Foundation Design - Programs for Isolated and Combined Footing Design with Estimate.xlsx

A

B

C

1

4

7

2

5

8

3

6

9

7

7

7

7

ISOLATED FOOTING DESIGN

C

D

4.5

E

G

F

I

H

10

13

16

19

22

25

11

14

17

20

23

26

12

15

18

21

24

27

4.5

4.5

4.5

4.5

4.5

J

1

31

28

7 2

32

29

7 33

30

4.5

4.5

3

COMBINED FOOTING DESIGN A

7

B

C

D

E

4

7

10

13

16

19

5

8

11

14

17

20

6

9

12

15

18

21

7

4.5

4.5

4.5

4.5

E

G

F

4.5

I

H

22

25

28

31

23

26

29

32

24

27

30

33

4.5

4.5

4.5

J

DEAD LOADS Component Ceiling Gypsum Board Mechanical Duct Allowance Wood Furring Suspension System Floor and Floor Finishes Cement Finish on Stone Concrete Fill Ceramic Quarry Tile Masonry For Plastering (both sides) LIVE LOADS Basic Floor Area

LOAD COMBINATIONS Load (kPa) 0.008 0.2 0.12 1.53 1.1 0.24

1.9

Load Case LC 1 LC 2 LC 3 LC 4 LC 5 LC 6 LC 7 LC 8

Modification F Dead 1.4 1.2 1.2 1.2 1.2 1.2 0.9 0.9

INATIONS Modification Factor Live Wind 1.4 1.2 1.2 0.8 1.2 1.3 1.2 1.2 0.9 0.9 1.3

Seismic

1 1 1

RESULT OF STRUCTURAL ANALYSIS IN THE GROUND FLOOR COLUMNS Column C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 C12 C13 C14 C15 C16 C17 C18 C19 C20 C21 C22 C23 C24 C25 C26 C27 C28 C29 C30 C31 C32 C33 Maximum Axial

Axial (kN) 1042.397 1580.476 1042.397 1582.783 2402.135 1582.783 1388.76 2105.535 1388.76 1197.474 1818.538 1197.474 1195.571 1812.648 1195.571 1195.831 1813.151 1195.831 1195.836 1813.161 1195.836 1195.846 1813.187 1195.846 1196.437 1814.215 1196.437 1186.426 1794.995 1186.426 854.569 1294.073 854.569 2402.135

Bending (X; kNm) 0.001 0 0.001 0.002 0.001 0.002 2 0.001 2 0.002 0 0.002 0.001 0 0.001 0.001 0 0.001 0.001 0 0.001 0.002 0 0.002 0.002 0 0.002 0.002 0 0.002 0.02 0 0.02 0.001

Bending (Z; kNm) -10.713 -10.733 -10.713 0.29 0.356 0.29 5.386 5.224 5.386 0.133 0.049 0.133 0.022 0.024 0.022 0.004 0.003 0.004 0.043 0.038 0.043 0.082 0.079 0.082 0.128 0.14 0.128 0.369 0.641 0.369 4.518 4.9 4.518 0.356

EXTERNAL INTERNAL

ISOLATED FOOTING - GOVERNING COLUMN FOOTING 5, INTERIOR Given qallowable γconcrete γsoil Φbar cc

150 23.56 18 25 75

kPa kN/m3 kN/m3 mm mm

fy f'c

415 20.7

MPa MPa

hsoil P column

1 2402.135 0.55

m kN m

Part 1. Determine the effective bearing pressure on the footing *Assume the thickness of the footing qeff = qa - γconcrete(tfooting) - γsoil(hsoil)

RESULTS ttrial qeff

Part 2. Determine the required dimension of footing qeff = P/Areq, solve for A Set a value of B, then solve for L L = A/B, pick a larger length

A B L L(used)

Part 3. Check the bearing pressure qu = P/(LB) *If qu < qa, dimension is ok, else, redesign Part 4. Solve Depthof the Footing Considering Wide Beam Shear Vu = qu*Acritical Acritical is the shaded area Vc = ᴓVn, ᴓ = .75 Vn = .17*sqrt(f'c)*B*d Vu = Vc, solve for d

qu SAFE d

Wide Beam Shear Vu x quB Vc k d

x L

B

d/2

Considering Punchig Shear Vu = qu*Acritical Acritical is the shaded area Vn = .33*sqrt(f'c)*bo*d, bo is perimeter of punching Vu = Vc, solve for d

Punching Shear Vu Vc k d dgoverning

Part 5. Determine the longitudinal and

Transverse Reinforcement Critical Area

Punching Area

Critical transverse reinforcement Area Reinforcement at Transverse Solve moment in the critical area M = qu*(Bx)(x/2) k = ᴓ*f'c*B*(d^2) Mu = kω*(1-.59ω) solve for ω, ᴓ = .9 ω = ρ*fy/f'c, solve for ρ ρmin = 1.4/fy *Get the greater value between ρ and ρmin As = ρ*B*d Abar = pi()*(ᴓbar^2)/4 N = As/Abar Reinforcement in Longitudinal M = quL*x*x/2 k = ᴓ*f'c*L*(d^2) Mu = kω*(1-.59ω) solve for ω, ᴓ = .9 ω = ρ*fy/f'c, solve for ρ

Punching Area

x

x M k ω ρ ρmin ρ(governing) As Abar N Longitudinal Reinforcement x M k ω ρ ρmin ρ(governing) As

ρmin = 1.4/fy *Get the greater value between ρ and ρmin As = ρ*L*d β = L/Bb

β γ As' As''

γ = 2/(β+1)

Abar N' N'' Ntotal

*For Middle Part As' = γAs *For Sides As'' = .5*(As - As') N = As/Abar Then Get the Total N Part 6. Solve for the Development Length (Ld/db) = (9*fy*α*β*γ*λ)/(10*sqrt(fc')*(c+ktr/db)) c = steel cover ktr = 0 α = 1.3 β=1 γ = .8 λ=1 solve for Ld

Development Length c ktr (c + ktr)/db *** α*β*γ*λ Ld

ERIOR

P

RESULTS 1 108.44

m MPa

22.1517428993 4 5.5379357248 5.6

mm2 m m m

107.2381696429 SAFE

mPa

de Beam Shear qu*B*x L/2-d 428.9526785714 kd 2274.8626332155 0.4005841333

kN m kN m

nching Shear qu*((L^2)-(c+d)) kd 4549.7252664309 0.4854776683

kN kN

0.4854776683

m

ansverse Reinforcement

0.1885620135

m

0.0235702517 30.366232092 1.0235702517 0.5759272769 -0.5208436367 0.4854776683 -1.0481427867

1.725 893.4816199219 24588916754.4749 0.0371510819 0.0018530781 0.003373494 0.003373494 6551.0239574124 490.8738521234 13

m kNm

mm2 mm2 pcs

ngitudinal Reinforcement 2.525 1367.4207106585 24588916754.4749 0.0575664623 0.0028713874 0.003373494 0.003373494 9171.4335403774

mm2

1.4 0.8333333333 7642.8612836479 1528.5722567296

mm2 mm2

490.8738521234 16 4 37

mm2 pcs pcs pcs

m kNm

velopment Length 75 0 3 2.5 1.04 853.7658369529

0.59 -1 0.0363367622 1.6577641724 0.0371510819

mm

mm

0.59 -1 0.0556112628 1.6373487919 0.0575664623

COMBINED FOOTINGGOVERNING GRID - B

1582.783

R

2402.135

1

2

z 7

0.5

7

Given Column 1 2 3

b (mm) 0.55 0.5 0.55

qallowable

150

γconcrete

23.56

γsoil Φbar cc

18 25 75

d (mm) 0.55 0.55 0.55

P (kN) 1582.783 2402.135 1582.783

Mx (kN-m) 0 0 0

Mz (kN-m) 0 0 0

fy

415

MPa

f'c

20.7

MPa

hsoil

1

m

kPa kN/m3 kN/m

3

mm mm

Solution RESULTS Part 1. Determine the effective soil pressure on the footing Assume a trial thickness of the footing (t) then solve for d d = t - cc Determine the effective upward soil pressure qeff = qallow - (t*γconcrete) - (h*γsoil)

ttrial d qeff R

Part 2. Determine the required dimensions of the footing. Determine the required Area of the footing

Areq z 100

R = P1 + P2 + P 3 qeff = R/Areq, solve for Areq Determine the position of the Resultant, L and W zR = .5P1 + 7.5P2 + 14.5P3, solve for z L = 2z Areq = WL, solve for W x = L - 14.5

Check the ultimate upward pressure q_u=P/A±((M_x∗.5W)/I_x )_X±((M_z∗.5L)/I_z )_Z

L W x Ix Iz

qu1 qu2 qu3 qu4 qu Rectangular Uniform Pressure

get the highest value of qu Determine the shape of the pressure

qu' VA VB VC VD VE VF VG VH Vmax (+) Vmax (-) |Vmax(-)| Vmax

Part 3. Draw the Shear and Moment diagram Determine the upward pressure (longutidnal face) qu' = qu*W *Determine the shear at the sections and get the maximum shear (Vmax) *Determine the maximum positive and negative moment (+Mu)

x1 x2 x3 x4 x5 x6 x7 x8 x9 x10

101

MA MB MC MD ME MF MG MH MI MJ MK ML MM Mmax (+) Mmax (-) |Mmax(-)| Mmax Part 4. Check the Wide Beam Shear of the Footing Vu = Vmax - qu'*d Vc = ᴓVn, ᴓ = .75 Vn = .17*sqrt(f'c)*W*d If Vu < Vc, the dimension of the footing is ok, else, change d Part 5. Check the Punching Shear at all Columns Vu = Pu - qu*.5d Vc = ᴓVn, ᴓ = .75 Vn = .33*sqrt(f'c)*bo*d, bo is perimeter of punching area If Vu < Vc, the dimension of the footing is ok, else, change d

Wide Beam Shear Vu Vn Vc Remarks on Wide Beam Shear Punching Shear At Leftmost Column Vu bo Vn Vc Remarks on Punching Shear At Middle Column Vu bo Vn Vc Remarks on Punching Shear At rightmost Column Vu

102

bo Vn Vc Remarks on Punching Shear Part 6. Determine the longitudinal and transverse reinforcement Reinforcement at Longitudinal Mu = Mmax k = ᴓ*f'c*b*(d^2) Mu = kω*(1-.59ω) solve for ω, ᴓ = .9 ω = ρ*fy/f'c, solve for ρ ρmin = 1.4/fy *Get the greater value between ρ and ρmin As = ρ*W*d Abar = pi()*(ᴓbar^2)/4 N = As/Abar Reinforcement at Transverse (per Column) q = P/W let j = .5*(W - c), where c is column dimension Mu = .5*q*(j^2) Solve for ω,ρ, until N using the same process in longitudinal

Longitudinal Mu k ω ρ ρmin ρ(governing) As Abar N Transverse At leftmost q Mu k ω ρ ρmin ρ(governing) As Abar N At middle q Mu k ω ρ ρmin ρ(governing) As Abar N

103

Part 7. Length Development of Longitudinal Bars (Ld/db) = (9*fy*α*β*γ*λ)/(10*sqrt(fc')*(c+ktr/db)) c = steel cover ktr = 0 α = 1.3 β=1 γ = .8 λ=1 solve for Ld

At rightmost q Mu k ω ρ ρmin ρ(governing) As Abar N

c ktr (c + ktr)/db *** α*β*γ*λ Ld

104

SHEAR DIAGRAM

MOMENT DIAGRAM

105

1582.783

3

x

RESULTS 800 725 113.152

mm mm kPa

5567.701

kN

49.2055023331 7.775

m2 m 106

15.55 3.1643409861 0.8

m m m

41058112154880.7 991501123159187

mm4 mm4

Negligible

113.152 Rectangular Uniform Pressure 358.051511254 0 179.025755627 -1206.8289131833 1120.5059099679 -1102.6033344052 1224.731488746 -161.1231800643 27 1224.731488746 -1206.8289131833 1206.8289131833 1224.731488746

