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hw11sol

# hw11sol - m PROBLEM 6.77 For the frame and loading shown...

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Unformatted text preview: , m PROBLEM 6.77 For the frame and loading shown, determine the force acting on member ABC ((1) at B, (b) at C. SOLUTION Note that BB is a two-force member FBD member ABC: (a) Q ZMC = 0: (16 MGFBDJ w (24 in.)(80 1b) = 0 3315 F3”: 1501b, FED = 150.0113 7 369° 4 (b) 1213 = o: Cy m§(1501b)= 0, CV = 901b x o: Cx—%(1501b)+801b=0, Cx=401b l E? C = 98.5 lb 466.0" 4 PROBLEM 6.84 Determine the components of the reactions at A and E when a 24-lb force directed vertically downward is applied (a) at B, (b) at D. SOLUTION (a) FBD AC: Note: CE is a two-force member 2% Q EMA : 0: (8111.)[5C‘i)+ (2 m.)(§@)_(6 in.)(24 1b) =0 2 5,, J5 J5 A . ‘ x 2“ FCE =14.4J§ 1b, so Ex =14.401b —~ 4 C. Ev =14.401b I 4 E / ' ——»2Fx=0: —Ax+14.41b=0 Ax=14401bw4 12:13,=0: Ay—241b+14.401b=0 Ay = 9.60 1b 1 4 (b) FBD CE: Note AC is a two-force member Q 2M5 = 0: (3 m.)[-5-FAC + LFAC) 4 (1 in.)(24 1b) = 0 . Jﬁ Jﬁ FAC =1.6ﬂ 1b, Ax = 6.40 lb «m 4 Ay = 1.600 lb 1 4 ———~>:P;=0: Ex—7:=7(1.6\/ﬁlb)=0, Ex=6.401b—~4 IZF, = 0: E), + %(1.6\/ﬁ1b)- 24 1b = 0 r Ey =22.41b I 4 1,51 1211—» 0 PROBLEM 6.104 «1th 2 ‘ ‘3‘“ g 7 7.211 The axis of the three-hinge arch ABC is a parabola with vertex at B. ‘ :: gm Knowing that P = 14 kips and Q = 21 kips, determine (a) the ‘ components of the reaction at A, (b) the components of the force exerted at B on segment AB. SOLUTION Members FBDs: E I:( EMA: 0: (12.8 ft)Bx- (32 ft)By- (20 ﬁ)(14 kips) = o 11:(' 2114C: 0: (7.2 ﬁ)Bx+ (24 11) By— (12 ft)(21kips)= 0 Solving: Bx = 27.5 kips, By = 2.25 kips, I: ——~ EEC: O: Ax— 27.5 kips = 0, Ax: 27.5 kips, (a) Ax: 27.5 kips ——> { ’( ZFy= 0: Ay— 14 kips — 2.25 kips = 0, Ay= 16.25 kips, Ay= 16.25 kips I 4 (b) Bx: 27.5 kips «—4 By: 2.25 kips l 4 —~—_—_H—___—_—__w—mw__l mm. —.1 PROBLEM 6.106 P Knowing that P = 411 1b and Q = 0, determine for the frame and loading shown (a) the reaction at D, (b) the force in member BF. I SOLUTION _l FBD Frame: (12 in.) Dx — (28 in.)(4111b)= 0 Dx = 9591b —~ 19.7 H I’ FBD DF: Note that BF and CE are two-force members. W'EF;=02 Solving: FBF = 357 1b, U ( ME 2 01: (22.5 in.)[~1%(357 113)] — (12 in.) Dy = 0 Dy=3151b(, so (a) D=10091b 418.1804 ...
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