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# 7 - j Prob 6.8-4 A simply supported beam AC of length L...

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Unformatted text preview: j.- Prob. 6.8-4. A simply supported beam AC of length L = " _ 4 m supports a concentrated load W = 1.0 kN that hangs from the beam at B, as shown in Fig. P6.8-4. The cross section . of the beam is 50 mm x 60 mm. Determine the ﬂexural - stress 0(x, y) and shear stress 1(x, y) at three levels—y = -. 0 mm, y = 10 mm, and y = 20 mm—on the cross section I just to the left of the load point B. P6.8-4 Qa/{fm r'n: ﬂfxalé/Jf/lrf dhc/fétm/ {/é’cr/ ﬁ/ AZ/ywm), zit/5740mm), C: (7’20MM}. Ezu/ﬂ‘én‘um/ (6/ n7 ago/V) (5/1/05 =0: 6’, = o, 75 AN (imam (7 =0, zfé/V M- —- 0. Um. M _.£_._._— 14. : 0. 75AA/ ﬂ/ymw/of Max ﬁfe} 54’ 1; ‘ 7? WI! 4’7 - l/E 7= 71 , 7"” [a -- 7.? am / =0): 7 N 1.___(b>(“/ZW4) N 1,} 0; =0 } [A : (M )4, :E—é—l’l :375épa '41:; IL E; ﬂ/ 5’ (wow )‘ — M (6/6 l 2 M 8 = (E) =' 6A1 ’2 AMA/n 4 V é . _.'= V(./>)( _ , a 7E - ff); ' 3’27 397:? p Iz ’ ﬂ/(H‘M‘ﬁh 0' - ”W, 7 =44 MM c ,. ‘ 3% ) ‘Prob. 6.844. As a set of wheels on one axle of a railway freight car passes directly over one of the cross ties, each wheel exerts a force through the rail and through the tie plate beneath the rail to a cross tie, as shown in Fig. P6.8- 14a. Each wheel load can be represented by a uniformly distributed load of p. = 32 kips/ft over the 1 ft width of the tie plate, as shown in Fig. P6.8-14b. Assume that the cross tie distributes this load as a uniform force-per-unit-length p. to the ballast on which the cross tie rests. (a) Solve for the distributed load 17.. (b) What is the maximum shear stress, 1..., in the wood cross tie and where does it occur? (c) Determine the maximum ﬂexural stress, am, and indicate where it occurs. Rail Tie plate (4) Uri/rile fr/ rate/429m far/”Zak m .’ *T {6‘0 I it (gm —(azL/#){2,c/)=o 141’ 6’ ilk/4f VM‘ = M A)“ Mutt : /é £9,qu /éJ Mac/mum {Aca/ .rﬂrrf. k) Mix/Mum //CK“r4/f//err, : ”(at its“! émmu 0;” T = 6A2. /6’r}1.)/g,f,,‘,. )2 = z 4053 As; 0;, = J/X'Yfé 743.4%) = 5,4/érf ' I' a? 'It‘ -.\: 21“ -:' .1 u'" . ' -. a '. -! 1‘: I; J‘- . _ iii '- é .. 3‘ L i‘.. i- E l . ,. a ' ' f. I' t r .“ ._ ' .g- --; § I: '-.'- 1 , 'F _ t . l . . . J. . -- '. E _“- ‘5. . _ )I I '. 3 1'- g : :t "I' '-" 11"?2': Prob. 6.10-10. A channel section is subjected to a vertical shear force V = 10 RN at the section shown in Fig. P.6.10-10. Determine the values of the horizontal shear stress 134 at the point designated A, and the vertical shear stress 1, at the point labeled B. Assume the thickness t, to be constant. 