김경종
435 lines
24 KiB
Markdown
435 lines
24 KiB
Markdown
<!-- source-page: 671 -->
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or, on multiplying the matrices of Eq. (16.2.21),
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$$
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[ \hat {m} ] = \rho A \int_ {0} ^ {L} \left[ \begin{array}{c c} \left(1 - \frac {\hat {x}}{L}\right) ^ {2} & \left(1 - \frac {\hat {x}}{L}\right) \frac {\hat {x}}{L} \\ \left(1 - \frac {\hat {x}}{L}\right) \frac {\hat {x}}{L} & \left(\frac {\hat {x}}{L}\right) ^ {2} \end{array} \right] d \hat {x} \tag {16.2.22}
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$$
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On integrating Eq. (16.2.22) term by term, we obtain the consistent-mass matrix for a bar element as
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$$
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[ \hat {m} ] = \frac {\rho A L}{6} \left[ \begin{array}{l l} 2 & 1 \\ 1 & 2 \end{array} \right] \tag {16.2.23}
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$$
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# Step 5 Assemble the Element Equations to Obtain the Global Equations and Introduce Boundary Conditions
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We assemble the element equations using the direct stiffness method such that interelement continuity of displacements is again satisfied at common nodes and, in addition, interelement continuity of accelerations is also satisfied; that is, we obtain the global equations
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$$
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\{F (t) \} = [ K ] \{d \} + [ M ] \{\ddot {d} \} \tag {16.2.24}
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$$
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where
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$$
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[ K ] = \sum_ {e = 1} ^ {N} [ k ^ {(e)} ] \quad [ M ] = \sum_ {e = 1} ^ {N} [ m ^ {(e)} ] \quad \{F \} = \sum_ {e = 1} ^ {N} \{f ^ {(e)} \} \tag {16.2.25}
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$$
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are the global stiffness, mass, and force matrices, respectively. Note that the global mass matrix is assembled in the same manner as the global stiffness matrix. Equation (16.2.24) represents a set of matrix equations discretized with respect to space. To obtain the solution of the equations, discretization in time is also necessary. We will describe this process in Section 16.3 and will later present representative solutions illustrating these equations.
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# 16.3 Numerical Integration in Time
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We now introduce procedures for the discretization of Eq. (16.2.24) with respect to time. These procedures will enable us to determine the nodal displacements at different time increments for a given dynamic system. The general method used is called direct integration. There are two classifications of direct integration: explicit and implicit. We will formulate the equations for three direct integration methods. The first, and simplest, is an explicit method known as the central difference method [3, 4]. The second and third, more complicated but more versatile than the central difference method, are implicit methods known as the Newmark-Beta (or Newmark’s) method [5] and the Wilson-Theta (or Wilson’s) method [7, 8]. The versatility of both Newmark’s and Wilson’s methods is evidenced by their adaptation in many commercially available computer programs. Wilson’s method is used in the Algor computer
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<!-- source-page: 672 -->
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<details>
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<summary>line</summary>
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| t | d(t) |
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|-------------|-------|
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| t_i - Δt | d_{i-1} |
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| t_i | d_{i+1} |
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| t_i + Δt | d_{i+1} |
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</details>
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Figure 16–5 Numerical integration (approximation of derivative at ti)
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program [16]. Numerous other integration methods are available in the literature. Among these are Houboldt’s method [8] and the alpha method [13].
