Lf Rd Retaining Wall
Transcript of Lf Rd Retaining Wall
1
1 INTRODUCTION
1.0 General
The LRFD design procedure for conventional gravity and cantilever retaining walls,
abutments and MSE walls, with a few exceptions, is identical to the ASD design procedure
utilized in the past. Generally, ultimate bearing capacity, resistance to sliding, overall stability,
wall foundation settlement, and lateral deflection limits are checked. Total as well as
differential settlements are major criteria for determining wall type. See excerpts in Appendix
C taken from FDOT’s Plans Preparation Manual – Volume I showing wall type selection
criteria based on anticipated settlement and the environmental classification of the site.
Therefore, the first step of a retaining wall design is to calculate the settlements based on the
fill heights.
1.1 Design Summary
In ASD design, all the uncertainties in the applied loads and ultimate geotechnical or
structural capacity are factored in safety factors or allowable stresses. Whereas, LRFD
separates the variability of these design components and resistance factors to the load and
material capacity, respectively. The key issues in the design of retaining walls and abutments
by LRFD is the application of maximum and minimum load factors for dead , earth and
surcharge loads. See Table 1 and Figures 1 through 3 below.
1.1.1 Dead or Permanent Loads
DC = dead load of structural component and nonstructural attachments (for conventional retaining walls not for MSE Walls)
DW = dead load for wearing surfaces and utilities EH = horizontal earth pressure load ES = earth surcharge load EV = vertical pressure from dead load of earth fill
1.1.2 Live or Transient Loads
LS = live load surcharge WA = water load and stream pressure
2
Table 3.4.1-1 Load Combinations and Load Factors.
Use One of These at a Time Load Combination
Limit State
DC DD DW EH EV ES EL
LL IM CE BR PL LS WA WS WL FR
TU CR SH TG SE EQ IC CT CV
STRENGTH I (unless noted)
γp 1.75 1.00 — — 1.00 0.50/1.20 γTG γSE — — — —
STRENGTH II γp 1.35 1.00 — — 1.00 0.50/1.20 γTG γSE — — — — STRENGTH III γp — 1.00 1.40 — 1.00 0.50/1.20 γTG γSE — — — — STRENGTH IV γp — 1.00 — — 1.00 0.50/1.20 — — — — — — STRENGTH V γp 1.35 1.00 0.40 1.0 1.00 0.50/1.20 γTG γSE — — — — EXTREME EVENT I
γp γEQ 1.00 — — 1.00 — — — 1.00 — — —
EXTREME EVENT II
γp 0.50 1.00 — — 1.00 — — — — 1.00 1.00 1.00
SERVICE I 1.00 1.00 1.00 0.30 1.0 1.00 1.00/1.20 γTG γSE — — — — SERVICE II 1.00 1.30 1.00 — — 1.00 1.00/1.20 — — — — — — SERVICE III 1.00 0.80 1.00 — — 1.00 1.00/1.20 γTG γSE — — — —
SERVICE IV 1.00 — 1.00 0.70 — 1.00 1.00/1.20 — 1.0 — — — —
FATIGUE—LL, IM & CE ONLY
— 0.75 — — — — — — — — — — —
Table 3.4.1-2 Load Factors for Permanent Loads, γp.
Load Factor Type of Load, Foundation Type, and Method Used to Calculate Downdrag Maximum Minimum
DC: Component and Attachments DC: Strength IV only
1.25 1.50
0.90 0.90
DD: Downdrag Piles, α Τοmlinson Method Piles, λ Method Drilled shafts, O’Neil and Reese (1999) Method
1.40 1.05 1.25
0.25 0.30 0.35
DW: Wearing Surfaces and Utilities 1.50 0.65 EH: Horizontal Earth Pressure
• Active • At-Rest
1.50 1.35
0.90 0.90
EL: Locked-in Erection Stresses 1.00 1.00 EV: Vertical Earth Pressure
• Overall Stability • Retaining Walls and Abutments • Rigid Buried Structure • Rigid Frames • Flexible Buried Structures other than Metal Box Culverts • Flexible Metal Box Culverts
1.00 1.35 1.30 1.35 1.95 1.50
N/A 1.00 0.90 0.90 0.90 0.90
ES: Earth Surcharge 1.50 0.75
3
1.1.3 Typical Application of Load Factors
a. Dead or permanent load
Figure 1 - Bearing resistance for both
conventional and MSE walls
Figure 2 - Sliding and Eccentricity for
conventional and MSE walls
Figure 3 - Live or transient loads
4
2.0 Conventional Retaining Walls 2.1 Overall Stability
Investigate Service 1 load combination using an appropriate resistance factor. In
general, the resistance factor, φ, may be taken as;
• 0.75 - where the geotechnical parameters are well defined, and slope does not
support or contain a structural element.
• 0.65 – where the geotechnical parameters are based on limited information or the
slope contains or supports a structural element.
2.2 Bearing Resistance
Investigate bearing resistance at the strength limit state using appropriate factored
loads and resistances as described below.
2.2.1 Wall Supported on Soils
The vertical stress shall be calculated assuming a uniformly distributed
pressure over an effective base area.
eBV
v 2−Σ
=σ (11.6.3.2-1)
Figure 4
5
2.2.2 Wall Supported on Rocks
If the resultant force is within the middle third of the base;
)61(max Be
BV
V +Σ
=σ (11.6.3.2-2)
)61(min Be
BV
V −Σ
=σ (11.6.3.2-3)
If the resultant force is outside of the middle third
of the base;
])2[(32
max eBV
V−
Σ=σ (11.6.3.2-4)
0min =Vσ (11.6.3.2-5)
Figure 5
2.2.3 Soil Bearing Resistance
nbR qq φ= (10.6.3.1.1-1)
γγγγ wmwqqmfcmn CBNgCNDgcNq 5.0++= (10.6.3.1.2a-1)
Where: g = gravitational acceleration (ft/sec2)
c = cohesion
Ncm = Ncscic, bearing capacity factor for cohesion
Nqm = Nqsqdqiq , surcharge bearing capacity factor
Nγm = Nγsγiγ, unit weight bearing capacity factor
γ = total (moist) unit weight of soil above or below the bearing depth of the
footing.
