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BRACED EXCAVATIONS

BRACED EXCAVATIONS. for deep, narrow excavations pipelines service cuts. Braced Excavations. 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

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BRACED EXCAVATIONS

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  1. BRACED EXCAVATIONS for deep, narrow excavations pipelines service cuts

  2. Braced Excavations • 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

  3. The rest is in Elastic Equilibrium Failure of the system usually occurs progressively: one strut fails, then another & so on The maximum deformation will be at the bottom Only the lower portion of the soil wedge will reach Plastic Equilibrium Therefore, Rankine’s Theory doesn’t apply

  4. For medium to dense sands: Since one strut failure means system failure, the pressure distibution assumed for design is conservative: Using measured strut loads various earth pressure distibutions have been documented an envelope based on field measurements.

  5. For clays: calculate Stability Number, where cu is the undrained shear strength of the clay: Shear Stress, τ (kPa) u 0 cu =τf σf σf σf Normal Stress, σn(kPa)

  6. For clays with SN  4:

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

  8. Example Find the strut loads 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 18.0 kPa 1. Find the equivalent active earth pressure on the piling

  9. Example Find the strut loads 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 18.0 kPa fixed to support 1.0 m 2. Split up A.E. distribution into tributary panels A 1.5 m 3. Determine height of each panel B 4. Label supports 2.0 m hinged 5. Since this arrangement is statically indeterminate, assume A is fixed support and the others are hinged C 1.5 m S

  10. Example Find the strut loads 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 18.0 kPa 1.0 m PA 6. Calculate thrust in top panel 45 kN/m A 1.5 m 1.25 m PB1 7. Top strut load = PA B PB2 18 kN/m 8. ΣMB = 0 2.0 m 18 kN/m PC1 9. Divide other panel thrusts and strut loads in half 13.5 kN/m C PC2 1.5 m 13.5 kN/m ps S

  11. Example Find the strut loads 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 18.0 kPa 1.0 m PA PA= 37.5 kN/m 45 kN/m A 10. ΣFH=0 down to B 1.5 m 1.25 m 37.5 - 45 +PB1 = 0 PB1 = 7.5 PB1 PB2 = 18 B PB2 18 kN/m 11. PB2 = 18 PB = 25.5 kN/m 12. PC1 = 18 2.0 m 18 kN/m PC = 31.5 kN/m 13. PC2 = 13.5 PC1 14. ps = 13.5 kPa/m 13.5 kN/m C PC2 1.5 m 13.5 kN/m ps S

  12. Example Find the strut loads 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 18.0 kPa 1.0 m PA PA= 37.5 kN/m 45 kN/m A PB = 25.5 kN/m 1.5 m 1.25 m PC = 31.5 kN/m PB1 B PB2 18 kN/m 15. Strut Loads for 5 m of wall: 2.0 m 18 kN/m Strut A Load = 37.5 kN/m x 5m = 187.5 kN PC1 Strut B Load = 25.5 kN/m x 5m = 127.5 kN 13.5 kN/m C PC2 Strut C Load = 31.5 kN/m x 5m = 157.5 kN 1.5 m 13.5 kN/m ps S

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