Braced Cuts in Deep Excavation by Prof.E.Saibaba Reddy & Eadala Rakesh Reddy

1. Prof. E.Saibaba Reddy
B.Tech, M.E.(Hons) Roorkee, Ph.D (Nottingham, UK)
Post Doc,(Halifax Canada), Post Doc (Birmingham UK)
&
Eadala Rakesh Reddy
B.Tech (JNTUH), M.Tech (VSSUT-Gold Medal),
(Ph.D)- Andhra University-DST-Inspire Fellow
Chief Consultant –EE Engineering Construction Services

2. INTRODUCTION
An excavation supported by suitable bracing system are called
braced cut. These excavation support systems are used to,
• Minimize the excavation area,
• Keep the sides of deep excavations stable, and
• Ensure that movements of soil will not cause damage to
neighboring structures or to utilities in the surrounding
ground.

3. Thedesign of braced cuts involves twodistinct but
interrelated features, namely
 Stability of excavation, ground movement, controlof
water into the excavation, effect of adjoining
structures and soon.
 Design of structural elements i.e sheet pile, strutsor
anchors and soforth.

5. TYPE I USE OF SOLDIER BEAMS
•Soldier beam is driven into the ground before excavation and is a
vertical steel or timber beam.
•Laggings, which are horizontal timber planks, are placed between
soldier beams as the excavation proceeds.
•When the excavation reaches the desired depth, wales and struts
(horizontal steel beams) are installed. The struts are horizontal
compression members.

6. TYPE II: USE OF SHEET
PILES
•Interlocking sheet piles are driven in to the soil before excavation.
•Wales and struts are inserted immediately after excavation reaches
the appropriate depth.

7. DIFFERENT TYPES OF
SHEETING AND BRACING
SYSTEMS

8.  Vertical Timber Sheeting: Vertical timber sheeting consisting of planks about
8 to 10 cm thick are driven around the boundary of the proposed excavation to
some depth below the base of the excavation. The soil between the sheeting is
then excavated. The sheeting is held in place by a system of wales and struts.
The wales are horizontal beams running parallel to the excavation wall. The
wales are supported by horizontal struts which extend from side to side of the
excavation. However, if the excavations are relatively wide, it becomes
economical to support the wales by inclined struts, known as rakers. For
inclined struts to be successful, it is essential that the soil at the base of the
excavation be strong enough to provide adequate reaction. If the soil can be
temporarily support itself an excavation of limited depth without an external
support, the timber sheeting can be installed in the open or in a partially
completed excavation. Vertical timber sheeting is economical up to a depth of 4
to 6 m.

9. VERTICAL TIMBER SHEETING

10. Steel Sheet Pile: In this method, the steel sheet piles are driven
along the sides of the proposed excavation. As the soil is excavated
from the enclosure, wales and struts are placed. The wales are
made of steel. The struts may be of steel or wood. As the
excavation progresses, another set of wales and struts is inserted.
The process is continued till the excavation is complete. It is
recommended that the sheet piles should be driven several meters
below the bottom of excavation to prevent local heaves. If the width
of a deep excavation is large, inclined bracing may be used.
Steel sheet pile

11. Soldier Beams: Soldier beams are H-piles which are driven at a
spacing of 1.5 to 2.5 m around the boundary of the proposed excavation. As
the excavation proceeds, horizontal timber planks called laggings are placed
between the soldier beams. When the excavation advances to a suitable
depth, wales and struts are inserted. The lagging is properly wedged
between the pile flanges or behind the back flange.
Soldier Beam

12. Tie Backs: In this method, no bracing in the form of struts or
inclined rakers is provided. Therefore, there is no hindrance to the
construction activity to be carried out inside the excavated area. The
tie back is a rod or a cable connected to the sheeting or lagging on
one side and anchored into soil (or rock) outside the excavation area.
Inclined holes are drilled into the soil (or rock), and the hole is
concreted. An enlargement or a bell is usually formed at the end of
the hole. Each tie back is generally prestressed the depth of
excavation is increased further to cope with the increased tension.

