
#1
Feb907, 01:20 AM

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Note: This article will assume the reader is familiar with the concepts of statics and soil mechanics (mainly the topics of shear strength in soils and effective pressure).
1.1 Introduction Retaining Walls A wall designed with the purpose of keeping a level difference of the soil on its both sides is called a Retaining Wall. These supporting elements can be considered either rigid or flexible. The rigid type are mainly those made of masonry, simple concrete or reinforced concrete. The flexible type are known as sheet pile walls. They are usually made of steel. Retaining walls can also be classified due to their primary function to hold the soil mass. They can be of gravity, cantilever, and counterfort. This is a retaining wall of the gravity type. It is made with masonry units of rocks and mortar. This wall is of the cantilever type with steel diagonal supports. It is made of precast concrete sheets. This is a gravity wall, but less rigid than the usual. It is known as gabion. It is a very popular solution because gabions are easy to install, economic, offer drainage, and can withstand ground movement better than its counterparts. 



#2
Feb907, 04:40 PM

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Considerations
There are a number of forces that act on the retaining wall. Some are relatively constant while others intensity may vary due to factors such as weather. These forces are: 1) Weight of the wall This force acts on the gravity centroid of the section. 2) Pressure of the retained soil 3) The pressures on the foundation They are usually considered as being linearly distributed in the form of a trapezoidal diagram. 4) The pressure of the soil against the front of the wall The soil on the front of the wall exerts a passive force (resistance) against the active force of the retained soil. This force is usually omitted due to the uncertainty of its magnitude. 5) The loads on the retained soil 6) Forces due to water If there's a body of water on the back of the wall, there'll be hydrostatic pressure acting on it. This can be prevented by installing adequate drainage. 7) Subpressures When the drainage under the wall is not adequate or is damaged, it can lead to storage of water in that zone. If the foundation is impervious the water will flow until it will emerge on the frontal part of the soil. If the foundation is pervious, the water will generate pressure against the wall. 8) Vibration They are produced by traffic, power plants, and others. Frequently, vibrations effects on retaining walls are neglected because of their little contribution. In some cases, engineers simply use the magnitude instead of the normal component of the pressures of the retained soil on the wall. In other words, making the angle of the resultant ([itex] \theta [/itex])with the horizontal zero. This is done to overdesign the wall in order to avoid problems due to vibrations. 9) Impact of Forces on the retained soil The effects are damped by the soil, therefore they are neglected. 10) Stresses due to frozen water If the drainage is not adequate, when the water freezes it will produce expansions of the retained soil. 11) Expansions due to changes of humidity on the retained soil On clay soils the expansions produce an increase of the pressures exerted by the retained soil on the wall. When the soil dries up, it contracts and the pressures decrease accordingly. If this process keeps repeating, it can be harmful for the wall. This effect is more intense on the surface, and then it decreases with depth. There are not safe methods to calculate the increases of pressure due to the humidity. However, they can be prevented by including coarsegrained soils stratums. This is a section of a retaining wall. The Green colored vector represents the weight of the wall acting on the section's centroid of gravity. The magenta colored vector (resultant) and distribution diagram is the force exerted by the soil on the foundation. The blue colored vector (resultant) and distribution diagram is the resistance of the soil against the front of the wall. The red colored vector (resultant)and distribution diagram is the active pressure of the retained soil on the wall. 



