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Assumptions:
The embedded depth can be determined by summarizing horizontal earth pressures and moments about the anchor.
å F_{x} = 0 [1]
åM_{o} = 0 [2]
The difficulty is that the lateral earth pressure is a function of embedded depth. Both equations are highly nonlinear. A trial and error method has to be used to determine the root.
For structural design, the sheet pile needs to be able to withstand
maximum moment and shear from lateral pressure. A structural
analysis needs to be done to determine maximum moment and shear.
The method for calculating active earth pressure is the same as that in cantilever sheet pile wall. The lateral forces H_{a1} is calculated as
H_{a1}=g K_{a} h^{2}/2+q K_{a} h
The depth a can be calculated as
a = p_{a} / g (K_{p}-K_{a})
The lateral forces H_{a2} can be calculated as
H_{a2}=p_{a}*a/2
The slope from point C to E in the figure above is g (K_{p}-K_{a}). The passive earth pressure at a depth Y below a is calculated as
P_{p} = g (K_{p}-K_{a})
Y
The passive lateral force
HCEF = g (K_{p}-K_{a}) Y^{2}/2
Derive equation for Y from åMo =
0
åM_{o} =
H_{a1}*y_{1} +
H_{a2}* y_{2} –
HCEF* y_{3} =
0
Where
y_{1 }= (2h/3-b)
y_{2} = (h+a/3-b)
y_{3} =
(h+a+2Y/3)
The equation needs to be determined by a trial and error process.
Determine anchor force T from å F_{x} =
0
å F_{x} = H_{a1}+ H_{a2}– HCEF-T = 0
Then,
T = H_{a1}+ H_{a2}– HCEF
The structural is the same as cantilever sheet piles in cohesionless soil.
Maximum moment locates at a distance y below T where shear stress equals to zero.
T-g K_{a} (y+b)^{2}/2=0
Solve for y, we have, y = -b+Ö2*T/(g K_{a})
The maximum moment is
Mmax = T y - g K_{a} (y+b)^{3}/6
The required section modulus is S = M_{max} / F_{b}
The sheet pile section is selected based on section modulus
The sheet pile design above is based on a unit width, foot or meter. The
tie back force T calculated from sheet pile design is force per
linearly width of sheet pile. The
top of sheet pile often supported with soldier beams and tie rods at
certain spacing.
Assume the spacing of tie rod is s, the tension in the rod is T times s. The required area of tie rod is
A = T s / F_{t}
Where F_{t} is
allowable tensile stress of steel and is equal to 0.6F_{y} in
AISC ASD design.
The soil beam is designed as a continuous beam that subjected to tie back force T. The maximum moment in the soldier beam is calculated from structural analysis. The required section modulus is equal to S = M_{max} / F_{b}.
Calculate lateral earth pressure at bottom of excavation, pa and Ha1.
p_{a} = g K_{a} H, H_{a1}=p_{a}*h/2
Calculate the length a, and Ha2.
a = p_{a} / g (K_{p}-K_{a}), H_{a2}=p_{a}*a/2
Assume a trial depth Y, calculate HCEF.
HCEF = g (K_{p}-K_{a}) Y^{2}/3
Let R = H_{a1}*y_{1} + H_{a2}* y_{2} – HCEF* y_{3}
y_{1} = (2h/3-b)
y_{2} = (h+a/3-b)
y_{3} = (h+a+2Y/3)
Substitute Y into R, if R = 0, the embedded depth, D = Y + a.
If not, assume a new Y, repeat step 3 to 4.
Calculate the length of sheet pile, L = h+F.S.*D, FS is from 1.2 to 1.4.
Calculate anchored force T = H_{a1}+ H_{a2}– HCEF
Calculate y = -b+Ö2*T/(g K_{a})
Calculate M_{max} = T y - g K_{a} (y+b)^{3}/6
Calculate required section modulus S= M_{max}/F_{b}.
Select sheet pile section.
Design tie rod
Design soldier beam.
