add documents
This commit is contained in:
김경종
+340
@@ -0,0 +1,340 @@
|
||||
<!-- source-page: 151 -->
|
||||
|
||||
yielding of the end sections followed by a further reduction when the central section becomes plastic resulting in a beam failure mechanism.
|
||||
|
||||
# 5.5 Elasto-plastic layered Timoshenko beams
|
||||
|
||||
# 5.5.1 Yielding of layered beams
|
||||
|
||||
In the ‘layered’ approach the beam or the plate is subdivided into a chosen number of layers, as shown in Fig. 5.8.
|
||||
|
||||
|
||||
|
||||
<details>
|
||||
<summary>text_image</summary>
|
||||
|
||||
(a) Layered beam
|
||||
Layer i
|
||||
</details>
|
||||
|
||||
|
||||
|
||||
<details>
|
||||
<summary>text_image</summary>
|
||||
|
||||
(b) Layered plate
|
||||
Layer i
|
||||
</details>
|
||||
|
||||
Fig. 5.8 Layered subdivision of beam and plate.
|
||||
|
||||
In the finite element solution it is assumed that as soon as the stress in the middle of the outer layers reaches the yield value, then the outer layers become plastic, while the rest of the layers remain elastic, as shown in
|
||||
|
||||
<!-- source-page: 152 -->
|
||||
|
||||
|
||||
Fig. 5.9 Yielding of layered beam.
|
||||
Fig. 5.9. Then, as plastification propagates, more layers become plastic, until the whole cross-section eventually becomes plastic.
|
||||
|
||||
# 5.5.2 Formation of the stiffness matrix in the layered approach
|
||||
|
||||
In the layered approach, we work in terms of stresses and not in terms of stress resultants as in the nonlayered approach. The state of stress at the middle of a layer is taken as representative for the entire layer.
|
||||
|
||||
Contributions to the stress resultants M and Q are found for each layer separately by integrating over the layer thickness only. The bending moments and shear forces are then found from the contributions of all the layers of the beam element.
|
||||
|
||||
The displacement field, stress-strain relationship and strain-displacement relationship are given in (5.1)-(5.10).
|
||||
|
||||
The virtual work expression is given by (5.11) and when we evaluate the bending moment M and shear force Q we use a mid-ordinate rule as follows:
|
||||
|
||||
$$
|
||||
M = E I \left(- \frac {d \theta}{d x}\right) \quad \text { and } \quad Q = G \hat {A} \epsilon_ {s} \tag {5.48}
|
||||
$$
|
||||
|
||||
where
|
||||
|
||||
$$
|
||||
E I = \sum_ {l} E _ {l} b _ {l} z _ {l} ^ {2} t _ {l} \tag {5.49}
|
||||
$$
|
||||
|
||||
and
|
||||
|
||||
$$
|
||||
G \hat {A} = \sum_ {l} G _ {l} b _ {l} t _ {l} \tag {5.50}
|
||||
$$
|
||||
|
||||
and where $b_{l}$ is the layer breadth
|
||||
|
||||
$t_{l}$ is the layer thickness
|
||||
|
||||
$z_{l}$ is the $z$ -coordinate at the middle of the layer
|
||||
|
||||
$E_{l}$ is the Young's modulus of the layer material
|
||||
|
||||
and $G_{l}$ is the Shear modulus of the layer material.
|
||||
|
||||
<!-- source-page: 153 -->
|
||||
|
||||
However, if the stress at the middle surface of a layer reaches the uniaxial yield stress of the layer material, the whole layer is considered to be plastic and $E_{l}$ is replaced by
|
||||
|
||||
$$
|
||||
E _ {l} \left(1 - \frac {E _ {l}}{E _ {l} + H ^ {\prime}}\right),
|
||||
$$
|
||||
|
||||
where $H'$ is the uniaxial strain hardening parameter. As mentioned before, the shear force–shear strain relationship is always elastic.
