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10.4 Moments of Inertia
            for an Area by Integration
When the boundaries for a planar area are
expressed by mathematical functions,
moments of inertia for the area can be
determined by the previous method
If the element chosen for integration has a
differential size in two directions, a double
integration must be performed to evaluate
the moment of inertia
Try to choose an element having a
differential size or thickness in only one
direction for easy integration
10.4 Moments of Inertia
               for an Area by Integration
Procedure for Analysis
  If a single integration is performed to determine
  the moment of inertia of an area bout an axis, it
  is necessary to specify differential element dA
  This element will be rectangular with a finite
  length and differential width
  Element is located so that it intersects the
  boundary of the area at arbitrary point (x, y)
  2 ways to orientate the element with respect to
  the axis about which the axis of moment of
  inertia is determined
10.4 Moments of Inertia
               for an Area by Integration
Procedure for Analysis
Case 1
  Length of element orientated parallel to the axis
  Occurs when the rectangular element is used to
  determine Iy for the area
  Direct application made since the element has
  infinitesimal thickness dx and
  therefore all parts of element
  lie at the same moment
  arm distance x from the y axis
10.4 Moments of Inertia
               for an Area by Integration
Procedure for Analysis
Case 2
  Length of element orientated perpendicular to
  the axis
  All parts of the element will not lie at the same
  moment arm distance from the axis
  For Ix of area, first calculate moment of inertia of
  element about a horizontal
  axis passing through the
  element’s centroid and x axis
  using the parallel axis theorem
10.4 Moments of Inertia
                for an Area by Integration
Example 10.1
Determine the moment of
inertia for the rectangular area
with respect to (a) the
   centroidal
x’ axis, (b) the axis xb passing
through the base of the
rectangular, and (c) the pole or
z’ axis perpendicular to the x’-y’
plane and passing through the
centroid C.
10.4 Moments of Inertia
              for an Area by Integration
Solution
Part (a)
  Differential element chosen, distance y’ from
  x’ axis
  Since dA = b dy’
                                  h/2
             I x = ∫ y ' dA = ∫
                       2                 2
                                        y ' dy
                   A          −h / 2

               1 3
             = bh
              12
10.4 Moments of Inertia
              for an Area by Integration
Solution
Part (b)
  Moment of inertia about an axis passing
  through the base of the rectangle obtained
  by applying parallel axis theorem
            I xb = I x + Ad 2
                                2
              1 3    h 1 3
            = bh + bh  = bh
             12      2  3
10.4 Moments of Inertia
             for an Area by Integration
Solution
Part (c)
  For polar moment of inertia about point C
                       1 3
               I y ' = hb
                      12
               JC = I x + I y'
                 1
               = bh(h 2 + b 2 )
                12
10.4 Moments of Inertia
            for an Area by Integration
Example 10.2
Determine the moment of
inertia of the shaded area
about the x axis
10.4 Moments of Inertia
              for an Area by Integration
Solution
  A differential element of area that is parallel
  to the x axis is chosen for integration
  Since element has thickness dy and
  intersects the curve at arbitrary point (x, y),
  the area
             dA = (100 – x)dy
  All parts of the element lie at the same
  distance y from the x axis
10.4 Moments of Inertia
                   for an Area by Integration
Solution
 I x = ∫ y 2 dA
          A

 = ∫ y 2 (100 − x)dy
    A

    200           y2 
 = ∫ y 100 −
          
              2
                      dy
    0
                 400 
                      
        200         1 200 4
 = 100∫ y dy −
             2
                        ∫0 y dy
       0           400
 = 107(10 6 )mm 4
10.4 Moments of Inertia
               for an Area by Integration
Solution
  A differential element parallel
  to the y axis is chosen for
  integration
  Intersects the curve at
  arbitrary point (x, y)
  All parts of the element do not
  lie at the same distance from
  the x axis
10.4 Moments of Inertia
                        for an Area by
                          Integration
Solution
  Parallel axis theorem used to determine moment
  of inertia of the element
  For moment of inertia about its centroidal axis,
            1 3
    Ix =      bh
           12
  For the differential element shown
    b = bx h = y
  Thus,
           1
    dI x = dxy3
          12
10.4 Moments of Inertia
                   for an Area by Integration
Solution
  For centroid of the element from the x axis
         ~ = y/2
         y

  Moment of inertia of the element
                                2
                         ~ 2 = 1 dxy 3 + ydx y  = 1 y 3dx
         dI x = dI x + dAy                   
                              12            2     3
  Integrating
                          1 3     100 1
         I x = ∫ dI x = ∫ y dx = ∫
                         A3       0 3
                                        (400 x )3 / 2 dx
               ( )
         = 107 106 mm 4
10.4 Moments of Inertia
               for an Area by Integration
Example 10.3
Determine the moment of inertia with respect
to the x axis of the circular area.
10.4 Moments of Inertia
                      for an Area by Integration
Solution
Case 1
  Since dA = 2x dy
   I x = ∫ y 2 dA
              A

