🟦 16.1 Introduction to internal forces (and moments)
We can subcategorize forces (and moments) into three groups:
applied forces and moments (loads)
reacting forces and moments (at supports and connections)
internal forces and moments (the effects of the fibers of the solid material, which stay connected under load and therefore must transfer force from each cross-sectional plane to its neighboring plane)
When do we include applied forces (and moments) on a FBD?
It's when they are applied to the portion of the member that is cut or disconnected from the context.
❏ Applied forces (and moments)
When do we include reacting forces (and moments) on a FBD?
It's when we disconnect a support from the context. For example, we remove a pin and replace it with x-direction and y-direction forces. These are the actions of the support on the body.
❏ Reacting forces (and moments)
When do we include internal forces (and moments) on a FBD?
It's whenever we pass a cutting plane through the solid material. For instance, beam AB has been cut at plane c-c.
If you don't cut through the structure, you do not show any internal forces (and moments) on the FBD.
❏ Internal forces (and moments)
Through this unit, we will use the generic term force to refer to both forces and moments. This may seem a little strange at first. Forces and moments aren't the same thing; they don't even share the same units.
We can portray a couple moment as a force couple (two forces, opposite in direction, equal in magnitude, and parallel).
Therefore, in the subsequent sections, you'll no longer see the parenthetical note "and moments" when internal forces are mentioned.
❏ Force couples and couple moments
🟦 16.2 Prior lessons that incorporated internal forces
This isn't our first foray into internal forces.
In the truss lesson, we learned to use the Method of Sections to slice through a truss and cut it into two pieces.
Then, we were able to use the equations of equilibrium to solve for unknown tension or compression forces.
The internal forces are shown in orange. They are the effects of everything cut away.
❏ Internal forces in a truss
In the cable lesson, we learned that we could cut a cable at any point and expose the internal tension in that particular fiber. Some of the key points were:
For straight-line segments, tension is constant in each segment of the cable.
For a cable supporting a uniform live load (uniformly distributed load), tension varies through the cable.
The cable is cut perpendicular to the tangent to the curve, to expose the normal force (which aligns to the tangent to the curve).
Over distance x1, the amount of load on the FBD is wx1. Over length L/2, the amount of load on the FBD is wL/2.
The internal forces are again shown in orange. They are the effects of everything cut away.
❏ Internal forces in a cable
🟦 16.3 The significance of internal forces (and moments)
In this flipbook, we will slice through a cantilever beam with two cuts. This allows us to extract and investigate a dx slice of the solid material.
We use the equations of equilibrium to solve for unknown forces (and moments) on the dx slice:
N = normal (perpendicular) force
V = in-plane (parallel) force
M = moment
We can imagine that the dx slice of solid material is replaced with springs. This helps us visualize the deformations that occur due to each internal force.:
normal force results in elongation (under tension) or shortening (under compression)
shear force results in angular warping
bending moment results in curvature
❏ Flipbook: effects of internal force
🟦 16.4 Force at a point and stress on a plane
After you finish Statics, you'll probably go on to study stress in a subsequent course. A simple definition for stress is force normalized over (divided by) area.
This flipbook is a preview of what you'll learn about stress after Statics.
The takeaway for this course is that internal forces are applied at a point while stresses lie in a plane. Force is a vector, while stresses are a vector field.
In Statics, we will cut through a plane to expose the force. That force results from integrating the stress distribution over the cross-sectional surface area.
❏ Flipbook: Force vs. Stress
For notation, lowercase letters indicate planes while uppercase letters continue to indicate points (or nodes, or particles).
Each point should be interpreted as the centroid of the cross-sectional area, if no other indication is provided.
🟦 16.5 2D terminology and sign conventions
❏ Normal force (N)
When the internal force is perpendicular (normal) to the cut plane, we call it a normal force and use the symbol N.
arrows that point away from the body create tension, which we define as positive
arrows that point toward the body create compression, which we define as negative
❏ Signs for normal force (N)
Say: "Normal? Positive = tension"
❏ Shear force (V)
When the internal force is parallel (in-plane) to the cut, we call it a shear force and use the symbol V.
This sign convention is notoriously tricky to apply:
Method 1: when the shear force on the left (negative x face) goes up (in the positive y direction), V is positive; when the shear force on the right (positive x face) goes down (in the negative y direction), V is positive
Method 2: when a shear force tends to rotate the body clockwise (yes, clockwise, that is not a typo), V is positive
Shear vectors are often drawn with half arrowheads in 2D drawings. This signifies that the force is located in the adjacent plane. We offset it slightly so that it's easy to see.
❏ Signs for shear force (V)
Say: "Shear? Positive = Up, Down"
❏ Moment or Bending Moment (M)
For a 2D (planar) problem in the xy plane, the internal moment will always be about the z-axis. We use the symbol M.
You can think about the sign convention in two ways:
Method 1: when the internal moment makes the beam "smile" at you (concave up), it's positive
Method 2: a clockwise moment on the negative x face is positive; a counterclockwise moment on the positive x face is positive
❏ Signs for bending moment (M)
Say: "Moment? Positive = Smile!"
