验证例题54.pdf

1Example 54 Title Tensile forces of cable members in a cable stayed bridge Description Given is a cable stayed bridge subjected to a uniformly distributed load.Determine the required initial tensile forces in the cable elements such that the horizontal displacement at the top of the main tower(Node A)and the vertical displacements of the nodes B and C are all zeros.Structural geometry and analysis model Verification Example 2Model Analysis Type 2-D static analysis Unit System m,tonf Dimension Length 160 m Height 90 m Element Main girder and main tower Beam element Cable Truss element Material Steel Modulus of elasticity E =2.1 107 tonf/m2 Poissons ratio =0.3 Section Property Cable Radius 0.25 m Main girder Box 3 20 0.03/0.05 m Main tower Bottom Box 10 2 0.05/0.05 m Top(Tapered section)I-End Box 10 2 0.05/0.05 m J-End Box 5 2 0.05/0.05 m Boundary Condition Node 1;Constrain Dx,Dy,and Dz.(Hinge support)Node 5;Constrain Dy,and Dz.(Roller support)Node 6;Constrain all DOFs(Fixed support)Example 54 3Load Case A unit tensile force is applied to each cable Load Case 1;A tensile force,1.0,is applied to the truss element.Load Case 2;A tensile force,1.0,is applied to the truss element.Load Case 3;A tensile force,1.0,is applied to the truss element.Load Case 4;A uniform load,=9 ton/m is distributed in the-Z direction.Composition of equations Constitute equations in order to find load combination factors.The number of unknown load combination factors and required limitations should be equal.Limitation;Set the horizontal displacement at the node 7 and the vertical displacements at the nodes 3 and 4 to zero.Verification Example 4Results Load Combination Factors Combined Z-displacements of the structure Example 54 5 Combined X-displacements of the structure Results of MIDAS/Gen Unit:m,tonf Load Case Load combination factor Limitation After combination 1 1496.47217 Vertical displacement at the node 3 0.0 2 565.86749 Vertical displacement at the node 4 0.0 3 722.65741 Horizontal displacement at the node 7 0.0