DESIGN OF RC PAD WITH SINGLE COLUMN FOOTING

Internal column

Column size = 400 mm 400 mm 0.4

Column spacing = 6 m 6 m

The unfactored column loads are given in the following table

DEAD IMPOSED WIND

Vertical Load N(KN) 610 480 0

0 0 42

38

95

105

suitable bearing stratum at 1000 mm below ground level.Medium dense silty sand

PROCEDURE

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

presumed allowable bearing capacity from BS 8004: 1986 150

Maximum Vertical Load = 1090 KN

= 0.1 m

Horizontal Shear Hx(KN)

Horizontal Shear Hy(KN)

Moment Mx(KNm)

Moment My(KNm)

STEP 1: select type and depth of foundation

STEP 2 : Select approximate size

KN/m2

Maximum eccentricity e

= 0.6 m <A(length of footing)

V/150 = 7.3

A = 3

Area of footing = 9

hale the foundation = 4.5

Determination of Minimum Thickness of pad

= C30 30

= 4.38

= 1600 mm

Total Factored Load from column

Nu = 1.4 DL + 1.6 IL = 1622 KN 1622000 N

= 430.71 mm

= 266.67 mm

= 300370.37

= 0.45 N/mm2

which corresponds to about 0.3% tension reinforcement

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

allowable bearing capacity = 190

6ex

m2

Assume a 3m x 3m foundation pad with area of 9 m2:

m2

m2

Assume grade of concrete N/mm2

Vmax = 0.8(fcu)0.5 or 5 N/mm2 whichever is less N/mm2

U0 = 2(Cx + Cy)

d > Nu/vmax U0 (OR) 1/2[(C12+4C2)0.5 - C1]

C1 = U0/6

C2 = Nu/12vc mm2

assume Vc

assume Vc=0.45 N/mm2

for fcu=30 N/mm2.

STEP 3: Calculate bearing capacity of soil

KN/m2

STEP 4: calculate column load combinations

Bearing pressure calculations

LC1:combined vertical column load,N = 1090 KN

LC3: N (vertical load) = 1090 KN

by inspection,wind in one direction only may be checked for a square foundation

Bending Moment and shear calculations

LC5: Nu = 1622 KN

= 1308 KN

= 50.4 KN

= 0

= 0

= 126 KNm

= 854 KN

= 58.8 KN

= 0

= 147 KNm

This step may be ignore since the foundations are not connected by groung beams and the differential settlements will have

little effect on the design of this foundation

LC1 = 1.0 DL+1.0 IL

LC3 = 1.0 DL+1.0 IL+1.0 WL

Hx=0;Hy=0;Mx=0;My=0

LC5: 1.4 DL+1.6 IL

LC6: 1.2 DL+1.2 IL+1.2WL

LC7: 1.4 DL+1.4 WL

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

LC 6: Nu

Hxu

Hyu

Mxu

My

LC7: Nu

Hxu

Hyu

My

STEP 5:Calculate approximate settlement

STEP 6: carry out analysis for bearing pressure

weight of soil = 18

weight of concrete = 24

weight of foundation = 108 KN

Thickness of ground sla = 150 mm 0.15 m

weight of back fill +ground slab = 113.4 KN

surcharge on ground slab = 5

weight of surcharge on half foundation = 22.5 KN

Eccentricity of surcharg = 0.75 m

weight of surcharge on full foundation = 45 KN

LC1:p(Total vertical Loa = 1356.4 KN

LC 3 : P = 1333.9 KN

= 42 KN

= 105 KNm

= 0

= 0

= 140.8 KNm

LC1:p = 150.7 < 190

= 0.106 m < 0.5 m

= 179.5 < 237.5

Note: 25% over stress on allowable bearing capacity may be allowed for combinations including wind

bearing pressures within allowable limits

Ignore passive resistance because horizontal movement of the foundation should be avoided