❏ The sign convention chant:
First, get a beat going. Then add in the lyrics:
Normal? Positive TENSION
Shear? Positive ⬆️⬇️
Moment? Positive SMILE 😊
ONE. MORE. TIME.
(repeat until you have it memorized)
Notes:
- These sign conventions can vary significantly by engineering discipline, by subfield of engineering, by software package, and by country / geographic region.
- Always make sure that you understand whatever sign convention is being used when you take a new class, work with a new person, use a new website, or open a new piece of software!
- The sign conventions explained here are fairly consistent within the United States in courses in Statics and Mechanics of Materials.
- The sign conventions used here are consistent with the Fundamentals of Engineering (FE) standardized exam that is required for future licensed (professional) engineers that wish to practice in the United States.
Now that you have the sign conventions memorized, make sure you can apply them correctly.
Check your understanding by working through this flipbook.
❏ Flipbook: how to apply the sign conventions
🟦 16.6 3D terminology and sign conventions
In this course, we only focus on solving internal forces in a 2D plane. Yet, it is also useful to think about internal forces in 3D space.
There are six internal forces and moments in 3D space:
N = normal force
V1 = shear in one direction
V2 = shear in the other direction
M1 = moment about one axis
M2 = moment about the other axis
T = torsion (twisting) about the longitudinal axis
The 3D sign conventions also vary quite a bit in different geographic areas, in different software programs, in different textbooks, etc. They are not a focus in this class.
❏ 3D internal forces
🟦 16.7 A friendly warning about sign errors
Sign errors on these types of problems are too common.
We need a deep and sophisticated understanding of how to use three different sets of sign conventions.
❏ Three sets of sign conventions to understand
Keep these three sets of sign conventions in mind as you work through the next section. The signs are the hardest part.
🟦 16.8 Solving for N, V, and M at a discrete cut plane
An example problem for solving for discrete values of N, V, and M is shown in the flipbook. The basic steps are:
Construct a global FBD.
Solve unknown reactions with the E.o.E.
- assume directions of reactions
- use signs for the E.o.E.
- interpret your results using signs for unknowns
Cut through the member.
- cut it into two pieces
- pick whichever piece looks easiest to solve
Draw the unknown internal forces.
- always draw these in the positive direction, using the signs for internal forces
Solve the unknown internal forces.
- use the signs for the E.o.E.
Interpret your results.
- use the signs for internal forces
❏ Flipbook: example problem
🟦 16.9 Cutting FBDs at "knife edges"
You can't cut a FBD directly through a force vector or through a moment vector. However, you can cut a differential distance dx to the left, and a differential distance dx to the right.
This idea is sometimes called a "knife edge." You can think of pin supports as "knife edges" (you'll calculate different internal forces when you cut to the left and to the right). This concept applies whenever there is a force vector or a moment vector on a body. You can cut just to the left or to the right, and solve for different values.
This example shows how you can cut on both sides of a single plane. It solves for the internal shear and moment just to the right of the pin-roller ... and also just to the left of the pin-roller.
🟦 16.10 Writing equations for V(x) and M(x) over a domain
Using cut planes and solving for N, V, and M at discrete planes is somewhat useful. But oftentimes we want to know more information, such as:
How does the internal shear force, V, vary with position x?
How does the bending moment, M, vary with position x?
What is the worst-case (maximum) shear force in this beam?
What is the worst-case (maximum) bending moment in this beam?
To answer these questions, we can write equations for V and M in terms of position variable x. We also need to specify the domain of x.
❏ Range and domain
Here is an example of how to write functions for V(x) and M(x) for a simply-supported beam that supports a uniformly distributed line load. There is only one domain (0<x<L).
❏ Flipbook: Example 1 of 2
Here is another example of the same concept. This time, there are two domains in play.
❏ Flipbook: Example 2 of 2
➜ Practice Problems
Problem 1.
Solve for the internal forces and moments at planes c-c and d-d.
Note: plane c-c is just to the left of the reaction at A (x=2.9999999'). Plane d-d is just to the right of the reaction at A.
Problem 2.
Solve for the internal forces and moments at plane c-c.
answer:
Problem 3.
Solve for the internal shear force and bending moment for the planes shown in the image. (Note: all of the normal forces are zero.)
Here are scrambled (unitless) answers, so that you can check your work:-6, -4, -2, 0, 0, 0, 2, 4, 6, 10, 10, 16, 16, 18Problem 4.
Solve for the internal shear force and bending moment for the planes shown in the image. (Note: all of the normal forces are zero.)
Here are scrambled (unitless) answers, so that you can check your work:0, 0, -4, -8, -8, -12, -32, -72Problem 5. Write the V(x) and M(x) equations for this cantilever beam. Please use the origin designated.
Answer:
Problem 6. Write the V(x) and M(x) equations for this simply-supported beam. Please use the origin designated.
Note:
The V(x) equation will be valid for the domain of 0 ≤ x ≤ L.
There will be two M(x) equations, one for each half of the beam.
Answer:
Fall 2024: This is the end of the problem set.
Author's note: add in some contextual examples, such as some pliers or nail clippers, and a gate subjected to fluid pressure.