# distributed loads handout - = 2 max L f r = F v • f(x...

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1 Slide Summary Sept. 22, 2004 Distributed forces Introduction to equilibrium problems Distributed forces (review) For distributed forces along & perp. to a straight line of length L • Let x = direction of line and y = direction of the force(s) • f(x) = “distributed force intensity” = force / length Resultants at x = 0: Location Q of single force resultant: •F rom : • We find: For discrete forces following similar rules: () () j F ˆ x f dx x d = v () j F ˆ 0 = = L x r dx x f v () k M ˆ 0 0 , = = L x r dx x xf v ( ) r OQ OQ OQ r OQ r z y x F k j i F r M v v v v + + = = ˆ ˆ ˆ 0 , () () = = = L x L x OQ dx x f dx x xf x 0 0 0 = OQ z anything y OQ = = = = F F i N i i N i i OA OQ F F x x 0 0 • “Q” is (2/3)L from where f(x) = 0 Location Q of single force resultant: Distributed forces: important results () j F ˆ 0 = = L x r dx x f v () k M ˆ 0 0 , = = L x r dx x xf v () ( ) = = L x L x OQ dx x f dx x xf x 0 0 For distributed forces applied along a line from x = 0 to L in (+/-) y-dir. f(x) = “distributed force intensity” …or for similar “discrete” forces: = = F F i N i i N i i OA OQ F F x x 0 0 Two special cases: • Constant f(x): () 0 f x f = j F ˆ 0 L f r = v 2 / L
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Unformatted text preview: = 2 / max L f r = F v • f(x) steadily increases or decreases from / to 0 Chapter 4: General Equilibrium Problems • Typical problem: Find a force, point of application of a force, couple, etc. • Steps: – Draw free body diagram(s) • Isolate subject(s) • Include all forces/ couples – “loads” – often, known – “reactions” – forces that constrain • Include as much information as possible: lines of action, points of application, etc. – See pp. 160-162 for reactions of common connections • Use action/reaction principle when applicable – Solve equilibrium equations: – Sometimes this will be enough For future discussion: static indeterminacy = ∑ = F N i i F v , = + ⊗ = ∑ ∑ ∑ = = = + C F i C F N i i N i i PA N i i P C F r M v v v v...
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