“
NEW WORDS Reinforced Earth A composite mat materia eriall havi aving facing panels, soil reinforcement stri strips ps and and sele select ct fill as basic components.
Highway Embankment A rais raised ed eart earthe hen n structure structure used for increasing the level of pavement from from the grou ground nd due due to a numb number er of factors.
Study and Design of Reinforced Earth Retaining Wall with steel Reinforcement for Highway Embankment TABLE OF CONTENTS Why highway embankment is needed to be strengthened? _____________________________________________3 What is done to strengthen the embankment? _________3 Which method is the best? _______________________________4 What is a Rei Reinforced nforced Earth? ______________________________5 Introduction to reinforced soil structure ________________6 The Reinforced Earth Retaining Wall _____________________6 Mechanism Of Reinforced Soil ____________________________7 Shear Box Analogy Concept ______________________________8 What can be used as a Reinforcement? ________________10 1. _________________________________________________ Geosynthetics _________________________________________________ Geosynthetics _______________________________ _______________________________________________ _________________________________10 _________________10 2. _____________________________________________ Reinforcing _____________________________________________ Reinforcing strips _______________________________ _______________________________________________ _________________________________10 _________________10
Stability of Retaining Walls ______________________________11 Metallic Strip Re Reinforcement inforcement ____________________________12 APPENDIX APPENDI X A (Review (Review of Shear Strength) _______________22 APPENDI X B (Rankine APPENDIX (Rankine’s ’s Theory of Lateral earth Pressure) _________________________________________________26 _________________________________________________26
Table Of Figures Figure a External stabilization..... stabilization........................... ......................................6 ................6 Figure b Internally stabilized reinforced soil.......................6 soil.......................6 Figure c (a) Element of unreinforced soil, (b) Element of reinforced soil............................................................7 soil............................................................7 Figure d Slopes showing failure surface.............................8 surface.............................8 Figure e shear box test on unreinforced and reinforced soil...............................10 soil ...............................10
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Figure Figure Figure Figure Figure
f External stability checks (after Transportation Transportation Research Board, 1995)...11 1995)...11 g Steel reinforced earth retaining wall..... wall........... ....................... ........................................ .......................12 12 h Analysis of a Reinforced Earth retaining Wall.......................................14 Wall.......................................14 i Understanding the equation of Lateral pressure due to Surcharge.........15 Surcharge .........15 j Calculation for Overturning..................................................................20 Overturning..................................................................20
PART 1
Introduction
Why highway embankment is needed to be strengthened? A highway embankment is a structure (usually made up of earth in Pakistan) which has a sufficient height to separate the pavement built over it from the ground level. The reasons why it is necessary to make it strong are given below: •
It has to bear the forces of water during the flood.
•
It has to bear huge loading of traffic with large number of repetitions.
• To avoid the Land sliding. •
Over-burden
•
Dead and Live Load Surcharge
•
Earth Pressure Pressure
•
Hydrostatic Pressure Pressure
•
Seismic Loads
•
Construction Loads etc…..
What is done to strengthen the embankment? Ther There e are are a large large number number of soluti solutions ons to the problem problem of streng strengthe thenin ning g an earthen embankment. These include: •
Giving slopes on both sides of embankment when embankment height is sufficiently high.
• •
Ability of Plant Roots to Strengthen Soil Influence of Short Polymeric Fibers Fibers on Crack Development in Clays
•
Recycled Recycled Plastic Soil Nails Provide Slope Stabilization Project
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• • • • •
In-situ densification of soils. Ground improvement and modification. Reinforced soil. Grouting. Grading and other soil improvement methods
Sheet piling Sheet piling piling may be composed of steel, steel, timber or concrete concrete piles, with each pile being linked to the next to form a continuous wall. Sheet pile walls are sufficiently watertight for most practical purposes. Grouting covers different injection techniques of special liquid or slurry materials called grouts into the ground for the purpose of improving the soil or rock. descri ribe bed d as the the excavat avatiion and and exchang hange e of Muckshift can be desc unsuitable soil regions by more qualified ones. It is a kind of large scale land clearance. Sandbagging Sand Sandba bags gs can be used sed for for pre prevent ventin ing g a leach ach ate discharge downstream of the embankment site. Geosynthetical structures Rock facing / rock riprap slope surface consists of rock or cobble fills, no special slope surface treatment is necessary. necessary. Downstream slopes with outer sand and gravel should be protected against erosion especially during flood.
