HomeMy WebLinkAbout2013-12-03 Planning Board Supplemental Materials (85) DESIGN CALCULATIONS
FOR
UNIVERSIRL
uja11TM
Precast Concrete Retaining Wall Produced by Concrete Systems, Inc.
The Arbors at North Andover
North Andover, MA
Revision 1
Pro Con, Inc. � ,OkA OF Aq�SgC
C] CHRISTOPHER M.
Uwall Scope of Engineering: C) S CTURAL
Cn
• Overturning 48517
• Sliding
• Applied bearing pressureAL
• Pullout _g ^
Analysis not in Uwall Scope of Engineering: ' 17 l
Y P
• Slip circle/ global stability
• Bearing capacity of foundation soils
• Settlement
• Scour
• Slope stability for temporary cut
• Site drainage (runoff must be diverted away from wall)
Design 1: pp. 1 —3
Design 2: pp. 1—8
Design 3: pp. 1 - 8
• 12' high
• 6:1 Batter
• Seismic analysis
• No geogrid
Pro Con, Inc. Uwall Calculations Page 1 of 3
The Arbors Design 1
HNorth Andover, MA
WALL PARAMETERS
H:= 12ft wall height from top of leveling pad L:= 3.83ft stem length (fromface)*
P:= 0 radians backslope angle w:= 0.16515 radians wall batter
WU := 5in depth of UWall face L := L—Wu L =3.413-ft
L •tan((3)•tan(w)
L := P - L =O-ft L := L + L L =3.413-ft
PP 1 —tan((3)•tan(w) PP p p pP R
h:= Lp•tan(p) h =O-ft height of backslope soil wedge
H
Hs:= Oft height of broken back slope I:= atan s I=0
(2-H)
5:= if(I>0,I,(3) =0
RETAINED FILL PROPERTIES
.5236 angle of internal friction lbf
125 ft3 soil unit weight
6:= .6667•(0 friction angle between soil & wall
cos( + w)
2
Ka '= 2
2 Ka= 0.235
cos(W)2•cos(w—S)•11 + (4 sin( + S)•sin(�— (3)
Cos(W—6)•cos(W+ R)
HORIZONTAL FORCES
Ps:= .5•Ka•-yr(H+ h)2 cos( — W)
Ps= 1.98 x 103•lbf Horizontal force due to retained soil
ft
ql •= 0 lbf live load surcharge qd:= 0 lbf dead load surcharge
ft2 ft2
Pq:= (ql + gd)Ia•(H+ h)•cos((�— w)
P —0 lbf Horizontal force due to surcharges behind wall
q ft
Pa:= Ps+ Pq Pa= 1.98 x 103•lbf ft Total Horizontal active force
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Pro Con, Inc. Uwall Calculations Page 2 of 3
The Arbors
North Andover, MA
Resisting Moment:
Wri:= L•ryr H•1 ft Wri=5.745 x 103•lbf Weight of reinforced soil mass
L•�yr h•lft
Wrp:= Wro=0•1bf Weight of soil due to backslope
2
L+ H•tan(w)
Xri:= Xri=2.915-ft Moment arm for soil mass
2
2•L
Xrp• 3
H•tan(w) + Wu+ P Xrp=4.692-ft Moment arm for backslope
L Moment arm for dead load
Xqp := L+ [(H+ h)•tan(w)] — P Xqp =4.123-ft surcharge
Mr:= Wri'Xn + WrO'Xr(3+ qd*l ft•LO•Xgp
Mr= 1.675 x 104•ft•1bf
Overturning Moment:
Ys.— H+ h Ys=4-ft
3
Yq .— H+ h Yq = 6•ft
2
Mo := (Ps'Ys+ Pq•Yq)•l ft Mo =7.921 x 103 ft.IV
M
FSot:= r FSot=2.114
Mo
Base Sliding: Cds := 1 Coefficient of direct sliding
Rs:= Cds'(gd•lft•Lp+ Wri+ Wrp)•tan(�)) Rs=3.317 x 103 IV
R
FSSl := s FSSI = 1.675
Pa•1 ft
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Pro Con, Inc. Uwall Calculations Page 3 of 3
The Arbors
North Andover, MA
Seismic:
Seismic Load
A:= .17 Am:= (1.45 —A)•A Am=0.218 PAE:= .375•Am•'yr H2
MSeis:= .5•PAE•.6•H•lft MSeis =5.288 x 103•lbf-ft Seis:= .5•PAE
Overturning
M
i'Sot'R r FSot= 1.268
M + M if(FSot>_ 1.125,"OK" ,"NG") _ "OK"
Sets o
Slidin
R
FSsI ;= s FSsj = 1.222 if(FS,l >_ 1.125,"OK" ,"NG") ="OK"
(Pa+ Seis)•1 ft
Soil Bearing Stress: ( 1
MD —[Wri.
