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