KN/m3

KN/m3

KN/m2

Hx=0;Hy=0;Mx=0;My=0

Hx

My

Hy

Mxx

Myy

STEP 7:calculate bearing pressure under foundation

KN/m2 KN/m2

LC2: ex = Myy/P , A/6

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

STEP:8 calculate sliding resistance of foundation

b = 17 from table

P = 831.4 KN For Dead Load only

= 254 KN > 63 KN

= 426 KN > 254

P = 1356.4 KN

= 1710 KN

= 42 KN

= 426 KN

= 0.89 < 1

= 1622 KN

= 2004 KN

= 0

= 0

= 0

= 0

= 1601 KN

= 0

= 168.93 KNm

= 1164 KN

= 0

= 173.5 KNm

Ps = P tang

PH = qA tan pi +cA

STEP 9: check combined sliding and bearing

Pv

Hx

PHx

P/Pv + Hx/PHx

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

LC5: Nu

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

Hxu

Hyu

Mxxu

Myyu

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

Mxxu

Myyu = My+Hxuh+M*yu

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

Mxxu

Myyu = My+Hxuh

= 222.7

= 215.4

= 140.3

= 167.9

= 90.8

LC 5: downward load on pad =

= self weight of pad +back fill+ surcharge 42.4

upward load on pad = 222.7

= pressure of ground on pad

= 42.4

= 222.7

Cantilever overhang at section 1 l 1300 mm 1.3 m

= 457.1 KNm

= 703.2 KN

assume d = 425 mm 0.425 m

= 473.3 KN

= 243.4 KN

LC6: pressure at section 1-1 = 182.9

= 177.9

= 37.5

= 35.5

STEP 11: calculate bearing pressure for bending moment and shear

LC5: p = Pu/AB KN/m2

LC6 : p1 = Pu/AB + 6 Myyu/A2B KN/m2

p2 = Pu/AB - 6 Myyu/A2B KN/m2

LC 7: p1 = Pu/AB + 6 Myyu/A2B KN/m2

p2 = Pu/AB - 6 Myyu/A2B KN/m2

STEP 12: calculate bending moments and shears in pad

pd

pd

pu KN/m2

pu

pd KN/m2

pu KN/m2

bending moment at section 1-1: M1=(pu-pd)Bl2/2

Shear at section 1; V1= (p

Shear at section 2: V2 = (p

Shear at section 3: V3 =(p

KN/m2

Pu/AB KN/m2

6Myyu/A2B KN/m2

pd 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

from SI report,total SO3 0.50%

class of exposure = 3

75 mm blinding concrete will be used

Minimum cover on blinding concr= 50 mm

Assume 16mm diameter HT type 2 deformed bars

effective depth of top la (symmetrical reinforcement in both directions)

d = 426 mm

dia of bars = 16 mm

Maximum bending moment on section 1-1 = 457.1 KNm

= 30

= 460

= 0.028

= 405 mm

= 2820

use 15 no.s 16 dia type HT bars in each direction

distribution of tension reinforcement

= 400 mm

STEP 13: Determine cover to reinforcement

STEP 14: calculate area of tensile reinforcement

Grade of concrete (fcu) N/mm2

Grade of steel (fy) N/mm2

K = M/fcubd2

z = d[0.5+sqrt(0.25-K/0.9)]

Ast = M/0.87fyd mm2

Cx

= 400 mm

Assuming dia of bars = 16 mm

= 442 mm

= 426 mm

= 2517 mm < =

= 2589 mm < =

= 1880

reinforcement over centr= 1678 mm 1.678 m

= 1726 mm 1.726 m

reinforcement = 1120

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 use=

all bars are HT type 2

see step 12 -LC5

= 0.55 < 4.38

= 0.37 < 0.9

= 15 n0.16 dia bars = 3015

Use larger d(442mm) for calculation of p

p = 0.23%

= 30

= 0.42

Cy

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

dy=(overall depth - cove - dia of bar - 0.5 x dia of bar)

1.5(Cy+3dy) ly

1.5(Cx+ 3 dx) lx

2/3 Ast mm2

Cy+3 dy

and Cx+ 3dx

mm2/m

use 11 no.16mm dia.bars at 175 mm centers (1149 mm2/m) over the central zone in each direction.

15(3105 mm2)

STEP 15:check shear stress

check v1 = V1/bd < 0.8(fcu)0.5 or 5 N/mm2 N/mm2 N/mm2

check v2 = V2/bd < 2 vc N/mm2 N/mm2

Ast mm2

from fig 11.3 for fcu N/mm2

vc N/mm2

no more shear checks are necessary

= 442 mm

= 426 mm

= 434 mm

= 1600 mm

= 6808 mm

= 2.34 < 4.38

= 1622 KN

= 180.3

= 2.9

= 0.37

= 0.42

Minimum tensile reinforcement = 1950 < 3015

no top tension in pad foundation

percentage reinforcement,p = 0.23%

Maximum spacing = 750 mm not exceeded

R = 0.15 say

T1 = 28

STEP 16:Check punching shear

dx

dy

d = 0.5 (dx + dy)

U0 = 2(Cx + Cy)

U1= (U0+12d)

v0 = Nu/U0d <(fcu)0.5 N/mm2 N/mm2

Nu

p1 = pu-pd KN/m2

A1 = (Cx +3.0 dx)(Cy+3.0 dy) m2

v1=Nu-p1A1/U1d N/mm2

vc N/mm2

STEP 17: check minimum reinforcement for flexure

mm2

STEP 18:Check spacing of reinforcement

STEP 19:Check early thermal cracking

alpha =

=

x = 250 mm assumed

= 148 mm

= 0.01 mm < 0.3 mm

Top reinforcement = 2625 over 3000 mm

Bottom reinforcement = 1050

servicebility limit state

Loading condition LC1

= 150.7

= 29.6

M = 307 KN m

x = 99 mm

z = 393 mm

= 259

=

=

=

= 66 mm

= 0.24 mm < 0.3 mm

12 x 10-6/C

er 4.032 x 10-5

acr

Wmax

STEP 20:check minimum reinforcement to distributr thermal cracking

mm2

mm2

STEP 21: check crack width due to flexure

pu KN/m2

pd KN/m2

fs N/mm2

Es 1.295 *10-3

Eh 1.588*10-3

emh 0.773*10-3

acr

Wmax

STEP 22: design mass concrete foundation

not required

Load combination LC13:

LC13 = 1.0 DL+0.5IL vertical loads only

P = 1093.9 KN

grass foundation pressu = 121.5

weight of soil removed = 162 KN

= 103.5

Foundation design over

STEP 23: calculate settlement

KN/m2

qn KN/m2

STEP 24: design connection of pad to column

DESIGN OF RC PAD WITH SINGLE COLUMN FOOTING

m 0.4 m

on plan

m B = 3 m

This step may be ignore since the foundations are not connected by groung beams and the differential settlements will have

Note: 25% over stress on allowable bearing capacity may be allowed for combinations including wind

Ignore passive resistance because horizontal movement of the foundation should be avoided

carry out analysis of bearing pressure for bending moment and shear

KN/m2

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

3000 mm

3000 mm

or 5 N/mm2

/m) over the central zone in each direction.

< 5

provided

N/mm2

mm2