mathcad - caract de backuprigiditate pentru o diafragma cu 2 siruri de goluri

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4. Determinarea sectiunilor active ale diafragmelor transversale 4.1.1. Sectiuni active montanti MONTANT 1, DT1 MONTANT 2,3, DT1 t 11 0.30 := m t 12 0.30 := m d gol11 1.5 := m d gol12 1.8 := m Δbdr 11 min 6 t 11 d gol11 , ( ) := Δbdr 12 min 6 t 12 d gol12 , ( ) := Δbdr 11 1.5 = Δbdr 12 1.8 = Δbstg 11 0 := Δbstg 12 0 := leff 11 Δbdr 11 t 11 + Δbstg 11 + := leff 12 Δbdr 12 t 12 + Δbstg 12 + := leff 11 1.8 = m leff 12 2.1 = m MONTANT 1, DT2 t 21 0.25 := m d gol21 1.725 := m Δbstg 21 min 6 t 21 d gol21 , ( ) := Δbstg 21 1.5 = Δbdr 21 1 := leff 21 Δbdr 21 t 21 + Δbstg 21 + := leff 21 2.75 = m MONTANT 2, DT2 t 22 0.25 := m d gol22 0.8 := m Δbstg 22 min 6 t 22 d gol22 , ( ) := Δbstg 22 0.8 = Δbdr 22 0.8 := leff 22 Δbdr 22 t 22 + Δbstg 22 + := leff 22 1.85 = m

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Page 1: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

4. Determinarea sectiunilor active ale diafragmelor transversale

4.1.1. Sectiuni active montanti

MONTANT 1, DT1 MONTANT 2,3, DT1

t11 0.30:= m t12 0.30:= m

dgol11 1.5:= m dgol12 1.8:= m

∆bdr11 min 6 t11⋅ dgol11, ( ):= ∆bdr12 min 6 t12⋅ dgol12, ( ):=

∆bdr11 1.5= ∆bdr12 1.8=

∆bstg11 0:= ∆bstg12 0:=

leff11 ∆bdr11 t11+ ∆bstg11+:= leff12 ∆bdr12 t12+ ∆bstg12+:=

leff11 1.8= m leff12 2.1= m

MONTANT 1, DT2

t21 0.25:= m

dgol21 1.725:= m

∆bstg21 min 6 t21⋅ dgol21, ( ):=

∆bstg21 1.5=

∆bdr21 1:=

leff21 ∆bdr21 t21+ ∆bstg21+:=

leff21 2.75= m

MONTANT 2, DT2

t22 0.25:= m

dgol22 0.8:= m

∆bstg22 min 6 t22⋅ dgol22, ( ):=

∆bstg22 0.8=

∆bdr22 0.8:=

leff22 ∆bdr22 t22+ ∆bstg22+:=

leff22 1.85= m

Page 2: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

MONTANT 3, DT2

t23 0.25:= m

dgolstg23 1.725:= m

∆bstg23 min 6 t23⋅ dgolstg23, ( ):=

∆bstg23 1.5=

dgoldr23 1:= m

∆bdr23 min 6 t23⋅ dgoldr23, ( ):=

∆bdr23 1=

leff23 ∆bdr23 t23+ ∆bstg23+:=

leff23 2.75= m

MONTANT 1, DT3 MONTANT 1.1, DT3

t31 0.25:= m t311 0.25:= m

dgolstg31 1.275:= m dgolstg311 1.5:= m

∆bstg31 min 6 t31⋅ dgolstg31, ( ):= ∆bstg311 min 6 t31⋅ dgolstg31, ( ):=

∆bstg31 1.275= m ∆bstg311 1.275= m

dgoldr31 1:= m dgoldr311 0.8:= m

∆bdr31 min 6 t31⋅ dgoldr31, ( ):= ∆bdr311 min 6 t311⋅ dgoldr311, ( ):=

∆bdr31 1= ∆bdr311 0.8=

leff31 ∆bdr31 t31+ ∆bstg31+:= leff311 ∆bdr311 t311+ ∆bstg311+:=

leff31 2.525= m leff311 2.325= m

Page 3: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

MONTANT 2, DT3 MONTANT 2.1, DT3

t32 0.25:= m t321 0.25:= m

dgolstg32 0:= m dgolstg321 1.225:= m

∆bstg32 0:= ∆bstg321 min 6 t321⋅ dgolstg321, ( ):=

∆bstg32 0=∆bstg321 1.225= m

dgoldr32 1.5:= m dgoldr321 0:= m

∆bdr32 min 6 t32⋅ dgoldr32, ( ):= ∆bdr321 min 6 t321⋅ dgoldr321, ( ):=

∆bdr32 1.5= ∆bdr321 0=

leff32 ∆bdr32 t32+ ∆bstg32+:= leff321 ∆bdr321 t321+ ∆bstg321+:=

leff32 1.75= m leff321 1.475= m

MONTANT 1, DT4MONTANT 2.1, DT3

t41 0.25:= mt3211 0.25:= m

dgolstg41 1.225:= mdgolstg3211 0:= m

∆bstg41 min 6 t41⋅ dgolstg41, ( ):=∆bstg3211 min 6 t3211⋅ dgolstg3211, ( ):=

∆bstg41 1.225= m∆bstg3211 0= m

dgoldr41 1.5:= mdgoldr3211 1.5:= m

∆bdr41 min 6 t41⋅ dgoldr41, ( ):=∆bdr3211 min 6 t3211⋅ dgoldr3211, ( ):=

∆bdr41 1.5= m∆bdr3211 1.5=

leff41 ∆bdr41 t41+ ∆bstg41+:=leff3211 ∆bdr3211 t3211+ ∆bstg3211+:=

leff41 2.975= mleff3211 1.75= m

Page 4: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

MONTANT 2, DT4

t42 0.25:= m

dgolstg42 1.925:= m

∆bstg42 min 6 t42⋅ dgolstg42, ( ):=

∆bstg42 1.5= m

dgoldr42 1.5:= m nuavemgol( )

