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Unformatted text preview: PROBLEM 3.98 ‘4‘ ﬂ) ‘ W, , Irmw
mg"??? mg? L Mil: A 5—m—long beam is subjected to a variety of loadings. (6:) Replace each
" loading with an equivalent forcecouple system at end B of the beam.
1 CEQE‘ZWLD u L: l—WXS: ,. (b) Which of the loadings are equivalent?
,E: in ‘lx > N.  1’ “ M w 662 »~ 7 $1
lf‘i”: C41 SOLUTION (a)a:2Fy: Ra=—400N—600N
or Ra =1000Nl 4
MB: Ma =(2 kNm)+(2 kNm)+(5 m)(400 N)
or Ma = 6.00 kNm ‘) 4
b: 211,: Rb=—1200N+200N
or Rb =1000N l 4
W5: Mb = (0.6 kNm) + (5 m)(1200 N)
or Mb = 6.60 kNm ‘) 4
c22F: RCEZOON—IZOON
or RC = 1000 N14
ME: MG =—(4kNm)—(1.6kNm)—(5 m)(200 N)
or M0 = 6.60 kNm ) 4
d:EF,: Rd=—800N*200N
or Rd =1000 N14
2MB: Md = —(1.6 kNm) + (462 kNm) + (5 m)(800 N) or Md = 6.60 kNvm ‘) 4 continued f:£F' )" PROBLEM 3.98 CONTINUED
Re : —500 N — 400 N or Re = 900 N l 4
Me = (3.8 kNm) + (0.3 kNm) + (5 m)(500 N) 01' Me =6.60 kNIn ) 4
Rf=400N—1400N or Rf =1000N1 4
M, = (8.6 kNm) — (0.8 kNvm) — (5 m)(400 N)
or Mf = 5.80 kNm U 4
Rg =—1200N+300N
or Rg : 900 N l 4
Mg 2 (0.3 kNm) + (0,3 kNm) + (5 m)(1200 N)
or ME = 6.60 kNm ) 4
R), = —250N—750N
or Rh =1000N ( 4
Mk = —(0.65 mm) + (6 kNm) + (5 m)(250 N)
or M), = 6.60 kNm 3 4 (b) The equivalent loadings are (b), (d), (h) 4 PROBLEM 3.102 The masses of two children sitting at ends A and B of a seesaw are 38 kg
and 29 kg, respectively. Determine where a third child should sit so that
the resultant of the weights of the three children will pass through C if she has a mass of (a) 27 kg, (b) 24 kg. B First WA : mAg = (38 kg) g
W3 = mBg = (29 kg)g
(:1) WC = Meg = (27 kg)g
For resultant weight to act at C, ZMC = 0
Then [(38 kg)g](2 m) — [(27 keglw) — [(29 kg)g](2 m) = o
. d = 76 ‘ 58 = 0.66667m
or d = 0.667 m {
(b) We = meg = (24 kg)g
For resultant weight to act at C, ZMC = 0
Then [(38 kg)g](2 m)  [(24 kg)g](d)  [(29 kg)g](2 111) = 0
_ _ 76 — 58 _
. d — 24 — 0.75 m or d = 0.750m 4 y PROBLEM 3.124 can kk A concrete foundation mat in the shape of a regular hexagon with 3m {no gm sides Supports four column loads as shown. Determine the magnitudes of
the additional loads that must be applied at B and F if the resultant of all
six loads is to pass through the center of the mat. SOLUTION Have: We FA(ZA) + FB(ZB) + FC(ZC) + FD(ZD) + FE(ZE) + FF(ZF) = Rlza)
(80 kN)(0) + F3 [(3 m)sin60°] + (40 kN)[(3 m) $11600] + (100 kN)(0)
+(60 kN)[—(3 m)sin60°] + FF[7(3 m) sin60°] = R(O)
FE — FF = 20 kN (1)
A150 W23 FA (xA) + FEW) + FC(XC) + FD(xD) + 35le) + FleF) 7" RUG)
(80 kN)[—(3 m)cos60a — 1.5 m] + F3 {—1.5 m) + (40 kN)(l.5 m)
+ (100 kN)[(3 m)cos60° +1.5 m] + (60 kN)(1‘5 m) + FF(—1.S m) = R(0)
F3 + FF = 140 kN (2)
Solving equations (1) and (2): FB = 80.0 kN 4
F; = 60.0 kN 4 ...
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 Fall '07
 Delph

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