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Chapter 4, Problem 4/068 Your answer is partially correct. Try again Determine the magnitudes of the pin reactions at A, B, a

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TE To FBD of Beam .1.3.1/ 2.45, 1.15. 2.6 4250 lb. (+ n) ΣΜ: = 0; Tø(1.15) . T,(2.45) = 0 1.15 T๖-2.45 Tg = 0 Tp +Tg = 4250FBD of member AEC Cy 6.2 Ar TE 1357.639(1.3) 7.5 -235.324 lb A,--235.324 + 1357.639 1122.315 lb. =By FBD of member ADB To 40 4.9 2.6 Ax 1122.315+2892.361 4014.676 lb. = By= (+ n) Σ MA-0; By(7.5)-Bx(7.5 tan 400-To(4.9)-0 401From Eqn. (ii)A -Cx0 Су- 2532.476 lb A (2532.476)(1122.315)7 (2532.476)2 + (4014.676)2 (2532.476) (235.324)7 2770.023 lb 4746

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