Engineering Mechanics: Statics. Appendix A: Area Moments of Inertia. Moment of Inertia. When forces are distributed continuously over an area, it is often necessary to calculate moment of these forces about some axis (in or perpendicular to the plane of area)
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Engineering Mechanics: Statics
Appendix A: Area Moments of Inertia
When forces are distributed continuously over an area, it is often necessary to calculate moment of these forces about some axis (in or perpendicular to the plane of area)
Frequently, intensity of the distributed force is proportional to the distance of the line of action from the moment axis, p = ky
dM = y(pdA) = ky2dA
I is a function of geometry only!
Bending moment in beam
Moment of inertia of area/
Second moment of area(I )
Torsion in shaft
Rectangular moment of inertia
Polar moment of inertia
-- Moment of inertia about x-axis
Determine the moments of inertia of the rectangular area about the centroidal x0- and y0-axes, the centroidal polar axis z0 through C, the x-axis, and the polar axis z through O.
-- Must remember!: for a rectangular area,
: for a circular area, - see sample problem A/3
For an area A with moment of inertia Ix and Iy
Visualize it as concentrated into a long narrow strip of area A a distance kxfrom the x-axis. The moment of inertia about x-axis is Ix. Therefore,
The distance kx = radius of gyration of the area about x-axis
Do not confused with centroid C!
Moment of inertia of an area about a noncentroidal axis
The axis between which the transfer is made must be parallel
One of the axes must pass through the centroid of the area
the centroid on x0-axis
Centroid of composite areas:
Part Area, A
Sum SA S S
The moment of inertia of a composite area about a particular axis is the sum of the moments of inertia of its component parts about the same axis.
I = SI + SAd2
The radius of gyration for the composite area cannot be added, k =I/A
Part Area, Adxdy Adx2Ady2
Sum SA SAdx2SAdy2S S
Calculate the moment of inertia and radius of gyration about the x-axis for the shaded area shown
Unsymmetrical cross section
Ixy = xydA
may be positive, negative or zero
Ixy= 0 when either the reference axes is an axis of symmetry
because x(-y)dA cancel x(+y)dA
Transfer of Axes
Ixy = (x0+dy)(y0+dx)dA
Ixy = Ixy + Adx dy
Determine the product of inertia of the area shown with respect to the x-y axes.
To calculate the moment of inertia of an area
about an inclined axes
Ix’ = y’2 dA = (ycosq– xsinq)2 dA
Iy’ = x’2 dA = (ysinq– xcosq)2 dA
-- expand & substitutesin2q = (1- cos 2q)/2
cos2q = (1+ cos 2q)/2
The angle which makes Ix’ and Iy’ either max or min
dIx’/dq= (Iy - Ix)sin 2q - 2Ixycos 2q = 0
The critical angle a:
called “Principal Axes of Inertia”
5. Angle 2a is found from AS and OS as
6. Imax = O + R and Imin = O - R
4. Radius of the circle is OA or OB
Determine the orientation of the principle axes of inertia through the centroid of the angle section and determine the corresponding maximum and minimum moments of inertia.