How to Use Partial Derivative Method Due to Flexural Strains?

Set-Up Placeholder

The first step is introducing a placeholder load at point \(E\). Since we are looking for its translation, we place a point load \(F\). If we are looking for rotation at a point, we introduce a placeholder couple instead.

The direction of \(F\) will correspond to our assumed direction translation at \(E\), downwards.

Reactions

Afterwards, we need to solve for the reactions of the beam example. In this case, we'll need to express it in terms of the placeholder load \(F\)

\(\sum{M_B}=0]\space{\circlearrowright_+}\)

\(R_A(9)-F(9-2)-81(3)=0\)

\(R_A=27+\frac{7}{9}F\)

\(\sum{F_v}=0]\space{\uparrow_+}\)

\(R_A-F-81+D_v=0\)

\(27+\frac{7}{9}F-F-81+D_v=0\)

\(D_v=54+\frac{2}{9}F\)

Formulate M, E, and I Equations

With a placeholder introduced, we formulate moment equations \(M\), \(E\), and \(I\) based on the real and placeholder loads.

Creating \(M\) expressions is similar to constructing moment equations. Its flexural rigidity \(EI\) varies per segment. We can see how we can formulate these expressions in the following section:

Segment AE

• \(M_{AE}=\left(27+\frac{7}{9} F\right)(x)\)
• \(E_{AE}=E\)
• \(I_{AE}=I\)

Segment EC

• \(M_{EC}=\left(27+\frac{7}{9} F\right)(x)-F(x-2)\)
• \(E_{EC}=E\)
• \(I_{EC}=I\)

Segment CD

• \(M_{CD}=\left(27+\frac{7}{9} F\right)(x)-F(x-2)\)
• \(E_{CD}=E\)
• \(I_{CD}=2I\)

Segment DB

• \(M_{DB}=\left(27+\frac{7}{9} F\right)(x)-F(x-2)-81(x-6)\)
• \(E_{DB}=E\)
• \(I_{DB}=2I\)

Summarize Equations

The following table shows the summary of each variable per segment.

Apply Partial Derivative Equations

At this point, we apply the partial derivative method to solve for the deflection at \(E\).

As the name suggests, we need to find the partial derivative of the moment equation with respect to the placeholder load:

Segment AE

\(\frac{\partial M}{\partial F}=\frac{7}{9} x\)

Segment EC

\(\frac{\partial M}{\partial F}=-\frac{2}{9} x+2\)

Segment CD

\(\frac{\partial M}{\partial F}=-\frac{2}{9} x+2\)

Segment DB

\(\frac{\partial M}{\partial F}=-\frac{2}{9} x+2\)

Afterward, we apply the partial derivative equation to solve for the deflection:

\(\Delta_{E_v}=\int \frac{M}{E I} \times \frac{\partial M}{\partial F} d x\)

\(\Delta_{E_v}=\frac{1}{E I} \int_0^2\left(27+\frac{7}{9} F\right)(x)\left(\frac{7}{9} x\right) d x+\frac{1}{E I} \int_2^5\left[\left(27+\frac{7}{9} F\right)(x)-F(x-2)\right]\left(-\frac{2}{9} x+2\right) d x\)

\(+\frac{1}{E(2 I)} \int_5^6\left[\left(27+\frac{7}{9} F\right)(x)-F(x-2)\right]\left(-\frac{2}{9} x+2\right) d x+\frac{1}{E(2 I)} \int_6^9[\left(27+\frac{7}{9} F\right)(x)\)

\(-F(x-2)-81(x-6)]\left(-\frac{2}{9} x+2\right) d x\)

Notice that the translation at \(E\) is in terms of two variables: the placeholder load \(F\) and \(x\). We cannot directly evaluate the integral because of it; however, we recall that the purpose of \(F\) is to represent the required deflection. So, we let \(F = 0\) to have:

\(\Delta_{E_v}=\frac{1}{E I} \int_0^2(27)(x)\left(\frac{7}{9} x\right) d x+\frac{1}{E I} \int_2^5(27)(x)\left(-\frac{2}{9} x+2\right) d x\)

\(+\frac{1}{E(2 I)} \int_5^6(27)(x)\left(-\frac{2}{9} x+2\right) d x+\frac{1}{E(2 I)} \int_6^9\left[(27(x)-81(x-6)]\left(-\frac{2}{9} x+2\right) d x\right.\)

\(\Delta_{E_v}=\frac{1001}{2 E I}=\frac{500.5}{EI}\)

The positive direction indicates that the translation at \(E) is downward, which was our initial assumption.