13. Use of Slurry Trenches: An alternative to use of sheeting and
bracing system, which is being increasingly used these days, is the
construction of slurry trenches around the area to be excavated and is
kept filled with heavy, viscous slurry of a bentonite clay-water mixture.
The slurry stabilizes the walls of the trench, and thus the excavation can
be done without sheeting and bracing. Concrete is then placed through
a tremie. Concrete displaces the slurry. Reinforcement can also be
placed before concreting, if required. Generally, the exterior walls of the
excavation are constructed in a slurry trench.
Slurry Trench

14. LATERAL EARTH PRESSURE
DISTRIBUTION

15. Lateral earth pressure is the pressure that soil exerts against a
structure in a sideways, mainly horizontal direction. Since most open
cuts are excavated in stages within the boundaries of sheet pile
walls or walls consisting of soldier piles and laggings and since
struts are inserted progressively as the excavation proceeds, the
walls are likely to deform (as shown in figure below). Little inward
movement can occur at the top of the cut after the first strut is
inserted
Typical pattern of deformation of vertical wall (Braced cuts)

16. EARTH PRESSURE DISTRIBUTION
• In Sand
• In Clay

17. IN SAND
FOLLOWING FIGURES SHOWS VARIOUS
RECOMMENDATIONS FOR EARTH PRESSURE DISTRIBUTION
BEHIND SHEETING THIS PRESSURE, PA MAY BE EXPRESSED AS
Terzaghi and Peck’s Peck’s earth
pressure distribution for loose
sand
0.8γHKa

18. 0.8γHKa
TERZAGHI AND PECK’S EARTH
PRESSURE DISTRIBUTION FOR DENSE
SAND
0.8γHKa
Tschebotarioff’s Peck’s earth pressure
distribution
H
H

19. 0.65γHKa
Thornburn’s
distribution
PECK, HANSEN AND
PECK’S EARTH
PRESSURE FOR
MOIST AND DRY SANDS
H
Where,, γ= unit weight
H= height of the cut
Ka= Rankine’s active pressure coefficient.

20. Cuts in Clay
The given figures represent the different earth pressure distribution
recommendations for clay. In clay braced cuts becomes unstable
due to bottom heave .To ensure the stability of braced system
γH/cb must be kept less than 6, where γH/cb is the undrained shear
strength of soil below base or excavation level.
For plastic clay by Peck

21. H
NEUTRAL EARTH PRESSURE
RATIO METHOD BY
TSCHEBOTARIOFF
H Peck, Hanson and Thornburn’s
diagram when γH/c ≤ 4

22. H
Peck, Hanson and Thornburn’s
diagram when γH/c > 4
As the most probable value of any individual strut load is about
25 percent lower than the maximum as obtained from Peck,
Hanson and Thornburn’s earth pressure distribution theories,
so among all the given earth pressure distribution profiles,
Peck, Hanson and Thornburn’s earth pressure distribution
theories are most widely and popularly used

23. PRESSURE ENVELOPE FOR CUTS IN LAYERED
SOIL
Sometimes, layers of both sand and clay are encountered when a braced
cut is being constructed. In this case, Peck (1943) proposed that an
equivalent value of cohesion should be determined according to the
formula,
cav = [γsKsHs tanФs’ + (H-HS)n’qu]2
The average unit weight of the layers may be expressed as,
γa H =[ γs Hs + ( H - HS ) γc ]

24. Where,
H = total height of cut
γs = unit weight of sand
Ks = lateral earth pressure coefficient for sand layer ( =1 )
Hs = height of sand layer
Фs’= effective angle of friction of sand
qu= unconfined compression strength of clay
n’= a coefficient of progressive failure (ranging from 0.5 to 1.0
average value 0.75)
γc = saturated unit weight of clay layer
•Once the average values of cohesion and unit weight are
determined, the pressure envelopes in clay can be used to
design the cuts

25. Similarly, when several clay layers are encountered in the cut, the
average undrained cohesion becomes
Cav = (c1H1 + c2H2 + ...... + cnHn)
The average unit weight is now,
γa =[ γ1 H1 + γ 2H2 + ... + γn Hn ]
Where,
c1,c2,...., cn = undrained cohesion in layer 1,2,...,n
H1 , H2 ,..., Hn = thickness of layers 1, 2, ... , n

26. LIMITATIONS OF THE PRESSURE ENVELOPE
When using the pressureenvelope justdescribed, following pointsare to
be noted:
 The pressure envelopes are sometimes referred to a apparent pressure
envelope. However, theactual pressuredistribution is a functionof the
constructionsequenceand relative flexibilityof wall.
 Theyapply toexcavation having depthsgreaterthanabout 6m
 Theyare based on theassumption thatwatertable is below the bottom
of thecut.
 Sand is assumed to bedrained with zero porewaterpressure.
 Clay is assumed to be undrained and porewaterpressure is not
considered

27. LOADS ON BRACES
Tributary Area Method
Equivalent Beam Method

28. Tributary Area Method
The load on a strut is equal to the load resulting
from pressure distribution over the tributary area over that strut.
For e.g Strut load PB in the fig. is the load on the tributary area 1-
2-3-4.

