HomeMy WebLinkAboutCT 04-14A; TRAILS END; RESPONSE TO PLAN REVIEW; 2017-11-16I
A ENGINEERING
I CONSULTANTS. INC.
1701 E. Edinger Avenue, Suite A9, Santa Ma, CA 92705
Tel.: (888) 220-5596 Fax: (714) 866-4171 E-Mail: abi@abiconsultants.com
Memo
To: Paul Hutêhins, DM1
From: Daya Bettadapura
CC:
Date: November 16, 2017
Re: Response to Plan Review comments from City of Fullerton
Keystone Retaining Wall Plan
Trails End Development, Carlsbad, CA
ABI Project No. 17201
The Keystone wall plan for the subject project, prepared by ABI, has been revised to incorporate
recommendations from project Geotechnical Engineer. The design calculations for the wall for its
internal and external stability (see attached) requires shorter geogrid reinforcement length relative to
that required for global stability. Based on the results of global stability analysis results (see attached)
the recommended geogrids length is 18 feet. ABI has also incorporated a drainage ditch at the top of
wall to drain the surface water drainage over the slope.
The attached plans are electronically stamped and signed. Wet signed plans will be sent via overnight
mail to Pacifica office.
If you have any questions regarding this memo, please contact me at (888) 220-5596.
RECEIVED
NOV 202017
LAND DEVELOPMENT
ENGINEERING
Unit Type:
Leveling Pad
Wall HI:
BackSlope:
Surcharge:
Results:
Factors ofSafelv:
Compaclll / 120.00 pcI
Crushed Stone
6.00 ft
26.60 deg. slope,
LL: 0 psfunform surcharge
Load Width: 0.00 ft
Sliding Overturning
1.57 3.82
'STONE
_RE1A1N1NGWMLSYS1MS RETAINING WALL DESIGN
KeyWall 2012 Version 3.7.2 Build 10
Project: Trails End
Project No: 17201
Case: Case 1
Design Method: NCI14 3rd Edition (parallelogram soil interface)
Desin Parameters
Soil Parameters: 4) deg c psf y pcf
Reinforced Fill 30 0 130
Retained Zone 30 0 130
Foundation Soil 20 100 130
Reinforced Fill Type: Sand, Silt or Clay
Unit Fill: Crushed Stone, 1 inch minus
Minimum Design Factors of Safety
sliding: 1.50 pullout: 1.50 uncertainties: 1.50
overturning: 2.00 shear: 1.50 connection: 1.50
bearing: 2.00 bending: 1.50
Date: 11/8/2017
Designer: JC-.
Design Preferences
Professional Mode
Reinforcing Parameters: MirafiXT Geogrids
Tult RFcr RFd RFid LTDS FS Tal Ci Cds
3XT 3500 1.58 1.10 1.05 1918 1.50 1279 0.80 0.80
Analysis: Case: Case 1
Wall Batter: 0.90 deg (Hinge fit N/A)
embedment: 0.50 ft
9.00 ft long
DL: 0 pf uniform surcharge
Load Width: 0. 00ft
Bearing Shear Bending
3.21 6.60 3.28
Calculated Bearing Pressure: 1122/1122 psf
Eccentricity at base: 0.58 ft
Reinforcing: (ft & lbs/fl)
Calc.
Layer Height Length Tension
4 4.67 6.0 126
3 3.33 6.0 224
2 2.00 6.0 331
1 0.67 6.0 428
Allow Ten Pk Conn Pullout
Reinf. Tyne Li Is.! FS
3XT 1279 ok 747 ok 3.51 ok
3XT 1279 ok 844 ok 5.35 ok
3XT 1279 ok 941 ok 6.76 ok
3XT 1279 ok 1038 ok 8.32 ok
Reinforcing Quantities (no waste included):
3XT 2.67 sy/ft
Date 11/8/2017 Case 1 Page 1
I I
DETAILED CALCULATIONS
Project: Trails End
Project No: 17201
Case: Case 1
Design Method: NCIvIA 3rd Edition (parallelogram soil interface)
Soil Parameters: 4b deg c nsf y ncf
Reinforced Fill 30 0 130
Retained Zone 30 0 130
Foundation Soil 20 100 130
Leveling Pad: Crushed Stone
Modular Concrete Unit: Compaclil
Depth: 1.00 ft in-Place Wt: 120 pcf
Date: 11/8/2017
Designer: JCW
Geometry
Internal Stability
(Broken geometry)
Height: 6.00 ft
BackSlope:
Angle: 26.6 deg
Height: 4.5/ ft
Batter: 0. 90deg
Surcharge:
DeadLoad: 0.00 psf
Live Load: 0.00 psf
Base width: 6. Oft
Earth Pressures:
mn2(c+)
ka -
sink stn(a—c) 1+
n(a-c)stn(c+/3)
External Stability
(Broken geometry)
Height: 8.52 ft
Angle: 26.6 deg
Height: 1.98 ft
Batter: 0. 90deg
Dead Load: 0.00 psf
Live Load.0.00 psf
Internal External
0 = 30 deg 4) = 30 deg
a = 90.90 deg a = 90.90 deg
= 26.60 deg = 26.60 deg
6 = 20.00 deg 6 = 30.00 deg
