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### Partial Differentials Notes_Part_18

Course: MATH 5587, Fall 2010
School: UCF
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Word Count: 820

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table, following we display a list of all relative vibrational frequencies (12.158) that are &lt; 6. Once the lowest frequency 0,1 has been determined -- either theoretically, numerically or experimentally -- all the higher overtones m,n = m,n 0,1 are simply obtained by rescaling. Relative Vibrational Frequencies of a Circular Disk n 1 2 3 4 . . . m 0 1 2 3 4 5 6 7 8 9 ... 1.000 1.593 2.136 2.653 3.155 3.647...

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table, following we display a list of all relative vibrational frequencies (12.158) that are < 6. Once the lowest frequency 0,1 has been determined -- either theoretically, numerically or experimentally -- all the higher overtones m,n = m,n 0,1 are simply obtained by rescaling. Relative Vibrational Frequencies of a Circular Disk n 1 2 3 4 . . . m 0 1 2 3 4 5 6 7 8 9 ... 1.000 1.593 2.136 2.653 3.155 3.647 4.132 4.610 5.084 5.553 . . . . . . . . . 2.295 2.917 3.500 4.059 4.601 5.131 5.651 . . . . . . . 3.598 4.230 4.832 5.412 5.977 . . . . . . . . 4.903 5.540 . . . . . . . . . Nodal Curves When a membrane vibrates, its individual atoms move up and down in a quasiperiodic manner. As such, there is little correlation between their motion at different locations. However, if the membrane is set to vibrate in a pure eigenmode, say un (t, x, y) = cos(n t) vn (x, y), (12.159) then all points move up and down at a common frequency n = n , which is the square root of the eigenvalue corresponding to the eigenfunction vn (x, y). The exceptions are the points where the eigenfunction vanishes: vn (x, y) = 0, (12.160) which remain stationary. The set of all points (x, y) that satisfy (12.160) is known as the nth nodal set of the domain . Scattering small particles (e.g., fine sand) over a membrane vibrating in an eigenmode will enable us to visualize the nodal set, because the particles will tend to accumulate along the stationary nodal curves. It can be shown that, in general, each nodal set consists of a finite system of nodal curves, [35], that partition the membrane into disjoint nodal regions. As the membrane vibrates, the nodal curves remain stationary, while each nodal region is entirely either above or below the equilibrium plane except, momentarily, when the entire membrane has zero displacement. Adjacent nodal regions, lying on the opposite sides of a nodal curve, move in opposing directions -- when one is up, its neighbor is down, then switching roles after the membrane becomes momentarily flat. Let us look at a couple of examples where the nodal curves can be readily determined. 1/22/12 470 c 2012 Peter J. Olver 1.000 1.593 2.136 2.295 2.653 2.917 3.155 3.500 3.598 Figure 12.11. Nodal Curves and Relative Vibrational Frequencies of a Circular Membrane. Example 12.15. Circular Drums. Since the eigenfunctions (12.147) for a disk are products of trigonometric functions in the angular variable and Bessel functions of the radius, the nodal curves for the normal modes of vibrations of a circular membrane are rays emanating from and circles centered at the origin. Consequently, the nodal regions are annular sectors. Pictures of the nodal curves for the first nine normal modes, by indexed their relative frequencies, are plotted in Figure 12.11. Representative displacements of the membrane in each of the first twelve modes can be found earlier in Figure 12.5. The dominant (lowest frequency) mode is the only one that has no nodal curves; it has the form of a radially symmetric bump where the entire membrane flexes up and down. The next lowest modes vibrate proportionally faster at a relative frequency 1,1 1.593. The most general solution with this vibrational frequency is a linear combination u1,1 + u1,1 of the two eigensolutions. Each such combination has a single diameter as a nodal curve, 1/22/12 471 c 2012 Peter J. Olver ==1 Figure 12.12. = 2, = 1 = 5, = 1 Some Nodal Curves for a