SlideShare a Scribd company logo
1 of 19
Download to read offline
LECTURE EIGHT
DESIGN OF COMBINED
PAD FOUNDATIONS
1
Copyright ©Structures Centre, 2020. All rights reserved
25th October, 2020
Time: 13:00-13:45 Hrs
Introduction
▪ Spread foundations supporting two or more columns are classified as combined
footings or combined bases.
▪ Combined footings can become a necessity for two reasons
▪ Instances where two columns are so closely spaced such that designing isolated pad footings
under each column will result in a base overlap.
▪ Instances where the external columns are very close to a property line such that
constructing isolated pad bases is impossible without a projection beyond the line.
2
Principles
▪ The design of combined bases can be done using two approach, the flexibility method
and the conventional rigid method. The latter is most common method.
▪ The conventional rigid approach makes two assumptions:
▪ The base is infinitely rigid and therefore the deflection of the footing does not influence the
pressure distribution.
▪ The footing is proportioned such that the resultant of all applied loads pass through the
centroid of the base.
▪ The design of combined bases in principle is the same procedure as for an isolated
pad foundation.
3
Steps for Design
▪ Determine the required size of the combined base using the permissible bearing
stress and the critical loading at the serviceability limit state.
▪ Proportion the footing for uniform pressure distribution by finding the point of
application of the resultant of the column loads and adjusting the geometry of the
footing appropriately.
▪ Determine the bearing pressures associated with the critical loading arrangement at
the ultimate limit state.
▪ Assume a suitable thickness for the footing, h and effective depth, d and verify
punching shear at both column faces
▪ Analyze the footing in the longitudinal and transverse direction.
▪ In the longitudinal/transverse direction, the footing is idealized as an inverted overhanging
beam subjected to the earth pressure at the ultimate limit state.
▪ Design the bending reinforcement in the longitudinal/transverse direction
▪ Verify shear at the critical sections including punching shear at the basic control
perimeter 2d from the column face.
▪ Verify detailing requirements
4
Worked Example
A combined footing is required in a proposed facility for two closely spaced
column. Design the footing completely using a presumed bearing capacity of
180kN/m2 using C20/25 concrete and reinforcing steel of 410Mpa bars. The design
data is presented as follows:
Table 1.0 Design Data
Figure 1.0: Column Layout
5
Column Size Gk (kN) Qk (kN)
C1 275x275 365 185
C2 300x300 525 290
Solution Cont’d
▪ Serviceability Limit State
𝑁 = 1.0𝐺 𝑘 + 1.0𝑄 𝑘
▪ For Column C1:
▪ 𝐺 𝑘 = 365𝑘𝑁 𝑄 𝑘 = 185𝑘𝑁
▪ 𝑁 = 1.0𝐺 𝑘 + 1.0𝑄 𝑘 = 365 + 185 = 550𝑘𝑁
▪ For Column C2:
▪ 𝐺 𝑘 = 525𝑘𝑁 𝑄 𝑘 = 290𝑘𝑁
▪ 𝑁 = 1.0𝐺 𝑘 + 1.0𝑄 𝑘 = 525 + 290 = 815𝑘𝑁
Assume 10% increase in axial actions to actions to account for footing self weight
𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑎𝑑 = 1 + 0.1 × 550 + 815 = 1501.5𝑘𝑁
𝐴𝑟𝑒𝑎 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 =
𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑎𝑑
𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
=
1501.5
180
= 8.34𝑚2
6
Solution Cont’d
▪ Serviceability Limit State Cont’d
Assume a width of 1.8m, length of base =
8.34
1.8
= 4.63
Provide a Base of 4.7m × 1.8m ( say 0.5m deep) 𝐴𝑟𝑒𝑎 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 = 8.46𝑚2
▪ Proportioning the Base
▪ Longitudinal Direction
Take moment about the centrelines of column C1
815 × 2.75 − 𝑅𝑒1 = 0
𝑅 = 815 + 550 = 1365𝑘𝑁
𝑒1 =
815 × 2.75
1365
= 1.64
𝑒2 = 2.75 − 1.64 = 1.11m
𝑦1 =
4.7
2
− 𝑒1 = 2.35 − 1.64 = 0.71
𝑦2 =
4.7
2
− 𝑒2 = 2.35 − 1.11 = 1.24
7
Solution Cont’d
▪ Ultimate Limit State
𝑁𝐸𝑑 = 1.35𝐺 𝑘 + 1.5𝑄 𝑘
▪ For Column C1
▪ 𝐺 𝑘 = 365𝑘𝑁 𝑄 𝑘 = 185𝑘𝑁