I X — 2 I + WrP•I XrO— 2 I + lftgd•Lp (XqP 2lJ
Wri+ WrR+ qd•1 ft-Lo
e =4.546.in eccentricity Check a<U6 6 =7.66-in
Be := L—2•e Be =3.072-ft effective base width
Wri + Wr(3+ (ql + gd)•LR•lft
Qa Be•lft
Qa= 1.87 x 103.lbf applied bearing pressure ft2
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Design 2
• 10.5' high
• 6:1 Batter
• Seismic analysis
• 3:1 Backslope
• Geogrid
Pro Con, Inc. Uwall Calculations Page 1 of 8
The Arbors Design 2
North Andover, MA
WALL PARAMETERS
H:= 10.5ft wall height from top of leveling pad L := 7ft reinforcement length
P:= 0.321 F radians backslope angle W:= 0 radians wall batter
Wu := 5in depth of UWall face L := L-Wu L =6.583-ft
L •tan((3)•tan(w)
L := P L =0•ft L := L + L L =6.583-ft
PP 1 -tan(p)•tan(w) PP R P PP
h:= Lp•tan((3) h =2.195-ft height of backslope soil wedge
H
Hs:= 6.5ft height of broken back slope I.= atan s I=0.3
(2-H)
(3:= if(I> 0,I,(d) (3=0.3
RETAINED FILL PROPERTIES
.5236 angle of internal friction - 125 lbf
3 soil unit weight
ft
8:= .6667• friction angle between soil & wall
cos( + W)2
Ka: 2 ' Ka=0.388
S)•sin(�- (d)
COS(W) cos(W— b) 1 +
HORIZONTAL FORCES
Ps:= .5-Ka'-If(H+ h)2•cos(q)- W)
Ps=3.382 x 103•Ibf Horizontal force due to retained soil
ft
ql 0 IV live load surcharge qd:= 0 lbf dead load surcharge
ft2 ft2
Pq := (qt + gd)Ka•(H+ h)•cos((� - w)
Pq =0 lbf Horizontal force due to surcharges behind wall
ft
Pa:= Ps+ P Pa=3.382 x 103.lbf ft Total Horizontal active force
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Pro Con, Inc. Uwall Calculations Page 2 of 8
The Arbors
North Andover, MA
Resisting Moment:
Wri := L•-Yr•H•1ft Wri =9.188 x 103•IV Weight of reinforced soil mass
L•ryr h•lft
Wrp:= 2 Wr(3=960.228•1bf Weight of soil due to backslope
L+ H•tan(w)
Xri:= Xri=3.5.ft Moment arm for soil mass
2
2•L
Xrp• 3
= H•tan(w) + Wu+ P Xro=4.806•ft Moment arm for backslope
L Moment arm for dead load
XqP := L+ [(H+ h)-tan(w)] - XqP =3.708-ft surcharge
Mr:= Wri'Xn + Wr(3•XrR+ qd.1 ft•L P'Xq(3
Mr=3.677 x 104•ft.IV
Overturning Moment:
Ys H+ h Ys=4.232-ft
3
Yq .- H+ h Yq =6.347-ft
2
Mo := (Ps'Ys+ Pq•Yq)•lft Mo = 1.431 x 104 ft.IV
M
