Area Moments of Inertia by Integration • Second moments or moments of inertia of an area with respect to the x and y axes, 2 2 I x   y dA I y   x dA

• Evaluation of the integrals is simplified by choosing dA to be a thin strip parallel to one of the coordinate axes

ME101 - Division III Kaustubh Dasgupta 1 Area Moments of Inertia Products of Inertia: for problems involving unsymmetrical cross-sections and in calculation of MI about rotated axes. It may be +ve, -ve, or zero • Product of Inertia of area A w.r.t. x-y axes: I xy   xy dA x and y are the coordinates of the element of area dA=xy

• When the x axis, the y axis, or both are an axis of symmetry, the product of inertia is zero. • Parallel axis theorem for products of inertia:

I xy  I xy  xyA - Ixy + Ixy

Quadrants

+ Ixy - Ixy ME101 - Division III Kaustubh Dasgupta 2 Area Moments of Inertia Rotation of Axes Product of inertia is useful in calculating MI @ inclined axes.  Determination of axes about which the MI is a maximum and a minimum

2 2 I x'   y' dA   y cos  xsin  dA 2 2 I y'   x' dA   xcos  ysin  dA

I x'y'   x' y'dA   xcos  ysin y cos  xsin dA 1 cos 2 1 cos 2 sin2   cos 2   2 2 sin cos 1/ 2sin 2 cos 2   sin2   cos 2 2 2 I x   y dA I y   x dA I x  I y I x  I y I x   cos 2  I xy sin 2 I xy   xy dA 2 2 Moments and product of inertia I  I I  I I  x y  x y cos 2  I sin 2 w.r.t. new axes x’ and y’ ? y 2 2 xy Note: x  xcos  ysin I  I I  x y sin 2  I cos 2 y  y cos  xsin xy' 2 xy

ME101 - Division III Kaustubh Dasgupta 3 Area Moments of Inertia Rotation of Axes Adding first two eqns: Ix’ + Iy’ = Ix + Iy = Iz  The Polar MI @ O

Angle which makes Ix’ and Iy’ either max or min can be found by setting the derivative

of either Ix’ or Iy’ w.r.t. θ equal to zero:

I  I I  I dIx' x y x y  Iy  Ix sin2  2Ixy cos2  0 I    cos2  I sin 2 x 2 2 xy d Ix  Iy Ix  Iy Denoting this critical angle by α I    cos2  I sin 2 y 2 2 xy 2Ixy Ix  Iy tan 2  Ixy'  sin 2  Ixy cos2  2 Iy Ix  two values of 2α which differ by π since tan2α = tan(2α+π)  two solutions for α will differ by π/2  one value of α will define the axis of maximum MI and the other defines the axis of minimum MI  These two rectangular axes are called the principal axes of inertia

ME101 - Division III Kaustubh Dasgupta 4 Area Moments of Inertia

Rotation of Axes 2Ixy 2Ixy tan 2   sin2  cos2 Iy  Ix Iy  Ix Substituting in the third eqn for critical value

of 2θ: Ix’y’ = 0

 Product of Inertia Ix’y’ is zero for the Ix  Iy Ix  Iy Principal Axes of inertia Ix   cos2  Ixy sin 2 2 2 Substituting sin2α and cos2α in first two eqns Ix  Iy Ix  Iy Iy   cos2  Ixy sin 2 for Principal Moments of Inertia: 2 2 I  I  x y    Ixy' sin 2 Ixy cos2 Ix  Iy 1 2 I   I  I 2 4I 2 max 2 2 x y xy Ix  Iy 1 I   I  I 2 4I 2 min 2 2 x y xy Ixy@  0

ME101 - Division III Kaustubh Dasgupta 5 Area Moments of Inertia Mohr’s Circle of Inertia :: Graphical representation of the MI equations

- For given values of Ix, Iy, & Ixy, corresponding values of Ix’, Iy’, & Ix’y’ may be determined from the diagram for any desired angle θ.

