7: Statically Indeterminate Beams and Plane Frames

7.2: Example, Beam Structure 1

Draw the complete free body, shear and moment diagrams for the beam shown in Fig. 7.2-1.

Figure

Figure 7.2-1: Beam Example 1

Following the steps of the detailed procedure of Section 7.1.3, we have the following.

Step 1: Determine Statical Determinacy

There are 4 reaction components and no internal conditions, so this structure is 1 degree statically indeterminate and so 1 redundant must be identified. The complete free body diagram is shown in Fig. 7.2-2.

Figure

Figure 7.2-2: Real Structure and Free Body Diagram

Step 2: Identify the redundants

Choose the vertical force at point c as the only redundant.

The verticals at a or b could have been chosen instead; there is no completely compelling reason to choose c. Experience has shown that the bending moment diagrams are slightly simpler this way, and thus slightly simpler to integrate, but the difference is not large.

Step 3: Analyze the primary structure

Fig. 7.2-3 shows the primary structure (with the redundant force removed).

Figure

Figure 7.2-3: Primary structure with real loads

That figure also shows the results of the analysis - i.e. of determining the reactions and bending moments due to the real loads acting on the primary structure. We will find it slightly more convenient to integrate later if we show the bending moment diagrams for each load separately. $M_{udl}$ are the bending moments due to the uniformly distributed load of 16 kN/m. $M_{cl}$ are the bending moments due to the concentrated load of 60 kN. The total bending moments are the sum of these two.

Step 4: Apply unit values of the redundants

Fig. 7.2-4 shows the statically determinate structure with a unit value of the redundant applied to point c. The unit value is applied in the same direction as was shown for that reaction earlier.

Figure

Figure 7.2-4: Unit value of the redundant

The figure also shows the corresponding reactions and bending moment diagram, $m_1$.

We note that it is probably better to show the reactions in the same direction as we have shown on the previous 3 free body diagrams. In this case, then, we label the value of $v_{b1}$, the center reaction, as -1.75 as it actually would act in the direction opposite to that shown.

Step 5: Compute displacements in the primary structure

Using the unit value of the redundant as the virtual load, the method of virtual work is used to compute the displacement at the redundant due to the real loads on the primary structure:

Figure

Step 6: Compute flexibility coefficients

The displacement due to the unit value of the redundant is calculated next, again using the method of virtual work. Note that the unit load is used both as a virtual load and as a real unit load.

Figure

Figure 7.2-4 Unit value of the redundant

The resulting displacement is called a flexibility coefficient.

Step 7: Write compatibility equations

The compatibility equations are illustrated graphically in Fig. 7.2-5:

Figure

Figure 7.2-5: Compatibility Relationships Illustrated

The specific equation used involves the known displacement (= 0) at point c:

\[\begin{split} 0 &= \Delta_{10} + V_c f_{11}\\ 0 &= -\frac{3136~kNm^3}{EI_0} + V_c \times \frac{120 m^3}{EI_0}\\ \end{split}\]

Step 8: Solve for the unknown redundant forces

Solving the above equation, we obtain:

\[\begin{split} V_c &= 26.133~kN~~~~~(\therefore\uparrow) \end{split}\]

Step 9: Use superposition to determine forces in the real structure

Now, superposition can be used to determine all of the other reactions, and the moment at b:

\[\begin{split} V_a &= V_{a0} + V_c v_{a1}\\ &= 34~kN + 26.133~kN \times 0.75\\ V_a &= 53.60~kN\\ \\ V_b &= V_{b0} + V_c v_{b1}\\ &= 154~kN + 26.133~kN \times -1.75\\ V_b &= 108.27~kN\\ \\ M_b &= -240~kNm + 26.133~kN \times 6~m\\ M_b &= -83.20~kNm \end{split}\]

The summary is shown in Fig. 7.2-6:

Figure

Figure 7.2-6: Example 1 Summary