The document describes Mohr's circle, which is used to analyze the principal moments of inertia for a given cross-sectional area. It presents equations to determine the radius and center of the Mohr's circle based on the area's moments of inertia (Ix, Iy) and product of inertia (Ixy). An example problem is shown where these values are used to construct the circle and determine the maximum and minimum moments of inertia and their corresponding principal axes.

1. 10.8 Mohr’s Circle for Moments
of Inertia
It can be found that
2 2
 Ix + I y   Ix − Iy 
 Iu −
  + I uv = 
2
 + I xy
2
 2  
 2 
 
In a given problem, Iu and Iv are variables and Ix,
Iy and Ixy are known constants
(I u − a )2 + I uv = R 2
2
When this equation is plotted on a set of axes
that represent the respective moment of inertia
and the product of inertia, the resulting graph
represents a circle

2. 10.8 Mohr’s Circle for Moments
of Inertia
The circle constructed is known as a Mohr’s
circle with radius
2
 Ix + Iy 
R= 
 2   + I xy
2
 
and center at (a, 0) where
a = (I x + I y ) / 2

3. 10.8 Mohr’s Circle for Moments
of Inertia
Procedure for Analysis
Determine Ix, Iy and Ixy
Establish the x, y axes for the area, with the
origin located at point P of interest and
determine Ix, Iy and Ixy
Construct the Circle
Construct a rectangular coordinate system such
that the abscissa represents the moment of
inertia I and the ordinate represent the product
of inertia Ixy

4. 10.8 Mohr’s Circle for Moments
of Inertia
Procedure for Analysis
Construct the Circle
Determine center of the circle O, which is located
at a distance (Ix + Iy)/2 from the origin, and plot
the reference point a having coordinates (Ix, Ixy)
By definition, Ix is always positive, whereas Ixy
will either be positive or negative
Connect the reference point A with the center of
the circle and determine distance OA (radius of
the circle) by trigonometry
Draw the circle

5. 10.8 Mohr’s Circle for Moments
of Inertia
Procedure for Analysis
Principal of Moments of Inertia
Points where the circle intersects the abscissa
give the values of the principle moments of
inertia Imin and Imax
Product of inertia will be zero at these points
Principle Axes
To find direction of major principal axis,
determine by trigonometry, angle 2θp1,
measured from the radius OA to the positive I
axis

6. 10.8 Mohr’s Circle for Moments
of Inertia
Procedure for Analysis
Principle Axes
This angle represent twice the angle from the x
axis to the area in question to the axis of
maximum moment of inertia Imax
Both the angle on the circle, 2θp1, and the angle
to the axis on the area, θp1must be measured in
the same sense
The axis for the minimum moment of inertia Imin
is perpendicular to the axis for Imax

7. 10.8 Mohr’s Circle for Moments
of Inertia
Example 10.10
Using Mohr’s circle, determine the principle
moments of the beam’s cross-sectional area
with respect to an axis
passing through the
centroid.

8. 10.8 Mohr’s Circle for Moments
of Inertia
Solution
Determine Ix, Iy and Ixy
Moments of inertia and the product of inertia
have been determined in previous examples
( )
I x = 2.90 109 mm4 I y = 5.60(109 )mm4
I xy = −3.00(109 )mm4
Construct the Circle
Center of circle, O, lies from the origin, at a
distance
(I x + I y )/ 2 = (2.90 + 5.60) / 2 = 4.25

9. 10.8 Mohr’s Circle for Moments
of Inertia
Solution
With reference point A (2.90, -3.00) connected to
point O, radius OA is determined using
Pythagorean theorem
OA = (1.35)2 + (− 3.00)2
= 3.29
Principal Moments of Inertia
Circle intersects I axis at
points (7.54, 0) and
(0.960, 0)

10. 10.8 Mohr’s Circle for Moments
of Inertia
Solution
( )
I max = 7.54 109 mm4 ( )
I min = 0.960 109 mm 4
Principal Axes
Angle 2θp1 is determined from
the circle by measuring CCW
from OA to the direction of the
positive I axis
−1
BA 
2θ p1 = 180 − sin 
o
 OA 

 
o −1 3.00  o
= 180 − sin   = 114.2
 3.29 

11. 10.8 Mohr’s Circle for Moments
of Inertia
Solution
The principal axis for Imax = 7.54(109) mm4
is therefore orientated at an angle θp1 =
57.1°,
measured CCW from the positive x axis
to the positive u axis
v axis is perpendicular
to this axis