Centroid s

CENTROIDS and

MOMENT OF INERTIA

AMET UNIVERSITY

Centroids: The concept of the centroid is nearly the same as the center of mass of an object in two dimensions, as in a very thin plate. The center of mass is obtained by breaking the object into very small bits of mass dM, multiplying these bits of mass by the distance to the x (and y) axis, summing over the entire object, and finally dividing by the total mass of the object to obtain the Center of Mass – which may be considered to be the point at which the entire mass of the object may be considered to "act". See Diagram 1.

The only difference between the center of mass and the centroid is that rather than summing the product of each bit of mass dM and the distance xi (and yi) to

an axis then dividing by the total mass, we instead divided the object into small bits of areas dA, and then take the sum of the product of each bit of area dA and the distance xi (and yi) to an axis then

divide by the total area of the object. This results in an Xct. and Yct location for

the Centroid (center of area) of the object. See Diagram 2

We will assume all our beams have uniform density and will not consider the case of non-uniform density beams. We will also point out that for any beam cross section (or object) which is symmetry, the centroid will simply be at the geometric center of the cross section. Thus for rectangular beam and I-beams, the centroid is located at the exact center of the beam. This is not the case for T-beams.Centroid of Composite Areas:Some objects or beams may be formed from several simple areas, such as rectangles, triangles, etc. (See Diagram 3) In this case the centroid of the compose area may be found by taking the sum of the produce of each simple area and the distance it's centroid is from the axis, divided by the sum of the areas. For the composite area shown in Diagram 3, the location of it's x - centroid would be given by: X ct = (A1 * x1 + A2 * x2 + A3 * x3 + A4 * x4)/(A1 +A2 +A3 + A4) where x1, x2, x3, and x4 are the distances from the centroid of each simple area to the y-axis as shown in the Diagram 3. The location of the y - centroid would be given in like manner, Y ct = (A1 * y1 + A2 * y2 + A3 * y3 + A4 * y4)/(A1 +A2 +A3 + A4)

Moment of InertiaA second quantity which is of importance when considering beam stresses is the Moment of Inertia. Once again, the Moment of Inertia as used in Physics involves the mass of the object. The Moment of Inertia is obtained by breaking the object into very small bits of mass dM, multiplying these bits of mass by the square of the distance to the x (and y) axis and summing over the entire object. See Diagram 4. For use with beam stresses, rather than using the Moment of Inertia as discussed above, we will once again use an Area Moment of Inertia. This Area Moment of Inertia is obtained by breaking the object into very small bits of area dA, multiplying these bits of area by the square of the distance to the x (and y) axis and summing over the entire object. See Diagram 5. The actual value of the moment of inertia depends on the axis chosen to calculate the moment of the inertia with respect to. Note that the moment of inertia of any object has its smallest value when calculated with respect to an axis passing through the centroid of the object.That is, for a rectangular object, the moment of inertia about an axis passing through the centroid of the rectangle is: I = 1/12 (base * depth3) with units of inches4., while the moment of inertia with respect to an axis through the base of the rectangle is: I = 1/3 (base * depth3) in4.

Parallel Axis Theorem: Moments of inertia about different axis may calculated using the Parallel Axis Theorem, which may be written: Ixx = Icc + Adc-x

2 This says that the moment of inertia about any axis

(Ixx) parallel to an axis through the centroid of the object is equal to the moment of inertia

about the axis passing through the centroid (Icc) plus the product of the area of the object and

the distance between the two parallel axis (Adc-x2).

We lastly take a moment to define several other concepts related to the Moment of Inertia.Radius of Gyration: rxx = (Ixx/A)1/2 The radius of gyration is the distance from an axis which,

if the entire area of the object were located at that distance, it would result in the same moment of inertia about the axis that the object has. Polar Moment of Inertia J = ∑ r2 dA The polar moment of inertia is the sum of the produce of each bit of area dA and the radial distance to an origin squared. In a case as shown in fig. 7, the polar moment of inertia in related to the x & y moments of inertia by: J = Ixx + Iyy.

All the summations shown above become integrations as we let the dM's and dA's approach zero.

