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# HW 7 - 352 CHAPTER 5 Stresses in Beams Problem 5.11-4 A b...

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352 CHAPTER 5 Stresses in Beams Problem 5.11-4 A box beam of wood is constructed of two 260 mm X 50 mm boards and two 260 mm X 25 mm boards (see figure). The boards are nailed at a longitudinal spacing s = 100 mm. If each nail has an allowable shear force F = 1200 N, what is the maximum allowable shear force V ? max z Solution 5.11-4 Wood box beam 2F 2(1200 N) All dimensions in millimeters. =---= = 24kN/m ill a ow S 100 mm b = 260 b = 260 - 2(50) = 160 l VQ V =falloj h = 310 hi = 260 f=~- ] max Q s = nail spacing = 100 mm F = allowable shear force ] = - I (bh 3 - b h 3 ) = 411.125 X 10 6 mm 4 12 1 1 3 for one nail = 1200 N f = shear flow between one flange Q = Qflange = Afd f = (260)(25)(142.5) = 926.25 X 10 3 mm and both webs fallow] (24 kN/m)(411.125 X 10 6 mm 4 ) V = ---- = -'-------'-----'----------,---,--------'- max Q 926.25 X 10 3 mm 3 = 10.7 kN - Problem 5.11-5 A box beam constructed of four wood boards of size 6 in. X I in. (actual dimensions) is shown in the figure. The boards are joined by screws for which the allowable load in shear is F = 250 Ib y ----f!f!t------'O 1 in. 1 in. ---1 J in. per screw. Calculate the maximum permissible longitudinal spacing smax of the screws if the shear force V is 1200 lb. 6 in. =--1 1 in. ~6in·-fl Solution 5.11-5 Wood box beam VQ 2F 2F] All dimension s in inches. f=---=- ,". Smax =- ] s VQ b = 6.0 b l = 6.0 - 2(1.0) = 4.0 h = 8.0 hi = 6.0 ] = ~ (bh 3 - blh~) = 184 in. 4 12 F = allowable shear force for one screw = 250 Ib V = shear force = 1200 Ib Q = Qtlange = Afd f = (6.0)(1.0)(3.5) = 21 in. 3 s = longitudinal spacing of the screws 2F1 2(250 Ib)(l84 in. 4 ) s --- f = shear flow between one flange and both webs max - VQ - (1200 Ib )(21 in. 3 ) = 3.65 in. -

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356 CHAPTER 5 Stresses in Beams Solution 5.11-10 Beam with cover plates t = 0.5 ~~\$3 ~w 16 x 77 1 d N.A. l~ t=0.5 t I. .j b =10 All dimensions in inches. Wide-flange beam (W 16 X 77): d = 16.52 in. I beam = 1110 in. 4 Cover plates: b = 10 in. t = 0.5 in. F = allowable load per bolt = 2.1 k V = shear force = 30 k S = spacing of bolts in the longitudinal direction Find smax MOMENT OF INERTIA ABOlIT THE NElITRAL AXIS 1= I beam + z[ /2 bt3 + (bt)(~ + ±YJ = 1110 in. 4 + z[ /2 (10)(0.5)3 + (10)(0.5)(8.51)2 ] = 1834 in. 4 FIRST MOMENT OF AREA OF A COVER PLATE d + t) Q = bt ( -2- = (10)(0.5)(8.51) = 42.55 in. 3 MAXIMUM SPACING OF BOLTS VQ 2F 2FI j=-=- s=- I S VQ 2(2.1 k) (1834 in. 4 ) S = = 6.03 Ill. - max (30 k) (42.55 in. 3 ) Problem 5.11-11 Two W 10 X 45 steel wide-flange beams are bolted together to form a built-up beam as shown in the figure. What is the maximum permissible bolt spacing s if the shear force W 10 x 45 V = 20 kips and the allowable load in shear on each bolt is F = 3.1 kips? (Note: Obtain the dimensions and properties of the W shapes from Table E-1.) W 10 x 45 Solution 5.11-11 Built-up steel beam All dimensions in inches. FIRST MOMENT OF AREA OF ONE BEAM W 10 X 45: II = 248 in. 4 d = 10.10 in. A = 13.3 in. 2 Q = A(~) = (13.3)(5.05) = 67.165 in. 3 V=20k F=3.1k Find maximum allowable bolt spacing smax' MAXIMUM SPACING OF BOLTS IN THE LONGITUDINAL DIRECTION MOMENT OF INERTIA OF BUILT-UP BEAM VQ 2F 2FI j = -1- = -;- s = VQ I=z[I I +A(~YJ =2[248+(13.3)(5.05)2] 2(3.1 k)(1l74.4in. 4 ) = 1174.4 in. 4 Smax = = 5.42 Ill. (20 k)(67.l65 in. 3 ) -
SECTION 5.12 Beams with Axial Loads 359 Problem 5.12-4 A rigid frame ABC is formed by welding two B steel pipes at B (see figure). Each pipe has cross-sectional area A = 11.31 X 10 3 mm 2 , moment of inertia 1 = 46.37 X 10 6 mm 4 , and outside diameter d = 200 mm.

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