Anchored sheet pile wall
Topics:
Design using free earth support method
Assumptions:
- Sheet
pile is rigid, and lateral deflection is small.
- The
lateral pressure distributes according to Rankine’s or Coulomb’s
theories
- The
tie back is strong, and sheet pile rotate about the tie rod anchor point at
failure.
- Bottom
of sheet pile is free to move.
The embedded depth can be determined by summarizing
horizontal earth pressures and moments about the anchor.
å
Fx = 0
[1]
åMo
= 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.
Design length of sheet pile
Theory
Calculating active earth pressure
The method for calculating active earth pressure is the
same as that in cantilever sheet pile wall. The lateral forces Ha1
is calculated as
Ha1=g
Ka h2/2+q Ka h
The depth a can be calculated as
a = pa / g
(Kp-Ka)
The lateral forces Ha2
can be calculated as
Ha2=pa*a/2
Calculating passive earth pressure
The slope from point C to E in the figure above is g
(Kp-Ka). The passive earth pressure at a depth Y below a is calculated as
Pp = g
(Kp-Ka) Y
The passive lateral force
HCEF = g
(Kp-Ka) Y2/2
Derive equation for Y from åMo = 0
åMo
= Ha1*y1
+ Ha2* y2
– HCEF* y3 = 0
Where
y1
= (2h/3-b)
y2 =
(h+a/3-b)
y3 = (h+a+2Y/3)
The equation needs to be determined by a trial and error
process.
Determine anchor force T from å
Fx = 0
å
Fx = Ha1+ Ha2– HCEF-T
= 0
Then,
T = Ha1+ Ha2–
HCEF
Design size of sheet pile
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
Ka (y+b)2/2=0
Solve for y, we have, y = -b+Ö2*T/(g
Ka)
The maximum moment is
Mmax = T y - g
Ka (y+b)3/6
The required section modulus is S = Mmax / Fb
The sheet pile section is selected based on section modulus
Design of tie rod and soldier beam
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 / Ft
Where Ft is allowable tensile stress of steel and is equal
to 0.6Fy 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 = Mmax
/ Fb.
Design procedure
-
Calculate
lateral earth pressure at bottom of excavation, pa and Ha1.
pa
= g
Ka H, Ha1=pa*h/2
-
Calculate
the length a, and Ha2.
a = pa / g
(Kp-Ka), Ha2=pa*a/2
-
Assume
a trial depth Y, calculate HCEF.
HCEF
= g
(Kp-Ka) Y2/3
-
Let
R = Ha1*y1 + Ha2* y2
– HCEF* y3
y1
= (2h/3-b)
y2
= (h+a/3-b)
y3
= (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 = Ha1+ Ha2–
HCEF
-
Calculate
y = -b+Ö2*T/(g
Ka)
-
Calculate
Mmax = T y - g
Ka (y+b)3/6
-
Calculate
required section modulus S= Mmax/Fb.
-
Select
sheet pile section.
-
Design
tie rod
-
Design
soldier beam.
Example 3. Design anchored sheet pile in cohesionless soil.
Given:
Depth of excavation, h = 10 ft
Unit weight of soil, g
= 115 lb/ft3
Internal friction angle, f
= 30 degree
Allowable design stress of sheet pile = 32 ksi
Yield strength of soldier beam,
Fy = 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:
Ka = tan (45-f/2)
= 0.333
Kp = tan (45-f/2)
= 3
The lateral earth pressure at bottom of excavation is
pa =
Ka g
h = 0.333*115*10 = 383.33 psf
The active lateral force above excavation
Ha1 = pa*h/2
= 383.33*10/2 = 1917 lb/ft
The depth a = pa / g
(Kp-Ka) = 383.3 / [115*(3-0.333)] =1.25 ft
The corresponding lateral force
Ha2 = pa*a/2 =
383.33*1.25/2 = 238.6 lb/ft
Assume Y = 2.85 ft
HCEF = g
(Kp-Ka) Y2/3 = 115*(3-0.333)*2.852/3
= 830.3 lb/ft
y1 = (2h/3-b) =
(2*10/3-2)=4.67 ft
y2 =
(h+a/3-b) =
(10+1.25/3-2)=8.42 ft
y3 = (h+a+2Y/3)
= (10+1.25+2*2.85/3) = 13.15 ft
R = Ha1*y1
+ Ha2* y2
– HCEF* y3
= 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 = Ha1+ Ha2–
HCEF = 1917+238.6-830.3 = 1326 lb/ft
Calculate location of maximum moment,
y = -b+Ö2*T/(g
Ka) = -2 ft + Ö2*1326/(115*0.333)
= 6.32 ft
Mmax = T y - g
Ka (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= Mmax/Fb = 4.7*12/32 =
1.8 in3/ft
Use PS28, S = 1.9 in3/ft
Design tie rod, the required cross section area,
A = T s / (0.6*Fy) = 1.326*12/(0.6*36) = 0.442 in3.
