BeerMOM ISM_C05-P001-P079BeerMOM ISM_C05-P080-P163CCHHAAPPTTEERR 55PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 681 BwAL 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: 022BLwLMALwL A   0: 022ALwLMBLwL B   Free body diagram for determining reactions: Over whole beam, 0xL Place section at x. Replace distributed load by equivalent concentrated load. 0: 02ywLFwxV   2LVw x  0: 022JwL xMxwxM    2()2wMLx x ()2wMxL x  Maximum bending moment occurs at .2Lx  2max8wLM PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 682 BPCALba 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: 0CPbMLAbPAL   0: 0APaMLCaPCL   From A to B, 0xa 0: 0yPbFVL  PbVL  0: 0JPbMMxL   PbxML  From B to C, axL 0: 0yPaFVL   PaVL  0: ( ) 0KPaMMLxL  ()PaL xML  At section B, 2PabML PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 683 Bw0AL PROBLEM 5.3 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 Free body diagram for determining reactions. Reactions: 000: 022   yA AwL wLFR R 020: 023 AAwL LMM 220033 AwL wLM Use portion to left of the section as the free body. Replace distributed load with equivalent concentrated load. 0010: 022  ywL wxFxVL 20022wL wxVL 200 00:1() 032 2 3      JMwL wL wx xxxML 2300 0326  wL wLx wxML PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 684 BwLA 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 Free body diagram for determining reactions. Reactions: 0: 0  yAFRwL ARwL 0: ( ) 02 AALMMwL 202AwLM Use portion to the right of the section as the free body. Replace distributed load by equivalent concentrated load. 0: ( ) 0   yFVwLx ()VwLx 0: ( ) 02  JLxMMwLx 2()2 wMLx PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 685 BPPCAaa 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 A to B: 0xa 0: 0 yFPV VP  0: 0JMPx M MPx From B to C: 2ax a 0: 0 yFPPV 2VP  0: ( ) 0  JMPxPxaM 2MPx Pa  PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 686 DABaaCLw w 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 Reactions: ADwa From A to B, 0xa 0:yF 0wa wx V ()Vwax  0:JM () 02xwax wx M  22xMwax  From B to C, axLa 0:yF 0wa wa V 0V   0:JM 02awax wa x M  212Mwa  From C to D, La x L 0: ( ) 0yFVwLxwa ()VwLxa  0: () ()02JLxMMwLxwaLx  21()()2MwaLx Lx PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 687 BACD E3 kN 2 kN 2 kN5 kN0.3 m 0.3 m0.3 m 0.4 mPROBLEM 5.7 Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment. SOLUTION Origin at A:  Reaction at A: 0: 3 2 5 2 0 2 kNyA AFR R  0: (3kN)(0.3m) (2kN)(0.6m) (5kN)(0.9m) (2kN)(1.3m) 0     AAMM  0.2 kN mAM  From A to C: 0: 2 kNyFV 10: 0.2 kN m (2 kN) 0MxM 0.2 2 Mx From C to D: 0: 2 3 0yFV 1kNV  20: 0.2 kN m (2 kN) (3 kN)( 0.3)