Statics

Lesson 5

Loads and Static Equivalency

🟦  5.1 An introduction to loads

Engineers use the word loads to talk about the demands placed on an engineered system.

In Statics, loads are the inputs into the system. We have a designed structure and we need it to support a certain amount of weight. Loads are the demand on the structure. Reactions are the forces (and moments) that the structure develops to keep the loads in static equilibrium. We just learned about reactions in the previous lesson, so that should be fresh on your mind.

The main types of loads are:

Line loads and area loads are vector fields. Vector calculus can be complex, but in Statics you only need to know a few basic things (all of which will be part of this text).

In a nutshell, a vector field is a function that describes a set of vectors with both location and magnitude/direction.

This wind map is a useful an introduction to vector fields. At each geographic point, the wind speed can be modeled as a vector with magnitude and direction. In Statics, our vector fields will be much simpler than the vector field visualized in the wind map.

❏ Types of loads


Line loads and area loads are both examples of distributed loads. The force has been distributed along a length or over a surface.

🟦  5.2 An introduction to static equivalency

Static equivalency is the idea that if two systems produce the same tendency to translate and the same tendency to rotate, then they are considered statically equivalent.

Consider identical boxes A and B. Box A has a load of 4N applied at the center, while Box B is subjected to four loads (four 1N forces at the corners).

These two scenarios are statically equivalent because:

We could say that these scenarios are not fully equivalent because the push forces occur at different locations.

❏ Statically equivalent:

Now, consider these two scenarios (Box C and D).

These scenarios are not statically equivalent. Box C only translates. Box D translates and rotates (about y) because the force is applied off-center.

Important: even though both boxes have the same tendency to translate (a net force of 4N in the positive x-direction), but they are not statically equivalent because they do not have the same net moment on the body.

These scenarios are neither statically equivalent or fully equivalent.

❏ Not statically equivalent:

🟦  5.3  Vector components and resultants are statically equivalent

One type of static equivalency that we have already discussed is the idea of force components and force resultants. 

In the flipbook, F1 is statically equivalent to the combination of F2 and F3. It is also statically equivalent to the combination of F4 and F5.

Resultant forces are always statically equivalent to their component forces and vice versa.

This flipbook emphasizes the fact that you can create components for a force in any convenient coordinate system. You're not limited to x being rightwards and y being upwards. Sometimes it's helpful to draw the bounding box and components in an entirely different coordinate system.

❏ Flipbook: Force components, resultant

🟦  5.4 Force couples and couple moments are statically equivalent

Another type of static equivalency previously discussed is the idea of force couples and couple moments. These produce a tendency to rotate (a net moment) without a tendency to translate (a net force).

For example, the following five systems (I, II, III, IV, and V) are all statically equivalent. They all produce a net moment of 10kN • 1m = 10 kN•m↺. None has a tendency to translate (or a net force).

Were you surprised that systems IV and V were statically equivalent? Couple moments are free vectors: you can move them anywhere you want to create a statically equivalent system. Analytically (mathematically), their location doesn't matter. That said, we do like to draw them where they are applied in real life, so that our FBD communicates the problem context effectively.

🟦  5.5  The equations of static equivalency

In Section 5.2, we considered different scenarios (Box A vs. Box B and Box C vs. Box D).

Let's use the word system for each scenario, and let's use Roman Numerals (I, II, etc.) to identify each system.

Here are the six equations of static equivalency. If these equations are satisfied, then we can state that Systems I and II are statically equivalent.

❏ The equations of static equivalency (3D format)

If you're working on a planar (2D) problem, only three of the six equations of static equivalency matter. 

If these three equations are satisfied, then Systems I and II are statically equivalent. 

In other words, they have the same tendency to translate and the same tendency to rotate.

❏ The equations of static equivalency (2D or planar format)

Here is how you could explain the equations of static equivalency to a friend: 

"if you have two systems of applied forces and/or moments, the systems are statically equivalent when they produce the identical tendency to translate and the identical tendency to rotate."

Here is how you could explain the equations of static equilibrium to a friend: 

"if the connections can create reactions that ensure the body neither translates nor rotates, then the net force and the net moment are both equal to zero, and we call that state static equilibrium."

🟦  5.6  The Principle of Transmissibility

The Principle of Transmissibility states that if you move a vector along its line of action, you will create a statically equivalent system.

Recall that force vectors have four attributes or parameters:

When we use the Principle of Transmissibility, the only parameter we are changing is the point of application.

❏ Four attributes of a force vector

Example 1.

Force vectors A and B are statically equivalent because they:

Vector A depicts a push and Vector B depicts a pull. That little detail means that they aren't fully equivalent -- only statically equivalent.

Vector C is not statically equivalent to A and B. It shares the same line of action, but has the opposite direction.

Vector D is not statically equivalent to A. It does not share the same line of action.

❏ Illustration: Example 1

Example 2.

Consider a box of weight W. 

