GROUP DESIGN PROJECT

SITE A: SHALLOW SHOUNDATION

Given Properties: Silty Clay/ Clayey Silt σ‘p = σ‘0 + 80 KPA Cr =0.01 Cc= 0.10 e0= 0.85 Height of tank = 18 m Tank thickness = 0.003 m Shallow Foundation Thickness = 0.8 m Shallow Foundation Width = 18.5 m GWT at 4.8m below ground surface Assumptions: ���� = 19.5��/�� ���� = 21��/�� ����� = 24��/�� GS OIL = 0.9 GS STEEL= 7.86 Square Footing 1 – Design Shallow Foundation:

Layer Thickness = 23m – 1.8m = 21.20m

Midheight of layer = 21.2m/2 = 10.60m

σ‘zo = (4.8m)x(19.5 KN/m3) + (9m)x(21 KN/m3 – 9.81 KN/m3) = 194.31 kPa

WOIL = π x (9m – 0.003m)2 x (20m – 0.003m) x (0.9)x(9.81 KN/m

3) = 44897.45 KN

WSTEEL = [π x (9m)2 x (20m) - π x (9m-0.003m)

2x(20m - 0.003m)] x (7.86)x(9.81 KN/m

3) = 320.40 KN

WFOOTING = (0.8m) x (18.5m)2 x (24 KN/m

3) = 6571.20 KN

WSOIL = [(18.5x18.5) – ((Π/4)(18)2 ] x (19.5 KN/m

3) = 1711.73 KN

FWIND = CfqCgCeA

From Table 1.1.3.1 get q value for 75 year return period in Sault Ste Marie

• Hourly wind Pressure (1/30 year) = 0.37

• Hourly wind Pressure (1/100 year) = 0.43

• Calculate q value at (1/75 year)

� q = 0.43 - { [ (0.06) x (-0.00333) ] ÷ (-0.0233) } = 0.421 kPa

q (kPa) 0.37 ? 0.43

Return Period 1/30

1/75

1/100

From table 4.1.7.1 (for Normal Importance, SLS)

Iw= 0.75 (Normal Importance, SLS) Assume open terrain:

Ce= (h/10)

0.2 = (

20m/10)

0.2 =1.15 > 0.9 ok

Calculate Area: A= h*d = (20m)(18m)=360 m

2

For Building as a whole:

Cg=2.0 For Cylinder Tank of moderately smooth:

Slenderness = (h÷d) = (20m/18m) = 1.11 Cf = ?

Slenderness = 7 Cf = 0.6

Slenderness = 6 Cf = 0.5

� Cf = { [ (1.1 – 1)x(0.6-0.5) ] ÷ (7 – 1) } + 0.5 = 0.502

Calculate FWIND:

FWIND = CfqCgCeA = (0.502)(0.421 kPa)(2.0)(1.15)(360 m2) = 175 kN

Calculate SnowLOAD:

S = Is[ ss (CbCwCsCa + Sr )

Ss = 2.8 KPA (base on 1 in 50 year ground snow load)

Sr =0.4 KPA (base on 1 in 50 year ground rain load)

IS = 1.15 (for high house toxic chemicals)

Cb = 0.8 ( for basic snow load factor)

Cw = 1.0 (Building is of high import)

CS = 1.0 (for flat roof)

Ca = 1.0 (for basic shape factor)

� S = 1.15 x [ 2.8 x (0.8 x 1 x 1 x1 + 0.4) ] = 3.036 KPA

� PS = 3.036 KPA x (18.5m)2 = 1039.07 KN

P+Wf = WOIL + WSTEEL + WFOOTING + WSOIL + PS = 44897.45 kN + 320.40 kN + 6571.20 kN + 1711.73 kN + 1039.07 = 54539.85 kN

Calculate Eccentricity

ΣMoment @ center = 0

� (175 kN) x [(20m/2) + 0.8m)] = 54539.85 eB

� eB= 0.035 m < (B/6) = (18.5m/6) = 3.083m � B’= B – 2eB = 18.5m – 2(0.035m) = 18.428m

qapp=P+Wf = (54539.85 kN) = 159.98 kPa A’ (18.5m)(18.428m) qmax = (P+Wf -uD) x (1+6e ) = [ (54539.85 kN) -0 ] x (1 + 6 x 0.035) = 161.17 KPA A L (18.5m x 18.5m) 18 .5m

� Choose bigger q value for FS calculation, thus use qmax = 161.17 KPA

σ‘ZD = (1.8m) x (19.5 KN/m3) = 35.1 kPa

Calculate qULT

qULT = c’Ncscdcic + σ‘ZDNqsqdqiq +0.5γ’B’Nγsγdγiγ

c’ = 191.64 kPa

γ’ = γ – γW�1 − ������ � = 21.0 KN/m

3 – 9.81 KN/m

3�1 − .!�".!"!. #! �= 12.787 KN/m

3

Φ’ = 0° Nc=5.1 Nq= 1.0 N�= 0.0

SHAPE FACTORS s$= 1 + (B’/L)(N%/N$) = 1 + (18.428m/18.5m)(1.0/5.1) = 1.20

s%= 1 + (B’/L)tanΦ’ = 1 + (18.428m/18.5m)tan(0o) = 1.0

s�= 1 - 0.4(B’/L) = 1 - 0.4(18.428m/18.5m) = 0.6

DEPTH FACTORS d$= 1 + 0.4(D/B’) = 1 + 0.4(1.8m/18.428m) = 1.04 d%= 1 +2(D/B’)tan Φ’(1 - sin 0’)

