Continuing from this example, let's learn how to use the double integration method to evaluate the beam's deflection.
The solution presented is in SI. The author will update the post soon to reflect English units.
Slope and Deflection Equations
To evaluate deflections using this method, we need the slope and deflection equations which we can obtain either by segmented analysis or the general moment equation.
Below are the equations using segment analysis:
Segment Slope Equations
\(\frac{d \Delta}{d x}=\theta=\frac{1}{E I}(27 x-877.92),\{0\leq{x}\lt {2}\}\)
\(\frac{d \Delta}{d x}=\theta=\frac{1}{E I}\left(-\frac{5}{12} x^4+\frac{10}{3} x^3+35 x^2-\frac{419}{3} x-704.59\right),\{2\leq{x}\lt {6.5}\}\)
\(\frac{d \Delta}{d x}=\theta=\frac{1}{E I}\left(-6 x^3+\frac{891}{8} x^2-\frac{1629}{4} x-372.75\right),\{6.5\leq{x}\lt {11}\}\)
\(\frac{d \Delta}{d x}=\theta=\frac{1}{E I}\left(45 x^2-1125 x+7567.88\right),\{11\leq{x}\lt {12.5}\}\)
Segment Deflection Equations
\(\Delta=\frac{1}{E I}\left(\frac{27}{2} x^2-877.92 x+1701.84\right),\{0\leq{x}\lt {2}\}\)
\(\Delta=\frac{1}{E I}\left(-\frac{1}{12} x^5+\frac{5}{6} x^4+\frac{35}{3} x^3-\frac{419}{6} x^2-704.59 x+1584.51\right),\{2\leq{x}\lt {6.5}\}\)
\(\Delta=\frac{1}{E I}\left(-\frac{3}{2} x^4+\frac{891}{24} x^3-\frac{1629}{8} x^2-372.75 x+1287\right),\{6.5\leq{x}\lt {11}\}\)
\(\Delta=\frac{1}{E I}\left(15 x^3-\frac{1125}{2} x^2+7567.88 x-35149.13\right),\{11\leq{x}\lt {12.5}\}\)
General Slope Equation
\(\frac{d \Delta}{d x}=\frac{1}{E I}(27\langle x\rangle+45\langle x-2\rangle^2-\frac{5}{12}\langle x-2\rangle^4+\frac{15}{2}\langle x-6.5\rangle^3+\frac{5}{12}\langle x-6.5\rangle^4\)
\(-6\langle x-6.5\rangle^3+6\langle x-11\rangle^3+\frac{1053}{8}\langle x-11\rangle^2-877.92)\)
General Deflection Equation
\(\Delta=\frac{1}{E I}(\frac{27}{2}\langle x\rangle^2+15\langle x-2\rangle^3-\frac{1}{12}\langle x-2\rangle^5+\frac{15}{8}\langle x-6.5\rangle^4+\frac{1}{12}\langle x-6.5\rangle^5-\frac{3}{2}\langle x-6.5\rangle^4\)
\(+\frac{3}{2}\langle x-11\rangle^4+\frac{351}{8}\langle x-11\rangle^3-877.92\langle x\rangle+1701.84)\)
Evaluating Deflections
Now that we have the equations, we'll illustrate how to find the deflection at specific points.
Which set of equations should I use? The segment equations or general equations? It doesn't matter because both groups are the same. The difference lies in how we evaluate the expression.
- If we have segment equations, we have a specific domain. We must only substitute \(x\) within the said domain.
- If we use general equations, we have explicitly defined terms. We must neglect particular terms if the value inside the brackets is negative.
Say we're interested in finding the translation at \(x = 1.8m\). Using the segment deflection equation, we substitute and evaluate the following:
\(\theta_{x=1.8}=\frac{1}{E I}(27(1.8)-877.92), 0 \leq x<2\)
\(\theta_{x=1.8}=\frac{1}{E I}(-829.32)\)
If we use the general equation, we have the following:
\(\Delta=\frac{1}{E I}(\frac{27}{2}\langle 1.8\rangle^2+15\langle 1.8-2\rangle^3-\frac{1}{12}\langle 1.8-2\rangle^5+\frac{15}{8}\langle 1.8-6.5\rangle^4+\frac{1}{12}\langle 1.8-6.5\rangle^5-\frac{3}{2}\langle 1.8-6.5\rangle^4\)
\(+\frac{3}{2}\langle 1.8-11\rangle^4+\frac{351}{8}\langle 1.8-11\rangle^3-877.92\langle 1.8\rangle+1701.84)\)
\(\theta_{x=1.8}=\frac{1}{E I}(-829.32)\)
Slope and Deflection Diagrams
Just like axial \(N\), shear \(V\), moment \(M\), and torsion \(T\) diagrams, we can graphically express the deflection characteristics of a structure.
The following shows the graphical representation of these components.
In this example, the flexural rigidity \(EI\) of the beam example is constant throughout its entire length.
Plotting the slope equations will lead you to the slope diagram \(\theta\). If you look closely, a relative maximum occurs at zero moment point \(x = 10.145\). It signifies a point where there is a change in curvature in the deflected shape.
Similarly, plotting the deflection equations will let you draw the deflection diagram. It is the actual shape of the deflected beam when subjected to the given load.
In structural design, we're usually interested in finding deflection at critical points, such as the point of maximum deflection. From calculus, a maximum occurs when the tangent drawn to it is zero; hence, the maximum deflection will appear at the zero rotation point \(\theta = 0\). If you look at the figure, it will occur when \(x = 6.365\) with a value of \(-2223.75 / EI\).
