How to Use the Area Moment Method Using Moment By Parts?

Main Solution

Draw Moment Diagram By Parts

The first step is to create the moment diagram of our example by superimposing the bending effects of cantilever parts. This article shows a more descriptive discussion of how to construct one.

Set Up M/EI Expressions

After getting the moment diagram by parts, we set up the \(M/EI\) expressions per segment. We divide the said moment diagram \(M\) equation with the flexural rigidity of the beam \(EI\).

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 translation \(\Delta\) is upward from the longitudinal axis (+x-axis), it has a positive sense. Otherwise, it is negative. 
• If rotation \(\theta\) is counterclockwise from the longitudinal axis (+x-axis), it has a positive sense. Otherwise, it is negative.

From these conventions, translations \(\Delta_{A_v}\) is positive while \(\Delta_{C_v}\) and \(\Delta_{E_v}\) are negative. For rotations, \(\theta_C\), \(\theta_D\), and \(\theta_E\) are positive while \(\theta_A\), and \(\theta_B\) are negative.

Later on, we can verify if these signs are indeed correct.

Derive Key Relationships

Next, we derive relationships for all deflection components based on our sketch.

Like our first approach, we draw a reference tangent at one of the supports (at \(B\)). Then, draw and label all translations \(\Delta\), rotation \(\theta\), and deviations \(t\). Afterward, we identify similar triangles in our drawing, which will be our basis for forming relationships

Translation

From triangles \(BAA^{\prime \prime}\), \(BCC^{\prime \prime}\), \(BDD^{\prime \prime}\), and \(BEE^{\prime \prime}\), we can derive the following:

Equation 1-1: \(\frac{A A^{\prime \prime}}{2}=\frac{C C^{\prime \prime}}{4.5}=\frac{D D^{\prime \prime}}{9}=\frac{E E^{\prime \prime}}{10.5}\)

From the same figure, we can express \(AA^{\prime \prime}\), \(CC^{\prime \prime}\), \(DD^{\prime \prime}\), and \(EE^{\prime \prime}\) the following:

Equation 1-2: \(A A^{\prime \prime}=\left(+\Delta_{A_v}\right)-\left(+t_{A / B}\right)\)

Equation 1-3: \(C C^{\prime \prime}=\left(-\Delta_{C_v}\right)+\left(+t_{C / B}\right)\)

Equation 1-4: \(D D^{\prime \prime}=+t_{D / B}\)

Equation 1-5: \(E E^{\prime \prime}=\left(+t_{E / B}\right)-\left(+\Delta_{E_v}\right)\)

We shall use these equations to solve for the translations after we get the values for deviation.

Rotation

If we look at triangle \(BDD^{\prime \prime}\), we can also derive a relationship for the clockwise rotation at \(B\):

Equation 2-1: \(-\theta_B=-\frac{D D^{\prime \prime}}{9}=-\frac{t_{D / B}}{9}\)

Later on, we shall use the first theorem to get other rotational components; hence, there is no need to derive relationships further for other angular deflections.

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 \(E\). We will use moment by parts \(M/EI\) for this approach.

Let's compute for \(t_{A/B}\), \(t_{C/B}\), \(t_{D/B}\), and \(t_{E/B}\):

\(t_{A / B}=\frac{1}{E I}(2 \times 27)\left(\frac{1}{2} \times 2\right)=\frac{54}{E I}\)

\(t_{C / B}=\frac{1}{E I}[\left(\frac{1}{2} \times 4.5 \times 405\right)\left(\frac{1}{3} \times 4.5\right)+(4.5 \times 27)\left(\frac{1}{2} \times 4.5\right)\)

\(-\left(\frac{1}{4} \times 4.5 \times 151.875\right)\left(\frac{1}{5} \times 4.5\right)]=\frac{1486.48}{E I}\)

\(t_{D / B}=\frac{1}{E I}[(4.5 \times 27)\left(\frac{4.5}{2}+4.5\right)+\left(\frac{1}{2} \times 4.5 \times 405\right)\left(\frac{4.5}{3}+4.5\right)-\left(\frac{1}{4} \times 4.5 \times 151.875\right)\left(\frac{4.5}{5}+4.5\right)\)

\(-\left(\frac{1}{3} \times 4.5 \times 364.5\right)\left(\frac{3}{4} \times 4.5\right)+\left(\frac{1}{2} \times 4.5 \times 1184.625\right)\left(\frac{2}{3} \times 4.5\right)-(4.5 \times 135)\left(\frac{1}{2} \times 4.5\right)\)

\(-\left(\frac{1}{2} \times 4.5\right)(540-135)\left(\frac{2}{3} \times 4.5\right)]=\frac{474579}{64 E I}\)

\(t_{E / B}=\frac{1}{E I}[(4.5 \times 27)\left(\frac{4.5}{2}+6\right)+\left(\frac{1}{2} \times 4.5 \times 405\right)\left(\frac{4.5}{3}+6\right)-\left(\frac{1}{4} \times 4.5 \times 151.875\right)\left(\frac{4.5}{5}+6\right)\)

\(-\left(\frac{1}{3} \times 4.5 \times 364.5\right)\left(\frac{3 \times 4.5}{4}+1.5\right)+\left(\frac{1}{2} \times 4.5 \times 1184.625\right)\left(\frac{2 \times 4.5}{3}+1.5\right)-\frac{1}{2}(6 \times 540)\left(\frac{2}{3} \times 6\right)]\)

\(t_{E / B}=\frac{9506.74}{E I}\)

After solving for the deviations, we interpret our results. Since our answers are positive, this indicates that \(A^{\prime \prime}\), \(C^{\prime \prime}\), \(D^{\prime \prime}\), and \(E^{\prime \prime}\) must be above the reference tangent. Comparing these results with our drawing, it matches our sketch. If it doesn't, we may need to resketch and reformulate our relationships in the previous step.

