STRAP FOOTINGS STRAP #1 Design assumptions 1. Strap does not provide bearing 2. Strap is ridge enough to transfer moment from one footing to the other. 3. Soil bearing pressure
=
100
kN/m2
A
B 3.05
3.05
2.75
2.75
150
5700
150
6000 5700 DL LL Pa LL
420 430 850 1276
N2
N1
1.525 Ra
~assume ~assume a footing footing width of e
= =
2952
3048 mm, the eccentricity eccentricity of footing footing
1524 mm 1.524 m
~the distance between footing reaction, L L = 2952 mm = 2.952 m ~the eccentric moment is
Rb 1.525
M
= =
Pa * e 1295 kNm
A is
420 430 850 1276
DL LL Pb LL
~the shear produced by M is, V
= M/L = 438.8 kNm/m
Ra = Pa + V
~reaction at footing A ,
Ra = ~soil bearing capacity
=
100
~req ~requi uired red foot footin ing g area area of A = = use
3.05
x
e
= =
kN/m2
Ra/so Ra/soil il bear bearin ing g capa capaci city ty 12.89 kN/m2
2.75
~assume a footing width of
1289 kN
=
8.388 kN/m2
3048 mm, the eccentricity of footing
1524 mm 1.524 m
~the distance between footing reaction, L L = 2952 mm = 2.952 m ~the eccentric moment is
M
= =
~the shear produced by M is, V
= M/L = 438.8 kNm/m
Rb = Pb + V
~reaction at footing B ,
Rb = ~soil bearing capacity
=
100
~req ~requi uired red foot footin ing g area area of B = = use
3.05
x
Pb * e 1295 kNm
2.75
1289 kN kN/m2
Rb/so Rb/soil il bear bearin ing g capa capaci city ty 12.89 kN/m2 =
8.388 kN/m2
3.05
3.05 STRAP
A 2.75
B 2.75
~Factored colum n load of A
= =
1.4*gk+1.6*qk 1276 kN
~Factored colum n load of B
= =
1.4*gk+1.6*qk 1276 kN
~factored eccentric moment, M ua
~Factored shear, Vua
= =
= =
Pa * e ~Mub 1945 kN ~Vub
M/L 658.7 kN
~Factored footing reaction at A = =
1276 + 1935 kN
~Factored footing pressure per linear foot of A
~Factored footing reaction at B = =
~Factored footing pressure per linear foot of B
= =
634.3 * -1086 kN
0.3
-
1276
At point 2:
Vu = =
634.3 * 657 kN
3.048
-
1276
At point 3:
Vu = =
634.3 * -1086 kN
0.3
-
1276
At point 4:
Vu
634.3 * 657 kN
3.048
-
1276
~Moment diagram At point 1: Mu = =
wl2 2 -184
kNm
At point 2:
Mu
=
-749.9 kNm
At point 3:
Mu
=
-813.1 kNm
At point 4:
Mu
= 1408.4 kNm
M/L 658.7 kN
1935 / 3.05 634.3 kN/m
658.7
~Shear diagram At point 1: Vu = =
= =
Pb * e 1945 kN
658.7
= =
1276 + 1935 kN
= =
= =
1935 / 3.05 634.3 kN/m
1276 kN
1276 kN Pub
Pua LOAD DIAGRAM (kN)
2.90
300
2.90
2748
-96
2748
634.3 kN/m
300
634.3 kN/m
+657
1 SHEAR DIAGRAM
2
+657
3
0
4 -1086
-1086
4 +1408.4
MOMENT DIAGRAM
0 -184
1 -749.9
2
3 -813.1
REINFORCED CONCRETE DESIGN OF STRAP FOOTING ~design footing strap as a reinforced concrete beam 2 ~yield strenght of rebar = 410 N/mm 2 ~strenght of concrete = 25 N/mm 1. Design footing strap * * * *
assume 375 x 1050 footing strap and the reinforcement is H 25 top cover 50 mm effective depth, d = 987.5 mm depth to comression. rebar d' = 87.5 mm
(b) design flexural reinforcement maximum factored moment at point 4, Mu = (K) k = M = 0.1541 > 2 f cu * b * d
