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InClassLecture20

# InClassLecture20 - 7-33 Determine all ferees acting on...

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Unformatted text preview: 7-33* Determine all ferees acting on member BED ef the linkage Shawn in Fig. P?-33. HDLUTIDH Member AC is 3 tun-farce member: therefereI the line ef aetien ef feree C313 kneun 33 3heun_en the freerbedy diagram fer member BEE: H + C 333 = 3 333 33” {3.31 - 33 333 33a 13.31 = 3 3 = 33.33 13 3 33.3 13 C 3 33.3 13 3 33" 333. + —+ Erx = 3 + 33 333 33D - 33.33 333 33” = 3 I B = 25.93 lb B Eﬁ.ﬂ 1h ‘* + T 331' = 33 - 33.33 313 33° + 33 313 33“ = 3 3? = 33.33 13 3 33.3 13 3 3 = J13ﬂ13 + 13?}2 = £133.33}2 + 133.33}2 = 33.33 13 E _ -1_3_ -13D.33_ _ 3 3B - 333 33 - 333 35*93 - 33.3 3 3 33.3 13 3 33.3” 333. *eﬂli Determine all forces acting en member AECD ef the frame ehewn in Fig. PT-Ql. Tim SDLUTIDN From a free-bed? diagram for the complete frame: + C EMA = 51121 - ?5{24ﬂ = 1:: 11m. F = lﬁﬂ.ﬂ 1b = 150.ﬂ lb T +—}EF =9. -?5=ﬂ H I A = f5.ﬂ lb = ?5.0 lb"* 1 + 1 EE = A? + liﬂ = u A = -15ﬂ.ﬂ lb = 150.0 lb l r e e A - {AH} + {Ar} H 2 2 = {?5.03 + {-150.ﬂ} = 15?.T1 1b _ ~I -lﬁﬂ.ﬂ _ n e an — ten 15-0 - 53.43 I s: 15H 1 53.4“” Ans. m... Member HE is e tee-farce member: therefore. the line of action of force H is kneen ee ehem'l en the L Ill! free-bed? diagram for member ABCD: ?*91 [Guntinuadi + c E”: = -3 333 33° 13} 3 731131 - 33131 = 3 3 = -334.3 13 E = 331 13'3 43.3“ Ans. + —1 E}? = ‘33 +1-133.31333 43° + 33 - T5 = 3 J! C = ﬂﬂﬂ 1h = Bﬂﬂ 1b —4 I! + T 33 = 3 ~ 1-433.31 sin 33“ + 133 = 3 TI" '5" 3? = -133.3 13 = -133.3 13 1 I'll— 3 = #13112 + [UTIE = #13331E + [-133.312 = 333.41 13 _ -1 -15['.' _ ﬂ 1::- SC — tan _§Eﬁ - 2E.5T C = 333 13'3 33.3” Ans. ...
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