Library of Congress Cataloging-in-Publication Data Budhu, M. Soil mechanics and foundations / Muni Budhu.—3rd ed. p. cm. Includes bibliographical. SOIL MECHANICS. AND FOUNDATIONS. MUNI BUDHU. Professor, Department of Civil Engineering & Engineering Mechanics. University of Arizona. MUNIRAM (Muni) BUDHU is Professor of Civil Engineering and Engineering ics topics including soil mechanics, foundation engineering.
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Soil Mechanics and Foundations 3rd Edition By Muni Budhu and Foundations 3rd Edition By Muni Budhu is available for free download in PDF format. Find all the study resources for Soil Mechanics and Foundations by Muniram Budhu; M. Budhu. by Muni Budhu · soil soil mechanics and foundations A companion lab manual is available from the Publisher: Soil Mechanics This book is intended -.
The influence factor for the vertical stress 3. You can program your calculator or usc a spreadsheet program to find It. You must be careful in the last te rm Ian - I in progra mming.
I method. The vertical stress increase under the center of the load is. The area.. Set the scale, shown on the chart, equal to the depth at which the increase 2.
We will call this the depth scale. Identify the point on the loaded area below which the stress is required. Let us say this point is A. Plot the loaded area using the depth scale with point A at the center of the chart. Count the number of segments Ns covered by the scaled loaded area. Calculate the increase in vertical stress i The essential points are: The increases in stresses below"a surjp ce load are found by assuriUng the soil is an elastic, semi-infinite 2.
Use the equation for a point load. Use Eq. The load on the slab is kN. Determine the vertica sl ss increase at a depth of 3 m 3 under the ceoter of the slab. Q neT, poiot C Fig. You shoultf ivide the area so that the point of inlerest is Find the centroid. The scale on the chart is set equal to the; depth. Mean stress causes volume changes. Equation 3. You only need to apply the equations gi ven in the pre T he negative sign for the radial displacement indi a es an expap l tQ.
The strains, in. The maximum lateral displacement applied a t e top is O. Calculate the principal, volumeteic, and dejiatOri1 ains. Strategy You ar? Let us rble 10 a hemispherica ole anH stack the other on top of it Fig. We wilt calt this loading "A". We can represent loading " A" by a Jin as shown in Fig. The line OA is called a load path or a force path. There are now two components offorce. If the frictional resistance at the contacts of the two marbles is less than the horizontal force, the top marble will slide relative to the bottom.
You should recall from your mechanics or physics course that the frictional resistance is fLFz Coulomb's law , where fL is the coefficient of friction at the contact between the two marbles. Our one-dimensional system now has two modes of instability or failure-one due to relative sliding and the other due to crushing of the marbles.
The force path for loading "B" is represented by OB in Fig. The ssential point or principle. We wiU call this load ing condition, loading "1. Conseq uently, we are going to usc the incremental form of the stress invaria nts. Point B in Fig. The implication is that the volume of our soil sample remains constant.
In Chapter 5, we will discuss drained and undrained load ing conditions in more detail. For loading " 2," the total stress path is AB. In this book, we will represent total stress paths by dashed lines.
If OU f soil were an isotropic, elastic material, then accordi to Eg. Remember the truss analogy we used for effective stresseb.. In this case, fip cannot be o.. T eslmplicattons or Egs. Since G E. The effective stress path for loading "2," assuming our soil sample be haves like all isotropic, elastic materi al, is represented by A B' Fig.
A stress path is a graphical representation of stresses in stress space. Th e response, stability, and f ailure of soils depend on stress paths.
Calcul ate the initial loading va. Ca lculate the incre;]. Co ect t e points id lltifyin effective stresses and do the same for total y. Two cylindrical specimens. A and B. Subsequently, the radial stress applied on speci men A was held constant and the axial stress was incrementally increased to kPa under undrained conditions. The adal stress on specimen B was held constant and the radial stress incrementally reduced to 50 kPa under drained conditions.
Plot the total and effective stress paths fo r each specimen assu nung the soH is a linea r, isotropic, clastic materia l. Calculate the maximum excess pore water pressure in specimen A. Strategy The loading condit ions on both specimens are axisymmet ric. The easiest approach is to write the mean stress and deviatoric st ress equations in terms of increments and make the necessary substitutions. Determine loading cond ition.
