1. Problem 9.190 For the 5 x 3 x -in. angle cross
section shown, use Mohr’s circle
to determine (a) the moments of
inertia and the product of inertia
with respect to new centroidal
axes obtained by rotating the x
and y axes 30o clockwise, (b) the
orientation of the principal axes
through the centroid and the
corresponding values of the
moments of inertia. 1 2
3 in
5 in
0.5 in
1.75 in x
0.5 in
0.75 in y
L5x3x 1 2

2. Solving Problems on Your Ownx
0.5 in
0.75 in
L5x3x 1 2
Solving Problems on Your Own
For the 5 x 3 x -in. angle cross section
shown, use Mohr’s circle to determine
(a) the moments of inertia and the
product of inertia with respect to new
centroidal axes obtained by rotating
the x and y axes 30o clockwise, (b) the
orientation of the principal axes through
the centroid and the corresponding
values of the moments of inertia. 1 2
1. Draw Mohr’s circle. Mohr’s circle is completely defined by the
quantities R and IAVE which represent, respectively, the radius of
the circle and the distance from the origin O to the center C of the
circle. These quantities can be obtained if the moments and
product of inertia are known for a given orientation of the axes.

3. Solving Problems on Your Ownx
0.5 in
0.75 in
L5x3x 1 2
Solving Problems on Your Own
For the 5 x 3 x -in. angle cross section
shown, use Mohr’s circle to determine
(a) the moments of inertia and the
product of inertia with respect to new
centroidal axes obtained by rotating
the x and y axes 30o clockwise, (b) the
orientation of the principal axes through
the centroid and the corresponding
values of the moments of inertia. 1 2
2. Use the Mohr’s circle to determine moments of inertia of
rotated axes. As the coordinate axes x-y are
rotated through an angle q, the associated
rotation of the diameter of Mohr’s circle is
equal to 2q in the same sense
(clockwise or counterclockwise). x y 2q
Ix , Iy
Ixy x’ y’

4. Solving Problems on Your Ownx
0.5 in
0.75 in
L5x3x 1 2
Solving Problems on Your Own
For the 5 x 3 x -in. angle cross section
shown, use Mohr’s circle to determine
(a) the moments of inertia and the
product of inertia with respect to new
centroidal axes obtained by rotating
the x and y axes 30o clockwise, (b) the
orientation of the principal axes through
the centroid and the corresponding
values of the moments of inertia. 1 2
2. Use the Mohr’s circle to determine the orientation of principal
axes, and the principal moments of inertia. Points A and B where
the circle intersects the horizontal axis
represent the principal moments of inertia.
The orientation of the principal axes is
determined by the angle 2qm.
Ixy
2qm x B A
Ix, Iy y

5. Ix = 9.45 in4, Iy = 2.58 in4 Ixy = (Ixy)1 + (Ixy)2
Problem Solution
3 in
5 in
0.5 in
1.75 in x
0.5 in
0.75 in
L5x3x 1 2
From Fig. 9.13A:
Ix = 9.45 in4, Iy = 2.58 in4
Product of inertia Ixy:
Ixy = (Ixy)1 + (Ixy)2
For each rectangle Ixy = Ix’y’ + x y A
and Ix’y’ = 0 (symmetry)
Thus: Ixy = S x y A y
0.5 in 2
A, in2 x, in y in x y A, in4
Ixy = S x y A = -2.81
1 in x
1.5 in 1
0.75 in

6. OC = Iave = (Ix + Iy) OC = (9.45 + 2.58) = 6.02 in4
Draw Mohr’s circle.
Problem Solution
The Mohr’s circle is defined by the
diameter XY where X(9.45, -2.81)
and Y(2.58, 2.81).
The coordinate of the center C and
the radius R are calculated by:
OC = Iave = (Ix + Iy)
OC = ( ) = 6.02 in4
R = [ (Ix - Iy)]2 + Ixy
R = [ ( )]2 + (-2.82)2
R = 4.44 in4 1 2 1 2 1 2
X (9.45, -2.81)
Ix, Iy (in4)
Ixy (in4)
Y (2.58, +2.81) C O
2qm F 1 2
tan 2qm = = FX CF
2.81
qm = 19.6o

7. Ix’ = OE = OC _ CE = 6.02 _ 4.44 cos 80.7o Ix’ = 5.30 in4y
Problem Solution y’
Use the Mohr’s circle to determine
moments of inertia of rotated axes.
(a) The coordinates of X’ and Y’ give the
moments and product of inertia with
respect to the x’y’ axes. x
30o
Ix’ = OE = OC _ CE = 6.02 _ 4.44 cos 80.7o
Ix’ = 5.30 in4
Iy’ = OD = OC + CD = cos 80.7o
Iy’ = 6.73 in4 x’
Ixy (in4)
6.02 in4 Y’ Y
R = 4.44 in4 E D O C
Ix, Iy (in4)
Ix’y’ = EX’ = _ 4.44 sin 80.7o
80.7o X
39.3o
Ix’y’ = _ 4.38 in4 X’
2q = 60o

8. y
Problem Solution b
Use the Mohr’s circle to determine
the orientation of principal axes, and
the principal moments of inertia. a
qm = 19.6o
(b) The principal axes are obtained by
rotating the xy axes through angle qm. x
tan 2qm = = FX CF
2.81
qm = 19.6o
X (9.45, -2.81)
Ix, Iy (in4)
Ixy (in4)
Y (2.58, +2.81) C O
2qm F
The corresponding moments of inertia
are Imax and Imin : _ _
Imax, min = OC + R =
Imin
Imax
Imax = in4
Imin = in4
a axis corresponds to Imax
b axis corresponds to Imin
R = 4.44 in4