I-I = 6.00 ft
ka = 0.481 ka = 0.399
Hinge Height: Hinge Ht= Not applicable
Date 11/8/2017 Case 1 Page 2
Pa 0.5H.(}i-ka - 2c) Pq := q-}Ika
P Pa-COS(6) P%:= Pq•co(6)
:= Pa sin(s) Pq, p sin(8)
Reactions are:
Area Force Arm-x
Wi 720.00 [0.547]
W2 36.76 [1.063]
W3 3826.48 [3.547]
W4 36.76 [6.031]
W5 820.19 [4.441]
Pa_h 1647.86 6.031
Sum V= 5440.19
Sum 11= 1647.86
L
4
pa
V12
Arm-y Moment
3.000 393.93
2.000 39.07
3.000 13573.02
4.000 221.71
6.841 3642.31
[2.841] -4681.92
Sum Mr = 17870.04
Sum Mo = -4681.92
Reinforcing Parameters: MirafiXT Cieogrids
Tult RFcr RFd REid LTDS FS Tal Ci Cds
3X7' 3500 1.58 1.10 1.05 1918 1.50 1279 0.80 0.80
Connection Parameters: Mirafi XT Geogrids
Frictional 1 Break Pt Frictional 2
3XT Tcl=Ntan(42.30) + 975 1300 Tc1=Ntan(8.10) +1973
Unit Shear Data
Shear = N tan (40. 00)
Inter- Unit ShearShear = N tan(34. 00) + 900.00
Calculated Reactions
effective sliding length = 6.00 ft
Date 11/8/2017 Case 1 Page 3
I
Calculate Sliding at Base
For Sliding, Vertical Force = Wl+W2+W3+W4+W5+W6+qd
The resisting force within therein. mass, Rf_1
The resisting force at the foundation, Rf_2
The driving forces, Df, are the sum of the external earth pressures:
Pa_h + Pql_h + Pqd_h
the Factor of Safety for Sliding is Rf_2/Df
Calculate Overturning:
Overturning moment: Mo = Sum Mo
Resisting moment: Mr = Sum Mr
Factor of Safety of Overturning: Mr/Mo
= 5440
= N tan(30)
= 3141
= N tan(20) + (6.00 x 100.00)
= 2580
= 1648
= 1.57
= -4682
= 17870
= 3.82
Date 11/8/2017 Case 1 Page 4
Calculate eccentricity at base: with Surcharge / without Surcharge
Sum Moments = 13188! 13188
Sum Vertical = 5440/5440
Base Length = 6.00
e = 0.576 / 0.576
Calculate Ultimate Bearing based on shear:
where:
Nq=6.40
Nc = 14.83
Ng = 5.39 (ref. Vesic(1973, 1975) eqns)
Qult= 3597 psf
Equivalent footing width, B'= L -2e = 4.85 / 4.85
Bearing pressure = sumV!B' = 1122 psf / 1122 psf [bearing is greatest without liveload]
Factor of Safety for bearing = Qultibearing= 3.21
Calculate Tensions in Reinforcing:
The tensions in the reinforcing layer, and the assumed load at the connection,
is the vertical area supported by each respective layer, Sv.Column [7] is 2c sqrt(ka)'.
Table of Results ppf
[i] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11]
Layer Depth zi hi ka/rho Pa (Pas+Pasd) c (5+6)cos(d)-7 Ti Tcl Tsc
4 1.33 1.00 0.513/42 133 0 0 126 126 747 N/A
3 2.67 2.67 0.513/42 237 0 0 224 224 844 N/A
2 4.00 4.00 0.505/45 350 0 0 331 331 941 N/A
1 5.33 5.33 0.490/47 453 0 0 428 428 1038 N/A
Calculate sliding on the reinforcing:
The shear value is the lessor of base-shear or inter -unit shear.
[1] [2] [3] [4] [5] [6] [7] [8] 191 [101 [11] [12]
Layer Depth zi Pas+Pasd ji E
4 1.33 1687 5.00 0.80 1008 1787 0.475 460 0 402 4.45
3 2.67 2554 5.00 0.80 1116 2295 0.446 780 0 682 3.37
2 4.00 3420 5.00 0.80 1224 2803 0.424 1172 0 1024 2.74
1 5.33 4287 5.00 0.80 1332 3312 0.406 1631 0 1425 2.32
Date 11/8/2017 Case 1 Page 5
1]
Calculate pullout of each layer
The FoS (R*/S*) of pullout is calculated as the individual
layer pullout (RI) divided by the tension (DI) in that layer.
The angle of the failure plane is: 40.90 degrees from vertical.
[1] [2] [3] [4] [5] [6] [7] [8]
Layer Depth zi Le SumV Ci IQi Ti FS P0
4 1.33 1.03 478 0.80 442 126 3.51
3 2.67 2.16 1297 0.80 1198 224 5.35
2 4.00 3.30 2422 0.80 2237 331 6.76
1 5.33 4.43 3853 0.80 3559 428 8.32
Check Shear & Bending at each layer
Bending on the top layer is the FOS of overturning of the Units
(Most surcharge loads need to be moved back from the face)
[1] [2] [3] [4] [5] [6] [7] [8] [9]
Layer Depth z Si DM Pv LM LIk DS RS FS_Sh
4 1.33 1.33 25 160 82 3.28 56 1008 18.00
3 2.67 1.33 31 240 198 6.39 98 1116 11.42
2 4.00 1.33 49 400 303 6.16 152 1224 8.05
1 5.33 1.33 66 560 408 6.19 202 1332 6.60
Date 11/8/2017 Case 1 Page 6