Section 4.6 Equivalent Transformations
Key Questions
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.
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.
Answer.
\(M_r = \Nm{130.7}\) clockwise.
Solution.
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*}