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# Sol HW9 - w Problem 6.1 3 6.13 Two steel plates of 12 X...

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Unformatted text preview: w Problem 6 .1 3 6.13 Two steel plates of 12 X 200~mm rectangular cross section are welded to the W 310 X 52 beam' as shown. Determine the largest allowable vertical shear if the % shearing stressin the beam is not to exceed 90 MPa. Wela 7_’é mm X '45.8 MM Caupcu [hie Mame-4f 67C Imer‘l‘ice- 12 mm X 200 mm 'Aom‘) aw» 'A‘a‘ao‘mm") .E comm 2400 *I 65 55. 34 GQYO o 0 2qu “ .55 65.3? TOP P)a‘\€ W 3| 0 “52 80+. PM? Jimemsioms 9.: 2A}: 3:2.73w/os m’ = 812.73‘K/O"m3 t= m mm : Zéx/o‘ m f r MEL V: 43.33}: : (ZQ‘L741(0")(7.6>‘Io'37(70><(0‘) 73 Q 8(2.73*lo" : 210x103 N 2:0 W A 6161 Problem 6.13 6.13 Two steel plates of 12 x 220-mm rectangular cross section are welded to the W250 x 58 beam as shown. Determine the largest allowable vertical shear if the shearing stress in the beam is not to exceed 90 MPa. Cmioagail‘c momem‘l' 0‘? in ew‘l’Ca. I = 2 A4“ + ‘2‘}. = mmsézno‘m” = ”reap/o" w." TM,“ occur“; cl” hed'l‘v‘v? “gr-‘5. t: 8.0 MM = 8.5¥10_!w GD £35.er W4 we ® TOP Plait C3) (’3 Top Wan e 326.805 . C3) Hakka/ea 50.625 2 Dt‘mcmsfous in mm: C) IQXXKO; (i) 13.5):205 5 ® 2.0x1\2.5’ Q = 2437 = view/03 Mm’ = 725.6:w 10" m3 va \/ = It’Z _ (179.342xzo" (8.0m?) ﬁowoé) If: ff 0. ‘ 725,611,404 2 H73” :03 N \/= 177.6: m A 6‘“ Problem 6.18 6.18 ‘For the beam and loading shown, determine the minimum required depth h, knowmg that for the grade of timber used, Han = 1750 psi and Ta" = 130 psi. 750 lb/ft Eh]? ﬂow? (750 ﬂé/Fi )(lsﬁﬂ : [bi/(>3 jg Readiion (ii A R3 = 61/03 )5? VW = a x «03 41 MW = ix (3 Paloma) ~— Q‘NOS 1"” .— 232 x103 IL-in Bevmi'mg: 5 =' ‘46 E51 ‘FW‘ reVYLMJJja¢ sec/Hon. Mma 288x103 64,4 - i750 h ___ [.353 t/(CXI:V.S7) 1.14.05 ih S:- = 1e4.57 .313 Show: IF 715 £9143 For NJ’an—Jajqr SeaHovr. y _ ﬂ A = a“, ' " ék - 7 j = ﬁk .Q .— A5 .— eiox.(§m&m=§w H4 T t v_Q_. r LVN w“ I5 \ZEH h ~ 3V1.” _ (3%me eggsm. V_ W723; ’ (2 53030} = TLe jqwjer Value o‘F 14 «‘s ‘Hle mfm‘mow. Nﬁuir‘cj Jep‘i'in. h? H.05Fn.‘ PTObIem 6-21 6.21 and 6.22 For the beam and loading shown, consider section n-n and determine the shearing stress at (a)‘point a, (b) point b. 500 mm 30 mm 30 mm 20 mm V Draw ‘Hme shear albagvmm. \VLM." qt) 1:“ ﬁokN X 283 -qDLN @ 1600 40 5* I600 40 54 *_5 — _ ‘Héwos Y‘ .24Ai‘— gqoo = 55mm I = 2AA“L 2i .— (411,000+ mayhema W? = 5.8!33x10" MML' = 5.9133* lo‘c’ M" (BX A '— (3O)(Z°\ : ECO wwvsz 3 -' 65 -' ’5 = 50 MM _ 9+ = A3 = 30x303m§ = 30r10‘m3 ’Z’ = \/ Q1. .- (QOXI03\3§30>¢!6‘) 5 b It (.5_<z:33x20“)(2oud3) ->IZO|* T5 = 23.11 MPO. “ = 23.2w 10‘ Pa Problem 6_ 25 6.25 through 6.28 A beam having the cross section shown is subjected to a vertical shear V. Determine (a) the horizontal line along Which the shearing stresS is maximum, (b) the constant k in the following expression for the maximum Shearing stress V = k— Tmax A where A is the cross-sectional area of the beam. A 7 20415” = U1 I:2(éu,3‘) : £5143 Fol~ a CU+ 05" jocvjiovx Y) ULeVQ— ys L) Ag>= Wm = ‘31? ~ .M9_. L6... b‘ehv 1- 2 ”i“ It ' Vere E3“ aéL‘gpL E’Wh) 2%)] dial (a) To ‘Fivwl ‘er IDCJ‘IOW or MMI‘MUM H 7:) Self g}; = O, a“ - 13'3- LHC-"J = o 1,? = :3; e—L‘Mrm ‘6’ b h,— he'J‘\'V‘J MCS . “ up) ’L’tthZ—ﬁ—[C’zékdﬂﬂ - 43%]; = mas—2% , krMQS‘ M Problem 617 6.17 For the beam and loading shown, determine the minimum required width b, 5 ' knowing that for the grade of timber used, (Tau = 12 MPa and Ta“ = 825 kPa. 2.4 \<N 4.8 kN 7.2 kN 2.4- m 4.8{44 72k“ A ’ ISOmm 0.5m 9:: MD = o ‘- , i 3 A “212.43 + (new) «051(sz = o A =' 2 m 'r ' . . M kNv Draw 1H6 skew» amai bet/winds) momew'i aingmws. ’ m MW: 11w = 7.2)403 N MM; 3.5 tau-m = 3.6xl05 NW ‘ . - M Bel/Linn: . 6' S 3 1 Sm“ t Egg”? %%':3%‘ 300x10" m3 = 300x103 mm pow a. rewiiavxiu’oaw sec/'i'iow) S = é‘E iﬁz __ GS _ (é'KaooxlO‘) _ b - MW. ~ W ~ 80 m4 Shem»? MMiw‘vM shearing s—(waﬁi Occuﬂs aJi‘ A ‘Hne nay-imaﬂ wxfs o‘P ioevmi‘in? ‘FDA ‘ T a W: ai‘xmjupaw sec/+800. h be .1 l-<——-+i I=ébh3 trb b 1 - 1.9— t vegm 3%. It (145% )(5) w, b: ELV -_- (33( 7.2*£033 3 27 3 x103m ...
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Sol HW9 - w Problem 6.1 3 6.13 Two steel plates of 12 X...

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