Upload
secsever-everedy
View
345
Download
12
Embed Size (px)
AllanBlock
TM
Mortarless Technology
Allan BlockEngineering Manual
Water Application
Roadway Application
Roadway Application
Slope Below Wall
The information and product applicationsillustrated in this manual have been carefullycompiled by the ALLAN BLOCK CORPORA-TION and to the best of our knowledge accu-rately represent ALLAN BLOCK product use.Final determination of the suitability of anyinformation or material for the use contem-plated and its manner of use is the soleresponsibility of the user.
Earth AnchorApplication:2 Terraces,Slope AboveWall
FORWARD
This manual presents the techniques that we use in our engineering practice todesign retaining walls. It is not intended as a textbook of soil mechanics orgeotechnical engineering. The methods we use have evolved over the course of nineyears and continue to evolve as our knowledge and experience grows. If any of theusers of this manual want to offer suggestions about ways to improve our design, wewould be very glad to hear them. The intended users of this manual are practicingengineers. When writing it, we assumed that the reader would already be familiarwith the basic principles of statics and soil mechanics. We encourage others tocontact a qualified engineer for help with the design of geogrid reinforced retainingwalls. The example problems in this manual are based on walls constructed with AllanBlock Retaining Wall System’s AB Stones.
AB Stones provide a setback of twelve degrees from vertical. We believe thata twelve degree setback maximizes the leverage achieved by a battered wall,while providing a finished retaining wall that fulfills the goal of more useable flatland. Allan Block also has developed products with three and six degree setbacks.The equations that follow can be used for each product by selecting theappropriate � angle. (� = 90 - Wall Batter)
AB Stones
AB Classic
AB Rocks
AB Three
AB Lite Stones
AB Lite Rocks
3º� 6º� 12º�
TABLE OF CONTENTS
Chapter One - Concepts & Definitions• Soil Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1• Retaining Wall Failure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2
Sliding Failure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2Overturning Failure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2Effects of Water on Wall Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2
• Types of Retaining Walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3• Forces Acting on Retaining Walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4• Soil States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4• Active and Passive Zones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5• Pressure Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5• Active Force on the Wall . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6• Two-Dimensional Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7• Calculating the Effective Unit Weight of the Wall Facing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7• Safety Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .8
Chapter Two - Basic Wall Design• Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .9• Simple Gravity Walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .9
Sliding Failure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .9Overturning Failure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .11
• Tieback Walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .12The Geogrid as a Tieback . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .14Earth Anchors as a Tieback . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .16
• Coherent Gravity Walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .17Length of Geogrid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .17External Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .18Internal Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .19Bearing Pressure on the Underlying Soil . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .22Attachment of the Geogrid to the Wall Facing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .25Mechanical Connection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .25
Chapter Three - Surcharges• Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .30• Surcharges on Simple Gravity Walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .31• Surcharges on Tieback Walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .33• Surcharges on Coherent Gravity Walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .35
External Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .35Internal Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .39
• Tiered Walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .42
Chapter Four - Sloped Backfill• Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43• Simple Gravity Walls with Sloped Backfill . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43• Tieback Walls with Sloped Backfill . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44• Coherent Gravity Walls with Sloped Backfill. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
External Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47Internal Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
TABLE OF CONTENTS
Chapter Five - Seismic Analysis• Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51• Pressure Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51• Dynamic Earth Force on the Wall . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54• Safety Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55• Simple Gravity Wall with Seismic Influence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
Safety Factor Against Sliding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57Safety Factor Against Overturning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
• Coherent Gravity Wall with Seismic Influence. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59Safety Factor Against Sliding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60Safety Factor Against Overturning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61Factor of Safety Geogrid Tensile Overstress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62Geogrid / Block Connection Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63Geogrid Pullout from the Soil . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63Top of the Wall Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
LIST OF FIGURES
Chapter One - Concepts & DefinitionsFigure 1-1 Forces Acting on Retaining Walls ..............................................................................4Figure 1-2 Relative Pressures for the Three Soil States..................................................................4Figure 1-3 Definition of Terms for Coulomb's Equation ................................................................5Figure 1-4 Effect of Wall Friction on Active Force .......................................................................6Figure 1-5 Active Pressure Distribution on a Retaining Wall .........................................................6
Chapter Two - Basic Wall DesignFigure 2-1 Schematic Diagram of Simple Gravity Retaining Wall..................................................10Figure 2-2 Diagram of Retaining Wall for Tieback Analysis .........................................................12Figure 2-3 Freebody Diagram of Retaining Wall for Tieback Analysis ...........................................13Figure 2-4 Coherent Gravity Wall for Example 2-3 .....................................................................17Figure 2-5 Typical Geogrid Reinforcement Embedment for 12 Degree AB System...........................17Figure 2-6 Typical Geogrid Reinforcement Embedment for 6 Degree AB System.............................17Figure 2-7 Typical Geogrid Reinforcement Embedment for 3 Degree AB System.............................17Figure 2-8 Freebody Diagram of a Coherent Gravity Wall ..........................................................19Figure 2-9 Calculating the Spacing of Geogrid Layers ................................................................20Figure 2-10 Freebody Diagram for Bearing Pressure Analysis ........................................................22Figure 2-11 Friction Attachment of Geogrid to Blocks....................................................................27
Chapter Three - SurchargesFigure 3-1 Effect of uniform Surcharge on a Retaining Wall .........................................................30Figure 3-2 Simple Gravity Retaining Wall with Surcharge............................................................31Figure 3-3 Freebody Diagram of a Simple Gravity Retaining Wall with Surcharge .........................32Figure 3-4a Tieback Retaining Wall with Surcharge......................................................................33Figure 3-4b Freebody Diagram of Tieback Wall with Surcharge.....................................................34Figure 3-5 Locations of Surcharge on Coherent Gravity Walls......................................................35Figure 3-6 Coherent Gravity Wall with Surcharge.......................................................................39Figure 3-7 Pressure Distributions Due to the Soil Weight and Surcharge ........................................39Figure 3-8 Retaining Wall with Three Tiers .................................................................................42Figure 3-9 Average Bearing Stress of Top Wall Applied as Surcharge to Second Wall ...................42
Chapter Four - Sloped BackfillFigure 4-1 Tieback Wall with Sloped Backfill..............................................................................44Figure 4-2 Coherent Gravity Wall with Sloped Backfill ................................................................47Figure 4-3 Failure Surface in a Coherent Gravity Wall ................................................................49Figure 4-4 Effect of Sloped Backfill on Spacing of Geogrid..........................................................49
Chapter Five - Seismic AnalysisFigure 5-1 Static Component of Active Pressure Distribution .........................................................54Figure 5-2 Dynamic Increment Component of the Active Pressure Distribution.................................55Figure 5-3 Dynamic Earth Force Pressure Distribution ..................................................................55Figure 5-4 Free Body Diagram of Simple Gravity Wall Under Seimic Influence ..............................56Figure 5-5 Free Body Diagram of Coherent Gravity Wall Under Seismic Influence..........................59
CHAPTER ONEConcepts & Definitions
Soil Characteristics
Soil can be described in many different ways. One way to describe it is by the average size of the particles thatmake up a soil sample. Sandy soil consists of relatively large particles, while clay soil consists mainly of smaller particles.Another way to describe soil is by the tendency of the particles to stick together -- a property called cohesion. Sand,such as is found at the beach, has very low cohesion. Even when it is wet, you can pick up a handful of sand and itwill pour out of your hand as individual particles. Clay, on the other hand, is much more cohesive than sand. A wetclay soil can be molded into a ball or rolled into a thread that resists being pulled apart.
Still another way to describe a soil is by its natural tendency to resistmovement. This property can be expressed by a number known as the coefficientof internal friction, or simply, the friction angle (PHI �). If you take a dry soilsample and pour it out onto a flat surface, it will form a cone-shaped pile. Theangle formed by the base of the cone and its sides is known as the angle ofrepose. The angle of repose of a soil is always smaller than the friction angle forthe same soil. However, the difference between the two angles is small and forthe design of retaining walls the angle of repose can be used to approximate thefriction angle. The larger the friction angle the steeper the stable slope that canbe formed using that soil.
Soil that consists mainly of sand has a larger friction angle than soil composedmainly of clay. This is due to the fact that sand particles are roughly spherical withirregular surfaces, while clay particles are flat and smooth. When subjected toexternal pressure, the clay particles tend to slide past one another. The surfaceirregularities of the sand particles tend to interlock and resist movement.
Clay soil has some characteristics that make it undesirable for use as backfillfor a retaining wall. First of all, clay soil is not readily permeable and retains thewater that filters into it. The added weight of the retained water increases theforce on the retaining wall. Secondly, once the clay becomes saturated, its cohesion decreases almost to zero. The shearstrength of the soil is the sum of the frictional resistance to movement and the cohesion of the soil. Once the cohesion is lostdue to soil saturation, the full force of the weight of water and most of the weight of the soil is applied to the wall. For thesereasons, clay soil is not a good choice for retaining wall backfill.
The preferred soil for backfill behind retaining walls is soil that contains a high percentage of sand and gravel. Sucha soil is referred to as a granular soil and has a friction angle of approximately 32º to 36º, depending on the degree
of compaction of the soil.The main reason forpreferring a granular soil forbackfill is that it allows waterto pass through it morereadily than a nongranular,or clayey soil does. Also, theshear strength of a granularsoil doesn't vary withmoisture content andtherefore its shear strength ismore predictable.
SANDLarge, spherical,angular surfaces
CLAYSmall, flat,
smooth surfaces
1
Typical Soil Properties
Soil Cohesion Cohesion Soil FrictionGroups Compacted Saturated Angle
Clean Gravel-Sand Mix
Sand-Silt Clay Mix
Inorganic Clays
0
1050 PSF(50 KPA)
1800 PSF(86 KPA)
0
300 PSF(14 KPA)
270 PSF(13 KPA)
36º
32º
27º
Retaining Wall FailureThere are two primary modes of retaining wall failure. The wall can fail by sliding too far forward and encroaching onthe space it was designed to protect. It can also fail by overturning -- by rotating forward onto its face.
Sliding FailureSliding failure is evident when the wallmoves forward, and occurs when thehorizontal forces tending to cause slidingare greater than the horizontal forcesresisting sliding. Generally, this willoccur when either the driving force isunderestimated or the resisting force isoverestimated. Underestimating thedriving force is the most common mistakeand usually results from: 1) neglectingsurcharge forces from other walls, 2)designing for level backfill when thebackfill is in fact sloped, 3) usingcohesive soils for backfill.
Overturning FailureOverturning failure is evident when the wall rotates about its bottom front edge (also called the toe of the wall). Thisoccurs when the sum of the moments tending to cause overturning is greater than the sum of the moments resistingoverturning. As with sliding failures, overturning failures usually result from underestimating the driving forces.
Effects of Water on Wall StabilityPerhaps the single most important factor in wall failure is water. Water contributes to wall failure in several differentways. If the soil used for backfill is not a free-draining granular soil, it will retain most of the water that filters into it. Theforce on a wall due to water can be greater than the force due to soil.
As the moisture content ofthe soil increases, the unitweight of the soil increasesalso, resulting in greaterforce on the wall. When thesoil becomes saturated, theunit weight of the soil isreduced because of thebuoyant force of the wateron the soil particles.However, the water exertshydrostatic pressure on thewall. Therefore, the totalforce on the wall is greaterthan it is for unsaturated soil,because the force on the
wall is the sum of the force exerted by the soil and the force exerted by thewater. The problem is even greater if the soil contains a high percentage ofclay. Saturated, high-clay-content soil loses its cohesion and the force on thewall increases. Good drainage is an essential for proper wall design.
OverturningSliding
2
Drainage
AB Geogrid Wall Below Grade Section
AB Geogrid Wall Above Grade Section
Some clay soils exhibit the characteristic of expanding when wet. Thisexpansion, coupled with contraction when the soil dries, can work to weakenthe wall and cause failure.
Another way in which water contributes to wall failure is by the action of the freeze-thaw cycle. Water trapped in the soil expands when it freezes causing increasedpressure on the wall. Water in contact with the wall itself can cause failure of thestructural materials. The freeze-thaw cycle is the basic mechanism by which rocksare turned into soil. The same thing can happen to a wall in contact with waterduring the winter months.
Several things can be done to reduce the likelihood of wallfailure due to water. First, use a free-draining granularmaterial for the back fill. Second, create a drain field in andaround the block cores and 6-12 inches (15-30 cm) deepbehind the wall using a material with large individual particles,such crushed limestone. Third, install a drain pipe at the bottomrear of the base and provide outlets as needed. Finally, directwater away from the top and bottom of the wall using swalesas required. All these measures will ensure that excess water isremoved from behind the wall before it can build up or freezeand cause damage.
Types of Retaining Walls•Gravity
A wall that relies solely on its weight to prevent failure is called a gravity wall. For a gravity wall, the primary factoraffecting the wall's resistance to overturning is the horizontal distance from the toe of the wall to the center of gravity ofthe wall. The greater this distance is, the less likely it is that the wall will overturn. For example, a wall four feet highand two feet thick will have a lower resistance to overturning than a wall two feet high and four feet thick, even if theweights are equal. Battering the retaining wall (sloping it into the backfill) also enhances stability bymoving the center of gravity back from the toe of the wall.
•Geogrid
Studies have shown that retaining walls reinforced with several layers of geogrid act as giant gravity walls. “Geogridreinforced soil masses create the same effect as having an extremely thick wall with the center ofgravity located well back from the toe of the wall.” For this reason, reinforced soil walls are more likely tofail by sliding than by overturning.
•Anchor
Tieback walls rely on mechanical devices embedded in the backfill to provide the force necessary to resist sliding andoverturning. Battering a tieback wall will shift its center of gravity and enhance its stability.
AB Gravity WallTypical Section
AB Geogrid WallTypical Section
AB Earth AnchorTypical Section
3
Forces Acting on Retaining WallsThe forces that act on a retaining wall can be divided into two groups:
• Those forces that tend to cause the wall to move
• Those forces that oppose movement of the wall (see Figure 1-1)
Included in the first group are the weight of the soil behind theretaining wall and any surcharge on the backfill. Typical surchargesinclude driveways, roads, buildings, and other retaining walls. Forcesthat oppose movement of the wall include the frictional resistance tosliding due to the weight of the wall, the passive resistance of the soilin front of the wall, and the force provided by mechanical restrainingdevices. When the forces that tend to cause the wall to move becomegreater than the forces resisting movement, the wall will move.
Soil States
The soil behind a retaining wall exists in one of three states:
1) the active state, 2) the passive state, 3) the at-rest state.
When a wall is built and soil is placed behind it and compacted, the soil is in the at-rest state. If the pressure on thewall due to the soil is too great, the wall will move forward. As the wall moves forward, the soil settles into a newequilibrium condition called the active state. The pressure on the wall due to the soil is lower in the active state than itis in the at-rest state (see Figure 1-2).
The passive state is achieved when a wall is pushed back into the soil.This could occur by building the retaining wall, placing andcompacting the soil, and then somehow forcing the retaining wall tomove into the backfill. Usually, the passive state occurs at the toe ofthe wall when the wall moves forward. The movement of the wallcauses a horizontal pressure on the soil in front of the wall. Thispassive resistance of the soil in front of the wall helps keep the wallfrom sliding. However, the magnitude of the passive resistance at thetoe of the wall is so low that it is usually neglected in determining thestability of the wall.
The occurrence of the passive state behind a retaining wall isextremely rare and it will most likely never be encountered behind anAllan Block wall. The at-rest condition occurs whenever a retainingwall is built. Some designers may prefer to take a conservativeapproach and design for the higher at-rest pressure rather than theactive pressure. However, this is not necessary since the amount ofwall movement required to cause the pressure to decrease from the at-rest level to the active level is very small. Studies of soil pressure onretaining walls have shown that the top of a retaining wall needs tomove only 0.001 times the height of the wall in order for the pressureto drop to the active value.
There are some applications where the wall cannot be allowed to move. These include bridge abutments and walls thatare rigidly connected to buildings. In cases such as these, the design should be based on the higher at-rest pressure;otherwise, the lower active pressure can be used. Designing on the basis of the active pressure will reduce the cost ofthe wall and give a more accurate model of the actual behavior of most retaining walls.
4
Figure 1-1. Forces Actingon the Retaining Walls
Figure 1-2. Relative Pressuresfor the Three Soil States
Active and Passive ZonesWhen the wall moves forward, a certain portion of the soil behind the wall movesforward also. The area containing the soil that moves with the wall is referred to asthe active zone. The area behind the active zone is called the passive zone. Theline that divides the two zones is called the failure plane. The failure plane can beestimated by drawing a line that begins at the bottom rear edge of the wall andextends into the backfill at an angle of 45º plus one-half the friction angle of the soil(45º � �/2) and intersects a vertical line three 0.3 the height of the wall. (H x 0.3)
The active zone for a geogrid reinforced soil mass includes the entire reinforcementzone and the area included in the theoretical failure surface. The origin of thetheoretical failure surface is located at the back bottom of the reinforced zone.
Pressure CoefficientsThe horizontal stress (��h) on a retaining wall due to the retained soil is directly proportional to the vertical stress (��v) onthe soil at the same depth. The ratio of the two stresses is a constant called the pressure coefficient:
� = (��h)(��v)
The pressure coefficient for the at-rest state can be calculated using theformula:
Ko = 1 — sin (�)
where: � is the friction angle of the soil.
The active pressure coefficient can be calculated using an equation thatwas derived by Coulomb in 1776. This equation takes into account theslope of the backfill, the batter of the retaining wall, and the effects offriction between the retained soil and the surface of the retaining wall.Figure 1-3 illustrates the various terms of Coulomb's equation.
The Coulomb equation for the active force on a retaining wall is:
Fa = (0.5) (�) (Ka) (H)2
where:
Fa = the active force on the retaining wall; it is the resultant forceof the active pressure on the retaining wall
H = distance from the bottom of the wall to the top of the wall
Ka = the active pressure coefficient
� = angle between the horizontal and the sloped back face of the wall
i = slope of the top of the retained soil
�w = angle between a line perpendicular to the wall face and the line of action of the active force
Ka = csc (�) sin (� - �) 2
sin (� + �w) + sin (� + �w) sin (� — i)sin (� — i)
[ ]
5
Figure 1-3. Definition forActive Force for Coulomb’s Equation
Theoretical Failure Plan
As the wall moves forward slightly, the soil enters the active stateby moving forward and downward. At the interface of the soiland the wall, this downward movement of the wall is resisted bythe friction between the soil and the wall. Figure 1-4 shows theresultant active force on a retaining wall and the effect of wallfriction on the direction of the force.
The magnitude of �w varies depending on the unit weight of thebackfill. For a loose backfill, �w is approximately equal to �.For a dense back-fill, however, �w < �. Since retaining wallbackfill is thoroughly compacted, the design method in thismanual assumes that �w = (0.66) �.
Active Force on the WallOnce the active pressure coefficient has been determined, the active force on the wall can be determined. Figure 1-5shows the active pressure distribution on a retaining wall. The active pressure distribution is triangular, which reflects thefact that soil pressure increases linearly with soil depth. The vertical pressure at any depth is given by:
Pv = (�) (d)Where:
� = the unit weight of the soil
d = the depth from the top of the retained soil mass.
As discussed previously, the horizontal pressure (Ph) is related to the vertical pressure (Pv) by the active pressurecoefficient:
Ka = (Ph)(Pv)
Ph = (Ka) (Pv) = (Ka) (�) (d)Since Ka and � are constants, the horizontal pressure increases linearly as the depth increases and the resulting pressuredistribution is triangular. The magnitude of the resultant force of a triangular pressure distribution is equal to the area ofthe triangle. The pressure at the base of the triangle is given by:
Ph = (Ka) (�) (H)The magnitude of the active force is:
Fa = (area of the triangle)
= (0.5) (base) (height)
= (0.5) (�) (Ka) (H) (H)
= (0.5) (�) (Ka) (H)2
The resultant force acts at a point above the base equal to one-third of theheight of the triangle. Adding a surcharge or slope above the wall hasthe effect of developing a rectangular pressure distribution. The resultantforce of a rectangular pressure distribution acts at a point above the baseequal to one-half of the height of the rectangle.
6
Figure 1-4. Effect of WallFriction on Active Force
Figure 1-5. Active PressureDistribution on a Retaining Wall
Two-Dimensional AnalysisA retaining wall is a three-dimensional object. It has height, length, and depth. In order to simplify the analysis, thelength of the wall is taken to be one foot (or one meter) and the wall is analyzed as a two-dimensional system. Becauseof this, the units for forces will always be pounds per foot (lb/ft) (kilograms per meter (kg/m)), and the units for momentswill be foot-pounds per foot (ft-lb/ft) (newton-meters per meter (N-m/m)).
Calculating the Effective Unit Weight of the Wall FacingThe effective unit weight of the wall facing is often needed for wall design. Allan Block’s unit weight is the sum of theblocks plus the granular in fill material and is calculated below. Concrete usually weighs more than soil. A typical unitweight for concrete is 135 lb/ft3 (2,163 kg/m3) while a typical unit weight for soil is 120 lb/ft3 (1,923 kg/m3).Depending on the size of the wall, this difference may be significant, and the design engineer should know how tocalculate the weight of the wall facing.
