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Foundation Engineering Lecture ##31 31

Braced Cuts and Excavations

L. Prieto-Portar 2008

Deep excavations for large utilities, heavy rail transit lines and deep basements for buildings typically require the shoring of vertical walls that may be 30 m deep or more. Underground transportation routes (such as subways) commonly follow existing street alignments, which permit the use of the cut-and-cover method of construction. In this method the lateral walls are placed into the ground first (with sheet piling, slurry walls, soil-cement mixing, etc.) followed by the open excavation of a limited length, say 300 m long. Shortly after excavation, a roof is placed over the job site that permits the restoration of traffic on the street above. For buildings, a method called top-down construction permits the simultaneous work of building the basements and foundations at the same time that the superstructure is erected. Both methods rely on the ability of engineers to effectively brace the deep walls that enclose the excavation. That is the subject of this lecture.

Some shallow temporary utility excavations are not anchored or braced.

This temporary excavation is protected from collapse by a movable earth shield. Temporary excavations kill many workers each year, and these movable shields are becoming very popular with engineering contractors. This photo shows a shield in a utility trench dug to place a drainage pipe at a depth of 15 feet. The pipe is corrugated polyethylene 36” diameter.

A larger movable shield wall is used to place a 108” diameter water main.

This is a very common setup for deeper temporary braced cuts. The walls are steel sheet piles, supported with three rows of horizontal walers, and laterally braced with WF struts.

Lateral Earth Pressures on Deep Walls. Deep walls with multiple braces deflect differently than shallow un-braced retaining walls, as illustrated in photos and figures of the next two slides. Retaining walls rotate about the toe and experience most of their movement at the top. Deep multiply-braced walls experience increasing movement with depth. This variation is complex, and a function of many factors, such as depth, soil parameters, speed of construction, etc. Since there is very little movement at the top of the wall, the lateral earth pressure h is essentially an at-rest o pressure. Further down, wall movement towards the excavation signifies that the soil behind the wall enters an increasing active a pressure. Karl Terzaghi proposed in 1943 a general wedge theory that estimates the total lateral force Fh on the wall by assuming that the failure surface is the arc of a logarithmic spiral, where the radius r = ro e( tan ) from the spiral centroid. The total lateral force F on the wall is located at a distance 0.33H from the toe (from the bottom of excavation), whereas braced walls total lateral force F may range from 0.3 to 0.6 of H. Their failure modes differ considerably. Few structures are actually analyzed via Terzaghi's formula, which may however be suitable for use with a computer code that will test a multitude of wedges to select the Fmax. A simpler analytical alternative was offered by Ralph Peck in 1969 via three typical apparent pressure envelopes.

The difference between the failure mechanisms of a retaining wall and a multi-braced wall.

The Terzaghi failure mechanism for a multi-braced wall.

In the event the backfill has layers of both granular and cohesive soils, Peck proposed a transformation of the granular layers into equivalent single cohesive layer via the formula,

cavg = 0.5H [

2 sandKaH sand

tan

sand +

(H – Hsand )n' qu ]

where qu is the unconfined compression strength of clay and n' is a coefficient of progressive failure (that ranges from 0.5 < n' < 1.0). A similar calculation for the average unit weight

avg

is,

avg = 1/H [ sHs + (H - Hs) c] where c is the saturated unit weight of clay. Once all the granular layers are transformed to equivalent cohesive layers, the overall value of cohesion for the entire backfill is found from, n Cavg = 1/H (c1H1 + c2H2 + .......... + cnHn) = 1/H ciHi i and for the unit weight avg, n avg = 1/H ( 1H1 + 2H2 + .......... + nHn) = 1/H iHi i

Figure 8 A similar calculation for the average unit weight γavg is = 1/H [ γs Hs + (H-Hs) γc] where γc is the saturated unit weight of the clay.

Once all the granular layer(s) are transformed to the equivalent cohesive layer(s), the overall value of cohesion for the entire backfill is found from, cavg = 1/H (c1h1 +c2H2 + ................................... cnHn) and for the unit weight, avg. γavg. = 1/H (γ1H1 + γ2H2 + ............................... γnHn)

A typical excavation to bury a utility may require several levels of horizontal struts. When the soil in back of the wall is a sand, the pressure pa will be uniform from top to bottom, with a value of pa = 0.65 H Ka.