Square Membrane. whose angle with the horizontal depends upon the ratio /. The two semicircular halves of the drum vibrate in opposing directions -- when the top half is up, the bottom half is down and vice versa. The next set of modes have two perpendicular diameters as nodal curves; the four quadrants of the drum vibrate in tandem, with opposite quadrants having the same displacements. Next in increasing order of vibrational frequency is a single mode, that has a circular nodal curve whose (relative) radius equals the ratio of the first two roots of the order zero Bessel function, 0,2 /0,1 .43565; see Exercise for a justification. In this case, the inner disk and the outer annulus vibrate in opposing directions. And so on ... Example 12.16. Rectangular Drums. For most rectangular drums, the nodal curves are relatively uninteresting. Since the normal modes (12.142) are separable products of trigonometric functions in the coordinate variables x, y, the nodal curves are equi-spaced straight lines parallel to the sides of the rectangle. The internodal regions are smaller rectangles, of identical size and shape, with adjacent rectangles vibrating in opposite directions. A more interesting collection of nodal curves occurs when the rectangle admits multiple eigenvalues -- so-called accidental degeneracies. Two of the eigenvalues (12.140) coincide, m,n = k,l , if and only if n2 k2 l2 m2 + 2 = 2 + 2 (12.161) a2 b a b where (m, n) = (k, l) are distinct pairs of positive integers. In such situations, the two eigenmodes happen to vibrate with a common frequency = m,n = k,l . Consequently, any linear combination of the eigenmodes, e.g., m x n y k x l y , , R, sin + sin sin a b a b is also a pure vibration, and hence qualifies as a normal mode. The associated nodal curves, cos( t) sin 0 y b, have a more intriguing geometry, which can change dramatically as , vary. 1/22/12 472 c 2012 sin m x n y k x l y sin + sin sin = 0, a b a b 0 x a, (12.162) Peter J. Olver
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UCF - MATH - 5587
For example, on a unit square R = 0 &lt; x, y &lt; 1 , an accidental degeneracy occurs whenever m2 + n2 = k 2 + l2 (12.163) for distinct pairs of positive integers (m, n) = (k, l). The simplest possibility arises whenever m = n, in which case we can merely reve
UCF - MATH - 5587
Chapter 9 Linear and Nonlinear Evolution EquationsIn this chapter, we analyze several of the most important evolution equations, both linear and nonlinear, involving a single spatial variable. Our first stop is to revisit the heat equation. We introduce
UCF - MATH - 5587
Chapter 3 Fourier SeriesJust before 1800, the French mathematician/physicist/engineer Jean Baptiste Joseph Fourier made an astonishing discovery. Through his deep analytical investigations into the partial differential equations modeling heat propagation
UCF - MATH - 5587
Chapter 8 Fourier TransformsFourier series and their ilk are designed to solve boundary value problems on bounded intervals. The extension of Fourier methods to the entire real line leads naturally to the Fourier transform, an extremely powerful mathemat
UCF - MATH - 5587
Chapter 6 Generalized Functions and Green's FunctionsBoundary value problems, involving both ordinary and partial differential equations, can be profitably viewed as the infinite-dimensional function space versions of finite dimensional systems of linear
UCF - MATH - 5587
Math 5587 September 8, 2011Homework #1Problems: Chapter 1: 1.1ae, 1.2b,d, 1.5a,e, 1.6, 1.12a, 1.16ad, 1.18, 1.19, 1.20, 1.24. Chapter 2: 2.1 2, 3c,e, 4, 6.Due: Thursday, September 15
UCF - MATH - 5587
Math 5587 September 20, 2011Homework #2Problems: Chapter 2: 2.2 2.3 2a, 3b, 9, 17, 26, 27. 2, 5, 14, 15.Due: Thursday, September 29 First Midterm: Tuesday, October 11 Will cover chapters 1 &amp; 2. You will be allowed to use one 8&quot; 11&quot; sheet of notes. Note
UCF - MATH - 5587
Math 5587 September 29, 2011Homework #3Problems: Chapter 2: 2.4 2, 3, 4c,d, 8, 11, 12.Also, in 2.4.8, determine the domain of influence of the point (0,2) and the domain of dependence of the point (3,-1). Discuss what these tell you about the solution.
