Design of Rectangular Combined Footing<Program Created by: Engr. Jeremy E Caballes>

P1 DL = 25 kN P2 DL = 25 kN

P1 LL = 31 kN P2 LL = 31 kN

M1 DL = 0.1 kN-m M2 DL = -0.1 kN-m

E = 200,000.00 MPa M1 LL = 0.3 kN-m M2 LL = -0.3 kN-m

f'c = 20.70 MPa

fy = 275.00 MPa

Wconc = 23.50 kN per cu m

Wsoil = 16.50 kN per cu m

qa = 144.00 kPa

OR

qe = kPa

D = 1.5 m

Assume T = 300 mm

0.3

Remarks: SAFE!

Distance from the Column Spacing = 2.1 m

Property Line = m

C1y = 400 mm C2y = 400 mm

0.4 0.4

Main bar Ø = 16 mm

0.016

Concrete cover = 75 mm

0.075

Temperature bar Ø = 12 mm

C1x = 400 mm C2x = 400 mm

Optional W = m 0.4 0.4

Computed dimensions: W = 0.39 m and L = 2.5 m

Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005

Solve for L to give a uniform pressure:

P1 = P1 DL + P2 LL

P1 = 56.000 kN

P2 = P2 DL + P2 LL

P2 = 56.000 kN

M1 = M1 DL + M2 LL

M1 = 0.400 kN-m

M2 = M2 DL + M2 LL

M2 = -0.400 kN-m x-factor = 2 1.7

X1 = 0.200 m

X2 = 2.300 m 0.5

P = P1 + P2 0.000 0.000 -1.5

P = 112.000 kN

P(Xc) = P1(X1) + P2(X2) + M1 +M2 0.000 0.400 -1.110 1.100 -1.100 4.200 2.100

Xc = 1.250 m ≤ X2, OK! 5.000 5.000 -1.110 1.100 -1.100 5.000 2.100

L = 2Xc 5.000 5.000 -1.890 0.500 -1.900 5.000 1.100

L = 2.500 m say 2.500 m 0.000 0.000 -1.890 0.500 -1.900 4.200 1.100

Determine the width, W: 0.000 0.000 -1.110 1.100 -1.100 4.200 2.100

qe = qa - qs - qc

qs = Wsoil x (D - T)

qs = 19.800 kPa 0.000 4.5

qc = Wconc x T 0.000 -6.250

qc = 7.050 kPa

Therefore: qe = 117.150 kPa 0.000 4.5 1.234 3.766 5.000

Thus: A = P/qe 0.000 -6.250 1.234 3.766 5.000

A = 0.956 sq m

W = A/L Pres. fac = 0.006

W = 0.390 m say 0.390 m 0.000 -5.000 0.000 -1.100

Therefore Use: L = 2.500 m 5.000 -5.000 0.800 -1.100

W = 0.390 m 5.000 -6.000 0.800 -1.900

Assume: A = LW 0.000 -6.000 0.000 -1.900

A = 0.975 sq m 0.000 -5.000 0.000 -1.100

Design loads:

Pu1 = 1.4(P1 DL) + 1.7(P1 LL)

Pu1 = 87.700 kN

Pu2 = 1.4(P2 DL) + 1.7(P2 LL)

Pu2 = 87.700 kN

Pu = Pu1 + Pu2

Pu = 175.400 kN

Mu1 = 1.4(M1 DL) + 1.7(M1 LL)

Mu1 = 0.650 kN-m

Mu2 = 1.4(M2 DL) + 1.7(M2 LL)

Mu2 = -0.650 kN-m

Mu = Mu1 + Mu2

Mu = 0.000 kN-m

Assume: qu = P/A qu = 179.897 kPa

qu = 179.897 kPa w1 = 70.16 kN/m kN/m

qu2 = 179.897 kPa w2 = 70.16 kN/m kN/m Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005

2222

Critical Section for One-Way or Direct Shear

70

.16

kN

/m

70

.16

kN

/m

Check for One-Way or Direct Shear:

d = T - 0.5Ø - cover d = 217 mm

(a) Exterior Column.: 0.000 4.500 1.017 3.983 5.000

Xa = X1-C1x/2-d Xa = 0.000 m 0.000 -6.250 1.017 3.983 5.000

Xb = X1+C1x/2+d Xb = 0.617 m

qua = 179.897 kPa

qub = 179.897 kPa

Vua = 0.000 kN 0 0.000 -0.883 3.983 -0.883

Vub = -44.411 kN 44.41128 1.017 -0.883 5.000 -0.883

(b) Interior Column: 1.017 -2.117 5.000 -2.117

Xc = X2-C2x/2-d Xc = 1.883 m 0.000 -2.117 3.983 -2.117

Xd = X2+C2x/2+d Xd = 2.500 m 0.000 -0.883 3.983 -0.883

quc = 179.897 kPa

qud = 179.897 kPa V Scale = 0.022805017

Vuc = 44.411 kN 0.000 -5.000

Vud = 0.000 kN 0.000 -5.000

0.4 -7

Governing Vu = 44.411 kN 1.234 -6.013

2.500 -5.000

ØVc = Ø(1/6)√(f'c)bd 3.766 -3.9872

ØVc = 54.548 kN SAFE! 4.6 -5.9872

5.000 -5

Check for Two-Way or Punching Shear: 5.000 -5.000

(a) Exterior Column.:

