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### Integration with variables Notes_Part_14

Course: MATH 5587, Fall 2010
School: UCF
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Word Count: 1026

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Applications 7.5. of Conformal Mapping. Let us now apply what we have learned about analytic/conformal maps. We begin with boundary value problems for the Laplace equation, and then present some applications in fluid mechanics. We conclude by discussing how to use conformal maps to construct Green's functions for the two-dimensional Poisson equation. Applications to Harmonic Functions and Laplace's Equation We are...

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Applications 7.5. of Conformal Mapping. Let us now apply what we have learned about analytic/conformal maps. We begin with boundary value problems for the Laplace equation, and then present some applications in fluid mechanics. We conclude by discussing how to use conformal maps to construct Green's functions for the two-dimensional Poisson equation. Applications to Harmonic Functions and Laplace's Equation We are interested in solving a boundary value problem for the Laplace equation on a domain R 2 . Our strategy is to map it to a corresponding boundary value problem on the unit disk D that we know how to solve. To this end, suppose we know a conformal map = g(z) that takes z to D. As we know, the real and imaginary parts of an analytic function F () defined on D are harmonic. Moreover, according to Proposition 7.31, the composition f (z) = F (g(z)) defines an analytic function whose real and imaginary parts are harmonic functions on . Thus, the conformal mapping can be regarded as a change of variables that preserves the property of harmonicity. In fact, this property does not even require the harmonic function to be the real part of an analytic function, i.e., we need not assume the existence of a harmonic conjugate. Proposition 7.37. If U (, ) is a harmonic function of , , and = + i = (x, y) + i (x, y) = g(z) is any analytic function, then the composition u(x, y) = U ((x, y), (x, y)) is a harmonic function of x, y. Proof : This is a straightforward application of the chain rule: U U u = + , x x x 2U 2u = x2 2 2u 2U = y 2 2 x y 2 (7.81) (7.82) u U U = + , y y y x y 2 2 U 2U +2 + x x 2 +2 2 U 2U + y y 2 U 2 U 2 + + , x2 x2 + U 2 U 2 + . y 2 y 2 2 2 Using the CauchyRiemann equations =- , x y = , y x for the analytic function = + i , we find, after some algebra, u = 1/19/12 2u 2u + 2 = x2 y x 2 + x 255 2 2U 2U + 2 2 = | g (z) |2 U, c 2012 (7.83) Peter J. Olver where | g (z) |2 = ( /x ) + ( /x ) . We conclude that whenever U (, ) is any harmonic function, and so solves the Laplace equation U = 0 (in the , variables), then u(x, y) is a solution to the Laplace equation u = 0 in the x, y variables, and is thus also harmonic. Q.E.D. This observation has immediate consequences for boundary value problems arising in physical applications. Suppose we wish to solve the Dirichlet problem u = 0 in , u=h on , (7.84) 2 2 on a simply connected domain C. Let = g(z) = p(x, y) + i q(x, y) be a oneto-one conformal mapping from the domain to the unit disk D, whose existence is guaranteed by the Riemann Mapping Theorem 7.33. (Although its explicit construction may be problematic.) Then the change of variables formula (7.82) will map the harmonic function u(x, y) on to a harmonic function U (, ) on D. Moreover, the boundary values of U = H on the unit circle D correspond to those of u = h on by the same change of variables formula: h(x, y) = H(p(x, y), q(x, y)), for (x, y) . on D. (7.85) We conclude that U (, ) solves the Dirichlet problem U = 0 in D, U =H (7.86) But we already know how to solve the Dirichlet problem (7.86) the on unit disk by the Poisson integral formula (4.116)! We conclude that the solution to the original boundary value