RETAINING WALL PROBLEMS
CANTILEVER RETAINING WALL For the retaining wall and the t he profile shown below, calculate: a. The safety safety factor for overturning (minimum required required F.S. F.S. =2.0) =2.0) b. The safety safety factor factor for sliding sliding (minimum required F.S. F.S. =1.5) Do not consider the passive resistance of the fill in front of the wall. c. If the overturning safety is not not satisfactory, satisfactory, extend the base to the right and satisfy satisfy the minimum requirement requirement of 2.0 . If the sliding is i s not satisfactory, design a key (location, thickness, depth) under the base slab and take the advantage advantage of passive resistance resistance of the foundation foundation soil. Calculate the vertical stress starting from the top level of the base but consider the passive resistance starting from the bottom level of the base slab (i.e. in the sand). Satisfy the minimum requirement of 1.5. Use a factor of safety of 2.0 with respect to passive resistance. q=30kPa
0.5m
SURCHARGE
GRANULAR BACKFILL
7m
1m
3 γ =20 =20 kN/m FILL 1m
c=0 φ=300 3 γ =18kN/m =18kN/m 1m BASE SLAB
5m MEDIUM DENSE SILTY SAND c=0,
3 3 φ=320, γ =20kN/m =20kN/m , f=0.5(base friction) , γ conc conc=24kN/m
1
Solution: q=30kPa
0.5m
GRANULAR BACKFILL c=0 φ=300 γ =18kN/m3
V
1m
γ =20 kN/m3 FILL
V
H
V
H
V
1m
V
5.0m
2
Ka=tan (45-φ /2) 2
For granular backfill⇒Ka=tan (45-30/2)=0.333 Active pressure, pa=(q+γ γz )Ka-2c√Ka
z=0⇒ pa=30x0.333=10 kN/m
2
z=8⇒ σa=(30+18x8)0.333=58 kN/m
Force(kN/m)
2
Arm,about toe(m)
Moment(kN.m/m)
V1=0.5x7x24=84
1.25
105
V2=0.5x7x1/2x24=42
1.67
70
V3=1x5x24=120
2.5
300
V4=0.5x7x1/2x18=31.5
1.83
57.75
V5=3x7x18=378
3.5
1323
ΣV=655.5
ΣMr=1855.75
H1=10x8=80
4
320
H2=(58-10)x8x1/2=192
8/3
512
ΣH=272.0
ΣMov=832
2
(F.S.)ov =
∑ Mr ∑ Mov
=
1855 .75 = 2.23 832
(F.S.)ov=2.23>2.0 O.K.
c=0(at the base)
(F.S.) sliding =
( F.S.) sliding
=
do not consider
∑ V. tan δ + ( 2 / 3cB) + Pp ∑H
∑ V. tan δ
=
∑H
655 .5 x 0.5 272
= 1.20
(F.S.)sliding=1.20>1.5 NOT O.K. DESIGN BASE KEY
Base key design:
Passive resistance at the base key;
FILL, γ =20kN/m
3
1m
P1
D
P2 65D 65
2
Kp=tan (45+32/2)=3.25 z=0⇒pp=1x20x3.25=65 kPa z=D⇒pp=(1x20+20xD)x3.25=65+65D kPa 2
Pp=P1+P2=65D+1/2x65D
2
Use F.S.=2.0 w.r.t. passive resistance ⇒ Pp=1/2(65D+1/2x65D )
(F.S.)sliding =
∑ V. tan δ + Pp ∑H
655.55 x0.5 + 1 / 2(65D + 1 / 2x65D 2 ) = = 1. 5 272
3
2
Then, 65D+32.5D =160.5
⇒
D=1.43m
If Pp is substituted below the equation;
(F.S.)sliding = Then,
∑ V. tan δ ∑ H − Pp
= 1.5
D=1.07m
Take D=1.43m as it is on safe side.
