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Engineering Statics: Open and Interactive

Section 4.6 Equivalent Transformations

An equivalent transformation occurs when a loading on an object is replaced with another loading that has the same external effect on the object. By external effect, we mean the response of the body that we can see from the outside, with no consideration of what happens to it internally. If the object is a free body, the external effect would be translation and rotation. In statics, since objects are not accelerating, the external effect really means the reactions at the supports required to maintain equilibrium. The external effects will be exactly the same before and after an equivalent transformation.
Equivalent transformation permits us to swap out one set of forces with another one without changing the fundamental physics of the situation. This is usually done to simplify or clarify the situation, or to give you an alternate way to think about, understand, and solve a mechanics problem.
You already know several equivalent transformations although we have not used this terminology before. Here are some transformations you have applied previously.

Vector Addition.

When you add forces together using the rules of vector addition, you are performing an equivalent transformation. You can swap out two or more components and replace them with a single equivalent resultant force.
Any number of concurrent forces can be added together to produce a single resultant force. By definition, the lines of action of concurrent forces all intersect at a common point. The resultant must be placed at this intersection point in order for this replacement to be equivalent. This is because before and after the replacement, the moment about the intersection point is zero. If the resultant was placed somewhere else, that would not be true.

Replacing a Force with its Component.

Resolving forces into components is also an equivalent transformation, as it is the inverse operation of vector addition. The components are usually orthogonal and in the coordinate directions, or in a given plane and perpendicular to it, but any combination of force components that add to the original force is equivalent.
A fixed ring is shown with two forces F1 and F2. This is equal to a fixed ring with the resultant force shown acting at an angle of theta from the horizontal, which would be pointing towards the coordinates (F1x+F2x, F1y+F2y). This is equal to a fixed ring with a horizontal force of F*cos(theta) and a vertical force of F*sin(theta).
Figure 4.6.1. Equivalent transformations of vectors
In this diagram,
\begin{equation*} \vec{F} = \vec{F_1} + \vec{F}_2 = (F; \theta) = \langle F \cos\theta, F\sin\theta \rangle\text{.} \end{equation*}
The effects of the force in the \(x\text{,}\) \(y\) and (in three dimensions the \(z\)) directions remain the same, and by Varignon’s theorem, we know that the moment these forces make about any point will also be the same.
An interesting special case occurs when two forces are equal and opposite and have the same line of action. When these are added together, they cancel out, so replacing these two forces with nothing is an equivalent transformation. The opposite is true as well, so you can make two equal and opposite forces spontaneously appear at a point if you wish.

Thinking Deeper 4.6.2. Internal Effects.

We made a point of saying that equivalent systems of force have the same \(external\) effect on the body. This implies that there may be some other effects that are not the same. As you will see in Chapter 8, we sometimes need to consider internal forces and moments. These are the forces inside a body that hold all the parts of the object to each other, otherwise, it would break apart and fail. Although the external effects are the same for all equivalent systems, the internal forces depend on the specifics of how the loads are applied.
Let’s imagine that you have gone off-roading and have managed to get your Jeep stuck in the mud. You have two basic options to get it out: you can \(pull\) it out using the winch on the front bumper, or you can ask your friend to \(push\) you out with his truck. Both methods (assuming that they apply forces with the same magnitude, direction and line of action) are statically equivalent, and both will equally move your vehicle forward.
The difference is what might happen to your vehicle. With one method there’s a danger that you will rip your front bumper off, with the other, you might damage your rear bumper. These are the internal effects and they depend on where the equivalent force is applied. These forces are necessary to maintain rigidity and hold the parts of the body together.

Sliding a force along its line of action.

Sliding a force along its line of action is an equivalent transformation because sliding a force does not change its magnitude, direction or the perpendicular distance from the line of action to any point, so the moments it creates do not change either. This transformation is called the “Principle of Transmissibility”.
A force vector is shown acting along a line of action. It is equivalent to a diagram where the same force vector has been shifted along the same line of action.
Figure 4.6.3. Sliding a vector along its line of action

Replacing a couple with couple-moment or vice-versa.