problem is given by the composition formula u(x, y) = U p(x, y), q(x, y) . In summary, the solution to the Dirichlet problem on a unit disk can be used to solve the Dirichlet problem on more complicated planar domains -- provided we are in possession of an appropriate conformal map. Example 7.38. According to Example 7.24, the analytic function + i = = z-1 x2 + y 2 - 1 2y = +i 2 + y2 z+1 (x + 1) (x + 1)2 + y 2 (7.87) maps the right half plane R = { x = Re z > 0 } to the unit disk D = { | | < 1 }. Proposition 7.37 implies that if U (, ) is a harmonic function in the unit disk, then u(x, y) = U 2y x2 + y 2 - 1 , 2 + y2 (x + 1) (x + 1)2 + y 2 (7.88) is a harmonic function on the right half plane. (This can, of course, be checked directly by a rather unpleasant chain rule computation.) To solve the Dirichlet boundary value problem u = 0, x > 0, u(0, y) = h(y), (7.89) on the right half plane, we adopt the change of variables (7.87) and use the Poisson integral formula to construct the solution to the transformed Dirichlet problem U = 0, 1/19/12 2 + 2 < 1, 256 U (cos , sin ) = H(), c 2012 (7.90) Peter J. Olver on the unit disk. The relevant boundary conditions are found as follows. Using the explicit form x+ iy = z = (1 + )(1 - ) 1 + - - | |2 1 - 2 - 2 + 2 i 1+ = = = 1- | 1 - |2 | 1 - |2 ( - 1)2 + 2 for the inverse map, we see that the boundary point = + i = e i on the unit circle D will correspond to the boundary point iy = 2 2 i sin = i cot = 2 2 + 2 ( - 1) 2 (cos - 1)2 + sin (7.91) on the imaginary axis R = { Re z = 0 }. Thus, the boundary data h(y) on R corresponds to the boundary data H() = h cot 1 2 on the unit circle. The Poisson integral formula (4.116) can then be applied to solve (7.90), from which we are able to reconstruct the solution (7.88) to the boundary value problem (7.88) on the half plane. Let's look at an explicit example. If the boundary data on the imaginary axis is provided by the step function u(0, y) = h(y) 1, 0, y > 0, y < 0, then the corresponding boundary data on the unit disk is a (periodic) step function H() = 1, 0, 0 < < , - < < 0. After some tedious algebra, we find that the corresponding solution in the right half plane is simply 1 1 1 y 1 u(x, y) = + ph z = + tan-1 , 2 2 x an answer that, in hindsight, we should have been able to guess. Remark : The solution to the preceding Dirichlet boundary value problem is not, in fact, unique, owing to the unboundedness of the domain. The solution that we pick out by using the conformal map to the unit disk is the one that remains bounded at . The unbounded solutions would correspond to solutions on the unit disk that have a singularity in their boundary data at the point -1; see Exercise . 1/19/12 257 c 2012 Peter J. Olver According to (4.119), the corresponding solution in the unit disk is 2 2 1 - 1 tan-1 1 - - , 2 + 2 < 1, > 0, 2 1 U (, ) = , 2 + 2 < 1, = 0, 2 2 2 - 1 tan-1 1 - - , 2 + 2 < 1, < 0. 2
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UCF - MATH - 5587
Figure 7.25.A NonCoaxial Cable.Example 7.39. A non-coaxial cable. The goal of this example is to determine the electrostatic potential inside a non-coaxial cylindrical cable, as illustrated in Figure 7.25, with prescribed constant potential values on th
UCF - MATH - 5587
0 Figure 7.29.15 Fluid Flow Past a Tilted Plate.30Note that = ( 1, 0 ), and hence this flow satisfies the Neumann boundary conditions (7.95) on the horizontal segment D = . The corresponding complex potential is (z) = z, with complex velocity f (z) = (
UCF - MATH - 5587
on the unit disk D for an impulse concentrated at the origin; see Section 6.3 for details. How do we obtain the corresponding solution when the unit impulse is concentrated at another point = + i D instead of the origin? According to Example 7.25, the lin
UCF - MATH - 5587
as long 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+
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