Df = footing embedment depth (ft)
B = footing width (ft)
Cwq,Cwγ= correction factors for water table
sc,sγ,sq = footing shape correction factors
dq = correction factor for shear resistance along the failure surface passing
through cohesionless materials above the bearing elevation.
ic,iγ,iq = load inclination factors
Figure 5
6
n
fq cBLV
Hi⎥⎥⎦
⎤
⎢⎢⎣
⎡
+−=
φcot1 (10.6.3.1.2a-7)
1
cot1
+
⎥⎥⎦
⎤
⎢⎢⎣
⎡
+−=
n
fcBLVHi
φγ (10.6.3.1.2a-8)
θθ 22 sin1
2cos
1
2
⎥⎥⎦
⎤
⎢⎢⎣
⎡
+
++
⎥⎥⎦
⎤
⎢⎢⎣
⎡
+
+=
LB
LB
BL
BL
n (10.6.3.1.2a-9)
All these factors can be found in Article 10.6.3.1.2 of the Interim AASHTO 2006 LRFD
Specifications.
2.3 Overturning or Eccentricity (Article 11.6.3.3)
For foundations on soil, the resultant of the reaction pressure distribution shall be
located within the middle one-half of the base width.
For foundations on rock, the resultant of the reaction pressure distribution shall be
located within the middle three-fourths of the base width.
2.4 Sliding
Check spread footing against sliding in accordance with the provisions of Article
10.6.3.4. Passive resistance of soil in front of walls used in stability computation shall be
carefully evaluated accounting for possible future excavations, either temporary or permanent,
and/or long term erosion, etc.
For clayey soils beneath the footing, QR, against failure by sliding is:
RR = φτR τ + φep Rep (11.6.3.4-1)
where:
φτ = Resistant factor for shear resistance between soil and foundation specified in
Table 10.5.5.2.2-1
Rτ= Nominal shear resistance between footing and foundation material (ton); and
Rep= Nominal passive resistance of foundation material available throughout the
design life of the footing (ton). [ pp2
p K H c 2 K H ½ P += γ ]
φep = 0.5, Table 10.5.5.2.2-1
If the soil beneath the footing is cohesionless, then:
7
Rτ= V*tan δ (11.6.3.4-2)
in which: tan δ = tan φf for concrete cast against soil,
= 0.8 tan φf for precast concrete footing (φ = 0.8 from Table 3, page 26)
V = total vertical force (tons)
φf = the internal friction angle of soil (o)
Example Problem 1:
Foundation soil properties:
γ2 = 110 pcf γ2’= 58 pcf φ2 = 33o
LS = 250 psf
The cantilever retaining wall is being considered for a grade separation between roadway
lanes in a non-seismic area. The wall will be backfilled with a free draining granular fill such
that the seasonal high water table will be below the bottom of the footing. The vehicular live
load surcharge, LS, on the backfill will be applied as shown in the figure.
Approach: To perform the evaluation, the following steps are taken:
The loads and resulting moments due to structure components, earth pressures and live
load surcharge are calculated; and
The appropriate load factors and combinations are determined and multiplied by the
unfactored loads and moments to determine the factored loading conditions.
Step 1: Calculate the Unfactored Loads
(A) Dead Load of Structural Components and Nonstructural Attachments (DC)
Figure 6
8
Assuming a unit weight of concrete, γc, equal to 150 pcf, W1 = B1 H1 γc =1’x15’x0.150 =2.25 kips/ft W2 = 0.5 B2 H1 γc=0.5x0.5’x15’x0.150=0.562 kips/ft W3 = B H2 γc = 9.5’x1.5’x0.150=2.14 kips/ft
(B) Vertical Earth Pressure (EV)
Unit Weight of Soil γ1 = 105 pcf
Weight of Soil on Footing
PEV = W4 = B3H1γ1 = 6.5’ x 15’ x 105= 10.24 kips/ft
(C ) Live Load Surcharge (LS)
A live load surcharge is applied when vehicle loads will be supported on the backfill within a
distance equal to H. The unit vertical component of LS is:
For a heel width B3 of 6.5’:
length wallof ftkips = )psf)(( = B p = P 3LSVLSV /625.1'5.6250
The active earth pressure coefficient ka for a wall friction angle, δ = 0 and a horizontal back
slope is:
k = ka = 0.33
ftpsfLS k = p /5.8225033.0 =×=Δ
Using a rectangular distribution, the live load horizontal earth pressure acting on the wall is:
length wallof ftkips = )psf)(( = H p = PLSH /36.1'5.165.82Δ
(D) Horizontal Earth Pressure (EH)
The lateral earth pressure is assumed to vary linearly with the depth of soil backfill as
given by:
p = kγ'z where k = 0.33
At the base of the footing (i.e., @ z = H):
p = (0.333) (105) (16.5’) = 577 psf
The horizontal earth pressure (triangular distribution) acting on the wall is:
PEH = ½ pH = (0.5) (577) (16.5’) = 4.76 kips/ft length of wall
Figure 7
9
(E) Summary of Unfactored Loads
Unfactored Vertical Loads and Resisting Moments
Item
V
(kips/ft)
Moment Arm About Toe (ft)
Moment About Toe (kip-ft/ft)
W1
2.25 2.5 5.625
W2
0.56 1.83 1.03
W3
2.14 4.75 10.165
PEV 10.24 6.25 64.00
PLSV 1.63 6.25 10.188
TOTAL 16.82
Unfactored Horizontal Loads and Overturning Moments
Item
H
(kip/ft)
Moment Arm About Toe (ft)
Moment About Toe (kip-ft/ft)
PLSH
1.36
8.25
11.22
PEH
4.76
5.5
26.18
Step 2: Determine the Appropriate Load Factors
In theory, structures could be evaluated for each of the limit states. However, depending on
the particular loading conditions and performance characteristics of a structure, only certain
controlling limit states need to be evaluated. For the example problem, each limit state will
be qualitatively assessed below relative to that limit state is applicable for the design problem:
• Strength I - Basic load combination related to the normal vehicular use of the bridge
without wind. (Applicable as a standard load case).
• Strength II - Load combination relating to the use of the bridge by Owner specified
special design vehicles and/or evaluation permit vehicles, without wind. (Not
10
applicable because special vehicle loading is not specified).
• Strength III - Load combination relating to the bridge exposed to wind velocity
exceeding 55 mph without live loads. (Not applicable because wall is not subjected to
other than standard wind loading).
