MultiPhysicsVault/.raw/FiniteElementProcedures/FiniteElementProcedures_020.md

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Thickness = 0.5 cm p1v p1 vS p1v 1 p2v p2 3 x us 1 cm Element m p2u 2 1 cm (a) Element layout

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v1 = U23 v3 = U15 v2 = U11 u1 = U22 u3 = U14 u2 = U10 Y X

(b) Local-global degrees of freedom
Figure E4.7 Pressure loading on element (m)

The vector of surface loads is (with p_{1} and p_{2} positive)

\mathbf {f} ^ {s} = \left[ \begin{array}{c} \frac {1}{2} (1 + x) p _ {1} ^ {u} + \frac {1}{2} (1 - x) p _ {2} ^ {u} \\ - \frac {1}{2} (1 + x) p _ {1} ^ {v} - \frac {1}{2} (1 - x) p _ {2} ^ {v} \end{array} \right]

To obtain \mathbf{R}_S^{(m)} we first evaluate

\mathbf {R} _ {s} = 0. 5 \int_ {- 1} ^ {+ 1} \mathbf {H} ^ {s ^ {T}} \mathbf {f} ^ {s} d x

to obtain

\mathbf {R} _ {s} = \frac {1}{3} \left[ \begin{array}{c} p _ {1} ^ {u} \\ p _ {2} ^ {u} \\ 2 (p _ {1} ^ {u} + p _ {2} ^ {u}) \\ - p _ {1} ^ {v} \\ - p _ {2} ^ {v} \\ - 2 (p _ {1} ^ {v} + p _ {2} ^ {v}) \end{array} \right]

Thus, corresponding to the global degrees of freedom given in Fig. E4.7, we have

\begin{array}{l} \mathbf {R} _ {S} ^ {(m) T} = \frac {1}{3} [ 0 \quad \dots \quad 0 \quad ; \quad p _ {2} ^ {u} \quad - p _ {2} ^ {v} \quad ; \quad 0 \quad 0 \quad ; \quad 2 (p _ {1} ^ {u} + p _ {2} ^ {v}) \quad - 2 (p _ {1} ^ {v} + p _ {2} ^ {v}) \quad ; \quad 0 \quad \dots \\ \dots 0 \mid p _ {1} ^ {u} - p _ {1} ^ {v} \mid 0 \dots 0 ] \\ \end{array}

U_{22} U_{23}\gets Assemblage degrees of freedom

The Assumption About Stress Equilibrium

We noted earlier that the analyses of truss and beam assemblages were originally not considered to be finite element analysis because the “exact” element stiffness matrices can be employed in the analyses. These stiffness matrices are obtained in the application of the principle of virtual displacements if the assumed displacement interpolations are in fact the exact displacements that the element undergoes when subjected to the unit nodal point displacements. Here the word “exact” refers to the fact that by imposing these displacements on the element, all pertinent differential equations of equilibrium and compatibility and the constitutive requirements (and also the boundary conditions) are fully satisfied in static analysis.

In considering the analysis of the truss assemblage in Example 4.5, we obtained the exact stiffness matrix of element 1. However, for element 2 an approximate stiffness matrix was calculated as shown in the next example.

EXAMPLE 4.8: Calculate for element 2 in Example 4.5 the exact element internal displacements that correspond to a unit element end displacement u_{2} and evaluate the corresponding stiffness matrix. Also, show that using the element displacement assumption in Example 4.5, internal element equilibrium is not satisfied.

Consider element 2 with a unit displacement imposed at its right end as shown in Fig. E4.8. The element displacements are calculated by solving the differential equation (see Example 3.22),

E \frac {d}{d x} \left(A \frac {d u}{d x}\right) = 0 \tag {a}

subject to the boundary conditions u|_{x=0} = 0 and u|_{x=80} = 1.0 . Substituting for the area A and integrating the relation in (a), we obtain

u = \frac {3}{2} \left(1 - \frac {1}{1 + x / 4 0}\right) \tag {b}

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u₂ = 1.0 cm 1 2 x 80 cm

Figure E4.8 Element 2 of bar analyzed in Example 4.5

These are the exact element internal displacements. The element end forces required to subject the bar to these displacements are

k _ {1 2} = - E A \left. \frac {d u}{d x} \right| _ {x = 0}

k _ {2 2} = E A \left. \frac {d u}{d x} \right| _ {x = L} \tag {c}

Substituting from (b) into (c) we have

k _ {2 2} = \frac {3 E}{8 0}; \quad k _ {1 2} = - \frac {3 E}{8 0}

Hence we have, using the symmetry of the element matrix and equilibrium to establish k_{21} and k_{11} ,

\mathbf {K} = \frac {3}{8 0} E \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \tag {d}

The same result is of course obtained using the principle of virtual displacements with the displacement (b).