The weight of a AB Stone unit is approximately 72 lb (33 kg). The unit weight of the concrete is 135 lb/ft3 (2,163 kg/m3).From these values, the volume of concrete for each Allan Block unit can be calculated:
Vc = (72 lb) = 0.53 ft3 = (33 kg) = 0.015 m3
(135 lb/ft3) (2,163 kg/m3)
The total volume occupied by each standard Allan Block unit, including the voids, is:
Vt = (1.5 ft) (0.635 ft) (0.97 ft) = (0.46 m) (0.19 m) (0.3 m)= 0.92 ft3 = 0.026 m3
Therefore, the volume of the voids is:Vv = Vt Vc
= 0.92 ft3 — 0.53 ft3 = 0.026 m3 — 0.015 m3
= 0.39 ft3 = 0.011 m3
The unit weight of the wall facing can now be calculated. Assuming that the voids arefilled with soil with a unit weight of 120 lb/ft3 (1,923 kg/m3), the unit weight of thewall facing is:
� = (weight of concrete) + (weight of soil)(volume of block)
�w = (0.53 ft3) (135 lb/ft3) + (0.39 ft3) (120 lb/ ft3) = 130 lb/ft3
(0.92 ft3)
= (0.015 m3) (2,163 kg/m3) + (0.011 m3) (1,923 kg/m3) = 2,061 kg/m3
0.026 m3
Once the unit weight of the wall facing is known, it is a simple matter to calculate the weight per linear foot of wall:
Ww = (unit weight of wall) (volume of wall)= (unit weight of wall) (wall height) (facing depth)
For a wall 6 feet (1.83 m) tall with a facing depth of 0.97 foot (0.3 m), the weight of the facing would be:
Ww = (130 lb/ft3) (6 ft) (0.97 ft) = (2,061 kg/m3) (1.83 m) (0.3 m)
= 757 lb/ft = 1,119 kg/m3
In general, the weight of the facing is:
Ww = (125 lb/ft2) (wall height) = (610 kg/m2) (wall height)
7
Total Unit Weight
Safety FactorsThe safety factors used in this design manual conform to the guidelines of the American Association of State Highwayand Transportation Officials (AASHTO). In the draft version of Guidelines for the Design of Mechanically Stabilized EarthWalls, they recommend using the following safety factors:
Sliding > 1.5Overturning > 2.0
These are the same values recommended by most governmental agencies. However, you should check your state andlocal building codes to make sure these safety factors are sufficient.
8
CHAPTER TWOBasic Wall Design Techniques
IntroductionOne way to classify retaining walls is by the type of reinforcement the walls require. If a wall is stable withoutreinforcement, it is referred to as a simple gravity wall. When the forces behind a wall are greater than a simple gravitysystem can provide, a tieback wall can often be built using a single layer of geogrid or anchors to tie the wall to the soil.When the forces behind a wall are greater than a tieback system can provide, a coherent gravity wall can be built byusing two or more layers of geogrid to stabilize the soil mass.
Simple Gravity WallsSimple gravity walls rely on the weight of the wall to counteract the force of the retainedsoil. Figure 2-1 is a diagram showing the forces acting on a simple gravity wall. Twomodes of failure must be analyzed, sliding and overturning.
Sliding FailureA simple gravity wall will not fail in sliding if the force resisting sliding, Fr, is greater thanor equal to the force causing sliding, Fh. The force resisting sliding is the frictionalresistance at the wall base. The minimum safety factor for sliding failure is 1.5.Therefore, Fr, must be greater than or equal to (1.5) Fh. The following example illustratesthe procedure for analyzing stability in sliding.
Example 2-1:Given:� = 30º Ka = 0.2197i = 0º H = 3.44 ft (1.05 m)� = 78º � = 120 lb/ft3 (1,923 kg/m3)�w = (0.666) (�) = 20ºUnit weight of wall = 130 lb/ft3 (2,061 kg/m3)
Find: The safety factor against sliding, SFS.The first step is to determine the total active force exerted by the soil on the wall:
Fa = (0. 5) (�) (Ka) (H)2
= (0.5) (120 lb/ft3) (0.2197) (3.44 ft)2 = 156 lb/ft= (0.5) (1,923 kg/m3) (9.81) (0.2197) (1.05m)2 = 2,295 N/m
9
Gravity Wall Tieback Wall Coherent Gravity Wall
Sliding
Free body Ex. 2-1
As explained in Chapter One, because of the effects of friction between the soil and the wall, the active force acts at anangle to a line perpendicular to the face of the wall. The active force can be resolved into a component perpendicularto the wall and a component parallel to the wall.
The degree of the angle between the active force and a line perpendicular to the face of the wall is �w. �w variesaccording to the unit weight of the soil. For very loose soil, �w approaches �; for compacted soil, �w can be aslow as (0.666) �. Since our wall designs involve compacting the backfill soil, we use the more conservative value of�w = (0.666) �. Thus, the horizontal component of the active force is:
Fh = (Fa) cos (�w)= (Fa) cos [ (0.666) (�) ]= (156 lb/ft) cos (20º) = (2,295 N/m) cos (20º)= 147 lb/ft = 2,157 N/m
Similarly, the vertical component of the active force is:
Fv = (Fa) sin (�w)= (Fa) sin [ (0.666) (�) ]= (156 lb/ft) sin (20º) = (2,295 N/m) sin (20º)= 53 lb/ft = 785 N/m
The weight of the wall facing must be determined before the frictionalresistance to sliding can be calculated:
Wf = (�w) (H) (d)= (130 lb/ft3) (3.44 ft) (0.97 ft) = (2061 kg/m3) (1.05 m) (0.3 m)= 434 lb/ft = 6,369 N/m
The maximum frictional resistance to sliding, Fr is calculated by multiplying the total vertical force, Vt , by the coefficient
of friction. The total vertical force is the sum of the weight of the wall and the vertical component of the active force.The coefficient of friction, Cf , is assumed to be equal to tan (�). Thus, the maximum frictional resistance is:
Fr = (Vt) (Cf)= (Vt) tan (�)= (Wf + Fv) tan (�)= (434 lb/ft + 53 lb/ft) tan (30º)= 281 lb/ft= (6,369 N/m + 785 N/m) tan (30º)= 4,130 N/m
Finally, the safety factor against sliding can be calculated:SFS = (Force resisting sliding) = Fr
(Force causing sliding) Fh= (281 lb/ft) = 1.9 > 1.5 OK
(147 lb/ft)
= (4,130 N/m) = 1.9 > 1.5 OK(2,157 N/m)
The safety factor against sliding is greater than 1.5. Therefore, the wall is stable and doesn't require reinforcement toprevent sliding failure. However, the wall must still be analyzed for overturning failure.
Figure 2-1. Schematic Diagram ofSimple Gravity Retaining Wall
10
Free body Ex. 2-1
Overturning FailureOverturning failure occurs when the forces acting on the wall cause it to rotate aboutthe bottom front corner of the wall (Point A in Figure 2-1). For stability, the momentsresisting overturning, Mr, must be greater than or equal to the moments causingoverturning, Mo. The minimum safety factor for overturning is 2.0. Therefore, Mrmust be greater than or equal to (2.0) Mo.
Find the safety factor against overturning, SFO, for Example 2-1.
Two forces contribute to the moment resisting overturning of the wall. These are theweight of the wall and the vertical component of the active force on the wall.Summing these moments about Point A:
Mr = (Wf) [ (X1) + (0.5) (H) tan (90º—�) ] + (Fv) [ (X2) + (0.333) (H) tan (90º—�) ]= (434 lb/ft) [ (0.49 ft) + (0.5) (3.44 ft) tan (90º—78º) ]+ (53 lb/ft) [ (0.97 ft) + (0.333) (3. 44 ft) tan (90º—78º) ]= 436 ft-lb/ft= (6,369 N/m) [ (0.149 m) + (0.5) (1.05 m) tan (90º—78º) ]+ (785 N/m) [ (0.3 m) + (0.333) (1.05 m) tan (90º—78º) ]= 1,954 N-m/m
(NOTE: The quantities (0. 5) (H) tan (90º—�) and (0.333) (H) tan (90º—�) accountfor the distance added to the moment arms because the wall is not vertical.)
The horizontal component of the active force is the only force that contributes tothe overturning moment. The active force is the resultant of the active pressuredistribution, which is triangular. For triangular pressure distributions, the verticalcentroid is located at one-third the height of the triangle. Therefore, the horizontalcomponent of the active force acts on the wall (0.333) H from the bottom of thewall. The moment causing overturning is given by:
Mo = (Fh) (�1) = (Fh) (0.333) (H)
= (147 lb/ft) (0.333) (3.44 ft) = 168 ft-lb/ft= (2,157 N/m) (0.333) (1.05 m) = 754 N-m/m
The safety factor against overturning is:
SFO = (Moment resisting overturning) = Mr(Moment causing overturning) Mo
= (436 ft-lb/ft) = 2.6 > 2.0 OK(168 ft-lb/ft)
= (1,954 N-m/m) = 2.6 > 2.0 OK(754 N-m/m)
The safety factor against overturning is greater than 2.0. Therefore, the wallis stable and doesn't require geogrid reinforcement to prevent overturning.As calculated previously, the safety factor against sliding is also greater than1.5 for this wall. This wall is adequate in both sliding and overturning andno geogrid reinforcement is required.
Overturning
11
Free body Ex. 2-1
Free body Ex. 2-1
Tieback WallsA simple gravity wall may be analyzed and found tobe unstable in either sliding or overturning. Whenthis occurs, the next logical step is to analyze the wallwith a single layer of geogrid or earth anchorsbehind it. The single layer of grid or earth anchor istreated as a restraining device or anchor. The forceon the wall due to the weight of the retained soil iscalculated exactly as it was in the simple gravity wallanalysis. However, the forces resisting failure in thisinstance are the frictional resistance due to the weightof the wall plus the friction force due to the weight of the soil on the grid orrestraining force of the anchor. Figure 2-2 is a schematic diagram of atieback wall and Figure 2-3 is a freebody diagram of the forces on the wall.
Example 2-2:Given:� = 30º �w = 20ºKa = 0.2197 H = 5.16 ft (1.57 m)
� = 78º � = 120 lb/ft3 (1,923 kg/m3)i = 0ºUnit weight of wall facing = 130 lb/ft3 (2,061 kg/m3)
Find: The safety factors against sliding, SFS, and overturning, SFO.
The first step in the analysis is to determine the weight of the wall facing:
Wf = (5.16 ft) (0.97 ft) (130 lb/ft3) = 651 lb/ft= (1.57 m) (0.3 m) (2,061 kg/m3) = 9,523 N/m
Next, the active force of the soil on the wall is calculated:
Fa = (0.5) (120 lb/ft3) (0.2197) (5.16 ft)2 = 351 lb/ft= (0.5) (1,923 kg/m3) (0.2197) (1.57 m)2 = 5,108 N/m
The horizontal and vertical components of the active force are:
Fh = (351 lb/ft) cos (20º) = 330 lb/ft= (5,108 N/m) cos (20º) = 4,800 N/m
Fv = (351 lb/ft) sin (20º) = 120 lb/ft= (5,108 N/m) sin (20º) = 1,747 N/m
The total vertical force due to the weight of the wall and the verticalcomponent of the active force is:
Vt = Wf + Fv= 651 lb/ft + 120 lb/ft= 771 lb/ft= 9,523 N/m + 1,747 N/m= 11,270 N/m
12
Figure 2-2. Diagram of retainingWall for Tieback Analysis
The force that resists sliding of the wall because of friction between thewall and the soil is:
Fr = (Vt) (Cf)= (771 lb/ft) tan (30º)= 445 lb/ft
= (11,270 N/m) tan (30º)= 6,507 N/m
The safety factor against sliding is:
SFS = Fr = (445 lb/ft) = 1.35Fh (330 lb/ft)
= Fr = (6,507 N/m) = 1.35Fh (4,800 N/m)
The safety factor against overturning is:
Mr = (Wf) [ (X1) + (0.5) (H) tan (90º—�) ] + (Fv) [ (X2) + (0.333) (H) tan (90º—�) ]= (651 lb/ft) [ (0.49 ft) + (0.5) (5.16 ft) tan (90º—78º) ]+ (120 lb/ ft) [ (0.97 ft) + (0.333) (5.16 ft) tan (90º—78º) ]= 836 ft-lb/ft
= (9,523 N/m) [ (0.149 m) + (0.5) (1.57 m) tan (90º—78º) ]+ (1,747 N/m) [ (0.3 m) + (0.333) (1.57 m) tan (90º—78º) ]= 3,726 N-m/m
Mo = (Fh) (�1)= (330 lb/ft) (0.333) (5.16 ft)= 567 ft-lb/ft
= (4,800 N/m) (0.333) (1.57m)= 2,510 N-m/m
SFO = Mr = (836 ft-lb/ft) = 1.47Mo (567 ft-lb/ft)
= Mr = (3,726 N-m/m) = 1.47Mo (2,510 N-m/m)
Without reinforcement, this wall isnot adequate with respect to eithersliding failure or overturning failure.Therefore, a tieback wall will berequired. A good rule of thumb is toplace the reinforcement as close aspossible to halfway between the topand bottom of the wall.
Figure 2-3. Freebody Diagram ofRetaining Wall for tieback Analysis
13
Geogrid as a TiebackA single layer of geogrid reinforcement acts as an anchor to keep theretaining wall from moving forward. The geogrid extends into the backfillsoil and the frictional resistance due to the weight of the soil on top of thegeogrid provides the restraining force. The relationship can be expressedas follows:
Fg = (Unit weight of soil) x (Depth to grid)x (2) x (Area of the grid in the passive zone)x (Coefficient of friction)
The following equation can be used to calculate the maximum potentialrestraining force:
Fg = (2) (dg) (�) (Le) (Ci) tan (�)
where:
Fg = the maximum potential restraining force.
The factor 2 is used since both the top and the bottom of the geogrid interactwith the soil.
dg = the depth from the top of the backfill to the layer of geogrid.
� = the unit weight of the backfill soil.
Le = the length of geogrid embedded in the passive zone of the soil.
Ci = the coefficient of interaction between the soil and the geogrid, a measure of the ability of the soil to hold the geogrid when a force is applied to it. Typical values of Ci are 0.9 for gravelly soil, 0.85 for sand or silty sands, and 0.75 for silts and clays.
tan(�) = the coefficient of friction (shear strength) between adjacent layers of soil.
The next step in the design process is to estimate the length of geogrid required. First, the depth to the geogrid, dg, mustbe specified. To complete Example 2-2, let dg = 2.30 ft (0.7 m). Another important assumption is that the geogrid willextend far enough into the passive zone to develop the full allowable design strength of the geogrid. In this case anaverage strength geogrid will be used, the full long-term allowable load is 1,250 lb/ft (18,249 N/m). A safety factorof 1.5 is applied to this value and the design strength is 833 lb/ft (12,161 N/m). The embedment length required togenerate that force can be calculated as follows:
Fg = (2) (dg) (�) (Le) (Ci) tan (�)Le = Fg
(2) (dg ) (�) Ci) tan (�)
= (833 lb/ft)(2) (2.3 ft) (120 lb/ft3) (0.85) tan (30º)
= 3.08 ft= (1,240 kg/m)
(2) (0.7 m) (1,923 kg/m3) (0.85) tan (30º)
= 0.94 m
14
Diagram Ex. 2-2
15
The total length of geogrid required per linear foot of wall is:
Lt = Lw + La + Lewhere:
Lt = total length of geogridLw = length of geogrid inside the Allan Block wall = 0.84 ft (0.26 m)
La = length of geogrid in the active zone
= (H — dg) [ tan (45—�/2) — tan (90—�) ]Le = length of geogrid embedded in the passive zone.
The estimated total length of geogrid required for the wall in Example 2-2 is:
Lt = (0.84 ft) + (5.16 ft — 2.3 ft) [ tan (30) — tan (12) ] + 3.08 ft= 4.96 ft= (0.26 m) + (1.57 m — 0.7 m) [ tan (30º) — tan (12º) ] + 0.94 m= 1.52 m
For the convenience of the workers installing the retaining wall, we round off the geogrid length to the nearest 0.5 ft (0.15 m). For this example, Lt = 5.0 ft (1.52 m).
With a total geogrid length of 5.0 ft (1.52 m) the actual embedment length is:
Le = Lt — (0.84 ft) — (H — dg) [ tan (45—�/2) — tan (90—�) ]= 5.0 ft — (0.84 ft) — (5.16 ft — 2.3 ft) [ tan (30) — tan (12) ]= 3.12 ft= Lt — (0.26 m) — (H — dg) [tan (45º — �/2) — tan (90º—�)]= 1.52 m — (0.26 m) — (1.57 m — 0.7m) [ tan (30º) — tan (12º) ]= 0.94 m
The maximum potential restraining force on the geogrid for an embedment length of 3.12 feet (0.95 m) is:
Fg = (2) (2.3 ft) (120 lb/ft3) (3.12 ft) (0.85) tan (30) = 845 lb/ft= (2) (0.7 m) (1,923 kg/m3) (0.95 m) (0.85 m) tan (30º) = 1,255 kg/m
However, the long-term allowable design load (LTADL) of the grid specified is only 833 lb/ft (12,161 N/m). Themaximum restraining force must be less than or equal to the LTADL. Therefore, let Fg = 833 lb/ft (12,161 N/m).
Finally, the safety factors against sliding and overturning can be recalculated:
SFS = Fr + LTAD = (445 lb/ft + 833 lb/ft) = 3.87 > 1.5 OKFh (330 lb/ft)
= (6,507 N/m + 12,161 N/m) = 3.87 > 1.5 OK4,800 N/m
SFO = Mr = (836 ft-lb/ft) + (833 lb/ft) (5.16 ft — 2.30 ft) = 5.7 > 2.0 OKMo (567 ft-lb/ft)
= (3,726 N-m/m) + (12,161 N/m) (1.57 m — 0.7m) = 5.7 > 2.0 OK2,510 N-m/m
Since SFS > 1.5 and SFO > 2.0, this retaining wall is stable with one layer of geogrid, 5.0 ft (1.52 m) long. At thispoint, you may want to go back, shorten the geogrid to optimize the design, and reanalyze the wall. However, werecommend using a geogrid length that is sufficient to develop the full long-term allowable design strength of the geogrid,833 lb/ft (12,161 N/m), even if it is not required for wall stability.
16
Earth Anchors as a TiebackA single row of earth anchors can be utilized to provide the additional tieback resistance. The earth anchors extendbeyond the failure plane and provide additional resistance to overturning and sliding. This additional force can beutilized in our calculations as follows:
Preloaded value of anchor.
Fe = 10,500 lbs. (4,763 kg) Fe = Preloaded value of installed anchor.
This is the preloaded value of the anchor for correction for design purposes we will use a weighted value and correc-tion for horizontal anchor spacing. For this example we will specify spacing of anchors on 8 foot (2.44 m) centersand Huesker 35/-20-20 geogrid (Diagram Ex.2-2). Therefore the additional force resisting sliding is:
Fr = (Wf + Fv) tan (30º)
Fr = (651 lb/ft + 120 lb/ft) tan (30º) = 445 lb/ft Fr = The maximum frictional resistance to sliding.
= (9,523 N/m + 1,747 N/m) tan (30º) = 6,507 N/m
Fwe = (0.67) Fe ÷ 8 ft = 879 lb/ft Fwe = Weighted design value of anchor.= (0.67) Fe ÷ (2.44 m) = 12,830 N/m
Fga = 833 lb/ft (12,161 N/m) Fga = The maximum potential restraining force.
� Fp = 888 lb/ft + 0.130 x N Fp = Grid pullout from block.
= 888 lb/ft + [0.130 x 1.9 (125 lb/ft)] = 919 lb/ft= 12,964 N/m + 0.130 x N= 12,964 N/m + [0.130 x 1.9 (1,825 N/m)] = 13,415 N/m
The resulting factor of safety with one row of earth anchors is:
SFS = Fr + * Fga = (445 lb/ft + 833 lb/ft) = (6,507 N/m + 12,161 N/m)Fh 330 lb/ft 4,800 N/m
= 3.87 > 1.5 OK = 3.87 > 1.5 OKThe safety factor against overturning is:
Mr = (Wf) [ (X1) + (0.5) (H) tan (90º — �) ] + (Fv) [(X2) + (0.333) (H) tan (90º — �) ] + *Fga (H/2)= (651 lb/ft) [ (0.49 ft) + (0.5) (5.16 ft) tan (90º — 78º) ]+ (120 lb/ft) [ (0.97 ft) + (0.333) (5.16 ft) tan (90º — 78º) ] + (833 lb/ft) (2.58 ft)= 2,985 ft-lb/ft= (9,523 N/m) [ (0.149 m) + (0.5) (1.57 m) tan (90º — 78º) ]+ (1,747 N/m) [ (0.296 m) + (0.333) (1.57 m) tan (90º — 78º) ] + (12,161 N/m) (0.786 m)= 13,278 N-m/m
* Use the least of Fwe, Fga, or Fp.
SFO = Mr = (2,985 ft-lb/ft) = 5.3 > 2.0 OK = (13,278 N-m/m) = 5.3 > 2.0 OKMo (567 ft-lb/ft) (2,510 N-m/m)
The anchor length requires a 3 foot (0.9 m) embedment into the passive zone. (Past the theoretical failure plane)
Lt = La + 3 ft= (5.16 ft 2.3 ft) [tan (30º) tan (12º) ] + 3.0 ft = 4 ft= (1.57 m — 0.7 m) [ tan (30º) — tan (12º) ] + 0.9 m = 1.22 m
Check to determine if the Fwe or the grid pullout from the block or rupture is the determining factor.Note: The pullout from the block can be eliminated as the governing factor by bonding the block to grid interface with
a construction grade adhesive. � Reference Table 2-1 Page 50
Diagram Ex. 2-2
Coherent Gravity WallsThe theory behind coherent gravity walls is that two or more layers ofgeogrid make the reinforced soil mass behave as a single unit. The wallfacing and reinforced soil mass are then treated as a unit and analyzed asa large simple gravity wall. The wall must be analyzed for stability insliding and overturning. In addition, the number of layers of geogridrequired, and their spacing, must be determined. Finally, the bearingpressure of such a large gravity wall must be checked to ensure that itdoesn't exceed the allowable bearing capacity of the soil.
Example 2-3:
Figure 2-4 is a schematic diagram of a coherent gravity wallwith two layers of geogrid. Figure 2-8 is a freebody diagramof the same wall. The subscripts r and o refer to the reinforcedsoil and the onsite soil, respectively. The values shown in thefigure will be used to analyze the stability of the wall.
Given:i = 0º (Slope above wall) �wo = 18º�wr = 20º �o = 27º�r = 30º � = 78ºKar = 0.2197 Kao = 0.2561H = 9.17 ft (2.8 m)
�o = 120 lb/ft3 (1,923 kg/m3)�r = 125 lb/ft3 (2,002 kg/m3)
Find: The safety factors against sliding, SFS, and overturning, SFO.
Length of Geogrid
The first step in analyzing the stability of the wall is to estimate the length of geogrid required. A rule of thumb is thatthe minimum reinforcement length is 50% of the wall height for the 12º block systems and 60% of the wall height for 3ºand 6º block systems.
Figure 2-4. Coherent Gravity Wall forExample 2-3
17
RetainedSoil
2.13 FT (0.65 m)
5.13 FT (1.56 m)
10.0 FT(3.05 m)
5.0 FT (1.52 m)Of Geogrid
1.1 FT (0.34 m) 6.13 FT (1.87 m)
6.0 FT (1.83 m) Of Geogrid
10.0 FT(3.05 m)
0.53 FT (0.16 m) 6.13 FT (1.87 m)
6.0 FT (1.83 m) Of Geogrid
10.0 FT (3.05 m)
Figure 2-5. Typical GeogridReinforcement Embedment for
12 degree AB System
Figure 2-6. Typical GeogridReinforcement Embedment for
6 degree AB System
Figure 2-7. Typical GeogridReinforcement Embedment for
3 degree AB System
External Stability
Once the length of the geogrid is known, the weight of the coherent gravity wall can be calculated. The weight of thestructure is the sum of the weights of the wall facing and the reinforced soil mass. The weight of the wall facing is equalto the unit weight of the wall facing times the height times the depth:
Wf = (130 lb/ft3) (9.17 ft) (0.97 ft) = 1,156 lb/ft= (2,061 kg/m3) (2.8 m) (0.3 m) = 16,983 N/m
The weight of the reinforced soil mass is equal to the unit weight of the backfill soil, times the height of the reinforcedsoil mass, times the depth (measured from back face of wall to the end of the geogrid):
Ws = (125 lb/ft3) (9.17 ft) (6.0 ft 0.84 ft) = 5,915 lb/ft= (2,002 kg/m3) (2.8 m) (1.83 m — 0.256 m) = 86,556 N/m
The total weight of the coherent gravity wall is:
Ww = Wf + Ws= (1,156 lb/ft) + (5,915 lb/ft) = 7,071 lb/ft= (16,983 N/m) + (86,556 N/m) = 103,539 N/m
(NOTE: Using a value of 125 lb/ft3 (2,002 kg/m3) for the unit weight of the wall facing would simplify the calculationsand result in a conservative design.)
The next step is to calculate the active force on the gravity wall. The properties of the retained soil are used to calculatedthe active force since it acts at the back of the reinforced soil zone. The active force is given by the equation:
Fa = (0.5) (�o) (Kao) (H)2
= (0.5) (120 lb/ft3) (0.2561) (9.17 ft)2
= 1,292 lb/ft= (0.5) (1,923 kg/m3) (0.2561) (2.8 m)2
= 18,938 N/m
The horizontal and vertical components of the active force are:
Fh = (Fa) cos (�wo)= (1,292 lb/ft) cos (18º) = (18,938 N/m) cos (18º)= 1,229 lb/ft = 18,011 N/m
Fv = (Fa) sin (�wo)= (1,292 lb/ft) sin (18º) = (18,938 N/m) sin (18º)= 399 lb/ft = 5,852 N/m
Next, the total vertical force is calculated:
Vt = Ww + Fv= (7,071 lb/ft) + (399 lb/ft) = 7,470 lb/ft
= (103,539 N/m) + (5,852 N/m) = 109,391 N/m
The force resisting sliding is calculated by multiplying the total vertical force by the coefficient of friction between thereinforced soil mass and the underlying soil:
Fr = (Vt) (Cf)= (7,470 lb/ft) tan (30º) = 4,313 lb/ft= (109,391 N/m) tan (30º) = 63,157 N/m
18
Diagram Ex. 2-2
The safety factor against sliding is:
SFS = Fr = (4,313 lb/ft) = 3.5 > 1.5 OKFh (1,229 lb/ft)
= Fr = (63,157 N/m) = 3.5 > 1.5 OKFh (18,011 N/m)
The safety factor against overturning is:
(NOTE: All moments are taken about Point A in Figure 2-8.)
�Mr = (Wf) [ (0.5) (X1) + (0.5) (H) tan (90º — �) ]+ (Ws) [ (0.5) (X2 — X1) + (X1) + (0.5) (H) tan (90º — �) ]+ (Fv) [ (X2) + (0.333) (H) tan (90º — �) ]= (1,156 lb/ft) [ (0.5) (0.97 ft) + (0.5) (9.17 ft) tan (90º — 78º) ]+ (5,915 lb/ft) [ (0.5) (6.13 ft — 0.97 ft) + (0.97 ft) + (0.5) (9.17 ft) tan (90º — 78º) ]+ (399 lb/ft) [ (6.13 ft) + (0.333) (9.17 ft) tan (90º — 78º) ]= 31,155 ft-lb/ft= (16,983 N/m) [ (0.5) (0.3 m) + (0.5) (2.8 m) tan (90º — 78º) ]+ (86,556 N/m) [ (0.5) (1.87 m — 0.3 m) + (0.3 m) + (0.5) (2.8 m) tan (90º — 78º)]+ (5,852 N/m) [ (1.87 m) + (0.333) (2.8 m) tan (90º — 78º) ]= 139,375 N-m/m
Mo = (Fh) (0.333) (H)= (1,229 lb/ft) (0.333) (9.17 ft)= 3,753 ft-lb/ft= (18,011 N/m) (0.333) (2.8 m)= 16,793 N-m/m
SFO = �Mr = (31,155 ft-lb/ft) = 8.3 > 2.0 OK�Mo (3,753 ft-lb/ft)
= �Mr = (139,375 N-m/m) = 8.3> 2.0 OK�Mo (16,793 N-m/m)
The minimum recommended safety factors for geogrid reinforcedretaining walls are 1.5 for sliding failure and 2.0 for overturning failure.Since both safety factors for this wall exceed the minimum values, the wallis adequate with respect to sliding and overturning. In cases where eitherof the safety factors is lower than required, the length of geogrid isincreased and the analysis is repeated. The process ends when bothsafety factors exceed the minimum recommended values.
Internal Stability
This part of the design consists of spacing the geogrid layers so that each layer is subjected to a force less than or equal to thelong-term allowable design load of the geogrid. Geogrid for this example has the long-term allowable design load of 1,250lb/ft (18,249 N/m). Applying a safety factor of 1.5 results in an allowable load of 833 lb/ft (12,161 N/m) per layer of grid.
The first step is to determine the minimum number of layers of geogrid required. (We recommend no more than 4-course spacingbetween each layer of geogrid reinforcement for a 12º system and no more than 3-course spacing for 3º and 6º systems.) Thisis done by dividing the total horizontal force at the back of the wall facing by the load each layer of grid can handle:
19
Figure 2-8. Freebody Diagram of a Coherent Wall
N = Fh = (0.5) (�r) (Kar) (H)2 cos (�wr)(833 lb/ft) (833 lb/ft)
= (0.5) (125 lb/ft3) (0.2197) (9.17 ft)2 cos (20º)(833 lb/ft)
= 1.30 layers= Fh = (0.5) (�r) (Kar) (H)2 cos (�wr)
(12,161 N/m) (12,161 N/m)= (0.5) (2002 kg/m3) (0.2197) (2.8 m)2 cos (20º)
(12,161 N/m)= 1.30 layers
Since it is impossible to put in a fraction of a layer, the number of layers mustalways be rounded up to the nearest whole number or use a higher strengthgrid. For this example, N, = 2 layers. As the number of layers of geogridincreases, it may become necessary to put in more than the minimum numberof layers. This is because the spacing of the layers is limited by the height of the individual Allan Block units.
The load on each layer of geogrid is equal to the average pressure on the wall section, Pavg, multiplied by the heightof the section, dh, (Figure 2-9). The pressure at any depth is given by:
Pi = (�r) (di) (Kar) cos (�wr)The load on each layer of grid is given by:
Fg = (Pavg) (dh)
where:
Pavg = (0.5) (P1 + P2)= (0.5) [ (�r) (d1) (Kar) cos (�wr)+ (�r) (d2) (Kar) cos (�wr) ]= (0.5) (�r) (Kar) cos (�wr) (d1 + d2)
Internal StabilityInternal stability is the ability of the reinforcement combined with the internalstrength of the soil to hold the soil mass together and work as a single unit.
Grid Rupture Bulging
Rupture occurs whenexcessive forces from theretained soil mass exceedthe ultimate tensilestrength of the geogrid.Increase grid strength
Pullout results when gridlayers are not embeddeda sufficient distancebeyond the failure plane.
Increase embedment length
Bulging occurs whenhorizontal forces betweenthe geogrid layers causeslocalized rotation of thewall.Increase number of grid layers
Pullout
Figure 2-9. Calculating the Spacingof Geogrid Layers
20
dh = d1 d2
d1 = distance from the top of the backfill to the bottom of the zone supported by the layerof geogrid.
d2 = distance from the top of the backfill to the top of the zone supported by the layer of geogrid.
21
To simplify the analysis, let:C = (0.5) (�r) (Kar) cos (�wr)then:Fg = [ C (d1) + C (d2) ] (d1 — d2)
= C (d1)2 — C (d2)2
The condition for internal stability is that:Fg < 833 lb/ft (12,161 N/m)
orC (d1)2 — C (d2)2 < 833 lb/ft (12,161 N/m)
where:C = (0.5) (125 lb/ft3) (0.2197) (cos 20º)
= 12.9 lb/ft3
= (0.5) (2,002 kg/m3) (0.2197) (cos 20º)= 207 kg/m3
d1 = 9.17 ft (2.8 m)Solving for d2:d2 = [ (d1)2 — ( (833 lb/ft)/C) ]0.5
= [ (9.17 ft)2 — ( (833 lb/ft)/(12.9 lb/ft3) ) ]0.5
= 4.42 ft= [ (d1)2 — ( (1,240 kg/m)/C) ]0.5
= [ (2.8 m)2 — ( (1,240 kg/m)/(207 kg/m3) ) ]0.5
= 1.36 m
Solving for dh:dh = (d1 — d2)
= 9.17 ft 4.42 ft = 4.75 ft= 2.8 m — 1.36 m = 1.44 m
Place the first layer of geogrid within dh so that theforce above the geogrid is equal to the forcebelow the geogrid. For a triangular pressure distri-bution, place the geogrid up one-third of the dis-tance from the bottom of dh to the top of dh. For arectangular pressure distribution, place the geogridup one-half of the distance from bottom to top ofdh. In general, the geogrid should be placedbetween one-third and one-half of the distancefrom the bottom of dh to the top of dh.
Geogrid can only be placed between the blocksforming the wall facing. For AB Stones standardunits, normally that means that the geogrid can only be placed at heights evenly divisible by 7.62 inches or 0.635 ft (194mm). Therefore, the first layer should be placed 1.905 ft (0.58 m), 3 blocks, up from the bottom.
This wall only requires two layers of geogrid. The second layer should be placed at one-third to one-half the distance betweenthe location of d2 and the top of the wall. In this case, the second layer of geogrid should be placed 5.73 ft (1.75 m), 10 blocks, from the bottom of the wall.
22
Bearing Pressure on the Underlying SoilAnother consideration in the design of a coherent gravity wall is theability of the underlying soil to support the weight of a giant gravitywall. Most undisturbed soils can withstand pressures between 2,500(120 kPa) and 4,000 (192 kPa) pounds per square foot.
Figure 2-10 is a freebody diagram of the coherent gravity wall inExample 2-3. It shows the forces acting on the wall. With thisinformation, the maximum bearing pressure can be calculated andcompared to the allowable bearing pressure.
The first step is to calculate the resultant vertical resisting force, R,exerted on the gravity wall by the soil:
R = �Fy = W + Fv
= (7,071 lb/ft + 399 lb/ft)= 7,470 lb/ft= 103,539 N/m + 5,852 N/m= 109,391 N/m
The next step is to locate the point of application of the resultant force. This is done by summing moments aroundPoint A, setting the result equal to zero, and solving for X.
�MA = (7,470 lb/ft) (X) + (1,229 lb/ft) (3.06 ft)— (7,071 lb/ft) (4.04 ft) — (399 lb/ft) (6.78 ft)
X = (28,567 ft-lb/ft) + (2,705 ft-lb/ft) — (3,761 ft-lb/ft) = 3.68 ft(7,470 lb/ft)
= (109,391 N/m) (X) + (18,011 N/m) (0.933 m)— (103,539 N/m) (1.23 m) — (5,852 N/m) (2.07 m)
= (127,353 N-m/m) + (12,144 N-m/m) — (16,804 N-m/m) = 1.12 m(109,391 N/m)
The eccentricity, e, of the resultant vertical force, is the distance from the centerline of bearing of the gravity wall to thepoint of application of the resultant force, R. In this case:
e = (0.5) (6.13 ft) – X= (0.5) (6.13 ft) – 3.68 ft = – 0.62 ft= (0.5) (1.87 m) — X= (0.5) (1.87 m) — 1.12 m = – 0.185 m
Assuming a linear bearing pressure distribution, the average bearing pressure occurs at the centerline of the wall. Itsmagnitude is:
�avg = R = (7,470 lb/ft) = 1,219 lb/sq ftL (6.13 ft)
= R = 109,391 N/m = 58 kPaL 1.87 m
Figure 2-10. Freebody Diagramfor Bearing Pressure Analysis
23
Next, the bearing pressure due to the moment about the centerline of bearing is calculated. This is done by finding themoment due to the resultant vertical force about the centerline of bearing (Point B) and dividing it by the section modulusof a horizontal section through gravity wall. The moment due to the eccentricity of the resultant force is:
MB = (R) (e)= (7,470 lb/ft) (—0.62 ft) = (109,391 N/m) (—0.188 m)= —4,631 ft-lb/ft = —20,566 N-m/m
The section modulus of a 1-foot wide section of the wall is given by:
S = (b) (d)2
6Where:
b = the width of the section = 1.0 ft (0.305 m)d = the depth of the section = L = 6.13 ft (1.87 m)
S = (1 ft/6) (L)2 = (0.305 m/6) (L)2
= (1 ft/6) (6.13 ft)2 = (0.305 m/6) (1.87 m)2
= 6.26 ft3 = 0.177 m3
The difference in stress due to the eccentricity is:
�mom = MBS
= (—4,631 ft-lb/ft) = (—20,566 N-m/m)(6.26 ft3) (0.177 m3)
= —740 lb/ft2 = —116 kPa
Finally, the maximum and minimum bearing pressures are calculated:
� = �avg � �mom
� = �avg + �mom= (1,219 lb/sq ft) + (—740 lb/sq ft) = (58 kPa) + (—116 kPa)= 479 lb/sq ft = —58 kPa = —5,914 kg/m2
� = �avg — �mom� = (1,219 lb/sq ft) — (—740 lb/sq ft) = (58 kPa) — (—116 kPa)
= 1,959 lb/sq ft = 174 kPa = 17,743 kg/m2
The maximum bearing pressure is greater than the allowable bearing pressure of 2,500 lb/sq ft (120 kPa). Therefore,the wall is unstable with respect to the allowable bearing capacity of the underlying soil.The procedure outlined above can be simplified by rearranging the equations as follows:
� = �avg � �mom� = R � M = R � (6)M = R � (6) (R) (e)
L S L L2 L L2
Note that the eccentricity can be negative as well as positive. A conservative assumption is that the maximum bearingpressure occurs at the toe of the wall.
When the maximum bearing pressure is greater than the allowable bearing pressure the underlying soil is not stable.Stabilizing the soil under the wall is accomplished by spreading the forces of the wall over a larger area. Engineersuse this concept in designing spread footings.
Diagram Ex. 2-3
Once the �max is determined, compare it to ultimate bearing (qf):
qf = (½) (�) (B) (N�) + (c) (Nc) + (D) (Nq)
(Craig, p. 303, Soil Mechanics, Fifth Edition)
Where:Nq = exp ( tan �) tan2 (45 + �/2)Nc = (Nc — 1) cot �N� = (Nq — 1) tan (1.4�)
� = �f = Unit weight of foundation soils
C = Cf = Cohesion of foundation soils
D = di = Depth of wall embedment = Buried block + Footing thickness (di).
The ultimate bearing (qf) should be designed to a factor of safety of 2.0
If FSbearing = qf < 2.0, then increase the size of the base.�max
The material in the base will always be a select crushed stone. Therefore, � = 36º.
tan (45 — �/2) = 0.5 ft/WW = 0.5 ft / tan (45 — 36º/2)W = 1.0 fttan (45 — �/2) = 0.15 m/WW = 0.15 m / tan (45 — 36º/2)W = 0.3 m
Therefore, the incremental base size is:
di = (di — 1) + 0.5 ft.= (di — 1) + 0.15 m
Bi = (Bi — 1) + (2) (W)= (Bi — 1) + (2) (1 ft)= (Bi — 1) + (2) (0.3 m)
The toe extension will be equal to the footing depth.
24
Original Base Size:
Increment to next size:
Increase Width by:
Base Footing Location:
Attachment of the Geogrid to the Wall FacingA logical question to ask is: What keeps the geogrid from slipping out from between the courses of Allan Block? Theanswer is that the weight of the Allan Blocks sitting on top of the geogrid creates friction between the blocks and thegeogrid. In addition, some of the material used to fill the voids in the Allan Blocks becomes wedged in the apertures ofthe geogrid. This is called interlocking and results in additional resistance to sliding.
Pullout tests were conducted at the University of Wisconsin-Platteville byKliethermes, et al. Two sets of tests were run. In the first set, the voidsof the Allan Blocks were filled with crushed limestone. In the second set,the voids were left empty.
When the voids were filled with crushed limestone, there was anapparent coefficient of friction (ACF) of about 3.0 between the geogridand the Allan Blocks. When the voids were left empty, the ACF wasabout 0.88. The surprising magnitude of the ACF for crushed limestoneis due to a significant amount of interlocking between the crushed rockand the geogrid.
The hollow core, pinless design of Allan Block raises questions on how the geogrid is attached to the wall facing. AllanBlock’s gravel filled hollow core provides a multi-point interlock with the grid. As wall heights increase, our exclusive"rock lock" connection, combined with the weight of the Allan Block units, provides a more uniform block-to-grid interlockthan any system on the market.
Pullout tests were conducted at the University of Washington. A total of ten geogrids and two geofabrics were tested. Eachproduct was tested three times under four loading conditions; 500 lbs. (226.8 kg), 1000 lbs. (453.6 kg), 1500 lbs. (680.4kg), and 2000 lbs. (907.18 kg) vertical load per lineal foot of wall. The data compiled was consistent. From a total of144 pullout tests, the results exhibited a uniform behavior based on grid strengths and normal loads applied. The test valuesincreased with added vertical loads. A typical pullout equation for service and ultimate loads takes the form X + Y * N.The variables X and Y are constant values as determined by testing. The normal (vertical) load N, is load applied to theblock. The location of the block to grid connection will be the determining factor for the amount of normal (vertical) loadapplied. Table 2-1 on page 50, shows the equations derived from the testing for each type of grid and fabric tested.
The maximum force in the geogrid occurs at the intersection of the failure plane - the boundary between the active andpassive zones of the retained soil. The force on the geogrid decreases as the horizontal distance from the failure planeincreases. At the back of the wall, the force on the geogrid is reduced to about two-thirds of the maximum force(McKittrick, 1979).
Mechanical Connection
A grouted / mechanical connection may be desirable in special circumstances such as for geogrid layers under highseismic loading or when barriers are attached. The hollow cores of the Allan Block provide for a cell to encapsulate thegeogrid placed between block courses. When a grouted connection is specified, a minimum of 3 inches (7.6 cm) ofgrout above and below the grid layers is required. Factors of safety for this connection are determined by comparingthe long-term allowable design strength (LTADS) of the geogrid to the applied load at the face.
FSmech= LTADS (Applied Load) (0.667)
Example 2-4
Given:H = 10.2 ft (3.1 m) LTADS = 1322 lb/ft (19,300 N/m)
Ao = 0.4 Fis = Fa = 45 lb/ft (657 N/m)d = 2 in. (5.1 cm) Fid = Fa + DFdyn + Pir = 1240 lb/ft (18,103 N/m)� = 12° Geogrid Length = 5.1 ft (1.6 m)� = 30° Geogrid Courses = 3, 7, 11, 15�s = 120 lb/ft3 (1,923 kg/m3) �w = 130 lb/ft3 (2,061 kg/m3)
25
Find the factors of safety for:
1. Static geogrid/block connection capacity2. Dynamic geogrid/block connection capacity3. Mechanical (grout) geogrid/block connection capacity
1. The static geogrid/block connection capacity factor of safety is determined by comparing the peak connectionstrength, which is a function of the normal load, to the applied load on each layer of geogrid.
FSconn = Fcs � 1.5(Fis)(0.667)
The peak connection strength (Fcs) is an equation of a line generated by comparing the maximum pullout force undervarious normal loads. The numbers in this example are based on testing done with Allan Block and Fortrac 35/20-20geogrid. The resulting equation for Fcs is:
Fcs = 888 lb/ft + 0.130(N) = 12,964 N/m + 0.130(N)
Where the normal load (N) is:
N = (H — grid elev) (�w) (d)= (10.2 ft — 9.525 ft) (130 lb/ft3) (0.97 ft) = 85 lb/ft = (3.1 m — 2.9 m) (2,061 kg/m3) (0.30 m) = (1,213 N/m)
Therefore, the peak connection strength (Fcs) is:
Fcs = 888 lb/ft + 0.130 (85 lb/ft) = 899 lb/ft = 12,964 N/m + 0.130 (1,213 N/m) = 13,122 N/m
The applied load (Fis) is equal to the active force acting on the wall:
Fis = Fa = 45 lb/ft (657 N/m)
FSconn = (899 lb/ft) = 29.9 � 1.5 , ok(45 lb/ft)(0.667)
= (13,122 N/m) (657 N/m)(0.667) = 29.9 � 1.5, ok
2. In a seismic condition, the applied load on each grid will increase due to the presence of the dynamic incrementforce (DFdyn) and the seismic inertial force (Pir).
FSconn = Fcs � 1.1(Fid)(0.667)
Fcs = 899 lb/ft (13,122 N/m)
Fid = Fa + DFdyn + Pir = 1,240 lb/ft (18,103 N/m)
Therefore:
FSconn = (899 lb/ft) = (13,122 N/m)
(1,240 lb/ft)(0.667) (18,103 N/m)(0.667)= 1.1 � 1.1, ok = 1.1 � 1.1, ok
In comparing the dynamic factor of safety with the static, we see a dramatic decrease. In such circumstances, amechanical connection is desirable.