|
||||
|
||||
# 5.5.3 Solution of nonlinear equations
|
||||
|
||||
Recall that the virtual work expression (5.11) has the form
|
||||
|
||||
$$
|
||||
\int_ {0} ^ {l} \int_ {- t / 2} ^ {t / 2} \int_ {b (- t / 2)} ^ {b (t / 2)} \left\{- z \frac {d (\delta \theta)}{d x} \sigma_ {x} + \delta \beta \tau_ {x z} \right\} d y d z d x - \int_ {0} ^ {l} \delta w q d x = 0. \tag {5.51}
|
||||
$$
|
||||
|
||||
The mid-ordinate rule is again used to evaluate the first two integrals in (5.51) so that we obtain
|
||||
|
||||
$$
|
||||
[ \delta \varphi ] ^ {T} [ \boldsymbol {p} _ {f} + \boldsymbol {p} _ {s} ] - [ \delta \varphi ] ^ {T} \boldsymbol {f} = 0 \tag {5.52}
|
||||
$$
|
||||
|
||||
where
|
||||
|
||||
$$
|
||||
\boldsymbol {p} _ {f} = \int_ {0} ^ {l} [ \boldsymbol {B} _ {f} ] ^ {T} \bar {M} d x
|
||||
$$
|
||||
|
||||
and
|
||||
|
||||
$$
|
||||
\boldsymbol {p} _ {s} = \int_ {0} ^ {l} [ \boldsymbol {B} _ {s} ] ^ {T} \bar {Q} d x
|
||||
$$
|
||||
|
||||
in which $B_{f}$ , $B_{s}$ and $\delta\varphi$ have been defined in (5.40), (5.41) and (5.43) respectively and in which
|
||||
|
||||
$$
|
||||
\overline {{{M}}} = \sum_ {l} b _ {l} \sigma_ {x l} z _ {l} t _ {l} \tag {5.53}
|
||||
$$
|
||||
|
||||
and
|
||||
|
||||
$$
|
||||
\bar {Q} = \sum_ {l} b _ {l} \tau_ {x z l} t _ {l}. \tag {5.54}
|
||||
$$
|
||||
|
||||
Note that $\sigma_{xl}$ and $\tau_{xzl}$ are the direct and shear stresses in the layer respectively. Since (5.52) is true for any arbitrary set of virtual displacements then
|
||||
|
||||
$$
|
||||
\boldsymbol {p} _ {f} + \boldsymbol {p} _ {s} - \boldsymbol {f} = 0. \tag {5.55}
|
||||
$$
|
||||
|
||||
Contributions to $p_{f}$ and $p_{s}$ may be evaluated separately from each element so that
|
||||
|
||||
$$
|
||||
\begin{array}{l} \boldsymbol {p} _ {f} ^ {(e)} = \int_ {x _ {1} ^ {(e)}} ^ {x _ {2} ^ {(e)}} [ \boldsymbol {B} _ {f} ^ {(e)} ] ^ {T} \bar {M} ^ {(e)} d x = \int_ {x _ {1} ^ {(e)}} ^ {x _ {2} ^ {(e)}} \left[ 0, \left(\frac {\bar {M}}{l}\right) ^ {(e)}, 0, - \left(\frac {\bar {M}}{l}\right) ^ {(e)} \right] ^ {T} d x \\ = [ 0, \bar {M} ^ {(e)}, 0, - \bar {M} ^ {(e)} ] ^ {T} \tag {5.56} \\ \end{array}
|
||||
$$
|
||||
|
||||
<!-- source-page: 154 -->
|
||||
|
||||
and
|
||||
|
||||
$$
|
||||
\begin{array}{l} \boldsymbol {p} _ {s} ^ {(e)} = \int_ {x _ {1} ^ {(e)}} ^ {x _ {2} ^ {(e)}} \left[ \boldsymbol {B} _ {s} ^ {(e)} \right] ^ {T} \bar {Q} ^ {(e)} d x = \int_ {x _ {1} ^ {(e)}} ^ {x _ {2} ^ {(e)}} \left[ - \frac {1}{l ^ {(e)}}, - \frac {1}{2}, \frac {1}{l ^ {(e)}}, - \frac {1}{2} \right] ^ {T} \bar {Q} ^ {(e)} d x \\ = \left[ - \bar {Q} ^ {(e)}, - \frac {(\bar {Q} l) ^ {(e)}}{2}, \bar {Q} ^ {(e)}, - \frac {(\bar {Q} l) ^ {(e)}}{2} \right] ^ {T}. \tag {5.57} \\ \end{array}
|
||||
$$
|
||||
|
||||
The complete sequence of nonlinear equation solving is very similar to the one adopted in Table 5.1 for the nonlayered beam. Step 5 is now written as:
|
||||
|
||||
5. For each element evaluate for each layer $\sigma_{xl}^{(e)}$ and $\tau_{xzl}^{(e)}$ . Check $\sigma_{xl}^{(e)}$ and adjust its value accordingly to account for any plastic behaviour. Evaluate the stress resultants $\bar{M}^{(e)}$ and $\bar{Q}^{(e)}$ and hence evaluate the residual force vector $[\psi^{(e)}]^{i+1} = p^{(e)} - f^{(e)}$ . Assemble $[\psi^{(e)}]^{i+1}$ into the global residual force vector $\psi^{i+1}$ .
