sh_exmp sheet pile

7/24/2019 Sh_exmp Sheet Pile

1/17

Civi lTech Software SHORING EXAMPLES 1

EXAMPLE 1DESIGN OF CANTILEVERED WALL, GRANULAR SOIL

Introduction:A sheet pile wall is required to support a 12 excavation. The soil is uniform as shown in the

figure. To take into account the friction between the wall and the soil, we use friction angle =

/2. Please note this value is applied only for passive pressure. Use factor of safety = 1.5 forpassive pressure.

Question:What is the embedment and maximum bending moment in the pile?

Calculation:Active pressure above cut:

= 35, Ka = 0.27, @ 12, P1=Ka(12) = (0.27)(125)(12) = 0.405 KsfActive pressure below cut:Below the cut, use pressure slope input.

The pressure slope Pa = Ka= (0.027)(125) = 0.034 KcfPassive pressure below cut:

Soil friction = 35, wall friction = /2, Kp = 6.74 (NAVFAC or Epres program)

The pressure slope Pp = Kp/F.S. = (6.74)(125)/1.50 = 0.56 KcfRunning Program:

A short result for input and output is presented.NOTE: 1 spacing for the sheet pile

Results:The program shows that the minimum embedment is 9.3 and the pile length is 21.3. Themaximum moment in the pile is 17 Kip-ft /ft. The maximum moment is at 15.9 from the groundsurface.

H = 12

Y0.56 kcf1

P1 = 0.405 ksf

0.034 kcf

1

Dredge Line

= 125 pcf

= 35o

= /2 (only for passive)X

EXAMPLE 1


2/17


EXAMPLE 2Alternative Input for Example 1

Introduction:The conditions are the same as example 1. In example 2, we use an alternative input to checkthe versatility of the program. Instead of inputting both active and passive pressures below the

cut, we input the net passive pressure below the cut. The net passive pressure is equal to Pp-Pa.

Question:What is the embedment and maximum bending moment in pile?

Calculation:Active pressure above cut: Same as Example 1.Net pressure below cut:

Active pressure slope is

Pa = Ka= (0.27)(125) = 0.034 KcfPassive pressure slope is

Pp = Kp/F.S. = (6.74)(125) /1.5 = 0.56 KcfNet passive pressure slope is

Pn = Pp - Pa = 0.56 - 0.034 = 0.526 KcfThe point of zero pressure is

e = P1/Pn = 0.405/0.526 = 0.77Running Program:

Input two active pressures in driving pressure input:One is from x = 0 to x = 12. Another is from x = 12 to x = 12 + 0.77There is no active pressure input below dredge line.The passive pressure starts Y = 0.77, with a net slope = 0.526 Kcf

Results:The same as example 1.

Alternative:Another alternative input is also presented in attached file, example 2A. In this input, the net

passive pressure slope, 0.526 Kcf is directly inputted from Y = 0 with starting pressure of -0.405Ksf. In this way, the calculation of zero pressure point, e, can be omitted.


3/17


H = 12

Y

0.526 kcf

1

P1 = 0.405 ksf

= 125 pcf

= 35o

=/2 (only for passive)X

EXAMPLE 2

e

Y = 0.77

H = 12

P5 = 1.752

P2 = 0.618

2 = 130 pcf

2 = 35o

= /2

EXAMPLE 3

1 = 125 pcf

1 = 30o

= /2

SOIL 1

SOIL 2

SURCHARGE, Ps = 0.24

P4 =

0.4

1

0.58

P1 = 0.495

P3 = 0.52615

10

1

0.041

1

0.035

1.2


4/17


EXAMPLE 3DESIGN OF SHEET PILE WALL WITH TWO SOIL AND SURCHARGE

LOAD

Introduction:

A sheet pile wall is required to support 12 excavation. F.S. = 1.5 is required for passivepressure. Soil conditions are shown in the figure. The surcharge load will be in a separatediagram.

Question: What is the embedment and maximum bending moment in pile?