#3
Feb907, 09:22 PM

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Note: a reader with a background in Mechanics of Materials will have a better understanding of this section.
Comments about the parameters of shear strength of a soil It is a well known fact for engineers that there are many theories of failure of a material. They can be divided by dynamic or kinematic criteriums. The first in function of the stresses and the last in function of the strains. Traditionally speaking soil mechanics has always studied the condition of the limit stress that causes the failure of soils either by fracture or plastic flow through the theory of MohrCoulomb. This theory states that the strenght of a material can be measured throught the maximum tangential stress, which is linked mathematically with the normal stress acting on the failure plane. It is interesting that the original theory of Coulomb which stated a linear relation between the normal stress and the shear stress has become a special case of the MohrCoulomb theory. This is because Mohr stated that the variation is generally represented with a curve. Classically speaking, it's that linear relation in Coulomb's theory that states the shear stress of a soil as in the following equation: [tex] \tau = \sigma \tan \phi + c [/tex] (1) where [itex] \tau [/itex] is the shear stress, [itex] \sigma [/itex] is the normal stress, [itex] \phi [/itex] is the "angle of internal friction" c is "cohesion" The graph made by equation (1) is called the line of failure. The values c and [itex] \phi [/itex] can be obtained throught the simple shear test, the triaxial stress test or others. These values or parameters are used to calculate the lateral foces by the soil, thus needed in the design of a retaining wall. The genesis of the classical theories of lateral pressure have their basis on equation (1), and the past interpretations of their parameters. The discrepancies with the results obtained experimentally and with equation (1) lead to modifications of the original equation by Coulomb. The studies made by Terzaghi in the 1920s proved that instead of the normal stress, it should be used an effective stress, due to the influences of water in the soil. Another study by Hvroslev in the 1930s acknowledged that the parameter c or "cohesion" in saturated clays was not constant, but it varied in function with the water content. This leads us to the equation: [tex] \tau = (\sigma  \sigma_{w}) \tan \phi + f(w) [/tex] (2) where [itex] \sigma_{w} [/itex] is the water stress, and f(w) is the cohesion in function of its water content. It is worth mentioning that the term "cohesion" is used even today, because of its tradition. It was discovered that there's no actual cohesion in the soils, and the strength of the soils is entirely due to friction. The parameter [itex] \phi [/itex] used to be known as the angle of repose of the soil. This was largely due to Rankine, however studies by Collin and Darwin proved that "internal friction angle" in some cases could drastically differ from the angle of repose. The reason is mainly due the behaviour of a loose coarsegrained soil and a compacted coarsegrained soil. The first will have a somewhat linear relation between the shear and the normal stresses (loose sand), while the compacted will have more of a curve linear between them (compacted sand). In the case of finegrained soils such as clay, the "internal angle of friction" will vary due the plasticity of the soil. There are a number of recommendations for calculating the "internal angle of friction" ([itex] \phi [/itex]), which go beyong the purpose of this article, however, they should be presented on any Geotechnical Engineering, Foundation Engineering and/or Soil Mechanics textbook. In practice, a structural engineer will only care about the values of c and [itex] \phi [/itex] obtained by the geotechnical engineer in order to calculate the lateral forces of the soil on the retaining wall. 



#4
Feb1007, 01:20 PM

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An Introduction to Retaining Walls Design.
1.2 Lateral Pressure of a soil
The estimation of the magnitude of the laterals forces acting on the retaining wall is not extremely accurate. The fact is, currently, engineers employ the classical theories of Rankine and Coulomb with certain modifications such as the ones made by Caquot and Kerisel (1948), Shields and Tolunay (1973), and Zhu and Qian (2000) to the Coulomb's theory. There are empirical methods such as the one developed by Terzaghi, which are considered pretty safe as long as we are in the range of its applicability, which is in fact limited. In this article, we will only focus on the theory by Rankine, and later on will discuss a bit more about the uncertainty in retaining walls. 