We note that the stiffness coefficient in (d) is smaller than the corresponding value obtained in Example 4.5 (3E/80 instead of 13E/240). The finite element solution in Example 4.5 overestimates the stiffness of the structure because the assumed displacements artificially constrain the motion of the material particles (see Section 4.3.4). To check that the internal equilibrium is indeed not satisfied, we substitute the finite element solution (given by the displacement assumption in Example 4.5) into (a) and obtain

E \frac {d}{d x} \left\{\left(1 + \frac {x}{4 0}\right) ^ {2} \frac {1}{8 0} \right\} \neq 0

The solution of truss and beam structures, using the exact displacements corresponding to unit nodal point displacements and rotations to evaluate the stiffness matrices, gives analysis results that for the selected mathematical model satisfy all three requirements of mechanics exactly: differential equilibrium for every point of the structure (including nodal point equilibrium), compatibility, and the stress-strain relationships. Hence, the exact (unique) solution for the selected mathematical model is obtained.

We may note that such an exact solution is usually pursued in static analysis, in which the exact stiffness relationships are obtained as described in Example 4.8, but an exact

solution is much more difficult to reach in dynamic analysis because in this case the distributed mass and damping effects must be included (see, for example, R. W. Clough and J. Penzien [A]).

However, although in a general (static or dynamic) finite element analysis, differential equilibrium is not exactly satisfied at all points of the continuum considered, two important properties are always satisfied by the finite element solution using a coarse or a fine mesh. These properties are (see Fig. 4.2)

1. Nodal point equilibrium
2. Element equilibrium.

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q - 1 q m - 1 Element m

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graph TD
    A["q"] -->|Sum of forces F(m) equilibrate externally applied loads| B["m-1"]
    B -->|Sum of forces F(m) equilibrate externally applied loads| C["m"]
    C -->|Sum of forces F(m) equilibrate externally applied loads| D["q-1"]
    D -->|Sum of forces F(m) equilibrate externally applied loads| A
    style A fill:#f9f,stroke:#333
    style B fill:#f9f,stroke:#333
    style C fill:#f9f,stroke:#333
    style D fill:#f9f,stroke:#333
    note right of A
        Forces F(m) are in equilibrium
    end

Figure 4.2 Nodal point and element equilibrium in a finite element analysis

Namely, consider that a finite element analysis has been performed and that we calculate for each finite element m the element nodal point force vectors

\mathbf {F} ^ {(m)} = \int_ {V ^ {(m)}} \mathbf {B} ^ {(m) T} \mathbf {\tau} ^ {(m)} d V ^ {(m)} \tag {4.29}

where \pmb{\tau}^{(m)} = \mathbf{C}^{(m)}\pmb{\epsilon}^{(m)} . Then we observe that according to property 1,

At any node, the sum of the element nodal point forces is in equilibrium with the externally applied nodal loads (which include all effects due to body forces, surface tractions, initial stresses, concentrated loads, inertia and damping forces, and reactions).

And according to property 2,

Each element m is in equilibrium under its forces \mathbf{F}^{(m)} .

Property 1 follows simply because (4.27) expresses the nodal point equilibrium and we have

\sum_ {m} \mathbf {F} ^ {(m)} = \mathbf {K U} \tag {4.30}

The element equilibrium stated in property 2 is satisfied provided the finite element displacement interpolations in \mathbf{H}^{(m)} satisfy the basic convergence requirements, which include the condition that the element must be able to represent the rigid body motions (see Section 4.3). Namely, let us consider element m subjected to the nodal point forces \mathbf{F}^{(m)} and impose virtual nodal point displacements corresponding to the rigid body motions. Then for each virtual element rigid body motion with nodal point displacements \widehat{u} , we have

\overline {{{\mathbf {u}}}} ^ {T} \mathbf {F} ^ {(m)} = \int_ {V ^ {(m)}} \left(\mathbf {B} ^ {(m)} \overline {{{\mathbf {u}}}}\right) ^ {T} \boldsymbol {\tau} ^ {(m)} d V ^ {(m)} = \int_ {V ^ {(m)}} \overline {{{\boldsymbol {\epsilon}}}} ^ {(m) T} \boldsymbol {\tau} ^ {(m)} d V ^ {(m)} = 0

because here \overline{\epsilon}^{(m)} = 0 . Using all applicable rigid body motions we therefore find that the forces \mathbf{F}^{(m)} are in equilibrium.