UCF - MATH - 5587
Math 5587 October 13, 2011Homework #4Problems: Chapter 3: 3.1 3.2 2b, 5. 1, 2g, 3a, 5, 6a,g, 15a,d, 16a,d, 24, 25, 34, 35, 41b, 52, 53.Due: Thursday, October 20
UCF - MATH - 5587
Math 5587 October 25, 2011Homework #5Problems: Chapter 3: 3.3 1, 2, 8. 3.4 2b, 3c, 7, 9 (just use one of the two methods). 3.5 2b,c,d, 4, 8, 11a,b,c. Due: Tuesday, November 1 Second Midterm: Thursday, November 17 Will cover chapters 3 &amp; 4. You will be a
UCF - MATH - 5587
Math 5587 November 3, 2011Homework #6Problems: Chapter 3: 3.5 13, 21c,e, 22b,c, 27b,d, 30, 31, 35a, 42. Chapter 4: 4.1 2, 4c, 10, 17a,b. Due: Thursday, November 10 Second Midterm: Thursday, November 17 Will cover chapters 3 &amp; 4. You will be allowed to u
UCF - MATH - 5587
Math 5587 November 10, 2011Homework #7Problems: Chapter 4: 4.2 3a, 4b,e, 8, 14a,d,e, 26. 4.3 4, 7, 10c, 11, 12a, 16, 24a, 29, 31. Due: Tuesday, November 22 Second Midterm: Thursday, November 17 Will cover chapters 3 &amp; 4. You will be allowed to use one 8
UCF - MATH - 5587
Math 5587 December 6, 2011Homework #8Problems: Chapter 4: 4.4 2a,e,f, 12a,e,f, 13, 17a,b. Chapter 6: 6.1 1b,d, 2d, 3, 5b, 8, 13, 19, 35. 6.2 2, 6. 6.3 1, 2, 6. Due: Tuesday, December 13 Final Exam: Take Home, to be handed out on Tuesday, December 13 and
UCF - MATH - 5587
Chapter 2 Linear and Nonlinear WavesOur exploration of the vast mathematical continent that is partial differential equations will begin with simple first order equations. In applications, first order partial differential equations are most commonly used
UCF - MATH - 5587
Chapter 5 Numerical Methods: Finite DifferencesAs you know, the differential equations that can be solved by an explicit analytic formula are few and far between. Consequently, the development of accurate numerical approximation schemes is essential for
UCF - MATH - 5587
Chapter 11 Numerical Methods: Finite ElementsIn Chapter 5, we introduced the first, the oldest, and in many ways the simplest class of numerical algorithms for approximating the solutions to partial differential equations: finite differences. In the pres
UCF - MATH - 5587
Chapter 10 A General Framework for Linear Partial Differential EquationsBefore pressing on to the higher dimensional forms of the heat, wave, and Laplace/ Poisson equations, it is worth taking some time to develop a general, abstract, linear algebraic fr
UCF - MATH - 5587
Chapter 12 Partial Differential Equations in SpaceAt last we have reached the ultimate rung of the dimensional ladder (at least for those of us living in a three-dimensional universe): partial differential equations in physical space. As in the one- and
UCF - MATH - 5587
Chapter 4 Separation of VariablesThere are three paradigmatic linear second order partial differential equations that have collectively driven the development of the entire subject. The first two we have already encountered: The wave equation describes v
UCF - MATH - 5587
Chapter 1 What are Partial Differential Equations?Let us begin by specifying our object of study. A differential equation is an equation that relates the derivatives of a (scalar) function depending on one or more variables. For example, d4 u du + u2 = c
University of Florida - CEG - 4012
Lecture 1 Review of Geostatic Stresses Unit WeightsYw = unit weight of water Ym = moist unit weight of unsaturated soil Ysat = unit weight of saturated soil y' = &quot;effective&quot; unit weight of soil = (Ysat - Yw) if soil saturated= Ym if soil not saturated=
University of Florida - CEG - 4012
Geostatic Stresses These self-weight stresses (ay, a'y, ah' a'h) are called geostatic stresses For a level surface there are no shear forces induced by the geostatic stresses, and therefore they are also principal stresses: a, =ay and a3 = ah Karl Terzag
University of Florida - CEG - 4012
using the effective weight concept:= 0&quot; h6'(100 pet + 4'(118-62.4 pet + 6'(126-62.4 pet+4'(120-62.4 pet 230 pst: 1434 pst= 600 pst + 222 pst + 382 pst +=-K (J'y : 0.5 (1434 pst) : 717 pst 14' (62.4 pet): 874 pst U(a' h + u): 717 pst + 874 pst: 15