1=30, Ka =0.33, P1= Ka(12) = (0.33)(125)(12) = 0.495 KsfActive pressure below cut:

pressure slope, Pa = Ka= (0.33)(125) = 0.041 Kcf

P2= Ka(15) = (0.33)(125)(15) = 0.618 Ksf

2= 35, Ka = 0.27, pressure slope Pa = Ka= (0.27)(130) = 0.035 Kcf

P3= Ka(15) = (0.27)(130)(15) = 0.526 Ksf

Passive pressure below cut: 1= 30, = 1/2, Kp = 4.8 (NAVFAC or Epres program)

pressure slope, Pp = Kp/F.S. = (4.8)(125) /1.5 = 0.4 KcfP4= Pp(3) = 0.4(3) = 1.2 Ksf

2= 35, = 2/2, Kp = 6.74 (NAVFAC or Epres program)

pressure slope, Pp = Kp/F.S.= (6.74)(130) /1.5 = (6.74)(130) /1.5 = 0.584 KcfP5= Pp(3) = 0.584(3) = 1.752 Ksf

Running Program:Use 1 spacing for the sheet pile and 0.24 Ksf for surcharge load.

Results:The program shows that the minimum embedment is 13.3 and the pile length is 25. Themaximum moment in the pile is 53.3 K-ft/ft at 17.7 from the ground surface.


5/17


EXAMPLE 4DESIGN OF SHEET PILE WALL WITH COHESIVE SOIL AND

SURCHARGE LOAD

Introduction :

In this example, the surcharge load is merged with the active pressure. It also can be separatedas example 3. Soil 2 below cut line is a cohesive material with cohesion = 500 pcf (F.S.included).

Question: What is the embedment and maximum bending moment in the pile?


1=30, Ka = 0.33, @ 0, P1= Ka q = (0.33)(300) = 0.1 Ksf.

P2= Ka (q + H) = 0.33 (300 + 122.5 x 12) = 0.485 KsfPassive pressure below cut:

2= 0, Ka = 0, @ 12, P3= 4c - v = 4(500) - (q + 1H) = 2000 - [300 + 122.5(12)] = 0.23Ksf

Running Program:For passive pressure input, use Top Pres.= 0.23 and Pres. Slope = 0

Results:The program shows that the minimum embedment is 41.58 and the pile length is 53.58. Themaximum moment in the pile is 43.2 Kip-ft/ft at 27.3 from the ground surface.

H = 12

P3 = 0.23

P2 = 0.485

2 = 110 pcf

2 = 0Cu = 500 psf

EXAMPLE 4

1 = 122.5 pcf1 = 30

o

=0Cu = 0

SOIL 1 (LOOSE SAND)

SOIL 2 (MED. CLAY)

P1 = 0.1

q = 300psf


6/17


EXAMPLE 5DESIGN OF SHEET PILE WALL WITH DIFFERENTIAL WATER LEVELS

Introduction:The water level is higher outside of the excavation base. Seepage is allowed to flow through thebottom of the pile. The water pressures of both sides are equal, therefore, the net pressure is

zero at the bottom of the pile.Question: What is the embedment and maximum bending moment in the pile?


= 35, Ka = 0.27, @ 5, P1= Ka(5) = (0.27)(125)(5) = 0.17 Ksf@ 12, P2= (0.27)(125 - 62.4)(12 - 5) + P1= 0.287 Ksf

Active pressure below cut:

Pa = Ka = (0.27)(125 - 62.4) = 0.017 KsfPassive pressure below cut:

= 35, Kp = 6.74, @ = /2, P1=0

Pp = Kp/F.S. = (6.74)(62.6) /1.5 = 0.281 KsfRunning Program:

Due to seepage, the net water pressure = 0 at the embedment bottom. Because theembedment depth is unknown, the value can be set 999 to let the program find embedment.(See Chapter 5).

