# HW23 - PROBLEM 7.83 Using the method of Sec 7.6 solve Prob...

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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} - 0.103kN-m MD = 0.103 I‘M-m; (0.5 m)(0.54kN) = 0.162 kN-m ME = 0.102 kN-m — (0.4m)(o.311:N) .—. -0.162kN-m 313': 0.162 kN-m + (0.3 m}(0.s4 kN) = 0 WI...“ = 0.162'kN-m = 102.0u-m< PROBLEM 7.67 Using tl-w 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 = Shun-Ding: Vjumps toll], = 9.6kipsatA,iscnmtm11to CLjnmps dnwn 12kips -~ to -2.4]dpsatC,ismtanttoD,andﬂaenislim-ar 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 kip-ﬂ, MislinearfrmnCtoD [% = 4.45pm] MD - 57.6kip-ft — 2.41am (a it) up = 50.4 513.11 His parabolic [% decreasingwid: V] to B. M, = 50.4kiP-ﬁ — %(2.4tips +14.4kips](6 ﬂ) - u = 0 mm = 57.6 kip-1H PROBLEM 174 Fmﬂlcbcam shown, drawﬁcshearamibmding-momentdiagmmsand detminethemaximum absolmevalue ofﬂnbendingmomentknawing 3 ﬂut(a)P=14kN,(b)P=20kN. 13F, -o:A,—(15kNm)(2m)—6kN+P-o A, =38kN-P A,-24kN1 Ay=18kNT __ 1L MA = D:(Sm)P—(3.5m)(6kN)—lm(]6kN/m)(2m)—MA = a .:_._; M4 = (Sm)P—53kN-m It‘aJ- MA =17kN-m ) (b) MA=47kN-m_) 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 [gt-g decreasingwithl’} Themaximmnocwrswhm 17-0. Mm - MA+-;+de,. I (0) MM =17kN-m+%(24kN)(l.5 m) = 351) ﬁlm d 1.5 11151111112! 4 (1:) MM = 47 mm +%(1sm)(1.125 m] = 57.125 kN-m Mm = mmmnsumu ME = Mm +%F’C(2m— III) (a) M5 = 35 kN-m - as kN)(0.5 m) = 33 kN-m (5} MC = 57.125 kN-m — £04 kN]{0.875 m) = 51 kN-m MispieoewiselinearalongC,D,B,with M, = 0 and Mo = (1.5 m)? (a) MD = Zlk‘N-m MD = 30mm SOLUTION (a) Replacing 1h: load at E with equivalmt form-mm]: at C. PROBLEM 7.79 Solve Prob. 7.78_assmningﬂ1a£thc 4-kN ﬂan-0e applied an? is directed ( m. = o: (s 111)!) — (s m](2 mm (3 m)(4 m) — 4kN-m — (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 =%4-kN—(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 kN-mat1.833 mﬁumAi' Jew-m- 4516,me 4kN-mt012kN-m. ThanMislinear [ﬂ=-EkN]toD. (it 3 MD =12kH-m—[%kﬂ](3m]= -4kN-m. Misagainl'mea: PROBLEM 7.31 For the beam and lending shnvm, (a) derive fhe equations of the shear and bending-moment curves. (1:) draw the shear and harming-moment diagram, (c) determine the magnimde and location of the maidmum bending moment. SOLUTION “Easel—em Notes: Atr=L,V=—woL AndY-uat [£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 V-oal—ngrzgz-o—pg-l M[¢=-J=-w[--% a- 12:: 2 2 M = Egg—manurofbemn‘ ...
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