김경종
42 KiB
42 KiB
where
[ \mathbf {K} ] = \sum_ {n = 1} ^ {\text { numel }} [ \mathbf {k} ] _ {n} \quad \text { and } \quad \{\mathbf {R} \} = \{\mathbf {P} \} + \sum_ {n = 1} ^ {\text { numel }} \{\mathbf {r} _ {e} \} _ {n} \tag {4.1-8}
Summations indicate assembly of element matrices by addition of overlapping terms, as in Section 2.7. Now \Pi_{p} is a function of d.o.f. \{D\} . Making \Pi_{p} stationary with respect to small changes in the D_{i} by use of convenient differentiation rules given in Appendix A, we write
\left\{\frac {\partial \Pi_ {p}}{\partial \mathbf {D}} \right\} = \{\mathbf {0} \} \quad \text { yields } \quad [ \mathbf {K} ] \{\mathbf {D} \} = \{\mathbf {R} \} \tag {4.1-9}
The latter matrix equation is a set of simultaneous algebraic equations to be solved for d.o.f. {D}. As noted in the latter part of Section 3.3, \partial \Pi_{p} / \partial D_{i} = 0 is a nodal equilibrium equation, [K] is a symmetric matrix, and K_{ij} = \partial^2\Pi_p / \partial D_i\partial D_j .
Some Particular Cases. The preceding derivation may have suggested that \{\epsilon\} must always be a 6 by 1 vector, as for a three-dimensional problem. However, no change in Eqs. 4.1-5 and 4.1-6 need be made if the problem is two-dimensional. For plane stress or plane strain, \{\epsilon\} is 3 by 1 and contains only \epsilon_x , \epsilon_y , and \gamma_{xy} , and matrix [E] is 3 by 3 (see Eq. 1.7-5, for example). For a symmetrically loaded solid of revolution, \{\epsilon\} is 4 by 1 and contains \epsilon_r , \epsilon_z , \epsilon_\theta , and \gamma_{zr} , and [E] is 4 by 4.
Formulas for [k] and \{\mathbf{r}_e\} applicable to a bar under axial load can be obtained by specialization of Eqs. 4.1-5 and 4.1-6. However, it is easier to rederive the formulas, using at the outset special forms applicable to a bar. Such a derivation has already been given in Section 3.9. In that development, note Eqs. 3.9-2 and 3.9-3 in particular. They state that the strain energy of a one-element structure is
U _ {e} = \int_ {V _ {e}} (\text { strain energy density }) d V = \frac {1}{2} \{\mathbf {d} \} ^ {T} [ \mathbf {k} ] \{\mathbf {d} \} \tag {4.1-10}
Equation 4.1-10 summarizes the argument that leads to Eq. 4.1-5: namely, that an element stiffness matrix is obtained by substitution of an interpolation scheme into a strain energy expression. Equation 4.1-6 can be explained similarly. Element nodal loads \{r_{e}\} are obtained by substitution of an interpolation scheme into an expression for work done by distributed loads that act on an element:
\Omega_ {e} = - \int_ {V _ {e}} (\text { work per unit volume }) d V = - \{\mathbf {d} \} ^ {T} \{\mathbf {r} _ {e} \} \tag {4.1-11}
(Work done by externally applied nodal loads \{\mathbf{P}\} is excluded from \{\mathbf{r}_e\} .)
In flexural problems, such as beam and plate bending, it is convenient to define [B] in such a way that the product [B]{d} represents curvatures rather than strains. Consider beam bending. Expressions for U_{e} and \Omega_{e} come from the first two integrals in Eq. 3.4-8. Noting that a scalar is its own transpose, we write w_{,xx}^{2} = w_{,xx}^{T}w_{,xx} and w = w^{T} . Hence
U _ {e} = \frac {1}{2} \int_ {0} ^ {L} w _ {, x x} ^ {T} E I w _ {, x x} d x \quad \text { and } \quad \Omega_ {e} = - \int_ {0} ^ {L} w ^ {T} q d x \tag {4.1-12}
where w is lateral displacement and L is the element length. Interpolating w from element nodal d.o.f. \{\mathbf{d}\} , we have
w = \lfloor \mathbf {N} \rfloor \{\mathbf {d} \} \quad \text { and } \quad w _ {, x x} = \lfloor \mathbf {B} \rfloor \{\mathbf {d} \}, \quad \text { where } \quad \lfloor \mathbf {B} \rfloor = \frac {d ^ {2}}{d x ^ {2}} \lfloor \mathbf {N} \rfloor \tag {4.1-13}
Therefore, in view of Eqs. 4.1-10 and 4.1-11, [k] and \{\mathbf{r}_e\} for a straight beam element are
\left\lfloor [ \mathbf {k} ] = \int_ {0} ^ {L} \left\lfloor \mathbf {B} \right\rfloor^ {T} E I \left\lfloor \mathbf {B} \right\rfloor d x\right) \quad \text { and } \quad \left\{\mathbf {r} _ {e} \right\} = \int_ {0} ^ {L} \left\lfloor \mathbf {N} \right\rfloor^ {T} q d x \tag {4.1-14}
Analogous expressions are written for flat plates in bending, where EI becomes a matrix of flexural rigidities and integration is over the area of the plate midsurface.