Loading is axisymmetric and both drained and undrained conditions are specified. Calculate initial stress invari ants for isotropic loading path. I Slep S: The mean Stress differe ce is. T he most important principle in soil mechanics is the principle of effective stress. Soil deformat ion is due to effective not total stresses. Applied surface Slresses are distribu ted such that their magnitudes decrease with depth and distance away from their points o f app1ication.
Stress paths provide a usefu l means through which the Illstory of loading of a soil can he followed. The elastic modulus o f the clay varies with de pth Fig.
Estimate the elastic settlement of the clay under the center of the foundation assumi ng 1 d rai ned con dition and 2 undrained condition. O ne opti'n': Beyond a depth oflB. J Solution 3. Find Ihe applieo--ve"rt;al surface stress.
In soil mechanics, the loading imF. Assuming all the settlement is elastic and to the medium clay, detennine the average stresses imposed on the medium clay under the cenler of the embankment using the elastic equations.
Assume the lateral strain is zero. Plot Mohr's circle of stress and determine a th. The effeHive size of a fine sand is 0. Determine the vertical effedive stress in the sand at depths of 1 m above and 3 m below the groundwater level for the following cases: The void ratio of the sand is 0. If a tube is inserted into the bottom sand layer Fig. The of,ti. The sand rests on a deep de. The existing high rise complex is founded on a concrete slab, m X m located at 2 m below ground surface, that transmits a load of MN to the soil.
Your office foundation is 50 m X 80 m and transmits a load of MN. You also intend to locate your foundation at 2 m below ground level.
The front of your building is aligned with the existing office complex and the side distance is 0. The lesser dimension of each building is the frontal dimension. The owners of the existing building are concerned about possible settlement of their building due to your building. You are invited to a meeting with your client, the owners of the existing building, and their technical staff.
You are expected to determine what effects your office building would have on the existing building. You only have one hour to make the preliminary calculations and you are expected to present the increase in stresses and the amount of settlement of the existing office complex due to the construction of your office building.
Prepare your analysis and presentation. Under load, all soi ls will set! Structures may settl uniformlr nonuniforml 1'he.. S3hapter 7. Primary consolidation is the change in volume of a fin e-grained soil caused by the expulsion of water from the voids and the transfer of load from the excess pore water pressure to the soil particles.
Drainage path, 1I. Wh at is the go e. How is the excess pore water pressure distributed within the soil when a load is applied and aftcr various elapsed times? What factors determine the consolidation settlement of soils? What are the ave rage degree of consolidation, time fa clor, modulus of volume compressibility, and compression and recompression indices? What is the difference between primary consolidation and secondary compression? What is the drainage path for single drainage and double drainage?
Why do we need to carry ou t consolida tion tests, how are they conducted, and what parameters are deduced from the test results? How is time rate of settlement and consolidation settlement calculated?
Are there significant differences between the calculated settlements and field settlements? Since the side wall of the con taine r is rigid, no radial displacement can occur.
Of wa er to drain. Before the valve was opened. The later time settlement response is called secon ary com ession or creep. The term consolidation is re or the process in which settlement of a soil occurs from changes in effective stresses resulting from decreases in excess pore water pressure.
The rate of settlement from secondary compression is very slow compared with primary consolidation. We have separated primary consolidation and secondary compression.
In reality, the distinction is not clear because secondary compression occurs as part of the primary consolidation phase especially in soft clays. The mechanics of consolidation is still not fully understood and to make estimates of settlement, it is convenient to separate primary consolidation and secondary compression. Because we allowed the soil to drain on 4.
When a load is applied to a saturated soil, all of the applied stress is supported initially by the pore water initial excess pore water pressure ,. The change in effective stress is zero If drainage of pore water is permitted, the initial excess pore water pressure decreases and soil settlement The change in effective stress is When t - co, the change in volume and the change in excess pore water pressure of the soil approach zero; that is, The change in vertical effective stress is Three types of graph are shown in Fig.
Figure 4. The segment AB in Figs. We are going to define two slopes for primary consolidation. The history of loading of a soil is locked in its fabric and the soil maintains a memory of the past maximum effective stress. To understand how the soil will respond to loads, we have to unlock its memory. If a soil were to be consolidated to stresses below its past maximum vertical effective stress, then settlement would be small because the soil fabric was permanently changed by a higher stress in 4.
