EML 6223: Mechanical Vibrations EGN 4323: Vibration Synthesis Analysis Isaac Elishakoff Lecture #16 Spring 2025 Topic 67: Galerkin’s Method Beam’s bending; governing differential equation is: π2 π2π€ πΈ π₯ πΌ π₯ = ππ¦ (π₯) ππ₯ 2 ππ₯ 2 The solution of this ODE may prove very difficult if EI is not constant. Question: C’mon, why do we need now a complication? Answer: In modern times, to save material, tapered beams, plates and shells are used. Example: An airplane’s wing is not of uniform thickness or form. L#16 - 2 Topic 67: Galerkin’s Method π π2π€ πΈπΌ π₯ = ΰΆ± ππ¦ π₯ ππ₯ + πΆ1 ππ₯ ππ₯ 2 π2π€ πΈπΌ π₯ = ΰΆ± ΰΆ± ππ¦ π§ ππ§ ππ¦ + πΆ1 π₯ + πΆ2 ππ₯ 2 π2π€ 1 πΆ1 π₯ + πΆ2 = ΰΆ΅ … ππ₯ππ¦ + ππ₯ 2 πΈ π₯ πΌ(π₯) πΈ π₯ πΌ(π₯) Boris Galerkin 1871-1945 In 1915 Galerkin, Russia suggested the following trick: Galerkin B.G., Sterzhni i plastiny. Ryady v nekotorykh voprosakh uprugogo ravnovesiya sterzhnei i plastin (Rods and Plates Series Occurring in Some Problems of Elastic Equilibrium of Rods and Plates), Vestnik Inzhenerov i Tekhnikov, Petrograd, Vol. 19, 897-908, 1915 (in Russian). (English Translation: 63-18925, Clearinghouse Fed. Sci. Tech. Info). L#16 - 3 Topic 67: Galerkin’s Method π€ π₯ = πΌ β π(π₯) is approximated π(π₯) must satisfy all boundary conditions Let us substitute πΌπ(π₯) into our differential equation. π2 π2π πΌ 2 πΈ π₯ πΌ π₯ = ππ¦ π₯ ππ₯ ππ₯ 2 ? Two things may happen: (a) Left Side = Right Side → You are done, you have the solution! (b) Note: Usually this doesn’t happen; an “error” is created π2 π2π πΌ 2 πΈ π₯ πΌ π₯ − ππ¦ π₯ = π π₯ ππ₯ ππ₯ 2 πππ πππππ L#16 - 4 Topic 67: Galerkin’s Method Let us make this error not to be seen from the π(π₯) direction. _____________________________________________________________________________ Orthogonality of error to π(π₯) error ⊥ π(π₯) inner product must = 0 πΏ ΰΆ± πππππ π π₯ ππ₯ = 0 0 L#16 - 5 Topic 67: Galerkin’s Method πΏ ΰΆ± 0 π2 π2π πΌ 2 πΈ π₯ πΌ π₯ − ππ¦ π₯ ππ₯ ππ₯ 2 π π₯ ππ₯ = 0 We don’t “see” the error from π(π₯) direction, as it were, πΏ π2 π2π πΌΰΆ± πΈ π₯ πΌ π₯ π π₯ ππ₯ − ΰΆ± ππ¦ π₯ π π₯ ππ₯ = 0 2 2 ππ₯ ππ₯ 0 π³ β«ππ π π π ππ πΧ¬β¬ πΆ= π π π(π) π³ π π π¬ π π° π π π π π β«πΧ¬β¬ π ππ π ππ L#16 - 6 Topic 68: Petrov’s Method Suggested in 1940; Galerkin method was suggested in 1915. Petrov G.I., Application of Galerkin’s Method to the Problem of Stability of Flow of a Viscous Fluid, PMM: Journal of Applied Mathematics and Mechanics, Vol. 4 (3), 3-12, 1940 (in Russian). ________________________________________________ ππ¦ (π₯) Example: Clamped Simply Supported Boundary conditions are arbitrary. In Galerkin’s method, we utilized the function π(π₯) as an approximating function. 