How to Use the Double Integration Method Using Segment Analysis?

There are two ways to solve this problem using the double integration method: segment analysis and general equation analysis. This post explains the former.

Set Up Equations

The first step in this method is to formulate the moment equations per segment which we can see here.

Equation 1-1: \(M_{AB}=27,\{0\leq{x}\lt {2}\}\)

Equation 1-2: \(M_{BC}=-\frac{5}{3} x^3+10 x^2+70 x-\frac{419}{3},\{2\leq{x}\lt {6.5}\}\)

Equation 1-3: \(M_{CD}=-18 x^2+\frac{891}{4} x-\frac{1629}{4},\{6.5\leq{x}\lt {11}\}\)

Equation 1-4: \(M_{DE}=90x-1125,\{11\leq{x}\lt {12.5}\}\)

Also, if there are any variations in \(E\) and \(I\), we have to express it in terms of \(x\); however, since the flexural rigidity is constant for this example, we disregard this effect.

Apply Bernoulli-Euler Beam Equation

Next, we apply the Bernoulli-Euler Beam Equation to each moment equation (as well as the functions of \(E\) and \(I\), if any):

Equation 2-1: \(\frac{d^2 \Delta}{d x^2}=\frac{1}{E I}(27),\{0\leq{x}\lt {2}\}\)

Equation 2-2: \(\frac{d^2 \Delta}{d x^2}=\frac{1}{E I}\left(-\frac{5}{3} x^3+10 x^2+70 x-\frac{419}{3}\right),\{2\leq{x}\lt {6.5}\}\)

Equation 2-3: \(\frac{d^2 \Delta}{d x^2}=\frac{1}{E I}\left(-18 x^2+\frac{891}{4} x-\frac{1629}{4}\right),\{6.5\leq{x}\lt {11}\}\)

Equation 2-4: \(\frac{d^2 \Delta}{d x^2}=\frac{1}{E I}(90 x-1125),\{11\leq{x}\lt {12.5}\}\)

Find General Solution

Afterward, we are ready to find the slope and deflection equations per segment.

General Slope Equations

Let's start by finding the general slope equation. We integrate each Bernoulli-Euler Beam Equation once to get such equations:

Equation 3-1: \(\frac{d \Delta}{d x}=\theta=\frac{1}{E I}\left(27 x+C_1\right),\{0\leq{x}\lt {2}\}\)

Equation 3-2: \(\frac{d \Delta}{d x}=\theta=\frac{1}{E I}\left(-\frac{5}{12} x^4+\frac{10}{3} x^3+35 x^2-\frac{419}{3} x+C_2\right),\{2\leq{x}\lt {6.5}\}\)

Equation 3-3: \(\frac{d \Delta}{d x}=\theta=\frac{1}{E I}\left(-6 x^3+\frac{891}{8} x^2-\frac{1629}{4} x+C_3\right),\{6.5\leq{x}\lt {11}\}\)

Equation 3-4: \(\frac{d \Delta}{d x}=\theta=\frac{1}{E I}\left(45 x^2-1125 x+C_4\right),\{11\leq{x}\lt {12.5}\}\)

General Deflection Equations

After getting the general slope equations, we integrate these again to get the deflection equations per segment:

Equation 4-1: \(\Delta=\frac{1}{E I}\left(\frac{27}{2} x^2+C_1 x+C_5\right),\{0\leq{x}\lt {2}\}\)

Equation 4-2: \(\Delta=\frac{1}{E I}\left(-\frac{1}{12} x^5+\frac{5}{6} x^4+\frac{35}{3} x^3-\frac{419}{6} x^2+C_2 x+C_6\right),\{2\leq{x}\lt {6.5}\}\)

Equation 4-3: \(\Delta=\frac{1}{E I}\left(-\frac{3}{2} x^4+\frac{891}{24} x^3-\frac{1629}{8} x^2+C_3 x+C_7\right),\{6.5\leq{x}\lt {11}\}\)

Equation 4-4: \(\Delta=\frac{1}{E I}\left(15 x^3-\frac{1125}{2} x^2+C_4 x+C_8\right),\{11\leq{x}\lt {12.5}\}\)

Find Particular Solution

After finding the general slope and deflection solutions, we must find the particular solution to our example. 

There are eight general expressions - one for rotation and one for translation for \(AB\), \(BC\), \(CD\), and \(DE\). In each of these, there are at most two constants of integration per segment - totaling eight constants (\(C_1\), \(C_2\), \(C_3\), \(C_4\), \(C_5\), \(C_6\), \(C_7\), and \(C_8\)) To find a particular solution, we need to solve for these eight constants. We can do this by establishing several conditions first.

