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1Foundation Design (Isolated Footing)
An internal column transferring the loads given in table 1 is considered for isolated footing design. The material to be used for are C25/30 and S400. If the column size is 600mm by 600 mm. and the soil has presumptive bearing capacity of 280 Kpa(Allowable), determine the size and reinforcement necessary for the isolated footing to safely transfer the load.
ETABS OUTPUT
Base^ Story Joint Label Load Case/Combo 33 Comb1-Ultimate Base 33 Comb2-Service
Fx Fy Fz Mx My Mz
Table 1: Joint Reaction from ETABS to an internal column
1.1 Material data
For C25/30 : 𝑓𝑐𝑘= 25 𝑀𝑝𝑎
For S400: 𝑓𝑦𝑘= 400 𝑀𝑝𝑎
Concrete design strength
fcd = αccfck γc
[ES EN 1992 −1 −1: 2015, Eqn. 3.15 ]
Where: the recommended value of αcc is 0.85.
fcd = 0.85 ∗25 1.5 = 14.17 𝑀𝑝𝑎
Steel design strength
fyd = 𝑓𝑦𝑘 𝛾𝑠
[ES EN 1992 −1 −1: 2015, Figure 3.8 ]
fyd =400 1.115 = 347.83 Mpa
1.2 Footing proportioning Factor of safety (F.S)
F. S
Ultimate load Service load
all 280 Kpa
ult .
* 280 Kpa 391.81 Kpa
Eccentricity at service load
y e ^ M^ ^ .8 173^ . 188
x 10
3 x (^) P 256 4
e M^ x .2 978^ . 161
x 10
3 y (^) P 2564. 125 Now, the stress at the four corners of the footing can be determined from:
P^ 1( 6 e^ x 6 e^ y (^) ) A B B
Figure 1 Stress distribution on isolated footing with eccentric loading.
Any plan dimension of footing, B and L, must satisfy two things:
a)The maximum stress must be less than or equal to the allowable bearing capacity.
b)The minimum stress must be greater than zero.
Condition a) Since the eccentricity is relatively equal in both axis’s, a square footing is ideal.
(^) P^ 1( 6 e
x (^) 6 e^ y^ ) all max (^) A B B
x 10
3 2564.1252 (^1) )
(
x 10
3
B^2 B B all
2 1 (
x 10
3 )^ all B^2 B
The average stress:
37 4.92^ ^371. 79
kpa 373. 356 Kpa avg (^) 4
Wide beam shear
According to ES EN 1992-1-1:2015, Article 6.2.1(8) the Critical section for shear is at distance d
from the face of supports.
Figure 2: Critical section for wide beam shear, plan.
Figure 3: Critical section for wide beam shear, Section.
The design shear force at section y-y is:
Vy y^ ^ avg B *^ (^ B^ ^ C^ x^ d^ ) 2 2
6.0^ d ) 1154. 304
d ) 1442. 88
V^ d y y ^ ( 372
)( )1.3 * (^ 1.
Wide beam shear resistance (the design value of shear resistance for members not requiring shear reinforcements) is given by:
Vrd,c = [Crd,c𝑘(100ρ 1 fck) 1 3 ] bwd [ES EN 1992 −1 −1: 2015, Eqn. 6.2 a]
Crd,c = 0. 𝛾𝑐
2 2 (^3) 0.3𝑓𝑐 0.3 ∗(25) 3 = 0.0017 < 0.
𝑘
𝑓𝑦𝑘 = 0.26 ∗^400
Vrd,c = [0.12(1 + √200𝑑)(100 ∗0.0017 ∗25) 3 ] 3100 ∗d KN = 602.57 ∗d ∗(1 + √200𝑑)
And the minimum Vrd,c
𝑉𝑟𝑑,𝑐(min) = (𝑣𝑚𝑖𝑛)bwd [ES EN 1992 −1 −1: 2015, Eqn. 6.2 b]
𝑣𝑚𝑖𝑛= 0.035𝑘
3 3
(min) = 0.175 (1 + √200𝑑)
Equating the shear capacity and the design shear force:
602.57 ∗d ∗(1 + √200𝑑) = 1442.88 −1154.304𝑑
Solving the nonlinear equation using excels goal seek
𝑑= 0.69354 𝑚=693.54 mm
Figure 4: Critical section for punching shear.
Check punching shear at critical perimeter 1(column face) VEd,red = VEd −∆VEd [𝐸𝑆 𝐸𝑁 1992 −1 −1: 2105, 𝐸𝑞𝑛. 6.48] 𝑉𝐸𝑑= VEd,red𝑢𝑑
[𝐸𝑆 𝐸𝑁 1992 −1 −1: 2105: 𝐸𝑞𝑛. 6.
