REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica:...

164
IPERΣ01.bhng.002 REAZIONI 781715 Bao Hong Da @ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18 39/40F 1/2F 19/40Fb 1/40F 1/2F A B 1/2F 1/2F Fb 1/2F 1/2F 1/2Fb C D 79/40F 1/2F 5/2Fb 79/40F 1/2F 21/40Fb E C 1/40F 5/2F 5/2Fb 1/40F 5/2F E B 59/40F 19/40Fb 59/40F 19/40Fb A C 1/2F 1/2F Fb 1/2F 1/2F F A 1/2F 1/2F 1/2F 1/2F G F 1/2F 1/2F 1/2Fb 1/2F 1/2F D G IPERΣ01.bhng.002 AZIONI INTERNE 781715 Bao Hong Da @ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18 -1/2 -1/2 -1/2 -1/2 1/40 59/40 0 -1/2 -1/2 -1/2 F 39/40 -1/40 -1/2 -79/40 5/2 0 -2/2 -1/2 1/2 -1/2 F -19/40 0 1 1/2 5/2 21/40 -5/2 0 19/40 19/40 1 0 0 0 1/2 0 Fb

Transcript of REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica:...

Page 1: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.bhng.002REAZIONI 781715 Bao Hong Da

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

39/40F

1/2F19/40Fb

1/40F

1/2F

A

B

1/2F

1/2FFb

1/2F

1/2F1/2Fb

C

D

79/40F

1/2F5/2Fb

79/40F

1/2F21/40Fb

E

C

1/40F

5/2F5/2Fb

1/40F

5/2F

EB

59/40F19/40Fb

59/40F19/40Fb

A C

1/2F

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Fb

1/2F

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F

A

1/2F

1/2F

1/2F

1/2F

GF

1/2F

1/2F1/2Fb

1/2F

1/2F

DG

IPERΣ01.bhng.002AZIONI INTERNE 781715 Bao Hong Da

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

-1/2

-1/2

-1/2

-1/2

1/40

59/40

0

-1/2-1/2 -1/2

F

39/4

0-1

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5/2

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F

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1

0

00 1/20

Fb

Page 2: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

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Page 3: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.bhng.002PROCEDIMENTO E RISULTATI 781715 Bao Hong Da

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAB = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXBA = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXCE = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXAC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXCA = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoAB = ∫

o

b(1/2 x/b - x2/b2 +1/2 x3/b3 ) Fb 1/EJ dx = [1/4 x2/b -1/3 x3/b2 +1/8 x4/b3 ]o

b Fb 1/EJ

= (1/4 b -1/3 b +1/8 b ) Fb 1/EJ = 1/24 Fb2/EJ

LXoBA = ∫

o

b(1/2 x2/b2 -1/2 x3/b3 ) Fb 1/EJ dx = [1/6 x3/b2 -1/8 x4/b3 ]o

b Fb 1/EJ

= (1/6 b -1/8 b ) Fb 1/EJ = 1/24 Fb2/EJ

LXoEC = ∫

o

b(5/2 x/b -3/2 x2/b2 ) Fb 1/EJ dx = [5/4 x2/b -1/2 x3/b2 ]o

b Fb 1/EJ

= (5/4 b -1/2 b ) Fb 1/EJ = 3/4 Fb2/EJ

LXoCE = ∫

o

b(1 +1/2 x/b -3/2 x2/b2 ) Fb 1/EJ dx = [ x +1/4 x2/b -1/2 x3/b2 ]o

b Fb 1/EJ

= ( b +1/4 b -1/2 b ) Fb 1/EJ = 3/4 Fb2/EJ

IPERΣ01.bhng.002PROCEDIMENTO E RISULTATI 781715 Bao Hong Da

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 756. mm2

Ju = 165564. mm4

Jv = 74844. mm4

yg = 39. mmN = 15.25 NTy = 1525. NMx = -930250. Nmmxm = 18. mmum = -3. mmvm = -39. mmσm = N/A-Mv/Ju = -219.1 N/mm2

xc = 21. mmyc = 16. mmvc = -23. mmσc = N/A-Mv/Ju = -129.2 N/mm2

τc = 4.569 N/mm2

σo = √σ2+3τ2 = 129.5 N/mm2

S* = 2976. mm3mm 0 18 24 42x

0

42

54

y

16σc,τc

σm

u

v

Page 4: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.bhng.002

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.bhng.002

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 5: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.bnst.003REAZIONI 843782 Benassai Tommaso