0.0710494165 0.4789505835 3.3705455088 3.1294544912 0.2520132361 0.2479867639 3.0794544912 3.4205455088 0.4860555251 0.0139444749

kPa kPa kPa kPa kPa

kN/m kN kN kN kN kN kN kN kN kN kN kN kN

mm mm mm mm mm mm mm mm mm mm

0.59 -1 0.0731690141 1.618281311 0.0766339432

107

0 49.2320827974 55.5919205367 -233.4137855305 -2267.2496721523 -513.9635459807 -372.772385734 -509.48790209 -2207.1962971523 -112.5714004823 185.072352974 183.9489639068 141.6541291399 185.072352974 -2267.2496721523 2267.2496721523 2267.2496721523

kNm kNm kNm kNm kNm kNm kNm kNm kNm kNm kNm kNm kNm kNm kNm kNm kNm

965.1441430868 1739.6232580824 1304.7174435618 emarks on Wide Beam Shear

kN kN kN OK

1379.00332 5.1 mm 5607.5363908761 4205.6522931571 Remarks on Punching Shear

kN

2213.27724 4.9 mm 5387.6330029986 4040.724752249 Remarks on Punching Shear

1379.00332

kN kN OK

0.59 2.5 0.0137911484 -0.0055236599

0.59 -1 0.0209303488 1.6737198508 0.0211954034

0.59 -1 0.0137911484 1.6810100262 0.0139052281

kN kN kN OK

kN

108

5.1 mm 5607.5363908761 kN 4205.6522931571 kN Remarks on Punching Shear OK

2267.2496721523 30986472894.7072 0.0766339432 0.0038224642 0.003373494 0.0038224642 8769.2954995252 490.8738521234 18

500.1935654139 427.3390465585 30986472894.7072 -0.0055236599 -0.0002755175 0.003373494 0.003373494 7739.2918092654 490.8738521234 16

759.1264691721 648.5576864325 30986472894.7072 0.0211954034 0.0010572165 0.003373494 0.003373494 7739.2918092654 490.8738521234 16

kNm

mm2 mm2 pcs

kPa kNm

mm2 mm2 pcs

kPa kNm

mm2 mm2 pcs

109

500.1935654139 427.3390465585 30986472894.7072 0.0139052281 0.0010572165 0.003373494 0.003373494 7739.2918092654 490.8738521234 16

kPa kNm

mm2 mm2 pcs

75 0 3 2.5 1.04

mm

853.7658369529

mm

110

111

112

113

114

115

116

117

APPENDIX B - FINAL ESTIMATE OF TRADEOFFS FINAL ESTIMATE OF VALUES FOR ISOLATED FOOTING ECONOMIC

Transverse Exterior Longitudinal Exterior Interior

Material Cement Sand Gravel

Quantity 6 18 9

Quantiity 3528.1558368 bags 147.006 cu. m 588.026 cu.m Rebars

Total Amount

L 4.9 4.9 5.6

Unit Price PHP 240 PHP 700 PHP 1,350 1050

Summary of Values B t 3 0.512779 3 0.513203 4 0.560478 Total

V 45.2271078 135.7935138 112.9923648 294.0129864

Amount PHP 846,757.40 PHP 102,904.55 PHP 793,835.06 PHP 1,080,450.00

PHP 2,823,947.01

CONSTRUCTABILITY Excavation (Assumed medium soil): Excavation Volume 294.0129864 cu.m. Item 1 cu.yd. 3/4 cu.yd 1/2 cu.yd

Duration .01hr/cu.yd 6 8.4 11.1

Total Duration 17.641 24.697 32.635

Duration hr/ton

Total Duration

5.12

100.47

15.7 0.0054

308.1 32.208

Excavation duration:

17.641

hrs

d N= L= Wt=

25 308.7 16.000 19.623

mm bars m tons

440.768

hrs

Rebarworks Item Fabricate cut &bend Place&tie Mesh

Rebar duration:

Formworks item Fabricate Erect Strip

Duration hr/sf 0.046 0.045 0.024

Total Duration 25.502 24.948 13.306

Area= 1 meter= Area=

0 sq.m. 3.281 ft 0.00 sq.ft

Formworks duration:

63.756

hrs

Duration hr/cu.yd. 0.57 0.7 0.94 0.82

Total Duration 167.587 205.809 276.372 241.091

Concreting duration:

167.587

hrs

Concreting: item Chute Buggies Crane Conveyor

TOTAL DURATION

689.752

SETTLEMENT COMPUTATION

qu B Es μs L/B αav

107.24 4 200000 0.4 1.4 1.1

KPa m Kpa

Se

8.836576

mm

hrs

FINAL ESTIMATE OF VALUES FOR COMBINED FOOTING ECONOMIC GRID A B C D E F G H I J K

Material Cement Sand Gravel

L 15.55 15.55 15.55 15.55 15.55 15.55 15.55 15.55 15.55 15.55 15.55

Summary of Values B t 2.775 0.8 3.164 0.8 2.39 0.8 2.39 0.8 2.39 0.8 2.39 0.8 2.39 0.8 2.39 0.8 2.39 0.8 2.37 0.8 1.71 0.8 Total

Quantiity 5191.017936 bags 216.292 cu. m 865.170 cu.m Rebars

Total Amount

Unit Price PHP 240 PHP 700 PHP 1,350 2850

N of Bars (Φ25) V 34.521 30 39.36016 34 29.7316 26 29.7316 26 29.7316 26 29.7316 26 29.7316 26 29.7316 26 29.7316 26 29.4828 25 21.2724 19 432.584828 290 Amount PHP 1,245,844.30 PHP 151,404.69 PHP 1,167,979.04 PHP 826,500.00

PHP 3,391,728.03

CONSTRUCTABILITY Excavation (Assumed medium soil): Excavation Volume 432.584828 cu.m. Item 1 cu.yd. 3/4 cu.yd 1/2 cu.yd

Duration Total hr/100cu.yd Duration (hrs) 6 25.955 8.4 36.337 11.1 48.017

Excavation duration:

25.955

hrs

Rebarworks Item Fabricate cut &bend Place&tie Mesh

Duration Total duration hr/ton &hr/sf

d N= L= Wt=

25 377 16.000 23.965

mm bars m tons

313.871

hrs

5.12

122.70

15.7 0.0054

376.3 24.165

Duration hr/sf 0.046 0.045 0.024

Total Duration (hrs) 19.134 18.718 9.983

Formworks duration:

47.834

hrs

Duration hr/cu.yd. 0.57 0.7 0.94 0.82

Total Duration (hrs) 246.573 302.809 406.630 354.720

Concreting duration:

197.259

hrs

Rebar duration:

Formworks item Fabricate Erect Strip

Area= 1 meter= Area=

415.94695 sq.m. 3.281 ft 4477.22 sq.ft

Concreting: item Chute Buggies Crane Conveyor

TOTAL DURATION

584.919

SETTLEMENT COMPUTATION

qu B Es μs L/B αav

113.152 3.164 200000 0.4 4.8988621997 2.1

KPa m Kpa

hrs

Se

5.9430146048

mm

S

TING

Bars (Φ25) 29 29 37 95

88.2 264.6 201.6 554.4

Nbars(Total) 174 522 333 1029

43.15125 49.2002 37.1645 37.1645 37.1645 37.1645 37.1645 37.1645 37.1645 36.8535 26.5905 415.94695

APPENDIX G - ISOLATED FOOTING DESIGN ISOLATED FOOTING - TRANSVERSE EXTERIOR COLUMN FOUNDATION DESIGN (2)

P Given qallowable

150

kPa

γconcrete

23.56

kN/m

γsoil Φbar cc

18 25 75

fy

415

MPa

f'c

20.7

MPa

hsoil P column

1 1580.476 0.55

m kN m

3

kN/m3 mm mm

Part 1. Determine the effective bearing pressure on the footing

RESULTS

*Assume the thickness of the footing qeff = qa - γconcrete(tfooting) - γsoil(hsoil) Part 2. Determine the required dimension of footing qeff = P/Areq, solve for A Set a value of B, then solve for L L = A/B, pick a larger length Part 3. Check the bearing pressure qu = P/(LB) *If qu < qa, dimension is ok, else, redesign Part 4. Solve Depthof the Footing Considering Wide Beam Shear Vu = qu*Acritical Acritical is the shaded area Vc = ᴓVn, ᴓ = .75 Vn = .17*sqrt(f'c)*B*d Vu = Vc, solve for d Considering Punchig Shear Vu = qu*Acritical Acritical is the shaded area Vn = .33*sqrt(f'c)*bo*d, bo is perimeter of punching Vu = Vc, solve for d

d x L

B

d/2

ttrial qeff

1 108.44

A B L L(used)

14.574658797 3 4.8582195992 4.9

qu

107.51537415 SAFE

Wide Beam Shear Vu qu*B*x x L/2-d quB 322.54612245 Vc kd k 1706.1469749 d 0.3458077603 Punching Shear Vu qu*((L^2)-(c+d)) Vc kd k 3412.2939498 d 0.4377785923 dgoverning

0.4377785923

103

Part 5. Determine the longitudinal and transverse reinforcement Critical Area Reinforcement at Transverse Solve moment in the critical area M = qu*(Bx)(x/2) k = ᴓ*f'c*B*(d^2) Mu = kω*(1-.59ω) solve for ω, ᴓ = .9 ω = ρ*fy/f'c, solve for ρ ρmin = 1.4/fy *Get the greater value between ρ and ρmin As = ρ*B*d Abar = pi()*(ᴓbar^2)/4 N = As/Abar Reinforcement in Longitudinal M = quL*x*x/2 k = ᴓ*f'c*L*(d^2) Mu = kω*(1-.59ω) solve for ω, ᴓ = .9 ω = ρ*fy/f'c, solve for ρ ρmin = 1.4/fy *Get the greater value between ρ and ρmin As = ρ*L*d β = L/Bb γ = 2/(β+1) *For Middle Part As' = γAs *For Sides As'' = .5*(As - As') N = As/Abar Then Get the Total N Part 6. Solve for the Development Length (Ld/db) = (9*fy*α*β*γ*λ)/(10*sqrt(fc')*(c+ktr/db)) c = steel cover ktr = 0 α = 1.3 β=1 γ = .8 λ=1

Punching Area

x

Transverse Reinforcement x 1.225 M 395.28363292 k 17495162300 ω 0.0229033751 ρ 0.0011424093 ρmin 0.003373494 ρ(governing) 0.003373494 As 4430.5303314 Abar 490.87385212 N 9 Longitudinal Reinforcement x 2.175 M 762.92237526 k 17495162300 ω 0.0447913136 ρ 0.0022341691 ρmin 0.003373494 ρ(governing) 0.003373494 As 7236.5328746 β 1.6333333333 γ 0.7594936709 As' 5496.1009174 As'' 1740.4319572 Abar N' N'' Ntotal

490.87385212 12 4 29

Development Length c 75 ktr 0 (c + ktr)/db 3 *** 2.5 α*β*γ*λ 1.04 Ld 853.76583695

104

solve for Ld

ISOLATED FOOTING - LONGITUDINAL EXTERIOR FOUNDATION DESIGN (1,3,31,33)

P Given qallowable

150

γconcrete

23.56

γsoil Φbar cc

18 25 75

kPa

fy

415

MPa

f'c

20.7

MPa

hsoil P column

1 1582.783 0.55

m kN m

kN/m3 kN/m

3

mm mm

Part 1. Determine the effective bearing pressure on the footing *Assume the thickness of the footing qeff = qa - γconcrete(tfooting) - γsoil(hsoil)

RESULTS

Part 2. Determine the required dimension of footing qeff = P/Areq, solve for A Set a value of B, then solve for L L = A/B, pick a larger length Part 3. Check the bearing pressure qu = P/(LB) *If qu < qa, dimension is ok, else, redesign Part 4. Solve Depthof the Footing Considering Wide Beam Shear Vu = qu*Acritical Acritical is the shaded area Vc = ᴓVn, ᴓ = .75 Vn = .17*sqrt(f'c)*B*d Vu = Vc, solve for d Considering Punchig Shear Vu = qu*Acritical Acritical is the shaded area Vn = .33*sqrt(f'c)*bo*d, bo is perimeter of punching Vu = Vc, solve for d

d x L

B

d/2

ttrial qeff

1 108.44

A B L L(used)

14.595933235 3 4.8653110783 4.9

qu

107.67231293 SAFE

Wide Beam Shear Vu qu*B*x x L/2-d quB 323.01693878 Vc kd k 1706.1469749 d 0.3462321782 Punching Shear Vu qu*((L^2)-(c+d)) Vc kd k 3412.2939498 d 0.4382026979

