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Chapter 6: Center of Gravity and Centroid. Chapter 6.1 C.G and Center of Mass. Chapter 6.1 C.G and Center of Mass..2. The same way you can find the centroid of the line and the volume. Example 6-2 (pg.251, sections 6.1-6.3). Locate the centroid of the area shown in figure. Solution I

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chapter 6 1 c g and center of mass 2
Chapter 6.1 C.G and Center of Mass..2

The same way you can find the centroid of the line and the volume.

example 6 2 pg 251 sections 6 1 6 3
Example 6-2 (pg.251, sections 6.1-6.3)

Locate the centroid of the area shown in figure

Solution I

Differential Element. A differential element of thickness dx is shown in the Figure. The element intersects the curve at the arbitrary point (x,y), and so it has a height of y.

Area and Moment Arms. The area of the element is dA=y dx, and its centroid is located at

con t example 6 2 pg 251 sections 6 1 6 3
Con’t Example 6-2 (pg. 251, Sections 6.1-6.3)

Integrations. Applying Equations 6-5 and integrating with respect to x yields

con t example 6 2 pg 251 sections 6 1 6 36
Con’t Example 6-2 (pg. 251, Sections 6.1-6.3)
  • Solution II
  • Differential Element. The differential element of thickness dy is shown in Figure. The element intersects the curve at the arbitrary point (x,y) and so it has a length (1-x).
  • Area and Moment Arms. The area of the element is dA = (1-x) dy, and its centroid is located at
con t example 6 2 pg 251 sections 6 1 6 37
Con’t Example 6-2 (pg. 251, Sections 6.1-6.3)
  • Integrations. Applying Equations 6-5 and integrating with respect to y, we obtain
example 6 6 pg 261 sections 6 1 6 3

Solution I The y axis is placed along the axis of symmetry so that To obtain we will establish the x axis (reference axis) thought the base of the area.l The area is segmented into two rectangles and the centroidal location for each is established.

Example 6-6 (pg. 261, Sections 6.1-6.3)

Locate the centroid C of the cross-sectional area for the T-beam

slide10

Con’t Example 6-6 (pg. 261, Sections 6.1-6.3)

Solution II

Using the same two segments, the x axis can be located at the top of the area. Here

The negative sign indicates that C is located below the origin, which is to be expected. Also note that from the two answers 8.55 in + 4.45 in =13.0 in., which is the depth of the beam as expected

slide11

Con’t Example 6-6 (pg. 261, Sections 6.1-6.3)

Solution III

It is also possible to consider the cross-sectional area to be one large rectangle less two small rectangles. Hence we have

problem 6 30 pg 264 sections 6 1 6 3
Problem 6-30 (pg. 264, Sections 6.1-6.3)

Determine the distance to the centroid of the shaded area.

Solution

6 6 moments of inertia for areas
6.6 Moments of Inertia For Areas
  • By definition moments of inertia with respect to any axis (i.e. x and y) are
  • Polar moment of inertia

Always positive value

Units:

slide14

Geometric Properties of An Area and Volume Geometric Properties of An Area and Volume (page 786-787)

slide15

Geometric Properties of An Area and Volume Geometric Properties of An Area and Volume (page 786-787)

con t 6 6 moments of inertia for areas
Con’t 6.6 Moments of Inertia For Areas
  • Example 6.14

Lets proof for rectangular area

Solution

Part (a)

example 6 15 pg 287 sections 6 8 6 9
Example 6-15 (pg. 287 Sections 6.8-6.9)

Solution I (CASE I)

Determine the moment of inertia of the shaded area about the x axis.

A differential element of area that is parallel to the x axis is chose for integration.

dA =(100-x) dy.

Limits of integration wrt y, y=0 to y=200 mm

6 7 parallel axis theorem
6-7 Parallel-Axis Theorem
  • If we know Moment of Inertia of a given axis, we can compute M.I about another parallel axis
con t example 6 15 pg 287 sections 6 8 6 9
Con’t Example 6-15 (pg. 287 Sections 6.8-6.9)

Solution II (CASE 2)

A differential element parallel to the y axis is chosen for integration.

Use the parallel-axis theorem to determine the moment of inertia of the element with respect to this axis.

For a rectangle having a base b and height h, the moment of inertia about its centroidal axis is

slide22

Con’t Example 6-15 (pg. 287 Sections 6.8-6.9)

Solution II (CASE 2)

For the differential element , b = dx and h = y, and thus

Since the centroid of the element is at from the x axis, the moment of inertia of the element about this axis is

example 6 17 pg 291 sections 6 8 6 9
Example 6-17 (pg. 291, Sections 6.8-6.9)
  • Determine the moment of inertia of the cross-sectional area of the T-beam about the centroidal axis.

In class workout

problem 6 87 pg 294 sections 6 8 6 9
Problem 6-87 (pg. 294, Sections 6.8-6.9)
  • Determine the moment of inertia of the shaded area with respect to a horizontal axis passing through the centroid of the section

Solution:

problem 6 92 pg 294 sections 6 8 6 9
Problem 6-92 (pg. 294, Sections 6.8-6.9)
  • Determine the moment of inertia of the beam’s cross-sectional area about the y axis

Solution:

problem 6 8 pg 257 sections 6 1 6 3
Problem 6-8 (pg. 257, Sections 6.1-6.3)

Determine the location of the centroid of the quarter elliptical plate.

Solution