Question 13.7: 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. 13.11(a) by the method of consistent deformations.
Degree of Indeterminacy The truss consists of ten members connected by six joints and is supported by three reaction components. Thus the degree of indeterminacy of the truss is equal to (m + r) – 2j = (10 + 3) – 2(6) = 1. The three reactions can be determined from the three equations of external equilibrium, so the truss is internally indeterminate to the first degree.
Primary Truss The axial force F_{CE} in the diagonal member CE is selected to be the redundant. The sense of F_{CE} is arbitrarily assumed to be tensile. The primary truss obtained by removing member CE is shown in Fig. 13.11(b). Next, the primary truss is subjected separately to the external loading and a unit tensile force in the redundant member CE, as shown in Fig. 13.11(b) and (c), respectively.
Compatibility Equation The compatibility equation can be expressed as
Δ_{CEO} + f_{CE,CE} F_{CE} = 0 (1)
in which Δ_{CEO} denotes the relative displacement between joints C and E of the primary truss due to the external loads, and the flexibility coefficient f_{CE,CE} denotes the relative displacement between the same joints due to a unit value of the redundant F_{CE}.
Deflections of Primary Truss The virtual work expression for Δ_{CEO} can be written as
Δ_{CEO} = ∑\frac{F_Ou_{CE}L}{AE} (2)
in which F_O and u_{CE} represent, respectively, the member forces due to the external loads and the unit tensile force in member CE. The numerical values of these forces are computed by the method of joints (Fig. 13.11(b) and (c)) and are tabulated in Table 13.3. Equation (2) is then applied as shown in Table 13.3, to obtain
Δ_{CEO} = -\frac{1860 kN-m}{AE}Next, the flexibility coefficient f_{CE,CE} is computed by using the virtual work expression (see Table 13.3):
f_{CE,CE} =∑\frac{u^2_{CE}L}{AE} = \frac{34.56 m}{AE}| TABLE 13.3 | ||||||
| Member | L (m) |
F_O (kN) |
u_{CE} (kN/kN) |
F_Ou_{CE}L (kN-m) |
u^2_{CE}L
(m) |
F = F_O + u_{CE}F_{CE} (kN) |
| AB | 6 | 150 | 0 | 0 | 0 | 150 |
| BC | 6 | 131.25 | -0.6 | -472.5 | 2.16 | 98.95 |
| CD | 6 | 131.25 | 0 | 0 | 0 | 131.25 |
| EF | 6 | -150 | -0.6 | 540 | 2.16 | -182.3 |
| BE | 8 | 200 | -0.8 | -1280 | 5.12 | 156.95 |
| CF | 8 | 150 | -0.8 | -960 | 5.12 | 106.95 |
| AE | 10 | -250 | 0 | 0 | 0 | -250 |
| BF | 10 | 31.25 | 1 | 312.5 | 10 | 85.07 |
| CE | 10 | 0 | 1 | 0 | 10 | 53.82 |
| DF | 10 | -218.75 | 0 | 0 | 0 | -218.75 |
| ∑ | -1,860 | 34.56 | ||||
f_{CE,CE} = \frac{1}{AE}∑u^2_{CE}L = \frac{34.56 m}{AE}
F_{CE} = -\frac{Δ_{CEO}}{f_{CE,CE}} = 53.82 kN (T)
Magnitude of the Redundant By substituting the values of Δ_{CEO} and f_{CE,CE} into the compatibility equation (Eq. (1)), we determine the redundant F_{CE} to be
-\frac{1860 }{AE} + (\frac{34.56 }{AE})F_{CE} = 0F_{CE} = 53.82 kN (T)
Reactions See Fig. 13.11(d). Note that the reactions due to the redundant F_{CE} are zero, as shown in Fig. 13.11(c).
Member Axial Forces The forces in the remaining members of the indeterminate truss can now be determined by using the superposition relationship:
F = F_O + u_{CE}F_{CE}The member forces thus obtained are shown in Table 13.3 and Fig. 13.11(d).
