The document discusses moments of inertia, which are integrals related to the distribution of mass of an object and how it resists rotational changes. It provides formulas for calculating the moment of inertia for basic shapes like rectangles and circles, and explains techniques like using differential area elements and the parallel axis theorem. Sample problems demonstrate calculating moments of inertia for composite shapes.

1. 12/2/2013
1
Area Moment of Inertia
Engineering Mechanics- Distributed Forces
• Encounter in the engineering problems involving deflection, bending, buckling in
statics
Moment of Inertia of an Area
• Consider distributed forces F whose magnitudes are proportional to the
elemental areas on which they act and also vary linearlywith the distance of
from a given axis.
A
A
2
M y dA

• The integral is already familiar from our study of centroids.
• The integral is known as the area moment of inertia, or more
precisely, the second moment of the area.
y dA
y2
dA
Engineering Mechanics- Distributed Forces

2. 12/2/2013
2
Moment of Inertia of an Area by Integration
• Second moments or moments of inertia of
an area with respect to the x and y axes,
  dAxIdAyI yx
22
• Evaluation of the integralsis simplified by
choosing dA to be a thin strip parallelto
one of the coordinate axes.
• For a rectangular area,
3
3
1
0
22
bhbdyydAyI
h
x  
• The formula for rectangularareas may also
be applied to strips parallelto the axes,
dxyxdAxdIdxydI yx
223
3
1 
Engineering Mechanics- Distributed Forces
Polar Moment of Inertia
• The polar moment of inertia is an important
parameter in problems involving torsion of
cylindrical shafts and rotations of slabs.
 dArJ 2
0
• The polar moment of inertia is related to the
rectangularmoments of inertia,
 
xy II
dAydAxdAyxdArJ

 
22222
0
Engineering Mechanics- Distributed Forces

3. 12/2/2013
3
Radius of Gyration of an Area
• Consider area A with moment of inertia
Ix. Imagine that the area is
concentrated in a thin strip parallel to
the x axis with equivalent Ix.
A
I
kAkI x
xxx  2
kx = radius of gyration with respect
to the x axis
• Similarly,
A
J
kAkJ
A
I
kAkI
O
OOO
y
yyy


2
2
222
yxO kkk 
Engineering Mechanics- Distributed Forces
Sample Problem
Determine the moment of
inertia of a triangle with respect
to its base.
SOLUTION:
• Adifferential strip parallel to the x axis is chosen for
dA.
dIx  y2
dA dA  l dy
• For similar triangles,
l
b

h  y
h
l  b
h  y
h
dA  b
h  y
h
dy
• Integrating dIx from y = 0 to y = h,
 
h
hh
x
yy
h
h
b
dyyhy
h
b
dy
h
yh
bydAyI
0
43
0
32
0
22
43 








 
12
3
bh
I x
Could a vertical strip have been
chosen for the calculation?
What is the disadvantage to that
choice?
Engineering Mechanics- Distributed Forces

4. 12/2/2013
4
Sample Problem
a) Determine the centroidal polar
moment of inertia of a circular
area by direct integration.
b) Using the result of part a,
determine the moment of inertia
of a circular area with respect to a
diameter of the area.
SOLUTION:
• An annular differential area element is chosen,
dJ O  u 2
dA dA  2 u du
JO  dJ O  u 2
2 u du 
0
r
  2 u 3
du
0
r

4
2
rJO


• From symmetry, Ix = Iy,
JO  Ix  Iy  2Ix

2
r4
 2I x
4
4
rII xdiameter


Why was an annular differential area chosen?
Would a rectangular area have worked?
Engineering Mechanics- Distributed Forces
Sample Problem
Engineering Mechanics- Distributed Forces

5. 12/2/2013
5
Determine the moment of inertia of the area
under the parabola about the x-axis
Engineering Mechanics- Distributed Forces
Sample Problem
Parallel Axis Theorem
• Consider moment of inertia I of an area A
with respect to the axis AA’
 dAyI 2
• The axis BB’ passes through the area centroid
and is called a centroidal axis.
 




dAddAyddAy
dAdydAyI
22
22
2
2
AdII  parallel axis theorem
Engineering Mechanics- Distributed Forces

6. 12/2/2013
6
Parallel Axis Theorem
• Moment of inertia IT of a circular area with
respect to a tangent to the circle,
 
4
4
5
224
4
12
r
rrrAdIIT




• Moment of inertia of a triangle with respect to a
centroidal axis,
IA A  IB B  Ad2
IB B  IA A  Ad2
 1
12
bh3
 1
2
bh 1
3
h 
2
 1
36
bh3
Engineering Mechanics- Distributed Forces
Sample Problem
Engineering Mechanics- Distributed Forces
Determine the moment of inertia of the
half circle with respect to the x-axis

7. 12/2/2013
7
Moments of Inertia of Composite Areas
• The moment of inertia of a composite area A about a given axis is
obtained by adding the moments of inertia of the component areas
A1, A2, A3, ... , with respect to the same axis.
Engineering Mechanics- Distributed Forces
Sample Problem
Determine the moment of inertia
of the shaded area with respect to
the x axis.
SOLUTION:
• Compute the moments of inertia of the
bounding rectangle and half-circle with
respect to the x axis.
• The moment of inertia of the shaded area is
obtained by subtracting the moment of
inertia of the half-circle from the moment
of inertia of the rectangle.
Engineering Mechanics- Distributed Forces

8. 12/2/2013
8
Sample Problem
SOLUTION:
• Compute the moments of inertia of the bounding
rectangle and half-circle with respect to the x axis.
Rectangle:
   463
3
13
3
1
mm102138120240  .bhIx
Half-circle:
Moment of inertia with respect to AA’,
  464
8
14
8
1
mm10762590  .rI AA 
  
 
23
2
2
12
2
1
mm1072.12
90
mm81.8a-120b
mm2.38
3
904
3
4






rA
r
a
Moment of inertia with respect to x’,
 
46
2362
mm10207
)238(107212107625

 
.
...AaII AAx
Moment of inertia with respect to x,
  
46
2362
mm10392
88110721210207

 
.
...AbII xx
Engineering Mechanics- Distributed Forces
Sample Problem
• The moment of inertia of the shaded area is obtained by
subtractingthe moment of inertia of the half-circle from
the moment of inertia of the rectangle.
46
mm109.45 xI
xI  46
mm102.138   46
mm103.92 
Engineering Mechanics- Distributed Forces