105

dgoverning Part 5. Determine the longitudinal and transverse reinforcement Critical Area Reinforcement at Transverse Solve moment in the critical area M = qu*(Bx)(x/2) k = ᴓ*f'c*B*(d^2) Mu = kω*(1-.59ω) solve for ω, ᴓ = .9 ω = ρ*fy/f'c, solve for ρ ρmin = 1.4/fy *Get the greater value between ρ and ρmin As = ρ*B*d Abar = pi()*(ᴓbar^2)/4 N = As/Abar Reinforcement in Longitudinal M = quL*x*x/2 k = ᴓ*f'c*L*(d^2) Mu = kω*(1-.59ω) solve for ω, ᴓ = .9 ω = ρ*fy/f'c, solve for ρ ρmin = 1.4/fy *Get the greater value between ρ and ρmin As = ρ*L*d β = L/Bb γ = 2/(β+1) *For Middle Part As' = γAs *For Sides As'' = .5*(As - As') N = As/Abar Then Get the Total N Part 6. Solve for the Development Length (Ld/db) = (9*fy*α*β*γ*λ)/(10*sqrt(fc')*(c+ktr/db)) c = steel cover ktr = 0 α = 1.3 β=1

Punching Area

x

0.4382026979

Transverse Reinforcement x 1.225 M 395.86062323 k 17529076205 ω 0.0228922786 ρ 0.0011418558 ρmin 0.003373494 ρ(governing) 0.003373494 As 4434.8224848 Abar 490.87385212 N 9 Longitudinal Reinforcement x 2.175 M 764.036003 k 17529076205 ω 0.0447693128 ρ 0.0022330717 ρmin 0.003373494 ρ(governing) 0.003373494 As 7243.5433918 β 1.6333333333 γ 0.7594936709 As' 5501.4253608 As'' 1742.1180309 Abar N' N'' Ntotal

490.87385212 12 4 29

Development Length c 75 ktr 0 (c + ktr)/db 3 *** 2.5 α*β*γ*λ 1.04 Ld 853.76583695

106

γ = .8 λ=1 solve for Ld

107

2)

RESULTS m MPa mm2 m m m mPa SAFE

kN m

0.1890494355

kN m

u*((L^2)-(c+d)) kN m

0.0315082392 23.299797674 1.0315082392 0.5846590632 -0.453639875 0.4377785923 -1.00457878

m

108

m kNm

0.59 -1 0.022593882 1.6720118791 0.0229033751

mm2 mm2 pcs

m kNm

0.59 -1 0.0436076192 1.6501239406 0.0447913136

mm2

mm2 mm2 mm2 pcs pcs pcs

mm

mm

109

3)

RESULTS m MPa mm2 m m m mPa SAFE

kN m

0.1893253884

kN m

u*((L^2)-(c+d)) kN m

0.0315542314 23.368751889 1.0315542314 0.5847096545 -0.454302047 0.4382026979 -1.005026658

110

m

m kNm

0.59 -1 0.0225830853 1.6720229756 0.0228922786

mm2 mm2 pcs

m kNm

0.59 -1 0.0435867808 1.6501459415 0.0447693128

mm2

mm2 mm2 mm2 pcs pcs pcs

mm

mm

111

112

APPENDDIX H - COMBINED FOOTING DESIGN GRID - A

1388.76

R

2105.535

1

2

z 7

0.5

7

Given Column 1 2 3

b (mm) 0.55 0.55 0.55

d (mm) 0.55 0.55 0.55

qallowable

150

kPa

γconcrete Φbar cc

23.56 25 75

kN/m mm mm

γsoil

18

P (kN) 1388.76 2105.535 1388.76

Mx (kN-m) 0 0 0

Mz (kN-m) 0 0 0

fy

415

MPa

f'c hsoil

20.7 1

MPa m

3

kN/m3

Solution RESULTS Part 1. Determine the effective soil pressure on the footing Assume a trial thickness of the footing (t) then solve for d d = t - cc Determine the effective upward soil pressure qeff = qallow - (t*γconcrete)

ttrial d qeff R

Part 2. Determine the required dimensions of the footing. Determine the required Area of the footing

Areq z 106

R = P1 + P2 + P 3 qeff = R/Areq, solve for Areq Determine the position of the Resultant, L and W zR = .5P1 + 7.5P2 + 14.5P3, solve for z L = 2z Areq = WL, solve for W x = L - 14.5

Check the ultimate upward pressure q_u=P/A±((M_x∗.5W)/I_x q_u=P/A±((M_x∗.5W)/I_x )_X±((M_z∗.5L)/I_z )_X±((M_z∗.5L)/I_z )_Z )_Z

L W x Ix Iz

qu1 qu2 qu3 qu4 qu Rectangular Uniform Pressure

get the highest value of qu Determine the shape of the pressure

qu' VA VB VC VD VE VF VG VH Vmax (+) Vmax (-) |Vmax(-)| Vmax

Part 3. Draw the Shear and Moment diagram Determine the upward pressure (longutidnal face) qu' = qu*W *Determine the shear at the sections and get the maximum shear (Vmax) *Determine the maximum positive and negative moment (+Mu)

x1 x2 x3 x4 x5 x6 x7 x8 x9 x10

107

MA MB MC MD ME MF MG MH MI MJ MK ML MM Mmax (+) Mmax (-) |Mmax(-)| Mmax Part 4. Check the Wide Beam Shear of the Footing Vu = Vmax - qu'*d Vc = ᴓVn, ᴓ = .75 Vn = .17*sqrt(f'c)*W*d If Vu < Vc, the dimension of the footing is ok, else, change d Part 5. Check the Punching Shear at all Columns Vu = Pu - qu*.5d Vc = ᴓVn, ᴓ = .75 Vn = .33*sqrt(f'c)*bo*d, bo is perimeter of punching area If Vu < Vc, the dimension of the footing is ok, else, change d

Wide Beam Shear Vu Vn Vc Remarks on Wide Beam Shear Punching Shear At Leftmost Column Vu bo Vn Vc Remarks on Punching Shear At Middle Column Vu bo Vn Vc Remarks on Punching Shear At rightmost Column Vu

108

bo Vn Vc Remarks on Punching Shear Part 6. Determine the longitudinal and transverse reinforcement Reinforcement at Longitudinal Mu = Mmax k = ᴓ*f'c*b*(d^2) Mu = kω*(1-.59ω) solve for ω, ᴓ = .9 ω = ρ*fy/f'c, solve for ρ ρmin = 1.4/fy *Get the greater value between ρ and ρmin As = ρ*W*d Abar = pi()*(ᴓbar^2)/4 N = As/Abar Reinforcement at Transverse (per Column) q = P/W let j = .5*(W - c), where c is column dimension Mu = .5*q*(j^2) Solve for ω,ρ, until N using the same process in longitudinal

Longitudinal Mu k ω ρ ρmin ρ(governing) As Abar N Transverse At leftmost q Mu k ω ρ ρmin ρ(governing) As Abar N At middle q Mu k ω ρ ρmin ρ(governing) As Abar N At rightmost

109

q Mu k ω ρ ρmin Part 7. Length Development of Longitudinal Bars (Ld/db) = (9*fy*α*β*γ*λ)/(10*sqrt(fc')*(c+ktr/db)) c = steel cover ktr = 0 α = 1.3 β=1 γ = .8 λ=1 solve for Ld

ρ(governing) As Abar N

c ktr (c + ktr)/db *** α*β*γ*λ Ld

110

GRID - C

1197.474

R

1818.538

1

2

z 7

0.5

7

Given Column 1 2 3

b (mm) 0.55 0.55 0.55

d (mm) 0.55 0.55 0.55

qallowable

150

kPa

γconcrete Φbar cc

23.56 25 75

kN/m mm mm

γsoil

18

3

P (kN) 1197.474 1818.538 1197.474

Mx (kN-m) 0 0 0

Mz (kN-m) 0 0 0

fy

415

MPa

f'c hsoil

20.7 1

MPa m

kN/m3

111

Solution RESULTS Part 1. Determine the effective soil pressure on the footing Assume a trial thickness of the footing (t) then solve for d d = t - cc Determine the effective upward soil pressure qeff = qallow - (t*γconcrete)

ttrial d qeff R

Part 2. Determine the required dimensions of the footing. Determine the required Area of the footing R = P1 + P2 + P 3 qeff = R/Areq, solve for Areq Determine the position of the Resultant, L and W zR = .5P1 + 7.5P2 + 14.5P3, solve for z L = 2z Areq = WL, solve for W x = L - 14.5

Check the ultimate upward pressure q_u=P/A±((M_x∗.5W)/I_x q_u=P/A±((M_x∗.5W)/I_x )_X±((M_z∗.5L)/I_z )_X±((M_z∗.5L)/I_z )_Z )_Z

Areq z L W x Ix Iz

qu1 qu2 qu3 qu4 qu Rectangular Uniform Pressure

get the highest value of qu Determine the shape of the pressure

qu' VA VB VC VD VE VF VG VH Vmax (+) Vmax (-) |Vmax(-)| Vmax

Part 3. Draw the Shear and Moment diagram Determine the upward pressure (longutidnal face) qu' = qu*W

x1 x2 112

*Determine the shear at the sections and get the maximum shear (Vmax) *Determine the maximum positive and negative moment (+Mu)

x3 x4 x5 x6 x7 x8 x9 x10

MA MB MC MD ME MF MG MH MI MJ MK ML MM Mmax (+) Mmax (-) |Mmax(-)| Mmax Part 4. Check the Wide Beam Shear of the Footing Vu = Vmax - qu'*d Vc = ᴓVn, ᴓ = .75 Vn = .17*sqrt(f'c)*W*d If Vu < Vc, the dimension of the footing is ok, else, change d Part 5. Check the Punching Shear at all Columns Vu = Pu - qu*.5d Vc = ᴓVn, ᴓ = .75 Vn = .33*sqrt(f'c)*bo*d, bo is perimeter of punching area If Vu < Vc, the dimension of the footing is ok,

Wide Beam Shear Vu Vn Vc Remarks on Wide Beam Shear Punching Shear At Leftmost Column Vu bo Vn Vc Remarks on Punching Shear

113

else, change d At Middle Column Vu bo Vn Vc Remarks on Punching Shear At rightmost Column Vu bo Vn Vc Remarks on Punching Shear Part 6. Determine the longitudinal and transverse reinforcement Reinforcement at Longitudinal Mu = Mmax k = ᴓ*f'c*b*(d^2) Mu = kω*(1-.59ω) solve for ω, ᴓ = .9 ω = ρ*fy/f'c, solve for ρ ρmin = 1.4/fy *Get the greater value between ρ and ρmin As = ρ*W*d Abar = pi()*(ᴓbar^2)/4 N = As/Abar Reinforcement at Transverse (per Column) q = P/W let j = .5*(W - c), where c is column dimension Mu = .5*q*(j^2) Solve for ω,ρ, until N using the same process in longitudinal

Longitudinal Mu k ω ρ ρmin ρ(governing) As Abar N Transverse At leftmost q Mu k ω ρ ρmin ρ(governing) As Abar N At middle q Mu

114

k ω ρ ρmin ρ(governing) As Abar N

Part 7. Length Development of Longitudinal Bars (Ld/db) = (9*fy*α*β*γ*λ)/(10*sqrt(fc')*(c+ktr/db)) c = steel cover ktr = 0 α = 1.3 β=1 γ = .8 λ=1 solve for Ld

At rightmost q Mu k ω ρ ρmin ρ(governing) As Abar N

c ktr (c + ktr)/db *** α*β*γ*λ Ld

115

GRID - D

1195.571

R

1812.648

1

2

z 7

0.5

7

Given Column 1 2

b (mm) 0.55 0.55

d (mm) 0.55 0.55

P (kN) 1195.571 1812.648

Mx (kN-m) 0 0

Mz (kN-m) 0 0 116

3

0.55

qallowable

150

kPa

γconcrete Φbar cc

23.56 25 75

kN/m mm mm

γsoil

18

0.55

1195.571

0

0

fy

415

MPa

f'c hsoil

20.7 1

MPa m

3

kN/m3

Solution RESULTS Part 1. Determine the effective soil pressure on the footing Assume a trial thickness of the footing (t) then solve for d d = t - cc Determine the effective upward soil pressure qeff = qallow - (t*γconcrete)

ttrial d qeff R

Part 2. Determine the required dimensions of the footing. Determine the required Area of the footing R = P1 + P2 + P 3 qeff = R/Areq, solve for Areq Determine the position of the Resultant, L and W zR = .5P1 + 7.5P2 + 14.5P3, solve for z L = 2z Areq = WL, solve for W x = L - 14.5

