Tablaestacas y Troqueles

SOCIEDAD MEXICANA DEINGENIERIA ESTRUCTURAL, A.C.

Curso: Diseño Estructural de Cimentaciones23 al 26 de septiembre, 2002

1

DISEÑO ESTRUCTURAL DE CIMENTACIONES

Diseño de elementos auxiliares de construcción: tablaestacas y troqueles.

PREAMBULO: El diseño ejecutivo de excavaciones profundas tiene una gran interacción entre el Ingeniero Estructural, el Ingeniero Geotécnico y el Ingeniero Constructor. Para poder estimar las cargas que se generan en los elementos de soporte de la excavación, se requiere conocer la distribución de presiones laterales que ejerce el terreno sobre estos elementos. Sin embargo, esta distribución de presiones varía en función de la separación que se establezca para los elementos de apoyo, de la rigidez de los mismos y del proceso constructivo que se siga. Dadas las incertidumbres que implica el comportamiento de los suelos, éste no es un problema que pueda resolverse con precisión, sino que su solución será más razonable que exacta. Con objeto de dar una idea de los procesos que se siguen en el diseño de elementos de apoyo de excavaciones profundas, en las notas siguientes se ha incluído el Capítulo 7 de un excelente libro de Ingeniería de Cimentaciones: “Principles of Foundation Engineering”; Braja M. Das; PWS Engineering; Boston, MA; 1984. Estas notas pretenden ser sólo una ayuda para entender el comportamiento y diseño de estos elementos, y se invita al lector a estudiarlas detenidamente y a profundizar en bibliografía adicional referente a este tema. SOCIEDAD MEXICANA DE INGENIERÍA ESTRUCTURAL Septiembre, 2002



2

Braced Cuts 7.1 Introduction Sometimes construction work requires ground excavations with vertical or near-vertical faces - for example, basements of buildings in developed areas or underground transportation facilities at shallow depths below the ground surface (cut-and-cover type of construction). The vertical faces of the cuts need to he protected by temporary bracing systems to avoid failure that may be accompanied by considerable settlement or by bearing capacity failure of nearby foundation(s). Figure 7.1 shows two types of braced cut commonly used in construction work. One type uses the soldier beam (Figure 7.la), which is driven into the ground before excavation. Soldier beams are vertical steel or timber beams. After the excavation is started, laggings. are placed between the soldier beams as the excavation proceeds. Laggings are horizontal timber planks. When the excavation reaches the desired depth, wales and struts are properly installed (wales and struts are horizontal steel beams). The struts act like horizontal columns. Figure 7. 11) shows another type of braced excavation. In this case, interlocking sheet piles are driven into tire soil before excavation. As the excavation proceeds, wales and struts are inserted immediately after reaching the appropriate depth. To design braced excavations (that is, to select wales, struts, sheet piles, and soldier beams), one must know the lateral earth pressure to which the braced cuts will be subjected. This topic is discussed in Section 7.2; subsequent sections cover the procedures of analysis and design of braced cuts.



3

Figure 7.1 Types of braced cut: (a) use of soldier beams: (b) use of sheet piles 7.2 Lateral Earth Pressure In Braced Cuts In Chapter 5 we learned that a retaining wall rotates about its bottom (Figure 7.2a). With sufficient yielding of the wall, the lateral earth pressure can he approximated to be equal to that obtained by Rankine´s or Coulomb´s theory.



4

In contrast to retaining walls, braced cuts show a different type of wall yielding (see Figure 7.2b). In this case, the deformation of the wall gradually increases with the depth of excavation. The variation of the amount of deformation will depend on several factors, such as the type of soil, the depth of excavation, and the workmanship. However, one can easily visualize that, with very little wall yielding at the top of the cut, the lateral earth pressure will be close to the at-rest pressure. At the bottom of the wall, with a much larger degree of yielding, the lateral earth pressure will be substantially lower than the Rankine active earth pressure. As a result, the distribution of lateral earth pressure will vary substantially in comparison to the linear distribution assumed in the case of retaining walls. A theoretical evaluation of the total lateral force, P, imposed on a wall can be made by using Terzaghi's general wedge theory (1943a) (Figure 7.3a), in which the failure surface is assumed to be the are of a logarithmic spiral, defined by the equation r = r0eθ tanφ (7.1) where φ = angle of friction of soil A detailed outline for the evaluation of P is beyond the scope of this text; readers should check a soil mechanics text for more information (for example, Das, 1979). However, a comparison of the lateral earth pressure for braced cuts in sand (with angle of wall friction δ = 0) with that for a retaining wall (δ = 0) is shown in Figure 7.3b. lf δ = 0, a retaining wall of height H will he subjected to a Rankine active earth pressure, and the resultant active force will intersect the wall at a distance of nH measured from the bottom of the wall. For this case, n = 1/3. In contrast, the value of n for a braced cut may vary from 0.33 to 0.5 or 0.6. The general wedge theory can also be used to analyze braced cuts in saturated clay (for example, see Das and Seeley, 1975). In any event, when choosing a lateral soil pressure distribution for design of braced cuts, one should keep in mind that the nature of failure in braced cuts



