- L 2 12 G K1 Wwwwww K2 Mwwwww M X For The Uniform Beam Shown Pivoted At The Centre Mass Of Beam M 24 Kg Length Of Be 1 (516.84 KiB) Viewed 8 times
L/2 12 G k1 WWWWWW k2 MWWWWW M X For the uniform beam shown, pivoted at the centre. Mass of beam m = 24 kg, length of be
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L/2 12 G k1 WWWWWW k2 MWWWWW M X For the uniform beam shown, pivoted at the centre. Mass of beam m = 24 kg, length of be
L/2 12 G k1 WWWWWW k2 MWWWWW M X For the uniform beam shown, pivoted at the centre. Mass of beam m = 24 kg, length of beam = 1m. Hanging mass M = 1 kg (a) Determine the flexibility matrix. (8 marks) (b) Formulate the matrix equation of motion in terms of flexibility matrix, mass and moment of inertia. (4 marks) (c) Use Matrix iteration to solve for the lowest natural frequency of the model. (8 marks) ki = 5 N/m, k2 = 10 N/m. As starting values use 1, 1 Use four iterations.