Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

mecmovies
ExAmpLE
A.6 The theory and procedures for determining principal
moments of inertia using Mohr’s circle are presented in an
interactive animation.
ExAmpLE A.8
Solve Example A.7 by means of Mohr’s circle.
y
Plan the Solution
The moments of inertia and the product of inertia determined in
Example A.5 will be used to construct Mohr’s circle for moments of inertia.
1.654 in.
SolutioN
From Example A.5, the moments of inertia and the product of inertia for the
unequal-leg angle shape are
8 in.
I
x
= 80.8 in.
4
x
I
I
y
xy
= 38.8 in.
4
=−32.3 in.
4
1 in.
2.654 in.
Moments of inertia are plotted along the horizontal axis, and products of
inertia are plotted on the vertical axis. Begin by plotting the point (I x , I xy ) and
labeling it x. Notice that since I xy has a negative value, point x plots below
the horizontal axis.
Next, plot the point (I y , −I xy ) and label this point y. Since I xy has a
negative value, point y plots above the horizontal axis.
Draw the circle diameter that connects points x and y. The center of the
circle is located where this diameter crosses the horizontal axis. Label the
circle center as C. Using the circle center C, draw the circle that passes
through points x and y. This is Mohr’s circle for moments of inertia. Points
on Mohr’s circle represent possible combinations of moment of inertia and
product of inertia.
The center of the circle is located midway between points x and y:
1 in. 5 in.
6 in.
80.8 + 38.8
C =
2
= 59.8
807