7: Statically Indeterminate Beams and Plane Frames
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7.5: Example - Elastic Supports
Figure 7.5-1: Elastic Supports
The beam shown in Fig. 7.5-1 has a fixed support on the left and on the right end is supported by a spring (or elastic support) that provides only a vertical force to the beam end.
It has 4 independent support forces and is thus 1 degree statically indeterminate. We choose the elastic support as the redundant and sketch a free body diagram of the real structure, labelling key forces and displacements.
Note that we have shown the force at b to be acting upward on the beam. To be consistent with this, that means the spring must be in compression to impart that force, and therefore the spring must shorten in order to develop compression. The displacement at point b should be shown downwards to be consistent with this.
Now we draw the primary structure (i.e., with the redundant removed) and determine the key reaction forces and the bending moment diagram.
\[\begin{aligned} \begin{split} H_{a0} &= 0\\ V_{a0} &= 24~kN/m \times 6~m = 144~kN\\ M_{a0} &= 24~kN/m \times 6~m \times \frac{6~m}{2} = 432~kN \end{split} \end{aligned}\]Next, we show the structure with a unit value of the redundant applied. In the real structure, we drew redundant force $V_b$ acting upwards; therefore we apply the unit load in the same direction, as shown. After drawing the reactions in the same direction as they are on the primary structure (and on the real structure)we determine the reactions due to the unit redundant:
\[\begin{aligned} \begin{split} h_{a1} &= 0\\ v_{a1} &= -1\\ m_{a1} &= -1 \times 6~m = -6~m\\ \end{split} \end{aligned}\]Finally, we sketch deflected shapes on all structures. Starting with the unit value of the redundant, plainly that will result in an upward displacement of point b, so that is what we draw.
We use the unit value of the redundant for two purposes. The second is as a virtual load so that we can compute the displacement in the primary structure. Drawing it the way we did (upwards) establishes a sign convention for displacements as being positive upwards. Therefore, that is what we show on the primary structure, even though we 'know' the displacement will be downward. The computations will tell us that eventually, but for now, sketching it this way makes it simpler to develop the superposition equations.
The displacement in the primary structure is:
\[\begin{aligned} \begin{split} \Delta_{10} &= \int \frac{m_1 M}{EI} dx\\ & = \frac{1}{EI}\times\frac{6~m}{4} \times 6~m \times -432~kNm\\ & = \frac{-3888~kNm^3}{EI} \end{split} \end{aligned}\]Substituting numeric values for the beam properties:
\[\begin{split} EI &= 200~kN/mm^2 \times 82.7\times10^6mm^4 \times10^{-6}m^2/mm^2 \\ &= 200\times82.7~kNm^2 \\ &= 16540~kNm^2\\ \Delta_{10} &= \frac{-3888~kNm^3}{16540~kNm^2}\\ \\ \Delta_{10} &= -0.23507~m~~~~~(\therefore\downarrow) \end{split}\]The displacement due to a unit value of the redundant is:
\[\begin{split} f_{11} &= \int \frac{m_1 m_1}{EI} dx\\ &= \frac{1}{EI} \times \frac{6~m}{3} \times 6~m \times 6~m\\ &= \frac{72~m^3}{EI}\\ &= \frac{72~m^3}{16540~kNm^2}\\ \\ f_{11} &= 0.0043531~m/kN\\ \end{split}\]The compatibility equation, noting that, due to the reasoning stated above, the real displacement at b is downward and therefore negative:
\[\begin{split} \Delta_{10} + V_b \times f_{11} &= -\frac{V_b}{K}\\ -0.23507~m + V_b\times 0.0043531~m/kN &= \frac{-V_b}{445~kN/m}\\ V_b\times 0.0066003~m/kN &= 0.23507~m\\ \\ V_b &= 35.62~kN\\ \end{split}\]Determine all other reactions in the real structure by superposition:
\[\begin{split} H_a &= 0\\ \\ V_a &= V_{a0} + V_b v_{a1} \\ &= 144~kN + 35.62~kN\times-1\\ V_a &= 108.38~kN~~~~~(\therefore\uparrow)\\ \\ M_a &= M_{a0} + V_b m_{a1}\\ &= 432~kNm + 35.62~kN\times -6~m \\ M_a &= 218.28~kNm ~~~~~(\therefore CCW)\\ \end{split}\]