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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 "Lil 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“ IIl
l<—6[t—1+4.5I'r 11511 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 kipﬂ — .r{2 kips)
—%x(2 kjpsm) + M = o M =[1kip1’1"’(}x2 + [2 kips)x—114 kipvft
M = —66kipﬁatC (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 kipﬂ — [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 kipﬂat!) [x = L5 R] (b) From diagrams: IVI = 14.00 kips on CD 4 Inuk M] = 14_0 kipHam 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.
 Fall '09
 K

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