For a factor of safety of 2 on the passive resistance, determine the required depth of penetration depth, D. (initial trial with D'=1.5m)
For a factor of safety of 2 on the passive resistance, determine the required depth of penetration depth, D. (initial trial with D'=1.5m)
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:Table 9.1 Properties of Some Sheet Pile Sections Produced by Bethlehem Steel Corporation
Section modulus
m³/m
of wall
326.4 x 10-
Section
designation Sketch of section
PZ-40
PZ-35
PZ-27
PZ-22
PSA-31
PSA-23
409 mm
(16.1 in.)
379 mm
(14.0 in.)
O
304.8 mm
(12 in.)
O
228.6 mm
(9 in.)
-Driving distance = 500 mm (19.69 in.)
12.7 mm (0.5 in.)
15.2 mm (0.6 in.)
12.7 mm
Driving distance = 575 mm (22.64 in.)
(in.)
12.7 mm (0.5 in.)
15.2 mm (0.6 in.)
9.53 mm ( in.)
9.53 mm
Driving distance = 457.2 mm (18 in.)
9.53 mm
9.53 mm (in.)
9.53 mm
05100
Driving distance 500 mm (19.7 in.)
- Driving distance 558.8 mm (22 in.) -
Driving distance 406.4 mm (16 in.) →
09/00
in.)
in.)
260.5 x 10-5 48.5
162.3 x 10-5
97 x 10-5
in³/ft
of wall
60.7
10.8 x 10-5
30.2
Moment of inertia
m¹/m
in"/ft
of wall
of wall
670.5 x 10-6 490.8
2.01
493.4 x 106 361.2
251.5 x 106 184.2
18.1 115.2 x 10-6 84.4
4.41 x 106 3.23
12.8 x 10-5 2.4 5.63 x 10-6 4.13

Transcribed Image Text:2. Design the anchored sheet pile wall supporting a loose sand fill as shown in the
following Figure. GWT is at the same height on both sides, and assume yw=10kN/m³.
Based on the log spiral solutions, the Ka for the loose sand is 0.3 while the Ka and Kp
for the dense sand are 0.2 and 13.125, respectively. Using the free earth support
method, do the following:
a) For a factor of safety of 2 on the passive resistance, determine the required depth of
penetration depth, D. (initial trial with D'=1.5m)
b) Determine the bending moment and the anchor load.
c) Select a sheet pile section from the Table 9.1 (E=210x10³ MN/m² and fair-210 MN/m²)
kN/m²
D
7.0m.
Yt = 16.5 kN/m³
o'= 30°
Loose sand fill:
Yt 19.5 kN/m3
o' = = 30°
Dense sand:
Yt = 21 kN/m³
$' = 40°
q=10
1.5m.
0.5m.
T
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