1408.4 kNm (K' ) 0.156 ==> compression bar is required As' =
lever arm, z
=
d
0.5 +
0.87*f y*(d -d' )
0.25 - K' 0.9 =
z As =
= 2
k'f cubd
771
2
(K -K' )f cu b d
-55.23
mm2
mm
+ As'
0.87*f y*z =
5130.58
provide 12 H
mm2
25 =
5890 mm2
check area of steel provide 0.4% < 100As Ac 100As
=
1.496
==>
area of steel provide within the lim it specified by code
Ac
(a) check direct shear from shear factored diagram, Vu (max) v
=
100As
V b*d
=
=
2.932 N/mm <
=
2.932 N/mm <
2
=
1086 kN
0.8 f cu 4.00
BS8110 - table 3.8,3.9 =======> OK BS8110 from table 3.9 1.5 0.72 1.591 vc
1.591
b*d vc
=
0.735
2 0.72 - vc 0.72-0.8 vc =
0.8 =
1.5-1.439 1.5-2 0.735
from table 3.9 BS 8110, this ultimate shear stress is not excessive, therefore h h = 1050 will be suitable
2. Design footing for colum A assume h
=
450
mm
dia. bar d d
= = =
25 mm 600 - cover -dia. main bar - dia. secondary bar 362.5 mm Ra = Pa + V
~reaction at footing A ,
Ra = use 2.75 x earth pressure
3.05
net upward pressure ~at column face
shear stress, vc
=
1289 kN 2
=
8.388 kN/m 1289 = 153.7 kN/m 8.388 153.7 - h x 24 x gF
=
138.5 kN/m
= =
2
N col. Perimeter x d
=
2.963
<
0.8 f cu
(
3.578 )
punching shear critical perimeter = column perimeter + 8(1.5d) = 5550 mm area within perimeter
= =
punching shear force V = =
punching shear stress v
= =
100As
=
100 2750
=
0.347
bd
vc
1608 362.5
2
2
(400+3h) - (4-pi)(1.5h) 2671388.2 mm2 138.5 8.388 792.1 kN
V perimeter x d 0.394 N/mm2 =
0.161
2.67
<
0.8 f cu
(
3.578 ) BS8110 from table 3.9 0.15 0.34 0.161 vc 0.25 0.34 - vc 0.34-0.4 vc =
from table 3.9 BS 8110, this ultimate shear stress is not excessive, therefore h h = 450 will be suitable
0.4 =
0.15-0.158 0.15-0.25 0.347
(b) bending reinforcement at column face which is the critical section M = Fl = 153.7 x 2.75 x 0.716 x = 108.3 kNm
0.358
the location of zero shear ia at X
= =
1.575*335/(335+402) 0.716 m from inside face of column
Mu = =
0.156f cubd
k
M 2 f cu * b * d
=
2
1409 kNm
As =
>
108.3 kNm
=
0.0000210
<
0.156 compression bar is not required
M 0.87 f y z
lever arm, z
=
d
0.5 +
z As = provide 100As
8 =
=
0.25 - k 0.9
281.6 mm
1078 mm H
16
0.161
>
As =
1608 mm
2
or H16 @ 150 c/c
0.13 as required by code
b*h
~check on shear stress earth pressure
=
1289 = 8.388
153.7 kN/m
2
V = 0.154 N/mm < bd therefore section is adequate in shear. v
=
3. Design footing for colum B assume h dia. bar d
= = =
450 mm 25 mm 600 - cover -dia. main bar - dia. secondary bar
0.347 N/mm
2
d
=
362.5 mm Rb = Pb + V
~reaction at footing B ,
Rb = use 2.75 x earth pressure
3.05
net upward pressure ~at column face shear stress, vc
=
1289 kN 2
=
8.388 kN/m 1289 = 153.7 kN/m 8.388 153.7 - h x 24 x gF
=
138.5 kN/m
= =
2
N col. Perimeter x d
=
2.963
<
3.578 N/mm
0.8 f cu
punching shear critical perimeter = column perimeter + 8(1.5d) = 5550 mm area within perimeter
= =
punching shear force V = =
punching shear stress v
= =
100As
=
100 2750