5

is much different from that in retaining walls. After observation of several braced cuts, Peck (1969) suggested using design pressure envelopes for braced cuts in sand and clay. Figure 7.4 shows Peck's pressure envelopes, to which the following guidelines apply: 1. Figure 7.4a is for braced cuts constructed in dry or moist sand. Note Kα. is the Rankine active earth pressure coefficient.



6

2. For cuts in clay, first calculate the value of γH/c (where c = undrained cohesion of the clay located on the sides of the cuts; φ = 0 concept). lf γH/c is less than or equal to 4, the pressure envelope shown in Figure 7.4c should be used. The value of ρα varies between 0.2γH and 0.4γH, with an average of 0.3γH. lf γH is greater than 4, the pressure envelope shown in Figure 7.4b should be used. In this case, ρα may he equal to γH[1 - (4c/γH)] or 0.3γH, whichever is greater. Peck's pressure envelopes are sometimes referred to as apparent pressure envelopes. Sometimes one encounters layers of both sand and clay when constructing a braced cut. In this case, Peck (1943) proposed that an equivalent value of cohesion φ = 0 concept) should be determined in the following manner (refer to Figure 7.5a):

Once the average values of cohesion and unit weight are determined, the pressure envelopes in clay (Figure 7.4b and c) can be used to design the cuts. In a similar manner, when a number of clay layers are encountered in the cut (Figure 7.5b), the average undrained cohesion can be expressed by the equation



7

7.3 Design of Various Components of a Braced Cut Struts In construction work, the struts should have a minimum vertical spacing of about 2.75 m or more. The struts are actually horizontal columns subject to bending. The load-carrying capacity of columns will depend on the slenderness ratio, l/r. The slenderness ratio can he reduced by providing vertical and horizontal supports at intermediate points. For cuts with large widths, it may be necessary to splice the struts. In the case of braced cuts in clayey soils, the depth of the first strut below the ground surface should he less than the depth of tensile crack, zc From Eq. (5. 1 1)

A simplified conservative procedure can be used to determine the strut loads. This procedure will vary depending on the engineers involved in the project, Following is a step-by-step outline of it (refer to Figure 7.6). 1. Draw the pressure envelope for the braced cut (see Figure 7.4). Also show the proposed strut levels. Figure 7.6a shows a pressure envelope for a



8

sandy soil; however, it could also be for a clay. Also, in this figure, the strut levels are marked A, B1, C, and D. The sheet piles (or soldier beams) can be assumed to be hinged at the strut levels, except for the top and bottom ones. In Figure 7.6a, the hinges are at the level of struts B and C. (Many designers also assume the sheet piles, or soldier beams, to be hinged at all strut levels, except for the top.) 2. Determine the reactions for the two simple cantilever beams (top and bottom) and all the simple beams in between. In Figure 7.6b, these reactions are A, B1, B2, C1, C2, and D. 3. The strut loads in Figure 7.6 can now be calculated as follows:



9

4. Knowing the strut loads at each level and the intermediate bracing conditions, one can now select the proper sections by using the steel construction manual. Sheet Piles In order to design the sheet piles, perform the following steps: 1. For each of the sections shown in Figure 7.6b, determine the maximum bending moment. 2. Determine the maximum value of the maximum bending moment (Mmax) obtained in Step 1. Note that the unit of this moment will he, for example kN-m/meter length of the wall. 3. Obtain the section modulus of the sheet piles:

all

MSσ

max=

where σall = allowable flexural stress of the sheet pile material 4. The sheet pile section can now be chosen from a table such as Table 6. 1. Wales 1. Wales can be treated as continuous horizontal members if they are spliced properly. Conservatively, they may also he treated as though they are pinned at the struts. For the section shown in Figure 7.6a, the maximum moments for the wales (assuming the they are pinned at the struts) are as follows:

where A, B1, B2, C1, C2, and D are the reactions under the struts per unit length of the wall (Step 2 of strut design). 2. Determine the section modulus of the wales:



10

The wales are sometimes fastened to the sheet piles at points that satisfy the lateral support requirements. Example 7.1 The cross section of a long braced cut is shown in Figure 7.7a. a. Draw the earth pressure envelope. b. Determine the strut loads at levels A, B, and C. c. Determine the section of the struts subjected to the largest load. d. Determine the sheet pile section required. e. Determine a design section for the wales at level B. Note: The struts are placed at 3 m center-to-center in the plan. Solution Part a Given: γ = 18 kN/m3 , c = 35 kN/m2 and H = 6 m.

