L J

The required nodal reactions are therefore X\ = —54 kN, X2 = 54 kN, X3 = 0 and Y3 = 162 kN.

Representing these reactions together with the applied forces on a sketch of the structure, Fig. 9.33, and considering force and moment equilibrium, gives

Hence, equilibrium is satisfied by the system of forces.

90 kN

90 kN

Fig. 9.33.

Example 9.2

Figure 9.34 shows the members and idealised support conditions for a roof truss. All three members of which are steel and have the same cross-sectional area such that AE = 12 MN throughout. Using the displacement based finite element method, treating the truss as a pin-jointed plane frame and each member as a rod:

(a) assemble the necessary terms in the structural stiffness matrix;

(b) hence, determine the nodal displacements with respect to the global coordinates, for the condition shown in Fig. 9.34.

(c) If, under load, the left support sinks by 5 mm, determine the resulting new nodal displacements, with respect to the global coordinates.

40 kN

40 kN

Fig. 9.34.

Solution

(a) Figure 9.35 shows suitable node, dof. and element labelling. Lack of symmetry prevents any advantage being taken to reduce the calculations. However, since both ends of the horizontal member are fixed it is redundant therefore and does not need to be considered further.

40 kN

40 kN

Fig. 9.35.

All three members have the same AE, hence

With reference to §9.7, the element stiffness matrix with respect to global coordinates is given by

cos2 a sin a cos a sin a symmetric — cos2 a — sin a cos a cos2 a

Evaluating the stiffness matrix for both elements:

sin2 a sin a cos a sin2 a

Element a

L(a) = 2m, a(a) = 60°, cosa(fl) = 1/2, sina(a) = >/(3)/2
0 0

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