PROBLEM 1.4 Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Knowing that d1 = 50 mm and d 2 = 30 mm, find the average normal stress at the midsection of (a) rod AB, (b) rod BC. SOLUTION (a) Rod AB P = 40 + 30 = 70 kN = 70 × 103 N A= σ AB = (b) π 4 d12 = π 4 (50) 2 = 1.9635 × 103 mm 2 = 1.9635 × 10 −3 m 2 P 70 × 103 = = 35.7 × 106 Pa A 1.9635 × 10−3 σ AB = 35.7 MPa Rod BC P = 30 kN = 30 × 103 N A= σ BC = π 4 d 22 = π 4 (30)2 = 706.86 mm 2 = 706.86 × 10 −6 m 2 P 30 × 103 = = 42.4 × 106 Pa A 706.86 × 10−6 σ BC = 42.4 MPa PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. This All rights reserved. material No part of this Manual may beinstructor displayed,use. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. is proprietary solely for authorized Notreproduced, authorized or fordistributed sale or distribution in or anybymanner. This without document be copied, scanned, duplicated, distributed, or posted in any form any means, themay priornotwritten permission of the publisher,forwarded, or used beyond the limited on distribution a website, intowhole or part. teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. 6 BeerMOM_ISM_C01.indd 6 12/12/2014 6:00:44 PM PROBLEM 1.7 0.4 m C 0.25 m 0.2 m B Each of the four vertical links has an 8 36-mm uniform rectangular cross section and each of the four pins has a 16-mm diameter. Determine the maximum value of the average normal stress in the links connecting (a) points B and D, (b) points C and E. E 20 kN D A SOLUTION Use bar ABC as a free body. M C 0 : (0.040) FBD (0.025 0.040)(20 103 ) 0 FBD 32.5 103 N Link BD is in tension. M B 0 : (0.040) FCE (0.025)(20 103 ) 0 FCE 12.5 103 N Link CE is in compression. Net area of one link for tension (0.008)(0.036 0.016) 160 106 m 2 For two parallel links, (a) BD A net 320 106 m 2 FBD 32.5 103 101.563 106 6 Anet 320 10 BD 101.6 MPa Area for one link in compression (0.008)(0.036) 288 106 m 2 For two parallel links, (b) CE A 576 106 m 2 FCE 12.5 103 21.701 106 A 576 106 CE 21.7 MPa PROPRIETARY MATERIAL. Copyright© ©2015 2015McGraw-Hill McGraw-Hill Education. Education. This This is authorized instructor use.use. PROPRIETARY MATERIAL. Copyright is proprietary proprietarymaterial materialsolely solelyforfor authorized instructor NotNot authorized forforsale document may may not notbebecopied, copied,scanned, scanned, duplicated, forwarded, distributed, or posted authorized saleorordistribution distributioninin any any manner. manner. This document duplicated, forwarded, distributed, or posted on on a website, in whole or part. a website, in whole or part. 99 BeerMOM_ISM_C01.indd 9 12/12/2014 6:00:45 PM PROBLEM 1.17 When the force P reached 8 kN, the wooden specimen shown failed in shear along the surface indicated by the dashed line. Determine the average shearing stress along that surface at the time of failure. SOLUTION Area being sheared: A = 90 mm × 15 mm = 1350 mm 2 = 1350 × 10−6 m 2 Force: P = 8 × 103 N Shearing stress: τ = P 8 × 103 − = 5.93 × 106 Pa A 1350 × 10−6 τ = 5.93 MPa PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc.This All rights reserved. material No part of this for Manual may beinstructor displayed,use. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. is proprietary solely authorized Notreproduced, authorized or fordistributed sale or distribution in any This document notwritten be copied, scanned, duplicated, forwarded, distributed, or posted in any form or bymanner. any means, without themay prior permission of the publisher, or used beyond the limited on adistribution website, into whole or part. teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. 