Question 7.13: Determine the deflection at point C of the beam shown in Fig...
Determine the deflection at point C of the beam shown in Fig. 7.21(a) by Castigliano’s second theorem.
This beam was previously analyzed by the moment-area, the conjugate-beam, and the virtual work methods in Examples 6.7, 6.13, and 7.9, respectively.
The 60-kN external load is already acting at point C, where the deflection is to be determined, so we designate this load as the variable P, as shown in Fig. 7.21(b). Next, we compute the reactions of the beam in terms of P. These are also shown in Fig. 7.21(b). Since the loading is discontinuous at point B, the beam is divided into two segments, AB and BC. The x coordinates used for determining the equations for the bending moment in the two segments of the beam are shown in Fig. 7.21(b). The equations for M (in terms of P) obtained for the segments of the beam are tabulated in Table 7.11, along with the partial derivatives of M with respect to P.
| TABLE 7.11 | ||||
| Segment | x Coordinate | M (kN-m) | \frac{∂M}{∂P}(kN-m/kN) | |
| Origin | Limits (m) | |||
| AB | A | 0-9 | (135-\frac{P}{3})x – x^2 | -\frac{x}{3} |
| CB | C | 0-3 | -Px | -x |
The deflection at C can now be determined by substituting P = 60 kN into the equations for M and ∂M/∂P and by applying the expression of Castigliano’s second theorem as given by Eq. (7.60):
Δ= \int_{0}^{L}{(\frac{∂M}{∂P}) (\frac{M}{EI}) dx} (7.60)
Δ_C = \int_{0}^{L}{(\frac{∂M}{∂P}) (\frac{M}{EI}) dx}Δ_C = \frac{1}{EI}\left[\int_{0}^{9}{(-\frac{x}{3}) (135x -\frac{60x}{3} – x^2) dx} + \int_{0}^{3}{(-x) (-60x) dx}\right]
= \frac{1}{EI}\left[\int_{0}^{9}{(-\frac{x}{3}) (115x – x^2) dx} + \int_{0}^{3}{(-x) (-60x) dx}\right]
=- \frac{8228.25 kN-m^3}{EI} = -\frac{8228.25}{200(10^6)800(10^{-6})} = -0.0514 m
The negative answer for Δ_C indicates that point C deflects upward in the direction opposite to that of P.
Δ_C = 51.4 mm ↑