FSot:= r FSot=2.57
Mo
Base Sliding: Cds := 1 Coefficient of direct sliding
Rs Cds'(gd'lft•Lp+ Wri + Wro)•tan((�) Rs=5.859 x 103 IV
R
FSsI := s FSsI= 1.733
Pa•1 ft
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Pro Con, Inc. Uwall Calculations Page 3 of 8
The Arbors
North Andover, MA
Seismic:
Seismic Load
A:_ .17 Am:_ (1.45 -A)•A Am=0.218 Pte:_ .375•Atn" H2
MSeis` .5•PA•.6•H•lft 3
MSeis =3.542 x 10 •lbf-ft Seis :_ .5•PAE
Overturning
Mr FS
r5ot.-- MSeis+ Mo of=2.06 if(FSot>_ 1.125,"OK" "NG") _ "OK"
Sliding
FS Rs FSsl = 1.486 if(FSsl > 1.125,"OK" ,"NG") _"OK"
sl (Pa+ Seis)•1 ft —
Soil Bearing Stress:
Mo — Wri' ' — + WrO' Xr.O—2) 2 J + lftgd•L�- XqO 2/J
e:=
Wri+ Wro+ qd•lft•Lo
e = 15.439•in eccentricity Check a<U6 L = 14-in
6
Be:= L —2•e Be =4.427-ft effective base width
Wri+ WrO+ (ql+ gd)•LO•lft
Qa Be•lft
Qa= 2.292 x 103•IV applied bearing pressure
ft2
Note: The eccentricity is slightly outside the kem, therefore, Qa is underestimated.
Recalculate for Qmax.
a:= L-�L + e) a=26.561-in
2
2•(Wri+ WrO+ ql + gd)•lft 3 2 lbf
Qmax�= 3-a Qmax= 3.056 x 10 ft Z
ft
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The Arbors
North Andover, MA
INFILL PROPERTIES
� := .5934 angle of internal friction Si:= .6667•(�friction angle between soil & wall
-yl .- 125 lbf infill unit weight
ft3
cos(0)i + w)2
Kai:= 2
rcos(w
Kai=0.32]
2 ni + S sin �i - p
cos(w) •cos(w- S) 1 +
HORIZONTAL FORCES DUE TO REINFORCED SOIL
Ps:= .5•Kal ryi•H2•cos((�i -w)
Ps= 1.832 x 103.IV Horizontal force due to reinforced soil
ft
q1 =0 1bf live load surcharge qd=0.IV dead load surcharge
ft ft2
Pq:= (ql + gd)Kai•H•cos(Oi -w)
Pq =0 lbf Horizontal force due to surcharges above zone
ft
Pa:= Ps+ Pq Pa= 1.832 x 103.lbf ft Total Horizontal active force
GEOGRID MATERIAL PROPERTIES
Tult:= 5475 llbt ultimate tensile strength
ft
RFD:= 1.2 durability reduction factor
RFID:= 1.1 installation damage reduction factor
RFCR:= 1.42 creep reduction factor
FSunc 1.1 factor of safety against uncertainties
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The Arbors
North Andover, MA
ALLOWABLE WORKING STRESS OF GEOGRID
T Tult 3lbf
T =2.655 x 10 .