Use ¾” diameter tie rod, A = 0.442 in3.
Design
soldier beam:
The
maximum moment of a continuous beams with 3 or more span is
M
= 0.1*T s2 = 0.1*1326*122
=19.1 kip-ft
Required
section modulus, S = M / (0.6*Fy) = 19.1*12/(0.6*36) = 6.4 in3.
Use
W6x15, S = 9.72 in3.
Theory
Calculating active earth pressure
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, pa
= g
h – 2C, where g
is unit weight of soil. The resultant force Ha=pa*h/2
Calculating passive earth pressure
For cohesive soil, friction angle, f
= 0, Ka = Kp = 1. The earth
pressure below excavation,
p1=
sp-sa=
2C-(gh-2C)
= 4C-gh
Assume the embedded depth is D, the resultant force below
bottom of excavation is
HBCDF = p1*D
Derive equation for D from åMo = 0
åMo
= Ha1*y1
– HBCDF* y3
= 0
Where
y1 =
2(h-d)/3-(b-d)
y3 =
h-b+D/2
The equation can be determined with a trial and error
process.
Determine anchor force T from å
Fx = 0
å
Fx = Ha1– HBCDF-T = 0
T = Ha1+ Ha2–
HCEF
Design size of sheet pile
Maximum moment locates at a distance y below T where shear
stress equals to zero.
T-g
Ka (y+b-d)2/2=0
Solve for y, we have, y = -b+d+Ö2*T/(g
Ka)
The maximum moment is
Mmax = T y - g
Ka (y+b-d)3/6
The required section modulus is S = Mmax / Fb
The sheet pile section is selected based on section modulus
Design of tie rod and soldier beam
Design of tie rod and soldier beam is the same as that of
anchored sheet pile in cohesionless soil.
Design procedure
-
Calculate
free standing height, d = 2C/g
-
Calculate
pa=g(h-d)
-
Calculate
Ha=pa*h/2
-
Calculate
p1=4C-gh,
-
Assume
a value of D, and calculate HBCDF = p1*D
-
Calculate
R= Ha*y1 – HBCDF* y3.
Where
y1
= 2(h-d)/3-(b-d)
y3
= 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 = Ha – HBCDF
-
Calculate
y = -b+d+Ö2*T/g
-
Calculate
Mmax = T y - g
(y+b-d)3/6
-
Calculate
required section modulus S= Mmax/Fb.
Select sheet pile section.
-
Design
tie rod
-
Design
soldier beam.
Example 4: Design anchored sheet pile in cohesive soil.
Given:
Depth of excavation, h = 15 ft
Unit weight of soil, g
= 115 lb/ft3
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, pa
= g(h-d)=115*(10-8.7)=150
psf
Total active force,
Ha=pa*h/2
= 150*10/2 = 750 lb/ft
p1=4C-gh
= 4*550-115*15 = 275 psf
Assume D = 11.5 ft,
HBCDF = p1*D = 3163 lb/ft
y1 =
2(h-d)/3-(b-d) =2 (15-8.7)/3-(2-8.7) = 10.9 ft
y3 =
h-b+D/2 =
15-2+11.5/2 = 18.75 ft
R= Ha*y1
– HBCDF* y3
= 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 = Ha
– 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= Mmax/Fb
= 22.47*12/32 = 8.4 in3/ft
Use
PDA 27 section modulus 10.7 in3/ft
Design tie rod
Cross
section of tie rod required, A = T*s/(0.6*Fy) = 2.275*12/(0.6*36) = 0.91 in2.
Diameter
of tie rod, d = Ö4*A/p
= 1.08 in
Use
1-1/8” diameter tie rod.
Design soldier beam
Maximum
moment in solider beam, Mmax =
0.1*T*s2 = 0.1*2275*122 = 32760 lb-ft
Required
section modulus, S= Mmax/Fb=
32.76*12/(0.6*36) = 13.1 in3.
Use
W 8x18, section modulus S = 15.2 in3.
Topics:
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