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## Q. 10.9

Determine the force in each member of the Warren truss shown in $Fig. 10.20(a)$.

## Verified Solution

This truss was analyzed in Example 4.4 without taking advantage of its
symmetry.

Symmetry This truss is symmetric with respect to the vertical s axis passing through member $CG$, as shown in Fig. 10.20(b). The truss is subjected to vertical loads only, so the horizontal reaction at support $A$ is zero $(A_x = 0)$. The half of the truss to the right of the axis of symmetry, $CEHG$, will be used for analysis.

Symmetric and Antisymmetric Components of Loading The symmetric and antisymmetric components of the given loading with respect to the axis of symmetry of the truss are determined by using the procedure described in Section 10.2. These loading components are shown in Fig. 10.20(b) and (c). Note that the sum of the two components is equal to the total loading given in Fig. 10.20(a).

Member Forces Due to the Symmetric Loading Component The substructure (right half of the truss) with symmetric boundary conditions is shown in Fig. 10.20(d). Note that the joints $C$ and $G$, which are located at the axis of symmetry, are supported by rollers that prevent their movements in the horizontal direction (perpendicular to the s axis). The symmetric component of loading (Fig. 10.20(b)) is applied to the substructure, with the magnitude of the 30-k concentrated load acting along the axis of symmetry reduced by half, as shown in Fig. 10.20(d). The reactions of the substructure are obtained by applying the equilibrium equations:

$+\uparrow \sum{F_y}=0$             $-15 – 18 + E_y = 0$          $E_y = 33 k \uparrow$

$+\circlearrowleft \sum{M_C}=0$             $-G_x(15) – 18(20) + 33(40) = 0$          $G_x = 64 k \longrightarrow$

$\longrightarrow \sum{F_x}=0$            $-C_x + 64 = 0$           $C_x = 64 k \longleftarrow$

The axial forces in the members of the substructure are determined by applying the method of joints. These member forces are also shown in Fig. 10.20(d).

The member axial forces in the left half of the truss can now be obtained by rotating the member forces in the right half (Fig. 10.20(d)) through $180°$ about the $s$ axis, as shown in Fig. 10.20(e).

Member Forces Due to the Antisymmetric Loading Component The substructure with antisymmetric boundary conditions is shown in Fig. 10.20(f ). Note that joints $C$ and $G$, located at the axis of symmetry, are supported by rollers to prevent their deflections in the vertical direction. Also, member $CG$, which is located along the axis of symmetry, is removed from the substructure, as shown in the figure. (The force in member $CG$ will be zero under antisymmetric loading.) The antisymmetric component of loading (Fig. 10.20(c)) is applied to the substructure, and its reactions and member axial forces are computed by applying the equilibrium equations and the method of joints (see Fig. 10.20(f )).

The member axial forces in the left half of the truss are then obtained by reflecting the negatives (i.e., the tensile forces are changed to compressive forces and vice versa) of the member forces in the right half to the left side of the axis of symmetry, as shown in Fig. 10.20(g).

Total Member Forces Finally, the total axial forces in members of the truss are obtained by superimposing the forces due to the symmetric and antisymmetric components of the loading, as given in Fig. 10.20(e) and (g), respectively. These member forces are shown in Fig. 10.20(h).                Ans.