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

3F

F2Fb

2F

F1/2Fb

A

B

4/5F4/5Fb

4/5F

C

D

4/5F

4F4Fb

4/5F

4F

ED

14/5F

F4Fb

14/5F

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E

A

FFb

F

F

C

1/5F4/5Fb

1/5F4/5Fb

C A

F

F1/2Fb

F

BG

FGF

IPERΣ01.bnst.003AZIONI INTERNE 843782 Benassai Tommaso

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

11

0

-4/5

1

-√2/2

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111

F

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0

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2

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0 40

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1

0

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1/2000

Fb

Page 6: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

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Page 7: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.bnst.003PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXCD = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXDC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEA = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXAE = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXCA = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXAC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoEA = ∫

o

b(-4 x/b +2 x2/b2 ) Fb 1/EJ dx = [-2 x2/b +2/3 x3/b2 ]o

b Fb 1/EJ

= (-2 b +2/3 b ) Fb 1/EJ = -4/3 Fb2/EJ

LXoAE = ∫

o

b(-2 +2 x2/b2 ) Fb 1/EJ dx = [-2 x +2/3 x3/b2 ]o

b Fb 1/EJ

= (-2 b +2/3 b ) Fb 1/EJ = -4/3 Fb2/EJ

IPERΣ01.bnst.003PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 864. mm2

Ju = 251424. mm4

Jv = 62208. mm4

yg = 33. mmN = -528. NTy = -2640. NMx = 1742400. Nmmxm = 18. mmum = -6. mmvm = -33. mmσm = N/A-Mv/Ju = 228.1 N/mm2

xc = 24. mmyc = 14. mmvc = -19. mmσc = N/A-Mv/Ju = 131.1 N/mm2

τc = 3.822 N/mm2

σo = √σ2+3τ2 = 131.2 N/mm2

S* = 4368. mm3mm 0 18 30 48x

0

48

54

y

14σc,τc

σm

u

v

Page 8: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.bnst.003

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.bnst.003

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 9: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.brts.004REAZIONI 811986 Beretta Stefano

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

3F

11/20F5/2Fb

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11/20F

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F

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IPERΣ01.brts.004AZIONI INTERNE 811986 Beretta Stefano

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

11/2

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1/20

Fb

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2

01/

3Xb/

EJ

BE

bx/

b0

0x2 /b

2

DF

b0

1/2F

b-F

x+1/

2qx2

00

00

FD

b0

-1/2

qx2

00

FG

b0

00

00

0G

F b

00

00

GE

√2b

0F

b-√2

/2F

x0

00

0

EC

b1

00

10

Xb/

EJ

CE

b-1

00

1

tota

li11

/12F

b2 /EJ

5/3X

b/E

J

iper

stat

ica

X=

WE

B-1

1/20

Fb

Svi

lupp

i di c

alco

lo ip

erst

atic

a

Page 11: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.brts.004PROCEDIMENTO E RISULTATI 811986 Beretta Stefano

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXCA = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXEB = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXBE = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXCE = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoAC = ∫

o

b(5/2 x/b - x2/b2 ) Fb 1/EJ dx = [5/4 x2/b -1/3 x3/b2 ]o

b Fb 1/EJ

= (5/4 b -1/3 b ) Fb 1/EJ = 11/12 Fb2/EJ

LXoCA = ∫

o

b(3/2 -1/2 x/b - x2/b2 ) Fb 1/EJ dx = [3/2 x -1/4 x2/b -1/3 x3/b2 ]o

b Fb 1/EJ

= (3/2 b -1/4 b -1/3 b ) Fb 1/EJ = 11/12 Fb2/EJ

IPERΣ01.brts.004PROCEDIMENTO E RISULTATI 811986 Beretta Stefano

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 1080. mm2

Ju = 276955. mm4

Jv = 116640. mm4

yg = 35.4 mmN = 577.5 NTy = -3150. NMx = 1863750. Nmmxm = 18. mmum = -6. mmvm = -35.4 mmσm = N/A-Mv/Ju = 238.8 N/mm2

xc = 24. mmyc = 15. mmvc = -20.4 mmσc = N/A-Mv/Ju = 137.8 N/mm2

τc = 4.76 N/mm2

σo = √σ2+3τ2 = 138.1 N/mm2

S* = 5022. mm3mm 0 18 30 48x

0

42

54

y

15σc,τc

σm

u

v

Page 12: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.brts.004

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.brts.004

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 13: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.brmm.006REAZIONI 829837 Bormolini Matteo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

