Centroids Moment of Inertia by swenthomasovelil

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									                       Lecture Notes



Module 2: Centroids & Moment of Inertia
                              Centroid
• Centroid or center of gravity is the point within an object
  from which the force of gravity appears to act.
• Centroid of 3D objects often (but not always) lies
  somewhere along the lines of symmetry.
   Hollowed pipes, L shaped section have centroid
   located outside of the material of the section

• The Centroid of any area can be found by taking Centroidal axis
                                                  or Neutral
  moments of identifiable areas (such as rectangles or
  triangles) about any axis. i.e A.y or A.x

• The moment of an area about any axis is equal to the
  algebraic sum of the moments of its component areas.
• The moment of any area is defined as the product of
  the area and the perpendicular distance from the
  Centroid of the area to the moment axis.
 First Moment of Area-Centroid
• Composite areas
  –   Subdivide
  –   Select coordinate axes
  –   Find the overall area
  –   Apply the composite
      centroid formula
              Moment of Inertia (I)
• Also known as the Second Moment of the Area is a term
  used to describe the capacity of a cross-section to resist
  bending.
• It is a mathematical property of a section concerned with a
  surface area and how that area is distributed about the
  reference axis. The reference axis is usually a centroidal
  axis.

                          where
Cross Section of Beam




                            Y



                        Y
             Internal Forces

• Naming those internal forces




            2011-2-11
                                 7
         Perpendicular Axis Theorem

                                        • For flat objects the
                                          rotational moment of inertia
                  Iy = (1/12) b3d
                                          of the axes in the plane is
                                          related to the moment of
                                          inertia perpendicular to the
Ix = (1/12) bd3                     d     plane.
                     Iz  Ix  Iy

                            b
                                                    Iz  Ix  Iy
              Iz = (1/12) bd(b2 + d2)
                                        Izz  Polar Moment of Inertia
             Parallel-Axis Theorem for an Area
• Used to calculate the moment of inertia about other axis.
• The moment of inertia of an area about an axis is equal to the
  moment of inertia of the area about a parallel axis passing through
  the area’s Centroid plus the product of the area and the square of
  the perpendicular distance between the axes.

      - Ix’: Moment of Inertia about new axis
      - Ix: Moment of Inertia about original
      axis
      - A: Area of shape                               A
      - d: Perpendicular distance from new to
      original axis                           x                          x



                                                   d
            I X '  I X  Ad 2                x'                         x'

2011-2-11
                                                                        10
Moment of Inertia - Parallel
    Axis Theorem

Parallel axis theorem:
Consider the moment
of inertia Ix of an area
A with respect to an
axis AA’. Denote by y
the distance from an
element of area dA to
                           I x   y dA
                                   2

AA’.
 Moment of Inertia - Parallel
     Axis Theorem
Consider an axis BB’
parallel to AA’ through the
Centroid C of the area,
known as the centroidal
axis. The equation of the
moment inertia becomes

    I x   y dA    y  d  dA
              2               2



         y dA  2 ydA  d
              2                   2
                                       dA
   Moment of Inertia - Parallel
       Axis Theorem
The first integral is the
moment of inertia about the
centroid.
            I x   y dA
                       2


The second component is the first moment area
about the centroid i.e if the area is located at the C.G
               yA   ydA  y  0
                      ydA  0
   Moment of Inertia - Parallel
       Axis Theorem
Modify the equation obtain
the parallel axis theorem.


Ix   y 2 dA  2  ydA  d 2  dA

    Ix  d A
            2
                Moments of Inertia and Radius of
                Gyration
• In general case                                   Polar Moment of
                                                    Inertia
                                                                IX
                                                         kX 
                                                                 A
                                                                Iy
                                                         ky 
                                                                A
                                                                JO
                                                         kO 
                                                                 A

• It is a mathematical property of a section concerned with a surface
  area and how that area is distributed about the reference axis. Look
  at IX of following shapes:

            x               x              x             x




                                                                     15
                  Moment of Inertia example
                   simple rectangular shape




         dA  bdy




   Centroid
or Neutral axis
                     Rectangular Area
 •      Moment of Inertia about centroidal axes for a rectangle          To
                                                                         Remember
          1         1
     IX    bh 3     (300mm )(150mm ) 3  562500 mm 4
         12        12                                                            1
                                                                         IX       bh 3
          1         1                                                           12
     IY  hb 3  hb 3  (150mm )(300mm ) 3  1125000 mm           4
                                                                                 1
         12        12                                                    IY       hb 3
       – Which one is bigger?                                                   12
                                                  New Unit!
                            y                     1. Standard sections (C-shapes,
                                                     Wide flange beams, Hollow steel
                                                     sections)

150mm       x                               x     2. Rectangular shapes:
     h




                                                     Calculate yourself, memorize
                                                     equations.
                            y                     3. Other shapes (circles, triangles,
                          300mm
                             b                       etc..): The equation or value will
                                                     be given to you.
                                                                                     17
       Moments of Inertia for
        Composite Areas
• If the moment of inertia of each simpler area (a part of the
  composite area) is known or can be determined about a common
  axis, then the moment of inertia of the composite area equals the
  algebraic sum of the moments of inertia of all its parts.

• Steps to follow:
   – Divide the composite area into smaller and simpler parts
   – Find the Centroid of each part
   – Calculate the moment of inertia (I) for each part about its centroidal axis
   – Use parallel-axis theorem to calculate the moment of inertia of each
     part about the given axis (normally the centroidal axis of composite area,
     if this is the case, very likely you need to find the Centroid of that
     composite area first)
   – Take the algebraic sum of the moments of inertia of all parts to get the
     moment of inertia of the composite area
                                                                             20
 Find the M.I of the I - section                    Y
      24 mm

                    6mm           X3                                        X3


                8 mm                      y3                               h3y
                                                            h2y
48 mm
                                      X                                         X
                                                                           X2
                                 X2
                                          y2                y      h1y
                         6mm
                                   X1                                       X1

        48 mm
                                   y1                   Y


  Ix = Ix1 + A1h1y2 + Ix2 + A2h2y2 + Ix3 + A3h3y2   Symmetric about Y axis hence
                                                    h1x, h2x, h3x = 0
                    C.G  x = 0 ,
                    y = (A1y1 + A2y2 + A3y3)/ (A1 + A2 + A3)

                    Ix1 , Ix2, Ix3= (b1d13 )/12, ……….

                    h1y = y-y1       h2y = y – y2       h3y = y – y3



                    Similarly find Iy.

                     Iy1 = (d1b13 )/12


Radius of Gyration along X & Y axis

								
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