Area Moments of Inertia

  • Define moment of inertia.

  • Understand importance of moment of inertia in rotational motion and bending.

  • Understand terms such as polar moment of inertia, radius of gyration.

Area Moment of Inertia is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis.
The area moment of inertia indicates the resistance of a beam to bending and deflection, around an axis that lies in the cross-sectional plane.
Beams with higher area moments of inertia are used in construction because they offer more resistance to rotation.
Polar Moment of Inertia of an area about a point is equal to the sum of moments of inertia of the area about any two perpendicular axes in the area and passing through the same point.
The polar moment of inertia is a measure of an object’s ability to resist torsion as a function of its shape.

\[J = I_x+ I_y\]

Radius of Gyration is the distance from a reference axis at which all of the area can be considered to be concentrated to produce the moment of inertia.

\[r_{x}=\sqrt{\dfrac{I_{x}}{A}}\]

where,
\(r_{x}\) = radius of gyration,
\(I_{x}\) = Moment of Inertia about x axis
A = Area of cross section

Solved Examples

Solved Example:

26-3-01

The SI unit for polar moment of inertia is:

Solution:
$J = \int \int r^2 dA$
The SI unit for polar moment of inertia, like the area moment of inertia, is meters to the fourth power (m$^4$), and inches to the fourth power (in$^4$) in imperial units.

Correct Answer: C

Solved Example:

26-3-02

Determine the radius of gyration k$_y$ of the area between parabola and x-axis.

26.3-02



Solution:

Consider an elemental strip parallel to y-axis.

Its area will be: $dA = y dx = 0.1(1600-x^2) dx$

Its moment of inertia will be:

\[dI =dA \times r^2 =0.1(1600-x^2)dx \times x^2\] The limits are 0 to 40 (half parabola) and then multiply by 2. \begin{align*} I_y &= \int_{0}^{40} 0.1(1600-x^2)dx \times x^2 \times 2\\ &= \left[\dfrac{160x^3}{3}-0.1\dfrac{x^5}{5} \right]_0^{40} \times 2\\ &=2730666\ mm^4 \end{align*} Again, the limits are 0 to 40 (half parabola) and then multiply by 2. \begin{align*} A &= \int_{0}^{40} y dx \times 2\\ &= \int_{0}^{40} 0.1(1600-x^2) dx \times 2\\ &= \left[160x-0.1\dfrac{x^3}{3} \right]_0^{40} \times 2\\ &= 8533~mm^2 \end{align*}

substitute the values of I,A in k$_y$ we get

\[k_y =\sqrt{\dfrac{I_y}{A}} =\sqrt{\dfrac{2730666}{8533}} =17.89\ mm\]

Correct Answer: B

Solved Example:

26-3-03

The second moment of area is an important value which is used to ____________. It can also be called moment of inertia.

Solution:

In structural engineering, the second moment of area of a beam is an important property used in the calculation of the beam's deflection and the calculation of stress caused by a moment applied to the beam.

Correct Answer: D

Solved Example:

26-3-04

Second moment of area is the product of:

Solution:
For a section component having an axis of symmetry that is parallel to either of the section reference axes, the product second moment of area is the product of the coordinates of its centroid multiplied by its area.

Correct Answer: A

Solved Example:

26-3-05

In a square plate, a circular hole is drilled at the center. Which of the following will most likely happen? (Prof. Prashant More's Strength of Materials Exam- Dec 2010)

Solution:

In a square plate, when a circular hole is drilled at the center, the mass is removed symmetrically about the existing center of gravity, so its CG will not change.
However, the MI will decrease to reflect the reduction in mass because of removal of material.

Correct Answer: A

Solved Example:

26-3-06

The moment of inertia of the area under the curve y=kx$^2$ shown in the following figure about x-axis is:

26-3-06



Solution:

26-3-06

Consider an elemental strip parallel to x-axis. Its area will be: \[dA = (a-x) dy = (a-x) (2kxdx)\] Its moment of inertia will be: \[dI =dA \times y^2 =(a-x) (2kxdx) \times (kx^2)^2 = 2k^3 x^5 (a-x)dx\] The limits are x=0 to x=a. \begin{align*} I_x &= 2k^3 \int_{0}^{a} \left[ax^5 - x^6 \right]dx\\ &= 2k^3 \left[\dfrac{ax^6}{6} - \dfrac{x^7}{7} \right]_0 ^a\\ &= 2k^3 \left[\dfrac{a^7}{6} - \dfrac{a^7}{7}\right] = \dfrac{k^3 a^7}{21} \end{align*} Using, $b = ka^2$ as it satisfies the condition y = kx$^2$ since the point (a,b) is on the curve. \[I_x = \dfrac{k^3a^6.a}{21} = \dfrac{b^3.a}{21} = \dfrac{ab^3}{21}\]

Correct Answer: D