Question 9.8: 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. 9-20a). 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. 9-20d). (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. 9-20b), 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. 9-20c). 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:
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\nu=-\frac{qx^{2}}{24EI}(6L^{2}-4Lx+x^{2}) \nu^{\prime}=-\frac{qx}{6EI}(3L^{2}-3Lx+x^{2})
\delta_{B}=\frac{qL^{4}}{8EI} \theta_{B}=\frac{qL^{3}}{6EI} |
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\nu=-\frac{Px^{2}}{6EI}(3L-x) \nu^{\prime}=-\frac{Px}{2EI}(2L-x)
\delta_{B}=\frac{PL^{3}}{3EI} \theta_{B}=\frac{PL^{2}}{2EI} |
or, since F = 2P/3,
\delta_{B}=\frac{qb^{4}}{8EI}+\frac{2Pb^3}{9EI} (9-56)
The angle of rotation \theta_{A} at support A (Fig. 9-20d) 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
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\nu=-\frac{Pbx}{6LEI}(L^{2}-b^{2}-x^{2}) \nu^{\prime}=-\frac{Pb}{6LEI}(L^{2}-b^{2}-3x^{2}) (0 ≤ x ≤ a)
\theta_{A}=\frac{Pab(L+b)}{6LEI} \theta_{B}=\frac{Pab(L+a)}{6LEI}
If a ≥ b, \delta_{C}=\frac{Pb(3L^{2}-4b^{2})}{48EI} If a ≤ b, \delta_{C}=\frac{Pa(3L^{2}-4a^{2})}{48EI}
If a ≥ b, x_{1}=\sqrt{\frac{L^{2}-b^{2}}{3}} and \delta_{\max}=\frac{Pb(L^{2}-b^{2})^{{3}/{2}}}{} |
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. 9-20a), the angle of rotation is
(\theta_{A})_{2}=\frac{P\left(\frac{2a}{3}\right)\left(\frac{a}{3}\right)\left(a+\frac{a}{3}\right)}{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} (9-57)
This example illustrates how the method of superposition can be adapted to handle a seemingly complex situation in a relatively simple manner.
