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MUHAMMAD ZAHID 12-FT-MSC-STR-07 FOUNDATION BEAM FOOTING Dr. Liaqat Ali Qureshi

Foundation Beam Footing

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7/27/2019 Foundation Beam Footing

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M U H A M M A D Z A H I D

1 2 - F T - M S C - S T R - 0 7

FOUNDATION BEAM FOOTING 

Dr. Liaqat Ali Qureshi

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FOUNDATION BEAM FOOTING 

To resist the contact pressure, in this type of footing, a slab strip is provided at the bottom for

more than one columns, which in turn is supported by a beam running along the column centre line.

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Following points may be consider in design of  beam footing………… 

1. The point of application of the resultant loadmust coincide with the centroid of the contactarea.

2. The soil pressure may approximately be assumedto be uniform.

3. Only one way shears in the beam and in theoverhanging slab are to be satisfied and there areno chances of two way shear failure.

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Following points may be consider in design of  beam footing………… 

4. The longitudinal section of beam is then considered with uniformly distributed load equal to qn,u X Bacting over the full length.

5. If the BMD do not change in sagging-hogging sensealong the length of the beam at mid-span & columnsection, this type of footing must be seriously analyzed.

6. The minimum width of foundation beam is to betaken equal to column width + 75mm on both side.

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Following points may be consider in design of  beam footing………… 

7. The depth of beam usually lies b/w  ℓ/10 & ℓ/8, ascompared to with ℓ/12 for superstructure due toheavy loads.

7. The beam may either be designed as rectangularsections as a conservative approach or the mid-spanmay be considered as T-beams.

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D E S I G N A C O M B I N E D F O O T I N G U S I N G F O U N D A T I O N B E A M F O R  T H E A R R A N G E M E N T O F C O L U M N S S H O W N I N F I G U R E . F C ’ = 2 0

M P A , F Y = 3 0 0 M P A A N D N E T A L L O W A B L E B E A R I N GC A P A C I T Y I S E Q U A L T O 8 0 K P A . T H E S I Z E O F A L L C O L U M N S I S 3 0 0

X 3 0 0 M M .  

Problem

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Length and loading for footing in problem

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Solution…… 

L = 19.3 m

qn,u = 80 kPa

P1, P5 = 300 kN

P2, P3, P4 = 650 kN Pu1, Pu5 = 400 kN

Pu2, Pu3, Pu4 = 875 kN

∑P = 2250 kN

∑Pu = 3425 kN f c’ = 20 Mpa

fy   = 300MPa

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Solution…… 

2,

,

2

,

,

m32.81LB

7.1651.13.19

875.31)(

875.3180

2550

   

 prov f  

 sayreq f  

an

req f  

 A

mm L

 A B

mq

 P  A

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kPa A

 P 

q  prov f  u 39.10481.32

3425,

figure in shown are

diagramsmoment bending and forceshear The/46.177mkN  Bqw uu

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7/27/2019 Foundation Beam Footing

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349.77kN0.252)(0.15177.46-469.55Vu 

force,shear design 

10390.81 beamsreinforcedsinglyfor d 

mm52575-h d 

600mm)(saymm8

l hbeam,of Depth 

mm450275300bbeamof WidthLet,

6

min

mm46020450205.0

5948/4750

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Bars#257Use 

mm31900A ,0135.0

1509.3 bd

Mum,-kN390.81Mu

00467.04.1

retangular  betosectionsallconsider DesignFor 

2s

2

min

  

  

 MPa

 f   y

7/27/2019 Foundation Beam Footing

http://slidepdf.com/reader/full/foundation-beam-footing 14/20kN 134.71 

1000/525x450x200.17x0.75V

Bars25# 4 Use mm 1890A 0800.0

 9582.1 bdMu

m,-kN 242.88Mu

cv

2

s

2

 

  

 MPa

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cc mm140s

 use

mm1411000x134.71)-(349.77

525x300x258x0.75 

f  A iv)

mm600 iii)

mm262.5 2

d ii)

mm491450x

300x258 

f  A i)

:valuesfour followingtheof least s

mm258 A

stirrups13#legged-2Using

yvv

yv

max

2

v

cu

w

V V 

b

 

 

35.035.0

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1000x134.71)-35.274(

525x300x142x0.75 

df A stirrups10#legged-2for s

kN274.351.1x177.46-469.55Vu

:is1.1m)(xsupportof edgefrom5of distanceaatforceShear 

yvv

max

cvuV V   

 

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mm175d

mm)250(saymm21875143hminmm,143d

0.625x104.391000

dx1000x200.17x0.75x8.0

xqVx0.8

m0.6252

0.45-1.7 

2

 b-B length,Cantilever 

Design

ucv

 

Slab

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(om0.625m0.4231000

1320

300x0.485 13#for 

cc mm250@13#Use

mmm 500 250x1000x0.0026,A

mmm 455 175x1000x0.0026A

0.0026 MPa,0.6658175x100010x20.39 

m-kN20.392

0.625x104.39 

2

m ins

2

s

2

6

2

2

u

u

bd  M 

 M 

  

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