Hence, a finite element analysis can be interpreted as a process in which

1. The structure or continuum is idealized as an assemblage of discrete elements connected at nodes pertaining to the elements.
2. The externally applied forces (body forces, surface tractions, initial stresses, concentrated loads, inertia and damping forces, and reactions) are lumped to these nodes using the virtual work principle to obtain equivalent externally applied nodal point forces.
3. The equivalent externally applied nodal point forces (calculated in 2) are equilibrated by the element nodal point forces that are equivalent (in the virtual work sense) to the element internal stresses; i.e., we have

\sum_ {m} \mathbf {F} ^ {(m)} = \mathbf {R}

4. Compatibility and the stress-strain material relationship are satisfied exactly, but instead of equilibrium on the differential level, only global equilibrium for the com-

plete structure, at the nodes, and of each element m under its nodal point forces \mathbf{F}^{(m)} is satisfied.

Consider the following example.

EXAMPLE 4.9: The finite element solution to the problem in Fig. E4.6, with P = 100, E = 2.7 \times 10^{6} , \nu = 0.30 , t = 0.1, is given in Fig. E4.9. Clearly, the stresses are not continuous between elements, and equilibrium on the differential level is not satisfied. However,

1. Show that \Sigma_{m}\mathbf{F}^{(m)} = \mathbf{R} and calculate the reactions.
2. Show that the element forces \mathbf{F}^{(4)} for element 4 are in equilibrium.

The fact that \Sigma_{m}\mathbf{F}^{(m)} = \mathbf{R} follows from the solution of (4.17), and \mathbf{R} consists of the sum of all nodal point forces. Hence, this relation can also be used to evaluate the reactions.

Referring to the nodal point numbering in Fig. E4.6(b), we find for node 1:

\text { reactions } R _ {x} = 1 0 0. 1 5

R _ {y} = 4 1. 3 6

for node 2:

\begin{array}{l} \text { reactions } R _ {x} = 2. 5 8 - 2. 8 8 = - 0. 3 0 \\ R _ {y} = 1 6. 7 9 + 5. 9 6 = 2 2. 7 4 \text {(because of rounding)} \\ \end{array}

for node 3:

\text { reactions } R _ {x} = - 9 9. 8 5

R _ {y} = 3 5. 9 0

for node 4:

\text { horizontal   force   equilibrium:   } - 4 2. 0 1 + 4 2. 0 1 = 0

\text { vertical   force   equilibrium: } - 2 2. 9 0 + 2 2. 9 0 = 0

for node 5:

\text { horizontal   force   equilibrium:   } - 6 0. 7 2 - 1 2. 0 4 + 4 4. 7 3 + 2 8. 0 3 = 0

\text { vertical   force   equilibrium: } - 3 5. 2 4 - 3 5. 0 4 + 1 9. 1 0 + 5 1. 1 8 = 0

for node 6:

\text { horizontal   force   equilibrium:   } 5 7. 9 9 - 5 7. 9 9 = 0

\text { vertical   force   equilibrium: } - 6. 8 1 + 6. 8 1 = 0

And for nodes 7, 8, and 9, force equilibrium is obviously also satisfied, where at node 9 the element nodal force balances the applied load P = 100.

Finally, let us check the overall force equilibrium of the model: horizontal equilibrium:

1 0 0. 1 5 - 0. 3 0 - 9 9. 8 5 = 0

vertical equilibrium:

4 1. 3 6 + 2 2. 7 4 + 3 5. 9 0 - 1 0 0 = 0

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1129 1123 ② -95.92 -101.6

radar

Dimension	Value
1	1.146
2	624.0
3	298.5
4	4
5	-324.3

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-95.92 -55.16 Element ① -972.1 -931.4

radar

Value
47.60
-79.76
③
-219.9
-347.3

(a) Exploded view of elements showing stresses \tau_{xx}^{(m)} . Note the stress discontinuities between elements and the nonzero stresses along the free edges

radar

Point	Value
1	338.7
2	319.7
3	-28.77
4	-47.79

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169.9 -914.9 ④ -16.96 -1102

other

Element ①

Label	Value
-28.77	107.1
-291.6	-155.8
(Top label)	107.1
(Bottom label)	-28.77
(Bottom label)	-291.6

radar

Point	Value
1	137.9
2	-286.7
3	③
4	57.64
5	-366.9

(b) Exploded view of elements showing stresses \tau_{\gamma \gamma}^{(m)}
Figure E4.9 Solution results for problem considered in Example 4.6 (rounded to digits shown)

(d) Exploded view of elements showing element nodal point forces equivalent (in the virtual work sense) to the element stresses. The nodal point forces are at each node in equilibrium with the applied forces (including the reactions)
Figure E4.9 (continued)

moment equilibrium (about node 2):

- 1 0 0 \times 4 + 1 0 0. 1 5 \times 2 + 9 9. 8 5 \times 2 = 0

It is important to realize that this force equilibrium will hold for any finite element mesh, however coarse the mesh may be, provided properly formulated elements are used (see Section 4.3).

