add documents
This commit is contained in:
김경종
+412
@@ -0,0 +1,412 @@
|
||||
<!-- source-page: 141 -->
|
||||
|
||||
|
||||
|
||||
<details>
|
||||
<summary>line</summary>
|
||||
|
||||
| Curvature | Bending Moment | Slope EI |
|
||||
| --------- | -------------- | -------- |
|
||||
| dεf | 0 | 0 |
|
||||
| dεf | M0 | (dεf)e |
|
||||
| dεf | (dεf)p | (dεf)p |
|
||||
</details>
|
||||
|
||||
Fig. 5.4 Moment curvature relationship for a Timoshenko beam.
|
||||
|
||||
$$
|
||||
H ^ {\prime} = \frac {d M}{(d \epsilon_ {f}) _ {p}}.
|
||||
$$
|
||||
|
||||
This can be interpreted as the slope of the strain-hardening portion of the moment-curvature curve after the removal of the elastic curvature component. Thus
|
||||
|
||||
$$
|
||||
H ^ {\prime} = \frac {d M}{d \epsilon_ {f} - (d \epsilon_ {f}) _ {e}} = \frac {(E I) _ {T}}{1 - [ (E I) _ {T} / E I ]}. \tag {5.32}
|
||||
$$
|
||||
|
||||
It is therefore possible to rewrite (5.31) as
|
||||
|
||||
$$
|
||||
d \epsilon_ {f} = \frac {d M}{E I} + \frac {d M}{H ^ {\prime}} = \frac {d M (H ^ {\prime} + E I)}{E I H ^ {\prime}} \tag {5.33}
|
||||
$$
|
||||
|
||||
and then the incremental moment–curvature relationship can be written in the form
|
||||
|
||||
$$
|
||||
d M = \frac {E I H ^ {\prime}}{(E I + H ^ {\prime})} d \epsilon_ {f}. \tag {5.34}
|
||||
$$
|
||||
|
||||
Thus during yielding the incremental stress–strain resultant relationship is
|
||||
|
||||
$$
|
||||
d M = E I \left(1 - \frac {E I}{E I \pm H ^ {\prime}}\right) d \epsilon_ {f}
|
||||
$$
|
||||
|
||||
$$
|
||||
d Q = G \hat {A} d \epsilon_ {s}. \tag {5.35}
|
||||
$$
|
||||
|
||||
<!-- source-page: 142 -->
|
||||
|
||||
The shear force/shear strain relationship is always elastic whereas the moment–curvature relationship is elasto-plastic. After yielding the flexural rigidity EI is replaced by
|
||||
|
||||
$$
|
||||
E I \left(1 - \frac {E I}{E I + H ^ {\prime}}\right).
|
||||
$$
|
||||
|
||||
If the hardening parameter $H'$ is equal to zero then the material behaviour is elasto-perfectly plastic and as mentioned in Section 3.5 for elasto-plastic axial bar elements this may lead to tangential stiffness matrices which are singular. This difficulty can also be avoided by use of the initial stiffness method in which the elastic element stiffnesses are employed at every stage of the computation thereby guaranteeing a positive definite assembled stiffness matrix.
|
||||
|
||||
# 5.4.3 Solution of nonlinear equations
|
||||
|
||||
Let us now generate the nonlinear equilibrium equations using the virtual expression (5.11). In order to do this we require the global rather than the element expressions for the lateral displacements, rotation, curvature and shear strain. At any point in the finite element mesh the lateral displacement and rotation can be obtained from the expression
|
||||
|
||||
$$
|
||||
\left[ \begin{array}{l} w \\ \theta \end{array} \right] = N \varphi \tag {5.36}
|
||||
$$
|
||||
|
||||
where the shape function matrix is
|
||||
|
||||
$$
|
||||
N = \left[ \begin{array}{l l l l l l} N _ {1}, & 0, & N _ {2}, & 0, & \dots , & N _ {n}, & 0 \\ 0, & N _ {1}, & 0, & N _ {2}, & \dots , & 0, & N _ {n} \end{array} \right] \tag {5.37}
|
||||
$$
|
||||
|
||||
and the vector of nodal displacements is
|
||||
|
||||
$$
|
||||
\varphi = [ w _ {1}, \theta_ {1}, w _ {2}, \theta_ {2}, \dots , w _ {n}, \theta_ {n} ] ^ {T} \tag {5.38}
|
||||
$$
|
||||
|
||||
where $w_{i}$ , $\theta_{i}$ and $N_{i}$ are the lateral displacement, rotation and global shape functions associated with node i.
