Rectangular Footing

CONCENTRIC COLUMNSI

Units: 1 English

INT

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SI)

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Design of Square/Rectangular Isolated Footing<Program Created by: Engr. Jeremy E. Caballes>

For SI Units put E = 200000 P DL = 8 kN

For English Units put E = 29000000 P LL = 8 kN

M DL =

E = 200000 MPa M LL =

f'c = 20.7 MPa

fy = 275 MPa

Wconc = 23.5 kN per cu m

Wsoil = 18 kN per cu m H DL =

qa = 144 kPa H LL =

OR

(OPTIONAL) qe = kPa

SAFE ! h =

Assume T = 200 mm

Shear: SAFE!!!

Bending: OK!

C2 = 250 mm

bar Ø = 16 mm

Concrete cover = 75 mm

Remarks: Standard Hook is Required...

Allow Clear Spacing = 25 C1 = 250

Remarks: OK!!! Computed Length, L = 0.96 m

(OPTIONAL) L = 1

0.96

TH

IS P

OR

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OT

AV

AIL

AB

LE

...


For SI Units put E = 200000 P DL = 185 kips

For English Units put E = 29000000 P LL = 150 kips

M DL =

E = 29000000 psi M LL =

f'c = 3000 psi

fy = 60000 psi

Wconc = 150 lbs per cu ft

Wsoil = 100 lbs per cu ft H DL =

qa = 4000 psf H LL =

OR

(OPTIONAL) qe = psf

SAFE ! h =

Assume T = 24 in

Shear: SAFE!!!

Bending: OK!

C2 = 18 in

bar Ø = 0.875 in

Concrete cover = 3 in


Allow Clear Spacing = 1 C1 = 18


(OPTIONAL) L =

0.96

CONCENTRIC COLUMN ECCENTRIC COLUMNUnits: 1

Design of Square/Rectangular Isolated Footing

INT

ER

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L S

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M (

SI)

UN

ITS

Design of Square/Rectangular Isolated Footing<Program Created by: Engr. Jeremy E. Caballes> <Program Created by: Engr. Jeremy E. Caballes>

For SI Units put E = 200000

For English Units put E = 29000000

3 kN-m

3 kN-m E =

f'c =

fy =

Wconc =

kN Wsoil =

kN qa =

D = 0.5 m (OPTIONAL) qe =

m

(Optional) Clear Edge Distance =

Width, S = 0.5 m

OR

L/S Ratio = 1.2

(OPTIONAL)

bar Ø =

Concrete cover =

mm

Allow Clear Spacing =

Computed Length, L = 0.96 m Remarks:

m

Design of Square/Rectangular Isolated Footing

TH

IS P

OR

TIO

N I

S N

OT

AV

AIL

AB

LE

...

Design of Square/Rectangular Isolated Footing<Program Created by: Engr. Jeremy E. Caballes> <Program Created by: Engr. Jeremy E. Caballes>

For SI Units put E = 200000

For English Units put E = 29000000

kips-ft

kips-ft E =

f'c =

fy =

Wconc =

kips Wsoil =

kips qa =

D = 5 ft (OPTIONAL) qe =

ft

(Optional) Clear Edge Distance =

Width, S = 0.5 ft

OR

L/S Ratio =

(OPTIONAL)

bar Ø =

Concrete cover =

Allow Clear Spacing =

Computed Length, L = 0.96 m Remarks:

ft

ECCENTRIC COLUMNSIEnglish




M DL = 1 kN-m

200000 MPa M LL = 1 kN-m

20.7 MPa

275 MPa

23.5 kN per cu m

18 kN per cu m H DL = 25 kN

144 kPa H LL = 222 kN

OR

kPa D = 3

SAFE ! h = m

Assume T = 300

Shear: SAFE!!!

Bending: OK!!! Computed offset: 0.32 m to make the pressure uniform

(Optional) Clear Edge Distance = offset = 0.32 m from the center

Width, S = 3

OR

C2 = 300 L/S Ratio =

(OPTIONAL)

16

75

25

OK!!!

C1 = 300 mm

Computed Length, L = 3.99 m

(OPTIONAL) L = m

3.99

AG30

You should input the length L of the footing.




M DL = kN-m

29000000 MPa M LL = kN-m

3000 MPa

60000 MPa

150 kN per cu m

100 kN per cu m H DL = kN

4000 kPa H LL = kN

OR

kPa D = 5

SAFE ! h = m

Assume T = 24 mm

Shear: SAFE!!!

Bending: OK! Computed offset: 0.32 m to make the pressure uniform

(Optional) Clear Edge Distance = m offset = m from the center

Width, S = 0.5

OR

C2 = 18 mm L/S Ratio =

(OPTIONAL)

0.875 mm

3 mm

1 mm

OK!!!

