Transmission And Substation Foundations - Technical Design Manual

SECTION 7: DESIGN EXAMPLES

Design Example 8

Ultimate Pile Capacity • P c = Compression/2 piles + 10,000 ft-lbs/4 ft spacing = (5000 lbs / 2) + 2,500 lbs = 5,000 lbs • P t = Tension/2 piles + 10,000 ft-lbs/4 ft spacing = (2000 lbs / 2) + 2,500 lbs = 3,500 lbs • Q t = (A 16 ) q h N q A 16 = Projected area of helical plates. Compression will have full helix area = A 16C = 1.281 ft 2 Tension will area will be full helix area minus pipe shaft = A 16T = 0.972 ft 2 Nq = Bearing Capacity Factor related to ɸ of residual soil (30°) = 0.5 (12 x Φ) Φ /54 = 13 • q h = γ’ x D h (Effective unit weight times depth of helix below ground line, ft) = (90 pcf – 62.4pcf) (5ft) + (120 pcf – 62.4pcf) (8.5ft) = 627 psf • QtC = (1.281 ft 2 ) (627 psf) (13) = 10,441 lbs (Ultimate Compression Capacity) • QtT = (0.972 ft 2 ) (627 psf) (13) = 7,922 lbs (Ultimate Tension Capacity)

Check Qt • Conduct Field Load Test (if required per specifications)

Estimate installation Torque P = 5000 lbs T = (P X FS)/Kt = (5,000 lb x 2)/5 = 2,000 ft-lb K t = empirical torque factor (default value =5 for the RS6625 series) NOTE: If during installation T = 2,000 ft-lb is not achieved then two options are available: (1) install piles deeper, or (2) change helix configuration to a larger combination, i.e, (16-18) Factor of Safety • Theoretical Ultimate Bearing Capacity FS = (QtC/P) = 10,441/ 5,000 = 2.09 (OK Compression) FS = (QtT/P) = 7,922 / 3,500 = 2.26 (OK Tension)

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