a-Basis Formulation. If displacements are initially expressed in terms of d.o.f. {a}, one usually replaces {a} by element nodal d.o.f. {d} before generating a stiffness matrix. The process of replacing {a} by {d} is explained in Sections 3.8 and 3.13. Sometimes it is convenient to delay the replacement of {a} by {d}. The procedure is as follows.
Displacements and strains, expressed in terms of \{a\} , are
\{\mathbf {u} \} = [ \mathbf {N} _ {a} ] \{\mathbf {a} \} \quad \text { and } \quad \{\boldsymbol {\epsilon} \} = [ \mathbf {B} _ {a} ] \{\mathbf {a} \}, \quad \text { where } \quad [ \mathbf {B} _ {a} ] = [ \partial ] [ \mathbf {N} _ {a} ] \tag {4.1-15}
(For example, [\mathbf{N}_a] = \lfloor 1 \quad \mathrm{s} \rfloor in Eq. 3.8-5.) But \{\mathbf{a}\} = [\mathbf{A}]^{-1}\{\mathbf{d}\} , so
[ \mathbf {N} ] = [ \mathbf {N} _ {a} ] [ \mathbf {A} ] ^ {- 1} \quad \text { and } \quad [ \mathbf {B} ] = [ \mathbf {B} _ {a} ] [ \mathbf {A} ] ^ {- 1} \tag {4.1-16}
Substitution of Eqs. 4.1-16 into Eqs. 4.1-5 and 4.1-6 yields the “d-basis” matrices [k] and \{r_{e}\} ,
[ \mathbf {k} ] = [ \mathbf {A} ] ^ {- T} [ \mathbf {k} _ {a} ] [ \mathbf {A} ] ^ {- 1} \quad \text { and } \quad \{\mathbf {r} _ {e} \} = [ \mathbf {A} ] ^ {- T} \{\mathbf {r} _ {e a} \} \tag {4.1-17}
where the “a-basis” matrices are defined as
[ \mathbf {k} _ {a} ] = \int_ {V _ {e}} [ \mathbf {B} _ {a} ] ^ {T} [ \mathbf {E} ] [ \mathbf {B} _ {a} ] d V \quad \text { and } \quad \{\mathbf {r} _ {e a} \} = \int_ {V _ {e}} [ \mathbf {B} _ {a} ] ^ {T} [ \mathbf {E} ] \{\boldsymbol {\epsilon} _ {0} \} d V - \dots \tag {4.1-18}
The a -basis matrices [\mathbf{k}_a] and \{\mathbf{r}_{ea}\} are converted to d -basis matrices [\mathbf{k}] and \{\mathbf{r}_e\} before global equations are assembled, as the relationship between d.o.f. \{\mathbf{a}\} in neighboring elements is complicated and unwieldy.
An example application appears in Eqs. 4.2-6 and 4.2-7.
4.2 OVERVIEW OF ELEMENT STIFFNESS MATRICES
In this section we outline the formulation of selected element stiffness matrices, with the intent of showing the conceptual simplicity of the process. Details of manipulations may be found in the text sections cited.
Bar. Figure 4.2-1 shows a straight bar whose nodal d.o.f. are axial displacements u_{1} and u_{2} . A linear axial displacement field, as used in Section 3.9, is appropriate,
u = \lfloor \mathbf {N} \rfloor \left\{ \begin{array}{l} u _ {1} \\ u _ {2} \end{array} \right\}, \quad \text { where } \quad \lfloor \mathbf {N} \rfloor = \left\lfloor \frac {L - x}{L} \frac {x}{L} \right\rfloor \tag {4.2-1}
Using Eq. 3.9-3, with \lfloor \mathbf{B} \rfloor = d\lfloor \mathbf{N} \rfloor / dx = \lfloor -1 - 1 \rfloor / L , we obtain
[ \mathbf {k} ] = \int_ {0} ^ {L} \left\lfloor \mathbf {B} \right] ^ {T} A E \left\lfloor \mathbf {B} \right\rfloor d x = \frac {A E}{L} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \tag {4.2-2}
where the latter expression is for a uniform bar, AE = constant. The integral expression for [k] does not demand that AE be independent of x.