=
0.347
bd
vc
1608 362.5
2
2
(400+3h) - (4-pi)(1.5h) 2 2671388.2 mm 138.5 8.388 792.10 kN
V perimeter x d 0.394 N/mm =
2.67
<
0.161
0.8 f cu
(
3.578 ) BS8110 from table 3.9 0.15 0.34 0.161 vc 0.25 0.34 - vc 0.34-0.4 vc =
from table 3.9 BS 8110, this ultimate shear stress is not excessive, therefore h h = 450 will be suitable
(b) bending reinforcement at column face which is the critical section M = Fl = 153.7 x 2.75 x 0.716 x
0.358
0.4 =
0.15-0.158 0.15-0.25 0.347
=
108.3 kNm
the location of zero shear ia at X
= =
1.575*335/(335+402) 0.716 m from inside face of column
Mu = =
0.156f cubd
k
M 2 f cu * b * d
=
2
1409 kNm
As =
>
108.3 kNm
=
0.0001560
<
0.156 compression bar is not required
M 0.87 f y z
lever arm, z
=
d
0.5 +
z As = provide 100As
8 =
=
0.25 - k 0.9
281.6 mm
1078 mm H
16
0.161
>
As =
1608 mm
2
or H16 @ 150 c/c
0.13 as required by code
b*h
~check on shear stress earth pressure
=
1289 = 8.388
153.7 kN/m
2
V = 0.154 N/mm < bd therefore section is adequate in shear. v
=
0.347 N/mm
2
STRAP FOOTINGS STRAP #3 Design assumptions 1. Strap does not provide bearing 2. Strap is ridge enough to transfer moment from one footing to the other. 3. Soil bearing pressure
=
100
kN/m2
A
B 3.05
3.05
3.35
3.35
150
5700
150
6096 5796 DL LL Pa LL
525 525 1050 1575
N2
N1
1.525 Ra
~assume a footing width of e
= =
3048
Rb 1.525
3048 mm, the eccentricity of footing
1524 mm 1.524 m
~the distance between footing reaction, L L = 3048 mm = 3.048 m ~the eccentric moment is
M = =
~the shear produced by M is, V
Pa * e 1600 kNm = M/L = 525
kNm/m
A is
525 525 1050 1575
DL LL Pb LL
Ra = Pa + V
~reaction at footing A ,
Ra = ~soil bearing capacity
=
100
~required footing area of A = = use
3.05
x
e
= =
kN/m2
Ra/soil bearing capacity 15.75 kN/m2
3.35
~assume a footing width of
1575 kN
=
10.22 kN/m2
3048 mm, the eccentricity of footing
1524 mm 1.524 m
~the distance between footing reaction, L L = 3048 mm = 3.048 m ~the eccentric moment is
M = =
~the shear produced by M is, V
= M/L = 525
Rb = =
100
~required footing area of B = = use
3.05
x
kNm/m
Rb = Pb + V
~reaction at footing B ,
~soil bearing capacity
Pb * e 1600 kNm
3.35
1575 kN kN/m2
Rb/soil bearing capacity 15.75 kN/m2 =
10.22 kN/m2
3.05
3.05 STRAP
A
B
3.35
3.35
~Factored column load of A
= =
1.4*gk+1.6*qk 1575 kN
~Factored column load of B
= =
1.4*gk+1.6*qk 1575 kN
~factored eccentric moment, M ua
=
Pa * e
~Mub
=
Pb * e
= ~Factored shear, V ua
= =
2400 kN ~Vub
M/L 787.5 kN
~Factored footing reaction at A = =
1575 + 2363 kN
~Factored footing pressure per linear foot of A
~Factored footing reaction at B = =
2363 / 3.05 774.6 kN/m
= =
774.6 * -1343 kN
0.3
-
1575
At point 2: Vu = =
774.6 * 786 kN
3.048
-
1575
At point 3: Vu = =
774.6 * -1343 kN
0.3
-
1575
At point 4: Vu
774.6 * 786 kN
3.048
-
1575
=
M/L 787.5 kN
787.5
~Shear diagram At point 1: Vu = =
~Moment diagram At point 1: Mu =
= =
2400 kN
787.5
