tutorial (13-16th April)

tutorial (13-16th April) - Problem 5.5-“ A compound beam...

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Unformatted text preview: Problem 5.5-“! A compound beam ABCD (see figure) is supported at points A. B, and D and has :1 splice (represented by the pin connection) at point C. The distance a = 6.0 ft and the beam is a W 16 X 57 wide-flange shape with an allowable bending stress of 10,800 psi. Find the allowable uniform load 9'le that may be placed on top of the beam. taking into account the weight of the beam itself. MW: 2 =61]th _. 20.,th I I gm = 5a: qrallow = gm — (weight of beam) fim— din—of DATA: a = 6 ft = 72 in. alum = [0,800 psi .114“! RE: 459:1 RD: 2% W 16 x 57 S = 92.2 in.3 34—1— Aunwnnua UNIFORM LOAD u 200,800 psi)(92.2 in?) gm" 502 in.)2 = 9221bift quill"W = 9221b/fi — 571blft = 8651blft +— Pin connection at point C. = 76.833 lbr‘in. Prnhlem 5.3-5 A steel beam of length L = 16 in. and cross-sectional __ 3;: 2393??!9; dimensions b = 0.6 in. and h = 2 in. (see figure) supports a uniform load 1 i i E I of intensity q = 240 lbfin., which includes the weight of the beam. _ Calculate the shear stresses in the beam (at the cross section of maximum shear force} at points located 1M in.. “2 in.. 3M in.. and l in. I I I ' ' ' b = D 6 in from the top surface of the beam. From these calculations. plot a graph |-— ' ' ' showing the distribution of shear stresses from top to bottom of the beam. Snlulinn 5.0-5 Shear stresses in a simple beam Distance from the yl 1- top surface (in) (in) (psi) 0 1.00 0 0.25 0.75 l050 0.50 0.50 1800 0.75 0.25 2250 1.00 (NA) 0 2400 v 9:2 54:]. (5—39): 1' = —_y31) GRAPH or SHEAR STRESS 'r L 3 V=%=l9201b f=%=0.4in.‘ o 1050 UNTts: pounds and inches 1800 2250 192“ (2)2 2 NA 1' 4400 ' = _. : 4 _ . . .. p5: T 2(o.43[ 4 n] (2 00m y?) 233:3 (r = psi; y. = in.) 1353? Shear SlfBSSES ill the WOIJS [If Beams Willi Flanges Problem 5.10-1 Il‘ll‘lluflll 5.19-5 A wide—flange beam (see figure) having the cross section described below is subjected to a shear force V. Using the dimensions of the cross section, calculate the moment of inertia and then determine the following quantities: (a) The maximum shear stress TM in the web. (13) The minimum shear stress 7min in the web. (c) The average shear stress 1' (obtained by dividing the shear 8"“ force by the area of the web) and the ratio Twirlm. (d) The shear force th carried in the web and the ratio th1 V. Note: Disregard the fillets at the junctions of the web and flanges and detemfine all quantities. including the moment of inertia. by considering Probe. 5.10-1 through 5.10-5 the cross section to consist of three rectangles. Prohle 5.10-1 Dimensions of cross section: b = 6 in., r = 0.5 in., h =12in.. h1=10.5in., and V = 30 kt Solution 5.10-1 Wide-flange beam (b) MINIMUM SHEAR STRESS IN THE wen (Eq. 5-48b) Vb Tmin = — h: " Hz) = Si ‘— b = 6.0 in. 8k< 1 p I = 0.5 in. . g, = 12 o in (c) AVERAGE SHEAR mess IN THE wen (Eq. 5-50) . I. v hl = 10.5 In. Tm, = __ = psi ‘— V = 30 k zhl 7"“ = 1.0M «— MOMENT OF INERTIA (Eq. 5-47) Tam I = Lam: _ bk? + m?) : 333 4 in 4 (d) Sam FORCE IN THE WEB (Eq. 5-49) ' ' If! [2 vm = f (2%,. + rm...) = 28.25 R 4—- (aJ MAXIMUM SHEAR STRESS IN THE wen (Eq. 548a) m -—— = 0.942 «— Tmln = _ + = psi 4— v BuilI-Up Beams Prnhlem 5.11-1 A prefabricated wood I-beam serving as a floor joist has the cross section shown in the figure. The allowable load in shear for the glued joints bflween the web and the flanges is 65 lbfin. in the longitudinal direction. Determine the maximum allowable shear force Vm for the beam. Solution 5.11-1 Wood l-beam All dimensions in inches. Find Vm baSed upon shear in the glued joints. Allowable load in shear for the glued joints is 65 lbfin. '. flu“ = 65 Ibfm. f=fl V’m =‘(fl 1 Q _bh"' (la—ma _ 1 3 1 , - 12 I2 r~ 12 (5)(9.5) 12 (4.375K3) = 170.57 in.4 Q = Qnmfl = Afdf= (5)(0.75)(4.375) = 16.406 in.3 Jun“! _ (651b/in.)(]70.57 in.‘) ' Q ' 16.4mm} Vm =6761b 4— ...
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This note was uploaded on 03/15/2012 for the course CIVL 000 taught by Professor Kk during the Spring '10 term at HKUST.

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