46.335

)7)(18( <==cHγ

So, the pressure envelope will he like the one in Figure 7.4c. This is plotted in Figure 7,7a with maximum pressure intensity, pα equal to 0.3γH = 0.3(18)(7) = 37.8 kN/m2

Part b For determination of the strut loads, refer to Figure 7.7b. Taking the moment about B1,



11

Figure 7.7 (Continued)

0275.1)8.37)(75.1(

375.175.1)75.1)(8.37(

21)5.2( =

+

+

=A



12

or A = 54.02 kN/m Also, Σ vertical forces = 0. Thus

21

(1.75)(37,8) + (37.8)(1.75) = A + B1

33.08 + 66.15 - A = B1 So B1 = 45.2 kN/m Due to symmetry B2 = 45.2 kN/m C = 54.02 kN/m Strut load at level Pa = 54.02 x horizontal spacing, s = 54.02 x 3 = 162.06 kN Pb = (B1 + B2)3 = (45.02 + 45.2)3 = 271.2 kN Pc = 54.02 x3 = 162.06 kN

Part C The struts at level B are subjected to the largest load-that is, PB = 271.2 kN. For the struts, effective length (KL: refer to the American lnstitute of Steel Construction, Manual of Steel Construction, 1980, pp. 3-29) with respect to x and y axes is 6 m. Accordingly, the section W 250 mm X 49 kg/m (in English units, it is section W 10 X 33) will be more than sufficient. (Note: Fy = 248.4 MN/m2.) Part d Refer to the left side of Figure 7.7b. For the maximum moment, the shear force should be zero. The nature of variation of the shear force is shown in Figure 7.7c. The location of point E can be given as

mBatreactionx 196.18.372.451

8.37===

The magnitude of moment a

=311

75.18.37)1(

21 xA

lmeterofwalmkN /6.3 −=



13

The magnitude of moment at

−=2196.1)196.18.37()196.12.45( xxE

lmeterofwalmkN /03.2703.2706.54 −=−= Because the loading on the left and right sections of Figure 7.7b are the same, the magnitude of moments at F and C (Figure 7.7e) will be the same as E and A, respectively. Hence, the maximum moment = 27.03 kN-m/meter of wall. The section modulus of the sheet piles,

mmxmkNxmkNMS

all/109.15

/1017003.27 35

23max −=−==

σof the wall

According to Table 6. 1, section PMA-22 can he used. Part e The reaction at level B has been calculated in Part b. Hence

mkNsBBM −=+=+= 7.1018

3)2.452.45(8

)2( 221

max

Section modulus, )10004.248(6.0

7.1016.0

7.1017.101xF

Syall

x ===σ

3310682.0 mx −= So, a section of W 310 mm x 54 kg/m (in English units, W 12 x 36) with an Sx = 0.754 x 10-3 m3 can be used. (Note: Lc = 2.1 m and Lu = 4.085 m – AISC Manuel.) 7.4 Stability of Braced Cuts Heave of the- Bottom of the Cut in Clay Braced cuts in clay may become unstable as a result of the heaving of the bottom of the excavation. Terzaghi (1943b) has analyzed the factor of safety of braced excavations against bottom heave. The failure surface for such a case is shown in Figure 7.8. The vertical load per unit length of the cut at the level of the bottom of the cut along the line bd and af is equal to



14

Q = γHB1 - cH (7.8)

where B1 = 0.7B

c = cohesion (φ = 0 concept)

This load Q can he treated like a load per unit length on a continuous foundation at the level of bd (and af) having a width of B1 = 0.7B. Based on Terzaghi´s bearing capacity theory, the net ultimate load-carrying capacity per unit length of this foundation can be given by the equation [Chapter 3; see Eqs. (3.3) and (3.34)] Qu = cNcB1 = 5.7cB1 Hence, from Eq. (7.8), the factor of safety against bottom heave can he given as

−=

−==

Bcc

HcHHBcB

QQFS u

7.0

7.517.51

1

γγ

(7.9) The preceding factor of safety [Eq. (7.9)] has been derived based on the assumption that the clay layer is homogeneous, at least up to a depth of 0.7B below the bottom of the cut. However, if a hard layer of rock or rock-like material is located at a depth D < 0.7B, the failure surface will he modified to some extent. In such a´ case, the factor of safety can be modified to the form

