Question 16.9: Calculate the forces in the members of the truss shown in Fi...
Calculate the forces in the members of the truss shown in Fig. 16.19(a). All members have the same cross sectional area, A, and Young’s modulus, E.
By inspection we see that the truss is both internally and externally statically indeterminate since it would remain stable and in equilibrium if one of the diagonals, \mathrm{AD} or \mathrm{BD}, and the support at \mathrm{C} were removed; the degree of indeterminacy is therefore
2. Unlike the truss in Ex. 16.18, we could not remove any member since, if BC or \mathrm{CD} were removed, the outer half of the truss would become a mechanism while the portion \mathrm{ABDE} would remain statically indeterminate. Therefore we select \mathrm{AD} and the support at \mathrm{C} as the releases, giving the statically determinate truss shown in Fig. 16.19(b); we shall designate the force in the member \mathrm{AD} as X_{1} and the vertical reaction at \mathrm{C} as R_{2}.
In this case we shall have two compatibility conditions, one for the diagonal \mathrm{AD} and one for the support at \mathrm{C}. We therefore need to investigate three loading cases: one in which the actual loads are applied to the released statically determinate truss in Fig. 16.19(b), a second in which unit loads are applied to the cut member AD (Fig. 16.19(\mathrm{c}) ) and a third in which a unit load is applied at \mathrm{C} in the direction of R_{2} (Fig. 16.19(\mathrm{~d}) ). By comparison with the previous example, the compatibility conditions are
\begin{array}{r} \Delta_{\mathrm{AD}}+a_{11} X_{1}+a_{12} R_{2}=0 & (\text{i})\\ v_{\mathrm{C}}+a_{21} X_{1}+a_{22} R_{2}=0 & (\text{ii}) \end{array}
in which \Delta_{\mathrm{AD}} and v_{\mathrm{C}} are, respectively, the change in length of the diagonal \mathrm{AD} and the vertical displacement of \mathrm{C} due to the actual loads acting on the released truss, while a_{11}, a_{12}, etc., are flexibility coefficients, which we have previously defined (see Ex. 16.7). The calculations are similar to those carried out in Ex. 16.8 and are shown in Table 16.2.
From Table 16.2
\begin{aligned} \Delta_{\mathrm{AD}} & =\sum_{j=1}^{n} \frac{F_{0, j} F_{1, j}\left(X_{1}\right) L_{j}}{A E}=\frac{-27.1}{A E} \quad \text { (i.e. AD increases in length) } \\ v_{\mathrm{C}} & =\sum_{j=1}^{n} \frac{F_{0, j} F_{1, j}\left(R_{2}\right) L_{j}}{A E}=\frac{-48.11}{A E} \quad \text { (i.e. C displaced downwards) } \\ a_{11} & =\sum_{j=1}^{n} \frac{F_{1, j}^{2}\left(X_{1}\right) L_{j}}{A E}=\frac{4.32}{A E} \\ a_{22} & =\sum_{j=1}^{n} \frac{F_{1, j}^{2}\left(R_{2}\right) L_{j}}{A E}=\frac{11.62}{A E} \\ a_{12} & =a_{21} \sum_{j=1}^{n} \frac{F_{1, j}\left(X_{1}\right) F_{1, j}\left(R_{2}\right) L_{j}}{A E}=\frac{2.7}{A E} \end{aligned}
Substituting in Eqs (i) and (ii) and multiplying through by AE we have
\begin{array}{r} -27.1+4.32 X_{1}+2.7 R_{2}=0 & (\text{iii})\\ -48.11+2.7 X_{1}+11.62 R_{2}=0 & (\text{iv}) \end{array}
Solving Eqs (iii) and (iv) we obtain
X_{1}=4.28 \mathrm{kN} \quad R_{2}=3.15 \mathrm{kN}
The actual forces, F_{\mathrm{a}, j}, in the members of the complete truss are now calculated by the method of joints and are listed in the final column of Table 16.2.
TABLE 16.2
| Member | { L}_{j} | F_{0,j} | F_{1,j}\left(X_{1}\right) | F_{1,j}\left(R_{2}\right) | F_{0,j}F^{}_{1,j}({X_{1})L_{j}} | F_{0,j}F_{1,j}\ (R_{2})L_{j} | F_{1,j}^{2}\ (X_{1})L_{j} | F_{1,j}^{2}\left(R_{2}\right)L_{j} | F_{1,j}(\mathbf{X}_{1})\,F_{1,j}(R_{2})L_{j} | F_{\mathbf{a},j} |
| AB | 1 | 10.0 | −0.71 | −2.0 | −7.1 | −20.0 | 0.5 | 4.0 | 1.41 | 0.67 |
| BC | 1.41 | 0 | 0 | −1.41 | 0 | 0 | 0 | 2.81 | 0 | −4.45 |
| CD | 1 | 0 | 0 | 1.0 | 0 | 0 | 0 | 1.0 | 0 | 3.15 |
| DE | 1 | 0 | −0.71 | 1.0 | 0 | 0 | 0.5 | 1.0 | −0.71 | 0.12 |
| AD | 1.41 | 0 | 1.0 | 0 | 0 | 0 | 1.41 | 0 | 0 | 4.28 |
| BE | 1.41 | −14.14 | 1.0 | 1.41 | −20.0 | −28.11 | 1.41 | 2.81 | 2.0 | −5.4 |
| BD | 1 | 0 | −0.71 | 0 | 0 | 0 | 0.5 | 0 | 0 | −3.03 |
| ∑ = −27.1 | ∑ = −48.11 | ∑ = 4.32 | ∑ = 11.62 | ∑ = 2.7 |