|
||||
|
||||
# 5.5.4 Overall structure of layered beam program TIMLAY
|
||||
|
||||
The overall structure of the layered beam program is exactly the same as that of the nonlayered beam program given in Fig. 5.5. Subroutine STIFBL replaces STIFFB and subroutine RFORBL replaces REFORB. Subroutine STIFBL calls a further new routine called LAYER. The master routine BEML has minor changes as shown in the next section.
|
||||
|
||||
# 5.5.5 Modified and new routines
|
||||
|
||||
Master BEML This routine is almost identical to routine BEAM described earlier.
|
||||
```txt
|
||||
MASTER BEML LYBM 1
|
||||
C**************************LYBM 2
|
||||
C LYBM 3
|
||||
C *** ELSTO-PLASTIC LAYERED TIMOSHENKO BEAM PROGRAM LYBM 4
|
||||
C LYBM 5
|
||||
C**************************LYBM 6
|
||||
COMMON/UNIM1/NPOIN,NELEM,NBOUN,NLAYR,NPROP,NNODE,IINCS,IITER, LYBM 7
|
||||
. KRESL,NCHEK,TOLER,NALGO,NSVAB,NDOFN,NINCS,NEVAB, LYBM 8
|
||||
. NITER,NOUTP,FACTO LYBM 9
|
||||
COMMON/UNIM2/PROPS(5,25),COORD(26),LNODS(25,2),IFPRE(52), LYBM 10
|
||||
. FIXED(52),TLOAD(25,4),RLOAD(25,4),ELOAD(25,4), LYBM 11
|
||||
. MATNO(25),STRES(25,2),PLAST(250),XDISP(52), LYBM 12
|
||||
. TDISP(26,2),TREAC(26,2),ASTIF(52,52),ASLOD(52), LYBM 13
|
||||
. REACT(52),FRESV(1352),PEFIX(52),ESTIF(4,4), LYBM 14
|
||||
. STRSL(250,2) LYBM 15
|
||||
CALL DATA LYBM 16
|
||||
CALL INITIAL LYBM 17
|
||||
DO 30 IINCS=1,NINCS LYBM 18
|
||||
CALL INCLOD LYBM 19
|
||||
DO 10 IITER=1,NITER LYBM 20
|
||||
CALL NONAL LYBM 21
|
||||
IF(KRESL.EQ.1) CALL STIFBL LYBM 22
|
||||
CALL ASSEMB LYBM 23
|
||||
IF(KRESL.EQ.1) CALL GREDUC LYBM 24
|
||||
```
|
||||
|
||||
<!-- source-page: 155 -->
|
||||
|
||||
<table><tr><td>IF(KRESL.EQ.2) CALL RESOLV</td><td>LYBM</td><td>25</td></tr><tr><td>CALL BAKSUB</td><td>LYBM</td><td>26</td></tr><tr><td>CALL RFORBL</td><td>LYBM</td><td>27</td></tr><tr><td>CALL CONUND</td><td>LYBM</td><td>28</td></tr><tr><td>IF(NCHEK.EQ.0) GO TO 20</td><td>LYBM</td><td>29</td></tr><tr><td>IF(IITER.EQ.1.AND.NOUTP.EQ.1) CALL RESULT</td><td>LYBM</td><td>30</td></tr><tr><td>IF(NOUTP.EQ.2) CALL RESULT</td><td>LYBM</td><td>31</td></tr><tr><td>10 CONTINUE</td><td>LYBM</td><td>32</td></tr><tr><td>WRITE(6,900)</td><td>LYBM</td><td>33</td></tr><tr><td>900 FORMAT(1H0,5X,'SOLUTION NOT CONVERGED')</td><td>LYBM</td><td>34</td></tr><tr><td>STOP</td><td>LYBM</td><td>35</td></tr><tr><td>20 CALL RESULT</td><td>LYBM</td><td>36</td></tr><tr><td>30 CONTINUE</td><td>LYBM</td><td>37</td></tr><tr><td>STOP</td><td>LYBM</td><td>38</td></tr><tr><td>END</td><td>LYBM</td><td>39</td></tr></table>
|
||||
|
||||
Subroutine STIFBL This routine calculates the element stiffness matrices for the elasto-plastic layered Timoshenko beam element.