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# Central Difference Method
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The central difference method is based on finite difference expressions in time for velocity and acceleration at time t given by
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$$
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\underline {{d}} _ {i} = \frac {\underline {{d}} _ {i + 1} - \underline {{d}} _ {i - 1}}{2 (\Delta t)} \tag {16.3.1}
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$$
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$$
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\ddot {\underline {{d}}} _ {i} = \frac {\dot {\underline {{d}}} _ {i + 1} - \dot {\underline {{d}}} _ {i - 1}}{2 (\Delta t)} \tag {16.3.2}
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$$
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where the subscripts indicate the time step; that is, for a time increment of $\Delta t , \underline { { d } } _ { i } =$ dðtÞ and $\underline { { d } } _ { i + 1 } = \underline { { d } } ( t + \Delta t )$ . The procedure used in deriving Eq. (16.3.1) is illustrated by use of the displacement/time curve shown in Figure 16–5. Graphically, Eq. (16.3.1) represents the slope of the line shown in Figure 16–5; that is, given two points at increments i 1 and i þ 1 on the curve, two Dt increments apart, an approximation of the first derivative at the midpoint i of the increment is given by Eq. (16.3.1). Similarly, using a velocity/time curve, we could obtain Eq. (16.3.2), or we can see that Eq. (16.3.2) is obtained simply by differentiating Eq. (16.3.1) with respect to time.
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It has been shown using, for instance, Taylor series expansions [3] that the acceleration can also be expressed in terms of the displacements by
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$$
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\ddot {\underline {{d}}} _ {i} = \frac {\underline {{d}} _ {i + 1} - 2 \underline {{d}} _ {i} + \underline {{d}} _ {i - 1}}{(\Delta t) ^ {2}} \tag {16.3.3}
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$$
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Because we want to evaluate the nodal displacements, it is most suitable to use Eq. (16.3.3) in the form
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$$
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\underline {{d}} _ {i + 1} = 2 \underline {{d}} _ {i} - \underline {{d}} _ {i - 1} + \ddot {\underline {{d}}} _ {i} (\Delta t) ^ {2} \tag {16.3.4}
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$$
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<!-- source-page: 673 -->
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Equation (16.3.4) will be used to determine the nodal displacements in the next time step i þ 1 knowing the displacements at time steps i and i 1 and the acceleration at time i.
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From Eq. (16.2.24), we express the acceleration as
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$$
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\ddot {\underline {{d}}} _ {i} = \underline {{M}} ^ {- 1} (\underline {{F}} _ {i} - \underline {{K}} \underline {{d}} _ {i}) \tag {16.3.5}
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$$
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To obtain an expression for $\underline { { d } } _ { i + 1 }$ , we first multiply Eq. (16.3.4) by the mass matrix $\underline { { M } }$ and then substitute Eq. (16.3.5) for $\ddot { d } _ { i }$ into this equation to obtain
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$$
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\underline {{M}} \underline {{d}} _ {i + 1} = 2 \underline {{M}} \underline {{d}} _ {i} - \underline {{M}} \underline {{d}} _ {i - 1} + (\underline {{F}} _ {i} - \underline {{K}} \underline {{d}} _ {i}) (\Delta t) ^ {2} \tag {16.3.6}
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$$
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Combining like terms of Eq. (16.3.6), we obtain
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$$
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\underline {{M}} \underline {{d}} _ {i + 1} = (\Delta t) ^ {2} \underline {{F}} _ {i} + [ 2 \underline {{M}} - (\Delta t) ^ {2} \underline {{K}} ] \underline {{d}} _ {i} - \underline {{M}} \underline {{d}} _ {i - 1} \tag {16.3.7}
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$$
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To start the computations to determine $\underline { { d } } _ { i + 1 } , \dot { \underline { { d } } } _ { i + 1 }$ , and $\ddot { \underline { d } } _ { i + 1 }$ , we need the displacement $\underline { d } _ { i - 1 }$ initially, as indicated by Eq. (16.3.7). Using Eqs. (16.3.1) and (16.3.4), we solve for $\underline { d } _ { i - 1 }$ as
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$$
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\underline {{d}} _ {i - 1} = \underline {{d}} _ {i} - (\Delta t) \underline {{\dot {d}}} _ {i} + \frac {(\Delta t) ^ {2}}{2} \underline {{\ddot {d}}} _ {i} \tag {16.3.8}
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$$
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The procedure for solution is then as follows:
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1. Given: $\underline { { d } } _ { 0 } , \dot { \underline { { d } } } _ { 0 }$ , and $\underline { { F } } _ { i } ( t )$ .
