How to Use the Conjugate Beam Method?

Free Body Diagram B'D'

Let's investigate $B^{\prime }{D}^{\prime }$. We aim to find the shear and moment of the conjugate beam at key points $B^{\prime }$, $C^{\prime }$, and $D^{\prime }$.

We first solve for the segment's endpoints for $B^{\prime }$ and $D^{\prime }$.

To start, let's investigate the moment. This free-body diagram's endpoints are internal hinges, meaning that the moment is equal to zero at $B^{\prime }$ and $D^{\prime }$.

Next, we solve for shear. We can solve it by applying equilibrium equations at $B^{\prime }$ and $D^{\prime }$. To start, let's solve for the shear at $B^{\prime }$, which is assumed to be upward (take a moment summation at $D^{\prime }$:

$\sum M_{D_R^{\prime }}=0]{\circlearrowright }_{+}$

$V_{B^{\prime }}(9)-\frac{27}{EI}(4.5)(\frac{4.5}{2}+4.5)-\frac{1}{2}\frac{405}{EI}(4.5)(\frac{4.5}{3}+4.5)+\frac{1}{4}\frac{151.875}{EI}(4.5)(\frac{4.5}{5}+4.5)$

$-\frac{1}{2}\frac{1184.625}{EI}(4.5)(\frac{2}{3}\times 4.5)+\frac{1}{3}\frac{364.5}{EI}(4.5)(\frac{3}{4}\times 4.5)$

$+\frac{135}{EI}(4.5)(\frac{1}{2}\times 4.5)+\frac{1}{2}\frac{540-135}{EI}(4.5)(\frac{2}{3}\times 4.5)=0$

$V_{B^{\prime }}=\frac{823.92}{EI}$

With the shear at $B^{\prime }$ solved, let's proceed with the shear at $D^{\prime }$. We'll assume that it is acting downward:

$\sum F_{v_L}=0]\uparrow$

$\frac{823.92}{EI}-\frac{27}{EI}(4.5)-\frac{1}{2}\frac{405}{EI}(4.5)+\frac{1}{4}\frac{151.875}{EI}(4.5)$

$-\frac{1}{2}\frac{1184.625}{EI}(4.5)+\frac{1}{3}\frac{364.5}{EI}(4.5)+\frac{135}{EI}(4.5)+\frac{1}{2}\frac{540-135}{EI}(4.5)-V_{D^{\prime }}=0$

$V_{D^{\prime }}=-\frac{637.9}{EI}$

Since the shear at $D^{\prime }$ is negative, we have the wrong assumption. Instead of acting downwards, it should be upward.

After solving for $B^{\prime }$ and $D^{\prime }$, we now move on to the remaining key point $C^{\prime }$. In the following figure, we place a section at said point (from $B^{\prime }$) to reveal $V_{C^{\prime }}$ and $M_{C^{\prime }}$. Now, we consider the left section and apply the equilibrium principle to solve for $V_{C^{\prime }}$ and $M_{C^{\prime }}$:

$\sum F_{v_L}=V_{C^{\prime }}]{\uparrow }_{+}$

$\frac{823.92}{EI}-\frac{27}{EI}(4.5)-\frac{1}{2}\frac{405}{EI}(4.5)+\frac{1}{4}\frac{151.875}{EI}(4.5)=V_{C^{\prime }}$

$V_{C^{\prime }}=-\frac{37.9}{EI}$

$\sum M_L=M_{C^{\prime }}]{\circlearrowright }_{+}$

$\frac{823.92}{EI}(4.5)-\frac{27}{EI}(4.5)\left(\frac{4.5}{2}\right)-\frac{1}{2}\frac{405}{EI}(4.5)\left(\frac{4.5}{3}\right)+\frac{1}{4}\frac{151.875}{EI}(4.5)\left(\frac{4.5}{5}\right)=M_C^{\prime }$

$M_{C^{\prime }}=\frac{2221}{EI}$

Free Body Diagram A'B'

Let's proceed to $A^{\prime }{B}^{\prime }$ and solve for $A^{\prime }$. Since we already know $V_{B^{\prime }}$ and \(M_{B^{\prime}\), we can directly use the equilibrium principle to solve these:

$\sum F_{v_R}=0]\uparrow$

$A_v^{\prime }-\frac{27}{EI}(2)-\frac{823.92}{EI}=0$

$A_v^{\prime }=\frac{877.9}{EI}$

$\sum M_{B_R^{\prime }}=0]{\circlearrowright }_{+}$

$\frac{877.9}{EI}(2)-M_{A^{\prime }}-\frac{27}{EI}(2)\left(\frac{2}{2}\right)=0$

$M_{A^{\prime }}=\frac{1701.8}{EI}$

Free Body Diagram D'E'

At this point, the only remaining point we haven't considered yet is $E^{\prime }$; hence, we now look at $D^{\prime }{E}^{\prime }$ and solve for the shear and moment at said point:

$\sum F_{v_L}=0]\uparrow$

$E_v^{\prime }+\frac{1}{2}\frac{135}{EI}(1.5)-\frac{637.9}{EI}=0$

$E_v^{\prime }=\frac{536.6}{EI}$

$\sum M_{D_L^{\prime }}=0]{\circlearrowright }_{+}$

$M_{E^{\prime }}-\frac{1}{2}\frac{135}{EI}(1.5)\left(\frac{1.5}{3}\right)-\frac{536.6}{EI}(1.5)=0$

$M_{E^{\prime }}=\frac{855}{EI}$