|
||||
|
||||
<table><tr><td></td><td>SUBROUTINE STIFBL</td><td>STBL</td><td>1</td></tr><tr><td>C**********</td><td>**********</td><td>STBL</td><td>2</td></tr><tr><td>C</td><td></td><td>STBL</td><td>3</td></tr><tr><td>C *** CALCULATES ELEMENT STIFFNESS MATRICES</td><td></td><td>STBL</td><td>4</td></tr><tr><td>C</td><td></td><td>STBL</td><td>5</td></tr><tr><td>C**********</td><td>**********</td><td>STBL</td><td>6</td></tr><tr><td></td><td>COMMON/UNIM1/NPOIN,NELEM,NBOUN,NLAYR,NPROP,NNODE,IINCS,IITER,</td><td>STBL</td><td>7</td></tr><tr><td></td><td>KRESL,NCHEK,TOLER,NALGO,NSVAB,NDOFN,NINCS,NEVAB,</td><td>STBL</td><td>8</td></tr><tr><td></td><td>NITER,NOUTP,FACTO</td><td>STBL</td><td>9</td></tr><tr><td></td><td>COMMON/UNIM2/PROPS(5,25),COORD(26),LNODS(25,2),IFPRE(52),</td><td>STBL</td><td>10</td></tr><tr><td></td><td>FIXED(52),TLOAD(25,4),RLOAD(25,4),ELOAD(25,4),</td><td>STBL</td><td>11</td></tr><tr><td></td><td>MATNO(25),STRES(25,2),PLAST(250),XDISP(52),</td><td>STBL</td><td>12</td></tr><tr><td></td><td>TDISP(26,2),TREAC(26,2),ASTIF(52,52),ASLOD(52),</td><td>STBL</td><td>13</td></tr><tr><td></td><td>REACT(52),FRESV(1352),PEFIX(52),ESTIF(4,4),</td><td>STBL</td><td>14</td></tr><tr><td></td><td>STRSL(250,2)</td><td>STBL</td><td>15</td></tr><tr><td></td><td>REWIND 1</td><td>STBL</td><td>16</td></tr><tr><td></td><td>DO 20 IELEM=1,NELEM</td><td>STBL</td><td>17</td></tr><tr><td></td><td>LPROP=MATNO(IELEM)</td><td>STBL</td><td>18</td></tr><tr><td></td><td>CALL LAYER(IELEM,EIVAL,SVALU)</td><td>STBL</td><td>19</td></tr><tr><td></td><td>HARDS=PROPS(LPROP,4)</td><td>STBL</td><td>20</td></tr><tr><td></td><td>NODE1=LNODS(IELEM,1)</td><td>STBL</td><td>21</td></tr><tr><td></td><td>NODE2=LNODS(IELEM,2)</td><td>STBL</td><td>22</td></tr><tr><td></td><td>ELENG=ABS(COORD(NODE2)-COORD(NODE1))</td><td>STBL</td><td>23</td></tr><tr><td></td><td>VALU1=0.5*SVALU</td><td>STBL</td><td>24</td></tr><tr><td></td><td>VALU2=SVALU/ELENG</td><td>STBL</td><td>25</td></tr><tr><td></td><td>VALU3=EIVAL/ELENG</td><td>STBL</td><td>26</td></tr><tr><td></td><td>VALU4=0.25*SVALU*ELENG</td><td>STBL</td><td>27</td></tr><tr><td></td><td>ESTIF(1,1)=VALU2</td><td>STBL</td><td>28</td></tr><tr><td></td><td>ESTIF(1,2)=VALU1</td><td>STBL</td><td>29</td></tr><tr><td></td><td>ESTIF(1,3)=-VALU2</td><td>STBL</td><td>30</td></tr><tr><td></td><td>ESTIF(1,4)=VALU1</td><td>STBL</td><td>31</td></tr><tr><td></td><td>ESTIF(2,2)=VALU3+VALU4</td><td>STBL</td><td>32</td></tr><tr><td></td><td>ESTIF(2,3)=-VALU1</td><td>STBL</td><td>33</td></tr><tr><td></td><td>ESTIF(2,4)=-VALU3+VALU4</td><td>STBL</td><td>34</td></tr><tr><td></td><td>ESTIF(3,3)=VALU2</td><td>STBL</td><td>35</td></tr><tr><td></td><td>ESTIF(3,4)=-VALU1</td><td>STBL</td><td>36</td></tr><tr><td></td><td>ESTIF(4,4)=VALU3+VALU4</td><td>STBL</td><td>37</td></tr><tr><td></td><td>DO 10 ISTIF=1,4</td><td>STBL</td><td>38</td></tr><tr><td></td><td>DO 10 JSTIF=ISTIF,4</td><td>STBL</td><td>39</td></tr><tr><td>10</td><td>ESTIF(JSTIF,ISTIF)=ESTIF(ISTIF,JSTIF)</td><td>STBL</td><td>40</td></tr><tr><td></td><td>WRITE(1) ESTIF</td><td>STBL</td><td>41</td></tr><tr><td>20</td><td>CONTINUE</td><td>STBL</td><td>42</td></tr><tr><td></td><td>RETURN</td><td>STBL</td><td>43</td></tr><tr><td></td><td>END</td><td>STBL</td><td>44</td></tr></table>
|
||||
|
||||
<!-- source-page: 156 -->
|
||||
|
||||
STBL 19 Call routine LAYER which evaluates approximate values of EI and exact values of $GA$ using a mid-ordinate rule. Note that certain layers may be plastic.
