lec36-20110408 (1) - fifiofiifié In : EUGJ :thy);...

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Unformatted text preview: fifiofiifié In : EUGJ :thy); JVZCI 395 e a to 10-37 Determine the moment of inertia of the beam’s cross-sectional area about the anis. a) (9 (3 ea @:®+® 67M FDA/EMT dim )5 AC ( Au )0 (AXLA)U bx (mm‘t) (Mn?) (Mm) (Ml/n4") (MW) 0 a f awaitloo) 0mm) {gate} (maxim? 00 F— 6 9‘ mm) =28“ :Alow woos moo) 2531:: it it» (atom? Crow) gets MW? mot) 2—S- _ Ix : 04302-006) :mb :[O’b SXPDHD) ‘ fist 3 a $2: (52's) (1:) (329m) 2( 6 M :wiszubé) :im 0 O 59“; w) b 5 G°°)(2S) J_ , ioO 9‘ (FZ>(AX‘°°) :23‘00 0 0 04101006) I : 0. {501 (I b“) Z (iD‘) SHAFT : Gram; ~Sau125 CIRCLE : Jo : (a) (*4) (r) (5* SauAR JD : (2) (rt) (05) 3 {6’ M ° SHAFT: JD _: Ego“: 2L“? ,Il Ag; 10-52 Determine the polar moment of inertia of the shaft’s cross-sectional area bout the centre 0. A? 6* Tamra/Es KXZ‘. a”; J (5&6 4 b‘ GDMPbNfl/T (Mm’r) (mm’) M 0) 3 $0“ (L {oh (I DQLCOD) ’X (LUMXG ) :50000 [D :[ocfi-q‘ ('0 ® Dom thgoo)ubo); ([ob) (:09 5 0 (Id!) :g'booo © 9 __ («Fl-i)([bo)(|°vh)3 " 000006) JD :a&3§3([o‘) zr/ovmo Z (flioné) $3060 MS 0 X106 kw: If — [0? m V) 10-53 Determine the radius of gyration kx for the column’s cross-sectional area. 41/; 10.5 PRODUCT OF INERTIA - DEFJNITION Ix}, = f xydA - UNIT: [LENGTH]4 ' CAN BE +IVE , -IVE OR ZERO 6’0 00 _10.5 PRODUCT OF INERTIA ' IXy :0, IF EITHER X OR YAXIS CODJCDDES 1WITH AN AXIS OF SWMETRY (1,933 ; (Ix 1='(Ix 4 (12:92:41an (11:33: (113 armsslwzwgwfio 351+aiyh+axgg+ax94=o - PARALLEL-AXIS THEOREM [xy = [1,317, + dx dyA 10-62 Determine the product of inertia of the shaded area with respect to the x and y axes. 43% : o + (M) (5:) (Mr): m (M : i [(2 (PM 33 13 «M K"‘(\/ 9< :— Marlmmm ‘“ \ L33 3 9J6! (3’43 fly) Ag]: (2) (Jag—(afrdr «YD—Nb _3L 4 M , . 5TH ...
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This note was uploaded on 04/27/2011 for the course ENGG 130 taught by Professor Lubell during the Spring '08 term at University of Alberta.

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lec36-20110408 (1) - fifiofiifié In : EUGJ :thy);...

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