Vibrations_rao_4thsi_ch10

Chapter 10\nVibration Measurement\nand Applications\n\n10.1\nVoltage sensitivity = V = 0.098 volt-meter/newton\nthickness = t = 2 mm = 2 / 1000 m\noutput voltage = 220 volts, pressure applied = Pz = ?\nE = v t Pz => 220 = (0.098)(1000) Pz\n. Pz = 1.1224 x 10^6 N/m^2\n\n10.2\nm = 0.5 kg, k = 10,000 N/m, c < 0\namplitude = Y = 4 / 1000 m\ntotal displacement of mass x = 12 / 1000 m\nrelative displacement = 3z = x - y = 3 / 1000 m\nZ = r^2 Y\n1 - r^2\nr = ωn / ω = 0.8165\nω3 = r ωn = 0.8165 (10000)^0.5 = 115.4705 rad/sec = 18.3777 Hz\n\n10.3\nZ = r^2 Y\n√((1 - r^2) + (2 s r)^2)\n= r^2 Y\n√((1 - r^2)^2 + (√2 r)²) = r^2 Y\n√(1 + r^2)\n\n10.4\nSpeed range:\n500 rpm = 52.36 rad/sec - 1500 rpm = 157.08 rad/sec\nx = X cos ωt ;\nZ = r^2 Y (E_1)\n\nCASE(i): Let Y = 0\nE0(E1) gives, for 2% error,\nZ / Y = |r^2 / |r^2 - 1|| = 1.02 => r = ω / ωn = 7.41414\n\n628 ωn = ω3 500 / 7.41414 = 70.0143 or 210.0428 rpm\n= 1.1669 or 3.5007 Hz = 7.3319 or 21.9957 rad/sec\n∴ ωn = 1.1669 Hz\n\nCASE (ii): Let Y = 0.6\nE0(E1) gives\nZ / Y = r^2 = E = 1.02\n=> 0.0404 r^4 - 0.5826 r^2 + 1.0404 = 0\nwhich gives r = 2.2562, 3.0546\nSince the quantity E attains maximum at\nr = 1 / √(1 - 2 r^2) = 1.8898 for Y = 0.6\nwe use r = 2.2562.\n∴ Maximum ωn = ω / r = 500 / (2.2562 × 60) = 3.6935 Hz\n\n10.5\nError factor for vibrometer is\nE = √(r^2) / √((1 - r^2) + (2 r Y)²)\nMaximum of E occurs at\nr* = 1 / √(1 - 2 Y^2)\nE |r=4 = |1 - 4²| = 1.0667\n\nPercent error = (E - 1) 100 = 6.67%. \n\n10.6\nError factor = E = √(r^2) / √((1 - r^2) + (2 r Y)²) \nE attains maximum at r* = √(1 - 2 r^2)\nFor Y = 0.67, r* = 3.1281 and E |r=r*, Y=0.67 = 1.0053\nPercent error = 0.53% 10.7\nSelect the vibrometer on the basis of lowest frequency being measured.\n(a) Y = 0:\nFrom E0 (10.19), Z / Y = 1.03 = r^2 / |r - 1| , r^2 = 1.03 / 0.03 = 34.3333\n=> r = ω / ωn = 2π(500) / 60 ωn = 5.9595\nωn = 8.9359 rad/sec = 1.4222 Hz\n\n(b) Y = 0.6:\nFrom E0 (10.19), Z / Y = 1.03 = r^2 / √((1 - r^2) + (2 rY)^2) = E\n(1.03)² {1 + r + 2 r² + 4 r³} = r^4\ni.e., 0.057404 r^4 - 0.56 r² + 1 = 0\ni.e., r² = 2.3535, 7.4018\ni.e., r = 1.5341, 2.7206\nBy selecting r = 2.7206,\nωn = 1000 π\n= 60 / 2.7206 = 19.2458 rad/sec\n= 3.06030 Hz\n\nThe quantity E attains maximum at\nr = 1 / √(1 - 2 r²) = 1.8898 for Y = 0.6. Hence we have\nto take r = 2.7206 to avoid peak of E.