Transmission And Substation Foundations - Technical Design Manual

SECTION 7: DESIGN EXAMPLES

Design Example 11

Buckling Example Using The Davisson Method

EQUATION 7-21

= 4 √ (30 x 10 6 x 0.396) / (15 x 1.5) = 26.96

R

l max

= (15 x 12) / 26.96 = 6.7

Project A three-helix Chance® Type SS150 1-1/2” square shaft helical pile is to be installed into the soil profile as shown in Figure 7-34. The top three feet is uncontrolled fill and is assumed to be soft clay. The majority of the shaft length (12 feet) is confined by soft clay with a k h = 15 pci. The helix plates will be located in stiff clay below 15 feet. The buckling model assumes a pinned-pinned end condition for the helical pile head and tip. Determine the critical buckling load using the Davisson method. Assumptions • k h is constant, i.e., it does not vary with depth. This is a conservative assumption because k h usually varies with depth, and in most cases increases with depth. • Pinned-pinned end conditions are assumed. In reality, end conditions are more nearly fixed than pinned, thus the results are generally conservative. • From Figure 7-33, U cr ≈ 2

P cr

= (2 x 30 x 106 x 0.396) / 26.96 2 = 32.69 kip

Chance Type SS150 Square Shaft Foundations Physical Properties, Table 7-2 Modulus of Elasticity (E p ) Moment of Inertia (I p ) Shaft Diameter (D)

30 x 10 6 psi

0.396 in 4

1.5 in

P cr

P cr

Model as

Foundation

3’

Soft Clay N = 3 K h = 15 pci Soft Clay N = 3

K h = 15 pci h

12’

15’

3.0

ft-p

ft-p

ft-p

ft-p

Sti Clay N ≥ 5 Stiff Clay

2.0

p-p

p-p

p-p

Legend f= Free p= pinned ft = fixed, traslating Note: Upper and condition listed first

Foundation Details Figure 7-35

ft-f

1.0

U cr =P cr R

2 /E

P I P

ft-p

f-f

R= 4 √ E P I P /K h d

I max =L/R

0

0

2

4

6

8

10 12

Poulos and Davis (1980) Figure 7-34

7-36 | www.hubbell.com/hubbellpowersystems