Question 10.6: A beam ABC (Fig.10-19a) rests on simple supports at points A......

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 we must select one redundant quantity for purposes of analysis.
The tensile force T in the cable is a suitable choice for the redundant. We can 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.
The deflection at point C of beam ABC (Fig. 10-19b) consists of two parts, a downward deflection (δ_C)_1 due to the uniform load and an upward deflection (δ_C)_2 due to the force T. At the same time, the lower end C of cable CD displaces downward by an amount (δ_C)_3. 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

(δ_C)_1\ -\ (δ_C)_2=(δ_C)_3                                 (w)

Having formulated this equation, we now turn to the task of evaluating all three displacements.
The deflection (δ_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). Using Eq. (9-59) of that example, and substituting a = L. we get

δ_C=\frac{qa}{24EI}(a+L)(3a^2+aL\ -\ L^2)                     (9-59)

(δ_C)_1=\frac{qL^4}{4E_bI_b}                                (x)

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 Problem 9.8-5. which gives the deflection (δ_C)_2 at the end of the overhang when the length of the overhang is a:

(δ_C)_2=\frac{Ta^2(L+a)}{3E_bI_b}

Again substituting a = L, we find

(δ_C)_2=\frac{2TL^3}{3E_bI_b}                         (y)

Finally, the elongation of the cable is

(δ_C)_3=\frac{Th}{E_cA_c}                               (z)

where h is the length of the cable and E_cA_c is its axial rigidity.
By substituting the three displacements (Eqs. x. y. and z) into the equation of compatibility (Eq. w), we get

\frac{qL^4}{4E_bI_b}\ -\ \frac{2TL^3}{3E_bI_b}=\frac{Th}{E_cA_c}

Now solving for the force T, we get

T=\frac{3qL^4E_cA_c}{8L^3E_cA_c+12hE_bI_b}                      (10-34)

With the force T known, we can find all reactions, shear forces, and bending moments 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.