The Construction of Deep Excavations. From shallow excavations (for utilities) to very deep (up to 40 m) for tall buildings, it is required to stabilize the areas immediately around them, against: 1) Lowered bearing capacity; 2) Additional settlements (∆H); 3) Lateral settlements (∆x and ∆y); and 4) Vibrations that cause damage. A stabilized excavation has three elements: 1) A wall; 2) A retention system; and 3) A controlling and a monitoring system. The wall could be built by: 1) Sheet piling (steel, plastic or reinforced concrete). 2) Soldier piles with or without lagging. 3) Drilled in-place piles. 4) Poured in-place concrete panel walls (slurry walls). The retention system could be: 1) Wales supported by horizontal struts or diagonal rakers. 2) Tiebacks or ground anchors. 3) Compression rings.

Beam or pile Strut

Wale

Lagging (2x6,2x8,etc) Strut

Lagging

Wale

Strut spacing is variable depending on tolerable lateral movements and wale strength.

Vertical strut spacing depends Wales on tolerable lateral movement and pile section

Strut

Strut

Plan view for a soldier pile and lagging wall

Elevation view for a soldier pile and lagging wall

Wall constructed of arch web piles.

Depth to first raker depends on the pile section and tolerable lateral movements.

Strut

Wale Wale

Space as required.

Spacing as for soldier beams For shallow excavations, wales and rakers may be omitted.

Rakers

Strut

Plan view for a sheet piling wall.

Elevation view for a sheet piling wall.

Peck’s Design Pressure Envelopes.

pa = γ H [1 – 4 c / γ H]

pa ≈ 0.3 γ H

The pressure against the walls in dense sands: pa = 0.65 γ H Ka where Ka = tan2 (45° - φ/2) A clay is considered soft to medium if: γ H / c > 4, then pa = γ H [1 – 4 c / γ H]

or

0.3 γ H

whichever is greater.

A clay is considered stiff if: γ H / c ≤ 4 , then pa ≈ 0.3 γ H. When there are several layers of sand and clays, the averaged properties are, cavg = 1/(2H) [ γs Ks (Hs)2 tan(φs) + (H - Hs)0.75 qu] γavg = 1/H [γs Hs + (H - Hs) γc] where γc is the saturated unit weight of the clay layer. For multiple clay layers N cavg = 1/H (ci Hi) i

and

γavg =

N 1/H (γi Hi) i

Analysis of Strut (or Anchor) Loads. The minimum vertical spacing for struts (and anchors) is between 2.0 to 3.0 m. The advantages of anchors over struts are (1) they permit to have uncluttered excavations, and (2) struts are axial and bending load limited by their slenderness ratios l / r. The location of the topmost brace in cohesive soils (strut or anchor) should be at a distance d1 from the top that is less than the depth Zc of tensile cracking of the clay soil. The determination of brace loads is as follows: 1. Establish a pressure envelope on wall; 2. Struts and anchors are assumed to be hinged at the wall, except for topmost and lowest levels (ie. A and D in Figure); 3. Separate the loads into 2 supported segments, as shown in Figure; 4. Calculate the strut bracing loads via, FA = A s FB = (B1 + B2)(s) FC = (C1 + C2)(s) FD = D(s) 5. Knowing the bracing loads, prepare a shear and a moment diagram to design the steel struts or anchors.

Analysis of wales. If struts are used, their capacity to properly carry the horizontal loads at each level is enhanced through the use of horizontal steel members against the wall called wales or walers. Their design is based on an analysis of their maximum bending moments Mmax, such that: for wales at level A: for wales at level B: for wales at level C: for wales at level D:

Mmax = [A s2] / 8 Mmax = [(B1 + B2) s2] / 8 Mmax = [(C1 + C2) s2] / 8 Mmax = [D s2] / 8

From these maximum moments, the section modulii are found for each level from, S = Mmax /

all

for steel

Analysis of the Retaining Wall. Each section of the wall in Figure 32.8 is analyzed to determine its maximum bending moment Mmax. Depending on the materials of the wall (steel sheet piling, cast-in-place concrete, drilled shafts, etc.), the section modulus S is then determined and hence the wall thickness, reinforcement, etc.