▪ 𝑁 𝐸𝑑,1 = (1.35 × 365) + (1.5 × 185) = 770.25𝑘𝑁
▪ For Column C2
▪ 𝐺 𝑘 = 365𝑘𝑁 𝑄 𝑘 = 185𝑘𝑁
▪ 𝑁 𝐸𝑑,2 = (1.35 × 525) + (1.5 × 290) = 1143.75𝑘𝑁
𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 =
𝐷𝑒𝑠𝑖𝑔𝑛 𝑎𝑥𝑖𝑎𝑙 𝑎𝑐𝑡𝑖𝑜𝑛
𝐴𝑟𝑒𝑎 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑
=
𝑁𝐸𝑑,1+ 𝑁 𝐸𝑑,2
𝐴
=
770.25 + 1143.75
8.46
= 226.24𝑘𝑁/𝑚2
8
Solution Cont’d
▪ Punching check at Column face
Assume cover to reinforcement is 50mm and reinforcement of 16mm
𝑑 = ℎ − 𝑐 +
𝜙
2
= 500 − 50 +
16
2
= 442𝑚𝑚
𝑉𝑒𝑟𝑖𝑓𝑦 𝑡ℎ𝑎𝑡 𝑁 𝐸𝑑 ≤ 𝑉𝑅𝑑,𝑚𝑎𝑥
▪ For Column C1:
▪ 𝑉𝑅𝑑,𝑚𝑎𝑥 = 0.2 1 −
𝑓 𝑐𝑘
250
𝑓𝑐𝑘 𝑝 𝑜 𝑑 = 0.2 1 −
20
250
20 × 4 × 275 × 442 × 10−3 = 1789.2𝑘𝑁
▪ Thus (𝑁 𝐸𝑑= 770.25𝑘𝑁) < 𝑉𝑅𝑑,𝑚𝑎𝑥 = 1789.2kN O. K
▪ For Column C2:
▪ 𝑉𝑅𝑑,𝑚𝑎𝑥 = 0.2 1 −
𝑓 𝑐𝑘
250
𝑓𝑐𝑘 𝑝 𝑜 𝑑 = 0.2 1 −
20
250
20 × 4 × 300 × 442 × 10−3 = 1951.9𝑘𝑁
▪ Thus (𝑁 𝐸𝑑= 1143.75𝑘𝑁) < 𝑉𝑅𝑑,𝑚𝑎𝑥 = 1951.9kN O. K
9
Solution Cont’d
▪ Structural Analysis
▪ Longitudinal Y-Direction
10
Solution Cont’d
▪ Structural Analysis
▪ Transverse Z-Direction
11
Solution Cont’d
▪ Flexural Design
▪ Longitudinal Direction
▪ Hogging in Spans
𝑀 𝐸𝑑 = 102.4kN. m
𝑘 =
𝑀 𝐸𝑑
𝑏𝑑2 𝑓𝑐𝑘
=
102.4 × 106
103 × 4422 × 20
= 0.026 < 0.168
𝑧 = 𝑑 0.5 + 0.25 − 0.882𝑘 ≤ 0.95𝑑
= 0.95𝑑 = 0.95 × 442 = 419.9𝑚𝑚
𝐴 𝑠 =
𝑀 𝐸𝑑
0.87𝑓𝑦𝑘 𝑧
=
102.4 × 106
0.87 × 410 × 419.9
= 683.7𝑚𝑚2/m
𝑇𝑟𝑦 𝑌12 − 150 𝑐𝑒𝑛𝑡𝑟𝑒𝑠 𝑇𝑜𝑝 𝐴 𝑠,𝑝𝑟𝑜 = 753𝑚𝑚2/𝑚
12
Solution Cont’d
▪ Flexural Design
▪ Longitudinal Direction
▪ Sagging at Column Interface
𝑀 𝐸𝑑 = 173.93kN. m
𝑘 =
𝑀 𝐸𝑑
𝑏𝑑2 𝑓𝑐𝑘
=
173.93 × 106
103 × 4422 × 20
= 0.045 < 0.168
𝑧 = 𝑑 0.5 + 0.25 − 0.882𝑘 ≤ 0.95𝑑
= 0.95𝑑 = 0.95 × 442 = 419.9𝑚𝑚
𝐴 𝑠 =
𝑀 𝐸𝑑
0.87𝑓𝑦𝑘 𝑧
=
173.93 × 106
0.87 × 410 × 419.9
= 1161.2𝑚𝑚2/m
𝑇𝑟𝑦 𝑌16 − 10 𝑐𝑒𝑛𝑡𝑟𝑒𝑠 𝐵𝑜𝑡𝑡𝑜𝑚 𝐴 𝑠,𝑝𝑟𝑜 = 1340𝑚𝑚2/𝑚
13
Solution Cont’d
▪ Flexural Design
▪ Transverse Direction
▪ Hogging in Span 𝑀 𝐸𝑑 = 0kN. m ( Hence Provide Nominal Reinforcement)
▪ Sagging at Column Interface
𝑀 𝐸𝑑 = 91.6kN. m
𝑘 =
𝑀 𝐸𝑑
𝑏𝑑2 𝑓𝑐𝑘
=
91.6 × 106
103 × 4422 × 20
= 0.023 < 0.168
𝑧 = 𝑑 0.5 + 0.25 − 0.882𝑘 ≤ 0.95𝑑
= 0.95𝑑 = 0.95 × 442 = 419.9𝑚𝑚
𝐴 𝑠 =
𝑀 𝐸𝑑
0.87𝑓𝑦𝑘 𝑧
=
91.6 × 106
0.87 × 410 × 419.9
= 611.57𝑚𝑚2/m
𝑇𝑟𝑦 𝑌12 − 150 𝑐𝑒𝑛𝑡𝑟𝑒𝑠 𝑁𝑒𝑎𝑟 𝐵𝑜𝑡𝑡𝑜𝑚 𝐴 𝑠,𝑝𝑟𝑜 = 753𝑚𝑚2
/𝑚
14
Solution Cont’d
▪ Punching Shear at Basic Control Perimeter 2.0d from column face
Punching at the basic control perimeter is not a critical check in combined bases, hence
the check is ignored in this example, however a check can be carried out using the same
procedures outlined in the design of isolated pad bases for each column.
▪ Transverse shear
The maximum shear force is used to carry out the check. By Inspection, this occurs in the
longitudinal direction.
▪ N:B The design shear force is taken at 1.0d from the supports.
𝑉𝐸𝑑 = 353.6 − 𝑝 1.0d = 353.6 − 226.24 × 0.442 = 253.6𝑘𝑁
𝑣 𝐸𝑑 =
𝑉 𝐸𝑑
𝑏𝑑
=
253.6×103
1000×442
= 0.57Mpa
15
Solution Cont’d
▪ Transverse shear Cont’d
𝑣 𝑅𝑑,𝑐 = 0.12𝑘(100𝜌𝑓𝑐𝑘)1/3
≥ 0.035𝑘3/2
𝑓𝑐𝑘
𝜌 =
𝐴 𝑠
𝑏𝑑
=
1340
1000×442
= 0.0030 ; 𝑘 = 1 +
200
𝑑
= 1 +
200
442
= 1.67 < 2
𝑣 𝑅𝑑,𝑐 = 0.12 × 1.67(100 × 0.0030 × 20)1/3 ≥ 0.035 × 1.67
3
2 20 = 0.36𝑀𝑝𝑎
𝑆𝑖𝑛𝑐𝑒 (𝑣 𝐸𝑑 = 0.57𝑀𝑝𝑎) > (𝑣 𝑅𝑑,𝑐= 0.42𝑀𝑝𝑎) 𝑁𝑜𝑡 𝑜𝑘
The section is not adequate in shear, hence the section depth can be increased to 600mm,
or reinforcement increased, or concrete grade increased and then redesigned.
16
Solution Cont’d
▪ Verify Minimum Area of Steel
𝐴 𝑠,𝑚𝑖𝑛 =
0.078𝑓𝑐𝑘
2/3
𝑓𝑦𝑘
𝑏𝑡d ≥ 0.0013𝑏𝑡 𝑑
𝐴 𝑠,𝑚𝑖𝑛 =
0.078(20)2/3
410
× 1000 × 442 ≥ 0.0013 × 1000 × 144
= 619.6𝑚𝑚2 < (753𝑚𝑚2, 1340𝑚𝑚2) o. k
▪ Summary of Reinforcement
12Y16-150mm Centers (Bottom)
31Y12-150mm Centers (Near Bottom)
12Y12-150mm Centers (Top)
31Y12-150mm Centers (Near Top).
17
Quiz
A combined footing is required in a proposed facility for two closely spaced
column. Design the footing completely using a presumed bearing capacity of
150kN/m2 using C30/37 concrete and reinforcing steel of 460Mpa bars. The design
data is presented as follows:
Table 1.0 Design Data
Figure 1.0: Column Layout
18
Column Size Gk (kN) Qk (kN)
C1 275x275 365 185
C2 300x300 525 290
▪Questions
19