|
||||
|
||||
The curvature and shear strain at any point within the entire finite element mesh is given as
|
||||
|
||||
$$
|
||||
- \frac {d \theta}{d x} = B _ {f} \varphi \quad \text { and } \quad \frac {d w}{d x} - \theta = B _ {s} \varphi \tag {5.39}
|
||||
$$
|
||||
|
||||
where
|
||||
|
||||
$$
|
||||
\boldsymbol {B} _ {f} = \left[ 0, - \frac {d N _ {1}}{d x}, 0, - \frac {d N _ {2}}{d x}, \dots , 0, - \frac {d N _ {n}}{d x} \right] \tag {5.40}
|
||||
$$
|
||||
|
||||
and
|
||||
|
||||
$$
|
||||
\boldsymbol {B} _ {s} = \left[ \frac {d N _ {1}}{d x}, - N _ {1}, \frac {d N _ {2}}{d x}, - N _ {2}, \dots , \frac {d N _ {n}}{d x}, - N _ {n} \right] \tag {5.41}
|
||||
$$
|
||||
|
||||
<!-- source-page: 143 -->
|
||||
|
||||
Virtual curvatures and shear strains are given as
|
||||
|
||||
$$
|
||||
- \frac {d (\delta \theta)}{d x} = \boldsymbol {B} _ {f} \delta \varphi \quad \text { and } \quad \frac {d (\delta w)}{d x} - \delta \theta = \boldsymbol {B} _ {s} \delta \varphi \tag {5.42}
|
||||
$$
|
||||
|
||||
respectively, where the vector of virtual nodal displacements is written as
|
||||
|
||||
$$
|
||||
\delta \varphi = [ \delta w _ {1}, \delta \theta_ {1}, \delta w _ {2}, \delta \theta_ {2}, \dots , \delta w _ {n}, \delta \theta_ {n} ] ^ {T}. \tag {5.43}
|
||||
$$
|
||||
|
||||
Thus the virtual work expression (5.11) can now be written as
|
||||
|
||||
$$
|
||||
\begin{array}{l} \int_ {0} ^ {l} [ \delta \varphi ] ^ {T} [ \boldsymbol {B} _ {f} ] ^ {T} M d x + \int_ {0} ^ {l} [ \delta \varphi ] ^ {T} [ \boldsymbol {B} _ {s} ] ^ {T} Q d x \\ - \int_ {0} ^ {l} [ \delta \varphi ] ^ {T} [ \bar {\mathbf {N}} ] ^ {T} q d x = 0 \tag {5.44} \\ \end{array}
|
||||
$$
|
||||
|
||||
where $\bar{\mathbf{N}} = [N_1, 0, N_2, 0, \dots, N_n, 0]$ . (5.45)
|
||||
|
||||
Since (5.44) must be true for any set of virtual displacements $\delta \varphi$ then we have
|
||||
|
||||
$$
|
||||
\left\{\int_ {0} ^ {l} \left[ \boldsymbol {B} _ {f} \right] ^ {T} M d x + \int_ {0} ^ {l} \left[ \boldsymbol {B} _ {s} \right] ^ {T} Q d x \right\} - \int_ {0} ^ {l} \left[ \bar {\mathbf {N}} \right] ^ {T} q d x = 0 \tag {5.46}
|
||||
$$
|
||||
|
||||
or
|
||||
|
||||
$$
|
||||
p - f = 0.
|
||||
$$
|
||||
|
||||
In fact this equation is identical to (5.22) when there is no plasticity.