4
GRAVITY RETAINING WALL
A gravity retaining wall is shown below. Use δ=2/3
φ and Coulomb active earth pressure
theory. Determine a. The factor of safety against overturning b. The factor of safety against sliding c. The pressure on the soil at the toe and heel, for both cases; -
considering the passive pressure, and
-
neglecting it.
Soil properties: I)
γ 1=18.5 kN/m3 ,
φ1=320,
c1= 0
II)
γ 2=18.0 kN/m3 ,
φ1=240,
c1= 30 kN/m
γ concrete=23.58
kN/m
5
3
2
Note: In Coulomb’s active earth pressure theory, the forces to be considered are only Pa(Coulomb) and weight of the wall i.e. the weight of the soil above the back face of the wall is not taken into account
Coulomb active forces;
Pa
1
= γ H 2K a 2
where
H =Height of the wall Ka = Coulomb’s active earth pressure coefficient
Ka
=
sin 2 (β + φ)
sin(φ + δ).sin(φ − α)
sin(β − δ).sin(α + β)
sin 2 β. sin(β − δ).1 +
With horizontal backfill;
α=0
With vertical retaining wall; β=90
δ: friction between the wall and adjacent soil
6
2
eqn 1
Solution:
For
α=0 β=750
⇒
Ka=0.4023 (use eqn 1)
φ=320 δ=(2/3)x32=21.30 1 Pa= .γ .H 2 .K a 2
=
1 2
x18.5x 6.52 x 0.4023 = 157.22 kN / m
Ph
= Pa . cos(15 + δ) =157.22 x cos 36.3 =126.65 kN / m
Pv
= Pa .sin(15 + δ) =157.22 x sin 36.3 = 93.15 kN / m
7
Force
Moment arm about pt. A
(kN/m)
Moment
(m)
(kN.m/m)
1 1) ( x1.53x 5.7) x 23.58 = 102.81 2
2.18
224.13
2) (0.6x 5.7) x 23.58 = 80.64
1.37
110.48
1 3) ( x 0.77x 5.7) x 23.58 = 51.75 2
0.81
41.92
4) (3.5x 0.8) x 23.58 = 66.02
1.75
115.54
Pv=
2.83
263.61
93.15
ΣV=394.37 ΣMoverturning= Ph x H/3
a) (F.S.)overt.=
∑ Mr ∑ Mo
=
ΣMresisting =755.68
= 126.65x 2.17 = 274.83 kN.m/m
755.68 274.83
=
2.75 > 2.0 O.K.
Note: if there is cohesionless soil at the base (c=0) ignore this term if passive pressure is considered
2
∑ V. tan δ + ( .c 2 .B) + Pp
b) (F.S.)sliding=
3 ∑H
δ=(2/3)xφ2
-
(F.S)sliding
Pp is ignored
=
2 2 394.37 x tan( x 24) + ( x30x3.5) 3 3 126.65
8
= 1.45
c. Pressure on soil at toe and heel
-If Pp is ignored
Σmnet=755.68-274.83=480.85 x=
∑ M net ∑V
=
480.85 394.37
= 1.22
⇒ e
=
B 2
− x = 0.53
ΣV x
2
qmax=215.05 kN/m (toe)
q max min
=
394.37 6 x 0.53 1± 3.5 3.5
A 2
qmin=10.30 kN/m (heel)
B/2
B/2
-If Pp is considered 2
2
Mres (due to Pp)=92.4x1.5 x0.5+0.5x63.99x(1/3)x1.5 =127.92 kN.m/m
Σmres=755.68+127.92
= 883.6 kN.m/m
Σmnet=883.6-274.83=608.77
x=
∑ M net ∑V
=
608.77 394.37
= 1.54
kN.m/m
⇒ e
=
3.5 2
− 1.54 = 0.21
ΣV x
2
qmax=153.24 kN/m (toe)
q max min
=
394.37 6x 0.21 1± 3.5 3.5
A 2
qmin=72.11 kN/m (heel)
9
B/2
B/2
REINFORCED EARTH WALL
A reinforced earth wall is to be constructed as shown in the figure below. The material that will be used as backfill shall have the following properties,
γ =17 kN/m3, φ=300, c=0. The
strips will be galvanised steel and will have a width of 75mm. The yield stress for strip 5
2
material is f y=3x10 kN/m .
a Design the reinforcements (i.e. determine the length and thickness) by using a factor of safety of 3.0 for both tie-breaking and pull-out.
b. Find the factor of safety along sliding on the base and calculate the base pressures for the foundation soil.
