Engineering Mechanics: Statics Appendix A: Area Moments of Inertia.
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Transcript of Engineering Mechanics: Statics Appendix A: Area Moments of Inertia.
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Engineering Mechanics: Engineering Mechanics: Statics Statics
Appendix A: Area Moments of Inertia
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Moment of InertiaMoment 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!
Hydrostatic pressure
Bending moment in beam
Torsion in shaft
2M k y dA
Moment of inertia of area/ Second moment of area (I )
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Rectangular moment of inertia
Polar moment of inertia
DefinitionsDefinitions
2
2
x
y
I y dA
I x dA
-- Moment of inertia about x-axis
2 x yzI r dA I I
• Notice that Ix, Iy, Iz involve the square of the distance from the inertia axis -- always positive!
• dimensions = L4 (ex. m4 or mm4)
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Sample Problem A/1Sample Problem A/1 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
3 3,
12 12x ybh hbI I
4
4x yrI I
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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 kx from 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
Radius of GyrationRadius of Gyration
2x xk A I
x xk I A
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Similarly,
Do not confused with centroid C!
Radius of GyrationRadius of Gyration
y yk I A
z zk I A 2 2 2z x yk k k
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Transfer of AxesTransfer of Axes 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
2 20( )x xdAdI y y d dA
2 20 0 2x x x xd dA d dA dI I y y dA
2x x xI I Ad
0Ay 0 0y and with the centroid on x0-axis
2
2
2
x x x
y y y
z z
I I Ad
I I Ad
I I AdParallel-axis theorems
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Composite AreasComposite Areas Centroid of composite areas:
i i
i
A xx
A
Part Area, A
Sum A
x y AyAx
Ax Ay
i i
i
A yy
A
100 mm
400 mm
400 mm
100 mm
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Composite AreasComposite Areas 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 = I + Ad2
o The radius of gyration for the composite area cannot be added, k = I/A
xIPart Area, A dx dy Adx2 Ady
2yI
Sum A Adx2 Ady
2
xI yI
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Example A/7Example A/7
Calculate the moment of inertia and radius of gyration about the x-axis for the shaded area shown
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Products of InertiaProducts of Inertia 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
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Sample Problem A/8 & A/10Sample Problem A/8 & A/10 Determine the product of inertia of the area shown with respect
to the x-y axes.
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Rotation of AxesRotation of Axes To calculate the moment of inertia of an area
about an inclined axes
Ix’ = y’2 dA = (ycos – xsin )2 dA
Iy’ = x’2 dA = (ysin – xcos )2 dA
-- expand & substitute sin2 = (1- cos 2)/2
cos2 = (1+ cos 2)/2
'
'
' '
cos2 sin22 2
cos2 sin22 2
sin2 cos22
x y x yx xy
x y x yy xy
x yx y xy
I I I II I
I I I II I
I II I
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Rotation of AxesRotation of Axes The angle which makes Ix’ and Iy’ either max or
min
dIx’/d = (Iy - Ix)sin 2 - 2Ixycos 2 = 0
The critical angle :
2tan2 xy
y x
I
I I
This equation gives two value of 2 [tan 2 = tan (2+) ] obtain two values for (differ by /2)
axis of minimum moment of inertia axis of maximum moment of inertia
called “Principal Axes of Inertia”
2 2max
2 2min
1( ) 4
2 21
( ) 42 2
x yx y xy
x yx y xy
I II I I I
I II I I I
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Mohr’s Circle of InertiaMohr’s Circle of Inertia
1. Draw x-axis as I and y-axis as Ixy
2. Plot point A at (Ix, Ixy) and B at (Iy, -
Ixy)
3. Find the center of the circle at O2 2R OS AS
R
S ImaxImin
6. Imax = O + R and Imin = O - R
tan2ASOS
5. Angle 2 is found from AS and OS as
4. Radius of the circle is OA or OB
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Sample Problem A/11Sample Problem A/11
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.