Pad Footing BS8110

Design Calculation Footing Pad Load by column = Column self wt. = Footing self wt. = Column size =

base, h 430 cover, c = 40 200 kN kN/m2

800 kN

l.l = 300 kN earth bearing presure = 60 kN assumed 400 x 400 mm

SLS design load = = ULS design load = =

1Gk+1Qk 1160 kN 1.4Gk + 1.6 Qk 1600 kN

Require area base =

35 460

Gk = load col+sw footing Gk = only load col

1160 200

Ult. Earth pressure' = assume bar Ø = d= h - c - Ø - 0.5 Ø

fcu = fy =

= =

5.8 m2 2.41 m x 2.5 m x

2.41 m 2.5 m

1.05

0.69

256 kN/m2

20 mm mm =

360 mm

Bending reinforcement

0.4 ep' x strip x width

M=

=

2. 5

0.4

352.8 kNm

2.5

K = M / f cu cubd^2

=

0.03 < K'

Z = d { 0.5 + ( √ 0.25 – K/0.9)} = 0.97d > 0.95d,

0.95 d

As = M / 0.87 f y.z Provide = As min =

=

########

As min < Asc < As max

vc =

(Min 0.24% for mild steel)

43000 mm m m2 O.K

Shear Transverse V= ep' *b *l

=

442

=

400/d =

2578 2578 mm2 mm2

1398 mm2

4bh/100 =

v= V/bd 100As/bvd =

342 mm

2827 mm2

0.13bh/100 =

As max =

=

kN

0.49 N/mm2

0.314111 < 3

O.K

1.111111 > 1

O.K

.'. Take larger

O.K

no shear reinf. Rqd.

0.49 0.49 N/mm N/mm2 2

Punching shear critical perimeter, u =

.'. v < vc

4 x [col width + 2(1.5d)] 5920 mm

= V= 2

ep' * (footing width - criticial section area) 1.5d

=

1039.258 kN kN

v = V/ud = vc =

0.49 N/m2 0.49 N/ N/mm2

max shear at column face ,

.'. v < Vc

O.K 2

V=

ep' * (footing width - column area) 1559.04 kN (also N for column calc.)

= at column face, max shear stress, v =

N/ud =

2.706667

<

0.8√(f cu cu)

or 5 N/mm2 N/mm2

4.732864 N/ N/mm2 Cracking spacing spacing at 3d @ 3d = 1080 mm

750 mm max

Clear spacing =

380 mm

.'. So take <

750 mm

.'. Ok

750 mm O.K

h > 200mm Bent up anchorage bond length , 38 * 20 mm rebar = 760 mm (from column face) (footing width-col width)/2-c-anchor length = 250 mm of space .'. no bent up needed

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