|
||||
|
||||
Unfortunately in elasto-plastic problems M is a nonlinear function and in general we can only predict the vector p approximately. Thus (5.46) is nonlinear and since p is only approximately known than p-f will equal a residual value $\psi(\varphi)$ which we attempt to reduce to zero in our solution procedure.
|
||||
|
||||
We evaluate contributions to p element by element and assemble in the usual manner. The contribution from element e has the form
|
||||
|
||||
$$
|
||||
\begin{array}{l} \boldsymbol {p} ^ {(e)} = \int_ {x _ {1} ^ {(e)}} ^ {x _ {2} ^ {(e)}} \left[ \begin{array}{c} 0 \\ \frac {1}{l ^ {(e)}} \\ 0 \\ - \frac {1}{l ^ {(e)}} \end{array} \right] M ^ {(e)} d x + \int_ {x _ {1} ^ {(e)}} ^ {x _ {2} ^ {(e)}} \left[ \begin{array}{c} - \frac {1}{l ^ {(e)}} \\ \frac {x ^ {(e)} - x _ {2} ^ {(e)}}{l ^ {(e)}} \\ \frac {1}{l ^ {(e)}} \\ \frac {x _ {1} ^ {(e)} - x ^ {(e)}}{l ^ {(e)}} \end{array} \right] Q ^ {(e)} d x \\ = \left[ - Q ^ {(e)}, M ^ {(e)} - \frac {(Q l) ^ {(e)}}{2}, Q ^ {(e)}, - M ^ {(e)} - \frac {(Q l) ^ {(e)}}{2} \right] ^ {T}. \tag {5.47} \\ \end{array}
|
||||
$$
|
||||
|
||||
\*The second integral evaluation is equivalent to using a 1-point Gauss rule.
|
||||
|
||||
<!-- source-page: 144 -->
|
||||
|
||||
|
||||
|
||||
<details>
|
||||
<summary>flowchart</summary>
|
||||
|
||||
```mermaid
|
||||
graph TD
|
||||
A["DATA"] --> B["Input data defining geometry, loading and boundary conditions, material properties, etc."]
|
||||
B --> C["INITAL"]
|
||||
C --> D["Input data for current increment."]
|
||||
D --> E["INCREM"]
|
||||
E --> F["Initialize accumulative arrays to zero. Update load vector."]
|
||||
F --> G["NONAL"]
|
||||
G --> H["Set indicator to identify type of solution algorithm."]
|
||||
H --> I{Is new element stiffness matrix required?}
|
||||
I -->|No| J["No"]
|
||||
I -->|Yes| K["STIFFB"]
|
||||
K --> L["Calculate the element stiffness matrices and store on disc."]
|
||||
L --> M["ASSEMB and GREDUC or RESOLV and BAKSUB"]
|
||||
M --> N["Assemble global stiffness matrix (or take previous one) and global load vector and solve the resulting equations for unknowns."]
|
||||
N --> O["REFORB"]
|
||||
O --> P["Calculate the residual force vector."]
|
||||
P --> Q["CONUND"]
|
||||
Q --> R["Has solution converged?"]
|
||||
R -->|No| S["No"]
|
||||
R -->|Yes| T["RESULT"]
|
||||
T --> U["Output the results."]
|
||||
U --> V["END"]
|
||||
V --> W["LOAD INCREMENT LOOP"]
|
||||
W --> X["ITERATION LOOP"]
|
||||
```
|
||||
</details>
|
||||
|
||||
Fig. 5.5 Overall structure of program TIMOSH.
|
||||
|
||||
<!-- source-page: 145 -->
|
||||
|
||||
Note that the appropriate value of bending moment $M^{(e)}$ is inserted in (5.47).
|
||||
|
||||
Table 5.1 shows the complete sequence of nonlinear equation solving which is very similar to the one adopted for the axially-loaded bars in Chapter 3.
|
||||
|
||||
1. Begin load increment.
|
||||
Set $f = f + \Delta f$ , iteration counter i = 0 and $\psi^{i} = \Delta f + \psi$ (that is, include equilibrium correction from previous increment).
