Rods - United States Naval Academy
MECHANICAL ENGINEERING DEPARTMENT UNITED STATES NAVAL ACADEMY EM423 - INTRODUCTION TO MECHANICAL VIBRATIONS CONTINUOUS SYSTEMS β LONGITUDINAL VIBRATION IN RODS SYMBOLS Ο mass density AX Cross-sectional area u Longitudinal deflection from equilibrium position x Distance along the rod P Internal longitudinal (axial) force. Varies with position along the rod. INTRODUCTION This theory is applicable to longitudinal vibration in slender rods. With this type of vibration, very small amplitudes of motion can produce very large forces. Typical problems include damage to thrust bearings in ship propulsion systems, noise and damage caused by engine support props in helicopters, and radiated noise problems in submarines. ASSUMPTIONS 1. The rod is homogeneous and isotropic. 2. The material is within the elastic limit, and obeys Hooke's Law. 3. The rod is thin compared with its length. 4. Plane sections remain plane. 5. The lateral deflection caused by changes in the length of the rod is small. Rods - 1 THEORY Consider a small element of a rod that is vibrating axially. The element is at position x along the rod, where we assume that the left end of the rod is at x = 0, with positive x to the right. At a certain instant in time the element is displaced an amount u from the equilibrium position. P P x P+ dx x u+ dx u x dx u We first apply strength of materials relationships to the element. The element changes length by an amount ππππ ππππ ππππ strain = and so: (change in length) ππππ = (original length) ππππ From Hooke's Law, E = (stress) / (strain), and (stress) = (force) / (area), so Hence Rearrange and differentiate πΈπΈ = ππ (stress) οΏ½ οΏ½π΄π΄π₯π₯ οΏ½ = ππππ (strain) οΏ½ οΏ½ ππππ ππ ππππ = ππππ π΄π΄π₯π₯ πΈπΈ ππ 2 π’π’ ππππ = π΄π΄π₯π₯ πΈπΈ 2 πππ₯π₯ ππππ We now apply Newtonβs 2nd Law (f=ma) to the element. Resolve forces along the rod. Rods - 2 οΏ½ππ + Combining the two results yields: ππππ ππ2 π’π’ πΏπΏπΏπΏοΏ½ β ππ = (Οπ΄π΄π₯π₯ πΏπΏπΏπΏ) 2 ππππ πππ‘π‘ 1 ππ2 π’π’ ππ 2 π’π’ = πππ₯π₯ 2 ππ 2 πππ‘π‘ 2 With πΈπΈ ππ = οΏ½ = wave velocity ππ This is the same wave equation as derived for string vibrations, but with a different wave velocity. Using the same method to solve the equation, firstly separate the variables. Hence: π’π’(π₯π₯, π‘π‘) = ππ(π₯π₯). πΊπΊ(π‘π‘) ππππ ππππ with ππ(π₯π₯) = οΏ½π΄π΄1 π π π π π π οΏ½ οΏ½ + π΄π΄2 ππππππ οΏ½ οΏ½οΏ½ ππ ππ and πΊπΊ(π‘π‘) = π π π π π π (ππππ) ππππ ππππ π’π’(π₯π₯, π‘π‘) = ππ(π₯π₯). πΊπΊ(π‘π‘) = οΏ½π΄π΄1 π π π π π π οΏ½ οΏ½ + π΄π΄2 ππππππ οΏ½ οΏ½οΏ½ . π π π π π π (ππππ) ππ ππ COMMON BOUNDARY CONDITIONS For a FREE end the internal stress is zero, since there is nothing for the end of the rod to react against, and thus no axial force can be generated. therefore: For a FIXED end the displacement is zero. stress = πΈπΈ ππππ =0 ππππ ππππ ππππ π’π’ = 0 Rods - 3 NATURAL FREQUENCIES OF A FREE-FREE ROD, length = L ππππ ππππ π’π’(π₯π₯, π‘π‘) = ππ(π₯π₯). πΊπΊ(π‘π‘) = οΏ½π΄π΄1 π π π π π π οΏ½ οΏ½ + π΄π΄2 ππππππ οΏ½ οΏ½οΏ½ . π π π π π π (ππππ) ππ ππ At x = 0, ππππ ππππ = 0 for all time. Therefore ππππ ππππ ππ(π₯π₯) = οΏ½π΄π΄1 π π π π π π οΏ½ οΏ½ + π΄π΄2 