#### Transcript Slide 1

```Engineering Mechanics:
Statics
Appendix A: Area Moments of Inertia
Moment of Inertia

Hydrostatic pressure

Bending
moment in
beam
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
M  k y2dA
Moment of inertia of area/
Second moment of area (I )

Torsion in shaft
I is a function of geometry only!
Definitions

Rectangular moment of inertia
Ix   y2dA
Iy   x2dA

-- Moment of inertia about x-axis
Polar moment of inertia
Iz   r 2dA  Ix  Iy
• 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 xaxis, and the polar axis z through O.
3
3
-- Must remember!: for a rectangular area, Ix  bh , Iy  hb
12
12
4
: for a circular area, Ix  Iy   r - see sample problem A/3
4


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,
kx2A  Ix

The distance kx = radius of gyration of the area about x-axis
 kx  Ix A

Similarly,
ky  Iy A
kz  Iz A

Do not confused with centroid C!
kz2  kx2  ky2
Transfer of Axes

Moment of inertia of an area about a noncentroidal axis
dIx  y2dA  (y0  dx )2 dA
Ix   dIx   y02dA  2dx  y0dA  dx2  dA
2
Iy  Iy  Ady 2
 Ay0 and y0  0 with
the centroid on x0-axis
Parallel-axis theorems

The axis between which the transfer is made must be parallel

One of the axes must pass through the centroid of the area
Composite Areas

Centroid of composite areas:
x
  Ai xi
y
 Ai
 Ai yi
 Ai
400 mm
100 mm
Part
Area, A
Sum
SA
x
y
Ax
Ay
S Ax
S Ay
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
o The radius of gyration for the composite area cannot be added,
k = I/A
Part
Sum
Area, A
SA
dx
dy
Ix
Iy
S Iy
Example A/7
Calculate the moment of inertia and radius of
shown
Table D/3
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
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
Ix’ =  y’2 dA =  (ycos q – xsin q )2 dA
Iy’ =  x’2 dA =  (ysin q – xcos q )2 dA
-- expand & substitute
Ix ' 
Iy ' 
Ix  Iy
2
Ix  Iy
Ix ' y ' 


2
Ix  Iy
2
Ix  Iy
2
Ix  Iy
2
sin2q = (1- cos 2q)/2
cos2q = (1+ cos 2q)/2
cos 2q  Ixy sin2q
cos 2q  Ixy sin2q
sin2q  Ixy cos 2q
2
Ix  Iy 

I

 x'
 
2


I 
x 'y '
2
I

x
 Iy 
4
2
 Ixy 2
Mohr’s Circle of Inertia
R
Imin
S
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
4.
Radius of the circle is OA or OB
R  OS2  AS2
Imax
5.
Angle 2 is found from AS and OS as
tan2 
6.
AS
OS
Imax = O + R and Imin = O - R
Rotation of Axes

The critical angle :
tan2 


2Ixy
Iy  Ix
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
Imax  Iave  R
Imin  Iave  R
called “Principal Axes of Inertia”
Sample Problem A/11
Determine the orientation of the principal axes of
inertia through the centroid of the angle section and
determine the corresponding maximum and minimum
moments of inertia.
```