C1 = 18 mm


(OPTIONAL) L = m

3.99

AG80

You should input the length L of the footing.

ECCENTRIC COLUMN


m

Computed offset: 0.32 m to make the pressure uniform

m


m

Computed offset: 0.32 m to make the pressure uniform

m




M DL = 3 kN-m

E = 200000 MPa M LL = 3 kN-m

f'c = 20.7 MPa

fy = 275 MPa


Wsoil = 18 kN per cu m H DL = 0 kN

qa = 144 kPa H LL = 0 kN

OR

(OPTIONAL) qe = 0 kPa D = 0.5 m

SAFE ! h = 0 m

Assume T = 200 mm

Shear: SAFE!!!

Bending: OK!

Width, S = 0.5 m

C2 = 250 mm OR

0.25 L/S Ratio = 1.2

(OPTIONAL)

bar Ø = 16 mm



Allow Clear Spacing = 25 mm C1 = 250 mm

0.25


(OPTIONAL) L =

0.96

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

Effective bearing capacity:

qe = qa - qsoil - qconc

qconc = 4.700 kPa

qsoil = 5.400 kPa

qe = 133.900 kPa 133.9 kPa

Determine the footing dimension: X Origin = 0 Y Origin = -1.5

P = P DL + P LL Scale = 0.001041667 Scale = 0.001389207

P = 16.000 kN qmin is positive qmin is negative

M = M DL + M LL + (H DL+H LL)(T+h) r is given S is given r is given S is given -0.13020833 0 -0.5 -0.6

M = 6.000 kN-m 133.900 133.900 1.000 0.478 -0.13020833 -0.29166667 -0.5 -0.53971963

0.000 -32.000 -0.750 0.130208333 -0.29166667 0.5 -0.75

-19.200 -72.000 -0.191 0.130208333 0 0.5 -0.6

e = M/P -43.200 -0.13020833 0 -0.5 -0.6

e = 0.375 m 0.755 0.862 0.951 1.069

L = 0.755 L = 0.951 -0.5 -0.29166667 -0.675 -1.26458333

q = qe = P(1+6e/L) -0.5 -0.5 1 -1.26458333

LS 0.5 -0.5

sq m 0.5 -0.29166667 0.5 -1.08333333

L = 0.755 m -0.5 -0.29166667 1 -1.08333333

S = 0.629 m

qmax = P(1+6e/L)/(LS) -0.5 -1.08333333 -1 -1.5

qmax = 133.900 kPa SAFE ! -0.5 -1.91666667 1 -1.5

qmin = P(1-6e/L)/(LS) 0.5 -1.91666667

qmin = -66.594 kPa SAFE ! 0.5 -1.08333333 -1 -1.08333333

Tension/Upliftment! -0.5 -1.08333333 -0.5 -1.08333333

Use: L = 0.951 m say 0.960 960 -0.13020833 -1.36979167 -1 -1.08333333

S = 0.793 m say 0.800 800 -0.13020833 -1.63020833 -1 -1.5

Design Loads: L/2 = 0.48 m 0.130208333 -1.63020833

Pu = 1.4DL + 1.7LL 0.5C1 + d = 0.242 m 0.130208333 -1.36979167 1 -1.08333333

Pu = 24.800 kN S/2 = 0.4 m -0.13020833 -1.36979167 1 -1.26458333

Mu = 1.4MDL + 1.7MLL + (1.4HDL+1.7HLL)(T+h) 1 -1.5

Mu = 9.300 kN-m 0.252083333 -0.11666667

Assume: eu = e 0.252083333 -2.05

eu = 0.375 m

qu max = Pu(1+6eu/L)/(LS) 0 -0.11666667 0 -2.25

qu max = 107.975 kPa qu max = 107.975 kPa 0 -2.25 0.5 -2.25

qu min = Pu(1-6eu/L)/(LS)