|
||||
Subroutine RFORBL This routine evaluates p for the layered beam using the mid-ordinate rule.
|
||||
|
||||
<table><tr><td>SUBROUTINE RFORBL</td><td>RFRL</td><td>1</td></tr><tr><td>C**********</td><td>RFRL</td><td>2</td></tr><tr><td>C</td><td>RFRL</td><td>3</td></tr><tr><td>C *** CALCULATES INTERNAL EQUIVALENT NODAL FORCES</td><td>RFRL</td><td>4</td></tr><tr><td>C</td><td>RFRL</td><td>5</td></tr><tr><td>C**********</td><td>RFRL</td><td>6</td></tr><tr><td>COMMON/UNIM1/NPOIN,NELEM,NBOUN,NLAYR,NPROP,NNODE,IINCS,IITER,</td><td>RFRL</td><td>7</td></tr><tr><td>KRESL,NCHEK,TOLER,NALGO,NSVAB,NDOFN,NINCS,NEVAB,</td><td>RFRL</td><td>8</td></tr><tr><td>NITER,NOUTP,FACTO</td><td>RFRL</td><td>9</td></tr><tr><td>COMMON/UNIM2/PROPS(5,25),COORD(26),LNODS(25,2),IFPRE(52),</td><td>RFRL</td><td>10</td></tr><tr><td>FIXED(52),TLOAD(25,4),RLOAD(25,4),ELOAD(25,4),</td><td>RFRL</td><td>11</td></tr><tr><td>MATNO(25),STRES(25,2),PLAST(250),XDISP(52),</td><td>RFRL</td><td>12</td></tr><tr><td>TDISP(26,2),TREAC(26,2),ASTIF(52,52),ASLOD(52),</td><td>RFRL</td><td>13</td></tr><tr><td>REACT(52),FRESV(1352),PEFIX(52),ESTIF(4,4),</td><td>RFRL</td><td>14</td></tr><tr><td>STRSL(250,2)</td><td>RFRL</td><td>15</td></tr><tr><td>DIMENSION STRAN(2)</td><td>RFRL</td><td>16</td></tr><tr><td>DO 15 IELEM=1,NELEM</td><td>RFRL</td><td>17</td></tr><tr><td>DO 10 IEVAB=1,NEVAB</td><td>RFRL</td><td>18</td></tr><tr><td>10 ELOAD(IELEM,IEVAB)=0.0</td><td>RFRL</td><td>19</td></tr><tr><td>DO 15 IDOFN=1,NDOFN</td><td>RFRL</td><td>20</td></tr><tr><td>15 STRES(IELEM,IDOFN)=0.0</td><td>RFRL</td><td>21</td></tr><tr><td>KLAYR=0</td><td>RFRL</td><td>22</td></tr><tr><td>DO 70 IELEM=1,NELEM</td><td>RFRL</td><td>23</td></tr><tr><td>LPROP=MATNO(IELEM)</td><td>RFRL</td><td>24</td></tr><tr><td>YOUNG=PROPS(LPROP,1)</td><td>RFRL</td><td>25</td></tr><tr><td>SHEAR=PROPS(LPROP,2)</td><td>RFRL</td><td>26</td></tr><tr><td>YIELD=PROPS(LPROP,3)</td><td>RFRL</td><td>27</td></tr><tr><td>HARDS=PROPS(LPROP,4)</td><td>RFRL</td><td>28</td></tr><tr><td>THKTO=PROPS(LPROP,5)</td><td>RFRL</td><td>29</td></tr><tr><td>NODE1=LNODS(IELEM,1)</td><td>RFRL</td><td>30</td></tr><tr><td>NODE2=LNODS(IELEM,2)</td><td>RFRL</td><td>31</td></tr><tr><td>ELENG=ABS(COORD(NODE2)-COORD(NODE1))</td><td>RFRL</td><td>32</td></tr><tr><td>WNOD1=XDISP(NODE1*NDOFN-1)</td><td>RFRL</td><td>33</td></tr><tr><td>WNOD2=XDISP(NODE2*NDOFN-1)</td><td>RFRL</td><td>34</td></tr><tr><td>THTA1=XDISP(NODE1*NDOFN)</td><td>RFRL</td