More Related Content

What's hot

Worked example extract_flat_slabs
Worked example extract_flat_slabsWorked example extract_flat_slabs
Worked example extract_flat_slabs
Luan Truong Van
 
Flow Chart - One Way Joist Construction
Flow Chart - One Way Joist ConstructionFlow Chart - One Way Joist Construction
Flow Chart - One Way Joist Construction
Qamar Uz Zaman
 
INDUSTRIAL BUILDING GANTRY GIRDER
INDUSTRIAL BUILDING  GANTRY GIRDERINDUSTRIAL BUILDING  GANTRY GIRDER
INDUSTRIAL BUILDING GANTRY GIRDER
Harsh Shani
 

What's hot (20)

Compression member
Compression memberCompression member
Compression member
 
Footing design
Footing designFooting design
Footing design
 
Lec 3 design problem of flat plate slab
Lec 3 design problem of flat plate slabLec 3 design problem of flat plate slab
Lec 3 design problem of flat plate slab
 
10 columns
10 columns10 columns
10 columns
 
Worked example extract_flat_slabs
Worked example extract_flat_slabsWorked example extract_flat_slabs
Worked example extract_flat_slabs
 
Koppolu abishek prying action
Koppolu abishek   prying actionKoppolu abishek   prying action
Koppolu abishek prying action
 
Square footing design
Square footing designSquare footing design
Square footing design
 
Singly reinforced beam design
Singly reinforced beam   designSingly reinforced beam   design
Singly reinforced beam design
 
Flow Chart - One Way Joist Construction
Flow Chart - One Way Joist ConstructionFlow Chart - One Way Joist Construction
Flow Chart - One Way Joist Construction
 
Column Interaction Diagram construction
Column Interaction Diagram constructionColumn Interaction Diagram construction
Column Interaction Diagram construction
 
Isolated Footing
Isolated FootingIsolated Footing
Isolated Footing
 
23-Design of Column Base Plates (Steel Structural Design & Prof. Shehab Mourad)
23-Design of Column Base Plates (Steel Structural Design & Prof. Shehab Mourad)23-Design of Column Base Plates (Steel Structural Design & Prof. Shehab Mourad)
23-Design of Column Base Plates (Steel Structural Design & Prof. Shehab Mourad)
 
Design of combined footing ppt
Design of combined footing pptDesign of combined footing ppt
Design of combined footing ppt
 
Design of two way slabs(d.d.m.)
Design of two way slabs(d.d.m.)Design of two way slabs(d.d.m.)
Design of two way slabs(d.d.m.)
 
