hw solutions – chapter 6

hw solutions – chapter 6 - T CD = 2.443 kN T...

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Homework Solutions – Chapter 6 6.1 T AB = -625 lb T AC = 375 lb T BC = 1000 lb T BD = -625 lb T CD = 375 lb 6.2 T AB = -500 lb T AC = 0 lb T BC = 625 lb T BD = -375 lb T CD = 625 lb T CE = 0 lb T DE = -500 lb 6.3 Safety Factor = 3.20 for 1000 lb load (weakest element – is in compression in this case: T AB and T BD ) 6.4 Largest force that may be applied is 3200 lb 6.5 Safety Factor = 8.00 for 1000 lb load (weakest element – is in compression in this case: T AB and T DE ) 6.6 Largest force that may be applied is 8000 lb 6.7 T AB = -2.175 kN T AC = 1.088 kN T BC = -1.288 kN T BD = -0.4434 kN
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Unformatted text preview: T CD = 2.443 kN T CE-0.7783 kN T DE = -2.443 kN 6.8 T DF = -3.657 kN T CE = 3.753 kN 6.9 T AB = 3.61 kN T AC = -2.00 kN T BC = 0 kN T BD = 3.61 kN T CD = -4.00 kN T CE 0 kN T CF = -2.00 kN T DE = 3.61 kN T EF = 3.61 kN 6.10 T AB = 4.00 kN T AC = 0 kN T BC = -7.21 kN T BD = 6.00 kN T CD = -5.00 kN T CE-6.00 kN T DE = 7.21 kN 6.17 By inspection: T BD = T DE = 0 T AB = T BC = P/2 T AD = -(Q+(P/2))(2/√2) T EF =T CE = P/2 T DF = -(Q+P)(2/√2) T CD = -(P/√2)...
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This note was uploaded on 11/08/2011 for the course ENGR 111 A taught by Professor Reddy during the Spring '10 term at Texas A&M.

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