Durchlaufträger
Mathcad in der Tragwerksplanung
Teil IV, Kap. 6
Zuletzt aktualisiert: 05.12.2002
Bearbeiter: Horst Werkle
Durchlaufträger
Symbolische Lösung der Dreimomentengleichung für den Dreifeldträger mit feldweise konstanter Streckenlast
Lösung durch symbolisches Invertieren der Matrix und Multiplikation mit dem Vektor der rechten Seite:
 − 2 ⋅ l ⋅ l ⋅ R + 2 ⋅ l 2 ⋅ L + 2 ⋅ l ⋅ l ⋅ R + 2 ⋅ l ⋅ l ⋅ L − l 2 ⋅ R − l ⋅ l ⋅ L 
2 2
3 1 1
3 2 2 2
2 2 3 3
  2 1 1
−1

2


l2
 2 ⋅ ( l1 + l2)
  −l1 ⋅ R1 − l2 ⋅ L2 
 4 ⋅ l1 ⋅ l2 + 4 ⋅ l1 ⋅ l3 + 3 ⋅ l2 + 4 ⋅ l2 ⋅ l3 

 ⋅
 vereinfachen → 
l2
2 ⋅ ( l2 + l3) 
2
2



 −l2 ⋅ R2 − l3 ⋅ L3 
 − −l2 ⋅ l1 ⋅ R1 − l2 ⋅ L2 + 2 ⋅ l1 ⋅ l2 ⋅ R2 + 2 ⋅ l1 ⋅ l3 ⋅ L3 + 2 ⋅ l2 ⋅ R2 + 2 ⋅ l2 ⋅ l3 ⋅ L3 

 4⋅l ⋅l + 4⋅l ⋅l + 3⋅l 2 + 4⋅l ⋅l 
1 3
2
2 3

 1 2








Die Lastglieder werden nun durch die entprechenden Ausdrücke für Gleichlasten ersetzt und das Ergebnis wieder vereinfacht:
 − 2 ⋅ l ⋅ l ⋅ R + 2 ⋅ l 2 ⋅ L + 2 ⋅ l ⋅ l ⋅ R + 2 ⋅ l ⋅ l ⋅ L − l 2 ⋅ R − l ⋅ l ⋅ L 
2 2
3 1 1
3 2 2 2
2 2 3 3
  2 1 1

 4⋅l ⋅l + 4⋅l ⋅l + 3⋅l 2 + 4⋅l ⋅l 
1 3
2
2 3
 1 2

 
2
2

 − −l2 ⋅ l1 ⋅ R1 − l2 ⋅ L2 + 2 ⋅ l1 ⋅ l2 ⋅ R2 + 2 ⋅ l1 ⋅ l3 ⋅ L3 + 2 ⋅ l2 ⋅ R2 + 2 ⋅ l2 ⋅ l3 ⋅ L3 

 4⋅l ⋅l + 4⋅l ⋅l + 3⋅l 2 + 4⋅l ⋅l 
1 3
2
2 3

 1 2
  −1 ⋅ l ⋅ l 3 ⋅ q − 2 ⋅ l 2 ⋅ L − 1 ⋅ l ⋅ l 3 ⋅ q − 2 ⋅ l ⋅ l ⋅ L + l 2 ⋅ R + l ⋅ l ⋅ L 


2 1 1
2 2
3 1 1
3 2 2 2
2 2 3 3
2


 2

2

2


 4 ⋅ l1 ⋅ l2 + 4 ⋅ l1 ⋅ l3 + 3 ⋅ l2 + 4 ⋅ l2 ⋅ l3 
 ersetzen , R = q1 ⋅ l1 → 
1

 1
4
3
2

2

  4 ⋅ l2 ⋅ l1 ⋅ q1 + l2 ⋅ L2 − 2 ⋅ l1 ⋅ l2 ⋅ R2 − 2 ⋅ l1 ⋅ l3 ⋅ L3 − 2 ⋅ l2 ⋅ R2 − 2 ⋅ l2 ⋅ l3 ⋅ L3 



 4⋅l ⋅l + 4⋅l ⋅l + 3⋅l 2 + 4⋅l ⋅l 


1 3
2
2 3

 1 2
1
  −1 ⋅ l l 3
3

2
2

2 ⋅ 1 ⋅ q1 − 2 ⋅ l2 ⋅ L2 − ⋅ l3 ⋅ l1 ⋅ q1 − 2 ⋅ l3 ⋅ l2 ⋅ L2 + l2 ⋅ R2 + l2 ⋅ l3 ⋅ L3 
2

 2

 4⋅l ⋅l + 4⋅l ⋅l + 3⋅l 2 + 4⋅l ⋅l 
1 3
2
2 3
 1 2

 1
3
2

2
  4 ⋅ l2 ⋅ l1 ⋅ q1 + l2 ⋅ L2 − 2 ⋅ l1 ⋅ l2 ⋅ R2 − 2 ⋅ l1 ⋅ l3 ⋅ L3 − 2 ⋅ l2 ⋅ R2 − 2 ⋅ l2 ⋅ l3 ⋅ L3 


 4⋅l ⋅l + 4⋅l ⋅l + 3⋅l 2 + 4⋅l ⋅l 

1 3
2
2 3

 1 2
© Vieweg Verlag und Bearbeiter









1 4
1
1

  −1 ⋅ l l 3
3
3

2

2 ⋅ 1 ⋅ q1 − ⋅ l2 ⋅ q2 − ⋅ l3 ⋅ l1 ⋅ q1 − ⋅ l3 ⋅ l2 ⋅ q2 + l2 ⋅ R2 + l2 ⋅ l3 ⋅ L3 
2
2
2


