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+<!-- source-page: 791 -->
+
+The solution of the equations in (b) is trivial because the coefficient matrix is diagonal.
+
+Calculating the solution to (b) for each time step we obtain
+
+<table><tr><td>Time</td><td> $\Delta t$ </td><td> $2\Delta t$ </td><td> $3\Delta t$ </td><td> $4\Delta t$ </td><td> $5\Delta t$ </td><td> $6\Delta t$ </td><td> $7\Delta t$ </td><td> $8\Delta t$ </td><td> $9\Delta t$ </td><td> $10\Delta t$ </td><td> $11\Delta t$ </td><td> $12\Delta t$ </td></tr><tr><td rowspan="2"> $^{t}U$ </td><td>0</td><td>0.0307</td><td>0.168</td><td>0.487</td><td>1.02</td><td>1.70</td><td>2.40</td><td>2.91</td><td>3.07</td><td>2.77</td><td>2.04</td><td>1.02</td></tr><tr><td>0.392</td><td>1.45</td><td>2.83</td><td>4.14</td><td>5.02</td><td>5.26</td><td>4.90</td><td>4.17</td><td>3.37</td><td>2.78</td><td>2.54</td><td>2.60</td></tr></table>
+
+The solution obtained is compared with the exact results in Example 9.7.
+
+Consider now case (2), in which $\Delta t = 28$ . Following through the same calculations, we find that
+
+$$
+^ {\Delta t} \mathbf {U} = \left[ \begin{array}{c} 0 \\ 3. 8 3 \times 1 0 ^ {3} \end{array} \right]; \quad^ {2 \Delta t} \mathbf {U} = \left[ \begin{array}{c} 3. 0 3 \times 1 0 ^ {6} \\ - 1. 2 1 \times 1 0 ^ {7} \end{array} \right]
+$$
+
+and the calculated displacements continue to increase. Since the time step $\Delta t$ is about 6 times larger than $T_{1}$ and 10 times larger than $T_{2}$ , we can certainly not expect accuracy in the numerical integration. But of particular interest is whether the calculated values decrease or increase. The increase in the values as observed in this example is a consequence of the time integration scheme not being stable. As pointed out above, the time step $\Delta t$ must not be larger than $\Delta t_{cr}$ for stability in the integration using the central difference method, where $\Delta t_{cr} = (1/\pi) T_{2}$ . In this case the time step $\Delta t$ is much larger, and the calculated response increases without bound. This is the typical phenomenon of instability. We shall see in Examples 9.2 to 9.4 that the response predicted using $\Delta t = 28$ with the unconditionally stable Houbolt, Newmark and Bathe methods is also very inaccurate but does not increase.
+
+We discussed above some advantages and disadvantages of the central difference method. Another disadvantage is that the solutions can contain significant spurious oscillations. An explicit scheme that shows less such errors has been presented by G. Noh and K.J. Bathe [A]. However, the effective use of conditionally stable methods is limited to certain problems. Therefore we consider in the following sections integration schemes which are unconditionally stable. With these methods, the time step $\Delta t$ is selected without a requirement such as (9.13), namely, only based on accuracy considerations, and in many cases $\Delta t$ can be orders of magnitude larger than (9.13) would allow. However, the integration methods discussed in the following are implicit; i.e., a triangularization of an effective stiffness matrix that includes K, is required for solution.
+
+# 9.2.2 The Houbolt Method
+
+The Houbolt integration scheme is somewhat related to the previously discussed central difference method in that standard finite difference expressions are used to approximate the acceleration and velocity components in terms of the displacement components. The following finite difference expansions are employed in the Houbolt integration method (see J. C. Houbolt [A]):
+
+$$
+{ } ^ { t + \Delta t } \ddot { \mathbf { U } } = \frac { 1 } { \Delta t ^ { 2 } } \left( 2 { } ^ { t + \Delta t } \mathbf { U } - 5 { } ^ { t } \mathbf { U } + 4 { } ^ { t - \Delta t } \mathbf { U } - { } ^ { t - 2 \Delta t } \mathbf { U } \right) \tag {9.14}
+$$
+
+<!-- source-page: 792 -->
+
+and $t + \Delta t\dot{\mathbf{U}} = \frac{1}{6\Delta t} (11^{t + \Delta t}\mathbf{U} - 18^{t}\mathbf{U} + 9^{t - \Delta t}\mathbf{U} - 2^{t - 2\Delta t}\mathbf{U})$ (9.15)
+
+which are two backward-difference formulas with errors of order $(\Delta t)^{2}$ .
+
+In order to obtain the solution at time $t + \Delta t$ , we now consider (9.1) at time $t + \Delta t$ (and not at time t as for the central difference method), which gives
+
+$$
+\mathbf {M} ^ {t + \Delta t} \ddot {\mathbf {U}} + \mathbf {C} ^ {t + \Delta t} \dot {\mathbf {U}} + \mathbf {K} ^ {t + \Delta t} \mathbf {U} = ^ {t + \Delta t} \mathbf {R} \tag {9.16}
+$$
+
+Substituting (9.14) and (9.15) into (9.16) and arranging all known vectors on the right-hand side, we obtain for the solution of ${}^{t+\Delta t}U$ ,
+
+$$
+\begin{array}{l} \left(\frac {2}{\Delta t ^ {2}} \mathbf {M} + \frac {1 1}{6 \Delta t} \mathbf {C} + \mathbf {K}\right) ^ {t + \Delta t} \mathbf {U} = ^ {t + \Delta t} \mathbf {R} + \left(\frac {5}{\Delta t ^ {2}} \mathbf {M} + \frac {3}{\Delta t} \mathbf {C}\right) ^ {t} \mathbf {U} \\ - \left(\frac {4}{\Delta t ^ {2}} \mathbf {M} + \frac {3}{2 \Delta t} \mathbf {C}\right) ^ {t - \Delta t} \mathbf {U} + \left(\frac {1}{\Delta t ^ {2}} \mathbf {M} + \frac {1}{3 \Delta t} \mathbf {C}\right) ^ {t - 2 \Delta t} \mathbf {U} \tag {9.17} \\ \end{array}
+$$
+
+As shown in (9.17), the solution of $^{t+\Delta t}\mathbf{U}$ requires knowledge of $^t\mathbf{U}$ , $^{t-\Delta t}\mathbf{U}$ , and $^{t-2\Delta t}\mathbf{U}$ . Although the knowledge of $^0\mathbf{U}$ , $^0\dot{\mathbf{U}}$ , and $^0\ddot{\mathbf{U}}$ is useful to start the Houbolt integration scheme, it is more accurate to calculate $^{\Delta t}\mathbf{U}$ and $^{2\Delta t}\mathbf{U}$ by some other means; i.e., we employ special starting procedures. One way of proceeding is to integrate (9.1) for the solution of $^{\Delta t}\mathbf{U}$ and $^{2\Delta t}\mathbf{U}$ using a different integration scheme, possibly a conditionally stable method such as the central difference scheme with a fraction of $\Delta t$ as the time step (see Example 9.2). Table 9.2 summarizes the Houbolt integration procedure for use in a computer program.
