Transmission And Substation Foundations - Technical Design Manual (TD06088E)

For this example the working loading is as follows: Loads • Shear: 2300 lbs • Shear applied to switch 10 feet above grade. Soil Profile: • Soil is a clay with a cohesion of 0.5 ksf.

Solution

P = Applied horizontal shear load: Use 2300 lbs. Include a Factor of Safety of 2 in the calculations, thus doubling the horizontal shear load; P = 2 x 2300 lbs = 4600 lbs C u = Cohesion of Clay: 500 psf

D = Diameter of foundation: Use D = 10.75” (10” nominal pipe size) e = Eccentricity; distance above grad to resolve load: Given e = 10 ft. L = Minimum Length of foundation based on above criteria.

Equation 7-11

F = P/ [9 (C u ) D]

= 4600 lbs/ [9 (500 psf) (10.75in/12)] = 1.141 ft

M POS

mAX = P (e +1.5D + 0.5F)

= 4600 lbs [10 + 1.5(10.75 in /12) + 0.5 (1.141 ft)] = 54,806 ft-lbs

M POS mAX = 2.25 D x g2 x Cu 54,806 ft-lb = 2.25 (10.75 in/12) g 2 (500 psf) g 2 = 54.38 g = 7.38 ft L = 1.5D + F + G = 1.5 (10.75 in/12) + 1.141ft + 7.38ft = 9.87 FT

DESIGN EXAMPLES

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