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1089_Mechanics Homework Mechanics of Materials Solution

# 1089_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: PROBLEM 9.79 B DJjeefl’l‘oh 4* C. E *~= 200 GPa. SOLUTION Wyn.” .- 9. 79 For the cantilever beam shown, determine the slope and deﬂection at end C', Use ‘ a“... ..,,.',.-s E UM+\$ : Forces in kN 3 ﬁemj'} LS in M . A C 1.5 . L—0-75 m—J‘-0-5 m‘J 8100 X 1 Load in 3 I 3 Con cedw+ed .9044 a." B Caseﬂl 6? Appendix: ID __¢pkfiea‘ +6 Pari‘fon AB. __ _.__(._3__)§g.7_5)§ _ _ b.8437: an E, c. 251 25.1? ‘ EI‘ <—--—0.7\$—--——-H4-O.\$ -——| .- _ __ PL3 _ _(3)(o.75)’ : _ 0.42187: 3’6 ‘ 351: ' 351: El Por‘h'on BC Pemeaa'ns 5+7“?th 9 ‘ :- 68‘ :' _ 0.89375- c. 51‘ . - .8 37 J...‘ = 33 40.6394 = * 2‘53“": Lowlimi II: Concen‘l'v‘ahal Pascal 0:} C. Case I 0-? Appendix D. e II___E_|=_1__‘§33§L2\$)‘ _ _ 2.34375“ 8 ' 251‘“ 25: ' EI -... PLS - Aﬂﬁf- _ I-_9§3I_2\$ 3* ' ‘35:: ' 3E1 ' Er I II 3.:87 By soperfosH-bn? 9‘ = 9,. + an = - T55: _ I u .. _ 2.7016375 y» ' ‘Yn + j» ' _EI Dark: E : Zoouo“ Pa, I = 2.53%!0" um“ = 2.53am" m“ 5: =(xoox10“)(2.53wd‘) = soeutoa N-m" = Soc kM-m‘ Siege 03 C c= — iégeli : ~6.3.oxto" m! = 6.30:16‘M-a- «- 5‘ r -M = ~5.53 x155 m = 5.53 m t and ...
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