김경종
22 KiB
where we define
[ B ] = \left[ - \frac {1}{L} \quad \frac {1}{L} \right] \tag {3.10.13}
The axial stress/strain relationship is given by
\{\sigma_ {x} \} = [ D ] \{\varepsilon_ {x} \} \tag {3.10.14}
where [ D ] = [ E ] ð3:10:15Þ
for the one-dimensional stress/strain relationship and E is the modulus of elasticity. Now, by Eq. (3.10.12), we can express Eq. (3.10.14) as
\{\sigma_ {x} \} = [ D ] [ B ] \{\hat {d} \} \tag {3.10.16}
Using Eq. (3.10.7) expressed in matrix notation form, we have the total potential energy given by
\pi_ {p} = \frac {A}{2} \int_ {0} ^ {L} \left\{\sigma_ {x} \right\} ^ {T} \left\{\varepsilon_ {x} \right\} d \hat {x} - \left\{\hat {d} \right\} ^ {T} \left\{P \right\} - \iint_ {S _ {1}} \left\{\hat {u} _ {s} \right\} ^ {T} \left\{\hat {T} _ {x} \right\} d S - \iint_ {V} \iint \left\{\hat {u} \right\} ^ {T} \left\{\hat {X} _ {b} \right\} d V \tag {3.10.17}
where \{ P \} now represents the concentrated nodal loads and where in general both \underline { { \sigma } } _ { x } and \underline { { \boldsymbol { \varepsilon } } } _ { x } are column matrices. For proper matrix multiplication, we must place the transpose on \left\{ \sigma _ { x } \right\} . Similarly, fu^g and \{ \hat { T } _ { x } \} in general are column matrices, so for proper matrix multiplication, fu^g is transposed in Eq. (3.10.17).
Using Eqs. (3.10.8), (3.10.12), and (3.10.16) in Eq. (3.10.17), we obtain
\begin{array}{l} \pi_ {p} = \frac {A}{2} \int_ {0} ^ {L} \{\hat {d} \} ^ {T} [ B ] ^ {T} [ D ] ^ {T} [ B ] \{\hat {d} \} d \hat {x} - \{\hat {d} \} ^ {T} \{P \} \\ - \iint_ {S _ {1}} \left\{\hat {d} \right\} ^ {T} \left[ N _ {S} \right] ^ {T} \left\{\hat {T} _ {x} \right\} d S - \iint_ {V} \left\{\hat {d} \right\} ^ {T} \left[ N \right] ^ {T} \left\{\hat {X} _ {b} \right\} d V \tag {3.10.18} \\ \end{array}
In Eq. (3.10.18), \pi _ { p } is seen to be a function of \{ \hat { d } \} ; that is, \pi _ { p } = \pi _ { p } ( \hat { d } _ { 1 x } , \hat { d } _ { 2 x } ) . However, ½B and [ D ] , Eqs. (3.10.13) and (3.10.15), and the nodal degrees of freedom \hat { d } _ { 1 x } and \dot { d } _ { 2 x } are not functions of x^. Therefore, integrating Eq. (3.10.18) with respect to x^ yields
\pi_ {p} = \frac {A L}{2} \{\hat {d} \} ^ {T} [ B ] ^ {T} [ D ] ^ {T} [ B ] \{\hat {d} \} - \{\hat {d} \} ^ {T} \{\hat {f} \} \tag {3.10.19}
where \{ \hat { f } \} = \{ P \} + \int _ { S _ { 1 } } \int _ { } ^ { } [ N _ { S } ] ^ { T } \{ \hat { T } _ { x } \} d S + \int _ { V } \int _ { } ^ { } \int _ { } ^ { } [ N ] ^ { T } \{ \hat { X } _ { b } \} d V ð3:10:20Þ
From Eq. (3.10.20), we observe three separate types of load contributions from concentrated nodal forces, surface tractions, and body forces, respectively. We define
these surface tractions and body-force matrices as
\{\hat {f} _ {s} \} = \iint_ {S _ {1}} \left[ N _ {S} \right] ^ {T} \{\hat {T} _ {x} \} d S \tag {3.10.20a}
\{\hat {f} _ {b} \} = \iint_ {V} \left[ N \right] ^ {T} \{\hat {X} _ {b} \} d V \tag {3.10.20b}
The expression for \{ \hat { f } \} given by Eq. (3.10.20) then describes how certain loads can be considered to best advantage.
