8. Slope Deflection

8.1 Introduction - Slope Deflection Equations

To use the method of slope-deflection, it is necessary express the end moments of a beam segment as a function of the end rotations, $\theta_a$ and $\theta_b$, and of the relative lateral displacement (i.e., translation) of the ends, $\Delta_{ab}$, and of the applied loads.

Figure 8.1-1: End moments as functions of displacements and loads

In other words, referring to Fig. 8.1-1, we wish to be able to write something like:

\[\begin{split} M_{ab} &= f_1(\theta_a,\theta_b,\Delta_{ab},P,w(x),\dots)\\ M_{ba} &= f_2(\theta_a,\theta_b,\Delta_{ab},P,w(x),\dots)\\ \end{split}\]

If we are able to do that, then we can treat joint displacements in beams and frames as unknowns, write expressions for member end moments in terms of those unknowns, and then write equilibrium equations involving the moments. This would allow us to solve for the unknown displacements, and then use those to calculate the member end moments.

Use Superposition to Simplify the Derivation

Figure 8.1-2: Superposition of three simpler cases

As seen in Fig. 8.1-2, we can make some headway by realizing that the most general case can be made up of the superposition of three simpler cases:

i. the beam segment with the applied loads and fixed ends - case i), plus i. a segment undergoing only end rotations - case ii), plus i. a segment undergoing only end translations - case iii).

We can then write the expressions for the end moments:

\[\begin{split} M_{ab} &= M^f_{ab} + M^r_{ab} + M^d_{ab}\\ M_{ba} &= M^f_{ba} + M^r_{ba} + M^d_{ba} \end{split}\]

Superposition Case ii)

Consider case ii) first. Using the method of virtual work, it is fairly straightforward to express the inverse relationship – the end rotations as functions of the end moments. The relevant details are shown in Fig. 8.1-3.

Figure 8.1-3: Case ii) - end moments as functions of end rotations

In that figure, we show the bending moments corresponding to the applied end moments, $M^r_{ab}$ and $M^r_{ba}$ separately; this simplifies the resulting integration.

Using a virtual moment of 1 applied to end a, the rotation of end a is determined thus:

\[\begin{split} \theta_a &= \int \frac{m_1 M_1}{E I} + \int \frac{m_1 M_2}{E I}\\ &= \frac{1}{EI}\frac{L}{3}\times 1 \times M^r_{ab} + \frac{1}{EI}\frac{L}{6}\times 1 \times -M^r_{ba}\\ \\ \theta_a &= \frac{1}{EI}(\frac{L}{3} M^r_{ab} - \frac{L}{6} M^r_{ba}) \end{split}\]

Similarly, the rotation of end b is given as:

\[\begin{split} \theta_b &= \int \frac{m_2 M_1}{E I} + \int \frac{m_2 M_2}{E I}\\ &= \frac{1}{EI}\frac{L}{6}\times -1 \times M^r_{ab} + \frac{1}{EI}\frac{L}{3}\times -1 \times -M^r_{ba}\\ \\ \theta_b &= \frac{1}{EI}(-\frac{L}{6} M^r_{ab} + \frac{L}{3} M^r_{ba}) \end{split}\]

These two relationships can be expressed in matrix form:

\[\begin{split} \left\{ \begin{array}{c} \theta_a\\ \theta_b \end{array} \right\} = \frac{1}{6EI} \left[ \begin{array}{cc} 2L & -L\\ -L & 2L \end{array} \right] \left\{ \begin{array}{c} M^r_{ab}\\ M^r_{ba} \end{array} \right\} \end{split}\]

from which, by inverting the 2x2 matrix and pre-multiplying both sides by that inverse, we get:

\[\begin{split} \left\{ \begin{array}{c} M^r_{ab}\\ M^r_{ba} \end{array} \right\} = \frac{6EI}{1} \frac{1}{(2L)^2 - (L)^2} \left[ \begin{array}{cc} 2L & L\\ L & 2L \end{array} \right] \left\{ \begin{array}{c} \theta_a\\ \theta_b \end{array} \right\} \end{split}\]

or, simplifying:

\[\begin{split} \left\{ \begin{array}{c} M^r_{ab}\\ M^r_{ba} \end{array} \right\} = \frac{EI}{L} \left\{ \begin{array}{c} 4 \theta_a + 2 \theta_b\\ 2 \theta_a + 4 \theta_b \end{array} \right\} \end{split}\]

The above expresses the end moments as a function of the end rotations.

Superposition Case iii)

Now consider case iii), end translation only:

Figure 8.1-4: Case iii) - end moments as a function of end translation

As we see in Fig. 8.1-4, the beam segment with a translation and no end rotation is equivalent to a segment with a rotation of $\Psi=-\Delta/L$ from the chord. This gives us:

\[\begin{split} \left\{ \begin{array}{c} M^d_{ab}\\ M^d_{ba} \end{array} \right\} = \frac{EI}{L} \left\{ \begin{array}{c} 4 \times -\Psi_{ab} + 2 \times -\Psi{ab}\\ 2 \times -\Psi_{ab} + 4 \times -\Psi{ab} \end{array} \right\} = \frac{EI}{L} \left\{ \begin{array}{c} -6\Psi_{ab}\\ -6\Psi_{ab} \end{array} \right\} \end{split}\]

Summary

Case i) end moments are called ‘fixed-end moments’ and they will be treated separately, below. For now, we will not develop these any further.

Using superposition of all three cases, we can now express the end moments in terms of displacements and applied loads:

\[\begin{split} M_{ab} &= M^f_{ab} + M^r_{ab} + M^d_{ab}\\ &= M^f_{ab} + \frac{EI}{L}(4\theta_a+2\theta_b) + \frac{EI}{L}(-6\Phi_{ab})\\ M_{ab} &= \frac{EI}{L}(4\theta_a + 2\theta_b - 6\frac{\Delta_{ab}}{L}) + M^f_{ab} \end{split}\] \[\begin{split} M_{ba} &= M^f_{ba} + M^r_{ba} + M^d_{ba}\\ &= M^f_{ba} + \frac{EI}{L}(2\theta_a+4\theta_b) + \frac{EI}{L}(-6\Phi_{ab})\\ M_{ba} &= \frac{EI}{L}(2\theta_a + 4\theta_b - 6\frac{\Delta_{ab}}{L}) + M^f_{ba} \end{split}\] \[\begin{split} M_{ab} &= \frac{EI}{L}(4\theta_a + 2\theta_b - 6\frac{\Delta_{ab}}{L}) + M^f_{ab}\\ M_{ba} &= \frac{EI}{L}(2\theta_a + 4\theta_b - 6\frac{\Delta_{ab}}{L}) + M^f_{ba}\\ \end{split}\]

where $M^f_{ab}$ and $M^f_{ba}$ are the fixed-end moments due to the applied loads.