1912 – 1987 Georgii Ivanovich Petrov (on His 100th Birthday), Fluid Mechanics, Vol. 47(3), 289-291, 2012. L#16 - 7 Topic 68: Petrov’s Method Note: For Galerkin’s method, as well as for Petrov method trial, approximate, comparison functions MUST satisfy all boundary conditions: conditions at x=0 & x=L. ______________________________________________________________________________ π€ ≈ πΌπ(π₯) π2 π2π€ πΈ π₯ πΌ π₯ = ππ¦ (π₯) ππ₯ 2 ππ₯ 2 error = difference between LHS & RHS π2 π 2 πΌπ(π₯) π π₯ = 2 πΈ π₯ πΌ π₯ − ππ¦ (π₯) ππ₯ ππ₯ 2 In Galerkin’s method, we demanded the inner product of two functions π(π₯) & π(π₯) to be zero. L#16 - 8 Topic 68: Petrov’s Method For vectors: π΄Τ¦ = π΄π₯ πΤ¦ + π΄π¦ πΤ¦ + π΄π§ π π΅ = π΅π₯ πΤ¦ + π΅π¦ πΤ¦ + π΅π§ π π΄Τ¦ β π΅ = π΄π₯ π΅π₯ + π΄π¦ π΅π¦ + π΄π§ π΅π§ Here, with two functions: πΏ ΰΆ± π π₯ π π₯ ππ₯ = 0 0 Petrov (1940): We don’t have to multiply by π(π₯). If I have another set of functions, say π π₯ , I have the freedom to multiply by π π₯ . L#16 - 9 Topic 68: Petrov’s Method This is like being invited to two weddings that take place at the same time and the same place. πΏ ΰΆ± π π₯ π π₯ ππ₯ = 0 0 πΏ π2 π 2 πΌπ π₯ ΰΆ± πΈ π₯ πΌ π₯ − ππ¦ π₯ 2 2 ππ₯ ππ₯ 0 π π₯ ππ₯ = 0 π³ β«ππ π π π ππ πΧ¬β¬ πΆ= π π π(π) π³ π π π¬ π π° π π π β«)π(π πΧ¬β¬ π ππ π ππ Once I found πΌ, then π€(π₯) ≈ πΌπ(π₯) We can write them down differently: πΏ πΏ ΰΆ± πφ π₯ π π₯ ππ₯ = 0 ΰΆ± ππ π₯ π π₯ ππ₯ = 0 0 0 L#16 - 10 Topic 68: Petrov’s Method With the Petrov method, we have a possibility of two versions. If we have two sets of functions π(π₯) & π(π₯) that each satisfies all boundary conditions, then we can perform two versions of the Galerkin method πΏ ΰΆ± ππ π₯ π π₯ = 0 πΌ 0 πΏ ΰΆ± ππ π₯ π π₯ = 0 (πΌπΌ) 0 and two versions of the Petrov method: πΏ ΰΆ± ππ π₯ π π₯ ππ₯ = 0 πΌπΌπΌ 0 πΏ ΰΆ± ππ π₯ π π₯ ππ₯ = 0 0 πΌπ L#16 - 11 Topic 68: Petrov’s Method Question: What is the relationship between the functions π(π₯) & π(π₯)? Answer: None. Question: What is the relationship between π€(π₯) & π(π₯)? Answer: π(π₯) is drawn from functions satisfying all boundary conditions. The expression "you can't dance at two weddings with one pair of legs" means you can only do one thing at a time. You must choose one of two things to do at that time. The expression "you can't have your cake and eat it too" means that you cannot use the same limited resource for two purposes. Of course, we can dance at two weddings, in special cases. L#16 - 12 Example – 40 [Galerkin’s One-Term Method] π₯ A S-C beam is subjected to a triangularly distributed load π = 2 πΏ ππ over a span