Loads calculated by Eqs. (3.10.20a) and (3.10.20b) are called consistent because they are based on the same shape functions ½N used to calculate the element stiffness matrix. The loads calculated by Eq. (3.10.20a) and (3.10.20b) are also statically equivalent to the original loading; that is, both \{ \hat { f } _ { s } \} and \{ \hat { f } _ { b } \} and the original loads yield the same resultant force and same moment about an arbitrarily chosen point.
The minimization of \pi _ { p } with respect to each nodal displacement requires that
\frac {\partial \pi_ {p}}{\partial \hat {d} _ {1 x}} = 0 \quad \text { and } \quad \frac {\partial \pi_ {p}}{\partial \hat {d} _ {2 x}} = 0 \tag {3.10.21}
Now we explicitly evaluate \pi _ { p } given by Eq. (3.10.19) to apply Eq. (3.10.21). We define the following for convenience:
\{U ^ {*} \} = \{\hat {d} \} ^ {T} [ B ] ^ {T} [ D ] ^ {T} [ B ] \{\hat {d} \} \tag {3.10.22}
Using Eqs. (3.10.10), (3.10.13), and (3.10.15) in Eq. (3.10.22) yields
\left\{U ^ {*} \right\} = \left[ \hat {d} _ {1 x} \quad \hat {d} _ {2 x} \right] \left\{ \begin{array}{c} - \frac {1}{L} \\ \frac {1}{L} \end{array} \right\} [ E ] \left[ - \frac {1}{L} \quad \frac {1}{L} \right] \left\{ \begin{array}{c} \hat {d} _ {1 x} \\ \hat {d} _ {2 x} \end{array} \right\} \tag {3.10.23}
Simplifying Eq. (3.10.23), we obtain
U ^ {*} = \frac {E}{L ^ {2}} \left(\hat {d} _ {1 x} ^ {2} - 2 \hat {d} _ {1 x} \hat {d} _ {2 x} + \hat {d} _ {2 x} ^ {2}\right) \tag {3.10.24}
Also, the explicit expression for \{ \hat { d } \} ^ { T } \{ \hat { f } \} is
\{\hat {d} \} ^ {T} \{\hat {f} \} = \hat {d} _ {1 x} \hat {f} _ {1 x} + \hat {d} _ {2 x} \hat {f} _ {2 x} \tag {3.10.25}
Therefore, using Eqs. (3.10.24) and (3.10.25) in Eq. (3.10.19) and then applying Eqs. (3.10.21), we obtain
\frac {\partial \pi_ {p}}{\partial \hat {d} _ {1 x}} = \frac {A L}{2} \left[ \frac {E}{L ^ {2}} (2 \hat {d} _ {1 x} - 2 \hat {d} _ {2 x}) \right] - \hat {f} _ {1 x} = 0 \tag {3.10.26}
and \frac { \hat { \sigma } \pi _ { p } } { \hat { \sigma } \hat { d } _ { 2 x } } = \frac { A L } { 2 } \left[ \frac { E } { L ^ { 2 } } ( - 2 \hat { d } _ { 1 x } + 2 \hat { d } _ { 2 x } ) \right] - \hat { f } _ { 2 x } = 0
In matrix form, we express Eqs. (3.10.26) as
\frac {\partial \pi_ {p}}{\partial \{\hat {d} \}} = \frac {A E}{L} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \left\{ \begin{array}{l} \hat {d} _ {1 x} \\ \hat {d} _ {2 x} \end{array} \right\} - \left\{ \begin{array}{l} \hat {f} _ {1 x} \\ \hat {f} _ {2 x} \end{array} \right\} = \left\{ \begin{array}{l} 0 \\ 0 \end{array} \right\} \tag {3.10.27}
or, because \{ \hat { f } \} = [ \hat { k } ] \{ \hat { d } \} , we have the stiffness matrix for the bar element obtained from Eq. (3.10.27) as
[ \hat {k} ] = \frac {A E}{L} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \tag {3.10.28}
As expected, Eq. (3.10.28) is identical to the stiffness matrix obtained in Section 3.1.
