Bscode Footing

DESIGN OF RC PAD WITH SINGLE COLUMN FOOTING Internal column Column size

=

400 mm

400 mm

Column spacing

=

6m

6m

0.4 m on plan

The unfactored column loads are given in the following table DEAD Vertical Load N(KN)

IMPOSED

WIND

610

480

0

0

0

42

Horizontal Shear H x(KN) Horizontal Shear H y(KN)

38

Moment Mx(KNm)

95

Moment My(KNm)

105

suitable bearing stratum at 1000 mm below ground level.Medium dense silty sand PROCEDURE STEP 1: select type and depth of foundation Type: Reinforced concrete pad with single RC column Depth: 1000 mm below finished ground level 1150 mm below finished floor level Depth selected from considerations of: * Frost action * Swelling of soil * Suitable bearing stratum STEP 2 : Select approximate size presumed allowable bearing capacity from BS 8004: 1986 Maximum Vertical Load V =

1090 KN

Maximum eccentricity ex =

0.1 m

6ex

0.6 m

=

150 KN/m2


V/150

=

7.3 m2

Assume a 3m x 3m foundation pad with area of 9 m 2: Area of footing

=

9 m2

hale the foundation

=

4.5 m2

A =

3m

Determination of Minimum Thickness of pad Assume grade of concrete

C30

=

Vmax = 0.8(fcu)0.5 or 5 N/mm=

4.38 N/mm2

U0 = 2(Cx + Cy)

1600 mm

=

30 N/mm2

Total Factored Load from column Nu = 1.4 DL + 1.6 IL

=

1622 KN

d rel="nofollow"> Nu/vmax U0 (OR) 1/2[(C=

430.71 mm

C1 = U0/6

=

266.67 mm

C2 = Nu/12vc

=

assume Vc assume Vc=0.45 N/mm

300370.37 mm2

= 2

1622000 N

0.45 N/mm2

which corresponds to about 0.3% tension reinforcement

for fcu=30 N/mm 2. choose overall depth of pad equal to 500 mm allowing for adequate cover overall depth

=

500 mm

0.5 m

Effective depth

=

450 mm

0.45 m

STEP 3: Calculate bearing capacity of soil allowable bearing capacity = STEP 4: calculate column load combinations Bearing pressure calculations LC1 = 1.0 DL+1.0 IL

190 KN/m2

LC3 = 1.0 DL+1.0 IL+1.0 WL LC1:combined vertical column load,N Hx=0;Hy=0;Mx=0;My=0 LC3: N (vertical load)

=

=

1090 KN

1090 KN

by inspection,wind in one direction only may be checked for a square foundation Bending Moment and shear calculations LC5: 1.4 DL+1.6 IL LC6: 1.2 DL+1.2 IL+1.2WL LC7: 1.4 DL+1.4 WL LC5: Nu

=

1622 KN

LC 6: Nu

=

1308 KN

Hxu

=

50.4 KN

Hyu

=

0

Mxu

=

0

My

=

126 KNm

LC7: Nu

=

854 KN

Hxu

=

58.8 KN

Hyu

=

0

My

=

Hxu=0;Hyu=0;Mxu=0;My=0

147 KNm

STEP 5:Calculate approximate settlement

This step may be ignore since the foundations are not connected by groung beams and the differential settlements will h little effect on the design of this foundation STEP 6: carry out analysis for bearing pressure weight of soil