University of Florida - CEG - 4012
S1 Stresses Changes Due to Surface Loads (Aerv) o Stresses within a soil mass will change as a result of surface loads. The change in total stress spreads and diminishes with distance from the load. Equations and charts are available to calculate both th
University of Florida - CEG - 4012
S2 Pyramid Approximation For Surface Loads: The &quot;2:1 melhod &quot;used for &quot;back of Ihe envelope&quot; solutions (seen on the PE exam). o A=LoadedArea =BxLQ = Tolal Load = q x B x LQr+-&quot;q=unit pressurewhere B = width (short sjde, alwill's) L _ length !loM-si
University of Florida - CEG - 4012
S3 Point Load on the Surface: Boussinesq (1883) - French Physicist(with the parameters as shown at right)Qo Assumotionc:. I. nALF-SPACE - semi-infinite 2. ELASTIC 3. HOMOGENEOUS - same properties throughout 4. ISOTROPIC - same properties in all directi
University of Florida - CEG - 4012
84Circular Surface Load:o The stress increase at a point (A) on the axis beneath the center of a circular loaded area is &quot;easily&quot; determined by integrating the Boussinesq point solution over the loaded area. (same assumptions as before)dav6=30z 5 2
University of Florida - CEG - 4012
S5IPta (m)z(m)~&quot;f.r~;,&lt;,iJa 0r (m)&lt;w, (&quot;,1,;r/a~I,.()OAGy=I x q(kPa)IS ':JI'llC.L.Surface12 12 12 12 12 12 12 1200 0 12 0 12 0 12 2400 1 0 1 0 1 2AB6 612 12 24 24 24C 0E FG0.5 0.5 1 1 2 2 2o'I ()h'f'ltt&quot;,o~f) (,((),
University of Florida - CEG - 4012
r~.lr0.20.3 0.4 0.50.6,86810_ Influence Chart for Vertical Stress Increase Beneath Circular Loaded Area '(From Perlott and Baron)Influence value, I(x 100) 2 3 4 5 60.8 1.020_. 30 40 50 60 I I I I I80 100I11-~V1lllll1Ui .-I'm'N-I'3,!)
University of Florida - CEG - 4012
S11~~trlp and Square FootingsStrips footings have fixed width and infinite (or relatively long) length and are often used for smaller buildings (1-2 story) and walls. Square footings are a special case of the rectangular footing.BqzInfluence charts
University of Florida - CEG - 4012
812o Boussinesq Influence Chart for Strip &amp; Square Footings48382B88I - 1-.a~ I ~/v 28'. l' 1-~'l.O; 1./0-11 VI VI I , /'0.' VI / I ~, l&quot; / / / \ IIV ~ v ~ rtf :\l/I-&quot; ./ /\ \III 11I\.q,.I'.'~) OAq~ ~ ~.-.~ ~ t'. ~0.4;1 ~l\.'~~8 8U!
University of Florida - CEG - 4012
813 Rectangular Surface Loads: (Boussinesq solution - the limn&quot; methodo This method determines the vertical stress ~ay at a point P under the corner of a rectangular loaded area using footing dimensions normalized to the depth z: m = Biz n=Uz (note that
University of Florida - CEG - 4012
o814 Influence Chart for Vertical Stress Increase Beneath Corner of a Rectangular Load (Perloff &amp; Baron)I PI JrI i J I1II i,i J:It- -_ 'j0.23 1i-I--TyO.22~mt!tJ0.21Ffj:fSffi Ff1':11+WM-10,0.20 H+J+.;+J+j 0.190.18=pl&lt; pflm,nl for sq
University of Florida - CEG - 4012
S 15a Example 3: Solution: Find the stress increase at Point T al a depth of 6 1t.qfor &lt;D m = Biz = ':J.)(&quot; 6&gt;~ n = Uz = ~), : cs 0 in chart find I = 0.0(, I for Q) m = BIz = 3/~ v.n) n = Uz = 'I!&quot; t.F) in chart find I = 0,10&lt;.&gt;= 1000 psf : 3'2'T4'
University of Florida - CEG - 4012
I\:._~B:.-_ _I-=-_ c.(j;)A_ _-,~ ,e: .,III~. - - - -,- . ' . - - . - 1&gt;I,pl; -.-- .~ p ,bt;'-~,:-JE~IL_-j. - _.\ ,~( 5'1ftt /'d-+ncfw_;.lj~ 0: = ~ [ &quot;I HP6-rae-tiP] x:~c. - - - 6 - l - - - _ . IiEp
University of Florida - CEG - 4012
S16 Embankment Load: a Another important &quot;shape&quot; is the embankment (e.g. highways, dams, etc.) If the embankment material is soil then the surface load (pressure) at full height is ~ q '1 H.ba=Chart on next page provides influence factors due to 1/2
University of Florida - CEG - 4012
517oInfluence Chart for Vertical Stress Increase Beneath and Embankment (PerloH and Baron)VALUE OF~05'0.4500'0020050102,,.20,e1.2~~1a35 680050.8v. v./45bit ~ LOl./.400.9/ 1/ .4 / / 1/21/40oe07.35\.-1/I I II I I II
University of Florida - CEG - 4012