|
||||
2. Evaluate the new tangential stiffness matrix $K_{T}$ if necessary.
|
||||
3. Solve $\psi^i = \mathbf{K}_T\Delta \varphi^i$
|
||||
4. Evaluate $\varphi = \varphi + \Delta \varphi^i$ .
|
||||
5. For each element evaluate $M^{(e)}$ and $Q^{(e)}$ . Check $M^{(e)}$ and adjust its value accordingly to account for any plastic behaviour. Evaluate the element residual force vector $[\psi^{(e)}]^{i+1} = \mathfrak{p}^{(e)} - \mathbf{f}^{(e)}$ and assemble into the global residual force vector $\psi^{i+1}$ .
|
||||
6. Check $\Delta\varphi^{i}$ for convergence.
|
||||
7. If solution has converged set $\psi = \psi^{i+1}$ and go to step 1, otherwise set $i = i+1$ and go to step 2.
|
||||
|
||||
Table 5.1 Solution procedure for elasto-plastic nonlayered Timoshenko beam analysis.
|
||||
|
||||
# 5.4.4 Overall program structure of TIMOSH
|
||||
|
||||
A modular approach is adopted for program TIMOSH. In fact the overall structure is identical to the structure in the programs of Chapter 3. Figure 5.5 shows the overall structure of TIMOSH. Routines DATA, INITIAL, INCREM, NONAL, ASSEMB, GREDUC, BAKSUB, CONUND, RESOLV and RESULT have already been described in Chapter 3. The only new routines are STIFFB, REFORB and, of course, the MASTER routine BEAM.
|
||||
|
||||
5.4.5 New routines for nonlayered elasto-plastic Timoshenko beam analysis Master BEAM The master calling routine BEAM simply organises the calling of the main routines as described in Fig. 5.5.
|
||||
|
||||
<table><tr><td></td><td>MASTER BEAM</td><td>EPBM</td><td>1</td></tr><tr><td>C</td><td>**********</td><td>EPBM</td><td>2</td></tr><tr><td>C</td><td></td><td>EPBM</td><td>3</td></tr><tr><td>C ***</td><td>ELSTO-PLASTIC NONLAYERED TIMOSHENKO BEAM PROGRAM</td><td>EPBM</td><td>4</td></tr><tr><td>C</td><td></td><td>EPBM</td><td>5</td></tr><tr><td>C</td><td>**********</td><td>EPBM</td><td>6</td></tr><tr><td>.</td><td>COMMON/UNIM1/NPOIN,NELEM,NBOUN,NLOAD,NPROP,NNODE,IINCS,IITER,</td><td>EPBM</td><td>7</td></tr><tr><td></td><td>KRESL,NCHEK,TOLER,NALGO,NSVAB,NDOFN,NINCS,NEVAB,</td><td>EPBM</td><td>8</td></tr><tr><td></td><td>NITER,NOUTP,FACTO</td><td>EPBM</td><td>9</td></tr><tr><td></td><td>COMMON/UNIM2/PROPS(5,4),COORD(26),LNODS(25,2),IFPRE(52),</td><td>EPBM</td><td>10</td></tr><tr><td></td><td>FIXED(52),TLOAD(25,4),RLOAD(25,4),ELOAD(25,4),</td><td>EPBM</td><td>11</td></tr><tr><td></td><td>MATNO(25),STRES(25,2),PLAST(25),XDISP(52),</td><td>EPBM</td><td>12</td></tr></table>
|
||||
|
||||
<!-- source-page: 146 -->
|
||||
|
||||
```csv
|
||||
TDISP(26,2),TREAC(26,2),ASTIF(52,52),ASLOD(52),
|
||||
