the relationship between nonlinear restoring forces p(t
+
.6.t) and applied load is (6.29) p(t+
.6.t) = f8+
A(t+
.6.t)fLwhere f8 is the static gravity loading applied before the displacement history and A(t+.6.t) is the unknown factor of the predetermined lateral forces, fL, to be applied at this step. Since this is a nonlinear problem, the restoring forces and load factor need to be linearized as follows:
(6.30) (6.31)
p(t
+
.6..t) = KT..O..x+
p(t) A(t+
.6.t)=
A(t)+
.6.A.Rewriting equation (6.29) for an iterative solution, the first iteration in step t
+
.6.tlS
(6.32)
(6.33) where K~
=
[KT(x(t))](6.34) A0(t
+
.6.t)=
A(t) (6.35) and p0(t+
.6.t)=
p(t).Solve equation (6.32) for Ax1 and .6.A1 such that the control displacement is satis- fied:
(6.36)
The displacements and load factor are updated as follows:
(6.37) (6.38)
x1(t
+
.6.t) = x(t)+
Ax1 A1(t+
.6.t)=
A(t)+
.6.A1.The stiffness matrix K} and restoring forces p1 ( t
+
.6.t) are updated using theseupdated displacements.
For the second iteration, solve (6.39)
for .6.x2 and .6..>.2 such that the control displacement is satisfied:
(6.40) x~(t
+
.6.t)-x~(t+
.6.t)=
0.Note that the right-hand side of equation (6.40) is zero. In the first iteration, the control displacement increment at dof q is satisfied by equation (6.36) and the corresponding factor .6..>.1 so that in future iterations the current displacement at the control dof should not change. This can be thought of as imposing the control displacement in the first iteration and iterating to reduce the residual forces until they are smaller than a specified tolerance as shown in figure 6.7. The solution keeps returning to the control displacement similar to the way in which the Newton- Raphson (NR) method (section 6.2) keeps returning to the applied load. The DC method can thus work past an ultimate load by reducing the applied load, allowing investigation of behavior in this region that the NR method cannot achieve.
The procedure for determining the load factor will now be discussed. Let the control displacement be represented as a function of iteration step k as follows:
(6.41) if k
=
1,if k 2: 2.
The equilibrium equations (6.32) for a general iteration step k are (6.42)
f
X· J
Figure 6. 7 Displacement control iterative method.
Separating these equations to solve for two separate loadings gives
{6.43) (6.44)
X
The right-hand side of equation (6.44) consists of all loads applied through all prior steps and iterations within the current step minus the restoring forces calculated at the previous iteration. This is simply the residual stresses from the previous step,
rk-1(t
+
ll.t), so the equation is rewritten as (6.45)Note that the solution to the original equation (6.42) can be obtained from the solutions of equations (6.43) and (6.45) by the following relationship:
(6.46)
Specifically, the displacement increments at the control dof q can be written (6.47)
This defines the factor !:J..>..k
(6.48)
Now, equations (6.43) and (6.45) for the first iteration become (6.49)
(6.50)
For simpler implementation of the DC method, it will be assumed that the residual force r0(t
+
11t)=
0, which means that the previous step converged exactly. This is a fair assumption for small tolerances. Under this assumption, the displacement increment ~x1 = 0 and thus at dof q the displacement increment 11x~ is also zero.The solution of equation (6.49) provides a displacement !1x~ which is scaled to the control displacement 11x~ by equation (6.48) for iteration k
=
1 as(6.51)
For iterations k
>
2, the control displacement 11x~ is zero, so equation (6.48) becomes(6.52)
The displacement control method can be summarized in the following steps. In the first iteration of a control step:
• Apply specified load f L and solve for
.6..x
1.• Calculate ~). 1 and adjust incremental and total displacements.
In the following iterations:
• Determine nodal forces and assemble in pk-l(t+~t). Determine rk-1(t+~t).
• Solve equation (6.43) for specified load fL.
• Solve equation (6.45) for residual forces rk-l(t
+
~t).• Calculate ~>. k and adjust incremental and total displacements, .6..xk and xk(t
+
~t).• Repeat until residual forces are less than tolerances.
Note that the incremental displacement for iterations k ~ 2 will be equal to zero at the control degree of freedom q. Even though it is zero at this dof, the other dofs may experience an increment. Similarly, the total displacement at the control dof will remain the same throughout a step, but total displacements must be updated at all the other dofs for the non-zero incremental changes.
To implement this approach for pushover analysis, the vertical gravity loading is applied first. This is the load f8 from equation (6.29). Once the solution is obtained for the gravity loading, the displacement control increments may be imposed. The load used for the first iteration of the first step is chosen to be proportional to the UBC static lateral force load distribution. Any fraction of the load can be used for f L since the factors ~). will adjust the load level to match the control displacement at the control degree of freedom.
The solution technique requires two solutions each iteration, but uses the same stiffness matrix for both (equations 6.43 and 6.45). For the first solution, Gaussian elimination with forward reduction and backward substitution is performed. For the second solution, only backward substitution is required, reducing the processing
time. In the IMNR method used in the program, elastic material properties are used in the tangent stiffness matrix for the first iteration and subsequent iterations use the tangent material properties to provide faster convergence. In the IMNR method, the objective is to minimize the number of iterations required to reach the displacement associated with a fixed load level. In the DC method, the objective is to minimize the number of iterations required to reach the load level associated with a fixed displacement. Because of the fixed displacements, the fraction of the predetermined lateral load (e.g., fL