of πΏ = 5π. The modulus of elasticity and area moment of inertia in the beam varies as a function of length as given: πΌ π₯ = 10−5 ⋅ π₯ 3 π4 , πΈ π₯ = 200 ⋅ 1 + sin ππ₯ πΏ πΊππ Find the displacement π€(π₯) of the beam by a single term Galerkin’s approximation, using MAPLE. Also, depict the displacement plot. Use π π₯ = 2π₯ 4 − 3π₯ 3 πΏ + π₯πΏ3 π as an approximation function in S-C beam. L#16 - 13 Example – 40 [Galerkin’s One-Term Method] Solution L#16 - 14 Example – 40 [Galerkin’s One-Term Method] Solution We got: We have: πΌ = 0.00016426 π π₯ = 2π₯ 4 − 3π₯ 3 πΏ + π₯πΏ3 ∴π€ π₯ =πΌ⋅π π₯ π€ π₯ = 0.00016426 ⋅ 2π₯ 4 − 3π₯ 3 πΏ + π₯πΏ3 L#16 - 15 Example – 41 [Petrov’s One-Term Method] Let us solve the same problem with additional approximation function: π π₯ = π₯ 5 − 2π₯ 3 πΏ2 + π₯πΏ4 L#16 - 16 Example – 41 [Petrov’s One-Term Method] Solution We got: We have: πΌ = 0.00015266 π π₯ = 2π₯ 4 − 3π₯ 3 πΏ + π₯πΏ3 ∴π€ π₯ =πΌ⋅π π₯ π€ π₯ = 0.00015266 ⋅ 2π₯ 4 − 3π₯ 3 πΏ + π₯πΏ3 L#16 - 17 Topic 69: Galerkin’s Method – 2 Term Approximation π2 ππ₯ 2 π2 π€ πΈπΌ ππ₯ 2 = ππ¦ (π₯) static displacement problem We said π€ ≈ πΌπ(π₯) (1 term approximation) In some problems, the 1 term approximation can result in higher errors. Therefore, there is a need for improvement. π€ ≈ πΌ1 π1 π₯ + πΌ2 π2 (π₯) (two-term approximation) πΌ1 and πΌ2 are unknown coefficients. π1 (π₯) & π2 (π₯) are known functions, chosen by the engineer himself/herself & satisfy ALL boundary conditions. We substitute into the equation: π2 π2π€ πΈπΌ 2 − ππ¦ π₯ = 0 ππ₯ 2 ππ₯ L#16 - 18 Topic 69: Galerkin’s Method – 2 Term Approximation We substitute into the equation: π2 π2π€ πΈπΌ 2 − ππ¦ π₯ = 0 ππ₯ 2 ππ₯ π2 π2 πΈπΌ 2 πΌ1 π1 π₯ + πΌ2 π2 (π₯) ππ₯ 2 ππ₯ − ππ¦ (π₯) This should be zero, but it is not. I call it error πΊ(π). _________________________________________________________________________ I should not see this error from the “direction” of π1 (π₯) & π2 (π₯). Neater notation (I): π π₯ , π1 (π₯) = 0 , = πππππ, πππ‘ πππππ’ππ‘ πΏ ππ, ΰΆ± π π₯ π1 π₯ ππ₯ = 0 0 L#16 - 19 Topic 69: Galerkin’s Method – 2 Term Approximation Neater notation (II): π π₯ , π2 (π₯) = 0 πΏ ππ, ΰΆ± π π₯ π2 π₯ ππ₯ = 0 0 ______________________________________________________________________________ πΏ π2 π2 (πΌ): ΰΆ± π1 (π₯) 2 πΈπΌ ππ₯ 2 πΌ1 π1 π₯ + πΌ2 π2 (π₯) ππ₯ 0 − ππ¦ (π₯) ππ₯ = 0 πΏ π2 π2 (πΌπΌ): ΰΆ± π2 (π₯) πΈπΌ 2 πΌ1 π1 π₯ + πΌ2 π2 (π₯) 2 ππ₯ ππ₯ 0 − ππ¦ (π₯) ππ₯ = 0 πΏ πΏ π2 π 2 π1 (π₯) π2 π 2 π2 (π₯) πΌ1 ΰΆ± π1 (π₯) 2 πΈπΌ ππ₯ + πΌ2 ΰΆ± π1 (π₯) 2 πΈπΌ ππ₯ − ΰΆ± ππ¦ π₯ π1 π₯ ππ₯ = 0 2 2 ππ₯ ππ₯ ππ₯ ππ₯ 0 0 L#16 - 20 Topic 69: Galerkin’s Method – 2 Term Approximation