Which method is the best? Most of these methods work in different problems in different situations and therefore none can be said as best. Another reason for this is that other than reinforcing the earth and making a composite material called reinforced earth, all the above mentioned methods require having slopes of embankment for an economical design. There can be a situation where one wants to have an embankment with angle of repose 90 degrees. This can be achieved by employing the reinforced earth concept for the embankment.
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PART 2
Reinforced Earth What is a Reinforced Earth? A Reinf einfor orce ced d Soil Soil Syst System em (RSS (RSS)) is a compo composi site te mater materia iall whic which h has has the the following basic components; •
Facing Panel (Commonly made of concrete, steel plate, wire mesh, block etc…)
•
Soil Reinforcement Strips (Galvanized steel, geotextiles, etc)
•
Select Fill (Cohesionless soil meeting specific defined requirements)
The frictional forces created when combining the select fill with the flexible meta metall llic ic or nonnon-me meta tall llic ic rein reinfor forci cing ng stri strips ps resu result lt in a robu robust st stru struct ctura urall material, commonly known as Reinforced Earth. The strips are attached to a front facing panel, which may be manufactured from concrete or steel. The faci facing ng mate materi rial al sele select cted ed is gene general rally ly depe depend nden entt on it havi having ng suff suffic icie ient nt durability to accommodate the design life of the structure, and also meet the aesthetic needs of the project. The Reinforced Earth monolithic mass acts cohesively and supports it’s own weight and any applied loads which may include all the forces as described in part 1 of this report. The forces induced in the steel strips can be precisely calculated and depend on ;•
Strip geometry
•
Strip frictional characteristics
•
Vertical soil pressure on the strip
•
Strength and stiffness characteristics on the strip
Importantly, the durability of the structure relies heavily on the ability of the soil soil rein reinfo forc rcem emen entt stri strip p to main mainta tain in a leve levell of tens tensil ile e stre streng ngth th in the the operational operational environment environment for the duration of the structure’s structure’s design life. The strip made up of steel, if used, is therefore designed to include a sacrificial 5 | Page
steel thickness, which predicts the amount of strip corrosion throughout the design life of the structure. This is achieved by controlling the environment in which the strip will be operating. The select fill, whilst having certain physical requirements that ensure it is activated in forming part of the structural mass, mass, is also also requir required ed to have have electr electroch ochemi emical cal charac character terist istics ics that that also also ensures that corrosion of the strip is not excessive or beyond the allowance made made in the the stri strip p desi design gn.. Furth urther ermo morre, the the stri strip p is coat coated ed with with zinc zinc galvanizing for further protection. The final length and frequency of the soil reinforcement strips is a function of the combinations of geometric and physical properties of the structure and the applie applied d design design loads. loads. Whilst Whilst the facing facing to the Reinfo Reinforc rced ed Earth Earth wall wall technically does not take on a structural role in support of the loads, it obviously forms an important part in the wall in preventing the erosion of back backfi fill ll,, supp suppor orti ting ng the the soil soil reinf einfor orce ceme ment nt and and weat weathe heri ring ng the the loca locall environment. Typically, for roads projects, concrete is the only economical material that can achieve the necessary 100 year design life without the need for any continuous maintenance or repair. The facing also forms the most most visu visual al aspe aspect ct of the the stru struct ctur ure e and and is ofte often n requ requir ired ed to have have some some aesthetic appeal, particularly in urban areas. Concrete can lend itself readily to the provi provisio sion n of archit architect ectural ural and aesthe aesthetic tic requi require remen ments. ts. The facing facing panels can however, often be a complex component to manufacture as each facing facing panel panel may have very very indivi individua duall charac racteristics with respect to its geometry, finish or cast-in inclusions.
