Question 10.4: A fixed-end beam AB (Fig. 10-15a) is loaded by a force P act......

This beam has four unknown reactions (a force and a moment at each support), but only two independent equations of equilibrium are available. Therefore, the beam is statically indeterminate to the second degree. In this example, we will select the reactive moments M_A and M_B as the redundants.
Equations of equilibrium. The two unknown force reactions (R_A and R_B) can be expressed in terms of the redundants (M_A and M_B) with the aid of two equations of equilibrium. The first equation is for moments about point B, and the second is for moments about point A. The resulting expressions are

R_A=\frac{Pb}{L}+\frac{M_A}{L}\ -\ \frac{M_B}{L}\quad \quad R_B=\frac{Pa}{L}\ -\ \frac{M_A}{L}+\frac{M_B}{L}                 (n,o)

removing the rotational restraints at the ends of the beam, we are left with a simple beam as the released structure (Figs. 10-15b, c, and d). The angles of rotation at the ends of the released structure due to the concentrated load P are denoted (θ_A)_1 and (θ_B)_1, as shown in Fig. 10- 15b. In a similar manner, the angles at the ends due to the redundant M_A are denoted (θ_A)_2 and (θ_B)_2, and the angles due to the redundant M_B are denoted (θ_A)_3 and (θ_A)_3.
Since the angles of rotation at the supports of the original beam are equal to zero, the two equations of compatibility are

θ_A=(θ_A)_1\ -\ (θ_A)_2\ -\ (θ_A)_3 = 0                           (p)

θ_B=(θ_B)_1\ -\ (θ_B)_2\ -\ (θ_B)_3 = 0                           (q)

in which the signs of the various terms are determined by inspection from the figures.
Force-displacement relations. The angles at the ends of the beam due to the load P (Fig. 10-15b) are obtained from Case 5 of Table G-2:

(θ_A)_1=\frac{Pab(L+b)}{6LEI}\quad \quad (θ_B)_1=\frac{Pab(L+a)}{6LEI}

in which a and h are the distances from the supports to point D where the load is applied.
Also, the angles at the ends due to the redundant moment M_A are (see Case 7 of Table G-2):

(θ_A)_2=\frac{M_AL}{3EI}\quad \quad (θ_B)_2=\frac{M_AL}{6EI}

Similarly, the angles due to the moment M_B are

(θ_A)_3=\frac{M_BL}{6EI}\quad \quad (θ_B)_3=\frac{M_BL}{3EI}

When the preceding expressions for the angles are substituted into the equations of compatibility (Eqs. p and q), we arrive at two simultaneous equations containing M_A and M_B as unknowns:

\frac{M_AL}{3EI}+\frac{M_BL}{6EI}=\frac{Pab(L+b)}{6LEI}                             (r)

\frac{M_AL}{6EI}+\frac{M_BL}{3EI}=\frac{Pab(L+a)}{6LEI}                             (s)

Solving these equations for the redundants, we obtain

M_A=\frac{Pab^2}{L^2}\quad \quad M_B=\frac{Pa^2b}{L^2}                                    (10-25a,b)

Substituting these expressions for M_A and M_B into the equations of equilibrium (Eqs. n and o), we obtain the vertical reactions:

R_A=\frac{Pb^2}{L^3}(L+2a)\quad \quad R_B=\frac{Pa^2}{L^3}(L+2b)                           (10-26a,b)

Thus, all reactions for the fixed-end beam have been determined.
The reactions at the supports of a beam with fixed ends are commonly referred to as fixed-end moments and fixed-end forces. They are widely used in structural analysis, and formulas for these quantities are occasionally listed in engineering handbooks.
Deflection at point D. To obtain the deflection at point D in the original fixed-end beam (Fig. 10-15a), we again use the principle of superposition. The deflection at point D is equal to the sum of three deflections: (1) the downward deflection (δ_D)_1 at point D in the released structure due to the load P (Fig. 10-15b); (2) the upward deflection (δ_D)_1 at the same point in the released structure due to the redundant M_A (Fig. 10-15c), and (3) the upward deflection (δ_D)_3 at the same point in the released structure due to the redundant M_B (Fig. 10-15d). This superposition of deflections is expressed by the following equation:

δ_D=(δ_D)_1\ -\ (δ_D)_2\ -\ (δ_D)_3                           (t)

in which δ_D is the downward deflection in the original beam.
The deflections appearing in Eq. (t) are obtained from the formulas given in Table G-2 of Appendix G (see Cases 5 and 7) by substituting x = a and simplifying. The results are

(δ_D)_1=\frac{Pa^2b^2}{3LEI}\quad \quad (δ_D)_2=\frac{M_Aab}{6LEI}(L+b)\quad \quad (δ_D)_3=\frac{M_Bab}{6LEI}(L+a)

Substituting the expressions for M_A and M_B from Eqs. (10-25a and b) into the last two expressions, we get

(δ_D)_2=\frac{Pa^2b^3}{6L^3EI}(L+b)\quad \quad (δ_D)_3=\frac{Pa^3b^2}{6L^3EI}(L+a)

Therefore, the deflection at point D in the original beam, obtained by substituting (δ_D)_1. (δ_D)_2, and (δ_D)_3 into Eq. (t) and simplifying, is

δ_D=\frac{Pa^3b^3}{3L^3EI}                                       (10-27)

The method described above for finding the deflection δ_D can be used not only to find deflections at individual points but also to find the equations of the deflection curve.
Concentrated load acting at the midpoint of the beam (Fig. 10-16). When the load P acts at the midpoint C the reactions of the beam (from Eqs. 10-25 and 10-26 with a = b = L /2) are

M_A=M_B=\frac{PL}{8}\quad \quad R_A=R_B=\frac{P}{2}                                  (10-28a,b)

Also, the deflection at the midpoint (from Eq. 10-27) is

δ_C=\frac{PL^3}{192EI}                               (10-29)

This deflection is only one-fourth of the deflection at the midpoint of a simple beam with the same load, which shows the stiffening effect of clamping the ends of the beam.
The above results for the reactions at the ends and the deflection at the middle (Eqs. 10-28 and 10-29) agree with those found in Example 10-2 by solving the differential equation of the deflection curve (see Eqs. 10-13. 10 -14, and 10-19).

R_A=R_B=\frac{P}{2}                             (10-13)

M_A=M_B=\frac{PL}{8}                              (10-14)

δ_{max}=-(v)_{x=L/2}=\frac{PL^3}{192EI}                      (10-19)