26
27
3. For a mechanical connection in a seismic condition, the factor of safety for geogrid/block connection is a compar-ison of the long term allowable design strength of the geogrid (with no creep reduction factor taken due to thetemporary nature of a seismic event) to the dynamic applied load on each grid.
FSconn = LTADS(Fid)(0.67)
= (1,322 lb/ft) (1.67) = (19,300 N/m) (1.67)
(1,240 lb/ft) (0.67) (18,103 N/m) (0.67)
= 2.6 � 1.1, ok = 2.6 � 1.1, ok
Example 2-5
Let's analyze the wall of Example 2-3 for pullout of the geogrid.Figure 2-11 shows the wall and some of the dimensions that will beneeded in the calculations.
Calculate the horizontal force on the bottom layer of geogrid:
P1 = (�) (Ka) (d1)= (125 lb/ft3) (0.2197) (9.17 ft)= 252 lb/ft2
= (2,002 kg/m3) (0.2197) (2.8 m)= 1,232 kg/m2
P2 = (�) (Ka) (d2)= (125 lb/ft3) (0.2197) (5.44 ft)= 149 lb/ft2
= (2,002 kg/m3) (0.2197) (1.66 m)= 730 kg/m2
Pavg = (0.5) (252 lb/ft2 + 149 lb/ft2)= 201 lb/ft2
= (0.5) (1,232 kg/m2 + 730 kg/m2)= 981 kg/m2
F1 = Pavg (dh)= (201 lb/ft2) (3.73 ft)= 750 lb/ft= (981 kg/m2) (1.14 m)= 10,971 N/m
The force on the geogrid at the back face of the wall will be approximatelytwo-thirds of F1:
Fw = (0.666) (F1) = (0.666) (750 lb/ft)= 500 lb/ft= (0.666) (F1) = (0.666) (10,971 N/m)
= 7,307 N/mThe force resisting pullout, caused by the weight of the aggregate-filled blocks above the bottom geogrid layer, is:
N = (120 lb/ft3) (0.97 ft) (7.45 ft) = 867 lb/ft= (1,923 kg/m3) (0.3 m) (2.27 m) = 12,847 N/m
Figure 2-11. Friction Attachment of Geogrid to Blocks
Diagram Ex. 2-3
28
Using Huesker 35/20-20 equation from Table 2-1:
P = Fr = 888 lb/ft + 0.130 (867 lb/ft) = 12,964 N/m + 0.130 (12,847 N/m)
= 1,000.71 lb/ft = 14,634 N/m
The safety factor against pullout for the bottom layer of geogrid is:
SFP = (1,000 lb/ft) = 2.0 = (14,634 N/m) = 2.0(500 lb/ft) = (7,300 N/m)
The horizontal force on the top layer of geogrid is:P2 = (�) (Ka) (d2) = (125 lb/ft3) (0.2197) (5.44 ft) = 149 lb/ft2
= (�) (Ka) (d2) = (2,002 kg/m3) (0.2197) (1.67 m) = 730 kg/m2
P3 = (�) (Ka) (d3) = (125 lb/ft3) (0.2197) (0 ft) = 0 lb/ft2
= (�) (Ka) (d3) = (2,002 kg/m3) (0.2197) (0 m) = 0 kg/m2
Pavg = (0.5) (149 lb/ft2 + 0 lb/ft2) = 75 lb/ft2
= (0.5) (730 kg/m2 + 0 kg/m2) = 365 kg/m2
F2 = (Pavg) (dh) = (75 lb/ft2) (5.44 ft) = 408 lb/ft= (Pavg) (dh) = (365 kg/m2) (1.67 m) = 6,012 N/m
The force on the geogrid at the back face of the wall will be approximately two-thirds of F2:
Fw = (0.666) (F2) = (0.666) (408 lb/ft) = 272 lb/ft= (0.666) (F2) = (0.666) (6,012 N/m) = 4,004 N/m
The force resisting pullout, caused by the weight of the aggregate filled blocks above the top geogrid layer, is:
N2 = W2 = (120 lb/ft3) (0.97 ft) (3.44 ft) = 400 lb/ft= W2 = (1,923 kg/m3) (0.3 m) (1.05 m) = 5,942 N/m
P = Fr = 888 lb/ft + 0.130 (400 lb/ft) = 940 lb/ft= Fr = 12,964 N/m + 0.130 (5,942 N/m) = 13,736 N/m
The safety factor against pullout for the top layer of geogrid is:
SFP = (940 lb/ft) = 3.45 = (13,736 N/m) = 3.45(272 lb/ft) (4,004 N/m)
At a certain depth, the force holding the geogrid between the blocks will be equal to or greater than the long-termallowable design load of the geogrid. Any layer of geogrid located below this critical depth can be assumed to be safefrom pullout failure. The critical depth will be different for each wall depending on the type of soil, the slope of thebackfill, and the presence of surcharges, if any.
The long-term allowable design load for Huesker geogrid is 1100 lb/ft (16,059 N/m). In a properly designed wall, themaximum tensile force in any layer of geogrid will be less than or equal to 1100 lb/ft (16,059 N/m). The tensile forceon the geogrid at the back face of the concrete blocks is approximately two-thirds of the maximum tensile force:
Fw = (0.666) (1100 lb/ft) = 732 lb/ft= (0.666) (16,059 N/m) = 10,695 N/m
The required pullout resistance for a single layer of geogrid is equal to the tensile force at the back face of the concreteblocks times a safety factor of 1.5. Therefore, the maximum required pullout resistance for any one layer of geogrid is:
Fr = 1100 lb/ft= 16,059 N/m
29
The following equation can be used to calculate the critical depth at which the pullout resistance equals the long-termallowable design load of geogrid:
Fr = (SFP) (Wf) (Wb) (dc)where:
SFP = the safety factor against pulloutWf = the unit weight of the wall facing
Wb = the width of the wall facing (from front to back)
dc = the depth at which the pullout resistance equals the
long-term allowable design load of the geogrid.
The critical depth for the wall in Example 2-3 can be determined byrearranging and solving the equation given above:
dc = F(Wf) (Wb) (SFP)
= (1,100 lb/ft) = 6.3 ft(120 lb/ft3) (0.97 ft) (1.5)
= (16,059 N/m) = 1.89 m(1,923 kg/m3) (0.3 m) (1.5) (9.81)
For the wall in Example 2-3, any geogrid layer placed 6.3 ft (1.89 m) ormore below the level of the backfill need not be checked for pullout failure.
Diagram Ex. 2-3
Chapter ThreeSurcharges
IntroductionA surcharge is an external load applied to the retained soil. Typicalsurcharges include: sidewalks, driveways, roads, buildings, and otherretaining walls. Retaining walls as surcharges will be dealt with in aseparate section entitled "Tiered Walls." In this chapter, we will showhow to apply the force due to surcharges to each of the three types ofretaining walls -- simple gravity walls, tieback walls, and coherentgravity walls.
The effect a surcharge has on a wall depends on the magnitude of thesurcharge and the location of the surcharge relative to the wall. Asurcharge located directly behind a wall will have a much greatereffect than one located ten or twenty feet behind the wall. Generally,in good soil if the distance from the back of the wall to the surchargeis greater than the height of the wall, the effect of the surcharge will beinsignificant. Keep in mind that the back of a coherent gravity wall islocated at the end of the geogrid furthest from the wall facing.
In order to properly determine the effects of a surcharge load, it isnecessary to determine how the stress within the soil varies withvertical and horizontal distance from the surcharge. There areseveral theories about how to calculate the stress at some pointwithin the soil and they range from relatively simple to extremelycomplex. The one that we have chosen to use is illustrated in Figure3-1. We assume that the force due to a surcharge load on theretained soil is transmitted downward through the soil at an angle of45º + �/2 to the horizontal. (� is the friction angle of the soil.) Theplane of influence can be approximated by drawing a line up fromthe bottom rear edge of the wall at an angle of 45º + �/2 until itintersects the top of the backfill. Any surcharge located between thefront of the wall and the point of intersection will have a measurableeffect on the wall. Surcharges located beyond the point of intersection will have a minimal effect on the wall and willbe neglected.
The nature of a surcharge can be defined as a live load or a dead load. Essentially, a live load is that which is transientin its influence on the wall structure and a dead load is that which is taken as a permanent influence on the wall structure.In our calculations for stability, a conservative approach is followed that does not assume the presence of the live loadweight and vertical forces.
The location of the live or dead load surcharge, be it the retained soil or the infill soil affects individual forces on the wallresulting in increased or decreased stability factors of safety. For example, a coherent gravity wall with a live loadsurcharge on the infill soil will act to decrease FOS overstress and also decrease FOS for sliding and overturning. If thelive load surcharge is acting on the retained soil, we see decreases in FOS sliding and overturning. As for a coherentgravity wall with a dead load surcharge on the infill soils, we see a decrease in FOS overstress and an increase in FOSsliding and overturning. If the dead load is on the retained soil, we see an increase in FOS sliding and overturning.
Another assumption we make in analyzing a surcharge load is that the stress within the soil due to the surcharge isconstant with depth. This assumption is fairly accurate for surcharges covering a large area and will result in an error onthe conservative side while greatly simplifying the analysis. More exact methods of analysis are available and can beused if desired.
30
Figure 3-1
31
Surcharges on Simple Gravity Walls
Example 3-1:
Figure 3-2 shows the simple gravity wall of Example 2-1 with a uniform surcharge of 120 lb/ft (586 kg/m2) behind it.Assume that the surcharge is a sidewalk running parallel to the face of the wall. The sidewalk is 4 feet wide (1.22 m)and is located right next to the back of the wall. The first step in the analysisis to calculate the pressure on the retaining wall due to the surcharge:
Pq = (q) (Ka)= (120 lb/ft2) (0.2197) = (586 kg/m2) (0.2197)= 26 lb/ft2 = 129 kg/m2
Again, because of the effects of friction between the wall and the soil, thepressure due to the surcharge has both a horizontal component and a verticalcomponent. Therefore, the next step in the analysis is to calculate thehorizontal and vertical components of the pressure:
Pqh = (Pq) cos (�w)= (26 lb/ft2) cos (20º) = (129 kg/m2) cos (20º)= 24 lb/ft2 = 121 kg/m2 = 1,189 Pa
Pqv = (Pq) sin (�w)= (26 lb/ft2) sin (20º) = (129 kg/m2) sin (20º)= 9 lb/ft2 = 44 kg/m2 = 433 Pa
Finally, the total surcharge forces on the wall are calculated:
Fqh = (Pqh) (H)= (24 lb/ft2) (3.44 ft) = (121 kg/m2) (1.05 m)= 83 lb/ft = 127 kg/m = 1,246 N/m
Fqv = (Pqv) (H)= (9 lb/ft2) (3.44 ft) = (44 kg/m2) (1.05 m)= 31 lb/ft = 46.2 kg/m = 453 N/m
Figure 3-3 is a freebody diagram showing the active forces on the wall. Now that the force and pressure distributiondue to the surcharge are known, the wall can be analyzed as described in Chapter Two. (The rest of the forces havealready been calculated in Example 2-1.) For a simple gravity wall, the horizontal force due to the surcharge is aforce that tends to cause both sliding and overturning. Therefore, it must be added to those forces when the safetyfactors are calculated.
Figure 3-2. Simple Gravity Retaining Wall with Surcharge
Assumptions:
1. Stress in Soil Due to Surcharge Does Not Vary with Depth.
2. Wall Friction is Neglected in this Example.
Figure 3-1. Effect of Uniform Surcharge on a Retaining Wall
CASE 1X = 0
Pq = (q) (Ka)Sliding Force:
Fs = (Pq) (H2) COS (�w)Overturning Moment:Mq = (0.5) (H2) (Fs)
CASE 20 < X < L2
Pq = (q) (Ka)Sliding Force:
Fs = (Pq) (H1) COS (�w)Overturning Moment:Mq = (0.5) (H1) (Fs)
CASE 3X > L2
Pq = 0Sliding Force:
Fs = 0Overturning Moment:
Mq = 0
The safety factor against sliding is:
SFS = Fr + (Fqv) (Cf)Fh + Fqh
= (281 lb/ft) + (31 lb/ft) tan (30º) = 1.29(147 lb/ft) + (83 lb/ft)
= (4,130 N/m) + (453 N/m) tan (30º) = 1.29
(2,157 N/m) + (1,246 N/m)(NOTE: Fr and Fh were calculated in Example 2-1).
The safety factor against overturning is:
�Mr = (Wf) [ (X1) + (0.5) (H) tan (90º — �) ]+ (Fv) [ (X2) + (0.333) (H) tan (90º — �) ]+ (Fqv) [ (X2) + (0.5) (H) tan (90º — �) ]= (434 lb/ft) [ (0.49 ft) + (0.5) (3.44 ft) tan (90º — 78º) ]+ (53 lb/ft) [ (0.97 ft) + (0.333) (3.44 ft) tan (90º — 78º) ]+ (31 lb/ft) [ (0.97 ft) + (0.5) (3.44 ft) tan (90º — 78º) ]= 477 ft-lb/ft= (6,369 N/m) [ (0.149 m) + (0.5) (1.05 m) tan (90º — 78º) ]+ (785 N/m) [ (0.3 m) + (0.333) (1.05 m) tan (90º — 78º) ]+ (453 N/m) [ (0.3 m) + (0.5) (1.05 m) tan (90º — 78º) ]= 2,140 N-m/m
�Mo = (Fh) (0.333) (H) + (Fqh) (0.5) (H)= (147 lb/ft) (0.333) (3.44 ft) + (83 lb/ft) (0.5) (3.44 ft)= 311 ft-lb/ft= (2,157 N/m) (0.333) (1.05 m) + (1,246 N/m) (0.5) (1.05 m)= 1,408 N-m/m
SFO = �Mr = (477 ft-lb/ft) = 1.53 = (2,140 N-m/m) = 1.53�Mo (311 ft-lb/ft) (1,408 N-m/m)
Notice that, with the surcharge on the backfill, the safety factors are much lower than the recommended minimum valuesof 1.5 for sliding and 2.0 for overturning. This illustrates that a surcharge can make the difference between a stablewall and an unstable one.
32
Figure 3-3. Freebody Diagramof a Simple Gravity Wall with
Surcharge
Surcharges on Tieback WallsJust as in the case of simple gravity walls, a surcharge on atieback wall adds a horizontal force that contributes to wallfailure. However, the surcharge also adds to the verticalforce on the geogrid and this helps resist wall failure. Thevertical force due to the surcharge is transmitted downthrough the soil and the full force of the surcharge is felt onthe geogrid. To be conservative and to simplify the analysis,we will assume that the surcharge is felt only by the portionof the geogrid lying directly beneath the surcharge.
Example 3-2:
Analyze the tieback wall of Example 2-2 with a surcharge of120 lb/sq ft (586 kg/m2) beginning 2.0 ft (0.61 m) behindthe front wall face and ending 6.0 ft (1.83 m) behind the front wall face. Figure 3-4a is a schematic diagram of thewall and Figure 3-4b is a freebody diagram showing the forces on the wall. The first step in the analysis is to calculatethe maximum potential restraining force on the geogrid. This force has two components -- the force due to the weight ofthe soil on top of the geogrid and the force due to the surcharge. These two components are represented by the termsin the square brackets:
Fg = (2) (Ci) [ (dg) (�) (Le) + (q) (Lq) ] tan (�)
where:
q = the surcharge, in lb/ft2 (Pa)Lq = the length of the geogrid in the passive zone that is underneath the surcharge.
The maximum potential restraining force on the geogrid is:
Fg = (2) (0.85) [ (2.30 ft) (120 lb/ft3) (3.12 ft) = (2) (0.85) [ (0.70 m) (1,923 kg/m3) (0.95 m)
+ (120 lb/ft2) (2.62 ft) ] tan (30º) + (586 kg/m2) (0.80 m) ] tan (30º)= 1,154 lb/ft = 16,827 N/m
This is the maximum potential restraining force that can be developed by the geogrid due to the weight of the soil andthe surcharge. The actual restraining force will vary to balance the force on the wall due to the weight of the soil andthe surcharge force. If the actual force is less than both the maximum potential restraining force and the long-termallowable design load (including safety factor), then the design is acceptable. Since the maximum potential restrainingforce is over twice the long-term allowable design load (LTADL) of the geogrid selected, check to see if a restrainingforce equal to the LTADL will provide acceptable factors of safety. If so, the design is adequate.
For this example, let Fg = 833 lb/ft (12,161 N/m), the LTADL of the selected geogrid, and calculate the factors ofsafety. The first step is to calculate the magnitude of the pressure on the wall due to the surcharge:
Pq = (q) (Ka)= (120 lb/ft2) (0.2197) = (5,747 Pa) (0.2197)= 26 lb/ft2 = 1,263 Pa
Next, calculate the horizontal and vertical components of the pressure on the wall due to the surcharge:
Pqh = (Pq) cos (�w)= (26 lb/ft2) cos (20º) = (1,150 Pa) cos (20º)= 24 lb/ft2 = 1,081 Pa
Pqv = (P) sin (�w)= (26 lb/ft2) sin (20º) = (1,245 Pa) sin (20º)= 9 lb/ft2 = 426 Pa
33
Figure 3-4a. Tieback Retaining Wall
Finally, the total surcharge forces on the wall are calculated:
Fqh = (Pqh) (Hq)= (24 lb/ft2) (2.33 ft)= 56 lb/ft= (1,081 Pa) (0.71 m)= 768 N/m
Fqv = (Pqv) (Hq)= (9 lb/ft2) (2.33 ft)= 21 lb/ft= (426 Pa) (0.71 m)= 302 N/m
Next, the safety factors can be calculated:
SFS = Fr + Fg + (Fqv) (Cf)Fh + Fqh
= (445 lb/ft) + (833 lb/ft) + (21 lb/ft) tan (30) = 3.34(330 lb/ft) + (56 lb/ft)
= (6,507 N/m) + (12,161 N/m) + (302 N/m) tan (30º) = 3.34(4,800 N/m) + (768 N/m)
�Mr = (Wf) [ (X1) + (0.5) (H) tan (90º — �) ] + (Fv) [(X2 + (0.333) (H) tan (90º — �)]+ (Fqv) [ (X2) + (0. 5) (Hq) tan (90º — �) ] + (Fg) (H — dg)= (651 lb/ft) [ (0.49 ft) + (0.5) (5.16 ft) tan (90º — 78º) ]+ (120 lb/ft) [ (0.97 ft) + (0.333) (5.16 ft) tan (90º — 78º) ]+ (21 lb/ft) [ (0.97 ft) + (0.5) (2.33 ft) tan (90º — 78º) ]+ (833 lb/ft) (5.16 ft 2.30 ft) = 3,244 ft-lb/ft= (9,523 N/m) [ (0.149 m) + (0.5) (1.57 m) tan (90º — 78º) ]+ (1,747 N/m) [ (0.296 m) + (0.333) (1.57 m) tan (90º — 78º) ]+ (302 N/m) [ (0.296 m) + (0.5) (0.71 m) tan (90º — 78º) ]+ (12,161 N/m) (1.57 m — 0.701 m) = 14,399 N-m/m
�Mo = (Fh) (0.333) (H) + (Fqh) (Hq)= (330 lb/ft) (0.333) (5.16 ft) + (56 lb/ft) (0.5) (2.33 ft) = 632 ft-lb/ft
= (4,800 N/m) (0.333) (1.57 m) + (768 N/m) (0.5) (0.71 m) = 2,782 N-m/mSFO = �Mr = (3,244 ft-lb/ft) = 5.13
�Mo (632 ft-lb/ft)
= �Mr = (14,399 N-m/m) = 5.13�Mo (2,782 N-m/m)
Since SFS > 1.5 and SFO > 2.0, the stability of this retaining wall is acceptable. Note that in some cases thesurcharge actually increases the stability of the wall. Therefore, it is important to check the stability of the wall withoutthe surcharge if it is possible that the surcharge may be removed. Also check the stability without the surcharge if it isa live-load surcharge, such as a parking lot or driveway.
Figure 3-4b. Tieback Retaining Wall
34
Surcharges on CoherentGravity WallsAnalyzing the effects of a surcharge on a coherent gravitywall is a two-part problem. First, the effect on the entirereinforced soil mass (external stability) must be analyzed.The surcharge will have an effect on both sliding failureand overturning failure. Second, the effect of thesurcharge on the individual layers of geogrid (internalstability) must be analyzed. The surcharge will affect thestress in each layer of geogrid and will influence thespacing of the layers.
External Stability
The effect of a surcharge on the external stability of a coherent gravity retaining wall is nearly identical to the effect ona simple gravity wall and depends on the location of the surcharge. Recall that the back of a coherent gravity wall islocated at the end of the geogrid farthest from the wall facing.
Figure 3-5 shows three possible locations of a surcharge. The surcharge in Location A contributes to the forces resistingboth sliding and overturning. Surcharges at location B contribute to the forces causing sliding and overturning. InLocation C, the surcharge contributes partly to the forces causing sliding and partly to the forces resisting sliding. In thesame manner, it also contributes both to the forces causing overturning and the forces resisting overturning.
Example 3-3:
Consider the coherent gravity wall analyzed in Example 2-3, but with a three-foot-wide surcharge of 120 lb/ft2 (5,748 Pa)on the backfill. Analyze the external stability of the wall with the surcharge in the three locations shown in Figure 3-5.
