# Braced Excavations - spin.mohawkc.on.ca

```BRACED EXCAVATIONS
 for deep, narrow excavations
 pipelines
 service cuts
Braced Excavations
1.
2.
3.
4.
5.
6.
7.
8.
drive in piling
excavate first portion
install wales and top struts
excavate next portion
install next wales and struts
excavate next portion
install next wales and struts
excavate last portion
The rest is in Elastic Equilibrium
Failure
the system
occurs
The maximum
Only
theoflower
deformation
portionusually
of the
will
soil
bewedge
at the will
bottom
progressively:
reach
Plastic
Equilibrium
Therefore, Rankine’s Theory doesn’t apply
one strut fails, then another & so on
Since one strut failure means system failure, the
For
medium
to dense
sands:
pressure
distibution
for design
Using
measured
strutassumed
earthis
conservative:
pressure distibutions have been documented
an envelope based on field measurements.
For clays: calculate Stability Number,
γH
SN 
cu
Shear Stress, τ (kPa)
where cu is the undrained shear strength of the
clay:
u  0
cu = τ f
σf
σf
σf
Normal Stress, σn(kPa)
For clays with SN  4:
For clays with SN > 4:
Ususally m = 1.0, however,
for soft or normally consolidated clay, m can
be as low as 0.4
Example
18.0 kPa
Excavation in sand
γ = 17 kN/m3
’ = 35
6 m deep,
braced at 1, 2.5 and 4.5 m depths
struts spaced at 5 m c-c
1. Find the equivalent active earth
pressure on the piling
1  sin35
Ka 
 0.271

1  sin35
pa  0.650.271176
pa  17.9673 18.0 kPa
Example
18.0 kPa
fixed to support
Excavation in sand
γ = 17 kN/m3
’ = 35
6 m deep,
braced at 1, 2.5 and 4.5 m depths
struts spaced at 5 m c-c
1.0 m
A
2. Split up A.E. distribution into
tributary panels
1.5 m
3. Determine height of each panel
B
4. Label supports
2.0 m
hinged
C
1.5 m
S
5. Since this arrangement is
statically indeterminate, assume A is
fixed support and the others are
hinged
Example
18.0 kPa
1.0 m
PA
A
1.5 m
45 kN/m
1.25 m
PB1
Excavation in sand
γ = 17 kN/m3
’ = 35
6 m deep,
braced at 1, 2.5 and 4.5 m depths
struts spaced at 5 m c-c
6. Calculate thrust in top panel
FTOP  18.02.5  45 kN/m
7. Top strut load = PA
B
PB2
18 kN/m
2.0 m
18 kN/m
PC1
PC2
C
1.5 m
13.5 kN/m
13.5 kN/m
ps
S
8. ΣMB = 0
451.25  PA 1.5  0
PA  37.5kN/m
9. Divide other panel thrusts and
Example
18.0 kPa
1.0 m
PA
A
1.5 m
Excavation in sand
γ = 17 kN/m3
’ = 35
6 m deep,
braced at 1, 2.5 and 4.5 m depths
struts spaced at 5 m c-c
PA= 37.5 kN/m
45 kN/m
10. ΣFH=0 down to B
1.25 m
37.5 - 45 +PB1 = 0
PB1
B
PB2
18 kN/m
2.0 m
18 kN/m
C
1.5 m
ps
12. PC1 = 18
13. PC2 = 13.5
PC1
PC2
11. PB2 = 18
S
13.5 kN/m
13.5 kN/m
14. ps = 13.5 kPa/m
PB1 = 7.5
PB2 = 18
PB = 25.5 kN/m
PC = 31.5 kN/m
Example
18.0 kPa
1.0 m
PA
A
1.5 m
45 kN/m
1.25 m
PB1
B
PB2
18 kN/m
Excavation in sand
γ = 17 kN/m3
’ = 35
6 m deep,
braced at 1, 2.5 and 4.5 m depths
struts spaced at 5 m c-c
PA= 37.5 kN/m
PB = 25.5 kN/m
PC = 31.5 kN/m
15. Strut Loads for 5 m of wall:
2.0 m
18 kN/m
PC1
PC2
Strut B Load = 25.5 kN/m x 5m = 127.5 kN
C
1.5 m
ps
Strut A Load = 37.5 kN/m x 5m = 187.5 kN
S
13.5 kN/m
13.5 kN/m
Strut C Load = 31.5 kN/m x 5m = 157.5 kN
```