Results:The program shows that the minimum embedment is 16.8 and the pile length is 28.8. Themaximum moment in the pile is 38.9 Kip-ft/ft at 20 from the ground surface.

1 = 125 pcf

1 = 35o

=/2 (Only for passive)H = 12Water @ 5Seepage @ bottomF. S. = 1.5

0.2811

Pw = 0.43

0.017

1 The depth is unknown,input 999 in program

H = 12

EXAMPLE 5

5P1 = 0.17

P2 = 0.28712


7/17


EXAMPLE 6DESIGN OF ANCHORED SHEET PILE WALL WITH THE SAME WATER

LEVEL

Introduction:

This example has a 36 deep excavation to which tieback anchors are applied. The water levelsare the same inside and outside of the excavation. Two layers of soil are presented in theexample.

Question: What is the embedment, anchor force, and maximum bending moment in the pile?


1= 34, Ka = 0.28, @ 10, P1= Ka(10) = 0.308 Ksf

@ 36, P2= P1+ Ka(36-10) = P1+ (0.28)(60)(26) = 0.745 KsfActive pressure below cut:

2= 34.5, Ka = 0.26, @ 36, P3= Ka[(10) + (26)] = 1.26 [110(10) + 60(26)]= 0.692 Ksf

Passive pressure below cut:

2= 34.5, Kp = 6.63, Pp = KpPn = Pp - Pa = (6.63 - 0.26)(65) = 0.414 KsfRunning Program:

Try input anchor level at 9 from the ground surface.Results:

The program shows that the minimum embedment is 7.5 and the pile length is 43.5. Themaximum moment in the pile is 68.7 Kip-ft/ft at 26 from the ground surface. The brace force =8.8 K/ft.

SAND BACK FILL

= 110 = 60 pcf

= 34o

= 0Ka = 0.28

T

0.4141

H = 36

EXAMPLE 6

10P1=0.308

P2=0.745

P3=0.692

MEDIUM SAND

= 65 pcf

Kp = 6.63

= 34.5o

/= 0.4

Ka = 0.26Kp - Ka = 6.37

9


8/17


EXAMPLE 7DESIGN OF ANCHORED SOLDIER PILE WALL WITH LAGGING

Introduction: A soldier pile wall is used in this example with one tieback anchor applied for each pile.



= 30, Ka = 0.33, @ 12, P1= Ka(12) = 0.485 KsfPassive pressure below cut:

C = 700 psf, = 0, Ka = 0, @ 12, P2= 4c - v = 4(700) - (122.5)(12) = 1.33 KsfRunning Program:

Try an anchor level = 5 from the pile top.Results:

The program shows that the minimum embedment is 1.3 and the pile length is 13.3. Themaximum moment in the pile is 13.2 Kip-ft/ft at 9.3 from the ground surface. Horizontal tiebackforce is 10.6 Kips per pile.

EXAMPLE 7

H = 12

P2 = 1.33

P1 = 0.485

2 = 110 pcf

2 = 0Cu = 700 psf

1 = 122.5 pcf

1 = 30o

=0Cu = 0Ka = 0.33

LOOSE SAND

MED. CLAY

Lagging spacing = 6

T5

Pile Dia. = 2


9/17


EXAMPLE 8DESIGN OF SHEET PILE WALL WITH TIEBACK ANCHOR

Introduction:This example has a limited surcharge load. The surcharge pressure from the Lpres program is0.42 Ksf as shown in the figure. The water pressure is similar to the condition in example 5.


Calculation:Surcharge:Ps = 0.42 KsfWater pressure:

Pw = w (16 - 8) = (62.4)(8) = 0.499 KsfActive pressure:

P1= Ka(8) = (0.27)(125)(8) = 0.27 Ksf

P2= P1+Ka= P1+ (0.27)(125 - 62.4)(7) = 0.388 Ksf

Pa = Kp= (0.27)(125-62.4) = 0.017 KcfPassive pressure: starting 2 below dredge line

P3= Kp[(1) +w (1)] = 8.95 [125 + 62.4] = 1.26 Ksf

Pp = Kp = (8.95)(125 - 62.4) = 0.56 KcfRunning Program:

Use 999 for the unknown depth. (See Example 5)Results:

The program shows that the minimum embedment is 5 and the pile length is 20. Themaximum moment in the pile is 17.3 Kip-ft/ft at 12.5 from the ground surface. T = 8.2 Kips.