F

F

F1/2Fb

AB

1/5F1/5Fb

1/5F

C D

2F

1/5F2Fb

2F

1/5F

E

D

F

11/5F2Fb

F

11/5F1/5Fb

EA

F

Fb

F

F

C

6/5F1/5Fb

6/5F1/5Fb

C

A

F

F1/2Fb

F

B

G

F

G

F

IPERΣ01.brmm.006AZIONI INTERNE 829837 Bormolini Matteo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

11

0

-1/5

1

√2/2

6/5

-1-1

-1

F

-10

1/5

2

-11/5

-√2/2

0

10

0

F

0-1/2

-1/50

-20

2-1/5

1

0

1/5

1/5-1/2

00

0

Fb

Page 14: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPE

RΣ0

1.br

mm

.006

PR

OC

ED

IME

NT

O E

RIS

ULT

AT

I 829

837

Bor

mol

ini M

atte

o

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

AB

C

D

E

F

G

F

W

X

X

q

0-1

/2

00

-20

20

1

0

00

-1/20 0 0

Mo

fless

ione

da

caric

hi a

sseg

nati

00

-10

0 0

0-1

0

0

11

0 0 0 0

Mx

fless

ione

da

iper

stat

ica

X=

1

IPE

RΣ0

1.br

mm

.006

PR

OC

ED

IME

NT

O E

RIS

ULT

AT

I 829

837

Bor

mol

ini M

atte

o

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

Qua

dro

cont

ribut

i PLV

per

iper

stat

ica

X=

WC

D

→M

x(x)

Mo(

x)M

xMo

MxM

x∫M

xMo/

EJd

x∫X

MxM

x/E

Jdx

AB

b0

-Fx+

1/2q

x20

00

0B

A b

01/

2Fb-

1/2q

x20

0

CD

b-1

+x/

b0

01-

2x/b

+x2 /b

2

01/

3Xb/

EJ

DC

bx/

b0

0x2 /b

2

ED

b0

-2F

b+2F

x0

00

0D

E b

02F

x0

0

EA

b-x

/b2F

b-2F

x-2

Fx+

2Fx2 /b

x2 /b2

-1/3

Fb2 /E

J1/

3Xb/

EJ

AE

b1-

x/b

-2F

x-2

Fx+

2Fx2 /b

1-2x

/b+

x2 /b2

FC

√2b

0F

b-√2

/2F

x0

00

0

CA

b1

00

10

Xb/

EJ

AC

b-1

00

1

BG

b0

-1/2

Fb+

Fx-

1/2q

x20

00

0G

B b

01/

2qx2

00

GF

b0

00

00

0F

G b

00

00

tota

li-1

/3F

b2 /EJ

5/3X

b/E

J

iper

stat

ica

X=

WC

D1/

5Fb

Svi

lupp

i di c

alco

lo ip

erst

atic

a

Page 15: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.brmm.006PROCEDIMENTO E RISULTATI 829837 Bormolini Matteo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXCD = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXDC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEA = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXAE = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXCA = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXAC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoEA = ∫

o

b(-2 x/b +2 x2/b2 ) Fb 1/EJ dx = [- x2/b +2/3 x3/b2 ]o

b Fb 1/EJ

= (- b +2/3 b ) Fb 1/EJ = -1/3 Fb2/EJ

LXoAE = ∫

o

b(-2 x/b +2 x2/b2 ) Fb 1/EJ dx = [- x2/b +2/3 x3/b2 ]o

b Fb 1/EJ

= (- b +2/3 b ) Fb 1/EJ = -1/3 Fb2/EJ

IPERΣ01.brmm.006PROCEDIMENTO E RISULTATI 829837 Bormolini Matteo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 612. mm2