Check the ultimate upward pressure q_u=P/A±((M_x∗.5W)/I_x q_u=P/A±((M_x∗.5W)/I_x )_X±((M_z∗.5L)/I_z )_X±((M_z∗.5L)/I_z )_Z )_Z

Areq z L W x Ix Iz

qu1 qu2 qu3 qu4 qu Rectangular Uniform Pressure

get the highest value of qu Determine the shape of the pressure

qu' VA VB VC VD VE VF VG 117

VH Vmax (+) Vmax (-) |Vmax(-)| Vmax Part 3. Draw the Shear and Moment diagram Determine the upward pressure (longutidnal face) qu' = qu*W *Determine the shear at the sections and get the maximum shear (Vmax) *Determine the maximum positive and negative moment (+Mu)

x1 x2 x3 x4 x5 x6 x7 x8 x9 x10

MA MB MC MD ME MF MG MH MI MJ MK ML MM Mmax (+) Mmax (-) |Mmax(-)| Mmax Part 4. Check the Wide Beam Shear of the Footing Vu = Vmax - qu'*d Vc = ᴓVn, ᴓ = .75 Vn = .17*sqrt(f'c)*W*d

Wide Beam Shear Vu Vn Vc

118

If Vu < Vc, the dimension of the footing is ok, else, change d Part 5. Check the Punching Shear at all Columns Vu = Pu - qu*.5d Vc = ᴓVn, ᴓ = .75 Vn = .33*sqrt(f'c)*bo*d, bo is perimeter of punching area If Vu < Vc, the dimension of the footing is ok, else, change d

Remarks on Wide Beam Shear Punching Shear At Leftmost Column Vu bo Vn Vc Remarks on Punching Shear At Middle Column Vu bo Vn Vc Remarks on Punching Shear At rightmost Column Vu bo Vn Vc Remarks on Punching Shear

Part 6. Determine the longitudinal and transverse reinforcement Reinforcement at Longitudinal Mu = Mmax k = ᴓ*f'c*b*(d^2) Mu = kω*(1-.59ω) solve for ω, ᴓ = .9 ω = ρ*fy/f'c, solve for ρ ρmin = 1.4/fy *Get the greater value between ρ and ρmin As = ρ*W*d Abar = pi()*(ᴓbar^2)/4 N = As/Abar Reinforcement at Transverse (per Column) q = P/W let j = .5*(W - c), where c is column dimension Mu = .5*q*(j^2) Solve for ω,ρ, until N using the same process

Longitudinal Mu k ω ρ ρmin ρ(governing) As Abar N Transverse At leftmost q Mu k ω ρ

119

in longitudinal

ρmin ρ(governing) As Abar N At middle q Mu k ω ρ ρmin ρ(governing) As Abar N

Part 7. Length Development of Longitudinal Bars (Ld/db) = (9*fy*α*β*γ*λ)/(10*sqrt(fc')*(c+ktr/db)) c = steel cover ktr = 0 α = 1.3 β=1 γ = .8 λ=1 solve for Ld

At rightmost q Mu k ω ρ ρmin ρ(governing) As Abar N

c ktr (c + ktr)/db *** α*β*γ*λ Ld

120

GRID - E

1195.831

R

1813.151

1

2

z 121

0.5

7

7

Given Column 1 2 3

b (mm) 0.55 0.55 0.55

d (mm) 0.55 0.55 0.55

qallowable

150

kPa

γconcrete

23.56

kN/m

Φbar cc

25 75

mm mm

γsoil

18

P (kN) 1195.831 1813.151 1195.831

Mx (kN-m) 0 0 0

Mz (kN-m) 0 0 0

fy

415

MPa

f'c

20.7

MPa

hsoil

1

m

3

kN/m3

Solution RESULTS Part 1. Determine the effective soil pressure on the footing Assume a trial thickness of the footing (t) then solve for d d = t - cc Determine the effective upward soil pressure qeff = qallow - (t*γconcrete)

ttrial d qeff R

Part 2. Determine the required dimensions of the footing. Determine the required Area of the footing R = P1 + P2 + P 3 qeff = R/Areq, solve for Areq Determine the position of the Resultant, L and W zR = .5P1 + 7.5P2 + 14.5P3, solve for z L = 2z Areq = WL, solve for W x = L - 14.5

Check the ultimate upward pressure q_u=P/A±((M_x∗.5W)/I_x )_X±((M_z∗.5L)/I_z )_Z

Areq z L W x Ix Iz

qu1 qu2 qu3 qu4 qu Rectangular Uniform Pressure 122

get the highest value of qu Determine the shape of the pressure

qu' VA VB VC VD VE VF VG VH Vmax (+) Vmax (-) |Vmax(-)| Vmax

Part 3. Draw the Shear and Moment diagram Determine the upward pressure (longutidnal face) qu' = qu*W *Determine the shear at the sections and get the maximum shear (Vmax) *Determine the maximum positive and negative moment (+Mu)

x1 x2 x3 x4 x5 x6 x7 x8 x9 x10

MA MB MC MD ME MF MG MH MI MJ MK ML MM

123

Mmax (+) Mmax (-) |Mmax(-)| Mmax Part 4. Check the Wide Beam Shear of the Footing Vu = Vmax - qu'*d Vc = ᴓVn, ᴓ = .75 Vn = .17*sqrt(f'c)*W*d If Vu < Vc, the dimension of the footing is ok, else, change d Part 5. Check the Punching Shear at all Columns Vu = Pu - qu*.5d Vc = ᴓVn, ᴓ = .75 Vn = .33*sqrt(f'c)*bo*d, bo is perimeter of punching area If Vu < Vc, the dimension of the footing is ok, else, change d

Wide Beam Shear Vu Vn Vc Remarks on Wide Beam Shear Punching Shear At Leftmost Column Vu bo Vn Vc Remarks on Punching Shear At Middle Column Vu bo Vn Vc Remarks on Punching Shear At rightmost Column Vu bo Vn Vc Remarks on Punching Shear

Part 6. Determine the longitudinal and transverse reinforcement Reinforcement at Longitudinal Mu = Mmax k = ᴓ*f'c*b*(d^2) Mu = kω*(1-.59ω) solve for ω, ᴓ = .9 ω = ρ*fy/f'c, solve for ρ ρmin = 1.4/fy

Longitudinal Mu k ω ρ ρmin ρ(governing) As Abar

124

*Get the greater value between ρ and ρmin As = ρ*W*d Abar = pi()*(ᴓbar^2)/4 N = As/Abar Reinforcement at Transverse (per Column) q = P/W let j = .5*(W - c), where c is column dimension Mu = .5*q*(j^2) Solve for ω,ρ, until N using the same process in longitudinal

N Transverse At leftmost q Mu k ω ρ ρmin ρ(governing) As Abar N At middle q Mu k ω ρ ρmin ρ(governing) As Abar N

Part 7. Length Development of Longitudinal Bars (Ld/db) = (9*fy*α*β*γ*λ)/(10*sqrt(fc')*(c+ktr/db)) c = steel cover ktr = 0 α = 1.3 β=1 γ = .8

At rightmost q Mu k ω ρ ρmin ρ(governing) As Abar N

c ktr

125

λ=1 solve for Ld

(c + ktr)/db *** α*β*γ*λ Ld

GRID - F

1195.836

1813.161

R 126

1195.836

1813.161

1

2

z 7

0.5

7

Given Column 1 2 3

b (mm) 0.55 0.55 0.55

d (mm) 0.55 0.55 0.55

qallowable

150

kPa

γconcrete Φbar cc

23.56 25 75

kN/m mm mm

γsoil

18

P (kN) 1195.836 1813.161 1195.836

Mx (kN-m) 0 0 0

Mz (kN-m) 0 0 0

fy

415

MPa

f'c hsoil

20.7 1

MPa m

3

kN/m3

Solution RESULTS Part 1. Determine the effective soil pressure on the footing Assume a trial thickness of the footing (t) then solve for d d = t - cc Determine the effective upward soil pressure qeff = qallow - (t*γconcrete)

ttrial d qeff R

Part 2. Determine the required dimensions of the footing. Determine the required Area of the footing R = P1 + P2 + P 3 qeff = R/Areq, solve for Areq Determine the position of the Resultant, L and W zR = .5P1 + 7.5P2 + 14.5P3, solve for z L = 2z

Areq z L W x Ix 127

Areq = WL, solve for W x = L - 14.5

Check the ultimate upward pressure q_u=P/A±((M_x∗.5W)/I_x )_X±((M_z∗.5L)/I_z )_Z

Iz

qu1 qu2 qu3 qu4 qu Rectangular Uniform Pressure

get the highest value of qu Determine the shape of the pressure

qu' VA VB VC VD VE VF VG VH Vmax (+) Vmax (-) |Vmax(-)| Vmax

Part 3. Draw the Shear and Moment diagram Determine the upward pressure (longutidnal face) qu' = qu*W *Determine the shear at the sections and get the maximum shear (Vmax) *Determine the maximum positive and negative moment (+Mu)

x1 x2 x3 x4 x5 x6 x7 x8 x9 x10

MA MB MC MD ME

128

MF MG MH MI MJ MK ML MM Mmax (+) Mmax (-) |Mmax(-)| Mmax Part 4. Check the Wide Beam Shear of the Footing Vu = Vmax - qu'*d Vc = ᴓVn, ᴓ = .75 Vn = .17*sqrt(f'c)*W*d If Vu < Vc, the dimension of the footing is ok, else, change d Part 5. Check the Punching Shear at all Columns Vu = Pu - qu*.5d Vc = ᴓVn, ᴓ = .75 Vn = .33*sqrt(f'c)*bo*d, bo is perimeter of punching area If Vu < Vc, the dimension of the footing is ok, else, change d

Wide Beam Shear Vu Vn Vc Remarks on Wide Beam Shear Punching Shear At Leftmost Column Vu bo Vn Vc Remarks on Punching Shear At Middle Column Vu bo Vn Vc Remarks on Punching Shear At rightmost Column Vu bo Vn Vc Remarks on Punching Shear

129

Part 6. Determine the longitudinal and transverse reinforcement Reinforcement at Longitudinal Mu = Mmax k = ᴓ*f'c*b*(d^2) Mu = kω*(1-.59ω) solve for ω, ᴓ = .9 ω = ρ*fy/f'c, solve for ρ ρmin = 1.4/fy *Get the greater value between ρ and ρmin As = ρ*W*d Abar = pi()*(ᴓbar^2)/4 N = As/Abar Reinforcement at Transverse (per Column) q = P/W let j = .5*(W - c), where c is column dimension Mu = .5*q*(j^2) Solve for ω,ρ, until N using the same process in longitudinal

Longitudinal Mu k ω ρ ρmin ρ(governing) As Abar N Transverse At leftmost q Mu k ω ρ ρmin ρ(governing) As Abar N At middle q Mu k ω ρ ρmin ρ(governing) As Abar N At rightmost q Mu k ω ρ

130

ρmin Part 7. Length Development of Longitudinal Bars (Ld/db) = (9*fy*α*β*γ*λ)/(10*sqrt(fc')*(c+ktr/db)) c = steel cover ktr = 0 α = 1.3 β=1 γ = .8 λ=1 solve for Ld

ρ(governing) As Abar N

c ktr (c + ktr)/db *** α*β*γ*λ Ld

131

GRID - G

1195.846

R

1813.187

1

2

z 7

0.5

7

Given Column 1 2 3

b (mm) 0.5 0.5 0.5

qallowable

150

γconcrete Φbar cc

23.56 25 75

γsoil

18

d (mm) 0.5 0.5 0.5

P (kN) 1195.846 1813.187 1195.846

Mx (kN-m) 0 0 0

Mz (kN-m) 0 0 0

fy

415

MPa

f'c hsoil

20.7 1

MPa m

kPa kN/m3 mm mm kN/m3

Solution RESULTS Part 1. Determine the effective soil pressure on the footing Assume a trial thickness of the footing (t) then solve for d d = t - cc Determine the effective upward soil pressure

ttrial d qeff 132

qeff = qallow - (t*γconcrete) R Part 2. Determine the required dimensions of the footing. Determine the required Area of the footing R = P1 + P2 + P 3 qeff = R/Areq, solve for Areq Determine the position of the Resultant, L and W zR = .5P1 + 7.5P2 + 14.5P3, solve for z L = 2z Areq = WL, solve for W x = L - 14.5

