• The basic techniques used for centroids are similar (not identical)

• For centroids we assume the force of gravity was uniform over the face. When the forces are not uniform, but are a function of the distance from the centroid, then we need to use moments of inertia.

I. Last lecture we discussed the forces in a beam, today we will relate those forces to the stresses in the beam. These stresses can be used to predict failure of the beam or to size the beam to prevent failure.

• The location of the neutral axis:

• Consider the basic equation below.

• These can be evaluated using integration.

Problem 13.1 Determine the value of Ix.

• We can use sums to find moments of inertias for given objects.

• Consider the fully symmetrical shape below.

• The basic application of the parallel axis theorem is shown below.

• The polar moment of inertia may be simple to calculate with rotational parts, but it can also be found using a simple relationship (note: the polar ‘J’ and the ‘o’ for the origin)

• Evaluate the polar moment of inertia ‘J’ for the shape below.

Problem 13.2 Find the Polar Moment of Inertia about the origin for the beam section shown below.

• When we considered the centroids/center of mass we assumed that the force acting on each chunk was the same (uniform gravity).

• If we have a force that varies over a surface linearly we use the second moment of inertia.

• Consider a simple case of a beam (not quite accurate). The material in the beam acts as small springs. As the beam is bent the material in the center is almost neutral, but towards the sides it is compressed or stretched.

• If we consider the beam above as a continuous force, instead of distributed springs it changes the solution (by analogy you might recognize that with centroids we could lump forces, but with second moments we cannot).

• Another example of a linearly varying force is hydrostatic pressure based on water depth. Basically we add the mass of the water directly above to find the pressure on an area.

• The general forms of moments resulting from linearly varying forces are,

• We generally calculate second moments of inertia using,

• as an example we can consider a triangle. The calculations will be to find the second moments of inertia about the x and y axis: not the centroid.

• After we have found the second moment of inertia about an axis, we can find it about another parallel axis using the parallel axis theorem.

• Consider the application of the parallel axis theorem to the triangle seen before, To find the moment of inertia about the y centroid, when all we have is the y moment of inertia about the x axis.

Problem 13.3 Find the Second Moment of Inertia about the centroid for the beam section shown below.

• NOTE: when using the parallel axis theorem, the centroid should always be used as a reference.

• Like centroids, we can calculate moments of inertia for simple areas using weighted sums (a slightly different technique to finding centroids). The basic steps are,

1. For each simple shape find the moment of inertia about some global axis. This may require the use of centroids and areas to move the axis.

2. Add moments of inertia (or subtract for cut out areas.

• As an example, let’s consider a problem (9.26 pg. 64 Beer and Johnston)

Problem 13.4 Find the Second Moment of Inertia about the x-y axes using the composite bodies method.

• When we want to find the maximum moment of inertia of an area we can start by finding the product of inertia.

• This value can be either positive or negative.

• The value will be zero if either x or y is an axis of symmetry. This will allow us to find the product of inertia if we know the area and location of the centroid.

• If we want to find the maximum and minimum moment of inertia we can use the derived relationship below. We can also find the axis of these maxima and minima,

• We can also shift the moment of inertia values to other arbitrary axes using the following relationships,

Problem 13.5 Find the Second Moment of Inertia for the beam section shown below, about a line that lies 45° between the positive x and y axes.

• If we plot out the values as we rotate the axis, we generate a plot called Mohr’s force circle.

13.1 Soustas-Little, R.W. and Inman, D.J., Engineering Mechanics Statics, Prentice-Hall, 1997.