PAD FOOTING ANALYSIS AND DESIGN (BS8110-1:1997) TEDDS calculation version 2.0.05.06
Pad footin d!tai"# Length of pad footing;
L = 1$00 mm 1$00 mm
idth of pad footing;
! = 1$00 mm 1$00 mm
"rea of pad footing; footing;
" = L × ! = 1%&&0 m 1%&&0 m2
Depth of pad footing;
h = $'0 mm $'0 mm
Depth of soil over pad footing;
hsoil = $000 mm $000 mm
Densit# of concrete;
ρconc = $% $%&m $% $%&m'
*o"+,n d!tai"# (olumn )ase length;
l " $00 mm " = $00 mm
(olumn )ase *idth;
) " = 100 mm 100 mm
(olumn eccentricit# in +;
e ,+" = 0 mm
(olumn eccentricit# in #;
e ,#" = 0 mm
Soi" d!tai"# Densit# of soil;
ρsoil = $0%0 $%&m $0%0 $%&m '
Design shear strength;
φ- = $'%0 deg $'%0 deg
Design )ase friction;
δ = 19% deg 19% deg
"llo*a)le )earing pressure; pressure;
,)earing = 100 $%&m 100 $%&m2
Aia" "oadin on .o"+,n Dead a+ial load on column;
," = 1%8 $% 1%8
%$/mposed a+ial load on column;
," = 9%0 $% 9%0
%$ind a+ial load on column;
," = 0%0 $% 0%0
%$Total a+ial load on column;
, " = 10%8 $% 10%8
%$ Fo+ndation "oad# Dead surcharge load;
1sur = 0%000 $%&m2
/mposed surcharge load;
1sur = 0%000 $%&m2
,ad footing self *eight;
1s*t = h × ρconc = '%900 $%&m2
Soil self *eight;
1soil = hsoil × ρsoil = &0%000 $%&m2
Total foundation load;
1 = " × 1sur 3 1sur 3 1s*t 3 1soil4 = %1
%$/oionta" "oadin on .o"+,n 2a#! Dead horiontal load in + direction;
+" = 0%$
%$/mposed horiontal load in + direction;
+" = $%8
%$ind horiontal load in + direction;
+" = 0%0
%$Total horiontal load in + direction;
+" = %0
%$Dead horiontal load in # direction;
#" = 0%0
%$/mposed horiontal load in # direction;
#" = 0%0
%$ind horiontal load in # direction;
#" = 0%0
%$Total horiontal load in # direction;
#" = 0%0
%$3o,!nt on .o"+,n 2a#! Dead moment on column in + direction;
7+" = 0%1$0 $%m
/mposed moment on column in + direction;
7+" = 1%&'0 $%m
ind moment on column in + direction;
7+" = 0%000 $%m
Total moment on column in + direction;
7+" = 1%'70 $%m
Dead moment on column in # direction;
7#" = 0%000 $%m
/mposed moment on column in # direction;
7#" = 0%000 $%m
ind moment on column in # direction;
7#" = 0%000 $%m
Total moment on column in # direction;
7#" = 0%000 $%m
*4!.5 #ta2i"it6 aain#t #"idin 8esistance to sliding due to )ase f riction friction = ma+9," 3 1sur 3 1s*t 3 1soil4 × ": 0 $%4 × tanδ4 = $%8 $% < p = 3 sinφ-44 & > sinφ-44 = $%&&
,assive pressure coefficient; Sta2i"it6 aain#t #"idin in di!.tion ,assive resistance of soil in + direction;
+pas = 0.5 ×
%$Total resistance to sliding in + direction;
+res = friction 3 +pas = ''%$
%$PASS - Resistance to sliding is greater than horizontal load in x direction *4!.5 #ta2i"it6 aain#t o!t+nin in di!.tion 7+?T = 7+" 3 +" × h = $%$8 $%m
Total overturning moment; !#toin ,o,!nt in di!.tion