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 \(\Delta_{E_v}\):

\(\frac{\left(+\Delta_{A_v}\right)-\left(+t_{A / B}\right)}{2}=\frac{t_{D / B}}{9}\)

\(\Delta_{A_v}=\frac{2}{9} \times \frac{474579}{64 E I}+\frac{54}{E I}\)

\(\Delta_{A_v}=\frac{1701.84}{E I}\)

\(\frac{\left(-\Delta_{C_v}\right)+\left(+t_{C / B}\right)}{4.5}=\frac{t_{D / B}}{9}\)

\(\Delta_{C_v}=-\left(\frac{4.5}{9} \times \frac{474579}{64 E I}-\frac{1486.48}{E I}\right)\)

\(\Delta_{C_v}=-\frac{2221.18}{E I}\)

\(\frac{t_{D / B}}{9}=\frac{\left(+t_{E / B}\right)-\left(+\Delta_{E_v}\right)}{10.5}\)

\(\Delta_{E_v}=\left(\frac{9506.74}{E I}-\frac{10.5}{9} \times \frac{474579}{64 E I}\right)\)

\(\Delta_{E_v}=\frac{855.62}{E I}\)

With translations solved, we verify the positioning of these translations and compare it with our sketch. 

Translation at \(A\) and \(E\) is positive, which means that \(A^{\prime \prime}\) and \(E^{\prime \prime}\) is above the +x-axis. On the other hand, \(C\) is negative, indicating that \(C^{\prime \prime}\) is below the +x-axis. These results confirm the accuracy of our sketch, and we may proceed with solving for the rotational components.

Again, if our answers do not conform to our initial sketch - meaning the signs of the results are opposite to that of our drawing, then we have the wrong sign assumption. The actual direction would be the opposite of what we have established.

Rotation

With translations solved, the only remaining thing to do is to solve for \(\theta_A\), \(\theta_B\), \(\theta_C\), \(\theta_D\), and \(\theta_E\). With \(t_{D/B}\) solved, we can use Equation 2-1 to find the rotation at \(B\):

\(-\theta_B=-\frac{474579}{9 \times 64 E I}\)

\(-\theta_B=-\frac{823.92}{E I}\)

Before solving for the other rotations, verify if the direction of \(B\) is clockwise by checking whether the corrected translations match. If it doesn't, we must change the sign to positive to indicate that the direction is counterclockwise.

After verifying, we use the first theorem of the area moment method to solve for the remaining rotations. Rather than using integration, we use the moment-by-parts diagram again. With \(\theta_B\) as our reference, let's first solve \(\theta_A\):

\(\theta_B-\theta_A=\frac{1}{E I}(4.5 \times 27)\)

\(\theta_A=-\frac{823.92}{E I}-\frac{54}{E I}=-\frac{877.9}{E I}\)

With \(\theta_A\) solved, let’s compute for the remaining rotations using \(\theta_B\) as reference again:

\(\theta_C-\theta_B=\frac{1}{E I}\left[\left(\frac{1}{2} \times 4.5 \times 405\right)+(4.5 \times 27)-\left(\frac{1}{4} \times 4.5 \times 151.875\right)\right]\)

\(\theta_C=\frac{861.89}{E I}-\frac{823.92}{E I}=\frac{37.8}{E I}\)

\(\theta_D-\theta_B=\frac{1}{E I}[\left(\frac{1}{2} \times 4.5 \times 405\right)+(4.5 \times 27)-\left(\frac{1}{4} \times 4.5 \times 151.875\right)-\left(\frac{1}{3} \times 4.5 \times 364.5\right)\)

\(+\left(\frac{1}{2} \times 4.5 \times 1184.625\right)-(4.5 \times 135)-\left(\frac{1}{2} \times 4.5\right)(540-135)]\)

\(\theta_D=\frac{1461.80}{E I}-\frac{823.92}{E I}=\frac{637.9}{E I}\)

\(\theta_E-\theta_B=\frac{1}{E I}[\left(\frac{1}{2} \times 4.5 \times 405\right)+(4.5 \times 27)-\left(\frac{1}{4} \times 4.5 \times 151.875\right)-\left(\frac{1}{3} \times 4.5 \times 364.5\right)\)

\(+\left(\frac{1}{2} \times 4.5 \times 1184.625\right)-\left(\frac{1}{2} \times 6 \times 540\right)]\)

\(\theta_E=\frac{1360.55}{E I}-\frac{823.92}{E I}=\frac{536.6}{E I}\)

After solving for the rotations, we interpret the results. \(\theta_A\) 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.

Deflection Matrix

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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.