#5
Feb1107, 02:38 PM

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Note: The derivations for Rankine equations (710) should be consulted on a Soil Mechanics, Geotechnical Engineering or a Foundation Engineering book.
The Theory The lateral pressure on the retaining wall will depend on three "soil states". The highest push against the wall will be by the passive state, and the lowest by the active. Lateral pressure at rest This "state" is referred to the original form the soil had, before a part of it was removed. This is shown on the picture: Now consider an element restricted (no displacement) and "small enough" (a particle) inside a soil at depth of z. The vertical effective pressure acting on it, will be given by: [tex] P_{v} = \gamma z [/tex] (3) where [itex] \gamma [/itex] is the specific weight of the material, and z is the depth. Under the vertical pressure the element is pressured laterally, thus originating an horizontal pressure on the element, which is given by the equation: [tex] P_{h} = K_{o} \gamma z [/tex] (4) where [itex] K_{o} [/itex], the constant of proportionality, is called the coefficient of pressure at rest, and its values are obtained experimentally. However, there are some empirical equations that get close enough to the real values. Empirical Relations: (Jaky, 1944) [tex] K_{o} \approx 1  \sin \phi [/tex] (5) Consolidated clays (Brooker & Ireland, 1965) [tex] K_{o} \approx 0.95  \sin \phi [/tex] (6) where [itex] \phi [/itex] in this case is the maximun angle of friction drained. There are more relations such as the one by Mayne and Kulhay (1982) for both sand and clay. In this state, the soil has no failed and the lateral forces will be higher than the active state. Active lateral pressure with Rankine's Theory If the horizontal pressures decrease, and the vertical pressures remain constant, on the element we mentioned, then the backfill or retained soil will fail due to shear stresses. Rankine thought the wall will yield and displace under the influence of the backfill forces, thus decreasing the pressure of the retained soil to values under the "rest state". This value will be the minimum pressure the wall will have to withstand, even if it displaces more. The generalized equation for active horizontal pressure by Rankine is: [tex] P_{h} = P_{v} K_{a}  2c \sqrt{K_{a}} + wK_{a} [/tex] (7) where [itex] P_{v} [/itex] is the vertical effective pressure [itex] K_{a} [/itex] is the active coefficiente of active pressure [itex] w [/itex] is the surcharge acting on the soil. Also note that: [tex] K_{a} = \tan ^{2} ( 45^{o}  \frac{\phi}{2}) [/tex] (8) This equation works for the case where the backfill is horizontal, for cases where the backfill is inclined, the equation will become much complex. Pasive lateral pressure with Rankine's Theory If the vertical pressures decrease, and the horizontal pressures remain constant, on the element we mentioned, then the backfill or retained soil will fail due to shear stresses. Rankine thought the wall will displace onto the backfill because of an strong enough exterior force, and in this way allowing the soil to develop the pasive pressure. In this case the wall will have to be designed with the maximum value of the lateral pressures. The generalized equation for pasive horizontal pressure by Rankine is: [tex] P_{h} = P_{v} K_{b} + 2c \sqrt{K_{b}} + wK_{b} [/tex] (9) where [itex] P_{v} [/itex] is the vertical effective pressure [itex] K_{b} [/itex] is the pasive coefficiente of active pressure [itex] w [/itex] is the surcharge acting on the soil. Also note that: [tex] K_{b} = \tan ^{2} ( 45^{o} + \frac{\phi}{2}) [/tex] (10) This equation works for the case where the backfill is horizontal, for cases where the backfill is inclined, the equation will become much complex. Addedum As it was discussed earlier, the lateral pressure theories have their basis on the equation (1). This is crucial in order to understand the active and pasive "soil states". If we draw the line of failure of a soil, equation (1), and the draw the corresponding Mohr circle for the "rest state", we will notice the soil will not fail. However, if we decrease the horizontal pressure and keep the vertical pressure constant (active state), the line of failure for this case Mohr circle will be its tangent. This means a wedge of the soil will fail. This can be achieved also for the inverted case (passive state). The planes of failure in the active state formed between the wall and the line of failure (wedge) will form [itex] \pm (45^{o} + \frac{\phi}{2}) [/itex] with the horizontal. In the case of the pasive state they will form [itex] \pm (45^{o}  \frac{\phi}{2}) [/itex] with the horizontal. The red lines represent the case of active state. The wall has a [itex] \Delta [/itex] displacement from its axis, therefore allowing the active failure state. In contrast, the blue lines represent the passive state. The lines of failure are believed to start right at the bottom of the foundation. 



#6
Feb1107, 09:46 PM

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The Reality, Part 1
As it was mentioned before, Rankine Equations (7) and (9) are: [tex] P_{h} = P_{v} K_{a}  2c \sqrt{K_{a}} + wK_{a} [/tex] (7) [tex] P_{h} = P_{v} K_{b} + 2c \sqrt{K_{b}} + wK_{b} [/tex] (9) If we take a good look at both equations, we will notice that equation (7) is the only who will allow negative pressure values, thus we will focus on this one. It is obvious that in Rankine equation (7) the horizontal pressure is linear and in function with the vertical effective pressure ([itex] Ph = f(Pv) [/itex]). The rest are considered "constants". The following horizontal pressure diagrams represent all the possible combinations allowable with said function. This will be done by changing the values of the parameters c and [itex] \phi [/itex] (water pressures are not considered). In cases (b) and (c), will have a compression distribution. In the case (a), the soill will have a tension distribution and a compresion distribution. As anyone who has taken a soil mechanics course knows the soil doesn't have the strength to work on tension. The parts that are on tension will produce vertical tearing on the soil. The depth of the fissures can be calculated with equation (7) by making Ph = 0. 