   = ∫ y 2 (2 x)dy
       A
        a
                  (
   = ∫ y 2 2 a 2 − y 2 dy
       −a
                         )
       πa 4
   =
        4
10.4 Moments of Inertia
              for an Area by Integration
Solution
Case 2
  Centroid for the element lies
  on the x axis
  Noting
             dy = 0
  For a rectangle,
                  1 3
          I x ' = bh
                 12
10.4 Moments of Inertia
                      for an Area by Integration
Solution
                   1
       dI x =        dx(2 y )
                             3

                  12
           2 3
       =     y dx
           3
Integrating with respect to x
        Ix = ∫
                  a2 2
                −a 3
                      (
                     a − x2      )
                                 3/ 2
                                        dx

           πa 4
       =
            4
10.4 Moments of Inertia
             for an Area by Integration
Example 10.4
Determine the moment of inertia of the
shaded area about the x
axis.
10.4 Moments of Inertia
                 for an Area by Integration
Solution
Case 1
  Differential element parallel to x axis chosen
  Intersects the curve at (x2,y) and (x1, y)
  Area, dA = (x1 – x2)dy
  All elements lie at the same distance y from
  the x axis
        I x = ∫ y dA = ∫ y (x1 − x2 )dy = ∫ y 2
             A
                 2
                        1

                        0
                            2
                                           1

                                           0
                                                  (        )
                                                      y − y dy
             2 7/2 1 4 1
        I x = y − y = 0.0357m 4
             7     4 0
10.4 Moments of Inertia
               for an Area by Integration
Solution
Case 2
  Differential element parallel
  to y axis chosen
  Intersects the curve at (x, y2)
  and (x, y1)
  All elements do not lie at the
  same distance from the x axis
  Use parallel axis theorem to
  find moment of inertia about
  the x axis
10.4 Moments of Inertia for
                     an Area by Integration
Solution
              1 3
    I x' =      bh
             12
  Integrating
                                                                      2
                           1                                y −y 
    dI x = dI x + dA~ 2 = dx( y2 − y1 ) + ( y2 − y1 )dx y1 + 2 1 
                                       3
                    y
                          12                                  2 

         (
       1 3
                     )  1
                             (
    = y2 − y13 dx = x 3 − x 6 dx
       3                3
                                 )
          1 1 3
               (
    I x = ∫ x − x 6 dx
          3 0
                         )
        1 4 1 71
    = x − x = 0.0357 m 4
       12       21 0

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6161103 10.4 moments of inertia for an area by integration

