7: Statically Indeterminate Beams and Plane Frames

7.3: Example, Beam Structure 2 - Alternate Redundants

Draw shear force and bending moment diagrams for the following structure.

Figure

Figure 7.3-1: Real structure

This example will illustrate the choice of internal bending moments as redundants.

1: Determine statical determinacy

The structure is one degree statically indeterminate. The complete free body diagram is shown in Fig. 7.3-2.

Figure

Figure 7.3-2: Real structure, free body diagram

2: Choose the redundants

For this example, we choose the internal bending moment at the center support, point b, as the redundant.

Releasing this redundant means releasing the constraint against rotational discontinuity at that point, and is essentially the same as inserting a hinge at that point.

3: Analyze the primary structure under real loads

Fig. 7.3-3 shows the primary structure under the real loads and the bending moments that result from this.

Figure

Figure 7.3-3: Determinate structure, real loads

An analysis using equilibrium would determine the following reactive forces on the primary structure:

\[\begin{split} H_{a0} &= 0\\ \\ V_{a0} &= wL/2\\ \\ V_{b0} &= wL\\ \\ V_{c0} &= wL/2 \end{split}\]

Also seen in the Fig. 7.3-3 is the deflected shape showing the rotational discontinuity, $\theta_{10}$, introduced by releasing the redundant.

4: Apply a unit value of the redundant

Fig. 7.3-4 shows the determinate structure with unit values of the redundants imposed. Note that internal bending moments always occur in equal and opposite pairs at a point, so we must apply an equal and opposite pair of unit moments.

Figure

Figure 7.3-4: Determinate structure, unit redundant

Equilibrium will tell us that:

\[\begin{split} h_{a1} &= 0\\ \\ v_{a1} &= 1/L\\ \\ v_{b1} &= -2/L~~~~~(\therefore\downarrow)\\ \\ v_{c1} &= 1/L \end{split}\]

5: Displacements in the primary structure

Using the unit value structure as the virtual system and integrating the product of the moment diagrams, taking advantage of symmetry about point b, we get a value for the relative rotation of the tangents to the elastic curve at point b in the primary structure:

\[\begin{split} \theta_{10} &= \int \frac{m_1 M}{E I} dx\\ &= 2 \times \frac{1}{EI} \times \frac{L}{3}\times 1\times\frac{wL^2}{8}\\ \\ \theta_{10} &= \frac{w L^3}{12 E I} \end{split}\]

6: Flexibility coefficients

The angle of rotation between the tangents in the released structure due to a unit value of the redundant is:

\[\begin{split} f_{11} &= \int \frac{m_1 m_1}{E i} dx\\ &= 2\times\frac{1}{EI}\times\frac{L}{3}\times 1\times 1\\ \\ f_{11} &= \frac{2 L}{3 E I} \end{split}\]

7: Compatibility equation

There is no rotational discontinuity in the real structure. In other words, $\theta_1=0$. Fig. 7.3-5 shows the graphical compatibilty ‘equation’ that expresses the real structure as the sum of the primary structure with real loads plus the released structure with the redundant applied.

Figure

Figure 7.3-5: Superposition of Structures

We can express that by superposition algebraically thus:

\[0 = \theta_{10} + M_b f_{11}\]

where $M_b$ is the currently unknown real bending moment in the real structure.

8: Solve for the unknown redundant

\[\begin{split} 0 &= \frac{w L^3}{12 E I} + M_b \frac{2 L}{3 E I}\\ \\ M_b &= -\frac{w L^2}{8}~~~~~(\therefore \mathrm{compr.~on~bottom}) \end{split}\]

9: Determine reactions by superposition

The reactions in the real structure can now be determined by superposition.

\[\begin{split} H_a &= 0\\ \\ V_a &= V_{a0} + M_b v_{a1}\\ &= \frac{w L}{2} - \frac{w L^2}{8}\times\frac{1}{L}\\ V_a &= \frac{3 w L}{8}\\ \\ V_b &= V_{b0} + M_b v_{a2}\\ &= w L + -\frac{w L^2}{8}\times -\frac{2}{L}\\ V_b &= \frac{5 w L}{4}\\ \\ V_c &= V_a\\ V_c &= \frac{3 w L}{8} \end{split}\]

Figure

Figure 7.3-6: Summary Results