Footing Design
Footing DesignFooting Design
Footing Design
 
INDUSTRIAL BUILDING GANTRY GIRDER
INDUSTRIAL BUILDING  GANTRY GIRDERINDUSTRIAL BUILDING  GANTRY GIRDER
INDUSTRIAL BUILDING GANTRY GIRDER
 
22-Design of Four Bolt Extended Endplate Connection (Steel Structural Design ...
22-Design of Four Bolt Extended Endplate Connection (Steel Structural Design ...22-Design of Four Bolt Extended Endplate Connection (Steel Structural Design ...
22-Design of Four Bolt Extended Endplate Connection (Steel Structural Design ...
 
Final Presentation
Final PresentationFinal Presentation
Final Presentation
 
Doubly reinforced beam design
Doubly reinforced beam   designDoubly reinforced beam   design
Doubly reinforced beam design
 
Flat slab
Flat slab Flat slab
Flat slab
 

Similar to Design of combined footings

Lecture-5-Slabs-and-Flat-Slabs-PHG-N-Rev13-15-Oct-16 (1).pdf
Lecture-5-Slabs-and-Flat-Slabs-PHG-N-Rev13-15-Oct-16 (1).pdfLecture-5-Slabs-and-Flat-Slabs-PHG-N-Rev13-15-Oct-16 (1).pdf
Lecture-5-Slabs-and-Flat-Slabs-PHG-N-Rev13-15-Oct-16 (1).pdf
ssuserd8a85b
 
Lecture-5-Slabs-and-Flat-Slabs-PHG-N-Rev13-15-Oct-16 (1).pdf
Lecture-5-Slabs-and-Flat-Slabs-PHG-N-Rev13-15-Oct-16 (1).pdfLecture-5-Slabs-and-Flat-Slabs-PHG-N-Rev13-15-Oct-16 (1).pdf
Lecture-5-Slabs-and-Flat-Slabs-PHG-N-Rev13-15-Oct-16 (1).pdf
ssuserd8a85b
 
140204-5-PTI EDC-130-Continuous Members-41.pdf
140204-5-PTI EDC-130-Continuous Members-41.pdf140204-5-PTI EDC-130-Continuous Members-41.pdf
140204-5-PTI EDC-130-Continuous Members-41.pdf
ephrem53
 
Combine-Footing-Design.pptx
Combine-Footing-Design.pptxCombine-Footing-Design.pptx
Combine-Footing-Design.pptx
ssuserd185de
 

Similar to Design of combined footings (20)

Stepped footing
Stepped footingStepped footing
Stepped footing
 
Design of Structures Chapter2.pptx
Design of Structures Chapter2.pptxDesign of Structures Chapter2.pptx
Design of Structures Chapter2.pptx
 
Lecture-5-Slabs-and-Flat-Slabs-PHG-N-Rev13-15-Oct-16.pdf
Lecture-5-Slabs-and-Flat-Slabs-PHG-N-Rev13-15-Oct-16.pdfLecture-5-Slabs-and-Flat-Slabs-PHG-N-Rev13-15-Oct-16.pdf
Lecture-5-Slabs-and-Flat-Slabs-PHG-N-Rev13-15-Oct-16.pdf
 
Lecture-5-Slabs-and-Flat-Slabs-PHG-N-Rev13-15-Oct-16 (1).pdf
Lecture-5-Slabs-and-Flat-Slabs-PHG-N-Rev13-15-Oct-16 (1).pdfLecture-5-Slabs-and-Flat-Slabs-PHG-N-Rev13-15-Oct-16 (1).pdf
Lecture-5-Slabs-and-Flat-Slabs-PHG-N-Rev13-15-Oct-16 (1).pdf
 
Lecture-5-Slabs-and-Flat-Slabs-PHG-N-Rev13-15-Oct-16 (1).pdf
Lecture-5-Slabs-and-Flat-Slabs-PHG-N-Rev13-15-Oct-16 (1).pdfLecture-5-Slabs-and-Flat-Slabs-PHG-N-Rev13-15-Oct-16 (1).pdf
Lecture-5-Slabs-and-Flat-Slabs-PHG-N-Rev13-15-Oct-16 (1).pdf
 
Design of Structures Chapter2.pdf
Design of Structures Chapter2.pdfDesign of Structures Chapter2.pdf
Design of Structures Chapter2.pdf
 
Analysis of two reinforced concrete multistored building
Analysis of two reinforced concrete multistored building Analysis of two reinforced concrete multistored building
Analysis of two reinforced concrete multistored building
 
Week 9 Lecture Material_watermark.pdf
Week 9 Lecture Material_watermark.pdfWeek 9 Lecture Material_watermark.pdf
Week 9 Lecture Material_watermark.pdf
 
DESIGN OF PARTIAL MAGNETIC REPULSION FOUNDATION BUILDING BY USING U-BOOTS AND...
DESIGN OF PARTIAL MAGNETIC REPULSION FOUNDATION BUILDING BY USING U-BOOTS AND...DESIGN OF PARTIAL MAGNETIC REPULSION FOUNDATION BUILDING BY USING U-BOOTS AND...
DESIGN OF PARTIAL MAGNETIC REPULSION FOUNDATION BUILDING BY USING U-BOOTS AND...
 
