Let's learn how to use the real 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 consider a simple beam shown. It only consists of an \(81 kN\) point load applied \(6m\) from the left end. Given already the reactions, we shall see how to use the general real work equation to solve for the vertical translation at \(D\).
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The first thing we have to do is to express \(M\), \(E\), and \(I\) as a function of position \(x\) per segment.
Creating \(M\) expressions is similar to constructing moment equations. Its flexural rigidity \(EI\) varies per segment but not per position \(x\). We can see how we can formulate these expressions in the following section:
The following table shows the summary of each variable per segment.
At this point, we apply the general real work equation to solve for the deflection. Using the table above as a reference:
\(P \times \Delta_{D_v}=\int \frac{M^2}{E I} d x\)
\(81^{k N} \times \Delta_{D_v}=\frac{1}{E I} \int_0^5(27 x)^2 d x+\frac{1}{E(2 I)} \int_5^6(27 x)^2 d x+\frac{1}{E(2 I)} \int_6^9[27 x-81(x-6)]^2 d x\)
\(\Delta_{D_v}=\frac{1347}{2 E I}\)
The positive sense of our answer indicates that the vertical translation matches the direction of the applied force. Meaning at point \(D\), the beam deflected downwards from its original position.
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