hw16_5.9_5.12

# hw16_5.9_5.12 - Problem 5.9 ‘ l 5.9 and 5.10 Draw the...

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Unformatted text preview: ' Problem 5.9 ‘ l 5.9 and 5.10 Draw the shear and bending-moment diagrams for the beam and ' loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment. ' Canupcizl'e rem/+1014: a} VA Mél E- T2erM 9\‘5-l-v~519u+eal foaal lay am equiv”?er ’ ‘ Lancew‘l'wwlej Joan? cat‘l’ew almuw'ng 'Hae ‘F‘Ne eody ACDB. +3543 =. 0; —"5A 4 (W601 + (New, = O A: 72 WT‘ 513 — one» — {3mm = o a = 48 {m 'r, 72—éO—G04—I-I'5 = 0 2M V =(72—80 my“; +37%? = at ,72x +£30>d§ + M =0 M =(72. - )5X2)1<N'w‘ .Ci-oD. 2m59¢<3m +TZF7 = O M :r ‘ 72 — Lso\<2\- V - o 2“ 9 V . V: 12w v . M (EB 4lZF7=o: V+ L43; 0 T7774 +C3>é)C2?(9C‘-D+M =0 'v=—43 kw M=<zxx+gowivw . i‘—>l _..__._______.___ '\ [+3 kw +§EMI1 o; -— M + Hg (51.x) FV‘OM‘lTlQQ alfaﬁv‘awxs‘ 5%” M: 2140 — 48% LCD MW : 72.0 W ‘3 =- 96.0 kN'M 4 Problem 5.12 5.11 and 5.12 Draw the shear and bending—moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment. 3 3 kN 92MB = o: (7002(3) — 1450 +(aao )(a)- 10004: 0 A = 2.55 W 1 ' 200 300 92 MA r O:' ~(300)(3> ‘ 450 ‘ 0007(3)” Woo ‘8 :0 Dimensions in mm B = 3_ 45 VN 1 vam At A. V: 2.55“; M=o‘ A +2, C, v = 2.551%! M A+ C. : VT? 430036255? + M = O 2'55 V M = 755 N. m C +0 E. v t __ 0515. N‘m ' 9.: A+ 3. 43M 193MB : O: . L________? . Lao—4200.1]. + M = o ’ M = €75 N-m M D". , 3 “*5” M ‘9"ZMD = o: 3 -C5003(2.55‘) +Qoo)(s) 300—31200 - £450 + M '= o 2'55 V M.- nas N-m E +0 ‘8“ v ': — 3.4: w m E,M v 592% :0: C -M + (3003(3).ng = o M = 1035 N-M. A+ B. 3.45. W, M = o (0.) Maximum M = 3-45 MW 4 (19‘) Maw-mom IM“ “25 N-m d ...
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hw16_5.9_5.12 - Problem 5.9 ‘ l 5.9 and 5.10 Draw the...

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