Location A:
The surcharge can be resolved into an equivalent vertical force, Q, of 360 lb/ft (5,256 N/m) that is located 2.5 ft (0.762 m)from the front face of the wall and acts at the center of the uniform surcharge. This force can be added to the forcesresisting sliding when calculating Fr:
Fr = (Ww + Fv + Q) (Cf)= [ (7,071 lb/ft) + (399 lb/ft) + (360 lb/ft) ] tan (30) = 4,521 lb/ft= [ (103,539 N/m) + (5,825 N/m) + (5,256 N/m) ] tan (30º) = 66,176 N/m
35
Figure 3-5. Locations of Surcharge on Coherent Gravity Walls
The new safety factor against sliding is:
SFS = Fr = (4,521 lb/ft) = 3.68 = (66,176 N/m) = 3.68
Fh (1,229 lb/ft) (18,011 N/m)
Q can also be added to the moments of the forces resisting overturning:
�Mr = (Ww) [ (X1) + (0.5) (H) tan (90º — �)]+ (Fv) [ (X2) + (0.333) (H) tan (90º — �) ]+ (Q) [ (X3) + (H) tan (90º — �) ]= (7,071 lb/ft) [ (3.0 ft) + (0.5) (9.17 ft) tan (90º — 78º) ]+ (399 lb/ft) [ (6. 13 ft) + (0. 333) (9.17 ft) tan (90º — 78º) ]+ (360 lb/ft) [ (2.5 ft) + (9.17 ft) tan (90º — 78º) ]= 32,411 ft-lb/ft= (103,539 N/m) [ (0.91 m) + (0.5) (2.8 m) tan (90º — 78º) ]+ (5,852 N/m) [ (1.87 m) + (0. 333) (2.8 m) tan (90º — 78º) ]+ (5,256 N/m) [ (0.762 m) + (2.8 m) tan (90º — 78º) ]= 144,212 N-m/m
�Mo = (Fh) (0.333) (H)= (1,229 lb/ft) (0.333) (9.17 ft) = 3,753 ft-lb/ft= (18,011 N/m) (0.333) (2.8 m) = 16,793 N-m/m
The new safety factor against overturning is:
SFO = �Mr = (32,411 ft-lb/ft) = 8.64 = (144,212 N-m/m) = 8.64�Mo (3,753 ft-lb/ft) (16,793 N-m/m)
Thus, the effect of a surcharge in Location A is to make the wall slightly more stable with respect to sliding and overturning.However, such a surcharge can have a detrimental effect on the internal stability of the wall. Also, the added force dueto the surcharge must be taken into account when calculating the bearing pressure on the underlying soil.
Location B:
A surcharge in this location has the same effect on the external stability of a coherent gravity wall as on a simple gravitywall. In this case, the surcharge results in a horizontal force with its point of application located at H/2 on the back ofthe reinforced soil mass. The magnitude of the force is:
Fq = (q) (Kao) (H)= (120 lb/ft2) (0.2561) (9.17 ft) = 282 lb/ft= (5,748 Pa) (0.2561) (2.8 m) = 4,122 N/m
The horizontal and vertical components of the force on the reinforced soil mass due to the surcharge are:
Fqh = (Fq) cos (�wo)= (282 lb/ft) cos (18º) = 268 lb/ft= (4,122 N/m) cos (18º) = 3,920 N/m
Fqv = (Fq) sin (�wo)= (282 lb/ft) sin (18º) = 87 lb/ft= (4,122 N/m) sin (18º) = 1,274 N/m
Notice that the pressure coefficient for the onsite soil is used. This is because the surcharge is located entirely outsidethe reinforced soil zone and the surcharge force is transmitted through the onsite soil.For Location B, the safety factors against sliding and overturning are:
36
SFS = Fr + (Fqv) (Cf)Fh + Fqh
= (4,313 lb/ft) + (87 lb/ft) tan (27º) = 2.91(1,229 lb/ft) + (268 lb/ft)
= (63,157 N/m) + (1,274 N/m) tan (27º) = 2.91(18,011 N/m) + (3,920 N/m)
SFO = �Mr + (Fqv) [ (X2) + (0.5) (H) tan (90º — �) ]�Mo + (Fqh) (0.5) (H)
= (31,155 ft-lb/ft) + (87 lb/ft) [ (6.13 ft) + (0.5) (9.17 ft) tan (90º — 78º) ](3,753 ft-lb/ft) + (268 lb/ft) (0.5) (9.17 ft)
= 6.38= (139,375 N/m) + (1,274 N/m) [ (1.87 m) + (0.5) (2.8 m) tan (90º — 78º) ]
(16,793 N-m/m) + (3,920 N/m) (0.5) (2.8 m)= 6.38
Location C:
With the surcharge at Location C, half of the surcharge is over the reinforced soil zone and half is not. Therefore, theeffects on the coherent gravity wall are a combination of the effects of a surcharge at Location A and a surcharge atLocation B. The part of the surcharge over the geogrid will contribute to the stability of the wall with respect to sliding andoverturning. The horizontal and vertical components of the force on the reinforced soil mass due to the surcharge are:
Fq = (q) (H) (Kao)= (120 lb/ft2) (9.17 ft) (0.2561) = 282 lb/ft= (5,748 Pa) (2.8 m) (0.2561) = 4,122 N/m
Fqh = (Fq) cos (�wo)= (282 lb/ft) cos (18) = 268 lb/ft= (4,122 N/m) cos (18º) = 3,920 N/m
Fqv = (Fq) sin (�wo)= (282 lb/ft) sin (18) = 87 lb/ft= (4,122 N/m) sin (18º) = 1,274 N/m
The force resisting sliding is:
Fr = [Ww + Fv + 0.5 (Q) + Fqv] (Cf)= [ (7,071 lb/ft) + (399 lb/ft) + (180 lb/ft) + (87 lb/ft) ] tan (30) = 4,467 lb/ft= [ (103,539 N/m) + (5,852 N/m) + (2,628 N/m) + (1,274 N/m) ] tan (30º) = 65,410 N/m
The force causing sliding is:
Fs = Fh + Fqh= (1,229 lb/ft) + (268 lb/ft) = 1,497 lb/ft= (18,011 N/m) + (3,920 N/m) = 21,931 N/m
37
The safety factor against sliding is:
SFS = (4,467 lb/ft) = 2.98(1,497 lb/ft)
= (65,410 N/m) = 2.98(21,931 N/m)
The sum of the moments resisting overturning is:
�Mr = (Ww) [ (X1) + (0.5) (H) tan (90º — �) ]+ (Fv) [ (X2) + (0.333) (H) tan (90º — �) ]+ (Fqv) [ (X2) + (0.5) (H) tan (90º — �) ]+ (0.5) (Q) [ (X3) + (H) tan (90º — �) ]= (7,071 lb/ft) [ (3.0 ft) + (0.5) (9.17 ft) tan (90º — 78º) ]+ (399 lb/ft) [ (6.13 ft) + (0.333) (9.17 ft) tan (90º — 78º) ]+ (87 lb/ft) [ (6.13 ft) + (0.5) (9.17 ft) tan (90º — 78º) ]+ (0.5) (360 lb/ft) [ (5.25 ft) + (9.17 ft) tan (90º — 78º) ]= 32,723 ft-lb/ft= (103,539 N/m) [ (0.91 m) + (0.5) (2.8 m) tan (90º — 78º) ]+ (5,852 N/m) [ (1.87 m) + (0.333) (2.8 m) tan (90º — 78º) ]+ (1,274 N/m) [ (1.87 m) + (0.5) (2.8 m) tan (90º — 78º) ]+ (0.5) (5,256 N/m) [ (1.6 m) + (2.8 m) tan (90º — 78º) ]= 145,609 N-m/m
The sum of the moments causing overturning is:
�Mo = (Fh) (0.333) (H) + (Fqh) (0.5) (H)= (1,229 lb/ft) (0.333) (9.17 ft) + (268 lb/ft) (0.5) (9.17 ft)= 4, 982 ft-lb/ft= (18,011 N/m) (0.333) (2.8 m) + (3,920 N/m) (0.5) (2.8 m)= 22,281 N-m/m
The safety factor against overturning is:
SFO = (32,723 ft-lb/ft) = 6.57(4,982 ft-lb/ft)
= (145,609 N-m/m) = 6.57(22,281 N-m/m)
38
39
Internal StabilityIn addition to its effects on sliding and overturning failure, asurcharge can also have an impact on the spacing of the geogridlayers. It does so by putting an additional load on some or all of thelayers of geogrid.
The first step in analyzing the effects of a surcharge on internalstability is to determine the horizontal soil stress within the reinforcedsoil zone. Once again, we will use the wall of Example 2-3 with asurcharge of 120 lb/sq ft (5,747 Pa), located as shown in Figure 3-6. The surcharge is 2 ft (0.61 m) wide.
Notice the diagonal lines connected to the beginning and end of thesurcharge pressure diagram. These lines are drawn at an angle of45º + �/2 to the horizontal and mark the limits of the zone ofinfluence of the surcharge within the soil. The horizontal stress dueto the surcharge will act only on the portion of the retaining walllocated in the area labeled “ZONE OF INFLUENCE.”
The magnitude of the horizontal surcharge stress is:
Pqh = (q) (Kar) cos (�wr)= (120 lb/ft2) (0.2197) cos (20º)= 25 lb/ft2
= (5,747 Pa) (0.2197) cos (20º)= 1,186 Pa
Figure 3-7 shows the wall facing with the two pressure distributionsthat affect it - one due to the soil weight and one due to the surcharge.The rectangular pressure distribution represents the effect of thesurcharge on the wall facing.
The addition of the surcharge stress makes calculating the gridspacing more complicated. Instead of solving a linear equation tofind the maximum allowable distance between two layers, thedesigner must either solve a quadratic equation or use a trial anderror method.
The quadratic equation is:
0 = [ (�r) (Kar) cos (�wr) ] (dh)2
+ (2) [ (d1) (�r) (Kar) cos (�wr) + (q) (Kar) cos (�wr) ] (dh) + 2 (Fga)
where:d1 = depth from the top of the wall to the bottom of the area reinforced by this layer of geogrid
Fga = the long-term allowable design strength of the geogrid.
The quadratic formula is:
0 = ax2 + bx + c
To use the quadratic formula to solve for x = dh, let:
z = (Kar) cos (�wr) b = (2) [ (d1) (a) + (q) (z) ]a = (�r) (z) c = (2) (Fga)
Figure 3-6. Coherent GravityWall with Surcharge
Figure 3-7. Pressure DistributionsDue to the Weight and Surcharge
Example 3-4:
Given the wall depicted in Figure 3-6, and using the data of Example 2-3, determine the geogrid spacing for the wall.Use the quadratic formula to determine dh for the first layer of geogrid:
a = (�r) (Kar) cos (�wr)= (125 lb/ft3) (0.2197) cos (20)= 26 lb/ft3
= (2,002 kg/m3) (0.2197) cos (20º)= 413 kg/m3
b = (—2) [ (d1) (�r) (Kar) cos (�wr) + (q) (Kar) cos (�wr) ]= (—2) [ (9.17 ft) (125 lb/ft3) (0.2197) cos (20)+ (120 lb/ft2) (0.2197) cos (20) ] = 523 lb/ft2
= (—2) [ (2.8 m) (2,002 kg/m3) (0.2197) cos (20)+ (586 kg/m2) (0.2197) cos (20) ] = 2,557 kg/m2 = 25.07 kPa
c = (2) (Fga)= (2) (833 lb/ft) = 1,666 lb/ft
= (2) (12,161 N/m) = 2,479 kg/m = 24,322 N/m
dh = —b � b2 — 4ac2a
= —(—523 lb/ft2) � (—523 lb/ft2)2 — (4) (26 lb/ft3) (1,666 lb/ft)(2) (26 lb/sq ft3)
= 16.15 or 3.97= —(—2,557 kg/m) � (—2,557 kg/m2)2 — (4) (413 kg/m3) (2,479 kg/m)
(2) (413 kg/m3)= 5.0 or 1.2
Since the wall is only 9.17 ft (2.8 m) tall, the first root, 16.15 (4.9 m), cannot be valid. Therefore,
dh = 3.97 ft (1.2 m)d2 = d1 — dh
= 9.17 ft — 3.97 ft = 2.8 m — 1.2 m= 5.2 ft = 1.6 m
The first layer of geogrid should be placed at a height equal to one half of dh:
hg = (0.5) (3.97 ft) = 1.99 ft = (0.5) (1.2 m) = 0.6 m
However, for Allan Block standard units, the geogrid can only be placed at heights that are even multiples of 7.62 inches(19.4 cm). In this case, the geogrid should be placed 3 blocks up, at a height of 1.905 ft (0.58 m).The next step in the analysis is to determine if more than one additional layer of geogrid is required. This is done bycalculating the total horizontal force on the wall above height d2 and comparing it to the allowable design strength ofthe geogrid. The horizontal component of the active force above d2 is:
Fh = (0.5) (�r) (Kar) (d2)2 cos (�wr)
= (0.5) (125 lb/ft3) (0.2197) (5.2 ft)2 cos (20º) = 349 lb/ft
= (0.5) (2,002 kg/m3) (0.2197) (1.6 m)2 cos (20º) = 5,190 N/m
40
The horizontal component of the surcharge force is calculated based on the height from d2 to the top of the zone of
influence depicted in Figure 3-7:
Qh = (q) (Kar) (d2 — hz) cos (�wr)= (120 lb/sq ft) (0.2197) (5.2 ft — 1.905 ft) cos (20º) = 82 lb/ft= (5,748 Pa) (0.2197) (1.6 m — 0.58 m) cos (20º) = 1,210 N/m
The total horizontal force on the wall above height d2 is:
Ft = Fh + Qh= (349 lb/ft) + (82 lb/ft) = 431 lb/ft= (5,190 N/m) + (1,210 N/m) = 6,400 N/m
Since the total horizontal force above height d2 is less than the allowable design strength of the geogrid, 833 lb/ft(12,161 N/m), then only one more layer of geogrid is required. In that case, the new dh is equal to the height of thewall above d2 (If more than one layer is required, set d1 = d2 and use the quadratic formula to determine dh for thenext layer of geogrid.) The geogrid should be placed at a height equal to:
hg = (H — d2) + (0.5) dh= (9.17 ft — 5.2 ft) + (0.5) (5.2 ft) = 6.57 ft= (2.8 m — 1.6 m) + (0.5) (1.6 m) = 2 m
Rounding down to the nearest 7.62 inches (19.4 cm): hg = 6.35 ft (1.9 m).
Another way to tell that you are working on the last layer of geogrid is to go ahead and do the analysis usingthe quadratic formula. If you calculate a negative number for the quantity [b2 4ac], you are working on thelast layer of geogrid.
To check the number of layers of geogrid required, calculate the total horizontal force on the wall facing and divide itby the allowable design strength of the geogrid. The horizontal component of the active force on the wall facing is:
Fh = (0.5) (125 lb/ft3) (0.2197) (9.17 ft)2 cos (20) = 1,085 lb/ft= (0.5) (2,002 kg/m3) (0.2197) (2.8 m)2 cos (20º) = 15,894 N/m
The horizontal component of the surcharge force on the wall facing is:
Qh = (120 lb/sq ft) (0.2197) (9.17 ft — 1.905 ft) cos (20º) = 180 lb/ft
= (5,748 Pa) (0.2197) (2.8 m — 0.58 m) cos (20º) = 2,634 N/m
The total horizontal force on the wall facing is:
Ft = (1,085 lb/ft) + (180 lb/ft) = 1,265 lb/ft= (15,894 N/m) + (2,634 N/m) = 18,528 N/m
The number of layers of geogrid required is:
N = (1,265 lb/ft) = 1.52(833 lb/ft)
= (18,528 N/m) = 1.52(12,161 N/m)
Therefore, 2 layers of geogrid are required.
41
Tiered WallsSometimes it is desirable to build two or more smaller walls at different elevations rather than one very tall wall. Suchan arrangement is called a tiered wall and an example is pictured in Figure 3-8. The analysis of tiered walls can becomevery complicated. We have decided upon a design method that we feel comfortable with and will briefly describe itbelow. However, you as an engineer must use your own judgement. If you are not comfortable with this design method,use your best engineering judgement or seek advice from a local expert.
You should also be aware that, as the number and walls increase, the threat of global instability increases also. A tieredwall consisting of three 5 ft (1.52 m) walls can have as great an impact on the underlying soil as a single 15 ft (4.6 m)wall. Again, if you are concerned about the global stability, you should do a global stability analysis or have someonedo one for you.
The first step in the design of a tiered wall is to decide how many tiers there will be and the height of each tier. Then,using the design procedures presented earlier, design the top retaining wall. Next, find the average bearing stress ofthe top wall on the underlying soil. This average bearing stress is then applied as a uniform surcharge to the retainedsoil mass of the second wall from the top. (See Figure 3-9.) The second wall is then analyzed using the proceduresdescribed earlier in this chapter.
The process is repeated until all of the tiers have been analyzed. As a final step, check the maximum soil bearingpressure of the bottom wall to make sure it doesn't exceed the allowable bearing pressure of the onsite soil.
42
Figure 3-8. Retaining Wallwith Three Tiers
Figure 3-9. Average Bearing Stress of TopWall Applied as Surcharge to Second Wall
CHAPTER FOURSloped Backfill
IntroductionSometimes it is not feasible or desirable to build a retaining wall that is tall enough toallow for a flat backfill. In that case, the backfill must be sloped. Sloped backfill is oneof the most significant factors contributing to the active force on the wall. The slope ofthe backfill must be taken into account when designing a geogrid-reinforced retainingwall. Also, it should be noted that the slope of the backfill cannot exceed the frictionangle of the soil. (This is not true if the cohesion of the soil is taken into account.However, the design procedures in this manual are based on the assumption that onlynoncohesive soils will be used as backfill.)
Simple Gravity Walls With Sloped BackfillAs discussed in Chapter One, Coulomb's equation for the active force on the wallincludes a term that changes the magnitude of the pressure coefficient as the slope of thebackfill changes. The active pressure coefficient of Coulomb's equation is given by:
where: i = the slope of the backfill.
csc (�) sin (� — �) 2sin (� + �w) + sin (� + �w) sin (� — i)
sin (� i)
Let's look at the wall in Example 2-1 and see what effect changing the backfill slope has on the active force.
Example 4-1:
Given:�w = 20º H = 3.44 ft (1.05 m)
� = 30º � = 120 lb/ft3 (1,923 kg/m3)� = 78º Unit weight of wall facing = 130 lb/ft3 (2,061 kg/m3)
The table below shows the effect increasing the backfill slope has on the active pressure coefficient and the active force.
Changing the slope of the backfill from 0º to 26º increased the active force by 67%. The wall in Example 2-1 wouldnot be stable if the back-fill had a slope of 26º. For simple gravity walls, the effect of the sloping backfill is automaticallytaken into account by using Coulomb's equation to calculate the active force.
43
[ ]
i
(degrees)
0
18
26
Ka
0.2197
0.2847
0.3662
Fa
1 lb/ft (1 N/m)
156 (2,277)
202 (2,949)
260 (3,796)
Ka =
Tieback Walls with Sloped BackfillJust as in the case of simple gravity walls, the effect of the sloping backfillis automatically taken into account when calculating the active forceusing Coulomb's equation. However, the sloped backfill also has aneffect on the maximum potential restraining force on the geogrid thatshould be taken into account.
Figure 4-1 shows the tieback wall of Example 2-2, but with the backfill slopedat 18. In Example 2-2, the maximum potential restraining force on the geogridwas calculated to be 845 lb/ft (12,336 N/m) (see page 2-12). Calculate themaximum potential restraining force on the geogrid for the wall in Figure 4-1 andcompare it to the value calculated in Example 2-2.
Example 4-2:
Given:i = 18º Ka = 0.2847 H = 5.16 ft (1.57 m)
�w = 20º dg = 2.3 ft (0.7 m) Lg = 5.0 ft (1.52 m)
� = 30º Hg = 2.86 ft (0.87 m) � = 120 lb/ft3 (1,923 kg/m3)
� = 78º Unit weight of wall facing = 130 lb/ft3 (2,061 kg/m3)
The maximum potential restraining force on the geogrid is given by:
Fg = (2) (Ci) (�) (davg) (Le) tan (�)
where:
davg = (0.5) (d1 + d2)d1 = depth from surface of backfill to the geogrid at the point where the failure plane intersects the geogridd2 = depth from surface of backfill to the geogrid at the back end of the geogrid layer.
Assuming that the backfill slope begins at the front of the wall facing, then L1 is given by:
L1 = 0.97 ft + Xa Xh= (0.97 ft) + (Hg) tan (45º — �/2) — (H) tan (90º — �)= (0.97 ft) + (2.86 ft) tan (45º — 15º) — (5.16 ft) tan (90º — 78º) = 1.52 ft= 0.3 m + Xa Xh= (0.3 m) + (Hg) tan (45º — �/2) — (H) tan (90º — �)= (0.3 m) + (0.87 m) tan (45º — 15º) — (1.57 m) tan (90º — 78º) = 0.47 m
Once L1 is known, d2 can be calculated:
d1 = dg + (L1) tan (i)= (2.3 ft) + (1.52 ft) tan (18º) = 2.79 ft= (0.7 m) + (0.46 m) tan (18º) = 0.85 m
L2 is given by:
L2 = Lg — (dg) tan (90º — �)= (5.0 ft) — (2.3 ft ) tan (90º — 78º) = 4.51 ft= (1.52 m) — (0.7 m) tan (90º — 78º) = 1.37 m
44
Figure 4-1. Tieback Wallwith Sloped Backfill
Once L2 is known, d2 can be calculated:
d2 = dg + (L2) tan (i)= (2.3 ft) + (4.51 ft) tan (18º) = 3.77 ft= (0.7 m) + (1.37 m) tan (18º) = 1.15 m
The length of geogrid embedded in the passive zone of the soil is:
Le = L2 — L1= 4.51 ft — 1.52 ft = 2.99 ft= 1.37 m — 0.47 m = 0.9 m
Now, davg can be calculated and the maximum potential restraining force on the geogrid can be determined:
davg = (0.5) (d1 + d2)= (0.5) (2.79 ft + 3.77 ft) = 3.28 ft= (0.5) (0.85 m + 1.15 m) = 1.0 m
Fg = (2) (0.85) (120 lb/ft3) (3.26 ft) (3.12 ft) tan (30) = 1,198 lb/ft= (2) (0.85) (1,923 kg/m3) (0.99 m) (0.9 m) tan (30º) = 16,497 N/m
The maximum potential restraining force on the geogrid for a flat back fill was 845 lb/ft (12,336 N/m). Increasing the slope ofthe backfill from 0º to 18º increased the magnitude of Fg by 42%. However, as pointed out in Chapter Two, the long-termallowable design strength of AB-260 geogrid is 833 lb/ft (12,161 N/m). Therefore, Fg should be set equal to 833 lb/ft(12,161 N/m) before calculating the safety factors for sliding and overturning. If the safety factors are too low, add one or morelayers of geogrid and analyze the wall as a coherent gravity wall.