Example #1. Analyze and design a wall and a multiple bracing system for the excavation shown below. Use a spacing s = 10 feet for the horizontal struts.

Clay φ = 0° γ = 112 pcf c = 700 pcf

Sheet piling

Step 1. Select the suitable pressure diagram. γH/c = (112 lb/ft3)(20ft) / 700 lb/ft2 = 3.2 < 4 ∴ the soil is a stiff clay ∴ pa = 0.3 γ H = 0.3(0.112 k/ft3)(20ft) = 0.672 ksf Step 2. Determine the loads in each strut. Draw the pressure diagram along the entire depth of the wall (draw sideways below for convenience).

Set B = B1 + B2

+ MB1 = 0 therefore -A(7) + 1/2 (5)(0.672)(5+5/3) + (5)(0.672)(5/2) = 0 ∴ A = 2.80 kip/ft Fv = 0

A + B1 = (5)(0.672) + 1/2(5)(0.672) ∴ B1 = 2.24 kips/ft

and by symmetry,

B2 = 2.24 kips/ft, and C = 2.80 kips/ft

Therefore, the loads (force) in each strut (spaced at 10 feet center-to-center) is, Fa = A s = (2.80 k/ft)(10 ft) = 28.0 kips Fb = (B1 + B2) s = (2)(2.24 k/ft)(10) = 44.8 kips Fc = C s = (2.80 k/ft)(10 ft) = 28.0 kips Step 3. Design the struts. Use the strut at B (the highest load) for the design with kL = 20 feet. One possible solution is a steel pipe with a 3” diameter and 0.6” thick walls, which provides a capacity of 60 kips. Alternatively, a W10 x 33 provides a capacity of 105 kips, therefore FS >2.

Step 4. Design the sheet piles. To find the location of the maximum moment find where the shear = 0,

x = 2.24 kips / (0.672 k-ft) = 3.33 ft.

Mmax occurs at E = (2.24 k)(3.33 ft) - (0.672 k-ft)(3.33 ft.)(3.33ft/2) = 3.73 k-ft./ft of wall. The required section modulus of sheet piling S = Mmax/ fall = (3.73 k-ft)(12 in/ft)/(20 k/in2) = 2.2 in3/ft of wall. Could choose a PSA28 which provides a S = 2.5 in3/ft of wall.

The PSA28 steel sheet pile cross-section. Step 5. Design the waler. At the reaction at B, Mmax = (B1 +B2)(s2) / 8 = (2)(2.24 k-ft)(10 ft)2 / 8 = 56 kip-ft. ∴ the required section modulus S = M max / fall = (56 k-ft)(12 in/ft) / (20 k/in2) = 33.6 in3. Suggest the use of a W12 x 36, which provide a S = 46 in3.

Example #2. Analyze and design a wall and a multiple bracing system for the excavation shown below. Use a spacing s = 3 m for the horizontal struts.

Step 1. Determine the lateral pressure pa on the wall. First check the ratio [ H] / c = [18 kN/m3 (7 m)] / 35 kN/m2 = 3.6 < 4 Therefore the soil behaves as a stiff clay, and its "apparent" pressure is, pa = 0.3 H = 0.3(18 kN/m3)(7 m) = 37.8 kN/m2. Step 2. Determine the strut loads A, B and C.

MB1 = 0 A(2.5) - 0.5(37.8)(1.75)(1.75 + 1.75/3) - (1.75)(37.8)(1.75/2) = 0 A = 54.0 kN/m = C

(by symmetry)

Fy = 0 0.5(1.75)(37.8) + (37.8)(1.75) - A - B1 = 0 ↓+ B1 = 45.2 kN/m = B2 (by symmetry) The loads in the struts, FA = 54.0 kN/m (3 m) = 162 kN on the A-level struts FB = 45.2 kN/m (2)(3 m) = 271 kN on the B-level struts FC = = 162 kN on the C-level struts Step 3. Design the struts. The B-level struts carry the largest loads (271 kN), and have an effective length KL = 6 m. Refer to AISC's Manual of Steel Construction and select a W250 mm x 49 kg/m for an fy = 248 MN/m2, or a steel pipe with a diameter = 75 mm (and a wall thickness t = 15 mm).

