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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)
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 )
Rectangular moment of inertia
Polar moment of inertia
Definitions
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)
Sample 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
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 Gyration
2x xk A I
x xk I A
Transfer 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 2
0 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 Ad
Parallel-axis theorems
Composite Areas
Centroid of composite areas:
i i
i
A xx
A
Part Area, A
Sum SA S S
x y AyAx
Ax Ay
i i
i
Ayy
A
100 mm
400 mm
400 mm
100 mm
Composite 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 = SI + SAd2
o Find the moment of inertia of the T-section
o The radius of gyration for the composite area cannot be added, k = I/A
xIPart Area, A dx dy Adx2 Ady
2 yI
Sum SA SAdx2 SAdy
2 S S xI yI
Products 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
xy xy x yI I Ad d
Sample Problem A/8 & A/10
Determine the product of inertia of the area shown with respect to the x-y axes.
Rotation of Axes
To calculate the moment of inertia of an area
about an inclined axes
Ix’ = y’2 dA = (ycos q – xsin q )
2 dA
Iy’ = x’2 dA = (ysin q – xcos q )
2 dA
-- expand & substitute sin2q = (1- cos 2q)/2
cos2q = (1+ cos 2q)/2
q q
q q
q q
'
'
' '
cos2 sin22 2
cos2 sin22 2
sin2 cos22
x y x y
x xy
x y x y
y xy
x y
x y xy
I I I II I
I I I II I
I II I
222 2
' ' ' 2 4
x yx y
x x y xy
I II II I I
Mohr’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 O
2 2R OS AS
R
S Imax Imin
6. Imax = O + R and Imin = O - R
tan2AS
OS
5. Angle 2 is found from AS and OS as
4. Radius of the circle is OA or OB
Rotation of Axes
The critical angle :
This equation gives two value of 2
[tan 2 = tan (2+) ]
2tan2
xy
y x
I
I I
obtain two values for (differ by /2)
axis of minimum moment of inertia
axis of maximum moment of inertia
called “Principal Axes of Inertia”
max
min
ave
ave
I I R
I I R