><td>35</td></tr><tr><td>THTA2=XDISP(NODE2*NDOFN)</td><td>RFRL</td><td>36</td></tr><tr><td>STRAN(1)=(THTA1-THTA2)/ELENG</td><td>RFRL</td><td>37</td></tr><tr><td>STRAN(2)=(WNOD2-WNOD1)/ELENG</td><td>RFRL</td><td>38</td></tr><tr><td>-0.5*(THTA1+THTA2)</td><td>RFRL</td><td>39</td></tr><tr><td>ZMIDL=-THKTO/2.0</td><td>RFRL</td><td>40</td></tr><tr><td>KOUNT=5</td><td>RFRL</td><td>41</td></tr><tr><td>DO 50 ILAYR=1,NLAYR</td><td>RFRL</td><td>42</td></tr><tr><td>KLAYR=KLAYR+1</td><td>RFRL</td><td>43</td></tr><tr><td>KOUNT=KOUNT+1</td><td>RFRL</td><td>44</td></tr><tr><td>BRDTH=PROPS(LPROP,KOUNT)</td><td>RFRL</td><td>45</td></tr><tr><td>KOUNT=KOUNT+1</td><td>RFRL</td><td>46</td></tr><tr><td>THICK=PROPS(LPROP,KOUNT)</td><td>RFRL</td><td>47</td></tr><tr><td>ZMIDL=ZMIDL+THICK/2.0</td><td>RFRL</td><td>48</td></tr><tr><td>STLIN=YOUNG*STRAN(1)*ZMIDL</td><td>RFRL</td><td>49</td></tr><tr><td>STCUR=STRSL(KLAYR,1)+STLIN</td><td>RFRL</td><td>50</td></tr><tr><td>PREYS=YIELD+HARDS*ABS(PLAST(KLAYR))</td><td>RFRL</td><td>51</td></tr><tr><td>IF(ABS(STRSL(KLAYR,1)).GE.PREYS) GO TO 20</td><td>RFRL</td><td>52</td></tr><tr><td>ESCUR=ABS(STCUR)-PREYS</td><td>RFRL</td><td>53</td></tr><tr><td>IF(ESCUR.LE.0.0) GO TO 40</td><td>RFRL</td><td>54</td></tr></table>
|
||||
|
||||
<!-- source-page: 157 -->
|
||||
|
||||
```csv
|
||||
RFACT=ESCUR/ABS(STLIN) RFRL 55
|
||||
GO TO 30 RFRL 56
|
||||
20 IF(STRSL(KLAYR,1).GT.0.0.AND.STLIN.LE.0.0) GO TO 40 RFRL 57
|
||||
IF(STRSL(KLAYR,1).LT.0.0.AND.STLIN.GE.0.0) GO TO 40 RFRL 58
|
||||
RFACT=1.0 RFRL 59
|
||||
30 REDUC=1.0-RFACT RFRL 60
|
||||
STRSL(KLAYR,1)=STRSL(KLAYR,1)+REDUC*STLIN+ RFRL 61
|
||||
• RFACT*YOUNG*(1.0-YOUNG/(YOUNG+HARDS))*STRAN(1)*ZMIDL RFRL 62
|
||||
PLAST(KLAYR)=PLAST(KLAYR)+RFACT*STRAN(1)*YOUNG/(YOUNG+HARDS) RFRL 63
|
||||
.*ZMIDL RFRL 64
|
||||
GO TO 45 RFRL 65
|
||||
40 STRSL(KLAYR,1)=STRSL(KLAYR,1)+STLIN RFRL 66
|
||||
45 STRSL(KLAYR,2)=STRSL(KLAYR,2)+STRAN(2)*SHEAR RFRL 67
|
||||
STRES(IELEM,1)=STRES(IELEM,1)+STRSL(KLAYR,1)* RFRL 68
|
||||
• BRDTH*THICK*ZMIDL RFRL 69
|
||||
STRES(IELEM,2)=STRES(IELEM,2)+STRSL(KLAYR,2)* RFRL 70
|
||||
• BRDTH*THICK RFRL 71
|
||||
ZMIDL=ZMIDL+THICK/2.0 RFRL 72
|
||||
50 CONTINUE RFRL 73
|
||||
ELOAD(IELEM,1)=ELOAD(IELEM,1)-STRES(IELEM,2) RFRL 74
|
||||
ELOAD(IELEM,2)=ELOAD(IELEM,2)+STRES(IELEM,1) RFRL 75
|
||||
• -0.5*ELENG*STRES(IELEM,2) RFRL 76
|
||||
ELOAD(IELEM,3)=ELOAD(IELEM,3)+STRES(IELEM,2) RFRL 77
|
||||
ELOAD(IELEM,4)=ELOAD(IELEM,4)-STRES(IELEM,1) RFRL 78