Step 4. Design the steel sheetpiling wall. Locate the point of zero shear, which corresponds to the location of the maximum moment,

The shear is zero at a distance x from strut B1 (ie. point E in figure), x = (FB1)/(pa) = (45.2 kN/m)/(37.8 kN/m2) = 1.2 m The maximum moment occurs at E, Mmax = - FB1 x + pa (x/2) = - (45.2 kN)(1.2 m) + (37.8 kN/m2)(1/2)(1.2 m)2 = - 54.2 + 27.2 = - 27 kN-m

The section modulus S, of the sheet-piling is, S = Mmax / all = [27 kN-m]/ [0.6(248 MN/m2)] S = 18 x 10-5 m3/m of wall Choose a PMA-22 with S = 29 x 10-5 m3/m. Step 5. Design the wale at the B-level. Mmax = [(B1 + B2) S2]/8 = [90.4 kN)(3 m)2]/8 = 101.7 kN-m The required section modulus S = Mmax /

all

= [101.7 KN-m]/[0.6(248 MN/m2)

S = 68 x 10-5 m3/m of wall The use of a W310 mm x 54 kg/m with Sx = 75 x 10-5 m3/m, shows an un-braced length Lu = 4.085 m (reference to the AISC Manual).

Notice the two struts bracing the opening of this utility tunnel.

A permanent strut braced retaining wall built for the underground Alameda railroad corridor from Long Beach to Los Angeles.

Anchors versus rakers support soldier pile walls.

This ventilation and access shaft for a transit tunnel shows the large steel compression rings that brace the steel bulkhead.

A 19 acre landslide in Manoa Valley, Hawaii, is stabilized with 1000 soil anchors, each 100 ft long, and vertical groundwater drains. The 120 home development was placed over a 10 degree slope, which started to move during a severe storm in 1988. Water from aquifers and broken water mains worsened the ground movement. The 100 ft anchors are held at the surface by concrete cap blocks set at 40 and 50 degrees from the hillside slope to form a Vshape beneath the ground, installed by Schnabel Foundation (CE Sept. 1998).

This retaining wall was placed underground using the soil mixing technique. This method was originated by the Japanese firm Seiko, and has been modified by the US firms Howard Baker and Schnabel. www.schnabel.com

Another example of large steel I-beams used as temporary raker supports of the wall.

An anchored slurry wall nest to the Hudson River, by Hoboken, New Jersey.

This historic industrial plant was preserved and protected against collapse into a new excavation with a soldier pile and wood lagging wall.

This motorway over the Alps, built from Salzburg to Klagenfurt had serious rock falls and sliding slopes near Flachau, Austria. It was stabilized by Bauer AG using over 100 anchors, each 60 m deep and carrying a load of 1,200 kN each. The anchors penetrated rock, gravel, sand, silt and clay strata.

The Stability of Braced Excavations. Bottom Heave. Relief of the loads upon deep clay layers during excavation creates large bulging at the center, and the possibility of “flow” or creep of the clay into the excavation.

The vertical load ”Q” per unit length of cut at the level of the bottom of the cut along the line bd and af, Q = γH (0.7B) - cH But,

Qult = cNc(0.7B) = 5.7c(0.7B)

Qu 5.7c(0.7 B) = FS(against heave) = Q γH(0.7B) - cH

Bjerrum - Eide

B ≈0 L

and increases to

For long cuts

Nc = 5.14

cNc FS = γH

Nc = 7.60

9

8.5

8

7.5

Nc

L/B= infinity L/B=3

7

L/B=2 L/B=1

6.5

6

5.5

5 0

0.5

1

1.5

2

2.5

H/B

3

3.5

4

4.5

5