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posted by  aash on 7/26/2008 12:21:10 PM  |  status: Live  

Determine the force in brace CE

Course Textbook Chapter Problem
N/A statics N/A N/A
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The 300 N door for a roof opening is hinged at corners A and B. The roof forms a 30° angle with the horizontal and the cover is maintained in a horizontal position by brace CE. The force in member CE acts along its axis. Determine the force in brace CE.

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posted by Juan Colorado on 9/3/2008 3:03:08 PM  |  status: Live
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aash's comment:
"Thanks man"
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The calculation is:
According to the diagram (next delivery)

TCE = TCE (CE / CE) = TCE(0.134 i + 0.500j + 0.0 k)/ (0.1342 + 0.5002 + 0.002 )1/2

TCE = 0.259TCE i + 0.966TCE j + 0.0 TCE k

W = -300N j

A = Ax i + Ay j + 0.0 k

B = Bx i + By j + 0.0 k

Forming the ecuations:

(I)...ΣFx=0: Ax + Bx + 0.259TCE= 0

(II)...ΣFy=0: Ay + By + 0.966TCE - 300  = 0

(III)...ΣFz =0:   Az = 0 

                  ΣMc = 0: (-0.5 i + 0.0j -0.4 k)X(-300j) + (-1.0 i)X(Axi + Ay) +

                                  + (-1.0 i + 0.0j -0.8 k)X(Bx i + By j) =

                               = 0.5(300)k - 0.4(300)i + (-Ay k+Az j)+ (-0.8Bx j - By k + 0.8By i)

separating:
(IV)...ΣMx = 0:      -0.4(300) + 0.8 By = 0
(V)...ΣMy = 0:      Az -0.8 Bx = 0
(VI)...ΣMz = 0:       0.50(300) - Ay - By = 0

Solving: By = 150N,  Ay = 0,  TCE = 155.3N,  Bx = 0,  Ax = -40.2N (on the contrary )

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posted by Juan Colorado on 9/3/2008 3:23:18 PM  |  status: Live
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aash's comment:
"thanks"
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