∆bdr42 min 6 t42⋅ dgoldr42, ( ):=

∆bdr42 1.5= m

leff42 ∆bdr42 t42+ ∆bstg42+:=

leff42 3.25= m

MONTANT 2, DT5MONTANT 1, DT5

t52 0.25:= mt51 0.25:= m

dgolstg52 1.725:= mdgolstg51 1.5:= m nuavemgol( )

∆bstg52 min 6 t52⋅ dgolstg52, ( ):=∆bstg51 min 6 t51⋅ dgolstg51, ( ):=

∆bstg52 1.5= m∆bstg51 1.5= m

dgoldr52 1:= mdgoldr51 1.95:= m

∆bdr52 min 6 t52⋅ dgoldr52, ( ):=∆bdr51 min 6 t51⋅ dgoldr51, ( ):=

∆bdr52 1= m∆bdr51 1.5= m

leff52 ∆bdr52 t52+ ∆bstg52+:=leff51 ∆bdr51 t51+ ∆bstg51+:=

leff52 2.75= mleff51 3.25= m

Page 5: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

MONTANT 1, DT6 MONTANT 1.1, DT6

t61 0.3:= m t611 0.3:= m

dgolstg61 1.725:= m dgolstg611 0:= m

∆bstg61 min 6 t61⋅ dgolstg61, ( ):= ∆bstg611 min 6 t611⋅ dgolstg611, ( ):=

∆bstg61 1.725= m ∆bstg611 0= m

dgoldr61 0:= m dgoldr611 1:= m

∆bdr61 min 6 t61⋅ dgoldr61, ( ):= ∆bdr611 min 6 t611⋅ dgoldr611, ( ):=

∆bdr61 0= m ∆bdr611 1= m

leff61 ∆bdr61 t61+ ∆bstg61+:= leff611 ∆bdr611 t611+ ∆bstg611+:=

leff61 2.025= m leff611 1.3= m

MONTANT 1.2, DT6 MONTANT 2, DT6

t612 0.25:= m t62 0.25:= m

dgolstg612 1.5:= m dgolstg62 0.8:= m

∆bstg612 min 6 t612⋅ dgolstg612, ( ):= ∆bstg62 min 6 t62⋅ dgolstg62, ( ):=

∆bstg612 1.5= m ∆bstg62 0.8= m

dgoldr612 0:= m dgoldr62 1.5:= m nuavemgol( )

∆bdr612 min 6 t612⋅ dgoldr612, ( ):= ∆bdr62 min 6 t62⋅ dgoldr62, ( ):=

∆bdr612 0= m ∆bdr62 1.5= m

leff612 ∆bdr612 t612+ ∆bstg612+:= leff62 ∆bdr62 t62+ ∆bstg62+:=

leff612 1.75= m leff62 2.55= m

Page 6: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

MONTANT 3, DT6

t63 0.25:= m

dgolstg63 1.25:= m

∆bstg63 min 6 t63⋅ dgolstg63, ( ):=

∆bstg63 1.25= m

dgoldr63 1:= m

∆bdr63 min 6 t63⋅ dgoldr63, ( ):=

∆bdr63 1= m

leff63 ∆bdr63 t63+ ∆bstg63+:=

leff63 2.5= m

MONTANT 1, DT7

t71 0.25:= m

dgolstg71 1.275:= m

∆bstg71 min 6 t71⋅ dgolstg71, ( ):=

∆bstg71 1.275= m

dgoldr71 1:= m

∆bdr71 min 6 t71⋅ dgoldr71, ( ):=

∆bdr71 1= m

leff71 ∆bdr71 t71+ ∆bstg71+:=

leff71 2.525= m

Page 7: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

MONTANT 2, DT7 MONTANT 2.1, DT7

t72 0.25:= m t721 0.25:= m

dgolstg72 1.5:= m dgolstg721 1.26:= m

∆bstg72 min 6 t72⋅ dgolstg72, ( ):= ∆bstg721 min 6 t72⋅ dgolstg72, ( ):=

∆bstg72 1.5= m ∆bstg721 1.5= m

dgoldr72 0.3:= m dgoldr721 1:= m

∆bdr72 min 6 t72⋅ dgoldr72, ( ):= ∆bdr721 min 6 t72⋅ dgoldr72, ( ):=

∆bdr72 0.3= m ∆bdr721 0.3= m

leff72 ∆bdr72 t72+ ∆bstg72+:= leff721 ∆bdr72 t72+ ∆bstg72+:=

leff72 2.05= m leff721 2.05= m

MONTANT 1, DT8 MONTANT 2, DT8

t81 0.3:= m t82 0.3:= m

dgolstg81 1.24:= m dgolstg82 2.725:= m

∆bstg81 min 6 t81⋅ dgolstg81, ( ):= ∆bstg82 min 6 t82⋅ dgolstg82, ( ):=

∆bstg81 1.24= m ∆bstg82 1.8= m

dgoldr81 0:= m dgoldr82 0:= m

∆bdr81 min 6 t81⋅ dgoldr81, ( ):= ∆bdr82 min 6 t82⋅ dgoldr82, ( ):=

∆bdr81 0= m ∆bdr82 0= m

leff81 ∆bdr81 t81+ ∆bstg81+:= leff82 ∆bdr82 t82+ ∆bstg82+:=

leff81 1.54= m leff82 2.1= m

Page 8: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

MONTANT 3, DT8

t83 0.3:= m

dgolstg83 1.225:= m

∆bstg83 min 6 t83⋅ dgolstg83, ( ):=

∆bstg83 1.225= m

dgoldr83 0:= m

∆bdr83 min 6 t83⋅ dgoldr83, ( ):=

∆bdr83 0= m

leff83 ∆bdr83 t83+ ∆bstg83+:=

leff83 1.525= m

4.1.2 Sectiuni active rigle

Pentru usa 1(m) x 2.1(m) Pentru usa 0.7(m) x 2.1(m)

