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Unformatted text preview: 5.41 Using the methods of Sec. 5.3, solve Prob. 5.1.
PROBLEM 5.41 5.1 through 5.6 Draw the shear and bendingmoment diagrams for the beam and
loading shown. SOLUTION 9§M¢=o LALon A:E_ b
L.
Dir/1pc LcaP=o 03%» M A“ v.—A=J;_’~2 M=o
A1L08_ O<x<a 5.42 Using the methods of Sec. 5.3, solve Prob. 5.2.
PROBLEM 5.42 5.1 through 5.6 Draw the shear and bending—moment diagrams for the beam and
loading shown. SOLUTION Maximum M occurs ad X= lj w{em v=ﬁf§=o a!
8 MWM': 5.47 Using the methods of Sec. 5.3, solve Prob. 5.13. 5.13 and 5.14 Draw the shear and bendingmoment diagrams for the beam and
loading shown, and determine the maximum absolute value (a) of the shear, (b) ofthe
bending moment. PROBLEM 5.47 1.3 kN L5 kN 1.2 kN SOLUTION 9M3 Dimensionsin mm (400)(L5) ‘300 C, + (2253(L2) 4 (’25 )(l.3 ) = O O C= 3.55 IN v (m Om: o 8: 0.85M)
Shear? ‘
A 1‘0 C V: I.5l(N c +0 D = 4.5+ 3.4:  2.15 w
DbE wax1.2 =03ng
E +~= B v: OAS1.8 = o.3sw Area: or shear 0954ij Mac 5)le = Gina»): use NM .
C +0 D 5m  '(2.Is'_)( 7‘5) = lCLZS’Nm
D haE Svay .~ (o.‘l$)(loo) = as Nm
E+o B SVJx = (—0.85Y1253 = IOé.25 Nm Banal [n3 momeﬁfs M570
M; : M"'+S:Vax = 0—}50 = —Iso~.m Mo = M.+ 5‘:va 2 450+ Iél.25 = ”.25 Nm
ME = MN gjva, = ”.25 + 95 = 106.2SNm
M3: ME+$DBVJx  wax10:1: = o Maximum [Vl2 2.1.5 kN «I MaﬁMum ”V”: [50 Mom N” 5.57 and 5.58 Determine (a) the equations of the shear and bendingmoment curves
PROBLEM 5'57 for the given beam and loading. (b) the maximum absolute value of the bending moment in the beam. SOLUTION L z  (HikélW
(2 X15) = (Ii 4—3:;NL
: W, 0.421é5 L J.
— w.(o.qzzesL1 wa goﬁazestf’ w,_(o.*tzzcs.t_)
2 J“ GL + 3 .— o.oe.4:5 w,L‘ 5.47 Using the methods of Sec. 5.3, solve Prob. 5.13. 5.13 and 5.14 Draw the shear and bending—moment diagrams for the beam and
loading shown, and determine the maximum absolute value (a) ofthe shear, (b) of the
bending moment. PROBLEM 5.47 1.5 PM 1.2 kN 1.5 kN SOLUTION 9M3=O Dimensionsin mm (400)(LS) ' 300 C + (22.5)(L2) + (I25)U3 ) = O V (W) Br 0.86”ka V=!.5kN
Ch, D v: 4.5+ 3.4! rats W
9195 V_=2.L§ 1.2 =0.¢?5ku
E +9 B V: 0351.? 2 4.35%: Ameaﬁ or Shear 0“}:ij A+oC 5)le = CJ..E)(loo) = —150 Mm .
C +00 .5le  ‘(2.us'_\( 73) = iCLZS’Nm
1)qu Svay: (om/5m») = 95 MW: E +9 8 SVJx =~ (0.85%:25) = IOé.25 Nm Benchn: Momen+5 M4?0
C
m . M‘HS‘Va‘x = 0—150 = 450M». Mo = M.+ S:volx = lSOHél.25’ = ”.25 Nm
ME = MN SDWA, : 11.25 + 95 = 10635an
Mr “#55le  ”925—1062? = o Maximum lv= 2.15 M: l MontiMUM [MI’ 150 Nm 5.46 Using the methods of Sec. 5.3, solve Prob. 5.6. BLEM 5.46
PRO 5.1 through 5.6 Draw the shear and bendingmoment diagrams for the beam and loading shown. SOLUTION Raqc'l’ions v—vc V : —w[.x.‘(Lo.)]
X x 
m MC: SLAVOL,‘ = SpaWE” Chatﬂow = VW[§JCLQ5X]L:
=  W[Zz:*(L"“Jx _§L_—is.ll, (13°an
2." ~W[§(LQ3X+%l1 M: réwa‘w[—¥ (L—a\x+%‘—5r] ...
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 Fall '09
 MOORTHY

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