DESIGN OF CIRCULAR OVERHEAD WATER TANK.pptx
DESIGN OF CIRCULAR OVERHEAD WATER TANK.pptxDESIGN OF CIRCULAR OVERHEAD WATER TANK.pptx
DESIGN OF CIRCULAR OVERHEAD WATER TANK.pptx
 
140204-5-PTI EDC-130-Continuous Members-41.pdf
140204-5-PTI EDC-130-Continuous Members-41.pdf140204-5-PTI EDC-130-Continuous Members-41.pdf
140204-5-PTI EDC-130-Continuous Members-41.pdf
 
Combine-Footing-Design.pptx
Combine-Footing-Design.pptxCombine-Footing-Design.pptx
Combine-Footing-Design.pptx
 
A Project on Transformer Design.pdf
A Project on Transformer Design.pdfA Project on Transformer Design.pdf
A Project on Transformer Design.pdf
 
Rectangular tank
Rectangular tankRectangular tank
Rectangular tank
 
ppt.pptx
ppt.pptxppt.pptx
ppt.pptx
 
RCC BMD
RCC BMDRCC BMD
RCC BMD
 
Design of torsion reinforcement
 Design of torsion reinforcement Design of torsion reinforcement
Design of torsion reinforcement
 
flat Slab
flat Slabflat Slab
flat Slab
 
design-of-mat-sd.pdf
design-of-mat-sd.pdfdesign-of-mat-sd.pdf
design-of-mat-sd.pdf
 
Roll pass design in continuous bar mills
Roll pass design in continuous bar millsRoll pass design in continuous bar mills
Roll pass design in continuous bar mills
 

Recently uploaded

Kuwait City MTP kit ((+919101817206)) Buy Abortion Pills Kuwait
Kuwait City MTP kit ((+919101817206)) Buy Abortion Pills KuwaitKuwait City MTP kit ((+919101817206)) Buy Abortion Pills Kuwait
Kuwait City MTP kit ((+919101817206)) Buy Abortion Pills Kuwait
jaanualu31
 
Introduction to Robotics in Mechanical Engineering.pptx
Introduction to Robotics in Mechanical Engineering.pptxIntroduction to Robotics in Mechanical Engineering.pptx
Introduction to Robotics in Mechanical Engineering.pptx
hublikarsn
 
Cybercrimes in the Darknet and Their Detections: A Comprehensive Analysis and...
Cybercrimes in the Darknet and Their Detections: A Comprehensive Analysis and...Cybercrimes in the Darknet and Their Detections: A Comprehensive Analysis and...
Cybercrimes in the Darknet and Their Detections: A Comprehensive Analysis and...
dannyijwest
 
Integrated Test Rig For HTFE-25 - Neometrix
Integrated Test Rig For HTFE-25 - NeometrixIntegrated Test Rig For HTFE-25 - Neometrix
Integrated Test Rig For HTFE-25 - Neometrix
Neometrix_Engineering_Pvt_Ltd
 

Recently uploaded (20)

Kuwait City MTP kit ((+919101817206)) Buy Abortion Pills Kuwait
Kuwait City MTP kit ((+919101817206)) Buy Abortion Pills KuwaitKuwait City MTP kit ((+919101817206)) Buy Abortion Pills Kuwait
Kuwait City MTP kit ((+919101817206)) Buy Abortion Pills Kuwait
 
Computer Graphics Introduction To Curves
Computer Graphics Introduction To CurvesComputer Graphics Introduction To Curves
Computer Graphics Introduction To Curves
 
Introduction to Artificial Intelligence ( AI)
Introduction to Artificial Intelligence ( AI)Introduction to Artificial Intelligence ( AI)
Introduction to Artificial Intelligence ( AI)
 
Compressing and Sparsifying LLM in GenAI Applications
Compressing and Sparsifying LLM in GenAI ApplicationsCompressing and Sparsifying LLM in GenAI Applications
Compressing and Sparsifying LLM in GenAI Applications
 
Augmented Reality (AR) with Augin Software.pptx
Augmented Reality (AR) with Augin Software.pptxAugmented Reality (AR) with Augin Software.pptx
Augmented Reality (AR) with Augin Software.pptx
 
School management system project Report.pdf
School management system project Report.pdfSchool management system project Report.pdf
School management system project Report.pdf
 
Worksharing and 3D Modeling with Revit.pptx
Worksharing and 3D Modeling with Revit.pptxWorksharing and 3D Modeling with Revit.pptx
Worksharing and 3D Modeling with Revit.pptx
 
fitting shop and tools used in fitting shop .ppt
fitting shop and tools used in fitting shop .pptfitting shop and tools used in fitting shop .ppt
fitting shop and tools used in fitting shop .ppt
 
Unsatisfied Bhabhi ℂall Girls Ahmedabad Book Esha 6378878445 Top Class ℂall G...
Unsatisfied Bhabhi ℂall Girls Ahmedabad Book Esha 6378878445 Top Class ℂall G...Unsatisfied Bhabhi ℂall Girls Ahmedabad Book Esha 6378878445 Top Class ℂall G...
Unsatisfied Bhabhi ℂall Girls Ahmedabad Book Esha 6378878445 Top Class ℂall G...
 