Ju = 149427. mm4

Jv = 27756. mm4

yg = 36.88 mmN = -104. NTy = 1040. NMx = -842400. Nmmxm = 12. mmum = -3. mmvm = -36.88 mmσm = N/A-Mv/Ju = -208.1 N/mm2

xc = 15. mmyc = 16. mmvc = -20.88 mmσc = N/A-Mv/Ju = -117.9 N/mm2

τc = 3.216 N/mm2

σo = √σ2+3τ2 = 118. N/mm2

S* = 2773. mm3mm 0 12 18 30x

0

42

54

y

16σc,τc

σm

u

v

Page 16: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.brmm.006

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.brmm.006

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 17: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.btts.007REAZIONI 808392 Bottini Stefano

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

109/40F

1/2F7/2Fb

69/40F

1/2F51/40Fb

A

B

29/40F

7/2F7/2Fb

29/40F

7/2F

A C

29/40F

1/2F29/40Fb

29/40F

1/2F

D

C

3/2F

1/2F2Fb

3/2F

1/2F1/2Fb

B

E

3/2F

1/2F

3/2F

1/2F

F G

3/2F

1/2F1/2Fb

3/2F

1/2F

E F

9/40F29/40Fb

9/40F29/40Fb

DB

3/2F

1/2F

Fb

3/2F

1/2F

G

D

IPERΣ01.btts.007AZIONI INTERNE 808392 Bottini Stefano

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1/2

1/2

29/40

1/2

1/2

-3/2 -3/2-3/2

-9/40

√2

F

109/

4069

/40

-7/2

-29/

40

3/2

1/2-1/2

1/2

0

-√2/2

F

-7/2

-51/

40

7/2 0

29/4

00

-2-1

/2

0 0-1/2

0

-29/40-29/40

1

0

Fb

Page 18: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPE

RΣ0

1.bt

ts.0

07P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 8

0839

2 B

ottin

i Ste

fano

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

A B

C

D

EF

G

F

W

X

X

q

q

Sch

ema

di c

alco

lo ip

erst

atic

o

-7/2 -2

7/2

0 00

-2 -1/2

00

-1/2

000

1

0

Mo

fless

ione

da

caric

hi a

sseg

nati

0-1

00

-10

0 0

00

001

1

0

0

Mx

fless

ione

da

iper

stat

ica

X=

1

IPE

RΣ0

1.bt

ts.0

07P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 8

0839

2 B

ottin

i Ste

fano

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

Quadro contributi PLV per iperstatica X=WBD

→Mx(x)Mo(x)MxMoMxMx∫MxMo/EJdx∫XMxMx/EJdx

AB b-x/b-7/2Fb+2Fx-1/2qx2

7/2Fx-2Fx2/b+1/2qx

3/bx

2/b

2

29/24Fb2/EJ1/3Xb/EJ

BA b1-x/b2Fb+Fx+1/2qx2

2Fb-Fx-1/2Fx2/b-1/2qx

3/b1-2x/b+x

2/b

2

AC b07/2Fb-7/2Fx0000

CA b0-7/2Fx00

DC b-1+x/b001-2x/b+x2/b

2

01/3Xb/EJCD bx/b00x

2/b

2

BE b0-2Fb+3/2Fx0000

EB b01/2Fb+3/2Fx00

FG b01/2Fx-1/2qx2

0000

GF b0-1/2Fx+1/2qx2

00

EF b0-1/2Fb+1/2Fx0000

FE b01/2Fx00

DB b10010Xb/EJ

BD b-1001

GD √2b0Fb-√2/2Fx0000

totali29/24Fb2/EJ5/3Xb/EJ

iperstatica X=WBD-29/40Fb

Svi

lupp

i di c

alco

lo ip

erst

atic

a

Page 19: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.btts.007PROCEDIMENTO E RISULTATI 808392 Bottini Stefano