tr1 0.25:= m tr2 0.25:= m

l0r1 1:= m l0r2 0.7:= m

∆bstr1 0.15 l0r1⋅:= ∆bstr2 0.15 l0r2⋅:=

∆bstr1 0.15= m ∆bstr2 0.105= m

∆bdrr1 ∆bstr1 0.15=:= m ∆bdrr2 ∆bstr2 0.105=:= m

lp1 ∆bstr1 tr1+ ∆bdrr1+:= lp2 ∆bstr2 tr2+ ∆bdrr2+:=

lp1 0.55= m lp2 0.46= m

4.2 Calculul caracteristicilor geometrice pentru montanti

DT1

tp 0.3:= m

lw1 11.3:= m

tp1 0.25:= m

t 0.3:= m

Am11 tp 2leff11 leff12+( )⋅ lw1 2 tp⋅− tp1−( ) t⋅+:=

Am11 4.845= m2

Page 9: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

DT2- 1 DT2 - 2

tp 0.25:= m tp 0.25:= m

lw21 3.35:= m lw22 2.025:= m

tp1 0.3:= m tp2 0.25:= m

t 0.25:= m t 0.25:= m

Am21 tp leff21( )⋅ lw21 tp1−( ) t⋅+:= Am22 tp leff22( )⋅ lw22 tp2−( ) t⋅+:=

Am21 1.45= m2

Am22 0.906= m2

DT2 - 3

tp 0.25:= m

lw23 4.23:= m

tp3 0.3:= m

t 0.25:= m

Am23 tp leff23( )⋅ lw23 tp3−( ) t⋅+:=

Am23 1.67= m2

DT3 - 1

t 0.25:= m

lw31 4.73:= m

tp1 0.3:= m

tp2 0.25:= m

Am31 tp1 leff31 leff311+( )⋅ lw31 tp1− tp2−( ) t⋅+:=

Am31 2.5= m2

DT3 - 2

t 0.25:= m

lw311 4.73:= m

tp1 0.25:= m

tp2 0.25:= m

tp3 0.3:=m

Am32 tp1 leff32 leff321+ leff321+( )⋅ lw311 tp1− tp2− tp3−( ) t⋅+:=

Am32 2.158= m2

Page 10: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

DT4 - 1

t 0.25:= m

lw41 3.525:= m

tp1 0.3:= m

tp2 0:= m

tp3 0:=m

Am41 tp1 leff41( )⋅ lw41 tp1− tp2− tp3−( ) t⋅+:=

Am41 1.699= m2

DT4 - 2

t 0.25:= m

lw42 0.5:= m

tp1 0.25:= m

mtp2 0:=

tp3 0:= m

Am42 tp1 leff42( )⋅ lw42 tp1− tp2− tp3−( ) t⋅+:=

Am42 0.875= m2

DT5 - 1

t 0.25:= m

lw51 0.8:= m

tp1 0.25:= m

mtp2 0:=

tp3 0:= m

Am51 tp1 leff51( )⋅ lw51 tp1− tp2− tp3−( ) t⋅+:=

Am51 0.95= m2

Page 11: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

DT5 - 2

t 0.25:= m

lw52 3.225:= m

tp1 0:= m

mtp2 0:=

tp3 0.3:= m

Am52 tp1 leff52( )⋅ lw52 tp1− tp2− tp3−( ) t⋅+:=

Am52 0.731= m2

DT6 - 1

t 0.25:= m

lw61 5.275:= m

tp1 0.3:= m

mtp2 0.25:=

tp3 0.25:= m

Am61 tp1 leff61 leff611+ leff612+( )⋅ lw61 tp1− tp2− tp3−( ) t⋅+:=

Am61 2.641= m2

DT6 -2

t 0.25:= m

lw62 0.8:= m

tp1 0.25:= m

tp2 0:= m

tp3 0:= m

Am62 tp1 leff62( )⋅ lw62 tp1− tp2− tp3−( ) t⋅+:=

Am62 0.775= m2

Page 12: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

DT6 - 3

t 0.25:= m

lw63 3.225:= m

tp1 0:= m

tp2 0:= m

tp3 0.3:= m

Am63 tp1 leff63( )⋅ lw63 tp1− tp2− tp3−( ) t⋅+:=

Am63 0.731= m2

DT7 -1

t 0.25:= m

lw71 2.825:= m

tp1 0.3:= m

tp2 0:= m

tp3 0:= m

Am71 tp1 leff71( )⋅ lw71 tp1− tp2− tp3−( ) t⋅+:=

Am71 1.389= m2

DT7 - 2

t 0.25:= m

lw72 5.875:= m

tp1 0.25:= m

tp2 0.3:= m

tp3 0:= m

Am72 tp1 leff72 leff721+( )⋅ lw72 tp1− tp2− tp3−( ) t⋅+:=

Am72 2.356= m2

Page 13: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

DT8

t 0.25:= m

lw8 10.3:= m

tp1 0.3:= m

tp2 0.25:= m

tp3 0.3:= m

Am8 tp1 leff81 leff82+ leff83+( )⋅ lw8 tp1− tp2− tp3−( ) t⋅+:=

Am8 3.912= m2

Aria activa la forta taietoare

DT1

t 0.3:= m

Av1 t lw1⋅:=

Av1 3.39= m2

DT2 1− DT2 2− DT2 3−

t 0.25:= m t 0.25:= m t 0.25:= m

Av21 t lw21⋅:= Av22 t lw22⋅:= Av23 t lw23⋅:=

Av21 0.838= m2

Av22 0.506= m2

Av23 1.058= m2

DT3 1− DT3 2−

t 0.25:= m t 0.25:= m

Av31 t lw31⋅:= Av311 t lw311⋅:=

Av31 1.183= m2

Av311 1.183= m2

DT4 1− DT4 2−

t 0.25:= m t 0.25:= m

Av41 t lw41⋅:= Av42 t lw42⋅:=

Av41 0.881= m2

Av42 0.125= m2

Page 14: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

DT5 1− DT5 2−

t 0.25:= m t 0.25:= m

Av51 t lw51⋅:= Av52 t lw52⋅:=

Av51 0.2= m2

Av52 0.806= m2

DT6 1− DT6 2− DT6 3−

t 0.25:= m t 0.25:= m t 0.25:= m

Av61 t lw61⋅:= Av62 t lw62⋅:= Av63 t lw63⋅:=

Av61 1.319= m2

Av62 0.2= m2

Av63 0.806= m2

DT7 1− DT7 2−

t 0.25:= m t 0.25:= m

Av71 t lw71⋅:= Av72 t lw72⋅:=

Av71 0.706= m2

Av72 1.469= m2

DT8

t 0.3:= m

