Statics

Lesson 5

Static Equivalency

🟦  5.1 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 is being pushed with 4N at the center, while Box B is being pushed with four 1N forces at the corners. These scenarios are statically equivalent because:

Statically equivalent:

Now, consider boxes C and D. 

They are not statically equivalent, because box D tends to rotate (about the y-axis) while box C does not.

They have the same tendency to translate (a net force of 4N in the positive x-direction), but since they do not have the same net moment, they are not statically equivalent.

Not statically equivalent:

🟦  5.2  Equivalent systems of vectors

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.

❏ Flipbook: Force components

🟦  5.3  The equations of static equivalency

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.

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

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

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

🟦  5.4  The Principle of Transmissibility

Recall that all vectors have four attributes:

The Principle of Transmissibility states that you can move a vector along its line of action to create a statically equivalent system.

Vectors A and B are statically equivalent because they share the same line of action (a-a). 

Vector A depicts a push and Vector B depicts a pull, yet the vectors are statically equivalent

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

Vectors D and E are not statically equivalent to A. They do not share the same line of action.

Now, consider a box. 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). The represent different actions on the box, yet they are statically equivalent.

Example

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

You create a statically equivalent system by sliding the vector along its line of action from C to B. The vertical 3 kip component causes a moment about A (-30 kip-feet). The horizontal 4 kip component is coincident with line AB and does not create a moment.

🟦  5.5 Equivalent systems with couple moments

Imagine placing your cell phone on your desk and applying the loads from system I, II, III, and IV. Which systems are equivalent?

Systems I and II have the same amount of force, but since the force is on a different line of action, they are not equivalent.

Systems II and III are combined into System IV.

Systems I and IV are statically equivalent. They both have:

You can move a force off of its line of action and create an equivalent system, provided that you introduce a couple moment to counteracts the rotation you caused.

You can compute the magnitude and sign of the moment by setting the tendency to rotate in System I equal to the tendency to rotate in System II. That is, the net moment about an axis in System I must equal the net moment about the same axis in System II.

You can do the moment calculation about any axis, but it's most intuitive when you sum moments about the destination of the force you're moving.

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.

Again, you want to ensure that both systems have the same tendency to rotate about an axis. 

You can do the moment summation about any axis, but it's most intuitive when you sum moments about the original location of the force you're moving.

🟦  5.6 Force couples and couple moments

Force couples and couple moments are statically equivalent. They produce a tendency to rotate (a net moment) without a tendency to translate (a net force).

The following five systems are all equivalent. They all produce a net moment of 10kN • 1m = 10 kNm.

Try to solve the unknowns on your own. The answer is below the image.

Answers:  P1 = 5 kN; P2 = 2 kN; M1 = M2 = 10 kN•m

Were you surprised that M1 = M2? Couple moments are called "free vectors." You can put them anywhere and create a statically equivalent system. Their location doesn't matter.

🟦  5.7 Combining multiple parallel forces into a single force

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).

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

🟦  5.8 Converting a line load into an equivalent force

One of the most common applications of static equivalency in Statics is to convert a line load to an equivalent point load.

You must solve for the magnitude of the equivalent force as well as its location. 

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.

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.

The force is located at the thirdpoint of the triangle, on the "heavy" side of the triangle.

🟦  5.9 Equivalent systems in 3D space

In this interactive model, you can explore a number of different equivalent systems in 3D space.

You can truly create an infinite number of equivalent systems!

➜ 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.

Problem 1.

Problem 2.

Problem 3.

Problem 4.

Problem 5.

Problem 6.

Problem 7.

Problem 8.

Problem 9.

Problem 10.