7+sur = " × 1sur 3 1s*t 3 1soil4 × L & 2 = 9%'8 $%m
1oundation loading;
7+a+ial = ,"4 × L & 2 > e ,+"4 = 1%08 $%m
"+ial loading on column; Total restoring moment;
7+res = 7+sur 3 7+a+ial = &0%7&& $%m PASS - Restoring moment is greater than overturning moment in x direction
*a".+"at! ad 2a#! !a.tion Total )ase reaction;
T = 1 3 , " = 7%9
%$Eccentricit# of )ase reaction in +;
e T+ = , " × e,+" 3 7+" 3 +" × h4 & T = 0 mm
Eccentricit# of )ase reaction in #;
e T# = , " × e,#" 3 7#" 3 #" × h4 & T = 0 mm
*4!.5 ad 2a#! !a.tion !..!nti.it6 a)seT+4 & L 3 a)se T#4 & ! = 0%0$' Base reaction acts within middle third of base *a".+"at! ad 2a#! !##+!# @ = T & " > 6 × T × eT+ & L × "4 > 6 × T × eT# & ! × "4 = &'%8 $%&m
2
@2 = T & " > 6 × T × eT+ & L × "4 3 6 × T × eT# & ! × "4 = &'%8 $%&m
2
1%'0& $%&m
2
@' = T & " 3 6 × T × eT+ & L × "4 > 6 × T × eT# & ! × "4 = @A = T & " 3 6 × T × eT+ & L × "4 3 6 × T × eT# & ! × "4 = 1%'0& $%&m2 7inimum )ase pressure;
@min = min@ @2 @' @A4 = &'%8 $%&m2
7a+imum )ase pressure;
@ma+ = ma+@ @2 @' @A4 = 1%'0& $%&m2 PASS - Maximum base pressure is less than allowable bearing pressure
Patia" #af!t6 fa.to# fo "oad# ,artial safet# factor for dead loads;
γ f = 1%&0
,artial safet# factor for imposed loads;
γ f = 1%0
,artial safet# factor for *ind loads;
γ f = 0%00
"ti,at! aia" "oadin on .o"+,n Bltimate a+ial load on column;
,u" = ," × γ f 3 ," × γ f 3 ," × γ f = 17%0
%$ "ti,at! fo+ndation "oad# Bltimate foundation load;
1u = " × 91sur 3 1s*t 3 1soil4 × γ f 3 1sur ×
γ f: = 9$%'
%$"ti,at! 4oionta" "oadin on .o"+,n Bltimate horiontal load in + direction;
+u" = +" ×
γ f 3 +" × γ f 3 +" × γ f = &%8
%$Bltimate horiontal load in # direction;
#u" = #" ×
γ f 3 #" × γ f 3 #" × γ f = 0%0
%$"ti,at! ,o,!nt on .o"+,n Bltimate moment on column in + direction;
7+u" = 7+" × γ f 3 7+" ×
γ f 3 7+" × γ f = $%&88 $%m
Bltimate moment on column in # direction;
7#u" = 7#" × γ f 3 7#" ×
γ f 3 7#" × γ f = 0%000 $%m
*a".+"at! +"ti,at! ad 2a#! !a.tion Bltimate )ase reaction;
Tu = 1u 3 ,u" = 109%'
%$Eccentricit# of ultimate )ase reaction in +;
e T+u = ,u" × e,+" 3 7+u" 3 +u" × h4 & Tu = & mm
Eccentricit# of ultimate )ase reaction in #;
e T#u = ,u" × e,#" 3 7#u" 3 #u" × h4 & T u = 0 mm
*a".+"at! +"ti,at! ad 2a#! !##+!# @u = Tu&" > 6 ×Tu×eT+u&L× "4 > 6×Tu×eT#u&!× "4 = %$0 $%&m 2 @2u = Tu&" > 6 ×Tu×eT+u&L× "4 3 6×Tu× eT#u&!× "4 = %$0 $%&m2 @'u = Tu&" 3 6 ×Tu×eT+u&L× "4 > 6×Tu×eT#u&!× "4 = 88%8'& $%&m
2
$%&m
2
@Au = Tu&" 3 6 ×Tu×eT+u&L× "4 3 6×Tu×eT#u&!× "4 = 88%8'& 7inimum ultimate )ase pressure;
@minu = min@u @2u @'u @Au4 = %$0 $%&m2
7a+imum ultimate )ase pressure;
@ma+u = ma+@u @2u @'u @Au4 = 88%8'& $%&m2
*a".+"at! at! of .4an! of 2a#! !##+! in di!.tion Left hand )ase reaction;
fuL = @u 3 @2u4 × ! & 2 = 7'%87 $%&m