• Strength IV - Load combination relating to very high dead load to live load force
effect ratios exceeding about 7.0 (e.g., for spans greater than 250 feet). (Applicable
because dead loads predominate).
• Strength V - Load combination relating to normal vehicular use of the bridge with
wind velocity of 55 mph (Not applicable because wind load not a design
consideration).
• Extreme Event I - Load combination including earthquake. (Not applicable because
problem does not include earthquake loading).
• Extreme Event II - Load combination relating to ice load or collision by vessels and
vehicles. (Not applicable because problem does not include ice or collision loading).
• Service I - Load combination relating to the normal operational use of the bridge with
55 mph wind. (Applicable for design loading).
• Service II - Load Combination intended to control yielding of steel structures and slip
of slip-critical connections due to vehicular live load. (Not applicable due to structure
type).
• Service III - Load combination relating only to tension in prestressed concrete
structures with the objective of crack control. (Not applicable due to structure type).
• Fatigue - Fatigue and fracture load combination relating to repetitive gravitational
vehicular live load and dynamic responses under a single design truck. (Not applicable
due to structure type).
Consequently, the applicable load factors and combinations the example problem are
summarized in the following tables. From the descriptions above, it is apparent that only the
Strength I, Strength IV and Service I Limit States apply to the retaining wall design. Strength
I-a and I-b represent the Strength I Limit State using maximum and minimum load factors,
respectively,
Load Factors
11
Group
DC
EV
LSv
LSH
EH
(active)
Probable USE
Strength I-a
0.90
1.00
1.75
1.75
1.50
BC/EC/SL
Strength I-b
1.25
1.35
1.75
1.75
1.50
BC (max. value)
Strength IV
1.50
1.35
-
-
1.50
BC (max. value)
Service I
1.00
1.00
1.00
1.00
1.00
Settlement
Notes: BC - Bearing Capacity; EC - Eccentricity; SL - Sliding
By inspection:
• Strength I-a (minimum vertical loads and maximum horizontal loads) will govern for
the case of sliding and eccentricity (overturning); and
• For the case of bearing capacity, maximum vertical loads will govern, and the factored
loads must be compared for Strength I-b and Strength IV.
Step 3: Calculate the Factored Loads and Factored Moments
Factored Vertical Loads
Group/
Item Units
W1
Kips/ft
W2
Kips/ft
W3
Kips/ft
PEV
Kips/ft
PLSV
Kips/ft
Total
Kips/ft V (Unf.) 2.25 0.56 2.14 10.24 1.63 16.82
Strength I-a 2.03 0.50 1.93 10.24 2.85 17.55
Strength I-b 2.81 0.70 2.68 13.82 2.85 22.86
Strength IV 3.38 0.84 3.21 13.82 - 21.25
Service I 2.25 0.56 2.14 10.24 1.63 16.82
Factored Horizontal Loads
12
Group/Item Units
PLSH Kips/ft
PEH Kips/ft
Total Kips/ft
H (Unf.) 1.36 4.76 6.12
Strength I-a 2.38 7.14 9.52
Strength I-b 2.38 7.14 9.52
Strength IV - 7.14 7.14
Service I 1.36 4.76 6.12
Factored Moments from Vertical Forces (Mv)
Group/Item
Units
W1
Kip-ft/ft
W2
Kip-ft/ft
W3
Kip-ft/ft
PEV
Kip-ft/ft
PLSV
Kip-ft/ft
Total
Kip-ft/ft Mv (Unf.) 5.63 1.03 10.17 64 10.19 91.01
Strength I-a 5.06 0.93 9.15 64 17.83 96.97
Strength I-b 7.04 1.29 12.71 86.40 17.83 125.25
Strength IV 8.45 1.55 15.25 86.40 --- 111.63
Service I 5.63 1.03 10.17 64 10.19 91.01
Factored Moments from Horizontal Forces (Mh)
Group/Item
Units
PLSH
Kip-ft/ft
PEH
Kip-ft/ft
Total
Kip-ft/ft Mh (Unf.) 11.23 26.18 37.41
Strength I-a 19.65 39.27 58.92
Strength I-b 19.65 39.27 58.92
Strength IV - 39.27 39.27
Service I 11.23 26.18 37.41
Step 4 Stability Analyses
13
a. Overturning or Eccentricity
e = B/2 -Xo B/2 =9.5 / 2 = 4.75’
Xo = (MVdl - MHtotal) / Vdl
emax = B/4 = 9.5 / 4 = 2.375’
Group/Item
Units
Vdl.
Kip /ft
Htotal
Kip /ft
MVdl
Kip-ft/ft
MHtotal
Kip-ft/ft
Xo Ft
e ft
emax Ft
Strength I-a
14.70 9.52 79.14 58.92 1.38 3.37 2.375
Strength I-b
20.01 9.52 107.43 58.92 2.42 2.33 2.375
Strength IV
21.25 7.14 111.63 39.27 3.41 1.34 2.375
Service I
15.19 6.12 80.82 37.41 2.86 1.89 2.375
For load case Strength 1-a, e is greater than emax , therefore, the design revised regard to
eccentricity.
b. Bearing Resistance
The adequacy for bearing resistance is based on a rectangular distribution of soil pressure
over the reduced effective area. For a rectangular distribution:
L’ = 1 foot (unit length of wall)
B’ = B – 2e
eB
VV 2−
Σ=σ (11.6.3.2-1)
MVtotal MHtotal X0 CDR
14
Group/Item Units
ΣV Kip /ft
Kips-ft/ft Kips-ft/ft e ft
B’ ft
σ ksf
Strength I-a 17.55 96.97 58.92 2.17 2.58 4.34 4.05 1.58
Strength I-b 22.86 125.25 58.92 2.90 1.85 5.80 3.94 1.62
Strength IV 21.25 111.63 39.27 3.41 1.34 6.81 3.12 2.05
Service I 16.82 91.01 37.41 3.19 1.56 6.31 2.64 2.42
Bearing resistance of soil
nbR qq φ= (10.6.3.1.1-1)
γγγγ wmwqqmfcmn CBNgCNDgcNq 5.0++= (10.6.3.1.2a-1)
c = 0, φ = 33o, γ2 = 110 pcf, and γ2’ = 58 pcf
Nq = 26.1, N γ = 35.2 [Table 10.6.3.1.2a-1]
Cwq= 1 (Dw = Df) [Table 10.6.3.1.2a-2]
Cw γ = 0.5
sq = 1 + (B/L)tanφf =1 (L>>B) [Table 10.6.3.1.2a-3]
sγ = 1 - 0.4(B/L) =1 [Table 10.6.3.1.2a-3]
dq = 1 [Table 10.6.3.1.2a-4]
For modest embedment of footing, it is usually to omission of the inclination factors.