REACT(52),FRESV(1352),PEFIX(52),ESTIF(4,4)
|
||||
CALL DATA
|
||||
CALL INITIAL
|
||||
DO 30 IINCS=1,NINCS
|
||||
CALL INCLOD
|
||||
DO 10 IITER=1,NITER
|
||||
CALL NONAL
|
||||
IF(KRESL.EQ.1) CALL STIFFB
|
||||
CALL ASSEMB
|
||||
IF(KRESL.EQ.1) CALL GREDUC
|
||||
IF(KRESL.EQ.2) CALL RESOLV
|
||||
CALL BAKSUB
|
||||
CALL REFORB
|
||||
CALL CONUND
|
||||
IF(NCHEK.EQ.0) GO TO 20
|
||||
IF(IITER.EQ.1.AND.NOUTP.EQ.1) CALL RESULT
|
||||
IF(NOUTP.EQ.2) CALL RESULT
|
||||
10 CONTINUE
|
||||
WRITE(6,900)
|
||||
900 FORMAT(1H0,5X,'SOLUTION NOT CONVERGED')
|
||||
STOP
|
||||
20 CALL RESULT
|
||||
30 CONTINUE
|
||||
STOP
|
||||
END
|
||||
EPBM 13
|
||||
EPBM 14
|
||||
EPBM 15
|
||||
EPBM 16
|
||||
EPBM 17
|
||||
EPBM 18
|
||||
EPBM 19
|
||||
EPBM 20
|
||||
EPBM 21
|
||||
EPBM 22
|
||||
EPBM 23
|
||||
EPBM 24
|
||||
EPBM 25
|
||||
EPBM 26
|
||||
EPBM 27
|
||||
EPBM 28
|
||||
EPBM 29
|
||||
EPBM 30
|
||||
EPBM 31
|
||||
EPBM 32
|
||||
EPBM 33
|
||||
EPBM 34
|
||||
EPBM 35
|
||||
EPBM 36
|
||||
EPBM 37
|
||||
EPBM 38
|
||||
.
|
||||
```
|
||||
|
||||
Subroutine STIFFB The purpose of this routine is to evaluate the element stiffness matrices and store them on disc prior to their use in the assembly and equation solving routines.
|
||||
|
||||
```csv
|
||||
SUBROUTINE STIFFB STFB 1
|
||||
C**************************STFB 2
|
||||
C STFB 3
|
||||
C *** CALCULATES ELEMENT STIFFNESS MATRICES STFB 4
|
||||
C STFB 5
|
||||
C**************************STFB 6
|
||||
COMMON/UNIM1/NPOIN,NELEM,NBOUN,NLOAD,NPROP,NNODE,IINCS,IITER, STFB 7
|
||||
KRESL,NCHEK,TOLER,NALGO,NSVAB,NDOFN,NINCS,NEVAB, STFB 8
|
||||
NITER,NOUTP,FACTO STFB 9
|
||||
COMMON/UNIM2/PROPS(5,4),COORD(26),LNODS(25,2),IFPRE(52), STFB 10
|
||||
FIXED(52),TLOAD(25,4),RLOAD(25,4),ELOAD(25,4), STFB 11
|
||||
MATNO(25),STRES(25,2),PLAST(25),XDISP(52), STFB 12
|
||||
TDISP(26,2),TREAC(26,2),ASTIF(52,52),ASLOD(52), STFB 13
|
||||
REACT(52),FRESV(1352),PEFIX(52),ESTIF(4,4) STFB 14
|
||||
REWIND 1 STFB 15
|
||||
DO 20 IELEM=1,NELEM STFB 16
|
||||
LPROP=MATNO(IELEM) STFB 17
|
||||
EIVAL=PROPS(LPROP,1) STFB 18
|
||||
SVALU=PROPS(LPROP,2) STFB 19
|
||||
HARDS=PROPS(LPROP,4) STFB 20
|
||||
NODE1=LNODS(IELEM,1) STFB 21
|
||||
NODE2=LNODS(IELEM,2) STFB 22
|
||||
ELENG=ABS(COORD(NODE2)-COORD(NODE1)) STFB 23
|
||||
IF(PLAST(IELEM).NE.0.0) EIVAL=EIVAL*(1.0-EIVAL/(EIVAL+HARDS)) STFB 24
|
||||
VALU1=0.5*SVALU STFB 25
|
||||
VALU2=SVALU/ELENG STFB 26
|
||||
VALU3=EIVAL/ELENG STFB 27
|
||||
VALU4=0.25*SVALU*ELENG STFB 28
|
||||
ESTIF(1,1)= VALU2 STFB 29
|
||||
ESTIF(1,2)= VALU1 STFB 30
|
||||
```
|
||||
|
||||
