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 \delta_B at the hinge and the angle of rotation \theta_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 \delta_B of the hinge) is readily found from Cases 1 and 4 of Table G-1, Appendix G, (available online):
\delta _B=\frac{qb^4}{8EI} +\frac{Fb^3}{3EI}
or, since F = 2P/3,
\delta _B=\frac{qb^4}{8EI} +\frac{2Pb^3}{9EI} (8-56)
The angle of rotation \theta_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
(\theta _A)_1=\frac{\delta _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 G-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
(\theta _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:
\theta _A=(\theta _A)_1 + (\theta _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.