a RFD'RFID'RFCR'FSunc a ft
P
Nmin` a Nmin=0.69 minimum number of geogrid layers
Ta
CHECK TENSION AT FIRST GEOGRID LAYER
E1 := 2ft E2 := 4-ft Act •- E2 + El Acl D1 =9 ft
2 2
Fgl := (ryi'D1 + ql + gd)'Kafcos(-�i —w)•Acl Fg1 = 897.107•IVft
if(Fg1 <Ta,"OK" ,"NG") ="OK"
CHECK TENSION AT SECOND GEOGRID LAYER
E3 := 6ft Ac2:= E3 -El D2 := H-Ac1 - Ac2
2 2
D2 =6.5 ft
Fg2 := (,Yi•D2 + ql + gd)•Kai-cos((i — w)•Ac2 Fg2 =431.941.IV
ft
if N2 <Ta,"OK" ,"NG")_ "OK"
CHECK TENSION AT TOP GEOGRID LAYER
Ac3 := (H_
E3 2 E2/ Ac3 = 5.5 ft D3 := A2 3 D3 =2.75 ft
IV
Fg3 := (7i'D3 + ql + gd)'Kai•cos(N - w)•Ac3 Fg3 = 502.546• ft
if(Fg3 <Ta,"OK" ,"NG") _ "OK"
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The Arbors
North Andover, MA
FAILURE PLANE ANGLE
+ tan (pi- tan(4)i- R + cot '4)i + w 1 + tan 6i -w •cot (�i + w
I + tan(6i —w)•(tan(Oj — 0) + cot(-�i + w)11
res=0.362 atan(res) =0.347 o,:= atan(res) + a,=0.871 radians
adeg:= a• 180 o.deg=49.909 degrees
7T
EMBEDMENT LENGTH
La3 := L—Wu—E3•tan(1.5708 —o) + E3-tan(w) La3= 1.533-ft
La2:= L—Wu—E2•tan(1.5708 —cx) + E2-tan(w) La2=3.216-ft
Lal := L— Wu—E1•tan(1.5708 —(i) + E1-tan(w) Lal =4.9-ft
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Pro Con, Inc. Uwall Calculations Page 7 of 8
The Arbors
North Andover, MA
SHEAR STRENGTH TRANSFERED TO GEOGRID(ANCHORAGE CAPACITY/PULLOUT)
Ci := .7 coefficient of interaction between soil &geogrid(from manufacturer)
E1 Lal depth of
d = H-E + -H•tan(w) + tan(p) d =9.79.779-ft1 1) (tan(a) 2 1 overburr
den
AC := 2•La1'Ci'(dli+ qd)"N AC=4.976 x 103.lbf ft anchorage capacity
FS:= AC FS= 5.547
Fgl
of
d = H-E + E2 -H•tan(w) + La2 tan(p) d =8.04-ft
depth
2 2) tan(o) 2 2 overburr
den
3lbf
AC := 2•La2•Ci•(d2•1i + gd)4i AC =2.685 x 10 •—ft anchorage capacity
FS:= AC FS=6.217
Fg2
of
d �H-E3) + E3 -H•tan(w) + La3J tan(p) d 6.301-ft depth
3 3 tan(a) 2 3 overburden
AC:= 2•La3' i �C• d3'1i + qd)4i x AC= 1.003 103. ft IVanchorage capacity
FS:= AC FS= 1.995
Fg3
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The Arbors
North Andover, MA
FAILURE PLANE ANGLE(RETAINED SOIL)
-tan( - (3) + tan(O - (3)•(tan(O- p) + cot(4)+ w))•(1 + tan(6- w)•cot(O + w))
res.- 1 + tan(8- w)•(tan(4- (3) + cot(O + w))
res=0.365 atan(res) = 0.35 ae:= atan(res) + 4) oe =0.873 radians
E2 -El a,deg:= ae• 180 a.deg=50.036 degrees
DL:= �
tan(ae)
RESISTING FORCE OF LOWEST GEOGRID(INTERNAL SLIDING)
Lps•tan((3)-tan(w)
Lps := L-Wu-DL Lpps;= I - tan(p)-tan(w)
LO1 := Lps+ Lpps Lel =4.907.ft
Wpri := Lps•�H-El) ryi Wpri = 5.214 x 103•lbf
ft
'ii•LRl•Lps•tan((3) IV from geogrid
WprO:= 2 WprO=465.858•ft Cds 1 manufacturer
Rps := cds'Od•L�1 + Wpri + WprO)•tan(0) Rps=3.831 x 103.lbf
ft
SLIDING FORCE AT LOWEST GEOGRID hl := LRl•tan((3) hl = 1.519ft
PShl :_ .5•Ka•-yr(H-El + hl)2•cos(O - w) PShl =2.106 x 103.lbf
ft
Pqhl (ql + gd)•Ka•(H-El + hl)•cos(O -w) Pqhl =0.Ibfbf
Pahl := PShl + Pqhl Pahl =2.106 x 103.lbf
ft
R
FSsll := ps FSs11 = 1.819 Factor of Safety against internal sliding
Pahl
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Design 3
• 11' high
• 6:1 Batter
• Seismic analysis
• Lower tiers with groundwater present
(saturated backfill 165 pst)
• Geogrid
Pro Con, Inc. Uwall Calculations Page 1 of 8
The Arbors Design 3
North Andover, MA
WALL PARAMETERS
H:= l lft wall height from top of leveling pad L:= 6ft reinforcement length
0:= 0 radians backslope angle W:= 0 radians wall batter
WU:= 5in depth of UWall face L := L-WU L = 5.583-ft
L tan(�)•tan(w)
L := P L =O•ft L := L + L L = 5.583-ft
PP 1 - tan(R)•tan(w) PP P PP
h := LP•tan(p) h=0•ft height of backslope soil wedge
H
Hs:= Oft height of broken back slope I:= atan s I= 0
(2-H)
0:= if(I>0,I, 0 71 = 125 Ibf infill unit weight
3
ft
RETAINED FILL PROPERTIES
�):= .5236 angle of internal friction = 165 lbf
ryr'• 3 unit weight saturated soil
8:= .6667• friction angle between soil & wall ft
cos( + w)2
Ka:= n(�('�'
(3) 2 1a=0.297
S)•si -
cos(w) cos(w-S) 1 +
HORIZONTAL FORCES
Ps:= .5•Ia•^lr(H+ h)2•cos((� - w)
Ps=2.57 x 103.lbf Horizontal force due to retained soil
ft
ql 0 lbf live load surcharge qd := 0 1bf dead load surcharge
ft2 ft2
P (ql + gd)Ka'(H+ h)•cos(�)- w)
P _0 IV0.— force due to surcharges behind wall
q
ft
Pa:= Ps+ P Pa=2.57 x 103.lbf ft Total Horizontal active force
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The Arbors
North Andover, MA
Resisting Moment:
Wri := L•-yi•H•l ft Wri= 8.25 x 103 lbf Weight of reinforced soil mass
L•ryr•h•1ft
WrP:= WrP=0•1bf Weight of soil due to backslope
2Xn
L+ H•tan(w)
;= Xri=3-ft Moment arm for soil mass
2
2•L
Xrp:= H•tan(w) + Wu+ P 3 Xro=4.139•ft Moment arm for backslope
L Moment arm for dead load