In scenario A, the weight of the box is counteracted with a tensile force above. 

In scenario B, the weight is held in equilibrium with a compressive force below.

Forces F1 and F2 are statically equivalent. They have the same magnitude, the same direction, and the same line of action.

F1 results from a cable above (a pull) and F2 results from a cable below (a push). They represent different actions on the box, yet they are statically equivalent systems. They are not fully equivalent because a push is not the same thing as a pull.

❏ Flipbook: Example 2

Example 3.

Let's say that you are asked to compute the moment of the 5 kip force about Point A.

It's kind of like playing chess. Instead of moving pieces around the board, you're moving vectors around and manipulating them.

For your first move, you create a statically equivalent system by sliding the vector along its line of action from C to B.

For your second move, you create another statically equivalent system by swapping out the 5 kip force into its x- and y- component forces.

After the vector moves to B, the vertical 3 kip component causes a clockwise moment about A (30 kip-feet ↻ due to the 10' moment arm). 

The horizontal 4 kip component is coincident with line AB and therefore does not contribute to the moment about A.

❏ Flipbook: Example 3


🟦  5.7 Equivalent systems with couple moments

❏ Introductory mini-experiment

Things get interesting once we learn to create equivalent systems with couple moments.

Let's begin with a mini-experiment. Place your cell phone (or a similar small object) on your desk. Apply the loads from systems I, II, III, and IV. Which systems seem to be statically equivalent?

You can use the equations of static equivalency as the basis of your reasoning process.

Systems I and II have the same net force, but since the forces lie on different lines of action, these systems are not statically equivalent.

System III has a tendency to rotate counterclockwise, but no tendency to translate. It's not statically equivalent to any of the other systems.

Systems I and IV are the only ones that are statically equivalent. They both have:

Inspect the different translations and rotations in the flipbook and make sure they all make sense to you. (They are based on the assumption that the load/s overcome the friction resistance between the phone and table.)

❏ Mini-flipbook: four cell phones on a desk

❏ Example 1

The key takeaway is that you can move a force off of its line of action and create an equivalent system, provided that you introduce a couple moment to counteract the rotation caused by the new moment arm.

Compute the magnitude and sign of the moment by setting the net moment in System I equal to the net moment in System II. 

That is, the tendency to rotate about an axis in System I must equal the tendency to rotate about the same axis in System II.

You can set up the moment equivalency calculation about any axis, but it's most intuitive when you sum moments about the destination of the force you're moving (in this example problem, you're moving the force from A to B, so the moment summation was set up about B).

An important reminder: the equations of static equivalency are not the same thing as the equations of static equilibrium.

❏ Example 1: moving a force to a new line of action

❏ Example 2

You can also combine a force and a moment into a single force, by moving the force off of its line of action to a new location.

We use the symbol x bar to solve for an unknown distance in the +x direction from the origin (shown as a solid circle).

Again, you must to ensure that both systems have the same tendency to rotate about any/all axis/axes.

When you're moving a force to get rid of a moment, it feels most intuitive to sum moments about the original location of the force you're moving. (Note: this is the opposite advice given in Example 1.)

Mathematically, you can do the moment summation about any axis.

That means you can always check your work by spot-checking moment equivalency about multiple axes.

❏ Example 2: moving a force to get rid of a moment

🟦  5.8 Equivalent systems for multiple parallel forces

We can use the same principle to combine multiple parallel forces into a single force.

The concept is the same as before: in order for two systems to be statically equivalent, they must have the same net force (tendency to translate) and the same net moment (tendency to rotate).

Note again that we commonly use the symbol x bar to solve for an unknown distance in the +x direction from an origin or anchor point (shown as a solid circle). 

In this solution, the origin (anchor point) was placed at A. This is was done strategically for two reasons.

First, summing moments about A is a good way to avoid sign errors, since the 2k force, the 3k force, and the resultant force all tend to rotate the body clockwise about that point. They'll all have the same sign.

Second, by choosing A, we can omit a term in the moment equivalency equation associated with the 1k force.

While learning this topic, do not try to do the math in your head. Get in the habit of writing out the equations of static equivalency when solving these types of problems.

❏ Combining multiple parallel forces into one force

🟦  5.9 A force distributed over an area is an area load

The image depicts two pushpin designs. Pushpin A is more expensive, and Pushpin B is quite a bit cheaper to manufacture.

Despite the savings in material, Pushpin B is a lousy design, right? It will be a little painful, while A will not. The additional surface area between the finger and Pushpin A is an important part of the design. 

In engineering applications, the idea of spreading a force over a surface is a powerful one.

This idea -- force per area -- is called an area load. Because the units (force per area) are the same as they are for pressure, we will use the symbol p for area loads. Here the difference between the two terms:

❏ Compare and constrast pushpin designs

The interaction between the bottom of the finger and the top of Pushpin A is most accurately represented as a vector field: a set of vectors that illustrate force per area over a surface. Here, we would model the area load vectors as a constant value.