I  I I  I 2I  x y  x y    xy Ix cos2 Ixy sin 2 tan 2  Ixy 2 2 Iy  Ix Ix  Iy Ix  Iy I    cos2  I sin 2 y 2 2 xy Ix  Iy I   sin 2  I cos2 x y' 2 xy

Ix  Iy 1 I   I  I 2 4I 2 I max 2 2 x y xy Ix  Iy 1 I   I  I 2 4I 2 min 2 2 x y xy Ixy@  0 • At the points A and B, I = 0 and I I  I 2  I 2  R2 x’y’ x’ x ave xy takes the maximum and minimum values 2 I  I  I  I  I  I  R x y  x y  2 max, min ave Iave  R     Ixy 2  2  ME101 - Division III Kaustubh Dasgupta 6 Ix  Iy Ix  Iy I    cos2  I sin 2 Area Moments of Inertia x 2 2 xy Ix  Iy Ix  Iy Iy   cos2  Ixy sin 2 2Ixy 2 2 Construction tan 2  Mohr’s Circle of Inertia: Iy  Ix Ix  Iy I   sin 2  I cos2 x y' 2 xy

Ix  Iy 1 I   I  I 2 4I 2 max 2 2 x y xy Ix  Iy 1 I   I  I 2 4I 2 min 2 2 x y xy Ixy@  0 Choose horz axis  MI Choose vert axis  PI

Point A – known {Ix, Ixy} Point B – known {Iy, -Ixy} Circle with dia AB Angle α for Area  Angle 2α to horz (same

sense)  Imax, Imin Angle x to x’ = θ  Angle OA to OC = 2θ  Same sense

Point C  Ix’, Ix’y’ Point D  Iy’ ME101 - Division III Kaustubh Dasgupta 7 Area Moments of Inertia Example: Product of Inertia SOLUTION: • Determine the product of inertia using direct integration with the parallel axis theorem on vertical differential area strips • Apply the parallel axis theorem to evaluate the product of inertia with respect to the centroidal axes.

Determine the product of inertia of the right triangle (a) with respect to the x and y axes and (b) with respect to centroidal axes parallel to the x and y axes.

ME101 - Division III Kaustubh Dasgupta 8 Area Moments of Inertia

Examples SOLUTION: • Determine the product of inertia using direct integration with the parallel axis theorem on vertical differential area strips  x   x  y  h1  dA  y dx  h1 dx  b   b  1 1  x  xel  x yel  y  h1  2 2  b 

Integrating dIx from x = 0 to x = b, 2 b  x  I  dI  x y dA  x 1 h2 1 dx xy  xy  el el  2   0  b  b 2 3 2 2 3 4 b 2  x x x  x x x   h    dx h      2 b 2  4 3b 2 0 2b   8b 0 I  1 b2h2 xy 24

ME101 - Division III Kaustubh Dasgupta 9 Area Moments of Inertia Examples SOLUTION

• Apply the parallel axis theorem to evaluate the product of inertia with respect to the centroidal axes. x  1 b y  1 h 3 3

With the results from part a,

Ixy  Ixy  xyA I  1 b2h2  1 b 1 h 1 bh xy 24 3 3 2 

I   1 b2h2 xy 72

ME101 - Division III Kaustubh Dasgupta 10 Area Moments of Inertia Example: Mohr’s Circle of Inertia SOLUTION:

• Plot the points (Ix , Ixy) and (Iy ,-Ixy). Construct Mohr’s circle based on the circle diameter between the points. • Based on the circle, determine the orientation of the principal axes and the The moments and product of inertia principal moments of inertia. with respect to the x and y axes are Ix = • Based on the circle, evaluate the 4 4 7.24x106 mm , Iy = 2.61x106 mm , and moments and product of inertia with 6 4 Ixy = -2.54x10 mm . respect to the x’y’ axes. Using Mohr’s circle, determine (a) the principal axes about O, (b) the values of the principal moments about O, and (c) the values of the moments and product of inertia about the x’ and y’ axes

