BeerMOM_GE_C05-P001-P060BeerMOM_GE_C05-P061-P082BeerMOM_GE_C05-P083-P116BeerMOM_GE_C05-P117-P163CCHHAAPPTTEERR 55PROPRIETARY MATERIAL. © 2013 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 5.1 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the equations of the shear and bending-moment curves. SOLUTION   Reactions: 0: 0CPbMLAbPALΣ= −= = 0: 0APaMLCaPCLΣ= −= = From A to B: 0 xa<< 0: 0yPbFVLΣ= −= PbVL=  0: 0JPbMMxLΣ= − = PbxML=  From B to C: axL<< 0: 0yPaFVLΣ= + = PaVL=−  0: ( ) 0KPaMMLxLΣ=−+ −= ()Pa L xML−=  At section B: 2PabML= PROPRIETARY MATERIAL. © 2013 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 5.2 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the equations of the shear and bending-moment curves. SOLUTION Reactions: 0: 022BLwLMALwL AΣ= −+⋅= = 0: 022ALwLMBLwL BΣ= −⋅= = Free body diagram for determining reactions. Over whole beam, 0 xL<< Place section at x. Replace distributed load by equivalent concentrated load. 0: 02ywLFwxVΣ= − −= 2LVw x=−  0: 022JwL xMxwxMΣ= − + += 2()2wMLxx=− ()2wMxLx=−  Maximum bending moment occurs at .2Lx = 2max8wLM = PROPRIETARY MATERIAL. © 2013 The reproduced, or distributed in any form or bydistribution to teachers and educators permittewithout permission. PFmb SOLUTION 0: 0yFwaV =−−= 0: ( )2JaMwaxM =−+ McGraw-Hill Companies, Inc. All rights reserved. No part of this May any means, without the prior written permission of the publisher, or ued by McGraw-Hill for their individual course preparation. A student usinPROBLEM 5.3 For the beam and loading shown, (a) draw the shmoment diagrams, (b) determine the equations obending-moment curves. From A to B (0 )xa<< : 0: 0yFwxV =−−= 0: ( ) 02JxMwxM=+=From B to C ():axL<< 0= M=anual may be displayed, used beyond the limited ng this manual is using it hear and bending-of the shear and Vwx=−  22wxM =−  Vwa=−  2awa x=−− PROPRIETARY MATERIAL. © 2013 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 5.4 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the equations of the shear and bending-moment curves. SOLUTION  010: 02ywxFxVLΣ= − ⋅−= 202wxVL=−  010: 023Jwx xMxMLΣ= ⋅⋅+= 306wxML=− At ,xL= 02wLV =− 0max||2wLV =  206wLM =− 20max||6wLM = PROPRIETARY MATERIAL. © 2013 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 5.5 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the equations of the shear and bending-moment curves. SOLUTION From C to D: Using entire beam ABCD as a free body, 0: ( ) 0DMAL P L a PaAP=−+ −+== 0: ( ) 0AMDL Pa P L aDP=−−−== Check: 0yF PPPPΣ=−−+= From A to B: 0 xa<< 0: 0yFPVΣ=−= VP=  0: 0JMPxMΣ=−+= MPx=  From B to C: axLa<<− 00yFPPVΣ=−−= 0V=  0()0JMPxPxaMΣ=−+ −+= MPa=  La xL−<< 0: 0yFVPΣ=+= VP=−  0: ( ) 0JMMPLxΣ=−+ −= ()MPL x=−  PROPRIETARY MATERIAL. © 2013 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 5.6 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the equations of the shear and bending-moment curves. SOLUTION  Calculate reactions after replacing distributed load by an equivalent concentrated load. Reactions are 1(2)2AD wL a== − From A to B: 0 xa<< 10: ( 2 ) 02yFwLaVΣ= − −= 1(2)2VwLa=−  10: ( 2 ) 02MwLaMΣ= − − + = 1(2)2MwLax=−  From B to C: axLa<<− 2xab−= Place section cut at x. Replace distributed load by equivalent concentrated load. 10: ( 2 ) ( ) 02yFwLawxaVΣ= − − −−= 2LVw x=−  10: ( 2 ) ( ) 022JxaMwLaxwxaM−=− − + − += 21[( 2 ) ( ) ]2MwLaxxa=−−− PROPRIETARY MATERIAL. © 2013 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 5.6 (Continued) From C to D: La x L−<< 10: ( 2 ) 02yFVwLaΣ= + − =