qu min = -43.392 kPa qu min = -43.392 kPa 0.5 -1.91666667 0 -2.05

0.5 -2.25 0.252083333 -2.05

0.5 -2.05

Critical Section for One-way Shear

qu min = -43.392 kPa

qs = 37.526 kPa

x = 0.238 m

y =

0

.17

4 m

S/2

= 0

.4

m

L/2 = 0.48 m

0.5C1 + d = 0.242 m

0.5

C2

+ d

=

0.2

26

m

qu max = 107.975 kPa


Check for One-Way or Direct Shear: -0.1828125 -1.6828125

(a) Transverse -0.1828125 -1.3171875

Eff. d = T - 0.5Ø - cover 0.117 0.1828125 -1.3171875

d = 117.000 mm 0.5C1 + d = 242 mm 0.1828125 -1.6828125

b = S -0.1828125 -1.6828125

b = 800.000 mm S/2 = 400 mm

Thus use: -0.1828125 -1.6828125

a = 0.685 m (compression zone) -0.1828125 -2.25


qu min = 0.000 kPa 0.1828125 -1.6828125

(Neglect Upliftment Pressure) 0.1828125 -2.25

x = L/2-C1/2-d

x = 0.238 m x = 0.238 m -0.1828125 -2.25

qs = 37.526 kPa qs = 37.526 kPa 0.1828125 -2.25

Vu = 0.5(qs+qumax) S xs 0.1828125 -1.6828125

Vu = 6.926 kN 1 -1.6828125

Vu = 6,926 N

ØVc = Ø(1/6)√(f'c)bd 0.1828125 -1.3171875

ØVc = 60,329 N SAFE! 1 -1.3171875

Check for d = 36.005 mm SAFE!

(b) Longitudinal 1 -1.3171875

Eff. d = T - 1.5Ø - cover 0.101 1 -1.6828125

d = 101.000 mm 0.5C2 + d = 0.226 m

b = L -1 -1.5

b = 960.000 mm 0.25 -1.5

y = S/2-C2/2-d

y = 0.174 m y = 0.174 m

Vu = 0.5(qumin + qumax) a y

Vu = 6.433 kN

Vu = 6,432.907 N

ØVc = Ø(1/6)√(f'c)bd

ØVc = 62,495 N SAFE!

Check for d = 21.254 mm SAFE!

Check for Two-Way or Punching Shear:

No failure for this condition!

Eff. d 101.000 mm C2 + d = 351 mm

C1 + d = 351 mm

ßc = 1.000

4 4

(1/3) 0.333333333 5.264330562

Critical Section for Two-way Shear

qu min = -43.392 kPa qu max = 107.975 kPa

S/2

= 0

.4

m

L/2 = 0.48 m

C1 + d = 351 mm

C2

+ d

= 3

51

mm

1343.076456

-76176.5462


Design of Reinforcement:

(a) Long Direction. √f`cor

1.4

Eff. d = T - 1.5Ø - cover 4fy fy

d = 101.000 mm = 0.004136114 = 0.005090909 0 -0.11666667

b = S 0.00509 0 -1.75

b = 800.000 mm

Thus use: -0.5 -0.83333333


qu min = 0.000 kPa

qu max = 107.975 kPa -0.5 -0.95

0 -0.95

x = L/2-C1/2

x = 0.355 m x = 0.355 m 0.5 -1.08333333

qm = 52.001 kPa 1 -1.08333333

Mu = 4.502 kN-m

Mu = 4,502,478 N-m -0.675 -1.36979167

Rn = Mu 1 -1.36979167

Øbd²

Rn = 0.613 MPa 1 -1.36979167

Act p = 0.00227 1 -1.5

Min p = 0.00509 1 -1.08333333

Use p = 0.00509

As = pbd 0.130208333 -0.11666667

As = 411.345 sq mm 0.130208333 -2.05

Ab = 201.062 sq mm

n = As/Ab 0.130208333 -2.05

n = 2.046 say 3 0.5 -2.05

Soc = (S - Ø - 2cover)/(n - 1)

Soc = 317.000 mm

Scl = Soc-Ø

Scl = 301.000 mm OK!

α = 1.00

β = 1.00

αβ = 1.00 ≤ 1.70 Use: 1.00

γ = 0.800 for ≤ 19 mm

λ = 1.00 Normal Weight Concrete

c = side cover = 83.000 mm

c = 0.5Soc = 158.500 mm Use: c = 83.000 mm

Ktr = 0.000

(c + Ktr)/db = 5.188 ≤ 2.5 Use: 2.500

ld/db = 9fyαβγλ

10√f'c(c+Ktr)/db Allow Ld = Ld x Mod Factor

ρmin = ρmin =

Use: ρmin =

Critical Section for Bending

qu min = -43.392 kPa

qs = 37.526 kPa


x = 0.355 m

y =

0

.27

5 m

S/2

= 4

00

mm

L/2 = 0.48 m

0.5

C1

= 0

.25

m

ld/db = 17.408 diameters Allow Ld = 189.940 mm

Ld = 278.522 mm Actual Ld = x - cover

Mod. Factor = (req As)/(prov As) Actual Ld = 280.000 mm OK !