−=

Dcc

HFS

/7.51

γ



15

Bjerrum and Eide (1956) also studied the problem of bottom heave for braced cuts in clay, and they proposed the following equation for the factor of safety

HcNFS c

γ=

The bearing capacity factor Nc varies witli the ratio of H/B and also L/B (where L = length of the cut). For infinitely long cuts (B/L = 0), Nc = 5.14 at H/B = 0 and increases to a value of Nc = 7.6 at H/B = 4. Beyond that-that is, for H/B > 4 - the value of Nc remains constant. For cuts square in plan (B/L = 1), Nc = 6.3 at H/B = 0, and Nc = 9 for H/B ≥ 4. In general, at any given H/B

+=LBNN squarecglerecc 16.084.0)()tan( (7.12)

Figure 7.9 shows the variation of the value of Nc for L/B = 1, 2, 3, and ∞. In any case, a factor of safety of 1.25 to 1.5 is desired. Stability of the Bottom of the Cut in Sand

The bottom of a cut in sand is generally stable. When the ground water table is encountered, the bottom of the cut is stable as long as the water level inside the excavation is higher than the



16

ground water level. lf the water level inside the cut is lowered below the ground water level by pumping, instability may be created as a result of the upward seepage of water into the cut. Section 7.5 discusses this problem in more detail. Lateral Yielding of Sheet Piles In braced cuts, some lateral movement of sheet pile walls may be expected (Figure 7. 10). Of course, the lateral yield will depend on several factors, the



17

most important of which is time elapsed after excavation that is required for the placement of wales and struts, Mana and Ciotigh (1981) analyzed the field records of several braced cuts in clay from the San Francisco, Oslo (Norway), Boston, Chicago, and Bowline Point (New York) areas. Under ordinary construction conditions, it was found that the maximum lateral wall yield [δH(max)] has a definite relationship with the factor of safety against heave. This is shown in Figure 7. 10. Note that the factor of safety against heave as plotted in Figure 7. 10 has been calculated by using Eqs. (7. 9) and (7. 10). In several instances, the sheet piles (or the soldier piles as the case may be) are driven to a certain depth below the bottom of the excavation. This is done to reduce the lateral yielding of the walls during the last stages of excavation. The lateral yielding of the walls will cause settlement of the ground surface surrounding the cut. The degree of lateral yielding, however, depends mostly on the soil type below the bottom of the cut. lf clay below the cut extends to a great depth and γH/c is less than about 6, extension of the sheet piles or soldier piles below the bottom of the cut will help considerably in reducing the lateral yield of the walls. However, under similar circumstances, if γH/c is about 8, the extension of sheet piles into the clay below the cut does not help to a great extent. In such circumstances, one nay expect a great degree of wall yielding that may result in the total collapse of the bracing systems. lf a hard soil layer is located below a clay layer at the bottom of the cut, the piles should be embedded in the stiffer layer. This will have a great effect in reducing the lateral yield. Ground Settlement



18

The lateral yielding of walls will generally induce ground settlement (δv) around a braced cut. This is generally referred to as ground loss. Based on several field observations, Peck (1969) has provided curves for prediction of ground settlement in various types of soil (see Figure 7.11). The magnitude of ground loss varies extensively; however, Figure 7. 11 can be used as a general guide. Based on the field data obtained from various exits in the areas of San Francisco, Oslo, and Chicago, Mana and Clough (1981) have provided a correlation between the maximum lateral yield of sheet piles [δH(max)] and the maximum ground settlement [δV(max)] This is shown in Figure 7.12. It can be seen that

δV(max) ≈ 0.5 to 1 δH(max) (7.13)

7.2 Example Refer to Example Problem 7.1. Determine the factor of safety against bottom heave using Eqs. (7.9) and (7.11)



19

Solution In Example Problem 7. 1, γ = 18 kN/m3, c = 35 kN/m2, and H = 7 m.

95.2

)6)(7.0(3518

)35)(7.5(71

7.0

7.51 =

−=

−=

Bcc

HFS

γ

Factor of Safety from Eq. (7.9)

HcNFS c

γ=

Factor of Safety from Eq. (7.11)



20

According to Figure 7.9, for H/B = 7/6 = 1. 16 and B/L ≈ 0, the value of Nc is equal to 6.46. Thus

79.1)7)(18(

)46.6)(35( ==FS

Tablaestacas y Troqueles

Documents

Transcript of Tablaestacas y Troqueles