Check the ultimate upward pressure q_u=P/A±((M_x∗.5W)/I_x q_u=P/A±((M_x∗.5W)/I_x )_X±((M_z∗.5L)/I_z )_X±((M_z∗.5L)/I_z )_Z )_Z

Areq z L W x Ix Iz

qu1 qu2 qu3 qu4 qu Rectangular Uniform Pressure

get the highest value of qu Determine the shape of the pressure

qu' VA VB VC VD VE VF VG VH Vmax (+) Vmax (-) |Vmax(-)| Vmax

Part 3. Draw the Shear and Moment diagram Determine the upward pressure (longutidnal face) qu' = qu*W *Determine the shear at the sections and get the maximum shear (Vmax) *Determine the maximum positive and negative moment (+Mu)

x1 x2 x3 x4 x5 x6 x7 x8 133

x9 x10

MA MB MC MD ME MF MG MH MI MJ MK ML MM Mmax (+) Mmax (-) |Mmax(-)| Mmax Part 4. Check the Wide Beam Shear of the Footing Vu = Vmax - qu'*d Vc = ᴓVn, ᴓ = .75 Vn = .17*sqrt(f'c)*W*d If Vu < Vc, the dimension of the footing is ok, else, change d Part 5. Check the Punching Shear at all Columns Vu = Pu - qu*.5d Vc = ᴓVn, ᴓ = .75 Vn = .33*sqrt(f'c)*bo*d, bo is perimeter of punching area If Vu < Vc, the dimension of the footing is ok, else, change d

Wide Beam Shear Vu Vn Vc Remarks on Wide Beam Shear Punching Shear At Leftmost Column Vu bo Vn Vc Remarks on Punching Shear At Middle Column Vu bo Vn Vc

134

Remarks on Punching Shear At rightmost Column Vu bo Vn Vc Remarks on Punching Shear Part 6. Determine the longitudinal and transverse reinforcement Reinforcement at Longitudinal Mu = Mmax k = ᴓ*f'c*b*(d^2) Mu = kω*(1-.59ω) solve for ω, ᴓ = .9 ω = ρ*fy/f'c, solve for ρ ρmin = 1.4/fy *Get the greater value between ρ and ρmin As = ρ*W*d Abar = pi()*(ᴓbar^2)/4 N = As/Abar Reinforcement at Transverse (per Column) q = P/W let j = .5*(W - c), where c is column dimension Mu = .5*q*(j^2) Solve for ω,ρ, until N using the same process in longitudinal

Longitudinal Mu k ω ρ ρmin ρ(governing) As Abar N Transverse At leftmost q Mu k ω ρ ρmin ρ(governing) As Abar N At middle q Mu k ω ρ ρmin ρ(governing) As

135

Abar N

Part 7. Length Development of Longitudinal Bars (Ld/db) = (9*fy*α*β*γ*λ)/(10*sqrt(fc')*(c+ktr/db)) c = steel cover ktr = 0 α = 1.3 β=1 γ = .8 λ=1 solve for Ld

At rightmost q Mu k ω ρ ρmin ρ(governing) As Abar N

c ktr (c + ktr)/db *** α*β*γ*λ Ld

136

GRID - H

1195.846

R

1813.187

1

2

z 7

0.5

7

Given Column 1 2 3

b (mm) 0.55 0.55 0.55

qallowable

150

γconcrete Φbar

23.56 25

d (mm) 0.55 0.55 0.55

kPa kN/m3 mm

P (kN) 1195.846 1813.187 1195.846

Mx (kN-m) 0 0 0

Mz (kN-m) 0 0 0

fy

415

MPa

f'c hsoil

20.7 1

MPa m 137

cc

75

γsoil

18

mm kN/m3

Solution RESULTS Part 1. Determine the effective soil pressure on the footing Assume a trial thickness of the footing (t) then solve for d d = t - cc Determine the effective upward soil pressure qeff = qallow - (t*γconcrete)

ttrial d qeff R

Part 2. Determine the required dimensions of the footing. Determine the required Area of the footing R = P1 + P2 + P 3 qeff = R/Areq, solve for Areq Determine the position of the Resultant, L and W zR = .5P1 + 7.5P2 + 14.5P3, solve for z L = 2z Areq = WL, solve for W x = L - 14.5

Check the ultimate upward pressure q_u=P/A±((M_x∗.5W)/I_x q_u=P/A±((M_x∗.5W)/I_x )_X±((M_z∗.5L)/I_z )_X±((M_z∗.5L)/I_z )_Z )_Z

Areq z L W x Ix Iz

qu1 qu2 qu3 qu4 qu Rectangular Uniform Pressure

get the highest value of qu Determine the shape of the pressure

qu' VA VB VC VD VE VF VG VH Vmax (+) Vmax (-) |Vmax(-)| Vmax

138

Part 3. Draw the Shear and Moment diagram Determine the upward pressure (longutidnal face) qu' = qu*W *Determine the shear at the sections and get the maximum shear (Vmax) *Determine the maximum positive and negative moment (+Mu)

x1 x2 x3 x4 x5 x6 x7 x8 x9 x10

MA MB MC MD ME MF MG MH MI MJ MK ML MM Mmax (+) Mmax (-) |Mmax(-)| Mmax Part 4. Check the Wide Beam Shear of the Footing Vu = Vmax - qu'*d Vc = ᴓVn, ᴓ = .75 Vn = .17*sqrt(f'c)*W*d If Vu < Vc, the dimension of the footing is ok, else, change d Part 5. Check the Punching Shear at all Columns Vu = Pu - qu*.5d Vc = ᴓVn, ᴓ = .75

Wide Beam Shear Vu Vn Vc Remarks on Wide Beam Shear Punching Shear At Leftmost Column Vu bo

139

Vn = .33*sqrt(f'c)*bo*d, bo is perimeter of punching area If Vu < Vc, the dimension of the footing is ok, else, change d

Vn Vc Remarks on Punching Shear At Middle Column Vu bo Vn Vc Remarks on Punching Shear At rightmost Column Vu bo Vn Vc Remarks on Punching Shear

Part 6. Determine the longitudinal and transverse reinforcement Reinforcement at Longitudinal Mu = Mmax k = ᴓ*f'c*b*(d^2) Mu = kω*(1-.59ω) solve for ω, ᴓ = .9 ω = ρ*fy/f'c, solve for ρ ρmin = 1.4/fy *Get the greater value between ρ and ρmin As = ρ*W*d Abar = pi()*(ᴓbar^2)/4 N = As/Abar Reinforcement at Transverse (per Column) q = P/W let j = .5*(W - c), where c is column dimension Mu = .5*q*(j^2) Solve for ω,ρ, until N using the same process in longitudinal

Longitudinal Mu k ω ρ ρmin ρ(governing) As Abar N Transverse At leftmost q Mu k ω ρ ρmin ρ(governing) As Abar N

140

At middle q Mu k ω ρ ρmin ρ(governing) As Abar N

Part 7. Length Development of Longitudinal Bars (Ld/db) = (9*fy*α*β*γ*λ)/(10*sqrt(fc')*(c+ktr/db)) c = steel cover ktr = 0 α = 1.3 β=1 γ = .8 λ=1 solve for Ld

At rightmost q Mu k ω ρ ρmin ρ(governing) As Abar N

c ktr (c + ktr)/db *** α*β*γ*λ Ld

141

GRID - I

1196.437

1

0.5

R

1814.215

2

z 7

7

Given

142

Column 1 2 3

b (mm) 0.55 0.55 0.55

d (mm) 0.55 0.55 0.55

qallowable

150

kPa

γconcrete Φbar cc

23.56 25 75

kN/m mm mm

γsoil

18

P (kN) 1196.437 1814.215 1196.437

Mx (kN-m) 0 0 0

Mz (kN-m) 0 0 0

fy

415

MPa

f'c hsoil

20.7 1

MPa m

3

kN/m3

Solution RESULTS Part 1. Determine the effective soil pressure on the footing Assume a trial thickness of the footing (t) then solve for d d = t - cc Determine the effective upward soil pressure qeff = qallow - (t*γconcrete)

ttrial d qeff R

Part 2. Determine the required dimensions of the footing. Determine the required Area of the footing R = P1 + P2 + P 3 qeff = R/Areq, solve for Areq Determine the position of the Resultant, L and W zR = .5P1 + 7.5P2 + 14.5P3, solve for z L = 2z Areq = WL, solve for W x = L - 14.5

Check the ultimate upward pressure q_u=P/A±((M_x∗.5W)/I_x q_u=P/A±((M_x∗.5W)/I_x )_X±((M_z∗.5L)/I_z )_X±((M_z∗.5L)/I_z )_Z )_Z

Areq z L W x Ix Iz

qu1 qu2 qu3 qu4 qu Rectangular Uniform Pressure

get the highest value of qu Determine the shape of the pressure

qu' VA VB VC VD 143

VE VF VG VH Vmax (+) Vmax (-) |Vmax(-)| Vmax Part 3. Draw the Shear and Moment diagram Determine the upward pressure (longutidnal face) qu' = qu*W *Determine the shear at the sections and get the maximum shear (Vmax) *Determine the maximum positive and negative moment (+Mu)

x1 x2 x3 x4 x5 x6 x7 x8 x9 x10

MA MB MC MD ME MF MG MH MI MJ MK ML MM Mmax (+) Mmax (-) |Mmax(-)| Mmax Part 4. Check the Wide Beam Shear of the Footing

Wide Beam Shear

144

Vu = Vmax - qu'*d Vc = ᴓVn, ᴓ = .75 Vn = .17*sqrt(f'c)*W*d If Vu < Vc, the dimension of the footing is ok, else, change d Part 5. Check the Punching Shear at all Columns Vu = Pu - qu*.5d Vc = ᴓVn, ᴓ = .75 Vn = .33*sqrt(f'c)*bo*d, bo is perimeter of punching area If Vu < Vc, the dimension of the footing is ok, else, change d

Vu Vn Vc Remarks on Wide Beam Shear Punching Shear At Leftmost Column Vu bo Vn Vc Remarks on Punching Shear At Middle Column Vu bo Vn Vc Remarks on Punching Shear At rightmost Column Vu bo Vn Vc Remarks on Punching Shear

Part 6. Determine the longitudinal and transverse reinforcement Reinforcement at Longitudinal Mu = Mmax k = ᴓ*f'c*b*(d^2) Mu = kω*(1-.59ω) solve for ω, ᴓ = .9 ω = ρ*fy/f'c, solve for ρ ρmin = 1.4/fy *Get the greater value between ρ and ρmin As = ρ*W*d Abar = pi()*(ᴓbar^2)/4 N = As/Abar Reinforcement at Transverse (per Column) q = P/W

Longitudinal Mu k ω ρ ρmin ρ(governing) As Abar N Transverse At leftmost q Mu

145

let j = .5*(W - c), where c is column dimension Mu = .5*q*(j^2) Solve for ω,ρ, until N using the same process in longitudinal

k ω ρ ρmin ρ(governing) As Abar N At middle q Mu k ω ρ ρmin ρ(governing) As Abar N

Part 7. Length Development of Longitudinal Bars (Ld/db) = (9*fy*α*β*γ*λ)/(10*sqrt(fc')*(c+ktr/db)) c = steel cover ktr = 0 α = 1.3 β=1 γ = .8 λ=1 solve for Ld

At rightmost q Mu k ω ρ ρmin ρ(governing) As Abar N

c ktr (c + ktr)/db *** α*β*γ*λ Ld

146

GRID - J

1186.426

R

1794.995

1

2 147

z 7

0.5

7

Given Column 1 2 3

b (mm) 0.55 0.55 0.55

qallowable

150

γconcrete Φbar cc

23.56 25 75

γsoil

d (mm) 0.55 0.55 0.55

P (kN) 1186.426 1794.995 1186.426

Mx (kN-m) 0 0 0

Mz (kN-m) 0 0 0

fy

415

MPa

f'c hsoil

20.7 1

MPa m

kPa kN/m3 mm mm

18

kN/m3

Solution RESULTS Part 1. Determine the effective soil pressure on the footing Assume a trial thickness of the footing (t) then solve for d d = t - cc Determine the effective upward soil pressure qeff = qallow - (t*γconcrete)

ttrial d qeff R

Part 2. Determine the required dimensions of the footing. Determine the required Area of the footing R = P1 + P2 + P 3 qeff = R/Areq, solve for Areq Determine the position of the Resultant, L and W zR = .5P1 + 7.5P2 + 14.5P3, solve for z L = 2z Areq = WL, solve for W x = L - 14.5

