BRACED CUTS (EXCAVATIONS) Design Steps and Equations To design a braced cut follow these steps (click on item for details): If the soil stratum is multi-layered go here first then perform steps (1) to (3)

If the soil stratum is single-layered continue with steps (1) to (3) (Click here for an EXAMPLE ON THE DESIGN OF BRACED CUTS)

1) Estimate lateral pressure on braced cut 2) Calculate the forces and moments on various components

a) Determine the load distribution b) Estimate forces on struts and select section c) Estimate the maximum moment on sheet pile and select

section d) Estimate the maximum moment on wales and select section

3) Check for bottom heave (in clays) Lateral Earth Pressure on Braced Cuts return to top

a) Sands: c = 0 pa = 0.65 γ H ka

where ⎟⎠⎞

⎜⎝⎛ φ

−=2

45tank 2a H

pa

b) Soft to Medium Clay: φ = 0 and 4ch>

γ

0.75H

pa

0.25H

pa = ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛

γ−γ

Hc4

1H

or 0.3 γ H

Whichever is higher

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0.5H

0.25H

0.25H

pa

c) Soft to Medium Clay: φ = 0 and 4ch≤

γ

pa = 0.2 γ H to 0.4 γ H with an average of 0.3 γ H

Forces and Moments on Various Components return to top

Load Distribution return to top

H

pa

x x

B

s

s

d5

d4

d3

d2

d1

D

C

B

A

pa

d1

d2

FA I FB1

+ = FB FB2

pa

d3

II FC1

+ = FC FC2

pad5

d4

III L FD

Section x-x (Plan)

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Note that the first strut A must be placed at a depth d1 < zc (depth of tension

crack) where zc = γc2 .

Forces on Struts and Selection of Section return to top(Designed as column, pined at both ends) 1) Draw the pressure diagram pa

2) Assume that the sheet pile is hinged at all levels of struts

3) Calculate FA , FB1, FB2, FC1, FC2, and FD which are the reaction in the load

distributions I, II and III.

4) The loads in the struts are calculated as:

PA = (FA) x s

PB = (FB1 + FB2) x s

PC = (FC1 + FC2) x s

PD = (FD) x s

Maximu Moment on Sheet Pile and selection of Section return to top

1) For each of the load distributions I, II and III find Mmax i.e. where the shear

is equal to zero.

2) The design moment for the sheet pile is the maximum of step (1)

3) Calculate the section modulus all

maxMSσ

= where σall = allowable stress for

sheet pile

4) Select the sheet pile section based on S in Step 3 (Table of Sheet Pile

Section Properties)

Maximu Moment on Wales and selection of Section return to top(Designed as beams pined at the struts)

( )8

sF 2A

At level A: MA, max =

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( ) ( )8

sFF 22B1B +

At level B: MB, max =

t level C: MC, max =

( ) ( )

8

sFF 22C1C +

A

D, max = ( )

8

sF 2D

At level D: M

Mmax is the maximum moment from levels A, B, C and D. The required minimun

section modulus all

Sσ

= where σmaxM

Check for Bottom Heave (in clays)

all = allowable stress for of the wales.

return to top(Click here for an Example on BOTTOM HEAVE)

Hard Stratum (ROCK)

H

D

B

Bottom Heave

q L

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Safety Factor against bottom heave SFH ≥ 1.5

SFH =

If D > 0.7 B

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−γB7.0

cc7.5

H1

If D ≤ 0.7 B

SFH = ⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−γB7.0

cc7.5

H1

R

SFH =

O

qHNc c

+γ

hichever is larger

the above equations c = cohesion

load

Nc =

W In

q = surcharge

⎥⎦⎤

⎢⎣⎡ +⎥

⎦

⎤⎢⎣

⎡ ⎛⎟⎠⎞

⎜⎝

×LB16.084.0

BH082.0exp805.6

For Strip: LB = 0 For Square:

LB = 1

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Multi-Layered Soil Stratum return to top(Click here for an Example on MULTI-LAYERED STRATUM)

:

eplace the layers by an equivalent soil with:

c =

:

: : : :

R

Weighted average cohesion:

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎜⎛ 21 n 1

⎟⎟⎠

⎞⎜⎜⎝

⎛+

⎟⎟

⎠

⎞

⎜⎝

φγ ∑∑==

jcjisi cH75.0tanHH

m

1j1isi2

Weighted average unit weight:

γ =

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎜⎛⎞⎛ mn1

⎟⎟⎠

⎞⎜⎝

γ+⎟⎟⎠

⎜⎜⎝

γ ∑∑== 1j

cj1i

si cjsi HHH

Where: γsi, Hsi and φi are the unit weight, height and internal friction angle of sand i

nd cj are the unit weight, height and cohesion of clay j.

Once th p

γcj, Hcj a

n = number of sand layers and m =number of clay layers. e multi-layered soil is replaced by a single layer, return to to

H

Sand 1

Sand 2

Clay 1

Clay 2

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