Given:
Depth of excavation, h = 10 ft
Unit weight of soil, g = 115 lb/ft^{3}
Internal friction angle, f = 30 degree
Allowable design stress of sheet pile = 32 ksi
Yield strength of soldier beam, F_{y }= 36 ksi
Location of tie rod at 2 ft below ground surface, spacing, s = 12
ft
Requirement: Design length of an anchored sheet pile, select
sheet pile section, and design tie rod
Solution:
Design length of sheet pile:
Calculate lateral earth pressure coefficients:
K_{a} = tan (45-f/2) = 0.333
K_{p} = tan (45-f/2) = 3
The lateral earth pressure at bottom of excavation is
p_{a} = K_{a} g h = 0.333*115*10 = 383.33 psf
The active lateral force above excavation
H_{a1} = p_{a}*h/2 = 383.33*10/2 = 1917 lb/ft
The depth a = p_{a} / g (K_{p}-K_{a}) = 383.3 / [115*(3-0.333)] =1.25 ft
The corresponding lateral force
H_{a2} = p_{a}*a/2 = 383.33*1.25/2 = 238.6 lb/ft
Assume Y = 2.85 ft
HCEF = g (K_{p}-K_{a}) Y^{2}/3 = 115*(3-0.333)*2.85^{2}/3 = 830.3 lb/ft
y_{1} = (2h/3-b) = (2*10/3-2)=4.67 ft
y_{2} = (h+a/3-b) = (10+1.25/3-2)=8.42 ft
y_{3} = (h+a+2Y/3) = (10+1.25+2*2.85/3) = 13.15 ft
R = H_{a1}*y_{1} + H_{a2}* y_{2} – HCEF* y_{3} = 1917*4.67+238.6*8.42-830.3*13.15 = 42.5 lb
R closes to zero, D = 2.85+1.25 = 4.1 ft
Length of sheet pile, L = 10 + 1.2* 4.1 = 14.9 ft Use 15 ft
Calculate anchor force,
T = H_{a1}+ H_{a2}– HCEF = 1917+238.6-830.3 = 1326 lb/ft
Calculate location of maximum moment,
y = -b+Ö2*T/(g K_{a}) = -2 ft + Ö2*1326/(115*0.333) = 6.32 ft
M_{max} = T y - g K_{a} (y+b)^{3}/6 = 1326*6.32 – 115*0.333*(6.32+2)^{3}/6 = 4.7 kip-ft/ft
The required section modulus S= M_{max}/F_{b} = 4.7*12/32 = 1.8 in^{3}/ft
Use PS28, S = 1.9 in^{3}/ft
Design tie rod, the required cross section area,
A = T s / (0.6*F_{y}) = 1.326*12/(0.6*36) = 0.442 in^{3}.
Use ¾” diameter tie rod, A = 0.442 in^{3}.
Design soldier beam:
The maximum moment of a continuous beams with 3 or more span
is
M = 0.1*T s^{2} =
0.1*1326*12^{2} =19.1
kip-ft
Required section modulus, S = M / (0.6*F_{y}) =
19.1*12/(0.6*36) = 6.4 in^{3}.
Use W6x15, S = 9.72 in^{3}.
Calculation of active earth pressure above excavation is the same as that of cantilever sheet pile in cohesive soil. The free-standing height of soil is d = 2C/g
The lateral earth pressure at bottom of excavation, p_{a} = g h
– 2C, where g is
unit weight of soil. The resultant force H_{a}=p_{a}*h/2
For cohesive soil, friction angle, f = 0, K_{a} = K_{p} = 1. The earth pressure below excavation,
p_{1}= s_{p}-s_{a}=
2C-(gh-2C) = 4C-gh
Assume the embedded depth is D, the resultant force below bottom of excavation is
HBCDF = p_{1}*D
Derive equation for D from åM_{o} =
0
åM_{o} =
H_{a1}*y_{1} –
HBCDF* y_{3} =
0
Where
y_{1} = 2(h-d)/3-(b-d)
y_{3} =
h-b+D/2
The equation can be determined with a trial and error process.