prov As = 603.186 sq mm

Mod. Factor = 0.682 Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005

(b) Short Direction. Ktr = 0.000

Eff. d = T - Ø/2 - cover (c + Ktr)/db = 4.688 ≤ 2.5 Use: 2.500

d = 117.000 mmld/db =

9fyαβγλ

b = L 10√f'c(c+Ktr)/db

b = 960.000 mm ld/db = 17.408 diameters

y = S/2-C2/2 Ld = 278.522 mm

y = 0.275 m y = 0.275 m Mod. Factor = (req As)/(prov As)

0.5C1 = 0.25 m prov As = 603.186 sq mm

Mu = 1.398 kN-m Mod. Factor = 0.948

Mu = 1,397,956 N-m Allow Ld = Ld x Mod Factor

Rn = Mu Allow Ld = 264.035 mm

Øbd² Actual Ld = S/2 - cover

Rn = 0.118 MPa Actual Ld = 200.000 mm Standard hook is required!

Act p = 0.00043

Min p = 0.00509 Check for Bearing Stress:

Use p = 0.00509 Act Pb = 1.4PDL+1.7PLL C1/C2 = 1.000

As = pbd Act Pb = 24,800 N L/S = 0.833

As = 571.811 sq mm A1= C1 x C2 S' = 0.800 m

Ab = 201.062 sq mm A1= 62,500 sq mm L' = 0.800 m

n = As/Ab A2= S' x L'

n = 2.844 say 3 A2= 640,000 sq mm

Nbw = 2N √(A2/A1) = 3.200 ≤ 2.0 Use: 2.000

ß + 1 All Pb = Ø 0.85 f'c A1 √(A2/A1)

ß = L/S All Pb = 1,539,563 N Dowels are not Required!

ß = 1.200

Nbw = 2.585 say 3

Band width = S

Band width = 800.000 mm

Soc = (Band Width)/(n-1)

Soc = 400.000 mm

Scl = Soc-Ø

Scl = 384.000 mm

Now = (n - Nbw)/2

Now = 0.000 say 0

Outer width = (L - Bandwidth - Ø - 2cover)/2

Outer width = -3.000 mm

Soc = (Outer Width-cover)/(Now)

Soc = mm

Scl = Soc-Ø

Scl = mm

0

α = 1.00

β = 1.00

αβ = 1.00 ≤ 1.70 Use: 1.00

γ = 0.80 for ≤ 20 mm

λ = 1.00 Normal Weight Concrete

c = side cover = 75.000 mm

c = 0.5Soc = 200.000 mm Use: c = 75.000 mm Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005

Details of Reinforcement: 0.96 x 0.8 (Using 16 mm Ø)

(250 mm x 250 mm) RC-Column/Pedestal Increase = -0.25

Dowels are not Req'd!!! -0.75 -0.44166667 -0.75 -0.441666667 -0.41666667 -1.16979167 0.5 -1.08333333

T = 200 mm 1 -0.44166667 -0.75 -0.530208333 -0.41666667 -1.83020833 0.5 -0.75

D = 0.5 m

d = 117 mm -0.75 -0.53020833 1 -0.441666667 0.416666667 -1.16979167 -0.5 -0.75

Edge dist. = 83 mm -0.5 -0.53020833 1 -0.65 0.416666667 -1.83020833 0.5 -0.75

S = 0.8 m

0 @ 0 mm -0.41354167 -0.53020833 -0.41354167 -0.563541667 -0.41666667 -1.83020833 -0.41354167 -2.3

L = 0.96 m 0.413541667 -0.53020833 0.413541667 -0.563541667 -0.41666667 -2.25 0.413541667 -2.3

3 @ 400 mm

3 @ 317 mm -0.13020833 0.25 -1 -0.129166667 0.416666667 -1.83020833 -0.5 -1.08333333

f'c = 20.7 MPa -0.13020833 -0.44166667 1 -0.129166667 0.416666667 -2.25 -0.5 -0.75