p1 SETILEMENT IN CLAY SUMMARY Strains due Surface Load: Surface loads cause a change in the stresses within the soil mass and induce two types of soil strain.zt zy/ t y'iIIa,tn/ 't xza,1&quot;Ez,. [ II/ !i1-,/I II.~Y'J- -'J,. IIztII-(
University of Florida - CEG - 4012
p2Immediate Settlement (Pi)-=-TJ].sn~ettlementis g,y,e.to,J;otatjonal strainJ&lt;;fLstortion) within tile soil - not a chan e in volume. (Note that if no shear strain, say a blanket loaa, then no Imme late settleiii8nt.)-QL PirOriginal DistortedImmedi
University of Florida - CEG - 4012
p3 ,lmmedlate-Set1lement Case I: B toading on the Surface aLan ElastiGJ:Ialf SpacJl (e.g. a thiclclayer of ciay)kv,~ ~t.'\C0 .JBIqI.,~'I IIBcircle, square, or rectangular footingL_._-'_. _ 1I IP, = C,qB(1-1&quot;)Ewhere~= =Poisson's ratio
University of Florida - CEG - 4012
p4 Ilmme(llate Sell ement Case II: tpading-on tb-e Sufface of a Com-p-ressible Soil (Clay) Onaerlain By a Rigid Bounaary (Roc or Dense Sand) BqHE,fl Rigid Boundary~Table P2 provides C's value under the center of flexible footings. (If the clay layer
University of Florida - CEG - 4012
p5 'Imme&lt;llatE! Settlement Case III:~oaaing-on the-Surface of a Stiff baye Underlain oy a Less Rigl&lt;'FEayer of Great Ihicknes.sFairly common for upper crust due to desiccation etc..HS.qE III &quot;Table P3 gives the C&quot;s values under the center of flex
University of Florida - CEG - 4012
p6~_Approximate Solutions: Using various combinations of Tables P1, P2 and P3 can get &quot;ballpark&quot; estimate of immediate settlement for many cases not covered. Always start withclosest initial approximation, then &quot;correct&quot; as necessary. Examples:1.Rigi
University of Florida - CEG - 4012
p7Table P1 Values of Shape and Rigidity Factor Cs at Various Points of Elastic Half-Space SurfaceMiddle of Short Side Middle of Long Side'-'Shape Circle (flexible) Circle (rigid) Square (flexible) Square (rigid) UB . 1.5 Rectangle (flexible) 2.0 UB= 3
University of Florida - CEG - 4012
p8SupplementThese are the ,?omplete interpolation cales for the Immediate Settlement Case II example on (p. p4). Since the boundary characteristics aren't defined we need to consider both cases (i.e., and u = 0). For each case, interpolate the C's for t
University of Florida - CEG - 4012
p9 o Review of Consolidation (Oedometer) Test Objective to determine the following parameters: Cc CR~~ ~-Lle I Ll log cr'v -M I Ll log cr'v Co (!1xpansion)~~ ~-M Ilog (p,/pil -M Ilog (p,/p,) Cs(~well)3.PI\:OflSO f,J.,4w\cr'c = Pc(F I L')4. (o
University of Florida - CEG - 4012
p10 4. Plot dial gage readings R vs. log time for each load and fit a smooth curve for analysis. Calculate Cv and Ca using the Casagrande construction:Ro-_.J-&quot;-a _:&quot;:a1 11c V -.:1 11T5~ (~/,J) &quot;l/cfw_su(rJ:1.O./~ ( fI/l) 2/i~d (7)1 Jo. t 1a Jf.
University of Florida - CEG - 4012
p11 6. Calculate void ratio e from R lOo for each load p and plot e-log p curve to find preconsoJidation pressure Pc using Casagrande constwction-:void ratio,e,\J, ~('tfJ(,lJ;(V (log P- a um curvalure (minimum radius) on reloaQ.Rortioll.ot. a) Find
University of Florida - CEG - 4012
p12 Obtain Field Curve from Lab Curveo Disturbance Effects- -. '&quot;'&quot;&quot;void ratio,ellc ~'&quot; lloU, -.: (;clJ yl&quot;/&quot; ,.~ .&quot; , ,'.&quot;. ,'A', , , , , , '. ' '. ,,curvature caused by stress removal, disturbance due to sampling and handling, and test a
University of Florida - CEG - 4012
p13oConstruction of NC Field Curve using Schmertmann Methodvoid ratio,er . itJ ~ffj&quot;11(OY-o.(tf l &lt;' ,;0~ 'tv'J [c.log PoLConstruction of OC Field Curve using Schmertmann Method~Recompression Slope CReoLab Virgin Compression void ratio,eo
University of Florida - CEG - 4012
p14Different ways to look at data from a Consolidation Test: (data from San Francisco Bay Mud, Holtz &amp; Kovacs, 1981)Stress, kPa Straint-t-t- -.- -0.35-~;:-0.40 +-~i-r-t-+-l-:l:J10 20 30 40 50 60 70 80 90 1000.000 0.013 0.031 0.138 0.235 0.315 0.