Beam. Figure 4.2-2 shows a four-d.o.f. straight beam element. Rotation \theta is assumed to be small, so that \theta \approx dw/dx . Four d.o.f. define a cubic lateral displacement field,
w = \lfloor \mathbf {N} \rfloor \left\lfloor w _ {1} \quad \theta_ {1} \quad w _ {2} \quad \theta_ {2} \right\rfloor^ {T} \tag {4.2-3}
where the four N_{i} are given in Fig. 3.13-2. The curvature field is w_{xx} = [\mathbf{B}]\{\mathbf{d}\} , where
\lfloor \mathbf {B} \rfloor = \frac {d ^ {2}}{d x ^ {2}} \lfloor \mathbf {N} \rfloor = \left\lfloor - \frac {6}{L ^ {2}} + \frac {1 2 x}{L ^ {3}} \quad - \frac {4}{L} + \frac {6 x}{L ^ {2}} \quad \frac {6}{L ^ {2}} - \frac {1 2 x}{L ^ {3}} \quad - \frac {2}{L} + \frac {6 x}{L ^ {2}} \right\rfloor \tag {4.2-4}
For constant EI , the element stiffness matrix given by Eq. 4.1-14 is
text_image
y L u₁ 1 A,E 2 u₂ x,u
Figure 4.2-1. Bar element with two d.o.f. ( u_{1} and u_{2} ).
text_image
z,w L w₁ θ₁ 1 E,I w₂ θ₂ 2 x
Figure 4.2-2. Standard four-d.o.f. beam element.
[ \mathbf {k} ] = \int_ {0} ^ {L} [ \mathbf {B} ] ^ {T} E I [ \mathbf {B} ] d x = \frac {E I}{L ^ {3}} \left[ \begin{array}{c c c c} 1 2 & 6 L & - 1 2 & 6 L \\ 6 L & 4 L ^ {2} & - 6 L & 2 L ^ {2} \\ - 1 2 & - 6 L & 1 2 & - 6 L \\ 6 L & 2 L ^ {2} & - 6 L & 4 L ^ {2} \end{array} \right] \tag {4.2-5}
This [k] operates on d.o.f. \{\mathbf{d}\} having the order shown in Eq. 4.2-3. A slightly modified form of Eq. 4.2-5 is able to account for transverse shear deformation [4.2,4.11].
The manipulations needed to obtain [k] are shortened by using the “a-basis” of Eqs. 4.1-17 and 4.1-18. With [X] and [A] given by Eqs. 3.13-1 and 3.13-3, we write
\left\lfloor \mathbf {B} \right\rfloor = \frac {d ^ {2}}{d x ^ {2}} \left\lfloor \mathbf {X} \right\rfloor [ \mathbf {A} ] ^ {- 1} = \left\lfloor 0 0 2 6 x \right\rfloor [ \mathbf {A} ] ^ {- 1} \tag {4.2-6}
[ \mathbf {k} ] = [ \mathbf {A} ] ^ {- T} \int_ {0} ^ {L} \left\lfloor 0 \quad 0 \quad 2 \quad 6 x \right] ^ {T} E I \left\lfloor 0 \quad 0 \quad 2 \quad 6 x \right\rfloor d x [ \mathbf {A} ] ^ {- 1} \tag {4.2-7}
and obtain the same [k] as given in Eq. 4.2-5.
The reader should understand the sign conventions for nodal moments and bending moment. Nodal moments M_1 and M_2 are positive when acting in the directions of \theta_1 and \theta_2 in Fig. 4.2-2. Bending moment M = EIw_{,xx} is positive when it creates tension on the bottom of the beam. Therefore, M = -M_1 at the left end and M = +M_2 at the right end.
Plane Frame. A plane frame member can deform both axially and in bending. Effectively, to obtain a plane frame element we superpose the bar and beam elements of Figs. 4.2-1 and 4.2-2 (and of Eqs. 4.2-2 and 4.2-5). Nodal d.o.f. are \{\mathbf{d}\} = \left[u_1 w_1 \theta_1 u_2 w_2 \theta_2\right]^T . If the element is uniform and lies along the x axis, its stiffness matrix is
[ \mathbf {k} ] = \frac {A E}{L} \left[ \begin{array}{c c c c c c} 1 & 0 & 0 & - 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ - 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right] + \frac {E I}{L ^ {3}} \left[ \begin{array}{c c c c c c} 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 2 & 6 L & 0 & - 1 2 & 6 L \\ 0 & 6 L & 4 L ^ {2} & 0 & - 6 L & 2 L ^ {2} \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & - 1 2 & - 6 L & 0 & 1 2 & - 6 L \\ 0 & 6 L & 2 L ^ {2} & 0 & - 6 L & 4 L ^ {2} \end{array} \right] \tag {4.2-8}
If a frame element is arbitrarily oriented in the plane, its stiffness matrix is easily determined from [k] of Eq. 4.2-8 by a coordinate transformation (see Section 7.5). Similarly, a plane truss element of arbitrary orientation can be obtained by coordinate transformation of Eq. 4.2-2 (again, see Section 7.5).
