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PROBLEM 6.1 Three boards, each of 1.5 × 3.5-in. rectangular cross section, are nailed together to form a beam that is subjected to a vertical shear of 250 lb. Knowing that the spacing between each pair of nails is 2.5 in., determine the shearing force in each nail. SOLUTION I = 1 3 1 (3.5)(4.5)3 = 26.578 in 4 bh = 12 12 A = (3.5)(1.5) = 5.25 in 2 y1 = 1.5 in. Q = Ay1 = 7.875 in 3 q= VQ (250)(7.875) = = 74.074 lb/in I 26.578 qs = 2 Fnail Fnail = qs (74.074)(2.5) = 2 2 Fnail = 92.6 lb PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 6.7 A columm is fabricated by connecting the rolled-steel members shown by bolts of 34 -in. diameter spaced longitudinally every 5 in. Determine the average shearing stress in the bolts caused by a shearing force of 30 kips parallel to the y axis. SOLUTION Geometry: f = d 2 + (t w )C s 10.0 + 0.303 = 5.303 in. 2 x = 0.534 in. = y1 = f − x = 5.303 − 0.534 = 4.769 in. Determine moment of inertia. Part C8 × 13.7 S 10 × 25.4 C8 × 13.7 Σ d (in.) 4.769 0 4.769 A(in 2 ) 4.04 7.4 4.04 Ad 2 (in 4 ) 91.88 0 91.88 183.76 I (in 4 ) 1.52 123 1.52 126.04 I = ΣAd 2 + ΣI = 183.76 + 126.04 = 309.8 in 4 Q = A y1 = (4.04)(4.769) = 19.267 in 3 VQ (30)(19.267) = = 1.8658 kip/in 309.8 I 1 1 = qs = (1.8658)(5) = 4.664 kips 2 2 q= Fbolt Abolt = τ bolt = π 4 2 = d bolt π 3 4 4 2 = 0.4418 in 2 4.664 Fbolt = = 10.56 ksi Abolt 0.4418 τ bolt = 10.56 ksi PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 6.8 The composite beam shown is fabricated by connecting two W6 × 20 rolled-steel members, using bolts of 58 -in. diameter spaced longitudinally every 6 in. Knowing that the average allowable shearing stress in the bolts is 10.5 ksi, determine the largest allowable vertical shear in the beam. SOLUTION W6 × 20: A = 5.87 in 2 , d = 6.20 in., I x = 41.4 in 4 y = Composite: 1 d = 3.1 in. 2 I = 2[41.4 + (5.87)(3.1)2 ] = 195.621 in 4 Q = A y = (5.87)(3.1) = 18.197 in 3 Bolts: d = Abolt = 5 in., τ all = 10.5 ksi, s = 6 in. 8 π 5 4 8 2 = 0.30680 in 2 Fbolt = τ all Abolt = (10.5)(0.30680) = 3.2214 kips Shear: q= 2Fbolt (2)(3.2214) = = 1.07380 kips/in s 6 q = VQ I V = Iq (195.621)(1.0780) = Q 18.197 V = 11.54 kips PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 6.31 Several wooden planks are glued together to form the box beam shown. Knowing that the beam is subjected to a vertical shear of 3 kN, determine the average shearing stress in the glued joint (a) at A, (b) at B. SOLUTION IA = 1 3 1 bh + Ad 2 = (60)(20)3 + (60)(20)(50) 2 12 12 = 3.04 × 106 mm 4 1 3 1 (60)(20)3 = 0.04 × 106 mm 4 bh = 12 12 1 3 1 (20)(120)3 = 2.88 × 106 mm 4 IC = bh = 12 12 IB = I = 2I A + I B + 2 I C = 11.88 × 106 mm 4 = 11.88 × 10−6 m 4 QA = Ay = (60)(20)(50) = 60 × 103 mm3 = 60 × 10−6 m3 t = 20 mm + 20 mm = 40 mm = 40 × 10−3 m (a) τA = VQA (3 × 103 )(60 × 10−6 ) = = 379 × 103 Pa It (11.88 × 10−6 )(40 × 10−3 ) QB = 0 (b) τB = VQB =0 It τ A = 379 kPa τB = 0 PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 6.33 The built-up wooden beam shown is subjected to a vertical shear of 8 kN. Knowing that the nails are spaced longitudinally every 60 mm at A and every 25 mm at B, determine the shearing force in the nails (a) at A, (b) at B. (Given: I x = 1.504 × 109 mm 4.) SOLUTION I x = 1.504 × 109 mm 4 = 1504 × 10−6 m 4 s A = 60 mm = 0.060 m sB = 25 mm = 0.025 m (a) QA = Q1 = A1 y1 = (50)(100)(150) = 750 × 103 mm3 = 750 × 10−6 m3 FA = q A s A = (8 × 103 )(750 × 10−6 )(0.060) VQ1s A = I 1504 × 10−6 FA = 239 N (b) Q2 = A2 y2 = (300)(50)(175) = 2625 × 103 mm3 QB = 2Q1 + Q2 = 4125 × 103 mm3 = 4125 × 10−6 m3 FB = qB sB = VQB sB (8 × 103 )(4125 × 10−6 )(0.025) = I 1504 × 10−6 FB = 549 N PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 6.37 Knowing that a given vertical shear V causes a maximum shearing stress of 75 MPa in an extruded beam having the cross section shown, determine the shearing stress at the three points indicated. SOLUTION τ = VQ It is proportional to Q/t. Qc = (30)(10)(75) Point c: = 22.5 × 103 mm3 tc = 10 mm Qc /tc = 2250 mm 2 Qb = Qc + (20)(50)(55) Point b: = 77.5 × 103 mm3 tb = 20 mm Qb /tb = 3875 mm 2 Qa = 2Qb + (120)(30)(15) Point a: = 209 × 103 mm3 ta = 120 mm Qa /ta = 1741.67 mm 2 τ m = τ b = 75 MPa (Q/t ) m occurs at b. τa Qa /ta τa 1741.67 mm 2 = = τb Qb /tb = τc Qc /tc 75 MPa τa = 2 3875 mm 2250 mm 2 τ a = 33.7 MPa τ b = 75.0 MPa τ c = 43.5 MPa PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.