Question 11.16: A load P is supported at B by three rods of the same materia...

The structure is statically indeterminate to the first degree. We consider the reaction at H as redundant and release rod B H from its support at H. The reaction \mathbf{R}_{H} is now considered as an unknown load (Fig. 11.50) and will be determined from the condition that the deflection y_{H} of point H must be zero. By Castigliano’s theorem y_{H}=\partial U / \partial R_{H}, where U is the strain energy of the three-rod system under the load \mathbf{P} and the redundant reaction \mathbf{R}_{H}. Recalling Eq. (11.72),

x_{j}={\frac{\partial U}{\partial P_{j}}}=\sum_{i=1}^{n}{\frac{F_{i}L_{i}}{A_{i}E}}{\frac{\partial F_{i}}{\partial P_{j}}}       (11.72)

we write

y_{H}=\frac{F_{B C}(B C)}{A E} \frac{\partial F_{B C}}{\partial R_{H}}+\frac{F_{B D}(B D)}{A E} \frac{\partial F_{B D}}{\partial R_{H}}+\frac{F_{B H}(B H)}{A E} \frac{\partial F_{B H}}{\partial R_{H}}   (11.90)

We note that the force in \operatorname{rod} B H is equal to R_{H} and write

F_{B H}=R_{H}  (11.91)

Then, from the free-body diagram of pin B (Fig. 11.51), we obtain

F_{B C}=0.6 P-0.6 R_{H} \quad F_{B D}=0.8 R_{H}-0.8 P     (11.92)

Differentiating with respect to R_{H} the force in each rod, we write

\frac{\partial F_{B C}}{\partial R_{H}}=-0.6 \quad \frac{\partial F_{B D}}{\partial R_{H}}=0.8 \quad \frac{\partial F_{B H}}{\partial R_{H}}=1  (11.93)

Substituting from (11.91), (11.92), and (11.93) into (11.90), and noting that the lengths B C, B D, and B H are, respectively, equal to 0.6 l, 0.8 l, and 0.5 l, we write

1.228 R_{H}-0.728 P=0

and, solving for R_{H},

R_{H}=0.593 P

Carrying this value into Eqs. (11.91) and (11.92), we obtain the forces in the three rods:

F_{B C}=+0.244 P \quad F_{B D}=-0.326 P \quad F_{B H}=+0.593 P