EXAMPLE 6-003 LINK GAP ELEMENT PROBLEM DESCRIPTION This example uses a single-bay, single-story rigid frame to test the gap link element. This link element carries compression loads only; it has zero stiffness when subjected to tension. The gap element is paced at the bottom of the righthand column in the frame. The frame is then loaded with a gravity load P (10 kips) at the center of the beam. Once the full load P is applied, a lateral load V (20 kips) is applied, pushing the frame from right to left. The compression load in the gap element after the full load P has been applied and the uplift at the gap after the full load V has been applied are compared with independent hand calculated results. The model is created in the XZ plane. Only the U x, U z and R y degrees of freedom are active for the analysis. The gap element is modeled as a single-joint link element at joint 2. This means that one end of the gap element is connected to the ground and the other end is connected to joint 2. The gap element is oriented such that its positive local 1 axis is parallel to the positive global Z axis. This is the default orientation of single joint link elements. Only U 1 degree of freedom properties are defined for the gap element. All three frame elements have identical properties. The compression stiffness of the gap element is chosen to be approximately 100 times the axial stiffness of the frame element directly above it, frame element 2. The initial gap element opening is zero inches. The loading is applied as follows. First the full load P is applied. Then, while the load P is maintained, the full load V is applied. The analysis is run several times using different types of load cases. Nonlinear static, nonlinear modal time history and nonlinear direct time history load cases are used. See the subsequent section titled Summary of Load Cases for more information. Two different models are used in this example. In Model A the linear effective stiffness of the gap element is set equal to zero. In Model B the linear effective stiffness of the gap element is set equal to the nonlinear stiffness of the gap element. EXAMPLE 6-003 - 1
The gap linear effective stiffness is only used for the linear load cases, which in this example are load cases P, V and MODAL. The gap linear effective stiffness is not used in the other load cases and thus has no direct effect on them; however, it does indirectly affect the nonlinear modal time history cases named NLMHIST1 and NLMHIST2 because those cases are solved using the modes from the load case named MODAL. Joint Mass kip-s 2 /in Joint DOF U x DOF U z 2 0 0.001 The lumped joint masses shown in the table to the right are used for the modal and modal time history analyses. 3 0.3 0.1 4 0.3 0.1 GEOMETRY, PROPERTIES AND LOADING 144" P = 10 k 3 3 4 V = 20 k Link Properties (Gap U 1 DOF) Linear K e (Model A) = 0 k/in Linear K e (Model B) = 200,000 k/in Linear C e = 0 k-sec/in Nonlinear K = 200,000 k/in Nonlinear Open = 0 in Z Y 144" X 1 1 2 Single-joint link element (gap) with compression stiffness defined for the U 1 degree of freedom 2 Frame Material Properties E = 29,900 k/in 2 ν =0.3 G= 11,500 k/in 2 Frame Section Properties A = 10 in 2 I = 100 in 4 A v =2 in 2 (shear area) Loading First apply full load P; then apply full load V Active Degrees of Freedom U x, U z, R y EXAMPLE 6-003 - 2
SUMMARY OF LOAD CASES The following table summarizes the load cases that are used in this example. Load Case P V MODAL NLMHIST1 NLMHIST2 NLSTAT1 NLSTAT2 Description A linear static load case with a 10 kip gravity load applied in the negative global Z direction at the center of frame element 3. A linear static load case with a 20 kip lateral load applied in the negative global X direction at joint 2. A ritz-type modal load case with starting loads defined for load P, load V and the link element. A nonlinear modal time history starting from zero initial conditions and using a ramp load to apply the load P. 20 second ramp load rise time. 99.9% modal damping for all modes. 20 output steps and a 2 second output time step (40 seconds of output total). Default values for all nonlinear parameters (tolerances). A nonlinear modal time history starting from the conditions at the end of NLMHIST1 and using a ramp load to apply the load V. 40 second ramp load rise time. 20 output steps and a 4 second output time step (80 seconds of output total). 99.9% modal damping for all modes. Default values for all nonlinear parameters (tolerances), except in Model B the Force Convergence Tolerance is reduced from the default 1E-05 to 1E-11. A nonlinear static case starting from zero initial conditions and applying the load P. This load case uses default nonlinear parameters (tolerances). A nonlinear static case starting from the conditions at the end of NLSTAT1 and applying the load V. This load case uses default nonlinear parameters (tolerances). EXAMPLE 6-003 - 3
Load Case NLSTAT3 NLDHIST1 NLDHIST2 NLDHIST3 Description A nonlinear static case starting from the conditions at the end of the direct integration time history NLDHIST1 and applying the load V. This load case uses default nonlinear parameters (tolerances). A nonlinear direct integration time history starting from zero initial conditions and using a ramp load to apply the load P. 20 second ramp load rise time. 400 output steps and a 0.1 second output time step (40 seconds of output total). Mass and stiffness proportional damping is used with damping set to 99.9% at periods of 1 and 3.4 seconds. Default values for all nonlinear parameters (tolerances) except for the Maximum Iterations per Substep and the Iteration Convergence Tolerance. The Maximum Iterations per Substep item is increased from the default 10 to 100. The Iteration Convergence Tolerance is reduced from the default 1E-04 to 1E-06. A nonlinear direct integration time history starting from the conditions at the end of NLDHIST1 and using a ramp load to apply the load V. 40 second ramp load rise time. 800 output steps and a 0.1 second output time step (80 seconds of output total). Mass and stiffness proportional damping is used with damping set to 99.9% at periods of 1 and 3.4 seconds. Default values for all nonlinear parameters (tolerances), except for the Maximum Iterations per Substep and the Iteration Convergence Tolerance. The Maximum Iterations per Substep item is increased from the default 10 to 100. The Iteration Convergence Tolerance is reduced from the default 1E-04 to 1E-06. A nonlinear direct integration time history that is the same as NLDHIST2, except it starts from the conditions at the end of nonlinear static case NLSTAT1 instead of nonlinear direct integration time history NLDHIST1. EXAMPLE 6-003 - 4