fy = 275 MPa 0.130208333 -0.44166667

Wconc = 23.5 kN per cu m 0.130208333 0.25 -1 -0.65 0.5 -1.16979167 0 -0.83333333

Wsoil = 18 kN per cu m -0.13020833 0.25 1 -0.65 1 -1.16979167 0 -2.25

qa = 144 kPa

bar Ø = 16 mm -0.5 -0.44166667 -1 -0.129166667 0.5 -1.83020833 -1 -1.91666667

Concrete cover = 75 mm -0.5 -0.65 -1 -0.65 1 -1.83020833 -0.5 -1.91666667

Dowels, not required! 0.5 -0.65

0.5 -0.44166667 0.413541667 -1.830208333 -0.41354167 -2.25 -1 -1.08333333

-0.5 -0.44166667 0.413541667 -2.25 -0.41666667 -2.25 -1 -1.91666667

0.416666667 -2.25

-0.41354167 -1.16979167 -0.41354167 -1.830208333 0.413541667 -2.25 -0.75 -1.5

-0.41354167 -1.83020833 -0.41354167 -2.25 0.75 -1.5

0.413541667 -1.83020833

0.413541667 -1.16979167 1 -1.169791667 -1 -1.08333333

-0.421875 -1.16145833 1 -1.830208333 -0.5 -1.08333333

3

400

Footing Reinforcement DetailProgrammed by: Engr. Jeremy E. Caballes

L = 0.96 m

3 @ 400 mm0 @ 0 mm 0 @ 0 mm

T = 200 mmD = 0.5 m

d =

11

7 m

m

Edge dist. = 83 mm

S =

0.8

m

3 @

31

7 m

m

(250 mm x 250 mm) RC-Column/Pedestal

f'c = 20.7 MPa

fy = 275 MPa


Wsoil = 18 kN per cu m

qa = 144 kPa

bar Ø = 16 mm


Dowels, not required!

Footing Reinforcement DetailProgrammed by: Engr. Jeremy E. Caballes

L = 0.96 m

3 @ 400 mm0 @ 0 mm 0 @ 0 mm

T = 200 mmD = 0.5 m

d =

11

7 m

m

Edge dist. = 83 mm

S =

0.8

m

3 @

31

7 m

m


f'c = 20.7 MPa

fy = 275 MPa



qa = 144 kPa

bar Ø = 16 mm






M DL = 1 kN-m

E = 200000 MPa M LL = 1 kN-m

f'c = 20.7 MPa

fy = 275 MPa


Wsoil = 18 kN per cu m H DL = 25 kN

qa = 144 kPa H LL = 222 kN

OR

(OPTIONAL) qe = kPa D = 3 m

SAFE ! h = 0 m

Assume T = 300 mm

0.3

Shear: SAFE!!!

Bending: OK!!! Computed offset: 0.32 m to make the pressure uniform

(Optional) Clear Edge Distance = m offset = 0.32 m from the center

Width, S = 3 m

OR

C2 = 300 mm L/S Ratio = 0

0.3

bar Ø = 16 mm


Allow Clear Spacing = 25 mm

Remarks: OK!!!

C1 = 300 mm

0.3


(OPTIONAL) L = 0 m

3.99


0.32

Effective bearing capacity:

qe = qa - qsoil - qconc

qconc = 7.050 kPa

qsoil = 48.600 kPa

qe = 88.350 kPa 88.35

Determine the footing dimension:

P = P DL + P LL

P = 238.000 kN q2 is positive q2 is negative X Origin = 0 X Origin = -2

M = M DL + M LL + (H DL+H LL)(T+h) r is given S is given r is given S is given Scale = 0.000250627 Scale = 0.007610631

M = 76.100 kN-m 88.350 88.350 1.000 1.796

M/P = 0.320 m 0.000 -238.000 0.000 -0.117794486 0 -0.5 -0.75

offset = 0.320 m 0.000 -456.600 0.000 -0.117794486 -0.42481203 -0.5 -1

e = M/P - offset 0.000 -0.042606516 -0.42481203 0.5 -1

e = 0.000 m 0.000 3.989 0.000 1.197 -0.042606516 0 0.5 -0.75

L = 3.989 L = 1.197 -0.117794486 0 -0.5 -0.75

A = P

qe -0.5 -0.42481203 -0.675 -1.91203008

A = 2.694 sq m -0.5 -0.5 1 -1.91203008

L = 3.989 m 0.5 -0.5

S = 3.000 m 0.5 -0.42481203 0.5 -1.62406015

q = P/A -0.5 -0.42481203 1 -1.62406015

q = 19.886 kPa SAFE !

