How to Use the Double Integration Method Using a General Equation?

General Moment Equation

The first step is to create a moment equation encompassing all bending effects of each load. This single equation allows us to find $M$ at any point.

Creating one is similar to forming any moment equation on any segment. To construct it, we have to:

1. Place the cutting plane at the last point of the beam. Doing this would divide the beam into two parts: the left with all loading conditions and the right part, which is the last joint.
2. Choose the left part and create a moment equation for all loads on the side.

The left side of the section has only five loads; we only consider these effects:

• $M_1=27$ is due to the $27kN•m$ couple at $A$
• $M_2=90⟨x-2⟩$ is due to the $90kN$ reaction at $B$
• $M_3=-\frac{5}{3}⟨x-2{⟩}^3+\frac{45}{2}⟨x-6.5{⟩}^2+\frac{5}{3}⟨x-6.5{⟩}^3$ is due to the $45kN/m$ uniform varying load between $B$ and $C$
• $M_4=-18⟨x-6.5{⟩}^2+18⟨x-11{⟩}^2$ is due to the $36kN/m$ uniform distributed load between $C$ and $D$
• $M_5=\frac{1053}{4}⟨x-11⟩$ is due to the $263.25kN$ reaction at $D$

The sum of these will lead us to the general moment equation:

$M_x=\sum _{i=1}^n=M_1+M_2+\cdots +M_n$

$M=27+90⟨x-2⟩-\frac{5}{3}⟨x-2{⟩}^3+\frac{45}{2}⟨x-6.5{⟩}^2+\frac{5}{3}⟨x-6.5{⟩}^3$

$-18⟨x-6.5{⟩}^2+18⟨x-11{⟩}^2+\frac{1053}{4}⟨x-11⟩$

One important thing to note is that this equation must be in this form.

The domain is from $0$ to $L$, which is the beam's length; however, we must note that the terms are explicitly defined.

Explicitly-Defined Terms

To explain what is meant by explicitly defined terms, let's test our equation.

Say that we are interested in finding the moment when $x=1$. One may insert $x=1$ into our general equation to get $M$ at that point, but doing so will lead to a wrong answer. The correct $M$ value is $27kN•m$ based on the moment equation of $AB$.

$27\ne 27+90⟨1-2⟩-\frac{5}{3}⟨1-2{⟩}^3+\frac{45}{2}⟨1-6.5{⟩}^2+\frac{5}{3}⟨1-6.5{⟩}^3$

$-18⟨1-6.5{⟩}^2+18⟨1-11{⟩}^2+\frac{1053}{4}⟨1-11⟩$

To explain what went wrong, let's look at our equation closer. Each term represents $M$ of a load as specified in $M_1$, $M_2$, $M_3$, $M_4$, and $M_5$. If we substitute $x=1$ into the equation and evaluate the result, we would consider the bending effects of ALL loading conditions, which is wrong.

To get the correct one, we should only consider terms relevant to $x=1$ and disregard others. To help us remember this condition, we have intentionally replaced the parenthesis with angle brackets for terms involving the position $x$.

If the value inside the brackets is negative, we don't consider it. So, evaluating the general equation with $x=1$ will lead to:

$M_{x=1}=27$

This result is similar to the moment equation at $AB$.

General Slope Equations

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

$\frac{d\mathrm{\Delta }}{dx}=\frac{1}{EI}(27⟨x⟩+45⟨x-2{⟩}^2-\frac{5}{12}⟨x-2{⟩}^4+\frac{15}{2}⟨x-6.5{⟩}^3+\frac{5}{12}⟨x-6.5{⟩}^4$

$-6⟨x-6.5{⟩}^3+6⟨x-11{⟩}^3+\frac{1053}{8}⟨x-11{⟩}^2+C_1)$

General Deflection Equations

After getting the slope equation, we integrate it again to get the general deflection equation:

$\mathrm{\Delta }=\frac{1}{EI}(\frac{27}{2}⟨x{⟩}^2+15⟨x-2{⟩}^3-\frac{1}{12}⟨x-2{⟩}^5+\frac{15}{8}⟨x-6.5{⟩}^4+\frac{1}{12}⟨x-6.5{⟩}^5$

$-\frac{3}{2}⟨x-6.5{⟩}^4+\frac{3}{2}⟨x-11{⟩}^4+\frac{351}{8}⟨x-11{⟩}^3+C_1⟨x⟩+C_2)$

Conditions

Conditions are constraints or facts that we know about the structure. The number of conditions we need is equal to the number of constants of integration. So, if we have two constants, we need at least two conditions.

Solving the Constants of Integration

With these two conditions, we are now ready to solve for $C_1$ and $C_2$.

Using algebra, we have:

$C_1=-877.92$

$C_2=1701.84$

Slope Equation

$\frac{d\mathrm{\Delta }}{dx}=\frac{1}{EI}(27⟨x⟩+45⟨x-2{⟩}^2-\frac{5}{12}⟨x-2{⟩}^4+\frac{15}{2}⟨x-6.5{⟩}^3+\frac{5}{12}⟨x-6.5{⟩}^4$

$-6⟨x-6.5{⟩}^3+6⟨x-11{⟩}^3+\frac{1053}{8}⟨x-11{⟩}^2-877.92)$

Deflection Equation

$\mathrm{\Delta }=\frac{1}{EI}(\frac{27}{2}⟨x{⟩}^2+15⟨x-2{⟩}^3-\frac{1}{12}⟨x-2{⟩}^5+\frac{15}{8}⟨x-6.5{⟩}^4+\frac{1}{12}⟨x-6.5{⟩}^5$

$-\frac{3}{2}⟨x-6.5{⟩}^4+\frac{3}{2}⟨x-11{⟩}^4+\frac{351}{8}⟨x-11{⟩}^3-877.92⟨x⟩+1701.84)$

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

From this example, we can see the benefits of using this method over the other. We only need one equation to make rather than per segment, and we have fewer conditions to formulate.