3—stress wave methods for deep foundations
Tiêu chuẩn Châu Âu EC6: Kết cấu gạch đá phần 3: Phương pháp đơn giản hóa (Eurocode6 BS EN 1996 3 Design of masonry structures part 3: Simplified calculation methods for unreinforced masonry structures)
... and ETAs) for products There is a need for consistency between the harmonised technical specifications for construction products and the technical rules for works4 Furthermore, all the information ... following parts: Part 1-1: General rules for reinforced and unreinforced masonry c) serve as a reference for the establishment of harmonised standards and guidelines for European technical approvals ... self-weight is given in Annex C for walls subject to a uniform lateral design load 22 EN 1996-3:2006 (E) Annex A (Informative) Simplified calculation method for unreinforced masonry walls of buildings...
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Báo cáo khoa học: Modified PCR methods for 3¢ end amplification from serial analysis of gene expression (SAGE) tags doc
... 2009 FEBS W.-J Xu et al Methods for 3¢ end amplification from SAGE tags A B Fig Schematic for GLGI (A) In this process, first-strand cDNA synthesized by oligo(dT) is used for PCR In the first cycle, ... only for the subsequent PCR but also for 5¢-RACE, 3¢-RACE and northern blotting, etc Finally, the two-step nested PCR principle can easily produce 3¢-end cDNA tag-specific frag- Methods for 3¢ ... approaches for transcriptome research and genome annotation Nat Methods 2, 495–502 FEBS Journal 276 (2009) 2657–2668 ª 2009 The Authors Journal compilation ª 2009 FEBS 2663 Methods for 3¢ end...
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Quantitative Methods for Business chapter 3 ppt
... It had a reputation for innovations, both technical and managerial As well as stock control it was the seedbed for breakthroughs in product quality 100 Quantitative methods for business Chapter ... The nature of the slope therefore depends on where we are along the horizontal scale, in other words, the value of x For the lower values of x the slope is positive, for the higher ones it is negative ... necessarily all involving x For the types of equation we shall look at you need only deal with the parts one at a time to reach the differential 90 Quantitative methods for business Chapter Example...
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Quantitative Methods for Ecology and Evolutionary Biology (Cambridge, 2006) - Chapter 3 doc
... log-likelihood function the ‘‘support for different values of p, given the data’’ for this very reason (Bayesian methods show how to use the support to combine prior and observed information) (a) (b) –67 –8 ... for discrete random variables too, but the distribution function has jumps in it Exercise 3.2 (E) What is the distribution function for the sum of two rolls of the fair die? We can also ask for ... Virtually all computer operating systems now provide random numbers that are uniformly distributed between and 1; for a uniform random number between and 1, the probability density is f(z) ¼ if z and...
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Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 3 pptx
... 3.2.1 sin x x has a removable discontinuity at x = The Heaviside function, 0 for x < 0, H(x) = 1/2 for x = 0, for x > 0, has a jump discontinuity at x = x has an infinite discontinuity at ... Uniform Continuity Consider a function f (x) that is continuous on an interval This means that for any point ξ in the interval and any positive there exists a δ > such that |f (x) − f (ξ)| < for ... 3.4: Piecewise Continuous Functions the function is said to be uniformly continuous on the interval A sufficient condition for uniform continuity is that the function is continuous on a closed interval...
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Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 3 ppt
... [ıθ]2π for n = −1 = for n = −1 ı2π for n = −1 for n = −1 We parameterize the contour and the integration z − z0 = ı2 + eıθ , 2π n ı2 + eıθ (z − z0 ) dz = C = (ı2+eıθ )n+1 n ı eıθ dθ 2π for ... z 1/2 , − log z + const Assuming that the above expression is non-singular, we have found a formula for writing the analytic function in terms of its real part, u(r, θ) With the same method, we ... cos t, y = sin t for ≤ t ≤ π π x2 dx + (x + y) dy = C cos2 t(− sin t) + (cos t + sin t) cos t dt π = − 10.2 Contour Integrals Limit Sum Definition We develop a limit sum definition for contour integrals...
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Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 1 potx
... (x), for x ≥ 0, for x ≤ The initial condition for y− demands that the solution be continuous Solving the two problems for positive and negative x, we obtain y(x) = e1−x , e1+x , for x > 0, for ... −1 sign x = for x < for x = for x > Since sign x is piecewise defined, we solve the two problems, y+ + y+ = 0, y− − y− = 0, y+ (1) = 1, y− (0) = y+ (0), for x > for x < 0, and define the ... xn F 1, y x (Just formally substitute 1/x for λ.) For example, x2 y + 2y , x+y xy , x cos(y/x) are homogeneous functions of orders 3, and 1, respectively Euler’s theorem for a homogeneous function...
