Ch 7. Centroid/Distributed Loads/Inertia Multimedia Engineering Statics Centroid: Line Area Vol Centroid: Composite Distributed Loads Area Moment of Inertia
 Chapter 1. Basics 2. Vectors 3. Forces 4. Moments 5. Rigid Bodies 6. Structures 7. Centroids/Inertia 8. Internal Loads 9. Friction 10. Work & Energy Appendix Basic Math Units Sections Search eBooks Dynamics Statics Mechanics Fluids Thermodynamics Math Author(s): Kurt Gramoll ©Kurt Gramoll

STATICS - CASE STUDY SOLUTION

Composite Parts - Moments of Inertia

Dimension Diagram

Object Split into Sections

After breaking the cross section into its composite parts, determine the area, the location of the centroid, and the centroidal moments of inertia for each part.

The centroid and the area of each part were found in the previous section, Centroid: Composite Parts. The moments of inertia for each part can be found from the tables in the Sections Appendix.

For the axis system as shown, the properties for part 1 are,

x1 = 1 cm    y1 = 3.5 cm     A1 = 6 cm2

Ix'1 = 1/12 2(3)3 = 54/12 cm4

Iy'1 = 1/12 3(2)3 = 24/12 cm4

The properties for part 2 are,

x2 = 8 cm    y2 = 1 cm     A2 = 32 cm2

Ix'2 = 1/12 16(2)3 = 128/12 cm4

Iy'2 = 1/12 2(16)3 = 8192/12 cm4

And the properties for part 3 are,

x3 = 15 cm    y3 = 7 cm     A3 = 20 cm2

Ix'3 = 1/12 2(10)3 = 2000/12 cm4

Iy'3 = 1/12 10(2)3 = 80/12 cm4

Part 4 is a triangle and its properties are,

x4 = 17.33 cm    y4 = 10.67 cm     A4 = 8cm2

Ix'4 = 1/36 4(4)3 = 256/36 cm4

Iy'4 = 1/36 4(4)3 = 256/36 cm4

Distances from Global Centroid

These can be used with the following equations to find the moments of inertia of the entire cross section with respect to the centroid of the cross section.

Here dxi and dyi are the distances from the centroid of the cross section (global centroid) to the centroid of any part i. These distances are given by the equations:

 Part x y 1 1 3.5 2 8 1 3 15 7 4 17.33 10.67 combined 10.616 4.218
Centroids for Each Part and Total
(from Centroid: Composite Parts)

Substituting x, y and the results of the above equations give,

dx1 = 10.616 - 1 = 9.616 cm
dy1 = 4.218 - 3.5 = 0.718 cm

dx2 = 10.616 - 1 = 2.616 cm
dy2 = 4.218 - 1 = 3.218 cm

dx3 = 10.616 - 15 = -4.384 cm
dy3 = 4.218 - 7 = -2.782 cm

dx4 = 10.616 - 17.33 = -6.714 cm
dy4 = 4.218 - 10.67 = -6.452 cm

Substituting each dxi, dyi, Ix'i, and Iy'i into the basic summation equations give

Ix = 1,011 cm4

Iy = 2,217 cm4

In the theory page, the polar moment of inertia was show to be equal to

Jz = Ix + Iy

Substituting for Ix and Iy gives the polar moment,

Jz = 3,228 cm4

Since this is not a symmetrical cross section, the product of inertia, Ixy is not zero. The produce of inertia is used for un-symmetrical bending which is not covered in this statics eBook.

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