Nqm = Nqsqdqiq=26.1x1x1x1=26.1
N γm = N γsγiγ=35.2x1x1=35.2
5.02.3534.4110.05.011.263110.0 ××××+×××=nq = 12.8 ksf/ft
qR = Фqn, use Ф= 0.5
qR = 0.5x12.8=6.4 ksf/ft
qR > σ in all loading cases , therefore, the footing design is adequate.
c. Check sliding
δφττ tan××= VR where: 8.0tantan
)9.0(
=
=+×=
τφ
φδ f
EVDCV
τR = 0.8 (17.55-2.85) (tan 33o) = 7.63 kips/ft < Htotal in load cases strength Ia & Ib.
15
Figure 8
The design has to be modified by either adding a key or lengthening the footing width.
3.0 Mechanically Stabilized Earth
(MSE) Walls Mechanically stabilized earth (MSE)
walls are composed of a reinforced soil
mass/volume, and a discrete modular
precast concrete facing which is vertical or
near vertical. The reinforced soil mass
consists of selected backfill. The tensile
reinforcements may be proprietary, and may
employ either metallic (i.e., strip or
mesh/grid-typed) or polymeric grid. Figure
8 shows the typical MSE wall configuration,
which is reproduced from the AASHTO
Specifications Figure 11.10.2-1.
MSE walls may be used where conventional gravity, cantilever, or counterfort
retaining walls are considered. They can accommodate larger total and differential
settlements than conventional gravity, cantilever, or counterfort retaining walls. However
there are some design limitations related to the rigid facing elements related to very high total
and differential settlement applications resulting in the need for two-phased construction
solutions.
MSE walls shall not be used in the following site conditions: (1) where utilities are to
be constructed within the reinforced soil mass; (2) where floodplain erosion or scour may
undermine the foundation support; and (3) in the case of steel reinforcements , where exposed
to surface or groundwater contaminated by acid mine drainage, industrial pollutants,
fertilizers or other aggressive environment that will cost long-term corrosions, (4) constructed
over existing storm sewer systems unless attention is given to including possible joint leakage
countermeasures.
16
Figure 9
The size of the reinforced soil mass is
based on consideration of external stability of
the system, geotechnical resistance, the
structural resistance within the reinforced soil
mass and panel units. The minimum soil
reinforcement length required for external
stability should be in accord with the FDOT’s
Structures Design Guidelines Section 3.13.2
in lieu of AASHTO Specifications 11.10.2.1.
The minimum required reinforcement length
for various conditions is as follows;
Walls in front of abutment on piling
L ≥ 8 feet and
L ≥ 0.7H1
Walls supporting abutment on spread footings
L ≥ 22 feet and
L ≥ 0.6(H1 + d) + 6.5 feet and
L ≥ 0.7H1
Where: H1 = mechanical height of wall and measured to the point
where potential failure plan intersects slope backfill
face, in feet.
L = reinforcement length required for external stability, in
feet.
d =fill height above wall
Reinforcement length shall be increased as required for surcharges and other external loads,
or for soft/weak foundation soils. Regardless of wall height, the minimum reinforcement
length shall be eight (8) feet.
3.1 Minimum Front Face Embedment
The AASHTO requirements for the minimum front face embedment are outlined in
Article 11.10.2.2. In addition to this FDOT’s Structures Design Guidelines also have the
following requirements;
17 Figure 10
1. Consider scour and bearing capacity when determining front face embedment depth;
2. Consult the District Drainage and Geotechnical Engineers to determine the elevation
of the top of leveling pad; and
3. The minimum front face embedment at a slope
should comply with Figure 9.
3.2 Other MSE Wall Design Issues Requirements according to the FDOT Structures
Design Guideline (Section 3.13.2)
A. Concrete Class and Cover
Table 2 (SDG Table 3.13.2.A)
Distance (D) from wall to a body of water with high chloride
content (≥ 2000 ppm) or any coal burning industrial facility,
pulpwood plant, fertilizer plant or similar industries
Concrete
Type
Concrete
Cover,
inches
D > 2,500 feet (low air contaminants) Class II 2
2,500 feet ≥ D ≥ 300 feet (moderate air contaminants) Class IV 2
D < 300 feet (extreme air contaminants) Class IV 3
B. Service Life
a. Permanent wall – 75 years
b. Abutment walls on spread
footings – 100 years
c. Temporary walls – for the length
of contract or three (3) years,
whichever is greater.
C. Acute Corner Walls (see Figure 10)
a. Define as two wall intersect at an angle
less than 70o ;
b. Design the acute nose wall as at-rest bin;
c. The bin should design with slip joints to
tolerate differential settlements from the
remainder of the structures;
d. The nose bin wall should be considered
18
Figure 11
as a facing element restrained by proper soil reinforcements; and
e. Design of facing connections, pullout and strength of reinforcing elements and
obstructions must conform to the general requirements of MSE wall design.
3.3 Stability Analyses
In LRFD analyses, it separates the
variability of the design components by applying
load and resistance factors to the load and material
capacity, respectively, similar to the conventional
retaining wall analyses.
3.3.1 Overall Stability (Article 11.10.4.3)
The provision of Article 11.6.2.3 shall apply.
Overall and compound stability of complex MSE
wall system shall also be investigated, especially
where the wall is located on sloping or soft ground where overall
stability may be inadequate. The long term strength of each backfill reinforcement layer
intersected by the failure surface should be considered as restoring forces in the limit
equilibrium slope stability analysis.
3.3.2 External Stability Analyses
The engineering properties of the retained soil mass used for MSE wall design
according to the FDOT’s Structures Design Guidelines, Section 3.13.12.G shall be
(1) Sand Backfill - friction angle, φ , of 30o, and unit weight, γ, of 105 pcf;
(2) Limerock backfill (in Dade and Monroe Counties), friction angle, φ , of 34o, and unit
weight, γ, of 115 pcf.
