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Unformatted text preview: PROBLEM 4.33 Rod ABC is bent in the shape of‘a circular arc of radius R. Knowing
that 6' = 50°, determine the reaction (a) at B, (b) at C. SOLUTION
For 6 = 50°
(a) From the £1311. of rod ABC
+) ZMD = 0: CAR) — P(R) = o
: P =P—u LEE; =0: P—Bsin50°={} B: .‘D 21.305411”
s1n50° or B = 1.3051“ in. 40.0”‘ From the f.b.d. of rod ABC +125, = o: Cy — P + (1.30541P)00550° = 0 Cy = 0.160900P or Cy = 0.1609P 1 Then C =1/C3 + Cf. =1/(1t')2 + (0.1609P)2 =1.01285P a = tan—'[C ] tan"'[0°1ffgp] = 9.1405° or C = 1.0131” A: 9.14°i PROBLEM 4.18
Determine the reactions at A and B when (a) h = 0, (b) h = 8 in. SOLUTION (a) (a) h = 0
From f.b.d. of plate +3 EMA = 0: (Bsill30°)(201n.)(401b)(10in.}= 9
B = 40119
01' B = 4D.01b:::.30°{ LEFT = 0: A,r — (4D 1b)cos30° =
AI = 34.641113 or A)r = 34.61%: “ +123? = o: A}. — 40 1b +(401b)sin3{]° = A}, = 2011:» or A}. = 29.0 119T A = ,lAf + A; = (34.641)2 + (29)3 = 39.999113 an" 0 =30.001°
34.641 or A : 40.0113430" *‘ (b) h : 8 in.
From mm. of plate
+3 EMA = 0: {Bsin30°)(20 in.) — (Ec0530°}(8 in.)
—(40 1b)(10 in.) = 0
B = 13921711:
or B =130.21b§:.30.0°4 unomrm: .93 002.22ch
+.. Mn n 9 hr I 3.3.»: 5:330 u a
ma. H IP13; E E. in H Ink 5 I...
L 3 n 9 Ar. I .3 a + :3”: 333”. u o kw u 13:5 5 E 3 n a; E h a :53% + 713.33“ "532.” :5; E "Lymﬂog SPE— E. b u 2mm Edﬂ HMumaA PROBLEM 4.51 Knowing that the tension in wire ED is 300 1b, determine the reaction at
ﬁxed support C for the frame shown. SOLUTION From f.h.d. of frame with T : 300 lb _+..):Fx =0: Cx—1001b+[%]300lb=0 CI = 415.38461b or c, =15.38461b— +1sz = o: Cy —lBOlb—[£]3ﬁﬂlb = o
13 . Cy : 456.921b or Cy = 456.921b 1
c = Jcﬁ + Ci = (15.38%)2 +(456.92)2 = 457.1311: C
a = tan‘[ ] = tan—(ﬂ) = 43.0720 _y_
CI —15.3846 or C = 4511b :25. 33.1°< +‘) EMC = a: MC + (1801b)(20in.} + (1001b)(16 in.) — [[%]3001b](16 in.) = o MC = —769.23lbin.
or MC = 7691hin. )1 ...
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 Fall '08
 C.Trujilliano
 Statics

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