Xq� := L+ [(H+ h)•tan(w)] - 2 XqP =3.208-ft surcharge
Mr:= Wri'Xn+ Wrp•Xrp+ gd•1ft•LR'XgP
Mr=2.475 x 104•ft•lbf
Overturning Moment:
YS = H+ h Ys=3.667 ft
3
Yq.- H+ h Yq = 5.5•ft
2
Mo := (Ps•Ys+ Pq•Yq)•lft Mo =9.424 x 103 ft•lbf
M
FSot:= r FSot=2.626
Mo
Base Sliding: Cds 1 Coefficient of direct sliding
Rs:= Cds'(gd'lft•Lp+ Wri+ Wrp)•tan((�) Rs=4.763 x 103 IV
R
FSSI := s FSsI = 1.853
Pa 1 ft
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The Arbors
North Andover, MA
Seismic:
Seismic Load
A:= .17 Am:= (1.45 —A)•A Am=0.218 PAE:= .375•Am•^yr H2
MSeis := .5•PA•.6•H•lft MSeis =5.376 x 103•lbf-ft Seis:= .5•PAE
Overturning
Mr
FSot:= MSeis + Mo FSot 1.67 if(FSot>_ 1.125,"OK" ,"NG") _ "OK"
Slidin
R
FSSI:= s FSsI= 1.407 if(FSs, >— 1.125,"OK" ,"NG") _ "OK"
�Pa+ Seis)•1ft
Soil Bearing Stress:
Mo —[Wri.(Xri — 2) + Wrp•(Xro— 2/ + lftgd•L�- NO L
C:=
.— J
Wri+ Wro+ qd•lft•Lo
e = 13.708•in eccentricity Check a<L/6 L = 12.in
6
Be := L —2•e Be=3.715-ft effective base width
Wn+ Wro+ (ql+ gd)•Lo.1ft
Qa Be•lft
Qa=2.221 x 103.lbf applied bearing pressure
2
ft
Note: The eccentricity is slightly outside the kem, therefore, Qa is underestimated.
Recalculate for Qmax.
a:= L— L + e V a=22.292-in 2 /
2'(Wri + Wr{3+ ql + qd)'I f•t 3 2 lbf
3-a
Qmax�= Qmax=2.961 x 10 ft 2
ft
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The Arbors
North Andover, MA
INFILL PROPERTIES
4)i := .5934 angle of internal friction 6i:= .6667•,Ofriction angle between soil & wall
cos(�i + W)2 Kai:= ?
Fn Kai=0.255
cos(w) cos(w— 6) 1 + —S) cos(w+ �i)
HORIZONTAL FORCES DUE TO REINFORCED SOIL
Ps:= .5-Kai ryi•H2•cos(�i —w)
Ps= 1.598 x 103.IV Horizontal force due to reinforced soil
ft
ql —0 lbf live load surcharge qd =0.IV dead load surcharge
ft ft2
P :_ (ql + gd)Kai•H•cos(�i —w)
P _0 Ibf Horizontal force due to surcharges above zone
q
ft
Pa:= Ps+ Pq Pa= 1.598 x 103.IV ft Total Horizontal active force
GEOGRID MATERIAL PROPERTIES
Tult:= 5475 lbf ultimate tensile strength
ft
RFD:= 1.2 durability reduction factor
RFID:= 1.1 installation damage reduction factor
RFCR:= 1.42 creep reduction factor
FSunc := 1.1 factor of safety against uncertainties
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The Arbors
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ALLOWABLE WORKING STRESS OF GEOGRID
T Tult 3lbf
T =2.655 x 10 .
