Second Moment, or Moment of Inertia, of an Area

Consider a beam of uniform cross section which is subjected to two equal and opposite couples applied at each end of the beam. Such a beam is said to be in pure bending, and it is shown in mechanics of materials that the internal forces in any section of the beam are distributed forces whose magnitudes:

ΔF = ky ΔA

vary linearly with the distance y between the element of area ΔA and an axis passing through the centroid of the section. This axis, represented by the x axis as shown:



is known as the neutral axis of the section. The forces on one side of the neutral axis are forces of compression, while those on the other side are forces of tension; on the neutral axis itself the forces are zero.

The magnitude of the resultant R of the elemental forces ΔF which act over the entire section is:

R = ∫ky dA = k ∫ y dA

The last integral obtained is recognized as the first moment Qx of the section about the x axis; it is equal to

 y
A and is thus equal to zero, since the centroid of the section is located on the x axis. The system of forces ΔF thus reduces to a couple. The magnitude M of this couple (bending moment) must be equal to the sum of the moments ΔMx = y ΔF = ky2 ΔA of the elemental forces. Integrating over the entire section:

M = ∫ky 2 dA = k ∫ y 2 dA

The last integral is known as the second moment, or moment of inertia, of the beam section with respect to the x axis and is denoted by Ix. The term second moment is more proper than the term moment of inertia, since, logically, the latter should be used only to denote integrals of mass (include reference). In engineering practice, however, moment of inertia is used in commection with areas as well as masses. It is obtained by multiplying each element of area dA by the square of its distance from the x axis and integrating over the beam section. Since each product y2 dA is positive, regardless of the sign of y, or zero (if y is zero), the integral Ix will always be positive.

Another example of a second moment, or moment of inertia, of an area is provided by the following problem from hydrostatics. A vertical circular gate used to close the outlet of a large reservoir is submerged under water as shown:



What is the resultant of the forces exerted by the water on the gate, and what is the moment of the resultant about the line of intersection of the plane of the gate and the water surface (x axis)?

If the gate were rectangular, the resultant of the forces of pressure could be determined from the pressure curve (put reference link in here). Since the gate is circular, however, a more general method must be used. Denoting by y the depth of an element of area ΔA and by γ the specific weight of water, the pressure at the element is p = γ y, and the magnitude of the elemental force exerted on ΔA is ΔF = p ΔA = γ y ΔA. The magnitude of the resultant of the elemental forces is thus:

R = ∫γ y dA = γ ∫ y dA

and can be obtained by computing the first moment:

Qx = ∫ y dA

of the area of the gate with respect to the x axis. The moment Mx of the resultant must be equal to the sum of the moments:

ΔMx = y ΔF = γ y2 ΔA of the elemental forces. Integrating over the area of the gate:

M = ∫γ y 2 dA = γ ∫ y 2 dA

Here again, the integral obtained represents the second moment, or moment of inertia, Ix of the area with respect to the x axis.