Here's what's kind of mind-blowing. Previously, whenever we drew a single vector to represent a force, we were actually drawing a resultant force. We can integrate area load over surface area to solve for the resultant force. If the distribution is constant, the integral simplifies to division:

❏ Point loads and area loads

The interaction between pushpin and cork is best modeled as a point load. The interaction between the finger and the pushpin is best modeled as an area load.

This is our first integral in this course, so it's a good time to refresh a few concepts from calculus.

The integral sign looks like a giant S because in addition to "area under the curve" it can also mean "to sum up the effects."

❏ Integral relationship between area loads and resultant forces

Don't panic: we will only tackle constant and linear area loads in this text.

Work through the sample calculation. Here, we are modeling the area load as a constant area load. This means that each particle of the solid material is experiencing the exact same magnitude of force per area.

Common units of force per area are:

❏ Sample calculation

🟦  5.10 A force distributed over a length is called a line load

❏ Introduction to line loads

Sometimes, it is also useful to take a force, and spread it out over a length. 

The result of this operation is called a line load.

This text uses the symbol w for line load. It's commonly used; ω (omega) is another common symbol for this concept.

You can think of line loads in terms of "force per distance." You may also find it useful to think of them as a "force intensity."

❏ Relationships between area loads, line loads, and point loads

The equation for converting a force to a line load (or vice versa) is:

This flipbook features a constant line load. A person is modeled as a point load and as an equivalent uniformly distributed line load.

We convert line loads to equivalent forces by taking the area under the curve.

❏ Flipbook: Converting line loads to point loads

❏ Static equivalency integrals

Let's say that you are given a non-constant line load, such as this w(x) function. 

You want to create an equivalent point load.

You will need to solve for the magnitude (and direction) of the equivalent force in addition to its location. 

We will use the symbol x bar to identify the location of the resultant force.

Work through the flipbook carefully. This is an important derivation.

❏ General form of the integrals for static equivalency

❏ Formulas for a constant load

Remember that a definite integral is simply the area under the curve. When your load function is a simple geometric shape (a rectangle, a triangle, etc.), don't integrate: just compute the area under the curve.

For a uniformly distributed load, the resultant force is equal to area under the curve: wL. 

The force is located at midspan, at L/2.

❏ Derivation of formulas for a constant load

❏ Formulas for a linear load

Here is the derivation for a load function that varies linearly. We often call it a triangle load.

The equivalent force is the area under the curve of the loading diagram: wL/2. Again, when you integrate, and you already know the area formula for the shape under the curve, use it.

The force is located at the thirdpoint of the triangle, on the "heavy" side of the triangle. Please commit this fact to memory.

❏ Derivation of formulas for a linear (or "triangular") load

🟦  5.11 Converting between area loads and line loads

❏ Converting an area load to a line load

We convert an area load to a line load by multiplying by the tributary width.

The tributary width tells you how much load you want to take into account. It's always oriented perpendicular to the direction of the line load. 

In the interactive, checkers are used to help make this concept feel real.

❏ Converting a line load to an area load

We convert line loads to area loads by the opposite operation: we divide by the tributary width.

You don't need to memorize these operations. Instead, write out the units. The units will make it crystal clear whether you need to multiply or to divide when you do these conversions.


❏ Be careful with units

A note of warning! In a 2D projection, a line load and an area load will appear identical. The only way to tell them apart is by inspecting the units. Line loads are force per distance and area loads are force per area. Be sure to put units on all of your diagrams and figures.

❏ Illustration of tributary (trib.) width

🟦  5.12 Equivalent systems in 3D space

Work carefully through this interactive model. You'll explore a number of different equivalent systems in 3D space.

You can truly create an infinite number of equivalent systems! The key is to use the equations of static equivalency.

➜ Practice Problems

When working these problems, be sure to draw System I (whatever is given) and System II (whatever is asked for). Then, when you set up the static equivalency equations, set them up exactly as demonstrated above.

Here are SCRAMBLED answers to problems 1-8 so you can check your work. They are all magnitudes; some of your answers will be negative.

FORCE ANSWERS: 2.33  7.50  14.4  16.8  17.5  22.5  27.0  30.0  34.0  48.0  58.5 

DISTANCE ANSWERS:  1.60  2.36  5.30  11.0 

Problem 1.

Problem 2.

Problem 3.

Problem 4.

Problem 5.

Note that in this problem, we want to use ybar instead of xbar. This is because we are measuring distance in the y direction.

Problem 6.

Problem 7.


(The solution for MD is 34 k-feet).

Problem 8.

(The solution for MA is 291 N-mm.)

Problem 9.


Here is the answer, so you can check your work:

FR = <20.8, 12.0> pounds

MA = 130 #-in

Problem 10.

This is a challenge problem! No answer is supplied intentionally.

Give this one a bit of thought, but if you're confused, that's OK. We'll talk about it in class.