ME211solution_3 - In case of using method of joints the...

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In case of using method of joints the supporting reactions are not required. Joint B: 3 0; cos45 100 0 5 xB C B A FF F ⎛⎞ ⎜⎟ ⎝⎠ = ------① 4 0; sin45 500 0 5 yB C B A F + = = = b ------② Solving Eqs. ① and ② yields 286 BA Fl b = (T) 384 BC b = (T) Joint C: 0; 384cos45 0 xC A =− ° 271 CA b = (C) B C 0; 384sin 45 0 yy FC ° 271.43 y Cl =
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First, find reaction forces at pin A from the entire structure. 0; 0 xx FA == 0; (12) 4(8) 8(6) 0 Gy MA =− + + = 6.667 y A kN = Then, consider equilibrium equations from the sectioned part of the structure. (Cutting line goes through the members which we are interested in.) 0; 6.667(4) (2) 0 CK MF + = J 13.3 KJ Fk N = (T) To find BC F , move BC F to point C and resolve to x and y components. 2 () 5 BC x BC F F ⎛⎞ = ⎜⎟ ⎝⎠ , 1 5 BC y BC F F = 0; 6.667(4) ( ) (2) 0 KB + = C x 14.9 BC N = (C) 0 AC K
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First, find the reaction forces at point E. ( ) 0 x E = 0; 1000(10) 1000(20) 1000(30) (40) 0 Ay ME =− + = lb 1500 y E = Next, find the force in GJ by taking moment about point C.
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This homework help was uploaded on 04/04/2008 for the course ME 211 taught by Professor Thouless during the Fall '08 term at University of Michigan.

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ME211solution_3 - In case of using method of joints the...

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