Locate the centroid of the plane area shown.







Center of GravityThe center of gravity of a body is the point where the equivalent resultant force caused by gravity is acting. Its coordinates are defined for an arbitrary set of axes as

where x, y, z are the coordinates of an element of weight dW, and W is the total weight of the body. In the general case, where = specific weight of the material and dV = elemental volume.

CentroidsIf is a constant, the center of gravity coincides with the centroid, which is a geometrical property of a body. Centroids of lines L, areas A, and volumes V are defined analogously to the coordinates of the center of gravity,

Determine by direct integration the centroid of the area shown. Express answer in terms of a and h.

Determine by direct integration the centroid of the area shown. Express answer in terms of a and h.

Determine by direct integration the centroid of the area shown. Expressanswer in terms of a and h.

Determine by direct integration the centroid of the area shown.

Determine by direct Integration the centroid of the area shown.

A homogeneous wire is bent into the shape shown. Determine by direct integration the x coordinate of its centroid.

The horizontal x axis is drawn through the centroid C of the area shown and divides the area into two component areas A1 and A2. Determine the first moment of each component area with respect to the x axis, and explain the results obtained.

The horizontal x axis is drawn through the centroid C of the area shown and divides the area into two component areas A1 and A2. Determine the first moment of each component area with respect to the x axis, and explain the results obtained.

Continued

Continued

A bronze bushing is mounted inside a steel sleeve. Knowing that the density of bronze is 8800 kg/m3 and of steel is 7860 kg/m3, determine the center of gravity of the assembly.

Moment Of Inertia or Second Moment of Area or Area Moment of Inertia

The Area Moment Of Inertia of a beams cross-sectional area measures the beams ability to resist bending. The larger the Moment of Inertia the less the beam will bend. The moment of inertia is a geometrical property of a beam and depends on a reference axis. The smallest Moment of Inertia about any axis passes through the centroid. The following are the mathematical equations to calculate the Moment of Inertia:

Ix and Iy Here y is the distance from the x axis to an infinitesimal area dA. x is the distance from the y axis to an infinitesimal area dA.

Parallel Axis Theorem or Transfer of Axis Theorem For Area Moments of Inertia where A is the Cross-sectional Area and d is the perpendicular distance between the centroidal axis and the parallel axis.

For Area Radius of Gyration : where k is the Radius of Gyration about an axis Parallel to the Centroidal axis. : is the Radius of Gyration about the Centroidal axis. : is the perpendicular distance between the centroidal axis and the parallel axis.

Parallel Axis TheoremThe moment of inertia of any object about an axis through its center of mass is the minimum moment of inertia for an axis in that direction in space. The moment of inertia about any axis parallel to that axis through the center of mass is given by

The expression added to the center of mass moment of inertia will be recognised as the moment of inertia of a point mass – the moment of inertia about a parallel axis is the center of mass moment plus the moment of inertia of the entire objecttreated as a point mass at the center of mass.

Polar Moment Of Inertia or Moment of Inertia about the z axis The Polar Area Moment Of Inertia of a beams cross-sectional area measures the beams ability to resist torsion. The larger the Polar Moment of Inertia the less the beam will twist. The following are the mathematical equations to calculate the Polar Moment of Inertia: Jz

Here y is the distance from the x axis to an infinitesimal area dA. x is the distance from the y axis to an infinitesimal area dA.

PERPENDICULAR AXIS THEOREM: The moment of inertia of a plane area about an axis normal to the plane is equal to the sum of the moments of inertia about any two mutually perpendicular axes lying in the plane and passing through the given axis. Jz = Ix+Iy

Find the moment of inertia about the x axis of the following figure: Calculating Ix:

Perpendicular Axis TheoremFor a planar object, the moment of inertia about an axis perpendicular to the plane is the sum of the moments of inertia of two perpendicular axes through the same point in the plane of the object.

The utility of this theorem goes beyond that of calculating moments of strictly planar objects. It is a valuable tool in the building up of the moments of inertia of three dimensional objects such as cylinders by breaking them up into planar disks and summing the moments of inertia of the composite disks.