ππππππ οΏ½ οΏ½οΏ½ ππ ππ ππππ ππ ππππ ππ ππππ = οΏ½π΄π΄1 οΏ½ οΏ½ ππππππ οΏ½ οΏ½ β π΄π΄2 οΏ½ οΏ½ π π π π π π οΏ½ οΏ½οΏ½ = 0 ππππ ππ ππ ππ ππ Therefore At x = L, ππππ ππππ π΄π΄1 = 0 = 0 for all time. Therefore ππππ ππ ππππ = οΏ½0 β π΄π΄2 οΏ½ οΏ½ π π π π π π οΏ½ οΏ½οΏ½ = 0 ππππ ππ ππ ππππ οΏ½ ππ This is only true for all time if π π π π π π οΏ½ = 0, so ππππ πΏπΏ = ππππ ππ with n = 1, 2, 3, β¦ Solving for the natural frequency in Hz: ππππ = ππππ ππ πΈπΈ = οΏ½ 2ππ 2πΏπΏ ππ with n = 1, 2, 3, β¦ Q: What are the first 3 natural frequencies of a steel rod that is 170 m long? Both ends are unrestrained. A: From the datasheet: πΈπΈ = 206 kN/mm2 = 206 × 109 N/m2 = 206 × 109 Pa ππ = 1738 kg/m3 ππππ = ππππ = 206 × 109 ππππ ππ πΈπΈ ππ οΏ½ = οΏ½ = = ππ × 32.02 Hz 2ππ 2πΏπΏ ππ 2 × 170 1738 n =1 gives f1 = 16 Hz n =2 gives f1 =48 Hz n =3 gives f1 = 80 Hz Rods - 4 NATURAL FREQUENCIES OF A FIXED-FREE ROD, length = L ππππ ππππ π’π’(π₯π₯, π‘π‘) = ππ(π₯π₯). πΊπΊ(π‘π‘) = οΏ½π΄π΄1 π π π π π π οΏ½ οΏ½ + π΄π΄2 ππππππ οΏ½ οΏ½οΏ½ . π π π π π π (ππππ) ππ ππ At x = 0, π’π’ = 0 for all time. Therefore ππππ ππππ ππ(π₯π₯) = οΏ½π΄π΄1 π π π π π π οΏ½ οΏ½ + π΄π΄2 ππππππ οΏ½ οΏ½οΏ½ = 0 ππ ππ Therefore At x = L, ππππ ππππ π΄π΄2 = 0 = 0 for all time. Therefore ππππ ππ ππππ = οΏ½π΄π΄1 οΏ½ οΏ½ ππππππ οΏ½ οΏ½ β 0οΏ½ = 0 ππππ ππ ππ ππππ οΏ½ ππ This is only true for all time if ππππππ οΏ½ = 0, so 1 ππππ πΏπΏ = οΏ½ππ β οΏ½ ππ 2 ππ Hence ππππ = ππππ οΏ½ππ β 1οΏ½2οΏ½ πΈπΈ = οΏ½ 2ππ 2πΏπΏ ππ with n = 1, 2, 3, β¦ with n = 1, 2, 3, β¦ Q: What are the first 3 natural frequencies of a steel rod that is 3 m long? One end is unrestrained, the other is fixed. A: From the datasheet: πΈπΈ = 206 kN/mm2 = 206 × 109 N/m2 = 206 × 109 Pa ππ = 1738 kg/m3 ππππ οΏ½ππ β 1οΏ½2οΏ½ πΈπΈ οΏ½ππ β 1οΏ½2οΏ½ 206 × 109 οΏ½ ππππ = = οΏ½ = = οΏ½ππ β 1οΏ½2οΏ½ × 1814.5 Hz 2ππ 2πΏπΏ ππ 2×3 1738 n =1 gives f1 = 907 Hz n =2 gives f1 =2722 Hz n =3 gives f1 = 4536 Hz Rods - 5 DISCUSSION 1. The solution for longitudinal vibrations in a rod is similar to flexural vibrations of a taught string, but with a different wave speed. 2. This means the same discussion applies to rod vibrations. The only difference is that, for graphical presentation, longitudinal displacements are usually drawn as if they were transverse. 3. The wave speed for longitudinal waves travelling in many common metals is almost the same. Calculate the wave speeds and fundamental natural frequencies for the following materials and complete the table. E (kN/mm2) Density (kg/m3) Longitudinal wave speed (m/s) Fundamental natural frequency for a 3m long rod (Hz) Steel 206 7843 Aluminum 70 2720 Brass 105 8410 Nickel Alloy 207 8580 Magnesium 45 1738 Rods - 6 EXAMPLES 1. A steel propulsion shaft on a ship is 25 m long. Its outside and inside diameters are 32 cm and 15 cm respectively. What are the first 2 natural frequencies of longitudinal vibration? Solve the problem 2 times. First, assume the gearbox and propeller are very light. Second, assume the gearbox is light, but the propeller is very heavy. L = 25 m; Ο = 7843 kg/m3; E = 206 kN/mm2 Solve for free-free boundary conditions. ππππ = So ππ1 = ππ2 = ππ πΈπΈ οΏ½ 2πΏπΏ ππ 