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hw01sol

# hw01sol - PROBLEM I 1.2 xn F:3 mi Determine the force in...

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Unformatted text preview: PROBLEM I 1.2 xn F :3 mi Determine the force in each member of the Gambrel roof truss shown. 3 m State whether each member is in tension or compression: SOLUTION FBD Truss: 6 kN By impaction of joints C and G, FCE = FAG and FBC = 0 4 3 fZFy=0:12kN—3kN—§FAB=O FAB=15.00kNC< -ZI§=OzFAC—%(15kN)=O FAC=12.00kNT< 10 4 F ——F =0 10.44 B” 5 ’35 2F; ,= 0: g—(15kN)— 3 3 F +—F =0 10.44 “D 5 BE 1213 =O:—:—(15kN)—6kN— Solving yields F30 = 11-93 W C 4 ——> 21th = 0: by symmetry FDF '=11.93kN C 4 3 12:ery = 0:405 —6kN+2m(11.93kN) = 0 F05 = 0.856 kN T 4 By symmetry: FEF = FEE FEF = 0.714 kN C 4 Fm = 15.00 kN c 4 170,, = 12.00 kN T 4 From above FCE = 12.00 kN T { F50 =12.00kN T 4 PROBLEM 2 I .~. . 1: W For the roof truss shown , determine the force nmn- ﬂ ‘ . I“ 111 each of the members located to the left of member GH. State 5.: whether each member is in tension or compression. SOLUTION FBD Truss: Q EMM = o: (3 ft)(300 11:) + (9 ﬁ)(300 lb) + (15 tt)(3001b) ' + (18 ﬁ)(300 1b) + (24 ﬁ)(3001b)+(30 ﬁ)(200 1b) —(30 ft)(Ay) = o A), = 890 1151 w H; = 0: Ax = o __69°1b=51_c;=h FAB=1794Ib c4 5 12 13 , Joint FBDs: F,,C =16561b T 4 200(1) gf—LZ EM fan ' ma lb By inspection ofjointF: FCF = O 4 ~% A 3-7911: M N EF 165611’=h=§!¢é1 FCE=17941bT< 12 13 5 ~ fti -- r I?” *2 - 5 2552/2 c ’1 ” ‘ 125,, = 0:5(1794115 — FED) + 690 lb —300 lb = o FED = 230311» FED = 2.81 Rips c 4 a 215;. = 0:17“ + %(1794 lb — 2808 lb): 0 FM = 936 lb T 4 ~. 2F = o: —6—F —936 1b —131794 lb = 0 .Y: J55: EH F”, = 432J3—7 1b Fm, = 2.63 kips T 4 . s I 1 12F). = 0:17,”; — E(1794115) — 73—;(432J3—7 1b) = o __.._.... M..__ PROBLEM 2 CONTINUED 1 J5 FD” ~3001b - 1122 lb = 0 # 2px = o; %(28081b+FDG)— FD” = 0 L J5 Solving: Foo = 1721 1b C 4 IEFy = o: %(28081b + FOG) + FDH = 1419 lb C { PROBLEM 3 For the given loading, determine the zero-force members in the truss By inspection of joint C: Then by inspection of joint B: By inspection of joint G: Then by insPection of joint F: Then by inspection of j oint E: By inspection of j 0th M: Then by inspection of joint N: By inspection of joint I : ...
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