+
+TABLE 9.2 Step-by-step solution using Houbolt integration method
+
+A. Initial calculations:
+
+1. Form stiffness matrix K, mass matrix M, and damping matrix C.   
+2. Initialize ${}^{0}U$ , ${}^{0}\dot{U}$ , and ${}^{0}\ddot{U}$ .   
+3. Select time step $\Delta t$ and calculate integration constants:
+
+$$
+a _ {0} = \frac {2}{\Delta t ^ {2}}; \quad a _ {1} = \frac {1 1}{6 \Delta t}; \quad a _ {2} = \frac {5}{\Delta t ^ {2}}; \quad a _ {3} = \frac {3}{\Delta t}; \quad a _ {4} = - 2 a _ {0};
+$$
+
+$$
+a _ {5} = \frac {- a _ {3}}{2}; \quad a _ {6} = \frac {a _ {0}}{2}; \quad a _ {7} = \frac {a _ {3}}{9}
+$$
+
+4. Use special starting procedure to calculate $\Delta tU$ and $2\Delta tU$ .   
+5. Calculate effective stiffness matrix $\hat{\mathbf{K}}$ : $\hat{\mathbf{K}} = \mathbf{K} + a_0\mathbf{M} + a_1\mathbf{C}$ .   
+6. Triangularize $\tilde{\mathbf{K}}$ : $\tilde{\mathbf{K}} = \mathbf{LDL}^T$ .
+
+B. For each time step:
+
+1. Calculate effective load at time $t + \Delta t$ :
+
+$$
+{ } ^ { t + \Delta t } \hat { \mathbf { R } } = { } ^ { t + \Delta t } \mathbf { R } + \mathbf { M } ( a _ { 2 } { } ^ { t } \mathbf { U } + a _ { 4 } { } ^ { t - \Delta t } \mathbf { U } + a _ { 6 } { } ^ { t - 2 \Delta t } \mathbf { U } ) + \mathbf { C } ( a _ { 3 } { } ^ { t } \mathbf { U } + a _ { 5 } { } ^ { t - \Delta t } \mathbf { U } + a _ { 7 } { } ^ { t - 2 \Delta t } \mathbf { U } )
+$$
+
+2. Solve for displacements at time $t + \Delta t$ :
+
+$$
+\mathbf {L D L} ^ {T} {} ^ {t + \Delta t} \mathbf {U} = ^ {t + \Delta t} \hat {\mathbf {R}}
+$$
+
+3. If required, evaluate accelerations and velocities at time $t + \Delta t$ :
+
+$$
+{ } ^ { t + \Delta t } \ddot { \mathbf { U } } = a _ { 0 } { } ^ { t + \Delta t } \mathbf { U } - a _ { 2 } { } ^ { t } \mathbf { U } - a _ { 4 } { } ^ { t - \Delta t } \mathbf { U } - a _ { 6 } { } ^ { t - 2 \Delta t } \mathbf { U }
+$$
+
+$$
+{ } ^ { t + \Delta t } \dot { \mathbf { U } } = a _ { 1 } { } ^ { t + \Delta t } \mathbf { U } - a _ { 3 } { } ^ { t } \mathbf { U } - a _ { 5 } { } ^ { t - \Delta t } \mathbf { U } - a _ { 7 } { } ^ { t - 2 \Delta t } \mathbf { U }
+$$
+
+<!-- source-page: 793 -->
+
+A basic difference between the Houbolt method in Table 9.2 and the central difference scheme in Table 9.1 is the appearance of the stiffness matrix K as a factor to the required displacements $t^{+\Delta t}U$ . The term $K^{t+\Delta t}U$ appears because in (9.16) equilibrium is considered at time $t + \Delta t$ and not at time t as in the central difference method. The Houbolt method is, for this reason, an implicit integration scheme, whereas the central difference method was an explicit procedure. With regard to the time step $\Delta t$ that can be used in the integration, there is no critical time step limit and $\Delta t$ can in general be selected much larger than given in (9.13) for the central difference method.
+
+A noteworthy point is that the step-by-step solution scheme based on the Houbolt method reduces directly to a static analysis if mass and damping effects are neglected, whereas the central difference method solution in Table 9.1 could not be used. In other words, if C = 0 and M = 0, the solution method in Table 9.2 yields the static solution for time-dependent loads.
+
+EXAMPLE 9.2: Use the Houbolt direct integration scheme to calculate the response of the system considered in Example 9.1.