πΏ πΏ π2 π 2 π1 (π₯) π2 π 2 π2 (π₯) πΌ1 ΰΆ± π2 (π₯) 2 πΈπΌ ππ₯ + πΌ2 ΰΆ± π2 (π₯) 2 πΈπΌ ππ₯ − ΰΆ± ππ¦ π₯ π2 π₯ ππ₯ = 0 2 2 ππ₯ ππ₯ ππ₯ ππ₯ 0 0 ______________________________________________________________________________ πΌ1 π΄11 + πΌ2 π΄12 = π΅1 α πΌ1 π΄21 + πΌ2 π΄22 = π΅2 πΏ π2 π 2 π1 (π₯) π΄11 = ΰΆ± π1 (π₯) 2 πΈπΌ ππ₯ 2 ππ₯ ππ₯ 0 πΏ π2 π 2 π2 (π₯) π΄12 = ΰΆ± π1 (π₯) 2 πΈπΌ ππ₯ 2 ππ₯ ππ₯ 0 πΏ π2 π 2 π1 (π₯) π΄21 = ΰΆ± π2 (π₯) 2 πΈπΌ ππ₯ 2 ππ₯ ππ₯ 0 L#16 - 21 Topic 69: Galerkin’s Method – 2 Term Approximation πΏ π2 π 2 π2 (π₯) π΄22 = ΰΆ± π2 (π₯) 2 πΈπΌ ππ₯ 2 ππ₯ ππ₯ 0 πΏ π΅1 = ΰΆ± ππ¦ π₯ π1 π₯ ππ₯ 0 πΏ π΅2 = ΰΆ± ππ¦ π₯ π2 π₯ ππ₯ 0 By Maple π΅1 π΄12 π΅ π΄22 πΌ1 = 2 π΄11 π΄12 π΄21 π΄22 π€ π₯ ≈ πΌ1 π1 π₯ + πΌ2 π2 (π₯) π΄11 π΅1 π΄ π΅2 πΌ2 = 21 π΄11 π΄12 π΄21 π΄22 approximation for π€(π₯) L#16 - 22 Topic 70: Use of the Galerkin Method in Vibrations Some students have used the Galerkin’s method in their project. πΈ π₯ πΌ π₯ = πΈ0 πΌ0 1 + ππ(π₯) Note: Both π and π(π₯) are given. Example: ππ₯ πΈ π₯ = πΈ0 1 + 0.1π ππ πΏ No exact solution exists for the natural frequency of such a beam. π = ππππ π‘πππ‘, π΄ = ππππ π‘πππ‘, πΌ = ππππ π‘πππ‘ L#16 - 23 Topic 70: Use of the Galerkin Method in Vibrations Governing differential equation: π2 π2π€ π2π€ πΈ π₯ πΌ(π₯) 2 + ππ΄ 2 = 0 ππ₯ 2 ππ₯ ππ‘ π€ π₯, π‘ = π π₯ sin(ππ‘) π2 π2π πΌ 2 πΈ(π₯) − ππ΄π2 π = 0 2 ππ₯ ππ₯ We approximate a 1 term approximation. π π₯ = πΌ1 π1 (π₯) π2 π 2 π1 (π₯) πππππ π₯ = πΌπΌ1 2 πΈ(π₯) − ππ΄πΌ1 π2 π1 (π₯) 2 ππ₯ ππ₯ L#16 - 24 Topic 70: Use of the Galerkin Method in Vibrations We demand error ⊥ π1 (π₯) π³ ΰΆ± πππππ π ππ π π π = π π πΏ π2 π 2 π1 π₯ ΰΆ± πΌπΌ1 2 πΈ π₯ ππ₯ ππ₯ 2 0 − ππ΄πΌ1 π2 π1 π₯ π1 π₯ ππ₯ = 0 πΏ π2 π 2 π1 π₯ πΌ1 ΰΆ± πΌ 2 πΈ π₯ ππ₯ ππ₯ 2 0 − ππ΄π2 π1 π₯ π1 π₯ ππ₯ = 0 πΌ1 ≠ 0, otherwise π(π₯) ≡ 0 But we need π(π₯) β’ 0 L#16 - 25 Topic 70: Use of the Galerkin Method in Vibrations πΏ πΏ π2 π 2 π1 (π₯) ΰΆ± π1 π₯ πΌ 2 πΈ(π₯) ππ₯ − ππ΄π2 ΰΆ± π12 π₯ ππ₯ = 0 2 ππ₯ ππ₯ 0 0 π2 = πΏ π2 β«Χ¬β¬0 π1 π₯ πΌ 2 ππ₯ π 2 π1 (π₯) πΈ(π₯) ππ₯ ππ₯ 2 πΏ ππ΄ β«Χ¬β¬0 π12 π₯ ππ₯ Final result for natural vibrations of elastic modulus, for 1st natural frequency. πΈ π₯ = πΈ0 ππ₯ 1 + ππ ππ πΏ ππ₯ π1 π₯ = π ππ πΏ Exact mode shape of uniform beam L#16 - 26 Topic 70: Use of the Galerkin Method in Vibrations π2 = πΏ ππ₯ π 2 β«Χ¬β¬0 π ππ πΏ πΌ 2 ππ₯ ππ₯ πΈ0 1 + ππ ππ πΏ πΏ ππ΄ β«Χ¬β¬0 π2 ππ₯ − 2 π ππ πΏ πΏ ππ₯ ππ₯ 2 π ππ πΏ ππ₯ π2 ππ₯ ππ₯ π2 ππ₯ ππ₯ 2 π2 ππ₯ πΈ 1 + ππ ππ π ππ = 2 πΈ0 π ππ + πΈ0 π π ππ = πΈ0 − 2 π ππ + πΈ0 π ππ₯ 2 0 πΏ πΏ ππ₯ πΏ πΏ πΏ πΏ Final Result: π2 = π12 |(1 + π β π πππ ππ’ππππ) πΈ = πΈ0 πΌ = πΌ0 When π = 0 If the beam is