Let's complete the example and see what effect the sloping backfill has on the safety factors. First, the active force onthe wall must be calculated using the new active pressure coefficient:
Fa = (0.5) (120 lb/ft3) (0.2847) (5.16 ft)2 = 455 lb/ft= (0.5) (1,923 kg/m3) (0.2847) (1.57 m)2 = 6,619 N/m
The horizontal and vertical components of the active force are:
Fh = (455 lb/ft) cos (20º) = 428 lb/ft= (6,619 N/m) cos (20º) = 6,220 N/m
Fv = (455 lb/ft) sin (20) = 156 lb/ft= (6,619 N/m) sin (20º) = 2,264 N/m
The force resisting sliding is:
Fr = (651 lb/ft + 156 lb/ft) tan (30º) = 466 lb/ft= (9,523 N/m + 2,264 N/m) tan (30º) = 6,805 N/m
The safety factor against sliding is:
SFS = (466 lb/ft + 833 lb/ft) = 3.04(428 lb/ft)
= (6,805 N/m + 12,161 N/m) = 3.04(6,220 N/m)
45
The safety factor against overturning is:
�Mr = (Wf) [ (X1) + (0.5) (H) tan (90º — �) ]+ (Fv) [ (X2) + (0.333) (H) tan (90º — �) ] + (Fg) (Hg)
= (651 lb/ft) [ (0.49 ft) + (0.5) (5.16) tan (90º — 78º) ]+ (156 lb/ft) [ (0.97 ft) + (0.333) (5.16 ft) tan (90º — 78º) ]+ (833 lb/ft) (2.86 ft)= 3,267 ft-lb/ft= (9,523 N/m) [ (0.15 m) + (0.5) (1.57 m) tan (90º — 78º) ]+ (2,264 N/m) [ (0.3 m) + (0.333) (1.57 m) tan (90º — 78º) ]+ (12,161 N/m) (0.87 m)= 14,528 N-m/m
Mo = (Fh) (0.333) (H)= (428 lb/ft) (0.333) (5.16 ft)= 735 ft-lb/ft= (6,220 N/m) (0.333) (1.57 m)= 3,252 N-m/m
SFO = (3,267 ft-lb/ft) = 4.44(735 ft-lb/ft)
= (14,528 N-m/m) = 4.44(3,252 N-m/m)
With the backfill sloped at 18º, the wall is still stable although the safety factors have been reduced by about 20%.
46
47
Coherent Gravity Walls With Sloped Backfill
One effect of a sloped backfill on a coherent gravity wall is to increase theweight of the wall and consequently, the resistance to sliding. Theincreased weight is due to the backfill soil that is located above the wallfacing and over the reinforced soil mass. In Figure 4-2, the area designat-ed Wt contains the soil that contributes the extra weight. The total weightof the wall can be calculated by adding the weight of the rectangular sec-tion, Wr to the weight of the triangular section, Wt:
Wr = (130 lb/ft3) (9.17 ft) (0.97 ft)
+ (125 lb/ft3) (9.17 ft) (6.0 ft — 0.84 ft)= 7,071 lb/ft= (2,061 kg/m3) (2.8 m) (0.3 m)+ (2,002 kg/m3) (2.8 m) (1.83 m — 0.26 m)= 103,319 N/m
Wt = (0.5) (6.0 ft) [ (6.0 ft) tan (18º) ] (125 lb/ft3)
= 731 lb/ft= (0.5) (1.83 m) [ (1.83 m) tan (18º) ] (2,002 kg/m3)= 10,685 N/m
Ww = (Wr) + (Wt)
= (7,071 lb/ft) + (731 lb/ft) = 7,802 lb/ft= (103,319 N/m) + (10,685 N/m) = 114,004 N/m
External StabilityThe external stability of the wall can be calculated as it was in Example 2-3, with three differences. First, the weight ofthe wall is greater, as shown above. Second, the height of the retaining wall is taken to be the height at the back of thereinforced soil mass, He. Third, the active force on the retained soil mass is greater because of the sloping backfill. Theincrease in the active force is automatically accounted for by using Coulomb's equation to calculate the active force.Calculate the safety factors for sliding and overturning of the wall in Figure 4-2. Compare these values to the safetyfactors in Example 2-3.
Example 4-3:
Given:
i = 18 �r = 30 H = 9.17 ft (2.8 m)
�wo = 18 � = 78 �o = 120 lb/ft3 (1,923 kg/m3)
�wr = 20 Kao = 0.3440 �r = 125 lb/ft3 (2,002 kg/m3)
�o = 27' Kar = 0.2847
The first step is to calculate the effective height, He at the rear of the coherent gravity wall:
He = (H) + (Lg) tan (i)
= (9.17 ft) + (6.0 ft) tan (18) = 11.12 ft= (2.8 m) + (1.83 m) tan (18º) = 3.39 m
Figure 4-2. Coherent GravityWall with Sloped Backfill
Next, the active force on the coherent gravity wall is calculated:
Fa = (0.5) (�o) (Kao) (He)2
= (0.5) (120 lb/ft3) (0.3440) (11.12 ft)2 = 2,552 lb/ft= (0.5) (1,923 kg/m3) (0.3440) (3.39 m)2 = 37,289 N/m
The horizontal component of the active force is:
Fh = (Fa) cos (�wo)= (2,552 lb/ft) cos (18º) = 2,427 lb/ft= (37,289 N/m) cos (18º) = 35,464 N/m
The vertical component of the active force is:Fv = (Fa) sin (�wo)
= (2,552 lb/ft) sin (18º) = 789 lb/ft= (37,289 N/m) sin (18º) = 11,523 N/m
The force resisting sliding is:Fr = (Ww + Fv) (Cf)
= (7,802 lb/ft + 789 lb/ft) tan (30) = 4,960 lb/ft= (114,004 N/m + 11,523 N/m) tan (30º) = 72,473 N/m
The safety factor against sliding is:
SFS = Fr = (4, 960 lb/ft) = 2.04 = (72,473 N/m) = 2.04Fh (2,427 lb/ft) (35,464 N/m)
The moment resisting overturning is:
�Mr = (Wf) [ (X1) + (0.5) (H) tan (90º — �) ] + (Wr) [ (X2) + (0.5) (H) tan (90º — �) ]+ (Wt) [ (X3) + (H) tan (90º — �) ] + (Fv) [ (X4) + (0.333) (He) tan (90º — �) ]
= (1,156 lb/ft) [ (0.49 ft) + (0.5) (9.17 ft) tan (90º — 78º) ]+ (7,071 lb/ft) [ (3.55 ft) + (0.5) (9.17 ft) tan (90º — 78º) ]+ (731 lb/ft) [ (4.08 ft) + (9.17 ft) tan (90º — 78º) ]+ (789 lb/ft) [ (6.13 ft) + (0.333) (11.12 ft) tan (90º — 78º) ]= 43,551 ft-lb/ft= (16,983 N/m) [ (0.149 m) + (0.5) (2.8 m) tan (90º — 78º) ]+ (103,319 N/m) [ (1.08 m) + (0.5) (2.8 m) tan (90º — 78º) ]+ (10,685 N/m) [ (1.24 m) + (2.8 m) tan (90º — 78º) ]+ (11,523 N/m) [ (1.87 m) + (0.333) (3.39 m) tan (90º — 78º) ]= 193,836 N-m/m
The moment causing overturning is:
Mo = (Fh) (0.333) (He)= (2,427 lb/ft) (0.333) (11.12 ft) = 8,987 ft-lb/ft= (35,464 N/m) (0.333) (3.39 m) = 40,034 N-m/m
The safety factor against overturning is:SFO = �Mr = (43,551 ft-lb/ft) = 4.85
�Mo (8,987 ft-lb/ft)
= �Mr = (193,836 N-m/m) = 4.85�Mo (40,034 N-m/m)
48
As calculated in Example 2-3, the same wall with a flat backfill had a safety factor against sliding of 3.5 and a safetyfactor against overturning of 8.3. Sloping the backfill cut the safety factors by 42% for sliding and 48% for overturning.
Internal StabilityStudies have shown that the failure plane for the soil inside the reinforced soil mass is not well represented by a straightline at an angle of 45º + �/2 to the horizontal. Instead, the failure surface looks more like the one depicted in Figure4-3. It begins at the bottom rear edge of the wall facing and extends upward at an angle of 45º + �/2 to the horizontal.The failure surface continues upward at that angle until it intersects a vertical line located behind the wall facing adistance equal to one-third the height of the wall.
When analyzing the loads on the individual layers of geogrid, the depth should be measured from the point where thefailure plane intersects the geogrid layer to the top of the backfill. However, to simplify the analysis, the depth can bemeasured from the point where the vertical portion of the assumed failure surface intersects the top of the back- fill. Thatpoint is shown as Point A in Figure 4-3. Doing so will result in a slightly conservative design.
Let's examine the effect of sloping backfill on the bottom layer of geogrid in the wall shown in Figure 4-4. The load ona layer of geogrid is given by:
F = (Pavg) (dh)
Suppose the wall in Figure 4-4 had a flat backfill, the load on the bottom layer of geogrid would be:
F1 = (Pavg) (dh)= (0.5) (P1 + P2) (d1 — d2)= (0.5) [ (�r) (Kar) (d1) cos (�wr) + (�r) (Kar) (d2) cos (�wr) ] (d1 — d2)
= (0.5) [ (125 lb/ft3) (0.2197) (9.17 ft) cos (20º)+ (125 lb/ft3) (0.2197) (4.42 ft) cos (20º) ] (9.17 ft — 4.42 ft) = 833 lb/ft= (0.5) [ (2002 kg/m3) (0.2197) (2.8 m) cos (20º)+ (2,002 kg/m3) (0.2197) (1.35 m) cos (20º) ] (2.8 m — 1.35 m) = 12,199 N/m
49
Figure 4-3. Failure Surface in aCoherent Gravity Wall
Figure 4-4. Effect of Sloped Backfillon Spacing of Geogrid Layers.
For the wall in Figure 4-4 with a backfill slope of 26º, Kar = 0.3662 and the load on the bottom layer of geogrid is:F1 = (Pavg) (dh)
= (0.5) (P3 + P4) (d3 — d4)= (0.5) [ (�r) (Kar) (d3) cos (�wr) + (�r) (Kar) (d4) cos (�wr) ] (d3 — d4)
= (0.5) [ (125 lb/ft3) (0.3662) (10.85 ft) cos (20)+ (125 lb/ft3) (0.3662) (6.1 ft) cos (20) ] (10.85 ft — 6.1 ft)= 1,732 lb/ft= (0.5) [ (2,002 kg/m3) (0.3662) (3.31 m) cos (20º)+ (2,002 kg/m3) (0.3662) (1.86 m) cos (20º) ] (3.31 m — 1.86 m)= 25,332 N/m
Increasing the slope of the backfill from 0º to 26º increased the load on the bottom layer of geogrid by more than 100%.Since the allowable design load on Huesker 35/20-20 is only 833 lb/ft (12,161 N/m), the load on the bottom layerof geogrid will have to be reduced. This can be done by adding another layer of geogrid between the existing layersand moving the top layer up. In some cases, it may be possible to reduce the load on the individual layers of geogridmerely by repositioning the layers of geogrid.
When designing a wall with a sloping backfill, start from the bottom of the wall and calculate the maximum dh as inExample 2-3. But this time, use the depth from Point A rather than the depth from the top of the wall facing.
50
GEOGRID REINFORCEMENT PRODUCT PULLOUT RESISTANCE EQUATIONSMIRIFI 3XT P = 551 lb/ft (8.04 kN/m) + 0.554 x N for N < 1056.20 lbs
P = 988 lb/ft (14.41 kN/m) + 0.141 x N for N > 1056.20 lbsMIRIFI 5XT P = 933 lb/ft (13.61 kN/m) + 0.287 x N MIRIFI 7XT P = 1091 lb/ft (15.92 kN/m) + 0.325 x N FORTRAC 35/20-20 P = 888 lb/ft (12.96 kN/m) + 0.130 x N FORTRAC 55/20-20 P = 431 lb/ft (6.29 kN/m) + 0.601 x N FORTRAC 80/30-20 P = 1214 lb/ft (17.71 kN/m) + 0.450 x NSYNTEEN (SYMPAFORCE) 35/30-25 P = 496 lb/ft (7.24 kN/m) + 0.625 x N for N < 1073.70 lbs
P = 1054 lb/ft (15.38 kN/m) + 0.105 x N for N > 1073.70 lbsSYNTEEN (SYMPAFORCE) 55/30-25 P = 395 lb/ft (5.76 kN/m) + 0.625 x N for N < 969.54 lbs
P = 777 lb/ft (11.34 kN/m) + 0.231 x N for N > 969.54 lbsSYNTEEN (SYMPAFORCE) 80/30-20 P = 989 lb/ft (14.43 kN/m) + 0.488 x N RAUGRID (LUCKENHAUS) 2/3-30 P = 340 lb/ft (4.96 kN/m) + 0.445 x NRAUGRID (LUCKENHAUS) 4/2-15 P = 830 lb/ft (12.11 kN/m) + 0.554 x N RAUGRID (LUCKENHAUS) 6/3-15 P = 1709 lb/ft (24.93 kN/m) + 0.194 x N TERRAGRID (WEBTEC) 35 X 30-25 P = 496 lb/ft (7.24 kN/m) + 0.625 x N for N < 1073.70 lbs
P = 1054 lb/ft (15.38 kN/m) + 0.105 x N for N > 1073.70 lbsTERRAGRID (WEBTEC) 55 X 30-25 P = 395 lb/ft (5.76 kN/m) + 0.625 x N for N < 969.54 lbs
P = 777 lb/ft (11.34 kN/m) + 0.231 x N for N > 969.54 lbsTERRAGRID (WEBTEC) 80/30-20 P = 989 lb/ft (14.43 kN/m) + 0.488 x N STRATTA 300 P = 827 lb/ft (12.07 kN/m) + 0.463 x N STRATTA 500 P = 1500 lb/ft (21.89 kN/m) + 0.241 x N
Table 2-1 Pullout Resistance Equations
CHAPTER FIVESeismic Analysis
IntroductionIn seismic design we take a dynamic force and analyze it as a temporary static load. The forces from seismic activityyield both a vertical and a horizontal acceleration. For our calculations, the vertical acceleration is assumed to be zero(Bathurst, 1998, NCMA Segmental Retaining Walls - Seismic Design Manual, 1998). Due to the temporary nature ofthe loading, the minimum recommended factors of safety for design in seismic conditions are 75% of the valuesrecommended for static design.
The lack of wall failure during the Northridge earthquake in Los Angeles, California and the Kobe earthquake in Japanproves that a soil mass reinforced with geogrid, which is flexible in nature, performs better than rigid structures in reallife seismic situations (Sandri, Dean, 1994, "Retaining Walls Stand Up to the Northridge Earthquake").
The following design uses the earth pressure coefficient method derived by Mononobe-Okabe (M-O) to quantify the loadsplaced on the reinforced mass and the internal components of the structure. Since the nature of segmental retaining wallsis flexible, an allowable deflection can be accepted resulting in a more efficient design while remaining within acceptedfactors of safety.
PRESSURE COEFFICIENTSThe calculation of the dynamic earth pressure coefficient is similar to the static earth pressure coefficient derived byCoulomb, with the addition by Mononobe-Okabe of a seismic inertia angle (�).
cos (� + � — �)2
cos (�) cos (�)2 cos (�w — � + �)
sin (� + �w) sin (� — i — �) 2
cos (�w — � + �) cos (� + i)
Where:
� = peak soil friction angle i = back slope angle� = block setback � = seismic inertia angle�w = angle between the horizontal and the sloped back face of the wall
The seismic inertia angle (�) is a function of the vertical and horizontal acceleration coefficients:
� = atan
Where:
Kv = vertical acceleration coefficient
Kh = horizontal acceleration coefficient
[ ][ ]
( )
1+
Kae =
Kh1 + Kv
51
The vertical acceleration coefficient (Kv) is taken to be zero based on the assumption that a vertical and horizontal peak accelerationwill not occur simultaneously during a seismic event (Bathhurst et al.). The horizontal acceleration coefficient (Kh) is based on the acceleration coefficient (Ao) and the allowable deflection (d) of thewall system. (See equations below) The acceleration coefficient(Ao) varies from 0 to 0.4 in our calculations and is defined as thefraction of the gravitational constant g experienced during a seismicevent. AASHTO provides recommendations for the accelerationcoefficient based on the seismic zone that the retaining wall isbeing designed for. The allowable deflection (d) represents the lateral deflection that theretaining wall can be designed to withstand during a seismic event.The amount of deflection allowed in design is based on engineer-ing judgement. A good approximation is the acceleration coefficient(Ao) multiplied by the wall height (H). The equation used to determine the horizontal acceleration coefficient (Kh) varies depending on the amount of deflec-tion allowed and whether it is calculated for the infill soils or the retained soils.
For Infill soils:
If d = 0, then: Kh = (1.45 — Ao) Ao
This equation, proposed by Segrestin and Bastic, is used in AASHTO / FHWA guidelines. It is assumed to be constantat all locations in the wall.
If d > 0, then: Kh = 0.67 Ao
Kh = 0.67 Ao
This is a standard equation for the horizontal acceleration coefficient based on the Mononobe-Okabe methodology(Mononobe, 1929; Okabe, 1926).
For Retained soils:
If d � 1, then: Kh = Ao
If d > 1, then: Kh = 0.67 Ao
Kh = 0.67 Ao
The following example illustrates the calculation of the dynamic earth pressure coefficient for the infill and retained soilswith a typical allowable deflection of 3 inches (7.6 cm).
( )(Ao) (1 in)d
0.25
( )(Ao) (1 in)d
0.25
( )(Ao) (2.5 cm)d
0.25
( )(Ao) (2.5cm)d
0.25
52
EXAMPLE 5-1
Given:
�i = 34° �r = 28°�wi = 2/3(34°) = 23° �wr = 2/3(28°) = 19°d = 3 in (7.6 cm) � = 12ºi = 0º Ao = 0.4
Find:The dynamic earth pressure coefficients (Kaei, Kaer) for the infill and retained soils.
cos (� + � — �)2
cos (�) cos(�)2 cos (�w — � + �)
sin (� + �w) sin (� — i — �) 2
cos (�w — � + �) cos (� + i)
The first step is to calculate the acceleration coefficents.
Kv = 0, based on the assumption that a vertical and horizontal peak acceleration will not occur simultaneously duringa seismic event.
To determine Kh, we must look at the allowable deflection (d). Since the allowable deflection is greater than zero, thefollowing equation is used:
Kh = 0.67 Ao Kh = 0.67 Ao
Kh = 0.67 (0.4) = 0.162 Kh = 0.67 (0.4) = 0.162
The seismic inertia angle (�) is:
� = atan = atan = 9.2º
Finally, the dynamic earth pressure coefficient for the infill is:
cos (34 + 12 — 9.2)2
cos (9.2) cos (12)2 cos (23 — 12 + 9.2)
sin (34 + 23) sin (34 — 0 — 9.2) 2
cos (23 — 12 + 9.2) cos (12 + 0)
[ ]Kae =
[ ]1+
[ ]Kaei = = 0.276
[ ]1 +
( )(Ao)(1 in)d
0.25
( )Kh1 + Kv
( )0.1621 + 0
( )(Ao)(2.5 cm)d
0.25
( )(0.4)(1 in)3 in
0.25 ( )(0.4)(2.5 cm)7.5 cm
0.25
53
The same process is followed in determining the dynamic earth pressure coefficient for the retained soil. Here again,the vertical acceleration coefficient (Kv) is equal to zero. With the allowable deflection greater than 1 inch (2.5 cm),the horizontal acceleration coefficient is the following:
Kh = 0.67 Ao Kh = 0.67 Ao
Kh = 0.67 (0.4) = 0.162 Kh = 0.67 (0.4) = 0.162
Next, the seismic inertia angle (�) can be calculated:
� = atan = atan = 9.2º
The dynamic earth pressure coefficient for the retained soil is:
cos (28 + 12 — 9.2)2
cos (9.2) cos (12)2 cos (19 — 12 + 9.2)
sin (28 + 19) sin (28 — 0 — 9.2) 2
cos (19 — 12 + 9.2) cos (12 + 0)
DYNAMIC EARTH FORCE ON THE WALLThe dynamic earth force is based on a pseudo-staticapproach using the Mononobe-Okabe (M-O) method.Figures 5-1, 5-2, and 5-3 illustrate the pressure distributionsfor the active force, dynamic earth increment, and thedynamic earth force. The magnitude of the dynamic earthforce is:
Fae = Fa + DFdyn
Where:
Fa = (0.5) (Ka) (�) (H)2
Fae = (0.5) (1 + Kv) (Kae)(�) (H)2
DFdyn = (0.5) (Kae - Ka) (�) (H)2
The magnitude of the resultant force (Fa) acts at 1/3 of the height of the wall and DFdyn acts at 6/10 of the height ofthe wall. The magnitude of the resultant force (Fae) acts at a ratio of the dynamic active earth force moment to the wallheight (m), multiplied by the height of the wall.
( )(Ao) (1 in)d
0.25 ( )(Ao) (2.5 cm)d
0.25
( )(0.4) (1 in)3 in
0.25 ( )(0.4) (2.5 cm)7.5 cm
0.25
( )Kh1 + Kv
( )0.1621 + 0
[ ]Kae = = 0.361
[ ]1+
54
Figure 5-1. Static Component of ActivePressure Distribution
Figure 5-2. Dynamic Increment Component of theActive Pressure Distribution
Figure 5-3. Dynamic Earth Force PressureDistribution
55
Safety Factors
The minimum accepted factors of safety for seismic design are taken to be 75% of the values recommended for staticdesign.
Sliding > 1.1
Overturning > 1.5
Note: The values 1.1 and 1.5 are based on 75% of the recommended minimum factors of safety for design of con-ventional segmental retaining walls. (Collin, 1996, P. 68, Design Manual for Segmental Retaining Walls).
SIMPLE GRAVITY WALL WITH SEISMIC INFLUENCEIn seismic analysis, the weight of a simple gravity wall must counteract the static and temporary dynamic forces of theretained soil. Figure 5-4 illustrates the forces on a simple gravity wall during a seismic event. In the following example,the same equilibrium principles apply as in a static gravity wall analysis with additional consideration for the seismicearth force and the allowed reductions in required factors of safety for sliding and overturning.