2= 1+ 2(1.8m/18.428m)tan0(1-sin0

0)2 = 1.0

d� = 1.0 LOAD INCLINATION FACTORS &= tan

-1(FWIND/(P + Wf)) = tan

-1(175 kN/54539.85 kN) = 0.184

i$=i% = (1- &/900)2 = (1 - (0.184)/90

o)2 = 0.996

i�= (1- &/ Φ’)2 = (1- (0.190)/0

o)2 = 1

qULT=(191.64 kPa)(5.1)(1.2)(1.04)(0.996) + (35.1 kPa)(1.0)(1.0)(1.0)(0.996)

+ 0.5(0)(12.787 kN/m3)(18.428m)(0.60)(1)(1)

= 1249.83 kPa

FS = qULT/qapp = 1249.83 kPa/161.17 kPa = 7 Satisfactory

Su = c‘ = 191.64 kPa

Eu = 300Su = 300(191.64 kPa) Eu = 57492 kPa

D/B’ = 1.8m/18.428m = 0.098

zh/B’ = (23m -1.8m)/ 18.428 m = 1.15

I1 =1.0

I2=0.36

σ‘ZD = (1.8m) x (19.5 KN/m

3) = 35.1 kPa

Distorsion Settlement

'd = (q - σ‘ZD)B’I1I2 = (159.98 kPa – 35.1 kPa)(18.428m)(1.0)(0.36) Eu 57492 kPa 'd = 14.41 mm

Site A: Shallow Foundations Settlement

Silty Clay Depth to Surface to Bottom of Footing

Use safety hammer type, efficiency mean value 0.57 (0.55-0.6) hf= 1.80 m (Treat the first 0.6m as Clay)

Water Depth 4.80 m Width 18.428 m

Cs = 1 Assume standard sampler Length 18.50 m

ϒsat = 21.00 KN/m3 B'/L = 0.996

ϒclay= 19.50 KN/m3

r= 0.85 (Concrete=Rigid)

Cc= 0.1

Cr= 0.01

eo= 0.85

σm= 80.0 kPa

W f = 54539.85 kN (su)avg = 191.64 kPa

q = 159.980 kPa Eu = 300su = 57492 kPa

σ'zD (kPa) = 35.1 kPa

Only consider layer below bottom of foundation in this case

Skempton and Bjerrum

LayerH

(m)

zf

(m)

σzo'

(kPa)Iσ

∆σ

(kPa)

σp'

(kPa)

σzo' + ∆σ

(kPa)OCR case Cc/(1+e0) Cr/(1+e0) δc (mm)

zh zh/BΨ Ψδc (mm)

1 0.8 0.70 48.75 1.00 124.87 128.75 173.62 Case2 0.054 0.005 6.32 1.10 0.06 0.90 5.69

2 1.1 1.65 67.28 1.00 124.60 147.28 191.88 Case2 0.054 0.005 7.53 2.20 0.12 0.93 7.00

3 1.5 2.95 92.63 0.98 122.98 172.63 215.60 Case2 0.054 0.005 8.52 3.70 0.20 0.95 8.09

4 1.5 4.45 109.83 0.95 118.24 189.83 228.06 Case2 0.054 0.005 7.13 5.20 0.28 0.96 6.85

5 1.5 5.95 126.61 0.88 110.41 206.61 237.02 Case2 0.054 0.005 5.58 6.70 0.36 0.97 5.41

6 1.5 7.45 143.40 0.81 100.56 223.40 243.95 Case2 0.054 0.005 3.96 8.20 0.44 0.98 3.88

7 1.5 8.95 160.18 0.72 89.95 240.18 250.14 Case2 0.054 0.005 2.43 9.70 0.53 0.98 2.38

8 2.0 10.70 179.76 0.62 77.94 259.76 257.70 Case1 0.054 0.005 1.47 11.70 0.63 0.98 1.44

9 2.5 12.95 204.94 0.51 64.26 284.94 269.20 Case1 0.054 0.005 1.64 14.20 0.77 0.99 1.63

10 2.0 15.20 230.12 0.42 53.00 310.12 283.12 Case1 0.054 0.005 1.19 16.20 0.88 0.99 1.18

11 1.5 16.95 249.70 0.37 45.82 329.70 295.52 Case1 0.054 0.005 0.83 17.70 0.96 0.99 0.82

12 2.0 18.70 269.28 0.32 39.81 349.28 309.09 Case1 0.054 0.005 1.04 19.70 1.07 0.99 1.03