of constant cross-section & uniform, then we get the exact solution π2 = π12 . Galerkin method hits the exact solution if π is small, small error like 2%, 3%, or 4%. L#16 - 27 Topic 70: Use of the Galerkin Method in Vibrations Two-term Galerkin approximation π2 π2π πΈπΌ − ππ΄π2 π = 0 2 2 ππ₯ ππ₯ π = πΌ1 π1 π₯ + πΌ2 π2 (π₯) πΌ1 , πΌ2 = unknown constants π1 (π₯) & π2 (π₯) are known functions, chosen by the investigator that is conducting the analysis Substitute into the differential equation & get an error: π2 π2 2 2 π (π₯) πππππ π₯ = πΌ1 πΈπΌπ (π₯) − ππ΄π π (π₯) + πΌ πΈπΌπ π₯ − ππ΄π 1 1 2 2 2 ππ₯ 2 ππ₯ 2 Two requirements of orthogonality: L#16 - 28 Topic 70: Use of the Galerkin Method in Vibrations πΏ ΰΆ± πππππ π1 π₯ ππ₯ = 0 (πΌ) 0 πΏ ΰΆ± πππππ π2 π₯ ππ₯ = 0 (πΌπΌ) 0 πππππ π₯ = πΌ1 π΄11 π₯ + πΌ2 π΄12 (π₯) π2 π΄11 = 2 πΈπΌπ1 (π₯) − ππ΄π2 π1 (π₯) ππ₯ π2 π΄12 = 2 πΈπΌπ2 (π₯) − ππ΄π2 π2 (π₯) ππ₯ L#16 - 29 Topic 70: Use of the Galerkin Method in Vibrations πΏ πΏ πΌ1 ΰΆ± π1 π₯ π΄11 π₯ ππ₯ + πΌ2 ΰΆ± π1 π₯ π΄12 π₯ ππ₯ = 0 0 0 πΏ πΏ πΌ1 ΰΆ± π2 π₯ π΄11 π₯ ππ₯ + πΌ2 ΰΆ± π2 π₯ π΄12 π₯ ππ₯ = 0 0 0 π11 , π12 , π21 , π22 πΌ1 π11 + πΌ2 π12 = 0 α πΌ1 π21 + πΌ2 π22 = 0 2 homogeneous equations for 2 unknowns Homogeneous set to have a nontrivial solution (πΌ12 + πΌ22 ≠ 0); we need the determinant in front of πΌ1 & πΌ2 to be zero. L#16 - 30 Topic 70: Use of the Galerkin Method in Vibrations π11 π21 π12 =0 π22 π11 π22 − π12 π21 = 0 The equation above will be reducible to the following equation. π1 π4 + π2 π2 + π3 = 0 Quadratic equation for π2 or bi-quadratic equation π2 = −π2 ± π2 − 4π1 π3 2π1 L#16 - 31 Topic 70: Use of the Galerkin Method in Vibrations −π2 − π2 − 4π1 π3 π1 = 2π1 Smallest/1st frequency −π2 + π2 − 4π1 π3 π2 = 2π1 Biggest/2nd frequency L#16 - 32 Example – 42 [Galerkin’s One Term Approximation] A S-C beam with a span of πΏ = 5π has following parameters: π₯ π΄ π₯ = 0.1 ⋅ 0.25 + πΏ π2 , πΌ π₯ = 10−8 ⋅ π₯ 3 π4 , ππ₯ π π₯ = 4000 ⋅ 2 + sin πΏ πΈ π₯ = 200 ⋅ 1 + sin ππ₯ πΏ ππ/π3 πΊππ Find the natural frequency ππ of the beam by a single term Galerkin’s approximation, using MAPLE. Use π π₯ = 2π₯ 4 − 3π₯ 3 πΏ + π₯πΏ3 π as an approximation function in S-C beam. Varying π΄, π, πΈ, πΌ w.r.t. π₯ L#16 - 33 Example – 42 [Galerkin’s One Term Approximation] Solution L#16 - 34 Example – 42 [Galerkin’s One Term Approximation] Solution ππ, ππ2 = 54.7193 ∴ ππ,πΊ1 = 7.3972 πππ π πππ The, natural frequency of the given beam is 7.3972 π ππ . L#16 - 35 Project Guidance L#16 - 36 Project Guidance L#16 - 37 Project Guidance L#16 - 38 Project Guidance L#16 - 39
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