206 × 109 1 οΏ½ = 102.5π»π»π»π» 7843 2 × 25 206 × 109 2 οΏ½ = 205.0π»π»π»π» 7843 2 × 25 Solve for fixed-free boundary conditions. ππππ = So ππ1 = οΏ½ππ β 1οΏ½2οΏ½ πΈπΈ οΏ½ 2πΏπΏ ππ οΏ½1 β 1οΏ½2οΏ½ 206 × 109 οΏ½ = 51.2π»π»π»π» 7843 2 × 25 οΏ½2 β 1οΏ½2οΏ½ 206 × 109 οΏ½ ππ2 = = 153.7π»π»π»π» 7843 2 × 25 2. What difference does it make to the previous question if you assume both the propeller and gearbox are very heavy? The solution is identical to that for a free-free rod. Why? Rods - 7 ASSIGNMENTS 1. Find, from first principles, the velocity of longitudinal waves along a thin steel rod. 2. The figure shows a schematic of a magnetostriction oscillator. The frequency of the oscillator is determined by L the length of the nickel alloy rod, which generates an alternating voltage in the surrounding coils equal to the frequency of the fundamental longitudinal vibration of the rod. Determine the full length, L, of the rod for a frequency of 20 kHz. 3. (extra credit) A uniform rod of length L and cross-sectional area Ax is fixed at the upper end and is loaded with a weight W=mg on the other end. Show that the natural frequencies of longitudinal vibration are determined from the equation: πΈπΈ π΄π΄π₯π₯ ππππππ πΈπΈ πππποΏ½ π‘π‘π‘π‘π‘π‘ οΏ½πππποΏ½ οΏ½ = ππ ππ ππ SOLUTIONS 1. Find, from first principles, the velocity of longitudinal waves along a thin steel rod. See the Course Handout. 2. The figure shows a schematic of a magnetostriction oscillator. The frequency of the oscillator is determined by the length of the nickel alloy rod, which generates an alternating voltage in the surrounding coils equal to the frequency of the fundamental longitudinal vibration of the rod. Determine the full length, L, of the rod for a frequency of 20 kHz. We consider half of the rod, with boundary conditions of fixed at one end (x = 0), and free at the other end (x = L/2). Note that in this situation we include only half of the total length of the rod. ππππ ππππ π’π’(π₯π₯, π‘π‘) = οΏ½π΄π΄1 π π π π π π οΏ½ οΏ½ + π΄π΄2 ππππππ οΏ½ οΏ½οΏ½ π π π π π π (ππππ) ππ ππ ππππ ππ ππππ ππ ππππ = οΏ½π΄π΄1 οΏ½ οΏ½ ππππππ οΏ½ οΏ½ β π΄π΄2 οΏ½ οΏ½ π π π π π π οΏ½ οΏ½οΏ½ π π π π π π (ππππ) ππππ ππ ππ ππ ππ Rods - 8 Boundary condition at x = 0, u = 0. The A2 = 0. At x = L/2, ππππ ππππ = 0 for all time. Therefore ππππ οΏ½ 2ππ This is only true if ππππππ οΏ½ From which = 0, so ππππ ππ ππππ = 0 = οΏ½π΄π΄1 οΏ½ οΏ½ ππππππ οΏ½ οΏ½ β 0οΏ½ ππππ ππ 2ππ ππππ πΏπΏ 1 = οΏ½ππ β οΏ½ ππ 2 2ππ Rearrange and solve for L with n =1. πΏπΏ = ππππ = with n = 1, 2, 3, β¦ ππππ οΏ½ππ β 1οΏ½2οΏ½ πΈπΈ = οΏ½ 2ππ πΏπΏ ππ οΏ½1 β 1οΏ½2οΏ½ πΈπΈ οΏ½1 β 1οΏ½2οΏ½ 206 × 109 οΏ½ οΏ½ = = 0.128 m = 128 mm 7843 ππ 20,000 ππ1 The total length of the rod must be 128.2 mm. 3. (extra credit) A uniform rod of length L and cross-sectional area Ax is fixed at the upper end and is loaded with a weight W=mg on the other end. Show that the natural frequencies are determined from the equation: πΈπΈ π΄π΄π₯π₯ ππππππ πΈπΈ πππποΏ½ π‘π‘π‘π‘π‘π‘ οΏ½πππποΏ½ οΏ½ = ππ ππ ππ The solutions to extra credit problems are available from your instructor. Rods - 9
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