# Hw4 answers - r PROBLEM 6.163 I a m For the frame and loading shown determine the components Of l the forces acting on member CF E at C and at F r

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Unformatted text preview: r; ; PROBLEM 6.163 I/ . a m For the frame and loading shown, determine the components Of _ l the forces acting on member CF E at C and at F. ' .. r ‘ " a :<.a__ ___'o_—_“~ 7—? \ 1 _ . - .- I) Q _ I O i F ' F Jﬂ— ‘ L3 "Li-l ill. 4 “NJ L SOLUTION _| FBD Frame: ;_ my = o: (13 in.)(4o lb] — [lOin.]A_‘. = o A, = 5211: —- .1" 2MB = 0: (4mm. -(6in.)(52 lb} = o a = 73 1b —n- onABF on CFE from above F\. = 78.0 lb ‘— 4 I; EM? = 0:(9in.)(401b)—(4in.)F_. —{4in.)(781b}= o _,. : 12.00 lb I «l (‘Zsz0rCr—F‘=0 r=781b c! = 73.0 lb —+ 4 ‘ 25.. = 0: —40 lb + a. + C_.. = 0 c3. = 40 lb — 12 1b = 28113 c". = 23.0 lb t 4 21:". PROBLEM 6.166] Rod CD is fitted with a collar at D that can be mowed along rod AB. which is bent in the shape ofa circular arc. For the position when 9 = 30", determine (a) the force in rod C D. (b) the reaction al B. SOLUTION FBD: (“J , I\_ my. = o: ([3 in.}(20 lb — 3.1.): o B}. = 20 lb 1 D-\ ‘ 2;}. = 0: -201b + Fmsinw — 20 lb = 0 Fm = 30.0 lb T <1 (5) —— 2F, 2 O: (801b]c0530° — B. = 0 Br = 69.232 lb —— 50 B = 72.1lb 716tlﬂ°4 PROBLEM 7.37 ‘/ '_-‘ Lllu 2 .jlh ‘ l 1 F or the beam and loading shown, {a} draw the shear and bending- _ B moment diagrams. (5] determine the maximum absolute values of the " shear and bending moment. 2 Lil“ II-l l<—6[t—-1+4.5I'r 1-1511 SOLUTION _ I_ ’f'P‘ {a} FBD Beam: 5- '* 4 ’ 5 121‘}: 0:11, + (a ﬁ}(2 kipslﬂ) —12kjps — Zlu'ps = o _//.._—” A). = 2 kjpsl _ #J: (321.11,T = 0: MA + [3 ﬂ](6 ﬁ](2 kipsm} gm — (10.5 ﬁ)(12 kips} - (12 11)(2 kips) = 11 um” ' M,I = 114 kip‘ﬁ ‘) 2",, I we -~ Along AC: L ’9 x M IW/qp/ “L 2WD, Zkrtpslﬁ‘i x TEF). = 0:2kips+x[2 kips!ft]—V =0 V=2kjps+(2 kipsfﬁ)x V =l4kipsatC [x=6ﬁ) QZMJ = o: 114 kip-ﬂ — .r{2 kips) —%x(2 kjpsm) + M = o M =[1kip1’1"’(}x2 + [2 kips)x—114 kipvft M = —66kip-ﬁatC (x=6ft} Along CD: .12 #705 K'IPS Z K151 #2 m tm‘:ﬂ:V—12Hps—25ps=0 V=14kips (21m = 0'. —M -—x.(12 Rips] — [1.5 R + x‘)(2 kips) = 0 PROBLEM 7.3? CONTINUED M = -3 kip-ﬂ — [I4kips}xl M = ~66 kipvﬁat C (x1: 4.5 a} AlongDB. 75 =5 Valor)" H MEET-=3 12F, =0: V—2kips= D V = +2kips (XML =0z—M—2kipx3=0 M = —[2 kips)xi M = —3 kip-ﬂat!) [x = L5 R] (b) From diagrams: IVI = 14.00 kips on CD 4 Inuk |M] = |14_0 kip-Ham 4 m an. PROBLEM 9.93/ Using Mohr’s circle, determine the moments ofinenia and the product of inertia of the area of Problem 9.73 with respect In new centroidal axes obtained by rotating the x and ‘1' axes through 30“ clockwise. SOLUTION From Problems 9.?3 and 9.81 I... = 133.24 x10“ mm" :1. = 51.34;? x 10“ mm“ = 162.86 x10“ mrn" E. = 103.68: x 10“ mm4 = 325.72 x10" mm‘ Now 2; _= 1“: +1} = 244.29 >< 10“ mm" '— 1—293 _. R= ' +."._" 2 J = 160.4405 x 10“ mmJ‘ From Problem 9.87 26,” = 595" Then a = 180 — 60° — 29m : 605° Then 1. = Tm + Room = 244.29 +160.4405cos60.5° = 323.29 x106 mm4 or JT-‘- = 323 x 10° mm" ‘ T, :m — Rcusa : 244.24 — 160.4405c03605" = [65.29 x 10" mm4 or I... = 155.3 x10“ mm4 4 I 1r. = Rsina = 160.44sin 605° = 1396 x 10“ mm" *1 .\ ...
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## This note was uploaded on 09/30/2010 for the course CIVL CIVL113 taught by Professor K during the Fall '09 term at HKUST.

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Hw4 answers - r PROBLEM 6.163 I a m For the frame and loading shown determine the components Of l the forces acting on member CF E at C and at F r

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