(3) Traffic surcharge present and locate within 0.5H1 of the back of the reinforced soil
volume should be included in the analysis.
(4) The Geotechnical Engineer of Record for the project is responsible for the external
stability analyses and design of the soil reinforcement length based on the analysis.
In performing the external stability analyses, engineers should assume the reinforced fill
19
volume as a rigid body similar to that of a gravity wall, which shall satisfy eccentricity,
sliding and bearing resistance criteria.
3.3.2.1 Overturning or Eccentricity
The provisions of Article 11.6.3.3 shall apply.
3.3.2.2 Sliding
The provisions of Article 10.6.3.3 shall apply. The coefficient of friction angle shall be
determined as;
• For discontinuous reinforcements, such as strips – the lesser of friction angle of either
reinforced backfill, φr, or foundation soil, φf;
• For continuous reinforcements, such as grids and sheets – the lesser of φr, φf, and ρ,
where ρ is the soil-reinforcement interface friction angle and a value of 2/3 φf may be
used. Other FDOT Structures Design Guideline Issues (Section 3.13.2)
3.3.2.3 Bearing Resistance
Effective width will be used to account for the effects of eccentricity as well as
inclination, B’=L-2e. Uniform bearing pressure shall be computed over the effective width.
nbR qq φ= (10.6.3.1.1-1)
γγγγ wmwqqmfcmn CBNgCNDgcNq 5.0++= (10.6.3.1.2a-1)
Where: Ncm = Ncscdcic
Nqm = Nqsqdqiq
N γm = N γsγiγ
Nc, Nq , N γ = bearing capacity factors
Cwq, Cw γ = correction factors to account for the location of the water table
g = gravitational acceleration (ft/sec2)
γ = total (moist) unit weight of soil above or below the bearing depth
Df = footing embedment depth (ft)
sq = 1 + (B/L)tanφf =1 (L>>B) [Table 10.6.3.1.2a-3]
sγ = 1 - 0.4(B/L) =1 [Table 10.6.3.1.2a-3]
dq = correction factor to account for the shearing resistance along the
failure surface passing through cohesionless material above the
bearing elevation [Table 10.6.3.1.2a-4]
20
D
Design Example 2a – MSE Wall External
Stability
Wall height
H = 20 feet, D = 2
Reinforced soil mass:
φr = 30o
γr = 105 pcf
Retained fill:
φf = 30o
γf = 105 pcf
Foundation soils: Figure 12 - MSE Example
φF = 30o
γF = 105 pcf (total/moist)
Traffic/live loads:
q = 250 psf/ft
ka = tan2(45 – φ/2) = 0.333
L = 14 feet (if the wall facing panels are thick, L will used and weight of panels should be
calculated as dead load, W)
Step 1 Calculate the unfactored loads
Vertical earth load of reinforced soil mass:
PEV = V1 = γr HL = 105 x 20 x 14 = 29,400 #/ft or 29.4 kips/ft
PLSV = q L = 250 x 14 = 3,500 #/ft or 3.5 kips/ft
PEH = F1 = 0.5 γf H2 ka = 0.5 x 105 x (20)2 x 0.3333 = 6,999 #/ft or 7 kips/ft
PLSH = F2 = qH ka = 250 x 20 x 0.3333 = 1,667 #/ft or 1.67 kips/ft
21
Unfactored Vertical Loads and Resisting Moments
Item
V
(kips/ft)
Moment Arm About Toe (ft)
Moment About Toe
(kip-ft/ft)
PEV 29.4 7.0 205.8
PLSV 3.5 7.0 24.5
TOTAL 32.9
Unfactored Horizontal Loads and Overturning Moments
Item
H
(kip/ft)
Moment Arm About Toe (ft)
Moment About Toe (kip-ft/ft)
PLSH
1.67
10.0
16.7
PEH
7
6.7
46.7
Step 2 Determine the appropriate load factors
Load Factors
Group
DC
EV
LSv
LSH
EH (active)
Probable USE
Strength I-a
0.90
1.00
1.75
1.75
1.50
BC/EC/SL
Strength I-b
1.25
1.35
1.75
1.75
1.50
BC (max. value)
Service I
1.00
1.00
1.00
1.00
1.00
Settlement
Notes: BC - Bearing Capacity; EC - Eccentricity; SL - Sliding
Step 3: Calculate the Factored Loads and Factored Moments
22
Factored Vertical Loads
Group/
Item Units
PEV
Kips/ft
PLSV
Kips/ft
Total
Kips/ft V (Unf.) 29.4 3.5 32.9
Strength I-a 29.4 6.13 35.5
Strength I-b 39.7 6.13 45.8
Service I 29.4 3.5 32.9
Factored Horizontal Loads
Group/Item
Units
PLSH
Kips/ft
PEH
Kips/ft
Total
Kips/ft H (Unf.) 1.67 7 8.7
Strength I-a 2.98 10.49 13.5
Strength I-b 2.98 10.49 13.5
Service I 1.67 7 8.7
Factored Moments from Vertical Forces (Mv)
Group/Item
Units
PEV
Kip-ft/ft
PLSV
Kip-ft/ft
Total
Kip-ft/ft Mv (Unf.)
205.8 24.5 230.3
Strength I-a
205.8 42.9 248.7
Strength I-b
277.8 42.9 320.7
Service I
205.8 24.5 230.3
23
Factored Moments from Horizontal Forces (Mh)
Group/Item
Units
PLSH
Kip-ft/ft
PEH
Kip-ft/ft
Total
Kip-ft/ft Mh (Unf.)
16.7 46.7 63.4
Strength I-a
29.23 69.93 99.16
Strength I-b
29.23 69.93 99.16
Service I
16.7 46.7 63.4
Step 4 Stability Analyses
a. Overturning or Eccentricity
e = B/2 -Xo B/2 =14 / 2 = 7’
Xo = (MV. Dead Load - MHtotal) / VDead Load
emax = B/4 = 14 / 4 = 3.5’
Group/Item
Units
VD.L.
Kip /ft
Htotal
Kip /ft
MV.D.L.