Example 5-2:
Given:
�i = �r = 30° � = 78°�w = 2/3(�) = 20° i = 0°� = (90 - �) = 12° d = 2 in. (5.1 cm)Kai = 0.2197 Ao = 0.4Kar = 0.2197 �wall facing = 130 lb/ft3 (2,061 kg/m3)
H = 2.20 ft (0.67 m) �soil = 120 lb/ft3 (1,923 kg/m3)Kaei = 0.5260 Kaer = 0.5260
Find: The safety factor against sliding (SFS) and overturning (SFO).
Note: The dynamic earth pressure coefficients Kaei and Kaer were determined by following the allowable deflection cri-teria established at the beginning of the section.
56
The first step is to determine the driving forces exerted by the soil on the wall:
Active earth force:
Fa = (0.5) (Ka) (�) (H)2
= (0.5) (0.2197) (120 lb/ft3) (2.20 ft)2 = 64 lb/ft = (0.5) (0.2197) (1,923 kg/m3) (0.67 m)2
= (95 kg/m) (9.81) = 930 N/m
Dynamic earth force:
Fae = (0.5) (1 + Kv) (Kae) (�) (H)2
= (0.5) (1 + 0) (0.5260) (120 lb/ft3) (2.20)2 = 153 lb/ft= (0.5) (1 + 0) (0.5260) (1,923 kg/m3) (0.67)2 = 2,227 N/m
Dynamic earth force increment:DFdyn = Fae — Fa
= 153 lb/ft — 64 lb/ft = 89 lb/ft = 2227 N/m — 930 N/m = 1,297 N/m
Resolving the active earth force and the dynamic earth force increment into horizontal and vertical components:
DFdynh = (DFdyn) cos (�w) Fah = (Fa) cos (�w)= (89 lb/ft) cos (20°) = 84 lb/ft = (64 lb/ft) cos (20°) = 60 lb/ft= (1,297 N/m) cos (20°) = 1219 N/m = (930 N/m) cos (20°) = 874 N/m
DFdynv = (DFdyn) sin (�w) Fav = (Fa) sin (�w)= (89 lb/ft) sin (20°) = 30 lb/ft = (64 lb/ft) sin (20°) = 22 lb/ft= (1,297 N/m) sin (20°) = 444 N/m = (930 N/m) sin (20°) = 318 N/m
The next step is to determine the resisting forces:
Weight of the wall facing:Wf = (�wall facing)(H)(d)
= (130 lb/ft3) (2.20 ft )(0.97 ft) = 277 lb/ft= (2,061 kg/m3) (0.67 m) (0.296m) = 4,010 N/m
Maximum frictional resistance to sliding:Fr = (Wf + Fav + DFdynv) tan (�)
= (277 lb/ft + 22 lb/ft + 30 lb/ft) tan (30°) = 190 lb/ft= (4,010 N/m + 318 N/m + 444 N/m) tan (30°) = 2,755 N/m
Figure 5-4. Free Body Diagram ofSimple Gravity Wall Under
Seismic Influence
Safety factor against sliding (SFS):
SFSseismic= (Force resisting sliding) = Fr
(Force driving sliding) Fh + DFdynh
= (190 lb/ft) = 1.3 � 1.1 ok(60 lb/ft + 84 lb/ft)
= (2,755 N/m) = 1.3 � 1.1 ok(874 N/m + 1,219 N/m)
The factor of safety of 1.3 shows that an AB gravity wall during an earthquake in a seismic zone 4 is stable and doesnot require reinforcement to prevent sliding. As a comparison, the factor of safety in a static condition is the following:
SFSstatic= (Force resisting sliding) = Fr = (Wf + Fav) tan �
(Force driving sliding) Fh Fh
= (277 lb/ft + 22 lb/ft) tan (30) = 2.9 � 1.5 ok(60 lb)
= (4,010 + N/m + 318 N/m) tan (30) = 2.9 � 1.5 ok(874 N/m)
Overturning Failure
In seismic analysis, the moments resisting overturning (Mr) must be greater than or equal to 1.5 times the moments caus-ing overturning (Mo).
The moments resisting overturning (Mr):
The weight of the wall, the vertical component of the active force, and the vertical component of the dynamic earth incre-ment force contribute to the moment resisting overturning failure of the wall.
Mr = (Wf ) (Wfarm) + (Fav) (Faarmv) + (DFdynv) (DFdynarmv) = (Wf ) [(X1) + (0.5)(H) tan (�)] + Fav [(L + s) + (0.333) (H) tan (�)] + DFdynv [(L + s) + (0.6)(H) tan (�)]
= (277 lb/ft) [(0.49 ft) + (0.5) (2.20) tan (12)] + (22 lb/ft) [(0) + (0.171 ft) + (0.333) (2.20) tan (12°)] + (30 lb/ft) [(0) + (0.171 ft) + (0.6) (2.20) tan (12)] = 221 ft-lb/ft= (4,010 N/m) [(0.149 m) +(0.5) (0.67 m) tan (12°)] + (318 N/m) [(0) + (0.053 m) + (0.333) (0.67 m) tan (12°)] + (444 N/m) [(0) + (0.053 m) + (0.6) (0.67 m) tan (12°)]= 976 N-m/m
Note : (s = setback per block, L = length of geogrid)
57
The moments causing overturning (Mo):
The horizontal components of the active and dynamic forces con-tribute to the moment causing overturning failure of the wall.
Mo = (Fah) (Faarmh) + (DFdynh) (DFdynarmh)= (Fah) (0.333)(H) + (DFdynh) (0.6)(H)= (60 lb/ft) (0.333) (2.2 ft) + (84 lb/ft) (0.6) (2.2 ft)= 155 ft-lb/ft
= (874 N/m) (0.333) (0.67 m) + (1219 N/m) (0.6) (0.67 m) = 685 N-m/m
Safety Factor Against Overturning (SFO):
SFOseismic= (Moments resisting overturning) = Mr � 1.5
(Moments driving overturning) Mo
= (221 ft-lb/ft) = 1.4 < 1.5, Not ok(155 ft-lb/ft)
= (976 N-m/m) = 1.4 < 1.5, Not ok(685 N-m/m)
This shows that the gravity wall is not adequate with respect to overturning failure. Geogrid reinforcement for this wallis needed to achieve proper factor of safety. Evaluating the wall under static conditions we see that the required factorsof safety are met.
Mr = (Wf) (Wfarm) + (Fav) (Faarmv) = (Wf) [(X1) + (0.5) (H) tan (�)] + (Fav) [(L + s + (0.333) (H) tan (�)]
= (277 lb/ft) [(0.49 ft) + (0.5) (2.20) tan (12º)] + (22 lb/ft) [(0) + (0.171 ft) + (0.333) (2.20) tan (12°)] = 208 ft-lb/ft
= (4010 N/m) [(0.149 m) + (0.5) (0.67 m) tan (12°)] + (318 N/m) [(0) + (0.053 m) + (0.333) (0.67 m) tan (12°)] = 915 N-m/m
Mo = (Fah) (Faarmh) = (Fah) (0.333) (H)= (60 lb/ft) (0.333) (2.2 ft) = (874 N/m) (0.333) (0.67 m)= 44 ft-lb/ft = 195 N-m/m
SFOstatic
= (Moments resisting overturning) = Mr � 2.0(Moments driving overturning) Mo
= (208 ft-lb/ft) = 4.7 � 2.0 ok = (915 N-m/m) = 4.7 � 2.0 ok(44 ft-lb/ft) (195 N-m/m)
58
COHERENT GRAVITY WALL WITH SEISMIC INFLUENCESeismic inertial force (Pir)In the external stability analysis of a geogrid reinforced retaining wall during a seismic event, a seismic inertial force(Pir) is introduced. The seismic inertial force is the sum of the weight components that exert a horizontal inertial forcewithin a reinforced soil mass during a seismic event. The three components exerting this inertial force are the block fac-ing, the reinforced soil mass, and the backslope angle.
Pir = Khr (Wf + Ws + Wi)
This force along with the dynamic earth increment forcecombine with the static earth forces from the retained soiland the weight forces from the wall structure to create theconditions during an earthquake.
Factor of safety against slidingCalculating the factor of safety against sliding for a coher-ent gravity wall follows the same stability criteria as a sim-ple gravity wall. The principle being that the forces resist-ing sliding must be 1.1 times the forces causing sliding.As can be seen below, the formula for calculating the fac-tor of safety sliding is the same as the gravity wall analy-sis with the addition of the seismic inertial force (Pir) andthe weight of the reinforced soil (Ws).
SFSseismic= Frseismic � 1.1
Fah + DFdynh + Pir
Where:
Frseismic
= (Fav + DFdynv + Wf + Ws) tan (�i)
Factor of safety against overturningThe factor of safety against overturning is computed in the same way as a simple gravity wall with the addition of theseismic inertial force (Pir) and the weight of the reinforced soil (Ws).
SFOseismic
Mr = (Wf) (Wfarm) + (Ws) (Wsarm) + (Fav) (Faarmv) + (DFdynv) (DFdynarmv) � 1.5Mo (Fah) (Faarmh) + (DFdynh) (DFdynarmh) + (Pir) (Hir)
Figure 5-5. Free Body Diagram of a CoherentGravity Wall Under Seismic Influence
59
Example 5-3:
Given:�i = �r = 30° � = 78° Fa = 1,360 lb/ft (19,855 N/m)
�w = 2/3(�) = 20° � = (90 — �) = 12° DFdyn = 716 lb/ft (10,453 N/m)
i = 0° Kai = 0.2197 Wf = 1,292 lb/ft (18,862 N/m)
d = 2 in (5.1 cm) Kar = 0.2197 Ws = 6,425 lb/ft (93,799 N/m)Ao = 0.4 Kaei = 0.3870 �wall facing = 130 lb/ft3 (2,061 kg/m3)H = 10.0 ft (3.05 m) Kaer = 0.3870 �soil = 120 lb/ft3 (1,923 kg/m3)Wi = 0 lb/ftFind:The safety factor against sliding and overturning.
Safety Factor Against SlidingBased on the given information, we must first determine the frictional resistance to sliding (Fr).
Fr = (Fav + DFdynv + Wf + Ws) tan (�)= [(1,360 lb/ft) sin (20°) + (716 lb/ft) sin (20°) + 1,292 lb/ft + 6,425 lb/ft ] tan (30°)= 4,865 lb/ft
= [(19,855 N/m) sin (20°) + (10,453 N/m) sin (20°) + 18,862 N/m + 93,799 N/m ] tan (30°)= 71,030 N/m
Next, the seismic inertial force is calculated:
Pir = Khr (Wf + Ws + Wi)
Since,
d = 2 in (5.1 cm)
Khr = (0.67) (Ao) = (0.67) (Ao)
= (0.67) (0.4) = (0.67) (0.4)
= 0.179 = 0.179
Pir = 0.179 (1,292 lb/ft + 6,425 lb/ft + 0 ) = 0.179 (18,862 N/m + 93,799 N/m + 0 )
= 1,381 lb/ft = 20,166 N/m
Finally, the safety factor against sliding can be calculated:
SFSseismic= (Forces resisting sliding) = Fr � 1.1
(Forces driving sliding) Fh + DFdynh + Pir
= (4,865 lb/ft) = 1.5 � 1.1 ok(1,360 lb/ft) cos 20º + (716 lb/ft) cos 20º + 1,381 lb/ft
= (71,030 N/m) = 1.5 � 1.1 ok(19,855 N/m) cos 20º + (10,453 N/m) cos 20º + 20,166 N/m
( )(Ao) (1 in)d
0.25
( )(0.4) (1 in)2 in
0.25
( )(Ao) (2.5cm)d
0.25
( )(0.4) (2.5cm)5.1 cm
0.25
60
Comparing the seismic SFS to the static SFS below, we again see much higher safety values for static.
SFSstatic= (Forces resisting sliding) = Fr = Fr
(Forces driving sliding) Fh (Fa) cos (�w)
= (4,865 lb/ft) = 3.8 � 1.5 ok(1,360 lb/ft) cos 20º
= (71,030 N/m) = 3.8 � 1.5 ok
(19,855 N/m) cos 20º
Safety Factor Against OverturningThe safety factor against overturning is equal to the moments resisting overturning divided by the moments driving over-turning (Mr / Mo) and must be greater than or equal to 1.5.
The moments resisting overturning (Mr):
Mr = (Wt) (Wtarm) + (Fav) (Faarmv) + (DFdynv) (DFdynarmv)
Where: Wt = Ws + Wf
= (Wt) [0.5 (L + s) + (0.5) (H) tan (�)] + Fav [(L + s) + (0.333) (H) tan (�)] + DFdynv [(L + s) + (0.6) (H) tan (�)]
= (7,717 lb/ft) [0.5 (6.1 ft + 0.171 ft) + (0.5) (10 ft) tan (12°)] + [(1,360 lb/ft) sin 20°] [6.1 ft + 0.171 ft + (0.333) (10 ft) tan (12°)] + [(716 lb/ft) sin (20°)] [6.1 ft + 0.171 ft + (0.6) (10 ft) tan (12°)]= 37,492 ft-lb/ft
= (112,661 N/m) [0.5 (1.86 m + 0.053 m) + (0.5) (3.05 m) tan (12°)] + [(19,855 N/m) sin 20°] [1.86 m + 0.053 m + (0.333) (3.05 m) tan (12°)] + [(10,453 N/m) sin (20°)] [1.86 m + 0.053 m + (0.6) (3.05 m) tan (12°)]= 166,966 N-m/m
The moments driving overturning (Mo):
Mo = (Fah) (Faarmh) + (DFdynh) (DFdynarmh)= (Fah) (0.333) (H) + (DFdynh) (0.6)(H)= [(1,360 lb/ft) cos (20°)] (0.333) (10 ft) + [(716 lb/ft) cos (20°)] (0.6) (10 ft)= 8,293 ft-lb/ft
= [(19,855 N/m) cos (20°)] (0.333) (3.05 m) + [(10,453 N/m) cos (20°)] (0.6) (3.05 m)= 36,925 N-m/m
Safety Factor Against Overturning (SFO):
SFOseismic = (Moments resisting overturning) = Mr � 1.5(Moments driving overturning) Mo
= (37,492 ft-lb /ft) = 4.5 � 1.5 ok = (166,966 N-m/m) = 4.5 � 1.5 ok(8,293 ft-lb/ft) (36,925 N-m/m)
61
Comparing the seismic (SFO) to the below static (SFO):
Mr = (Wt) (Wtarm) + Fav
Where: Wt = Ws + Wf= (Wt) [0.5 (L + s) + (0.5) (H) tan (�)] + (Fav) [(L + s) + (0.333) (H) tan (�)]
= (7,717 lb/ft) [0.5 (6.1 ft + 0.171 ft) + (0.5) (10 ft) tan (12°)] + [(1,360 lb/ft) sin 20°] [(6.1 ft + 0.171 ft) + (0.333) (10 ft) tan (12°)] = 35,644 ft-lb/ft
= (112,661 N/m) [0.5 (1.86 m + 0.053 m) + (0.5) (3.05 m) tan (12°)] + [(19,855 N/m) sin 20°] [(1.86 m + 0.053 m) + (0.333) (3.05 m) tan (12°)] = 158,736 N-m/m
Mo = (Fah) (Faarmh) = (Fah) (0.333) (H)
= [(1,360 lb/ft) cos (20°)] (0.333) (10 ft) = 4,256 ft-lb/ft
= [(19,855 N/m) cos (20°)] (0.333) (3.05 m) = 18,950 N-m/m
SFOstatic= (Moments resisting overturning) = Mr
(Moments driving overturning) Mo
= (35,644 ft-lb /ft) = 8.4 � 1.5 ok = (158,736 N-m/m) = 8.4 � 1.5 ok(4,256 ft-lb/ft) (18,950 N-m/m)
Internal StabilityThe factor of safety checks for the internal stability of a geogrid reinforced retaining wall under seismic conditions includethe geogrid overstress, geogrid / block connection strength, geogrid pullout from the soil, and localized or top of thewall stability. These calculations are identical to those for a static stability analysis with the exception of the seismicforces introduced which affect the tensile loading on the geogrid.
Factor of Safety Geogrid Tensile Overstress
In order to calculate the factor of safety for geogrid tensile overstress, the tensile force on each grid must first be deter-mined. In a seismic event, the sum of the active force (Fa), the dynamic earth force increment (DFdyn), and the seismicinertial force (Pir) represent the tensile force on each layer of geogrid.
Fid = Fa + DFdyn + Pir
Where:
Fa = (Ka) cos (�w) (�) (Ac) (0.5)
DFdyn = 0.8 - 0.6 (Kae) cos (�w) (�) (Ac)
and,
Pir = (Kh) (�) (Ac)
( )[ ](He) — (grid) (H)(He)
62
The variable Ac in the above equations represents the amount of area influencing each geogrid layer. The (grid) (H)term in the dynamic earth force increment equation refers to the elevation of the geogrid. Once the tensile force isdetermined for each grid, we calculate the factor of safety against geogrid tensile overstress, which is equal to thelong term allowable design strength of the geogrid divided by the tensile force acting on that grid.
FSoverstressed
= LTDS � 1.0Fid
In the calculation of the factor of safety geogrid tensile overstress for a seismic event, we do not take a reduction ofthe geogrid ultimate strength for long-term creep. This is due to the short-term loading during a seismic event.
Geogrid / Block Connection CapacityThe factor of safety for connection strength is equal to the peak connection strength divided by the tensile force on thatlayer of grid multiplied by 2/3. We take the additional 2/3rds reduction on the tensile force due to the reality thatsome of the tensile force is absorbed by the soil in the influence area.
FSconn = Fcs � 1.1Fid (0.667)
Geogrid Pullout from the SoilThe factor of safety for geogrid pullout from the soil is equal to the pullout capacity of the geogrid divided by the ten-sile force on each geogrid.
FSpullout = Fp � 1.1 Fid
where,
Fp = 2 (Ci) tan (�) [(He) — (grid) (�) (Le)]
The above pullout capacity equation takes into account the geogrid interaction coefficient (Ci) and is calculated basedon the length of geogrid embedded beyond the line of maximum tension (Le).
Localized Stability, Top of the Wall Stability
To determine local or top of the wall stability (SFS and SFO), the wall parameters and soils forces in the unreinforcedportion of the retaining wall are focused on. The unreinforced height of the wall (Ht) is simply the total height of thewall minus the elevation at which the last grid layer is placed. The local weight of the facing is:
Wf = (Ht) (t) (�wall)
The local sliding resistance (Fr) is an equation based on the Allan Block shear strength, which was developed throughempirical test data and is a function of the normal load acting at that point and is the following:
Fr = 805 lb/ft + (Wf) tan (56º)= 11,752 N/m + (Wf) tan (56º)
63
The soil and surcharge forces are as follows:
Active Force: Fa = (0.5) (Ka) (�) (Ht)2
Dynamic Force: Fae = (0.5) (1 + Kv) (Kae) (�) (Ht)2
Dynamic Earth Force Infrement: DFdyn = Fae - FaSeismic Inertial Force: Pir = (Kh) (Wf)
Finally, the safety factor equations are:
SFSlocalstatic
= Fr � 1.5(Fa) cos (�w)
SFSlocalseismic
= Fr � 1.1(Fa + DFdyn + Pir) cos (�w)
SFOlocalstatic
= Wf [(Ht/2) tan � + t/2] + (Fa) sin (�w) [(Ht/3) tan � + t] � 1.5(Fa) cos (�w) (Ht/3)
SFOlocalseismic
= Wf [(Ht/2) tan � + t/2] + (Fa) sin (�w) [(Ht/3) tan � + t] + (DFdyn) sin (�w) (0.6 Ht + t)(Fa) cos (�w) (Ht/3) + (DFdyn) cos (�w) (0.6 Ht) + Pir (Ht/2)
� 1.1
64
REFERENCESAmerican Association of State Highway and Transportation Officials -- Task Force 27. "Guidelines for the Design ofMechanically Stabilized Earth Walls." Draft version. Washington: 1987.Kliethermes, J., K. Buttry, E. McCullough, and R. Wetzel. "Modular Concrete Retaining Wall and GeogridPerformance and Laboratory Modeling." University of Wisconsin-Platteville, 1990.Leshchinsky, D. and E.B. Perry. "A Design Procedure for Geotextile Reinforced Walls." Geotechnical Fabrics Report.St. Paul: July/August, 1987.McKittrick, D.P. "Reinforced Earth: Application of Theory and Research to Practice." Reinforced Earth TechnicalServices, Report 79-1. Arlington, VA: The Reinforced Earth Company, 1978.Minnesota Department of Transportation. "Walls." Section 9-4.0 in Road Design Manual -- Part II. St. Paul: 1985.Peck, Ralph. "Earth Retaining Structures and Slopes." Chapter 13 in Soil Science, edited by T.W. Lambe and R.V.Whitman. New York: John Wiley & Sons, Inc., 1969.Sowers, G.B., and G.F. Sowers. "Problems in Earth Pressure." Chapter 8 in Introductory Soil mechanics andFoundations. New York: The Macmillan Company, 1970.R. F. Craig, Chapman & Hall, “Soil Mechanics” Fifth Edition, 1992Braja M. Das “Principles of Geotechnical Engineering” Third Edition, Chapter 10, 1994
65
Sample CalculationsExample S-1:
Given:
i = 0º H = 3.0 ft (0.91 m)� = 36º � = 120 lb/ft3 (1,923 kg/m3)� = 90 — 3 = 87º �w = 130 lb/ft3 (2,061 kg/m3)
�w = (0.666) (36) = 24º
csc (�) sin (� — �) 2sin (� + �w) + sin (� + �w) sin (� — i)
sin (� i)
csc (87) sin (87 — 36) 2sin (87 + 24) + sin (36 + 24) sin (36 — 0)
sin (87 — 0)
0.7782124 2
0.966219657 + 0.713957656 = 0.2145
Find: The safety factor against sliding, SFS.
