Question 11.S-P.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.
Castigliano’s Theorem. Since no vertical load is applied at joint C, we introduce the dummy load Q as shown. Using Castigliano’s theorem, and denoting by F_{i} the force in a given member i caused by the combined loading of P and Q, we have, 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. Considering in sequence the equilibrium of joints E, C, B, and D, we determine the force in each member caused by load Q.
Joint E: F_{C E}=F_{D E}=0
Joint C: F_{A C}=0 ; F_{C D}=-Q
Joint B: F_{A B}=0 ; F_{B D}=-\frac{3}{4} Q
The force in each member caused by the load P was previously found in Sample Prob. 11.4. The total force in each member under the combined action of Q and P is shown in the following table. Forming \partial F_{i} / \partial Q for each member, we then compute \left(F_{i} L_{i} / A_{i}\right)\left(\partial F_{i} / \partial Q\right) as indicated in the table.
| Member | F_{i} | \partial F _{i} / \partial Q | L_{i}, m | A_{i}, m ^{2} | \left(\frac{ F _{i} L_{i}}{ A _{i}}\right) \frac{\partial F _{i}}{\partial Q } |
| AB | 0 | 0 | 0.8 | 500 × 10^{-6} | 0 |
| AC | +15P/8 | 0 | 0.6 | 500 × 10^{-6} | 0 |
| AD | +5P/4 + 5Q/4 | \frac{5}{4} | 1.0 | 500 × 10^{-6} | +3125P +3125Q |
| BD | -21P/8 – 3Q/4 | -\frac{3}{4} | 0.6 | 1000 × 10^{-6} | +1181P + 338Q |
| CD | -Q | -1 | 0.8 | 1000 × 10^{-6} | + 800Q |
| CE | +15P/8 | 0 | 1.5 | 500 × 10^{-6} | 0 |
| DE | -17P/8 | 0 | 1.7 | 1000 × 10^{-6} | 0 |
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 Q}=\frac{1}{E}(4306 P+4263 Q)
Since the load Q is not part of the original loading, we set Q = 0 . Substituting the given data, P = 40 kN and E = 73 GPa, we find
y_{C}=\frac{4306\left(40 \times 10^{3} N \right)}{73 \times 10^{9} Pa }=2.36 \times 10^{-3} m y_{C}=2.36 mm \downarrow
