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ENCE 461Foundation Anal ysis and Desi gn

Retaining WallsSheet Piling Overview; Cantilever Walls

Overview of Sheet Pilin g as a Retainin g Wall

� Sheet piling is an “in-situ” type of retaining wall

� Do not rely on their mass to retain the soil, as opposed to a gravity wall

� In-situ walls rely on their flexural strength to retain soil, supported either by their own penetration into the soil or by an anchoring system

� Other types of in-situ walls

� Soldier pile walls – use H-beams to hold timber or concrete lagging to retain soil on a temporary or permanent basis

� Slurry walls – bentonite slurry is injected into a trench after which reinforcement and concrete are placed into the trench, forming a wall

Materials for Sheet Pilin g� Steel

� Cold formed� Hot rolled

� Aluminium� Extruded

� Vinyl� Extruded

� Fibreglass

� Pultruded

� Concrete

� Wood

Steel Sheet Piles

� Hot rolled� Panel and interlocks rolled in one operation� “Traditional” form of steel sheet piling

� Cold formed� Form rolled cold from steel plate� Common with lighter sheet pile profiles� Interlocks more prone to breakage

Aluminium, Vin yl and Fibre glass Sheetin g

� Made for lightweight and light load applications

� Common substitute for wood or concrete walls

� Require special handling in setting and driving

� Vinyl sheets can be obtained in various colours, but is subject to long term creep

Concrete and Wood Sheetin g

� Concrete Sheeting � Wood Sheeting

Sections of Sheet Pilin g

� Z-shaped sheeting– Popular in north America– Usually drive two at a time with split clamp– Wall stiffness developed with each sheet without assumed

assistance from the interlocks

� U-shaped sheeting (Larssen, etc.)– Very popular in Europe– Usually driven one at a time– Wall stiffness developed with two sheets and load transferred

using the interlocks (European practice; U.S. practice does not assume this load transfer)

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Sections of Sheet Pilin g� Flat-web sheeting

– Almost exclusively usedfor cellular cofferdams

– Can be driven singlyor two at a time

� Arched shaped– Used for shallower wall

construction– Used in cold formed steel and

aluminium sheeting

Transitional Sections

Interlock St yles

� Hot rolled and extruded sections

� Ball and socket� Single or double jaw� Double hook� Thumb and finger

� One point contact� Three point contact

� Cold formed sections� Hook and grip

Cantilever and Anchored Walls

� Cantilever Walls

� Walls which have no additional supports, and which rely on the lateral earth pressures in the lower portion of the wall to support the earth in the upper portion

� Limited in height and soil type

� Almost exclusively done with steel piling

� Anchored Walls

� Walls which have additional supports buried in the soil

� These are usually referred to as tiebacks

Anchored Walls

Deep-Seated Failure of Sheet Pile Walls

Rotational Failure due to Inadequate Penetration of

Sheet Pile Walls

Flexural Failure of Sheet Pile Walls

Anchora ge Failure in

Sheet Pile Walls

Active and Passive Pressures

Design of Cantilever WallsSource: British Steel Piling Handbook, http://www.corusconstruction.com

Simplified Method of Desi gn

� Eliminate the “bottom triangle” which fixes the pile toe to the wall; F3 replaces the forces at the toe

� Use the force triangles for resultant forces “F1” and F2

� Increase the penetration by 0.2 * OC to compensate for simplification (not a factor of safety!)

Example of Cantilever Wall Design

� Given

� Cantilever Sheet Pile as shown

� Find

� Necessary penetration to prevent overturning

� Suitable sheet piling for bending moment

� Assume

� Rankine earth pressure conditions

Factor of Safet y and Earth Pressure Considerations

� Two methods of incorporating factor of safety

� Divide the passive earth pressure coefficient by a factor of 1.5 – 2 (Coduto)

� Increase the toe length by 20 – 40% (toe length being distance from excavation level to toe of pile) (PBSSPDM)

� Use Coduto's method

� Ka = tan2 (45 – �/2) = tan2 (45 – 35/2) = 0.271

� Kp = tan2 (45 + �/2)/F = tan2 (45 + 35/2)/1.5 = 2.46

Earth Pressure Dia gram(Lateral and Vertical)

260 psf @ 8'

457 psf @ 20'

Hydrostatic Pressures

For this problem, they are balanced, and do not need to be taken into consideration. They can be a serious factor.