Determine anchor force T from å F_{x} =
0
å F_{x} = Ha1– HBCDF-T = 0
T = H_{a1}+ H_{a2}– HCEF
Maximum moment locates at a distance y below T where shear stress equals to zero.
T-g K_{a} (y+b-d)^{2}/2=0
Solve for y, we have, y = -b+d+Ö2*T/(g K_{a})
The maximum moment is
Mmax = T y - g K_{a} (y+b-d)^{3}/6
The required section modulus is S = M_{max} / F_{b}
The sheet pile section is selected based on section modulus
Design of tie rod and soldier beam is the same as that of anchored
sheet pile in cohesionless soil.
Calculate free standing height, d = 2C/g
Calculate p_{a}=g(h-d)
Calculate H_{a}=p_{a}*h/2
Calculate p_{1}=4C-gh,
Assume a value of D, and calculate HBCDF = p_{1}*D
Calculate R= Ha*y_{1} – HBCDF* y_{3}.
Where
y_{1} = 2(h-d)/3-(b-d)
y_{3} = h-b+D/2
If R is not close to zero, assume a new D, repeat steps 5 and 6
The design length of sheet pile is L=h+D*FS, FS=1.2 to 1.4.
Calculate anchored force T = H_{a} – HBCDF
Calculate y = -b+d+Ö2*T/g
Calculate M_{max} = T y - g (y+b-d)^{3}/6
Calculate required section modulus S= M_{max}/F_{b}. Select sheet pile section.
Design tie rod
Design soldier beam.
Given:
Depth of excavation, h = 15 ft
Unit weight of soil, g = 115 lb/ft^{3}
Cohesion of soil, C = 500 psf
Internal friction angle, f = 0 degree
Allowable design strength of sheet pile = 32 ksi
Yield strength of soldier beam, Fy = 36 ksi
Location of tie rod at 2 ft below ground surface, spacing =12 ft.
Requirement: Design length of sheet pile and select sheet
pile section
Solution:
Design length of sheet pile:
The free standing height, d = 2C/g = 2*500/115 = 8.7 ft
The lateral pressure at bottom of sheet pile, p_{a} = g(h-d)=115*(10-8.7)=150 psf
Total active force, H_{a}=p_{a}*h/2 = 150*10/2 = 750 lb/ft
p_{1}=4C-gh = 4*550-115*15 = 275 psf
Assume D = 11.5 ft,
HBCDF = p_{1}*D = 3163 lb/ft
y_{1} = 2(h-d)/3-(b-d) =2 (15-8.7)/3-(2-8.7) = 10.9 ft
y_{3} = h-b+D/2 = 15-2+11.5/2 = 18.75 ft
R= H_{a}*y_{1} – HBCDF* y_{3} = 5438*10.9-3163*18.75 = -36 lb Close to zero
The length of sheet pile, L = 15 + 1.2*11.5 = 28.8 ft Use 29 ft
Anchored force per foot of wall, T = H_{a} – HBCDF = 5438 – 3163 = 2275 lb/ft
Calculate location of maximum moment,
y = -b+d+Ö2*T/g = -2+8.7+Ö2*2275/115 = 13 ft
Maximum moment,
Mmax = T y - g (y+b-d)^{3}/6 = 2275*13 – 115*(13+2-8.7)^{3}/6 = 24770 lb-ft/ft
Required section modulus of sheet pile, S= M_{max}/F_{b} = 22.47*12/32 = 8.4 in^{3}/ft
Use PDA 27 section modulus 10.7 in^{3}/ft
Cross section of tie rod required, A = T*s/(0.6*F_{y})
= 2.275*12/(0.6*36) = 0.91 in^{2}.
Diameter of tie rod, d = Ö4*A/p =
1.08 in
Use 1-1/8” diameter tie rod.
Maximum moment in solider beam, Mmax =
0.1*T*s^{2} =
0.1*2275*12^{2} =
32760 lb-ft
Required section modulus, S=
M_{max}/F_{b}=
32.76*12/(0.6*36) = 13.1 in^{3}.
Use W 8x18, section modulus S = 15.2 in^{3}.
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