24 - MSE Retaining Walls 01. Pull out resistance at rods. 02. Design the length length L of geotextiles for a 16 ft wall.
MSE Walls-01: Pull out resistance at rods. (Revision: Aug-09)
Find the pull out resistance at road #4: lb γ = 1 2 0 3 φ = 3 0 Interaction Coefficient = 0.7 ft o
Failure Plane MSE Wall
2'
2'
2'
2'
2'
6 0 °
Footing
Solution:
ψ
=
45o
K a
=
K a
=
φ −
2
1 − s in φ 1 + s in φ 2
o
=
tan (6 0 )
2
ta n ( 4 5
=
+
φ 2
)
=
2
ta n (ψ )
0 .3 3 3
The force at H/3 from the bottom of the wall is:
F1
=
1 2
γ H 2 K a
F1
=
1 2
(1 2 0
lb ft 3
2
)(1 0 ft ) (0 .3 3 3)
F 1
=
1998
lb ft
Assuming a triangular distribution of the forces on the wall:
MSE Wall
2'
2'
2'
2' 3.333'
2'
Footing
1 9 9 8 lb / ft ) 6 . 6 6 7 ft
F ro d 4
=
8 ft (
F ro d 4
=
2397.48
lb ft
**MSE Walls-02: Design the length L of geotextiles for a 16 ft wall. (Revised: Aug.-09)
Determine the length L of a geotextile-reinforcing for the 16 foot high temporary retaining wall shown below. Also determine the required vertical spacing of the reinforcing layers S v and the required lap length l l. The geotextile chosen has an allowable strength of σ G of 80 lb/inch, and the available granular backfill has a unit 3 weight of γ = 110 lb/ft and the angle of internal friction is φ = 36°. Comment: These geotextile walls are usually used for temporary civil works, such as detour roadways, temporary abutments or excavation walls, etc. If the wall must become permanent, then the face is stabilized and the plastic geotextile is protected from ultraviolet light through a layer of shotcrete.
16 feet
Solution: The active pressure on the wall is, a
= K a
v
= K a ( z ) where K a = tan2 ( 45° − / 2) = tan2 ( 45° − 36° / 2 ) = 0.26
610
Step 1: Find the vertical spacing S V of each layer of geotextile, where the Factor of Safety (FS ) is generally chosen between 1.3 to 1.5 for temporary walls. Permanent walls should use at least FS ≥ 2. For this problem choose FS = 1.5.
S V =
G a
FS
=
G
(
zK a ) FS
At a depth of z = 8 feet from the top, SV =
G
(
zK a ) FS
=
( 80 x12 lb / ft ) = 2.8 (110 )(8 )( 0.26 )(1.5 )
feet ≈ 34 inches
At a depth of z = 12 feet from the top, SV =
G
(
zK a ) FS
=
( 80 x12 lb / ft ) = 1.9 . . 110 12 0 26 1 5 ( )( )( )( )
feet ≈ 22 inches
At a depth of z = 16 feet from the top, SV =
G
(
zK a ) FS
=
( 80 x12 lb / ft ) = 1.4 (110 )(16 )( 0.26 )(1.5)
feet ≈ 17 inches
611
Notice how the spacing becomes denser the deeper we go below the surface; choose S V = 20 inches from z = 0 to z = 8 feet; below z = 8 feet use S V = 16 inches throughout, as shown in the figure. Step 2: Find the length of each layer of geotextile L , which is composed of two parts, l R which is the length of the geotextile within the Rankine failure zone, and l e which is the effective length of the geotextile beyond the failure zone (see the first figure, page 408). Again use FS = 1.5, and the angle φF is the soil-to-geotextile angle of friction, which is usually assumed to be 2/3 φ of the soil. Other values can be used, and a few are shown in this table,
Therefore, the length L of the geotextile layer is,
SV K a ( FS ) SV ( 0.26 )(1.5 ) ( H − z ) ( H − z) + = + L = l R + l e = tan ( 45° − / 2 ) tan ( 45° − 36° / 2 ) 2 tan F 2 ( 0.445 ) L = ( 0.51)( H − z ) + 0.438 S V From this equation, we can now prepare a table with the required lengths. (0.51)(H-z) (0.438S V ) S V z
L
inches
feet
feet
feet
feet
feet
16 56 76 96 112 144 176
1.33 4.67 6.34 8.00 9.34 12.00 14.67
1.67 1.67 1.67 1.67 1.33 1.33 1.33
7.48 5.78 4.93 4.08 3.40 2.04 0.68
0.731 0.731 0.731 0.731 0.582 0.582 0.582
8.21 6.51 5.66 4.81 3.99 2.66 1.26
Based on this table, use L = 8.5 feet for z ≤ 8 feet and L = 4.0 feet for z > 8 feet. Step 3: Find lap length l l for the geotextile, but never smaller than 3 feet ,
612
l l =
SV
4
a v
( FS )
tan
F
=
SV ( 0.26 )(1.5)
4 tan ⎡⎣( 23 ) ( 36° ) ⎤⎦
= 0.219 S V
For a depth of z=16 inches,
⎛ 20 inches ⎞ ⎟ = 0.365 feet < 3 feet in / ft 12 ⎝ ⎠
l l = 0.219SV = ( 0.219 ) ⎜
Therefore, use l l = 3 feet
Comment: These MSE problems commonly use Rankine’s active pressure coefficient. However, the actual value of K must depend on the degree of restraint of the type of reinforcement, as shown in this figure:
613
