Continuing from this example, let's learn how to use the area moment method to determine the maximum deflection.
The solution presented is in SI. The author will update the post soon to reflect English units.
In this example, we are to find the maximum deflection between the roller \(B\) and hinge support \(D\) using the area-moment method.
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To start, we would need to locate where between \(B\) and \(D\) is \(\Delta_{max}\) positioned.
Since there are two loading conditions between \(B\) and \(D\), we don't know if \(\Delta_{max}\) is between \(B\) and \(C\) or \(C\) and \(D\). What we will do here is to assume first where it might be, then compute for \(x\). Afterward, we assess whether our assumption is correct.
Using the figure above, let's first assume that the maximum translation occurs between points \(B\) and \(C\). For ease, we'll consider the moment-by-parts diagram.
Recall that \(\Delta_{max}\) will occur at the point of zero rotation (when \(\theta = 0\), \(\Delta = \Delta_{max}\)). With any computed rotation, say \(\theta_B\), we use the first theorem of the area moment method to verify our assumption.
\(\theta_{\Delta_{\max }}-\theta_B=\int_{x_1}^{x_2} \frac{M}{E I} d x\)
\(0-\left(-\frac{823.92}{E I}\right)=\frac{1}{E I}\left[27(x-2)+\frac{1}{2}(405)(x-2)\right.\left.-\frac{1}{4}(151.875)(x-2)\right]\)
\(x=6.3\)
With \(x\) solved, we evaluate if this assumption is correct. Since it is within the limits of segment \(BC \left(2 \leq x \lt 6.5\right)\), we can say that the maximum deflection occurs at \(x = 6.3\). If it is outside of the limits, then our guess is wrong, and we would need to solve for \(x\) again with the new assumption that it is between other segments.
Now that we know where \(\Delta_{max}\) is, we now solve for it. Solving for maximum translation is the same as solving any vertical translation. From the figure, we can derive relationships for \(\Delta_{max}\):
Equation 1-1: \(\frac{F F^{\prime \prime}}{4.3}=\frac{D D^{\prime \prime}}{9}\)
Equation 1-2: \(F F^{\prime \prime}=\left(-\Delta_{\max }\right)+\left(+t_{F / B}\right)\)
Equation 1-3: \(D D^{\prime \prime}=\left(+t_{D / B}\right)\)
Afterward, we solve for \(t_{F/B}\). We'll use the second theorem to solve for the deviation of \(F\) with reference tangent at \(B\):
\(t_{F / B}=\frac{1}{E I}\left[(4.3 \times 27)\left(\frac{1}{2} \times 4.3\right)+\left(\frac{1}{2} \times 4.3 \times 387\right)\left(\frac{1}{3} \times 4.3\right)\right.\left.-\left(\frac{1}{4} \times 4.3 \times 132.512\right)\left(\frac{1}{5} \times 4.3\right)\right]\)
\(t_{F / B}=\frac{1319}{E I}\)
Since \(t_{F/B}\) is positive, it confirms our sketch that \(F^{\prime \prime}\) must be above the reference tangent drawn at \(B\).
The only thing left to do is to apply Equations 1-1 to 1-3 to solve for the maximum deflection:
\(\frac{\left(-\Delta_{\max }\right)+\left(+t_{F / B}\right)}{4.3}=\frac{\left(+t_{D / B}\right)}{9}\)
\(\Delta_{\max }=-\left(\frac{4.3}{9} \times \frac{474579}{64 E I}-\frac{1319}{E I}\right)\)
\(\Delta_{\max }=-\frac{2223.86}{E I}\)
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