Transmission And Substation Foundations - Technical Design Manual

SECTION 7: DESIGN EXAMPLES

Design Example 6

Helical Pile Foundation for New Substation Construction

CHANCE® Helical Pile Selection RS2875.203 with 8-10-12 helix configuration

EQUATION 7-7

Purpose This design example is intended to assist with the design of new construction substation structures that require deep piles beneath a concrete cap for compression capacity. This example will show how to calculate the bearing capacity of four piles beneath a 10 feet square concrete cap holding an oil filled transformer. After the loads for the structure have been determined, it is possible to design the piles. For this example the working load is as follows:

Ultimate Pile Capacity • Q t = (A 8 + A 10 + A 12 ) c N c A 8 , A 10 , A 12 = Projected area of helical plates. A 8 = 0.34ft 2 A 10 = 0.53 ft 2 A 12 = 0.77 ft 2 N c = Bearing Capacity Factor = 9.0 C = N/8 = 16/8 = 2 ksf • Q t = (1.64ft2)(2,000 psf) ( 9.0) • Q t = 29,520 lb (installed depth is over 20 ft) Check Qt

• Compression: 56 kip (Load includes weight of concrete cap)

• Conduct Field Load Test (if required per specifications) EQUATION 7-8

• Shear loads are assumed to be taken by up by passive pressure and fiction along the bottom of the concrete cap.

Estimate installation Torque

P = 56,000lb/4 Piles = 14,000 lb T = (P X FS)/K t = (14,000 lb x 2)/9 = 3,150 ft-lb K t = empirical torque factor (default value =9 for the RS2875 series) The rated installation torque of the RS2875.203 series is 7000 ft-lb, which greater than the required estimated installation torque of 3,150 ft-lb. (OK) NOTE: If during installation T = 3,150 ft-lb is not achieved then two options are available: (1) add piles if spacing allows, or (2) change helix configuration to a larger combination, i.e, (10-12-14) (3) Install Deeper EQUATION 7-9

10 FT SQ.

Figure 7-24

12” DIA 10” DIA 8” DIA

22.5 FT

Factor of Safety

• Theoretical Ultimate Capacity FS = (Qt/P) = 29,520/14,000 = 2.1 (OK) • Torque Correlation: FS = (T x Kt)/P FS = (3,150 x 9)/14,000 = 2.03 (OK)

Figure 7-25

Soil Profile

Figure 7-26

7-28 | www.hubbell.com/hubbellpowersystems