Question 10.6: A beam ABC (Fig. 10-19a) rests on simple supports at points ......
A beam ABC (Fig. 10-19a) rests on simple supports at points A and B and is supported by a cable at point C. The beam has total length 2L and supports a uniform load of intensity q. Prior to the application of the uniform load, there is no force in the cable nor is there any slack in the cable.
When the uniform load is applied, the beam deflects downward at point C and a tensile force T develops in the cable.
Find the magnitude of this force.
Use a four-step problem-solving approach.
1. Conceptualize:
Redundant force: The structure ABCD, consisting of the beam and cable, has three vertical reactions (at points A, B, and D). However, only two equations of equilibrium are available from a free-body diagram of the entire structure.
Therefore, the structure is statically indeterminate to the first degree, and one redundant quantity must be selected for purposes of analysis.
The tensile force T in the cable is a suitable choice for the redundant.
Release this force by removing the connection at point C, thereby cutting the structure into two parts (Fig. 10-19b). The released structure consists of the beam ABC and the cable CD as separate elements, with the redundant force T acting upward on the beam and downward on the cable.
2. Categorize:
Equation of compatibility: The deflection at point C of beam ABC (Fig. 10-19b) consists of two parts: a downward deflection (\delta_C)_1 due to the uniform load and an upward deflection (\delta_C)_2 due to the force T. At the same time, the lower end C of cable CD displaces downward by an amount (\delta_C)_3, which is equal to the elongation of the cable due to the force T. Therefore, the equation of compatibility, which expresses the fact that the downward deflection of end C of the beam is equal to the elongation of the cable, is
\quad\quad\quad\quad (\delta_{C})_{1}-(\delta_{C})_{2}=(\delta_{C})_{3}\quad\quad(a)
Having formulated this equation, now turn to the task of evaluating all three displacements.
3. Analyze:
Force-displacement relations: The deflection (\delta_C)_1 at the end of the overhang (point C in beam ABC) due to the uniform load can be found from the results given in Example 9-9 of Section 9.5 (see Fig. 9-21). Use Eq. (9-68) of that example and substitute a = L to get
\quad\quad\quad\quad (\delta_{C})_{1}={\frac{q L^{4}}{4E_{b}I_{b}}}\quad\quad(b)
where E_bI_b is the flexural rigidity of the beam.
The deflection of the beam at point C due to the force T can be taken from the answer to Problems 9.8-5 or 9.9-3. Those answers give the deflection (\delta_C)_2 at the end of the overhang when the length of the overhang is a:
\quad\quad\quad\quad(\delta_{C})_{2}=\frac{T a^{2}(L+a)}{3E_{b}I_{b}}
Now substitute a =L to obtain the desired deflection:
\quad\quad\quad\quad(\delta_{C})_{2}={\frac{2T L^{3}}{3E_{b}I_{b}}}\quad\quad(c)
Finally, use the force-displacement relation (Eq. 2-3) to find the elongation of the cable
\quad\quad\quad\quad \delta = \frac{PL}{EA}\quad\quad (2-3)
\quad\quad\quad\quad(\delta_{C})_{3}=\frac{T h}{E_{c}A_{c}}\quad\quad(d)
where h is the length of the cable and E_cA_c is its axial rigidity.
Force in the cable: Substitute the three displacements [Eqs. (b), (c), and (d)] into the equation of compatibility [Eq. (a)] to get
\quad\quad\quad\quad \frac{q L^{4}}{4E_{b}I_{b}}-\frac{2T L^{3}}{3E_{b}I_{b}}=\frac{T h}{E_{c}A_{c}}
Solve for the force T to find
\quad\quad\quad\quad T={\frac{3q L^{4}E_{c}A_{c}}{8L^{3}E_{c}A_{c}+12h E_{b}I_{b}}}\quad\quad(10-39)
4. Finalize: With the force T known, all reactions, shear forces, and bending moments can be found by means of free-body diagrams and equations of equilibrium.
This example illustrates how an internal force quantity (instead of an external reaction) can be used as the redundant.