However, if the soil were to be consolidated beyond its past maximum effective stress, settlement would be large for stresses beyond its past maximum effective stress because the soil fabric would now undergo further change from a current loading that is higher than its past maximum effective stress. The preconsolidation stress defines the limit of elastic be 0 s that are lower than the preconsolidation stress, the soil will URL and we can reasonably assume that the soil will behave like a stresses greater than the preconsolidation stress soil elastoplastic material.
Path AB Fig. The NCL is approximately a straight line in a plot of e versus log and is defined by a slope, Cc , called the compression index. A normally consolidated soil would behave like an elastoplastic material.
That is, part of the settlement under the load is recoverable while the other part is permanent. An overconsolidaled soil willjollow paths such as CDE Fig. For stresses below the preconsolidillion stress, an ovucpnsolidilted soli would approximately behave lib an elastic malerlal and selllement would be small.
However,jor stresses greater Ilian the preeonsolidalion SITtss, an overconsofidaled soil will b,thav , lik e an eluslOplastlc material, similar 10 U normall , cons lidaled soil. F OCR.! In this case, consolidation occurs along the URL and 4. Calculate the primary consolidation settlement. You can also calculate the primary consolidation settleITJ4lt -Il! Below the soft clay is a deposit of coarse sand. The groundwater table was observed at 3 m below ground level.
The building will impose a vertical stress increase of kPa at the middle of the clay layer. Estimate the primary consolidation settlement of the clay.
Calculate the increase of stress at the mid-depth of the clay layer. You do not need to calculate l1a z for this problem. Calculate a;inCY;in Step 4: All other soil values given in Example 4. Determine the primary con olidation settlement of the clay. Strategy Since the soil is overconsolidated, you will have 00 c the preconsolidation stress is less than or greater than the vertical effective stress and the applied vertical stress at This check will determine the appropriate equation to use unit weight of the sand is unchanged but the clay h s chan Solution 4.
FO Step 3: Determine the primary consolidation settlement of the clay. Strategy Since the soil is overconsolidated, you will have to check whether the preconsolidation stress is less than or greater than the sum of the current vertical effective stress and the applied vertical stress at the center of the clay.
This check will determine the appropriate equation to use. From Example 4. Calculate the preconsolidation stress. Calculate Step 1: J" fin. J; e assume a rough base, we can use the influence values specified by Milovic and To nier or if we assume a smooth base we can use the values specified by So!
The clay layer is 10 m thick, so it is best to subdivide the clay layer into sublayers s:: I Foundation: Find the vertical stress increase at the center of the clay layer below the foundatioD.
Divide the clay layer into five sublayers, each of thickness 2. The groundwater level is at the surface. Determine and plot the variation of water content and overconsolidation ratio with depth up to 50 m.
Strategy The overconsolidation state lies on the unloadinglreloading line Fig. Identify what given data is relevant to finding the equation for the unloading! Here you are given the slope, C" so you need to use the other data to find the complete question. You can find the coordinate of one point on the unloading!
The table below shows the calculated values and the resu lts, which are ploued in Fig. This is a characteristic of real soils. So far, we have only considered how to determine the fin imary consolidation settlement. This settlement might take months or years occur, depending essentially on the permeability of the soil, the soil thickness d ainage c nditions, and the magnitude of the applied stress.
Geotechnical e ine s hav to know the magnitude of the final primary consolidation settlemeflit an also t e rate of settlement so that the settlement at any given time can be ev ated. The next section deals with a theory to determine the settl Several assumptions are made in developing this the How that many of the observations we made in Section theory. At any depth , the change in vertical effective stress is equal to the change in excess pore water pressure at that depth.
The change in flow is then aq 3z dz dA. The rate of change in volume of water expelled, which is equal to the rate of change of volume of the soil, must equal e change in flow. Rewriting Eq. It is a common equation in many branches of engineering.
For example, the heat diffusion equation commonly used in mechanical engineering is similar to Eq. In the derivation of Eq. This is usually not the case because as the soil consolida are reduced and k z decreases. The consequence of k z and m v not being constants is thaf , is n a onstant.
