How to Use the Area Moment Method Using Integration?
Let's explore how to compute the deflection of our beam example using the Area Moment Method. In this post, let's use integration to calculate the slope and translation at a certain point.
Continuing from this example, let's learn how to use the area moment method to determine the deflection at critical points. We shall use integration for this example. For the alternative solution, click here.
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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The first step is to set up the \(M/EI\) expressions per segment. You can do this analytically by dividing the \(M\) equation with flexural rigidity \(EI\) or drawing the \(M/EI\) diagram.
Sketch the Deflected Shape
The next step is to draw an approximate deflected shape of the structure. Our basis in sketching is the structure's compatibility requirements and support conditions.
This approximate sketch will be our basis for deriving the critical relationships for the deflection components required.
Part of sketching is also assigning a sign convention to all our deflection components based on our drawing:
If the translation \(\Delta\) is upward from the longitudinal axis (+x-axis), it has a positive sense. Otherwise, it is negative.
If the rotation \(\theta\) is counterclockwise from the longitudinal axis (+x-axis), it has a positive sense. Otherwise, it is negative.
Based on our sketch, translations \(\Delta_{A_v}\) and \(\Delta_{E_v}\) are positive while \(\Delta_{C_v}\) is negative. For rotations, \(\theta_{C}\) and \(\theta_{D}\) are positive while \(\theta_{A}\), \(\theta_{B}\), and \(\theta_{E}\) are negative.
Remember that this is only an assumption. Later, we will be able to verify if the directions are correct.
Derive Key Relationships
Next, we derive relationships for all deflection components based on our sketch. We can create geometric relationships among the different elements and deviations.
Let's draw a reference tangent at one of the supports. In this case, the roller at \(B\).
Using the exact figure, we identify all key points \(A\), \(B\), \(C\), \(D\), \(E\) and their corresponding position in the deflected shape \(A^{\prime}\), \(B^{\prime}\), \(C^{\prime}\), \(D^{\prime}\), \(E^{\prime} \), as well on the reference tangent \(A^{\prime \prime}, B^{\prime \prime}, C^{\prime \prime}, D^{\prime \prime}, E^{\prime \prime}\). Next, identify similar triangles \(BAA^{\prime \prime}\), \(BCC^{\prime \prime}\), \(BDD^{\prime \prime}\), and \(BEE ^{\prime \prime}\). We draw and label all deviations, translations \(\Delta\), and rotations \(\theta\) of said triangles.
Translation
From our figure, we can derive these relationships:
Notice that we have expressed it into their respective deviations and deflections (with the appropriate sign convention).
Rotation
Similarly, we can create a relationship for the assumed clockwise rotation \(\theta_{B}\) by taking the ratio of a triangle's opposite and adjacent sides. Let's use triangle \(BDD^{\prime \prime}\) to form such a relationship:
We can derive relationships for rotation at other critical points based on our sketch; however, we will derive only one relation to show the use of the first theorem later.
Solve for Deviations
After formulating relationships among deviations and deflections, it's time to solve for deviations \(t\) of Equations 1-2 to 1-5.
We apply the second theorem to solve for tangential deviations among \(A\), \(C\), \(D\), and \(E\). For this approach, we will integrate math \(M/EI\) expressions.
Using the \(M/EI\) diagram we set up earlier, let's compute for \(t_{A/B}\), \(t_{C/B}\), \(t_{D/B}\), and \(t_{E/B}\):
\(t_{A / B}=\int_0^2 \frac{1}{E I}(27 x) d x=\frac{54}{E I}\)
\((12.5-x) d x+\int_{11}^{12.5} \frac{1}{E I}(90 x-1125)(12.5-x) d x=\frac{9506.74}{E I}\)
With deviations solved, it's important to check the sign convention. In this case, all deviations are positive. Recall that if a deviation is positive, its corresponding point in the deflected shape is above the reference tangent. Looking back at our figure, the directional sense of our deviations matches our drawing.
We may have an incorrect sketch if our answers contradict our drawing. If so, we may need to redraw and reformulate relationships.
Solve for Deflection Components
We can now compute for the deflection components with deviations solved using our earlier derived relationships.
Translation
Using Equations 1–1 to 1–5, let's solve for \(\Delta_{A_v}\), \(\Delta_{C_v}\), and \(\Delta_{E_v}\):
\(\Delta_{E_v}=-\left(\frac{10.5}{9} \times \frac{474579}{64 E I}-\frac{9506.74}{E I}\right)\)
\(\Delta_{E_v}=\frac{855.62}{E I}\)
With translations solved, the only thing we need to do is to verify the position of these deflections through their signs. For \(\Delta_{A_v}\) and \(\Delta_{E_v}\), our answers are positive, which means that translation is upwards at these points. On the other hand, \(\Delta_{C_v}\) is negative, which means point \(C\) deflected downwards.
Initially, we assumed that the translation at E is downward; however, our computations show that the point deflected upwards. The solution will correct itself if we have defined the relationships properly (Equation 1-1 to 1-5).
Rotation
Let us proceed with the remaining deflection components: the rotations at \(A\), \(B\), \(C\), \(D\), and \(E\).
Since we know the value of \(t_{D/B}\), we can use Equation 2-1 to solve for the slope at \(B\).
\(-\theta_B=-\frac{474579}{9 \times 64 E I}\)
\(-\theta_B=-\frac{823.92}{E I}\)
Remember that we have assumed the rotation at \(B\) to be negative; however, since we have solved the translations at all critical points, we can verify the directional sense of the rotation. We can see in the figure that the slope at \(B\) is indeed clockwise.
Verifying the sign convention is important because we will use the first theorem of the area moment method to solve for the remaining rotation components \(\theta_A\), \(\theta_C\), \(\theta_D\), and \(\theta_E\); From \(\theta_B\), we can choose any other point to solve for the missing rotations. For example, to find the slope at \(A\):
\(\theta_B-\theta_A=\frac{1}{E I} \int_0^2 27 d x\)
Next, we solve for the rotation at \(C\). You can choose any other solved rotation like \(\theta_A\) as your reference; however, let's still use \(\theta_B\) as our basis.
\(\theta_C-\theta_B=\frac{1}{E I} \int_2^{6.5}\left(-\frac{5}{3} x^3+10 x^2+70 x-\frac{419}{3}\right) d x\)
With angular deflections solved, verifying these components' directions is the only thing left to do. \(\theta_A\) and \(\theta_B\) are negative, which means the rotation is clockwise, while \(\theta_C\), \(\theta_D\), and \(\theta_E\) are positive indicating a counterclockwise rotation. We make sure these are correct by comparing them to our sketch.
Deflection Matrix
At this point, we have solved all 15 deflection components. It is best to summarise all our results using a deflection matrix. The sign convention described in the table is as follows:
A positive-sense rotation means that the point rotated counterclockwise from the local longitudinal axis (+x-axis), while negative means the opposite.
A positive-sense translation means the point deflected upward from the local longitudinal axis (+x-axis), while negative means the opposite.
