Question 7.16: Use Castigliano’s second theorem to determine the horizontal...
Use Castigliano’s second theorem to determine the horizontal and vertical components of the deflection at joint B of the truss shown in Fig. 7.24(a).
This truss was previously analyzed by the virtual work method in Example 7.3.
| TABLE 7.13 | ||||||
| Member | L (m) |
F (kN) |
\frac{∂F}{∂P_1} (kN/kN) |
\frac{∂F}{∂P_2} (kN/kN) |
For P_1 = 0 and P_2 = 84 kN | |
| (∂F/∂P_1)FL
(kN-m) |
(∂F/∂P_2)FL (kN-m) |
|||||
| AB | 4 | -15 + P_1 +0.43P_2 | 1 | 0.43 | 84.48 | 36.32 |
| BC | 3 | -15 + 0.43P_2 | 0 | 0.43 | 0 | 27.24 |
| AD | 5.66 | -28.28 – 0.61P_2 | 0 | -0.61 | 0 | 274.55 |
| BD | 4 | P_2 | 0 | 1 | 0 | 336.00 |
| CD | 5 | 25 – 0.71P_2 | 0 | -0.71 | 0 | 122.97 |
| ∑(\frac{∂F}{∂P})FL | 84.48 | 797.08 | ||||
= \frac{84.48}{EA} kN-m
= \frac{84.48}{200(10^6)(0.0012)} = 0.00035 m
Δ_{BH} = 0.35 mm →
Δ_{BV} =\frac{1}{EA}∑(\frac{∂F}{∂P_2})FL
= \frac{797.08}{EA} kN-m
= \frac{797.08}{200(10^6)(0.0012)} = 0.00332 m
Δ_{BV} = 3.32 mm ↓
As shown in Fig. 7.24(b), a fictitious horizontal force P_1 (= 0) is applied at joint B to determine the horizontal component of deflection, whereas the 84-kN vertical load is designated as the variable P_2 to be used for computing the vertical component of deflection at joint B. The member axial forces, in terms of P_1 and P_2, are then determined by applying the method of joints. These member forces F, along with their partial derivatives with respect to P_1 and P_2, are tabulated in Table 7.13. Note that the tensile axial forces are considered as positive and the compressive forces are negative. Numerical values of P_1 = 0 and P_2 = 84 kN are then substituted in the equations for F, and the expression of Castigliano’s second theorem, as given by Eq. (7.59) is applied, as shown in the table, to determine the horizontal and vertical components of the deflection at joint B of the truss.
Δ =∑(\frac{∂F}{∂P})\frac{FL}{EA} (7.59)
