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Structure I Course Code: ARCH 208 Dr. Aeid A. Abdulrazeg.

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Presentation on theme: "Structure I Course Code: ARCH 208 Dr. Aeid A. Abdulrazeg."— Presentation transcript:

1 Structure I Course Code: ARCH 208 Dr. Aeid A. Abdulrazeg

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6 COMPOSITE BODIES Today’s Objective: We will be able to determine:
The location of the center of gravity. The location of the center of mass. The location of the centroid using the method of composite bodies.

7 1. A composite body in this section refers to a body made of ____.
READING QUIZ 1. A composite body in this section refers to a body made of ____. A) carbon fibers and an epoxy matrix B) steel and concrete C) a collection of “simple” shaped parts or holes D) a collection of “complex” shaped parts or holes Answers: 1. C 2. C 2. The composite method for determining the location of the center of gravity of a composite body requires _______. A) integration B) differentiation C) simple arithmetic D) All of the above. Dr. Aeid A. Abdulrazeg

8 The I-beam is commonly used in building structures.
APPLICATIONS The I-beam is commonly used in building structures. When doing a stress analysis on an I - beam, the location of the centroid is very important. How can we easily determine the location of the centroid for a given beam shape? Dr. Aeid A. Abdulrazeg

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

10 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. 4. Sum the columns to get x, y, and z. Use formulas like x = (  xi Ai ) / (  Ai ) or x = (  xi Wi ) / (  Wi ) This approach will become clear by doing examples! Dr. Aeid A. Abdulrazeg

11 Find: The centroid of the part.
EXAMPLE Given: The part shown. Find: The centroid of the part. Plan: Follow the steps for analysis. a b c d Solution: 1. This body can be divided into the following pieces: rectangle (a) + triangle (b) + quarter circular (c) – semicircular area (d) Dr. Aeid A. Abdulrazeg

12 EXAMPLE (continued) a b c d
Steps 2 & 3: Make up and fill the table using parts a, b, c, and d. 39.83 76.5 28.0 - 2/3 – 9 0 (3) / (3 ) 4(1) / (3 ) 3 7 – 4(3) / (3 ) 0  / 4 –  / 2 Rectangle Triangle Q. Circle Semi-Circle A y ( in3) A x ( in3) y (in) x (in) Area A (in2) Segment Dr. Aeid A. Abdulrazeg

13 · EXAMPLE (continued) 
4. Now use the table data and these formulas to find the coordinates of the centroid. x = (  x A) / (  A ) = in3/ 28.0 in2 = in y = (  y A) / ( A ) = in3 / 28.0 in2 = in Dr. Aeid A. Abdulrazeg

14 CONCEPT QUIZ Answers: 1. C 2. C
1. Based on the typical centroid information available in handbooks, what are the minimum number of segments you will have to consider for determing the centroid of the given area? 1, 2, 3, or 4 3cm 1 cm Answers: 1. C 2. C Dr. Aeid A. Abdulrazeg

15 GROUP PROBLEM SOLVING Given: Two blocks of different materials are assembled as shown. The densities of the materials are: A = lb / ft3 and B = lb / ft3. 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 segments.

16 GROUP PROBLEM SOLVING (continued)
Weight = w =  (Volume in ft3) wA = (0.5) (6) (6) (2) / (12)3 = lb wB = (6) (6) (2) / (12)3 = lb 56.25 53.12 29.17 19.79 6.25 50.00 3.125 12.5 16.67 2 3 1 4 A B wz (lb·in) w y (lb·in) w x (lb·in) z (in) y (in) x (in) w (lb) Segment

17 GROUP PROBLEM SOLVING (continued)
~ x = ( x w) / ( w ) = /19.79 = 1.47 in y = ( y w) / ( w ) = / = 2.68 in z = ( z w) / ( w ) = / = in Dr. Aeid A. Abdulrazeg

18 ATTENTION QUIZ Answers: 1. B 2. D 2cm 4cm x y
1. A rectangular area has semicircular and triangular cuts as shown. For determining the centroid, what is the minimum number of pieces that you can use? A) Two B) Three C) Four D) Five Answers: 1. B 2. D 2. For determining the centroid of the area, two square segments are considered; square ABCD and square DEFG. What are the coordinates (x, y ) of the centroid of square DEFG? A) (1, 1) m B) (1.25, 1.25) m C) (0.5, 0.5 ) m D) (1.5, 1.5) m ~ A 1m y E F G C B x D Dr. Aeid A. Abdulrazeg


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