Strap Footing sample.pdf

DESIGN OF STRAP FOOTING AS PER IS 456: 2000

1. Data Axial service load on col 1= Axial service load on col 2= Size of column 1 = Size of column 2 = SBC of soil = Space between columns = Materials

500 kN 800 kN 350 mm 400 mm 200 kN/m2 5 m c/c M20 grade concrete Fe 415 HYSD bars

2. Ultimate stresses fck =

x x

350 mm 400 mm

20 N/mm2 415 N/mm2

fy = 3. Size of footings Service load on column 1 = Service load on column 2 = Self wt. of footing (10%) = Total load = W =

500 kN 800 kN 130 kN 1430 kN 1430 Footing area required = 200 = 7.15 m2 Let L1 and L2 be the lengths of footings under columns 1 and 2 respectively and B be the width of footing Then B(L1 + L2) = 7.15 Assuming B = 1.5 m L1 + L 2 = 4.77 m The centroid of loads from centre of column 2 = x = 1.923 m If x is also the distance of CG of areas from centre of column 2, we have B x L1 (l+b1/2-L1/2) x= B(L1 + L2) Substituting the values we get, 0.5

L1

2

-5.175

L 12

+

9.1666667

Solving we get, L1 =

2.269 m

L2 = Under column 1, adopt footing of size Under column 2, adopt footing of size Net area of footing provided =

2.498 m

Upward soil pressure = Factored soil pressure = P u = = 4. Factored moments

1.5 m x 1.5 m x 7.2 m2 180.6 kN/m2 270.8 kN/m2 0.271 N/mm2

2.4 m 2.4 m <

200 kN/m2

Assuming width of strap beam = 500 mm Cantilever projection of slab beyond beam = 2 Maximum service moment = M = 0.5 p L = Factored moment = M u = 1.5 M = 5. Depth of footing (a) From moment considerations Mu = 0.138 fck b d2 Therefore,

33.85 kNm

√

Effective depth = d =

Mu

0.138 fck b 111 mm

= IS 456-2000

0.5 m 22.57 kNm

(b) From shear considerations VU = pu*(cantilever projection - d)

clause 22.6.2.1

0.28 N/mm2

Assuming shear strength of concrete τc = (for M 20 grade concrete with reinforcement percentage p t = 0.15)

VU /bd 246 mm 200 mm and overall depth = D =

τc = ∴d = Hence adopt effective depth = d = 6. Reinforcement in footing

Mu = 0.87 fy Ast d (1 - Ast fy / bdfck)

IS 456-2000

∴ Ast =

clause G-1.1

Hence provide IS 456-2000 clause 26.5.2.1

Ast = But minimum reinforcement = ∴ Area of distribution bars = Hence provide Ast = Working shear force at a distance of V= Factored shear force = V u =

494 mm2 12 mm diameter bars @ 200 mm c/c 2 565 mm 0.12% of total cross-sectional area 300 mm2 10 mm diameter bars @ 200 mm c/c 2 393 mm 200 mm from the face of column is 54.17 kN 81.25 kN 0.406 N/mm2

Nominal shear stress = τv = VU/bd = IS 456-2000 clause 40.2.1.1 Table 19

100 Ast bd

=

0.283

Permissible shear stress = k s τc = τv 0.413 N/mm2 > SAFE 7. Design of strap beam Load on strap beam = 271 kN/m The strap beam is analysed for maximum bending moment and shear force with column forces at 1 and 2 as reactions. Maximum positive working BM occurs at a distance of 1.496 m from column 1 and has a magnitude of Mmax (positive) = 374 kNm Factored Mmax (positive) = Maximum negative BM at 2 =

561 kNm 195 kNm

250

Factored Mmax (negative) at 2 = Adopt width of strap beam = b =

292.5 kNm 500 mm

√

Effective depth = d =

Mu

IS 456-2000

0.138 fck b = 638 mm Depth required from shear consideration will be larger. Hence adopt effective depth = d = 950 mm and overall depth = D = Mu = 0.87 fy Ast d (1 - Ast fy / bdfck)

clause G-1.1

Substituting the values we get, 14.984

L1

2

L1

-342997.5

+

1000 5 561057692

Solving we get, The tension reinforcement for max BM is given by 1773 mm2 Hence provide 6 bars of ∴ Ast = 2281 mm2 The tension reinforcement for max negative BM is given by Ast = 887 mm2 Hence provide 3 bars of ∴ Ast = 1140 mm2 Maximum working shear force = 405 kN Factored shear force = V u = 608 kN 100 Ast

clause 40.4 Table 19

bd Permissible shear stress = k s τc = Balance shear force = V us =

22 mm diameter

1.280 N/mm2

Nominal shear stress = τv = VU/bd = IS 456-2000

22 mm diameter

=

0.480 0.470 N/mm2 384 kN

>

τv

Also, Vus = 0.87 fy Asv d / Sv Adopt

10 mm diameter 4-legged stirrups ∴ Spacing = Sv =

280 mm

Provide shear reinf

=0

SAFE

mm

2

Ast/bd < or =

τc (N/mm ) 0.15 0.25 0.5

0.28 0.36 #VALUE! 0.48 0.283 0.376

mm

=0

ide shear reinforcement

0.15 0.25

τc (N/mm2) 0.28 0.36

0.5

0.48

0.75

0.56

1

0.62

1.25

0.67

Ast/bd < or =

#VALUE! 0.48

0.470