Set Up General Equations
In this method, we will analyze the beam as a whole. To start, we would need to formulate general \(M\), \(E\), and \(I\) equations - expressions that would include all bending effects, changes in material, and cross-section properties for the beam.
Since we have assumed that \(E\) and \(I\) are constant for the whole beam, we'll only concentrate on creating a general moment \(M\) 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:
- 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.
- 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 \(27 kN•m\) couple at \(A\)
- \(M_2=90\langle x-2\rangle\) is due to the \(90kN\) reaction at \(B\)
- \(M_3=-\frac{5}{3}\langle x-2\rangle^3+\frac{45}{2}\langle x-6.5\rangle^2+\frac{5}{3}\langle x-6.5\rangle^3\) is due to the \(45 kN/m\) uniform varying load between \(B\) and \(C\)
- \(M_4=-18\langle x-6.5\rangle^2+18\langle x-11\rangle^2\) is due to the \(36 kN/m\) uniform distributed load between \(C\) and \(D\)
- \(M_5=\frac{1053}{4}\langle x-11\rangle\) is due to the \(263.25 kN\) reaction at \(D\)
The sum of these will lead us to the general moment equation:
\(M_x=\sum_{i=1}^n=M_1+M_2+\cdots+M_n\)
\(M=27+90\langle x-2\rangle-\frac{5}{3}\langle x-2\rangle^3+\frac{45}{2}\langle x-6.5\rangle^2+\frac{5}{3}\langle x-6.5\rangle^3\)
\(-18\langle x-6.5\rangle^2+18\langle x-11\rangle^2+\frac{1053}{4}\langle x-11\rangle\)
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 \(AB\).
\(27 \neq 27+90\langle 1-2\rangle-\frac{5}{3}\langle 1-2\rangle^3+\frac{45}{2}\langle 1-6.5\rangle^2+\frac{5}{3}\langle 1-6.5\rangle^3\)
\(-18\langle 1-6.5\rangle^2+18\langle 1-11\rangle^2+\frac{1053}{4}\langle 1-11\rangle\)
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 \(AB\).
Apply Bernoulli-Euler Beam Equation
With the general equation set-up, we can apply the Bernoulli-Euler Beam Equation. If we have any functions of \(E\) and \(I\), we should also consider it here:
\(\frac{d^2 \Delta}{d x^2}=\frac{1}{E I}(27+90\langle x-2\rangle-\frac{5}{3}\langle x-2\rangle^3+\frac{45}{2}\langle x-6.5\rangle^2+\frac{5}{3}\langle x-6.5\rangle^3\)
\(-18\langle x-6.5\rangle^2+18\langle x-11\rangle^2+\frac{1053}{4}\langle x-11\rangle)\)
Find General Solution
Afterward, we are ready to find the general slope and deflection equations.
General Slope Equations
Let's start by finding the general slope equation. We integrate the Bernoulli-Euler Beam Equation once to get it:
\(\frac{d \Delta}{d x}=\frac{1}{E I}\left(27\langle x\rangle+45\langle x-2\rangle^2-\frac{5}{12}\langle x-2\rangle^4+\frac{15}{2}\langle x-6.5\rangle^3+\frac{5}{12}\langle x-6.5\rangle^4\right.\)
\(\left.-6\langle x-6.5\rangle^3+6\langle x-11\rangle^3+\frac{1053}{8}\langle x-11\rangle^2+C_1\right)\)
General Deflection Equations
After getting the slope equation, we integrate it again to get the general deflection equation:
\(\Delta=\frac{1}{E I}\left(\frac{27}{2}\langle x\rangle^2+15\langle x-2\rangle^3-\frac{1}{12}\langle x-2\rangle^5+\frac{15}{8}\langle x-6.5\rangle^4+\frac{1}{12}\langle x-6.5\rangle^5\right.\)
\(\left.-\frac{3}{2}\langle x-6.5\rangle^4+\frac{3}{2}\langle x-11\rangle^4+\frac{351}{8}\langle x-11\rangle^3+C_1\langle x\rangle+C_2\right)\)
Find Particular Solution
In this example, we need to solve only two constants of integration \(C_1\) and \(C_2\). As we did in segment analysis, we create conditions first.
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.
Boundary Conditions
Boundary conditions deal with restrictions at the supports. For example, fixed supports in ideal cantilevers or restrained beams do not permit deflection.