Determine the moment of inertia and the radius of gyration of the shaded area with respect to the x axis.

Determine by direct integration the moment of inertia of the shaded area with respect to y axis.

At

Then

We know the M.I. about Y axis of a rectangle is (base x height 3)

Determine by direct integration the moment of inertia of the shaded area with respect to the y axis.


By observation

Then

Now


Have

At

or

Then

At

or

Then

Determine by direct integration the moment of inertia of the shaded area with respect to the x axis.


At


Center of Mass for ParticlesThe center of mass is the point at which all the mass can be considered to be "concentrated" for the purpose for the purpose of calculating the "first moment", i.e., mass times distance. For two masses this distance is calculated from

For the more general collection of N particles this becomes

and when extended to three dimensions:

This approach applies to discrete masses even if they are not point masses if the position xi is

taken to be the position of the center of mass of the ith mass.

The Mass Moment of Inertia of a solid measures the solid's ability to resist changes in rotational speed about a specific axis. The larger the Mass Moment of Inertia the smaller the angular acceleration about that axis for a given torque. The mass moment of inertia depends on a reference axis, and is usually specified with two subscripts. This helps to provide clarity during three-dimensional motion where rotation can occur about multiple axes.

Following are the mathematical equations to calculate the Mass Moment of Inertia:

x is the distance from the yz-plane to an infinitesimal area dA.

y is the distance from the zx-plane to an infinitesimal area dA.

z is the distance from the xy-plane to an infinitesimal area dA.

I r mO 2

m

r

O

O

Mass moment of inertia for a particle: The mass moment of inertia is one measure of the distribution of the mass of an object relative to a given axis. The mass moment of inertia is denoted by I and is given for a single particle of mass m as

where O-O is the axis around which one is evaluating the mass moment of inertia, and r is the perpendicular distance between the mass and the axis O-O.

As can be seen from the above equation, the mass moment of inertia has the units of mass times length squared.

Mass moments of inertia naturally appear in the equations of motion, and provide information on how difficult (how much inertia there is) it is rotate the particle around given axis.

The mass moment of inertial should not be confused with the area moment of inertia which has units of length to the power four.

Mass moment of inertia for a rigid body: When calculating the mass moment of inertia for a rigid body, one thinks of the body as a sum of particles, each having a mass of dm. Integration is used to sum the moment of inertia of each dm to get the mass moment of inertia of body.

dm

m

r

O

O

I r dm r dmm

2 2

The equation for the mass moment of inertia of the rigid body is the integration over mass can be replaced by integration over volume, area, or length. For a fully three dimensional body using the density one can relate the element of mass to the element of volume. In this case the density has units of mass per length cubed and the relation is given as and the equation for the mass moment of inertia becomes

dm dV

I r dVV

2

The integral is actually a triple integral ∫ ∫ ∫ . If the coordinate system used is rectangular then dV=dxdydz . If the coordinates uses are cylindrical coordinates then dV rdrd dz

For a two dimensional body like a plate or a shell one can use density per unit area (units of mass per length squared) to change the integration using the relation where A is the surface area and dA differential element of area. For example, for rectangular coordinates dA=dxdy and for polar coordinates . After this substitution one gets the equation to calculate the mass moment of inertia as .

If the body is a rod like object then one can use the relation to get where l is a coordinate along the length of the rod and the density is in units of mass per unit length.

dm dA

dA rdrd I r dA

A

2

dm dl

I r dll

2

Sometime in place of the mass moment of inertia the radius of gyration k is provided. The mass moment of inertia can be calculated from k using the relation where m is the total mass of the body. One can interpret the radius of gyration as the distance from the axis that one could put a single particle of mass m equal to the mass of the rigid body and have this particle have the same mass moment of inertia as the original body.

I mk 2

O

O

CM

d

r

r’

m

Parallel-axis theorem: The moment of inertia around any axis can be calculated from the moment of inertia around parallel axis which passes through the center of mass. The equation to calculate this is called the parallel axis theorem and is given as

where d is the distance between the original axis and the axis passing through the center of mass, m is the total mass of the body, and is the moment of inertia around the axis passing through the center of mass.