+
+First, we consider the case $\Delta t = 0.28$ . We then have, following Table 9.2, and showing three digits,
+
+$$
+a _ {0} = 2 5. 5; \quad a _ {1} = 6. 5 5; \quad a _ {2} = 6 3. 8; \quad a _ {3} = 1 0. 7;
+$$
+
+$$
+a _ {4} = - 5 1. 0; \qquad a _ {5} = - 5. 3 6; \qquad a _ {6} = 1 2. 8; \qquad a _ {7} = 1. 1 9
+$$
+
+To start the integration we need $\Delta^{r}U$ and $2\Delta^{r}U$ . Let us here use simply the values calculated with the central difference method in Example 9.1, i.e.,
+
+$$
+{ } ^ { \Delta \prime } \mathbf { U } = \left[ \begin{array} { c } 0 . 0 \\ 0 . 3 9 2 \end{array} \right] ; \quad { } ^ { 2 \Delta \prime } \mathbf { U } = \left[ \begin{array} { c } 0 . 0 3 0 7 \\ 1 . 4 5 \end{array} \right]
+$$
+
+Next we calculate $\hat{K}$ and obtain
+
+$$
+\hat {\mathbf {K}} = \left[ \begin{array}{c c} 6 & - 2 \\ - 2 & 4 \end{array} \right] + 2 5. 5 \left[ \begin{array}{l l} 2 & 0 \\ 0 & 1 \end{array} \right] = \left[ \begin{array}{c c} 5 7 & - 2 \\ - 2 & 2 9. 5 \end{array} \right]
+$$
+
+For each time step we need $^{t+\Delta t}\hat{R}$ , which is in this case
+
+$$
+^ {t + \Delta t} \hat {\mathbf {R}} = \left[ \begin{array}{l} 0 \\ 1 0 \end{array} \right] + \left[ \begin{array}{l l} 2 & 0 \\ 0 & 1 \end{array} \right] (6 3. 8 ^ {\prime} \mathbf {U} - 5 1. 0 ^ {t - \Delta t} \mathbf {U} + 1 2. 8 ^ {t - 2 \Delta t} \mathbf {U})
+$$
+
+Solving $\hat{\mathbf{K}}^{t + \Delta t}\mathbf{U} = {}^{t + \Delta t}\hat{\mathbf{R}}$ for 12 time steps, we obtain
+<table><tr><td>Time</td><td> $\Delta t$ </td><td> $2\Delta t$ </td><td> $3\Delta t$ </td><td> $4\Delta t$ </td><td> $5\Delta t$ </td><td> $6\Delta t$ </td><td> $7\Delta t$ </td><td> $8\Delta t$ </td><td> $9\Delta t$ </td><td> $10\Delta t$ </td><td> $11\Delta t$ </td><td> $12\Delta t$ </td></tr><tr><td rowspan="2">‘U</td><td>0</td><td>0.0307</td><td>0.167</td><td>0.461</td><td>0.923</td><td>1.50</td><td>2.11</td><td>2.60</td><td>2.86</td><td>2.80</td><td>2.40</td><td>1.72</td></tr><tr><td>0.392</td><td>1.45</td><td>2.80</td><td>4.08</td><td>5.02</td><td>5.43</td><td>5.31</td><td>4.77</td><td>4.01</td><td>3.24</td><td>2.63</td><td>2.28</td></tr></table>
+
+The solution obtained is compared with the exact results in Example 9.7.
+
+Next we consider the case $\Delta t = 28$ in order to observe the unconditional stability of the Houbolt operator. To start the integration we use the exact response at times $\Delta t$ and $2\Delta t$ (see Example 9.7),
+
+$$
+^ {\Delta t} \mathbf {U} = \left[ \begin{array}{l} 2. 1 9 \\ 2. 2 4 \end{array} \right]; \quad^ {2 \Delta t} \mathbf {U} = \left[ \begin{array}{l} 2. 9 2 \\ 3. 1 2 \end{array} \right]
+$$
+
+<!-- source-page: 794 -->
+
+It is interesting to compare $\hat{K}$ with K,
+
+$$
+\hat {\mathbf {K}} = \left[ \begin{array}{l l} 6 & - 2 \\ - 2 & 4 \end{array} \right] + 0. 0 0 2 5 5 \left[ \begin{array}{l l} 2 & 0 \\ 0 & 1 \end{array} \right] = \left[ \begin{array}{l l} 6. 0 0 5 1 & - 2. 0 0 0 \\ - 2. 0 0 0 & 4. 0 0 2 5 5 \end{array} \right]
+$$
+
+where it is noted that $\hat{\mathbf{K}}$ is almost equal to $\mathbf{K}$ . The displacement response over 12 time steps is given in the following table:
+<table><tr><td>Time</td><td> $\Delta t$ </td><td> $2\Delta t$ </td><td> $3\Delta t$ </td><td> $4\Delta t$ </td><td> $5\Delta t$ </td><td> $6\Delta t$ </td><td> $7\Delta t$ </td><td> $8\Delta t$ </td><td> $9\Delta t$ </td><td> $10\Delta t$ </td><td> $11\Delta t$ </td><td> $12\Delta t$ </td></tr><tr><td rowspan="2">‘U</td><td>2.19</td><td>2.92</td><td>1.00</td><td>1.00</td><td>1.00</td><td>1.00</td><td>1.00</td><td>1.00</td><td>1.00</td><td>1.00</td><td>1.00</td><td>1.00</td></tr><tr><td>2.24</td><td>3.12</td><td>3.00</td><td>3.00</td><td>3.00</td><td>3.00</td><td>3.00</td><td>3.00</td><td>3.00</td><td>3.00</td><td>3.00</td><td>3.00</td></tr></table>
+
+The static solution is $^{t}U=\begin{bmatrix}1.0\\ 3.0\end{bmatrix}$
+
+Therefore, the displacement response very rapidly approaches the static solution.
+
+# 9.2.3 The Newmark Method
+
+The Newmark integration scheme uses the following assumptions (see N. M. Newmark [A]):
+
+$$
+{ } ^ { t + \Delta t } \dot { \mathbf { U } } = { } ^ { t } \dot { \mathbf { U } } + \left[ ( 1 - \delta ) { } ^ { t } \ddot { \mathbf { U } } + \delta { } ^ { t + \Delta t } \ddot { \mathbf { U } } \right] \Delta t \tag {9.18}
+$$
+
+$$
+{ } ^ { t + \Delta t } \mathbf { U } = { } ^ { t } \mathbf { U } + { } ^ { t } \dot { \mathbf { U } } \Delta t + \left[ \left( \frac { 1 } { 2 } - \alpha \right) { } ^ { t } \ddot { \mathbf { U } } + \alpha { } ^ { t + \Delta t } \ddot { \mathbf { U } } \right] \Delta t ^ { 2 } \tag {9.19}
+$$
+
+where $\alpha$ and $\delta$ are parameters that can be determined to obtain integration accuracy and stability. When $\delta=\frac{1}{2}$ and $\alpha=\frac{1}{6}$ , we have the linear acceleration method. Newmark originally proposed as an unconditionally stable scheme the constant-average-acceleration method (also called trapezoidal rule, TR), in which case $\delta=\frac{1}{2}$ and $\alpha=\frac{1}{4}$ . The acceleration assumptions in Figures 9.1 and 9.2 are integrated in $\tau$ to obtain the schemes.
