BRACED BRACED EXCAVATIONSEXCAVATIONS

for deep, narrow excavationsfor deep, narrow excavations pipelinespipelines service cutsservice cuts

Braced ExcavationsBraced Excavations1.1. drive in pilingdrive in piling2.2. excavate first portionexcavate first portion3.3. install wales and top strutsinstall wales and top struts4.4. excavate next portionexcavate next portion5.5. install next wales and strutsinstall next wales and struts6.6. excavate next portionexcavate next portion7.7. install next wales and strutsinstall next wales and struts8.8. excavate last portionexcavate last portion

The rest is in Elastic EquilibriumThe rest is in Elastic Equilibrium

The maximum deformation will be at the The maximum deformation will be at the bottombottomTherefore, Rankine’s Theory doesn’t applyTherefore, Rankine’s Theory doesn’t apply

Only the lower portion of the soil wedge will Only the lower portion of the soil wedge will reach Plastic Equilibriumreach Plastic Equilibrium

Failure of the system usually occurs Failure of the system usually occurs progressively:progressively:

one strut fails, then another & so onone strut fails, then another & so on

For medium to dense sands:For medium to dense sands:Using measured strut loads various earth Using measured strut loads various earth pressure distibutions have been pressure distibutions have been documenteddocumented

Since one strut failure means system Since one strut failure means system failure, the pressure distibution assumed failure, the pressure distibution assumed for design is conservative:for design is conservative:

an envelope based on field an envelope based on field measurements.measurements.

For clays: calculate Stability Number,For clays: calculate Stability Number,

where cwhere cuu is the undrained shear strength of is the undrained shear strength of the clay:the clay:

ucH

SNγ

Normal Stress, σn(kPa)

cu = τf u 0

Shear

Str

ess

, τ

(kPa)

σf σf σf

For clays with SN For clays with SN 4: 4:

For clays with SN > 4:For clays with SN > 4:Ususally m = 1.0, however,Ususally m = 1.0, however,for soft or normally consolidated clay, m for soft or normally consolidated clay, m can be as low as 0.4can be as low as 0.4

ExampleExample

Excavation in sandγ = 17 kN/m3

’ = 356 m deep,braced at 1, 2.5 and 4.5 m

depthsstruts spaced at 5 m c-c

0.271sin351sin351

Ka

6170.2710.65pa

kPa 18.017.9673pa

18.0 kPa

1. Find the equivalent active earth pressure on the piling

Find the strut loads

ExampleExample

Excavation in sandγ = 17 kN/m3

’ = 356 m deep,braced at 1, 2.5 and 4.5 m

depthsstruts spaced at 5 m c-c1.0 m

1.5 m

2.0 m

1.5 m

AA

18.0 kPa

BB

CC

SS

fixed to support

hinged

2. Split up A.E. distribution into tributary panels

3. Determine height of each panel

4. Label supports

5. Since this arrangement is statically indeterminate, assume A is fixed support and the others are hinged

Find the strut loads

ExampleExample

Excavation in sandγ = 17 kN/m3

’ = 356 m deep,braced at 1, 2.5 and 4.5 m

depthsstruts spaced at 5 m c-c1.0 m

1.5 m

2.0 m

1.5 m

AA

18.0 kPa

BB

CC

SS

6. Calculate thrust in top panel

8. ΣMB = 0

7. Top strut load = PA

9. Divide other panel thrusts and strut loads in half

Find the strut loads

kN/m 452.518.0FTOP 1.25 m

45 kN/mPPAA

01.5P1.2545 A kN/m 37.5PA

PPB1B1

PPB2B2

PPC1C1

PPC2C2

18 kN/m

18 kN/m

ppss13.5 kN/m

13.5 kN/m

ExampleExample

Excavation in sandγ = 17 kN/m3

’ = 356 m deep,braced at 1, 2.5 and 4.5 m

depthsstruts spaced at 5 m c-c1.0 m

1.5 m

2.0 m

1.5 m

AA

18.0 kPa

BB

CC

SS

10. ΣFH=0 down to B

11. PB2 = 18

Find the strut loads

1.25 m

45 kN/mPPAA

PPB1B1

PPB2B2

PPC1C1

PPC2C2

18 kN/m

18 kN/m

ppss13.5 kN/m

13.5 kN/m

PA= 37.5 kN/m

37.5 - 45 +PB1 = 0 PB1 = 7.5

PB2 = 18

12. PC1 = 18 PB = 25.5

kN/m13. PC2 = 13.5

PC = 31.5 kN/m

14. ps = 13.5 kPa/m

ExampleExample

Excavation in sandγ = 17 kN/m3

’ = 356 m deep,braced at 1, 2.5 and 4.5 m

depthsstruts spaced at 5 m c-c1.0 m

1.5 m

2.0 m

1.5 m

AA

18.0 kPa

BB

CC

SS

Find the strut loads

1.25 m

45 kN/mPPAA

PPB1B1

PPB2B2

PPC1C1

PPC2C2

18 kN/m

18 kN/m

ppss13.5 kN/m

13.5 kN/m

PA= 37.5 kN/m

PB = 25.5 kN/m

PC = 31.5 kN/m

15. Strut Loads for 5 m of wall:

Strut A Load = 37.5 kN/m x 5m = 187.5 kN

Strut B Load = 25.5 kN/m x 5m = 127.5 kN

Strut C Load = 31.5 kN/m x 5m = 157.5 kN