Center of Mass and Centroids :: Guidelines Centroids of Lines, Areas, and Volumes

1.Order of Element Selected for Integration 2.Continuity 3.Discarding Higher Order Terms 4.Choice of Coordinates 5.Centroidal Coordinate of Differential Elements xdL ydL zdL x   y   z   L L L x dA y dA z dA x   c y   c z   c A A A x dV y dV z dV x   c y   c z   c V V V ME101 - Division III Kaustubh Dasgupta 1 Center of Mass and Centroids Theorem of Pappus: Area of Revolution - method for calculating surface area generated by revolving a plane curve about a non-intersecting axis in the plane of the curve

Surface Area Area of the ring element: circumference times dL dA = 2πy dL If area is revolved Total area, A  2 ydL through an angle θ<2π  θ in radians  yL   ydL  A  2 yL or A   y L y is the y-coordinate of the centroid C for the line of length L Generated area is the same as the lateral area of a right circular cylinder of length L and radius y Theorem of Pappus can also be used to determine centroid of plane curves if area created by revolving these figures @ a non-intersecting axis is known ME101 - Division III Kaustubh Dasgupta 2 Center of Mass and Centroids Theorem of Pappus: Volume of Revolution - method for calculating volume generated by revolving an area about a non-intersecting axis in the plane of the area Volume

Volume of the ring element: circumference times dA dV = 2πy dA If area is revolved Total Volume, V  2  ydA through an angle θ<2π θ in radians  yA   ydA  V  2 yA or V  yA y is the y-coordinate of the centroid C of the revolved area A Generated volume is obtained by multiplying the generating area by the circumference of the circular path described by its centroid.

Theorems of Pappus can also be used to determine centroid of plane areas if volume created by revolving these figures @ a non-intersecting axis is known

ME101 - Division III Kaustubh Dasgupta 3 Area Moments of Inertia

• Previously considered distributed forces which were proportional to the area or volume over which they act. - The resultant was obtained by summing or integrating over the areas or volumes - The moment of the resultant about any axis was determined by computing the first moments of the areas or volumes about that axis

• Will now consider forces which are proportional to the area or volume over which they act but also vary linearly with distance from a given axis. - the magnitude of the resultant depends on the first moment of the force distribution with respect to the axis. - The point of application of the resultant depends on the second moment of the distribution with respect to the axis.

ME101 - Division III Kaustubh Dasgupta 4 Area Moments of Inertia  • Consider distributed forces F whose magnitudes are proportional to the elemental areas A on which they act and also vary linearly with the distance of from a given axis. • Example: Consider a beam subjected to pure bending. Internal forces vary linearly with distance from the neutral axis which passes through the section centroid.  F  kyA

R  k y dA  0  y dA  Qx  first moment M  k y2dA  y2dA  second moment

First Moment of the whole section about the x-axis = yA = 0 since the centroid of the section lies on the x-axis.

Second Moment or the Moment of Inertia of the beam

section about x-axis is denoted by Ix and has units of (length)4 (never –ve)

ME101 - Division III Kaustubh Dasgupta 5 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 6 Area Moments of Inertia by Integration

• For a rectangular area, h I  y2dA  y2bdy  1 bh3 x   3 0

• The formula for rectangular areas may also be applied to strips parallel to the axes,

dI  1 y3dx dI  x2dA  x2 y dx x 3 y

ME101 - Division III Kaustubh Dasgupta 7 Area Moments of Inertia Polar Moment of Inertia 2 J0  I z   r dA

• The polar moment of inertia is an important parameter in problems involving torsion of cylindrical shafts and rotations of slabs

• The polar moment of inertia is related to the rectangular moments of inertia,

2 2 2 2 2 J 0  I z   r dA   x  y dA   x dA   y dA

 I y  I x

Moment of Inertia of an area is purely a mathematical property of the area and in itself has no physical significance.