Electromagnetic relays used for power system .pptx
Electromagnetic relays used for power system .pptxElectromagnetic relays used for power system .pptx
Electromagnetic relays used for power system .pptx
 
Design For Accessibility: Getting it right from the start
Design For Accessibility: Getting it right from the startDesign For Accessibility: Getting it right from the start
Design For Accessibility: Getting it right from the start
 
Introduction to Robotics in Mechanical Engineering.pptx
Introduction to Robotics in Mechanical Engineering.pptxIntroduction to Robotics in Mechanical Engineering.pptx
Introduction to Robotics in Mechanical Engineering.pptx
 
Theory of Time 2024 (Universal Theory for Everything)
Theory of Time 2024 (Universal Theory for Everything)Theory of Time 2024 (Universal Theory for Everything)
Theory of Time 2024 (Universal Theory for Everything)
 
Computer Networks Basics of Network Devices
Computer Networks  Basics of Network DevicesComputer Networks  Basics of Network Devices
Computer Networks Basics of Network Devices
 
Cybercrimes in the Darknet and Their Detections: A Comprehensive Analysis and...
Cybercrimes in the Darknet and Their Detections: A Comprehensive Analysis and...Cybercrimes in the Darknet and Their Detections: A Comprehensive Analysis and...
Cybercrimes in the Darknet and Their Detections: A Comprehensive Analysis and...
 
Max. shear stress theory-Maximum Shear Stress Theory ​ Maximum Distortional ...
Max. shear stress theory-Maximum Shear Stress Theory ​  Maximum Distortional ...Max. shear stress theory-Maximum Shear Stress Theory ​  Maximum Distortional ...
Max. shear stress theory-Maximum Shear Stress Theory ​ Maximum Distortional ...
 
Danikor Product Catalog- Screw Feeder.pdf
Danikor Product Catalog- Screw Feeder.pdfDanikor Product Catalog- Screw Feeder.pdf
Danikor Product Catalog- Screw Feeder.pdf
 
Introduction to Serverless with AWS Lambda
Introduction to Serverless with AWS LambdaIntroduction to Serverless with AWS Lambda
Introduction to Serverless with AWS Lambda
 
Databricks Generative AI Fundamentals .pdf
Databricks Generative AI Fundamentals  .pdfDatabricks Generative AI Fundamentals  .pdf
Databricks Generative AI Fundamentals .pdf
 
Integrated Test Rig For HTFE-25 - Neometrix
Integrated Test Rig For HTFE-25 - NeometrixIntegrated Test Rig For HTFE-25 - Neometrix
Integrated Test Rig For HTFE-25 - Neometrix
 

