Question 13.10: Determine the moments at the supports of the fixed beam show...
Determine the moments at the supports of the fixed beam shown in Fig. 13.15(a) by the method of consistent deformations. Also, draw the bending moment diagram for the beam.
Degree of Indeterminacy The beam is supported by six support reactions; thus, its degree of indeterminacy is i = 6 – 3 = 3. However, since the beam is subjected only to vertical loading, the horizontal reactions A_x and C_x must be zero. Therefore, to analyze this beam, we need to select only two of the remaining four reactions as the redundants.
Primary Beam The moments M_A and M_C at the fixed supports A and C, respectively, are selected as the redundants. The restraints against rotation at ends A and C of the fixed beam are then removed to obtain the simply supported primary beam shown in Fig. 13.15(b). Next, the primary beam is subjected separately to the external load P and the unit values of redundants M_A and M_C, as shown in Fig. 13.15(b), (c), and (d), respectively.
Compatibility Equations Noting that the slopes of the actual indeterminate beam at the fixed supports A and C are zero, we write the compatibility equations:
θ_{AO} + f_{AA} M_A + f_{AC} M_C = 0 (1)
θ_{CO} + f_{CA} M_A + f_{CC} M_C = 0 (2)
Slopes of the Primary Beam The slopes at ends A and C of the primary beam due to the external load P and due to the unit value of each of the redundants obtained by using either the deflection formulas or the conjugate-beam method are
θ_{AO} = -\frac{Pb(L^2 – b^2)}{6EIL}θ_{CO} = -\frac{Pa(L^2 – a^2)}{6EIL}
f_{AA} = f_{CC} = \frac{L}{3EI}
f_{CA} =\frac{L}{6EI}
By applying Maxwell’s law,
f_{AC} =\frac{L}{6EI}Magnitudes of the Redundants By substituting the expressions for slopes into the compatibility equations (Eqs. (1) and (2)), we obtain
-\frac{Pb(L^2 – b^2)}{6EIL} + ( \frac{L}{3EI})M_A + (\frac{L}{6EI})M_C = 0 (1a)
-\frac{Pa(L^2 – a^2)}{6EIL} + ( \frac{L}{6EI})M_A + (\frac{L}{3EI})M_C = 0 (2a)
which can be simplified as
2M_A + M_C = \frac{Pb(L^2 – b^2)}{L^2} (1b)
M_A + 2M_C =\frac{Pa(L^2 – a^2)}{L^2} (2b)
To solve Eqs. (1b) and (2b) for M_A and M_C, we multiply Eq. (1b) by 2 and subtract it from Eq. (2b):
M_A = -\frac{P}{3L^2}[a(L^2 – a^2) – 2b(L^2 – b^2)]= -\frac{P}{3L^2}[a(L – a)(L + a) – 2b(L – b)(L + b)]
= -\frac{Pab}{3L^2}[(L + a) – 2(L + b)]
=\frac{Pab^2}{L^2}
M_A =\frac{Pab^2}{L^2} \circlearrowleft
By substituting the expression for M_A into Eq. (1b) or Eq. (2b) and solving for M_C, we obtain the following.
M_C =\frac{Pa^2b}{L^2} \circlearrowrightBending Moment Diagram The vertical reactions A_y and C_y can now be determined by superposition of the reactions of the primary beam due to the external load P and due to each of the redundants (Fig. 13.15(b) through (d)). Thus
A_y = \frac{Pb}{L} + \frac{1}{L}(M_A – M_C) = \frac{Pb^2}{L^3}(3a + b)C_y = \frac{Pa}{L} – \frac{1}{L}(M_A – M_C) = \frac{Pa^2}{L^3}(a + 3b)
The bending moment diagram of the beam is shown in Fig. 13.15(e).
The moments at the ends of beams whose ends are fixed against rotation are usually referred to as fixed-end moments. Such moments play an important role in the analysis of structures by the displacement method, to be considered in subsequent chapters. As illustrated here, the expressions for fixed-end moments due to various loading conditions can be conveniently derived by using the method of consistent deformations. The fixed-endmoment expressions for some common types of loading conditions are given inside the back cover of the book for convenient reference.