<!-- source-page: 147 -->
|
||||
|
||||
<table><tr><td>ESTIF(1,3)=-VALU2</td><td>STFB</td><td>31</td></tr><tr><td>ESTIF(1,4)=VALU1</td><td>STFB</td><td>32</td></tr><tr><td>ESTIF(2,2)=VALU3+VALU4</td><td>STFB</td><td>33</td></tr><tr><td>ESTIF(2,3)=-VALU1</td><td>STFB</td><td>34</td></tr><tr><td>ESTIF(2,4)=-VALU3+VALU4</td><td>STFB</td><td>35</td></tr><tr><td>ESTIF(3,3)=VALU2</td><td>STFB</td><td>36</td></tr><tr><td>ESTIF(3,4)=-VALU1</td><td>STFB</td><td>37</td></tr><tr><td>ESTIF(4,4)=VALU3+VALU4</td><td>STFB</td><td>38</td></tr><tr><td>DO 10 ISTIF=1,4</td><td>STFB</td><td>39</td></tr><tr><td>DO 10 JSTIF=ISTIF,4</td><td>STFB</td><td>40</td></tr><tr><td>10 ESTIF(JSTIF,ISTIF)=ESTIF(ISTIF,JSTIF)</td><td>STFB</td><td>41</td></tr><tr><td>WRITE(1) ESTIF</td><td>STFB</td><td>42</td></tr><tr><td>20 CONTINUE</td><td>STFB</td><td>43</td></tr><tr><td>RETURN</td><td>STFB</td><td>44</td></tr><tr><td>END</td><td>STFB</td><td>45</td></tr></table>
|
||||
|
||||
STFB 15 Rewind disc ready for writing element stiffnesses.
|
||||
|
||||
STFB 16–38 For each element evaluate the upper triangular portion of the element stiffness matrix $K^{(e)}$ . Note that if plasticity has taken place the elastic EI is replaced by the elasto-plastic $(EI)_{T}$ .
|
||||
|
||||
STFB 39-41 Obtain using symmetry the lower triangular portion of $K^{(e)}$ .
|
||||
|
||||
STFB 42 Write all element stiffness matrices on to disc.
|
||||
|
||||
Subroutine REFORB This routine evaluates the equivalent nodal forces.
|
||||
|
||||
<table><tr><td>SUBROUTINE REFORB</td><td>RFRB</td><td>1</td></tr><tr><td>C**********</td><td>RFRB</td><td>2</td></tr><tr><td>C</td><td>RFRB</td><td>3</td></tr><tr><td>C *** CALCULATES INTERNAL EQUIVALENT NODAL FORCES</td><td>RFRB</td><td>4</td></tr><tr><td>C</td><td>RFRB</td><td>5</td></tr><tr><td>C**********</td><td>RFRB</td><td>6</td></tr><tr><td>COMMON/UNIM1/NPOIN.NELEM,NBOUN,NLOAD,NPROP,NNODE,IINCS,IITER,</td><td>RFRB</td><td>7</td></tr><tr><td>KRESL,NCHEK,TOLER,NALGO,NSVAB,NDOFN,NINCS,NEVAB,</td><td>RFRB</td><td>8</td></tr><tr><td>NITER,NOUTP,FACTO</td><td>RFRB</td><td>9</td></tr><tr><td>COMMON/UNIM2/PROPS(5,4),COORD(26),LNODS(25,2),IFPRE(52),</td><td>RFRB</td><td>10</td></tr><tr><td>FIXED(52),TLOAD(25,4),RLOAD(25,4),ELOAD(25,4),</td><td>RFRB</td><td>11</td></tr><tr><td>MATNO(25),STRES(25,2),PLAST(25),XDISP(52),</td><td>RFRB</td><td>12</td></tr><tr><td>TDISP(26,2),TREAC(26,2),ASTIF(52,52),ASLOD(52),</td><td>RFRB</td><td>13</td></tr><tr><td>REACT(52),FRESV(1352),PEFIX(52),ESTIF(4,4)</td><td>RFRB</td><td>14</td></tr><tr><td>DO 10 IELEM=1,NELEM</td><td>RFRB</td><td>15</td></tr><tr><td>DO 