Av8 t lw8⋅:=

Av8 3.09= m2

Pozitia centrului de greutate

DT1

yg11 5.7:= m

DT2 1− DT2 2− DT2 3−

yg12 2.49:= m yg22 0.974:= m yg23 1.3:= m

DT3 1− DT3 2−

yg31 3.31:= m yg32 2.71:= m

DT4 1− DT4 2−

yg41 2.72:= m yg42 0.18:= m

Page 15: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

DT5 1− DT5 2−

yg51 0.82:= m yg52 0.89:= m

DT6 1− DT6 2− DT6 3−

yg61 2.77:= m yg62 0.79:= m yg63 1.07:= m

DT7 1− DT7 2−

yg71 2.22:= m yg72 2.69:= m

Momentele de inertie ale montantilor

DT1

Im116340883270.8287

108

63.409=:= m4

DT2 1−

Im21173230248.0657

108

1.732=:= m4

DT2 2−

Im2251396433.5249

108

0.514=:= m4

DT2 3−

Im23328650286.1803

108

3.287=:= m4

DT3 1−

Im311273263548.5361

108

12.733=:= m4

DT3 2−

Im321091323645.7214

108

10.913=:= m4

DT4 1−

Im41254832088.9325

108

2.548=:= m4

Page 16: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

DT4 2−

Im4271582031.2499

108

0.716=:= m4

DT5 1−

Im5171621093.7500

108

0.716=:= m4

DT5 2−

Im52207530458.8246

108

2.075=:= m4

DT6 1−

Im61899480639.1248

108

8.995=:= m4

DT6 2−

Im6236764678.9627

108

0.368=:= m4

DT6 3−

Im63191367359.6916

108

1.914=:= m4

DT7 1−

Im71137221820.9449

108

1.372=:= m4

DT7 2−

Im721228005955.8220

108

12.28=:= m4

DT8

Im84580874773.2824

108

45.809=:= m4

Page 17: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

Pozitia centrului de greutate pentru intreaga diafragma

DT1 DT3 DT5 DT7

yg1 5.7:= m yg3 6.56:= m yg5 2.5:= m yg7 5.06:= m

DT4DT2 DT6 DT8

yg4 2.79:= myg2 4.56:= m yg6 6.85:= m yg7 5.16:= m

Momente de inertie pentru intreaga diafragma

DT1

Im1x6340883270.8287

108

63.409=:= m4

DT2

Im2x10574258361.5741

108

105.743=:= m4

DT3

Im3x10998181186.4256

108

109.982=:= m4

DT4

Im4x1238297604.0524

108

12.383=:= m4

DT5

Im5x1258843236.0877

108

12.588=:= m4

DT6

Im6x10724143987.1588

108

107.241=:= m4

DT7

Im7x5605907220.3889

108

56.059=:= m4

DT8

Im8x4580874773.2824

108

45.809=:= m4

Page 18: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

4.3 Calculul caracteristicilor geometrice pentru rigla

Aria activa la compresiune excentrica

Pentru usa 1(m) x 2.1(m)

lp1 0.55= m

hpl 0.13:= m

t 0.25:= m

hr 0.93:= m

Ar1 lp1 hpl⋅ t hr hpl−( )⋅+ 0.272=:= m2

Pentru usa 0.7(m) x 2.1(m)

lp2 0.46= m

hp2 0.13:= m

t 0.25:= m

hr 0.93:= m

Ar1 lp2 hp2⋅ t hr hpl−( )⋅+ 0.26=:= m2

Aria activa la forta taietoare

Av hr t⋅ 0.233=:= m2

Pozitia centrului de greutate

Pentru usa 1(m) x 2.1(m)

ygp1 0.52:= m

Pentru usa 0.7(m) x 2.1(m)

ygp2 0.51:= m

Moment de inertie

Pentru usa 1(m) x 2.1(m)

Ip12215029.4642

108

0.022=:= m4

Pentru usa 0.7(m) x 2.1(m)

Ip2209614

108

2.096 103−

×=:= m4

Page 19: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

CALCULUL CARACTERISTICILOR DE RIGIDITATE PENTRU O DIAFRAGMA CU 2

SIRURI DE GOLURI

CADRU REAL

Rigiditatea montantilor la distorsiune

Im21 1.732= m4

Im22 0.514= m4

Im23 3.287= m4

He 3.13:= m

km1

Im21

He

0.553=:= m3Em

km2

Im22

He

0.164=:= m3Em

km3

Im23

He

1.05=:= m3Em

Rigiditatea riglei la distorsiune

L12 5:= m

l0 0.7:= m

a 0.35 hr⋅ 0.325=:=

lrd l0 2 a⋅+ 1.351=:= m

μ1

130 Ip2⋅( )Av lrd

2⋅

+

0.871=:=

kr1

6 Ip2⋅( )L12

L12

lrd

3

⋅ μ⋅ 0.111=:= m3Er⋅

Page 20: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

L23 6:= m

l0 1:= m

a 0.35 hr⋅ 0.325=:=

lrd l0 2 a⋅+ 1.651=:= m

μ1

130 Ip1⋅( )Av lrd

2⋅

+

0.488=:=

kr2

6 Ip1⋅( )L23

L23

lrd

3

⋅ μ⋅ 0.519=:= m3Er⋅

CADRU ECHIVALENT

Rigiditatea montantilor la distorsiune

kme

km1 km2+ km3+( )2

0.884=:= m3

Rigiditatea riglei la distorsiune

kre kr1 kr2+ 0.63=:= m3

λkre

kme

0.713=:=

L13 11:= m

Am1 1.59:= m2

Am3 1.81:= m2

γ 1Im21 Im22+ Im23+( )