ME101 - Division III Kaustubh Dasgupta 8 Area Moments of Inertia

Radius of Gyration of an Area • Consider area A with moment of inertia

Ix. Imagine that the area is concentrated in a thin strip parallel to

the x axis with equivalent Ix. I I  k 2 A k  x x x x A

kx = radius of gyration with respect to the x axis • Similarly, I I  k 2 A k  y y y y A J J  I  k 2 A  k 2 A k  k  O O z O z O z A 2 2 2 2 kO  kz  kx  k y Radius of Gyration, k is a measure of distribution of area from a reference axis Radius of Gyration is different from centroidal distances

ME101 - Division III Kaustubh Dasgupta 9 Area Moments of Inertia

Example: Determine the moment of inertia of a triangle with respect to its base. SOLUTION: • A differential strip parallel to the x axis is chosen for dA. 2 dIx  y dA dA  l dy • For similar triangles, l h  y h  y h  y  l  b dA  b dy b h h h

• Integrating dIx from y = 0 to y = h, h h 2 2 h  y b 2 3 Ix   y dA   y b dy   hy  y dy 0 h h 0 3 4 h b  y y  bh3  h   h 3 4 I x  0 12

ME101 - Division III Kaustubh Dasgupta 10 Area Moments of Inertia

Example: a) Determine the centroidal polar moment of inertia of a circular area by direct integration. b) Using the result of part a, determine the moment of inertia of a circular area with respect to a diameter. SOLUTION: • An annular differential area element is chosen,

2 dJO  u dA dA  2 u du r r 2 3 JO   dJO  u 2 u du  2 u du 0 0  J  r4 O 2

• From symmetry, Ix = Iy,  J  I  I  2I r4  2I O x y x 2 x

 I  I  r4 diameter x 4

ME101 - Division III Kaustubh Dasgupta 11 Area Moments of Inertia Parallel Axis Theorem • Consider moment of inertia I of an area A with respect to the axis AA’ I   y2dA

• The axis BB’ passes through the area centroid and is called a centroidal axis.

I  y2dA  y  d 2 dA I   y dA   y  d  dA 2 2   y2dA  2d  ydA  d 2 dA Parallel Axis theorem:    MI @ any axis = • Second term = 0 since centroid lies on BB’ MI @ centroidal axis + Ad2 (∫y’dA = ycA, and yc = 0 The two axes should be 2 Parallel Axis theorem parallel to each other. I  I  Ad

ME101 - Division III Kaustubh Dasgupta 12 Area Moments of Inertia Parallel Axis Theorem

• Moment of inertia IT of a circular area with respect to a tangent to the circle, I  I  Ad 2  1  r 4   r 2 r 2 T 4    5  r 4 4

• Moment of inertia of a triangle with respect to a centroidal axis, 2 I AA  I BB  Ad 2 I  I  Ad 2  1 bh3  1 bh 1 h BB AA 12 2 3   1 bh3 36 • The moment of inertia of a composite area A about a given axis is obtained by adding the

moments of inertia of the component areas A1, A2, A3, ... , with respect to the same axis.

ME101 - Division III Kaustubh Dasgupta 13 Area Moments of Inertia: Standard MIs

ME101 - Division III Kaustubh Dasgupta 14 Area Moments of Inertia Example: Determine the moment of inertia of the shaded area with respect to the x axis.

SOLUTION: • Compute the moments of inertia of the bounding rectangle and half-circle with respect to the x axis. • The moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle.

ME101 - Division III Kaustubh Dasgupta 15 Area Moments of Inertia

Example: Solution SOLUTION: • Compute the moments of inertia of the bounding rectangle and half-circle with respect to the x axis. Rectangle: 1 3 1 3 6 4 I x  3 bh  3 240120 138.210 mm Half-circle: 4r 490 a    38.2 mm moment of inertia with respect to AA’, 3 3 I  1r4  1 904  25.76106mm4 b  120 - a  81.8 mm AA 8 8

1 2 1 2 Moment of inertia with respect to x’, A  2 r  2  90 2 6 3 2  12.72103 mm 2 I x  I AA  Aa  25.7610  12.7210 38.2  7.20106 mm 4

moment of inertia with respect to x, 2 6 3 2 I x  I x  Ab  7.2010  12.7210 81.8  92.3106 mm 4

ME101 - Division III Kaustubh Dasgupta 16 Area Moments of Inertia

Example: Solution

• The moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle.

6 4 6 4 I x  138.210 mm  92.310 mm

6 4 Ix  45.910 mm

ME101 - Division III Kaustubh Dasgupta 17 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 18