Sol HW8 - Problem 5.1 7 5.17 For’the beam and loading...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Problem 5.1 7 5.17 For’the beam and loading shown, determine the maximum normal stress due ' ’ - to bending on a transverse section at C. 25 kips 25 ldps Read/om :11“ A_ +DEMB : O - 45.0 R +025 )(257 + 00,0)(Q5) +- (3L15,)<7.5)(5) =‘ 0 9,544,375 (gigs get/31M? Morse-4+ 651' C. 92M¢=O ~(2.5)qu.2753 +' M = o A c M M= It7.i275 kip-H: —- 1.40625“ 1.03 10rd" 25. For WléX77 rowed? she/V seaflon S? 13L) {W3 ‘ 4‘37; V Norma/Q 5419.35 64’ C. 3 g: 154‘ = l-qfi’ffsxzo : 5: [0-97 ks! «é Problem 5.21 5.21 and 5.22 Draw the shear and bending-moment diagrams for the beam and loading shown and determine the maximum normal stress due to bending. j kips . 25 kips . kips ‘ I :9 E M B :7 O B (IliC25)- \oc +(2\(2s)+(21(25)= o $12><35 C = 52.5 16,0: 1&2ft 2ft 9 2 MC ___. O mam—(mus) - (9X25) + In B = o B = ‘6'” Shear ‘ A +0 C V = "‘ ‘(C‘es C:+ +9 D’ V = 27. ‘5 Mrs 3* +9 E‘ V: 2.5 kr‘PS Ew‘l'h) B v = “22.5 16195 Behalf n3 Momen+s . 15’ x A+xc A c3“ 92mm ‘ +11 mash-M = 0 V Mr'25 2- M MD be»? 92%“: T (3X25) - (21(52. 5) 52.5 V + M = o , M 3‘ 30 A+ E MC E ‘3 m 22.5 max “4' = [+5 1' 540 Ripvin = o -M + (21(22.5\ : o M= “5 W” For SIX X 35 roiled she} See'i‘iom :3 32?.Z in; M Norma! s+ress 6 = is". = fl =' 14.14 ksi A Problem 5.65 5.65 and 5.66 For the beam and loading shovvn, design the cross section of the beam, knowing that the grade of timber used has an allowable normal stress of 12 -?.HA + (L6)(L87+(o.8)(3.£) :' o A: 7.4 kid +DZM$ = O 40.3)03) - (1.4)(3.<) + 24'! D = o D r 3 “N Coms‘l'ruc‘} Shear amt; bemdlnfl momevx‘f' diafiv‘ama \Mimme ’ 2:4 kU-m :‘ 9.43110, N.“ 6‘.“ r -, :Qx/o‘ Pa 2.4m»? - 3 7T2???“— : 200 “’0 5"“ T RCMDNID’3 mm: 5 = e;- bh‘ = «$0403 if“ = 200x10? ht : e—L—Woa : 3ox/os Mml 40 h r [73.2 mm p bl 5. 85 Determine the allowable value of P for the loading shown, knowing that the ro em 5.85 . . . . . . . allowable normal stress 1s + 8ksn 1n tenslon and — 18 k51 1n compressmn. Rear/‘hova. B = D = l-SPl Skew Jig. \mw‘- A ‘lb 8 V: - P 8* o c‘ —- ~P+ 1.5 P = 0.5,? C‘luD' VEO.5P—P=—O.SP 0" +0 E V=‘0-5P+L~5P *— P Areas. A'l‘c'B (ION/P) = -10? B to C (603095133 .- 30 P C 'l‘o 2D (goX-OJ?) : -30 P D *0 E (“73093 = [OP Bevmhvxax momed‘S- MA: 0 MB=O—IOP = —/019 MC=-IOP+30P= 20? rx, . Mp: 20P-3OP :"/0? _w[? _|DP Me 7‘l0 P + '0 P 1‘ D I Lafaes’l loosf-llxm Leualirw, Mame-4+ I 2019 ‘ Lad/363+ neZcJ‘lvz bangean Mag)de .‘ -]O P a) l CEVI‘l'V‘ol?) Mall Momen‘l‘ o'y'lvxer'l’ic... 772/0) COMP, - l8 P: 7.8? Rips Bel. Teasiom 8 r “e P "" 8.57 kt!” 3.7.25. 80+- Comp ‘* I? r ““ P 3 3.3.3 ‘0?” Shad/55+ voice 0? ? ls "He flows/Lbs \lé/PUQ 9': 7.01.. L595 4 ...
View Full Document

Page1 / 4

Sol HW8 - Problem 5.1 7 5.17 For’the beam and loading...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online