Solution_05

# Solution_05 - CIV-100F Section B/E Solutions Fall 2011...

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CIV-100F Solutions Section B/E Fall 2011 Assignment 5 Due on Monday, October 17 th 2011 Draw a FBD if necessary; Pay attention to significant figures; A ruler is always required. 1- Four signs are mounted on a frame spanning a highway, and the magnitudes of the horizontal wind forces acting on the signs are as shown. For a=0.3m and b=3.6m, determine: (a) an equivalent force-couple system at B, (b) an equivalent force-couple system at C, and (c) a single force of resultant. Solution : (a) ∑∑ = = = kN F Z Y X 405 , 0 , 0 k B ) 405 ( = m M = + + + = 357 ) 50 )( 65 . 1 ( ) 160 )( 9 . 0 ( ) 105 )( 5 . 1 ( ) 90 )( 3 . 0 ( = + + + = 8 . 1530 ) 50 )( 75 . 6 ( ) 160 )( 35 . 4 ( ) 105 )( 65 . 1 ( ) 90 )( 6 . 3 ( 0 = m kN j i M B + = ) 1531 357 ( (b) = = = 405 , 0 , 0 C ) 405 ( = m kN j i j j i j M M B C = + = + = ) 1811 357 ( 3 . 3341 ) 8 . 1530 357 ( ) 5 . 1 75 . 6 )( 405 ( (c)Assume B is the origin of our coordinate, m F M x Y B 78 . 3 = = m F M y X B 881 . 0 = = 2- Two 150-mm-diameter pulleys are mounted on line shaft AD. The belts at B and C lie in vertical planes parallel to the y-z plane. Replace the belt forces shown with an equivalent force- couple system at A. Solution : 1.5m x y F R

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CIV-100F Solutions Section B/E Fall 2011 Assignment 5 Due on Monday, October 17 th 2011 Draw a FBD if necessary; Pay attention to significant figures; A ruler is always required. Finding the resultant of these four forces is actually a 2D problem. In y-z plane we have: F (145kN)(cos20° ) (240kN)(cos10° ) G (240kN)(sin10° ) 215kN (145kN)(sin20° ) (155kN)(cos10° ) E (155kN)(sin10° ) H y F1 F2 F3 F4 z kN F Z 41 . 339 ) 10 )(cos 155 ( ) 10 )(cos 240 ( 0 ) 20 )(sin 145 ( 4 3 2 1 = ° ° + ° = + + + = Y 87 . 419 ) 10 )(sin 155 ( ) 10 )(sin 240 ( 215 ) 20 )(cos 145 ( 4 3 2 1 = ° ° ° = + + + = = 0 X Then we have the resultant R is kN k j R ) 339 419 ( = Next, we take moment of each force about point A. m kN k j i k j i F r M A H F A = ° ° ° ° = × = ) 675 . 30 158 . 11 875 . 10 ( ) 20 )(sin 145 ( ) 20 )(cos 145 ( 0 ) 20 )(cos 075 . 0 ( ) 20 )(sin 075 . 0 ( 225 . 0 1 / 1 / m k i j r M A = = × = ) 375 . 48 125 . 16 ( 0 215 0 075 . 0 0 225 . 0 2 / 2 / G + = ° ° ° ° = × = ) 754 . 18 36 . 106 000 . 18 ( ) 10 )(cos 240 ( ) 10 )(sin 240 ( 0 ) 10 )(sin 075 . 0 ( ) 10 )(cos 075 . 0 ( 45 . 0 3 / 3 / E + = ° °
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## This note was uploaded on 12/10/2011 for the course CIV 100 taught by Professor Nahrvar during the Spring '08 term at University of Toronto- Toronto.

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Solution_05 - CIV-100F Section B/E Solutions Fall 2011...

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