= =
1575 + 2363 kN
~Factored footing pressure per linear foot of B
= =
=
2363 / 3.05 774.6 kN/m
wl2 2
-228
kNm
At point 2: Mu
=
At point 3: Mu
=
At point 4: Mu
= 1706.7 kNm
-964.7 kNm -964.69
kNm
1575 kN
1575 kN Pub
Pua
LOAD DIAGRAM (kN)
2.90
300
2748
2.90
0
2748
300
774.6 kN/m
774.6 kN/m
+786
1
SHEAR DIAGRAM
2
+786
3
0
4 -1343 -1343
4 +1706.7
MOMENT DIAGRAM
0 -228
1
3 -964.7
2
-964.69
REINFORCED CONCRETE DESIGN OF STRAP FOOTING ~design footing strap as a reinforced concrete beam 2 ~yield strenght of rebar = 410 N/mm 2 ~strenght of concrete = 25 N/mm 1. Design footing strap * * * *
assume 375 x 1050 footing strap and the reinforcement is H 25 top cover 50 mm effective depth, d = 987.5 mm depth to comression. rebar d' = 87.5 mm (b) design flexural reinforcement Mu = maximum factored moment at point 4, (K) k = M = 0.1867 > 2 f cu * b * d
1706.7 kNm (K' ) 0.156 ==> compression bar is required As' =
lever arm, z
=
d
0.5 +
0.87*f y*(d -d' )
0.25 - K' 0.9 =
z As =
= 2
k'f cubd
0.87*f y*z
697.5 mm + As'
(K -K' )f cu b d
873.9
mm2
=
6606.19
provide 14 H
mm2
25 =
6872 mm2
check area of steel provide 0.4% < 100As Ac 100As
=
1.745
==>
area of steel provide within the limit specified by code
Ac
(a) check direct shear from shear factored diagram, Vu (max) v
=
V b*d
100As = b*d vc
=
=
=
3.626 N/mm <
=
3.626 N/mm <
1343 kN
0.8 f cu
2
BS8110 - table 3.8,3.9
4.00
=======> OK BS8110 from table 3.9 1.5 0.72 1.856 vc
1.856
0.777
2 0.72 - v c 0.72-0.8 vc =
0.8 =
1.5-1.439 1.5-2 0.777
from table 3.9 BS 8110, this ultimate shear stress is not excessive, therefore h h = 1050 will be suitable
2. Design footing for colum A assume h dia. bar d d
= = = =
450 mm 25 mm 600 - cover -dia. main bar - dia. secondary bar 362.5 mm Ra = Pa + V
~reaction at footing A ,
Ra = use 2.75 x earth pressure
3.05
net upward pressure ~at column face
shear stress, v c
=
=
1575 kN 2
=
8.388 kN/m 1575 = 187.8 kN/m 8.388 187.8 - h x 24 x gF
=
172.7 kN/m
= =
2
N col. Perimeter x d 3.621
<
0.8 f cu
(
3.578 )
punching shear critical perimeter = column perimeter + 8(1.5d) = 5550 mm area within perimeter
= =
punching shear force V = =
punching shear stress v
= =
100As = bd
vc
100 3350
=
2
(400+3h) - (4-pi)(1.5h) 2671388.2 mm2 172.7 8.388 987.2 kN
2.67
V perimeter x d 0.491 N/mm2
1608 362.5
=
2
<
0.132 <
0.8 f cu 0.15
0.340
(
3.578 ) BS8110 from table 3.9 0.15 0.34 0.132 vc 0.25 0.34 - v c 0.34-0.4 vc =
0.4 =
0.15-0.158 0.15-0.25 0.329
from table 3.9 BS 8110, this ultimate shear stress is not excessive, therefore h h = 450 will be suitable
(b) bending reinforcement at column face which is the critical section M = Fl = 187.8 x 3.35 x 0.716 x = 161.2 kNm
0.358
the location of zero shear ia at X
= =
1.575*335/(335+402) 0.716 m from inside face of column
Mu = =
0.156f cubd
k
M 2 f cu * b * d
=
1717 kNm
As =
> =
161.2 kNm 0.0000210
M 0.87 f y z
lever arm, z
=
d
0.5 +
0.25 - k
<