a RFD'RFID'RFCR°FSunc a ft
P
Nmin a Nmin=0.602 minimum number of geogrid layers
Ta
CHECK TENSION AT FIRST GEOGRID LAYER
E1 := 2ft E2 := 4-ft - E2 + E1 Ael
Acl :- D1 := H- — D1 =9.5 ft
2 2
Fgl := (ryi-Dl + ql + qd)-Kafcos(�i - w)•Acl Fg1 =752.931.lbf
ft
if(Fgl <Ta,"OK" "NG") = "OK"
CHECK TENSION AT SECOND GEOGRID LAYER
E3 -El Act
E3 := 6ft Ac2:= 2 D2 := H-Acl - 2
D2 =7 ft
Fg2 := (ryi•D2 + ql + gd)•Kai•cos(�)i - w)•Ac2 Fg2 =369.861.lbf
ft
if(Fg2 <Ta,"OK" ,"NG") = "OK"
CHECK TENSION AT TOP GEOGRID LAYER
Ac3 := CH- E3 2 E2 Ac3 = 6-ft D3 := Ac3 D3 -3 ft
Fg3 := (�i•D3 + ql + gd)•Kai•cos((�i - w)•Ac3 Fg3 =475.536•lbf
—
ft
if(Fg3 <Ta,"OK" ,"NG") _ "OK"
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The Arbors
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FAILURE PLANE ANGLE
—tan(cpi — 13)+ tan(4)i — (3 • tan (�i -- + cot Oi + w I + tan $i — w•cot(4)i + w
I + tan(Si -w)•(tan(4)i - p)+ cot((�i + w))
res=0.453 atan(res) =0.425 oc:= atan(res) + a=0.949 radians
a.deg:= a• 180 a.deg=54.353 degrees
7
EMBEDMENT LENGTH
La3 := L-Wu-E3•tan(1.5708 -a) + E3•tan(w) Lai = 1.28-ft
La2:= L-Wu-E2•tan(1.5708 -o) + E2-tan(w) La2=2.715-ft
Lal := L-Wu-El•tan(1.5708-oc) + El•tan(w) Lal =4.149-ft
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SHEAR STRENGTH TRANSFERED TO GEOGRID(ANCHORAGE CAPACITY/PULLOUT)
Ci := .7 coefficient of interaction between soil &geogrid(from manufacturer)
d = H—E + —H•tan(w) + tan(p) d =9
tan(E1 Lal ft depth of
1 1) ( cx) 2 1 overburden
AC:= 2•La1'Ci'(dl'-ji + gd)4i AC =3.878 x 103.IV anchorage capacity
ft
FS:= AC FS=5.15
Fgl
of
d = (H—E2) + E2 —Man(w) + I a21 tan(p) d 7-ft depth
2 2 tan((.) 2 2 overburden
AC := 2•La2' i•C• (d2'ryi + qd)4i AC = 1.973 x 103.IVanchorage capacity
ft
FS:= AC FS=5.335
Fg2
of
d = H—E + E3 —H•tan(w) + La3 tan(p) d =5-ft depth
3 ( 3) tan(a) 2 3 overburden
).
AC:= 2•La3'Ci'(d3'^fi + gd)•gi AC =664.709•IV anchorage capacity
ft
FS:= AC FS= 1.398
Fg3
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FAILURE PLANE ANGLE(RETAINED SOIL)
—tan( — 0) + tan( — 0)•(tan((�— 0) + cot( + w))•(1 + tan(6— w)-cot( + w))
res:= 1 + tan(6 — w)•(tan(�)— 0) + cot( + w))
res=0.487 atan(res) =0.454 ae:= atan(res) + ae=0.977 radians
E2 —El adeg:= ote• 180 odeg= 55.984 degrees
DL:= �
tan(o e)
RESISTING FORCE OF LOWEST GEOGRID(INTERNAL SLIDING)
Lps•tan(0)•tan(w)
Lps := L—Wu —DL Lpps' 1 —tan(0)-tan(w)
LO1 := Lps + Lpps LOl =4.233-ft
Wpri := Lps•(H—El)--Ii Wpri =4.763 x 103.lbf
ft
ryi•L0l•Lps•tan(0) lbf from geogrid
WprO 2 WprO 0 ft Cds .— 1 manufacturer
Rps := Cds•(gd•L01 + Wpri + WprO)•ta+' ) =3.212 x 103.lbf
Rs ft
SLIDING FORCE AT LOWEST GEOGRID hl := LOl•tan(0) hl =0 ft
PShl .5'Ka'1r(H—El + hl)2•cos(4— w) PShl = 1.721 x 103.lbf
ft
Pqhl (ql + gd)•Ka•(H—El + hl)•cos((� — w) Pghl =0.lbf ft
P P + P P = 1.721 x 103.lbf
Pahl '= PShl qhl Pahl ft
FSsll :=: Rps FSsll = 1.867 Factor of Safety against internal sliding
Pahl
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