Constant-Strain Triangle. This element, shown in Fig. 4.2-3, is one of the earliest finite elements [1.8]. It can be used to solve problems of plane stress and plane strain. However, it is not a very good element for this purpose. We introduce it here primarily because it serves as a good example in subsequent discussions of why elements behave as they do.
text_image
y,v v₃ u₃ 3 u₂ 2 v₂ u₁ 1 v₁ x,u
Figure 4.2-3. Constant-strain triangle (six d.o.f.).
flowchart
graph TD
A["1"] --> B["2"]
B --> C["3"]
C --> D["4"]
D --> E["5"]
E --> F["6"]
F --> G["7"]
G --> H["8"]
H --> I["9"]
I --> J["10"]
J --> K["11"]
K --> L["12"]
L --> M["13"]
M --> N["14"]
N --> O["15"]
O --> P["16"]
P --> Q["17"]
Q --> R["18"]
R --> S["19"]
S --> T["20"]
T --> U["21"]
U --> V["22"]
V --> W["23"]
W --> X["24"]
X --> Y["25"]
Y --> Z["26"]
Z --> A["4"]
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style AN1 fill:#f9f,stroke:#380
style AT1 fill:#f9f,stroke:#380
style AU1 fill:#f9f,stroke:#380
style AV1 fill:#f9f,stroke:#380
style AZ1 fill:#f9f,stroke:#380
style AS1 fill:#f9f,stroke:#380
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style BI1 fill:#f9f,stroke:#380
style AC1 fill:#f9f,stroke:#380
style AB1 fill:#f9f,stroke:#380
style AD1 fill:#f9f,stroke:#380
style BN1 fill:#f9f,stroke:#380
style BH1 fill:#f9f,stroke:#380
style BI1 fill:#f9f,stroke:#380
style AC1 fill:#f9f,stroke:#380
style AD1 fill:#d9f9f,stroke:#333
style AB1 fill:#f9f,stroke:#380
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style AD1 fill:#f9f,stroke:#380
style BN1 fill:#f9f,stroke:#380
style BH1 fill:#f9f,stroke:#380
style BI1 fill:#f9f,stroke:#380
style Ac1 fill:#f9f,stroke:#380
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style AB1 fill:#f9f,stroke:#380
style AC1 fill:#f9f,stroke:#380
style AD1 fill:#f9f,stroke:#380
style BN1 fill:#f9f,stroke:#380
style BH1 fill:#d9f9f,stroke:#333
style BI1 fill:#f9f,stroke:#380
style AC1 fill:#f9f,stroke:#380
style AD1 fill:#f9f,stroke:#380
style BN1 fill:#f9f,stroke:#380
style BH1 fill:#f9f,stroke:#380
style BI1 fill:#f9f,stroke:#380
style AC1 fill:#d9f9f,stroke:#333
style AD1 fill:#f9f,stroke:#380
style AB1 fill:#f9f,stroke:#380
style AC1 fill:#f9f,stroke:#380
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style BN1 fill:#f9f,stroke:#380
style BH1 fill:#f9f,stroke:#380
style BE1 fill:#f9f,stroke:#380
style AC1 fill:#f9f,stroke:#380
style AD1 fill:#f9f,stroke:#380
style BN1 fill:#f9f,stroke:#380
style BH1 fill:#f9f,stroke:#380
style BI1 fill:#f9f,stroke:#380
style AC1 fill:#e9f9f,stroke:#333
style AD1 fill:#f9f,stroke:#380
style AB1 fill:#f9f,stroke:#380
style AC1 fill:#f9f,stroke:#380
style AD1 fill:#f9f,stroke:#380
style BN1 fill:#f9f,stroke:#380
style BH1 fill:#f9i9f,stroke:#333
style BI1 fill:#f9i9f,stroke:#380
style AC1 fill:#f9i9f,stroke:#380
style AD1 fill:#f9i9f,stroke:#380
style BN1 fill:#f9i9f,stroke:#380
style BH1 fill:#f9i9f,stroke:#380
style BI1 fill:#f9i9f,stroke:#380
style AC1 fill:#f9i9f,stroke:#380
style AD1 fill:#f9i9f,stroke:#380
style BN1 fill:#f9i9f,stroke:#380
style BH1 fill:#f9i9f,stroke:#380
style BI1 fill:#f9f,stroke:#380
style AC1 fill:#f9f,stroke:#380
style AD1 fill:#f9f,stroke:#380
style BN1 fill:#f9f,stroke:#380
style BH1 fill:#f9i9f,stroke:#380
style BI1 fill:#f9i9f,stroke:#380
style AC1 fill:#f9i9f,stroke:#380
style AD1 fill:#f9i9f,stroke:#380
style BN1 fill:#f9i9f,stroke:#380
</details>
Figure 4.2-4. Plane rectangular bilinear element (eight d.o.f.).