+
+![](images/page-794_5c172cda8eca75c815c151fdbd64fd7f87a10d1e0c66d487c16f2cdba59f28e2.jpg)
+
+<details>
+<summary>text_image</summary>
+
+t + \Delta t \dot{U}
+\tau
+t + \Delta t
+</details>
+
+Figure 9.1 Linear acceleration method
+
+![](images/page-794_26a356719baa14847a9b4ad56a07bb89ffc582e19935b731362ecf0daf196043.jpg)
+
+<details>
+<summary>text_image</summary>
+
+t\dot{U} = \frac{1}{2}(t\dot{U} + t+\Delta t\dot{U})
+t \tau t + \Delta t
+</details>
+
+Figure 9.2 TR scheme
+
+In addition to (9.18) and (9.19), for solution of the displacements, velocities, and accelerations at time $t + \Delta t$ , the equilibrium equations (9.1) at time $t + \Delta t$ are also used:
+
+$$
+\mathbf {M} ^ {t + \Delta t} \ddot {\mathbf {U}} + \mathbf {C} ^ {t + \Delta t} \dot {\mathbf {U}} + \mathbf {K} ^ {t + \Delta t} \mathbf {U} = ^ {t + \Delta t} \mathbf {R} \tag {9.20}
+$$
+
+<!-- source-page: 795 -->
+
+Considering the trapezoidal rule, which is mostly used, solving for $^{++\Delta t}\dot{\mathbf{U}}$ and $^{++\Delta t}\ddot{\mathbf{U}}$ from (9.18) and (9.19) in terms of $^{++\Delta t}\mathbf{U}$ , we solve for each time step
+
+$$
+\left(\frac {4}{\Delta t ^ {2}} \mathbf {M} + \frac {2}{\Delta t} \mathbf {C} + \mathbf {K}\right) ^ {t + \Delta t} \mathbf {U} = ^ {t + \Delta t} \mathbf {R} + \mathbf {M} \left(\frac {4}{\Delta t ^ {2}} ^ {\prime} \mathbf {U} + \frac {4}{\Delta t} ^ {\prime} \dot {\mathbf {U}} + ^ {\prime} \ddot {\mathbf {U}}\right) + \mathbf {C} \left(\frac {2}{\Delta t} ^ {\prime} \mathbf {U} + ^ {\prime} \dot {\mathbf {U}}\right) \tag {9.21}
+$$
+
+and then calculate $^{+\Delta t}\ddot{\mathbf{U}}$ and $^{+\Delta t}\dot{\mathbf{U}}$ . The complete solution procedure using the Newmark method is given in Table 9.3 and a problem is solved in Example 9.3.
+
+TABLE 9.3 Step-by-step solution using Newmark integration method
+
+# A. Initial calculations:
+
+1. Form stiffness matrix K, mass matrix M, and damping constant C.   
+2. Initialize $^{0}U$ , $^{0}\dot{U}$ , and $^{0}\ddot{U}$ .   
+3. Select time step $\Delta t$ and parameters $\alpha$ and $\delta$ and calculate integration constants:
+
+$$
+\delta \geq 0. 5 0;
+$$
+
+$$
+\alpha \geq 0. 2 5 (0. 5 + \delta) ^ {2}
+$$
+
+4. $a_0 = \frac{1}{\alpha\Delta t^2};$
+
+$$
+a _ {1} = \frac {\delta}{\alpha \Delta t};
+$$
+
+$$
+a _ {2} = \frac {1}{\alpha \Delta t};
+$$
+
+$$
+a _ {3} = \frac {1}{2 \alpha} - 1;
+$$
+
+$$
+a _ {4} = \frac {\delta}{\alpha} - 1;
+$$
+
+$$
+a _ {s} = \frac {\Delta t}{2} \left(\frac {\delta}{\alpha} - 2\right);
+$$
+
+$$
+a _ {6} = \Delta t (1 - \delta);
+$$
+
+$$
+a _ {7} = \delta \Delta t
+$$
+
+5. Form effective stiffness matrix $\hat{\mathbf{K}}:\hat{\mathbf{K}} = \mathbf{K} + a_0\mathbf{M} + a_1\mathbf{C}$ .   
+6. Triangularize $\hat{\mathbf{K}}:\hat{\mathbf{K}} = \mathbf{LDL}^T$
+
+# B. For each time step:
+
+1. Calculate effective loads at time $t + \Delta t$ :
+
+$$
+^ {t + \Delta t} \hat {\mathbf {R}} = ^ {t + \Delta t} \mathbf {R} + \mathbf {M} \left(a _ {0} ^ {\prime} \mathbf {U} + a _ {2} ^ {\prime} \dot {\mathbf {U}} + a _ {3} ^ {\prime} \ddot {\mathbf {U}}\right) + \mathbf {C} \left(a _ {1} ^ {\prime} \mathbf {U} + a _ {4} ^ {\prime} \dot {\mathbf {U}} + a _ {5} ^ {\prime} \ddot {\mathbf {U}}\right)
+$$
+
+2. Solve for displacements at time $t + \Delta t$ :
+
+$$
+\mathbf {L D L} ^ {T} {} ^ {t + \Delta t} \mathbf {U} = ^ {t + \Delta t} \hat {\mathbf {R}}
+$$
+
+3. Calculate accelerations and velocities at time $t + \Delta t$ :
+
+$$
+{ } ^ { t + \Delta t } \ddot { \mathbf { U } } = a _ { 0 } \left( { } ^ { t + \Delta t } \mathbf { U } - { } ^ { t } \mathbf { U } \right) - a _ { 2 } { } ^ { t } \dot { \mathbf { U } } - a _ { 3 } { } ^ { t } \ddot { \mathbf { U } }
+$$
+
+$$
+^ {t + \Delta t} \dot {\mathbf {U}} = ^ {t} \dot {\mathbf {U}} + a _ {6} ^ {t} \ddot {\mathbf {U}} + a _ {7} ^ {t + \Delta t} \ddot {\mathbf {U}}
+$$
+
+EXAMPLE 9.3: Calculate the displacement response of the system considered in Examples 9.1 and 9.2 using the Newmark method. Use $\alpha=0.25$ , $\delta=0.5$ .
