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Class 9 Notes

# Class 9 Notes - PROBLEM 3.98 Q “ it A 5-m-long beam is...

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Unformatted text preview: PROBLEM 3.98 Q “ - * it! A 5-m-long beam is subjected to a variety of loadings. (a) Replace each " loading with an equivalent force-couple system at end B of the beam. ,,,,,, a. .......... "gm; R) (b) Which of the loadings are equivalent? (A > H] \ - m < SOLUTION (a) a: my: Ra = —400N—600N or Ra =1000N ) 4 MB: Ma = (2 kN-m)+ (2 kN-m)+ (5 m)(400 N) or Ma = 6.00 kN-m 3 4 b12F: R,,=—1200N+200N or Rb =1000N l 4 2MB: Mb = (0.6 kNm) + (5 m)(1200 N) or Mb = 6.60 kN-m 3 4 RC =200N—1200N or RC =1000Nl 4 MB: ML. = —(4 kNm) — (1.6 kN-m) —(5 m)(200 N) or M6 = 6.60 kN-m ) 4 d:>:F,: Rd=—8OON—200N . 0r,Rdt=1000 N l ‘ 2MB: Md = —(1.6 kN-m) + (4.2 kNm) + (5 m)(800 N) or Md = 6.60 kN-m 3 4 PROBLEM 3.98 CONTINUED Re=—500N—400N orRe=900Nl4 Me = (3.8 kN-m) + (0.3 kN-m) + (5 m)(500 N) or Me = 6.60 kN-m ‘) 4 Rf=400N-—1400N or Rf =1000Nl 4 (0.8 kN-m) — (5 m)(400 N) or Mf = 5.80 kN~m ‘) 4 Rg=—1200N+300N Mg = (0.3 kN-m)+ R), = —250 N — 750 M), Vii-(0.65 kN-m) or Rg = 900 N l 4 (0.3 kN-m) + (5 m)(1200 N) or Mg = 6.60 kN-m V) 4 N orRh=1000Nl4 «.1 3 or M), = 6:60 kN-m§ 4 (b) The equivalent loadings are (1)), (@X‘ a + ,(6 kN-m) ,(5 m) (250 N) PROBLEM 3.1 1 5 A machine component is subjected to the forces shown, each of which is parallel to one of the coordinate axes. Replace these forces with an equivalent force-couple system at A. 30 mm SOLUTION For equivalence 2F: FB+FC+FD =RA RA = —(240 N)j — (125 N)k — (300 N)i + (150 N)k RA = —(300 N)i — (240 N)j + (25 N)k 4 Also for equivalence EMA: rB/AXFB+rC/AXFc+rD/AXFD=MA i j k i j k i j k or MA = 0 0.12 m 0 + 0.06 m 0.03 m —0.075 m + 0.06 m 0.08 m —0.75 m 0 —240 N —125 N —300 N 0 0 0 0 150 N = [—(15 N.m)i] + [(22.5 N-m)j + (9 N-m)k] + [(12 N-m)i — (9 N-m)j] or MA = —(3 N-m)i + (13.5 N-m)j + (9 N-m)k 4 PROBLEM 3.123 A concrete foundation mat in the shape of a regular hexagon with 3-m 100 kx sides supports four column loads as shown. Determine the magnitude and the point of application of the resultant of the four loads. \$0 kN SOLUTION \$0 in) H Have: 2F: FA + FC + FD + FE = R R=—(80kN)j—(40kN)j—(100kN)j—(60kN)j = —(280 kN)j or R = 280 kN 4 HaVei XMXI FA(ZA) + FC(ZC) + FD(ZD) + FE(ZE) = Rlzo) (80 kN)(O) + (40 kN)[(3 m)sin60°] + 60 kN(O) +(6o kN)[—(3 m)sm60°] = (280 kN)ZG ZG = —0.135577 m or 20 = —O.1§56 m 4 mm) + me) + Fax,» + mm = Rec) (30 kN)[—(3 m)cos60° — 1.5 m] + (40 kN)(1.5 m) + 60 kN(1.5 m) -~ + (100 kN)[(3 m)cos60° +1.5 m] = (280 kN)xG or x0 = 0.750 m 4 ...
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