Introduction Introduction to reinforced soil structure Reinforced soil structures are fundamentally different from conve convent ntio iona nall eart earth h retai etaini ning ng syst system ems s which which are exter external nally ly stabil stabiliz ized ed in that thatFigure a External stabilizationthey they utilize a different mechanism for support and are internally stabilized. An externally stabilized system uses an external structural wall against which stabil stabilizi izing ng force forces s are are mobiliz mobilized, ed, for examp example, le, gravity gravity retai retainin ning g walls walls and excavations supported with strutting (fig A) An internally internally stabilize stabilized d system system involves involves reinforc reinforcement ements s installed installed within within & extending beyond the potential failure mass (tied back, reinforced soil walls, and soil-nailed excavations). With this system, the interactions between the reinforcements and soil (to mobilize the tensile capacity of closely spaced 6 | Page
rein reinfor forci cing ng elem elemen ents ts)) elim elimin inat ate e the the need for a structural wall or a support (figure B)
The Reinforced Earth Retaining Wall According to Reinforced Earth company,
Figure b Internally stabilized reinforced soil
®
Reinforced Reinforced Earth retaining retaining walls are gravity structures consisting of alter alternat nating ing layers layers of granul granular ar backfi backfill ll and reinforcing strips with a modular precast concrete facing. They are used exte extens nsiv ivel ely y in tran transp sport ortat atio ion n and and othe otherr civi civill engi engine neer erin ing g appli applica cati tions ons.. Because of its high load-carrying capacity, Reinforced Earth is ideal for very high or heavy-loaded retaining walls. The inherent flexibility of the composite material makes it possible to build on compr compress essibl ible e founda foundatio tion n soils soils or unstab unstable le slopes slopes.. These These perfor performan mance ce advantages combined with low materials volume and a rapid, predictable and easy construction process make Reinforced Earth an extremely costeffective solution over conventional retaining structures.
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Mechanism Of Reinforced Soil The The mech mechan anis ism m of reinf einfor orce ced d soil soil can can be expla xplain ined ed in simp simple le term terms s cons consid ider erin ing g an elemen elementt of cohe cohesi sionl onles ess s soil soil show shown n in figure figure C(a) C(a) . If a vertical stress is applied on the soil it deforms both laterally and vertically and reaches a new equilibrium. If reinforcement in the form of plane sheet is introduced in the sample before the application of vertical stress on the sample,
Figure c (a) Element of unreinforced soil, (b) Element of reinforced soil
deform deformati ations ons are restrai restrained ned due to the interact interaction ion b/w the soil and the reinforcement reinforcement to some extent as in figure C (b). introduction of reinforcement generates generates inward inward lateral lateral stress, stress, which resists resists the shear shear stresses stresses that are gene genera rate ted d when when a vert vertic ical al str stress ess (sig (sigma ma 1) is appl applie ied. d. If ther there e is no significant deformation, the lateral stress is equal to Ko times the vertical stress and this condition prevails at higher vertical stresses also.
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The shear stress at the interface of soil and reinforcement generates strains in the reinforcement and tensile force is mobilized in the reinforcement. If the reinforc reinforcement ement force force exceeds exceeds the tensile tensile capacity capacity of sheet sheet reinforc reinforcement ement,, rupture failure occurs. Secondly, it is likely that a slip occurs b/w soil & reinf reinfor orcem cement ent if deform deformati ation on are are high high or interf interface ace is smooth. smooth. These These two conditions viz tesile failure & pullout failure need to be examined to ensure stability of reinforced reinforced soil structures.
Shear Box Analogy Concept The These se shea shearr box box test test simu simula late tes s the the mech mechan anis ism m and beha behavio viorr of both both unreinforced and reinforced soils. Figure on the left shows slope of unreinforced soil and on the right shows slope of reinforced soil.
