Question 11.14: A load P is supported at B by two rods of the same material ...
A load P is supported at B by two rods of the same material and of the same cross-sectional area A (Fig. 11.47). Determine the horizontal and vertical deflection of point B.
We apply a dummy horizontal load Q at B (Fig. 11.48). From Castigliano’s theorem we have
x_{B}=\frac{\partial U}{\partial Q} 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{P^{2} L}{2 A E} (11.14)
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 BC and BD, 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. 11.49), we obtain
F_{B C}=0.6 P+0.8 Q F_{B D}=-0.8 P+0.6 Q (11.85)
Differentiating these expressions with respect to Q and P, we write
\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 (11.86)
Substituting from (11.85) and (11.86) into both (11.83) and (11.84), making Q = 0 , and noting that BC = 0.6l and BD = 0.8l , we obtain the horizontal and vertical deflections of point B under the given load P:
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}
Referring to the directions of the loads Q and P, we conclude that
x_{B}=0.096 \frac{P l}{A E} \leftarrow 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.