In practice, Cu is assumed to be a constant and this assumpti n ' eas9'nable only if the stress changes are small enough such tha and not change significan tl y. The essential point is: We need to know the excess pore water pressure at a desired time because we have to determine the lerticai effectIve stress to cal'Culate the primary consolidatIOn ettl menlo fro F in programmin ensional consolidation the top and bottom eme.
J ecificatl 0 t e initial distribution of excess pore water pressures at the bOu'n ries, we can obtain solutions for the spatial variation of excess pore water pressu wit time and depth. Various distributions of pore water pressures within a soil ayer are possible.
Two of these are shown in Fig. One of these is a uniform distribution of initial excess pore water pressure with depth Fig. This may occur in a thin layer of fine-grained soils.
The other Fig. This may occur in a thick layer of fine- grained soils. Le s examine Eq. The m imum xcess pore water pressure occurs at the center of the soil layer because tn ainage path there is the longest, as obtained earlier in our experiment in Section 4. We now defin e a parameter, U z , called the degree of consolidation or consolidation ratio, which gives us the amount of consolidation completed at a particular time and depth.
There are special conditions that apply to boundary nodes. Calculate the excess pore water pressure at interior nodes using Eg. If the boundary is permeable, then the excess pore water pressure is zero at all nodes on this boundary.
An ideal soil isotropic, homogeneous, satur. Strains are assumed to be small. Excess pore water pressure dissipation depen ls on the time, soil thickness, drainage conditions, and per.
The average degree of e6ns Iidatio. The initial FO n the increment in applied stress and the degree of at he initial change in excess pore water pressure is lie ertical stress. From the data given decide on the n this case is Eg. From Eg. Calculate the vertical total stress and total excess pore water pressure.
Vertical total stress: Calculate the current vertical effective stress. Ca lculate the initial vertical effective Stress. Same as Step 2 above. The initial excess pore water pressure from an applied oad at i e i' Jaial exc 5S pore watef! Divide' ' he d7 ,'O five lay"" ti.. The bottom bou ndary is impermeable, therefore Eg. The top boundary is pervious, therefore the excess pore water pressure is zero at all times greater than ze ro.
You are given the distribution of initial excess pore water pressure as I:! For example, at r w 2, column 1 node 2 , I:! The inii ex ess pore water pressures are listed in column 1; see the table low.
FO 5 We have only described primary consolidation settlement. The other part of the total consolidation settlement is secondary compression, which will be discussed next.
The first part is primary consolidation, which occurs at early times. The second part is secondary compression, or creep, w. The physical reason for secondary compression in soils are not fully understood. One plausible planatiom is the expulsion of water from micropores; another is viscous fon of he soil structure.
We can make a plot of void ratio versus the logan experimental data in Section 4. Pri assumed to end at the intersection of the projectio of the the curve.
The secondary compression index is 4. This negative excess pore water pressure can cause water to flow into the soil and increase the soil's water content. Consequently, the tinal void ratio calculated from the final water content would be erroneous.
The data obtained from the one-dimensional consolidation est are as follows: Initial height of the soil, Ho, which is fixed by the hei fl. Current height of the soil at various time intervals un settlement data. Water content at the beginning and at the en of the soil at the end of the test. You now have to use these data to deter We will start with finding Cu' FO 4.
At early times, the theoretical relationship between U and Tv is given by Eq. You should note that 0 is below t e initi tion is achieved 1. Plot 2. The theoretical early time s ttle ent ponse in a plot of logarithm of times versus displacement gau ding's a parabola Section it a p arabola and a correc4.
The experimental early time curve is R'O tion is often required. The procedure, with reference dary 2. You will recall Fig. Calculate e for each loading step using Eq. We will call Fig. You wiJl now determine the preconsolidation stress using a method proposed by Casagrande F ed in practice is to project the straight mpres 'o curve to intersect the backward projec.
Degradation of the soil from its intact condition caused by sampling, transportation, handling, and sample preparation usually does not produce the ideal curve shown in Fig. The slope of this The FO C, without correction 0.
You should note that Fig. Three examples and their solutions are present how to find various consolidation soil parameters as discusse examples are intended to illustrate the determination of the com how to use them to make predictions.
The third examg. Use o n in Fig. C, Step 3: Preconsolidation effective stress: The void ratio at kPa. You know the in- FO Q28 te the modulus of volume recompressibility. Drainage was permitted from the top and bottom boundaries. Solution 4. Make a plot of settlement decreas shown in Fig. Follow the procedures out From Fig.
