ANCHORED SHEET PILE WALLDesign using free earth support method

Assumptions:1. Sheet pile is rigid, and lateral deflection is small.2. The lateral pressure distributes according to Rankine’s or Coulomb’s theories3. The tie back is strong, and sheet pile rotate about the tie rod anchor point at failure.4. 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.

Anchored sheet pile wall in cohesionless soilDesign length of sheet pile

TheoryCalculating active earth pressureThe method for calculating active earth pressure is the same as that in cantilever sheet pile wall. The lateral forces Ha1 is calculated asHa1=g Ka h2/2+q Ka hThe depth a can be calculated asa = pa / g (Kp-Ka)The lateral forces Ha2 can be calculated asHa2=pa×a/2

Calculating passive earth pressureThe 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 asPp = g (Kp-Ka) YThe passive lateral forceHCEF = g (Kp-Ka) Y2/2

Derive equation for Y from åMo = 0åMo = Ha1×y1 + Ha2× y2 – HCEF× y3 = 0Wherey1 = (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 = 0å Fx = Ha1+ Ha2– HCEF-T = 0Then,T = Ha1+ Ha2– HCEF

Design size of sheet pileThe 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=0Solve for y, we have, y = -b+Ö2×T/(g Ka)The maximum moment isMmax = T y - g Ka (y+b)3/6The 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 isA = 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 procedure1. Calculate lateral earth pressure at bottom of excavation, pa and Ha1.

pa = g Ka H, Ha1=pa×h/22. Calculate the length a, and Ha2.

a = pa / g (Kp-Ka), Ha2=pa×a/23. Assume a trial depth Y, calculate HCEF.

HCEF = g (Kp-Ka) Y2/34. 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.

5. Calculate the length of sheet pile, L = h+F.S.×D, FS is from 1.2 to 1.4.6. Calculate anchored force T = Ha1+ Ha2– HCEF7. Calculate y = -b+Ö2×T/(g Ka)8. Calculate Mmax = T y - g Ka (y+b)3/69. Calculate required section modulus S= Mmax/Fb.10. Select sheet pile section.11. Design tie rod12. Design soldier beam.

Example 3. Design anchored sheet pile in cohesionless soil.Given:Depth of excavation, h = 10 ftUnit weight of soil, g = 115 lb/ft3

Internal friction angle, f = 30 degreeAllowable design stress of sheet pile = 32 ksiYield strength of soldier beam, Fy = 36 ksiLocation 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) = 0.333Kp = tan (45-f/2) = 3The lateral earth pressure at bottom of excavation ispa = Ka g h = 0.333×115×10 = 383.33 psfThe active lateral force above excavationHa1 = pa×h/2 = 383.33×10/2 = 1917 lb/ftThe depth a = pa / g (Kp-Ka) = 383.3 / [115×(3-0.333)] =1.25 ftThe corresponding lateral forceHa2 = pa×a/2 = 383.33×1.25/2 = 238.6 lb/ftAssume Y = 2.85 ftHCEF = g (Kp-Ka) Y2/3 = 115×(3-0.333)×2.852/3 = 830.3 lb/fty1 = (2h/3-b) = (2×10/3-2)=4.67 fty2 = (h+a/3-b) = (10+1.25/3-2)=8.42 fty3 = (h+a+2Y/3) = (10+1.25+2×2.85/3) = 13.15 ftR = Ha1×y1 + Ha2× y2 – HCEF× y3 = 1917×4.67+238.6×8.42-830.3×13.15 = 42.5 lbR closes to zero, D = 2.85+1.25 = 4.1 ftLength of sheet pile, L = 10 + 1.2× 4.1 = 14.9 ft Use 15 ftCalculate anchor force,T = Ha1+ Ha2– HCEF = 1917+238.6-830.3 = 1326 lb/ftCalculate location of maximum moment,y = -b+Ö2×T/(g Ka) = -2 ft + Ö2×1326/(115×0.333) = 6.32 ftMmax = T y - g Ka (y+b)3/6 = 1326×6.32 – 115×0.333×(6.32+2)3/6 = 4.7 kip-ft/ftThe required section modulus S= Mmax/Fb = 4.7*12/32 = 1.8 in3/ftUse PS28, S = 1.9 in3/ftDesign 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 isM = 0.1×T s2 = 0.1×1326×122 =19.1 kip-ftRequired section modulus, S = M / (0.6×Fy) = 19.1×12/(0.6×36) = 6.4 in3.Use W6x15, S = 9.72 in3.