Kip-ft/ft
MHtotal
Kip-ft/ft
Xo ft
e ft
emax ft
CDR
Strength I-a 29.4 13.5 205.8 99.16 3.63 3.37 3.5 0.96
Strength I-b 39.7 13.5 277.8 99.16 4.49 2.51 3.5 0.72
Service I 29.4 8.7 205.8 63.4 4.84 2.16 3.5 0.62
In all loading cases, e < emax (or CDR < 1), thus the design is adequate with regard to
eccentricity.
b. Bearing resistance
B’ = B – 2e
eB
VV 2−
Σ=σ
Eccentricity to calculate reduced footing area:
e2 = B/2 -Xo B/2 =14 / 2 = 7’
Xo = (MV total - MHtotal) / Vtotal
24
emax = B/4 = 14 / 4 = 3.5’
Group/Item Units
Vtotal
Kip /ft
Htotal
Kip /ft
MVtotal
Kip-ft/ft
MHtotal
Kip-ft/ft
Xo ft
e2 ft
ema
x ft
B’ ft
qR σv
Ksf/ft CDR
Strength I-a 35.5 13.5 248.7 99.16 4.21 2.78 3.5 8.44 4.85 4.21 1.15
Strength I-b 45.8 13.5 320.7 99.16 4.83 2.16 3.5 9.68 5.25 4.73 1.11
Service I 32.9 8.7 230.3 63.4 5.07 1.92 3.5 10.1
6 5.41 3.24 1.67
CDR is > 1.0 in all cases; therefore design is adequate with regards to bearing resistance.
nbR qq φ= (10.6.3.1.1-1)
γγγγ wmwqqmfcmn CBNgCNDgcNq 5.0++= (10.6.3.1.2a-1)
c = 0, φ = 30o, γ2 = 105 pcf, and γ2’ = 50 pcf
Nq = 18.4, N γ = 22.4 [Table 10.6.3.1.2a-1]
Cwq= 1 (Dw = Df) [Table 10.6.3.1.2a-2]
Cw γ = 0.5
sq = 1 + (B/L)tanφf =1 (L>>B) [Table 10.6.3.1.2a-3]
sγ = 1 - 0.4(B/L) =1 [Table 10.6.3.1.2a-3]
dq = 1 [Table 10.6.3.1.2a-4]
For modest embedment of footing, it is usually to omission of the inclination factors.
Nqm = Nqsqdqiq=18.4x1x1x1=18.4
N γm = N γsγiγ=22.4 x1x1=22.4
5.04.22))18.2)(2(14(105.05.014.182105.0 ××−××+×××=nq = 9.55 ksf/ft
qR = Фqn, use Ф= 0.55 (FDOT recommended)
qR = 0.55x12.1=5.25 ksf/ft
qR > σv in all loading cases , therefore, the footing design is adequate.
c. Check sliding
δφττ tan××= VR where: 9.0
,tantan=
=
τφ
φδ
andf
τR = 0.9x29.4x tan 30o = 15.3 kips/ft > Htotal in all load cases
OK (CDR = 15.3/13.5) = 1.13
Table 3 [Table 11.5.6-1 Resistance Factors for Permanent Retaining Walls.]
25
WALL-TYPE AND CONDITION RESISTANCE FACTOR
Nongravity Cantilevered and Anchored Walls Bearing resistance of vertical elements Article 10.5 applies Passive resistance of vertical elements 1.00 Pullout resistance of anchors(1) • Cohesionless (granular) soils
• Cohesive soils • Rock
0.65 (1) 0.70 (1) 0.50 (1)
Pullout resistance of anchors(2) • Where proof test are conducted 1.0(2) Tensile resistance of anchor tendon
• Mild steel (e.g., ASTM A 615 bars) • High strength steel (e.g., ASTM A 722
bars)
0.90(3) 0.80(3)
Flexural capacity of vertical elements 0.90
Mechanically Stabilized Earth Walls Bearing resistance Article 10.5 applies Sliding Article 10.5 applies Tensile resistance of metallic reinforcement and connectors
Strip reinforcements(4) • Static loading • Combined static/earthquake loading
Grid reinforcements(4) (5) • Static loading • Combined static/earthquake loading
0.75 1.00
0.65 0.85
Tensile resistance of geosynthetic reinforcement and connectors
• Static loading • Combined static/earthquake loading
0.90 1.20
Pullout resistance of tensile reinforcement
• Static loading • Combined static/earthquake loading
0.90 1.20
Prefabricated Modular Walls
Bearing Article 10.5 Applies Sliding Article 10.5 Applies Passive resistance Article 10.5 Applies
(1) Apply to presumptive ultimate unit bond stresses for preliminary design only in Article C11.9.4.2. (2) Apply where proof test are conducted to a load of 1.0 or greater times the factored design load on the anchor. (3) Apply to maximum proof test load for the anchor. For mild steel apply resistance factor to Fy. For high-strength steel apply the resistance factor to guaranteed ultimate tensile strength. (4) Apply to gross cross-section less sacrificial area. For sections with holes, reduce gross area in accordance with Article 6.8.3 and apply to net section less sacrificial area. (5) Applies to grid reinforcements connected to a rigid facing element, e.g., a concrete panel or block. For grid reinforcements connected to a flexible facing mat or which are continuous with the facing mat, use the resistance factor for strip reinforcements.
Table 4
26
Resistance Factors for Geotechnical Resistance of Shallow Foundations at the Strength Limit
State. (Table 10.5.5.2.2-1, AASHTO 2006)
Method/Soil/Condition Resistance Factor
Theoretical method (Munfakh et al., 2001) in clay 0.5
Theoretical method (Munfakh et al., 2001) in sand, CPT 0.5
Theoretical method (Munfakh et al., 2001) in sand, SPT 0.45
Semi-emperical methods (Meyerhof,1957), all soils 0.45
Footing on rock 0.45
Bearing
Resistance φb
Plate load test 0.55
Precast concrete placed on sand 0.90
Cast-in-place concrete on sand 0.80
Cast-in-place or Precast concrete on clay 0.85 φτ
Soil on soil 0.90
Sliding
φep Passive earth pressure component of sliding resistance 0.50
3.3 Internal Stability
It is titled as ‘Safety against Structural Failure’ in AASHTO Section 11.10.6. In this
section, it requires engineers to evaluate the pullout resistance and rupture of soil
reinforcements.
3.4.1 Maximum reinforcement loads Simplified Method approach should be used to calculate the maximum reinforcement loads. )( HrVPH k σσγσ Δ+= (11.10.6.2.1-1)
where: γP = load factor for EV
kr = horizontal pressure coefficient
σv = vertical pressure due the reinforced soil mass and any surcharge loads above it.