Figure 5: pictorial representation of VEd and ΔVEd for control perimeter 1
V p V
Ed ,
red ^ P
d
( V
Ed
) [ avg^ (
C
x )^ *^ (
C
y )](^ V
Ed
KN
Punching shear resistance is given by (since ES EN is not clear about this verification Euro code is used):
With θ=22⁰
cot(𝜃) + tan (𝜃) [ES EN 1992 −1 −1: 2015, Eqn. 6.9(with b replaced with u)] 𝑣 1 = 𝑣= 0.6 [1 −𝑓𝑐𝑘250] = 0.6 [1 −25 250] = 0.
𝑧= 0.9𝑑= 0.9 ∗660 = 594 𝑚𝑚 [0.9𝑑 𝑖𝑠 𝑎𝑠𝑠𝑢𝑚𝑒𝑑 𝑓𝑜𝑟 𝑣𝑎𝑙𝑢𝑒 𝑓𝑜𝑟 𝑧]
cot(22) = 2.47 ≈2.5 and tan(22) = 0.404 ≈0.
𝑉𝑟𝑑,𝑚𝑎𝑥= cot(𝜃) + tan (𝜃) = 1 ∗2400 ∗594 ∗0.54 ∗14.17^ ∗^10 −^3 = 3761.52^ 𝐾𝑁
𝑉𝑟𝑑,𝑚𝑎𝑥> 𝑉𝑝(𝑉𝐸𝑑,𝑟𝑒𝑑)θ=22⁰ and control perimeter 1 is ok! Control perimeter 2(at 1.5d distance from the column face) Thiszcontrol zperimeterziszfollowingESzENz1992-1-:2105.zArticlez6.4.2(2)].zAndzsomezbooks zusezd zinstead zofz‘1.5d’zwhichzis zveryzconservative.
Shearz VEd,redz=zVEdz−∆VEdz z z z z z z z z z z z z z z z z[𝐸𝑆z𝐸𝑁z 1992 z− 1 z−1:z2105,z𝐸𝑞𝑛.z6.48]
V p VzEd^ z ,^ red P
zd
(z V
zEd
) [ avg^ (z C x 3
z d
) * (z C y 3 z d )]
VzEd )
𝑽𝒑=z3587.939z−373.36z ∗(3z∗0.66z+z0.6)z∗(3z∗0.66z+z0.6)z=z1102. z𝐾𝑁 Resistancez
Vrd,cz=zCrd,c𝑘(100ρ 1 fck) 1 z 3 𝑢𝑑≥𝑉𝑚𝑖𝑛zudz z[ESzENz 1992 z− 1 z−1:z2015,zEqn.z6.47(zwithzbz=zu)]
Crd,cz=z0. 𝛾𝑐
z=z0.18z1.5z=z0.
𝑘=z 1 z+z√ 𝑑
=z 1 z+z√200z 660 z=z1.55z<z 2 z𝑜𝑘!
𝜌 1 z=z𝜌𝑚𝑖𝑛=z0.26z𝑓𝑐𝑡𝑚 𝑓𝑦𝑘
=z0.26z∗
2 2 z0.3z∗(25)z 3
3 z0.3𝑓𝑐 𝑘
𝑓𝑦𝑘 =z0.26z∗^400 =z0.0017z<z0.
𝑢=z 2 z∗(3𝑑+z𝑐𝑥)z+z 2 z∗(3𝑑+z𝑐𝑦) 𝑢=z 2 z∗(3z∗0.66z+z0.6)z+z 2 z∗(3z∗0.66z+z0.6)z=z7.68z𝑚=z10,320z𝑚𝑚
Isolatedzfootingz Pzazgzez z|z 8 Afz
ISOLATED zFOOTINGzDESIGNzES zENz 1992 - 1 - 1:2015z
Designzmoment zperzmeterzwidthz
M 930. z 89
KN m (^) 300. z 288
KN m / m zsd (^) 1.
Reinforcement zcalculation:z
sd f
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z 288
.0z 047 cd bd^ .z^17 *^1000 *^660
2
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K (^) z .0 97 5
( from designz chart)
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Msd yd
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z 83
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𝜌𝑚𝑖𝑛=z0.26z𝑓𝑐𝑡𝑚 𝑓𝑦𝑘
=z0.26z∗
(^2 ) 3 z0.3𝑓𝑐𝑘
=z0.26z∗0.3z∗(25) z 400
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Az mim ^ .0z^0017 *^100 0
mm^2
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z 51 Checkzmaximumzspacingzofzreinforcement. z
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3 z d 3 z*
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Isolatedzfootingz Pzazgzez z|z 10 Afz
