Question 16.8: Determine the forces in the members of the truss shown in Fi...

The truss in Fig. 16.18(a) is clearly externally statically determinate but, from Eq. (16.5),

n_{\mathrm{s}}=M-2N+3      (16.5)

has a degree of internal statical indeterminacy equal to 1(M=6, N=4). We therefore release the truss so that it becomes statically determinate by ‘cutting’ one of the members, say BD, as shown in Fig. 16.18(b). Due to the actual loads ( P in this case) the cut ends of the member BD will separate or come together, depending on whether the force in the member (before it was cut) was tensile or compressive; we shall assume that it was tensile.

We are assuming that the truss is linearly elastic so that the relative displacement of the cut ends of the member BD (in effect the movement of B and D away from or towards each other along the diagonal BD) may be found using, say, the unit load method as illustrated in Exs 15.6 and 15.7. Thus we determine the forces F_{\mathrm{a}, j}, in the members produced by the actual loads. We then apply equal and opposite unit loads to the cut ends of the member BD as shown in Fig. 16.18(c) and calculate the forces, F_{1, j} in the members. The displacement of B relative to \mathrm{D}, \Delta_{\mathrm{BD}}, is then given by

\Delta_{\mathrm{BD}}=\sum\limits_{j=1}^{n} \frac{F_{\mathrm{a}, j} F_{1, j} L_{j}}{A E}          (see Eq. (viii)

\Delta_{\mathrm{C}}=\sum\limits_{j=1}^{n} \frac{F_{j} F_{1, j} L_{j}}{E_{j} A_{j}}     (viii)

in Ex. 15.7)

The forces, F_{\mathrm{a}, j}, are the forces in the members of the released truss due to the actual loads and are not, therefore, the actual forces in the members of the complete truss. We shall therefore redesignate the forces in the members of the released truss as F_{0, j}. The expression for \Delta_{\mathrm{BD}} then becomes

\Delta_{\mathrm{BD}}=\sum\limits_{j=1}^{n} \frac{F_{0, j} F_{1, j} L_{j}}{A E}      (i)

In the actual structure this displacement is prevented by the force, X_{\mathrm{BD}}, in the redundant member BD. If, therefore, we calculate the displacement, a_{\mathrm{BD}}, in the direction of \mathrm{BD} produced by a unit value of X_{\mathrm{BD}}, the displacement due to X_{\mathrm{BD}} will be X_{\mathrm{BD}} a_{\mathrm{BD}}. Clearly, from compatibility

\Delta_{\mathrm{BD}}+X_{\mathrm{BD}} a_{\mathrm{BD}}=0        (ii)

from which X_{\mathrm{BD}} is found. Again, as in the case of statically indeterminate beams, a_{\mathrm{BD}} is a flexibility coefficient. Having determined X_{\mathrm{BD}}, the actual forces in the members of the complete truss may be calculated by, say, the method of joints or the method of sections.

In Eq. (ii), a_{\mathrm{BD}} is the displacement of the released truss in the direction of BD produced by a unit load. Thus, in using the unit load method to calculate this displacement, the actual member forces \left(F_{1, j}\right) and the member forces produced by the unit load \left(F_{l, j}\right) are the same. Therefore, from Eq. (i)

a_{\mathrm{BD}}=\sum\limits_{j=1}^{n} \frac{F_{1, j}^{2} L_{j}}{A E}      (iii)

The solution is completed in Table 16.1.

From Table 16.1

\Delta_{\mathrm{BD}}=\frac{2.71 P L}{A E} \quad a_{\mathrm{BD}}=\frac{4.82 L}{A E}

Substituting these values in Eq. (i) we have

\frac{2.71 P L}{A E}+X_{\mathrm{BD}} \frac{4.82 L}{A E}=0

from which

X_{\mathrm{BD}}=-0.56 P       (i.e. compression)

The actual forces, F_{\mathrm{a}, j}, in the members of the complete truss of Fig. 16.18(a) are now calculated using the method of joints and are listed in the final column of Table 16.1.

We note in the above that \Delta_{\mathrm{BD}} is positive, which means that \Delta_{\mathrm{BD}} is in the direction of the unit loads, i.e. B approaches \mathrm{D} and the diagonal \mathrm{BD} in the released structure decreases in length. Therefore in the complete structure the member BD, which prevents this shortening, must be in compression as shown; also a_{\mathrm{BD}} will always be positive since it contains the term F_{1, j}^{2}. Finally, we note that the cut member BD is included in the calculation of the displacements in the released structure since its deformation, under a unit load, contributes to a_{\mathrm{BD}}.