|
||||
• -0.5*ELENG*STRES(IELEM,2) RFRL 79
|
||||
70 CONTINUE RFRL 80
|
||||
RETURN RFRL 81
|
||||
END RFRL 82
|
||||
```
|
||||
|
||||
Subroutine LAYER This routine evaluates EI and $GA\hat{A}$ using the mid-ordinate rule. Note that certain layers may be plastic and therefore have a modified E.
|
||||
|
||||
```txt
|
||||
SUBROUTINE LAYER(IELEM,EIVAL,SVALU) LAYR 1
|
||||
C******************************************************************************************
|
||||
C LAYR 2
|
||||
C LAYR 3
|
||||
C *** CALCULATES INTEGRATED VALUES FOR EI AND GA THROUGH DEPTH LAYR 4
|
||||
C LAYR 5
|
||||
C******************************************************************************************
|
||||
COMMON/UNIM1/NPOIN,NELEM,NBOUN,NLAYR,NPROP,NNODE,IINCS,IITER, LAYR 7
|
||||
. KRESL,NCHEK,TOLER,NALGO,NSVAB,NDOFN,NINCS,NEVAB, LAYR 8
|
||||
. NITER,NOUTP,FACTO LAYR 9
|
||||
COMMON/UNIM2/PROPS(5,25),COORD(26),LNODS(25,2),IFPRE(52), LAYR 10
|
||||
. FIXED(52),TLOAD(25,4),RLOAD(25,4),ELOAD(25,4), LAYR 11
|
||||
. MATNO(25),STRES(25,2),PLAST(250),XDISP(52), LAYR 12
|
||||
. TDISP(26,2),TREAC(26,2),ASTIF(52,52),ASLOD(52), LAYR 13
|
||||
. REACT(52),FRESV(1352),PEFIX(52),ESTIF(4,4), LAYR 14
|
||||
. STRSL(250,2) LAYR 15
|
||||
EIVAL=0.0 LAYR 16
|
||||
SVALU=0.0 LAYR 17
|
||||
LPROP=MATNO(IELEM) LAYR 18
|
||||
KLAYR=(IELEM-1)*NLAYR LAYR 19
|
||||
SHEAR=PROPS(LPROP,2) LAYR 20
|
||||
HARDS=PROPS(LPROP,4) LAYR 21
|
||||
THKTO=PROPS(LPROP,5) LAYR 22
|
||||
ZMIDL=-THKTO/2.0 LAYR 23
|
||||
KOUNT=5 LAYR 24
|
||||
DO 10 ILAYR=1,NLAYR LAYR 25
|
||||
KLAYR=KLAYR+1 LAYR 26
|
||||
YOUNG=PROPS(LPROP,1) LAYR 27
|
||||
IF(PLAST(KLAYR).NE.0.0) YOUNG=YOUNG*(1.0-YOUNG/(YOUNG+HARDS)) LAYR 28
|
||||
```
|
||||
|
||||
<!-- source-page: 158 -->
|
||||
|
||||
KOUNT=KOUNT+1 LAYR 29
|
||||
BRDTH=PROPS(LPROP,KOUNT) LAYR 30
|
||||
KOUNT=KOUNT+1 LAYR 31
|
||||
THICK=PROPS(LPROP,KOUNT) LAYR 32
|
||||
ZMIDL=ZMIDL+THICK/2.0 LAYR 33
|
||||
EIVAL=EIVAL+YOUNG*BRDTH*THICK*ZMIDL*ZMIDL LAYR 34
|
||||
SVALU=SVALU+SHEAR*BRDTH*THICK LAYR 35
|
||||
ZMIDL=ZMIDL+THICK/2.0 LAYR 36
|
||||
10 CONTINUE LAYR 37
|
||||
RETURN LAYR 38
|
||||
END LAYR 39
|
||||
|
||||
# 5.5.6 Examples of layered elasto-plastic Timoshenko beam analysis
|
||||
|
||||
The third example considered in this chapter is the elasto-plastic analysis of the simple beam of Example 5.1. The layered solution is adopted in this case. A typical input data listing is provided in Appendix IV.