The constant-strain triangle is based on a complete linear polynomial for both x- and y-direction displacements:
$$
u = a _ {1} + a _ {2} x + a _ {3} y \tag {4.2-9a}
$$
$$
v = a _ {4} + a _ {5} x + a _ {6} y \tag {4.2-9b}
$$
Strains are $\epsilon_x = u_{,x}, \epsilon_y = v_{,y}$ , and $\gamma_{xy} = u_{,y} + v_{,x}$ . Thus
$$
\epsilon_ {x} = a _ {2} \quad \epsilon_ {y} = a _ {6} \quad \gamma_ {x y} = a _ {3} + a _ {5} \tag {4.2-10}
$$
We see that strains are independent of $x$ and $y$ within the element; hence the name "constant-strain triangle."
The algebra of generating the element stiffness matrix is most easily carried out in area coordinates, as described in Chapter 5. Strain-displacement matrix [B] is 3 by 6 and contains only constants, which depend on the $x$ and $y$ coordinates of the three nodes. Hence, if material property matrix [E] and element thickness $t$ are constant over the element,
$$
\underset {6 \times 6} {[ \mathbf {k} ]} = \iint_ {6 \times 3} [ \mathbf {B} ] ^ {T} \underset {3 \times 3} {[ \mathbf {E} ]} \underset {3 \times 6} {[ \mathbf {B} ]} t d x d y = A t [ \mathbf {B} ] ^ {T} [ \mathbf {E} ] [ \mathbf {B} ] \tag {4.2-11}
$$
where $A$ is the area of the triangle.
The behavior of the constant-strain triangle is illustrated by numerical examples in Fig. 5.5-2.
Plane Rectangular Bilinear Element. This element, shown in Fig. 4.2-4, is based on the bilinear displacement field
$$
u = a _ {1} + a _ {2} x + a _ {3} y + a _ {4} x y \tag {4.2-12a}
$$
$$
v = a _ {5} + a _ {6} x + a _ {7} y + a _ {8} x y \tag {4.2-12b}
$$
In terms of nodal d.o.f., the displacement field $\{\mathbf{u}\} = [\mathbf{N}]\{\mathbf{d}\}$ is
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$$
\left\{ \begin{array}{l} u \\ v \end{array} \right\} = \left[ \begin{array}{c c c c c c c c} N _ {1} & 0 & N _ {2} & 0 & N _ {3} & 0 & N _ {4} & 0 \\ 0 & N _ {1} & 0 & N _ {2} & 0 & N _ {3} & 0 & N _ {4} \end{array} \right] \left\{ \begin{array}{l} u _ {1} \\ v _ {1} \\ u _ {2} \\ \vdots \\ v _ {4} \end{array} \right\} \tag {4.2-13}
$$
Shape functions $N_{i}$ of this element are presented as Eqs. 3.12-10. Each $N_{i}$ has the form $(a \pm x)(b \pm y)/4ab$ . (With an adequate understanding of shape functions, the reader should have no difficulty in choosing the proper algebraic signs for each of the $N_{i}$ by inspection.) The strain-displacement matrix is
$$
[ \mathbf {B} ] _ {3 \times 8} = \left[ \begin{array}{c c} \partial / \partial x & 0 \\ 0 & \partial / \partial y \\ \partial / \partial y & \partial / \partial x \end{array} \right] [ \mathbf {N} ] = \frac {1}{4 a b} \left[ \begin{array}{c c c c} - (b - y) & 0 & (b - y) & \dots \\ 0 & - (a - x) & 0 & \dots \\ - (a - x) & - (b - y) & - (a + x) & \dots \end{array} \right] \tag {4.2-14}
$$
We see that $\epsilon_x$ depends on $y$ , $\epsilon_y$ depends on $x$ , and $\gamma_{xy}$ depends on both $x$ and $y$ . The element stiffness matrix is
$$
\underset {8 \times 8} {[ \mathrm{k} ]} = \int_ {- b} ^ {b} \int_ {- a} ^ {a} \underset {8 \times 3} {[ \mathrm{B} ] ^ {T}} \underset {3 \times 3} {[ \mathrm{E} ]} \underset {3 \times 8} {[ \mathrm{B} ]} t d x d y \tag {4.2-15}
$$
where $t$ is the element thickness. The integrand involves polynomials in $x$ and $y$ and is easily evaluated.
If elements are rectangular, the bilinear element and the four-node plane isoparametric element of Section 6.3 are identical. Numerical examples that use isoparametric elements appear in Table 6.14-1.