Anchored sheet pile wall in cohesive soil.

TheoryCalculating active earth pressureCalculation 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/gThe 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 pressureFor cohesive soil, friction angle, f = 0, Ka = Kp = 1. The earth pressure below excavation,p1= sp-sa= 2C-(gh-2C) = 4C-ghAssume the embedded depth is D, the resultant force below bottom of excavation isHBCDF = p1*D

Derive equation for D from åMo = 0åMo = Ha1×y1 – HBCDF× y3 = 0Wherey1 = 2(h-d)/3-(b-d)

y3 = h-b+D/2The equation can be determined with a trial and error process.Determine anchor force T from å Fx = 0å Fx = Ha1– HBCDF-T = 0T = Ha1+ Ha2– HCEFDesign size of sheet pileMaximum moment locates at a distance y below T where shear stress equals to zero.T-g Ka (y+b-d)2/2=0Solve for y, we have, y = -b+d+Ö2×T/(g Ka)The maximum moment isMmax = T y - g Ka (y+b-d)3/6The required section modulus is S = Mmax / Fb

The sheet pile section is selected based on section modulusDesign 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

1. Calculate free standing height, d = 2C/g2. Calculate pa=g(h-d)3. Calculate Ha=pa×h/24. Calculate p1=4C-gh,5. Assume a value of D, and calculate HBCDF = p1×D6. Calculate R= Ha×y1 – HBCDF× y3.

Wherey1 = 2(h-d)/3-(b-d)y3 = h-b+D/2If R is not close to zero, assume a new D, repeat steps 5 and 6

7. The design length of sheet pile is L=h+D×FS, FS=1.2 to 1.4.8. Calculate anchored force T = Ha – HBCDF9. Calculate y = -b+d+Ö2×T/g10. Calculate Mmax = T y - g (y+b-d)3/611. Calculate required section modulus S= Mmax/Fb. Select sheet pile section.12. Design tie rod13. Design soldier beam.

Example 4: Design anchored sheet pile in cohesive soil.Given:Depth of excavation, h = 15 ftUnit weight of soil, g = 115 lb/ft3

Cohesion of soil, C = 500 psfInternal friction angle, f = 0 degreeAllowable design strength of sheet pile = 32 ksiYield strength of soldier beam, Fy = 36 ksiLocation of tie rod at 2 ft below ground surface, spacing =12 ft.

Requirement: Design length of sheet pile and select sheet pile sectionSolution:Design length of sheet pile:The free standing height, d = 2C/g = 2×500/115 = 8.7 ftThe lateral pressure at bottom of sheet pile, pa = g(h-d)=115×(10-8.7)=150 psfTotal active force, Ha=pa×h/2 = 150×10/2 = 750 lb/ftp1=4C-gh = 4×550-115×15 = 275 psfAssume D = 11.5 ft,HBCDF = p1×D = 3163 lb/fty1 = 2(h-d)/3-(b-d) =2 (15-8.7)/3-(2-8.7) = 10.9 fty3 = h-b+D/2 = 15-2+11.5/2 = 18.75 ftR= Ha×y1 – HBCDF×y3 = 5438×10.9-3163×18.75 = -36 lb Close to zeroThe length of sheet pile, L = 15 + 1.2*11.5 = 28.8 ft Use 29 ftAnchored force per foot of wall, T = Ha – HBCDF = 5438 – 3163 = 2275 lb/ftCalculate location of maximum moment,y = -b+d+Ö2×T/g = -2+8.7+Ö2×2275/115 = 13 ftMaximum moment,Mmax = T y - g (y+b-d)3/6 = 2275×13 – 115×(13+2-8.7)3/6 = 24770 lb-ft/ftRequired section modulus of sheet pile, S= Mmax/Fb = 22.47×12/32 = 8.4 in3/ftUse PDA 27 section modulus 10.7 in3/ftDesign tie rodCross 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 inUse 1-1/8” diameter tie rod.Design soldier beamMaximum moment in solider beam, Mmax = 0.1×T×s2 = 0.1×2275×122 = 32760 lb-ftRequired section modulus, S= Mmax/Fb= 32.76×12/(0.6×36) = 13.1 in3.Use W 8x18, section modulus S = 15.2 in3.