Finally, instead of the cumbersome process of explicitly evaluating \pi _ { p } , we can use the matrix differentiation as given by Eq. (2.6.12) and apply it directly to Eq. (3.10.19) to obtain
\frac {\partial \pi_ {p}}{\partial \{\hat {d} \}} = A L [ B ] ^ {T} [ D ] [ B ] \{\hat {d} \} - \{\hat {f} \} = 0 \tag {3.10.29}
where [ D ] ^ { T } = [ D ] has been used in writing Eq. (3.10.29). The result of the evaluation of A L [ B ] ^ { T } [ D ] [ B ] is then equal to [ \hat { k } ] given by Eq. (3.10.28). Throughout this text, we will use this matrix differentiation concept (also see Appendix \mathbf { A } ) , which greatly simplifies the task of evaluating [ \hat { k } ] .
To illustrate the use of Eq. (3.10.20a) to evaluate the equivalent nodal loads for a bar subjected to axial loading traction \hat { T } _ { x } . , we now solve Example 3.12.
Example 3.12
A bar of length L is subjected to a linearly distributed axial loading that varies from zero at node 1 to a maximum at node 2 (Figure 3–28). Determine the energy equivalent nodal loads.
text_image
T̂ₓ = Cx̂ (force/length) 1 L 2 x̂
Figure 3–28 Element subjected to linearly varying axial load
Using Eq. (3.10.20a) and shape functions from Eq. (3.10.9), we solve for the energy equivalent nodal forces of the distributed loading as follows:
\{\hat {f} _ {0} \} = \left\{ \begin{array}{l} \hat {f} _ {1 x} \\ \hat {f} _ {2 x} \end{array} \right\} = \iint_ {S _ {1}} [ N ] ^ {T} \{\hat {T} _ {x} \} d S = \int_ {0} ^ {L} \left\{ \begin{array}{c} 1 - \frac {\hat {x}}{L} \\ \frac {\hat {x}}{L} \end{array} \right\} \{C \hat {x} \} d \hat {x} \tag {3.10.30}
= \left\{ \begin{array}{c} \frac {C \hat {x} ^ {2}}{2} - \frac {C \hat {x} ^ {3}}{3 L} \\ \frac {C \hat {x} ^ {3}}{3 L} \end{array} \right\} _ {0} ^ {L}
= \left\{ \begin{array}{l} \frac {C L ^ {2}}{6} \\ \frac {C L ^ {2}}{3} \end{array} \right\} \tag {3.10.31}
where the integration was carried out over the length of the bar, because \hat { T } _ { x } is in units of force/length.
Note that the total load is the area under the load distribution given by
F = \frac {1}{2} (L) (C L) = \frac {C L ^ {2}}{2} \tag {3.10.32}
Therefore, comparing Eq. (3.10.31) with (3.10.32), we find that the equivalent nodal loads for a linearly varying load are
\hat {f} _ {1 x} = \frac {1}{3} F = \text { one - third of the total load } \tag {3.10.33}
\hat {f} _ {2 x} = \frac {2}{3} F = \text { two - thirds of the total load }
In summary, for the simple two-noded bar element subjected to a linearly varying load (triangular loading), place one-third of the total load at the node where the distributed loading begins (zero end of the load) and two-thirds of the total load at the node where the peak value of the distributed load ends.
We now illustrate (Example 3.13) a complete solution for a bar subjected to a surface traction loading.
Example 3.13
For the rod loaded axially as shown in Figure 3–29, determine the axial displacement and axial stress. Let E = 3 0 \times 1 0 ^ { 6 } psi, \overset { } { A } \overset { = } 2 \mathrm { i n } . ^ { 2 } , and L = 6 0 in. Use (a) one and (b) two elements in the finite element solutions. (In Section 3.11 one-, two-, four-, and eight-element solutions will be presented from the computer program Algor [9].
text_image
T_x = -10x lb/in. 60 in. x
Figure 3–29 Rod subjected to triangular load distribution
(a) One-element solution (Figure 3–30).