Figure d Slopes showing failure surface
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What can be used as a Reinforcement? 1. Geos Geosyn ynth thet etic ics s Figure 2 shear box test on unreinforced and reinforced soil ASTM
has defined a geosynthetic as a planar product manufactured from a polymeric material used with soil, rock, earth, or other geotechnical-related material as an integral part of a civil engineering project, structure, or system. Geotextiles: A geotextile is a permeable geosynthetic made of textile materials Geotextile in fabric form are being used as a basal reinforcement of embankment and fills on the ground Geogrids: Geogrids are primarily used for reinforcement; reinforcement; they are formed by a regular network of tensile elements with apertures of sufficient size to interlock with surrounding fill material. These are made up of High Density Polyethylene Polyethylene (HDPE) and interconnected longitudinal and transverse Member. Member. These are made from sheets of polymer by punching holes and stretching the sheets in one or two direction. Geocomposits: Geotextiles and related products such as nets and grids can be combined with geomembranes and other synthetics to take advantage of the best attributes of each component. These products are called geocomposites. 10 | P a g e
2. Rein Reinfo forc rcin ing g strip strips s
Reinforced Reinforced members are composed of thin wide steel or aluminum strips called ties. The ties. The flexibility of reinforcing strips and their tensile strengths are essential elements. The reinforcing strips are made of mild galvanized steel, stainless steel or aluminum alloy. alloy. Bolts and nuts for fixing the ties are made of the same material as that of the reinforcing strips. The durability of the strips depends on the chemical and electro-chemical behavior of these metals when in contact with soil particles
Works Cited Introduction to Soil Reinforcement and Geosynthetics [Book] / auth. Babu Visakumar. Visakumar. - to be added : to be added, to be added. - Vol. to be added.
PART 3
Stability of Retaining
Walls A retaining wall may fail in any of the following ways (see figures): •
It may overturn about its toe.
•
It may slide along its base.
•
It may fail due to the loss of bearing capacity of the soil supporting the base.
•
It may undergo deep-seated shear failure.
•
It may go through excessive settlement.
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Figure f External stability checks (after Transportation Transportation Research Board, 1995)
When a weak soil layer is located at a shallow s hallow depth—that is, within a depth of 1.5 times the width of the base slab of the retaining wall—the possibility of excessive settlement settlement should be considered. In some cases, the use of lightweight backfill material behind the retaining wall may solve the problem.
Retaining Walls
PART 4
with Metallic Strip Reinforcement
As discussed before, the reinforced earth retaining wall can deploy steel and geotextiles as reinforcements. reinforcements. In this part we will be looking at the design of reinforced earth wall using steel as reinforcement. A basic model of steel reinforced earth retaining wall is shown in figure (f).
