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• Manoj Jaiswal

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Client: (Client Name Here) Project: (Project Name Here) Description: (Description of what is being calculated)

Sheet: _____ of _____ Date: mm/dd/yy Job No: ######## By: (Author) Chkd By: _____

Cantilever Sheet Pile Wall in Cohesionless Soil These calculations are in accordance with IBC 2006 and ASCE 7-05. They are based on the procedures outlined in ACI 350.3-06, with modifications as required per ASCE 7-05.

Reference

The resulting forces are Strength Level Forces.

Task : 1) Design length of a cantilever sheet 2) Select sheet pile section

Cs - Cohesion of Soil

C  0

(For Cohesionless soil, C is zero) Note : The calculations are done for 1ft strip width.

Calculations Page 1 of 7

PF - 6.6.#

Client: (Client Name Here) Project: (Project Name Here) Description: (Description of what is being calculated)

Sheet: _____ of _____ Date: mm/dd/yy Job No: ######## By: (Author) Chkd By: _____

Design Parameters

Reference

h = Height of Pile above dredge line

h  12  ft

D = Height of Pile below dredge line L1 = Height of Pile above water table L2 = Height of Pile above dredge line upto water-table L3 = Height of Pile above dredge line L4 = Height of Pile below dredge line D - L3  = Unit weight of homogeneous Soil

γ  115pcf

sat = Specific unit weight of Soil

γsat  52.6pcf

ϕ = Angle of internal soil friction

ϕ  30° Ka  tan( 45°) 

ϕ

2

Ka = Active earth pressure Coefficient

Kp  tan( 45°) 

ϕ

2

Kp = Passive earth pressureCoefficient

K'p = Coefficient

K'p = Kp/SF

 

Kp = 1/ Ka

 

SF = 1.5 - 2.0 K'p 

Kp 1.5

  0.333

2 

  3

2 

2

Surcharge: q = Soil surcharge

q  0psf

The active earth pressure at bottom of excavation is; Pa = γ.h.Ka + q.Ka

Pa  γ  h  Ka  q  Ka Pa  460 psf

The lateral force Ha1 is calculated as Ha1 = γ.h.Ka.h/2 + q.Ka h

Ha1  γ  Ka 

h

2

2

 q  Ka  h

3

Ha1  2.76  10 plf Below the bottom of excavation, the sheet pile is subjected to active pressure on the earth side and passive pressure on the excavation side. Since the passive pressure is larger than active pressure, the lateral pressure on the earth side decreases. Calculations Page 2 of 7

PF - 6.6.#

Client: (Client Name Here) Project: (Project Name Here) Description: (Description of what is being calculated)

Sheet: _____ of _____ Date: mm/dd/yy Job No: ######## By: (Author) Chkd By: _____

At depth "a" below the dredge line, the earth pressure is zero. The depth "a" is calculated as

a 

Pa

γ   Kp  Ka

Reference

a  1.5 ft When the sheet pile rotates away from the earth side, there are active on the earth side and passive pressure on the other side i.e, excavation side The corresponding lateral force

Ha2  Pa 

a 2

Ha2  345 plf Assume Trial Depth Y

(choose arbitrary value for "Y")

Y  10.55ft

The pressure at the bottom of sheet pile on the excavation side P1 P1  γ   Kp  Ka  Y  3.235  10 psf 3

3

P1  3.235  10 psf

The pressure at the bottom of sheet pile on the earth side P2 3

P2  γ  Kp  ( h  a  Y)  q  Kp  γ  Ka  ( a  Y)  7.835  10 psf 3

P2  7.835  10 psf

Derive the depth Z from ΣFx = 0 Summarize lateral forces, Hence solving the equation for Z

ΣFx = Ha1  Ha2  Hp1  Hp2 = 0 Z 

P1  Y  2  Ha1  Ha2 P1  P2

Z  2.522 ft

Derive the depth of embedment D = Y + a Verify the assumed depth Y and Z calculated above from ΣMO = 0 Both P1 and P2 are function of Y, hence to determine Y, take moment about bottom of sheet pile "O" and equate the resultant "R" to zero or closed to zero The depth Y can be determined from a trial and error process. h R  Ha1    a  3

2 2 P1  Y 2a Z    Y  Ha2    Y    P1  P2   3.908 lbf 6 6   3 

Check_R 

Close to "zero", Hence OK

"OK" if R = 0 "Revise Y" otherwise

Calculations Page 3 of 7

PF - 6.6.#

Client: (Client Name Here) Project: (Project Name Here) Description: (Description of what is being calculated)

Sheet: _____ of _____ Date: mm/dd/yy Job No: ######## By: (Author) Chkd By: _____

The embedment depth "D"

Reference

D  Y  a  12.05 ft

D  12.05 ft

Fs  1.2

Lp  h  Fs  D

The design depth of sheet pile (Lp) Fs is factor of safety from 1.2 to 1.4

Lp  26.46 ft

Selection of Sheet Pile Section : (Based on maximum moment and shear) Find maximum shear force The maximum shear force is usually located at D where lateral earth pressure change from active to passive. 3 Vmax  Ha1  Ha2 Vmax  3.105  10 plf Find maximum moment which occurs at the point of zero shear: The maximum moment locates at where shear stress equals to zero between C and D shown in figure above Assume the maximum moment located at a distance "y" below point C, then

 Ha1  Ha2 

γ   Kp  Ka  y

h ΣM max  Ha1    a  3

2

2



y 

2   Ha1  Ha2 γ   Kp  Ka

3 γ   Kp  Ka  y 2a    y  Ha2    y  6   3 

y  4.5 ft

ΣM max  24.84

kip  ft ft

Determine minimum section modulus: Allowable Bending stress

σallow  32ksi

Required Section Modulus

S 

ΣM max σallow

3

S  9.32 

in

Ssp  5.5

in

ft 3

Select sheet pile section modulus per foot of wall

check 

"OK" if Ssp  S

ft

check  "FAILS"

"FAILS" otherwise

Calculations Page 4 of 7

PF - 6.6.#