1. Under the guidance of
Prof. Kalyan Kumar Chattopadhyay
Yogesh Kr Pandey
Exam roll: 110904034
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
3. The design of braced cuts involves two distinct but
interrelated features, namely
Stability of excavation, ground movement, control of
water into the excavation, effect of adjoining
structures and so on.
Design of structural elements i.e sheet pile, struts or
anchors and so forth.
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
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.
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.
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.
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.
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
Typical pattern of deformation of vertical wall (Braced cuts)
17. In Sand
Following figures shows various recommendations for earth
pressure distribution behind sheeting This pressure, pa may be
Terzaghi and Peck’s Peck’s earth
pressure distribution for loose
Peck, Hansen and Thornburn’s
Peck’s earth pressure distribution
for moist and dry sands
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
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
cav = [γsKsHs
2tanФs’ + (H-HS)n’qu]
The average unit weight of the layers may be expressed as,
γa H =[ γs Hs + ( H - HS ) γc ]
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 ]
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 pressure envelope just described, following points are to
The pressure envelopes are sometimes referred to a apparent pressure
envelope. However, the actual pressure distribution is a function of the
construction sequence and relative flexibility of wall.
They apply to excavation having depths greater than about 6m
They are based on the assumption that water table is below the bottom
of the cut.
Sand is assumed to be drained with zero pore water pressure.
Clay is assumed to be undrained and pore water pressure is not
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-
29. Equivalent Beam Method:
In this method entire depth is split into segments of
simply supported beams and reactions can then be determined by
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
Case 2: When a hard stratum is met within a depth equal to
The anchorage load block of
soil a b c d in Fig. (a) of width
(assumed) at the level of the
bottom of the cut per unit
length may be expressed as
The vertical pressure q per
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. Where, B’=T (thickness of clay below the base of excavation) or B/ (whichever
In 2000, Chang suggested few revisions, with his
modification, equation takes the form
B’=T if T B/
B’ = B/ if T> B/
35. Case: 2
Replacing 0.75B by D in Eq, the
factor of safety is represented by
For a cut in soft clay with a
constant value of cu below the
bottom of the cut, D in Eq.
becomes large, and Fs
approaches the value
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):
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 use of.
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
41. NC (rect) = (0.84 + 0.16B/L) NC (sq)
L = length of excavation.
B = width of excavation
The factor of safety for bottom heave may be expressed as
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.
Draw the pressure envelope of the braced cut and also show the
proposed strut level. Strut levels are marked A, B, C and D.
Determine the reactions for two simple cantilever beams (top and
bottom) and all the simple beams between. These reactions are
A, B1,B2,C1,C2 and D.
The strut loads may be calculated 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 struts at
levels A,B,C and D respectively
Knowing the strut loads at each level and intermediate bracing
condition allows selections for the proper selection from the steel
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.
49. Design of Braced Sheeting in Cuts
Sheet piles are used to retain the sides of the
cuts in sands and clays. The sheet piles are kept
in position by wales and struts. The first brace
location should not exceed the depth of the
potential tension cracks.
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)
1. The sheet pile is considered as continuous beam
supported on wales either cantilevered at
top, fixed, partially fixed, hinged, or
cantilevered at the bottom depending upon the
amount of penetration below the excavation
2. Bending moment and shearing force diagram
are then obtained using moment distribution
3. Section of the sheet pile is then designed in the
conventional way for the maximum bending
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
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
γ = 18.5KN/m3; c=0 ; Ф = 38o
(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
(d) The maximum bending moments for determining the section of
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 = 61.33 kN.
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 sheet pile
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
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
γ = 18 kN/m2; c= 35 kN/m2 and H= 7m
57. (a) Given: γ = 18 kN/m2; c= 35 kN/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 = 54.02 kN/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,
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