The ramp times for the load cases are selected using the rule of thumb that the ramp loading time should be approximately 10 times the period of interest. The period of interest is assumed to be the period of the first mode. The following table shows the first mode period obtained for the two models. Model Gap Effective Stiffness for Modal (Linear) Analysis k/in Mode 1 Period sec A 0 3.42 B 200,000 1.90 The load P is applied first and it causes the gap element to always be in compression. Thus the 1.90 sec period, which is the period when the gap stiffness is 200,000 k/in, is taken as the period of interest for application of the load P. The ramp rise time for this load is chosen as 20 seconds, which is approximately 10 times the period of interest. The load V is applied after load P and it causes the gap element to eventually uplift. Thus the 3.42 sec period, which is the period when the gap stiffness is 0 k/in, is taken as the period of interest for application of the load V. The ramp rise time for this load is chosen as 40 seconds, which is approximately 10 times the period of interest. Load case NLMHIST2 in Model B requires 200 output steps (800 seconds total) instead of the 20 output steps (80 seconds total) required in Model A. This occurs because in Model B the modes are calculated using the gap effective stiffness of 200,000 k/in and the 99.9% damping is applied to those modes. When the gap opens and the gap effective stiffness becomes zero, the period lengthens. Because the modal damping coefficient is held constant, the percent of critical damping increases in proportion to the period lengthening, thus over-damping the system. This over-damping that initiates when the gap opens is the reason that it takes so long for analysis to reach its final value in case NLMHIST2 for Model B. Load case NLMHIST2 in Model B also requires a force convergence tolerance of 1E-11 rather than the default 1E-05. The following equation, which is equation of motion used to solve nonlinear modal time history analyses, helps explain the rationale for the change in the convergence tolerance. EXAMPLE 6-003 - 5
... K L u(t) + K N u(t) + Cu(t) +Mu(t) + r N (t) = r(t) + K N u(t) In the preceding equation K L is the stiffness of all the linear elements, including the linear degrees of freedom of the link elements; K N is the linear effective stiffness matrix for all link element nonlinear degrees of freedom; C is the proportional damping matrix; M is the diagonal mass matrix; r is the vector of applied loads; and r N is the vector of forces from the nonlinear degrees of freedom of the link elements that is computed by iteration at time t. In the preceding equation, if the K N term is very large compared to the other terms, the relative force convergence tolerance needs to be very small to capture an accurate representation of r N. This is why the 1E-11 force convergence tolerance is required for load case NLMHIST2 in Model B. The nonlinear direct integration load cases need the maximum number of iterations per substep set to 100 rather than the default 10 and the iteration convergence tolerance set to 1E-06 rather than the default 1E-04. Reducing the iteration convergence tolerance to 1E-06 eliminates the slight fluctuations (high frequency chatter) in the results. For this particular example, increasing the maximum allowed number of iterations per substep to 100 decreases the running time of the problem by several orders of magnitude. Using the default 10 maximum iterations per substep causes the program to have to cut the time steps from the initially specified 0.1 second to as low as approximately 1E-07 second to solve the problem. When 100 maximum iterations per substep are allowed, the program does not have to reduce the time step size below 0.1 second to solve the problem. TECHNICAL FEATURES OF TESTED Gap element links Force-controlled nonlinear static analysis Nonlinear modal time history analysis Nonlinear direct time history analysis Frame point loads Joint force loads Joint mass assignments Ramp loading for time histories EXAMPLE 6-003 - 6
RESULTS COMPARISON Independent results are hand calculated using the unit load method described on page 244 in Cook and Young 1985. The results are presented separately for Model A and Model B Results for Model A (Gap K e = 0 k/in) Output Parameter Load Case Independent Percent Difference Link force after full load P is applied NLMHIST1-4.534 0% NLSTAT1-4.534-4.534 0% kip NLDHIST1-4.534 0% Link deformation after full load V is applied inch NLMHIST2 3.917 0% NLSTAT2 3.917 0% NLSTAT3 3.917 3.917 0% NLDHIST2 3.917 0% NLDHIST3 3.917 0% EXAMPLE 6-003 - 7
Results for Model B (Gap K e = 200,000 k/in) Output Parameter Load Case Independent Percent Difference Link force after full load P is applied NLMHIST1-4.534 0% NLSTAT1-4.534-4.534 0% kip NLDHIST1-4.534 0% Link deformation after full load V is applied inch NLMHIST2 3.917 0% NLSTAT2 3.917 0% NLSTAT3 3.917 3.917 0% NLDHIST2 3.917 0% NLDHIST3 3.917 0% COMPUTER FILE: Example 6-003a, Example 6-003b CONCLUSION The results show an acceptable comparison with the independent results both for Model A where the gap linear effective stiffness, K e, is 0 k/in and for Model B where the gap linear effective stiffness, K e, is 200,000 k/in. The comparison is exact when sufficiently small convergence tolerances are used. This example illustrates that solution of nonlinear problems using gap elements can be sensitive to the convergence and iteration tolerances that are used. For this verification example, it is easy to determine if the tolerances used are sufficient because the hand calculated results were available. In other situations it is helpful to run the analysis two or more times using tolerances that are different by an order of magnitude or more and verify that the results are the same. This is a good check that the tolerances used are sufficient. EXAMPLE 6-003 - 8
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