6 - Prob. 6.24. Couples M, are applied to a steel strap of...

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Unformatted text preview: Prob. 6.24. Couples M, are applied to a steel strap of length L = 30 in. and thickness 1: = i in. to bend it into the form of a circular are, as shown in Fig. P62-2. The deflection 6 at the midpoint of the are, measured from a line joining the tips, is 3.5 in. (a) Determine the radius of curvature, p, of the strap, measured from the center of curvature C to the heutral axis of the strap (which passes through the center of the cross section of the strap). (b) Calculate the maximum extensional strain, em in the curved strap. 4} fat/1w 05 («n/a 14m: h/z. QWW ll/z {i/IONA- at“ = /0 6w = /. 75200” 5/. 6—451. P6.2-2 A [’/0/Z¢) ¢=E7 f=/’(/’Cor¢) vol/ear 5,2” 3.570, =/0 [/- (al{':;;")] JE/ur [Ar algae 744/10: 710/ / =j/,‘f4/h. ——> J=£faaz,h. I4: = ——‘—“ = A ‘idlél/o" 7,7. Prob. 6.3-7. A timber beam consists of four planks fastened together with screws to form a box section 5.5 in. wide and 8.5 in. deep, as shown in Fig. P6.3-7. If the flexural stress at point B in the cross section is 900 psi ('1‘), (a) determine the flexural stress at point A in the cross section; (b) determine the stress at point C in the cross section; and (c) determine " ' 5 5 m the total force on the top plank. NA . __-_ - A 2.0 in. " J [/l 1.5 m. 0.75 in. . 2.5 in. 1-5 in- l.5 In. P6.3-7 duejco s mmeb’lj, cwbro'ld isin middle, of cross—Seceionfio 7T: ‘4. in. __-M93 53' I Ii O’A=_ I @3015 =%)fiwfii = " FLOO g? 9mm . 05:ng = (3.412353 = (%‘F2Mfioofii 0: = 67W ; 0'4 1.5m. 553m. 1.5m m 2.15m. O’(Lj=2.?5 in) =(2.25in)‘loo P5; = — H . -l'+ooEsi+Iloo§' ' . 2 ‘(L.5m)(5..5m) “(Krakow 0' Fm = ' “00 psi (L5in)(5.5in) + = -— ' _U_5_kuzi Prob. 63-29. For beam A15 in Fig. P6348, (a) sketch shear and moment diagrams, and (b) determine the maximum flexural stress in the beam. /4) 2164/ doc/moman/a/‘z I’de Ega/%/n’amr 475w), =0: ~//é,;o)/7/f) $5104!) —/z Lad/2,44 f a £44 ,A/ =0 3} ’ / [ff/kl “Mime =0: (/éq'o mm '(Z£I}I)/Z;4/J My Mm - H123 'fl‘ =0 0'25 fl} = / 7f/4/ryf fl] MdX/I‘YH/m 7481mm / {fled -MJ 4;" T 6- : Mil /Z = /7. 3/44 m. t : (é Mufti/e ’M/HMJ a” mam 0', ¢ ' = //I W“ 6;” = W5 A; I’l’l’ol). 6.3-35. Determine the maximum uniform distributed load. w, that can be applied to the beam with overhang shown in Fig. P6.3-35. The allowable stress (magnitude) in tension or compression is 150 MPa, and the beam is a W310 x 97 (see Table D2 of Appendix D). P6.3-35, P6.3-36, and P6341 " beam Worm ?(éM)B =0: AJHml‘ w(lpm¥)ml= o A3: W(i.5m) '1 A C 19(EMlA‘o: woomYSml " Bfi‘imk'o B B.j = w(4.5ml ‘1 A B l K—tj—Zm—fiklmak—‘JZM—H —2w — — — —" Mn!th = - :FromTalole, D2, al= 502mm and "1:222:00") mm“ __ = M =‘NmaLZJrfld ZIz z. 25‘] o WMAx ‘ (Zmzlg zgxmmzzzugbl mm‘l (2m2l(508 mm} = /08./Z kNgm UAHOW WW: Wm w = 1.5 kips/fi I"Prob. 6.4-9. From Table D.1 of Appendix D, select the lightest wide-flange steel beam that may be used for the application shown in Fig. P6.4-9. The allowable stress (mag- nitude) in tension or compression is 0.... = 20 ksi. P = 20 kips P6.4«9 ft/cc/ fifé/rr/ r/a/ Jam . Z ,‘I’én'um,’ f“ [fl/kr/ 4 26:0: {Db/'4, - fob/’4; 5% =o T4 = - /o A}, “(Z/WC to: MC *[3041241)//0//)- [mud/12 fl) =0 M .- —60 MW MW *1 leyll ’ 60 A,;.,£/ 19//0wd!/c ' {/Z/crf 09/1» : I 304,th fl’3 0' fit /%/rf/M¢m £7 mfg/Ia 51/! a/ fl/wJ/M a «Zr/pa . "(If/")5 ‘0 I' ‘ l’I’I'ol). 6.4-15. An inverted structural steel tee beam is sup- ‘1 [“Av ported by a pin at A and rests on a toilet at B, as shown in _f Fig. P6.4-15a. What is the maximum load P that can be hung from the beam at C if the cross section of the tee has the 554i" dimensions shown in Fig. P6.4-15b, with t, = 0.420 in., and l the allowable bending stress (in tension and compression) is 03... = 20 ksi? L5.02in.-J17 wonm Ml/mum /0¢a/, +(’(2’M)6 =0! H} (10/!) 1‘ F/fff)=0 l9; 5' ’ {P /lI/'u) @(w/H « P//5’fl)=0 45/.{P MW = /M//a//:/ r 5? 4/474! 7, e an!» #276,514): 3.6%. 7’, = $70.4: )=0.z/ ,2. fl. = {a.¢zrz.)/6.57rrh.l — [.7465th '71=/0,4z/h.)/£ozlh,)=z, watt/2‘ é'fi‘h'L. ’ I I a: '2. " fl; 1* l ’ 7: 7 z: =Z/7f5’m- (~— ' - E was 1 L— -—11 '1‘ - aflM. 5”“ é%m%Z;%mc£finz = Mm Jam (57’ AM/ We Ms. W) 4/1th [ : 27, 267/34 p _. (Mind/w.sz {57/102 1%,) {mm 5 /. 67/5, IP= [67Z/Ikr| [44/44/144 Mic/x) . ' f L : o’in */M,.;,)(4,o7,>.,s{Mi-m) , I ,t I: [M3, ; /£oZ/h./(o,~fl/}.J/Z./B'Oll/y] IL z wow/27‘ r 0’. 20m" = M, 2457 ,a‘ ...
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6 - Prob. 6.24. Couples M, are applied to a steel strap of...

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