|
||||
|
||||
The results for both nonlayered and layered solutions to this beam problem are reproduced in Fig. 5.10.
|
||||
|
||||
The last example to be considered here is the layered solution of the clamped I-beam given in Example 5.1.
|
||||
|
||||
Again, both nonlayered and layered solution results are illustrated in Fig. 5.11.
|
||||
|
||||
From Figs. 5.10 and 5.11 it is obvious that the layered solution is more realistic. Yielding takes place gradually through the layers, resulting in smoother curves representing the load-displacement relationship.
|
||||
|
||||
# 5.6 Problems
|
||||
|
||||
5.1 Derive the main expressions for the elasto-plastic analysis of Timoshenko beams using elements with
|
||||
|
||||
(i) Quadratic shape functions
|
||||
|
||||
$$
|
||||
N _ {1} ^ {(e)} = \frac {(x ^ {(e)} - x _ {2} ^ {(e)}) (x ^ {(e)} - x _ {3} ^ {(e)})}{(x _ {1} ^ {(e)} - x _ {2} ^ {(e)}) (x _ {1} ^ {(e)} - x _ {3} ^ {(e)})}
|
||||
$$
|
||||
|
||||
$$
|
||||
N _ {2} ^ {(e)} = \frac {\big (x ^ {(e)} - x _ {1} ^ {(e)} \big) \big (x ^ {(e)} - x _ {3} ^ {(e)} \big)}{\big (x _ {2} ^ {(e)} - x _ {1} ^ {(e)} \big) \big (x _ {2} ^ {(e)} - x _ {3} ^ {(e)} \big)}
|
||||
$$
|
||||
|
||||
$$
|
||||
N _ {3} ^ {(e)} = \frac {\left(x ^ {(e)} - x _ {1} ^ {(e)}\right) \left(x ^ {(e)} - x _ {2} ^ {(e)}\right)}{\left(x _ {3} ^ {(e)} - x _ {1} ^ {(e)}\right) \left(x _ {3} ^ {(e)} - x _ {2} ^ {(e)}\right)} \tag {5.58}
|
||||
$$
|
||||
|
||||
<!-- source-page: 159 -->
|
||||
|
||||
|
||||
|
||||
<details>
|
||||
<summary>line</summary>
|
||||
|
||||
| Central deflection (mm) | Nonlayered solution (KN) | Layered solution (KN) |
|
||||
| ----------------------- | ------------------------ | --------------------- |
|
||||
| 0 | 0 | 0 |
|
||||
| 5 | 600 | 550 |
|
||||
| 10 | 1250 | 1150 |
|
||||
| 15 | 1275 | 1200 |
|
||||
| 20 | 1280 | 1230 |
|
||||
| 25 | 1285 | 1250 |
|
||||
</details>
|
||||
|
||||
Fig. 5.10 Load-deflection diagrams for simply supported beam.
|
||||
|
||||
<!-- source-page: 160 -->
|
||||
|
||||
|
||||
|
||||
<details>
|
||||
<summary>line</summary>
|
||||
|
||||
| Layer number | Cross-section (mm) | Applied load intensity (KN/mm) |
|
||||
| ------------ | ------------------- | ------------------------------ |
|
||||
| 2 | 200 | 0.45 |
|
||||
| 3 | 200 | 0.44 |
|
||||
| 4 | 200 | 0.43 |
|
||||
| 5 | 200 | 0.42 |
|
||||
| 6 | 200 | 0.41 |
|
||||
</details>
|
||||
Reference in New Issue
Block a user