0.156 compression bar is not required
0.9 z
8
100As = b*h
281.6 mm
1605 mm
As = provide
=
H
16
0.132
As =
>
2
1608 mm
or H16 @ 150 c/c
0.13 as required by code
~check on shear stress earth pressure
=
1575 = 8.388
187.8 kN/m
2
V = 0.155 N/mm < bd therefore section is adequate in shear. v
=
0.34
3. Design footing for colum B assume h dia. bar d d
= = = =
450 mm 25 mm 600 - cover -dia. main bar - dia. secondary bar 362.5 mm Rb = Pb + V
~reaction at footing B ,
Rb = use 2.75 x earth pressure
3.05
net upward pressure ~at column face shear stress, v c
=
1575 kN 2
=
8.388 kN/m 1575 = 187.8 kN/m 8.388 187.8 - h x 24 x gF
=
172.7 kN/m
= =
2
N col. Perimeter x d
=
3.621
<
3.578 N/mm
0.8 f cu
punching shear critical perimeter = column perimeter + 8(1.5d) = 5550 mm area within perimeter
= =
punching shear force V = =
2
(400+3h) - (4-pi)(1.5h) 2 2671388.2 mm 172.7 8.388 987.18 kN
2
2.67
N/mm
2
punching shear stress v
= =
100As = bd
vc
100 3350
=
V perimeter x d 0.491 N/mm
1608 362.5
=
<
0.132 <
0.8 f cu 0.15
(
3.578 ) BS8110 from table 3.9 0.15 0.34 0.132 vc
0.340
0.25 0.34 - v c 0.34-0.4 vc =
0.4 =
0.15-0.158 0.15-0.25 0.329
from table 3.9 BS 8110, this ultimate shear stress is not excessive, therefore h h = 450 will be suitable
(b) bending reinforcement at column face which is the critical section M = Fl = 187.8 x 3.35 x 0.716 x = 161.2 kNm
0.358
the location of zero shear ia at X
= =
1.575*335/(335+402) 0.716 m from inside face of column 2
Mu = =
0.156f cubd
k
M f cu * b * d
=
1717 kNm
As =
>
161.2 kNm
=
0.0001560
<
0.156 compression bar is not required
M 0.87 f y z
lever arm, z
=
d
z As = provide 100As = b*h
8
0.5 +
=
0.25 - k 0.9
281.6 mm
1605 mm H 0.132
16 >
As =
2
1608 mm
or H16 @ 150 c/c
0.13 as required by code
~check on shear stress earth pressure
=
1575 = 8.388
187.8 kN/m
2
V = 0.155 N/mm < bd therefore section is adequate in shear. v
=
0.34
N/mm
2
STRAP FOOTINGS STRAP #3a Design assumptions 1. Strap does not provide bearing 2. Strap is ridge enough to transfer moment from one footing to the other. 3. Soil bearing pressure
=
100
kN/m2
A
B 3.05
3.05
3.05
3.05
150
5700
150
6096 5796 DL LL Pa LL
525 525 1050 1575
N2
N1
1.525 Ra
~assume a footing width of e
= =
3048
525 525 1050 1575
DL LL Pb LL
Rb 1.525
3048 mm, the eccentricity of footing
A is
1524 mm 1.524 m
~the distance between footing reaction, L L = 3048 mm = 3.048 m ~the eccentric moment is
M = =
~the shear produced by M is, V
Pa * e 1600 kNm = M/L = 525
kNm/m
5706/STRAP#3/19
Ra = Pa + V
~reaction at footing A ,
Ra = ~soil bearing capacity
=
100
~required footing area of A = = use
3.05
x
e
= =
kN/m2
Ra/soil bearing capacity 15.75 kN/m2
3.05
~assume a footing width of
1575 kN
=
9.303 kN/m2
3048 mm, the eccentricity of footing
1524 mm 1.524 m
~the distance between footing reaction, L L = 3048 mm = 3.048 m ~the eccentric moment is
M = =
~the shear produced by M is, V
= M/L = 525
Rb = =