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Figure 0 Steel reinforced earth retaining wall
It can be seen that it consists of: •
Granular soil as backfill
• Thin, wide steel reinforcing strips placed at regular intervals and •
A cover or skin on the front face of the wall
Calculation of active horizontal and vertical earth pressures
Figure (g) shows a retaining wall with granular backfill having a unit weight of γ 1 and a friction angle of ф1. Below the base of the retaining wall, the in-situ soil has been excavated and recompacted, with granular soil used as backfill. Below the backfill the in-situ soil has the unit weight of γ 2, 2, friction angle of ф 2 and cohesion of c2’. A surcharge having intensity of q per unit area lies atop the retaining wall, which has reinforcement reinforcement ties at depths z= 0, Sv, 2Sv, . . . , NSv. NSv. The height of the wall is NSv = H. According to Rankine active pressure theory (see Appendix B for details),
Where = Rankine active pressure at any depth z For dry granular soilswith no surcharge at the top, = 0, = and
K a = tan
2
(45 – ф1’/2). Thus,
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When surcharge is added at the top, as shown then =
(1) +
Due to soil
(2)
Due to Surchar
According to Laba and Kennedy Kennedy (1986),
(For z≤2b’) And
(For z›2b’) Also when a surcharge is added at the top the lateral pressure at any depth is = + (2) Due to soil
14 | P a g e
Due to Surchar
Figure h Analysis of a Reinforced Earth retaining Wall
According to Laba and Kennedy Kennedy (1986)
= M[2q/π(β-
↑ (In radians) Where
M = [1.4 – (0.4 /0.14H)] ≥ 1 Tie Force
The tie force per unit length of wall developed at any depth z (figure f) is T = active earth pressure at depth z x area of the wall to be b e supported by the tie
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Figure i Understanding the equation of Lateral pressure due to Surcharge
T = ) Factor of Safety against Tie Failure
The reinforcement reinforcement ties at each level, and thus the walls, could fail by either, (a) Tie breaking or (b) Tie pullout
Yield or breaking strength of each tie / maximum
force in any tie
/) Where = width of each tie = thickness of each tie = yield or breaking strength of the tie material A factor of safety of about 2.5 – 3 is generally recommended for ties at all levels. Reinforcing Reinforcing strips at any depth z will fail by pullout if the frictional resistance developed along the surfaces of the ties is less than the force to which the ties are being subjected. The effective length of the ties along 16 | P a g e
which frictional resistance is developed may be conservatively taken as the length that extends beyond the limits of the Rankine active failure zone , which is the zone ABC in figure(). Line BC makes makes an angle 45 + ф1’/2 with the horizontal. Now the maximum friction force that can be realized for a tie at depth z is
Where = effective length = effective vertical pressure at a depth z
фu = soil -
tie friction angle
thus the factor of safety against tie pullout at any depth z is
FS(P) = FR / T ) Total length of Tie
The total length of ties at any depth is
L = lr + le l
Where r = length within the rankine failure zone
le = effective length } + )} The Design Procedure with summary of steps Following is a step-by-step procedure for the design of reinforced-earth reinforced-earth retaining wall General
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Determine the height of the wall, H, and the properties of the granular .)backfill material, such as the unit weight () and the angle of friction (ф1 Obtain the soil-tie friction angle, (фu) FS(P) |
and the required value of FS (B) and
Exhibit 1
Given Data And general Requirements Height of the Embankment 8 cohesion 52 Unit Weight of Granular Backfill ( Angle of Friction of Backfill (
)
')
16.6 30
kN/m3 degrees degrees