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Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 2 potx
... ln t + ct , A = −2 For positive A, the solution is bounded at the origin only for c = For A = 0, there are no bounded solutions For negative A, the solution is bounded there for any value of c ... α→∞ α − 1 for x = 0, = for x > lim y(x) = lim α→∞ This behavior is shown in Figure 14.10 The first graph plots the solutions for α = 1/128, 1/64, , The second graph plots the solutions for α = ... lim x→∞ c + βx β =0 = lim αx x→∞ α eαx e For β = λ = 1, the solution is y= e−x +c e−αx (c + x) e−x α−1 828 for α = 1, for α = 1 Figure 14.9: The Solution for a Range of α The solution which satisfies...
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Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 3 pptx
... ··· 856 0 0 λ 1 λ 0 Jordan Canonical Form A matrix J is in Jordan canonical form if all the elements are zero except for Jordan blocks Jk along the diagonal J1 · · · 0 ... method for computing functions of matrices Example 15.3.3 Consider the matrix exponential function A= −3 eA for our old friend: 1 −1 In Example 15.3.2 we showed that the Jordan canonical form ... the information in part (i), write down one solution x(1) (t) of the system (15.2) There is no other solution of a purely exponential form x = xi eλt (c) To find a second solution use the form...
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Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 4 pot
... of independent variable τ = log t, y(τ ) = x(t), will transform (15.4) into a constant coefficient system dy = Ay dτ Thus all the methods for solving constant coefficient systems carry over directly ... greater than one, we will have solutions of the form, xitα , xitα log t + ηtα , xitα (log t)2 + ηtα log t + ζtα , ., analogous to the form of the solutions for a constant coefficient system, xi eατ , ... 1 We have an uncoupled equation for x1 x1 = x1 t x1 = c1 t 898 x1 x2 (15.5) By substituting the solution for x1 into (15.5), we obtain an uncoupled equation for x2 (c1 t + x2 ) t x2 − x2 =...
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Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 5 pdf
... order equation for y, but note that it is a first order equation for y We can solve directly for y d dx 3/2 x y =0 y = c1 exp − x3/2 exp Now we just integrate to get the solution for y y = c1 ... equations have the form d F (x, y, y , y , ) = f (x) dx If you can write an equation in the form of an exact equation, you can integrate to reduce the order by one, (or solve the equation for first order) ... +ay = 0, has the solution y = cxa Thus for the second order equation we will try a solution of the form y = xλ The substitution 940 y = xλ will transform the differential equation into an algebraic...
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Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 6 ppsx
... put in a simpler form by exchanging the dependent and independent variables Thus a differential equation for y(x) can be written as an equation for x(y) Solving the equation for x(y) will give ... a x da = ln x a=0 xa − x−a a Clearly y1 is a solution for all a For a = 0, y2 is a linear combination of xa and x−a and is thus a solution For a = 0, y2 is one half the derivative of xa evaluated ... linear equation for u which when solved will give us an implicit solution for y (See Exercise 18.4.) 984 Result 18.1.1 The Bernoulli equation y + p(t)y = q(t)y α , α = can be transformed to the...
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Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 7 pps
... Equation Normal Form Hint 19.1 Transform the equation to normal form Transformations of the Independent Variable Integral Equations Hint 19.2 Transform the equation to normal form and then apply ... 19.5 Transform the equation to normal form 1034 19.7 Solutions The Constant Coefficient Equation Normal Form Solution 19.1 y + 2+ x y + 24 + 12x + 4x2 y = To transform the equation to normal form we ... the form G(x|ξ) = c1 (x − a), c2 (x − b), for x ≤ ξ for x ≥ ξ Imposing continuity of G(x|ξ) at x = ξ and a unit jump of G(x|ξ) at x = ξ, we obtain G(x|ξ) = (x−a)(ξ−b) , b−a (x−b)(ξ−a) , b−a for...
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Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 8 ppsx
... form G(x|ξ) = c1 + c2 x d1 + d2 x for x < ξ for x > ξ Applying the two boundary conditions, we see that c1 = and d1 = −d2 The Green function now has the form G(x|ξ) = cx d(x − 1) for x < ξ for ... e−x for x < ξ for x > ξ Since the solution must be bounded for all x, the Green function must also be bounded Thus c2 = d1 = The Green function now has the form c ex for x < ξ G(x|ξ) = −x de for ... = H(x) = −∞ for x < for x > The integral of the Heaviside function is the ramp function, r(x) x H(t) dt = r(x) = −∞ x for x < for x > The derivative of the delta function is zero for x = At x...