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2. If $\ddot { d } _ { 0 }$ is not initially given, solve Eq. (16.3.5) at $t = 0$ for $\ddot { d } _ { 0 } ;$ that is,
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$$
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\ddot {\underline {{d}}} _ {0} = \underline {{M}} ^ {- 1} (\underline {{F}} _ {0} - \underline {{K}} \underline {{d}} _ {0})
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$$
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3. Solve Eq. (16.3.8) at $t = - \Delta t$ for $\underline { { d } } _ { - 1 } ;$ that is,
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$$
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\underline {{d}} _ {- 1} = \underline {{d}} _ {0} - (\Delta t) \dot {\underline {{d}}} _ {0} + \frac {(\Delta t) ^ {2}}{2} \ddot {\underline {{d}}} _ {0}
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$$
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4. Having solved for $\underline { d } _ { - 1 }$ in step 3, now solve for $\underline { { d } } _ { 1 }$ using Eq. (16.3.7) as
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$$
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\underline {{d}} _ {1} = \underline {{M}} ^ {- 1} \{(\Delta t) ^ {2} \underline {{F}} _ {0} + [ 2 \underline {{M}} - (\Delta t) ^ {2} \underline {{K}} ] \underline {{d}} _ {0} - \underline {{M}} \underline {{d}} _ {- 1} \}
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$$
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5. With $\underline { { d } } _ { 0 }$ initially given, and $\underline { { d } } _ { 1 }$ determined from step 4, use Eq. (16.3.7) to obtain
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$$
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\underline {{d}} _ {2} = \underline {{M}} ^ {- 1} \{(\Delta t) ^ {2} \underline {{F}} _ {1} + [ 2 \underline {{M}} - (\Delta t) ^ {2} \underline {{K}} ] \underline {{d}} _ {1} - \underline {{M}} \underline {{d}} _ {0} \}
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$$
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6. Using Eq. (16.3.5), solve for $\ddot { d } _ { 1 }$ as
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$$
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\ddot {\underline {{d}}} _ {1} = \underline {{M}} ^ {- 1} (\underline {{F}} _ {1} - \underline {{K}} \underline {{d}} _ {1})
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$$
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7. Using the result of step 5 and the boundary condition for $\underline { { d } } _ { 0 }$ given in step 1, determine the velocity at the first time step by Eq. (16.3.1) as
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$$
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\dot {\underline {{d}}} _ {1} = \frac {\underline {{d}} _ {2} - \underline {{d}} _ {0}}{2 (\Delta t)}
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$$
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8. Use steps 5–7 repeatedly to obtain the displacement, acceleration, and velocity for all other time steps.
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<!-- source-page: 674 -->
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<details>
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<summary>flowchart</summary>
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```mermaid
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graph TD
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A["START"] --> B["Input the boundary and initial conditions d₀ and ḋ₀, the number of time steps, and the size of the time step or increment Δt"]
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B --> C["Evaluate the initial acceleration from ḋ₀ = M⁻¹(F₀ - Kd₀)"]
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C --> D["Solve Eq. (16.3.8) for d₋₁"]
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D --> E["Solve Eq. (16.3.7) for ḋ₁"]
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E --> F["DO i = 1, Total number of time steps"]
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F --> G["Solve Eq. (16.3.7) for ḋᵢ₊₁"]
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G --> H["Solve Eq. (16.3.5) for ḋᵢ"]
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H --> I["Solve Eq. (16.3.1) for ḋᵢ"]
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I --> J["Output the displacements ḋᵢ, velocities ḋᵢ, and accelerations ḋᵢ for a given time step i"]
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J --> K["END"]
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```
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</details>
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Figure 16–6 Flowchart of the central difference method
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Figure 16–6 is a flowchart of the solution procedure using the central difference equations. Note that the recurrence formulas given by equations such as Eqs. (16.3.1) and (16.3.2) are approximate but yield sufficiently accurate results provided the time step Dt is taken small in relation to the variations in acceleration. Methods for determining proper time steps for the numerical integration process are described in Section 16.5.