8ight hand )ase reaction;
fu8 = @'u 3 @Au4 × ! & 2 = 10%$' $%&m
Length of )ase reaction;
L+ = L = 1$00 mm
8ate of change of )ase pressure;
(+ = f u8 > f uL4 & L+ = $'%$& $%&m&m
*a".+"at! ad "!nt4# in di!.tion Left hand length;
LL = L & 2 3 e ,+" = 00 mm
8ight hand length;
L8 = L & 2 > e ,+" = 00 mm
*a".+"at! +"ti,at! ,o,!nt# in di!.tion Bltimate moment in + direction;
7+ = f uL × LL2 & 2 3 ( + × LL' & 6 > 1u × LL2 & 2 × L4 3 +u" × h 3 7+u" = &%90 $%m
*a".+"at! at! of .4an! of 2a#! !##+! in 6 di!.tion Top edge )ase reaction;
fuT = @2u 3 @Au4 × L & 2 = 91%$'0 $%&m
!ottom edge )ase reaction;
fu! = @u 3 @'u4 × L & 2 = 91%$'0 $%&m
Length of )ase reaction;
L# = ! = 1$00 mm
8ate of change of )ase pressure;
(# = f u! > f uT4 & L# = 0%000 $%&m&m
*a".+"at! ad "!nt4# in 6 di!.tion Top length;
LT = ! & 2 > e ,#" = 00 mm
!ottom length;
L! = ! & 2 3 e ,#" = 00 mm
*a".+"at! +"ti,at! ,o,!nt# in 6 di!.tion Bltimate moment in # direction;
7# = f uT × LT2 & 2 3 ( # × LT' & 6 > 1u × LT2 & 2 × !4 = $%'&' $%m
3at!ia" d!tai"# (haracteristic strength of concrete;
f cu = $' %&mm2
(haracteristic strength of reinforcement;
f# = &0 %&mm2
(haracteristic strength of shear reinforcement;
f #v = $'0 %&mm2
%ominal cover to reinforcement;
cnom = '0 mm
3o,!nt d!#in in di!.tion Diameter of tension reinforcement;
φ+! = 1 mm
Depth of tension reinforcement;
d+ = h > c nom > φ+! & 2 = 19$ mm
D!#in fo,+"a fo !.tan+"a 2!a,# (." %&%&%&) <+ = 7+ & ! × d+2 × f cu4 = 0%00& <+- = 0.56 K x K x ! compression reinforcement is not re"uired + = d+ × min90.5 3 √0.25 > <+ & 0.C4: 0.C54 = 18$ mm
Lever arm; "rea of tension reinforcement re@uired;
"s+re@ = 7+ & 0.F × f # × +4 = '' mm2
7inimum area of tension reinforcement;
"s+min = 0.00' × ! × h = 90 mm2
Tension reinforcement provided;
No% 1 dia% 2a# 2otto, ($$' .!nt!#)
"rea of tension reinforcement provided;
"s+!prov = %+! × π × φ+!2 & A = 1$0 mm2
PASS - #ension reinforcement provided exceeds tension reinforcement re"uired 3o,!nt d!#in in 6 di!.tion Diameter of tension reinforcement;
φ#! = 1 mm
Depth of tension reinforcement;
d# = h > c nom > φ+! > φ#! & 2 = 17 mm
D!#in fo,+"a fo !.tan+"a 2!a,# (." %&%&%&) <# = 7# & L × d#2 × f cu4 = 0%00 <#- = 0.56 K $ K $ ! compression reinforcement is not re"uired # = d# × min90.5 3 √0.25 > <# & 0.C4: 0.C54 = 17 mm
Lever arm; "rea of tension reinforcement re@uired;
"s#re@ = 7# & 0.F × f # × #4 = ' mm2
7inimum area of tension reinforcement;
"s#min = 0.00' × L × h = 90 mm2
Tension reinforcement provided;
No% 1 dia% 2a# 2otto, ($$' .!nt!#)
"rea of tension reinforcement provided;
"s#!prov = %#! × π × φ#!2 & A = 1$0 mm2
PASS - #ension reinforcement provided exceeds tension reinforcement re"uired *a".+"at! +"ti,at! #4!a fo.! at d fo, i4t fa.! of .o"+,n Bltimate pressure for shear;