Check the ultimate upward pressure

Areq z L W x Ix Iz

qu1 qu2 qu3 148

q_u=P/A±((M_x∗.5W)/I_x q_u=P/A±((M_x∗.5W)/I_x )_X±((M_z∗.5L)/I_z )_X±((M_z∗.5L)/I_z )_Z )_Z

qu4 qu Rectangular Uniform Pressure

get the highest value of qu Determine the shape of the pressure

qu' VA VB VC VD VE VF VG VH Vmax (+) Vmax (-) |Vmax(-)| Vmax

Part 3. Draw the Shear and Moment diagram Determine the upward pressure (longutidnal face) qu' = qu*W *Determine the shear at the sections and get the maximum shear (Vmax) *Determine the maximum positive and negative moment (+Mu)

x1 x2 x3 x4 x5 x6 x7 x8 x9 x10

MA MB MC MD ME MF MG MH MI MJ

149

MK ML MM Mmax (+) Mmax (-) |Mmax(-)| Mmax Part 4. Check the Wide Beam Shear of the Footing Vu = Vmax - qu'*d Vc = ᴓVn, ᴓ = .75 Vn = .17*sqrt(f'c)*W*d If Vu < Vc, the dimension of the footing is ok, else, change d Part 5. Check the Punching Shear at all Columns Vu = Pu - qu*.5d Vc = ᴓVn, ᴓ = .75 Vn = .33*sqrt(f'c)*bo*d, bo is perimeter of punching area If Vu < Vc, the dimension of the footing is ok, else, change d

Wide Beam Shear Vu Vn Vc Remarks on Wide Beam Shear Punching Shear At Leftmost Column Vu bo Vn Vc Remarks on Punching Shear At Middle Column Vu bo Vn Vc Remarks on Punching Shear At rightmost Column Vu bo Vn Vc Remarks on Punching Shear

Part 6. Determine the longitudinal and transverse reinforcement Reinforcement at Longitudinal Mu = Mmax k = ᴓ*f'c*b*(d^2) Mu = kω*(1-.59ω)

Longitudinal Mu k ω ρ ρmin

150

solve for ω, ᴓ = .9 ω = ρ*fy/f'c, solve for ρ ρmin = 1.4/fy *Get the greater value between ρ and ρmin As = ρ*W*d Abar = pi()*(ᴓbar^2)/4 N = As/Abar Reinforcement at Transverse (per Column) q = P/W let j = .5*(W - c), where c is column dimension Mu = .5*q*(j^2) Solve for ω,ρ, until N using the same process in longitudinal

ρ(governing) As Abar N Transverse At leftmost q Mu k ω ρ ρmin ρ(governing) As Abar N At middle q Mu k ω ρ ρmin ρ(governing) As Abar N

Part 7. Length Development of Longitudinal Bars (Ld/db) = (9*fy*α*β*γ*λ)/(10*sqrt(fc')*(c+ktr/db)) c = steel cover ktr = 0

At rightmost q Mu k ω ρ ρmin ρ(governing) As Abar N

151

α = 1.3 β=1 γ = .8 λ=1 solve for Ld

c ktr (c + ktr)/db *** α*β*γ*λ Ld

GRID - K

152

854.569

R

1294.073

1

2

z 7

0.5

7

Given Column 1 2 3

b (mm) 0.55 0.55 0.55

qallowable

150

γconcrete Φbar cc

23.56 25 75

γsoil

18

d (mm) 0.55 0.55 0.55

P (kN) 854.569 1294.073 854.569

Mx (kN-m) 0 0 0

Mz (kN-m) 0 0 0

fy

415

MPa

f'c hsoil

20.7 1

MPa m

kPa kN/m3 mm mm kN/m3

Solution RESULTS Part 1. Determine the effective soil pressure on the footing Assume a trial thickness of the footing (t) then solve for d d = t - cc Determine the effective upward soil pressure qeff = qallow - (t*γconcrete)

ttrial d qeff R

Part 2. Determine the required dimensions of the footing. Determine the required Area of the footing R = P1 + P2 + P 3 qeff = R/Areq, solve for Areq

Areq z L W 153

Determine the position of the Resultant, L and W zR = .5P1 + 7.5P2 + 14.5P3, solve for z L = 2z Areq = WL, solve for W x = L - 14.5

Check the ultimate upward pressure q_u=P/A±((M_x∗.5W)/I_x q_u=P/A±((M_x∗.5W)/I_x )_X±((M_z∗.5L)/I_z )_X±((M_z∗.5L)/I_z )_Z )_Z

x Ix Iz

qu1 qu2 qu3 qu4 qu Rectangular Uniform Pressure

get the highest value of qu Determine the shape of the pressure

qu' VA VB VC VD VE VF VG VH Vmax (+) Vmax (-) |Vmax(-)| Vmax

Part 3. Draw the Shear and Moment diagram Determine the upward pressure (longutidnal face) qu' = qu*W *Determine the shear at the sections and get the maximum shear (Vmax) *Determine the maximum positive and negative moment (+Mu)

x1 x2 x3 x4 x5 x6 x7 x8 x9 x10

MA MB

154

MC MD ME MF MG MH MI MJ MK ML MM Mmax (+) Mmax (-) |Mmax(-)| Mmax Part 4. Check the Wide Beam Shear of the Footing Vu = Vmax - qu'*d Vc = ᴓVn, ᴓ = .75 Vn = .17*sqrt(f'c)*W*d If Vu < Vc, the dimension of the footing is ok, else, change d Part 5. Check the Punching Shear at all Columns Vu = Pu - qu*.5d Vc = ᴓVn, ᴓ = .75 Vn = .33*sqrt(f'c)*bo*d, bo is perimeter of punching area If Vu < Vc, the dimension of the footing is ok, else, change d

Wide Beam Shear Vu Vn Vc Remarks on Wide Beam Shear Punching Shear At Leftmost Column Vu bo Vn Vc Remarks on Punching Shear At Middle Column Vu bo Vn Vc Remarks on Punching Shear At rightmost Column Vu bo Vn

155

Vc Remarks on Punching Shear Part 6. Determine the longitudinal and transverse reinforcement Reinforcement at Longitudinal Mu = Mmax k = ᴓ*f'c*b*(d^2) Mu = kω*(1-.59ω) solve for ω, ᴓ = .9 ω = ρ*fy/f'c, solve for ρ ρmin = 1.4/fy *Get the greater value between ρ and ρmin As = ρ*W*d Abar = pi()*(ᴓbar^2)/4 N = As/Abar Reinforcement at Transverse (per Column) q = P/W let j = .5*(W - c), where c is column dimension Mu = .5*q*(j^2) Solve for ω,ρ, until N using the same process in longitudinal

Longitudinal Mu k ω ρ ρmin ρ(governing) As Abar N Transverse At leftmost q Mu k ω ρ ρmin ρ(governing) As Abar N At middle q Mu k ω ρ ρmin ρ(governing) As Abar N At rightmost q Mu

156

k ω ρ ρmin Part 7. Length Development of Longitudinal Bars (Ld/db) = (9*fy*α*β*γ*λ)/(10*sqrt(fc')*(c+ktr/db)) c = steel cover ktr = 0 α = 1.3 β=1 γ = .8 λ=1 solve for Ld

ρ(governing) As Abar N

c ktr (c + ktr)/db *** α*β*γ*λ Ld

157

1388.76

3

x

RESULTS 800 725 113.152

mm mm kPa

4883.055

kN

43.1548271352 7.775

m2 m 158

15.55 2.7752300408 0.8

m m m

27697853325982.1 869578757362884

mm4 mm4

Negligible

113.152 Rectangular Uniform Pressure 314.022829582 0 157.011414791 -1059.0360289389 982.1123633441 -950.7100803859 1090.4383118971 -125.6091318328 39 1090.4383118971 -1059.0360289389 1059.0360289389 1090.4383118971

0.0710139054 0.4789860946 3.3724810083 3.1275189917 0.2794678847 0.2205321153 3.0275189917 3.4724810083 0.4931888757 0.0068111243

kPa kPa kPa kPa kPa

kN/m kN kN kN kN kN kN kN kN kN kN kN kN

mm mm mm mm mm mm mm mm mm mm

0.59 -1 0.0732506 1.6181916 0.0767236

159

0 43.1781390675 48.7531359461 -204.8786298232 -1990.6680771758 -454.8805430064 -317.6461106484 -422.4771631833 -1861.6235751662 31.6395892283

kNm kNm kNm kNm kNm kNm kNm kNm kNm kNm

300.5356117483 300.1078420418 267.1354449357 300.5356117483 -1990.6680771758 1990.6680771758 1990.6680771758

kNm kNm kNm kNm kNm kNm kNm

862.7717604502 1525.7062203045 1144.2796652284 emarks on Wide Beam Shear

kN kN kN OK

1184.98032 5.1 mm 5607.5363908761 4205.6522931571 Remarks on Punching Shear

kN

1901.75532 5.1 mm 5607.5363908761 4205.6522931571 Remarks on Punching Shear

1184.98032

kN kN OK

0.59 2.5 0.0113973 -0.004564 -4.232724

0.59 -1 0.0172797 1.6774557 0.0174595

0.59 -1 0.0113973 1.6834403 0.0114749

kN kN kN OK

kN

160

5.1 mm 5607.5363908761 kN 4205.6522931571 kN Remarks on Punching Shear OK

1990.6680771758 27176145306.8088 0.0767236102 0.0038269367 0.003373494 0.0038269367 7699.9565312264 490.8738521234 16

500.4125710515 309.7334092838 27176145306.8088 -4.2327243188 -0.2111262492 0.003373494 0.003373494 6787.6108227961 490.8738521234 14

758.688457897 469.5948428212 27176145306.8088 0.0174595234 0.0008708726 0.003373494 0.003373494 6787.6108227961 490.8738521234 14

kNm

mm2 mm2 pcs

kPa kNm

mm2 mm2 pcs

kPa kNm

mm2 mm2 pcs

161

500.4125710515 309.7334092838 27176145306.8088 0.011474941 0.0008708726 0.003373494 0.003373494 6787.6108227961 490.8738521234 14

kPa kNm

mm2 mm2 pcs

75 0 3 2.5 1.04

mm

853.7658369529

mm

162

1197.474

3

x

163

RESULTS 800 725 113.152

mm mm kPa

4213.486

kN

37.237397483 7.775 15.55 2.394687941 0.8

m2 m m m m

17794915454990.7 750341317074231

mm4 mm4

Negligible

113.152 Rectangular Uniform Pressure 270.9637299035 0 135.4818649518 -912.9620836013 848.3021607717 -821.2057877814 940.0584565916 -108.3854919614 34 940.0584565916 -912.9620836013 912.9620836013 940.0584565916

0.0710720166 0.4789279834

kPa kPa kPa kPa kPa

kN/m kN kN kN kN kN kN kN kN kN kN kN kN

mm mm 164

3.3693147195 3.1306852805 0.2794632927 0.2205367073 3.0306852805 3.4693147195 0.4931423867 0.0068576133

mm mm mm mm mm mm mm mm

0 37.2575128617 42.0719975397 -176.5495472669 -1714.577840556 -386.694296463 -268.1596389532 -358.7126491961 -1603.1207958615 27.5585244373

kNm kNm kNm kNm kNm kNm kNm kNm kNm kNm

259.3498599045 258.9782270096 230.5270353698 259.3498599045 -1714.577840556 1714.577840556 1714.577840556

kNm kNm kNm kNm kNm kNm kNm

743.6097524116 1316.4999778552 987.3749833914 emarks on Wide Beam Shear

kN kN kN OK

993.69432 5.1 mm 5607.5363908761 4205.6522931571 Remarks on Punching Shear

kN kN kN OK

0.59 -1 0.0731172 1.6183383 0.076577

0.59 2.5 0.0090706 -0.003631 -4.233657

0.59 -1 0.013775 1.6810264 0.0138888

0.59 -1 0.0090706

165

1.6857956 0.0091197

1614.75832 5.1 mm 5607.5363908761 4205.6522931571 Remarks on Punching Shear

993.69432 5.1 mm 5607.5363908761 4205.6522931571 Remarks on Punching Shear

1714.577840556 23449727226.9521 0.076576954 0.0038196216 0.003373494 0.0038196216 6631.43122261 490.8738521234 14