ΔσH= horizontal stress at reinforcement level resulting from any applicable concentrated horizontal surcharge loads.
27
Figure 13 – Live Load and Dead Load
Surcharges for Internal Stability Analysis
for Horizontal Backfill
Figure 14 – Live Load and Dead Load
Surcharges for Internal Stability Analysis
for Slope Backfill
Figure 15 – Variation of Coefficient of Active
Lateral Earth Pressure Multiplier Factors with
Depth
3.4.2 Maximum tension force
The maximum tension force applied to the reinforcements per unit width of the wall
will be;
VH ST σ=max (11.10.6.2.1-2)
Where: σH = factored horizontal soil stress at the reinforcement
28
Sv = vertical spacing between reinforcements (< 2.7 ft.)
Figure 16 Potential Failure Surface for Internal Stability Design of MSE Walls
3.4.3 Pullout Resistance Length
The potential failure surface for inextensible and extensible wall system and the active
and resistant zones are shown in Figure 16. The pullout resistance length shall be determined
using the following equation;
cv
e CRFTLασφ *
max≥ (11.10.6.32-1)
Where: Le = length of reinforcement in resistant zone (ft.) >= 3’
29
Tamx = factored tension load in reinforcement (kips/ft)
ø = resistance factor for reinforcement pullout
F* = pullout friction factor, need not be reduced for properly placed and
compacted, saturated backfill (per FDOT Structures Design Guidelines
section 3.13).
α = scale effect correction
factor
σv = unfactored vertical stress
at reinforcement level
C = overall reinforcement
surface area geometry factor based on the gross perimeter of the
reinforcement, equal to 2 for strip, grid and sheet, i.e., two sides
Rc = reinforcement
coverage ratio
Figure 17 Default Values for Pullout
Friction Factor, F*
Soil reinforcements, in general,
should be designed laying perpendicular to the back of the wall facing panels. If this not
feasible because of obstructions, such as piles or other miscellaneous structures, the clearance
between the wall panels and obstructions shall be increased such that the soil reinforcement is
skewed no more than 15 degree off the perpendicular as per SDG. The pullout resistance
should be separately with the consideration of the skewed angles.
30
3.3.4 Reinforcement Strength
The reinforcement strength shall be checked at every level within the wall, both at the
boundary between the active and resistance zones (i.e., zone of maximum stress), and at the
connection of the reinforcement to the wall face.
• Zone of maximum stress
cal RTT φ=max
Where: Tmax = factor load to the reinforcement (kips/ft)
Ø = resistance factor for reinforcement tension
Tal = nominal long term reinforcement design strength (kips/ft)
For steel reinforcement; bFA
T ycal = , and
For geosynthetic reinforcement; RFTTal
max= RF=RFID x RFCR x RFD
Rc = reinforcement coverage ratio.
• Stress at the connection with the wall face
caco RTT φ=
Where: To = factor load at the reinforcement/facing connection (kips/ft)
Ø = resistance factor for reinforcement tension in connectors
Tac = nominal long term reinforcement/facing connection design strength (kips/ft)
Rc = reinforcement coverage ratio.
3.3.4 Design Life Consideration for Soil Reinforcement
In SDG section 3.13, Corrosion Rate for non-corrosive environments (low and moderate air
contaminants):
(A) Zinc (first 2 years) 0.59mils/year
(B) Zinc (subsequent years to depletion) 0.16mils/year
(C) Carbon Steel (after depletion of zinc) 0.48mils/year
(D) Carbon Steel (75 to 100 years) 0.28mils/year
Example 3 – MSE Wall Internal Stability
Wall height
31
PEV1
PVT
1.25
e
PaH
γp σv=quniform
PLS
L
H = 20 feet, Df = 2
Reinforced soil mass:
Assume 40mm x 5 mm ribbed steel strips will be used
φr = 30o
γr = 105 pcf
Traffic/live loads:
q = 250 psf/ft
ka = tan2(45 – φ/2) = 0.333
Step 1 – Calculate the factored horizontal force acting on the reinforcement
For this example assume steel strips reinforcements are used, at vertical spacing of Sv
= 2.5’ beginning at 1.25’ below the top of the wall. The vertical pressure at the reinforcement
is the vertical earth pressure uniformly distributed over an adjusted reinforcement length of
L’=L – e.
The load case that produces the largest uniform stress is Strength Ib, i.e. γEV = 1.35
and γEH = 1.5. Live load surcharge is not considered in the internal stability calculations for
vertical pressure. The factored vertical pressure as well as horizontal earth pressure is
calculated at each layer or
reinforcement strip determines the
eccentricity of each layer.
riEVEVi ZP γγ=
= 1.35*Zi*0.105
= 0.1418Zi
Zi= depth from the top of the
wall to the i reinforcement.
At level 1 reinforcement, where L=14’
)(1 ZiLP rEVET ×××= γγ = 1.35*14* 0.105*1.25= 2.48 kips/ft
21LPM ETV ×= =2.48*7 = 17.36 kips-ft/ft
)21()
21
31( 22
LSiariiaEHH qZKlshZZKM ××××+××××××= γγγ
= (1.5*0.333*0.333*1.25*0.5*1.252*0.105) +(1.75*0.333*0.5*1.252*0.25]
= 0.131 K-ft/ft
D
0.3
0 5
32
EV
HV
PMMLe −
−=2
= 7 - 48.2
131.036.17 − = 0.05’
eL
Pq TV
uniformVP 2−==σγ = 0.178
Layer,i Zi (ft) PVT, (kip/ft) MV(kips-ft/ft) Mh (kip-ft/ft) e, ft quniform, (ksf) 1 1.25 2.48 17.36 0.13 0.05 0.178 2 3.75 7.44 52.08 1.49 0.20 0.547 3 6.25 12.4 86.8 4.98 0.4 0.940 4 8.75 17.35 121.55 11.43 0.66 1.369 5 11.25 22.31 156.28 21.66 0.97 1.851 6 13.75 27.26 191.01 36.49 1.34 2.409 7 16.25 32.25 225.74 56.74 1.76 3.077 8 18.75 37.21 260.47 83.22 2.24 3.906
Since the wall is only 20 feet height, the multipliers of lateral coefficient of earth pressure
vary linearly from 1.7 at the top of the wall to 1.2 at 20 feet below the top of the wall.
