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Example : Design the struts force given that the soil is soft and normally consolidated, cu = 24kPa, γ =20k!m "  and #o = $m% &m 2m 2m &m 4m

Example Braced Excavation Design

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Page 1: Example Braced Excavation Design

7/23/2019 Example Braced Excavation Design

http://slidepdf.com/reader/full/example-braced-excavation-design 1/3

Example :

Design the struts force given that the soil is soft and normally consolidated, c u = 24kPa,

γ =20k!m" and #o = $m%

&m

2m

2m

&m

4m

Page 2: Example Braced Excavation Design

7/23/2019 Example Braced Excavation Design

http://slidepdf.com/reader/full/example-braced-excavation-design 2/3

SOLUTION

Stability against heave :

OK q H 

c N heave FS 

q

 B

 Ho Nc

 B Ho

 so

uc

 s

%%%%%%%%'%&'2%&$20

24$%()*

0

)'%&2%0&*$2%0&$

,'%&4$

>=×

=+

=

=

×+=   

   +=

==

γ  

Determine the lateral pressure Diagram

$4%0$%&4%0)*4%0

'24

$20

=×==

=

heave

u

o

 FOS m

c

 H γ  

+aximum ateral Pressure is : γ   H o – mcu = 20 x $ - 0%$ x 24 = &04%$ kPa

Pressure Diagram

.alculate all forces on the strut at each level%

/ll loads are per meter length all%

&04%$ kPa

&%'m

4%'m

Page 3: Example Braced Excavation Design

7/23/2019 Example Braced Excavation Design

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evel &

mkN  A

 A M  B

!(%&"(

('%0$%&04'%&'%&"

'%&'%&$%&04

2

&20&

=∴

××+ 

  

   +×××−==∑

evel 2

mkN C  B   !$%&042

2$%&04&2   =

×==

evel "

:0∑   = x F  .2 = &04%$ x & = &04%$ k!m

.alculate the forces at each strut:

/ = &"(% k!m

1 = 1&  12 = 3%& &04%$ = 202%( k!m

. = .& .2 = &04%$ &04%$ = 203%2 k!m

&%'m

&%'m

/

1&

&m

2m

&04%$ kPa

2m

12

.&

&04%$ kPa

2m

&m

2m

.2

D