100
~required footing area of B = = use
3.05
x
kNm/m
Rb = Pb + V
~reaction at footing B ,
~soil bearing capacity
Pb * e 1600 kNm
3.05
1575 kN kN/m2
Rb/soil bearing capacity 15.75 kN/m2 =
9.303 kN/m2
3.05
3.05 STRAP
A
B
3.05
3.05
~Factored column load of A
= =
1.4*gk+1.6*qk 1575 kN
~Factored column load of B
= =
1.4*gk+1.6*qk 1575 kN
~factored eccentric moment, M ua
=
Pa * e
~Mub
=
Pb * e
5706/STRAP#3/20
= ~Factored shear, V ua
= =
2400 kN ~Vub
M/L 787.5 kN
~Factored footing reaction at A = =
1575 + 2363 kN
~Factored footing pressure per linear foot of A
~Factored footing reaction at B = =
= =
774.6 * -1343 kN
0.3
-
1575
At point 2: Vu = =
774.6 * 786 kN
3.048
-
1575
At point 3: Vu = =
774.6 * -1343 kN
0.3
-
1575
At point 4: Vu
774.6 * 786 kN
3.048
-
1575
=
M/L 787.5 kN
2363 / 3.05 774.6 kN/m
787.5
~Shear diagram At point 1: Vu = =
~Moment diagram At point 1: Mu =
= =
2400 kN
787.5
= =
1575 + 2363 kN
~Factored footing pressure per linear foot of B
= =
=
2363 / 3.05 774.6 kN/m
wl2 2
-228
kNm
At point 2: Mu
=
At point 3: Mu
=
At point 4: Mu
= 1706.7 kNm
-964.7 kNm -964.69
kNm
5706/STRAP#3/21
1575 kN
1575 kN Pub
Pua
LOAD DIAGRAM (kN)
2.90
300
2.90
2748
0
2748
774.6 kN/m
300
774.6 kN/m
+786
1
SHEAR DIAGRAM
2
+786
3
0
4 -1343 -1343
4 +1706.7
MOMENT DIAGRAM
0 -228
1
3 -964.7
2
-964.69
REINFORCED CONCRETE DESIGN OF STRAP FOOTING ~design footing strap as a reinforced concrete beam 2 ~yield strenght of rebar = 410 N/mm 2 ~strenght of concrete = 25 N/mm 1. Design footing strap * * * *
assume 375 x 1050 footing strap and the reinforcement is H 25 top cover 50 mm effective depth, d = 987.5 mm depth to comression. rebar d' = 87.5 mm (b) design flexural reinforcement Mu maximum factored moment at point 4, (K)
=
1706.7 (K' )
kNm
5706/STRAP#3/22
k
=
M f cu * b * d
=
0.1867
>
0.156 ==> compression bar is required As' =
lever arm, z
=
d
0.5 +
=
697.5 mm
0.87*f y*(d -d' )
0.25 - K' 0.9 =
z As =
k'f cubd
2
(K -K' )f cu b d
873.9
mm2
+ As'
0.87*f y*z =
6606.19
provide 14 H
mm2
25 =
6872 mm2
check area of steel provide 0.4% < 100As Ac 100As
=
1.745
==>
area of steel provide within the limit specified by code
Ac
(a) check direct shear from shear factored diagram, Vu (max) v
=
V b*d
100As = b*d vc
=
=
=
3.626 N/mm <
=
3.626 N/mm <
2
1343 kN
0.8 f cu 4.00
BS8110 - table 3.8,3.9 =======> OK BS8110 from table 3.9 1.5 0.72 1.856 vc
1.856
0.777
2 0.72 - v c 0.72-0.8 vc =
0.8 =
1.5-1.439 1.5-2 0.777
from table 3.9 BS 8110, this ultimate shear stress is not excessive, therefore h h = 1050 will be suitable
2. Design footing for colum A assume h dia. bar d d
= = = =
450 mm 25 mm 600 - cover -dia. main bar - dia. secondary bar 362.5 mm
~reaction at footing A ,
Ra = Pa + V Ra =
use 3.05
x
3.05
=
1575 kN 2
9.303 kN/m
5706/STRAP#3/23
earth pressure
=
net upward pressure
=
1575 = 169.3 kN/m 9.303 169.3 - h x 24 x gF
=
154.2 kN/m
~at column face
shear stress, v c