 2


2
2


 4 ⋅ l1 ⋅ l2 + 4 ⋅ l1 ⋅ l3 + 3 ⋅ l2 + 4 ⋅ l2 ⋅ l3 
 ersetzen , L = q2 ⋅ l2 → 
2

 1
4
1 4
3
2


  4 ⋅ l2 ⋅ l1 ⋅ q1 + 4 ⋅ l2 ⋅ q2 − 2 ⋅ l1 ⋅ l2 ⋅ R2 − 2 ⋅ l1 ⋅ l3 ⋅ L3 − 2 ⋅ l2 ⋅ R2 − 2 ⋅ l2 ⋅ l3 ⋅ L3 



 4⋅l ⋅l + 4⋅l ⋅l + 3⋅l 2 + 4⋅l ⋅l 


1 3
2
2 3


 1 2
Seite 1


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





Dateiname:
TIV_10_Dreimomentengleichung_symbolisch_lösen.mcd
Durchlaufträger
Mathcad in der Tragwerksplanung
Teil IV, Kap. 6
1 4
1
1

  −1
3
3
3
2
  ⋅ l2 ⋅ l1 ⋅ q1 − ⋅ l2 ⋅ q2 − ⋅ l3 ⋅ l1 ⋅ q1 − ⋅ l3 ⋅ l2 ⋅ q2 + l2 ⋅ R2 + l2 ⋅ l3 ⋅ L3 
2
2
2
2




 4⋅l ⋅l + 4⋅l ⋅l + 3⋅l 2 + 4⋅l ⋅l 
1 3
2
2 3
 1 2

 1
1
3
4
2

  4 ⋅ l2 ⋅ l1 ⋅ q1 + 4 ⋅ l2 ⋅ q2 − 2 ⋅ l1 ⋅ l2 ⋅ R2 − 2 ⋅ l1 ⋅ l3 ⋅ L3 − 2 ⋅ l2 ⋅ R2 − 2 ⋅ l2 ⋅ l3 ⋅ L3 



 4⋅l ⋅l + 4⋅l ⋅l + 3⋅l 2 + 4⋅l ⋅l 

1 3
2
2 3

 1 2
  −1 ⋅ l ⋅ l 3 ⋅ q − 1 ⋅ l 4 ⋅ q − 1 ⋅ l ⋅ l 3 ⋅ q − 1 ⋅ l ⋅ l 3 ⋅ q + l ⋅ l ⋅ L 

2 1 1
2 2
3 1 1
3 2 2 2 3 3
4
2
2

 2

 4⋅l ⋅l + 4⋅l ⋅l + 3⋅l 2 + 4⋅l ⋅l 
1 3
2
2 3
 1 2

 1
1
1
3
4
3

  4 ⋅ l2 ⋅ l1 ⋅ q1 − 4 ⋅ l2 ⋅ q2 − 2 ⋅ l1 ⋅ l2 ⋅ q2 − 2 ⋅ l1 ⋅ l3 ⋅ L3 − 2 ⋅ l2 ⋅ l3 ⋅ L3 



 4⋅l ⋅l + 4⋅l ⋅l + 3⋅l 2 + 4⋅l ⋅l 

1 3
2
2 3

 1 2
© Vieweg Verlag und Bearbeiter
Zuletzt aktualisiert: 05.12.2002
Bearbeiter: Horst Werkle
1 4
1
1


  −1
3
3
3

  ⋅ l2 ⋅ l1 ⋅ q1 − ⋅ l2 ⋅ q2 − ⋅ l3 ⋅ l1 ⋅ q1 − ⋅ l3 ⋅ l2 ⋅ q2 + l2 ⋅ l3 ⋅ L3 
2
4
2
2




2


2


 4 ⋅ l1 ⋅ l2 + 4 ⋅ l1 ⋅ l3 + 3 ⋅ l2 + 4 ⋅ l2 ⋅ l3 
 ersetzen , R = q2 ⋅ l2 → 
2

 1
4
1 4
1
3
3


  4 ⋅ l2 ⋅ l1 ⋅ q1 − 4 ⋅ l2 ⋅ q2 − 2 ⋅ l1 ⋅ l2 ⋅ q2 − 2 ⋅ l1 ⋅ l3 ⋅ L3 − 2 ⋅ l2 ⋅ l3 ⋅ L3 




 4⋅l ⋅l + 4⋅l ⋅l + 3⋅l 2 + 4⋅l ⋅l 


1 3
2
2 3


 1 2

  −1 ⋅ l ⋅ l 3 ⋅ q − 1 ⋅ l 4 ⋅ q − 1 ⋅ l ⋅ l 3 ⋅ q − 1 ⋅ l ⋅ l 3 ⋅ q + 1 ⋅ l ⋅ l 3 ⋅ q 

2 1 1
2 2
3 1 1
3 2 2
2 3 3
4
2
2
4


 2
2


2


 4 ⋅ l1 ⋅ l2 + 4 ⋅ l1 ⋅ l3 + 3 ⋅ l2 + 4 ⋅ l2 ⋅ l3 
 ersetzen , L = q3 ⋅ l3 → 
3

 1
4
1 4
1
1
1
3
3
3
3


  4 ⋅ l2 ⋅ l1 ⋅ q1 − 4 ⋅ l2 ⋅ q2 − 2 ⋅ l1 ⋅ l2 ⋅ q2 − 2 ⋅ l1 ⋅ l3 ⋅ q3 − 2 ⋅ l2 ⋅ l3 ⋅ q3 




 4⋅l ⋅l + 4⋅l ⋅l + 3⋅l 2 + 4⋅l ⋅l 


1 3
2
2 3


 1 2
Seite 2
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
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

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











Dateiname:
TIV_10_Dreimomentengleichung_symbolisch_lösen.mcd