q2 = P(1-6e/L)/(LS) -0.5 -1.62406015 -1 -2

q2 = 19.886 kPa SAFE ! offset = 0.32 m -0.5 -2.37593985 1 -2

L/2 + offset = 2.315 m 0.5 -2.37593985

L/2 - offset = 1.675 m 0.5 -1.62406015 -1 -1.62406015

Use: L = 3.989 m say 3.990 3990 mm -0.5 -1.62406015 -0.5 -1.62406015

S = 3.000 m say 3.000 3000 mm

Design Loads: 0.5C1 + d = 0.367 m -0.117794486 -1.96240602 -1 -1.62406015

Pu = 1.4DL + 1.7LL S/2 = 1.5 m -0.117794486 -2.03759398 -1 -2

Pu = 393.200 kN -0.042606516 -2.03759398

Mu = 1.4MDL + 1.7MLL + (1.4HDL+1.7HLL)(T+h) -0.042606516 -1.96240602 1 -1.62406015

Mu = 126.820 kN-m -0.117794486 -1.96240602 1 -1.91203008

Assume: eu = e 1 -2

eu = 0.000 m 0.011779449 -0.24981203

qu = Pu/A 0.011779449 -2.75

qu = 32.849 kPa qu = 32.849 kPa

qu min = Pu(1-6eu/L)/(LS) -0.080200501 -0.24981203 -0.0802005 -2.75

qu2 = 32.849 kPa qu = 32.849 kPa -0.080200501 -2.75 0.5 -2.75

0.5 -0.75 -0.0802005 -2.75

0.5 -2.75 0.011779449 -2.75

0.5 -2.75


Critical Section for One-way Shear

Pressure Diagram

xs = 1.948 m

y =

1

.14

9 m

S/2

= 1

.5

m

qu = 32.849 kPaqu = 32.849 kPa

offset = 0.32 mL/2 - offset = 1.675 m

0.5C1 + d = 0.367 m

0.5

C2

+ d

=

0.3

51

m

Check for One-Way or Direct Shear:

(a) Transverse -0.14298246 -2.06278195

Eff. d = T - 0.5Ø - cover 0.217 -0.14298246 -1.93721805

d = 217.000 mm -0.01741855 -1.93721805

b = S -0.01741855 -2.06278195

b = 3,000.000 mm -0.14298246 -2.06278195

Thus use:


a = 3,990.000 mm -0.14298246 -2.75

qu1 = 32.849 kPa

qu2 = 32.849 kPa -0.01741855 -2.06278195

xs = a - (L/2 - offset + 0.5C1 + d) -0.01741855 -2.75

xs = 1.948 m xs = 1.948 m

qs = 32.849 kPa qs = 32.849 kPa -0.14298246 -2.75

-0.01741855 -2.75

Vu = qu xs S

Vu = 191.968 kN -0.01741855 -2.06278195

Vu = 191.968E+3 N 1 -2.06278195

ØVc = Ø(1/6)√(f'c)bd -0.01741855 -1.93721805

ØVc = 419.598E+3 N SAFE! 1 -1.93721805

(b) Longitudinal

Eff. d = T - 1.5Ø - cover 0.201 1 -1.93721805

d = 201.000 mm 1 -2.06278195

b = L

b = 3,990.000 mm -1 -2

L/2 - offset = 1,675.000 m 0.169799499 -2

y = S/2 - C2/2 - d 0.5C2 + d = 0.351 m

y = 1.149 m y = 1.149 m

Vu = qu y L

Vu = 150.596 kN

Vu = 150.596E+3 N

ØVc = Ø(1/6)√(f'c)bd

ØVc = 516.918E+3 N SAFE!

Check for Two-Way or Punching Shear:

Eff. d = 201.000 mm C2 + d = 501 mm

bo = 2(c1+d)+2(c2+d) C1 + d = 501 mm

bo = 2,004.000 mm

Vu = qu[LS-(c1+d)(c2+d)] 5.18920409 d/2 = 100.5 mm

Vu = 384.955 kN ßc = 1.000 1566.61586 C1/2 = 150 mm

Vu = 384.955E+3 N 4 4 -390243.609 (C1 + d)/2 = 250.5 mm

ØVc = Ø((1/3))√(f'c)bod (1/3) 0.333333333333 L/2 - offset = 1675 mm

ØVc = 610.883E+3 N SAFE! Difference = -1,424.500 mm

Check for d = 162.082 mm SAFE! Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005

Critical Section for Two-way Shear

Pressure Diagram

qu = 32.849 kPa qu = 32.849 kPa

S/2

= 1

.5

m

C1 + d = 501 mm

C2

+ d

= 5

01

mm

L/2 - offset = 1.675 m offset = 0.32 m

S/2

= 1

.5

m

Design of Reinforcement:

Eff. d = T - 1.5Ø - cover 0.201 √f`cor

1.4

d = 201.000 mm 4fy fy

b = S = 0.004136114 = 0.005090909

b = 3,000.000 mm 0.00509

Thus use: a-a b-b c-c (BW) c-c (OW) c-c (OW)

a = 3.990 m (compression zone) Mu = 230.954 114.591 119.435 119.435 kN-m 119.435 kN-m

qu2 = 32.849 kPa Mu = 230.954E+6 114.591E+6 119.435E+6 119.435E+6 N-mm 119.435E+6 N-mm

qu1 = 32.849 kPa Rn = 2.117 1.050 0.706 0.706 MPa 0.706 MPa

Act p = 0.00823 0.00394 0.00262 0.00262 0.003

xa = a - (L/2 - offset + 0.5C1) 0.00509 0.00509 0.00509 0.00509 0.005

xa = 2.165 m Use p = 0.00823 0.00509 0.00509 0.00509 0.005

qa = 32.849 kPa As = 4,962 3,070 3,784 624 sq mm 624.289 sq mm

xb = L/2 - offset - C1/2 n = 25 16 19 4 4

xb = 1.525 m Width = S S BW OW OW

qb = 32.849 kPa Width = 3,000.000 3,000.000 3,000.000 990 990.000

Soc = 118.000 188.000 166.000 206.000 mm 206.000 mm

yc = S/2-C2/2 Scl = 102.000 172.000 150.000 190.000 mm 190.000 mm

yc = 1.350 m Remarks: OK! OK! OK! OK! OK!

α = 1.00 1.0 1.0 1.0 1.00

Eff. d = T - 0.5Ø - cover β = 1.00 1.0 1.0 1.0 1.00

d = 217.000 mm αβ = 1.00 1.0 1.0 1.0 1.00

b = L αβ ≤ 1.70 Use: 1.00 1.0 1.0 1.0 1.00

b = 3,990.000 mm γ = 0.80 0.80 0.80 0.8 0.80

λ = 1.00 1.0 1.0 1.0 Normal Wt 1.00 Normal Wt

Rn = 0.706 MPa c = side cover = 75.000 75.0 75.0 75.0 mm 75.000 mm

Act p = 0.00262 c = 0.5Soc = 59.000 94.000 83.000 103.000 mm 103.000 mm

Min p = 0.00509 Use: c = 59.000 75.000 75.000 75.000 mm 75.000 mm

Use p = 0.00509 Ktr = 0.000 0.000 0.000 0.000 0.000

As = pbd (c + Ktr)/db = 3.688 4.688 4.688 4.688 4.688

As = 4,407.862 sq mm (c + Ktr)/db ≤ 2.5 Use: 2.500 2.500 2.500 2.500 2.500

Ab = 201.062 sq mmLd/db =

9fyαβγλ17.408 17.408 17.408 17.408 diameters 17.408 diameters

n = As/Ab 10√f'c(c+Ktr)/db

n = 21.923 say 22 Ld = 278.522 278.522 278.522 278.522 mm 278.522 mm

Nbw = 2n prov As = 5,027 3,217 3,820 804 sq mm 804.248 sq mm

ß + 1 Mod. Factor = (req As)/(prov As) = 0.987 0.954 0.990 0.776 0.776

ß = L/S Allow Ld = Ld x Mod Factor 274.922 265.780 275.854 216.200 mm 216.200 mm

ß = 1.330 Actual Ld = xa - cover xb - cover y - cover y - cover y - cover

Nbw = 18.818 say 19 Actual Ld = 2,090.000 1,450.000 1,275.000 1275.000 mm 1,275.000 mm

Now = 3.000 say Remarks: No Hooks! No Hooks! No Hooks! No Hooks! No Hooks!

Offset = 320.000 mm

L - 2cover = 3,840.000 mm

Left OW = L/2 - offset - S/2 - 0.5Ø - cover Right OW = L - BW - Left OW - Ø - 2cover

Left OW = 92.000 mm Right OW = 732.000 mm

n = 0.447 say 0 n = 3.553 say 3

Soc = 0.000 mm Soc = 244.000 mm

Scl = 0.000 mm Scl = 228.000 mm


ρmin = ρmin =

Use: ρmin =

ρmin =

Check for Bearing Stress: xa = 2.165 m -0.0802005 -0.24981203

Act Pb = 1.4PDL+1.7PLL qa = 32.849 kPa -0.0802005 -2.25

Act Pb = 393,200 N C1/C2 = 1.000 230.954495

A1= C1 x C2 L/S = 0.752 xb = 1.525 m 0 -1.25

A1= 90,000 sq mm S' = 3.000 qb = 32.849 kPa 0 -2.25

A2= S' x L' L' = 3.000 114.5909461

A2= 9,000,000 sq mm 0.5C2 = 0.15 m -0.5 -1.25

√(A2/A1) = 10.000 ≤ 2.0 Use: 2.000 yc = 1.35 m -0.5 -1.62406015

All Pb = Ø 0.85 f'c A1 √(A2/A1) 119.4345

All Pb = 2,216,970 N -0.5 -1.25

Dowels are not Required! -0.0802005 -1.25

0 -1.25

-0.675 -1.96240602

1 -1.96240602

1 -1.62406015

1 -1.96240602

1 -2

-0.04260652 -0.24981203

-0.04260652 -2.75

-0.04260652 -2.75

0.5 -2.75

-0.11779449 -0.24981203

-0.11779449 -2.75

-0.5 -1

-0.5 -2.75

-0.5 -2.75

-0.11779449 -2.75

0.5 -1.62406015

1 -1.62406015


Critical Section for Bending

Pressure Diagram

qu = 32.849 kPaqu = 32.849 kPa

S/2

= 1

.5

m yc

=

1.3

5 m

xa = 2.165 m

Details of Reinforcement: 3.99 m x 3 m (Using 16 mm Ø)

