By Ranjan Ganguli

This e-book addresses the answer of rotating beam free-vibration difficulties utilizing the finite point procedure. It offers an advent to the governing equation of a rotating beam, earlier than outlining the answer techniques utilizing Rayleigh-Ritz, Galerkin and finite aspect tools. the opportunity of enhancing the convergence of finite aspect equipment via a really apt collection of interpolation features, that are towards the matter physics, is additionally addressed. The booklet bargains a priceless advisor for college kids and researchers engaged on rotating beam difficulties – very important engineering constructions utilized in helicopter rotors, wind generators, gasoline generators, steam generators and propellers – and their functions. it might probably even be used as a textbook for specialised graduate classes on complicated purposes of finite aspect analysis.

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Q1 . K18 ⎪ ⎪ ⎪ ⎪ . K28 ⎥ q ⎪ 2 ⎥⎪ ⎪ ⎪ . ⎥ . ⎪ ⎥⎪ ⎨ . ⎥ ⎥ . ⎥ . ⎥⎪ . ⎪ ⎪ ⎪ . ⎥ . ⎪ ⎥⎪ ⎪ . ⎦⎪ . ⎪ ⎪ ⎩ . K88 q8 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ + ⎧ Q1 ⎪ ⎪ ⎪ ⎪ Q ⎪ 2 ⎪ ⎪ ⎪ . ⎪ ⎪ ⎨ . = . ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ . ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ . ⎪ ⎪ ⎪ ⎩ ⎭ ⎪ Q8 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ Or in matrix form, Here, KTOT is singular because rigid body modes are present. These are zero frequency modes and prevent the matrix from being inverted. However, this equation is valid for a free−free rotating beam.

14. The mass matrix is given by, ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ mij = m ⎢ ⎢ ⎢ ⎢ ⎣ 13 l 35 11 2 l 210 9 l 70 11 2 l 210 1 3 l 105 13 2 l 420 9 l 70 13 2 l 420 13 l 35 −13 2 l 420 −1 3 l 140 −11 2 l 210 −13 2 ⎤ l 420 ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ −11 2 ⎥ ⎥ l 210 ⎥ ⎦ −1 3 l 140 l3 105 For example, l m13 = mH1 H3 dx 0 l = m 2 x l 3 −3 0 =m 9 l 70 Fig. 14 Element on the blade x l 2 +1 −2 x l 3 +3 x l 2 dx 44 1 Introduction The stiffness matrix is given by, ⎡ 12 l3 ⎢ ⎢ ⎢ 6 ⎢ l2 ⎢ kij = EI ⎢ ⎢ −12 ⎢ l3 ⎢ ⎣ 6 l2 ⎡ −mi 3 x 5 i + 6 l2 −12 l3 4 l −6 l2 −6 l2 12 l3 2 l −6 l2 ⎢ ⎢ ⎢ l l2 ⎢ 10 xi + 28 ⎢ 2⎢ ⎢ ⎢ −3 6l ⎢ 5 xi − 35 ⎢ ⎣ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥+ −6 ⎥ ⎥ l2 ⎥ ⎦ ⎢ ⎢ ⎢ 1 2 Ai ⎢ ⎢ 10 ⎢ 2 ⎢ −6 ⎢ 5l ⎢ ⎣ 4 l 6 5l 1 10 + l2 28 −3 x 5 i 3 − 6l 35 2 l x 30 i + l 105 −l x 10 i − l 28 −l x 10 i − l2 28 3 x 5 i + 6l 35 −l2 x 60 i − l3 140 2 −l2 70 ⎡ 2 l l x 10 i 6l 35 ⎤ 6 l2 1 10 −6 5l 2l 15 −1 10 −1 10 6 5l −l 30 −1 10 2l 15 ⎤ ⎥ ⎥ l ⎥ −l x − 140 ⎥ 60 i ⎥ ⎥ ⎥ ⎥ l2 ⎥ 70 ⎥ ⎦ l3 l2 x + 30 i 105 3 N Ai = 2 mj xj+1 − xj2 j=i To understand the origin of the Ai term, consider, (Fig.

K18 ⎪ ⎪ ⎪ ⎪ . K28 ⎥ q ⎪ 2 ⎥⎪ ⎪ ⎪ . ⎥ . ⎪ ⎥⎪ ⎨ . ⎥ ⎥ . ⎥ . ⎥⎪ . ⎪ ⎪ ⎪ . ⎥ . ⎪ ⎥⎪ ⎪ . ⎦⎪ . ⎪ ⎪ ⎩ . K88 q8 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ + ⎧ Q1 ⎪ ⎪ ⎪ ⎪ Q ⎪ 2 ⎪ ⎪ ⎪ . ⎪ ⎪ ⎨ . = . ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ . ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ . ⎪ ⎪ ⎪ ⎩ ⎭ ⎪ Q8 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ Or in matrix form, Here, KTOT is singular because rigid body modes are present. These are zero frequency modes and prevent the matrix from being inverted. However, this equation is valid for a free−free rotating beam.

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