#7
Feb1207, 10:11 PM

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The Reality, Part 2
The horizontal pressure distributions, shown in part 1, will allow us to understand which diagrams will occur for types of soils classified by the parameters c and [itex] \phi [/itex]. I) Soils with [itex] \phi \neq 0 [/itex] and [itex] c = 0 [/itex] This soils are clean sands, clean gravels, rocks (even thought they are not part of the traditional soil definition) and the combinations of such materials. These work quite good with Rankine's Theory. The horizontal pressure diagrams for these soils will be of the type (b) and (c) in the picture of part 1. II) Soils with [itex] \phi = 0 [/itex] and [itex] c \neq 0 [/itex] This soils are tipycally preconsolidated clays. Rankine's Theory is not recommended to be used for these soils. The reason is that the values obtained from Rankine's equation (7) can be even equal to 0. However, the lateral pressure can be calculated like it was at the "rest state" without developing the "active state". The horizontal pressure diagrams for these soils will be of the type (a) in the picture of part 1. III) Soils with [itex] \phi \neq 0 [/itex] and [itex] c \neq 0 [/itex] Rankine's Theories have given satisfactory results for these soils (combination of the previous soils). The horizontal pressure diagrams for these soils will be of the type (a), (b), and (c) in the picture of part 1. Wall Displacement ([itex] \Delta [/itex]) in the active and passive state In order to achieve the active state in type I soils the displacement must be approximately 0.001 to 0.004 of the wall's height. In type II soils it shouls vary from 0.01 to 0.04 of the wall's height. For the passive state condition the displacement on the following soils should be: Loose sand  0.01*Height Dense sand  0.005*Height Firm Clay  0.01*Height Soft Clay  0.05*Height. 



#8
Feb1307, 10:46 PM

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...And the examples!, Part 1
Problem Statements: 1) For the retaining wall shown in the picture, suppose the wall displaces enough so the active state is developed. Calculate the active force of Rankine per length of the wall and the position of the resultant line of action. [ENGLISH UNITS] 2) For the retaining wall shown in the picture, suppose the wall displaces enough so the active state is developed. Calculate the active force of Rankine per length of the wall and the position of the resultant line of action. [ENGLISH UNITS] Soil Data: [itex] \gamma_{sample} = 100 pcf [/itex] [itex] \gamma_{saturated} = 115 pcf [/itex] [itex] \phi = 30^{o} [/itex] [itex] c = 0 [/itex] 3) For the retaining wall shown in the picture, suppose the wall displaces enough so the active state is developed. Calculate the active force of Rankine per length of the wall and the position of the resultant line of action. [METRIC UNITS] Soil Data: [itex] \gamma_{sample} = 1.6 \frac{To}{m^3} [/itex] [itex] \gamma_{saturated} = 1.8 \frac{To}{m^3} [/itex] [itex] \phi = 28^{o} [/itex] [itex] c = 2 \frac{To}{m^2} [/itex] 



#9
Feb1607, 11:48 PM

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... And the examples, Part 2
1) We will assume the retaining wall is prismatic (constant cross section), and therefore will only analyze a foot of it. We start by calculating the effective vertical pressure. [tex] P_{v} = \gamma z [/tex] At the top: [tex] P_{v} = (100 \frac{lb}{ft^3} )(0 ft) = 0 [/tex] At the bottom: [tex] P_{v} = (100 \frac{lb}{ft^3} )(19 ft) = 1900 \frac{lb}{ft^2} [/tex] The figure formed by the vertical effective pressure is a triangular prism. In cross section a triangle as shown on the picture in bluegreen color. Now we calculate the horizontal pressure (active state) according to Rankine equation (7). [tex] P_{h} = P_{v} K_{a}  2c \sqrt{K_{a}} + wK_{a} [/tex] (7) In our case, we don't have a surcharge (w = 0) and c is 0, therefore the equation reduces to: [tex] P_{h} = P_{v} K_{a} [/tex] We proceed to calculate [itex] K_{a} [/itex] (coefficient of active pressure) by using Rankine's equation (8). [tex] K_{a} = \tan ^{2} ( 45^{o}  \frac{\phi}{2}) [/tex] (8) [tex] K_{a} = \tan ^{2} ( 45^{o}  \frac{30^{o}}{2}) [/tex] we end up with: [tex] K_{a} = 0.33 [/tex] Therefore our equation for the horizontal pressure on this case is: [tex] P_{h} = 0.33P_{v} [/tex] Now we calculate the horizontal pressure at the top: [tex] P_{h} = 0.33 (0) = 0 [/tex] and the horizontal pressure at the bottom: [tex] P_{h} = 0.33 (1900 \frac{lb}{ft^2} ) = 627 \frac{lb}{ft^2} [/tex] The figure formed by the horizontal pressure is a triangular prism. In cross section a triangle as shown on the picture in magenta color. Now from our knowledge from Statics we calculate the resultant of the horizontal pressure triangle: [tex] \vec{R} = \frac{1}{2} (627 \frac{lb}{ft^2}) (19 ft) (1 ft) [/tex] [tex] \vec{R} = 5956.5 lb [/tex] or [tex] \vec{R} = 5.96 kip [/tex] Now we need to calculate the position of our resultant (from the bottom), which from Statics is: [tex] \bar{y} = \frac{1}{3} (19 ft) = 6.33 ft [/tex]. Solution: [tex] \vec{R} = 5.96 kip [/tex] [tex] \bar{y} = 6.33 ft [/tex] 



