hwsoln26 - 11.11[11.4 Determine the maximum hbfinai stress...

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Unformatted text preview: 11.11. [11.4] Determine the maximum hbfinai stress 7” anc- the maximum shear stress 1-,, at a section 5 ft from the right support for the simply supported beam shown in Fig. £11.11. 100 1b Find Reactions EM1=D (100)(5) + (40)(1) + R2(18) - 100 = 0 R2 = -24.4 1b ‘ EM 0 2 (100)(23) + (40)(19) - R1(18) - 100 = 0 R = 164.41b 1 At position 5 ft. from fight support: v = — 24.4'1b M = -(24.4)(5) = -122.0 ftjlb 52%4flligligliil = 4.570 psi (l—z—Hzxmnz) “— T = 1%§§11%31131 = 274.5 psi ‘ (IQ-Manse -_-_.__ !l(txy)maxl= 2.23 x 107 Pa E11.14. [ElLfi For a shear force of 45,000 N for the section lshown m Fig. P.11.14, what is the maximum shear stress 1- ;away from areas of stress concentration (at comers)? WWII/l 75mm ; \\\\ iOOIA‘. hymn] =H(45,nno)[(_.100) ( .025/2)( mm + .025/4) 4' (2)(.075/2)(.025/2) ‘ (.075/4)jyfi(%5)(.100)(.100)3 — (%§)(.075)(.u75)3](.025)}|' 1)’ 11.27. [11.4] Three 50 mm X 100 mm wooden plank-s are glued together (see Fig. P.11.27) and used as a simply sup- ported beam of 3 m to support a uniform load of 750 Mm. What maximum shear stress do you estimate the glue will be subject to for these conditions? H100 unnai “bunk. = t vmax (750)(3)/2 (at Supper s) = 1,125 N - ;_ a 12Z — 12 (.050)(.1oo) + 2[%E (.100)(.050)3 + (.100)(.oso)(.075)2] = 6.25 x 10‘5 m4 Assume $01 id beam 1?. a: 1 125 .100 .059 .075 ' yx (6.25x10'5)(.050y = 1.350 x 105 Pa — = 5 P HkTyX)g1uel 1 350 x 10 a F1nd5 Izz-abaat centroidal axis _____—._...—_.——_—— - 1. IZZ — (12)(-025)(-075)3 + (.025)(.075) — (.075/2 + .025 - .0375)1 + (%E) (.075)(.025)3 + (.075)(.025) (.0375 - .025/2)2 _ -6 4 I22 - 3.320 x 10 m £§§202,§2119_22§1 [TyXI=|(1292)[(.025)(.075)(.072/2+.025 7 - .03753/(3.320x10‘5)(.025) EE&XI= 7.296 x 105 Pa idF = (7.296 x 105)(.025){dx) . —-L— It. ‘ ' dMl “’4 f=§£= (7.296x105)(.025) ‘d—x _ ‘ x _ 4500 — 1400 _ nf = 7.296 x 105 .025 VI,“ax - — 1292 N ( } Find centroida1 axis (take moments about n'= Ju— (7,296 x 105)(.025) = 20.27 NAILS ‘base). ? 900 m (.025)(.075)’(.025+.075/2) 1/" ~_— 4934 m . + ('075H'025H'025/2) Nai'ls are 49.34 mm apart = §[(.075)<.025)12 " y = .0375 In ...
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hwsoln26 - 11.11[11.4 Determine the maximum hbfinai stress...

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