Design of combined footings

  • 1. LECTURE EIGHT DESIGN OF COMBINED PAD FOUNDATIONS 1 Copyright ©Structures Centre, 2020. All rights reserved 25th October, 2020 Time: 13:00-13:45 Hrs
  • 2. Introduction ▪ Spread foundations supporting two or more columns are classified as combined footings or combined bases. ▪ Combined footings can become a necessity for two reasons ▪ Instances where two columns are so closely spaced such that designing isolated pad footings under each column will result in a base overlap. ▪ Instances where the external columns are very close to a property line such that constructing isolated pad bases is impossible without a projection beyond the line. 2
  • 3. Principles ▪ The design of combined bases can be done using two approach, the flexibility method and the conventional rigid method. The latter is most common method. ▪ The conventional rigid approach makes two assumptions: ▪ The base is infinitely rigid and therefore the deflection of the footing does not influence the pressure distribution. ▪ The footing is proportioned such that the resultant of all applied loads pass through the centroid of the base. ▪ The design of combined bases in principle is the same procedure as for an isolated pad foundation. 3
  • 4. Steps for Design ▪ Determine the required size of the combined base using the permissible bearing stress and the critical loading at the serviceability limit state. ▪ Proportion the footing for uniform pressure distribution by finding the point of application of the resultant of the column loads and adjusting the geometry of the footing appropriately. ▪ Determine the bearing pressures associated with the critical loading arrangement at the ultimate limit state. ▪ Assume a suitable thickness for the footing, h and effective depth, d and verify punching shear at both column faces ▪ Analyze the footing in the longitudinal and transverse direction. ▪ In the longitudinal/transverse direction, the footing is idealized as an inverted overhanging beam subjected to the earth pressure at the ultimate limit state. ▪ Design the bending reinforcement in the longitudinal/transverse direction ▪ Verify shear at the critical sections including punching shear at the basic control perimeter 2d from the column face. ▪ Verify detailing requirements 4
  • 5. Worked Example A combined footing is required in a proposed facility for two closely spaced column. Design the footing completely using a presumed bearing capacity of 180kN/m2 using C20/25 concrete and reinforcing steel of 410Mpa bars. The design data is presented as follows: Table 1.0 Design Data Figure 1.0: Column Layout 5 Column Size Gk (kN) Qk (kN) C1 275x275 365 185 C2 300x300 525 290
  • 6. Solution Cont’d ▪ Serviceability Limit State 𝑁 = 1.0𝐺 𝑘 + 1.0𝑄 𝑘 ▪ For Column C1: ▪ 𝐺 𝑘 = 365𝑘𝑁 𝑄 𝑘 = 185𝑘𝑁 ▪ 𝑁 = 1.0𝐺 𝑘 + 1.0𝑄 𝑘 = 365 + 185 = 550𝑘𝑁 ▪ For Column C2: ▪ 𝐺 𝑘 = 525𝑘𝑁 𝑄 𝑘 = 290𝑘𝑁 ▪ 𝑁 = 1.0𝐺 𝑘 + 1.0𝑄 𝑘 = 525 + 290 = 815𝑘𝑁 Assume 10% increase in axial actions to actions to account for footing self weight 𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑎𝑑 = 1 + 0.1 × 550 + 815 = 1501.5𝑘𝑁 𝐴𝑟𝑒𝑎 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 = 𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑎𝑑 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 1501.5 180 = 8.34𝑚2 6
  • 7. Solution Cont’d ▪ Serviceability Limit State Cont’d Assume a width of 1.8m, length of base = 8.34 1.8 = 4.63 Provide a Base of 4.7m × 1.8m ( say 0.5m deep) 𝐴𝑟𝑒𝑎 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 = 8.46𝑚2 ▪ Proportioning the Base ▪ Longitudinal Direction Take moment about the centrelines of column C1 815 × 2.75 − 𝑅𝑒1 = 0 𝑅 = 815 + 550 = 1365𝑘𝑁 𝑒1 = 815 × 2.75 1365 = 1.64 𝑒2 = 2.75 − 1.64 = 1.11m 𝑦1 = 4.7 2 − 𝑒1 = 2.35 − 1.64 = 0.71 𝑦2 = 4.7 2 − 𝑒2 = 2.35 − 1.11 = 1.24 7