#10
Feb2207, 12:20 AM

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...And the examples!, Part 3
2) We will assume the retaining wall is prismatic (constant cross section), and therefore will only analyze a foot of it. We start by calculating the effective vertical pressure. However, it should be noticed that there will be additional pressure due to water presence. First we calculate the effective pressure for the dry soil above the water table: [tex] P_{v} = \gamma_{sample} z [/tex] At the top [tex] P_{v} = (100 \frac{lb}{ft^3} )(0 ft) = 0 [/tex] At the water table [tex] P_{v} = (100 \frac{lb}{ft^3} )(10 ft) = 1000 \frac{lb}{ft^2}[/tex] Now we need to use the inmersed specific weight of the soil: [tex] \gamma' = \gamma_{saturated}  \gamma_{water} [/tex] [tex] \gamma' = 115 \frac{lb}{ft^3}  62.4 \frac{lb}{ft^3} = 52.6 \frac{lb}{ft^3} [/tex] At the bottom: [tex] P_{v} = \gamma' (z  10 ft) + 1000 \frac{lb}{ft^2} [/tex] [tex] P_{v} = (52.6 \frac{lb}{ft^3}) (11 ft) + 1000 \frac{lb}{ft^2} = 1578.6 \frac{lb}{ft^2} [/tex] The figure formed by the vertical effective pressure is a triangular prism coupled with a trapezoidal prism. In cross section a triangle and a trapezoid as shown on the picture in bluegreen color. Now we need to calculate the hydrostatic pressure. Notice that any value of z above the water table will give us 0 hydrostatic pressure, so it's quite obvious where should we start. At the water table: [tex] P_{v} = \gamma_{water} (z10) [/tex] [tex] P_{v} = 62.4 \frac{lb}{ft^3} (0) = 0 [/tex] At the bottom: [tex] P_{v} = 62.4 \frac{lb}{ft^3} (11 ft) = 686.4 \frac{lb}{ft^2} [/tex] The figure formed by the hydrostatic pressure is a triangular prism. In cross section a triangle as shown on the picture in cyan color. Now we calculate the horizontal pressure (active state) according to Rankine equation (7). [tex] P_{h} = P_{v} K_{a}  2c \sqrt{K_{a}} + wK_{a} (7)[/tex] In our case, we don't have a surcharge (w = 0) and c is 0, therefore the equation reduces to: [tex] P_{h} = P_{v} K_{a} [/tex] We proceed to calculate [itex] K_{a} [/itex] (coefficient of active pressure) by using Rankine's equation (8). [tex] K_{a} = \tan ^{2} ( 45^{o}  \frac{\phi}{2}) [/tex](8) [tex] K_{a} = \tan ^{2} ( 45^{o}  \frac{30^{o}}{2}) [/tex] we end up with: [tex] K_{a} = 0.33 [/tex] Therefore our equation for the horizontal pressure on this case is: [tex] P_{h} = 0.33P_{v} [/tex] Now we calculate the horizontal pressure at the top: [tex] P_{h} = 0.33 (0) = 0 [/tex] Now at the water table: [tex] P_{h} = 0.33 (1000 \frac{lb}{ft^2}) = 330 \frac{lb}{ft^2} [/tex] Now at the bottom: [tex] P_{h} = 0.33 (1578.6 \frac{lb}{ft^2})= 520.94 \frac{lb}{ft^2} [/tex] The effects of the Hydrostatic pressure must be acknowledged therefore at the bottom the total horizontal pressure is: [tex] P_{h} = P_{Rankine@Bottom} + P_{water@bottom} [/tex] [tex] P_{h} = 520.94 \frac{lb}{ft^2} + 686.4 \frac{lb}{ft^2} = 1207.34 \frac{lb}{ft^2} [/tex] The figure formed by the horizontal pressure is a triangular prism coupled with a trapezoidal prism. In cross section a triangle and a trapezoid as shown on the picture in magenta color. Now from our knowledge from Statics we calculate the resultant of the horizontal pressure diagram: [tex] \vec{R} = \frac{1}{2} (330 \frac{lb}{ft^2}) (10 ft) (1 ft) + \frac{1}{2} (330 \frac{lb}{ft^2}+1207.34 \frac{lb}{ft^2}) (11 ft) (1 ft) = 10105.37 lb [/tex] [tex] \vec{R} = 10.11 kip [/tex] Now we need to calculate the position of our resultant (from the bottom), which from Statics is: [tex] \bar{y} = \frac{(\frac{1}{3} (10 ft) + 11 ft)\frac{1}{2} (330 \frac{lb}{ft^2}) (10 ft) (1 ft) + (330 \frac{lb}{ft^2}) (11 ft) (1 ft)\frac{1}{2} (11 ft) + \frac{1}{2} (1207.34 \frac{lb}{ft^2} 330 \frac{lb}{ft^2})(11 ft) (1 ft)\frac{1}{3} (11 ft)}{10105.37 lb} = 6.07 ft [/tex] Solution: [tex] \vec{R} = 10.11 kip [/tex] [tex] \bar{y} = 6.07 ft [/tex] 