+
+Consider the case $\Delta t = 0.28$ . Following the steps of calculations given in Table 9.3, we have
+
+$$
+{ } ^ { 0 } \mathbf { U } = \left[ \begin{array} { c } 0 \\ 0 \end{array} \right] ; \quad { } ^ { 0 } \dot { \mathbf { U } } = \left[ \begin{array} { c } 0 \\ 0 \end{array} \right] ; \quad { } ^ { 0 } \ddot { \mathbf { U } } = \left[ \begin{array} { c } 0 \\ 1 0 \end{array} \right]
+$$
+
+The integration constants are (showing three digits)
+
+<!-- source-page: 796 -->
+
+$$
+a _ {0} = 5 1. 0; \quad a _ {1} = 7. 1 4; \quad a _ {2} = 1 4. 3; \quad a _ {3} = 1. 0 0;
+$$
+
+$$
+a _ {4} = 1. 0 0; \quad a _ {5} = 0. 0 0; \quad a _ {6} = 0. 1 4; \quad a _ {7} = 0. 1 4
+$$
+
+Thus the effective stiffness matrix is
+
+$$
+\hat {\mathbf {K}} = \left[ \begin{array}{l l} 6 & - 2 \\ - 2 & 4 \end{array} \right] + 5 1. 0 \left[ \begin{array}{l l} 2 & 0 \\ 0 & 1 \end{array} \right] = \left[ \begin{array}{l l} 1 0 8 & - 2 \\ - 2 & 5 5 \end{array} \right]
+$$
+
+For each time step we need to evaluate
+
+$$
+{ } ^ { \prime + \Delta t } \hat { \mathbf { R } } = \left[ \begin{array} { c } 0 \\ 1 0 \end{array} \right] + \left[ \begin{array} { c c } 2 & 0 \\ 0 & 1 \end{array} \right] \left( 5 1 { } ^ { \prime } \mathbf { U } + 1 4 . 3 { } ^ { \prime } \dot { \mathbf { U } } + 1 . 0 { } ^ { \prime } \ddot { \mathbf { U } } \right)
+$$
+
+Then
+
+$$
+\hat {\mathbf {K}} ^ {t + \Delta t} \mathbf {U} = ^ {t + \Delta t} \hat {\mathbf {R}}
+$$
+
+and
+
+$$
+^ {t + \Delta t} \ddot {\mathbf {U}} = 5 1. 0 \left(^ {t + \Delta t} \mathbf {U} - ^ {t} \mathbf {U}\right) - 1 4. 3 ^ {t} \dot {\mathbf {U}} - 1. 0 ^ {t} \ddot {\mathbf {U}}
+$$
+
+$$
+^ {t + \Delta t} \dot {\mathbf {U}} = ^ {t} \dot {\mathbf {U}} + 0. 1 4 ^ {t} \ddot {\mathbf {U}} + 0. 1 4 ^ {t + \Delta t} \ddot {\mathbf {U}}
+$$
+
+Performing these calculations, we obtain
+<table><tr><td>Time</td><td> $\Delta t$ </td><td> $2\Delta t$ </td><td> $3\Delta t$ </td><td> $4\Delta t$ </td><td> $5\Delta t$ </td><td> $6\Delta t$ </td><td> $7\Delta t$ </td><td> $8\Delta t$ </td><td> $9\Delta t$ </td><td> $10\Delta t$ </td><td> $11\Delta t$ </td><td> $12\Delta t$ </td></tr><tr><td rowspan="2"> $^tU$ </td><td>0.00673</td><td>0.0505</td><td>0.189</td><td>0.485</td><td>0.961</td><td>1.58</td><td>2.23</td><td>2.76</td><td>3.00</td><td>2.85</td><td>2.28</td><td>1.40</td></tr><tr><td>0.364</td><td>1.35</td><td>2.68</td><td>4.00</td><td>4.95</td><td>5.34</td><td>5.13</td><td>4.48</td><td>3.64</td><td>2.90</td><td>2.44</td><td>2.31</td></tr></table>
+
+The solution we obtained is compared with the exact results in Example 9.7.
+
+Considering next the solution using $\Delta t = 28$ , ideally the static solution would rapidly be obtained, as when using the Houbolt method, see Example 9.2. However, we find that the solution is quite inaccurate with the displacements being 0.894 (instead of 1) and 1.45 (instead of 3) at $12\Delta t$ . The static solution is more closely obtained if the initial accelerations are set to zero (but the error for the static solution is then larger after 12 steps than after the first step where the error is very small).
+
+# 9.2.4 The Bathe Method
+
+The Bathe method uses two sub-steps for each time integration step $\Delta t$ . In the first sub-step the Newmark trapezoidal rule is used and in the second sub-step, the 3-point Euler backward method is employed, see K. J. Bathe [F]. Although different size sub-steps can be used, see K. J. Bathe and M.M.I. Baig [A], we present here, for clarity, the scheme with equal size sub-steps (the approach of integrating is the same when different sub-step sizes are used.). For the first sub-step we use
+
+$$
+{ } ^ { t + \Delta t / 2 } \dot { \mathbf { U } } = { } ^ { t } \dot { \mathbf { U } } + \left[ \frac { \Delta t } { 4 } \right] ( { } ^ { t } \ddot { \mathbf { U } } + { } ^ { t + \Delta t / 2 } \ddot { \mathbf { U } } ) \tag {9.22}
+$$
+
+$$
+{ } ^ { t + \Delta t / 2 } \mathbf { U } = { } ^ { t } \mathbf { U } + \left[ \frac { \Delta t } { 4 } \right] \left( { } ^ { t } \dot { \mathbf { U } } + { } ^ { t + \Delta t / 2 } \dot { \mathbf { U } } \right) \tag {9.23}
+$$
+
+and for the second sub-step we use
+
+<!-- source-page: 797 -->
+
+$$
+{ } ^ { t + \Delta t } \dot { \mathbf { U } } = \frac { 1 } { \Delta t } { } ^ { t } \mathbf { U } - \frac { 4 } { \Delta t } { } ^ { t + \Delta t / 2 } \mathbf { U } + \frac { 3 } { \Delta t } { } ^ { t + \Delta t } \mathbf { U } \tag {9.24}
+$$
+
+$$