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Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 9 doc
... (ξ) W (ξ) u1 (ξ)u2 (x) W (ξ) G(x|ξ) = for x < ξ, for x > ξ The solution for u is b u= G(x|ξ)f (ξ) dξ a Thus if there is a unique solution for v, the solution for y is b y=v+ G(x|ξ)f (ξ) dξ a 1099 ... constant How does the solution for λ = differ from that for λ = 1? The λ = case provides an example of resonant forcing Plot the solution for resonant and non-resonant forcing Hint, Solution Exercise ... solution, W (x) is nonzero on [a, b] The Green function has the form c u1 c u2 G(x|ξ) = for x < ξ, for x > ξ The continuity and jump conditions for Green function gives us the equations c1 u1 (ξ) − c2...
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Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 10 ppt
... function for this problem satisfies L[G(x|ξ)] = δ(x − ξ), G(0|ξ) = G (0|ξ) = Since the Green function must satisfy the homogeneous boundary conditions, it has the form G(x|ξ) = cx + d/x for x < ξ for ... e−k(x+ξ) 2k 2k −k(ξ−x) − 2k e + 2k e−k(x+ξ) , for x < ξ = 1 − 2k e−k(x−ξ) + 2k e−k(x+ξ) , for ξ < x = − k e−kξ sinh(kx), for x < ξ − k e−kx sinh(kξ), for ξ < x 1160 G(x; ξ) = − e−kx> sinh(kx< ) ... −k(x+ξ) e − e 2k 2k −k(ξ−x) − 2k e − 2k e−k(x+ξ) , for x < ξ 1 − 2k e−k(x−ξ) − 2k e−k(x+ξ) , for ξ < x − k e−kξ cosh(kx), for x < ξ − k e−kx cosh(kξ), for ξ < x G(x; ξ) = − e−kx> cosh(kx< ) k The Green...
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Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 3 pdf
... term in the sum is the leading order behavior For a few examples, • For sin x ∼ x − x3 /6 + x5 /120 − · · · as x → 0, the leading order behavior is x • For f (x) ∼ ex (1 − 1/x + 1/x2 − · · · ) as ... differential equation for y d2 s e = x es dx2 s + (s )2 es = x es s + (s )2 = x The Dominant Balance Now we have a differential equation for s that appears harder to solve than our equation for y However, ... (x) n=0 For fixed x = x0 , N f (x0 ) − an (x0 ) → as N → ∞ n=0 For an asymptotic series we have a quite different behavior If g(x) is asymptotic to fixed N , ∞ n=0 bn (x) as x → x0 then for N g(x)...
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Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 3 potx
... b2 − ac c This form is useful if a vanishes Another canonical form for hyperbolic equations is uσσ − uτ τ = K(σ, τ, u, uσ , uτ ) 1687 (36.5) We can transform Equation 36.3 to this form with the ... characteristic equations for ξ are √ b − b2 − ac d dy = , ξ(x, y(x)) = dx a dx Solving the differential equation for y(x) determines ξ(x, y) We just write the solution for y(x) in the form F (x, y(x)) ... equation for ξ is ξ(x, y(x)) = const, we then have ξ = F (x, y) Upon solving for ξ and ψ we divide Equation 36.2 by β(ξ, ψ) to obtain the canonical form Note that we could have solved for ξy /ξx...
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Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 3 ppsx
... −1 < for |β/α| < = for |β/α| = > for |β/α| > β Thus we see that α x may be a minimum for |β/α| ≥ and may be a maximum for |β/α| ≤ Jacobi Condition Jacobi’s accessory equation for this ... Gu = for (x, y) ∈ Γ 2089 Gu h ds = 0, Γ For the given integrand this is, (2pux , 2puy ) · n + 2σu = for (x, y) ∈ Γ, p u · n + σu = for (x, y) ∈ Γ We can also denote this as p ∂u + σu = for (x, ... − y − 2xy dx, y(0) = = y(1) For this problem take an approximate solution of the form y = x(1 − x) (a0 + a1 x + · · · + an xn ) , 2066 and carry out the solutions for n = and n = • (y )2 + y...
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Sensors and Methods for Robots 1996 Part 3 docx
... model 3ARG-D ($1,100) has an RS-232 output for connection to a computer Technical specifications for the 3ARG-D are given in Table 2.1 Specifications for the 3ARG-A are similar A more detailed ... routinely provided in the form of declination tables printed directly on the maps or charts for any given locale Instruments which measure magnetic fields are known as magnetometers For application to ... appropriate for true heading: Ht = Hi ± CFdev ± CFdec (2.9) where Ht = true heading Hi = indicated heading CFdev = correction factor for compass deviation CFdec = correction factor for magnetic...
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