A couple, defined as “two equal and opposite forces with different lines of action,” produces a pure turning action that is equivalent to a concentrated moment, called the couple-moment. Couples and couple-moments have no translational effect. Couple-moments are free vectors, which means that they are not bound to any point. Their external effect is on the entire body and is the same regardless of where it is applied.
This means that you are free to swap out a couple for its couple-moment, or swap a couple-moment for a couple that has the same moment, and you may put the replacement anywhere on you please and it will still be equivalent.
The diagram shows a series of equivalent transformations of a couple.
Six equivalent diagrams are shown representing a couple-moment. Two equal and opposite forces in parallel of magnitude F separated by a perpendicular distance d = two equal/opposite parallel forces of magnitude (F/2) separated by a perpendicular distance 2d = two equal/opposite parallel forces at a different angle separated by a perpendicular distance d = two equal/opposite parallel forces of magnitude 2F separated by a perpendicular distance of (d/2) = a Moment of magnitude M = Fd acting counter-clockwise at the middle of the object = a Moment of M = Fd acting counter-clockwise at the edge of the object.
Figure 4.6.4. Equivalent transformations of couples
Concentrated moments are free vectors, which you may draw the circular arrow anywhere you like on the body. In other words, moving a concentrated moment from one point to another is an equivalent transformation. Remember though, this equivalence only applies to the external effects. What happens inside the body definitely does depend on the specific point where the moment is applied.

Adding moments to produce a resultant moment.

If more than one couple-moment or concentrated moment acts on an object the situation may be simplified by adding them together to produce one resultant moment, \(\vec{M}_R\text{.}\) The standard rules of vector addition apply.
In two-dimensional problems moments are either clockwise or counter-clockwise, so they may be considered scalar values and added algebraically. Give counter-clockwise moments a positive sign and clockwise moments a negative sign according to the right-hand rule sign convention. If this is done, the sign of the resultant moment will indicate the direction of the net moment. You can use the right-hand rule to establish the direction of the moment vector, which will point into or out of the page.
\begin{equation*} M_R = \Sigma M \end{equation*}

Example 4.6.5. Equivalent Moment.

Two concentrated moments and a couple act on the object shown. Given: \(M_1 = \Nm{400}\text{,}\) \(M_2 =\Nm{200}\text{,}\) \(F = \N{40}\text{,}\) and \(d=\m{2}\text{.}\)
Replace these with a single, equivalent concentrated moment, and give the magnitude and direction of your result.
An object is shown with a Force couple where two equal/opposite parallel forces of magnitude 40 N are shown. The distance between them is shown as d = 2 m, but the forces act at an angle of 60 degrees from the distance vector. M1 = 400 N*m in clockwise direction, and M2 = 2200 N*m in counter-clockwise direction.
\(M_r = \Nm{130.7}\) clockwise.
First, replace the couple with an equivalent couple, \(M_3\text{,}\) the magnitude of which is
\begin{align*} M_3 \amp = F d_\perp\\ \amp = F d \sin \ang{60}\\ \amp = \Nm{69.3} \end{align*}
By observation, this is a counter-clockwise moment, as is \(M_2\text{.}\) \(M_1\) is clockwise. Summing the scalar magnitudes gives the resultant moment. The signs of the terms are assigned according to the sign convention: positive if counter-clockwise, negative if clockwise.
\begin{align*} M_R \amp = \Sigma M\\ \amp = M_1 + M_2 + M_3\\ \amp = - \Nm{400} + \Nm{200} + \Nm{69.3}\\ \amp = -\Nm{130.7}\\ \vec{M_R} \amp = \Nm{130.7} \text{ clockwise} \end{align*}

Resolving a moment into components.

For three-dimensional moment vectors, another potential equivalent transformation is to resolve a moment vector into components. These may be orthogonal components in the \(x\text{,}\) \(y\text{,}\) and \(z\) directions, or components in a plane and perpendicular to it, or components in some other rotated coordinate system.