+{ } ^ { t + \Delta t } \ddot { \mathbf { U } } = \frac { 1 } { \Delta t } { } ^ { t } \dot { \mathbf { U } } - \frac { 4 } { \Delta t } { } ^ { t + \Delta t / 2 } \dot { \mathbf { U } } + \frac { 3 } { \Delta t } { } ^ { t + \Delta t } \dot { \mathbf { U } } \tag {9.25}
+$$
+
+With (9.22) and (9.23) and the finite element equilibrium equations at time $t + \Delta t/2$ , that is, (9.16) but at time $t + \Delta t/2$ , and then (9.24) and (9.25) with the finite element equilibrium equations at time $t + \Delta t$ , that is (9.16), we obtain
+
+$$
+\left(\frac {1 6}{\Delta t ^ {2}} \mathbf {M} + \frac {4}{\Delta t} \mathbf {C} + \mathbf {K}\right) ^ {t + \Delta t / 2} \mathbf {U} = ^ {t + \Delta t / 2} \hat {\mathbf {R}} \tag {9.26}
+$$
+
+with ${}^{t+\Delta t/2}\hat{\mathbf{R}}={}^{t+\Delta t/2}\mathbf{R}+\mathbf{M}\left(\frac{16}{\Delta t^{2}}{}^{t}\mathbf{U}+\frac{8}{\Delta t}{}^{t}\dot{\mathbf{U}}+{}^{t}\ddot{\mathbf{U}}\right)+\mathbf{C}\left(\frac{4}{\Delta t}{}^{t}\mathbf{U}+{}^{t}\dot{\mathbf{U}}\right)$ (9.27)
+
+and $\left(\frac{9}{\Delta t^2}\mathbf{M} + \frac{3}{\Delta t}\mathbf{C} + \mathbf{K}\right)^{t + \Delta t}\mathbf{U} = ^{t + \Delta t}\hat{\mathbf{R}}$ (9.28)
+
+with $t^{+\Delta t/2}R$ in general best interpolated linearly from the load values at times t and $t + \Delta t$ (see G. Noh and K.J. Bathe [A]) and
+
+$$
+{ } ^ { t + \Delta t } \hat { \mathbf { R } } = { } ^ { t + \Delta t } \mathbf { R } + \mathbf { M } \left( \frac { 1 2 } { \Delta t ^ { 2 } } { } ^ { t + \Delta t / 2 } \mathbf { U } - \frac { 3 } { \Delta t ^ { 2 } } { } ^ { t } \mathbf { U } + \frac { 4 } { \Delta t } { } ^ { t + \Delta t / 2 } \dot { \mathbf { U } } - \frac { 1 } { \Delta t } { } ^ { t } \dot { \mathbf { U } } \right) + \mathbf { C } \left( \frac { 4 } { \Delta t } { } ^ { t + \Delta t / 2 } \mathbf { U } - \frac { 1 } { \Delta t } { } ^ { t } \mathbf { U } \right) \tag {9.29}
+$$
+
+The complete solution scheme is given in Table 9.4. While it appears that the solution requires twice the computational effort as the trapezoidal rule, in fact, the accuracy and stability characteristics are such that larger time steps can be used and the scheme is in general quite effective, in particular for nonlinear solutions, see K.J. Bathe [F], K.J. Bathe and G. Noh [A] and G. Noh, S. Ham, and K.J. Bathe [A].
+
+Table 9.4 Step-by-step solution using the Bathe integration method
+
+# A. Initial calculations:
+
+1. Form stiffness matrix K, mass matrix M, and damping matrix C   
+2. Initialize ${}^{0}U$ , ${}^{0}\dot{U}$ and ${}^{0}\ddot{U}$ .   
+3. Select time step $\Delta t$ and calculate integration constants:
+
+$$
+\begin{array}{l} a _ {0} = \frac {1 6}{\Delta t ^ {2}}; \quad a _ {1} = \frac {4}{\Delta t}; \quad a _ {2} = \frac {9}{\Delta t ^ {2}}; \quad a _ {3} = \frac {3}{\Delta t}; \\ a _ {4} = 2 a _ {1}; \quad a _ {5} = \frac {1 2}{\Delta t ^ {2}}; \quad a _ {6} = - \frac {3}{\Delta t ^ {2}}; \quad a _ {7} = - \frac {1}{\Delta t} \\ \end{array}
+$$
+
+4. Form effective stiffness matrices $\hat{K}_{1}$ and $\hat{K}_{2}$ :
+
+$$
+\hat {\mathbf {K}} _ {1} = \mathbf {K} + a _ {0} \mathbf {M} + a _ {1} \mathbf {C}; \quad \hat {\mathbf {K}} _ {2} = \mathbf {K} + a _ {2} \mathbf {M} + a _ {3} \mathbf {C}
+$$
+
+5. Triangularize $\hat{\mathbf{K}}_1$ and $\hat{\mathbf{K}}_2$ : $\hat{\mathbf{K}}_1 = \mathbf{L}_1\mathbf{D}_1\mathbf{L}_1^r$ ; $\hat{\mathbf{K}}_2 = \mathbf{L}_2\mathbf{D}_2\mathbf{L}_2^r$
+
+<!-- source-page: 798 -->
+
+# B. For each time step:
+
+First sub-step
+
+1. Calculate effective loads at time $t + \Delta t/2$ :
+
+$$
+^ {t + \Delta t / 2} \hat {\mathbf {R}} = ^ {t + \Delta t / 2} \mathbf {R} + \mathbf {M} \left(a _ {0} ^ {\prime} \mathbf {U} + a _ {4} ^ {\prime} \dot {\mathbf {U}} + ^ {\prime} \ddot {\mathbf {U}}\right) + \mathbf {C} \left(a _ {1} ^ {\prime} \mathbf {U} + ^ {\prime} \dot {\mathbf {U}}\right)
+$$
+
+2. Solve for displacements at time $t + \Delta t/2$ :
+
+$$
+\mathbf {L} _ {1} \mathbf {D} _ {1} \mathbf {L} _ {1} ^ {T} {} ^ {t + \Delta t / 2} \mathbf {U} = ^ {t + \Delta t / 2} \hat {\mathbf {R}}
+$$
+
+3. Calculate velocities at time $t + \Delta t/2$ :
+
+$$
+{ } ^ { t + \Delta t / 2 } \dot { \mathbf { U } } = a _ { 1 } \left( { } ^ { t + \Delta t / 2 } \mathbf { U } - { } ^ { t } \mathbf { U } \right) - { } ^ { t } \dot { \mathbf { U } }
+$$
+
+4. If required, evaluate accelerations at time $t + \Delta t/2$ :
+
+$$