15,21“ij 125 mm 125 mm P6.10-10 5 % 2(l2mml(‘=lla mml‘lssmmH 22bmm(ltamm5(‘3mml 71:2 ' 202 mmX‘Hpmml + 22b mmﬂtomrm =_\$_O§5me_ z. l2mm§bmm13 ' 7?. I "' 2———-[ + l2 mmFHommXE ham-702] - +[zalp n42_____m_____(llpmm\3 _ 2] [Z 22L, mm(llp mmﬂzmm —7O 2 +[th 5 GQL)mﬂﬂq Q». = ZAA/QA/ = l2 mm('-llpmm3(58mm‘ 77L) 4 anomalommy? mm “7—0 T I'2mm 'illlmm A Jr. = IZ-Qj'lllnsl [KHZ]; ”0W" VQA (lOkNll2. om loﬂmm?‘ K—H50mmT If It». = WIT—m 500" mm‘l—(uomrm - Stages; mg Er ﬂint BI _ ’H, 1‘ = Anal-«1‘5: |2mm (Momma—7k“ m3: ) =20.lﬁl§§lg”l Win43 , 05 BD I—“pmmwl \an (lOkNlzoJeBglo‘ﬁnnh’l3 —>' *— TB= In. = 2,04usuowmmmzmmw =L205LMEm l2 Mm =‘Z. 20 MP4, Prob. 6.10- 13. A rectangular box beam has the relative di- jected to a transverse shear force V, what IS the ratio (r...),l (1...). for the beam m orientations 2(a) and (b)? In each case, indicate where 7“ occurs. I = COHSI (a) (17) P6.1 0-1 3 12!: :(zotﬁ to: (181:33 4 = I2. ‘ 12 = 5I4o£ W - maximum Shear stress occurs at NA\thre (th- 2A4ﬁ’\- 2[(OL2(?/B]+ Iotzh. 5b = I515 t3 V max _Zf_‘L_ (Mix: @3239; = [2519112 (Qmaxib t 2A3}; = 2(utZX3t) + I2+2(5.5+) 285\$: 61V m L» NA (EMAﬁ It, (2:) ‘ #84 t2 ) 131m; ﬁor both boxes maxim ‘8) W (Tm... a = %_ﬁ_= [05 57 atmes occw’ 6d: vro , or the neutral (Tm "’ 0ch 1‘5 /? T’l’yrob. 6.11-1I Four 2 X 6 boards (nominal dimensions) are nailed together to form the box beam shown in Fig. P6.11-10. If the nails are spaced at regular intervals of Ax = 6 in. along the beam, and if each nail has an allowable force in shear of Vm = 300 lb, what is the maximum vertical shear force. V, for this built-up beam? (d = 5§_in., and t = 1! in.) c:/‘£.f/9.I /- K5,», :4 [-— Mala/3 _ 4mm? ,1? I a / /z ) )’ /z )/ ‘l’ and f a ark. 5J2 A752, 2.3%, , I é M _ In '://0,é7// _..l H l“ J—z’ . . 3 z‘ ‘1 2‘ and = J// T ) = Ari/n. ){malfz/z.) —— m. m m V -1 , Z 17“” 7/,w 4)“, = Z Vw/f_ z (foo/A )Maam) Z 1141/ ’ AK _ : f I / 7/1..“ 57‘de - //é,5-Dl'n ”(é/IL) =670,6j£4 K = é7/ a I’li'rob. 6.11-16. Two 2 x 8 boards are attached together by screws that are spaced at regular intervals Ax = 12 in. along the length of the beam to form the inverted tee beam shown in Fig. P6.11.16b. Each screw has an allowable shear-force capacity of V. =- 480 lb. If the beam supports a single concen- trated load P at the center and is simply supported, what is the maximum allowable load 1’? P6.11-16 :3 241 ' ' ' ' ' 77 ‘ 2A = _ 7.5ir’1..(1.5m)+ 1.5in.(:.5.‘n) gill]... 1.5]n.(|.5in.)3 _ . _ 2] I‘ IS? 15i+);l.51n.(l.5m)(’2.25m3 I 1‘ ‘ . . _ Z_ *[ h 12 In + l5in.(‘-l.5m.)(2,25 m)2 7L = I] \$.35 ‘34 05 = A; q; = l5 in.(‘+.5in)(2.25 in) = 2.5. 325 ‘n?’ WZML= o: AjL’ [3("3 — o A.’ =1“ Win/ﬁrm Phil—C51. = ([3) Cf?“ ->vm= P"? = VWQS ,M V ds = 23? I ‘ 2:1: _ 2V5]; _ 2(‘480llélllb8ﬂ‘5 m") '> PM" HQ? ‘ l2in.(25.5l25in’5) = 555.55 llg ...
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7 - j Prob 6.8-4 A simply supported beam AC of length L...

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