19 BeerMOM_ISM_C01.indd 19 12/12/2014 6:00:48 PM PROBLEM 1.25 P A 16 mm 750 mm 750 mm 50 mm B Knowing that 40° and P 9 kN, determine (a) the smallest allowable diameter of the pin at B if the average shearing stress in the pin is not to exceed 120 MPa, (b) the corresponding average bearing stress in member AB at B, (c) the corresponding average bearing stress in each of the support brackets at B. C 12 mm SOLUTION Geometry: Triangle ABC is an isoseles triangle with angles shown here. Use joint A as a free body. Law of sines applied to force triangle: P FAB FAC sin 20 sin110 sin 50 P sin110 FAB sin 20 (9)sin110 24.727 kN sin 20 PROPRIETARY MATERIAL. Copyright©©2015 2015McGraw-Hill McGraw-Hill Education. Education. This This is authorized instructor use.use. PROPRIETARY MATERIAL. Copyright is proprietary proprietarymaterial materialsolely solelyforfor authorized instructor NotNot authorized forforsale document may maynot notbebecopied, copied,scanned, scanned, duplicated, forwarded, distributed, or posted authorized saleorordistribution distributioninin any any manner. manner. This document duplicated, forwarded, distributed, or posted on on a website, in whole or part. a website, in whole or part. 28 28 BeerMOM_ISM_C01.indd 28 12/12/2014 6:00:52 PM PROBLEM 1.25 (Continued) (a) Allowable pin diameter. d2 FAB F 2FAB AB 2 where FAB 24.727 103 N 2 AP d2 24d 2 FAB (2)(24.727 103 ) 131.181 106 m 2 (120 106 ) d 11.4534 103 m (b) 11.45 mm Bearing stress in AB at A. Ab td (0.016)(11.4534 103 ) 183.254 106 m 2 b (c) FAB 24.727 103 134.933 106 Pa 6 Ab 183.254 10 134.9 MPa Bearing stress in support brackets at B. A td (0.012)(11.4534 103 ) 137.441 106 m 2 b 1 2 FAB A (0.5)(24.727 103 ) 89.955 106 Pa 6 137.441 10 90.0 MPa PROPRIETARY MATERIAL. Copyright© ©2015 2015McGraw-Hill McGraw-Hill Education. Education. This This is authorized instructor use.use. PROPRIETARY MATERIAL. Copyright is proprietary proprietarymaterial materialsolely solelyforfor authorized instructor NotNot authorized forforsale document may may not notbebecopied, copied,scanned, scanned, duplicated, forwarded, distributed, or posted authorized saleorordistribution distributioninin any any manner. manner. This document duplicated, forwarded, distributed, or posted on on a website, in whole or part. a website, in whole or part. 29 29 BeerMOM_ISM_C01.indd 29 12/12/2014 6:00:52 PM P' 150 mm 45 45 P 75 mm PROBLEM 1.30 Two wooden members of uniform rectangular cross section are joined by the simple glued scarf splice shown. Knowing that the maximum allowable shearing stress in the glued splice is 620 kPa, determine (a) the largest load P that can be safely applied, (b) the corresponding tensile stress in the splice. SOLUTION 90 45 45 A0 (150)(75) 11.25 103 mm 2 11.25 103 m 2 620 kPa 620 103 Pa P sin 2 2 A0 (a) P 2 A0 (2)(11.25 103 )(620 103 ) sin2 sin 90 13.95 103 N (b) P 13.95 kN P cos 2 (13.95 103 )(cos 45)2 A0 11.25 103 620 103 Pa 620 kPa PROPRIETARY MATERIAL. Copyright©©2015 2015McGraw-Hill McGraw-Hill Education. Education. This This is authorized instructor use.use. PROPRIETARY MATERIAL. Copyright is proprietary proprietarymaterial materialsolely solelyforfor authorized instructor NotNot authorized forforsale document may maynot notbebecopied, copied,scanned, scanned, duplicated, forwarded, distributed, or posted authorized saleorordistribution distributioninin any any manner. manner. This document duplicated, forwarded, distributed, or posted on on a website, in whole or part. a website, in whole or part. 34 34 BeerMOM_ISM_C01.indd 34 12/12/2014 6:00:54 PM A w 908 B PROBLEM 1.38 Link BC is 6 mm thick, has a width w 25 mm, and is made of a steel with a 480-MPa ultimate strength in tension. What was the safety factor used if the structure shown was designed to support a 16-kN load P? 480 mm C D P SOLUTION Use bar ACD as a free body and note