Increase = -0.2 Delta y = 0.75

(300 mm x 300 mm) RC-Column/Pedestal -0.75 -0.27481203 -0.75 -0.27481203 0.5 -1.60513784 -0.5 -0.87406015

Dowels, not required! 0.9 -0.27481203 -0.75 -0.321177945 0.9 -1.60513784 -0.5 -1.62593985

T = 300 mm 0.5 -1.62593985

D = 3 m -0.75 -0.32117794 0.9 -0.27481203 0.5 -0.87406015

d = 217 mm -0.5 -0.32117794 0.9 -0.35 -0.5 -0.87406015

Edge dist. = 83 mm

S = 3 m -0.47919799499 -0.32117794 -0.479197995 -0.329197995 -0.479197995 -2 -0.11779449 -1.21240602

b 16 @ 188 mm 0.479197994987 -0.32117794 0.479197995 -0.329197995 -0.456140351 -2 -0.11779449 -1.28759398

Center 19 @ 166 mm 0.295739348 -2 -0.04260652 -1.28759398

Right 3 @ 244 mm -0.11779448622 0.25 -1 0.401879699 0.479197995 -2 -0.04260652 -1.21240602

L = 3.99 m -0.11779448622 -0.27481203 1 0.401879699 -0.11779449 -1.21240602

offset = 0.32 m -0.04260651629 -0.27481203

L/2 - offset = 1.675 m -0.04260651629 0.25 -1 -0.35 -1 -0.87406015 -1 -1.62593985

L/2 = 1.995 m -0.11779448622 0.25 0.9 -0.35 -1 -1.62593985 -0.5 -1.62593985

a 25 @ 118 mm

f'c = 20.7 MPa -0.5 -0.27481203 -1 0.401879699 -0.75 -1.25 -1 -0.87406015

fy = 275 MPa -0.5 -0.35 -1 -0.35 0.75 -1.25 -0.5 -0.87406015

Wconc = 23.5 kN per cu m 0.5 -0.35

Wsoil = 18 kN per cu m 0.5 -0.27481203 0.479197995 -1.62593985 0 -0.55

qa = 144 kPa -0.5 -0.27481203 0.479197995 -2 0 -1.77593985

bar Ø = 16 mm

Concrete cover = 75 mm -0.47919799499 -0.89486216 -0.479197995 -1.62593985 0.295739348 -0.87406015

Left -0.47919799499 -1.60513784 -0.479197995 -2 0.295739348 -2

0.479197994987 -1.60513784

0.479197994987 -0.89486216 0.9 -0.894862155 -0.5 -0.55

-0.48120300752 -0.89486216 0.9 -1.605137845 -0.080200501 -0.55

0 -0.55

-0.45614035088 -0.87406015 0.5 -0.87406015 0.5 -0.55

-0.45614035088 -2 0.5 -0.4

0.295739348371 -0.87406015 -0.5 -0.4 -0.8 -0.89486216

0.295739348371 -2 0.5 -0.4 -0.5 -0.89486216

-0.45614035088 -1.62593985 -0.080200501 -0.55 -0.8 -1.60513784

-0.45614035088 -2 -0.080200501 -1.77593985 -0.5 -1.60513784

0.295739348371 -1.62593985 -0.5 -0.87406015 -0.8 -0.89486216

0.295739348371 -2 -0.5 -0.4 -0.8 -1.60513784

0.5 -0.89486216

0.9 -0.89486216


Footing Reinforcement DetailBy: Engr. Jeremy E. Caballes

L = 3.99 m

19 @ 166 mm 3 @ 244 mm

T = 300 mmD = 3 m

d =

21

7 m

m

Edge dist. = 83 mmS

= 3

m

25

@ 1

18

mm



f'c = 20.7 MPa

fy = 275 MPa



qa = 144 kPa

bar Ø = 16 mm


L/2 - offset = 1.675 m L/2 = 1.995 moffset = 0.32 m

16

@ 1

88

mm

Documents

Rectangular Footing