Sheet pile presentation

Sheet pile design simple way A to Z
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2. Course Instructor Md. Hishamur Rahman Faculty, Department Of Civil Engineering, IUBAT 2

3. Group: 03 group members GROUP MEMBERS 3 SL# NAME ID 01 NOOR E JANNAT 13106166 02 NUR AHMED ZUBAIR SHATU 13206064 03 MINTU MIAH 13206095 04 MD. ABDUL ALIM 13206097 05 MD. MAHADI NAWAZ 13206109 06 S. M. MEHEDI HASAN 13206112 07 MD. SHAMIM REZA 13206010 08 PIAS ROY CHOWDHURY 13306089 09 MD. OSMAN GONI 13306120 10 ATIQUR RAHMAN 13306127 11 MOTIUR RAHMAN 13306008

4. 4 SHEET PILE

5. Introduction Use of sheet piles Advantages Disadvantages Types of sheet piles Construction methods Design of sheet pile in cohesive soil 5

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7. Sheet piling is an earth retention and excavation support technique that retains soil, using  sheet sections with interlocking edges. Sheet piles are installed in sequence to design depth along the planned excavation perimeter or seawall alignment. Sheet pile  is act as a temporary supportive wall that been driven into a slope or excavation to support the soft soils collapse from higher ground to lower ground. 7

8. * Retaining walls * Bridge abutments * Tunnels * Pumping station * Water treatment plants * Basements * Underground car parks * Port facilities * Locks and dams * Waterfront structures * Piled foundations * Excavations and trenches * Cofferdams * Ground water diversion * Barrier for ground water treatment systems * Containment walls * Flood protection * Coastal protection * Tunnel cut and cover * Bulkheads and seawalls * Weir walls * Slope stabilization * Landfill

9. 1. Provides high resistance to driving stresses. 2. Light weight. 3. Can be reused on several projects. 4. Long service life above or below water with modest protection. 5. Easy to adapt the pile length by either welding or bolting. 6. Joints are less apt to deform during driving 9

10. 1. Sections can rarely be used as part of the permanent structure. 2. Installation of sheet piles is difficult in soils with boulders or cobbles. In such cases, the desired wall depths may not be reached. 3. Excavation shapes are dictated by the sheet pile section and interlocking elements. 4. Sheet pile driving may cause neighborhood disturbance. 5. Settlements in adjacent properties may take place due to installation vibrations 10

11. Several types of sheet piles are commonly used in construction Wooden sheet piles Precast concrete sheet piles Steel sheet piles Aluminum sheet piles 11

12. 12 Various types of wooden and concrete sheet piles

13. Sheet pile may be divided into two basic categories. Cantilever Anchored Construction methods generally can be divided into two categories. Backfilled structure Dredged structure 13

14. Construction methods Sequence for backfilled structure : step 1 : Dredged the situ soil in front and back of the proposed structure. step 2 : Drive the sheet piles Step 3 : Backfill up to the level of the anchor and place the anchor system. Step 4 : Backfill up to the top of the wall 14

15. Construction methods Sequence for dredged structure Step 1 : Drive the sheet piles Step 2 : Backfill up to the level of the anchor and place the anchor system. Step 3 : Backfill up to the top of the wall Step 4 : Dredged the front side of the wall 15

16. Design of sheet pile in cohesive soil

17. 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/  The lateral earth pressure at bottom of excavation, p a =  h – 2C, where  is unit weight of soil. The resultant force H a =p a *h/2

18. Calculating passive earth pressure For cohesive soil, friction angle,  = 0, K a = K p = 1. The earth pressure below excavation, p 1 =  p -  a = 2C-(  h-2C) = 4C-  h Assume the embedded depth is D, the resultant force below bottom of excavation is HBCDF = p 1 *D 18

19. Derive equation for D from  M o = 0  M 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. 19

20. Determination of anchor force: Determine anchor force T from  F x = 0  F x = Ha1– HBCDF-T = 0 T = H a1 + H a2 – HCEF Design size of sheet pile : Maximum moment locates at a distance y below T where shear stress equals to zero. T-  K a ( y+b -d) 2 /2=0 Solve for y, we have, y = - b+d +  2*T/(  K a ) The maximum moment is 20

21. 5. Mmax = T y -  K a ( y+b -d) 3 /6 6. The required section modulus is S = M max / F b 7. 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 cohesion less soil 21 Design of sheet pile in cohesive soil

22. Calculate free standing height, d = 2C/  Calculate p a =  (h-d) Calculate H a =p a *h/2 Calculate p 1 =4C-  h, 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 22

23. 10. If R is not close to zero, assume a new D, repeat steps 5 and 6 11. The design length of sheet pile is L= h+D *FS, FS=1.2 to 1.4. 12. Calculate anchored force T = H a – HBCDF 13. Calculate y = - b+d +  2*T/  14. Calculate M max = T y -  ( y+b -d) 3 /6 15. Calculate required section modulus S= M max / F b . Select sheet pile section. 16. Design tie rod 17. Design soldier beam. 23

24. 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 24

25. SOLUTION Design length of sheet pile: Calculate lateral earth pressure coefficients: K a = tan (45-  /2) = 0.333 K p = tan (45-  /2) = 3 The lateral earth pressure at bottom of excavation is p a = K a  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 /  ( 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 25

26. Assume Y = 2.85 ft HCEF =  ( 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 26

27. Calculate anchor force, T = H a1 + H a2 – HCEF = 1917+238.6-830.3 = 1326 lb / ft Calculate location of maximum moment, y = -b+  2*T/(  K a ) = -2 ft +  2*1326/(115*0.333) = 6.32 ft M max = T y -  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 2 . Use1@ ¾” diameter tie rod, A = 0.442 in 2 . 27

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