Therefore, Kr = [1.7 - Zi * (1.7 – 1.2)/20]*Ka = (1.7 - 0.025Zi)*Ka
σH=γp σvKr , and
Tmax = σHsvi
Summary of Factored Horizontal Load at Reinforcements
Layer Zi (ft) γp σv (ksf) Kr σH, (ksf) svi, ft Tmax (kips)
1 1.25 0.178 0.556 0.099 2.5 0.25 2 3.75 0.547 0.535 0.293 2.5 0.73 3 6.25 0.940 0.514 0.483 2.5 1.21 4 8.75 1.369 0.493 0.675 2.5 1.69 5 11.25 1.851 0.472 0.875 2.5 2.19 6 13.75 2.409 0.452 1.088 2.5 2.72 7 16.25 3.077 0.431 1.326 2.5 3.31 8 18.75 3.906 0.410 1.601 2.5 4.00
Step 2 Check pullout resistances
A
TCRF
TLcv
emax
*max =≥
ασφ= length of reinforcement in resistant zone (ft.) >= 3’
Tmax = factored tension load in reinforcement (kips/ft)
ø = 0.9 from Table 2 [AASHTO Table 11.5.6-1]
F* = varies from 1.2 + log Cu (assume Cu = 4) at the top to
tan φr= tan 30 = 0.577 at 20 feet below the top for ribbed steel
strip, i.e. 1.8 to 0.577 or at each reinforcement level F*i = 1.8 –
33
0.06*Zi
α = 1 for steel reinforcements
σv = unfactored vertical stress at reinforcement level
C = 2 (for strip)
Rc = b/Sh = (40/25.4)/ 40 = 0.03937 [ Sh = 40” is predetermined]
Layer Zi (ft) σv (ksf)/ft F* A Tmax /ft(kips) Le, ft 1 1.25 0.131 1.724 0.016 0.25 15.4 2 3.75 0.394 1.575 0.045 0.73 16.2 3 6.25 0. 656 1.425 0.07 1.21 17.3 4 8.75 0.919 1.275 0.091 1.69 18.6 5 11.25 1.181 1.125 0.108 2.19 20.3 6 13.75 1.444 0.975 0.121 2.72 22.4 7 16.25 1.706 0.825 0.13 3.31 25.4 8 18.75 1.969 0.675 0.134 4.00 29.9
A = ø F* α σv C Rc
L = La + Le
Active zone, La
Z =0 - 0.5 H or 0’ - 10’ La= 0.3 H = 6’
Z = 0.5H – H or 10’ – 20’ La varies from 6’ - 0’
Layer Zi (ft) Le, ft La, ft L, ft 1 1.25 15.4 6 21.4 2 3.75 16.2 6 22.2 3 6.25 17.3 6 23.3 4 8.75 18.6 6 24.6 5 11.25 20.3 5.25 25.6 6 13.75 22.4 3.75 26.2 7 16.25 25.4 2.25 27.7 8 18.75 29.9 0.75 30.7
The longest L is at Z8,
L = 29.9 + .75 = 30.7’ say 31’
Thus internal stability controls the design of reinforcement length at all layers. Adjust
the lengths to whole number (rounding up), which resulting in longer Le .
34
Layer Zi (ft) A Actual Le, ft Actual
Resistance*, (kips)/ft
Tmax (kips)/ft
1 1.25 0.016 16.00 0.26 0.25 2 3.75 0.045 17.00 0.77 0.73 3 6.25 0.07 18.00 1.26 1.21 4 8.75 0.091 19.00 1.73 1.69 5 11.25 0.108 20.75 2.24 2.19 6 13.75 0.121 23.25 2.81 2.72 7 16.25 0.13 25.75 3.35 3.31 8 18.75 0.134 30.25 4.05 4.00
* Resistance = (Actual Le, ft) x A, where A= cvCRF ασφ *
Since the actual resistance at each reinforcement level is larger than that of the
maximum tensile stress required for pullout, therefore the design is adequate.
Step 3 – Check Reinforcement Strength
The engineer should also check the tensile stress at the connection and the
design life based on the corrosion rate. Assume they are OK.
4.0 Nongravity Cantilever Walls (AASHTO 3.11.5.6). 4.1 Granular Soils or Rock
For permanent walls the simplified lateral earth pressure distributions as shown in
Figures 17 through 19 may be used. For temporary walls with discrete vertical elements,
Figures 17 & 18 may be used to determine passive resistance and Figures 20 & 21 may be
used to determine the active earth pressure due to the retained soil.
35
Figure 17 Figure 18
Figure 19 Figure 20
36
Figure 21
Figure 23
Figure 22
37
4.2 Cohesive Soils
If walls will support or are supported by cohesive soils for temporary applications,
wall must be designed based on total stress methods of analysis and undrained shear strength
parameters. Figures 20 – 23 may be used with the restrictions of;
• Ratio of total overburden pressure to undrained shear strength, Ns ( ;u
ss S
HN γ= γs =
total unit weight; H = total excavation depth, and Su = average undrained shear
strength of soil. AASHTO 3.11.5.7.2)
• The active earth pressure shall not be less than 0.25 times the effective overburden
pressure at any depth, or 0.035 ksf/ft of wall height, whichever is greater.
A portion of negative loading at top of wall due to cohesion is ignored and hydrostatic
pressure in a tension crack should be considered.
4.3 Nongravity Cantilevered Walls Design
Article 11.8 outlines the design and analysis. In C11.8.4.1 particularly it states the
procedures to determine the value x using factored loads as well as resistance factor. However,
at the end, the design embedment length shall also be extended by 20%, which is the same as
the design method of the old. With the combination of maximum load factor, the reduced
resistance factor and the 20% increase, we found that too much conservatism is built into the
design. Thus, we think the old method should be used because there aren’t many case
histories of failure.
Appendix
Changes in FDOT’s Structures Design Guidelines and Plans Preparation Manual are in the
Appendix.
i. Attachment A - Structural Manual Vol. 1; Structures Design Guidelines – Section 3.13
ii. Attachment C – Plans Preparation Manual – Vol. 1; Section 30.2.3, 30.2.4 and
Flowchart and Table of FDOT Wall Types