What is Conjugate Moment?

Verifying the Theorem

Let's focus on the same simple beam as our Theorem 1 example, but this time, say we are to find translation \(\Delta_x\) that is \(x\) units from the origin.

To solve for the translation, we need to use the area moment method and derive a relationship among \(\Delta_x\), \(t_{x/A}\), and \(t_{B/A}\). Looking closely at the figure, we can do that precisely by taking the ratio and proportion of similar triangles:

\(\frac{\Delta_x+t_{x/A}}{x}=\frac{t_{B/A}}{L}\)

Equation 1: \(\Delta_x=t_{B/A}\left( \frac{x}{L} \right)-t_{x/A}\)

From the exact figure, we now investigate the conjugate beam and solve for the moment at said position \(x\). Place a cutting plane at a distance \(x\) and solve for the moment:

\(\sum{M_L}=M_x]\circlearrowright_+\)

\(M_x=R_{A^{\prime}}(x)+A_2\left( \frac{x}{4} \right)-A_1\left( \frac{x}{3} \right)\)

Equation 2: \(M_x=R_{A^{\prime}}(x)-\left[ A_1\left( \frac{x}{3} \right) - A_2\left( \frac{x}{4} \right) \right]\)

In Equation 2, we have grouped all terms involving area \(A\). From the second theorem of the area moment method, this grouping corresponds to the tangential deviation at \(x\) from the tangent at \(A\); hence, we can rewrite Equation 2 as:

Equation 3: \(M_x=R_{A^{\prime}}(x)-t_{x/A}\)

Earlier, when we were verifying the first theorem, we solved for the reaction at \(A^{\prime}\):

\(R_{A^{\prime}}=\frac{t_{B/A}}{L}\)

Substituting this to Equation 3, we have:

Equation 4: \(M_x=t_{B/A}\left( \frac{x}{L} \right)-t_{x/A}\)

Equations 1 and 4 are the same. It means that if you take the moment of the conjugate beam at point \(x\), you get the translation at the actual beam at said point.

‍Summary

Let's summarize:

The fictional moment of the conjugate beam at any point is the translation (slope) of the actual beam at said point.

\(M^{\prime}=\Delta\)