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hw17-10sol - PAN 3 fl Sowflwg'6—49 Determine the...

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Unformatted text preview: PAN 3* fl, Sowflwg '6—49; Determine the maximum tensile and compressive bending stress in the'beam if it is subjected to a moment of M=4kip-ft. Section Properties: _ _ 2 yA ’ ” “EA— . W_ . J21. £~ . “71;.“ ’ 4(0.5) + 2[(3)(o.5)] + 10(05) ' 340‘“ M“. T i’ W a” 2, INA = 115006153) + 4(o.5)(3.4o — 0.25)2 [ 1 3 2 + 2 E (0.5)(3 ) + o.5(3)(3.4o — 2) ] 1 + E ((1.5)(103) + 0.5(10)(5.5 — 3.40)2 = 91.73 in“ . , . . . - Mc Maxtmum Bendlng Stress: Applying the flexure fonnula amax = ——I— 4 103 12 10.5 — 3.40 (cram = —(——)~(—)(————) = 3715.12 psi = 3.72 ksi Ans. 91.73 4 103 12 3.40 (mm = ~(—)7(1-%£——2 = 1779.07 psi = 1.78 ksi Ans, L l QNE‘N? 6:110“) : (LAY \Uv ("\Lwi‘m J knwv‘?u\;1;~n A £W12’<A\{W\ 19210644 QJBGK (re/$34} mm = 1m (WW :H "W, WM 5g WWW) (Malt: \-15 :1; ?\ND ~_ M My“. Kw JAM, 0'9“" Wis Wk ATM v 5536??) — TUILSTO“ ' 3171.“ MM 4h . 41? avg MM. Uofig—aqu‘sn M g: <5 -3112; —— 4.11% .‘x axm % \m f. 4:1“ ow. - (Acmkm sowmn; x? —— \1 :- 1311‘ :‘AJC WW Y—aK/‘WW . . M (3.11%?) a v Mb , um . _’ W MAW " :1." Tu? {S cm 3’4} 2’1 \“V‘ > 131,113, L. J, Lawfrvy-z«:M- in“) ' M g 11 \«g {13 MUM EDT)“; M4. \354 m (50W 6? ans/34;“ My; 2 M (4.1mm “wk #40 17mm“ M E i4 L; (131.2;fl 4.112.; . \V\ {_ \3‘C6 \Lr‘m :MM s \m w : mm m gw 35500 g1!!! “74:an .1 \ T3005 TDUI’JUM (0’ (fl (maxi) GWEN : ___‘L—__—’______ RA T \‘BTV‘ * \SV “ 0—wa : 11” PW: 61%“:th soLu‘rwJ: 5Y‘MAFXZ)‘ at: (\(90 $4.} " KBOWQV) LIA '150U'7K/503‘W id [\2‘5 :5L1.5U}, T MW: SL325» db. — 150w mama/i ’RA: (137.51}; It ems «315L353 : Womb?) F615 Vb!” :\%1.5(\53=1is\2,5 ‘ _ _ _ X —SL1.5 L50): #3525 wow?) “05‘“ MCUb'NW _ ._ “X No 0' z —— mm 1 \ r 61M“) 2 15(4 4, v3 i m WM...) > '5 4 L“? 6615 LOW/‘3 r _ J 'VVQ'HOOOPW 3 . i— 1 } ‘o‘q‘wmafl . v : O.C\%\51\« 0L: L‘WM 82114} A: 1.0 ‘m Beams—Internal Stresses ~ Normal Stress VIII The figure below shows a simply supported beam. The labeled points are evenly spaced along the a length of the beam and at the top of the beam. Points A and E are located directly above the supports Rank these points, from greatest to least, on the basis of the absolute value of normal stress. mm, ¢ mm m :3 4—50; w— W 345 Greatest 1 Q 2 ' C’ 3 E Q A :5 E Least Or, the normal stress is the same at every point. Please carefully explain your reasoning. » (Tm 2 Eff l A’stmfi 1% ¢ wvilall mania/LN“ W ”HM w ('5 ’ V” {WM ”W (”W‘ ALW‘L” )l WV 0'” VAN an, GYLDMJ‘ ‘m‘ ”50.4li 4—. VAN £ MAM] mtg/41% f. 9M» QM“ film ‘ MB 4 M° 4 M9 41% max >01“; >0‘mm’flflum: (/5) How sure are you of your ranking? (Circle one) , Basically Guessed Sure V Very Sure . 1 2 3 4 5 6 7 - 8 we? 9 10 73 ...
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