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Unformatted text preview: PROBLEM 7.83 Using the method of Sec. 7.6, solve Prob. 7.3 6. ; (1.1 m)({}.54 kN) — (0.9 may
+ (0.40:)(135 m)  (0.3 m)(0.s4 kN)  0 CJ.=1.08kNT
12F, =D:—0.54kN+].08kN].35kN+E—0.54kN =0 I:=1.35kN1 Shear Ding: .‘ Vis piecewise constant, [%=0everywhme] Withdjscoﬂﬁmﬁﬁcsli each memo force. (equal to the force) lV‘m = 310104 Moment Ding: Mispieoewiselinearstaﬂingwith MA = 0
MC = 0 — 02 m(0.s4kN}  0.103kNm
MD = 0.103 I‘Mm; (0.5 m)(0.54kN) = 0.162 kNm
ME = 0.102 kNm — (0.4m)(o.311:N) .—. 0.162kNm
313': 0.162 kNm + (0.3 m}(0.s4 kN) = 0 WI...“ = 0.162'kNm = 102.0um< PROBLEM 7.67 Using tlw method of Sec. 7.6, solve Prob. 7.38. SOLUTION
(a) math) m Beam:
Q m, = o: (3 n)[ %](a n) + (9 mm kips) — (15 up, = o h (F  A = 9.6M
: II! a y P! T
Hf = ShunDing: Vjumps toll], = 9.6kipsatA,iscnmtm11to CLjnmps dnwn 12kips
~ to 2.4]dpsatC,ismtanttoD,andﬂaenislimar
JV
._...—= B
(It mm)”
1’, = —2.4 Rips — [ikipsfﬁXE it)
= —l4.4kips Moment Ding:
Mis linearfmmd to C MC  9.6 ldps[6 a) = 57.6 kipﬂ, MislinearfrmnCtoD [% = 4.45pm] MD  57.6kipft — 2.41am (a it)
up = 50.4 513.11 His parabolic [% decreasingwid: V] to B. M, = 50.4kiPﬁ — %(2.4tips +14.4kips](6 ﬂ)  u
= 0 mm = 57.6 kip1H PROBLEM 174 Fmﬂlcbcam shown, drawﬁcshearamibmdingmomentdiagmmsand
detminethemaximum absolmevalue ofﬂnbendingmomentknawing
3 ﬂut(a)P=14kN,(b)P=20kN. 13F, o:A,—(15kNm)(2m)—6kN+Po
A, =38kNP
A,24kN1
Ay=18kNT
__ 1L MA = D:(Sm)P—(3.5m)(6kN)—lm(]6kN/m)(2m)—MA = a
.:_._; M4 = (Sm)P—53kNm
It‘aJ MA =17kNm )
(b) MA=47kNm_)
SharDiags: 1:, = Ay. Than Visli1:u:ar[ﬂ = 16 mm] m c.
 5 of: VC = FA —(16kNIm)(2m) = VA —32kN
Vc=—8kN
Vc=—l4kN
tabla—(mm);
x1 =].5m x1 =1.125 m Visoonsmnﬁ‘mCIoDdecreasesbyékNatDandismmmmm
Ma: .P} PROBLEM 7.74 CONTINUED Moment Diags: MA = M1 reaction'l'henMispanbolic [gtg decreasingwithl’} Themaximmnocwrswhm 170. Mm  MA+;+de,. I (0) MM =17kNm+%(24kN)(l.5 m) = 351) ﬁlm d
1.5 11151111112! 4 (1:) MM = 47 mm +%(1sm)(1.125 m] = 57.125 kNm
Mm = mmmnsumu ME = Mm +%F’C(2m— III)
(a) M5 = 35 kNm  as kN)(0.5 m) = 33 kNm (5} MC = 57.125 kNm — £04 kN]{0.875 m) = 51 kNm
MispieoewiselinearalongC,D,B,with M, = 0 and
Mo = (1.5 m)?
(a) MD = Zlk‘Nm
MD = 30mm SOLUTION (a)
Replacing 1h: load at E with
equivalmt formmm]: at C. PROBLEM 7.79 Solve Prob. 7.78_assmningﬂ1a£thc 4kN ﬂan0e applied an? is directed ( m. = o: (s 111)!) — (s m](2 mm (3 m)(4 m)
— 4kNm — (1.5 m)(8 kNImXS m) = 0 D=2kﬂr
3 22
TH, =D:Ay+?kN—(8Wm)(3m)+4kN—2Hl=0 44
Ay“s—mt Shear Ding: VA =Ay=131kN,thenVislinear[%=—Bk}€hn]toc vc =%4kN—(Sl:me)(3m) =—%m V=U=¥kN—(8kﬂfm)xl at xl=%m. AtC, F’imreasestltho —% EN. AtD, Vimrmes ~23? kN toZkN. PROBLEM 7.79 CONTINUED Mumm Ding: MA = 0. Than His parabolic [33% (lemming with V]. MaxMoccurs wheteV=0. Mm =13.44 kNmat1.833 mﬁumAi' Jewm 4516,me 4kNmt012kNm. ThanMislinear [ﬂ=EkN]toD.
(it 3 MD =12kHm—[%kﬂ](3m]= 4kNm. Misagainl'mea: PROBLEM 7.31 For the beam and lending shnvm, (a) derive fhe equations of the shear
and bendingmoment curves. (1:) draw the shear and harmingmoment
diagram, (c) determine the magnimde and location of the maidmum
bending moment. SOLUTION “Easel—em Notes: Atr=L,V=—woL AndYuat [£1.2[5 L L AlsoVismaxwhere w=o[£=l]
L 4 l
Vm=§HbL at»! Moment: PROBLEM 7.81 CONTINUED ”,4I_;{L,_,.) PROBLEM 7.83 I Be“ AB' Which 1i“ 0“ th= grounds supports the parabolic load
[III III} 51‘0“ Assuming ﬂlcwwardrcacﬁon of the gromdtohe uniformly
1 L l I distributed, (a) write the equations of the shear and bendingmmment
was, (b) dewunine the maximum bending moment. SOLUTION
TSP: o: wEL— $47? “’“(Lx— x2)dx= 4W0 2w“ ng_=2 ’“ ”T 2
W=%=°d€=%“°“°m=‘"°[%[§]Hm 01' w=4wu[—%+g—:‘] 'V=V(”)'.E4WL["%+€;z]d§= 0+4mL[%¢+§§’§E] V = §wnL(§3§’ +26) <1
M=Mu + gm=o+3woﬁﬁ [g—sghzgﬂdg
2 1
= gwmﬁof —c’+ Hr] Mk2 —2E +54) 4 MaxMoecurswhu'e Voal—ngrzgzo—pgl M[¢=J=w[% a 12:: 2 2
M = Egg—manurofbemn‘ ...
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 Spring '08
 Jenkins

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