bscode footing
TRANSCRIPT
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