29. Equivalent Beam Method:
In this method entire depth is split into segments of
simply supported beams and reactions can then be determined by
standard process.

30. STABILITY OF BRACED CUTS

31. Heaving in Clay Soil
The danger of heaving is greater if the bottom of the cut is
soft clay. Even in a soft clay bottom, two types of failure are
possible. They are
Case 1: When the clay below the cut is homogeneous at
least up to a depth equal 0.7 B where B is the width of the
cut.
Case 2: When a hard stratum is met within a depth equal to
0.7 B.

32. CASE:
1
 The anchorage load block of
soil a b c d in Fig. (a) ofwidth
(assumed) at the level of the
bottom of the cut per unit
length may be expressedas
 The vertical pressure qper
unit length of a
horizontal, ba, is

33. THE BEARING CAPACITY QU PER UNIT AREA AT LEVEL AB IS
QU = NCC =5.7C
Where, Nc =5.7
The factor of safety against heaving is
Because of the geometrical condition, it has been found that the width
cannot exceed 0.7 B . Substituting this value for
,
For excavations of limited length L, the factor of safety can be modified to

34. (WHICHEVE
RWhere, B’=T (thickness of clay below the base of excavation) or B/
is smaller)
In 2000, Chang suggested few revisions, with his
modification, equation takes the form
B’=T if T B/
B’ = B/ if T> B/
B’’= B’

35. CASE: 2
 Replacing 0.75B by D in Eq, the
factor of safety is representedby
 For a cut in soft clay with a
constant value of cu belowthe
bottom of the cut, D in Eq.
becomes large, and Fs
approaches thevalue

36. NS IS TERMED AS STABILITY NUMBER. THE
STABILITY NUMBER IS A USEFUL INDICATOR OF
POTENTIAL SOIL MOVEMENTS. THE SOIL
MOVEMENT IS SMALLER FOR SMALLER VALUES OF
NS.

37. Heaving in Cohesionless Soil
A bottom failure in cohesionless soils may occur because of a piping, or
quick, condition if the hydraulic gradient h/L is too large. A flow net
analysis may be used to estimate when a quick condition may occur.
Possible remedies are to drive the piling deeper to increase the length of
the flow path L of Fig or to reduce the hydraulic head h by less pumping
from inside the cell. In a few cases it may be possible to use a surcharge
inside the cell.
Fig. (a) condition for piping or quick, conditions ;(b) conditions for blow in

38. In Fig. 16(b ) the bottom of the excavation may blow in if the
pressure head hw indicated by the piezometer is too great, as follows
(SF = 1.0):
γwhw= γshs
This equation is slightly conservative, since the shear, or wall
adhesion, on the walls of the cofferdam is neglected. On the other
hand, if there are soil defects in the impervious layer, the blow-in
may be local; therefore, in the absence of better data, the equality
as given should be used. The safety factor is defined as

39. BJERRUM AND EIDE (1956) METHOD OF ANALYSIS
THIS METHOD OF ANALYSIS IS APPLICABLE IN THE
CASES :
 where the braced cuts are rectangular, square or
circular in plan or the depth of excavation exceeds the
width of the cut.
 In this analysis the braced cut is visualized
as a deep footing whose depth and horizontal
dimensions are identical to those at the bottom of the
braced cut. The theory of Skempton for computing Nc
(bearing capacity factor) for different shapes of footing
is made useof.

40. Figure gives values of Nc as a function of H/B for
long, circular or square footings. For rectangular
footings, the value of Nc may be computed by the
expression:

41. NC (rect) = (0.84 + 0.16B/L) NC (sq)
Where,
L = length of excavation.
B = width of excavation
The factor of safety for bottom heave may be expressed as
Where,
q, is the uniform surcharge load.

42. DESIGN OF VARIOUS COMPONENTS OF
BRACING
Struts: The strut is a compression member whose load-carrying
capacity depends upon slenderness ratio, l/r. The effective length ‘l’ of
the member can be reduced by providing vertical and horizontal
supports at intermediate points. The load carried by a strut can be
determined from the pressure envelope. The struts should have a
minimum vertical spacing of about 2.5 m. In the case of braced cuts in
clayey soils, the depth of the first strut below the ground surface
should be less than the depth of tensile crack (Zc), which is equal to ,
Zc= (2c/ γ)
While calculating the load carried by various struts, it is generally assumed
that the sheet piles (or soldier beams) are hinged at all the strut levels expect
for the top and bottom struts.