L132

1

Am1

1

Am3

+

⋅+ 1.054=:=

n 4:=

α n λ γ⋅⋅ 3.467=:=

1 α< 10< diafragma cu goluri mijlocii

a 0.8 0.6− 0.2=:=0.6 .................... 0.176

b 0.205 0.176− 0.029=:=0.713 ...................ηs c 0.713 0.6− 0.113=:=0.8 .....................0.205

Page 21: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

a ........................b

c ...................... .x

xc b⋅( )

a0.016=:=

ηs 0.176 x+ 0.192=:=

Moment de inertie global

I0γ

γ 1−Im21 Im22+ Im23+( )⋅ 107.952=:= m

4

Anet Am1 Am3+ 3.4=:= m4

ηs 0.192=

νs 10.43:=

n 4:=

He 3.13:= m

H n He⋅ 12.52=:= m

Ies

ηs I0⋅

1 νs

ηs I0⋅

Anet H2

⋅+

:=

Ies 14.767= m4

Page 22: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

6.3 . Distributia incarcarilor orizontale la diafragme transversale

Fb 1178:= kN

e0 0.58:= m

L 26:= m

ea 0.05 L⋅:=

ea 1.3= m

ls1 e0 ea+:=

ls1 1.88= m

Mts1 Fb ls1⋅:=

Mts1 2.215 103

×= kNm

ls2 ea e0−:=

ls2 0.72= m

Mts2 Fb ls2⋅:=

Mts2 848.16= kNm

Page 23: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

7. Diagrama de efort axial din incarcari gravitationale

7.1. Calculul suprafetelor elementare aferente

S1 0.878:= m2

S2 4.70:= m2

S3 1.13:= m2

S4 2.53:= m2

S5 5.58:= m2

S7 1.62:= m2

S8 0.845:= m2

S9 0.845:= m2

S10 1.62:= m2

S11 7.96:= m2

S12 2.53:= m2

S13 1.12:= m2

S14 7.11:= m2

S15 0.845:= m2

S16 4.47:= m2

S17 1.53:= m2

S6 0.882:= m2

S18 0.657:= m2

S19 0.660:= m2

S20 0.848:= m2

S21 6.38:= m2

S22 1.53:= m2

7.2. Incarcarea transmisa la plansee

Diafragma 1-DT1 -Diafragma plina

Trama 1

gpc 4.56:= kN /mp qk1 3:= kN /mp gt 8.08:= kN /mpgpr 5.45:= kN / mp

qkt 1.13:= kN /mpgpd 2.42:= kN /mp

gprbaie 5.49:= kN / mp

Apc1 S3 S4+ 2.94+ 2.97+ 9.57=:= mp

Aprbaie S7 2.37+ 3.99=:= mp

Apr S1 S8+ 1.48+ 3.203=:= mp

gmediu1

gpc Apc1⋅( ) gpr Apr⋅( )+ gprbaie Aprbaie⋅( )+

Apc1 Aprbaie+ Apr+:=

gmediu1 4.951= kN / mp

gmediuTRAMA1 gmediu1 gpd+ 0.3 qk1⋅+ 8.271=:= kN /mp

nivel curent

i1 S1 S2+ S3+ S4+ S5+ S7+ S8+( ) gmediuTRAMA1⋅ 142.955=:= kN

planseu terasa

Qtr1 S1 S2+ S3+ S4+ S5+ S7+ S8+( ) gt 0.4 qkt⋅+( )⋅ 147.459=:= kN

Page 24: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

Trama 2

gpc 4.56:= kN /mp qk1 3:= kN /mp gt 8.08:= kN /mpgpr 5.45:= kN / mp

qkt 1.13:= kN /mpgpd 2.42:= kN /mp

gprbaie 5.49:= kN / mp

Apc2 S12 S13+ S9+ S15+ 5.12+ 5.12+ 1.69+ 17.27=:= mp

Aprbaie2 S10 2.54+ 4.16=:= mp

Apr2 0 0=:= mp

gmediu2

gpc Apc2⋅( ) gpr Apr2⋅( )+ gprbaie Aprbaie2⋅( )+

Apc2 Aprbaie2+ Apr2+:=

gmediu2 4.741= kN / mp

gmediuTRAMA2 gmediu2 gpd+ 0.3 qk1⋅+ 8.061=:= kN /mp

nivel curent

i2 S9 S10+ S11+ S12+ S13+ S14+ S15+( ) gmediuTRAMA2⋅ 177.574=:= kN

planseu terasa

Qtr2 S9 S10+ S11+ S12+ S13+ S14+ S15+( ) gt 0.4 qkt⋅+( )⋅ 187.96=:= kN

Greutatea proprie pe nivel

gzidarie 1.479:= kN /mp

γbeton 25:=

Azidarie1 12.9 15.63+ 3.96 2⋅( )+ 4.91+ 41.36=:= mp

volumul centurii din beton armat

Vc1 11.2 0.17⋅ 0.3⋅( ) 2 1.5 0.17⋅ 0.3⋅( )+ 1.8 0.17⋅ 0.3⋅( )+ 0.816=:= mc

volumul buiandrugilor

Vb1 0.3 0.33⋅ 0.4⋅ 0.04=:= mc

volumul stalpisorilor din beton armat

Vs1 0.25 0.25⋅ 2.73⋅( ) 3⋅ 0.512=:= mc

GDT1 Azidarie1 gzidarie⋅ Vc1 Vb1+ Vs1+( ) γbeton⋅+:=