The first step is to determine the total active force exerted by the soil on the wall:
Fa = (0. 5) (�) (Ka) (H)2 = (0.5) (120 lb/ft3) (0.2145) (3.0 ft)2 = 116 lb/ft= (0.5) (�) (Ka) (H)2 = (0.5) (1,923 kg/m3) (0.2145) (0.91 m)2 = 1,675 N/m
Fh = (Fa) cos (�w) = (116 ft/lb) cos (24º) = 106 lb/ft= (Fa) cos (�w) = (1,693 N/m) cos (24º) = 1,547 N/m
Fv = (Fa) sin (�w) = (116 ft/lb) sin (24º) = 47 lb/ft= (Fa) sin (�w) = (1,693 N/m) sin (24º) = 689 N/m
Wf = (�w) (H) (d) = (130 lb/ft3) (3.0 ft) (0.97 ft) = 378 lb/ft= (�w) (H) ( d) = (2,061 kg/m3) (0.91 m) (0.3 m) = 5,520 N/m
Fr = (Vt) (Cf) = (Wf + Fv) tan (�) = (378 lb/ft + 47 lb/ft) tan (36º) = 309 lb/ft= (Vt) (Cf) = (Wf + Fv) tan (�) = (5,520 N/m + 689 N/m) tan (36º) = 4,511 N/m
SFS = Fr = 309 lb/ft = 2.92 > 1.5 OK Fh 106 lb/ft
= Fr = 4,511 N/m = 2.92 > 1.5 OKFh 1,547 N/m
66
[ ][ ][ ]
�a =
�a =
�a =
67
Find: The safety factor against overturning, SFO.
�Mr = (Wf) [ (x1) + (0.5) (H) tan (90º — �) ]+ (FV) [ (x2) + (0.333) (H) tan (90º — �) ]= (378 lb/ft) [ (0.49 ft) + (0.5) (3.0 ft) tan (90º — 87º) ]+ (47 lb/ft) [ (0.97 ft) + (0.333) (3.0 ft) tan (90º — 87º) ]= 263 ft-lb/ft= (5,520 N/m) [ (0.15 m) + (0.5) (0.9 m) tan (90º — 87º) ]+ (689 N/m) [ (0.3 m) + (0.333) (0.9 m) tan (90º — 87º) ]
= 1,176 N-m/m
Mo = (Fh) (0.333) (H)= (106 lb/ft) (0.333) (3.0 ft) = 106 ft-lb/ft
= (1,547 N/m) (0.333) (0.9 m) = 464 N-m/m
SFO = �Mr = (263 ft-lb/ft) = 2.48 > 2.0 OK�Mo (106 ft-lb/ft)
= �Mr = (1,176 N-m/m) = 2.48 > 2.0 OK�Mo (464 N-m/m)
Example S-2:
Given:� = 36º � = 120 lb/ft3 (1,923 kg/m3)H = 3.0 ft (0.9 m) �w = 130 lb/ft3 (2,061 kg/m3)� = 90 - 12 = 78º q = 250 lb/ft2 (11,974 Pa)i = 0º �w = (.666) (36) = 24º
csc (�) sin (� — �) 2sin (� + �w) + sin (� + �w) sin (� — i)
sin (� i)
csc (78) sin (78 — 36) 2
sin (78 + 24) + sin (36 + 24) sin (36 — 0)sin (78 — 0)
0.684079382 20.989013448 + 0.72139389 = 0.1599
[ ][ ][ ]
Ka =
Ka =
Ka =
Find: The safety factor against sliding, SFS.
The first step is to determine the total active force exerted by the soil on the wall:
Fa = (0.5) (�) (Ka) (H)2 = (0.5) (120 lb/ft3) (0.1599) (3.0 ft)2 = 86 lb/ft= (0.5) (�) (Ka) (H)2 = (0.5) (1,923 kg/m3) (0.1599) (0.9 m)2 = 1,222 N/m
Fh = (Fa) cos (�w) = (86 lb/ft) cos (24º) = 79 lb/ft= (Fa) cos (�w) = (1,222 N/m) cos (24º) = 1,116 N/m
Fv = (Fa) sin (�w) = (86 lb/ft) sin (24º) = 35 lb/ft= (Fa) sin (�w) = (1,222 N/m) sin (24º) = 497 N/m
Wf = (�w) (H) (d) = (130 lb/ft3) (3.0 ft) (0.97 ft) = 378 lb/ft= (�w) (H) (d) = (2,061 kg/m3) (0.9 m) (0.3 m) = 5,459 N/m
Fr = (Vt) (Cf) = (Wf + Fv) tan (�) = (378 lb/ft + 35 lb/ft) tan (36º) = 300 lb/ft= (Vt) (Cf) = (Wf + Fv) tan (�) = (5,459 N/m + 497 N/m) tan (36º) = 4,327 N/m
Pq = (q) (Ka) = (250 lb/ft2) (0.1599) = 40 lb/ft2
= (q) (Ka) = (11,974 N/m2) (0.1599) = 1,916 Pa
Pqh = (Pq) cos (�w) = (40 lb/ft2) cos (24º) = 37 lb/ft2
= (Pq) cos (�w) = (1,916 Pa) cos (24º) = 1,750 Pa
Pqv = (Pq) sin (�w) = (40 lb/ft2) sin (24º) = 16 lb/ft2
= (Pq) sin (�w) = (1,916 Pa) sin (24º) = 779 Pa
Fqh = (Pqh) (H) = (37 lb/ft2) (3.0 ft) = 111 lb/ft= (Pqh) (H) = (1,772 Pa) (0.9 m) = 1,595 N/m
Fqv = (Pqv) (H) = (16 lb/ft2) (3.0 ft) = 48 lb/ft= (Pqv) (H) = (766 Pa) (0.9 m) = 690 N/m
SFS = Fr + (Fqv) (Cf) = 300 lb/ft + (48 lb/ft) tan (36º) = 1.76 > 1.5 OKFh + Fqh 79 lb/ft + 111 lb/ft
= Fr + (Fqv) (Cf) = 4,327 N/m + (690 N/m) tan (36º) = 1.76 > 1.5 OKFh + Fqh 1,116 N/m + 1,595 N/m
68
Find: The safety factor against overturning, SFO.
�Mr = (Wf) [ (X1) + (0.5) (H) tan (90º — �) ]+ (Fv) [ (X2) + (0.333) (H) tan (90º — �) ]+ (Fqv) [ (X2) + (0.5) (H) tan (90º — �) ]
�Mr = (378 lb/ft) [ (0.49 ft) + (0.5) (3.0 ft) tan (90º — 78º) ]+ (35 lb/ft) [ (0.97 ft) + (0.333) (3.0 ft) tan (90º — 78º) ]+ (48 lb/ft) [ (0.97 ft) + (0.5) (3.0 ft) tan (90º — 78º) ]= 409 ft-lb/ft= (5,459 N/m) [ (0.15 m) + (0.5) (0.9 m) tan (90º — 78º) ]+ (497 N/m) [ (0.3 m) + (0.333) (0.9 m) tan (90º — 78º) ]+ (690 N/m) [ (0.3 m) + (0.5) (0.9 m) tan (90º — 78º) ]= 1,795 N-m/m
Mo = (Fh) (0.333) (H) + (Fqh) (0.5) (H)= (79 lb/ft) (0.333) (3.0 ft) + (111 lb/ft) (0.5) (3.0 ft) = 245 ft-lb/ft= (1,116 N/m) (0.333) (0.9 m) + (1,595 N/m) (0.5) (0.9 m) = 1,052 N-m/m
SFO = �Mr = (409 ft-lb/ft) = 1.7 > 2.0 NOT OK�Mo (245 ft-lb/ft)
= �Mr = (1,795 N-m/m) = 1.7 > 2.0 NOT OK�Mo (1,052 N-m/m)
Example S-3:
Given:� = 27º i = 0º � = 120 lb/ft3 (1,923 kg/m3)
H = 9.0 ft (2.74 m) Ci = 0.75 �w = 130 lb/ft3 (2,061 kg/m3)
� = 90 - 12 = 78º �w = (.666) (27) = 18º q = 250 lb/ft2 (11,974 Pa)
csc (�) sin (� — �) 2
sin (� + �w) + sin (� + �w) sin (� — i)sin (� i)
csc (78) sin (78 — 27) 2
sin (78 + 18) + sin (27 + 18) sin (27 — 0)sin (78 — 0)
0.794507864 20.997257186 + 0.572880034 = 0.256
69
[ ][ ]
[ ]Ka =
Ka =
Ka =
Find: The safety factor against sliding, SFS.
The first step is to determine the total active force exerted by the soil on the wall:
Fa = (0.5) (�) (Ka) (H)2 = (0.5) (120 lb/ft3) (0.256) (9.0 ft)2 = 1,244 lb/ft= (0.5) (�) (Ka) (H)2 = (0.5) (1,923 kg/m3) (0.256) (2.74 m)2 = 18,128 N/m
Fh = (Fa) cos (�w) = (1,244 lb/ft) cos (18º) = 1,183 lb/ft= (Fa) cos (�w) = (18,128 N/m) cos (18º) = 17,241 N/m
Fv = (Fa) sin (�w) = (1,244 lb/ft) sin (18º) = 384 lb/ft= (Fa) sin (�w) = (18,128 N/m) sin (18º) = 5,602 N/m
Wf = (�w) (H) (d) = (130 lb/ft3) (9.0 ft) (0.97 ft) = 1,135 lb/ft= (�w) (H) (d) = (2,061 kg/m3) (2.74 m) (0.3 m) = 16,620 N/m
Fr = (Vt) (Cf) = (Wf + Fv) tan (�) = (1,135 lb/ft + 384 lb/ft) tan (27º) = 774 lb/ft= (Vt) (Cf) = (Wf + Fv) tan (�) = (16,620 N/m + 5,602 N/m) tan (27º) = 11,323 N/m
SFS = Fr = 774 lb/ft = 0.65 > 1.5 NOT OK (Need Geogrid)Fh 1,183 lb/ft
= Fr = 11,323 N/m = 0.65 > 1.5 NOT OK (Need Geogrid)Fh 17,241 N/m
Analyze with single layer of grid.
Fg = 2 (dg) (�) (Le) (Ci) tan (�)
Find Le.Le = 833 lb/ft = 2.05 ft
2 (4.44 ft) (120 lb/ft3) (0.75) tan (27º)= 12,161 N/m = 0.648 m
2 (1.35 m) (18,865 N/m) (0.75) tan (27º)
Lt = Lw + La + Le = 0.85 + (H — dg) [ tan (45º — (��2)) — tan (90º — �) ] + 2.05 ft= 0.85 ft + (9 ft — 4.44 ft) [ tan (45º — 13.5º) — tan (90º — 78º)] + 2.05 ft= 4.72 ft= Lw + La + Le = 0.85 + (H — dg) [ tan (45º — (��2)) — tan (90º — �) ] + 0.624 m= 0.259 m + (2.74 m — 1.35 m) [ tan (45º — 13.5º) — tan (90º — 78º) ] + 0.624 m= 1.439 m
Actual Embedment Length.
Le = (Lt — Lw — La)= 5 ft — 0.85 ft — (9 ft — 4.44 ft) (0.4) = 2.33 ft= 1.524 m — 0.259 m — (2.74 m — 1.35 m) (0.4) = 0.71 m
Maximum potential restraining force with Le = 2.33.
70
Fg = 2 (4.44 ft) (120 lb/ft3) (2.33 ft) (0.75) tan (27º) = 949 lb/ft but max is 833 lb/ft= 2 (1.35 m) (1,923 kg/m3) (0.71 m) (0.75) tan (27º) = 13,820 N/m but max is 12,161 N/m
SFS = Fr + Fg = 774 lb/ft + 833 lb/ft = 1.36 > 1.5 NOT OK (Needs More Geogrid)Fh 1,183 lb/ft
= Fr + Fg = 11,323 N/m + 12,161 N/m = 1.36 > 1.5 NOT OK (Needs More Geogrid)
Fh 17,241 N/m
Lmin = 0.3 (H) + 0.85 ft + 2.4 ft = 0.3 (9 ft) + 0.85 ft + 2.4 ft = 5.94 ft Round up to Lg = 6 ft= 0.3 (H) + 0.256 m + 0.732 m = 0.3 (2.74 m) + 0.256 m + 0.732 m = 1.82 m
Ws = (�r) (H) (Lg — 0.85 ft) = (125 lb/ft3) (9.0 ft) (6 ft — 0.85 ft) = 5,794 lb/ft= (�r) (H) (Lg — 0.256 m) = (2,002 kg/m3) (2.74 m) (1.82 m — 0.256 m) = 84,163 N/m
Ww = Wf + Ws = 1,135 lb/ft + 5,794 lb/ft = 6,929 lb/ft= Wf + Ws = 16,620 N/m + 84,163 N/m = 100,783 N/m
Vertical Force; Solve using onsite soil
Vt = Ww + Fv = 6,929 lb/ft + 384 lb/ft = 7,313 lb/ft= Ww + Fv = 100,783 N/m + 5,602 N/m = 106,385 N/m
Fr = (Vt) (Cf) = (7,313 lb/ft) tan (27º) = 3,726 lb/ft= (Vt) (Cf) = (106,385 N/m) tan (27º) = 54,206 N/m
Pressure on the retaining wall due to the surcharge
Pq = (q) (Ka) = (250 lb/ft2) (0.256) = 64 lb/ft2
= (q) (Ka) = (11,974 Pa) (0.256) = 3,065 PaFind the horizontal and vertical components of the pressure.
Pqh = (Pq) cos (�wo) = (64 lb/ft2) cos (18º) = 61 lb/ft2
= (Pq) cos (�wo) = (3,065 Pa) cos (18º) = 2,915 Pa
Pqv = (Pq) sin (�wo) = (64 lb/ft2) sin (18º) = 20 lb/ft2
= (Pq) sin (�wo) = (3,065 Pa) sin (18º) = 947 Pa
Finally, the total surcharge forces on the wall are calculated:
Fqh = (Pqh) (H) = (61 lb/ft2) (9.0 ft) = 549 lb/ft= (Pqh) (H) = (2,915 Pa) (2.74 m) = 7,987 N/m
Fqv = (Pqv) (H) = (20 lb/ft2) (9.0 ft) = 180 lb/ft= (Pqv) (H) = (947 Pa) (2.74 m) = 2,598 N/m
Find the safety factor against sliding:
SFS = Fr + (Fqv) tan � = 3,726 lb/ft + 180 lb/ft (tan 27º) = 2.2 > 1.5 OKFh + Fqh 1,183 lb/ft + 549 lb/ft
= Fr + (Fqv) (Fqv) = 54,206 N/m + 2,598 N/m (tan 27º) = 2.208 > 1.5 OKFh + Fqh 17,241 N/m + 7,987 N/m
71
Find the safety factor against overturning:�Mr = (Wf) [ (0.5) (X1) + (0.5) (H) tan (90º — �) ]
+ (Ws) [ (0.5) (X2 X1) + (X1) + (0.5) (H) tan (90º — �) ]+ (Fv) [ (X2) + (0.333) (H) tan (90º — �) ]+ (Fqv) [ (X2) + (0.5) (H) tan (90º — �) ]
�Mr = (1,135 lb/ft) [ (0.5) (0.97 ft) + (0.5) (9.0 ft) tan (90º — 78º) ]+ (5,794 lb/ft) [ (0.5) (6.13 ft — 0.97 ft) + (0.97 ft) + (0.5) (9.0 ft) tan (90º — 78º) ]+ (384 lb/ft) [ (6.13 ft) + (0.333) (9.0 ft) tan (90º — 78º) ]+ (180 lb/ft) [ (6.13 ft) + (0.5) (9.0 ft) tan (90º — 78º) ]= 31,621 ft-lb/ft= (16,620 N/m) [ (0.5) (0.297 m) + (0.5) (2.74 m) tan (90º — 78º) ]+ (84,163 N/m) [ (0.5) (1.87 m — 0.297 m) + (0.297 m) + (0.5) (2.74 m) tan (90º — 78º) ]+ (5,602 N/m) [ (1.87 m) + (0.333) (2.74 m) tan (90º — 78º) ]+ (2,598 N/m) [ (1.87 m) + (0.5) (2.74 m) tan (90º — 78º) ]= 140,184 N-m/m
Mo = (Fh) (0.333) (H) + (Fqh) (0.5) (H)= (1,183 lb/ft) (0.333) (9.0 ft) + (549 lb/ft) (0.5) (9.0 ft) = 6,016 ft-lb/ft= (17,241 N/m) (0.333) (2.74 m) + (7,987 N/m) (0.5) (2.74 m) = 26,665 N-m/m
SFO = �Mr = (31,621 ft-lb/ft) = 5.3 > 2.0 OK�Mo (6,016 ft-lb/ft)
= �Mr = (140,184 N/m) = 5.3 > 2.0 OK= �Mo (26,665 N/m)
Internal Stability:
�r = 30º�r = 125 lb/ft3 (2,002 kg/m3)
�wr = 0.666 (30º) = 20º
csc (78) sin (78 — 30) 2
sin (78 + 19.98) + sin (30 + 19.98) sin (30 — 0)sin (78 — 0)
0.759747 20.995147 + 0.625671 = 0.2197
72
[ ][ ]
Kar =
Kar =
Pqh = (q) (Kar) cos (�wr) = (250 lb/ft2) (0.2197) cos (20º) = 52 lb/ft2
= (q) (Kar) cos (�wr) = (11,974 Pa) (0.2197) cos (20º) = 2,472 Pa
Quadratic equation = — b � b2 — 4ac2a
z = (Kar) cos (�wr) = (0.2197) cos (20º) = 0.2065a = (�r) (z) = (125 lb/ft3) (0.2065) = 26 lb/ft3
= (�r) (z) = (2,002 kg/m3) (0.2065) = 413 kg/m3
b = 2 [ (d1) (a) � (q) (z) ] = 2 [ (9.0 ft) (26 lb/ft3) � (250 lb/ft2) (0.2065) ]
= 571 lb/ft2
= 2 [ (d1) (a) � (q) (z) ] = 2 [ (2.74 m) (413 kg/m3) � (11,974 Pa) (0.2065) ]
= 27,148 Pa
c = (2) (Fga) = (2) (833 lb/ft) = 1,666 lb/ft= (2) (Fga) = (2) (12,161 N/m) = 24,322 N/m
dh = (571 lb/ft2) � (571 lb/ft2)2 4 (26 lb/ft3) (1,666 lb/ft)2 (26 lb/ft3)
= (571 lb/ft2) � (391 lb/ft2) = 18.5 or 3.5 The wall is only 9 ft (2.74 m) tall52 lb/ft3 so 18.51 (5.64 m) is not valid.
= (2,767 kg/m2) � (2,767 kg/m2)2 4 (413 kg/m3) (2,479 kg/m)2 (413 kg/m3)
= (2,767 kg/m2) � (1,887 kg/m2) = 5.6 or 1.1826 kg/m3
d2 = d1 dh = 9.0 ft — 3.5 ft = 5.5 ft= d1 dh = 2.74 m — 1.07 m = 1.67 m
The first layer of geogrid is placed at 1/2 dh.
hg = 1/2 dh = 1/2 (3.5 ft) = 1.75 ft= 1/2 dh = 1/2 (1.07 m) = 0.53 m
Analysis to determine if more than one additional layer of geogrid is required;
Fh = 0.5 (�r) (Kar) (d2)2 cos (�wr) = 0.5 (125 lb/ft3) (0.2197) (5.5 ft)2 cos (30º) = 360 lb/ft= 0.5 (�r) (Kar) (d2)2 cos (�wr) = 0.5 (2,002 kg/m3) (0.2197) (1.67 m)2 cos (30º)
= 5,211 N/m
Qh = (q) (Kar) (d2 hg) cos (�wr) = (250 lb/ft2) (0.2197) (5.5 ft 1.75 ft) cos (20º)= 194 lb/ft= (q) (Kar) (d2 hg) cos (�wr) = (11,974 Pa) (0.2197) (1.67 m — 0.53 m) cos (20º)= 2,818 N/m
73
Ft = Fh � Qh = 360 lb/ft + 194 lb/ft = 554 lb/ft= Fh � Qh = 5,211 N/m + 2,818 N/m = 8,029 N/m
Ft = 554 lb/ft < 833 lb/ft Only one more layer of geogrid is required.
= 8,029 N/m < 12,161 N/m Only one more layer of geogrid is required.
hg = (H d2) � 0.5 (dh) = (9.0 ft — 5.5 ft) + 0.5 (5.5) = 6.25 ft= (H d2) � 0.5 (dh) = (2.74 m — 1.67 m) + 0.5 (1.67 m) = 1.9 m
Check number of layers of geogrid required.
Fh = 0.5 (�r) (Kar) (H)2 cos (�wr) = 0.5 (125 lb/ft3) (0.2197) (9.0 ft)2 cos (20º)= 1,045 lb/ft= 0.5 (�r) (Kar) (H)2 cos (�wr) = 0.5 (2002 kg/m3) (0.2197) (2.74 m)2 cos (20º)= 15,220 N/m
Qh = (q) (Kar) (H hg) cos (�wr) = (250 lb/ft2) (0.2197) (9.0 ft 1.75 ft) cos (20º)= 374 lb/ft= (q) (Kar) (H hg) cos (�wr) = (11,974 N/m2) (0.2197) (2.74 m 0.53 m) cos (20º) = 5,463 N/m
Ft = Fh � Qh = 1,045 lb/ft + 374 lb/ft = 1,419 lb/ft= Fh � Qh = 15,220 N/m + 5,463 N/m = 20,683 N/m
Ft = N = 1,419 lb/ft = 1.7 = 2 LayersLTADS 833 lb/ft
= N = 20,683 N/m = 1.7 = 2 Layers12,161 N/m
Because the maximum spacing of geogrid is 3 ft (0.91 m), the total number of grid will need to be three.
Layer 1 = 1.905 ft (0.58 m) = 3 blocks from bottom.Layer 2 = 5.08 ft (1.55 m) = 8 blocks from bottom.Layer 3 = 8.255 ft (2.52 m) = 13 blocks from bottom.
74
Slope Surcharge
Roadway Application
Multi Terraces
Live Surcharge On Wall
Surcharge Above Wall, Slope Below Wall
Roadway Surcharge
Allan Block Retaining Wall Systems
Allan Block Corporation • 7400 Metro Blvd., #185, Edina, MN • 952-835-5309 • 952-835-0013 - FaxUS Patents #4,909,010 & 5,484,236 • Canadian Patent #2,012,286 & 2,133,675 • Taiwan Patent #NI-090824 AustralianPatent #133,306 & 682,394 • Int’l. & Other Patents Pending • ICBO #5087 • Copyright © 1999 • engman1099
Visit our web siteallanblock.com