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We will now illustrate the central difference equations as they apply to the following example problem.
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<!-- source-page: 675 -->
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Determine the displacement, velocity, and acceleration at 0.05-s time intervals up to 0.2 s for the one-dimensional spring-mass oscillator subjected to the time-dependent forcing function shown in Figure 16–7. [Guidelines regarding appropriate time intervals (or time steps) are given in Section 16.5.] This forcing function is a typical one assumed for blast loads. The restoring spring force versus displacement curve is also provided. [Note that Figure 16–7 also represents a one-element bar with its left end fixed and right node subjected to $F ( t )$ when a lumped mass is used.]
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Because we are considering the single degree of freedom associated with the mass, the general matrix equations describing the motion reduce to single scalar equations. We will represent this single degree of freedom by d.
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The solution procedure follows the steps outlined in this section and in the flowchart of Figure 16–6.
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Figure 16–7 Spring-mass oscillator subjected to a time-dependent force
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# Step 1
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At time $t = 0$ , the initial displacement and velocity are zero; therefore,
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$$
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d _ {0} = 0 \quad \dot {d} _ {0} = 0
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$$
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# Step 2
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The initial acceleration at t ¼ 0 is obtained as
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$$
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\ddot {d} _ {0} = \frac {2 0 0 0 - 1 0 0 (0)}{3 1 . 8 3} = 6 2. 8 3 \mathrm {in. / s^ {2}}
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$$
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where we have used $\underline { { F } } ( 0 ) = 2 0 0 0$ lb and $\underline { { K } } = 1 0 0 ~ \mathrm { { l b / i n } }$ .
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<!-- source-page: 676 -->
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# Step 3
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The displacement $d _ { - 1 }$ is obtained as
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$$
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d _ {- 1} = 0 - 0 + \frac {(0 . 0 5) ^ {2}}{2} (6 2. 8 3) = 0. 0 7 8 5 \text { in. }
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$$
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# Step 4
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The displacement at time $t = 0 . 0 5 \mathrm { ~ s ~ }$ is
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$$
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\begin{array}{l} d _ {1} = \frac {1}{3 1 . 8 3} \left\{\left(0. 0 5\right) ^ {2} (2 0 0 0) + \left[ 2 (3 1. 8 3) - (0. 0 5) ^ {2} (1 0 0) \right] 0 - (3 1. 8 3) (0. 0 7 8 5) \right\} \\ = 0. 0 7 8 5 \text { in. } \\ \end{array}
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$$
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# Step 5
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Having obtained $d _ { 1 }$ , we now determine the displacement at time $t = 0 . 1 0 \mathrm { ~ s ~ a s ~ }$
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$$
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\begin{array}{l} d _ {2} = \frac {1}{3 1 . 8 3} \left\{\left(0. 0 5\right) ^ {2} (1 5 0 0) + \left[ 2 (3 1. 8 3) - (0. 0 5) ^ {2} (1 0 0) \right] (0. 0 7 8 5) - (3 1. 8 3) (0) \right\} \\ = 0. 2 7 4 \text { in. } \\ \end{array}
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$$
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# Step 6
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The acceleration at time $t = 0 . 0 5 \mathrm { ~ s ~ }$ is
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$$
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\ddot {d} _ {1} = \frac {1}{3 1 . 8 3} [ 1 5 0 0 - 1 0 0 (0. 0 7 8 5) ] = 4 6. 8 8 \mathrm{in.} / \mathrm{s} ^ {2}
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$$
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# Step 7
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The velocity at time $t = 0 . 0 5 \mathrm { ~ s ~ }$ is
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$$
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\dot {d} _ {1} = \frac {0 . 2 7 4 - 0}{2 (0 . 0 5)} = 2. 7 4 \mathrm{in./s}
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$$
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# Step 8
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Repeated use of steps 5–7 will result in the displacement, acceleration, and velocity for additional time steps as desired. We will now perform one more time-step iteration of the procedure.
