Question 11.14: A load P is supported at B by two rods of the same material ...

We apply a dummy horizontal load \mathbf{Q} at B (Fig. 11.45). From Castigliano’s theorem we have

x_{B}=\frac{\partial U}{\partial Q} \quad y_{B}=\frac{\partial U}{\partial P}

Recalling from Sec. 11.4 the expression (11.14) for the strain energy of a rod, we write

U=\frac{F_{B C}^{2}(B C)}{2 A E}+\frac{F_{B D}^{2}(B D)}{2 A E}

where F_{B C} and F_{B D} represent the forces in B C and B D, respectively. We have, therefore,

x_{B}=\frac{\partial U}{\partial Q}=\frac{F_{B C}(B C)}{A E} \frac{\partial F_{B C}}{\partial Q}+\frac{F_{B D}(B D)}{A E} \frac{\partial F_{B D}}{\partial Q}      (11.83)

and

y_{B}=\frac{\partial U}{\partial P}=\frac{F_{B C}(B C)}{A E} \frac{\partial F_{B C}}{\partial P}+\frac{F_{B D}(B D)}{A E} \frac{\partial F_{B D}}{\partial P}     (11.84)

From the free-body diagram of pin B (Fig. 11.46), we obtain

F_{B C}=0.6 P+0.8 Q \quad F_{B D}=-0.8 P+0.6 Q      (11.85)

Differentiating these expressions with respect to Q and P, we write

\begin{array}{rlrl} \frac{\partial F_{B C}}{\partial Q} =0.8 & \frac{\partial F_{B D}}{\partial Q}=0.6 \\ \frac{\partial F_{B C}}{\partial P}=0.6 & \frac{\partial F_{B D}}{\partial P}=-0.8 \end{array}        (11.86)

Substituting from (11.85) and (11.86) into both (11.83) and (11.84), making Q=0, and noting that B C=0.6 l and B D=0.8 l, we obtain the horizontal and vertical deflections of point B under the given load \mathbf{P} :

\begin{aligned} x_{B} & =\frac{(0.6 P)(0.6 l)}{A E}(0.8)+\frac{(-0.8 P)(0.8 l)}{A E}(0.6) \\ & =-0.096 \frac{P l}{A E} \\ y_{B} & =\frac{(0.6 P)(0.6 l)}{A E}(0.6)+\frac{(-0.8 P)(0.8 l)}{A E}(-0.8) \\ & =+0.728 \frac{P l}{A E} \end{aligned}

Referring to the directions of the loads \mathbf{Q} and \mathbf{P}, we conclude that

x_{B}=0.096 \frac{P l}{A E} \leftarrow \quad y_{B}=0.728 \frac{P l}{A E} \downarrow

We check that the expression obtained for the vertical deflection of B is the same that was found in Example 11.09.