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### Integration with variables Notes_Part_18

Course: MATH 5587, Fall 2010
School: UCF
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Word Count: 1265

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long as as n = -1. Therefore, we can use the Fundamental Theorem of Calculus (which works equally well for real integrals of complex-valued functions), to evaluate n+1 1 -1 = n = 2 k + 1 odd, 0, 2 t + i (t - 1) 2 z n dz = = , n = 2 k even. n+1 P t = -1 n+1 Thus, when n 0 is a positive integer, we obtain the same result as before. Interestingly, in this case the complex integral is well-defined even when n is a...

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long as as n = -1. Therefore, we can use the Fundamental Theorem of Calculus (which works equally well for real integrals of complex-valued functions), to evaluate n+1 1 -1 = n = 2 k + 1 odd, 0, 2 t + i (t - 1) 2 z n dz = = , n = 2 k even. n+1 P t = -1 n+1 Thus, when n 0 is a positive integer, we obtain the same result as before. Interestingly, in this case the complex integral is well-defined even when n is a negative integer because, unlike the real line segment, the parabolic path does not go through the singularity of z n at z = 0. The case n = -1 needs to be done slightly differently, and integration of 1/z along the parabolic path is left as an exercise for the reader -- one that requires some care. We recommend trying the exercise now, and then verifying your answer once we have become a little more familiar with basic complex integration techniques. Finally, let us try integrating around a semi-circular arc, again with the same endpoints -1 and 1. If we parametrize the semi-circle S + by z(t) = e i t , 0 t , we find z n dz = S+ 0 zn dz dt = dt t=0 e i nt i e i t dt = 0 i e i (n+1)t dt 0 0, 2 , n = 2 k even. n+1 This value is the negative of the previous cases -- but this can be explained by the fact that the circular arc is oriented to go from 1 to -1 whereas the line segment and parabola both go from -1 to 1. Just as with line integrals, the direction of the curve determines the sign of the complex integral; if we reverse direction, replacing t by - t, we end up with the same value as the preceding two complex integrals. Moreover -- again provided n = -1 -- it does not matter whether we use the upper semicircle or lower semicircle to go from -1 to 1 -- the result is exactly the same. However, the case n = -1 is an exception to this "rule". Integrating along the upper semicircle S + from 1 to -1 yields S+ e i (n+1)t = n+1 1 - e i (n+1) = = - n+1 -1 = n = 2 k + 1 odd, dz = z i dt = i , 0 (7.111) whereas integrating along the lower semicircle S - from 1 to -1 yields the negative S- dz = z - 0 i dt = - i . (7.112) Hence, when integrating the function 1/z, it makes a difference which direction we go around the origin. Integrating z n for any integer n = -1 around an entire circle gives zero -- irrespective of the radius. This can be seen as follows. We parametrize a circle of radius r by z(t) = re i t for 0 t 2 . Then, by the same computation, 2 2 z dz = C 0 n (r e n i nt )(r i e ) dt = 0 it ir n+1 i (n+1)t e r n+1 i (n+1)t e dt = n+1 2 = 0, t=0 (7.113) 1/19/12 267 c 2012 Peter J. Olver provided n = -1. The circle on the integral sign serves to remind us that we are integrating around a closed curve. The case n = -1 remains special. Integrating once around the circle in the counter-clockwise direction yields a nonzero result dz = z 2 i dt = 2 i . 0 (7.114) C Let us note that a complex integral does not depend on the particular parametrization of the curve C. It does, however, depend upon its orientation: if we traverse the curve in the reverse direction, then the complex integral changes its sign: f (z) dz = - f (z) dz. C (7.115) -C Moreover, if we chop up the curve into two non-overlapping pieces, C = C1 C2 , with a common orientation, then the complex integral can be decomposed into a sum over the pieces: f (z) = C1 C2 C1 f (z) dz + C2 f (z) dz. (7.116) For instance, the integral (7.114) of 1/z around the circle is the difference of the individual semicircular integrals (7.111, 112); the lower semicircular integral acquires a negative to sign flip its orientation so as to agree with that of the entire circle. All these facts are immediate consequences of the basic properties of line integrals, or can be proved directly from the defining formula (7.108). Note: In complex integration theory, a simple closed curve is often referred to as a contour , and so complex integration is sometimes referred to as contour integration. Unless explicitly stated otherwise, we always go around contours in the counter-clockwise direction. Further experiments lead us to suspect that complex integrals are usually pathindependent, and hence evaluate to zero around closed contours. One must be careful, though, as the integral (7.114) makes clear. Path independence, in fact, follows from the complex version of the Fundamental Theorem of Calculus. Theorem 7.47. Let f (z) = F (z) be the derivative of a single-valued complex function F (z) defined on a domain C. Let C be any curve with initial point and final point . Then f (z) dz = C C F (z) dz = F () - F (). (7.117) Proof : This follows immediately from the definition (7.108) and the chain rule: b F (z) dz = C a F (z(t)) dz dt = dt b a d F (z(t)) dt = F (z(b)) - F (z(a)) = F () - F (), dt Q.E.D. c 2012 Peter J. Olver where = z(a) and = z(b) are the endpoints of the curve. 