Chance Technical Design Manual

BUCKLING: Check for buckling on the SS5 square shaft with 2 ft (0.61 m) of exposed shaft above grade. Assume a pin-pin (K = 1) connection.

7 ft (2.1 m)

EULER’S EQUATION:

EQUATION 8-27

60 psf (143 Pa) Dead load and Live load

2 EI/[KL

2

P crit = p

u ]

p 2 [30x10 6 ][.396]/[1 x 24] 2

P crit =

P crit = 203,354 lb (904 kN)

500 lb (2.22 kN) Wind load

The critical load is greater than the ultimate vertical load so buckling is not a concern.

2 ft (0.61 m) Above Grade

TORQUE:

EQUATION 8-28

22°

Torque required = Required load/K t

Soft to Medium Clay SPT N=6 Cohesion = 750 psf (36 kN/m 2 ) Unit weight = 92 pcf (14 kN/m 3 )

where

DESIGN EXAMPLES

K t = 10 (33) for square shaft Torque required = 5,204 lb / 10 Torque required = 520 ft∙lb (705 N∙m) This does not exceed the SS5 torque rating of 5,700 ft∙lb (7,730 N∙m).

15 ft (4.5 m)

5 ft (1.5 m) Spacing

HELICAL PILES FOR BOARDWALKS WITH LATERAL SUPPORT, FIGURE 8-6

DESIGN EXAMPLE 8: HELICAL TIEBACK ANCHORS IN CLAY

STRUCTURE TYPE • Cast concrete retaining wall • Height (H) = 18 ft, thickness = 2’-0 • nH = 0.25H = 4.5 ft, mH = 0.63H = 11.3 ft • Residual soils: stiff clay with N = 28. No ground water ta ble (GWT) present. • Tieback installation angle = 15° Structural Design Loads (See Figure 4-6 in Section 4)

AFP

nH

TIA

5(THD)

mH

THD

DL N

H

5(THD)

AFPA

• DL N /ft = (12 x H2) / cos(15°) • DL N /ft = (12 x 182)/ cos(15°) • DL N /ft = 4,025 lb/lin ft • DL M /ft = (18 x H2) / cos(15°) • DL M /ft = (18 x 182)/ cos(15°) • DL M /ft = 6,040 lb/lin ft

THD

DL M

Tieback Installation Angle (TIA) Top Helix Diameter (THD) Assumed Failure Plane (AFP) Assumed Failure Plane Angle (AFPA)

HELICAL TIEBACK ANCHOR FIGURE 8-7

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