#11
Feb2407, 12:15 AM

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Note: Problem #3 is left for the reader to do. HINT: A soil doesn't work in tension, disregard the tension part of the diagram for the resultant's magnitude and position.
1.3 Designing a retaining wall An engineer must know the basic parameters, specific weight ([itex] \gamma [/itex]), internal friction angle ([itex] \phi [/itex]), and cohesion (c), of the soil mass retained by the wall and the soil under the slab of the base. The purpose of knowing such values is to obtain the lateral pressure distribution that dictates the appropiate design. The design process of a retaining wall is divided in two parts. The first one deals with the stability, and the second one with the strength. This article will only focus on the stability of rigid retaining walls. 



#12
Mar1107, 05:11 PM

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Stability of retaining walls
A retaining wall may fail due to overturning with respect to its heel, sliding along its base, fail due to loss of bearing capacity of the soil ("foundation" failure), failure due to shear of the soil at a deeper depth, and excessive settlement of the soil of the foundation. 



#13
Mar2407, 10:11 PM

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Dimensioning
The design process of a retaining wall is iterative . The engineer must assign the dimensions of the wall, and then check for its stability. If the checks for stability fail, the dimensions change and are checked again. In the case the check for stability is adequate, then they must be checked for strength (steel, reinforced concrete, ...). The general dimensions for retaining walls has been studied, and usually help out as an initial guide for checking for stability and strength. This is the dimension guide for a cantilever retaining wall. There are guides for gravity walls and counterfort walls, too. D represents the depth, the distance to the base of the slab must be at least 0.6 m (approx 2 feet). The reason the top of the wall has a dimension of 0.3 m or 12 inches is to allow proper placement of the concrete. 



#14
Nov1707, 04:38 PM

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I fixed the images problem, My imagecave account got deleted. I'll continue this article when i've more time.




#15
Aug208, 11:46 PM

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Here is this article broken in two parts with the corresponding pictures.




#16
Oct609, 04:10 AM

P: 3

A retaining wall is a structure that holds back soil or rock from a building, structure or area. Retaining walls prevent downslope movement or erosion and provide support for vertical or nearvertical grade changes. Cofferdams and bulkheads, structures that hold back water, are sometimes also considered retaining walls. Retaining walls are generally made of masonry, stone, brick, concrete, vinyl, steel or timber. Once popular as an inexpensive retaining material, railroad ties have fallen out of favor due to environmental concerns.




#17
Jul111, 04:21 PM

P: 2

thanks a lot very helpfull




#18
Jul211, 11:05 AM

P: 1

what settings do i need to see the jpegs on this site. there is no use to be here if i cant see the diagrams



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