+^ {t + \Delta t / 2} \ddot {\mathbf {U}} = a _ {1} \left(^ {t + \Delta t / 2} \dot {\mathbf {U}} - ^ {t} \dot {\mathbf {U}}\right) - ^ {t} \ddot {\mathbf {U}}
+$$
+
+Second sub-step
+
+1. Calculate effective loads at time $t + \Delta t$ :
+
+$$
+{ } ^ { t + \Delta t } \hat { \mathbf { R } } = { } ^ { t + \Delta t } \mathbf { R } + \mathbf { M } ( a _ { 5 } { } ^ { t + \Delta t / 2 } \mathbf { U } + a _ { 6 } { } ^ { t } \mathbf { U } + a _ { 1 } { } ^ { t + \Delta t / 2 } \dot { \mathbf { U } } + a _ { 7 } { } ^ { t } \dot { \mathbf { U } } ) + \mathbf { C } ( a _ { 1 } { } ^ { t + \Delta t / 2 } \mathbf { U } + a _ { 7 } { } ^ { t } \mathbf { U } )
+$$
+
+2. Solve for displacements at time $t + \Delta t$ :
+
+$$
+\mathbf {L} _ {2} \mathbf {D} _ {2} \mathbf {L} _ {2} ^ {T} {} ^ {t + \Delta t} \mathbf {U} = ^ {t + \Delta t} \hat {\mathbf {R}}
+$$
+
+3. Calculate velocities and accelerations at time $t + \Delta t$ :
+
+$$
+^ {t + \Delta t} \dot {\mathbf {U}} = - a, ^ {t} \mathbf {U} - a _ {1} ^ {t + \Delta t / 2} \mathbf {U} + a _ {3} ^ {t + \Delta t} \mathbf {U}
+$$
+
+$$
+{ } ^ { t + \Delta t } \ddot { \mathbf { U } } = - a _ { 7 } { } ^ { t } \dot { \mathbf { U } } - a _ { 1 } { } ^ { t + \Delta t / 2 } \dot { \mathbf { U } } + a _ { 3 } { } ^ { t + \Delta t } \dot { \mathbf { U } }
+$$
+
+EXAMPLE 9.4: Calculate the displacement response of the system considered in Examples 9.1 to 9.3 using the Bathe method.
+
+First we consider the case $\Delta t = 0.28$ . Following the steps of calculations in Table 9.4, we have
+
+$$
+{ } ^ { 0 } \mathbf { U } = \left[ \begin{array} { c } 0 \\ 0 \end{array} \right] ; \quad { } ^ { 0 } \dot { \mathbf { U } } = \left[ \begin{array} { c } 0 \\ 0 \end{array} \right] ; \quad { } ^ { 0 } \ddot { \mathbf { U } } = \left[ \begin{array} { c } 0 \\ 1 0 \end{array} \right]
+$$
+
+Then (listed to three digits)
+
+$$
+a _ {0} = 2 0 4; a _ {1} = 1 4. 3; a _ {2} = 1 1 4; a _ {3} = 1 0. 7; a _ {4} = 2 8. 6; a _ {5} = 1 5 3; a _ {6} = - 3 8. 3; a _ {7} = - 3. 5 7
+$$
+
+Thus the effective stiffness matrices are
+
+$$
+\hat {\mathbf {K}} _ {1} = \left[ \begin{array}{l l} 4 1 4 & - 2 \\ - 2 & 2 0 8 \end{array} \right]; \quad \hat {\mathbf {K}} _ {2} = \left[ \begin{array}{l l} 2 3 5 & - 2 \\ - 2 & 1 1 8 \end{array} \right]
+$$
+
+For each time step, in the first sub-step, we first evaluate
+
+$$
+\hat {\mathbf {K}} _ {1} ^ {\prime + \Delta t / 2} \mathbf {U} = \left[ \begin{array}{l} 0 \\ 1 0 \end{array} \right] + \left[ \begin{array}{l l} 2 & 0 \\ 0 & 1 \end{array} \right] (2 0 4 ^ {\prime} \mathbf {U} + 2 8. 6 ^ {\prime} \dot {\mathbf {U}} + ^ {\prime} \ddot {\mathbf {U}})
+$$
+
+and then calculate
+
+$$
+^ {t + \Delta t / 2} \dot {\mathbf {U}} = 2 0 4 (^ {t + \Delta t / 2} \mathbf {U} - ^ {t} \mathbf {U}) - ^ {t} \dot {\mathbf {U}}
+$$
+
+<!-- source-page: 799 -->
+
+In the second sub-step, we evaluate
+
+$$
+\hat {\mathbf {K}} _ {2} ^ {t + \Delta t} \mathbf {U} = \left[ \begin{array}{l} 0 \\ 1 0 \end{array} \right] + \left[ \begin{array}{l l} 2 & 0 \\ 0 & 1 \end{array} \right] (1 5 3 ^ {t + \Delta t / 2} \mathbf {U} - 3 8. 3 ^ {t} \mathbf {U} + 1 4. 3 ^ {t + \Delta t / 2} \dot {\mathbf {U}} - 3. 5 7 ^ {t} \dot {\mathbf {U}})
+$$
+
+and then calculate
+
+$$
+{ } ^ { t + \Delta t } \dot { \mathbf { U } } = 3 . 5 7 ^ { t } \mathbf { U } - 1 4 . 3 ^ { t + \Delta t / 2 } \mathbf { U } + 1 0 . 7 ^ { t + \Delta t } \mathbf { U }
+$$
+
+$$
+^ {t + \Delta t} \ddot {\mathbf {U}} = 3. 5 7 ^ {t} \dot {\mathbf {U}} - 1 4. 3 ^ {t + \Delta t / 2} \dot {\mathbf {U}} + 1 0. 7 ^ {t + \Delta t} \dot {\mathbf {U}}
+$$
+
+With these calculations we obtain
+<table><tr><td>Time</td><td> $\Delta t$ </td><td> $2\Delta t$ </td><td> $3\Delta t$ </td><td> $4\Delta t$ </td><td> $5\Delta t$ </td><td> $6\Delta t$ </td><td> $7\Delta t$ </td><td> $8\Delta t$ </td><td> $9\Delta t$ </td><td> $10\Delta t$ </td><td> $11\Delta t$ </td><td> $12\Delta t$ </td></tr><tr><td rowspan="2"> $^{t}U$ </td><td>0.00458</td><td>0.0445</td><td>0.183</td><td>0.486</td><td>0.979</td><td>1.62</td><td>2.28</td><td>2.81</td><td>3.03</td><td>2.83</td><td>2.21</td><td>1.28</td></tr><tr><td>0.373</td><td>1.38</td><td>2.73</td><td>4.04</td><td>4.97</td><td>5.31</td><td>5.06</td><td>4.38</td><td>3.55</td><td>2.85</td><td>2.46</td><td>2.40</td></tr></table>
+
+Considering next the solution using $\Delta t = 28$ , the solution is immediately close to the static solution and exactly equal to the static solution from $4\Delta t$ onwards, as is much desired.
