Question 11.SP.5: For the truss and loading of Sample Prob. 11.4, determine th......
For the truss and loading of Sample Prob. 11.4, determine the vertical deflection of joint C.
STRATEGY: Add a dummy load associated with the desired vertical deflection at joint C. The truss is then analyzed to determine the member forces, first by drawing a free-body diagram of the truss to find the reactions and then by using equilibrium at each joint to find the member forces. Use Eq. (11.59) to get the deflection in terms of the dummy load Q.
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} (11.59)
MODELING and ANALYSIS:
Castigliano’s Theorem. We introduce the dummy vertical load Q as shown in Fig. 1. Using Castigliano’s theorem where the force F_i in a given member i is caused by the combined load of P and Q and since E = constant,
y_C=\sum\left(\frac{F_i L_i}{A_i E}\right) \frac{\partial F_i}{\partial Q}=\frac{1}{E} \sum\left(\frac{F_i L_i}{A_i}\right) \frac{\partial F_i}{\partial Q} (1)
Force in Members. Since the force in each member caused by the load P was previously found in Sample Prob. 11.4, we only need to determine the force in each member due to Q. Using the free-body diagram of the truss with load Q, we draw a free-body diagram (Fig. 2) to determine the reactions. Then, considering in sequence the equilibrium of joints E, C, B and D and using Fig. 3, we determine the force in each member caused by load Q.
\begin{array}{ll} \text { Joint } E: & F_{C E}=F_{D E}=0 \\ \text { Joint } C: & F_{A C}=0 ; F_{C D}=-Q \\ \text { Joint } B: & F_{A B}=0 ; F_{B D}=-\frac{3}{4} Q \end{array}
The total force in each member under the combined action of Q and P is shown in the following table. Form \partial F_i / \partial Q for each member, then compute \left(F_i L_i / A_i\right)\left(\partial F_i / \partial Q\right) , as indicated.
\sum\left(\frac{F_i L_i}{A_i}\right) \frac{\partial F_i}{\partial Q}=4306 P+4263 Q
Deflection of C. Substituting into Eq. (1), we have
y_C=\frac{1}{E} \sum\left(\frac{F_i L_i}{A_i}\right) \frac{\partial F_i}{\partial O}=\frac{1}{E}(4306 P+4263 Q)
Since load Q is not part of the original load, set Q = 0. Substituting P = 40 kN and E = 73 GPa gives
y_C=\frac{4306\left(40 \times 10^3 N \right)}{73 \times 10^9 Pa }=2.36 \times 10^{-3} m \quad y_C=2.36 mm \downarrow
| \left(\frac{ F _i L _i}{ A _i}\right) \frac{\partial F _i}{\partial Q } | A_i, m^2 | L_i, m | \partial F_i / \partial Q | F_i | Member |
| 0 | 500 \times 10^{-6} | 0.8 | 0 | 0 | AB |
| 0 | 500 \times 10^{-6} | 0.6 | 0 | +15P/8 | AC |
| +3125P + 3125Q | 500 \times 10^{-6} | 1.0 | \frac{5}{4} | +5P/4 + 5Q/4 | AD |
| +1181P + 338Q | 1000 \times 10^{-6} | 0.6 | -\frac{3}{4} | −21P/8 − 3Q/4 | BD |
| + 800Q | 1000 \times 10^{-6} | 0.8 | -1 | -Q | CD |
| 0 | 500 \times 10^{-6} | 1.5 | 0 | +15P/8 | CE |
| 0 | 500 \times 10^{-6} | 1.7 | 0 | -17P/8 | DE |