University of Florida - CEG - 4012
p15Consolidation Settlement: Recall that the settle. e esive soii under load IS mo':!ly due to consoli atlon:P=p;lfl,.le~p==-J-Time._._- .-wherepPitotal settlementimmediate settlement1-:;:]:.p, p,= (primary) consolidation settlement = seco
University of Florida - CEG - 4012
p16 o Graphical presentation of Fluid-Filled Cylinder with Spring Analogy: Depends on Opening Size Depends on Soil Properties~. .;.z._.~. .Pressure, - -PSPRINGpForce,o,&quot;, .ITime-AnalogyPWATER, ,,,- ., . -~cr'v_!i.crvSoil lIu (exces
University of Florida - CEG - 4012
p17NC Magnitude of Consolidation, pc: Normally consolidated (NC) clay (saturated)for NC Clay . cr'o = de (or Pc) i.e. the present effective overburden pressure, a'o. is the maximum pressure that the soilhas ever experienced,(j'C(or Pc)=preconsolida
University of Florida - CEG - 4012
p18 o Phase Diagrams: Initial StateVvo = ~'&quot; Av = HAIFinal StateIv = HAllH177:777.777771-t-VVF=H.rASubtracting:&quot;'e = (e, _ eo) = (H VFAlso:Hs-H,) = (- &quot;'H)Hsor(A)H= Hs + Hv = Hs + ~: ) = Hs (1 + eo)(1orH -s - 1+ eo( H)(8)Combinin
University of Florida - CEG - 4012
p19NC Settlement Example 1:Find Pc under center of mat foundation Sand Ym = 15.7 kNlm' Y,ot = 19.1 kN/m' 3 'water = 9.81 kN/m NC Clay (OCR = 1) ,&quot;I = 18.6 kNlm' Cc = 0.28, eo = 0.90Solution: calculate pc for the clay layer. Use point P at mid-height of
University of Florida - CEG - 4012
p20 NC Settlement Example 2: Find pc under tank center: Circular Tank, 30 ft Diameter, 20 ft High, Filled with Oil, SG = 0.91, , ~D1(b&quot;,-&quot; 0,'11) =1136pst Sand: 'i'Mo'&quot; = 100 pct, Ys&quot; = 1 05 pet Smectite Clay: Cc= 0.4, NC, eo= 1.4, Y= 120 pcSub-Iaye.!&quot; ~
University of Florida - CEG - 4012
p21 OC Magnitude of Consolidation, pc: Over-consolidated (OC) ciay (saturated)I for OC Clay cr'o &lt; cr'c (or Pc) i.e. the present effective overburden pressure, cr'o, is less than the maximum past pressure the soil has experienced, cr'c (or Pc) = preconso
University of Florida - CEG - 4012
p22Highly Over-consolidated (HOC):I (cr'O + !J.crv ) S; cr' cVirgin Curve with Slope Cc (= Compression index) Void Ratio,e6_81&quot;]:f\t\i . l 'rfl;Sh.f&lt;./ fl&quot;j't. ( .\$1.1(.RE.I~.Jeo eF-f~:~_-:_:-:_'!-;&quot;:~\~I t.(;.\(~lth_l'rtH'&quot; ~.)~.I4r I&quot;
University of Florida - CEG - 4012
p23Lightly Over-consolidated (LOC):Virgin Curve with Slope Cc (= Compression index) Void Ratio,efl.-c.kJ '-I 51'/e c.(:L,(~f&quot;&lt;~):! - ~Jt-&gt; )f -1-1-I ~f~_c _,'II'-~-'-~-t.ov!.ILog Vertical Effective Stress, o'v0'0Consolidation occurs in tw
University of Florida - CEG - 4012
p24OC Settlement Example 1:Find Pc under center of mat foundationSolution: calculate Pc for the clay layer. Use point Pat mid-height of layer to represent stress change in layer.(Pc in sand = 0)OC Clay (OCR = 1.4) 3 'Ysat = 18.6 kN/m Cc = 0.28, eo =