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Example. To demonstrate the matrix displacement method, the rigid frame shown in Fig. 3.82a. will be analyzed. The two-dimensional frame has three joints, or nodes, A, B, and C, and hence a total of nine possible degrees of freedom (Fig. 3.82b). The displacements at node A are not restrained. Nodes B and C have zero displacement. For both AB and AC, modulus of elasticity E = 29,000 ksi, area A = 1 in2, and moment of inertia I = 10 in4. Forces will be computed in kips; moments, in kip-in.

At each degree of freedom, the external forces must be balanced by the member forces. This requirement provides the following equations of equilibrium with reference to the global coordinate system:

At the free degree of freedom at node A, 2FxA = 0, 2FyA = 0, and 2MzA = 0:

At the restrained degrees of freedom at node B, 2FxB = 0, 2FyB = 0, and 2MzA = 0:

At the restrained degrees of freedom at node C, 2FxC = 0, 2FyC = 0, and 2MzC = 0:

RxC - |
FxCA |
= 0 |
(3.152a) |

RyC - |
FyCA |
= 0 |
(3.152b) |

Mzc - |
MzCA |
= 0 |
(3.152c) |

where subscripts identify the direction, member, and degree of freedom.

Member force components in these equations are then replaced by equivalent displacement relationships with the use of Eq. (3.148). With reference to the global coordinates, these relationships are as follows: For member AB with a = 0°

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