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

Info icon This preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
Image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
Image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 8
Image of page 9
This is the end of the preview. Sign up to access the rest of the document.

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‘ ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern