Question 13.16: Determine the reactions and the force in each member of the ...
Determine the reactions and the force in each member of the truss shown in Fig. 13.23(a) due to a temperature increase of 45°C in member AB and a temperature drop of 20°C in member CD. Use the method of consistent deformations.
Degree of Indeterminacy i = (m + r) – 2j = (6 + 3) – 2(4) = 1. The truss is internally indeterminate to the first degree.
Primary Truss The axial force F_{AD} in the diagonal member AD is selected to be the redundant. The primary truss obtained by removing member AD is shown in Fig. 13.23(b). Next, the primary truss is subjected separately to the prescribed temperature changes and a 1-kN tensile force in the redundant member AD, as shown in Fig. 13.23(b) and (c), respectively.
Compatibility Equation The compatibility equation can be expressed as
\Delta_{A D O}+f_{A D, A D} F_{A D}=0 (1)
in which \Delta_{A D O} denotes the relative displacement between joints A and D of the primary truss due to temperature changes and the flexibility coefficient f_{A D, A D} denotes the relative displacement between the same joints due to a unit value of the redundant F_{A D}.
Deflections of Primary Truss As discussed in Section 7.3, the virtual work expression for \Delta_{A D O} can be written as
\Delta_{A D O}=\sum \alpha(\Delta T) L u_{A D}in which the product α(ΔT)L equals the axial deformation of a member of the primary truss due to a change in temperature ΔT, and u_{AD} represents the axial force in the same member due to a 1-kN tensile force in member AD. The numerical values of these quantities are tabulated in Table 13.6, from which \Delta_{A D O} is determined to be
\Delta_{A D O} = -1.92 mmNext, the flexibility coefficient f_{A D, A D} is computed by using the virtual work expression (see Table 13.6)
f_{A D, A D}=\sum \frac{u_{A D}^{2} L}{A E}=0.0479 mmMagnitude of the Redundant By substituting the values of \Delta_{A D O} and f_{A D, A D} into the compatibility equation (Eq. (1)), we obtain
-1.92 + (0.0479)F_{AD} = 0F_{AD} = 40.084 kN (T)
| TABLE 13.6 | |||||||
| Member | L (m) |
A (m²) |
ΔT (°C) |
u_{AD} (kN/kN) |
(ΔT)Lu_{AD} (°C · m) |
u^2_{AD}L/A
(1/m) |
F=u_{AD}F_{AD} (kN) |
| AB | 8 | 0.005 | 45 | -0.8 | -288 | 1,024 | -32.067 |
| CD | 8 | 0.005 | -20 | -0.8 | 128 | 1,024 | -32.067 |
| AC | 6 | 0.005 | 0 | -0.6 | 0 | 432 | -24.05 |
| BD | 6 | 0.005 | 0 | -0.6 | 0 | 432 | -24.05 |
| AD | 10 | 0.003 | 0 | 1.0 | 0 | 3,333.333 | 40.084 |
| BC | 10 | 0.003 | 0 | 1.0 | 0 | 3,333.333 | 40.084 |
| ∑ | -160 | 9,578.667 | |||||
f_{A D, A D}=\frac{1}{E} \sum\frac{u_{AD}^{2}L}{A}=\frac{9,578.667}{200\left(10^{6}\right)} = 47.893\left(10^{-6}\right) m / kN =0.0479 mm / kN
F_{A D}=-\frac{\Delta_{A D O}}{f_{A D, A D}}=40.084 kN \ \ (T)
Reactions Since the truss is statically determinate externally, its reactions due to the temperature changes are zero.
Member Axial Forces The forces in the members of the primary truss due to the temperature changes are zero, so the forces in the members of the indeterminate truss can be expressed as
F=u_{A D} F_{A D}The member forces thus obtained are shown in Table 13.6 and Fig. 13.23(d).