=

18 KN/m3

weight of concrete

=

24 KN/m3

weight of foundation

=

108 KN

Thickness of ground slab =

150 mm

weight of back fill +ground slab

=

surcharge on ground slab

=

113.4 KN 5 KN/m2

weight of surcharge on half foundation Eccentricity of surcharge=

=

22.5 KN

0.75 m

weight of surcharge on full foundation LC1:p(Total vertical Load) =

0.15 m

=

45 KN

1356.4 KN

Hx=0;Hy=0;Mx=0;My=0 LC 3 : P

=

1333.9 KN

Hx

=

42 KN

My

=

105 KNm

Hy

=

0

Mxx

=

0

Myy

=

140.8 KNm

STEP 7:calculate bearing pressure under foundation LC1:p

=

LC2: ex = Myy/P , A/6

=

150.7 KN/m2 0.11 m

p1 = P/AB +6Myy/A2B < 1.25 = x 190(bearing capacity) 179.5 KN/m2

<

190 KN/m2

<

0.5 m

<

237.5 KN/m2

Note: 25% over stress on allowable bearing capacity may be allowed for combinations including wind bearing pressures within allowable limits STEP:8 calculate sliding resistance of foundation Ignore passive resistance because horizontal movement of the foundation should be avoided b

=

17

P

=

831.4 KN

Ps = P tang

=

254 KN

from table For Dead Load only >

63 KN

PH = qA tan pi +cA

=

426 KN

>

254

STEP 9: check combined sliding and bearing P

=

1356.4 KN

Pv

=

1710 KN

Hx

=

42 KN

PHx

=

426 KN

P/Pv + Hx/PHx

=

0.89 <

1

STEP 10: carry out analysis of bearing pressure for bending moment and shear LC5: Nu

=

1622 KN

Pu = Nu +1.4(foundation +back on KN back fill) = fill) +1.6(surcharge 2004 Hxu

=

0

Hyu

=

0

Mxxu

=

0

Myyu

=

0

LC 6:Pu = Nu +1.2(foundation = +back fill+surcharge) 1601 KN Mxxu

=

0

Myyu = My+Hxuh+M*yu

=

168.93 KNm

LC 7: Pu = Nu +1.4(foundation = +back fill)

1164 KN

Mxxu

=

0

Myyu = My+Hxuh

=

173.5 KNm

STEP 11: calculate bearing pressure for bending moment and shear LC5:

p = Pu/AB

=

222.7 KN/m2

LC6 :

p1 = Pu/AB + 6 Myyu/A =

p2 = Pu/AB - 6 Myyu/A2B LC 7:

215.4 KN/m2

=

140.3 KN/m2

p1 = Pu/AB + 6 Myyu=

167.9 KN/m2

p2 = Pu/AB - 6 Myyu/A2B

=

90.8 KN/m2

STEP 12: calculate bending moments and shears in pad pd

LC 5:

downward load on pad

=

pd

=

self weight of pad +back fill+ surcharge

upward load on pad

=

pu

pu

=

pressure of ground on pad

pd

=

42.4 KN/m2

pu

=

222.7 KN/m2

222.7 KN/m2

Cantilever overhang at section 1-1 l

1300 mm

bending moment at section 1-1: M1=(p =

457.1 KNm

Shear at section 1; V1= (pu = assume d

1.3 m

703.2 KN

=

425 mm

Shear at section 2: V2 = (p=

473.3 KN

Shear at section 3: V3 =(p =

243.4 KN

LC6: pressure at section 1-1

42.4 KN/m2

0.43 m

182.9 KN/m2

=

Pu/AB

=

177.9 KN/m2

6Myyu/A2B

=

37.5 KN/m2

pd

=

35.5 KN/m2

Bending Moment,M1

=

428.6 KNm

The shears at sections 1,1 and 3 need not be checked.By inspection they will be less critical than LC5

LC7 need not be checked .By inspection it will not be critical STEP 13: Determine cover to reinforcement from SI report,total SO3 class of exposure

0.50%

=

3

75 mm blinding concrete will be used Minimum cover on blinding concrete=

50 mm

Assume 16mm diameter HT type 2 deformed bars effective depth of top layer (symmetrical reinforcement in both directions) d

=

426 mm

dia of bars

=

16 mm

STEP 14: calculate area of tensile reinforcement Maximum bending moment on section 1-1

=

Grade of concrete (f cu)

=

30 N/mm2

Grade of steel (f y)