Soil-Tie Soil-Tie Friction Angle A ngle ( u') Factor of Safety
20
against Tie-Breaking Tie-Breaking FS(B) Factor of Safety
3
FS(P) Horizontal Tie Spacing ( SH) vertical Tie Spacing ( SV)
3 1
metres
0.5
metres
against Tie Pull-out
Width of Reinforcing Strip ( ) Load per unit area q Span of Load a Distance of Load from Wall b M
0.075 10 3 0.1 0.543
n/a n/a
metres metres metres
1.32
n/a radians
Angle Lateral Pressure due
0.1357
radians
to Surcharge 'a(2) Lateral Pressure due
7.033141
Angle
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Units metres kN/m2
to Soil Only 'a(1) Total Lateral Pressure 'a
137.4331
Tie Force T Thickness of ties
68.71657 0.011453
130.4
kN/m2 kN/m2 kN/m2 kN metres
Length within the rankine failure zone lr Effective length of Ties le Length of Ties Ka
L
0
metres
4.625899
metres
4.625899 0.333
metres
Internal Stability (see the attached Exhibit 2 and the result of internal stability analysis in succeeding pages)
Assume values for horizontal and vertical tie spacing. Also, assume the width of reinforcing strip, w, to be used. Calculate from given Eqs. Calculate the tie forces at various levels. For the known values of FS(B), calculate the thickness of ties, t, required to resist the tie breakout:
)/ The convention is to keep the magnitude of t the same at all levels, so in Eq. should equal (max). For the known values of Φu and FS(P), determine the length L of the ties at various levels. The magnitudes of S v , S H , t , , and L may be changed to obtain the most economical design. External Stability 1) Che Check for for overturning, using Figure () as a guide. Taking the moment about B yields the overturning moment for the unit length of the wall:
Mo = Pa z’ Here,
Pa = active force =
Calculation Calculation shown for the above equation
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soil only 8
⌠ ⌠ ⌡
5.478h ⋅ dh
→ 175. 296= 175.296
1.4 − .96 ⋅ 1.66dH → 2. 324⋅ H + −11.382857142857142857 ⋅ ln) H .14H
⌠ 2 1.4 − .04 ⋅ 20 ⋅ ) 1.159085553 1.159085553− sin) 1.159085553 1.159085553⋅ c os) 2⋅ 0 . 099668652 dH → 1.995 1.99557 5799 9915 1587 8760 6089 896 6 3 .14H π ⌠ ⌡1 1.4 − .04 ⋅ 20 ⋅ ) 0.947871788 0.947871788− sin) 0.947871788 0.947871788⋅ c os) 2⋅ 0. 04 9958 396 dH → 1.14 1.1424 2454 5413 1308 0846 4663 6395 9587 876 =61.142 .14H π ⌡2 ⌠ 7 4 1.4 − .04 ⋅ 20 ⋅ )0.460226744 0.460226744− sin) 0.460226744 ⋅ c os) 2⋅ 0. 01 6665 124 dH → 0.140 0.14089 8975 7585 8563 6346 4640 4070 7028 28 = 0.141 ⌠ .14H π .04 20 ⌡6 1.4 − ⋅ ⋅ )0.768469142 0.768469142− sin )0.768469142 ⋅ c os ) 2⋅ 0. 033320996 dH → 0.629 0.62900 0088 8839 3954 5415 15334 33476 7651 51 = 0.629 .14H π ⌡3
5 ⌠ 8⌠ 20 1.4− −.04.04 ⋅ 20 1.4 0.634315275 )0.634315275 0.63431527 H→ 0.3 6093 9389 8949 4903 0397 9726 2636 3661 ⋅ ⋅ )0.402613328 ⋅ ) 0.634315275 ⋅01.4022844979443794dHd→ =10.361 − s−insin =611 ) 0.402613328 0.40261 3328 ⋅ c5o⋅sc)o2s⋅)02. 0 0.094 0.00.360 9492 9271 7136 3684 8413 1385 8521 2133 3336 36 0.095 π .14H ⌡ .14H π ⌡7 4
⌠ 6 1.4 − ⌡5
⋅ 20 ⋅ )0.534998393 0.534998393− sin )0.534998393 0.534998393⋅ c os ) 2⋅ 0. 019997334 dH → 0.219 0.21938 3869 6930 3099 9958 5897 9765 6569 693 =30.219 .14H π .04
………… /{[ 20 | P a g e
Exhibit 2
Overturning Active force
175.3
4.5
179.8
soil
surchar ge
total
Resisting Moment W1 x1
1726.4 6.5 11221. 6 11269. 6
W1x1 MR
kn m kn-m kn-m
Overturning Moment z'
2.67 480.06 6
Mo
m
Figure j Calculation for Overturning
23.475 11
FOS
>
3
2) The check check for sliding sliding can can be done done by using using
………… /{ Sliding Numerat 639.27 or 69 FOS
Where k
≈
21 | P a g e
2/3
3.56 > 3
3) Check Check for ultimate ultimate bearing bearing Capacity Capacity can can be done by
The vertical stress at z = H is
So the factor of safety against bearing capacity failure is
Ultimate Bearing Capacity qu = 3805.2 Vertical stress at 8 m = 142.5 FOS = 26.7 > 5
Generally, minimum values of 3, 3, and = 3 to 5 are recommended.