Xa = X1-C1x/2-d/2 Xa = 0.000 m

Xb = X1+C1x/2+d/2 Xb = 0.509 m

qua = 179.897 kPa

qub = 179.897 kPa

x1 = 0.509 m

y1 = 0.617 m

Vu = Pu1 - 0.50(qua+qub)x1y1

Vu = 31.258 kN ßc = 1.000

bo = 1.634 m 4 4

ØVc = Ø(1/3)√(f'c)bod (1/3) 0.3333333333

ØVc = 457.083 kN SAFE!

(b) Interior Column:

Xc = X2-C2x/2-d/2 Xc = 1.992 m

Xd = X2+C2x/2+d/2 Xd = 2.500 m

quc = 179.897 kPa

qud = 179.897 kPa

x2 = 0.508 m 2.251

y2 = 0.617 m

Vu = Pu2 - 0.50(quc+qud)x2y2

Vu = 31.258 kN ßc = 1.000

bo = 1.634 m 4 4

ØVc = Ø(1/3)√(f'c)bod (1/3) 0.3333333333

ØVc = 457.083 kN SAFE!

Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005

3

222

43

2

Critical Section for Two-Way or Punching Shear

70

.16

kN

/m

70

.16

kN

/m

Design of Reinforcement:

(a) Long direction. (d = T - 0.5Ø - cover) d = 217.000 mm

0.000 4.500 0.800 2.500 4.200 5.000

Xa = X1 - d/2 Xa = 0.000 m 0.000 -6.250 0.800 2.500 4.200 5.000

Xb = X1 + d/2 Xb = 0.400 m

Xc = X[V=0] Xc = 1.250 m 0 0.01

Xd = X2 - d/2 Xd = 2.100 m 0 6 0.01

Xe = X2 + d/2 Xe = 2.500 m 350.8

-438.5

Mua = 0.000 kN-m Mua = 0.000E+0 N-mm

Mub = -11.277 kN-m Mub = -11.277E+6 N-mm

Muc = -36.623 kN-m Muc = -36.623E+6 N-mm

Mud = -11.277 kN-m Mud = -11.277E+6 N-mm

Mue = 0.000 kN-m Mue = -63.949E-9 N-mm

4√f`cor

1.4

fy fy

= 0.004 = 0.0050909091

0.00509

Ab = πd² / 4

Ab = 201.062 sq mm

Mu = 0.000 -11.277 -36.623 -11.277 0.000

Section a b c d e

Mu = 0.000E+0 11.277E+6 36.623E+6 11.277E+6 63.949E-9

b = S = 390.000 390.000 390.000 390.000 390.000

Rn = 0.00000 0.68230 2.21575 0.68230 0.00000

act p = 0.00000 0.00253 0.00864 0.00253 0.00000

Use: p = 0.00509 0.00509 0.00864 0.00509 0.00509

As = 430.844 430.844 731.267 430.844 430.844

n = 3 3 4 3 3

Soc = 112 112 74 112 112

Scl = 96.000 96.000 58.000 96.000 96.000

Location = Bottom Top Top Top Bottom

112.000 112.000

112.000 74.000 112.000

Min Spacing: Bottom Top n

a-b 112.000 3

c 74.000 4

d-e 112.000 3

Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005

ρmin = ρmin =

Use ρmin =

2 222

Critical Section for Bending (Long Direction)

70

.16

kN

/m

70

.16

kN

/m

(a) Short direction. (d = T - 0.5Ø - cover) d = 217.000 mm 0.000 4.500 1.126 3.875 5.000

d factor = 0.75 0.217 m 0.000 -6.250 1.126 3.875 5.000

b1 = C1x + 0.75d b2 = C2x + 0.75d 0.000 -1.110 3.875 -1.110

b1 = 562.750 mm b2 = 562.750 mm 1.126 -1.110 5.000 -1.110

d1 = W/2 - C1y/2 d2 = W/2 - C2y/2 1.126 -1.100 5.000 -1.100

d1 = -5.000 mm d2 = -5.000 mm 0.000 -1.100 3.875 -1.100

0.000 -1.110 3.875 -1.110

Xa = 0.000 m

Xb = 0.563 m

Xc = 1.937 m

Xd = 2.500 m

M1 = qu(b1)(d1²/2) M2 = qu(b2)(d2²/2)