GDT1 95.358= kN

Page 25: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

Diafragma 2 -DT2 -Diafragma cu goluri

Trama 3

gpc 4.56:= kN /mp qk1 3:= kN /mp gt 8.08:= kN /mpgpr 5.45:= kN / mp

qkt 1.13:= kN /mpgpd 2.42:= kN /mp

gprbaie 5.49:= kN / mp

Apc3 S17 2.79+ 4.32=:= mp

Aprbaie3 0 0=:= mp

Apr3 S6 S18+ 1.44+ 2.979=:= mp

gmediu3

gpc Apc3⋅( ) gpr Apr3⋅( )+ gprbaie Aprbaie3⋅( )+

Apc3 Aprbaie3+ Apr3+:=

gmediu3 4.923= kN / mp

gmediuTRAMA3 gmediu3 gpd+ 0.3 qk1⋅+ 8.243=:= kN /mp

nivel curent

i3 S16 S17+ S18+ S6+( ) gmediuTRAMA3⋅ 62.146=:= kN

planseu terasa

Qtr3 S16 S17+ S18+ S6+( ) gt 0.4 qkt⋅+( )⋅ 64.323=:= kN

Page 26: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

Trama 4

gpc 4.56:= kN /mp qk1 3:= kN /mp gt 8.08:=gpr 5.45:= kN / mp

qkt 1.13:= kN /mpgpd 2.42:= kN /mp

gprbaie 5.49:= kN / mp

Apc4 S19 S20+ S22+ 1.62+ 4.52+ 9.178=:= mp

Aprbaie4 0 0=:= mp

Apr4 0 0=:= mp

gmediu4

gpc Apc4⋅( ) gpr Apr4⋅( )+ gprbaie Aprbaie4⋅( )+

Apc4 Aprbaie4+ Apr4+:=

gmediu4 4.56= kN / mp

gmediuTRAMA4 gmediu4 gpd+ 0.3 qk1⋅+ 7.88=:= kN /mp

nivel curent

i4 S19 S20+ S21+ S22+( ) gmediuTRAMA4⋅ 74.214=:= kN

planseu terasa

Qtr4 S19 S20+ S21+ S22+( ) gt 0.4 qkt⋅+( )⋅ 80.354=:= kN

Greutatea proprie pe nivel

gzidarie 1.479:= kN /mp

γbeton 25:=

Azidarie2 8.62 1.86+ 2.48+ 9.92+ 2 4.04 2.6+( )⋅+ 1.93+ 1.93+ 40.02=:=

volumul centurii din beton armat

Vc2 11.2 0.17⋅ 0.3⋅( ) 2 2.75 0.17⋅ 0.3⋅( )+ 1.85 0.17⋅ 0.3⋅( )+ 0.946=:= mc

volumul buiandrugilor

Vb2 0.25 0.63⋅ 1.5⋅ 0.25 0.63⋅ 1.8⋅+ 0.25 0.33⋅ 0.175⋅+ 0.25 0.33⋅ 0.4⋅+ 2 0.25⋅ 0.63⋅ 0.4⋅+ 0.693=:= mc

volumul stalpisorilor din beton armat

Vs2 0.25 0.25⋅ 2.73⋅( ) 3⋅ 0.512=:= mc

GDT2 Azidarie2 gzidarie⋅ Vc2 Vb2+ Vs2+( ) γbeton⋅+:=

GDT2 112.967= kN

Page 27: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

Rezultanta eforturilor axiale la nivelul ,,0''

n 4:= niveluri

NGDT1 Qtr1 Qtr2+ n 1−( ) i1 i2+( )⋅+ n GDT1⋅+ 1.678 103

×=:= kN

NGDT2 Qtr3 Qtr4+ n 1−( ) i3 i4+( )⋅+ n GDT2⋅+ 1.006 103

×=:= kN

Page 28: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

9. Determinarea eforturilor din incarcari orizontale (seism)

MEd ,VEd,NEd pentru o diafragma plina (DT1) si o diafragma cu goluri (DT2)

9.1 Diafragma plina

DT1

F1 28.89:= kN

F2 57.78:= kN

F3 86.67:= kN

F4 115.56:= kN

F5 144.45:= kN

he 3.13:= m

Ms1 F5 he⋅ 452.128=:= kNm

Ms2 F5 2⋅ he⋅ F4 he⋅+ 1.266 103

×=:= kNm

Ms3 F5 3⋅ he⋅ F4 2⋅ he+ F3 he⋅+ 2.351 103

×=:= kNm

Ms4 F5 4⋅ he⋅ F4 3⋅ he+ F3 2⋅ he+ F2 he⋅+ 3.617 103

×=:= kNm

Ms5 F5 5⋅ he⋅ F4 4⋅ he+ F3 3⋅ he+ F2 2⋅ he+ F1 he⋅+ 4.973 103

×=:= kNm

Ts1 F5 144.45=:= kN

Ts2 F5 F4+ 260.01=:= kN

Ts3 F5 F4+ F3+ 346.68=:= kN

Ts4 F5 F4+ F3+ F2+ 404.46=:= kN

Ts5 F5 F4+ F3+ F2+ F1+ 433.35=:= kN

Page 29: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

9.2 Diafragma cu goluri

Cadru inlocuitor

Montanti

qs 23.93:= kN / m

H 12.52:= m

ms43

n 4=λ 0.713=γ 1.054=

pentru λ = 0.6

1................. 2.9

1.054.................γ43

1.1 ................2.62

a 1.1 1− 0.1=:= a .......................b

c .......................xb 2.9 2.62− 0.28=:=c 1.054 1− 0.054=:=

xc b⋅( )