  • 1. 10.4 Moments of Inertia for an Area by Integration When the boundaries for a planar area are expressed by mathematical functions, moments of inertia for the area can be determined by the previous method If the element chosen for integration has a differential size in two directions, a double integration must be performed to evaluate the moment of inertia Try to choose an element having a differential size or thickness in only one direction for easy integration
  • 2. 10.4 Moments of Inertia for an Area by Integration Procedure for Analysis If a single integration is performed to determine the moment of inertia of an area bout an axis, it is necessary to specify differential element dA This element will be rectangular with a finite length and differential width Element is located so that it intersects the boundary of the area at arbitrary point (x, y) 2 ways to orientate the element with respect to the axis about which the axis of moment of inertia is determined
  • 3. 10.4 Moments of Inertia for an Area by Integration Procedure for Analysis Case 1 Length of element orientated parallel to the axis Occurs when the rectangular element is used to determine Iy for the area Direct application made since the element has infinitesimal thickness dx and therefore all parts of element lie at the same moment arm distance x from the y axis
  • 4. 10.4 Moments of Inertia for an Area by Integration Procedure for Analysis Case 2 Length of element orientated perpendicular to the axis All parts of the element will not lie at the same moment arm distance from the axis For Ix of area, first calculate moment of inertia of element about a horizontal axis passing through the element’s centroid and x axis using the parallel axis theorem
  • 5. 10.4 Moments of Inertia for an Area by Integration Example 10.1 Determine the moment of inertia for the rectangular area with respect to (a) the centroidal x’ axis, (b) the axis xb passing through the base of the rectangular, and (c) the pole or z’ axis perpendicular to the x’-y’ plane and passing through the centroid C.
  • 6. 10.4 Moments of Inertia for an Area by Integration Solution Part (a) Differential element chosen, distance y’ from x’ axis Since dA = b dy’ h/2 I x = ∫ y ' dA = ∫ 2 2 y ' dy A −h / 2 1 3 = bh 12
  • 7. 10.4 Moments of Inertia for an Area by Integration Solution Part (b) Moment of inertia about an axis passing through the base of the rectangle obtained by applying parallel axis theorem I xb = I x + Ad 2 2 1 3 h 1 3 = bh + bh  = bh 12 2 3
  • 8. 10.4 Moments of Inertia for an Area by Integration Solution Part (c) For polar moment of inertia about point C 1 3 I y ' = hb 12 JC = I x + I y' 1 = bh(h 2 + b 2 ) 12
  • 9. 10.4 Moments of Inertia for an Area by Integration Example 10.2 Determine the moment of inertia of the shaded area about the x axis
  • 10. 10.4 Moments of Inertia for an Area by Integration Solution A differential element of area that is parallel to the x axis is chosen for integration Since element has thickness dy and intersects the curve at arbitrary point (x, y), the area dA = (100 – x)dy All parts of the element lie at the same distance y from the x axis
  • 11. 10.4 Moments of Inertia for an Area by Integration Solution I x = ∫ y 2 dA A = ∫ y 2 (100 − x)dy A 200  y2  = ∫ y 100 −  2 dy 0  400   200 1 200 4 = 100∫ y dy − 2 ∫0 y dy 0 400 = 107(10 6 )mm 4
  • 12. 10.4 Moments of Inertia for an Area by Integration Solution A differential element parallel to the y axis is chosen for integration Intersects the curve at arbitrary point (x, y) All parts of the element do not lie at the same distance from the x axis
  • 13. 10.4 Moments of Inertia for an Area by Integration Solution Parallel axis theorem used to determine moment of inertia of the element For moment of inertia about its centroidal axis, 1 3 Ix = bh 12 For the differential element shown b = bx h = y Thus, 1 dI x = dxy3 12
  • 14. 10.4 Moments of Inertia for an Area by Integration Solution For centroid of the element from the x axis ~ = y/2 y Moment of inertia of the element 2 ~ 2 = 1 dxy 3 + ydx y  = 1 y 3dx dI x = dI x + dAy   12 2 3 Integrating 1 3 100 1 I x = ∫ dI x = ∫ y dx = ∫ A3 0 3 (400 x )3 / 2 dx ( ) = 107 106 mm 4
  • 15. 10.4 Moments of Inertia for an Area by Integration Example 10.3 Determine the moment of inertia with respect to the x axis of the circular area.
  • 16. 10.4 Moments of Inertia for an Area by Integration Solution Case 1 Since dA = 2x dy I x = ∫ y 2 dA A = ∫ y 2 (2 x)dy A a ( = ∫ y 2 2 a 2 − y 2 dy −a ) πa 4 = 4
  • 17. 10.4 Moments of Inertia for an Area by Integration Solution Case 2 Centroid for the element lies on the x axis Noting dy = 0 For a rectangle, 1 3 I x ' = bh 12
  • 18. 10.4 Moments of Inertia for an Area by Integration Solution 1 dI x = dx(2 y ) 3 12 2 3 = y dx 3 Integrating with respect to x Ix = ∫ a2 2 −a 3 ( a − x2 ) 3/ 2 dx πa 4 = 4
  • 19. 10.4 Moments of Inertia for an Area by Integration Example 10.4 Determine the moment of inertia of the shaded area about the x axis.
  • 20. 10.4 Moments of Inertia for an Area by Integration Solution Case 1 Differential element parallel to x axis chosen Intersects the curve at (x2,y) and (x1, y) Area, dA = (x1 – x2)dy All elements lie at the same distance y from the x axis I x = ∫ y dA = ∫ y (x1 − x2 )dy = ∫ y 2 A 2 1 0 2 1 0 ( ) y − y dy 2 7/2 1 4 1 I x = y − y = 0.0357m 4 7 4 0
  • 21. 10.4 Moments of Inertia for an Area by Integration Solution Case 2 Differential element parallel to y axis chosen Intersects the curve at (x, y2) and (x, y1) All elements do not lie at the same distance from the x axis Use parallel axis theorem to find moment of inertia about the x axis
  • 22. 10.4 Moments of Inertia for an Area by Integration Solution 1 3 I x' = bh 12 Integrating 2 1  y −y  dI x = dI x + dA~ 2 = dx( y2 − y1 ) + ( y2 − y1 )dx y1 + 2 1  3 y 12  2  ( 1 3 ) 1 ( = y2 − y13 dx = x 3 − x 6 dx 3 3 ) 1 1 3 ( I x = ∫ x − x 6 dx 3 0 ) 1 4 1 71 = x − x = 0.0357 m 4 12 21 0