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1356_Mechanics Homework Mechanics of Materials Solution

1356_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: rod. Work oi: aim-upped «19431.1 = 23 G3éxlo '5' 11.63 The 2-kg block D is dropped fi-om the position shown onto the end of a 16-min- PROBLEM "'63 diameter rod. Knowing that E = 200 GPa, determine (a) the maximum deflection of end A, (b) the maximum bending moment in the rod, (c) the maximum normal stress in the UM = U.“ UBC = “League“ P...2 H]. U... = 4i PM}... = 2.2340 Mo3 3.: Equa/i‘a‘uj work and emewfly O. 7849 +1?.C? 5M3 3/: 1- 87szs’x10'3? y... - 35: ZfiSXIO‘ = (a) Jg=i'{3_'7325‘xto +1! (2 7282*“: 5) +(LHCBS'J. 243x10“) 3% = (4.468: 2103 )(zZesngJ‘) r tos.c: N (in) MM = -(Ios.c:)(o.c) =- - 64“.! NW. 4 (c) 6 - M — —————-—-——(‘;‘*-")(8"’°“) = rszmlo‘ P4 SOLUTION I t ”11%)" ’ E if"): 3.2I7ow103Mm“: 3.217owd‘ m“ ‘3 “g =8mm= awed-m tween OUev' AB M =-P...X Mh:_pha , °‘ 33x2 _ 1930.3 UM“ so 251' ”a” ’ 651: = (0433 P: (G 36200 no" )(3. am: No“) 2': 55'. 953x10“ P...‘ 3:} symmei‘ry 5F Evelina momemi chafing.“ Bar, = UM3 = 55.753 Ho" P: fioéslo" P.“ PM: 4.1mm x :03 y”, m3(h+yh\?(2)(q.81)(0.090 + 5.“) r 0.7948 + I162 3., 2.2340 x103 3/: II = 23.6MM -l stunnerq '-' i57.6 MP6». -“ ...
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