Question

Question 3 (Numerical integration, efficient schemes) The length L of the main supporting cable of a suspension bridge can be calculated by: 4h2 1/2 ral (1 +- L = -x²) dx Joq44 Where a is half the length of the bridge And h is the distance from the deck to the top of the tower where the cable is attached. Determine the length of a bridge with a=80 m and h=18m. Determine the cable length (with associated error of estimate) using: a) Richardson extrapolation using initial 2 and 4 segment estimates b) The two-point Gauss quadrature formula and c) Adaptive quadrature using initial 2 and 4 segment estimates d) MATLAB's Quad function e) Comment on the practical implication of your solution

Answer 1

Two point Gaun quadrature famula: (n=1) (2 = 48 (-1/2) + wf (ta) where w, z wy - where we = W₂ = 1 Men h = 18m 4 = 8000 het x = (-0)€ +(2+0) xafta @ x=0; & = -1 dx= adre @ x=a; & = 1 4bras +41 au 1-3 (1 + (**) *a)* )= (ht (a +3mm 7°)"} 1-5 (640x423)"? (it has a low va)*)"] Last (0-12030) - CHP (2-47 532 475 L-voſ (1.80704337)4 (1:55756723] L-tot-cous1147 +1.061108] L = 82.624902m Span for the limits oto a total length of the cable is equals to a L= 165.249804m for the timits at a

Richardson Extrapolation: I 3Chy) 1 1 (hu) In Ilha) - IChr) 4* for k=1 422 a h=18m a = 80m 932) ya + floxy). 80 20. 400 * IChs) 39 (40)+ altai) + tx)+ 0) + & 607 I Cha) = (40)+2 6,)} + fran] ho = 80-0 = 20 No=0 263678 +0.050825 * 40 ? (he) = 10/1 + 240.01263478 +86.050635 1.0063082 3+ 1.625 + 4055417578 + 1.09658561 hot 800 - 40 OT *](hy) - 10/1+6-1734516 +1.09657561] 10/1+6.1734516+ 1.0965 Ilhas = 82.7003721 Similarly 20h) = 30/1+2 (1025) +1.0765356] = 82.931412. m from richardson Carapolation method: 1 - 18 2.7003721) - 5 (92-931H2) 110-267143 - 27.643904 = 82.623259m for limit o to a Total length of the cable y at L = 165.346518m