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

2. Braced Excavations 1.d rive in piling 2.e xcavate first portion 3.i nstall wales and top struts 4.e xcavate next portion 5.i nstall next wales and struts 6.e xcavate next portion 7.i nstall next wales and struts 8.e xcavate last portion

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

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

5. For clays: calculate Stability Number, where cu is the undrained shear strength of the clay: Normal Stress, σ n (kPa) c u = τ f  u  0  S h e a r S t r e s s, τ ( k P a ) σfσf σfσf σfσf

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 Excavation in sand γ = 17 kN/m 3  ’ = 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 Find the strut loads

9. Example Excavation in sand γ = 17 kN/m 3  ’ = 35  6 m deep, braced at 1, 2.5 and 4.5 m depths struts spaced at 5 m c-c 1.0 m 1.5 m 2.0 m 1.5 m A 18.0 kPa B C S 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

10. Example Excavation in sand γ = 17 kN/m 3  ’ = 35  6 m deep, braced at 1, 2.5 and 4.5 m depths struts spaced at 5 m c-c 1.0 m 1.5 m 2.0 m 1.5 m A 18.0 kPa B C S 6. Calculate thrust in top panel 8. ΣM B = 0 7. Top strut load = P A 9. Divide other panel thrusts and strut loads in half Find the strut loads 1.25 m 45 kN/m PAPAPAPA P B1 P B2 P C1 P C2 18 kN/m 18 kN/m pspspsps 13.5 kN/m 13.5 kN/m

11. Example Excavation in sand γ = 17 kN/m 3  ’ = 35  6 m deep, braced at 1, 2.5 and 4.5 m depths struts spaced at 5 m c-c 1.0 m 1.5 m 2.0 m 1.5 m A 18.0 kPa B C S 10. ΣF H =0 down to B 11. P B2 = 18 Find the strut loads 1.25 m 45 kN/m PAPAPAPA P B1 P B2 P C1 P C2 18 kN/m pspspsps 13.5 kN/m P A = 37.5 kN/m 37.5 - 45 +P B1 = 0P B1 = 7.5 P B2 = 18 12. P C1 = 18 P B = 25.5 kN/m 13. P C2 = 13.5 P C = 31.5 kN/m 14. p s = 13.5 kPa/m

12. Example Excavation in sand γ = 17 kN/m 3  ’ = 35  6 m deep, braced at 1, 2.5 and 4.5 m depths struts spaced at 5 m c-c 1.0 m 1.5 m 2.0 m 1.5 m A 18.0 kPa B C S Find the strut loads 1.25 m 45 kN/m PAPAPAPA P B1 P B2 P C1 P C2 18 kN/m pspspsps 13.5 kN/m P A = 37.5 kN/m P B = 25.5 kN/m P C = 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