\n\n10.8\nS = t / 1000 m , ω = 4000 rpm = 418.88 rad/sec\nωn = √(g / S) = √(9.81 / 10) = 31.3209 rad/sec\nr = ω / ωn = 418.88 / 31.3209 = 13.3738\nError factor = r² / |1 - r²|\n= r² / |1 - 13.3738²|\n= 1.0056 10.9\nY = \\frac{r^2}{\\sqrt{(1-r^2)+(2rY)^2}} \\quad (E_1)\nMaximum of \\frac{Z}{Y} \\text{ occurs when } r = r^* = \\frac{1}{\\sqrt{1-2Y^2}} \\quad (see Eq. (3.84))\nFor \\text{ } Y=0.5, \\text{ } r^* = \\frac{1}{\\sqrt{0.5}} = 1.4142, \\text{ } (r^*)^2 = 2\n\\frac{Z}{Y} = \\frac{2}{\\sqrt{(1-2)^{2}+(2x0.5x1.4142)^{2}}} = \\frac{2}{\\sqrt{3}} = 1.1547\nWhen error is one percent, \\frac{Z}{Y} = 1.01 \\text{ or } \\frac{Y}{Z} = 0.9901\nEq. (E_1) can be rewritten as:\n\\left|\\frac{Z}{Y}\\right| = \\frac{(1-r^{2})^{2} + 4r^{3}r^{2}}{r^4}\n\\Rightarrow \\ r^{*} \\left(1 - \\left|\\frac{Z}{Y}\\right|^2 \\right) - r^2(2 + 4r^{2}) + 1 = 0 \\quad (E_2)\nFor \\frac{Z}{Y} = 0.9901 \\text{ and } Y = 0.5, (E_2) becomes \n0.0197 r^4 - r^2 + 1 = 0\n\\Rightarrow \\ r^{*} = 1.0203, 49.7411\n\\frac{Z}{Y} = 1.01 \\to 1.0527\nLowest frequency for one percent accuracy = 7.0527 \\text{ (5)} = 35.2635 Hz 10.10\nFrequency range > 100 Hz, \\text{ maximum error } = 2\\% \\nk = 4000 \\text{ N/m}, c = 0 \\Rightarrow T=0, m = ? \\nFor vibrometer with T=0, \\frac{Z}{Y} = \\frac{r^{2}}{\\sqrt{(r^{2}-1)^{2}+(2rY)^{2}}} \\text{ where } r^2 = \\frac{r^{2}}{|1-r|} = 1.02 \\text{ since } r \\text{ must be greater than one for higher frequencies}\\nor \\text{ } r^{2}=51, i.e., r=7.1414 10.18\n√( r² ) x 5 = 5.0294\nRecord indicated by vibrometer is given by\nx(t) = 21.0994 sin( 50t + 8.4934°) + 10.1331 sin (8πt + 4.0675°) + 5.0294 sin (12πt + 2.6905°) mm\nx(t) = -20 (50) sin 50t - 5 (150)² sin 150t mm/sec²\n = -50.000 sin 50t - 112500 sin 150t mm/sec²\nω_n = 100 rad/sec, ω₁ = ω₁ = √-1² = 80 ⇒ Y = 0.6\nr₁ = ω₁ / ω_n = 0.5; r₂ = ω₂ / ω_n = 1.5\nφ₁ = tan⁻¹( (2 Y r₁) / (1 - r₁²) ) = tan⁻¹( (2 × 0.25 × 0.5) / (1 - 0.25))\nφ₂ = tan⁻¹( (2 Y r₂)/(r - r₁) ) = tan⁻¹( (2 × 0.25 × 1.5)/(1 - 0.25)\n= 38.6589°\n50.00\n = 52.057.9206\n\n√( (r₁ - r₂)² + (2 Y r₁)² )\n\nr² = 112.500 = 51.335.6229\noutput of the accelerometer is given by\nx(t) = - 52.057.9206 sin (50t - 38.6589°)\n\t\t- 51.335.6229 sin (150t + 55.2222°) mm/sec² (E₃)\nIt can be seen that E₁ (E₃) is substantially different from E₂ (E₂). We have, from Eq. (10.19),\nZ = \n r² - 0.8312² \n | 1 - r¹|² \nY = Z/0.1232² = 40.5844 mm\nMax displacement of foundation = Y₁ = 40.5844 mm\nMax velocity of foundation = 6Y = 4249.9984 mm/sec\nMax acceleration of foundation = C²Y = 445059.8291 mm/sec²\n10.16\nMaximum speed = 1000 rpm = 50 Hz\nfor accelerometer, \n\t\t\t 1 \n\t( - ( r̃ - r²)² + (2Y r)² )\n = ——— \n\t\t √( - ( r̃ - r²)² + (2Y r)² )\n\nHere C = 20 N/m-s², Y = ——— = 0.8192 \n\t\t\t\t 2\nto100 rpm (E₁) gives \n\n\t\t ∴ = 24.3962 N-m\n\t\t\t\t \n 1/2m = 2.0/3.819 = 0.25\n\nk = m ω² => .01941 kg = 19.41 grams\nk = m = .01941 (100 × 2π) = 7622.7967 N/m 10.17\nG₁ = 2π(0.5) = rad/sec; ω₁ = ω₁/ω_n = 4; ω₂ = C₄/ω_n = 8; ω₃ = C₃/ω_n = 12\n\nφ₁ = tan⁻¹( (2 Y r₁) / (1 - r₁²) ) = tan⁻¹( (2(0.218)4) / .1 = -3.4934°\nφ₂ = tan⁻¹( (2 Y r₂) / (1 - r₁²) ) = tan⁻¹( (2 × 0.28 × 8) /(1 - .64)\n = -4.0675°\nφ₃ = tan⁻¹( (2 Y r₃) / (1 - r₁²) ) = tan⁻¹( (2 × 0.28 × 12) /(1 - .144)\n = -2.6905°\n\nr²\n√( - ( r̃ - r₁)² + (2 Y r)² )\n\n x20 = 21.0994\n\n√( ( r - r₁)² + (2 Y r₁²) )\nr² = x10 = 10.1331 10.19\nFor given beam,\nA = (1.6 × 25) mm² = 40 × 10⁻⁶ m²,\nI = 1/12 (0.025) (0.0016)³ = 8.53333 × 10⁻¹² m⁴,\ne = 0.05 m to 0.25 m.\nFor a cantilever beam, Fig. 8.15 gives ω₀ = (β₁, β₂) ² (E I / (ρ A e²))^{1/2}\nwhere (β₁, β₂)² = (1.875101)² = 3.516015\n(β₂)² = (4.694091)² = 22.03449\n(β₃)² = (7.854757)² = 61.69721\n(β₄)² = (10.995514)² = 120.90192\nFor spring steel, E = 200 × 10⁹ Pa, ρ = 7800 kg/m³\nE I / (ρ A e²)^{1/2} = (200 × 10⁹) (8.53333 × 10⁻¹²)\n(7800) (40 × 10⁻⁶) e² = 2.33882\nω₀ = (β₁, β₂)² / e²\nThe first four frequencies are given below:\n\n e = 0.05 m \n ω₁ = 3289.33 ω₂ = 20613.88 ω₃ = 57719.47 ω₄ = 113107.13\n e = 0.25 m \n ω₁ = 131.573 ω₂ = 824.6996 ω₃ = 2308.78 ω₄ = 4524.29\n\nHence, the range of frequencies that can be measured is given by ω > 131.573 rad/sec.\nHowever, for first mode only (which is easiest to excite), the range of frequencies is\n131.573 rad/sec ≤ ω ≤ 3289.33 rad/sec. 10.20\n\nk X R\nF0 = - 1 - r2\n(1 - r2)2 + (2 s) r2 N D (assume)\n\ndN dr - N dD dr = 0 ie, D dN dr - N dD dr = 0\n\nor {\n(1 - r2)2 + (2 s) r2}(-2 z r) - (1 - r2)2[2 (1 - r2) + 2 (2 s) r2] = 0\n\nThis equation can be simplified as\nr4 - 2 r2 + (1 - 4 s2) = 0\n\nand its solution is given by\nr2 = 1 ± 2 s or r = √1 + 2 s ; √1 - 2 s\n\nSince\n\nk X R\nF0 = - 1\n4 s (1 - s)\n(1)\n\nand\n\nk X R\nF0 = 1\n4 s (1 - s)\n(2)\n\nwe note that r = R1 = √1 - 2 s corresponds to a maximum and r = R2 = √1 + 2 s corresponds to a minimum of X R.\n\n637