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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 loads various 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 Find the strut loads 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 0.650.271176 pa 17.9673 18.0 kPa Example Find the strut loads 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 Find the strut loads 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 18.02.5 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 = 0 451.25 PA 1.5 0 PA 37.5kN/m 9. Divide other panel thrusts and strut loads in half Example Find the strut loads 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 Find the strut loads 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