43. Determination of strut load ;(a) section and plan of cut;
(b) Method for determining strut loads

44. STEPS
 Draw the pressureenvelopeof the braced cutand also show the
proposed strut level. Strut levels are marked A, B, C and D.
 Determine the reactions fortwosimple cantilever beams (topand
bottom) and all the simple beams between. These reactions are
A, B1,B2,C1,C2 and D.
 Thestrut loads may becalculated as
PA= (A) (s)
PB= (B1+B2) (s)
Pc= (C1+C2) (s)
PD= (D) (s)
Where PA,PB,PC,PD are the loads to be taken by individual strutsat
levels A,B,C and Drespectively
 Knowing the strut loads at each level and intermediate bracing
conditionallowsselections forthe properselection from thesteel
construction manual.

45. WALES
They are considered as horizontal beams pinned at strut levels. The
maximum bending moment will depend upon the span “s” and loads on the
struts. As the strut loads are different at various levels, maximum bending
moment would also be different.
At level A, Mmax : (A)(s2)/8
At level B, Mmax : (B1 +B2)(s2)/8
At level C, Mmax : (C1 + C2)(s2)/8
At level D, Mmax : (D)(s2)/8
once the maximum bending moment has been computed, the section modulus
is computed as,
S= (Mmax)/( σall)
σall = Allowable bearing stress.

46. SHEET PILES
Sheet piles act as vertical plates supported at strut levels.
The maximum bending moments in various sections such as
A, B, C, D is determined. Once the maximum bending moments
have been computed, the section modulus of the sheet pile can be
computed and the section chosen.

47. DESIGN OF BRACED SHEETING IN CUTS

48. EARTH PRESSURE DISTRIBUTION BEHIND SHEETING
FOR DRY OR MOIST SAND FOR SOFT TO MEDIUM CLAY FOR STIFF CLAY

49. SHEET PILES ARE KEPT IN POSITION BY WALES
AND STRUTS. THE FIRST BRACE LOCATION
SHOULD NOT EXCEED THE DEPTH OF THE
potential tension cracks.
2
h0
2c
tan 45
Since the formation of cracks will increase the
lateral pressure against the sheeting and if the
cracks are filled with water, the pressure will be
increased even more. The sheeting of a cut is
flexible and is restrained against deflection at the
first series of struts. The deflection, therefore, is
likely to be as shown in Fig. (a). The pressure
distribution on sheet pile walls to retain sandy soil
and clay soil are shown in Figs. (b) and (c)
respectively.
a
b
c

50. DESIGN:
1. The sheet pile is considered as continuous beam
top, fixed, partially fixed, hinged,
supported on wales either cantilevered at
or
cantilevered at the bottom depending upon the
amount of penetration below the excavation
line.
2. Bending moment and shearing force diagram
are then obtained using moment distribution
method.
3. Section of the sheet pile is then designed in the
conventional way for the maximum bending
moment.
A fast way of designing sheeting is to assume
conditions as shown in Fig. (d). The top is
treated as a cantilever beam including the first
two struts. The remaining spans between struts
are considered as simple beams with a hinge or
cantilever at the bottom.
Struts are designed as columns subjected to an
axial force. The wales as continuous members
or simply supported members pinned at the
d

51. EXAMPLE

52. #1 A long trench is excavated in medium dense sand for the foundation of a
multi-storey building. The sides of the trench are supported with sheet pile
walls fixed in place by struts and wales as shown in figure below. The soil
properties are:
γ = 18.5KN/m3; c=0 ; Ф = 38o
Determine:
(a) The pressure distribution on the walls with respect to depth.
(b) Strut loads. The struts are placed horizontally at distances L = 4
m centre to centre.
(c) The maximum bending moment for determining the pile wall
section.
(d) The maximum bending moments for determining the section of
the wales.

