김경종
461 lines
30 KiB
Markdown
461 lines
30 KiB
Markdown
<!-- source-page: 291 -->
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8.49 For the frame shown, geometry is symmetric but loads P are antisymmetric with respect to the y axis.
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(a) By considering a reflection and a load reversal, show that vertical support reactions are equal in magnitude and horizontal support reactions are zero.
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(b) If half the frame is analyzed, what support condition should be used at the point where the y axis crosses the frame?
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8.50 By superposing results from symmetric and antisymmetric loadings in Fig. 8.15-1b, and using formulas from beam theory, determine (a) the deflection at midspan, and (b) the rotation at the left end.
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# Section 8.16
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8.51 The three-node truss shown carries radial loads P and contains three identical bars, each of axial stiffness $k = AE/L$ . Use cyclic symmetry methods to determine the radial displacement of a typical node.
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<details>
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<summary>text_image</summary>
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P
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3
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1
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2
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P
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L
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</details>
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Problem 8.51
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<details>
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<summary>text_image</summary>
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y,v
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4
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3
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1
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2
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x,u
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</details>
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Problem 8.52
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8.52 The plane structure shown consists of four identical triangular areas, which form a square outer boundary and a square opening. Imagine that there exists an 8 by 8 stiffness matrix $[k']$ for one triangle, which operates on d.o.f. $u_{i}$ and $v_{i}$ at nodes i = 1, 2, 3, 4.
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(a) Construct a transformation matrix $[\mathbf{T}_a]$ for the operation $[\mathbf{k}] = [\mathbf{T}_a]^T [\mathbf{k}'][\mathbf{T}_a]$ , in which $[\mathbf{k}]$ operates on d.o.f. suitable for exploitation of cyclic symmetry.
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(b) Construct a 4 by 8 constraint transformation matrix [T] appropriate to this exercise (see Eq. 8.16-3).
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(c) What is the appropriate transformation matrix if parts (a) and (b) are to be done as a single transformation?
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8.53 A long, uniform beam is supported and loaded in a repetitive pattern, as shown. Use cyclic symmetry methods to determine nodal d.o.f. $w_{1}$ , $\theta_{1}$ , and $\theta_{2}$ in terms of P, a, E, and I.
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<details>
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<summary>text_image</summary>
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z,w
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P
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1
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2
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P
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x
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a
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2a
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a
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2a
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</details>
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Problem 8.53
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<!-- source-page: 292 -->
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# CONSTRAINTS
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Constraints enforce a relationship among d.o.f. Procedures for imposing a constraint include transformation, Lagrange multipliers, and penalty functions. Naturally arising constraints, constraint counting, and integration rules for incompressible materials are also discussed.
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# 9.1 CONSTRAINTS. TRANSFORMATIONS
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Constraints. A constraint either prescribes the value of a d.o.f. (as in imposing a support condition) or prescribes a relationship among d.o.f. In common terminology, a single-point constraint sets a single d.o.f. to a known value (often zero), and a multipoint constraint imposes a relationship between two or more d.o.f. Thus support conditions in the three-bar truss of Fig. 2.2-1 invoke three single-point constraints. Rigid links and rigid elements, discussed in Section 7.8, each invoke a multipoint constraint.
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Figure 9.1-1 shows an example in which constraints could be imposed. In a typical frame, axial deformation of a member can usually be ignored; only bending deformation is significant. Accordingly, in Fig. 9.1-1, one could impose the single-point constraints $v_{A} = 0$ and $v_{B} = 0$ , and the multipoint constraint $u_{A} = u_{B}$ , after which the active d.o.f. consist of only $\theta_{A}$ , $\theta_{B}$ , and either $u_{A}$ or $u_{B}$ . (Failure to impose the constraint $u_{A} = u_{B}$ invites numerical difficulty; see Section 18.2.) Special-purpose computer programs for tall buildings may incorporate constraints of this type by allowing only three d.o.f. per floor, these being the rotation $\theta_{z}$ of a floor about a vertical z axis and the horizontal displacement components u and v.
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For each equation of constraint, one d.o.f. can be eliminated from the vector of structural d.o.f. $\{D\}$ . However, doing so may involve appreciable manipulation and typically increases the bandwidth (or the frontwidth) of the structural equations. The Lagrange multiplier method of treating constraints, discussed subsequently, adds to the number of equations but requires less manipulation.
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Transformation Equations. Constraint equations that couple d.o.f. in $\{D\}$ can be written in the form
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$$
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[ \mathbf {C} ] \{\mathbf {D} \} = \{\mathbf {Q} \} \tag {9.1-1}
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$$
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where $[C]$ and $\{Q\}$ contain constants. There are more d.o.f. in $\{D\}$ than constraint equations, so $[C]$ has more columns than rows. We now consider the common case $\{Q\} = \{0\}$ . Let Eq. 9.1-1 be partitioned so that
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$$
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\left[ \begin{array}{l l} \mathbf {C} _ {r} & \mathbf {C} _ {c} \end{array} \right] \left\{ \begin{array}{l} \mathbf {D} _ {r} \\ \mathbf {D} _ {c} \end{array} \right\} = \{\mathbf {0} \} \tag {9.1-2}
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$$
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<!-- source-page: 293 -->
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<details>
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<summary>text_image</summary>
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u_A
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v_A
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θ_A
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A
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B
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v_H
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θ_H
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u_B
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C
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D
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</details>
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Figure 9.1-1. A three-element plane frame, fixed at nodes C and D. D.o.f. at nodes A and B are shown.
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where $\{D_{r}\}$ and $\{D_{c}\}$ are, respectively, d.o.f. to be retained and d.o.f. to be eliminated or “condensed out.” Because there are as many d.o.f. $\{D_{c}\}$ as there are independent equations of constraint in Eq. 9.1-2, matrix $[C_{c}]$ is square and nonsingular. Solution for $\{D_{c}\}$ yields
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$$
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\{\mathbf {D} _ {c} \} = [ \mathbf {C} _ {r c} ] \{\mathbf {D} _ {r} \}, \quad \text { where } \quad [ \mathbf {C} _ {r c} ] = - [ \mathbf {C} _ {c} ] ^ {- 1} [ \mathbf {C} _ {r} ] \tag {9.1-3}
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$$
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We now write as one relation the identity $\{\mathbf{D}_r\} = \{\mathbf{D}_r\}$ and Eq. 9.1-3:
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$$
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\left\{ \begin{array}{l} \mathbf {D} _ {r} \\ \mathbf {D} _ {c} \end{array} \right\} = [ \mathbf {T} ] \{\mathbf {D} _ {r} \}, \quad \text { where } \quad [ \mathbf {T} ] = \left[ \begin{array}{l} \mathbf {I} \\ \mathbf {C} _ {r c} \end{array} \right] \tag {9.1-4}
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$$
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With the transformation matrix [T] now defined, the familiar transformations $\{\mathbf{R}\} = [\mathbf{T}]^T\{\mathbf{R}'\}$ and $[\mathbf{K}] = [\mathbf{T}]^T[\mathbf{K}'][\mathbf{T}]$ of Eqs. 7.4-2 and 7.4-4 can be applied to the structural equations $[\mathbf{K}']\{\mathbf{D}'\} = \{\mathbf{R}'\}$ , which are partitioned as
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$$
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\left[ \begin{array}{l l} \mathbf {K} _ {r r} & \mathbf {K} _ {r c} \\ \mathbf {K} _ {c r} & \mathbf {K} _ {c c} \end{array} \right] \left\{ \begin{array}{l} \mathbf {D} _ {r} \\ \mathbf {D} _ {c} \end{array} \right\} = \left\{ \begin{array}{l} \mathbf {R} _ {r} \\ \mathbf {R} _ {c} \end{array} \right\} \tag {9.1-5}
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$$
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The condensed system is
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$$
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\left[ \mathbf {K} _ {r r} + \mathbf {K} _ {r c} \mathbf {C} _ {r c} + \mathbf {C} _ {r c} ^ {T} \mathbf {K} _ {c r} + \mathbf {C} _ {r c} ^ {T} \mathbf {K} _ {c c} \mathbf {C} _ {r c} \right] \left\{\mathbf {D} _ {r} \right\} = \left\{\mathbf {R} _ {r} + \mathbf {C} _ {r c} ^ {T} \mathbf {R} _ {c} \right\} \tag {9.1-6}
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$$
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After Eq. 9.1-6 is solved for $\{\mathbf{D}_r\}$ , Eq. 9.1-3 yields $\{\mathbf{D}_c\}$ . If $\{\mathbf{Q}\} \neq \{\mathbf{0}\}$ in Eq. 9.1-1, additional terms appear on the right-hand side of Eq. 9.1-6.
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If Eq. 9.1-2 simply sets certain d.o.f. $\{D_{c}\}$ to zero, then $[C_{r}]=[0]$ and $[C_{c}]=[I]$ , hence $[C_{rc}]=[0]$ , and Eq. 9.1-6 is equivalent to discarding rows and columns associated with $\{D_{c}\}$ . Otherwise, the choice of which d.o.f. to place in $\{D_{c}\}$ is not unique, so the choice of $[C_{c}]$ is not unique. One might then define $[C_{c}]$ to be the last c linearly independent columns of [C].
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It is possible to avoid the reordering, partitioning, and matrix multiplications implied by Eq. 9.1-6 by applying individual constraint equations serially and retaining all d.o.f. of $\{D_{r}\}$ and $\{D_{c}\}$ in the transformed equations [6.1]. The transformed coefficient matrix may not be positive definite.
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<!-- source-page: 294 -->
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<details>
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<summary>text_image</summary>
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y
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1
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2
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3
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x₁u
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P
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P
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P
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L
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L
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L
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</details>
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Figure 9.1-2. Three identical bar elements, each of axial stiffness $k = AE/L$ .
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$$
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\left[ \begin{array}{c c c} 2 k & - k & 0 \\ - k & 2 k & - k \\ 0 & - k & k \end{array} \right] \left\{ \begin{array}{l} u _ {1} \\ u _ {2} \\ u _ {3} \end{array} \right\} = \left\{ \begin{array}{l} P \\ P \\ P \end{array} \right\} \tag {9.1-7}
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$$
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Imagine that the constraint $u_{2} = u_{3}$ is to be imposed. With the choice $D_{c} = u_{3}$ , Eqs. 9.1-2 and 9.1-3 become
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$$
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[ 0 \quad 1 \quad | - 1 ] \left\{ \begin{array}{l} u _ {1} \\ u _ {2} \\ u _ {3} \end{array} \right\} = 0 \quad \text { and } \quad [ \mathbf {C} _ {r c} ] = [ 0 \quad 1 ] \tag {9.1-8}
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$$
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The transformation matrix of Eq. 9.1-4 and the reduced system of Eq. 9.1-6 are
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$$
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[ \mathbf {T} ] = \left[ \begin{array}{l l} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{array} \right] \quad \text { and } \quad \left[ \begin{array}{c c} 2 k & - k \\ - k & k \end{array} \right] \left\{ \begin{array}{l} u _ {1} \\ u _ {2} \end{array} \right\} = \left\{ \begin{array}{l} P \\ 2 P \end{array} \right\} \tag {9.1-9}
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$$
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Equation 9.1-9 yields $u_{1} = 3P / k$ and $u_{2} = 5P / k$ . Hence, Eq. 9.1-8 yields $u_{3} = 5P / k$ .
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In Section 8.16, we note that two different nodes can be forced to have the same d.o.f. in $\{D\}$ by giving them the same node number. (Actual nodal coordinates are still used in the generation of element matrices.) Thus a node whose d.o.f. would all appear in $\{D_{c}\}$ can be assigned a node number associated with $\{D_{r}\}$ instead of using the transformation, Eq. 9.1-6. Any externally applied loads on d.o.f. $\{D_{c}\}$ must be transferred to d.o.f. in $\{D_{r}\}$ . In applying this method to the foregoing example problem, one assigns the number 2 to the rightmost two nodes. This causes addition of the four coefficients in [k] of the right element, for a sum of zero at node 3, effectively removing the right element (but not its load) from the structure, and producing Eq. 9.1-9 upon assembly of the remaining two elements.
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The condensed system in Eq. 9.1-6 is different from the system obtained by static condensation, Eq. 8.1-3. In Eq. 8.1-3, condensed d.o.f. are related to retained d.o.f. by equilibrium equations already present in the system $[K]\{D\}=\{R\}$ . In Eq. 9.1-6, condensed d.o.f. $\{D_{c}\}$ are related to retained d.o.f. $\{D_{r}\}$ by supplementary equations of constraint that replace certain equilibrium equations. Accordingly, constraints may appear to falsify certain equilibrium equations. Figure 9.1-3 is a case in point. The original system, and the system that results from the constraint $v_{1}=v_{2}$ , are respectively
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$$
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\left[ \begin{array}{l l} k & 0 \\ 0 & k \end{array} \right] \left\{ \begin{array}{l} v _ {1} \\ v _ {2} \end{array} \right\} = \left\{ \begin{array}{l} P \\ 0 \end{array} \right\} \quad \text { and } \quad (2 k) v _ {1} = P \tag {9.1-10}
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$$
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<!-- source-page: 295 -->
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<details>
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<summary>text_image</summary>
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y,v
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L
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P
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Rigid bar
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1
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2
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x
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k
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k
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(a)
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</details>
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<details>
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<summary>text_image</summary>
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P
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1
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P/2
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2
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P/2
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(b)
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</details>
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Figure 9.1-3. (a) A rigid bar supported by two springs. (b) External and elastic forces applied to the bar if the constraint $v_{1} = v_{2}$ is imposed. Forces of constraint are not shown.
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Hence, $v_{1} = v_{2} = P/2k$ , and forces carried by the springs are $kv_{1} = kv_{2} = P/2$ . Net forces applied to the bar, Fig. 9.1-3b, satisfy equilibrium of y-direction forces but not moment equilibrium. Of course, the condensed structure is not that of Fig. 9.1-3b; it is a single spring of stiffness 2k, loaded by force P.
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# 9.2 LAGRANGE MULTIPLIERS
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Lagrange's method of undetermined multipliers is used to find the maximum or minimum of a function whose variables are not independent but have some prescribed relation. In structural mechanics the function is potential energy $\Pi_p$ and the variables are d.o.f. in $\{\mathbf{D}\}$ . System unknowns become $\{\mathbf{D}\}$ and the Lagrange multipliers.
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The theory is easy to describe. We write the constraint equation (Eq. 9.1-1) as the homogeneous equation $[C]\{D\} - \{Q\} = \{0\}$ and multiply its left-hand side by a row vector $\{\lambda\}^{T}$ that contains as many Lagrange multipliers $\lambda_{i}$ as there are constraint equations. Next we add the result to the potential expression, Eq. 4.1-7:
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$$
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\Pi_ {p} = \frac {1}{2} \{\mathbf {D} \} ^ {T} [ \mathbf {K} ] \{\mathbf {D} \} - \{\mathbf {D} \} ^ {T} \{\mathbf {R} \} + \{\lambda \} ^ {T} ([ \mathbf {C} ] \{\mathbf {D} \} - \{\mathbf {Q} \}) \tag {9.2-1}
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$$
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The expression in parentheses is zero, so we have added nothing to $\Pi_{p}$ . Next we make $\Pi_{p}$ stationary by writing the equations $\{\partial\Pi_{p}/\partial\mathbf{D}\}=\{\mathbf{0}\}$ and $\{\partial\Pi_{p}/\partial\boldsymbol{\lambda}\}=\{\mathbf{0}\}$ , following differentiation rules stated in Appendix A. The result is
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$$
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\left[ \begin{array}{l l} \mathbf {K} & \mathbf {C} ^ {T} \\ \mathbf {C} & \mathbf {0} \end{array} \right] \left\{ \begin{array}{l} \mathbf {D} \\ \boldsymbol {\lambda} \end{array} \right\} = \left\{ \begin{array}{l} \mathbf {R} \\ \mathbf {Q} \end{array} \right\} \tag {9.2-2}
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$$
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The lower partition of Eqs. 9.2-2 is Eq. 9.1-1, the equation of constraint. Equations 9.2-2 are solved for both $\{D\}$ and $\{\lambda\}$ . The $\lambda_{i}$ may be interpreted as forces of constraint (see the following example problem).
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Strict partitioning—that is, $\{D\}$ followed by $\{\lambda\}$ in Eq. 9.2-2—increases bandwidth to the maximum. If instead the $D_{i}$ and $\lambda_{i}$ are interlaced, bandwidth can be much less, although not as small as when the $\lambda_{i}$ are absent. However, in a Gauss elimination solution with pivoting on the diagonal, a zero pivot appears if a constraint equation is processed before any of the d.o.f. to which it is coupled. Otherwise, the null submatrix fills in and the solution proceeds normally if the stiffness matrix [K] is by itself positive definite.
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<!-- source-page: 296 -->
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The Lagrange multiplier method is more attractive than the transformation method of Section 9.1 if there are few constraint equations that couple many d.o.f. However, Lagrange multipliers are active at the structure level, but transformation equations can be applied at either the structure level or element by element. The latter has the appeal of disposing of constraints at an early stage, when the matrices are small and more manageable.
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Example. Again we solve the example problem of Fig. 9.1-2. The constraint equation is Eq. 9.1-8. Equation 9.2-2 assumes the form
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$$
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\left[ \begin{array}{c c c c} 2 k & - k & 0 & 0 \\ - k & 2 k & - k & 1 \\ 0 & - k & k & - 1 \\ 0 & 1 & - 1 & 0 \end{array} \right] \left\{ \begin{array}{l} u _ {1} \\ u _ {2} \\ u _ {3} \\ \lambda \end{array} \right\} = \left\{ \begin{array}{l} P \\ P \\ P \\ 0 \end{array} \right\} \tag {9.2-3}
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$$
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The solution of Eq. 9.2-3 is
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$$
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\left\lfloor u _ {1} \quad u _ {2} \quad u _ {3} \quad \lambda \right\rfloor = \left\lfloor \frac {3 P}{k} \quad \frac {5 P}{k} \quad \frac {5 P}{k} \quad - P \right\rfloor \tag {9.2-4}
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$$
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The result $\lambda = -P$ can be regarded as the force of constraint applied through the now rigid link 2-3. The algebraic sign of $\lambda$ is not significant: had we written [C] = [0 - 1 1] in Eq. 9.1-8, we would obtain $\lambda = +P$ but the same values of $u_{1}, u_{2}$ , and $u_{3}$ .
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# 9.3 PENALTY FUNCTIONS
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If the constraint equation $[\mathbf{C}]\{\mathbf{D}\} = \{\mathbf{Q}\}$ , Eq. 9.1-1, is written in the form
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$$
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\{\mathbf {t} \} = [ \mathbf {C} ] \{\mathbf {D} \} - \{\mathbf {Q} \} \tag {9.3-1}
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$$
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then $\{\mathbf{t}\} = \{\mathbf{0}\}$ implies satisfaction of the constraints. The usual potential $\Pi_p$ of a structural system can be augmented by a penalty function $\{\mathbf{t}\}^T[\alpha]\{\mathbf{t}\} / 2$ , where $[\alpha]$ is a diagonal matrix of "penalty numbers" $\alpha_i$ . Thus
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$$
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\Pi_ {p} = \frac {1}{2} \{\mathbf {D} \} ^ {T} [ \mathbf {K} ] \{\mathbf {D} \} - \{\mathbf {D} \} ^ {T} \{\mathbf {R} \} + \frac {1}{2} \{\mathbf {t} \} ^ {T} [ \alpha ] \{\mathbf {t} \} \tag {9.3-2}
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$$
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If $\{\mathbf{t}\} = \{\mathbf{0}\}$ the constraints are satisfied and we have added nothing to $\Pi_p$ . If $\{\mathbf{t}\} \neq \{\mathbf{0}\}$ the penalty of constraint violation becomes more prominent as $[\alpha]$ increases.
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Next we substitute Eq. 9.3-1 into Eq. 9.3-2 and write the minimum condition $\{\partial \Pi_p / \partial \mathbf{D}\} = \{\mathbf{0}\}$ . Thus, from Eqs. 9.3-1 and 9.3-2,
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$$
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\left([ \mathbf {K} ] + [ \mathbf {C} ] ^ {T} [ \alpha ] [ \mathbf {C} ]\right) \{\mathbf {D} \} = \{\mathbf {R} \} + [ \mathbf {C} ] ^ {T} [ \alpha ] \{\mathbf {Q} \} \tag {9.3-3}
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$$
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in which $[\mathbf{C}]^T [\alpha ][\mathbf{C}]$ can be called the penalty matrix. If $[\alpha ] = [\mathbf{0}]$ , the constraints are ignored., As $[\alpha ]$ grows, $\{\mathbf{D}\}$ changes in such a way that the constraint equations are more nearly satisfied. The analyst is responsible for selecting appropriate numerical values of the $\alpha_{i}$ .
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Preferably, for a reason that will subsequently be explained, penalty numbers
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<!-- source-page: 297 -->
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$\alpha_{i}$ are dimensionless. Equations 9.3-1 can easily be written in such a way that the $\alpha_{i}$ are dimensionless if d.o.f. coupled by the constraint equation are all of the same type, for example, all translations or all rotations. If d.o.f. are different types, some types can be redefined to agree with the others (e.g., $L\theta_{i}$ can replace $\theta_{i}$ ); however, the labor of making such a change may outweigh its benefits.
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The method of imposing a prescribed zero or nonzero d.o.f. $D_{i}$ by adding large numbers to $K_{ij}$ and $R_{i}$ is a penalty method. For example, in Fig. 2.10-6 the penalty matrix added to [K] contains a single coefficient—namely, the large spring stiffness $k_{s}$ . In this example, $\alpha$ is dimensionless if we define $k_{s} = \alpha k$ and the constraint as $\sqrt{k} v_{1} = 0$ , where $k$ is a spring stiffness whose magnitude is approximately the same as a typical $K_{ij}$ already present in [K].
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Example. Imagine that the constraint $u_{1} = u_{2}$ is to be imposed on the structure of Fig. 9.3-1.
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There is no unique way to write the constraint relation. We will write [C] in such a way that the penalty numbers are dimensionless. Thus for Eq. 9.3-1 we elect to write
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$$
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[ \mathbf {C} ] = \left\lfloor \sqrt {k} - \sqrt {k} \right\rfloor \quad \{\mathbf {D} \} = \left\{ \begin{array}{l} u _ {1} \\ u _ {2} \end{array} \right\} \quad \{\mathbf {Q} \} = \{\mathbf {0} \} \tag {9.3-4}
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$$
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where $k = AE / L$ . With but one constraint, $[\alpha] = \alpha$ , a scalar. Equation 9.3-3 becomes
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$$
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\left(\left[ \begin{array}{c c} 2 k & - k \\ - k & k \end{array} \right] + \alpha \left[ \begin{array}{c c} k & - k \\ - k & k \end{array} \right]\right) \left\{ \begin{array}{l} u _ {1} \\ u _ {2} \end{array} \right\} = \left\{ \begin{array}{l} P \\ P \end{array} \right\} \tag {9.3-5}
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$$
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which has the solution
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$$
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u _ {1} = \frac {2 P}{k} \quad \text { and } \quad u _ {2} = \frac {3 + 2 \alpha}{1 + \alpha} \frac {P}{k} \tag {9.3-6}
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$$
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If $\alpha = 0$ , then $u_{2} = 3P / k$ , as expected. As $\alpha$ becomes large, $u_{2}$ approaches the value $2P / k$ , which is correct for the constrained system. Note that $u_{1} - u_{2} = -P / k(1 + \alpha)$ and $\{\mathbf{t}\} = t = \sqrt{k}(u_{1} - u_{2})$ , so that the coefficient of $\alpha$ in the penalty function approaches zero as $\alpha$ approaches infinity.
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The symbolic manipulations that produce Eq. 9.3-6 from Eq. 9.3-5 obscure a difficulty that may arise if the manipulations are done numerically. The second square matrix in Eq. 9.3-5 is recognized as the stiffness of a bar element that spans nodes 1 and 2 (i.e., a “constraint” bar in parallel with the bar already there).
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<details>
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<summary>text_image</summary>
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y
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1
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2
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x,u
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P
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P
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L
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L
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</details>
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Figure 9.3-1. Two identical bar elements, each of axial stiffness $k = AE/L$ .
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As $\alpha$ grows, the structure becomes the error-prone case of a stiff region supported by a flexible region (see Section 18.2).
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If constraints do not couple all d.o.f. in $\{D\}$ , then $[C]$ has more columns than rows, and $[C]^{T}[\alpha][C]$ is certain to be a singular matrix. In some important problems, discussed in Section 9.4, constraints do couple all d.o.f. in $\{D\}$ , and singularity of $[C]^{T}[\alpha][C]$ is not guaranteed. However, we want this matrix to be singular, as the following argument illustrates.
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For simplicity let all $\alpha_{i}$ in $[\alpha]$ be the same number, say $\alpha$ . In addition, let $\{\mathbf{Q}\} = \{\mathbf{0}\}$ in Eq. 9.3-3. Then, as $\alpha$ becomes large, Eq. 9.3-3 becomes
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$$
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[ \mathbf {C} ] ^ {T} [ \mathbf {C} ] \{\mathbf {D} \} \approx \frac {1}{\alpha} \{\mathbf {R} \} \tag {9.3-7}
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$$
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Equation 9.3-7 shows that if $[C]^{T}[C]$ is nonsingular, then as $\alpha$ grows the solution vector $\{D\}$ approaches zero. In other words, the mesh “locks.” Only if $[C]^{T}[C]$ is singular can $\{D\}$ be nonzero. Then the number of independent nonzero $\{D\}$ 's that satisfy Eq. 9.3-7 is equal to the difference between the order of $[C]^{T}[C]$ and its rank. The practical significance of this argument is discussed in subsequent sections.
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In comparison with Lagrange multipliers, penalty functions have the advantage of introducing no new variables. However, the penalty matrix may significantly increase the bandwidth (or wave front) of the structural equations, depending on how d.o.f. are numbered and what d.o.f. are coupled by the constraint equation. Implementation of a penalty function can be as easy as assigning a high modulus to an element already in the structure. Penalty functions have the disadvantage that penalty numbers must be chosen in an allowable range: large enough to be effective but not so large as to provoke numerical difficulties $[9.2,9.3]$ .
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# 9.4 NATURALLY ARISING PENALTY FORMULATIONS. NUMERICAL INTEGRATION AND CONSTRAINTS
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In Section 9.3, constraints are imposed by explicitly adding a penalty matrix to an existing stiffness matrix [K]. In some situations [K] already contains a contribution that can be identified as a penalty matrix. Thus some applications of penalty methods arise naturally in the sense that large stiffnesses with respect to particular deformations can be interpreted as penalty numbers. Here we discuss two of these applications—transverse shear in beams and material incompressibility—which can easily lead to “locking” difficulties unless care is taken in the choice of element type and integration rule. Physical interpretation of the constraints makes it possible to understand the reasons for locking and leads to guidelines that can be used to avoid the difficulty. The guideline presented subsequently is called “constraint counting.” It does not provide a rigorous guarantee of success but can be quite effective in practice. In the present section we identify the constraints precisely, and in the next section we show how to count them.
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Mindlin Beam Element. The beam element shown in Fig. 9.4-1a allows transverse shear deformation. The element has four d.o.f., as is usual. However, rotational d.o.f. $\theta_{1}$ and $\theta_{2}$ are not values of dw/dx at the nodes, as is the case with the standard beam element of Eq. 4.2-5. A plane initially normal to the midsurface
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<details>
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<summary>text_image</summary>
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z,w
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w₁
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θ₁
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1
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x,u
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t
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2
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w₂
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θ₂
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L
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</details>
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(a)
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<details>
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<summary>text_image</summary>
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w
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θ
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u = -zθ
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w_{r,r}
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z
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w
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x,u
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</details>
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(b)
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Figure 9.4-1. (a) A “Mindlin” beam element. (b) Displacements and rotations.
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remains plane but not necessarily normal. In Fig. 9.4-1a, $\theta$ represents the rotation of a line that was initially normal to the undeformed longitudinal axis of the beam. We will interpolate $\theta$ and w independently, so that the beam can represent transverse shear deformation. This element is called a Mindlin beam element [9.4]. It can model constant bending moment but not linearly varying bending moment. The Mindlin beam element generalizes to Mindlin plate elements, discussed in Section 11.3. The standard beam element, also known as an Euler beam element, can model linearly varying bending moment but does not account for transverse shear deformation. $^{1}$
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Axial normal strain $\epsilon_{x}$ and transverse shear strain $\gamma_{zx}$ in a Mindlin beam element are, from Fig. 9.4-1b,
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$$
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\epsilon_ {x} = u _ {, x} = - z \theta_ {, x} \quad \text { and } \quad \gamma_ {z x} = w _ {, x} - \theta \tag {9.4-1}
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$$
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where $\theta$ is a small angle of rotation. Thus $\gamma_{zx}$ is taken as constant over the depth t. Nonzero stresses are assumed to consist only of axial normal stress $\sigma_{x}$ and transverse shear stress $\tau_{zx}$ . Accordingly, strain energy in the Mindlin beam element is $U = U_{b} + U_{s}$ , where $U_{b}$ and $U_{s}$ are strain energies of bending and shear, respectively:
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$$
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U _ {b} = \frac {1}{2} \int_ {V _ {e}} \frac {\sigma_ {x} ^ {2}}{E} d V = \frac {1}{2} \int_ {V _ {e}} E \epsilon_ {x} ^ {2} d V = \frac {1}{2} \frac {E b t ^ {3}}{1 2} \int_ {0} ^ {L} \theta_ {, x} ^ {2} d x \tag {9.4-2a}
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$$
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$$
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U _ {s} = \frac {1}{2} \int_ {V _ {e}} \frac {\tau_ {z x} ^ {2}}{G} d V = \frac {1}{2} \int_ {V _ {e}} G \gamma_ {z x} ^ {2} d V = \frac {1}{2} \frac {G b t}{1 . 2} \int_ {0} ^ {L} (w _ {, x} - \theta) ^ {2} d x \tag {9.4-2b}
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$$
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Here E = elastic modulus, G = shear modulus, b is the width of the beam, and 1.2 is the “form factor” that accounts for a parabolic distribution of $\tau_{zx}$ over a rectangular cross section. $^{2}$
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$^{1}$ The standard beam element, Eq. 4.2-5, can be modified to account for transverse shear deformation, while retaining the ability to model both constant and linearly varying bending moment [4.2,4.11].
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$^{2}$ Let a beam have a rectangular cross section of dimensions b by t. If P is the transverse shear force, then $\tau_{zx} = (3P/2bt^{3})(t^{2} - 4z^{2})$ , where z = 0 at the neutral axis. Equation 9.4-2b then yields $U_{s} = 1.2(P^{2}L/btG)/2$ . This result suggests the view that a uniform stress $\tau_{zx} = P/bt$ acts over a modified area A = bt/1.2, so that the same $U_{s}$ results. A uniform stress $\tau_{zx} = G\gamma_{zx}$ is provided by Eq. 9.4-1.
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Displacement w and rotation $\theta$ are interpolated linearly between nodes:
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$$
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w = \frac {L - x}{L} w _ {1} + \frac {x}{L} w _ {2} \quad \text { and } \quad \theta = \frac {L - x}{L} \theta_ {1} + \frac {x}{L} \theta_ {2} \tag {9.4-3}
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$$
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In the usual way, from Eqs. 9.4-2 and 9.4-3 we obtain a 4 by 4 element stiffness matrix $[k]=[k_{b}]+[k_{s}]$ , where $[k_{b}]$ resists bending strain $\epsilon_{x}$ and $[k_{s}]$ resists shear strain $\gamma_{zx}$ . With this notation, for a structure,
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$$
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([ \mathbf {K} _ {b} ] + [ \mathbf {K} _ {s} ]) \{\mathbf {D} \} = \{\mathbf {R} \} \tag {9.4-4}
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$$
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D.o.f. $w_{i}$ and $\theta_{i}$ in $\{\mathbf{D}\}$ are coupled by terms in $[\mathbf{K}_s]$ but not by terms in $[\mathbf{K}_b]$ .
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Deflections $\{\mathbf{D}\}$ of a thin beam should be governed by only $[\mathbf{K}_b]$ because transverse shear deformation is negligible. In other words, if $Gbt / 1.2$ becomes much larger than $Ebt^3 / 12$ in Eq. 9.4-2, then $[\mathbf{K}_s]$ , which arises from $U_s$ , should enforce the constraint $\gamma_{zx} = 0$ . But, as a beam becomes slender, $[\mathbf{K}_s]$ grows in relation to $[\mathbf{K}_b]$ . So $[\mathbf{K}_s]$ acts as a penalty matrix that causes Eq. 9.4-4 to yield $\{\mathbf{D}\} = \{\mathbf{0}\}$ — unless $[\mathbf{K}_s]$ is singular. In other words, unless $[\mathbf{K}_s]$ is singular, the computed deflection of a very slender beam is almost zero. A singular $[\mathbf{K}_s]$ can enforce the $\gamma_{zx} = 0$ constraint without locking. As discussed subsequently, a singular $[\mathbf{K}_s]$ can be achieved by reduced integration.
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As a simple example of locking, let the beam element of Fig. 9.4-1a be fixed at the left end and loaded by a transverse force P at the right end. Then $w = w_{2}x/L$ , $\theta = \theta_{2}x/L$ , and $\Pi_{p} = U - Pw_{2}$ . For simplicity let $\nu = 0$ , so that E = 2G. Then the equilibrium equations $\partial\Pi_{p}/\partial w_{2} = 0$ and $\partial\Pi_{p}/\partial\theta_{2} = 0$ yield
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$$
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w _ {2} = \frac {1 2 (t / L) ^ {2} + 2 0}{1 2 (t / L) ^ {2} + 5} \left(1. 2 \frac {P L}{G A}\right) \tag {9.4-5}
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$$
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where A = bt. For a very short beam, we obtain $w_{2} \approx 1.2PL/GA$ , which is the correct expression for the portion of end deflection that is due to transverse shear deformation. For a slender beam, L >> t, we obtain $w_{2} \approx 4.8PL/GA$ . This value includes no bending deformation and is far too small; that is, the beam locks as L/t increases.
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Incompressible Materials. As Poisson's ratio $\nu$ approaches 0.5, a material becomes incompressible. Values of $\nu$ near 0.5 occur in rubberlike materials and in materials that flow, such as fluids and plastic solids. Unless the problem is one of plane stress, the value $\nu = 0.5$ is forbidden because denominators become zero in material property matrices [E] (Eqs. 1.7-3 and 1.7-4, for example). It is tempting to approximate incompressibility by using (say) $\nu = 0.49$ . But, near $\nu = 0.5$ , stresses are strongly dependent on $\nu$ . Also, structural equations become ill conditioned as $\nu$ approaches 0.5, for reasons explained next.
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The shear modulus G and bulk modulus B of an isotropic material are
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$$
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G = \frac {E}{2 (1 + \nu)} \quad B = \frac {E}{3 (1 - 2 \nu)} \tag {9.4-6}
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$$
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In terms of $G$ and $B$ , the material property matrix [E] of Eq. 1.7-3 is
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