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 BH from its support at H. The reaction R _{H} is now considered as an unknown load (Fig. 11.53) 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 P and the redundant reaction R _{H}. Recalling Eq. (11.72), we write

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)

\begin{aligned}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}}\end{aligned}                        (11.90)

We note that the force in rod BH is equal to R _{H} and write

F_{B H}=R_{H}                          (11.91)

Then, from the free-body diagram of pin B (Fig. 11.54), we obtain

F_{B C}=0.6 P-0.6 R_{H}                        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             \frac{\partial F_{B D}}{\partial R_{H}}=0.8              \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 BC, BD, and BH are, respectively, equal to 0.6l, 0.8l, and 0.5l, we write

Setting y_{H}=0, we obtain

and, solving for R_{H},

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

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