13 1.5 20.45 288.87 0.28 34.78 368.87 323.65 Case1 0.054 0.005 0.73 21.20 1.15 0.99 0.72

Sum 46.12 mm

Total Settlement 60.53 mm

Deep Foundations Site A (Part I):

Silty Clay Weight of Stucture

21.00 KN/m2 Assume moist unit weight clay Weight of Oil= 44897 kN

19.50 KN/m2 Assume dry unit weight clay Weight of Steel Tank = 320.4 kN

Cc = 0.1 Weight of Cap = 4928.4 kN

e = 0.85 Snow Load = 1039 kN

Cr = 0.01 Total Weight = 51184.8 kN

Su>75kPa --> α = 0.5

Pile Bearing and Quantity Pile

Choosing Pile PP 20x1.000 pg 445

Nc = 9 Su= 232.09 KPa Area of Pile = 59.69 in2

q't = 2088.765 FS 3 0.03850005 m2

Pult= 1561.6 KPa Circumference of Pile = 0.69556157 m

Pall=Pult/FS= 520.54 KPa Diameter = 0.22140412

# of piles = Total weight/Pall = 98.33

Efficiency Factors:

Choose 22x22 grid (or 484piles) 22 Ge = 0.674

Ge = 0.203 <- governs

(Pall)group= 51226.6 KN <- good

FS= 3.00

Settlement

LayerH

(m)

zf

(m)

σzo'

(kPa)

∆σ

(kPa)

σp'

(kPa)Case Cc/(1+e0) Cr/(1+e0) δc (mm)

1 4.000 2.000 267.2 143.422 347.2 OC-II 0.054 0.005 18.21

2 4.000 6.000 311.9 97.679 391.9 OC-II 0.054 0.005 6.29

3 4.000 10.000 356.7 70.781 436.7 OC-I 0.054 0.005 1.70

4 4.000 14.000 401.5 53.637 481.5 OC-I 0.054 0.005 1.18

5 4.000 18.000 446.2 42.044 526.2 OC-I 0.054 0.005 0.85

6 4.000 22.000 491.0 33.840 571.0 OC-I 0.054 0.005 0.63

7 4.000 26.000 535.7 27.823 615.7 OC-I 0.054 0.005 0.48

8 4.000 30.000 580.5 23.279 660.5 OC-I 0.054 0.005 0.37

9 4.000 34.000 625.3 19.763 705.3 OC-I 0.054 0.005 0.29

Total Settlement 30.0 mm

Bearing

Lower

Depth

Upper

Depth

Depth

(m)Sublayers N N60

Su

(Kpa)α

fs

(kPa)

As

(m2)FsAs (kN)

0.0 0.5 0.25 1 0 0.0 0 1 0.00 0.35 0.00

0.5 1.5 1.00 2 22 21.9 146.3 0.5 73.15 0.70 50.88

1.5 2.1 1.80 3 25 24.9 166.25 0.5 83.13 0.42 34.69

2.1 2.9 2.50 4 25 24.9 166.25 0.5 83.13 0.56 46.25

2.9 4 3.45 5 28 27.9 186.2 0.5 93.10 0.77 71.23

4.0 5.5 4.75 6 28 27.9 186.2 0.5 93.10 1.04 97.14

5.5 7 6.25 7 27 26.9 179.55 0.5 89.78 1.04 93.67

7.0 8.5 7.75 8 29 28.9 192.85 0.5 96.43 1.04 100.60

8.5 10 9.25 9 29 28.9 192.85 0.5 96.43 1.04 100.60

10.0 11.5 10.75 10 30 29.9 199.5 0.5 99.75 1.04 104.07

11.5 13.5 12.50 11 30 29.9 199.5 0.5 99.75 1.39 138.76

13.5 16 14.75 12 30 29.9 199.5 0.5 99.75 1.74 173.46

16.0 18 17.00 13 31 30.9 206.15 0.5 103.08 1.39 143.39

18.0 19.5 18.75 14 30 29.9 199.5 0.5 99.75 1.04 104.07

19.5 21.5 20.50 33 32.9 32.8 218.79 0.5 109.39 1.39 152.18

21.5 23 22.25 35 34.9 34.8 232.09 0.5 116.04 1.04 121.07

SUM: 1481.20 KN

Center to Center Spacing

(Footing width/Spaces)18.5/23 = 0.80 m

Due to pilemovement in Claysoil DuringInstillation the Top1.5m is Ignored inthe Sum of FSAS

Oil Storage Tank’s Foundation Design Project Inline Consultant Engineering

3/24/2014 11

APPENDIX C: Site Schematics

1.8m

18.0 m

20.0

m

18.5 m

Water TableSite A: 4.8 mSite B: 3.4 m

Shallow Foundations

0.8

m

Pile Groups PP 20x1.0S

Dia.18.0m Site A: -Grid 22x22 Piles -Spacing: 0.8m o/cSite B: -Grid 12x12 Piles -Spacing: 1.46m o/c

18.0 mDeep Foundations

0.6

m

20.0

m

Water TableSite A: 4.8 mSite B: 3.4 m

18.5 m