Let's explore how to compute the deflection of our beam example using the Conjugate Beam Method. We shall use this to illustrate how to use the method to find the slope and translation at a point.
Continuing from this example, let's learn how to use the conjugate beam method to determine the deflection at critical points.
The solution presented is in SI. The author will update the post soon to reflect English units.
In this example, we are to find the deflection components at all critical points of the beam - \(A\), \(B\), \(C\), \(D\), and \(E\). It includes the slope and horizontal and vertical deflection components of said points. At the end of the solution, we have to show a deflection matrix - a table showing the deflection at each critical location.
Inspection Analysis
Let's identify the number of deflection components we need to solve. In a 2D structure, there are at most three per joint (rotation, horizontal translation, vertical translation). With five joints, we have a total of 15 components.
From inspection, we can already solve the value of some components:
Since no horizontal loads are acting on the beam that causes it to move along the x-axis, horizontal translation at all critical points is zero (5 inspected, ten remaining)
Vertical movement at supports is not allowed at \(B\) and \(D\); hence, vertical translation at these points is also zero (7 inspected, eight remaining)
From 15 components, we would need to solve only 8 using the area moment method.
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For this example, we will consider the moment-by-parts method (for ease in solution later on). After getting it, we divide it with the beam's flexural rigidity \(EI\). As a result, we'll have an \(M/EI\) diagram.
Construct the Conjugate Beam
The next step is to construct the conjugate beam of our example:
We create a similar structure with the same length \(L\).
Then, we load this beam with the \(M/EI\) diagram.
We change the original supports with their corresponding artificial constraints. Free ends \(A\) and \(E\) will become fixed supports, while the internal supports \(B\) and \(D\) will become hinges.
Solve for Deflection Components
After creating the conjugate beam, we can already proceed to find the deflection at any point. The solution is similar to finding the \(V\) and \(M\) of a beam using equations or diagrams; hence, if you're well-versed in the determinate analysis of beams, the following section is straightforward.
Our conjugate beam is a compound beam with internal connections at \(B^{\prime}\) and \(D^{\prime}\); hence, we would need to create free body diagrams for each part: \(A^{\prime}B^{\prime}\), \(B^{\prime}D^{\prime}\), and \(D^{\prime}E^{\prime}\).
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 \(B'\), \(C'\), and \(D'\).
We first solve for the segment's endpoints for \(B'\) and \(D'\).
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 \(D^{\prime}\):
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 \(M_{C^{\prime}}\):
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:
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:
With shear and moment solved at all critical points of the conjugate beam, the only remaining thing to do is to apply the theorems of the conjugate beam method.
The sign for each deflection component tells us the direction of the deflection. We base this convention on a separate post discussing how to interpret the conjugate shear and moment with the directional sense of actual slope and translation.
Positive-directed vertical translation means the point moved above, while negative means it moved below.
Positive rotation means the point rotated counterclockwise, while negative means it moved clockwise.