Solid Rectangular Trilinear Element. This element, shown in Fig. 4.2-5, is a simple generalization of the plane bilinear element. Its x-direction displacement is
$$
u = a _ {1} + a _ {2} x + a _ {3} y + a _ {4} z + a _ {5} x y + a _ {6} y z + a _ {7} z x + a _ {8} x y z \tag {4.2-16}
$$
<details>
<summary>text_image</summary>
2c
y,v
3
4
8
7
2b
2
z,w
1
6
x,u
5
2a
</details>
(a)
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v_i
i
u_i
w_i
</details>
(b)
Figure 4.2-5. (a) Solid rectangular trilinear element (24 d.o.f.). (b) D.o.f. at a typical node i.
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and similarly for $v$ and $w$ , for a total of 24 d.o.f. In terms of nodal d.o.f., the displacement field $\{\mathbf{u}\} = [\mathbf{N}]\{\mathbf{d}\}$ is
$$
\left\{ \begin{array}{l} u \\ v \\ w \end{array} \right\} = \left[ \begin{array}{c c c c c c c} N _ {1} & 0 & 0 & N _ {2} & 0 & 0 & \dots \\ 0 & N _ {1} & 0 & 0 & N _ {2} & 0 & \dots \\ 0 & 0 & N _ {1} & 0 & 0 & N _ {2} & \dots \end{array} \right] \left\{ \begin{array}{c} u _ {1} \\ v _ {1} \\ w _ {1} \\ u _ {2} \\ \vdots \\ w _ {8} \end{array} \right\}. \tag {4.2-17}
$$
where each $N_{i}$ has the form
$$
\frac {(a \pm x) (b \pm y) (c \pm z)}{8 a b c} \tag {4.2-18}
$$
The signs are all negative for $N_2$ , all positive for $N_8$ , and so on. The strain-displacement matrix is $[\mathbf{B}] = [\partial][\mathbf{N}]$ , where $[\partial]$ is given in Eqs. 1.5-6. The element stiffness matrix is
$$
\underset {2 4 \times 2 4} {[ \mathbf {k} ]} = \int_ {- c} ^ {c} \int_ {- b} ^ {b} \int_ {- a} ^ {a} \underset {2 4 \times 6} {[ \mathbf {B} ] ^ {T}} \underset {6 \times 6} {[ \mathbf {E} ]} \underset {6 \times 2 4} {[ \mathbf {B} ]} d x d y d z \tag {4.2-19}
$$
Again the integrations are straightforward.
Note that on any face (e.g., z = c), Eqs. 4.2-16 and 4.2-17 yield forms used for the bilinear element, Eqs. 4.2-12 and 4.2-13, respectively.
A possible application of solid elements is shown in Fig. 4.2-6.
Remark. According to terminology discussed in Section 3.11, the bar, triangular, bilinear, and trilinear elements are all $C^{0}$ elements, and the beam is a $C^{1}$ element. The frame element is $C^{0}$ in axial deformation and $C^{1}$ in bending.
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<summary>natural_image</summary>
3D wireframe model of a curved surface with grid lines, no text or symbols present
</details>
Figure 4.2-6. Quarter of a railway wheel, modeled by solid elements of a type not restricted to rectangular shape. Hidden lines are removed. (Courtesy of Algor Interactive Systems, Inc., Pittsburgh, Pennsylvania.)
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# 4.3 CONSISTENT ELEMENT NODAL
# LOADS $\{\mathbf{r}_e\}$
In this section we consider the element nodal load vector $\{\mathbf{r}_e\}$ , which is stated in Eq. 4.1-6. Equation 4.1-6 converts loads distributed throughout an element or write surface to discrete loads at element nodes.
Any of the four integrals in Eq. 4.1-6 may vanish. For example, the surface integral is zero unless the element has an edge on the structure boundary and traction is applied to that edge. Then we integrate over only that edge of the element. All four integrals vanish if externally applied nodal loads $\{P\}$ make up the entire load vector $\{R\}$ .
