PowerPoint PresentationHomework #9Homework #10Slide 4Slide 5Slide 6Slide 7Relations Between w, V, and MSlide 9Slide 10Slide 11Slide 12Slide 13Slide 14ME221 Lecture 27 1ME 221 StaticsLecture #27Section 7.4ME221 Lecture 27 2Homework #9Chapter 5 problems:–54, 56, 62, 64, 69, 71 & 73•Due TodayME221 Lecture 27 3Homework #10Chapter 7 problems:–5, 8, 19, 21, 24, 26 & 35•Due Wednesday, November 13ME221 Lecture 27 4Last Lecture:Internal Forces in Structures•Reviewed internal/external forces•Found internal forces•Started shear & moment diagramsME221 Lecture 27 5Generate a shear / bending diagram as follows: 2. Take a section on each side of an applied force or moment and inside a distributed load (take a new section whenever there is a change in the load or shape of the beam)- draw a FBD and sum forces / moments3. Repeat 2 along the length of the beam1. Find reaction forcesw(x) distributed loadV(x) shear forceM(x) momentShear and Moment Diagrams using Sectioning MethodME221 Lecture 27 6VMVMSign Convention Positive Shear and Positive MomentME221 Lecture 27 7Positive ShearMMPositive MomentEffect of External ForcesME221 Lecture 27 8Relations Between w, V, and MIn balancing forces, we can come up with differential equations relating w, V, and M. These are as follows:    ,dM x dV xV x w xdx dx This means you can integrate the shear diagram to obtain the moment diagram.dxM+dMMVV+dVw(x) 0)()(0 dxxwdVVVF0)(0 dMMVdxMM Thus,ME221 Lecture 27 9dxM+dMMVV+dVw(x)Shear and Moment Diagrams Using Integration Methodxj x j+1 dxxdVxw)()( 11)(jVjVjxjxdVdxxw jjVVV1Area under the load intensity diagrambetween xj and x j+11 - Distributed Load:ME221 Lecture 27 10 jjMMM1Area under the shear diagrambetween xj and x j+1Similarly: dxxdMxV)()( 11)(jMjMjxjxdMdxxVME221 Lecture 27 11 2 - Concentrated Load:PM+dMMVV+dVP0,  MPV3 - Concentrated Moment:CM+dMMVV+dVCCMV  ,0ME221 Lecture 27 12Notes•Concentrated force will cause a jump in the shear diagram by an amount equal to the applied load•Concentrated moment will cause a jump in the moment diagram by an amount equal to the negative of the applied momentConnecting points in the shear and moment diagramsxj+1xjxj+1xjNegative decreasing slopexjxj+1xjxj+1Negative increasing slopePositivedecreasing slopePositiveincreasing slopeME221 Lecture 27 13Developing Shear and Bending Diagrams1. Show FBD and statics for each section2. Determine equation for V(x) and M(x)3. Draw shear and bending diagrams indicating linear or parabolic4. Label end points of diagram as well as every region endpointME221 Lecture 27 1420 lb/in125 lb125 lb9 in. 12 in. 12 in.