@su = @u 3 (+ × L & 2 3 e ,+" 3 l " & 2 3 d +4 & ! 3 @ Au4 & 2 @su = 8'%'' $%&m2
"rea loaded for shear;
"s = ! × min' × L & 2 > e T+4 L & 2 > e ,+" > l " & 2 > d +4 = 0%70
2
m
Bltimate shear force;
Gsu = "s × @su > 1u & "4 = 7%87&
%$S4!a #t!##!# at d fo, i4t fa.! of .o"+,n (." %'%'%$) Design shear stress;
vsu = Gsu & ! × d+4 = 0%0& %&mm2
Fo, BS 8110:Pat 1:1997 - Ta2"! %8 vc = 0.FC %&mm 2 × min' 900 × "s+!prov & ! × d+4:&'4 ×
Design concrete shear stress;
ma+A00 mm & d +4&A 0.6F4 × minf cu & %&mm 2 A04 & 254&' & .25 = 0%1$ %&mm2 vma+ = min0. %&mm 2 × √f cu & %&mm 24 5 %&mm24 = &%000
"llo*a)le design shear stress; %&mm2
PASS - v su v c - %o shear reinforcement re"uired *a".+"at! +"ti,at! +n.4in #4!a fo.! at fa.! of .o"+,n Bltimate pressure for punching shear;
@pu" = @u39L&23e,+">l "&243l "4&2:×(+&!>9!&23e,#">) "&243 ) "4&2:×(#&L = 7%0&$ $%&m2
"verage effective depth of reinforcement;
d = d+ 3 d#4 & 2 = 18& mm
"rea loaded for punching shear at column;
"p" = l "4×) "4 = 0%0$0 m2
Length of punching shear perimeter;
up" = 2×l "432×) "4 = 00 mm
Bltimate shear force at shear perimeter;
Gpu" = ,u" 3 1u & " > @ pu"4 × "p" = 1%70
%$Effective shear force at shear perimeter;
Gpu"eff = Gpu"×93.5×a)s7+u"4&Gpu"×) "44: = '&%0'0
%$P+n.4in #4!a #t!##!# at fa.! of .o"+,n (." %7%7%$) vpu" = Gpu"eff & up" × d4 = 0%&90 %&mm2
Design shear stress;
vma+ = min0.%&mm 2 × √f cu & %&mm 24 5 %&mm 24 = &%000
"llo*a)le design shear stress; %&mm2
PASS - &esign shear stress is less than allowable design shear stress *a".+"at! +"ti,at! +n.4in #4!a fo.! at !i,!t! of 1%' d fo, fa.! of .o"+,n Bltimate pressure for punching shear;
@pu".5d = @u39L&23e,+">l "&2>.5×d43l "32×.5×d4&2:×(+&!> 9!&23e,#">) "&2>.5×d43) "32×.5×d4&2:×(#&L = 7%0&$ $%&m2
"verage effective depth of reinforcement;
d = d+ 3 d#4 & 2 = 18& mm
"rea loaded for punching shear at column;
"p".5d = l "32×.5×d4×) "32×.5×d4 = 0%&90 m2
Length of punching shear perimeter;
up".5d = 2×l "32×.5×d432×) "32×.5×d4 = $808 mm
Bltimate shear force at shear perimeter;
Gpu".5d = ,u" 3 1u & " > @ pu".5d4 × "p".5d = 11%189
%$Effective shear force at shear perimeter;
Gpu".5deff = Gpu".5d×93.5×a)s7+u"4&Gpu".5d×) "32×.5×d44: = 1%91
%$P+n.4in #4!a #t!##!# at !i,!t! of 1%' d fo, fa.! of .o"+,n (." %7%7%$) Design shear stress;
vpu".5d = Gpu".5deff & up".5d × d4 = 0%0 %&mm2
Fo, BS 8110:Pat 1:1997 - Ta2"! %8 Design concrete shear stress;
vc = 0.FC %&mm 2 × min' 900 × "s+!prov & ! × d+4 3 "s#!prov & L × d#44 & 2:&'4 × ma+00 mm & d + 3 d#44&A 0.6F4 × minf cu & %&mm2 A04 & 254 &' & .25 = 0%$8 %&mm2
"llo*a)le design shear stress;
vma+ = min0.%&mm 2 × √f cu & %&mm 24 5 %&mm 24 = &%000
%&mm2 PASS - v puA'()d v c - %o shear reinforcement re"uired