T

0.561

H = 15

EXAMPLE 8

10

P1 = 0.27

P2 = 0.39

P3 =1.26

= 62.6 pcf

= 125 pcf

= 35o

= /2

5

1718

q

Ps = 0.42

8

16

Pw = 0.4990.017

1

The depth is unknown,input 999 in program


10/17


EXAMPLE 9DESIGN OF COFFERDAM WITH TWO BRACES

Introduction : A cofferdam is designed for a bridge pier excavation. Two braces are used herein.

Question: What is the embedment, brace force, and maximum bending moment in the pile?

Calculation:

Surcharge:Ps = 0.42 KsfWater pressure:

Pw = w (16 - 8) = (62.4)(8) = 0.499 KsfActive pressure:

P1= Ka(8) = (0.27)(125)(8) = 0.27 Ksf

P2= P1+Ka= P1+ (0.27)(125 - 62.4)(7) = 0.388 Ksf

Pa= Kp= (0.27)(125-62.4) = 0.017 KcfPassive pressure: starting 2 below dredge line

P3= Kp[(1) + w (1)] =8.95 [125 + 62.4] = 1.26 Ksf

Pp = Kp= (8.95)(125 - 62.4) = 0.56 KcfRunning Program:

The condition is the same as in example 8. The maximum movement is significantly reduceddue to two levels of bracing.

Results:The program shows that the minimum embedment is 4.2 and the pile length is 19.2. Themaximum moment in the pile is 7.8 Kip-ft/ft at 14.1 from the ground surface. T1= 5.3 Kips/ ftand T2=2.9 Kips/ ft.

T1

0.51

H = 15

EXAMPLE 9

P1 = 0.27

P2 = 0.39

P3 =1.26

= 62.6 pcf

= 125 pcf

= 35o

= /2

1718

q

Ps = 0.42

8

16

0.01

1

T2

16

10

The depth is unknown,input 999 in program

10

5

Pw = 0.499

Pw = 0


11/17


EXAMPLE 10DESIGN OF COFFERDAM WITH THREE BRACES

Introduction:A cofferdam is designed to support an excavation for a bridge pier. Due to a railroad 6 from thewall, surcharge pressure of 0.42 Ksf is applied. No seepage is allowed below the pile bottom.

The water pressures at both sides are not equal. The outside pressure is higher than inside.After subtraction of the inside pressure, the net hydraulic pressure is constant below 36.



= 35, Ka = 0.27, @ 8, P1= Ka(8) = (0.27)(125)(8) = 0.27 Ksf

@ 35, P2= Ka(35 - 8) + P1= (0.27)(62.6)(27) + 0.27 = 0.724 KsfWater pressure:@ 36, Pw = (36 - 8)(62.4) = 1.747 KsfSurcharge load: (from Lpres program)from 0 to 10, Ps = 0.42 Ksfat 36, Ps = 0

Active pressure below cut:active pressure slope: Pa = Ka = (0.27)(62.6) = 0.017 KcfNet water pressure:Pw = 1.747 Ksf constantPassive pressure below cut:

= 35, = /2, Kp = 8.95

@ 37, P3= Kp(2) = 8.95(62.6)(2) = 0.843 Ksf

passive pressure slope: Pp = Kp= 8.95(62.6) = 0.56 KcfRunning Program:

The soil below the excavation base is disturbed. Therefore the passive pressure of the top 2soil below the dredge line is ignored. The input pressure diagram is shown as follows.