Conditions

Conditions are constraints or facts that we know about the structure. In this example, we're asking ourselves: what do we know about the deflected shape of the beam?

The number of conditions required is equal to the number of constants of integration. So, if we have eight constants, we need at least eight conditions.

We'll discuss two types of conditions: continuity and boundary.

Continuity Conditions

Let's draw an approximate deflected shape of our beam. Consider two adjacent segments, such as \(AB\) and \(BC\), and their common point \(B(2,0)\). At this point, the rotation \(\theta\) of \(AB\) and \(BC\) must be the same; so, at \((2, 0)\), Equations 3-1 and 3-2 must be equal:

\(\theta_{A B}=\theta_{B C}, x=2\)

\(\frac{1}{E I}\left[27(2)+C_1\right]=\frac{1}{E I}\left[-\frac{5}{12}(2)^4+\frac{10}{3}(2)^3+35(2)^2-\frac{419}{3}(2)+C_2\right]\)

Continuity Condition 1: \(C_1-C_2=-\frac{520}{3}\)

Similarly, the translation \(\Delta\) of \(AB\) and \(BC\) is also the same at point \(B\); Or, Equations 4-1 and 4-2 must be alike.

\(\Delta_{A B}=\Delta_{B C}, x=2\)

\(\frac{1}{E I}\left[\frac{27}{2}(2)^2+C_1(2)+C_5\right]=\frac{1}{E I}\left[-\frac{1}{12}(2)^5+\frac{5}{6}(2)^4+\frac{35}{3}(2)^3-\frac{419}{6}(2)^2+C_2(2)+C_6\right]\)

Continuity Condition 2: \(2 C_1-2 C_2+C_5-C_6=-\frac{688}{3}\)

The deflection components at the meeting point must be the same to ensure a continuous deflected shape. There should be no abrupt deflection changes at intersection points between segments (except for special instances).

With this in mind, let's continue establishing conditions. At \(C (6.5, 0)\), the rotation for \(BC\) and \(CD\) are equal (3-2 = 3-3):

\(\theta_{B C}=\theta_{C D}, x=6.5\)

Continuity Condition 3: \(C_2-C_3=-331.84\)

Likewise, the translation at \(C\) for both \(BC\) and \(CD\) is also equal (4-2 = 4-3):

\(\Delta_{B C}=\Delta_{C D}, x=6.5\)

Continuity Condition 4: \(6.5 C_2-6.5 C_3+C_6-C_7=-1859.44\)

The only remaining possible condition is the shared joint between segments \(CD\) and \(DE\). Similarly, the rotation and translation of said segments are equal at \((11, 0)\): (3-3 = 3-4, 4-3 = 4-4)

\(\theta_{C D}=\theta_{D E}, x=11\)

Continuity Condition 5: \(C_3-C_4=-\frac{63525}{8}\)

\(\Delta_{C D}=\Delta_{D E}, x=11\)

Continuity Condition 6: \(11 C_3-11 C_4+C_7-C_8=-\frac{203643}{4}\)

With all possible adjacent segments considered, we have created six conditions; however, we still need two conditions to complete the eight.

Boundary Conditions

The other condition we would need to set up deals with the restrictions of the deflected shape at the supports. For example, fixed supports in ideal cantilevers or restrained beams do not permit any form of deflection.

Look at our roller support at point \(B (2, 0)\). One restriction is that there should be no vertical translation at said point. Using this fact, we can create another condition by using 4-2 to express the following:

\(\Delta_{B_v}=0, x=2\)

\(0=\frac{1}{E I}\left[\frac{27}{2}(2)^2+C_1(2)+C_5\right]\)

Boundary Condition 1: \(2 C_1+C_5=-54\)

You can also use 4-3 to express the roller condition since both segments \(AB\) and \(BC\) share the constraint simultaneously.

\(0=\frac{1}{E I}\left(-\frac{1}{12}(2)^5+\frac{5}{6}(2)^4+\frac{35}{5}(2)^3-\frac{419}{6}(2)^2+C_2(2)+C_6\right)\)

Boundary Condition 2: \(2 C_2+C_6=\frac{526}{3}\)

Let's move on to the hinge support at point \(D (11, 0)\). Likewise, there shouldn't be any translation in this position. We can use either segment \(CD\) or \(DE\) to express this constraint (4-3 or 4-4):

\(\Delta_{D_v}=0, x=11\)

Boundary Condition 3: \(11 C_3+C_7=-\frac{11253}{4}\)

Boundary Condition 4: \(11 C_4+C_8=\frac{96195}{2}\)

We have made four conditions in terms of boundary constraints. Since we already have six conditions from continuity, we only need another two.

