Chapter 10 - Moments of Inertia of Areas
Consider the beam shown below
dA
Neutral Axis
Pure Bending
dF = k y dA
Internal forces are distributed forces whose magnitudes vary linearly with the distance y
from an axis passing through the centroid of the section.
Total resultant:
R   kydA  k  ydA
1st moment, Qx
The moment:
dM  ydF  ky2 dA
M   ky2 dA  k  y 2 dA
2nd moment, Qxx
Moment of inertia
Moment of inertia of an area by integration
y
y
x
x
dy
y
x
x
dx
a
dA = (a - x)dy
dA = ydx
Rectangular moments of inertia
I x   y 2 dA
Lecture21.doc
I y   x 2 dA
1
Polar moment of inertia
J o   r 2 dA
J o   r 2 dA   ( x 2  y 2 )dA   y 2 dA   x 2 dA
Jo  Ix  I y
Radius of gyration of an area
y
Has moment of inertia
A
x
Concentrate this area in a thin strip
y
If a strip is to have same moment of inertia,
the strip should be placed as shown.
A
kx
x
By def: I x  k x A
2
Thus:
kx 
Ix
A
I y  ky A
ky 
Iy
J o  ko A
ko 
Jo
A
2
2
A
or
ko  k x  k y
2
2
2
Radius of gyration - the distance, from an axis which passes through the CG, at which all
the area or mass of an object could be concentrated without changing its moment of
inertia.
Lecture21.doc
2
Parallel- Axis Theorem
Consider:
dA
y'
B
B'
centroidal axis
C
y
d
A
A'
Moment of inertia with respect to AA'
I   y 2dA
y  y  d
I   ( y   d ) 2 dA
I   y  2 dA  2d  y dA  d 2  dA
MOI, I W.R.T.
Centroidal Axis
I  I  Ad 2
1ST moment
W.R.T. BB.
Centroid, C, is
located on BB'
total area
Parallel axis theorem
The moment of inertia of an area W.R.T. to any given axis AA' is equal to the MOI I of
the area with respect to the centroidal axis plus Ad 2 . Product of area, A, and the square
of the distance d between the 2 axes.
J o  J c  Ad 2
The parallel axis theorem can be applied only if one of the two parallel axes passes
through the centroid of the area.
Lecture21.doc
3
Example
1). Given:
y
 x x2 
y  4h  2 
a a 
h
x
a
Find: Moment of inertia of shaded area with respect to the y-axis.
y
a
x
y
h
x
dx
dA = ydx
3
a
a x
 x x2 
 x4
 a3 a3 
x4 
x5  a
I y   x 2 dA  x 2 4h  2 dx  4h    2 dx  4h  2   4h  
0
0
5 
a a 
 a a 
 4a 5a  0
 4
Iy 
1 3
ha
5
Lecture21.doc
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