ME101 - Division III Kaustubh Dasgupta 11 Area Moments of Inertia Example: Mohr’s Circle of Inertia SOLUTION:

• Plot the points (Ix , Ixy) and (Iy ,-Ixy). Construct Mohr’s circle based on the circle diameter between the points. OC  I  1 I  I  4.925106 mm4 ave 2  x y  CD  1 I  I  2.315106 mm4 2  x y  R  CD2  DX 2  3.437 106 mm4

• Based on the circle, determine the orientation of the 6 4 Ix  7.2410 mm principal axes and the principal moments of inertia. 6 4 I y  2.6110 mm DX tan 2m  1.097 2m  47.6 m  23.8 6 4 CD Ixy  2.5410 mm

ME101 - Division III Kaustubh Dasgupta 12 6 4 Area Moments of Inertia OC  Iave  4.92510 mm 6 4 Example: Mohr’s Circle of Inertia R  3.437 10 mm • Based on the circle, evaluate the moments and product of inertia with respect to the x’y’ axes. The points X’ and Y’ corresponding to the x’ and y’ axes are obtained by rotating CX and CY counterclockwise through an angle θ 2(60o) = 120o. The angle that CX’ forms with the horz is f = 120o - 47.6o = 72.4o.

o Ix'  OF  OC  CXcos  Iave  Rcos72.4 6 4 Ix  5.9610 mm o I y'  OG  OC  CYcos  Iave  Rcos72.4

6 4 I y  3.8910 mm o Ixy'  FX  CYsin  Rsin 72.4

6 4 Ixy  3.2810 mm

ME101 - Division III Kaustubh Dasgupta 13 Mass Moment of Inertia • Application in rigid body dynamics - Measure of distribution of mass of a rigid body w.r.t. the axis (constant property for that axis)

I = ∫ r2 dm

r = perpendicular distance of the mass element dm from the axis O-O

r2Δm :: measure of the inertia of the system

ME101 - Division III Kaustubh Dasgupta 14 Mass Moment of Inertia • About individual coordinate axes

ME101 - Division III Kaustubh Dasgupta 15 Mass Moment of Inertia • Parallel Axis Theorem

ME101 - Division III Kaustubh Dasgupta 16 Mass Moment of Inertia Moments of Inertia of Thin Plates • For a thin plate of uniform thickness t and homogeneous material of density r, the mass moment of inertia with respect to axis AA’ contained in the plate is 2 2 I AA   r dm  rt r dA  rt I AA,area

• Similarly, for perpendicular axis BB’ which is also contained in the plate,

IBB  rt IBB,area

• For the axis CC’ which is perpendicular to the plate,

ICC  rt JC,area  rt I AA,area  I BB,area 

 I AA  I BB

ME101 - Division III Kaustubh Dasgupta 17 Mass Moment of Inertia Moments of Inertia of Thin Plates

• For the principal centroidal axes on a rectangular plate,

I  rt I  rt 1 a3b  1 ma 2 AA AA,area 12  12

I  rt I  rt 1 ab3  1 mb 2 BB BB,area 12  12

I  I  I  1 m a2  b2 CC AA,mass BB,mass 12  

• For centroidal axes on a circular plate, I  I  rt I  rt 1  r 4  1 mr 2 AA BB AA,area 4  4

ME101 - Division III Kaustubh Dasgupta 18 Mass Moment of Inertia Moments of Inertia of a 3D Body by Integration • Moment of inertia of a homogeneous body is obtained from double or triple integrations of the form I  r r 2dV

• For bodies with two planes of symmetry, the moment of inertia may be obtained from a single integration by choosing thin slabs perpendicular to the planes of symmetry for dm.

• The moment of inertia with respect to a particular axis for a composite body may be obtained by adding the moments of inertia with respect to the same axis of the components.

ME101 - Division III Kaustubh Dasgupta 19 Mass Moment of Inertia MI of some common geometric shapes

ME101 - Division III Kaustubh Dasgupta 20