Continuing from this example, let's learn how to use the virtual work method to solve deflections. This example presents a case of flexural strains.
The solution presented is in SI. The author will update the post soon to reflect English units.
Let's use the virtual work method for this beam example to get the deflection at \(C\) - rotation and translation.
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The first step is to represent each required deflection component with a unit load and assume its direction. For this example, we'll consider the following:
We apply these unit loadings to each virtual structure, as seen below.
The next step is to find the explicitly defined \(M\), \(E\), and \(I\) equations of the superimposed beam and the bending moment equations of our virtual structures \(m_v\) and \(m_{\alpha}\).
For this illustrative problem, we will not yet consider variations in the beam's flexural rigidity \(EI\); hence, we will assume it is constant.
We place a cutting plane between points \(A\) and \(B\) and consider the left section. The \(M\) and \(m\)-equations for this segment \(\left(0 \le x \lt 2 \right)\) are the following:
\(M_{A B}=27\)
\(m_{v_{A B}}=0\)
\(m_{\alpha_{A B}}=0\)
To determine \(M\) and \(m\) for \(BC\) \(\left(2 \le x \lt 6.5 \right)\), we place a section between said points and again investigate the left part:
\(M_{B C}=-\frac{5}{3} x^3+10 x^2+70 x-\frac{419}{3}\)
\(m_{v_{B C}}=\frac{1}{2}(x-2)\)
\(m_{\alpha_{B C}}=-\frac{1}{9}(x-2)\)
Next, we consider segment \(CD\) \(\left(6.5 \le x \lt 11 \right)\) and investigate the \(M\) and \(m\) equations for it. Like the previous segments, we place a section between \(C\) and \(D\). We can still choose either the left or right part in formulating the equations (depending on what is convenient). For this instance, we will still use the left section:
\(M_{C D}=-18 x^2+\frac{891}{4} x-\frac{1629}{4}\)
\(m_{v_{C D}}=\frac{1}{2}(x-2)-1(x-6.5)\)
\(m_{\alpha_{C D}}=-\frac{1}{9}(x-2)+1\)
Finally, we determine the \(M\) and \(m\)-equations for segment \(DE\) \(\left(11 \le x \lt 12.5 \right)\):
\(M_{D E}=90 x-1125\)
\(m_{v_{D E}}=\frac{1}{2}(x-2)-1(x-6.5)+\frac{1}{2}(x-11)\)
\(m_{\alpha_{D E}}=-\frac{1}{9}(x-2)+1+\frac{1}{9}(x-11)\)
After finding \(M\) and \(m\)-equations for all segments, we can summarise our results using a table.
The summarized table is for the deflection at point \(C\). If we want to find the deflection components at another point, say at \(B\), we need to set up another set of equations and table for it.
The remaining thing to do is to apply the virtual work equations to solve for \(\Delta_{C_v}\) and \(\theta_C\). Let's begin with the vertical translation at \(C\):
\(1 \times \Delta_{C_v}=\int_{x_1}^{x_2} \frac{M m_v}{E I} d x\)
\(1 \times \Delta_{C_v}=\frac{1}{E I} \int_0^2(27)(0) d x+\frac{1}{E I} \int_2^{6.5}\left(-\frac{5}{3} x^3+10 x^2+70 x-\frac{419}{3}\right)\left[\frac{1}{2}(x-2)\right] d x\)
\(+\frac{1}{E I} \int_{6.5}^{11}\left(-18 x^2+\frac{891}{4} x-\frac{1629}{4}\right)\left[\frac{1}{2}(x-2)-1(x-6.5)\right] d x+\frac{1}{E I} \int_{11}^{12.5}(90 x-1125)\)
\(\left[\frac{1}{2}(x-2)-1(x-6.5)+\frac{1}{2}(x-11)\right] d x\)
\(\Delta_{C_v}=\frac{142155}{64 E I}\)
\(\Delta_{C_v}=\frac{2221.17}{E I}\)
Next, we solve for the rotation at C:
\(1 \times \theta_C=\int_{x_1}^{x_2} \frac{M m_\alpha}{E I} d x\)
\(1 \times \Delta_{C_v}=\frac{1}{E I} \int_0^2(27)(0) d x+\frac{1}{E I} \int_2^{6.5}\left(-\frac{5}{3} x^3+10 x^2+70 x-\frac{419}{3}\right)\left[-\frac{1}{9}(x-2)\right] d x\)
\(+\frac{1}{E I} \int_{6.5}^{11}\left(-18 x^2+\frac{891}{4} x-\frac{1629}{4}\right)\left[-\frac{1}{9}(x-2)+1\right] d x+\frac{1}{E I} \int_{11}^{12.5}(90 x-1125)\)
\(\left[-\frac{1}{9}(x-2)+1+\frac{1}{9}(x-11)\right] d x\)
\(\theta_C=-\frac{1215}{32 E I}\)
\(\theta_C=-\frac{37.97}{E I}\)
With translation and rotation at \(C\) solved, the only thing left to do is to verify our assumptions for directions:
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