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

ExAmpLE A.7
Determine the principal moments of inertia for the unequal-leg angle shape
considered in Example A.5. Indicate the orientation of the principal axes.
1.654 in.
y
Plan the Solution
Using the moments of inertia and the product of inertia determined in Example
A.5, Equation (A.14) will give the magnitudes of I p1 and I p2 , and
Equation (A.13) will define the orientation of the principal axes.
SolutioN
From Example A.5, the moments of inertia and the product of inertia for
the unequal-leg angle shape are
I
I
I
x
y
xy
= 80.8 in.
= 38.8 in.
4
4
=−32.3 in.
4
8 in.
1 in.
1 in. 5 in.
6 in.
2.654 in.
x
The principal moments of inertia can be calculated from Equation (A.14):
y
I
p1, p2
Ix + Iy ⎛ Ix − Iy
= ±
2 ⎝
⎜
2
80.8 + 38.8
=
2
= 59.8 ± 38.5
±
= 98.3 in. ,21.3 in.
4 4
2
⎞
I
⎠
⎟ +
2
xy
⎛ 80.8 − 38.8⎞
⎝
⎜
2 ⎠
⎟
2
+ ( −32.3)
2
Ans.
8 in.
1.654 in.
p2
axis
28.5°
p1 axis
x
The orientation of the principal axes is found from Equation (A.13):
1 in.
2.654 in.
2Ixy
2( −32.3)
tan2θ p =− =−
= 1.538095
I − I 80.8 − 38.8
x
y
∴ 2θ p = 56.97°
1 in. 5 in.
6 in.
Therefore, θ p = 28.5°. Since the denominator of this expression (i.e., I x − I y ) is positive,
the value obtained for θ p gives the orientation of the p1 axis relative to the x axis. The
positive value indicates that the p1 axis is rotated 28.5° counterclockwise from the x
axis.
The orientation of the principal axes is shown in the sketch.
A.5 mohr’s circle for principal moments of Inertia
The use of Mohr’s circle for determining principal stresses was discussed in Section 12.9.
A comparison of Equations (12.5) and (12.6) with Equations (A.12) and (A.16) suggests
that a similar procedure can be used to obtain the principal moments of inertia for an area.
Figure A.8 illustrates the use of Mohr’s circle for moments of inertia. Assume that
I x is greater than I y and that I xy is positive. Moments of inertia are plotted along the
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