Compute Net PressuresN et Effective Stress

Active Pressure

Passive Pressure

F1

“Region of F1”

F2

“Region of F2”

C

O

Steps to Solve Problem

� Find location of Point O

� Find magnitude, location of F1

� Determine equation for magnitude, location of F2

� Check to make sure F3 > 0

� Sum moments around Point C to determine location of Point C

� Since F3 is assumed to act at Point C, it does not enter into the

calculations

� Increase OC by 20% to determine penetration of sheet piling

� Determine point and magnitude of maximum moment

� Size sheet pile based on maximum moment

Location of Point “O”

O

� Pressure at dredge line = 457 psf

� Slope of line below dredge line = �K

p –

�Ka = ��K

p – K

a) =

(123 – 62.4)(2.46-0.271) = 132.7 psf/ft

� O (or z) = 20 + 457/132.7 = 23.44'

Magnitude and Location of F 1

� F1a = (260)(8)/2 = 1040 lb/ft

� zF1a

= 2*8/3 = 5.33'

� F1b

= (457+260)(12)/2 = 4302 lb/ft

� zF1b

= 8'+6.55'=14.55'

� F1c = (457)(3.44)/2 = 786 lb/ft

� zF1c

= 20 + 3.44/3 = 21.15'

� F1 = 1040 + 4302 + 786 = 6128 lb/ft

� z1 = ((1040)(5.33) + (4302)(14.55) +

(786)(21.15))/6128 = 13.84'

8'

20'

23.44'

F1a

F1b

F1c

z�L

3�q1�q2��q1�2q2�

(distance from Point 1 for trapezoidal load)

Magnitude and Location of F 2

� Pressure at C = ��Kp –

Ka)(C - 23.44) = (123

– 62.4)(2.46-0.271) = 132.7 (C-23.44) psf

� F2 = (132.7) (C –

23.44)2 /2

� z2 = 23.44 + 2 (C-

23.44) /3

23.44'

F2

C

Determine Point C b y Summin g Moments

M c�F 1�C�z1��F 2�C�z2��0

0�6128�C�13.84���132.7��C�23.44�2

2�C�23.44

3�

C�43.66'C�O�43.66�23.44�20.22'�C�O� '�1.2�20.22�24.26'

Ltot�23.44�24.26�47.7'�F 1�F 2�F 3

F 2�27127lb� ftz2�36.92'

27127�6128�20999�0, soF 3OK

Addition for Simplified Method

Toe penetration of 27.7'

Determination of Maximum Moment

� Maximum moment takes place at point of zero shear

� Zero shear takes place at the point where the active and passive forces are equal, i.e., F

1 = F

2'

C

F1

F2'

Mmax

Determination of Maximum Moment

� F2' = (132.7)(z

Mmax –

23.44)2/2 = F1 = 6128

� zMmax

= 33.05'

� Mmax

= 6128 (33.05 -13.84) – 6128 ((33.05-23.44)/3) = 98,089 ft-lb/ft of wall

� From table, AZ-26 has adequate strength

C

F1

F2'

Mmax

Use of ChartsAssume 120 pcf� = 8/20 = 0.4K

p/K

a = 2.46/0.271 = 9.1

Kp/K

a

D/H = 1.25

Moment Ratio = 0.75

Use of Charts

Assume 120 pcf� = 8/20 = 0.4K

p/K

a = 2.46/0.271 = 9.1

D/H = 1.25Moment Ratio = 0.75

� Compute Depth

� D = (1.25)(20) = 25' (simplified method gives 27.7', book gives 34')

� Since we used a reduce Kp, we don't need to add an

additional factor of safety; otherwise, proceed as stated in chart

� M = (0.75)(60)(0.271)(20)3 = 97,560 ft-lb/ft of wall (simplified method gives 98,089 ft-lb/ft of wall)

Note use of submerged weight!

SPW 911

Cantilever Piles in Cla y

� Two step analysis

� Short term, where ��������������

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Questions