500.0542991353 212.7026966233 23449727226.9521 -4.2336567903 -0.2111727604 0.003373494 0.003373494 5856.8873738469 490.8738521234 12

759.4050017294 323.0199039912

kN kN kN OK

kN kN kN OK

kNm

mm2 mm2 pcs

kPa kNm

mm2 mm2 pcs

kPa kNm

166

23449727226.9521 0.0138888066 0.000692767 0.003373494 0.003373494 5856.8873738469 490.8738521234 12

500.0542991353 212.7026966233 23449727226.9521 0.0091196521 0.000692767 0.003373494 0.003373494 5856.8873738469 490.8738521234 12

mm2 mm2 pcs

kPa kNm

mm2 mm2 pcs

75 0 3 2.5 1.04

mm

853.7658369529

mm

167

1195.571

3

x

168

RESULTS 800 725 113.152

mm mm kPa

4203.79

kN

37.1517074378 7.775 15.55 2.3891773272 0.8

m2 m m m m

17672349892930.6 748614644810373

mm4 mm4

Negligible

113.152 Rectangular Uniform Pressure 270.340192926 0 135.170096463 -911.7137974277 845.4974565916 -818.463437299 938.7478167203 -108.1360771704

kPa kPa kPa kPa kPa

kN/m kN kN kN kN kN kN kN 169

34 938.7478167203 -911.7137974277 911.7137974277 938.7478167203

kN kN kN kN kN

0.0710141339 0.4789858661 3.3724685462 3.1275314538 0.2794678666 0.2205321334 3.0275314538 3.4724685462 0.4931886929 0.0068113071

mm mm mm mm mm mm mm mm mm mm

0 37.1717765273 41.9712701944 -176.3777412379 -1713.740793737 -391.580848955 -273.4361637493 -363.684907717 -1602.6468078046 27.2393254019

kNm kNm kNm kNm kNm kNm kNm kNm kNm kNm

258.7292297251 258.3609557074 229.9752354502 258.7292297251 -1713.740793737 1713.740793737 1713.740793737

kNm kNm kNm kNm kNm kNm kNm

742.7511768489 1313.4704712222 985.1028534166

kN kN kN

0.59 -1 0.07325 1.6181922 0.076723

0.59 2.5 0.0090437 -0.003621 -4.233668

0.59 -1 0.0137115 1.681091 0.0138243

170

emarks on Wide Beam Shear

OK

991.79132 5.1 mm 5607.5363908761 4205.6522931571 Remarks on Punching Shear

kN

1608.86832 5.1 mm 5607.5363908761 4205.6522931571 Remarks on Punching Shear

991.79132 5.1 mm 5607.5363908761 4205.6522931571 Remarks on Punching Shear

1713.740793737 23395765126.4034 0.0767230327 0.0038269079 0.003373494 0.0038269079 6628.7921423916 490.8738521234 14

500.4111609442 211.5846753037 23395765126.4034 -4.2336675552 -0.2111732973

kN kN OK

0.59 -1 0.0090437 1.6858228 0.0090925

kN kN kN OK

kN kN kN OK

kNm

mm2 mm2 pcs

kPa kNm

171

0.003373494 0.003373494 5843.4096074613 490.8738521234 12

758.6912781116 320.7911019253 23395765126.4034 0.0138242579 0.0006895473 0.003373494 0.003373494 5843.4096074613 490.8738521234 12

500.4111609442 211.5846753037 23395765126.4034 0.0090924942 0.0006895473 0.003373494 0.003373494 5843.4096074613 490.8738521234 12

mm2 mm2 pcs

kPa kNm

mm2 mm2 pcs

kPa kNm

mm2 mm2 pcs

75 0 3 2.5 1.04

mm

853.7658369529

mm

172

1195.831

3

173

x

RESULTS 800 725 113.152

mm mm kPa

4204.813

kN

37.1607483739 7.775 15.55 2.3897587379 0.8

m2 m m m m

17685254829054.8 748796821556034

mm4 mm4

Negligible

113.152 Rectangular Uniform Pressure

kPa kPa kPa kPa kPa

174

270.4059807074 0 135.2029903537 -911.9047202572 845.7341543408 -818.6935562701 938.945318328 -108.162392283 34 938.945318328 -911.9047202572 911.9047202572 938.945318328

kN/m kN kN kN kN kN kN kN kN kN kN kN kN

0.0710162326 0.4789837674 3.3723541094 3.1276458906 0.2794677005 0.2205322995 3.0276458906 3.4723541094 0.493187014 0.006812986

mm mm mm mm mm mm mm mm mm mm

0 37.1808223473 41.9816258503 -176.4121533762 -1714.0449687673 -391.4664926045 -273.2888029421 -363.5629892283 -1602.9200798606 27.2552374598

kNm kNm kNm kNm kNm kNm kNm kNm kNm kNm

258.7930563641 258.4246019293 230.031973955

kNm kNm kNm

0.59 -1 0.0732452 1.6181975 0.0767177

0.59 2.5 0.009047 -0.003622 -4.233666

175

258.7930563641 -1714.0449687673 1714.0449687673 1714.0449687673

kNm kNm kNm kNm

742.9009823151 1313.7901066683 985.3425800012 emarks on Wide Beam Shear

kN kN kN OK

992.05132 5.1 mm 5607.5363908761 4205.6522931571 Remarks on Punching Shear

kN

1609.37132 5.1 mm 5607.5363908761 4205.6522931571 Remarks on Punching Shear

992.05132 5.1 mm 5607.5363908761 4205.6522931571 Remarks on Punching Shear

1714.0449687673 23401458528.72 0.0767177295 0.0038266434 0.003373494 0.0038266434 6629.9469712154 490.8738521234

kN kN OK

0.59 -1 0.0137173 1.6810851 0.0138302

0.59 -1 0.009047 1.6858194 0.0090958

kN kN kN OK

kN kN kN OK

kNm

mm2 mm2

176

14

500.398212192 211.7129925566 23401458528.72 -4.2336662398 -0.2111732317 0.003373494 0.003373494 5844.8316118974 490.8738521234 12

758.717175616 321.0049113688 23401458528.72 0.0138301548 0.0006898415 0.003373494 0.003373494 5844.8316118974 490.8738521234 12

500.398212192 211.7129925566 23401458528.72 0.0090958128 0.0006898415 0.003373494 0.003373494 5844.8316118974 490.8738521234 12

75 0

pcs

kPa kNm

mm2 mm2 pcs

kPa kNm

mm2 mm2 pcs

kPa kNm

mm2 mm2 pcs

mm

177

3 2.5 1.04 853.7658369529

mm

1195.836 178

1195.836

3

x

RESULTS 800 725 113.152

mm mm kPa

4204.833

kN

37.1609251273 7.775 15.55 2.3897701046 0.8

m2 m m m m

17685507187563.3

mm4 179

748800383173740

Negligible

113.152 Rectangular Uniform Pressure 270.407266881 0 135.2036334405 -911.9083697749 845.7388649518 -818.6981382637 938.949096463 -108.1629067524 34 938.949096463 -911.9083697749 911.9083697749 938.949096463

mm4

kPa kPa kPa kPa kPa

kN/m kN kN kN kN kN kN kN kN kN kN kN kN

0.0710162792 0.4789837208 3.3723515654 3.1276484346 0.2794676968 0.2205323032 3.0276484346 3.4723515654 0.4931869766 0.0068130234

mm mm mm mm mm mm mm mm mm mm

0 37.1809991961 41.981828688 -176.4128032958 -1714.0506124739

kNm kNm kNm kNm kNm

0.59 -1 0.0732451 1.6181976 0.0767176

180

-391.4636939711 -273.2853476385 -363.5600406752 -1602.9251090173 27.2555734727

kNm kNm kNm kNm kNm

258.7943065179 258.4258483119 230.0330852894 258.7943065179 -1714.0506124739 1714.0506124739 1714.0506124739

kNm kNm kNm kNm kNm kNm kNm

742.9038279743 1313.7963556506 985.347266738 emarks on Wide Beam Shear

kN kN kN OK

992.05632 5.1 mm 5607.5363908761 4205.6522931571 Remarks on Punching Shear

kN

1609.38132 5.1 mm 5607.5363908761 4205.6522931571 Remarks on Punching Shear

992.05632 5.1 mm 5607.5363908761 4205.6522931571 Remarks on Punching Shear

kN kN OK

0.59 2.5 0.0090471 -0.003622 -4.233666

0.59 -1 0.0137174 1.681085 0.0138303

0.59 -1 0.0090471 1.6858194 0.0090959

kN kN kN OK

kN kN kN OK

181

1714.0506124739 23401569836.6831 0.0767176116 0.0038266375 0.003373494 0.0038266375 6629.9683181298 490.8738521234 14

500.3979243336 211.7154868715 23401569836.6831 -4.2336662143 -0.2111732304 0.003373494 0.003373494 5844.8594125706 490.8738521234 12

758.7177513327 321.0091215613 23401569836.6831 0.0138302713 0.0006898473 0.003373494 0.003373494 5844.8594125706 490.8738521234 12

500.3979243336 211.7154868715 23401569836.6831 0.0090958771 0.0006898473

kNm

mm2 mm2 pcs

kPa kNm

mm2 mm2 pcs

kPa kNm

mm2 mm2 pcs

kPa kNm

182

0.003373494 0.003373494 5844.8594125706 490.8738521234 12

mm2 mm2 pcs

75 0 3 2.5 1.04

mm

853.7658369529

mm

183

1195.846

3

x

RESULTS 800 725 113.152

mm mm kPa 184

4204.879

kN

37.1613316601 7.75 15.5 2.3975052684 0.75

m2 m m m m

17800375450720.7 744000827610943

mm4 mm4

Negligible

113.152 Rectangular Uniform Pressure 271.282516129 0 135.6412580645 -924.563483871 838.7728709677 -838.7728709677 924.563483871 -135.6412580645 0 924.563483871 -924.563483871 924.563483871 924.563483871

0.0639693696 0.4360306304 3.4081204239 3.0918795761 0.25 0.25 3.0918795761 3.4081204239

kPa kPa kPa kPa kPa

kN/m kN kN kN kN kN kN kN kN kN kN kN kN

mm mm mm mm mm mm mm mm

0.59 -1 185

0.4360306304 0.0639693696

mm mm

0 33.9103145161 38.248757404 -163.3202419355 -1738.8320882298 -442.139733871 -337.293125 -442.139733871 -1738.8320882298 -163.3202419355

kNm kNm kNm kNm kNm kNm kNm kNm kNm kNm

38.248757404 33.9103145161 -4.938272013533E-012 38.248757404 -1738.8320882298 1738.8320882298 1738.8320882298

kNm kNm kNm kNm kNm kNm kNm

727.8836596774 1318.0488274336 988.5366205752 emarks on Wide Beam Shear

kN kN kN OK

1006.98824 4.9 mm 5387.6330029986 4040.724752249 Remarks on Punching Shear

kN kN kN OK

0.0740643 1.6172963 0.0776189

0.59 2.5 0.0095619 -0.003828 -4.23346

0.59 -1 0.0144981 1.680291 0.0146243

0.59 -1 0.0095619 1.6852988 0.0096164

1624.32924 kN 4.9 mm 5387.6330029986 kN 4040.724752249 kN

186

Remarks on Punching Shear

OK

1006.98824 4.9 mm 5387.6330029986 4040.724752249 Remarks on Punching Shear

kN

1738.8320882298 23477315605.7888 0.0776189197 0.0038715943 0.003373494 0.0038715943 6729.5716183526 490.8738521234 14

498.7876422071 224.4872494671 23477315605.7888 -4.2334599254 -0.2111629409 0.003373494 0.003373494 5863.777945584 490.8738521234 12