  • 8. Solution Cont’d ▪ Ultimate Limit State 𝑁𝐸𝑑 = 1.35𝐺 𝑘 + 1.5𝑄 𝑘 ▪ For Column C1 ▪ 𝐺 𝑘 = 365𝑘𝑁 𝑄 𝑘 = 185𝑘𝑁 ▪ 𝑁 𝐸𝑑,1 = (1.35 × 365) + (1.5 × 185) = 770.25𝑘𝑁 ▪ For Column C2 ▪ 𝐺 𝑘 = 365𝑘𝑁 𝑄 𝑘 = 185𝑘𝑁 ▪ 𝑁 𝐸𝑑,2 = (1.35 × 525) + (1.5 × 290) = 1143.75𝑘𝑁 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = 𝐷𝑒𝑠𝑖𝑔𝑛 𝑎𝑥𝑖𝑎𝑙 𝑎𝑐𝑡𝑖𝑜𝑛 𝐴𝑟𝑒𝑎 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 = 𝑁𝐸𝑑,1+ 𝑁 𝐸𝑑,2 𝐴 = 770.25 + 1143.75 8.46 = 226.24𝑘𝑁/𝑚2 8
  • 9. Solution Cont’d ▪ Punching check at Column face Assume cover to reinforcement is 50mm and reinforcement of 16mm 𝑑 = ℎ − 𝑐 + 𝜙 2 = 500 − 50 + 16 2 = 442𝑚𝑚 𝑉𝑒𝑟𝑖𝑓𝑦 𝑡ℎ𝑎𝑡 𝑁 𝐸𝑑 ≤ 𝑉𝑅𝑑,𝑚𝑎𝑥 ▪ For Column C1: ▪ 𝑉𝑅𝑑,𝑚𝑎𝑥 = 0.2 1 − 𝑓 𝑐𝑘 250 𝑓𝑐𝑘 𝑝 𝑜 𝑑 = 0.2 1 − 20 250 20 × 4 × 275 × 442 × 10−3 = 1789.2𝑘𝑁 ▪ Thus (𝑁 𝐸𝑑= 770.25𝑘𝑁) < 𝑉𝑅𝑑,𝑚𝑎𝑥 = 1789.2kN O. K ▪ For Column C2: ▪ 𝑉𝑅𝑑,𝑚𝑎𝑥 = 0.2 1 − 𝑓 𝑐𝑘 250 𝑓𝑐𝑘 𝑝 𝑜 𝑑 = 0.2 1 − 20 250 20 × 4 × 300 × 442 × 10−3 = 1951.9𝑘𝑁 ▪ Thus (𝑁 𝐸𝑑= 1143.75𝑘𝑁) < 𝑉𝑅𝑑,𝑚𝑎𝑥 = 1951.9kN O. K 9
  • 10. Solution Cont’d ▪ Structural Analysis ▪ Longitudinal Y-Direction 10
  • 11. Solution Cont’d ▪ Structural Analysis ▪ Transverse Z-Direction 11
  • 12. Solution Cont’d ▪ Flexural Design ▪ Longitudinal Direction ▪ Hogging in Spans 𝑀 𝐸𝑑 = 102.4kN. m 𝑘 = 𝑀 𝐸𝑑 𝑏𝑑2 𝑓𝑐𝑘 = 102.4 × 106 103 × 4422 × 20 = 0.026 < 0.168 𝑧 = 𝑑 0.5 + 0.25 − 0.882𝑘 ≤ 0.95𝑑 = 0.95𝑑 = 0.95 × 442 = 419.9𝑚𝑚 𝐴 𝑠 = 𝑀 𝐸𝑑 0.87𝑓𝑦𝑘 𝑧 = 102.4 × 106 0.87 × 410 × 419.9 = 683.7𝑚𝑚2/m 𝑇𝑟𝑦 𝑌12 − 150 𝑐𝑒𝑛𝑡𝑟𝑒𝑠 𝑇𝑜𝑝 𝐴 𝑠,𝑝𝑟𝑜 = 753𝑚𝑚2/𝑚 12
  • 13. Solution Cont’d ▪ Flexural Design ▪ Longitudinal Direction ▪ Sagging at Column Interface 𝑀 𝐸𝑑 = 173.93kN. m 𝑘 = 𝑀 𝐸𝑑 𝑏𝑑2 𝑓𝑐𝑘 = 173.93 × 106 103 × 4422 × 20 = 0.045 < 0.168 𝑧 = 𝑑 0.5 + 0.25 − 0.882𝑘 ≤ 0.95𝑑 = 0.95𝑑 = 0.95 × 442 = 419.9𝑚𝑚 𝐴 𝑠 = 𝑀 𝐸𝑑 0.87𝑓𝑦𝑘 𝑧 = 173.93 × 106 0.87 × 410 × 419.9 = 1161.2𝑚𝑚2/m 𝑇𝑟𝑦 𝑌16 − 10 𝑐𝑒𝑛𝑡𝑟𝑒𝑠 𝐵𝑜𝑡𝑡𝑜𝑚 𝐴 𝑠,𝑝𝑟𝑜 = 1340𝑚𝑚2/𝑚 13
  • 14. Solution Cont’d ▪ Flexural Design ▪ Transverse Direction ▪ Hogging in Span 𝑀 𝐸𝑑 = 0kN. m ( Hence Provide Nominal Reinforcement) ▪ Sagging at Column Interface 𝑀 𝐸𝑑 = 91.6kN. m 𝑘 = 𝑀 𝐸𝑑 𝑏𝑑2 𝑓𝑐𝑘 = 91.6 × 106 103 × 4422 × 20 = 0.023 < 0.168 𝑧 = 𝑑 0.5 + 0.25 − 0.882𝑘 ≤ 0.95𝑑 = 0.95𝑑 = 0.95 × 442 = 419.9𝑚𝑚 𝐴 𝑠 = 𝑀 𝐸𝑑 0.87𝑓𝑦𝑘 𝑧 = 91.6 × 106 0.87 × 410 × 419.9 = 611.57𝑚𝑚2/m 𝑇𝑟𝑦 𝑌12 − 150 𝑐𝑒𝑛𝑡𝑟𝑒𝑠 𝑁𝑒𝑎𝑟 𝐵𝑜𝑡𝑡𝑜𝑚 𝐴 𝑠,𝑝𝑟𝑜 = 753𝑚𝑚2 /𝑚 14
  • 15. Solution Cont’d ▪ Punching Shear at Basic Control Perimeter 2.0d from column face Punching at the basic control perimeter is not a critical check in combined bases, hence the check is ignored in this example, however a check can be carried out using the same procedures outlined in the design of isolated pad bases for each column. ▪ Transverse shear The maximum shear force is used to carry out the check. By Inspection, this occurs in the longitudinal direction. ▪ N:B The design shear force is taken at 1.0d from the supports. 𝑉𝐸𝑑 = 353.6 − 𝑝 1.0d = 353.6 − 226.24 × 0.442 = 253.6𝑘𝑁 𝑣 𝐸𝑑 = 𝑉 𝐸𝑑 𝑏𝑑 = 253.6×103 1000×442 = 0.57Mpa 15
  • 16. Solution Cont’d ▪ Transverse shear Cont’d 𝑣 𝑅𝑑,𝑐 = 0.12𝑘(100𝜌𝑓𝑐𝑘)1/3 ≥ 0.035𝑘3/2 𝑓𝑐𝑘 𝜌 = 𝐴 𝑠 𝑏𝑑 = 1340 1000×442 = 0.0030 ; 𝑘 = 1 + 200 𝑑 = 1 + 200 442 = 1.67 < 2 𝑣 𝑅𝑑,𝑐 = 0.12 × 1.67(100 × 0.0030 × 20)1/3 ≥ 0.035 × 1.67 3 2 20 = 0.36𝑀𝑝𝑎 𝑆𝑖𝑛𝑐𝑒 (𝑣 𝐸𝑑 = 0.57𝑀𝑝𝑎) > (𝑣 𝑅𝑑,𝑐= 0.42𝑀𝑝𝑎) 𝑁𝑜𝑡 𝑜𝑘 The section is not adequate in shear, hence the section depth can be increased to 600mm, or reinforcement increased, or concrete grade increased and then redesigned. 16
  • 17. Solution Cont’d ▪ Verify Minimum Area of Steel 𝐴 𝑠,𝑚𝑖𝑛 = 0.078𝑓𝑐𝑘 2/3 𝑓𝑦𝑘 𝑏𝑡d ≥ 0.0013𝑏𝑡 𝑑 𝐴 𝑠,𝑚𝑖𝑛 = 0.078(20)2/3 410 × 1000 × 442 ≥ 0.0013 × 1000 × 144 = 619.6𝑚𝑚2 < (753𝑚𝑚2, 1340𝑚𝑚2) o. k ▪ Summary of Reinforcement 12Y16-150mm Centers (Bottom) 31Y12-150mm Centers (Near Bottom) 12Y12-150mm Centers (Top) 31Y12-150mm Centers (Near Top). 17
  • 18. Quiz A combined footing is required in a proposed facility for two closely spaced column. Design the footing completely using a presumed bearing capacity of 150kN/m2 using C30/37 concrete and reinforcing steel of 460Mpa bars. The design data is presented as follows: Table 1.0 Design Data Figure 1.0: Column Layout 18 Column Size Gk (kN) Qk (kN) C1 275x275 365 185 C2 300x300 525 290