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Unformatted text preview: PROBLEM 3.98 Q “  * it! A 5mlong beam is subjected to a variety of loadings. (a) Replace each " loading with an equivalent forcecouple 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 kNm)+ (2 kNm)+ (5 m)(400 N)
or Ma = 6.00 kNm 3 4
b12F: R,,=—1200N+200N
or Rb =1000N l 4
2MB: Mb = (0.6 kNm) + (5 m)(1200 N)
or Mb = 6.60 kNm 3 4
RC =200N—1200N
or RC =1000Nl 4
MB: ML. = —(4 kNm) — (1.6 kNm) —(5 m)(200 N)
or M6 = 6.60 kNm ) 4
d:>:F,: Rd=—8OON—200N .
0r,Rdt=1000 N l ‘
2MB: Md = —(1.6 kNm) + (4.2 kNm) + (5 m)(800 N)
or Md = 6.60 kNm 3 4 PROBLEM 3.98 CONTINUED Re=—500N—400N orRe=900Nl4 Me = (3.8 kNm) + (0.3 kNm) + (5 m)(500 N) or Me = 6.60 kNm ‘) 4 Rf=400N—1400N or Rf =1000Nl 4
(0.8 kNm) — (5 m)(400 N) or Mf = 5.80 kN~m ‘) 4 Rg=—1200N+300N Mg = (0.3 kNm)+
R), = —250 N — 750 M), Vii(0.65 kNm) or Rg = 900 N l 4
(0.3 kNm) + (5 m)(1200 N)
or Mg = 6.60 kNm V) 4
N orRh=1000Nl4 «.1 3 or M), = 6:60 kNm§ 4
(b) The equivalent loadings are (1)), (@X‘ a +
,(6 kNm) ,(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 forcecouple 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 Nm)j + (9 Nm)k] + [(12 Nm)i — (9 Nm)j] or MA = —(3 Nm)i + (13.5 Nm)j + (9 Nm)k 4 PROBLEM 3.123 A concrete foundation mat in the shape of a regular hexagon with 3m
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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This note was uploaded on 10/05/2009 for the course CE statics taught by Professor Epstein during the Spring '09 term at UConn.
 Spring '09
 Epstein

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