Mechanics of materials represents one of the main topics the PE exam will test you on. You are almost guaranteed to come across at least a few basic concepts in the morning portion regardless of what afternoon session you choose. Strength of materials and statics is a potentially huge area of study but the main focus areas – and the best place to start studying for the exam – are as follows.
Bending stresses for a beam under simple loading are defined as:
Where σ is the bending stress, M is the moment about the neutral axis at the point of interest, c is the distance from the neutral axis to the point of stress, and I is the second moment of area about the neutral axis. Let’s tackle each of these concepts individually:
Neutral Axis: In this context of beam loading, the neutral axis is considered to be the centroidal axis which corresponds to the point at which there is no bending stress. Note this is only true if there are no axial loads on the beam. For this simple loading scenario, the stress profile can be shown as:
Second Moment of Area: The second moment of area is a geometrical property with the units of length to the 4th power [L4]. It can also be known as the moment of inertia or area moment of inertia. In general, moment of inertia is a second integral over a given area but for the purposes of study for the PE exam we will only consider moment of inertia calculations for composite areas as this is the level of complexity that will most likely be given on the exam.
Torsional stress is the other type of “basic” stress configuration that can likely be on the test. Torsional stresses can either be considered alone or along with bending stresses in which case the stress is known as “combined stress”. Torsional stress for a circular solid shaft is defined as:
Where T is the torque applied to the member under load, R is the radius of the shaft, and J is the polar moment of inertia of the shaft cross section. Polar moment on inertia for a circle is defined as:
For a circular shaft, the stress distribution looks like figure MM1-2:
It’s important to note here that in this configuration, maximum stress is located on the surface of the shaft. This is important because many strength of materials problems are concerned with the maximum stress occurring in an element under load.
Combined loading and principal and maximum shear stresses are topics that go hand in hand. Principal stresses are just maximum normal stresses which are a product of two or more stresses acting at a given point. If there is only one load on a member and therefore only one stress acting at a point of interest that stress will be the principal stress. Let’s take a look at how combined loading and stresses are used to determine both the maximum normal (principal stress) and shear stress.
The key to determining combined stresses is in understanding the stress profile for a given geometry. See figures MM1-1 and MM1-2 for examples of simple stress profiles. Understanding what direction a stress acts on a member at a given point will allow you to set up the correct directions and magnitudes which are needed to determine the principal stress. Consider a cylindrical bar with the following loading:
There is a torque, T, which act on the bar as well as a downward force, M, which creates a moment in the bar as well as a normal stress on point A. Per Figures MM1-1 and MM1-2, these stresses act on the element as:
Completing the stress distribution to achieve equilibrium gives:
Understanding how these stresses act on a given point allows for the proper calculation of the maximum or principal stress. Principal stress is determined using Eq MM1-4:
The maximum shear stress is the square root term of Eq. MM1-4:
Using these formulas along with the appropriate failure theory criteria yields the Factor of Safety which is another very popular exam question and will be covered later in another section.