Transmission And Substation Foundations - Technical Design Manual (TD06088E)

Cohesionless Soil (See Figure 7-32) Assumed values:

• Applied shear load at the groundline (V) = 460 lbs. • Applied moment at the groundline (M) = 8600 ft lbs. • Foundation diameter is 6” nominal Schedule 40. Use 6.625” as the actual pipe size in calculations. Cableway openings are 2.5” wide by 12” high. The allowable moment capacity of this foundation shaft size and cableway opening is 10,860 ft-lbs. • The required length (L) will be determined using the Broms method. • ф = 30° • γ = 100 lbs/ft3 • Factor of Safety = 2.

Foundation in Cohesionless Soil Figure 7-32

Equation 7-16

= V (FS) = 460 (2) = 920 lbs

DESIGN EXAMPLES

V F

Equation 7-17

= M (FS)

V M = 8600 (2) = 17,200 ft-lbs Broms equation for cohesionless soil requires a trial and error solution. For the trial and error solution, start by assuming the foundation diameter (D) is 6.625” and the length (L) is 6 feet:

≤ L 3 - ( 2V F L / K P g

Equation 7-18

0

D ) – ( 2VM / K P g

D )

= 6 3 - [ 2 x 920 x 6) / (3 x 100 {6.625/12})] - [(2 x 17200) / (3 x 100 x {6.625/12})] = - 58.35

where:

0

> - 58.35

= tan 2 (45 + j /2 ) = 3.0

K P

g = Effective unit weight of soil = 100 lbs/ft 3 The 6 foot length is too short so we will try a 7 foot length and repeat the calculation: 0 = 7 3 - [2 x 920 x 7) / (3 x 100 {6.625/12})] - [(2 x 17200) / (3 x 100 x {6.625/12})] = 57.53 0 < 57.53 A 7 foot long SLF will be adequate. The maximum moment in the foundation shaft can be determined with the following equation: M MAX = V ( H + 0.54 x ( V / g DK P ) 0.5 ) Equation 7-19 = 460 (18.69565 + 0.54 x ( 460/100 x (6.625/12) x 3) 0.5 ) = 9013.968 ft-lbs This is less than the allowable moment capacity of 10,860 ft-lbs, therefore a 6” diameter by 7’ long SLF is adequate for the applied load in the sandy soil.

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