-
Design life for structure 50 yrs.
-
Corrosion=0.025 mm/yr
-
Use Rankine Earth Pressure Theory and take the friction angle between soil and reinforcement as 20
0
q=20 kN/m
2
2.0 m st
1 reinforcement
4
6.0 m
8
6.0 m
6.4 m
12
16
10
Sv =0.75m Sh = 1.00m
Solution:
2.0 m Lwedge
Le
1
6.67 kPa
6.0 m
8
6.0 m
6.4 m
0
45+φ /2=60
φ=300
⇒
16 74.64 kPa
Ka = 1/3
a) Design of reinforcement
As far as the tie breaking is concerned, bottom reinforcement (16) is the most critical one since the lateral pressure is maximum at that level.
pa=(γ z+q)Ka-2c(Ka)
0.5
sv
t w
( F.S.) breaking
T
=
w.t.f y Tmax
= 3.0
= S v .S h .(γ z + q )K a
Tmax
1
= 0.75x1.0 x (12x17 + 20) x = 56 kN 3
11
( F.S.) breaking
=
0.075 x t x3x10 5 56
= 3.0
⇒
t
= 7.46 mm
corrosion rate ⇒ 0.025mm/yr x 50 =1.25mm
t=7.46+1.25 = 8.71mm USE tdesign =9mm
•
As far as tie pull-out is concerned,
σv=(γ z+q)
σvtanφµ
t w Le
Frictional resistance is available on both surface (top and bottom)
( F.S.) pull− out
=
(F.S.) pull−out
2( γ z + q ) tan φ µ L e w ( γ z + q )K a S v S h
=
2 tan φ µ L e w
=
2x 0.075xL e x tan 20 1 x 0.75x1.0 3
Friction angle between soil and reinforcement
K a SvSh
= 3.0
⇒
Le
= 3.0
= 13.7m
Since first reinforcement (1) is the most critical one when the pull-out criterion is concerned, tan(45-φ/2)=
L wedge
+2
12.4 − 0.75
⇒
L wedge
= 4.72m
•
Total tie length L=Lwedge+Le=13.7+4.72=18.42m for upper 6m of the wall
•
For lower 6m of the wall , L=20.42m
12
b) (F.S.)sliding and Base Pressure 18.42m q=20 kPa 6.67 kPa
6.0m W1 H1
H2
6.4m W2
76.93 kPa
A
20.42m
Forces (kN/m)
Moment arm, about A
Moment
W1=18.42x6.0x17=1878.8
11.21
21061
W2 = (18.42+2)x6.4x17 = 2221.7
10.21
22684
Load = 20x18.42
11.21
4130
= 368.4
ΣFv = 4469 H1 = 6.67x12.4
ΣMr = 47875
kN
= 82.7
12.4 /2
H2 = (76.93-6.67)x12.4x(1/2) = 435.7
ΣFh = 518
12.4 /3
512.7 1800
ΣMov = 2313
kN
(FS)sliding = (ΣFv . tan δ) / ΣFh (Note:
δ = friction angle between soil and a different material = (2/3).φ = 20ο ; but as there
is soil at the bottom of the soil block in our concern, the internal friction angle used in FS formula.)
13
φ will
be
(FS)sliding = (4469 x tan30) / 518 = 4.98
X = ΣMnet / ΣFv = (47875 – 2313) / 4469 =10.2 m e = B/2 – X = (18.42 + 2)/2 – 10.2 = 0.01
z
0
qmax = ΣFv / B = 4469 / 20.42 = 218.85 kPa
14
⇒
no eccentricity
15