=
2
N col. Perimeter x d
=
3.621
<
0.8 f cu
(
3.578 )
punching shear critical perimeter = column perimeter + 8(1.5d) = 5550 mm area within perimeter
= =
punching shear force V = =
punching shear stress v
= =
100As = bd
vc
=
100 3050
2
(400+3h) - (4-pi)(1.5h) 2671388.2 mm2 154.2 9.303 1023 kN
2.67
V perimeter x d 0.508 N/mm2
1608 362.5
=
2
<
0.145 <
0.340
0.8 f cu 0.15
(
3.578 ) BS8110 from table 3.9 0.15 0.34 0.145 vc 0.25 0.34 - v c 0.34-0.4 vc =
0.4 =
0.15-0.158 0.15-0.25 0.337
from table 3.9 BS 8110, this ultimate shear stress is not excessive, therefore h h = 450 will be suitable
(b) bending reinforcement at column face which is the critical section M = Fl = 169.3 x 3.05 x 0.716 x = 132.3 kNm
0.358
the location of zero shear ia at
5706/STRAP#3/24
X
= =
1.575*335/(335+402) 0.716 m from inside face of column
Mu = =
0.156f cubd
k
M 2 f cu * b * d
1563 kNm
=
As =
>
132.3 kNm
=
0.0000210
<
0.156 compression bar is not required
M 0.87 f y z
lever arm, z
=
d
0.5 +
z
8
100As = b*h
281.6 mm
1317 mm
As = provide
=
0.25 - k 0.9
H
16
0.145
As =
>
2
1608 mm
or H16 @ 150 c/c
0.13 as required by code
~check on shear stress earth pressure
=
1575 = 9.303
169.3 kN/m
2
V = 0.153 N/mm < bd therefore section is adequate in shear. v
=
0.34
N/mm
2
3. Design footing for colum B assume h dia. bar d d
= = = =
450 mm 25 mm 600 - cover -dia. main bar - dia. secondary bar 362.5 mm
~reaction at footing B ,
Rb = Pb + V Rb =
use 3.05 x earth pressure
3.05
net upward pressure ~at column face
1575 kN 2
=
9.303 kN/m 1575 = 169.3 kN/m 9.303 169.3 - h x 24 x gF
=
154.2 kN/m
= =
2
5706/STRAP#3/25
shear stress, v c
=
N col. Perimeter x d
=
3.621
<
3.578 N/mm
0.8 f cu
punching shear critical perimeter = column perimeter + 8(1.5d) = 5550 mm area within perimeter
= =
punching shear force V = =
punching shear stress v
= =
100As = bd
vc
=
100 3050
2
(400+3h) - (4-pi)(1.5h) 2 2671388.2 mm 154.2 9.303 1022.66 kN
2.67
V perimeter x d 0.508 N/mm
1608 362.5
=
2
<
0.145 <
0.8 f cu 0.15
0.340
(
3.578 ) BS8110 from table 3.9 0.15 0.34 0.145 vc 0.25 0.34 - v c 0.34-0.4 vc =
0.4 =
0.15-0.158 0.15-0.25 0.337
from table 3.9 BS 8110, this ultimate shear stress is not excessive, therefore h h = 450 will be suitable
(b) bending reinforcement at column face which is the critical section M = Fl = 169.3 x 3.05 x 0.716 x = 132.3 kNm
0.358
the location of zero shear ia at X
= =
1.575*335/(335+402) 0.716 m from inside face of column 2
Mu = =
0.156f cubd
k
M f cu * b * d
=
As =
1563 kNm
> =
132.3 kNm 0.0001560
<
0.156 compression bar is not required
M
5706/STRAP#3/26
0.87 f y z lever arm, z
=
d
0.5 +
z As = provide 100As = b*h
8
=
0.25 - k 0.9
281.6 mm
1317 mm H
16
0.145
>
~check on shear stress earth pressure
As =
2
1608 mm
or H16 @ 150 c/c
0.13 as required by code
=
1575 = 9.303
169.3 kN/m
2
V = 0.153 N/mm < bd therefore section is adequate in shear. v
=
0.34
N/mm
2
5706/STRAP#3/27