Results:The program shows that the minimum embedment is 8.5 and the pile length is 43.5. Themaximum moment in the pile is 32.4 Kip-ft/ft at 25.2 from the ground surface. T1= 13.6 Kips/ft, T2= 12.6 Kips/ ft, and T3= 16.1 Kips/ ft.


12/17


T1

0.561

H = 35

EXAMPLE 10

P1 = 0.27

P2 = 0.46

P3 =1.26

= 62.6 pcf

= 125 pcf

= 35o

= /2

37

36

Surcharge q

Ps = 0.42

8

36

Pw = 1.7470.017

1

T2

36

10

19

10

T331

Cancel out


13/17


EXAMPLE 11DESIGN OF TIEBACK WALL

Introduction:The tieback wall has an 8 spacing of soldier pile supported by timber lagging (3 x 12). Theactive pressure diagram is a trapezoid. The surcharge load of 2 of soil is added to the active

diagram. The soldier pile has a 2 diameter shaft. Below the dredge line, the active pressure isacting on one diameter. The passive pressure is acting on two diameters of pile due to thearching effect.

Question: What is the embedment, tieback force, and maximum bending moment in the pile?

Calculation:The soil report of this project requires 20(H + 2) for active pressure and ignoring the top 2 ofpassive pressure. The passive resistance is 400 pcf including F.S.

Above base 8 spacingActive pressure P2 = P3 = 20(H + 2) = 20(30 + 2) = 0.64 Ksf

Below baseActive pressure P4 = 30(Z + 2) = 30(30 + 2) = 0.96 Ksf

Active pressure slope Pa = 0.03 KcfActing 1 diameter of pile. Diameter of pile = 2.

Passive pressure slope Pp = 0.4 KcfActing 2 diameter of pile. Diameter of pile = 2.

Running Program:Use 3 pressure programs to input the active pressure

Results:The program shows that the minimum embedment is 6.7 and the pile length is 36.7. Themaximum moment in the pile is 49.7 Kip-ft at 27.1 from the ground surface. T 1= 44.7Kips, T2= 45.1Kips, T3= 42.9Kips.

P4 = 0.96

Acting 2 Dia. of Pile = 4 Acting 1 Dia. Of Pile = 2

0.41

H = 30

EXAMPLE 11

P1 = 0.04

0.4.Y

Surcharge = 2 (Height of Soil)

P2 = 0.64

1

32

0.2 H = 6

Lagging Spacing = 8

0.03 X + 2

P3 = 0.640.2 H = 6

22

14

5.5

22

T1

T2

T3


14/17


EXAMPLE 12DESIGN OF ANCHORED AND RACKING BRACED WALL

Introduction:The original design using one level of tieback is not adequate to support the wall. An additionalraking brace is added to help the stability of the wall.


Calculation:Active pressure above exc. 20(H) = 20(30) = 0.6 Ksf Acting 8Active below exc. 30(Y) = 30(30) = 0.6 Ksf Acting 2 (diameter of pile)Passive below exc. 400Y Acting 4 (2 diameter of pile)Diameter of pile = 2 Lagging spacing = 8Surcharge as shown Ps = 0.36

Running Program:The program only calculates the horizontal bracing force. The user has to calculate the verticalforce and the total force based on the horizontal force calculations.

Results:The program shows that the minimum embedment is 5.8 and the pile length is 35.8. Themaximum moment in the pile is 128.6 Kip-ft at 24 from the ground surface. T 1= 61.3 Kips, T2=

71.1 Kips. The force is in brace = 50 /Sin 45= 70.7 Kips.

0.4

1

H = 30

EXAMPLE 12

P1 = 0.6

Ps = 0.36

0.0675

Surcharge

P2 = 0.6

0.03

1

36

0.2 H = 6

Lagging Spacing = 8

45o

P3 = 0.9

0.2 H = 6

30

6

17

4

T2

T1


15/17


EXAMPLE 13DESIGN OF BRACED CUT WITHOUT EMBEDMENT

Introduction :For trench excavation, sometimes the supporting system only uses steel plates, swale, andbraces. The steel plates do not penetrate into the ground. The program also can solve these

types of problems. The program also can be applied for trench box excavation for pipelines.This example uses three level of braces for a 20 cut. To check the stability problem, the heaveprogram can be used.

