Question 8.8: A compound beam ABC has a roller support at A, an internal h...
A compound beam ABC has a roller support at A, an internal hinge at B, and a fixed support at C (Fig. 8-20a). Segment AB has length a and segment BC has length b. A concentrated load P acts at distance 2a/3 from support A and a uniform load of intensity q acts between points B and C.
Determine the deflection δ_B at the hinge and the angle of rotation θ_A at support A (Fig. 8-20d). (Note: The beam has constant flexural rigidity EI.)
For purposes of analysis, we will consider the compound beam to consist of two individual beams: (1) a simple beam AB of length a, and (2) a cantilever beam BC of length b. The two beams are linked together by a pin connection at B.
If we separate beam AB from the rest of the structure (Fig. 8-20b), we see that there is a vertical force F at end B equal to 2P/3. This same force acts downward at end B of the cantilever (Fig. 8-20c). Consequently, the cantilever beam BC is subjected to two loads: a uniform load and a concentrated load. The deflection at the end of this cantilever (which is the same as the deflection δ_B of the hinge) is readily found from Cases 1 and 4 of Table H-1, Appendix H, available online:
δ_B=\frac{qb^4}{8EI}+\frac{Fb^3}{3EI}
or, since F = 2P/3,
δ_B=\frac{qb^4}{8EI}+\frac{2Pb^3}{9EI} (8-56)
The angle of rotation θ_A at support A (Fig. 8-20d) consists of two parts:
(1) an angle BAB^′ produced by the downward displacement of the hinge, and (2) an additional angle of rotation produced by the bending of beam AB (or beam AB^′) as a simple beam. The angle BAB^′ is
(θ_A)_1=\frac{δ_B}{a}=\frac{qb^4}{8aEI}+\frac{2Pb^3}{9aEI}
The angle of rotation at the end of a simple beam with a concentrated load is obtained from Case 5 of Table H-2. The formula given there is
\frac{Pab(L + b)}{6LEI}
in which L is the length of the simple beam, a is the distance from the left-hand support to the load, and b is the distance from the right-hand support to the load. Thus, in the notation of our example (Fig. 8-20a), the angle of rotation is
(θ_A)_2=\frac{P(\frac{2a}{3})(\frac{a}{3})(a + \frac{a}{3})}{6aEI}=\frac{4Pa^2}{81EI}
Combining the two angles, we obtain the total angle of rotation at support A:
θ_A=(θ_A)_1+(θ_A)_2=\frac{qb^4}{8aEI}+\frac{2Pb^3}{9aEI}+\frac{4Pa^2}{81EI} (8-57)
This example illustrates how the method of superposition can be adapted to handle a seemingly complex situation in a relatively simple manner.