+
+The scheme as presented here requires the use of two effective stiffness matrices, see (9.26) and (9.28). If, however, the first sub-step is $\gamma\Delta t$ , with $\gamma=2-\sqrt{2}$ , the same matrices are obtained, see K.J. Bathe and M.M.I. Baig [A]. Hence, in linear analysis it is most effective to use the sub-steps $\gamma\Delta t$ and $(1-\gamma\Delta t)$ , also because the accuracy characteristics are quite close to those when using equal-size sub-steps, see K.J. Bathe and G. Noh [A].
+
+The use of different matrices is of course not a disadvantage in nonlinear analysis since new tangent stiffness matrices are in any case established during the Newton-Raphson iterations, see Section 9.5.2. Considering nonlinear solutions, values of $\gamma$ other than 0.5 could also be used but the optimal value is problem-dependent. The optimal sub-step size for a specific problem solution should be measured on the best accuracy and least computational effort (in terms of total number of required Newton-Raphson iterations for the complete response evaluation), and the value of $\gamma$ thus obtained would ideally be independent of the total number of time steps used. Considering the solutions of two problems with different physical characteristics shows that different optimal values of $\gamma$ are obtained for such cases, see K.J. Bathe [K].
+
+The Bathe method was initially proposed for nonlinear dynamic analyses with large deformations and long time durations, that is, cases, in which the trapezoidal shows an unstable behavior, see K.J. Bathe and M.M.I. Baig [A] and K.J. Bathe [F]. However, it was then found in practical analyses that the method is also effective in linear solutions. In structural dynamics, spurious oscillations (due to artificially stiff elements) are cut out of the solutions, see K.J. Bathe and G. Noh [A], and in wave propagations, modes that cannot be resolved spatially are automatically not included in the response prediction, see G. Noh, S. Ham, and K.J. Bathe [A].
+
+# 9.2.5 The Coupling of Different Integration Operators
+
+So far we have assumed that all dynamic equilibrium equations are solved using the same time integration scheme. As discussed in Section 9.4, the choice of which method to use for an effective solution depends on the problem to be analyzed. However, for certain kinds
+
+<!-- source-page: 800 -->
+
+of problems it may be advantageous to use different time integration schemes to solve for the response in different regions of the total element assemblage, specifically, an explicit scheme and an implicit technique may be employed coupled in one solution. This is particularly the case when the stiffness and mass characteristics (i.e., the characteristic time constants) of the total element assemblage are quite different in different parts of the element assemblage. An example is the analysis of fluid-structure systems, in which the fluid is very flexible when measured on the stiffness of the structure. Here the explicit time integration of the fluid response using the conditionally stable central difference method and an implicit unconditionally stable time integration of the structural response (using, for example, the Newmark method) may be a natural choice (see Section 9.4). The reasons are that, first, the physical phenomenon to be analyzed may be a wave propagation in the fluid and a structural vibration of the structure, and second, the critical time step size for an explicit time integration of the fluid response is usually much larger than the time step required for explicit time integration of the structural response. The result may be that by proper choice of the finite element idealizations of the fluid and the structure, the explicit time integration of the fluid response and implicit time integration of the structural response can be performed with a time step that is relatively large but small enough to yield a stable and accurate solution. Of course, it may also be efficient to use different time step sizes for the explicit and implicit integrations, with one time step size being a multiple of the other.
+
+The use of a combination of operators for the integration of dynamic response raises the questions of which methods to choose and how to couple them. There are a large number of possibilities, but in general the selection of the schemes depends on their stability and accuracy characteristics, including the effects due to the operator coupling, and the overall effectiveness of the resulting time integration. We demonstrate the use of explicit-implicit time integration in the analysis of the simple problem considered in Examples 9.1 to 9.4.
+
+EXAMPLE 9.5: Solve for $U_{1}$ and $U_{2}$ of the simple system considered in Example 9.1 using the explicit central difference method for $U_{1}$ and the implicit trapezoidal rule (Newmark's method with $\alpha = \frac{1}{4}, \delta = \frac{1}{2}$ ) for $U_{2}$ .
+
+In the explicit integration we consider the equilibrium at time t to calculate the displacement for time $t + \Delta t$ . For degree of freedom 1, we have
+
+$$
+2 ^ {\prime} \ddot {U} _ {1} + 6 ^ {\prime} U _ {1} - 2 ^ {\prime} U _ {2} = 0 \tag {a}
+$$
+
+In the implicit integration we consider the equilibrium at time $t + \Delta t$ to calculate the displacement for time $t + \Delta t$ . Thus, we have for degree of freedom 2,
+
+$$
+{ } ^ { t + \Delta t } \ddot { U } _ { 2 } - 2 { } ^ { t + \Delta t } U _ { 1 } + 4 { } ^ { t + \Delta t } U _ { 2 } = 1 0 \tag {b}
+$$
+
+For (a) we now use the central difference method,
+
+$$
+^ \prime \ddot {U} _ {1} = \frac {{} ^ {t + \Delta t} U _ {1} - 2 ^ {t} U _ {1} + {} ^ {t - \Delta t} U _ {1}}{(\Delta t) ^ {2}} \tag {c}
+$$
+
+and for (b) we use the trapezoidal rule,
+
+$$
+^ {t + \Delta t} \dot {U} _ {2} = ^ {t} \dot {U} _ {2} + \frac {\Delta t}{2} (^ {t} \ddot {U} _ {2} + ^ {t + \Delta t} \ddot {U} _ {2}) \tag {d}
+$$
+
+$$
+^ {t + \Delta t} U _ {2} = ^ {t} U _ {2} + ^ {t} \dot {U} _ {2} \Delta t + \frac {(\Delta t) ^ {2}}{4} (^ {t} \ddot {U} _ {2} + ^ {t + \Delta t} \ddot {U} _ {2})
+$$