1/19/12 268 For example, when n = -1, the function f (z) = z n is the derivative of the single1 z n+1 . Hence valued function F (z) = n+1 z n dz = C n+1 n+1 - n+1 n+1 whenever C is (almost) any curve connecting to . The only restriction is that, when n < 0, the curve is not allowed to pass through the singularity at the origin z = 0. In contrast, the function f (z) = 1/z is the derivative of the complex logarithm log z = log | z | + i ph z, which is not single-valued on all directly. However, if our curve is does not include the origin, 0 complex logarithm to evaluate the of C \ {0}, and so Theorem 7.47 cannot be applied contained within a simply connected subdomain that C, then we can use any single-valued branch of the integral dz = log - log , z C where , are the endpoints of the curve. Since the common multiples of 2 i cancel, the answer does not depend upon which particular branch of the complex logarithm is chosen as long as we are consistent in our choice. For example, on the upper semicircle S + of radius 1 going from 1 to -1, S+ dz = log(-1) - log 1 = i , z where we use the branch of log z = log | z | + i ph z with 0 ph z . On the other hand, if we integrate on the lower semi-circle S - going from 1 to -1, we need to adopt a different branch, say that with - ph z 0. With this choice, the integral becomes S- dz = log(-1) - log 1 = - i , z thus reproducing (7.111, 112). Pay particular attention to the different values of log(-1) in the two cases! Cauchy's Theorem The preceding considerations suggest the following fundamental theorem, due in its general form to Cauchy. Before stating it, we introduce the convention that a complex function f (z) is to be called analytic on a domain C provided it is analytic at every point inside and, in addition, remains (at least) continuous on the boundary . When is bounded, its boundary consists of one or more simple closed curves. In general, as in Green's Theorem 6.13, we orient so that the domain is always on our left hand side. This means that the outermost boundary curve is traversed in the counter-clockwise direction, but those around interior holes take on a clockwise orientation. Our convention is depicted in Figure 7.32. 1/19/12 269 c 2012 Peter J. Olver
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UCF - MATH - 5587
Figure 7.32.Orientation of Domain Boundary.Theorem 7.48. If f (z) is analytic on a bounded domain C, then f (z) dz = 0.(7.118)Proof : If we apply Green's Theorem to the two real line integrals in (7.109), we find u dx - v dy = - u v - x y = 0,v dx +
UCF - MATH - 5587
Proof : Note that the integrand f (z) = 1/(z - a) is analytic everywhere except at z = a, where it has a simple pole. If a is outside C, then Cauchy's Theorem 7.48 applies, and the integral is zero. On the other hand, if a is inside C, then Proposition 7.
UCF - MATH - 5587
0 Figure 7.36.15 Kutta Flow Past a Tilted Airfoil.30which remains asymptotically 1 at large distances. By Cauchy's Theorem 7.48 coupled with formula (7.123), if C is a curve going once around the disk in a counter-clockwise direction, then i 1 dz = - 2
UCF - MATH - 5587
is analytic in the disk | z | 2 since its only singularity, at z = 3, lies outside the contour C. Therefore, by Cauchy's formula (7.135), we immediately obtain the integral ez dz = z2 - 2 z - 3 f (z) i dz = 2 i f (-1) = - . z+1 2eCCNote: Path independe
UCF - MATH - 5587
Chapter 12 Dynamics of Planar MediaIn previous chapters we studied the equilibrium configurations of planar media - plates and membranes - governed by the two-dimensional Laplace and Poisson equations. In this chapter, we analyze their dynamics, modeled
UCF - MATH - 5587
In this manner, we arrive at the basic conservation law relating the heat energy density and the heat flux vector w. As in our one-dimensional model, cf. (4.3), the heat energy density (t, x, y) is proportional to the temperature, so (t, x, y) = (x, y) u(
UCF - MATH - 5587
for the diffusion equation. See [35; p. 369] for a precise statement and proof of the general theorem. Qualitative Properties Before tackling examples in which we are able to construct explicit formulae for the eigenfunctions and eigenvalues, let us see w
UCF - MATH - 5587
Theorem 12.1. Suppose u(t, x, y) is a solution to the forced heat equation ut = u + F (t, x, y), for (x, y) , 0 &lt; t &lt; c,where is a bounded domain, and &gt; 0. Suppose F (t, x, y) 0 for all (x, y) and 0 t c. Then the global maximum of u on the set cfw_ (t, x
UCF - MATH - 5587
so there are no non-separable eigenfunctions . As a consequence, the general solution to the initial-boundary value problem can be expressed as a linear combination u(t, x, y) =m,n = 1cm,n um,n (t, x, y) =m,n = 1cm,n e- m,n t vm,n (x, y)(12.41)of
UCF - MATH - 5587
Let us start with the equation for q(). The second boundary condition in (12.50) requires that q() be 2 periodic. Therefore, the required solutions are the elementary trigonometric functions q() = cos m or sin m , where = m2 , (12.53)with m = 0, 1, 2, .