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Repeating step 5 for the next time step, we have
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$$
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\begin{array}{l} d _ {3} = \frac {1}{3 1 . 8 3} \left\{\left(0. 0 5\right) ^ {2} (1 0 0 0) + \left[ 2 (3 1. 8 3) - (0. 0 5) ^ {2} (1 0 0) \right] (0. 2 7 4) \right. \\ - (3 1. 8 3) (0. 0 7 8 5) \} = 0. 5 4 6 \text { in. } \\ \end{array}
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$$
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Repeating step 6 for the next time step, we have
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$$
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\ddot {d} _ {2} = \frac {1}{3 1 . 8 3} [ 1 0 0 0 - 1 0 0 (0. 2 7 4) ] = 3 0. 5 6 \mathrm {in. / s^ {2}}
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$$
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<!-- source-page: 677 -->
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Table 16–1 Results of the analysis of Example 16.1
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<table><tr><td>t (s)</td><td>F(t) (lb)</td><td>di(in.)</td><td>Q (lb)</td><td>ddi(in./s2)</td><td>di(in./s)</td><td>di(exact)</td></tr><tr><td>0</td><td>2000</td><td>0</td><td>0</td><td>62.83</td><td>0</td><td>0</td></tr><tr><td>0.05</td><td>1500</td><td>0.0785</td><td>7.85</td><td>46.88</td><td>2.74</td><td>0.0718</td></tr><tr><td>0.10</td><td>1000</td><td>0.274</td><td>27.40</td><td>30.56</td><td>4.68</td><td>0.2603</td></tr><tr><td>0.15</td><td>500</td><td>0.546</td><td>54.64</td><td>13.99</td><td>5.79</td><td>0.5252</td></tr><tr><td>0.20</td><td>0</td><td>0.854</td><td>85.35</td><td>-2.68</td><td>6.07</td><td>0.8250</td></tr><tr><td>0.25</td><td>0</td><td>1.154</td><td>115.4</td><td>-3.63</td><td>5.91</td><td>1.132</td></tr></table>
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Finally, repeating step 7 for the next time step, we obtain
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$$
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\dot {d} _ {2} = \frac {0 . 5 4 6 - 0 . 0 7 8 5}{2 (0 . 0 5)} = 4. 6 8 \mathrm{in./s}
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$$
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Table 16–1 summarizes the results obtained through time $t = 0 . 2 5 \mathrm { ~ s ~ }$ . In Table 16–1, $Q = k d _ { i }$ is the restoring spring force. Also, the exact analytical solution for displacement based on the equation in Reference [14] is given by
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$$
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y = \frac {F _ {0}}{k} (1 - \cos \omega t) + \frac {F _ {0}}{k t _ {d}} \left(\frac {\sin \omega t}{\omega} - t\right)
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$$
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where $F _ { 0 } = 2 0 0 0$ lb, k ¼ 100 lb/in., td ¼ 0:2 s, and
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$$
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\omega = \sqrt {\frac {k}{m}} = \sqrt {\frac {1 0 0}{3 1 . 8 3}} = 1. 7 7 \mathrm{rad/s}
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$$
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# Newmark’s Method
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We will now outline Newmark’s numerical method, which, because of its general versatility, has been adopted into numerous commercially available computer programs for purposes of structural dynamics analysis. (Complete development of the equations can be found in Reference [5].) Newmark’s equations are given by
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$$
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\dot {\underline {{d}}} _ {i + 1} = \dot {\underline {{d}}} _ {i} + (\Delta t) [ (1 - \gamma) \ddot {\underline {{d}}} _ {i} + \gamma \ddot {\underline {{d}}} _ {i + 1} ] \tag {16.3.9}
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$$
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$$
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\underline {{d}} _ {i + 1} = \underline {{d}} _ {i} + (\Delta t) \dot {\underline {{d}}} _ {i} + (\Delta t) ^ {2} [ (\frac {1}{2} - \beta) \ddot {\underline {{d}}} _ {i} + \beta \ddot {\underline {{d}}} _ {i + 1} ] \tag {16.3.10}
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$$
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where $\beta$ and $\gamma$ are parameters chosen by the user. The parameter $\beta$ is generally chosen between 0 and $\textstyle { \frac { 1 } { 4 } } ,$ and $\gamma$ is often taken to be ${ \frac { 1 } { 2 } } .$ For instance, choosing $\begin{array} { r } { \gamma = \frac { 1 } { 2 } } \end{array}$ and $\beta = 0$ , it can be shown that Eqs. (16.3.9) and (16.3.10) reduce to the central difference Eqs. (16.3.1) and (16.3.2). If $\begin{array} { r } { \gamma = \frac { 1 } { 2 } } \end{array}$ and $\begin{array} { r } { \beta = \frac { 1 } { 6 } } \end{array}$ are chosen, Eqs. (16.3.9) and (16.3.10) correspond to those for which a linear acceleration assumption is valid within each time interval. For $\begin{array} { r } { \gamma = \frac { 1 } { 2 } } \end{array}$ and $\begin{array} { r } { \beta = \frac { 1 } { 4 } } \end{array}$ , it has been shown that the numerical analysis is stable; that is, computed quantities such as displacement and velocities do not become unbounded regardless of the time step chosen. Furthermore, it has been found $[ 5 ]$ that a time step of approximately $\textstyle { \frac { 1 } { 1 0 } }$ of the shortest natural frequency of the structure being analyzed usually yields the best results.
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<!-- source-page: 678 -->
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To find $\underline { { d } } _ { i + 1 }$ , we first multiply Eq. (16.3.10) by the mass matrix $\underline { { M } }$ and then substitute Eq. (16.3.5) for $\ddot { d } _ { i + 1 }$ into this equation to obtain
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$$
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\underline {{M}} \underline {{d}} _ {i + 1} = \underline {{M}} \underline {{d}} _ {i} + (\Delta t) \underline {{M}} \dot {\underline {{d}}} _ {i} + (\Delta t) ^ {2} \underline {{M}} (\frac {1}{2} - \beta) \ddot {\underline {{d}}} _ {i} ] + \beta (\Delta t) ^ {2} [ \underline {{F}} _ {i + 1} - \underline {{K}} \underline {{d}} _ {i + 1} ] \tag {16.3.11}
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$$
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Combining like terms of Eq. (16.3.11), we obtain
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$$
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(\underline {{M}} + \beta (\Delta t) ^ {2} \underline {{K}}) \underline {{d}} _ {i + 1} = \beta (\Delta t) ^ {2} \underline {{F}} _ {i + 1} + \underline {{M}} \underline {{d}} _ {i} + (\Delta t) \underline {{M}} \underline {{\dot {d}}} _ {i} + (\Delta t) ^ {2} \underline {{M}} (\frac {1}{2} - \beta) \ddot {\underline {{d}}} _ {i} \tag {16.3.12}
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$$
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Finally, dividing Eq. (16.3.12) by $\beta ( \Delta t ) ^ { 2 }$ , we obtain
|
||
|
||
$$
|
||
\underline {{K}} ^ {\prime} \underline {{d}} _ {i + 1} = \underline {{F}} _ {i + 1} ^ {\prime} \tag {16.3.13}
|
||
$$
|
||
|
||
where
|
||
|
||
$$
|
||
\underline {{K}} ^ {\prime} = \underline {{K}} + \frac {1}{\beta (\Delta t) ^ {2}} \underline {{M}} \tag {16.3.14}
|
||
$$
|
||
|
||
$$
|
||
\underline {{{F}}} _ {i + 1} ^ {\prime} = \underline {{{F}}} _ {i + 1} + \frac {\underline {{{M}}}}{\beta (\Delta t) ^ {2}} \left[ \underline {{{d}}} _ {i} + (\Delta t) \underline {{{\dot {d}}}} _ {i} + \left(\frac {1}{2} - \beta\right) (\Delta t) ^ {2} \underline {{{\ddot {d}}}} _ {i} \right]
|
||
$$
|
||
|
||
The solution procedure using Newmark’s equations is as follows:
|
||
|
||
1. Starting at time $t = 0 , \underline { { d } } _ { 0 }$ is known from the given boundary conditions on displacement, and $\dot { d } _ { 0 }$ is known from the initial velocity conditions.
|
||
2. Solve Eq. (16.3.5) at $t = 0$ for $\ddot { d } _ { 0 }$ (unless $\ddot { d } _ { 0 }$ is known from an initial acceleration condition); that is,
|
||
|
||
$$
|
||
\ddot {\underline {{d}}} _ {0} = \underline {{M}} ^ {- 1} (\underline {{F}} _ {0} - \underline {{K}} \underline {{d}} _ {0})
|
||
$$
|
||
|
||
3. Solve Eq. (16.3.13) for $\underline { { d } } _ { 1 }$ , because $\underline { { F } } _ { i + 1 }$ is known for all time steps and $\underline { { d } } _ { 0 } , \underline { { d } } _ { 0 }$ , and $\ddot { d } _ { 0 }$ are now known from steps 1 and 2.
|
||
|
||
4. Use Eq. (16.3.10) to solve for $\ddot { d } _ { 1 }$ as
|
||
|
||
$$
|
||
\ddot {\underline {{d}}} _ {1} = \frac {1}{\beta (\Delta t) ^ {2}} \left[ \underline {{d}} _ {1} - \underline {{d}} _ {0} - (\Delta t) \dot {\underline {{d}}} _ {0} - (\Delta t) ^ {2} \left(\frac {1}{2} - \beta\right) \ddot {\underline {{d}}} _ {0} \right]
|
||
$$
|
||
|
||
5. Solve Eq. (16.3.9) directly for $\dot { d } _ { 1 }$ .
|
||
|
||
6. Using the results of steps 4 and 5, go back to step 3 to solve for $\underline { { d } } _ { 2 }$ and then to steps 4 and 5 to solve for $\ddot { d } _ { 2 }$ and $\dot { \underline { d } } _ { 2 }$ . Use steps 3–5 repeatedly to solve for $\underline { { d } } _ { i + 1 } , \dot { \underline { { d } } } _ { i + 1 }$ , and $\ddot { d } _ { i + 1 }$ .
|
||
|
||
Figure 16–8 is a flowchart of the solution procedure using Newmark’s equations. The advantages of Newmark’s method over the central difference method are that Newmark’s method can be made unconditionally stable (for instance, if $\begin{array} { r } { \beta = \frac { 1 } { 4 } } \end{array}$ and $\begin{array} { r } { \gamma = \frac { 1 } { 2 } ) } \end{array}$ and that larger time steps can be used with better results because, in general, the difference expressions more closely approximate the true acceleration and displacement time behavior [8] to [11]. Other difference formulas, such as Wilson’s and Houboldt’s, also yield unconditionally stable algorithms.
|
||
|
||
We will now illustrate the use of Newmark’s equations as they apply to the following example problem.
|
||
|
||
<!-- source-page: 679 -->
|
||
|
||
|
||
|
||
<details>
|
||
<summary>flowchart</summary>
|
||
|
||
```mermaid
|
||
graph TD
|
||
A["START"] --> B["Input the boundary and initial conditions d̃₀ and d̃₀, the number of time steps, the size of the time step or increment Δt, and the values of β and γ"]
|
||
B --> C["Evaluate the initial acceleration from d̃₀ = M⁻¹(F₀ - Kd₀)"]
|
||
C --> D["DO i = 1, Total number of time steps"]
|
||
D --> E["Solve Eq. (16.3.13) for d̃ᵢ₊₁; that is, solve K′d̃ᵢ₊₁ = F′ᵢ₊₁"]
|
||
E --> F["Solve Eqs. (16.3.10) and (16.3.9) for d̃ᵢ₊₁ and d̃ᵢ₊₁"]
|
||
F --> G["Output the displacements d̃ᵢ, velocities d̃ᵢ, and accelerations d̃ᵢ for a given time step i"]
|
||
G --> H["END"]
|
||
```
|
||
</details>
|
||
|
||
Figure 16–8 Flowchart of numerical integration in time using Newmark’s equations
|
||
|
||
# Example 16.2
|
||
|
||
Determine the displacement, velocity, and acceleration at 0.1-s time increments up to a time of 0.5 s for the one-dimensional spring-mass oscillator subjected to the timedependent forcing function shown in Figure 16–9, along with the restoring spring force versus displacement curve. Assume the oscillator is initially at rest. Let $\begin{array} { r } { \beta = \frac { 1 } { 6 } } \end{array}$ and $\begin{array} { r } { \gamma = \frac { 1 } { 2 } . } \end{array}$ , which corresponds to an assumption of linear acceleration within each time step.
|
||
|
||
Because we are again considering the single degree of freedom associated with the mass, the general matrix equations describing the motion reduce to single scalar equations. Again, we represent this single degree of freedom by d.
|
||
|
||
The solution procedure follows the steps outlined in this section and in the flowchart of Figure 16–8.
|
||
|
||
# Step 1
|
||
|
||
At time t ¼ 0, the initial displacement and velocity are zero; therefore,
|
||
|
||
$$
|
||
d _ {0} = 0 \qquad \dot {d} _ {0} = 0
|
||
$$
|
||
|
||
<!-- source-page: 680 -->
|
||
|
||
|
||
Figure 16–9 Spring-mass oscillator subjected to a time-dependent force
|
||
|
||
# Step 2
|
||
|
||
The initial acceleration at $t = 0$ is obtained as
|
||
|
||
$$
|
||
\ddot {d} _ {0} = \frac {1 0 0 - 7 0 (0)}{1 . 7 7} = 5 6. 5 \mathrm {in. / s^ {2}}
|
||
$$
|
||
|
||
where we have used $\underline { { F } } _ { 0 } = 1 0 0$ lb and $\underline { { K } } = 7 0 ~ \mathrm { { l b / i n } }$ .
|
||
|
||
# Step 3
|
||
|
||
We now solve for the displacement at time t ¼ 0:1 s as
|
||
|
||
$$
|
||
K ^ {\prime} = 7 0 + \frac {1}{(\frac {1}{6}) (0 . 1) ^ {2}} (1. 7 7) = 1 1 3 2 \mathrm{lb/in}.
|
||
$$
|
||
|
||
$$
|
||
F _ {1} ^ {\prime} = 8 0 + \frac {1 . 7 7}{(\frac {1}{6}) (0 . 1) ^ {2}} \left[ 0 + (0. 1) (0) + \left(\frac {1}{2} - \frac {1}{6}\right) (0. 1) ^ {2} (5 6. 5) \right] = 2 8 0 \mathrm{lb}
|
||
$$
|
||
|
||
$$
|
||
d _ {1} = \frac {2 8 0}{1 1 3 2} = 0. 2 4 8 \text { in. }
|
||
$$
|
||
|
||
# Step 4
|
||
|
||
Solve for the acceleration at time $t = 0 . 1 \mathrm { ~ s ~ a s }$
|
||
|
||
$$
|
||
\ddot {d} _ {1} = \frac {1}{\left(\frac {1}{6}\right) (0 . 1) ^ {2}} \left[ 0. 2 4 8 - 0 - (0. 1) (0) - (0. 1) ^ {2} \left(\frac {1}{2} - \frac {1}{6}\right) (5 6. 5) \right]
|
||
$$
|
||
|
||
$$
|
||
\ddot {d} _ {1} = 3 5. 4 \mathrm {in. / s^ {2}}
|
||
$$
|