We have described the consolidation test of a small sample of soil and the soil consolidation parameters that can be obtained. What is the relationship between this small test sample and the soil in the field? Can you readily calculate the settlement of the soil in the field based on the results of your consolidation test? The next section provides the relationship between the small test sam e and the soil in the field.
FO settlement fr Strategy You are given all the data to directly use Eq. For part 1 there is double drainage in the field and the lab, so the drainage path is one-half the soil thickness.
For part 2 , there is single drainage in the field, so the drainage path is equal to the soil thickness. Calculate the drainage path. Q1 m; ". Ca lcula te the field time uSing Eq. Step I: Calculat e the d ra in age path. Ca lculate fi eld time us ing Eq. In the next settion. You sho d be ut ious in using these relationships be- betweeq,.
LG, Naga raj and Murt hy. OO9 wll. Sometimes, we may have to build structures on a site for which the calculated settlement of the soil is intolerable. One popular method to reduce the consolidation settlement to tolerable limits is to preload the soil and use sand drains to speed up the drainage of the excess pore water pressure. Next, we will discuss sand drains. The d iameter of sa nd drai ns ranges from about to mm. The diameter required must only be large enough to d rain the pore wate r and prevent prematu re clogging from fi nes in the soil to be drained.
Filter fab rics are now commonly used at the interface of the natural soil and the backjin to prevent clogging. Richart repOrle solutions 10 Eq. Richart 1? You now have to find a value of T, such that the degree of consolidation matches Vr calculated in Step 6.
You do this by iteration since you do not know either T, or n. Select a value of n from the values shown in Fig. Determine Vr from the curve corresponding to the selected value of n. Compare this value of V, with Vr calculated in Step 6.
You may interpolate from Fig. You can also plot a graph of n versus V r and find the n value corresponding to the desired V , Step 6. Calculate the spacing, s, depending on the grid you desire. Below the soft clay is a stiff overconsolidated clay 20 m thick. The calculated settlemen t cannot be tolerated and It was decided that the soi l should be preeo solida ted 1 re: The da n The given va lues are: From Fig. Find U,. From Eq. Determine T,. Step 8: Determine n.
During unload ing or reloadi ng, the soil stresses m ust adjust I e in eqllli ium with the applied stress. This means that stress 01 oJ vertically but also horizon tally. From Chapter 3. Various eq uations have been suggested linking K': When an increment of ve rtica l stress is applied to a soil, the instantaneous i nitial excess po re water pressure is equal to the vert ical stress increment.
Wit h time, the ini tial excess pore water pressure decreases, the vertical eHective stress increases by the amount o f decrease of the in itial excess pore water pressure , and settlement 4. The consolidation settlement is made up of two pa rIs-th e ea rly time response called primary consolidation and a later time response called secondary comptession.
Soils retain a memory of the past maximum effective stress, which may be erased by loading to a higher stress level. If the current vert ical effect jve st ress on a soil wa s never exceeded in the paSI a normally consolidated 11 , '"I would behave elastoplastically when stressed. To find time for a given degree of consolidation, you need to find Cu from the data. Find Cu using the root time method. Use the data from the kPa load step t Vtime curve as depicted in Fig.
Equation 4. Follow the procedures in Sectio Step 4: FO R Step 5: Divide the clay layer into three sublayers of 1. Soils will,! When you complete this chapter, you "1' The highway rou te will pass through a terrai n Ihat is relallve!
Bulk unit weight 'Y is the weight density, that is, the weight of a soil per unit volume. Saturated unit weight 'Ysat is the weight of a saturated soil per unit volume. Dry unit weight 'Yd is the weight of a dry soil per unit volume. Effective unit weight 'Y' is the weight of soil solids in a submerged soil per unit volume. Relative density Dr is an index that quantifies the degree of packing between the loosest and densest state of coarse-grained soils.
Liquid limit wLd is the water content at which a soil changes from a plastic state to a liquid state. Plastic limit WPL is the water content at which a soil changes fro , to a plastic state. Shrinkage limit wsd is the water content at which a soil c a to a semisolid state without further change in volun. Groundwater is water under gravity in excess of th pores. Today Updates.
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