UCF - MATH - 5587
15 10 5 -4 -2 -5 -10 -15 2 4Figure 12.3.The Gamma Function.Thus, at integer values of x, the gamma function agrees with the elementary factorial. A few other values can be computed exactly. One important case is when x = 1 . Using 2 the substitution t
UCF - MATH - 5587
Remark : The definition of a singular point assumes that the other coefficients do not both vanish there, i.e., either q(x0 ) = 0 or r(x0 ) = 0. If all three functions happen to vanish at x0 , we can cancel any common factor (x - x0 )k , and hence, withou
UCF - MATH - 5587
we find that the only non-zero coefficients un are when n = 3 k +1. The recurrence relation u3 k+1 = u3 k-2 (3 k + 1)(3 k) yields u3 k+1 = 1 . (3 k + 1)(3 k)(3 k - 2)(3 k - 3) 7 6 4 3The resulting solution isx3 k+1 . (3 k + 1)(3 k)(3 k - 2)(3 k - 3) 7 6
UCF - MATH - 5587
two different Frobenius expansions. Usually, this expectation is valid, but there is an important exception, which occurs when the indices differ by an integer. The general result is summarized in the following list: (i ) If r2 - r1 is not an integer, the
UCF - MATH - 5587
We have thus found the series solution (-1)k xm+2k . 22k k(k - 1) 3 2 (r + k)(r + k - 1) (r + 2)(r + 1) k=0 k=0 (12.93) So far, we not paid attention to the precise values of the indices r = m. In order to continue the recurrence, we need to ensure that t
UCF - MATH - 5587
where h0 = 0, while = limkhk = 1 +1 1 1 + + + , 2 3 k (12.102)hk - log k .5772156649 . . .is known as Euler's constant. All Bessel functions of the second kind have a singularity at the origin x = 0; indeed, by inspection of (12.101), we find that th
UCF - MATH - 5587
of the Bessel boundary value problem (12.5455) are the squares of the roots of the Bessel function of order m. The corresponding eigenfunctions are wm,n (r) = Jm (m,n r) , n = 1, 2, 3, . . . , m = 0, 1, 2, . . . , (12.112)defined for 0 r 1. Combining (12
UCF - MATH - 5587
t=0t = .04t = .08t = .12 Figure 12.6.t = .16 Heat Diffusion in a Disk.t = .212.5. The Fundamental Solution of the Heat Equation.As we learned in Section 4.1, the fundamental solution to the heat equation measures the temperature distribution result
UCF - MATH - 5587
for the planar heat equation is given by the linear superposition formula u(t, x, y) = 1 4 t f (, ) e- [ (x-)2+(y-)2 ]/(4 t)d d.(12.125)We can interpret the solution formula (12.125) as a two-dimensional convolution u(t, x, y) = F (t, x, y) f (x, y)
UCF - MATH - 5587
Vibration of a Rectangular Drum Let us first consider the vibrations of a membrane in the shape of a rectangle R= 0 &lt; x &lt; a, 0 &lt; y &lt; b ,with side lengths a and b, whose edges are fixed to the (x, y)plane. Thus, we seek to solve the wave equation utt = c2
UCF - MATH - 5587
A table of their values (for the case c = 1) can be found in the preceding section. The Bessel roots do not follow any easily discernible pattern, and are not rational multiples of each other. This result, known as Bourget's hypothesis, [142; p. 484], was
UCF - MATH - 5587
following table, 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 o
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