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HW12-Solution

# HW12-Solution - 73 AC AB F F 128 kN AB F 128.0 kN AB F T...

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HW12 - Solution Terry Cooke: Cal Poly San Luis Obispo

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Chapter 6, Solution 9 Free body: Truss 0: 0 x x F A Due to symmetry of truss and load 1 total load 21kN 2 y A H Free body: Joint A : 15.3 kN 37 35 12 AC AB F F 47.175 kN 44.625 kN AB AC F F 47.2 kN AB F C 44.6 kN AC F T Free body: Joint B : From force polygon: 47.175 kN, 10.5 kN BD BC F F 10.50 kN BC F C 47.2 kN BD F C Free body: Joint C : 3 0: 10.5 0 5 y CD F F 17.50 kN CD F T 4 0: (17.50) 44.625 0 5 x CE F F 30.625 kN CE F 30.6 kN CE F T
PROBLEM 6.9 (Continued) Free body: Joint E : DE is a zero-force member 0 DE F Truss and loading symmetrical about c L

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Chapter 6, Solution 28 Free body: Truss 0: 0 x x F H 0: 48(16) (4) 0 192 kN H M G G 0: 192 48 0 y y F H 144 kN 144 kN y y H   H Zero-Force Members : Examining successively joints C , B , E , D , and F , we note that the following are zero-force members: BC , BE , DE , DG , FG Thus, 0 BC BE DE DG FG F F F F F Also note: AB BD DF FH F F F F (1) AC CE EG F F F (2) Free body: Joint A : 48 kN 8 3 73
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Unformatted text preview: 73 AC AB F F 128 kN AB F 128.0 kN AB F T 136.704 kN AC F 136.7 kN AC F C From Eq. (1): 128.0 kN BD DF FH F F F T From Eq. (2): 136.7kN CE EG F F C Free body: Joint H 144 kN 9 145 GH F 192.7 kN GH F C Also 144 kN 8 9 FH F 128.0 kN (Checks) FH F T Chapter 6, Solution 34 ( a ) Truss of Problem 6.23 : Joint : IJ FB I F : Joint : GJ FB J F : Joint : GH FB G F The zero-force members, therefore, are , , GH GJ IJ ( b ) Truss of Problem 6.28 : Joint : BC FB C F : Joint : BE FB B F : Joint : DE FB E F : Joint : DG FB D F : Joint : FG FB F F The zero-force members, therefore, are , , , , BC BE DE DG FG...
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