756.2807155859 340.3760704969 23477315605.7888 0.0146242658 0.0007294513 0.003373494 0.003373494 5863.777945584

kN kN OK

kNm

mm2 mm2 pcs

kPa kNm

mm2 mm2 pcs

kPa kNm

mm2

187

490.8738521234 12

mm2 pcs

498.7876422071 224.4872494671 23477315605.7888 0.0096164397 0.0007294513 0.003373494 0.003373494 5863.777945584 490.8738521234 12

kPa kNm

mm2 mm2 pcs

75 0 3 2.5 1.04

mm

853.7658369529

mm

188

1195.846

3

x

189

RESULTS 800 725 113.152

mm mm kPa

4204.879

kN

37.1613316601 7.775 15.55 2.3897962482 0.8

m2 m m m m

17686087621243.2 748808574894464

mm4 mm4

Negligible

113.152 Rectangular Uniform Pressure 270.4102250804 0 135.2051125402 -911.9152636656 845.7511993569 -818.7101768489 938.9562861736 -108.1640900322 34 938.9562861736 -911.9152636656 911.9152636656 938.9562861736

kPa kPa kPa kPa kPa

kN/m kN kN kN kN kN kN kN kN kN kN kN kN

190

0.0710164883 0.4789835117 3.3723401672 3.1276598328 0.2794676802 0.2205323198 3.0276598328 3.4723401672 0.4931868094 0.0068131906

mm mm mm mm mm mm mm mm mm mm

0 37.1814059486 41.9823020918 -176.4138856109 -1714.0581219912 -391.4470946141 -273.2670317447 -363.5430590032 -1602.93101757 27.2567963022

kNm kNm kNm kNm kNm kNm kNm kNm kNm kNm

258.7972237729 258.428752492 230.0356788585 258.7972237729 -1714.0581219912 1714.0581219912 1714.0581219912

kNm kNm kNm kNm kNm kNm kNm

742.9088729904 1313.81072831 985.3580462325 emarks on Wide Beam Shear

kN kN kN OK

0.59 -1 0.0732446 1.6181982 0.0767171

0.59 2.5 0.0090472 -0.003622 -4.233666

0.59 -1 0.0137177 1.6810847 0.0138306

992.06632 kN 5.1 mm

191

5607.5363908761 4205.6522931571 Remarks on Punching Shear

kN kN OK

1609.40732 5.1 mm 5607.5363908761 4205.6522931571 Remarks on Punching Shear

kN

992.06632 5.1 mm 5607.5363908761 4205.6522931571 Remarks on Punching Shear

1714.0581219912 23401825844.9984 0.0767170834 0.0038266111 0.003373494 0.0038266111 6629.9952009231 490.8738521234 14

500.3966346013 211.7209582793 23401825844.9984 -4.2336661603 -0.2111732278 0.003373494 0.003373494 5844.9233541191 490.8738521234 12

0.59 -1 0.0090472 1.6858192 0.009096

kN kN OK

kN kN kN OK

kNm

mm2 mm2 pcs

kPa kNm

mm2 mm2 pcs

192

758.7203307974 321.019336252 23401825844.9984 0.0138305625 0.0006898618 0.003373494 0.003373494 5844.9233541191 490.8738521234 12

500.3966346013 211.7209582793 23401825844.9984 0.0090960134 0.0006898618 0.003373494 0.003373494 5844.9233541191 490.8738521234 12

kPa kNm

mm2 mm2 pcs

kPa kNm

mm2 mm2 pcs

75 0 3 2.5 1.04

mm

853.7658369529

mm

193

1196.437

3

x

194

RESULTS 800 725 113.152

mm mm kPa

4207.089

kN

37.1808629101 7.775 15.55 2.3910522772 0.8

m2 m m m m

17713988638270.5 749202133650974

mm4 mm4

Negligible

113.152 Rectangular Uniform Pressure 270.5523472669 0 135.2761736334 -912.3570353698 846.233221865

kPa kPa kPa kPa kPa

kN/m kN kN kN kN 195

-819.1779871383 939.4122700965 -108.2209389068 34 939.4122700965 -912.3570353698 912.3570353698 939.4122700965

kN kN kN kN kN kN kN kN

0.0710190311 0.4789809689 3.3722015151 3.1277984849 0.279467479 0.220532521 3.0277984849 3.4722015151 0.4931847751 0.0068152249

mm mm mm mm mm mm mm mm mm mm

0 37.2009477492 42.0045391434 -176.4962892283 -1714.8221777256 -391.398683119 -273.151350552 -363.4790438907 -1603.6319780472 27.2823757235

kNm kNm kNm kNm kNm kNm kNm kNm kNm kNm

258.9342902929 258.5655152733 230.1575188103 258.9342902929 -1714.8221777256 1714.8221777256 1714.8221777256

kNm kNm kNm kNm kNm kNm kNm

0.59 -1 0.0732388 1.6182046 0.0767107

0.59 2.5 0.0090545 -0.003625 -4.233663

0.59 -1

196

743.261818328 1314.5012408573 985.875930643 emarks on Wide Beam Shear

kN kN kN OK

992.65732 5.1 mm 5607.5363908761 4205.6522931571 Remarks on Punching Shear

kN

1610.43532 5.1 mm 5607.5363908761 4205.6522931571 Remarks on Punching Shear

992.65732 5.1 mm 5607.5363908761 4205.6522931571 Remarks on Punching Shear

1714.8221777256 23414125374.9296 0.0767106584 0.0038262907 0.003373494 0.0038262907 6632.9242407379 490.8738521234 14

500.3809458377 212.0034936828

kN kN OK

0.0137298 1.6810724 0.0138429

0.59 -1 0.0090545 1.6858118 0.0091034

kN kN kN OK

kN kN kN OK

kNm

mm2 mm2 pcs

kPa kNm

197

23414125374.9296 -4.2336632295 -0.2111730816 0.003373494 0.003373494 5847.9953285118 490.8738521234 12

758.7517083247 321.4710998504 23414125374.9296 0.0138428517 0.0006904748 0.003373494 0.003373494 5847.9953285118 490.8738521234 12

500.3809458377 212.0034936828 23414125374.9296 0.0091034071 0.0006904748 0.003373494 0.003373494 5847.9953285118 490.8738521234 12

mm2 mm2 pcs

kPa kNm

mm2 mm2 pcs

kPa kNm

mm2 mm2 pcs

75 0 3 2.5 1.04

mm

853.7658369529

mm

198

1186.426

3 199

x

RESULTS 800 725 113.152

mm mm kPa

4167.847

kN

36.8340550764 7.775 15.55 2.3687495226 0.8

m2 m m m m

17222911368961 742213883550077

mm4 mm4

Negligible

kPa kPa kPa 200

Negligible

113.152 Rectangular Uniform Pressure 268.0287459807 0 134.0143729904 -904.9958167203 837.1910321543 -810.3881575563 931.7986913183 -107.2114983923 34 931.7986913183 -904.9958167203 904.9958167203 931.7986913183

kPa kPa

kN/m kN kN kN kN kN kN kN kN kN kN kN kN

0.0709405027 0.4790594973 3.3764878965 3.1235121035 0.2794737094 0.2205262906 3.0235121035 3.4764878965 0.4932475979 0.0067524021

mm mm mm mm mm mm mm mm mm mm

0 36.8539525723 41.6074760653 -175.1659444534 -1703.0196552269 -395.5314942926 -278.5450526747 -367.9009998392 -1593.0102012879 26.6832348875

kNm kNm kNm kNm kNm kNm kNm kNm kNm kNm

0.59 -1 0.0734195 1.6180058 0.0769094

0.59 2.5 0.0089283 -0.003574 -4.233714

201

256.4869679776 256.1250004019 227.981982074 256.4869679776 -1703.0196552269 1703.0196552269 1703.0196552269

kNm kNm kNm kNm kNm kNm kNm

737.4778504823 1302.2401126298 976.6800844723 emarks on Wide Beam Shear

kN kN kN OK

982.64632 5.1 mm 5607.5363908761 4205.6522931571 Remarks on Punching Shear

kN

1591.21532 5.1 mm 5607.5363908761 4205.6522931571 Remarks on Punching Shear

982.64632 5.1 mm 5607.5363908761 4205.6522931571 Remarks on Punching Shear

1703.0196552269 23195728020.378 0.076909424 0.003836205 0.003373494

kN kN OK

0.59 -1 0.013508 1.6812978 0.0136174

0.59 -1 0.0089283 1.6859394 0.0089758

kN kN kN OK

kN kN kN OK

kNm

202

0.003836205 6588.0813586621 490.8738521234 13

500.8659584658 207.0986716913 23195728020.378 -4.2337137961 -0.2111756038 0.003373494 0.003373494 5793.4476275525 490.8738521234 12

757.7816830685 313.3285010549 23195728020.378 0.0136174325 0.000679231 0.003373494 0.003373494 5793.4476275525 490.8738521234 12

500.8659584658 207.0986716913 23195728020.378 0.0089758447 0.000679231 0.003373494 0.003373494 5793.4476275525 490.8738521234 12

mm2 mm2 pcs

kPa kNm

mm2 mm2 pcs

kPa kNm

mm2 mm2 pcs

kPa kNm

mm2 mm2 pcs

203

75 0 3 2.5 1.04

mm

853.7658369529

mm

204

854.569

3

x

RESULTS 800 725 113.152

mm mm kPa

3003.211

kN

26.5413868071 7.775 15.55 1.7068415953

m2 m m m 205

0.8

m

6443603388369.55 534814473619188

mm4 mm4

Negligible

113.152 Rectangular Uniform Pressure 193.1325401929 0 96.5662700965 -651.7798327974 603.5816784566 -584.2684244373 671.0930868167 -77.2530160772 24 671.0930868167 -651.7798327974 651.7798327974 671.0930868167

kPa kPa kPa kPa kPa

kN/m kN kN kN kN kN kN kN kN kN kN kN kN

0.0709717714 0.4790282286 3.3747799938 3.1252200062 0.2794712248 0.2205287752 3.0252200062 3.4747799938 0.4932225829 0.0067774171

mm mm mm mm mm mm mm mm mm mm

0 26.5557242765

kNm kNm

0.59 -1 0.0733475 1.6180851 0.0768302

206

29.9824639003 -126.1280054662 -1225.9347755141 -282.772007074 -198.4301515972 -262.8541516077 -1146.6244149032 19.3260011254

kNm kNm kNm kNm kNm kNm kNm kNm

184.825133938 184.5633459807 164.2844292604 184.825133938 -1225.9347755141 1225.9347755141 1225.9347755141

kNm kNm kNm kNm kNm kNm kNm

531.0719951768 938.3506234492 703.7629675869 emarks on Wide Beam Shear

kN kN kN OK

650.78932 5.1 mm 5607.5363908761 4205.6522931571 Remarks on Punching Shear

kN

1090.29332 5.1 mm 5607.5363908761 4205.6522931571 Remarks on Punching Shear

kN kN OK

0.59 2.5 0.0050111 -0.002005 -4.235283

0.59 -1 0.0075882 1.6872927 0.0076225

0.59 -1 0.0050111 1.6898893 0.005026

kN kN kN OK

650.78932 kN 5.1 mm 5607.5363908761 kN

207

4205.6522931571 Remarks on Punching Shear

1225.9347755141 16714064970.1891 0.076830192 0.0038322529 0.003373494 0.0038322529 4742.2603348902 490.8738521234 10

500.67270586 83.7551885985 16714064970.1891 -4.2352827624 -0.2112538631 0.003373494 0.003373494 4174.5643837188 490.8738521234 9

kN OK

kNm

mm2 mm2 pcs

kPa kNm

mm2 mm2 pcs

758.1681882801 126.8304000909 16714064970.1891 0.0076225245 0.0003802078 0.003373494 0.003373494 4174.5643837188 490.8738521234 9

kPa kNm

mm2 mm2 pcs

500.67270586 83.7551885985

kPa kNm

208

16714064970.1891 0.005025964 0.0003802078 0.003373494 0.003373494 4174.5643837188 490.8738521234 9

mm2 mm2 pcs

75 0 3 2.5 1.04

mm

853.7658369529

mm

209

210

211

212

213

214

215

216

217

218

219

220

221

222

223

224

225

226

227

228

229

230

231

232

233

234

235

236

237

238

239

240

241

242

243

244

245

246

247

248

249

250

251

252

253

254

255

256

257

258

259

260

261