We get each per support; in this case, one for the roller at \(B\) and another for the hinge at \(D\).

We can use any of the combinations:

• Boundary Conditions 1 and 3
• Boundary Conditions 1 and 4
• Boundary Conditions 2 and 3
• Boundary Conditions 2 and 4

In our example, we will use Boundary Conditions 1 and 3. 

Solving the Constants of Integration

With eight conditions created, we are ready to solve for the constants simultaneously. 

We can solve systems of linear equations algebraically, but for many unknowns, it is preferable to solve them using matrices. 

\(\begin{bmatrix}1 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\2 & -2 & 0 & 0 & 1 & -1 & 0 & 0 \\0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 \\0 & 6.5 & -6.5 & 0 & 0 & 1 & -1 & 0 \\0 & 0 & 1 & -1 & 0 & 0 & 0 & 0 \\0 & 0 & 11 & -11 & 0 & 0 & 1 & -1 \\2 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\0 & 0 & 11 & 0 & 0 & 0 & 1 & 0\end{bmatrix}
\begin{bmatrix}C_1 \\C_2 \\C_3 \\C_4 \\C_5 \\C_6 \\C_7 \\C_8\end{bmatrix}
=\begin{bmatrix}-\frac{520}{3} \\-\frac{688}{3} \\-331.84 \\-1859.44 \\-\frac{63525}{8} \\-\frac{203643}{4} \\-54 \\-\frac{11253}{4}\end{bmatrix}\)

In the end, we'll have the following results:

\(\begin{bmatrix}C_1 \\C_2 \\C_3 \\C_4 \\C_5 \\C_6 \\C_7 \\C_8\end{bmatrix}=\begin{bmatrix}-877.92 \\-704.59 \\-372.75 \\7567.88 \\1701.84 \\1584.51 \\1287 \\-35149.13\end{bmatrix}\)

Summarize Equations

With constants of integration solved, the final step is to summarise the slope and deflection equations per segment by substituting constants of integration to the general slope and deflection equations:

Slope Equations

\(\frac{d \Delta}{d x}=\theta=\frac{1}{E I}(27 x-877.92),\{0\leq{x}\lt {2}\}\)

\(\frac{d \Delta}{d x}=\theta=\frac{1}{E I}\left(-\frac{5}{12} x^4+\frac{10}{3} x^3+35 x^2-\frac{419}{3} x-704.59\right),\{2\leq{x}\lt {6.5}\}\)

\(\frac{d \Delta}{d x}=\theta=\frac{1}{E I}\left(-6 x^3+\frac{891}{8} x^2-\frac{1629}{4} x-372.75\right),\{6.5\leq{x}\lt {11}\}\)

\(\frac{d \Delta}{d x}=\theta=\frac{1}{E I}\left(45 x^2-1125 x+7567.88\right),\{11\leq{x}\lt {12.5}\}\)

Deflection Equations

\(\Delta=\frac{1}{E I}\left(\frac{27}{2} x^2-877.92 x+1701.84\right),\{0\leq{x}\lt {2}\}\)

\(\Delta=\frac{1}{E I}\left(-\frac{1}{12} x^5+\frac{5}{6} x^4+\frac{35}{3} x^3-\frac{419}{6} x^2-704.59 x+1584.51\right),\{2\leq{x}\lt {6.5}\}\)

\(\Delta=\frac{1}{E I}\left(-\frac{3}{2} x^4+\frac{891}{24} x^3-\frac{1629}{8} x^2-372.75 x+1287\right),\{6.5\leq{x}\lt {11}\}\)

\(\Delta=\frac{1}{E I}\left(15 x^3-\frac{1125}{2} x^2+7567.88 x-35149.13\right),\{11\leq{x}\lt {12.5}\}\)

The equations above are the slope and deflection equations of the beam example per segment. With this, you can determine the rotation or translation at any beam point.

Slope and Deflection Diagrams

We can represent these equations using a graph. If we plot the slope \(\theta\) against position \(x\) and deflection \(\Delta\) against \(x\), we get the slope and deflection diagrams of the structure.

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These diagrams assume the beam example's flexural rigidity \(EI\) is constant.