APPENDIX A Review of Shear Strength Sources of stresses in the ground Geostatic stresses
σ τ xσ
zx xz z =
γ z
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Geostatic stresses are those that occur due to the weight of the soil above the point being evaluated. They are caused by gravity acting on soil or rock so the direct result is the vertical normal stress σz. This vertical normal stress indirectly produces horizontal normal stresses and shear stresses.
Exhibit 3: Three dimensional soil element (left), 2D soil element (middle), Induced horizontal stress in an unconfined soil element (right)
Horizontal stresses are either direct (e.g. braking forces from wheels of large truck) or induced induced (see figure (right)). (right)). Unlike the figure, figure, real real soils in the field are not unconfined and the adjacent elements of soil or rock also wish to expand but in opposite direction. These opposing forces may cancel each other (i.e. (i.e. there may be no horizontal horizontal strain) or the horizontal horizontal strain may be much less than would occur in an unconfined unconfined sample. sample. Either Either way, the result will be the formation of horizontal stresses in the ground. If ground surface is horizontal, the geostatic shear stresses on horizontal and vertical planes are all equal to zero.
τxz = τzx = τyz = τzy = τxy = τyx
(1)
Induced stresses
These are the stresses due to external loads. Equation (1) will not be true for induced stresses.
Effective stress The compressive stress, σ, is carried partially by the solid particles & partially by the pore water. It is called the total stress because it is sum of the stresses carried by these two phases in the soil.
u = γ w zw Where γ w = unit weight of water
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zw = depth from the groundwater table to the point where pressure head is to be computed. The particle contact forces in an element of soil that is initially above the groundwater table will decrease if the groundwater rises above a bove that element. This is because of the buoyancy force of water. water.
Vertical effective stress = σz = (σz) geostatic + (σz) induced - u
Horizontal effective stress = σx’ = σx – u σy’ = σy – u
Coefficient of lateral earth pressure The ratio of horizontal to vertical effective stresses is defined as the coefficient of lateral earth pressure, K.
Mohr’s circle It is used for computation of stresses acting on planes other than horizontal and vertical planes. It describes the 2D stresses at a point po int in a material. Each point on the circle represents the normal & shear stresses acting on one side of an element oriented at a certain angle. For example, in this mohr’s circle, points A & B represent (σz, τzx) and (σx,
τ
xz),
which are the stresses acting on an element aligned with the x and z axes, while points C & D represent represent the stresses on an element oriented as shown.
Principal stresses If the soil element is rotated to a certain angle, shear stresses will be zero on all four sides. The planes on each side of this element are represented represented by points E & F and are known as principal planes. Stresses acting along these planes are known as principle stresses. 24 | P a g e
σ
= major principal stress (greatest normal stress acting on any plane)
σ
= minor principal stress (smallest normal stress acting on any plane)
1
3
these two stresses act on right angle to each other. As told earlier, when the ground surface is level, the geostatic shear stresses on vertical & shear planes ar all zero. Therefore, these are the principal planes and the geostatic principal stresses act vertically and horizontally. horizontally. If σx >
σz then σ3 = σz and σ1 = σx
If σz >
σx then σ3 = σx and σ1 = σz
Mohr’s circle for effective stress principle of effective stress applies only to normal stresses, not to shear stresses because the pore water can’t carry a static shear stress.
Shear failure in soils In soil, shear failure occurs when the shear stresses b/w the particles are such that they slide or roll pass each other. The shear stress primarily depends upon interactions b/w the particles, not on their internal strength. Shear strength = frictional strength + cohesive strength
Frictional strength The force that resists sliding is equal to the normal force multiplied by the coefficient of friction, µ. For soils, ф’ is used i.e. ф’ = tan-1 µ Frictional shear strength = effective stress acting on shear plane tan ф
The value of ф depends on both the frictional properties of the individual particles and interlocking b/w particles. It also depends upon mineralogy, shape gradation, void ratio & organic material in soil. Pore water pressure reduces the frictional shear strength.