a0.151=:=

γ43 2.9 x− 2.749=:=

pentru λ = 0.8

1................. 2.83

1.054.................γ43

1.1 ................2.54

a 1.1 1− 0.1=:= a .......................b

c .......................xb 2.54 2.83− 0.29−=:=c 1.054 1− 0.054=:=

xc b⋅( )

a0.157−=:=

γ431 x 2.83+ 2.673=:=

0.6 ........................ 2.749

0.713 .................... ms430.8 .........................2.987

a 0.8 0.6− 0.2=:=

b γ431 γ43− 0.075−=:=

c 0.713 0.6− 0.113=:=

Page 30: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

xc b⋅( )

a0.043−=:=

ms43 x γ43+ 2.706=:=

Mme43

qs H2

100ms43⋅ 101.511=:= kNm

ms34

n 4=λ 0.713=γ 1.054=

pentru λ = 0.6

1................. 0.23−1.054.................γ34

1.1 ................0.51

a 1.1 1− 0.1=:= a .......................b

c .......................xb 0.51 0.23−( )− 0.74=:=c 1.054 1− 0.054=:=

xc b⋅( )

a0.4=:=

γ34 0.23− x+ 0.17=:=

pentru λ = 0.8

1................. 0.3

1.054.................γ43

1.1 ................0.58

a 1.1 1− 0.1=:= a .......................b

c .......................xb 0.58 0.3− 0.28=:=c 1.054 1− 0.054=:=

xc b⋅( )

a0.151=:=

γ341 0.3 x+ 0.451=:=

0.6 ........................ 0.17

0.713 .................... ms340.8 .........................0.451

a 0.8 0.6− 0.2=:=

b γ341 γ34− 0.282=:=

c 0.713 0.6− 0.113=:=

Page 31: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

xc b⋅( )

a0.159=:=

ms34 x γ34+ 0.329=:=

Mme34

qs H2

100ms34⋅ 12.33=:= kNm

Page 32: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

ms32

n 4=λ 0.713=γ 1.054=

pentru λ = 0.6

1................. 3.47

1.054.................γ32

1.1 ................2.9

a 1.1 1− 0.1=:= a .......................b

c .......................xb 2.9 3.47− 0.57−=:=c 1.054 1− 0.054=:=

xc b⋅( )

a0.308−=:=

γ32 3.47 x+ 3.162=:=

pentru λ = 0.8

1................. 3.54

1.054.................γ43

1.1 ................2.95

a 1.1 1− 0.1=:= a .......................b

c .......................xb 2.95 3.54− 0.59−=:=c 1.054 1− 0.054=:=

xc b⋅( )

a0.319−=:=

γ321 3.54 x+ 3.221=:=

0.6 ........................ 3.162

0.713 .................... ms320.8 .........................3.221

a 0.8 0.6− 0.2=:=

b γ321 γ32− 0.059=:=

c 0.713 0.6− 0.113=:=

xc b⋅( )

a0.033=:=

ms32 x γ32+ 3.196=:=

Mme32

qs H2

100ms32⋅ 119.87=:= kNm

Page 33: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

ms23

n 4=λ 0.713=γ 1.054=

pentru λ = 0.6

1................. 2

1.054.................γ23

1.1 ................2.57

a 1.1 1− 0.1=:= a .......................b

c .......................xb 2.57 2− 0.57=:=c 1.054 1− 0.054=:=

xc b⋅( )

a0.308=:=

γ23 2 x+ 2.308=:=

pentru λ = 0.8

1................. 1.93

1.054.................γ43

1.1 ................2.52

a 1.1 1− 0.1=:= a .......................b

c .......................xb 2.52 1.93− 0.59=:=c 1.054 1− 0.054=:=

xc b⋅( )

a0.319=:=

γ231 1.93 x+ 2.249=:=

0.6 ........................ 2.52

0.713 .................... ms320.8 .........................2.249

a 0.8 0.6− 0.2=:=

b γ231 γ23− 0.059−=:=

c 0.713 0.6− 0.113=:=

xc b⋅( )

a0.033−=:=

ms23 γ23 x− 2.341=:=

Mme23

qs H2

100ms23⋅ 87.821=:= kNm

Page 34: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

ms21

n 4=λ 0.713=γ 1.054=

pentru λ = 0.6

1................. 2.13

1.054.................γ21

1.1 ................1.3

a 1.1 1− 0.1=:= a .......................b

c .......................xb 1.3 2.13− 0.83−=:=c 1.054 1− 0.054=:=

xc b⋅( )

a0.448−=:=

γ21 2.13 x+ 1.682=:=

pentru λ = 0.8

1................. 2.56

1.054.................γ21

1.1 ................1.66

a 1.1 1− 0.1=:= a .......................b

c .......................xb 2.56 1.66− 0.9=:=c 1.054 1− 0.054=:=

xc b⋅( )

a0.486=:=

γ211 2.56 x− 2.074=:=

0.6 ........................ 1.682

0.713 .................... ms210.8 .........................3.054

a 0.8 0.6− 0.2=:=

b γ211 γ21− 0.392=:=

xc b⋅( )

a0.106=:=

ms21 x γ21+ 1.788=:=

Mme21

qs H2

100ms21⋅ 67.057=:= kNm

Page 35: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

ms12

n 4=λ 0.713=γ 1.054=

pentru λ = 0.6

1................. 4.9

1.054.................γ12

1.1 ................5.73

a 1.1 1− 0.1=:= a .......................b

c .......................xb 5.73 4.9− 0.83=:=c 1.054 1− 0.054=:=

xc b⋅( )

a0.448=:=

γ12 4.9 x+ 5.348=:=

pentru λ = 0.8

1................. 4.47

1.054.................γ12

1.1 ................5.37

a 1.1 1− 0.1=:= a .......................b

c .......................xb 5.37 4.47− 0.9=:=c 1.054 1− 0.054=:=

xc b⋅( )

a0.486=:=

γ121 4.47 x+ 4.956=:=

0.6 ........................ 5.348

0.713 .................... ms210.8 .........................4.956

a 0.8 0.6− 0.2=:=

b γ121 γ12− 0.392−=:=

c 0.713 0.6− 0.113=:=

xc b⋅( )

