PowerPoint PresentationHomework #5Slide 3Distributed LoadsContact Distributed LoadCenter of GravitySlide 7CG in Discrete SenseDiscrete EquationsSlide 10Continuous EquationsSlide 12Slide 13Slide 14Slide 15Slide 16Slide 17ME221 Lecture 13 1ME 221 StaticsLecture #13Sections 4.1 – 4.2ME221 Lecture 13 2Homework #5•Chapter 3 problems:–48, 51, 55, 57, 61, 62, 65, 71, 72 & 75–May use MathCAD, etc. to solve–Due Friday, October 3ME221 Lecture 13 3Distributed Forces (Loads);Centroids & Center of Gravity•The concept of distributed loads will be introduced•Center of mass will be discussed as an important application of distributed loading–mass, (hence, weight), is distributed throughout a body; we want to find the “balance” pointME221 Lecture 13 4Distributed LoadsTwo types of distributed loads exist:–forces that exist throughout the body•e. g., gravity acting on mass•these are called “body forces”–forces arising from contact between two bodies•these are called “contact forces”ME221 Lecture 13 5Contact Distributed Load•Snow on roof, tire on road, bearing on race, liquid on container wall, ...ME221 Lecture 13 6Center of Gravityxyzw5(x5,y5,z5)˜ ˜ ˜xyzw3(x3,y3,z3)˜ ˜ ˜w1(x1,y1,z1)˜ ˜ ˜w2(x2,y2,z2)˜ ˜ ˜w4(x4,y4,z4)˜ ˜ ˜The weights of the n particles comprise a system of parallel forces. We can replace them with an equivalent force w located at G(x,y,z), such that: x w=x1w1+x2w2+x3w3+x4w4~ ~ ~ ~yzxME221 Lecture 13 7Orniiniiiniiniiiniiniiiwwzzwwyywwxx111111~,~,~Where are the coordinates of each point. Point G is called the center of gravity which is defined as the point in the space where all the weight is concentrated.zyx~,~,~ME221 Lecture 13 8CG in Discrete SenseWhere do we hold the bar to balance it?20 10???? ??Find the point where the system’s weight may be balanced without the use of a moment.ME221 Lecture 13 9Discrete EquationsDefine a reference framexyzdwrdwdwxxwwxxiii~~ME221 Lecture 13 10. . . . . .; ;i i i i i ii i ic m c m c mm x m y m zx y zM M M    Mass center is defined byThe total mass is given by MiiM mCenter of MassME221 Lecture 13 11Continuous EquationsTake our volume, dV, to be infinitesimal. Summing over all volumes becomes an integral.. . . . . .1 1 1; ;c m c m c mV V Vx xdV y ydV z zdVV V V    1VM dVVNote that dm = dV . Center of gravity deals with forces andgdm is used in the integrals.ME221 Lecture 13 12If  is constantdvdvzzdvdvyydvdvxx~,~,~•These coordinates define the geometric center of an object (the centroid)dAdAzzdAdAyydAdAxx~,~,~•In case of 2-D, the geometric center can be defined using a differential element dAME221 Lecture 13 13If the geometry of an object takes the form of a line (thin rod or wire), then the centroid may be defined as:dLdLzzdLdLyydLdLxx~,~,~ME221 Lecture 13 14Procedure for Analysis1-Differential elementSpecify the coordinate axes and choose an appropriate differential element of integration.•For a line, the differential element is dl•For an area, the differential element dA is generally a rectangle having a finite height and differential width.•For a volume, the element dv is either a circular disk having a finite radius and differential thickness or a shell having a finite length and radius and differential thickness.ME221 Lecture 13 152- SizeExpress the length dl, dA, or dv of the element in terms of the coordinate used to define the object.3-Moment ArmDetermine the perpendicular distance from the coordinate axes to the centroid of the differential element.4- EquationSubstitute the data computed above in the appropriate equation.ME221 Lecture 13 16xySymmetry conditions•In the case where the shape of the object has an axis of symmetry, then the centroid will be located along that line of symmetry.In this case, the centoid is located along the y-axis•The centroid of some objects may be partially or completely specified by using the symmetry conditionsME221 Lecture 13 17In cases of more than one axis of symmetry, the centroid will be located at the intersection of these