10 IEVAB=1,NEVAB</td><td>RFRB</td><td>16</td></tr><tr><td>10 ELOAD(IELEM,IEVAB)=0.0</td><td>RFRB</td><td>17</td></tr><tr><td>DO 70 IELEM=1,NELEM</td><td>RFRB</td><td>18</td></tr><tr><td>LPROP=MATNO(IELEM)</td><td>RFRB</td><td>19</td></tr><tr><td>EIVAL=PROPS(LPROP,1)</td><td>RFRB</td><td>20</td></tr><tr><td>SVALU=PROPS(LPROP,2)</td><td>RFRB</td><td>21</td></tr><tr><td>YIELD=PROPS(LPROP,3)</td><td>RFRB</td><td>22</td></tr><tr><td>HARDS=PROPS(LPROP,4)</td><td>RFRB</td><td>23</td></tr><tr><td>NODE1=LNODS(IELEM,1)</td><td>RFRB</td><td>24</td></tr><tr><td>NODE2=LNODS(IELEM,2)</td><td>RFRB</td><td>25</td></tr><tr><td>ELENG=ABS(COORD(NODE2)-COORD(NODE1))</td><td>RFRB</td><td>26</td></tr><tr><td>WNOD1=XDISP(NODE1*NDOFN-1)</td><td>RFRB</td><td>27</td></tr><tr><td>WNOD2=XDISP(NODE2*NDOFN-1)</td><td>RFRB</td><td>28</td></tr><tr><td>THTA1=XDISP(NODE1*NDOFN)</td><td>RFRB</td><td>29</td></tr><tr><td>THTA2=XDISP(NODE2*NDOFN)</td><td>RFRB</td><td>30</td></tr><tr><td>STRAN=(THTA1-THTA2)/ELENG</td><td>RFRB</td><td>31</td></tr><tr><td>STLIN=STRAN*EIVAL</td><td>RFRB</td><td>32</td></tr><tr><td>STCUR=STRES(IELEM,1)+STLIN</td><td>RFRB</td><td>33</td></tr><tr><td>PREYS=YIELD+HARDS*ABS(PLAST(IELEM))</td><td>RFRB</td><td>34</td></tr><tr><td>IF(ABS(STRES(IELEM,1)).GE.PREYS) GO TO 20</td><td>RFRB</td><td>35</td></tr></table>
|
||||
|
||||
<!-- source-page: 148 -->
|
||||
|
||||
```txt
|
||||
ESCUR=ABS(STCUR)-PREYS RFRB 36
|
||||
IF(ESCUR.LE.0.0) GO TO 40 RFRB 37
|
||||
RFACT=ESCUR/ABS(STLIN) RFRB 38
|
||||
GO TO 30 RFRB 39
|
||||
20 IF(STRES(IELEM,1).GT.0.0.AND.STLIN.LE.0.0) GO TO 40 RFRB 40
|
||||
IF(STRES(IELEM,1).LT.0.0.AND.STLIN.GE.0.0) GO TO 40 RFRB 41
|
||||
RFACT=1.0 RFRB 42
|
||||
30 REDUC=1.0-RFACT RFRB 43
|
||||
STRES(IELEM,1)=STRES(IELEM,1)+REDUC*STLIN+ RFRB 44
|
||||
RFACT*EIVAL*(1.0-EIVAL/(EIVAL+HARDS))*STRAN RFRB 45
|
||||
PLAST(IELEM)=PLAST(IELEM)+RFACT*STRAN*EIVAL/(EIVAL+HARDS) RFRB 46
|
||||
GO TO 50 RFRB 47
|
||||
40 STRES(IELEM,1)=STRES(IELEM,1)+STLIN RFRB 48
|
||||
50 STRES(IELEM,2)=STRES(IELEM,2)+(SVALU/ELENG)*(WNOD2-WNOD1) RFRB 49
|
||||
-0.5*SVALU*(THTA1+THTA2) RFRB 50
|
||||
ELOAD(IELEM,1)=ELOAD(IELEM,1)-STRES(IELEM,2) RFRB 51
|
||||
ELOAD(IELEM,2)=ELOAD(IELEM,2)+STRES(IELEM,1) RFRB 52
|
||||
-0.5*ELENG*STRES(IELEM,2) RFRB 53
|
||||
ELOAD(IELEM,3)=ELOAD(IELEM,3)+STRES(IELEM,2) RFRB 54
|
||||
ELOAD(IELEM,4)=ELOAD(IELEM,4)-STRES(IELEM,1) RFRB 55
|
||||
-0.5*ELENG*STRES(IELEM,2) RFRB 56
|
||||
70 CONTINUE RFRB 57
|
||||
RETURN RFRB 58
|
||||
END RFRB 59
|
||||
```
|
||||
|
||||
RFRB 15-17 Zero space for storing $\pmb{p}$ .