a0.222−=:=

ms12 γ12 x+ 5.127=:=

Mme12

qs H2

100ms12⋅ 192.301=:= kNm

Page 36: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

ms10

n 4=λ 0.713=γ 1.054=

pentru λ = 0.6

1................. 1.6−1.054.................γ10

1.1 ................-2.6

a 1.1 1− 0.1=:= a .......................b

c .......................xb 2.6− 1.6−( )− 1−=:=c 1.054 1− 0.054=:=

xc b⋅( )

a0.54−=:=

γ10 1.6− x+ 2.14−=:=

pentru λ = 0.8

1................. 0.75−1.054.................γ10

1.1 ................1.05

a 1.1 1− 0.1=:= a .......................b

c .......................xb 1.05 0.75−( )− 1.8=:=c 1.054 1− 0.054=:=

xc b⋅( )

a0.972=:=

γ101 0.75− x+ 0.222=:=

0.6 ........................ -2.14

0.713 .................... ms210.8 .........................0.222

a 0.8 0.6− 0.2=:=

b γ101 γ10− 2.362=:=

c 0.713 0.6−:=

xc b⋅( )

a1.335=:=

ms10 x γ21+ 3.016=:=

Mme10

qs H2

100ms10⋅ 113.144=:= kNm

Page 37: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

ms01

n 4=λ 0.713=γ 1.054=

pentru λ = 0.6

1................. 9.41

1.054.................γ01

1.1 ................10.4

a 1.1 1− 0.1=:= a .......................b

c .......................xb 10.4 9.41− 0.99=:=c 1.054 1− 0.054=:=

xc b⋅( )

a0.535=:=

γ01 9.41 x+ 9.945=:=

pentru λ = 0.8

1................. 8.56

1.054.................γ01

1.1 ................9.67

a 1.1 1− 0.1=:= a .......................b

c .......................xb 9.67 8.56( )− 1.11=:=c 1.054 1− 0.054=:=

xc b⋅( )

a0.599=:=

γ011 8.56 x+ 9.159=:=

0.6 ........................ 9.945

0.713 .................... ms010.8 .........................9.159

a 0.8 0.6− 0.2=:=

b γ011 γ01− 0.785−=:=

c 0.713 0.6− 0.113=:=

xc b⋅( )

a0.444−=:=

ms01 x γ01+ 9.501=:=

Mme01

qs H2

100ms01⋅ 356.385=:= kNm

Page 38: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

Cadrul real (2 siruri) de goluri

Mm143

km1

kme

Mme43⋅ 63.566=:= kNm Mm134

km1

kme

Mme34⋅ 7.721=:= kNm

Mm243

km2

kme

Mme43⋅ 18.86=:= kNm Mm234

km2

kme

Mme34⋅ 2.291=:= kNm

Mm343

km3

kme

Mme43⋅ 120.596=:= kNm Mm334

km3

kme

Mme34⋅ 14.648=:= kNm

Mm132

km1

kme

Mme32⋅ 75.062=:= kNm Mm123

km1

kme

Mme23⋅ 54.993=:= kNm

Mm232

km2

kme

Mme32⋅ 22.271=:= kNm Mm223

km2

kme

Mme23⋅ 16.316=:= kNm

Mm332

km3

kme

Mme32⋅ 142.407=:= kNm Mm323

km3

kme

Mme23⋅ 104.333=:= kNm

Mm121

km1

kme

Mme21⋅ 41.991=:= kNm Mm112

km1

kme

Mme12⋅ 120.418=:= kNm

Mm221

km2

kme

Mme21⋅ 12.458=:= kNm Mm212

km2

kme

Mme12⋅ 35.727=:= kNm

Mm321

km3

kme

Mme21⋅ 79.665=:= kNm Mm312

km3

kme

Mme12⋅ 228.456=:= kNm

Mm110

km1

kme

Mme10⋅ 70.85=:= kNm Mm101

km1

kme

Mme01⋅ 223.167=:= kNm

Mm210

km2

kme

Mme10⋅ 21.021=:= kNm Mm201

km2

kme

Mme01⋅ 66.212=:= kNm

Mm310

km3

kme

Mme10⋅ 134.416=:= kNm Mm301

km3

kme

Mme01⋅ 423.39=:= kNm

Page 39: Mathcad - Caract de Backuprigiditate Pentru o Diafragma Cu 2 Siruri de Goluri

M4r1 Mm143− 63.566−=:= kNm

M4r2sup Mm243−kr1

kr1 kr2+⋅ 3.324−=:= kNm

M4r2inf Mm243−kr2

kr1 kr2+⋅ 15.535−=:= kNm

M4r3 Mm343− 120.596−=:= kNm

M3r1 Mm134 Mm132+( )− 82.783−=:= kNm

M3r2sup Mm234 Mm232+( )−kr1

kr1 kr2+⋅ 4.329−=:= kNm

M3r2inf Mm234 Mm232+( )−kr2

kr1 kr2+⋅ 20.232−=:= kNm

M3r3 Mm334 Mm332+( )− 157.055−=:= kNm

M2r1 Mm123 Mm121+( )− 96.984−=:= kNm

M2r2sup Mm223 Mm221+( )−kr1

kr1 kr2+⋅ 5.072−=:= kNm

M2r2inf Mm223 Mm221+( )−kr2

kr1 kr2+⋅ 23.703−=:= kNm

M2r3 Mm323 Mm321+( )− 183.997−=:= kNm

M1r1 Mm112 Mm110+( )− 191.269−=:= kNm

M1r2sup Mm212 Mm210+( )−kr1

kr1 kr2+⋅ 10.003−=:= kNm

M1r2inf Mm212 Mm210+( )−kr2

kr1 kr2+⋅ 46.745−=:= kNm

M1r3 Mm312 Mm310+( )− 362.872−=:= kNm