ME 221 Statics Lecture #14 Sections 4.1 – 4.2

1. ME 221 Statics Lecture #14Sections 4.1 – 4.2 Lecture 14

2. Homework #5 • Chapter 3 problems: • 48, 51, 55, 61, 62, 65, 71, 72 & 75 • May use MathCAD, etc. to solve • Due Friday, October 4 Lecture 14

3. Distributed 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” point Lecture 14

4. Distributed Loads Two 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” Lecture 14

5. Contact Distributed Load • Snow on roof, tire on road, bearing on race, liquid on container wall, ... Lecture 14

6. z z x y z y y ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ w1(x1,y1,z1) w2(x2,y2,z2) w5(x5,y5,z5) w4(x4,y4,z4) w3(x3,y3,z3) x x ~ ~ ~ ~ x w=x1w1+x2w2+x3w3+x4w4 Center of Gravity 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: Lecture 14

7. 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 Or Lecture 14

8. 20 10 ?? ?? ?? CG in Discrete Sense Where do we hold the bar to balance it? Find the point where the system’s weight may be balanced without use of a moment. Lecture 14

9. y r dw z x Discrete Equations Define a reference frame Lecture 14

10. Center of Mass The total mass is given by M Mass center is defined by Lecture 14

11. Continuous Equations Take our volume, dV, to be infinitesimal. Summing over all volumes becomes an integral. Note that dm = rdV . Center of gravity deals with forces and gdm is used in the integrals. Lecture 14

12. If r is constant • These coordinates defines the geometric center of an object (the centroid) • In case of 2-D, the geometric center can be defined using a differential element dA Lecture 14

13. If the geometry of an object takes the form of a line (thin rod or wire), then the centroid may be defined as: Lecture 14

14. Procedure for Analysis 1-Differential element • Specify 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 the 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. Lecture 14

15. 2- Size Express the length dl, dA, or dv of the element in terms of the coordinate used to define the object 3-Moment Arm Determine the perpendicular distance from the coordinate axes to the centroid of the differential element 4- Equation Substitute the data computed above in the appropriate equation. Lecture 14

16. Symmetry conditions y x • The centroid of some objects may be partially or completely specified by using the symmetry 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 Lecture 14

17. In cases of more than one axis of symmetry, the centroid will be located at the intersection of these axes. Lecture 14