Question

Question: The shear stresses in (kPa) of four specimens taken at various depths in a clay stratum are given by Depth, meter 1.8 3.1 4.1 6.6 Stress ,kPa 14.4 28.7 m 43.1 (a) Find the value of m if the Newton's third divided difference is -0.73. (10 marks) (b) Hence, estimate the shear stress at a depth of 4.5 meter. (5 marks) (c) Using the value of m obtained in (a), (i) Approximate the shear stress at a depth of 4.5 meter using second-order Lagrange interpolation. (5 marks) (ii) Estimate the second-order derivative using most appropriate numerical differentiation at a depth of 6.6 meter. (5 marks)

Answer 1

Shear Staedes at various depths in clay Stuatur, are given by • Derth (vetes) 1.8 3.1 11.1 66 Stoess (kla) 14.4 28.2 43.1 Given that Newton's pod dinises difference is -0.73 il. (x0, 21, 22,X3] = -0.73 (0 (70, X3, 43,13) = (10, 19,X7] - (24,42,43] 10-13 - 2 Now (xo, Xg, X2 Yo-Y (y-72) (10.84]-[X9,X2]. = Xox X-A2 Xo - X2 Xo-X2 2 f. 8-307 14-4-28.7 - 28-7-m 1.8 - 3-1 3.1-4.1 1.8-4.1 -14.3 (m -28.7) -1-3 - 2.3

11-m+28.7 -2.3 (voids, 123= M - 39.2 3 2.3 (1,12, 13] = (11,12}-{ 13,73} Y-3 Yo-Y2 X-2 72 73 X2 X3 (28-2-m) - (m-43:1) 3.1-4.1 4:1-66 3.1 - 6.6 X-13 m- 28.7 + (m-43-1) 2.5 -3.5 =(2.5)m ~71.75 m-43-1 - 2.5x3.5 13,5)m ~114.85 2.5x3.5 (1,1,1)= 45.34 – (1-4) 3.5 Put (3 84 in @ we get [X0,X.X2,83] = M–39.2 - (45.94-41-4)m) 2.3 3.5 1.8 - 6.6

= (3-5)m - 138.95 - 105.66+0.22) (-3)(3-5) *(-4.8) (vo, Yo13 23= (672)m – 241.61 -38.64 -0.73x63864) =(G72)m – 244-61 28.207 = (6-22)m-244.61 28-2077 244.61 6-72 40-66=m m=h0.66 f(1) = Yo + (1-4)[10, X1] + (x-7)(x-X3) [70,W2,X2] +(x-Y-XX-X3 ) (X-X2) (40,X9,X2,133 f(4.5) = 14.4+(x-1-8)(y->) +(x-48)(1–31) (40-66–34) 7.-11 2.3 +(x1-8)(x-3-1)(1-4.1) (-0.73) f(4.5) = 14.4 +(4.5-1-8)(11) +(4+5-1-8X4.5-3-1)(0.9 2.3 +(4.5-1.8 X(4.573-1)(4.5-4.1)60.73)

= 14.4 + (2.7)*11 + (2.7)x(1.4)*(0.39) + R-7)8(1.4) xo.4 x 60.73) = 14.4 + 29.2 +1:42 - 1.103 f(4.5) E44.462 Owe know that and onder Lagrange intes polation is f(x)= y (x-x1)(x-13) (7.-1) (1.-X2) (1-7)(x-x2) + YA (-xo)(x1 -72). 1 (1-1)(x-1) (12-7)(x2-x) f(4.5)= 34.5x (4.5-3-1)(4.5 -4.1) +28.7x(4.5–1-8 34.5-4-1) (1-8 - 3-1)(1-8-4.1) 3.9-1-8)(3-1-4-1) + 40.66(405–1-8)(4.5-3-1) (4-1 - 1-8)(4-1-3-1) f(4-5) = 14.4X 1.4x0.4 f 28.7 x 2.7% 0.4 -1-3x6-23) 1.3 X(-1) + 40.66X(2.7) 1.4 2-3 X 1

2.99 44 8.06 + 30.996 + 153-69 (-1-3) 2:3 = 2.69 - 23.84 + 66-82 f(4.5) = 45.62 I we 2nd know that, ooder derivative using Newton's divided difference is given by f"(1) = 2(40, 91,72] + 2[(x-xo)+(1-1)+(**)] [X0, X.X2,13] put 1= 6.6 f"(66) = 2 X(0:39) + 2[(66–1-8) +(66–3-1)+(66-4-1)]*(-0.73) = 0.78 + 2 [4.8 +3.5 +2.5) (073) = 0.78 - 9:46)[10.8] = 0.78 - 15.768 f"(66)= -14.988

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