54. (a) For a braced cut in sand use the apparent pressure envelope given
in Fig. 20.28 b. The equation for pa is
pa = 0.65 γ H KA = 0.65 x 18.5 x 8 tan2 (45 - 38/2) = 23 kN/m2
b) Strut loads
The reactions at the ends of struts A, B and C are represented by RA, RB
and Rc respectively
For reaction RA , take moments about B
RA x3 = 4x23x4/2
or RA = 184/3 = 61.33 kN
RB1 = 23 x 4 - 61.33 = 30.67 kN
Due to the symmetry of the load distribution,
RB1 = RB2 = 30.67 kN, and RA = Rc = 61.33kN.
Now the strut loads are (for L = 4 m)
Strut A, PA= 61.33 x 4 = 245 kN
Strut B, PB = (RB1 + RB2) x 4 = 61.34 x 4 = 245 kN
Strut C, Pc = 245 kN

55. (c)Moment of the pile wall section
To determine moments at different points it is necessary to draw a diagram
showing the shear force distribution.
Consider sections DB1 and B2E of the wall in Fig. (b). The distribution of
the shear forces are shown in Fig. (c) along with the points of zero shear.
The moments at different points may be determined as follows
MA = 0.5 x 1 x 23 = 11.5 kN- m
Mc = 0.5x 1 x 23 = 1 1.5 kN- m
Mm = 0.5 x 1.33 x 30.67 = 20.4 kN- m
Mn =0.5 x 1.33 x 30.67 = 20.4 kN- m
The maximum moment Mmax = 20.4kN-m. A suitable section of sheetpile
can be determined as per standard practice.
(d) Maximum moment for wales
The bending moment equation for wales is
Mmax = (RL2)/8
Where R = maximum strut load = 245 kN
L = spacing of struts = 4 m
Mmax = (245 x 42)/8 = 490 kN-m
A suitable section for the wales can be determined as per standard
practice.

56. #2 The cross section of a long braced cut is shown in Figure
a. Draw the earth-pressure envelope.
b. Determine the strut loads at levels A, B, and C.
c. Determine the section modulus of the sheet pile section required.
d. Determine a design section modulus for the wales at level B.
e.Calculate the factor of safety against heave (L= 20m, T= 1.5m and q= 0)
(Note: The struts are placed at 3 m, centre to centre, in the plan.) Use σall = 170 x
103 kN/m2
γ = 18 kN/m2; c= 35 kN/m2 and H= 7m

57. (a) Given: γ = 18 kN/m2; c= 35kN/m2 AND H= 7M. SO ,
Thus , the pressure envelope will be like the one as shown in previous figure
and the maximum pressure intensity Pa,
Pa = 0.3 γ H = (0.3) (18) (7) = 37.8 kN/m2
(b)To calculate the strut load, examine fig. b, taking moments abount
B1, we have MB1 = 0,
RA x 2.5 – 0.5 x 37.8 x 1.75 x (1.75 + 1.75/3 ) – 1.75 x 37.8 x (1.75/2) = 0
RA = 54.02 kN/m
Also vertical forces = 0 . Thus,
0.5 x 1.75x 37.8 + 37.8 x 1.75 = RA + RB1
Therefore, RB1 = 45.2 kN/m
Due to symmetry,
RB2 = 45.2kN/m
RC = 54.02kN/m

58. Hence the horizontal strut loads ,
Pa = RA x Horizontal spacing(s) = 54.02 x 3 = 162.06 kN
Pb = (RB1+ RB2 ) x s = (45.02 +45.02) x 3 = 271.2 kN
Pc =RC x s = 54.02 x 3 = 162.06kN
(c) Location of the point of maximum moment, i.e. shear force is zero,
37.8(x) = 45.2
Therefore, x= 1.196m from B
Moment at A= 0.5x1x (37.8/1.75 x1) x 0.33 = 3.6 kN-m/m of wall
Moment at E (point of maximum moment)
= 45.2x 1.196 – 37.8 x 1.196x 1.196/2 = 27.03kN-m/m of wall.
Therefore section modulus of sheet pile
S= (Mmax / σall) = (27.03)/(170x 103)= 15.9 x 10-5m3/m of wall

59. (d)the reaction at level B has been calculated in part b. Hence,
Mmax =(RB1 + RB2)S2 / 8 = (45.2 + 45.2)32 / 8 = 101.7 kN-m
And section modulus s = = (Mmax / σall) = (101.7)/(170x 103)= 0.598x 10-3 m3
(e)Factor of safety against heave is given by the equation,
Where,
L=20m; c=35kN/m2 ;Nc = 5.7 ; γ = 18 kN/m2 ; H= 7m ; B=3m ;q=0
B/ = 3/ = 2.12m
B’=T = 1.5m
B’’= = 1.5x = 2.12m

60. Braced cut showing arrangement of sheet piles, wales and struts

61. The wall is supported by “rakers,” or inclined struts. The
bottom ends of the rakers are raced against the central
part of the building foundation slab.