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CHAPTER 6 4 y a, id e Ri o E VCS Cotas $.1 Three full-size 50 x 100-mm boards are nailed together to form a beam that is PROBLEM 6,1 Subjecied to a vertical shear of 1500 N. Enowing that the allowable shearing force L in each nail is 400 N, determine the largest longitudinal spacing s that can bo nsed (Ds between ench pair of nails, Lo a J SOLUTION T= pbh'= (o ist= ze jaguios mm 5 = 28.128 x)0* m* “| 4! P . ny A À = (loo)(50) = 5000 me | Zi J = SO mm 7 FA !J A Ay 260010! mm! = ASo wo mê | Lob | gs Ve - (isoo tiago ter) 12.883 x/0* Ním LJ a » 1! GS = 2 Fr ss = $X40or GO xto? m — - 60 mm mt, F 13. 253x165 A | [| £ fa rod la PROBLEM 6.3 6.3 A aquare box beam is made of tuo 3 x 3.5-in. planks and two É x Sin Planks nailed together as shown. Knowing that the spacing betweca nails is s = 1.25 im. and that the vertical shear in the beam is = 250 Tb, determine (a) the shearing force in each nail, (b) the maximum shearing stress in the beam. SOLUTION E? 2 I=Ybhi-dbh HEKSP - flaskasy = api in? to AMD) 3.75 im Ko 25-F - 2u2Sim GQ, Ay, - laica! YB - tason aa) 45 4 SS) . 50,84 4b/in = S — (o s4Xr2s) Fa = da = CECHLOS , gs db = 6 Q,= R+4 QUIsAGNo grs) [ 7.969 + 2297 = |0.26€ in* tr QN sin . VA. QsoXioge) . . Tui ve. Gases) O tHtesi a y PROBLEM 6.4 6.4 A square box beam is made of two + x 3.S-in. planks and two i * Sin ç plunks nailed together às shown. Knowing that te spacing between nails is s = 2 t in. and that the allowahie shearing force in each nail is 75 Ib, determine (a) the largest aliowable vertical shear in the beam, (b) the corresponding maximum shearing stress in the beam. SOLUTION s 1= E bah - E bh, [1 - EXP - (asas) = 39.578 io! K A 03 AG) 3.75 mt a QUELALEICA pm We 25-G 7 2025 in % E Ê + Q,= Ay - (a.78%2.128)" 7.969 in! . ps E Fres (rs) - ”Y Ge = Er = Ego = 75 Ab lin + - VYQ . Igm (ELO À $ T Viu * Age - ELSDOS - 572 4h a 1) . tw Q:Q+ ataIsÃiNo2s) [1º Os Ls = 19644 2297 = 10.266 in? PR Noms E LUG) ns in J — - VQ: (374)M(10.266) + . Tx E 7575 AS) = 644 pai nal ls ca ca o ls lo Ca cd id CJ 6.5 The beam shown has been reinforced by attaching to it two 12 x 175-mm PROBLEM 6,6 pintes, using bolts of 18-mm diameter spaced longitudinally every 125 mm Knowing that the average allowabie shearing stress in the bolts is 85 MPa, determine the largest permissible vertical shearing force. 6.6 Solve Prob. 6.5, assuming that the reinforcing plates are only 9 mm thick. SOLUTION Cafco dude moment oP inertia Part “A QD] 4 Gm] AGE (OS mm] T Cota) Tpplde | 1578]! i37.5] 249.77? 0.08 W250 744.8] 5720 (o) o 71.1 ad. plate 1578] 139.5] 29.97? col. z 54.555 THlal 1 dE 7 = 375 mm “Te TAdtA EI isocêrio mm - 1g0.68»/0* m' Ge Atum * USTS)N37S) = 216.56 niotmm = 216.56 2150 Am Taro Elia) = assento t mt Fm o Thin - (esmo MASEMINOS) = 263210! N $: AF (a XaLés xii) = 346 1X Nm .I EI CAIA « gs Ya Vez a .4 TA 20% nto N = 209 kN
825 EPk Design is not acceptoble. — : - Edy 3 2Sxlo? a Resign A = E Ta * TRasrios US. IS wo wm El] Us ago! am h = a = AS Igulo? Tão = E79 tm h=37%mm at O Ls PROBLEM 6.16 6.16 Knowing that the allowabie shearing stress for the Limber used is 130 nsi, check whether the design obtained for the beam indicated is acceptable and, if not, redesign the cross section of the beam. Consider the beam of (a) Prob. 5.77, (b) Prob. 5.78. [&) SOLUTION Vas T AGIR 36 lips trom sodotimm to PROBLEM 5.77 a = G6iwn As ata ads in” For a rectangular sectron Tom * $ Vire 3 ão - . . Tum ET É = OMS ks = 265 ps € I32 par Design is acceptable . E ni » (4) SOLUTION F= E fo From solotion to PROBLEM 5.78 tosa —— Was > 1000 4b b= 2.95 in. A = babe able N7.90 int Iv mas For «a rectangular cross seckhun Tomas T 2 A . 3 lo0o . « Toa * não 26.2 ps < 130 ps Design is acceptable. LI . - O O O CS 6.18 Determine the average shearing stress in the web of the beam indicated and PROBLEM 6.18 check whether the design obtained earlier for that beam is acceptable, knowing that the allowabje shearing stress tor the steel used is 14,5 ksi Consider the beam ot(a) Prob. 5.83, (b) Prob. 5.84 24 kips (a) SOLUTION 2.7 Kipadt os From the soluticoa to PROBLEM 5.83 vias 7 48 kipr The selected section is W 27x 8% For that section to= O 460 mm dz R6.T) in. Amb = ds to guto Kage. 11) * 12.29 in” Vie Wi MB ga ks < 10.5 doi ee O Aedo MR Design is acceptable. iskpt (b) SOLUTION From the sofution to TROBLEM 5.84 Nico EUNMoS+ LE) 18 kipa The sefected sechon is Wigx50 tu = 0.855 in. dz Gin. For that section t.d = (oassMr a) = 6.37 nt Ameb E |V hmm Tu? - É = Ad 6.39 2.82 Ms; = 14.5 dese Design is acceptabte E. + Cd — a o N a] a oa &1P A simply supported timber beam AB of rectangular cross section carries à PROBLEM 6.19 single concentrated load P at its midpoint C. (4) Show that the ratio 5, /G, Of lhe meximum values of the shearing and normal stresses in the beam is equal to 4/2L, where h and É are, respectively, the depth and the length of the beam. (5) Determine the depth h and width b of the beam, knowing that = 2.m, P = 40 kN, É, = 960 kPa, and 0,= 12 MPa. SOLUTION Recctims RarRgr P/2 V Dar Rr É 42 (0) À = bh For rechang ular section * 65) E z É Mm . + For retangular section - PR MM Mia E EL M PL/4 O ss bh! fo rectangular section DAR x Bo. O sa - tb) Sofving Porh hz der - (2a aco nto”) = 820 416º m = 320 mm a Solving equation (3) fe bh 2 3P (Xdono) º “+ bEqhe * Wiiodo Nana) + 177NO m = 977 mm a 