Let's continue with our analysis. We know the vertical translation at the roller \(B (2, 0)\) is restricted. We can use this to create one condition.
\(\Delta_{B_v}=0, x=2\)
Before evaluating, it's essential to note that the terms are explicitly defined. We must disregard some terms if the values inside the brackets are negative.
\(0=\frac{1}{E I}\left(\frac{27}{2}\langle 2\rangle^2+15\langle 2-2\rangle^3-\frac{1}{12}\langle 2-2\rangle^5+\frac{15}{8}\langle 2-6.5\rangle^4+\frac{1}{12}\langle 2-6.5\rangle^5\right.\)
\(\left.-\frac{3}{2}\langle 2-6.5\rangle^4+\frac{3}{2}\langle 2-11\rangle^4+\frac{351}{8}\langle 2-11\rangle^3+C_1\langle 2\rangle+C_2\right)\)
Boundary Condition 1: \(2 C_1+C_2=-54\)
For the other condition, we look at \(D (11, 0)\) or the hinge support. Since vertical translation is not allowed, we use this to form the other condition:
\(\Delta_{D_v}=0, x=11\)
\(0=\frac{1}{E I}\left(\frac{27}{2}\langle 11\rangle^2+15\langle 11-2\rangle^3-\frac{1}{12}\langle 11-2\rangle^5+\frac{15}{8}\langle 11-6.5\rangle^4+\frac{1}{12}\langle 11-6.5\rangle^5\right.\)
\(\left.-\frac{3}{2}\langle 11-6.5\rangle^4+\frac{3}{2}\langle 11-11\rangle^4+\frac{351}{8}\langle 11-11\rangle^3+C_1\langle 11\rangle+C_2\right)\)
Boundary Condition 2: \(11 C_1+C_2=\frac{509139}{64}\)
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:
\(C_1=-877.92\)
\(C_2=1701.84\)
Summarize Equations
With constants of integration solved, the final step is to summarize the equations by substituting constants of integration:
Slope Equation
\(\frac{d \Delta}{d x}=\frac{1}{E I}\left(27\langle x\rangle+45\langle x-2\rangle^2-\frac{5}{12}\langle x-2\rangle^4+\frac{15}{2}\langle x-6.5\rangle^3+\frac{5}{12}\langle x-6.5\rangle^4\right.\)
\(\left.-6\langle x-6.5\rangle^3+6\langle x-11\rangle^3+\frac{1053}{8}\langle x-11\rangle^2-877.92\right)\)
Deflection Equation
\(\Delta=\frac{1}{E I}\left(\frac{27}{2}\langle x\rangle^2+15\langle x-2\rangle^3-\frac{1}{12}\langle x-2\rangle^5+\frac{15}{8}\langle x-6.5\rangle^4+\frac{1}{12}\langle x-6.5\rangle^5\right.\)
\(\left.-\frac{3}{2}\langle x-6.5\rangle^4+\frac{3}{2}\langle x-11\rangle^4+\frac{351}{8}\langle x-11\rangle^3-877.92\langle x\rangle+1701.84\right)\)
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.
General and Segment Equations
Let's recall the beam limits per segment.
- AB: \(0 \leq {x} \lt 2\)
- BC: \(2 \leq {x} \lt 6.5\)
- CD: \(6.5 \leq {x} \lt 11\)
- DE: \(11 \leq {x} \lt 12.5\)
We can use these limits to find the segment equation from the general expression.
As an example, let's evaluate the general slope equation with \(BC: 2 \leq {x} \lt 6.5\) as our limits:
\(\frac{d \Delta}{d x}=\frac{1}{E I}\left(27\langle x\rangle+45\langle x-2\rangle^2-\frac{5}{12}\langle x-2\rangle^4+\frac{15}{2}\langle x-6.5\rangle^3+\frac{5}{12}\langle x-6.5\rangle^4\right.\)
\(\left.-6\langle x-6.5\rangle^3+6\langle x-11\rangle^3+\frac{1053}{8}\langle x-11\rangle^2-877.92\right)\)
\(\frac{d \Delta}{d x}=\theta=\frac{1}{E I}\left(-\frac{5}{12} x^4+\frac{10}{3} x^3+35 x^2-\frac{419}{3} x-704.59\right)\)
This result is similar to the segmented slope equation.
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.
These diagrams assume the beam example's flexural rigidity \(EI\) is constant.
