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# hw14 - PROBLEM 5.63 ‘ll Determine the reactions atmebeam...

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Unformatted text preview: PROBLEM 5.63 ' ‘ll - Determine the reactions atmebeam supportsforthegivenloading. _ - mom mm .- 4‘ _ B SOLUTION m R, - (100 lwuq'h) = 4001b xi = \$200 Maﬁa 1%) = 600 lb ' kin - (zoo mum) = 3061b ' Amie: A, = o +j ml =0: - (211K400 lb) — (411](6UD'II3) —(12 ﬂ){8001b) +.(10ﬂ)3y = o By=8001b 13:300th 1 . +1..sz=0: Ay +snnIb—4oalb—somh—300=n -._A,=1oomb A=muomf1 PROBLEM 5. 7-3- AgradnbeamABsnppertsmreeconcmﬁ‘atedloadsandrestsmsoﬂand thetapofalargarock.’1hesailexertsanupwarddistributedlcniand theruckfertsacuncenmudluangasshownaningﬂth= 4kN 1w,“ detanuncmevalmofwgmdﬂgeorreapondirgto -. SOLUTION RI"=.[1_2m)(wA mm) .+ 1.2 16,. EN Rn -—3'%[1.8m)[%w1 1mm] = 11.45 19‘ HI Rm=(1.8m)[21wAkNlm]=0.9wde a.) £1.10— 11. — 06m][(1de)kN]+(06m)[[04de )kwm] +(03 (“1.)[03- 9 wA]kam]— (1.2 m)(4 mu.) _— (0.3 m)(_1__s mm) + (0.7 m)(24kNIm)- - o wd=5.567kH/m ll _ _ . 1114:6571:le +1261; =:o R,+L1.2m__)[6.6jrk14;1n]+Lo.45m)(6.6TkN;m) +(D.9m)(667kN_}m]:f 2411:11— 1111:11- 41:}: RR swim: H Rn ... 29.0kN4' ‘ l SDLU'HON 2 gr“; gig-rm»: -' PROBLEM 5.7.5 ' Thccrosssuctianofammrete damisaashown. Fundamsecﬁonof mitwidﬂl.detennhle(a)mereactionforceaexertadbyﬂwgrwudnnthe basaABofﬂaednm,(b)lhepointofapplicaﬁonof1h=raulhntofﬂne rwcﬁonfomesofparta,(c)ﬂ1emsuﬂmtofﬂlepmmeﬁxcﬁmd byﬂnwatermtheﬁneBCofmedam. In the following problems.use r: 52.4 lbfﬂ.’ ﬁnrﬂle speciﬁc wcighl of ﬁ'eshwatﬁandk:1501b/ﬁ3forﬂmespeciﬁcweightofmncmtciflm. customary-units are-used. With 81 min, use p =103 kgfma for ﬂu demityofﬁeshwaterandpc=z40 x 10’me formedensityof commie. (Sea the ﬁoomoteonpage 222 fmhowtodemnﬁncﬂmspedﬁc . weighxofgmaterialgivmiﬁdemity.) - Theﬁ'eebodyshownconsistsofal-mthicksocﬁmofmedammdthc uiangtﬂarsecﬁonBCDofthnwaterbchindthechm. §l=6m X3 =(9+3)m= 12m .- j, =(15+2)m =17m X" '="(15 +4]m =19m W = ng sathat W. = [wokym3)(9.81nys1)[%(9 m)(15 m](l 2.1)]: lségm' - W; = (2400 m3)(9.81ws1)[(6m)(1s m)(1m)] = 2543 km 53; (2400 kym’)(9.slmxsl)[%(5 m)(1sm)(1 111)] = 12711:}: W; = [2400 kyma)(9.81mfsz)[%[6 m)(1s m)(1m)] =- 529.7kN Also P = éAp = %[(1s m)(1m)][(1o3 kgfm3](9.8]m!sz)[18m)] -1ss9kN Then LEF,=0: H—1589kN=u or H=1589kN 'H=lss9kN--d ﬁzz-“go: P"—1589kN-2543kN—1271kN—529.7kN or r=s933kN V=s.93MNf< PROBLEM 5.75 CONTINUED (bj Ham 3.3 maﬁa X(5933kN)+ (6m)(1539kN) — (6 111-)(1589 kN) — (12 m)(2543 kN) — (17m)(1271kN) -—| (19 m)[529.7) = 0 or X = 10.48111 X =10.48md ' . to theright ofA . (c) Cmsider water section BCD as the free bod-y. Have 2!? =0 Pelt-3894a“. e8 Mﬁzmm 411' =1675kN a: 18.43° or R =1675kN 718.43“! Altcrmxive whtiontopart (c) Comiderthe faoeBCofthedam. BC = 62 +18? = 13.973711: m9 =' i 9 £13.43" 1 _ R = (pgyg =[ 000 kym‘)(9.31m1)(13m) :17“ mm: H Ap =§[(13.97 m)[1m)](176.6 man?) ' .=15751:N R = 16751131 7 18.43” ...
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hw14 - PROBLEM 5.63 ‘ll Determine the reactions atmebeam...

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