Initial Strain and Initial Stress. Terms in Eq. 4.1-6 that contain $\{\epsilon_{0}\}$ and $\{\sigma_{0}\}$ produce nodal loads that account for heating or cooling of the element, swelling (due perhaps to irradiation), and initial lack of fit. These nodal loads are self-equilibrating; that is, $\{r_{e}\}$ produces zero resultant force and zero resultant moment. The familiar bar element provides a convenient example (Fig. 4.3-1). Here
$$
\lfloor \mathbf {N} \rfloor = \left\lfloor \frac {L - x}{L} \quad \frac {x}{L} \right\rfloor \quad \text { and } \quad \lfloor \mathbf {B} \rfloor = \frac {1}{L} \lfloor - 1 \quad 1 \rfloor \tag {4.3-1}
$$
Imagine that the bar is $\Delta L$ units too long, so that the initial strain is $\epsilon_{0} = \Delta L/L$ . Also, let the bar be heated $T^{\circ}$ so that (with expansion prohibited) the initial stress caused by heating is $\sigma_{0} = -E\alpha T$ . The first two integrals in Eq. 4.1-6 yield
$$
\left\{\mathbf {r} _ {e} \right\} _ {2 \times 1} = \int_ {0} ^ {L} \frac {1}{L} \left\{ \begin{array}{c} - 1 \\ 1 \end{array} \right\} E \frac {\Delta L}{L} A d x - \int_ {0} ^ {L} \frac {1}{L} \left\{ \begin{array}{c} - 1 \\ 1 \end{array} \right\} (- E \alpha T) A d x \tag {4.3-2a}
$$
hence
$$
\left\{\mathbf {r} _ {e} \right\} _ {2 \times 1} = \left\{ \begin{array}{c} - F \\ F \end{array} \right\}, \quad \text { where } \quad F = E A \left(\frac {\Delta L}{L} + \alpha T\right) \tag {4.3-2b}
$$
Forces $F$ are shown in Fig. 4.3-1b. It is a worthwhile exercise for the reader to show that the same forces $F$ are produced by regarding the lack of fit as an initial stress and the temperature effect as an initial strain.
Although forces $F$ sum to zero, they act to deform the bar an amount $FL/AE$ if axial deformation of the bar is uninhibited. This deformation creates a stress $\sigma = F/A$ , which is exactly canceled by the initial stress $\sigma = -E\epsilon_0 + \sigma_0$ . Thus we obtain axial strain without axial stress, which is entirely correct for an unrestrained bar.
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y
L
1
2
x,u
</details>
(a)
(b)
Figure 4.3-1. (a) Bar element. (b) Nodal forces caused by heating and initial lack of fit (from Eq. 4.3-2).
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In this example, nodal forces could easily be deduced by direct physical argument. Elements more complicated than bars and beams often require formal use of Eq. 4.1-6.
Mechanical Loads. Loads $\{r_{e}\}$ produced by body forces $\{F\}$ and surface tractions $\{\Phi\}$ are given by the latter two integrals in Eq. 4.1-6. These loads are called work-equivalent loads for the following reason: work done by nodal loads $\{r_{e}\}$ in going through nodal displacements $\{d\}$ is equal to work done by distributed loads $\{F\}$ and $\{\Phi\}$ in going through the displacement field associated with the element shape function. To show this we argue as follows. Work W done by loads $\{r_{e}\}$ during small nodal displacements $\{d\}$ is $W = \{d\}^{T}\{r_{e}\}$ . Taking for example the surface integral in Eq. 4.1-6, and substituting the displacement field $\{u\}^{T} = \{d\}^{T}[N]^{T}$ , we have
$$
W = \{\mathbf {d} \} ^ {T} \left\{\mathbf {r} _ {e} \right\} = \int_ {S _ {e}} \left\{\mathbf {d} \right\} ^ {T} [ \mathbf {N} ] ^ {T} \left\{\boldsymbol {\Phi} \right\} d S = \int_ {S _ {e}} \left\{\mathbf {u} \right\} ^ {T} \left\{\boldsymbol {\Phi} \right\} d S \tag {4.3-3}
$$
The latter integral sums the work of force increments $\{\Phi\}$ dS in going through displacements $\{u\}$ , where $\{u\}$ are field displacements created by $\{d\}$ via shape functions [N]. (See Eqs. 4.3-8 and 4.3-9 for an illustrative example.)
Loads $\{r_{e}\}$ calculated by Eq. 4.1-6 are also called consistent because they are based on the same shape functions as used to calculate the element stiffness matrix. Finally, loads $\{r_{e}\}$ calculated by Eq. 4.1-6 are statically equivalent to the original distributed loading; that is, both $\{r_{e}\}$ and the original loading have the same resultant force and the same moment about an arbitrarily chosen point. That this is true may be seen by considering the work equivalence of the two loadings during a rigid-body translation and a small rigid-body rotation about an arbitrarily chosen point.
We define inconsistent loading or lumping as the conversion of a distributed load to nodal loads that are inconsistent with Eq. 4.1-6, but are statically equivalent to the distributed load in that they provide the same resultant force. Typically, lumping is achieved by (a) computing the total force on an element caused by distributed loading, then assigning the same fraction of the total force to each element node, and (b) ignoring any nodal moments that would be present in the consistent vector $\{r_{e}\}$ . As will be seen in subsequent examples, the consistent method generally does not allot the total force on an element equally to element nodes. Lumping can yield poor answers in a coarse mesh, produce locally poor answers in a fine mesh, and lead to failure of the patch test (the patch test is discussed in Section 4.6).