|
||||
|
||||
RFRB 18-57 For each element evaluate $p^{(e)}$ and assemble into $p$ .
|
||||
|
||||
# 5.4.6 Examples of nonlayered elasto-plastic Timoshenko beam analysis
|
||||
|
||||
Two numerical examples are considered. The first example, Example 5.1, involves the yielding of a rectangular simple beam under uniformly distributed load. The beam material has the following properties:
|
||||
|
||||
$$
|
||||
E = 2 1 0 \cdot 0 \mathrm{kN} / \mathrm{mm} ^ {2}
|
||||
$$
|
||||
|
||||
$$
|
||||
\nu = 0 \cdot 3
|
||||
$$
|
||||
|
||||
$$
|
||||
\sigma_ {0} = 0 \cdot 2 5 \mathrm{kN} / \mathrm{mm} ^ {2}
|
||||
$$
|
||||
|
||||
$$
|
||||
H ^ {\prime} = 0 \cdot 0
|
||||
$$
|
||||
|
||||
and the beam proportions are:
|
||||
|
||||
$$
|
||||
b = 1 5 0 \mathrm{mm}
|
||||
$$
|
||||
|
||||
$$
|
||||
t = 3 0 0 \mathrm{mm}
|
||||
$$
|
||||
|
||||
$$
|
||||
l = 3 0 0 0 \mathrm{mm}
|
||||
$$
|
||||
|
||||
Typical input data is provided in Appendix IV.
|
||||
|
||||
The problem, finite element idealisation employed and the results are illustrated in Fig. 5.6, which shows that the beam fails as soon as a plastic hinge forms at the centre of the beam. Note that the beam material is assumed to have no strain hardening.
|
||||
|
||||
The second example considered, Example 5.2, is the clamped I beam shown in Fig. 5.7. The beam has the same material properties as those of Example 5.1.
|
||||
|
||||
The dimensions and finite element discretisation of the beam are given in Fig. 5.7; the load-displacement relationship at the beam centre is also provided. It is seen that there is an initial loss of stiffness corresponding to the
|
||||
|
||||
<!-- source-page: 149 -->
|
||||
|
||||
|
||||
|
||||
<details>
|
||||
<summary>line</summary>
|
||||
|
||||
| Central deflection (mm) | Applied load (KN) |
|
||||
| ----------------------- | ----------------- |
|
||||
| 0 | 0 |
|
||||
| 10 | 1260 |
|
||||
| 10 @ 300 mm | 1260 |
|
||||
| 3000 mm | 1260 |
|
||||
| 300 mm | 1260 |
|
||||
| 150 mm | 1260 |
|
||||
| 300 mm | 1260 |
|
||||
</details>
|
||||
|
||||
<!-- source-page: 150 -->
|
||||
|
||||
|
||||
Fig. 5.7 Nonlayered elasto-plastic clamped beam.
|
||||
Reference in New Issue
Block a user