FAUZIAH MAT Applied Mechanics Division School of Mechatronic Engineering

1. Chapter 8 Centre of Gravity & Centroid FAUZIAH MAT Applied Mechanics Division School of Mechatronic Engineering Universiti Malaysia Perlis (UniMAP) fauziah@unimap.edu.my

2. Today’s Objective: • Students will be able to determine: • The location of the center of gravity (CG), • The location of the center of mass, and • The location of the centroid using the method of composite bodies. COMPOSITE BODIES • In-Class Activities: • Check homework, if any • Applications • Method of Composite Bodies • Group Problem Solving

3. The I-beam (top) or T-beam (bottom) shown are commonly used in building various types of structures. When doing a stress or deflection analysis for a beam, the location of its centroid is very important. APPLICATIONS How can we easily determine the location of the centroid for different beam shapes?

4. The compressor is assembled with many individual components. APPLICATIONS (continued) In order to design the ground support structures, the reactions at blocks A and B have to be found. To do this easily, it is important to determine the location of the compressor’s center of gravity (CG). If we know the weight and CG of individual components, we need a simple way to determine the location of the CG of the assembled unit.

5. Consider a composite body which consists of a series of particles(or bodies) as shown in the figure. The net or the resultant weight is given as WR = W. Similarly, we can sum moments about the x- and z-axes to find the coordinates of the CG. Summing the moments about the y-axis, we get x WR = x1W1 + x2W2 + ……….. + xnWn where x1 represents x coordinate of W1, etc.. ~ ~ ~ ~ CG / CM OF A COMPOSITE BODY – By replacing the W with a M in these equations, the coordinates of the center of mass can be found.

6. Many industrial objects can be considered as composite bodies made up of a series of connected “simple” shaped parts or holes, like a rectangle, triangle, and semicircle. CONCEPT OF A COMPOSITE BODY Knowing the location of the centroid, C, or center of gravity, CG, of the simple shaped parts, we can easily determine the location of the C or CG for the more complex composite body.

7. This can be done by considering each part as a “particle” and following the procedure as described in Section 9.1. This is a simple, effective, and practical methodof determining the location of the centroid or center of gravity of a complex part, structure or machine. CONCEPT OF A COMPOSITE BODY(continued)

8. 4. Sum the columns to get x, y, and z. Use formulas like x = (  xi Ai ) / (  Ai ) or x = (  xiWi ) / (  Wi ) This approach will become straightforward by doing examples!   STEPS FOR ANALYSIS 1. Divide the body into pieces that are known shapes. Holes are considered as pieces with negative weight or size. 2. Make a table with the first column for segment number, the second column for weight, mass, or size (depending on the problem), the next set of columns for the moment arms, and, finally, several columns for recording results of simple intermediate calculations. 3. Fix the coordinate axes, determine the coordinates of the center of gravity of centroid of each piece, and then fill in the table.

9. Given: The part shown. Find: The centroid of the part. Plan:Follow the steps for analysis. EXAMPLE Solution: 1. This body can be divided into the following pieces: rectangle (a) + triangle (b) + quarter circular (c) – semicircular area (d). Note the negative sign on the hole!

10. Steps 2 & 3: Make up and fill the tableusing parts a, b, c, and d. Note the location of the axis system.     Segment Area A(m2) x(m) y(m) x A( m3) y A( m3) RectangleTriangle Q. CircleSemi-Circle 184.59  / 4–  / 2 37– 4(3) / (3 )0 1.51 4(3) / (3 )4(1) / (3 ) 274.59 - 2/3 5431.5– 9 0  28.0 76.5 39.83 EXAMPLE (continued)

11. C 4. Now use the table data results and the formulas to find the coordinates of the centroid.  x = (  x A) / (  A ) = 76.5 m3/ 28.0 m2 = 2.73 m y = (  y A) / (  A ) = 39.83 m3 / 28.0 m2 = 1.42 m  EXAMPLE (continued)   Area A x A y A 28.0 76.5 39.83

12. GROUP PROBLEM SOLVING Given: Two blocks of different materials are assembled as shown. The densities of the materials are: A = 30 kN / m3 and B = 80 kN / m3. Find: The center of gravity of this assembly. Plan: Follow the steps for analysis. Solution: 1. In this problem, the blocks A and B can be considered as two pieces (or segments).

13.       Segment w (kN) x (m) y (m) z (m) xW (kN·m) yW (kN·m) zW (lb·in) A B 1.08 5.76 0.4 0.1 0.1 0.3 0.2 0.3 0.432 0.576 0.108 1.728 0.216 1.728  6.84 1.008 1.836 1.944 GROUP PROBLEM SOLVING (continued) Weight = w =  (Volume in m3) wA = 30 (0.5) (0.6) (0.6) (0.2) = 1.08 kN wB = 80 (0.6) (0.6) (0.2) = 5.76 kN

14. GROUP PROBLEM SOLVING (continued) Table Summary    • W (kN) xw y wz w • (kN·m) (kN·m) (kN·m) • 6.84 1.008 1.836 1.944 Substituting into the Center of Gravity equations: ~ x = ( x w) / ( w ) = 1.008 / 6.84 = 0.147 m y = ( y w) / ( w ) = 1.836 / 6.84 = 0.268 m z = ( z w) / ( w ) = 1.944 / 6.84 = 0.284 m ~ ~

15. End of the Lecture Let Learning Continue