Cohesive strength Cohesion may be true or apparent. True cohesion is cementation and electrostatic forces. Apparent cohesion is negative pore water pressures, etc. 25 | P a g e
APPENDIX B
Rankin’s theory for Active Earth Pressures Active Earth Pressure Assumption
Frictionless wall
Before the wall moves the stress condition is given by circ le “a”
State of Plast ic equilibrium equilibrium represented by circle “b”. This is the “Rankine’s active state”
Rankine’s active earth pressure is giv en by
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∆L
A'
A
σ a'
B'
σ o'
B
27 | P a g e
z
With geometrical manipulations we get:
'
'
σ a = σ o
'
1 −si sin n φ ′ 1 +si sin n φ ′
σ a = γz tan
2
cos co sφ ′ 1 +si sin n φ ′
(4 5 −φ ) −2c' tan (45 −φ ) ′
′
2
For For cohesi cohesionl onless ess soil, soil,
' a
−2c'
' 0
2
’=0 ’=0
c
φ '
2
σ =σ tan ) 4 5 −
2
Rankine’s Rankine’s Active Pressure Pressure Coefficient, Coefficient, Ka
The Rankine’s Rankine’s active pressure coefficient coefficient is given given by: σ a' φ ′ 2 K a = ' = tan ( 45 − 2 ) σ o The angle between the failure planes /slip planes and major principal plane (horizontal) is:
±
( 45 + ) φ ′ 2
σ a' The variation of
28 | P a g e
With depth:
The slip planes:
Lateral Earth Pressure Pressure Distribution against Retaining Walls There are three different cases considered: Horizontal backfill Cohesionless soil Partially Partially submerged cohesionless soil with surcharge Cohesive soil 29 | P a g e
Sloping backfill Cohesionless soil Cohesive soil Walls with Friction
Active Case
Horizontal backfill with Cohesionless soil
σ a
P a
= K γ z a
= 1 K γ H 2 2
a
Horizontal backfill with Cohesionless, partially submerged soil
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σ a' = K a [q + γ H 1+γ ' ) z − H 1 ]
Horizontal backfill with cohesive soil
The
depth at which the active pressure becomes equal to zero (depth of tension crack) is z 0
=
2c ' γ
K a
For the undrained condition, φ = 0, then K a becomes 1 c=cu. Therefore, z 0
(tan245° = 1) and
= 2c
u
γ
Tensile Tensile crack is taken taken into account when finding the total active force. force. i.e., consider only the pressure p ressure distribution below the crack
Active total pressure force will be 31 | P a g e
P a
=
1 2
K a γ H
2
−2
'
K a c H
+
2c ' 2 γ
Active total pressure force when φ = 0
P a =
1 2
2
H − 2cu H + γ
2cu2 γ
Sloping backfill cohesionless soil
Earth pressure acts an angle of α to the horizontal
σ a' P a
= K γ z a
= 1 K γ H 2 2
a
This force acts H/3 from bottom and inclines α to the horizontal cos co sα − cos co s2 α − cos co s2 φ ′ K a = cos co sα ⋅ 2 2 cos co sα + cos co s α − cos co s φ ′
Sloping backfill cohesive soil
σ a'
= γ zK = γ zK " cos co sα a
a
Depth to the tensile crack is given by
z 0 =
"
K a
=
K a
cos co sα
2c' 1 + sin si n φ ' γ
1 − sin φ '
Friction walls Rough retaining retaining walls with granular backfill. Angle of friction between the wall and the backfill is δ'
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Case 1: Positive wall friction in the th e active case (+δ´)
Wall AB
A’B
causes a downward motion of soil relative to wall.
Causes downward shear on the wall (fig. b) Pa will be inclined δ’ to the normal drawn to the back face of the retaining wall Failure surface is BCD (advanced studies): BC curve & CD straight Rankine’s active state exists in the zone ACD
Case 2: Negative wall friction in the active case (-δ’) - Wall is forced to a downward motion relative to the backfill
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