M1 = 1,265 N-mm M2 = 1,265 N-mm

Mu = 1,265 1,265

Section b1 x d1 b2 x d2

Mu = 1.265E+3 1.265E+3

b = W = 562.750 562.750

Rn = 0.00005 0.00005

act p = 0.00000 0.00000

Use: p = 0.00509 0.00509

As = 621.685 621.685

n = 4 4

Soc = 137 137

Scl = 121.000 121.000

Location = Bottom Bottom

Temperature/Shrinkage Reinforcement: Ø = 12 mm

Ast = 0.002bh

Ast = 600.000 sq mm

n = 5.305 say 6

Soc = 200.000 mm

Scl = 188.000 mm > 25 mm OK!

n = 0 to 0 m

n = 5 0.563 m to 1.937 m

n = 2 2.5 m to 2.5 m

n = 2 W

Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005

1435

2222

Critical Section for Bending (Short Direction)

qu

= 1

79

.89

7 k

Pa

qu

= 1

79

.89

7 k

Pa

f'c = 20.7 MPa Increase = -3.5

fy = 275 MPa 0.000 0.600 0.000 2.000 4.200 2.000

Wconc = 23.5 kN per cu m 5.000 0.600 0.800 2.000 5.000 2.000

Wsoil = 16.5 kN per cu m 5.000 0.000 0.800 0.600 5.000 0.600

qa = 144 kPa 0.000 0.000 0.000 0.600 4.200 0.600

T = 300 mm 0.000 0.600 0.000 2.000 4.200 2.000

Edge Dist. = 0.2 m

Col Dist. = 2.1 m

Edge Dist. = 0.2 m 0.000 -3.110 0.000 -3.100 4.200 -3.100

Ext. Col: 400 mm x 400 mm 5.000 -3.110 0.800 -3.100 5.000 -3.100

Int. Col: 400 mm x 400 mm 5.000 -3.890 0.800 -3.900 5.000 -3.900

Main bar Ø = 16 mm 0.000 -3.890 0.000 -3.900 4.200 -3.900

Concrete cover = 75 mm 0.000 -3.110 0.000 -3.100 4.200 -3.100

Temperature bar Ø = 12 mm

L = 2.5 m 0.166 -3.110 1.1255 3.8745 5 5.000 0

W = 0.39 m 0.166 -6.500 1.1255 3.8745 5 5.000 -3.110

D = 1.5 m

d = 217 mm Edge Dist. = 0.166

0.166 -3.276 -1 -3.110 0.166 -6.5 -0.500 -3.276

4-16 @ 137 mm 4.834 -3.276 -1 -3.890 0.166 -6.5 -0.500 -3.724

5-12 @ 200 mm 4.834 -3.724 1.1255 -6.5

4-16 @ 137 mm 0.166 -3.724 6.000 -3.276 3.8745 -6.5

0.166 -3.110 6.000 -3.724 5 -6.5

Bottom 3-16 @ 112 mm 4.834 -6.5

Top 4-16 @ 74 mm 0.166 -3.724 0.166 0.166

Bottom 3-16 @ 112 mm 0.166 -6.5 4.834 0.166 0.000 -1

0.400 -1

4.834 -3.724 0.000 0.434 4.6 -1

4.834 -6.500 5.000 0.434 5.000 -1

Optional W = m -1 -3.110 -1 3.000 0.4 -1.000

Computed dimenComputed dimension 0.000 -3.110 6.000 3.000 0.4 -3.500

-1 -3.890 -1 0.000 4.6 -1.000

0 -3.890 6.000 0.000 4.6 -3.500

-0.500 -3.276 -1 3.000 -0.5 0.600

6.000 -3.276 -1 0.000 -0.5 0.166

-0.500 -3.724 6.000 0.600 -0.5 0.600 -0.5 0.166

6.000 -3.724 6.000 0.000 6.000 0.600 0.000 0.166

-0.500 -3.724 0.000 -0.5 -0.5 0.166

0 -0.5

6.000 -3.724 5.000 -0.5 0.000 0.166

Reinforcement Detail of Combined Rectangular FootingBy: Engr. Jeremy E. Caballes

L = 2.5 m

W =

0.3

9 m

T =

30

0

mmD

= 1

.5 m

Col Dist. = 2.1 m

Ext. Col: 400 mm x 400 mm

4-16 @ 137 mm 4-16 @ 137 mm

5-12 @ 200 mm

3-1

6 @

11

2 m

m

4-1

6 @

74

mm

Bo

tto

m

To

p

Edge Dist. = 0.2 m

3-1

6 @

11

2 m

m

Bo

tto

m

f'c = 20.7 MPa

fy = 275 MPa

Wconc = 23.5 kN per cu m

Wsoil = 16.5 kN per cu m

qa = 144 kPa

d =

21

7 m

m

Main bar Ø = 16 mm

Concrete cover = 75 mm

Temperature bar Ø = 12 mm

Int. Col: 400 mm x 400 mm