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAB = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXBA = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXDC = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXCD = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXDB = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXBD = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoAB = ∫

o

b(7/2 x/b -2 x2/b2 +1/2 x3/b3 ) Fb 1/EJ dx = [7/4 x2/b -2/3 x3/b2 +1/8 x4/b3 ]o

b Fb 1/EJ

= (7/4 b -2/3 b +1/8 b ) Fb 1/EJ = 29/24 Fb2/EJ

LXoBA = ∫

o

b(2 - x/b -1/2 x2/b2 -1/2 x3/b3 ) Fb 1/EJ dx = [2 x -1/2 x2/b -1/6 x3/b2 -1/8 x4/b3 ]o

b Fb 1/EJ

= (2 b -1/2 b -1/6 b -1/8 b ) Fb 1/EJ = 29/24 Fb2/EJ

IPERΣ01.btts.007PROCEDIMENTO E RISULTATI 808392 Bottini Stefano

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 792. mm2

Ju = 225759. mm4

Jv = 30240. mm4

yg = 31.36 mmN = 377. NTy = -1820. NMx = 1565200. Nmmxm = 12. mmum = -6. mmvm = -31.36 mmσm = N/A-Mv/Ju = 217.9 N/mm2

xc = 18. mmyc = 13. mmvc = -18.36 mmσc = N/A-Mv/Ju = 127.8 N/mm2

τc = 2.606 N/mm2

σo = √σ2+3τ2 = 127.9 N/mm2

S* = 3879. mm3mm 0 12 24 36x

0

48

54

y

13σc,τc

σm

u

v

Page 20: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.btts.007

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.btts.007

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 21: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.clnl.008REAZIONI 832133 Calonico Lorenzo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

5/2F

1/2F3/2Fb

3/2F

1/2F1/2Fb

A

B

13/20F

1/2F13/20Fb

13/20F

1/2F

C

D

13/20F

7/2F7/2Fb

13/20F

7/2F

ED

53/20F

1/2F7/2Fb

53/20F

1/2F17/20Fb

E

A

1/2F

1/2F

Fb

1/2F

1/2F

F

C

3/20F13/20Fb

3/20F13/20Fb

C A

1/2F

1/2F1/2Fb

1/2F

1/2F

BG

1/2F

1/2F

1/2F

1/2F

GF

IPERΣ01.clnl.008AZIONI INTERNE 832133 Calonico Lorenzo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1/2

1/2

1/2

-13/20

1/2

0

3/20

1/21/21/2

F

5/2

3/2

-13/

20

-7/2

53/2

0

-√2/2

0

-1/2-1/21/2

F

-3/2

1/2

13/2

00 7/20

-7/2

-17/

20

1

0

-13/20 -13/20

1/2000

Fb

Page 22: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPE

RΣ0

1.cl

nl.0

08P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 8

3213

3 C

alon

ico

Lore

nzo

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

A

B

CDE

FG

F

X

X

q

q

-3/21/2

00

7/2

0

-7/2 -3/2

1

0

00 1/

20

00

Mo

fless

ione

da

caric

hi a

sseg

nati

0 0

10

00

0 1

0

0

-1-1

00

00

Mx

fless

ione

da

iper

stat

ica

X=

1

IPE

RΣ0

1.cl

nl.0

08P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 8

3213

3 C

alon

ico

Lore

nzo

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

Qua

dro

cont

ribut

i PLV

per

iper

stat

ica

X=

WC

A

→M

x(x)

Mo(

x)M

xMo

MxM

x∫M

xMo/

EJd

x∫X

MxM

x/E

Jdx

AB

b0

-3/2

Fb+

5/2F

x-1/

2qx2

00

00

BA

b0

-1/2

Fb+

3/2F

x+1/

2qx2

00

CD

b1-

x/b

00

1-2x

/b+

x2 /b2

01/

3Xb/

EJ

DC

b-x

/b0

0x2 /b

2

ED

b0

7/2F

b-7/

2Fx

00

00

DE

b0

-7/2

Fx

00

EA

bx/

b-7

/2F

b+2F

x-7

/2F

x+2F

x2 /bx2 /b

2

-13/

12F

b2 /EJ

1/3X

b/E

JA

E b

-1+

x/b

3/2F

b+2F

x-3

/2F

b-1/

2Fx+

2Fx2 /b

1-2x

/b+

x2 /b2

FC

√2b

0F

b-√2

/2F

x0

00

0

CA

b-1

00

10

Xb/

EJ

AC

b1

00

1

BG

b0

1/2F

b-1/

2Fx

00

00

GB

b0

-1/2

Fx

00

GF

b0

-1/2

Fx+

1/2q

x20

00

0F

G b

01/

2Fx-

1/2q

x20

0

tota

li-1

3/12

Fb2 /E

J5/

3Xb/

EJ

iper

stat

ica

X=

WC

A13

/20F

b

Svi

lupp

i di c

alco

lo ip

erst

atic

a

Page 23: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.clnl.008PROCEDIMENTO E RISULTATI 832133 Calonico Lorenzo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXCD = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXDC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEA = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXAE = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXCA = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXAC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoEA = ∫

