Transmission And Substation Foundations - Technical Design Manual

SECTION 7: DESIGN EXAMPLES

Design Example 9

Broms’ method.

EQUATION 7-18

• j = 30° • g = 100 lb/ft3 • Factor of Safety = 2.

0 ≤ L 3 - ( 2V

F L / K P g D ) – ( 2V M / K P g D )

=

63 - [ 2 x 920 x 6) / (3 x 100 {6.625/12})] - [(2 x 1 7200) / (3 x 100 x {6.625/12})]

EQUATION 7-16

=

- 58.35

V F

=

V (FS)

where

=

460 (2)

0

> - 58.35

=

920 lb

2 (45 + j /2 ) = 3.0

K P

= tan

EQUATION 7-17

g = Effective unit weight of soil = 100 lb/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: EQUATION 7-19

V M

=

M (FS)

=

8600 (2)

= 17,200 ft∙lb 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:

V ( H + 0.54 x ( V / g D KP )

M MAX

=

0.5 )

=

460 (18.69565 + 0.54 x ( 460/100 x (6.625/12) x 3) 0.5 )

=

9013.968 ft∙lb

This is less than the allowable moment capacity of 10,860 ft∙lb, therefore a 6” diameter by 7’ long SLF is adequate for the applied load in the sandy soil.

7-34 | www.hubbell.com/hubbellpowersystems