that member BC is a two-force member. M A 0: (480) FBC (600) P 0 FBC Ultimate load for member BC: 600 (600)(16 103 ) 20 103 N P 480 480 FU U A FU (480 106 )(0.006)(0.025) 72 103 N Factor of safety: F.S. FU 72 103 FBC 20 103 F.S. 3.60 PROPRIETARY MATERIAL. Copyright©©2015 2015McGraw-Hill McGraw-Hill Education. Education. This This is authorized instructor use.use. PROPRIETARY MATERIAL. Copyright is proprietary proprietarymaterial materialsolely solelyforfor authorized instructor NotNot authorized forforsale document may maynot notbebecopied, copied,scanned, scanned, duplicated, forwarded, distributed, or posted authorized saleorordistribution distributioninin any any manner. manner. This document duplicated, forwarded, distributed, or posted on on a website, in whole or part. a website, in whole or part. 42 42 BeerMOM_ISM_C01.indd 42 12/12/2014 6:00:56 PM PROBLEM 1.46 Three steel bolts are to be used to attach the steel plate shown to a wooden beam. Knowing that the plate will support a 110-kN load, that the ultimate shearing stress for the steel used is 360 MPa, and that a factor of safety of 3.35 is desired, determine the required diameter of the bolts. SOLUTION For each bolt, Required: P= 110 = 36.667 kN 3 PU = (F. S.) P = (3.35)(36.667) = 122.83 kN τU = d = PU P 4P = π U 2 = U2 A πd d 4 4PU πτU = (4)(122.83 × 103 ) = 20.8 × 10−3 m 6 π (360 × 10 ) d = 20.8 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. This All rights reserved. material No part of this Manual may beinstructor displayed,use. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. is proprietary solely for authorized Notreproduced, authorized or fordistributed sale or distribution in or anybymanner. This without document be copied, scanned, duplicated, distributed, or posted in any form any means, themay priornotwritten permission of the publisher,forwarded, or used beyond the limited on distribution a website, intowhole or part. teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. 52 BeerMOM_ISM_C01.indd 52 12/12/2014 6:00:59 PM PROBLEM 1.55 Top view 200 mm 180 mm 12 mm In the structure shown, an 8-mm-diameter pin is used at A, and 12-mm-diameter pins are used at B and D. Knowing that the ultimate shearing stress is 100 MPa at all connections and that the ultimate normal stress is 250 MPa in each of the two links joining B and D, determine the allowable load P if an overall factor of safety of 3.0 is desired. 8 mm A B C B A C B 20 mm P 8 mm 8 mm D D 12 mm Front view Side view SOLUTION Statics: Use ABC as free body. M B 0 : 0.20FA 0.18P 0 M A 0 : 0.20FBD 0.38P 0 Based on double shear in pin A, A FA 2 U A 4 10 FA 9 10 P FBD 19 P d2 4 (0.008)2 50.266 106 m 2 (2)(100 106 )(50.266 106 ) 3.351 103 N 3.0 F .S . 10 P FA 3.72 103 N 9 Based on double shear in pins at B and D, A FBD 2 U A 4 d2 4 (0.012) 2 113.10 106 m 2 (2)(100 106 )(113.10 106 ) 7.54 103 N 3.0 F .S. 10 P FBD 3.97 103 N 19 Based on compression in links BD, for one link, A (0.020)(0.008) 160 106 m 2 2 U A (2)(250 106 )(160 106 ) 26.7 103 N F .S . 3.0 10 P FBD 14.04 103 N 19 Allowable value of P is smallest, P 3.72 103 N FBD P 3.72 kN PROPRIETARY MATERIAL. Copyright©©2015 2015McGraw-Hill McGraw-Hill Education. Education. This authorized instructor use.use. PROPRIETARY MATERIAL. Copyright This isis proprietary proprietarymaterial materialsolely solelyforfor authorized instructor NotNot authorized forforsale document may maynot notbebecopied, copied,scanned, scanned, duplicated, forwarded, distributed, or posted authorized saleorordistribution distributioninin any any manner. manner. This document duplicated, forwarded, distributed, or posted on on a website, in whole or part. a website, in whole or part. 61 62 BeerMOM_ISM_C01.indd 62 12/12/2014 6:01:03 PM