Concentrated Loads Not at Nodes. We denote by $\{p\}$ a concentrated force that has components $p_{x}$ , $p_{y}$ , and $p_{z}$ . The contribution of $\{p\}$ to $\{r_{e}\}$ can be evaluated from the surface integral in Eq. 4.1-6 by regarding a concentrated force as a large traction $\{\Phi\}$ acting on a small area dS. Thus $\{p\} = \{\Phi\}$ dS. The integral of $[N]^{T}\{\Phi\}$ dS is simply $[N]^{T}\{p\}$ where the concentrated force acts and is zero elsewhere. Thus, if there are n concentrated forces applied to an element, Eq. 4.1-6 yields
$$
\text { Owing to concentrated forces } \quad \{\mathbf {p} \} _ {i}, \quad \{\mathbf {r} _ {e} \} = \sum_ {i = 1} ^ {n} [ \mathbf {N} ] _ {i} ^ {T} \{\mathbf {p} \} _ {i} \tag {4.3-4}
$$
where $[N]_{i}$ is the value of [N] at the location of $\{p\}_{i}$ .
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<details>
<summary>text_image</summary>
y
q = \frac{L - x}{L} q_1 + \frac{x}{L} q_2
F
1
2
x_1 u
\frac{2L}{3}
L
</details>
(a)
<details>
<summary>text_image</summary>
N₁ = \frac{L - x}{L}
1
\frac{1}{3}
\frac{2L}{3}
L
</details>
(b)
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N₂ = x/L
2/3
2L/3
L
</details>
(c)
Figure 4.3-2. (a) Linearly varying distributed load $q$ and concentrated force $F$ on a bar. (b,c) Shape functions for axial displacement $u$ .
Similarly, if $n$ concentrated moments $\{\mathbf{m}\} = \left\lfloor m_x - m_y - m_z \right]^T$ act at $n$ nonnodal locations on an element,
Owing to concentrated moments $\{\mathbf{m}\}_{i}, \{\mathbf{r}_e\} = \sum_{i=1}^{n} [\mathbf{N}']_i^T \{\mathbf{m}\}_{i}$ (4.3-5)
where $[N']$ contains derivatives of [N] and is evaluated at the location of $\{m\}_{i}$ . Derivatives are needed to calculate rotations $[N']_{i}\{d\}$ , through which moments $\{m\}_{i}$ act in doing work equal to $\{d\}^{T}\{r_{e}\}$ . An example of loads that result from Eq. 4.3-5 appears in Fig. 4.3-6c.
Example. Bar Element. The bar element in Fig. 4.3-2 carries a distributed axial load $q$ , having dimensions of force per unit length, which varies linearly from intensity $q_{1}$ at $x = 0$ to intensity $q_{2}$ at $x = L$ . In addition, a concentrated axial force $F$ acts at $x = 2L / 3$ . Consistent nodal loads $\{\mathbf{r}_e\}$ are required.
2L/3. Consistent nodal loads $[t_e]$ are required. To account for force $F$ we use Eq. 4.3-4. To account for load $q$ we can use the last integral in Eq. 4.1-6, writing the force increment as $q dx$ instead of $\{\Phi\} dS$ . Thus
$$
\left\{\mathbf {r} _ {e} \right\} _ {2 \times 1} = \int_ {0} ^ {L} \left\lfloor \mathbf {N} \right] ^ {T} q d x + \left\lfloor \mathbf {N} \right\rfloor_ {2 L / 3} F \tag {4.3-6}
$$
from which, with $\lfloor \mathbf{N}\rfloor = \left\lfloor \frac{L - x}{L} \frac{x}{L} \right\rfloor$ and $q$ as given in Fig. 4.3-2a,
$$
\left\{\mathbf {r} _ {e} \right\} = \frac {L}{6} \left\{ \begin{array}{l} 2 q _ {1} + q _ {2} \\ q _ {1} + 2 q _ {2} \end{array} \right\} + \left\{ \begin{array}{l} F / 3 \\ 2 F / 3 \end{array} \right\} \tag {4.3-7}
$$
We see that if $q_{1} = q_{2} = q$ , a constant, then the total load $qL$ is equally distributed to element nodes. The fraction of force $F$ allocated to each node is dictated by Eq. 4.3-4 and is shown graphically as an ordinate of the appropriate shape function in Fig. 4.3-2. (In civil engineering parlance, $N_{1}$ and $N_{2}$ are influence lines for reactions at the nodes.)
The concept of work-equivalent loads is discussed in connection with Eq. 4.5.9.16 show that $\{\mathbf{r}_e\}$ of eq 4.3-7 is indeed a set of work-equivalent loads, we must show that
$$
\left\lfloor u _ {1} \quad u _ {2} \right\rfloor \left\{\mathrm{r} _ {e} \right\} = \int_ {0} ^ {L} u q d x + F u _ {2 L / 3} \tag {4.3-8}
$$