = 1000 kN, Qk = 525 kN

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

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

NA = 1075 kN

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

2.0𝐻= 200 H = 3.46 mm

Footing A size: 2.00 × 3.50 m

𝑅𝐴 + 𝑅𝐵 = 𝑁𝐴 + 𝑁𝐵 +𝑊𝐴 +𝑊𝐵 +𝑊𝑅𝑅𝐴 + 𝑅𝐵 = 1075 + 1525 + 108 + 153 + 38𝑅𝐵 = 2898 − 𝑅𝐴𝑅𝐵 = 2898 − 1382 RA = 1516 kN

𝑅𝐵𝑆2= 200

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

Footing Self Weight

NA = 1075 kN6 m

NB = 1525 kN

WA = 123 kN WB = 132 kNWR = 38 kN

2.75 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

Shear Force & Bending Moment Diagram

1493 kN

2138 kN

0.3 0.3

276.5 −165

2.0×3.5= 253 × 3.5 = 885 kN/m 276.5 −

2.75×2.75= 253 × 2.75 = 695 kN/m

5.7= 9.1 kN/m

1.2251.2252.7751.7

2621.40

492 522

-313 BMD (kNm)

SFD (kN)

DESIGN OF FOOTING A

Main Reinforcement

1.6 m1.6 m

w = 506 kN/m

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

𝐾 =𝑀𝐸𝑑

647×106

30×2000×6522= 0.025 Kbal = 0.167

𝑧 = 𝑑 0.25 −𝐾

1.134= 0.98𝑑 0.95d

647×106

0.87×500×0.95×652= 𝟐𝟒𝟎𝟐mm2/m As,min

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)

(i) Vertical Shear

Design shear force, VEd = 253 0.948 2.0 = 479.4 kN

0.948d=

(i) Vertical Shear

𝑘 = 1 +200

𝑑= 1 +

652= 1.55 2.0

= 0.12 × 1.55 100 × 0 × 30 1/3 3500 × 652 = 0 N = 0 kN

= 0.035 × 1.553/2 30 3500 × 652 = 484194 𝑁 = 484.2 kN

VEd (479.4 kN) Vmin (484.2 kN) OK

Cracking

h = 700 mm 200 mm

= 0.59𝑓𝑦𝑘

2815= 219 N/mm2

13= 146mm 200 mm

Cracking OK

Max bar spacing

DESIGN OF FOOTING B

Main Reinforcement

1.225 m1.225 m

w = 695.75 kN/m

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

𝐾 =𝑀𝐸𝑑

522×106

30×2750×6362= 0.016 Kbal = 0.167

𝑧 = 𝑑 0.25 −𝐾

1.134= 0.99𝑑 0.95d

522×106

0.87×500×0.95×636= 𝟏𝟗𝟖𝟓mm2/m As,min = 2623 mm2

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)

(i) Vertical Shear

Design shear force, VEd = 253 0.589 2.75 = 409.5 kN

0.589d=

(i) Vertical Shear

𝑘 = 1 +200

𝑑= 1 +

636= 1.56 2.0

= 0.12 × 1.56 100 × 0 × 30 1/3 2750 × 636 = 0 N = 0 kN

= 0.035 × 1.563/2 30 2750 × 636 = 653774 𝑁 = 653.8 kN

VEd (409.5 kN) Vmin (653.8 kN) OK

Cracking

h = 700 mm 200 mm

= 0.54𝑓𝑦𝑘

3016= 204 N/mm2

14= 189mm 200 mm

Cracking OK

Max bar spacing

DESIGN OF TIE BEAM

Design similar to beam design. Refer to RC1.

Moment Design for Flexural

Reinforcement

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

Load from Structure

Lower Density

Medium Density

High Density

Pile Cap

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

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

Determination of Piles Number & Spacing

• Pile load Single pile capacity• Piles are usually arranged symmetrically with respect to

the column axis

Determination of Piles Number & SpacingFoundation Subjected to Axial Load Only

Load applied at centroid of the group of piles:

𝑭𝒂 =𝑵+𝑾

N = Axial load from columnW = Self weight of pile capn = Number of piles

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

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

Design of Pile Cap

Design for Shear

/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

• 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

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

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 1500

Critical shear section

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

Main Reinforcement (continued)

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

(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 ×

2×988= 𝟕𝟓𝟏 kN

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 1500

Critical section

𝑘 = 1 +200

𝑑= 1 +

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

(ii) Punching Shear at Perimeter 2.0d from Column Face

Since pile spacing 3p

No punching shear check in necessary

𝑉𝑅𝑑,𝑚𝑎𝑥 = 0.5 0.6 1 −𝑓𝑐𝑘250

𝑓𝑐𝑘1.5𝑢𝑑

= 0.5 0.6 1 −30

1.54 × 500 × 988 = 10428 kN VEd,max = 8475 kN

Cracking

h = 1100 mm 200 mm

= 0.50𝑓𝑦𝑘

7855= 205 N/mm2

15= 142mm 200 mm

Cracking OK

Max bar spacing

Detailing

5H16 Binders

500 mm spun pile

400 1500

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

• 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

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

𝐹𝑎=(3000 + 200)

600= 5.3 Use n = 6

Size of Pile Cap

Service load = 4200

1.40= 3000 kN

𝐹𝑎=(3000 + 200)

600= 5.3 Use n = 6

Size of Pile Cap

Service load = 4200

1.40= 3000 kN

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

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

Critical shear section

900300

Analysis

Service moment, Mservice = 75

1.40= 53.6 kNm

Maximum service load per pile, 𝐹 =𝑁+𝑊

𝑛+𝑀𝑥

=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

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

𝑧 = 𝑑 0.25 −𝐾

1.134= 0.98𝑑 0.95d

1009×106

0.87×500×0.95×1017= 𝟐𝟒𝟎𝟏mm2

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

Transverse Bar

𝐾 =𝑀𝐸𝑑

525×106

30×2400×10012= 0.007 Kbal = 0.167

𝑧 = 𝑑 0.25 −𝐾

1.134= 0.9𝑑 0.95d

525×106

0.87×500×0.95×1001= 𝟏𝟐𝟔𝟗mm2

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

(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 ×

2×1017= 𝟒𝟑𝟐 kN

Note:Bar extend beyond critical section at = 315 + 1001 = 1316 mm 𝑙𝑏𝑑 + 𝑑 = 36∅ + 𝑑 = 36 × 16 + 1001 = 1577 mm Asl = 0 mm2

Critical section

av = 610

300 900

Critical section

𝑘 = 1 +200

𝑑= 1 +

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

(ii) Punching Shear at Perimeter 2.0d from Column Face

Since pile spacing 3p

No punching shear check in necessary

𝑉𝑅𝑑,𝑚𝑎𝑥 = 0.5 0.6 1 −𝑓𝑐𝑘250

𝑓𝑐𝑘1.5𝑢𝑑

= 0.5 0.6 1 −30

1.54 × 400 × 1017 = 8456 kN VEd,max = 4200 kN

Cracking

h = 1100 mm 200 mm

= 0.55𝑓𝑦𝑘

2413= 238 N/mm2

17= 131mm 200 mm

11= 121mm 200 mm

Cracking OK

Max bar spacing

Detailing

900 900300

4H12 Binders

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