Now consider element 4:

horizontal equilibrium:

0 - 5 7. 9 9 + 2 8. 0 3 + 2 9. 9 7 = 0 \text {(because of rounding)}

vertical equilibrium:

- 1 0 0 + 6. 8 1 + 5 1. 1 8 + 4 2. 0 1 = 0

moment equilibrium (about its local node 3):

- 1 0 0 \times 2 + 5 7. 9 9 \times 2 + 4 2. 0 1 \times 2 = 0

Hence the element nodal forces are in equilibrium.

Element Local and Structure Global Degrees of Freedom

The derivations of the element matrices in Example 4.6 and 4.7 show that it is expedient to first establish the matrices corresponding to the local element degrees of freedom. The construction of the finite element matrices, which correspond to the global assemblage degrees of freedom [used in (4.19) to (4.25)] can then be directly achieved by identifying the global degrees of freedom that correspond to the local element degrees of freedom. However, considering the matrices \mathbf{H}^{(m)} , \mathbf{B}^{(m)} , \mathbf{K}^{(m)} , and so on, corresponding to the global assemblage degrees of freedom, only those rows and columns that correspond to element degrees of freedom have nonzero entries, and the main objective in defining these specific matrices was to be able to express the assemblage process of the element matrices in a theoretically elegant manner. In the practical implementation of the finite element method, this elegance is also present, but all element matrices are calculated corresponding only to the element degrees of freedom and are then directly assembled using the correspondence between the local element and global assemblage degrees of freedom. Thus, with only the element local nodal point degrees of freedom listed in \hat{u} , we now write (as in Example 4.6)

\mathbf {u} = \mathbf {H} \hat {\mathbf {u}} \tag {4.31}

where the entries in the vector \mathbf{u} are the element displacements measured in any convenient local coordinate system. We then also have

\epsilon = \mathbf {B} \hat {\mathbf {u}} \tag {4.32}

Considering the relations in (4.31) and (4.32) , the fact that no superscript is used on the interpolation matrices indicates that the matrices are defined with respect to the local element degrees of freedom. Using the relations for the element stiffness matrix, mass

matrix, and load vector calculations as before, we obtain

\boxed {\mathbf {K} = \int_ {V} \mathbf {B} ^ {T} \mathbf {C B} d V} \tag {4.33}

\boxed {\mathbf {M} = \int_ {V} \rho \mathbf {H} ^ {T} \mathbf {H} d V} \tag {4.34}

\boxed {\mathbf {R} _ {B} = \int_ {V} \mathbf {H} ^ {T} \mathbf {f} ^ {B} d V} \tag {4.35}

\boxed {\mathbf {R} _ {S} = \int_ {S} \mathbf {H} ^ {S ^ {T}} \mathbf {f} ^ {S} d S} \tag {4.36}

\boxed {\mathbf {R} _ {I} = \int_ {V} \mathbf {B} ^ {T} \boldsymbol {\tau} ^ {I} d V} \tag {4.37}

where all variables are defined as in (4.19) to (4.25), but corresponding to the local element degrees of freedom. In the derivations and discussions to follow, we shall refer extensively to the relations in (4.33) to (4.37). Once the matrices given in (4.33) to (4.37) have been calculated, they can be assembled directly using the procedures described in Example 4.11 and Chapter 12.

In this assemblage process it is assumed that the directions of the element nodal point displacements \hat{u} in (4.31) are the same as the directions of the global nodal point displacements U. However, in some analyses it is convenient to start the derivation with element nodal point degrees of freedom \tilde{u} that are not aligned with the global assemblage degrees of freedom. In this case we have

\mathbf {u} = \tilde {\mathbf {H}} \tilde {\mathbf {u}} \tag {4.38}

and

\tilde {\mathbf {u}} = \mathbf {T} \hat {\mathbf {u}} \tag {4.39}

where the matrix T transforms the degrees of freedom \hat{u} to the degrees of freedom \tilde{u} and (4.39) corresponds to a first-order tensor transformation (see Section 2.4); the entries in column j of the matrix T are the direction cosines of a unit vector corresponding to the jth degree of freedom in \hat{u} when measured in the directions of the \tilde{u} degrees of freedom. Substituting (4.39) into (4.38), we obtain

\mathbf {H} = \tilde {\mathbf {H}} \mathbf {T} \tag {4.40}