Composite bodies: If a body is composed of several bodies, to calculate the moment of inertia about a given axis –

simply calculate the moment of inertia of each part around the given axis and then add them to get the mass moment of inertia of the total body.

I I md 2

I

Calculate the mass moment of inertia of the triangular plate about the y-axis. Assume the plate is made of a uniform material and has a mass of m.

The mass moment of inertia about the y-axis is given by

The element of area in rectangular coordinate system is given by

A

yy dArdmrI 22

B

x

y

h

a

dydxdxdydA

yh

ax

hy

0

0yh

a

hz y

aThe domain of the triangle is defined by

The distance from the y-axis is x. Therefore, r=x. The mass moment of inertia about the y-axis can be written as

12

33

0

33

3

0 0

22

ha

dyyh

a

dxdyxdArI

hy

y

hy

y

yh

ax

xA

yy

For a uniform plate the density can be calculated using the total mass and total area of the plate so that

ah

m

A

m

2

1

612

23 mahaI yy

Therefore, the moment of inertia in terms of the total mass of the cone can be written as

a

h

z

y

x

Calculate the mass moment of inertia of the cone about the z-axis. Assume the cone is made of a uniform material of density (mass per unit volume).

The mass moment of inertia about the z-axis is given by

V

zz dVrdmrI 22

B

The element of volume in a cylindrical coordinate system is given by

The domain of the cone in cylindrical coordinates is defined by

dzrdrddV

zh

a

hz z

a

r

z

20

0

0

zh

ar

hz

Therefore, the mass moment of inertia about the z-axis can be written as

102

4

4

0

44

4

0

2

0

44

4

0

2

0 0

32

hadzz

h

a

dzdzh

a

dzdrdrdVrI

hz

z

hz

z

hz

z

zh

ar

rV

zz

ha

m

V

m

2

3

1

10

3

10

24 mahaI zz

For a uniform cone the density can be calculated using the total mass and total volume of the cone so that

Therefore, the moment of inertia in terms of the total mass of the cone can be written as

Moment of Inertia: CylinderThe expression for the moment of inertia of a solid cylinder can be built up from the moment of inertia of thin cylindrical shells. Using the general definition for moment of inertia:

The mass element can be expressed in terms of an infinitesimal radial thickness dr by

Substituting gives a polynomial form integral:

Moment of Inertia: Hollow Cylinder

The expression for the moment of inertia of a hollow cylinder or hoop of finite thickness is obtained by the same process as that for a solid cylinder. The process involves adding up the moments of infinitesimally thin cylindrical shells. The only difference from the solid cylinder is that the integration takes place from the inner radius a to the outer radius b:

Moment of Inertia: Cylinder About Perpendicular AxisThe development of the expression for the moment of inertia of a cylinder about a diameter at its end (the x-axis in the diagram) makes use of both the parallel axis theorem and the perpendicular axis theorem. The approach involves finding an expression for a thin disk at distance z from the axis and summing over all such disks.

Obtaining the moment of inertia of the full cylinder about a diameter at its end involves summing over an infinite number of thin disks at different distances from that axis. This involves an integral from z=0 to z=L. For any given disk at distance z from the x axis, using the parallel axis theorem gives the moment of inertia about the x axis.

Note that it is the sum of the expressions for a thin disk about a diameter plus the expression for a thin rod about its end. If you take the limiting case of R=0 you get the thin rod expression, and if you take the case where L=0 you get the thin disk expression.

Now expressing the mass element dm in terms of z, we can integrate over the length of the cylinder.

Moment of Inertia: Thin DiskThe moment of inertia of a thin circular disk is the same as that for a solid cylinder of any length, but it deserves special consideration because it is often used as an element for building up the moment of inertia expression for other geometries, such as the sphere or the cylinder about an end diameter.

The moment of inertia about a diameter is the classic example of the perpendicular axis theorem

For a planar object:

Moment of Inertia: SphereThe expression for the moment of inertia of a sphere can be developed by summing the moments of infinitesimally thin disks about the z axis.

The moment of inertia of a thin disk is

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