text_image
T_x = -10x -600 1 ① 2
Figure 3–30 One-element model
From Eq. (3.10.20a), the distributed load matrix is evaluated as follows:
\{F _ {0} \} = \int_ {0} ^ {L} [ N ] ^ {T} \{T _ {x} \} d x \tag {3.10.34}
where T _ { x } is a line load in units of pounds per inch and \underline { { \hat { f } } } _ { 0 } = \underline { { F } } _ { 0 } as \underline { x } = \hat { \underline { x } } . . Therefore, using Eq. (3.1.4) for ½N in Eq. (3.10.34), we obtain
\{F _ {0} \} = \int_ {0} ^ {L} \left\{ \begin{array}{c} 1 - \frac {x}{L} \\ \frac {x}{L} \end{array} \right\} \{- 1 0 x \} d x \tag {3.10.35}
or \left\{ \begin{array} { c c } { { F _ { 1 x } } } \\ { { F _ { 2 x } } } \end{array} \right\} = \left\{ \begin{array} { c c } { { \displaystyle \frac { - 1 0 L ^ { 2 } } { 2 } + \frac { 1 0 L ^ { 2 } } { 3 } } } \\ { { \displaystyle \frac { - 1 0 L ^ { 2 } } { 3 } } } \end{array} \right\} = \left\{ \begin{array} { c c } { { \displaystyle \frac { - 1 0 L ^ { 2 } } { 6 } } } \\ { { \displaystyle \frac { - 1 0 L ^ { 2 } } { 3 } } } \end{array} \right\} = \left\{ \begin{array} { c c } { { \displaystyle \frac { - 1 0 \left( 6 0 \right) ^ { 2 } } { 6 } } } \\ { { \displaystyle \frac { - 1 0 \left( 6 0 \right) ^ { 2 } } { 3 } } } \end{array} \right\}
or ð3:10:36Þ
Using Eq. (3.10.33), we could have determined the same forces at nodes 1 and 2—that is, one-third of the total load is at node 1 and two-thirds of the total load is at node 2.
Using Eq. (3.10.28), we find that the stiffness matrix is given by
k ^ {(1)} = 1 0 ^ {6} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right]
The element equations are then
1 0 ^ {6} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} \\ 0 \end{array} \right\} = \left\{ \begin{array}{c} - 6 0 0 0 \\ R _ {2 x} - 1 2, 0 0 0 \end{array} \right\} \tag {3.10.37}
Solving Eq. 1 of Eq. (3.10.37), we obtain
d _ {1 x} = - 0. 0 0 6 \text { in. } \tag {3.10.38}
The stress is obtained from Eq. (3.10.14) as
\begin{array}{l} \{\sigma_ {x} \} = [ D ] \{\varepsilon_ {x} \} \\ = E [ B ] \{d \} \\ = E \left[ - \frac {1}{L} \quad \frac {1}{L} \right] \left\{ \begin{array}{l} d _ {1 x} \\ d _ {2 x} \end{array} \right\} \\ = E \left(\frac {d _ {2 x} - d _ {1 x}}{L}\right) \\ = 3 0 \times 1 0 ^ {6} \left(\frac {0 + 0 . 0 0 6}{6 0}\right) \\ = 3 0 0 0 \mathrm{psi} (T) \tag {3.10.39} \\ \end{array}
(b) Two-element solution (Figure 3–31).