Question: What is the bracing force?

Calculation:

Active pressure: = 30 Ka = 0.33

P1= 0.65 KaH = (0.65)(0.33)(120) 20 = 0.51 KsfRunning Program:

In the program output on page 2, check the bottom which will indicate the shoring wall withoutbracing.

Results:Brace @ 3.7 T1 = 2.64k/ ft

Brace @ 10 T2 = 2.87k/ ftBrace @ 16.3 T3 = 2.64k/ ftMaximum moment in plate: 2 Kip-ft/ft

EXAMPLE 13

P1= 0.51

0.2 H = 4

= 120pcf = 30

o

0.2 H = 4

H = 20


16/17


EXAMPLE 14DESIGN OF SHEET PILE WALL, POUR SEAL AT BOTTOM

Introduction:Sometimes the bottom of the excavation is sealed to prevent water and heave. In this case, twoways can be used for inputting the concrete seal. 1) Use a brace at the level of the seal. Note:

the depth of the brace cannot be equal to the depth of the excavation. 0.5 to 1.0 is requiredabove the excavation level as the input of this example. 2) In the passive pressure input, use alarge value of Pres. Top and Pres. Slope. However, if the value is too large, the program cannotfunction. Trial and error is required.

Question: What is the embedment and maximum bending moment in the pile?

Calculation:Active: P1= 650 psf

Water: Pw= w(25 - 10) = 62.4 (15) = 936psfPassive: Pp = 300 pcfConcrete seal was poured at bottom of excavation, then dewater to excavation base.Sheet pile spacing (width) = 1

Running Program:

Results:The program shows that the minimum embedment is 3.7 and the pile length is 28.7. Themaximum moment in pile is 11.7 Kip-ft/ft at 18 from the ground surface. T1=5.8 Kips/ ft,T2=5.8 Kips/ ft, and T3=4.2 Kips/ ft. T3 is the concrete seal. If the seal is 2 thick, thecompression stress is 175 psi.

T1

0.3

1

25

EXAMPLE 14

CONCRETE SEAL, 2 THICK

P1= 0.65

10

Cancel out

Pw = 0.936

T2

T3

4

12

24


17/17


EXAMPLE 15DESIGN OF DEEP EXCAVATION USING MULTI-TIEBACK WALL

Introduction:This is a real project in Seattle area. The excavation is 60 below ground. The lagging spacing is6. 7 levels of tiebacks are used. Vertical spacings of tiebacks need to be adjusted for a

minimum moment in piles and smaller reaction force in tieback. The program proves easy inputinterface, so the user can change the depth of each tieback and run the program several timesto get an optimum result.

Question: What is the embedment, tieback force and maximum bending moment in the pile?

Calculation:

Active: Above Base P1 = 0.65 KaH = (0.65)(0.28)(125)(60) = 1.365 ksf

Below Base P2 = KaH = (0.28)(125)(60)= 2.1 ksf

Pa = Ka= (0.28)(125) = 0.035 kcf

Passive: Below Base Pp = Kp= 6.74 (130) = 0.876 kcf

Surcharge 0-10, Ps1 = 0.64 ksf; >10, Ps2 = 0.24 ksfRunning Program:

Results:The output results are shown in the following:

= 125 pcf

= 34o

ka = 0.28

13.3

0.8761

H = 60

EXAMPLE 15

Lagging Spacing = 6

P1= 1.365

0.035

Ps2 = 0.24

Ps1= 0.64

50

= 130 pcf

= 35o

=/2kp = 6.74

42.6

5.5

35.3

20.6

28.0

2

Acting 2 Dia. of pile = 4. Acting 1 Dia. of pile = 2

Surcharge

48

12

P2 = 2.1

701

sh_exmp sheet pile

Documents