Question 9.5: Before the distributed load q0 is applied, there is a gap s ......
Before the distributed load q_0 is applied, there is a gap s between the ends of the cantilevers. Determine the reactions provided by the walls at A and C when the load is applied, assuming that the load is more than large enough to close the gap. Both beams have the same E and I.
Given: Cantilever that will act as a flexible support for another cantilever when gap is closed.
Find: Reaction forces at the walls.
Assume: Hooke’s law applies; long slender beams with constant cross-section; the load causes the gap to close.
If the load is not sufficient for the gap to close, then each beam may be analyzed separately. The problem is more interesting, and statically indeterminate, when the gap does close. When the beams are in contact, they exert equal and opposite forces F_B on each other, so the beams are supported by three unknown forces, R_A, \ R_C, and F_B and two moment reactions M_A and M_C but there are only four equilibrium equations available, two from each beam:
\begin{aligned} & R_A=F_B, \quad M_A=F_B L_1, \\ & R_C=-F_B+q_0 L_2, \quad M_C=-F_B L_2+\frac{1}{2} q_0 L_2^2 . \\ & \end{aligned}To plan our implementation of the force method, we start by constructing the compatibility relationship to become the fifth needed equation. Since the beams end up touching, we know that the difference between the tip deflections must be s. Also, using superposition we can say that the deflection of the tip of the upper beam is the sum of the deflections due to the applied load and the unknown reaction force F_B. Note that all terms are drawn and will be computed using the convention of positive deflection being up. So
\Delta_{1 / F_B}+s=\Delta_{2 / F_B}+\Delta_{2 / q_0} .Table 9.1 gives us the information we need to calculate the tip deflections. For the lower beam,
w_1(x)=\frac{-F_B L_1^3}{6 E I}\left[3\left(\frac{x}{L_1}\right)^2-\left(\frac{x}{L_1}\right)^3\right] ,
so
\Delta_{1 / F_B}=w_1\left(L_1\right)=\frac{-F_B L_1^3}{3 E I} .Similarly for the upper beam, considering x going right to left, the point reaction force causes deflection
\Delta_{2 / F_B}=w_2\left(L_2\right)=\frac{F_B L_2^3}{3 E I}And for the distributed load,
w_2(x)=\frac{-q_0 L_2^4}{24 E I}\left[6\left(\frac{x}{L_2}\right)^2-4\left(\frac{x}{L_2}\right)^3+4\left(\frac{x}{L_2}\right)^4\right],so
\Delta_{2 / q_0}=w_2\left(L_2\right)=\frac{-q_0 L_2^4}{8 E I}Now enforcing compatibility
\Delta_{1 / F_B}+s=\Delta_{2 / F_B}+\Delta_{2 / q_{0^{\prime}}}becomes
-\frac{F_B L_1^3}{3 E I}=\frac{F_B L_2^3}{3 E I}-\frac{q_0 L_2^4}{8 E I}-s .Solving for the unknown force, we obtain
F_B=\frac{q_0 L_2^4 / 8 E I+s}{L_1^3 / 3 E I+L_2^3 / 3 E I}=\frac{3 q_0 L_2^4+24 s E I}{8\left(L_1^3+L_2^3\right)} .Then from the equilibrium relations
\begin{array}{ll} R_A=\frac{3 q_0 L_2^4+24 s E I}{8\left(L_1^3+L_2^3\right)}, & M_A=\frac{\left(3 q_0 L_2^4+24_s E I\right) L_1}{8\left(L_1^3+L_2^3\right)}, \\ R_C=-\frac{3 q_0 L_2^4+24 s E I}{8\left(L_1^3+L_2^3\right)}+q_0 L_2, & M_C=-\frac{\left(3 q_0 L_2^4+24 s E I\right) L_2}{8\left(L_1^3+L_2^3\right)}+\frac{1}{2} q_0 L_2^2. \end{array}