text_image
-300 -600 1 ① 2 ② 3
Figure 3-31 Two-element model
We first obtain the element forces. For element 2, we divide the load into a uniform part and a triangular part. For the uniform part, half the total uniform load is placed at each node associated with the element. Therefore, the total uniform part is
(3 0 \text { in. }) (- 3 0 0 \text { lb / in. }) = - 9 0 0 0 \text { lb }
and using Eq. (3.10.33) for the triangular part of the load, we have, for element 2,
\left\{ \begin{array}{l} f _ {2 x} ^ {(2)} \\ f _ {3 x} ^ {(2)} \end{array} \right\} = \left\{ \begin{array}{l} - \left[ \frac {1}{2} (9 0 0 0) + \frac {1}{3} (4 5 0 0) \right] \\ - \left[ \frac {1}{2} (9 0 0 0) + \frac {2}{3} (4 5 0 0) \right] \end{array} \right\} = \left\{ \begin{array}{l} - 6 0 0 0 \mathrm{lb} \\ - 7 5 0 0 \mathrm{lb} \end{array} \right\} \tag {3.10.40}
For element 1, the total force is from the triangle-shaped distributed load only and is given by
\frac {1}{2} (3 0 \text { in. }) (- 3 0 0 \text { lb / in. }) = - 4 5 0 0 \text { lb }
On the basis of Eq. (3.10.33), this load is separated into nodal forces as shown:
\left\{ \begin{array}{l} f _ {1 x} ^ {(1)} \\ f _ {2 x} ^ {(1)} \end{array} \right\} = \left\{ \begin{array}{l} \frac {1}{3} (- 4 5 0 0) \\ \frac {2}{3} (- 4 5 0 0) \end{array} \right\} = \left\{ \begin{array}{l} - 1 5 0 0 \mathrm{lb} \\ - 3 0 0 0 \mathrm{lb} \end{array} \right\} \tag {3.10.41}
The final nodal force matrix is then
\left\{ \begin{array}{l} F _ {1 x} \\ F _ {2 x} \\ F _ {3 x} \end{array} \right\} = \left\{ \begin{array}{c} - 1 5 0 0 \\ - 6 0 0 0 - 3 0 0 0 \\ R _ {3 x} - 7 5 0 0 \end{array} \right\} \tag {3.10.42}
The element stiffness matrices are now
\underline {{k}} ^ {(1)} = \underline {{k}} ^ {(2)} = \frac {A E}{L / 2} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] = (2 \times 1 0 ^ {6}) \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \tag {3.10.43}
The assembled global stiffness matrix is
\underline {{K}} = (2 \times 1 0 ^ {6}) \left[ \begin{array}{c c c} 1 & - 1 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 1 & 1 \end{array} \right] \frac {\mathrm{lb}}{\mathrm{in.}} \tag {3.10.44}
The assembled global equations are then
(2 \times 1 0 ^ {6}) \left[ \begin{array}{r r r} 1 & - 1 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 1 & 1 \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} \\ d _ {2 x} \\ d _ {3 x} = 0 \end{array} \right\} = \left\{ \begin{array}{c} - 1 5 0 0 \\ - 9 0 0 0 \\ R _ {3 x} - 7 5 0 0 \end{array} \right\} \tag {3.10.45}
where the boundary condition d_{3x}=0 has been substituted into Eq. (3.10.45). Now, solving equations 1 and 2 of Eq. (3.10.45), we obtain
d _ {1 x} = - 0. 0 0 6 \text { in. } \tag {3.10.46}
d _ {2 x} = - 0. 0 0 5 2 5 \mathrm{in}.
The element stresses are as follows:
Element 1
\begin{array}{l} \sigma_ {x} = E \left[ - \frac {1}{3 0} \quad \frac {1}{3 0} \right] \left\{ \begin{array}{l} d _ {1 x} = - 0. 0 0 6 \\ d _ {2 x} = - 0. 0 0 5 2 5 \end{array} \right\} \\ = 7 5 0 \mathrm{psi} (T) \tag {3.10.47} \\ \end{array}
Element 2
\begin{array}{l} \sigma_ {x} = E \left[ - \frac {1}{3 0} \quad \frac {1}{3 0} \right] \left\{ \begin{array}{l} d _ {2 x} = - 0. 0 0 5 2 5 \\ d _ {3 x} = 0 \end{array} \right\} \\ = 5 2 5 0 \mathrm{psi} (T) \tag {3.10.48} \\ \end{array}
3.11 Comparison of Finite Element Solution to Exact Solution for Bar
We will now compare the finite element solutions for Example 3.13 using one, two, four, and eight elements to model the bar element and the exact solution. The exact solution for displacement is obtained by solving the equation
u = \frac {1}{A E} \int_ {0} ^ {x} P (x) d x \tag {3.11.1}
where, using the following free-body diagram,
text_image
10x lb/in. x P(x)
we have
P (x) = \frac {1}{2} x (1 0 x) = 5 x ^ {2} \mathrm{lb} \tag {3.11.2}
Therefore, substituting Eq. (3.11.2) into Eq. (3.11.1), we have
\begin{array}{l} u = \frac {1}{A E} \int_ {0} ^ {x} 5 x ^ {2} d x \\ = \frac {5 x ^ {3}}{3 A E} + C _ {1} \tag {3.11.3} \\ \end{array}