=

460 N/mm2

K = M/fcubd2

=

0.03

z = d[0.5+sqrt(0.25-K/0.9)] = Ast = M/0.87fyd

=

405 mm 2820 mm2

use 15 no.s 16 dia type HT bars in each direction distribution of tension reinforcement Cx

=

400 mm

Cy

=

400 mm

Assuming dia of bars

=

16 mm

dx =(overall depth - cover =-0.5 x dia of bar)

442 mm

457.1 KNm

dy=(overall depth - cove - dia bar) = of bar - 0.5 x dia of 426 mm 1.5(Cy+3dy)

=

2517 mm

<

ly

=

3000

1.5(Cx+ 3 dx)

=

2589 mm

<

lx

=

3000

2/3 Ast

=

1880 mm2

reinforcement over central = and Cx+ 3dx

=

reinforcement

=

Cy+3 dy

1678 mm

1.68 m

1726 mm

1.73 m

1120 mm2/m

use 11 no.16mm dia.bars at 175 mm centers (1149 mm 2/m) over the central zone in each direction. Use 2 no.16mm dia bars on each side outside the central zone Total number of 16mm dia bars on each side outside the central zone Total number of 16 mm bars used =

15(3105 mm2)

all bars are HT type 2 STEP 15:check shear stress see step 12 -LC5 check v1 = V1/bd < 0.8(fcu)0.5 =

0.55 N/mm2

<

4.38 N/mm2

check v2 = V2/bd < 2 vc

=

0.37 N/mm2

<

0.9 N/mm2

Ast

=

15 n0.16 dia bars

Use larger d(442mm) for calculation of p p

=

0.23%

from fig 11.3 for f cu

=

30 N/mm2

vc

=

0.42 N/mm2

no more shear checks are necessary STEP 16:Check punching shear dx

=

442 mm

=

3015 mm2

or

dy

=

426 mm

d = 0.5 (dx + dy)

=

434 mm

U0 = 2(Cx + Cy)

=

1600 mm

U1= (U0+12d)

=

6808 mm

v0 = Nu/U0d <(fcu)0.5

=

2.34 N/mm2

Nu

=

1622 KN

p1 = pu-pd

=

180.3 KN/m2

A1 = (Cx +3.0 dx)(Cy+3.0 dy=

<

4.38 N/mm2

<

2.9 m2

v1=Nu-p1A1/U1d

=

0.37 N/mm2

vc

=

0.42 N/mm2

STEP 17: check minimum reinforcement for flexure Minimum tensile reinforcement

=

1950 mm2

<

no top tension in pad foundation STEP 18:Check spacing of reinforcement percentage reinforcement,p Maximum spacing

=

=

0.23% 750 mm

not exceeded

STEP 19:Check early thermal cracking R

=

0.15 say

T1

=

alpha

=

12 x 10-6/C

er

=

4.032 x 10-5

x

=

250 mm

acr

=

148 mm

28

assumed

3015 mm2

Wmax

=

0.01 mm

<

0.3 mm

STEP 20:check minimum reinforcement to distributr thermal cracking Top reinforcement

=

2625 mm2

Bottom reinforcement

=

1050 mm2

over

3000 mm

STEP 21: check crack width due to flexure servicebility limit state Loading condition LC1 pu

=

150.7 KN/m2

pd

=

29.6 KN/m2

M

=

307 KN m

x

=

99 mm

z

=

393 mm

fs

=

259 N/mm2

Es

=

1.295 *10-3

Eh

=

1.588*10-3

emh

=

0.773*10-3

acr

=

66 mm

Wmax

=

0.24 mm

<

0.3 mm

STEP 22: design mass concrete foundation not required STEP 23: calculate settlement Load combination LC13: LC13

=

P

=

1.0 DL+0.5IL 1093.9 KN

vertical loads only

grass foundation pressure= weight of soil removed

=

qn

=

121.5 KN/m2 162 KN 103.5 KN/m2

STEP 24: design connection of pad to column Foundation design over

COLUMN FOOTING

0.4 m

B=

3m

fferential settlements will have

mm mm

5 N/mm2

5 N/mm2

provided