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Unformatted text preview: MAE 101: Homework 4 Solution February 3, 2011 Problem 7.49 Determine the reactions at A and B . x B A L /2 L /2 y Let us break the load into two parts and use the area analogy. A B y x L 1 L 2 L 2 L 2 For Load L 1 ω ( x ) = 2 ω L x for (0 ≤ x ≤ L/ 2) For Load L 2 ω ( x ) = ω for L 2 ≤ x ≤ L Load 1 L 1 = Z L/ 2 2 ω L xdx = 2 ω L x 2 2 L/ 2 L 1 = ω L L 2 4 = Lω 4 using the area analogy, load L 1 acts 2/3 of the distance from the origin to L/2. Thus x 1 = L/ 3 Load 2 1 L 2 = Z L L/ 2 ω dx = ω x L L/ 2 = ω L 2 And from the area analogy, L 2 acts half way between L/2 and L. x 2 = 3 L 4 . x B A L /2 L /2 Now we can find the support reactions A x A y B y L 2 L 3 L 4 L 2 L 4 3 X F x : A x = 0 X F y : A y + B y Lω 4 Lω 2 = 0 X M A : B y L 2 Lω 4 L 3 Lω 2 3 L 4 = 0 Solving the third eqn. B y = Lω 6 + 3 Lω 4 = 11 12 Lω From the second eqn, A y + B y = 3 4 Lω Hence A y = 3 4 lω B y = Lω 6 A x = 0 A y = Lω / 6 B y = 11 Lω / 12 2 Problem 7.53 The aerodynamic lift of the wing is described by the distributed load w = 300 p 1 . 04 x 2 N/m . The mass of the wing is 27 kg, and its center of mass is located 2 m from the wing root R ....
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This note was uploaded on 11/06/2011 for the course MAE 101 taught by Professor Orient during the Spring '08 term at UCLA.
 Spring '08
 ORIENT
 Statics

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