Now, applying the boundary condition at x ¼ L, we obtain
u (L) = 0 = \frac {5 L ^ {3}}{3 A E} + C _ {1}
or
C _ {1} = - \frac {5 L ^ {3}}{3 A E} \tag {3.11.4}
Substituting Eq. (3.11.4) into Eq. (3.11.3) makes the final expression for displacement
u = \frac {5}{3 A E} (x ^ {3} - L ^ {3}) \tag {3.11.5}
line
| Axial coordinate in inches | One element | Two elements | Four elements | Eight elements |
|---|---|---|---|---|
| 0 | -0.006 | -0.006 | -0.006 | -0.006 |
| 20 | -0.005 | -0.005 | -0.005 | -0.005 |
| 40 | -0.003 | -0.003 | -0.003 | -0.003 |
| 60 | 0 | 0 | 0 | 0 |
Figure 3–32 Comparison of exact and finite element solutions for axial displacement (along length of bar)
Substituting A ¼ 2 in.2, E = 3 0 \times 1 0 ^ { 6 } psi, and L = 6 0 in. into Eq. (3.11.5), we obtain
u = 2. 7 7 8 \times 1 0 ^ {- 8} x ^ {3} - 0. 0 0 6 \tag {3.11.6}
The exact solution for axial stress is obtained by solving the equation
\sigma (x) = \frac {P (x)}{A} = \frac {5 x ^ {2}}{2 \mathrm{in} ^ {2}} = 2. 5 x ^ {2} \mathrm{psi} \tag {3.11.7}
Figure 3–32 shows a plot of Eq. (3.11.6) along with the finite element solutions (part of which were obtained in Example 3.13). Some conclusions from these results follow.
-
The finite element solutions match the exact solution at the node points. The reason why these nodal values are correct is that the element nodal forces were calculated on the basis of being energyequivalent to the distributed load based on the assumed linear displacement field within each element. (For uniform cross-sectional bars and beams, the nodal degrees of freedom are exact. In general, computed nodal degrees of freedom are not exact.)
-
Although the node values for displacement match the exact solution, the values at locations between the nodes are poor using few elements (see one- and two-element solutions) because we used a linear displacement function within each element, whereas the exact solution, Eq. (3.11.6), is a cubic function. However, because we use increasing
line
| Axial coordinate in inches | One element | Two elements | Four elements | Eight elements |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 |
| 10 | 0 | 0 | 0 | 0 |
| 20 | 0 | 0 | 0 | 0 |
| 30 | 0 | 0 | 0 | 0 |
| 40 | 0 | 0 | 0 | 0 |
| 50 | 0 | 0 | 0 | 0 |
| 60 | 0 | 0 | 0 | 0 |
Figure 3–33 Comparison of exact and finite element solutions for axial stress (along length of bar)
numbers of elements, the finite element solution converges to the exact solution (see the four- and eight-element solutions in Figure 3–32).
- The stress is derived from the slope of the displacement curve as
\sigma = E \varepsilon = E ( d u / d x ). Therefore, by the finite element solution, because u is a linear function in each element, axial stress is constant in each element. It then takes even more elements to model the first derivative of the displacement function or, equivalently, the axial stress. This is shown in Figure 3–33, where the best results occur for the eightelement solution. - The best approximation of the stress occurs at the midpoint of the element, not at the nodes (Figure 3–33). This is because the derivative of displacement is better predicted between the nodes than at the nodes.
- The stress is not continuous across element boundaries. Therefore, equilibrium is not satisfied across element boundaries. Also, equilibrium within each element is, in general, not satisfied. This is shown in Figure 3–34 for element 1 in the two-element solution and element 1 in the eight-element solution [in the eight-element solution the forces are obtained from the Algor computer code [9]]. As the number of elements used increases, the discontinuity in the stress decreases across element boundaries, and the approximation of equilibrium improves.
Finally, in Figure 3–35, we show the convergence of axial stress at the fixed end ðx ¼ LÞ as the number of elements increases.