o

b(-7/2 x/b +2 x2/b2 ) Fb 1/EJ dx = [-7/4 x2/b +2/3 x3/b2 ]o

b Fb 1/EJ

= (-7/4 b +2/3 b ) Fb 1/EJ = -13/12 Fb2/EJ

LXoAE = ∫

o

b(-3/2 -1/2 x/b +2 x2/b2 ) Fb 1/EJ dx = [-3/2 x -1/4 x2/b +2/3 x3/b2 ]o

b Fb 1/EJ

= (-3/2 b -1/4 b +2/3 b ) Fb 1/EJ = -13/12 Fb2/EJ

IPERΣ01.clnl.008PROCEDIMENTO E RISULTATI 832133 Calonico Lorenzo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 936. mm2

Ju = 248849. mm4

Jv = 52704. mm4

yg = 33.46 mmN = -344.5 NTy = -1855. NMx = 1688050. Nmmxm = 12. mmum = -6. mmvm = -33.46 mmσm = N/A-Mv/Ju = 226.6 N/mm2

xc = 18. mmyc = 14. mmvc = -19.46 mmσc = N/A-Mv/Ju = 131.6 N/mm2

τc = 2.762 N/mm2

σo = √σ2+3τ2 = 131.7 N/mm2

S* = 4446. mm3mm 0 12 24 36x

0

42

54

y

14σc,τc

σm

u

v

Page 24: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.clnl.008

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.clnl.008

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 25: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.cmna.009REAZIONI 867174 Comana Alberto

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1/2F

109/40F7/2Fb

1/2F

69/40F51/40Fb

A B

7/2F

29/40F7/2Fb

7/2F

29/40F

A

C

1/2F

29/40F29/40Fb

1/2F

29/40F

DC

1/2F

3/2F2Fb

1/2F

3/2F1/2Fb

B E

1/2F

3/2F

1/2F

3/2F

F

G

1/2F

3/2F1/2Fb

1/2F

3/2F

E

F

9/40F29/40Fb

9/40F29/40Fb

D

B

1/2F

3/2F

Fb

1/2F

3/2F

G

D

IPERΣ01.cmna.009AZIONI INTERNE 867174 Comana Alberto

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

-1/2 -1/2

-29/

40

-1/2

-1/2

3/2

3/2

3/2

9/40

-√2

F

109/40 69/40

-7/2

-29/40

3/2

1/2

-1/2

1/20

-√2/2

F

-7/2 -51/40

7/2

0

29/400

-2 -1/2

00

-1/2

0-29/

40-2

9/40

1

0

Fb

Page 26: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPE

RΣ0

1.cm

na.0

09P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 8

6717

4 C

oman

a A

lber

to

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

AB

C

D

E

F

G

F

W

X

X

q

q

Sch

ema

di c

alco

lo ip

erst

atic

o

-7/2

-2

7/2 0

00

-2-1

/2

0 0-1/2

0

00

1

0

Mo

fless

ione

da

caric

hi a

sseg

nati

0-1

0 0

-10

00

0 00 0

11

0

0

Mx

fless

ione

da

iper

stat

ica

X=

1

IPE

RΣ0

1.cm

na.0

09P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 8

6717

4 C

oman

a A

lber

to

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

Quadro contributi PLV per iperstatica X=WBD

→Mx(x)Mo(x)MxMoMxMx∫MxMo/EJdx∫XMxMx/EJdx

AB b-x/b-7/2Fb+2Fx-1/2qx2

7/2Fx-2Fx2/b+1/2qx

3/bx

2/b

2

29/24Fb2/EJ1/3Xb/EJ

BA b1-x/b2Fb+Fx+1/2qx2

2Fb-Fx-1/2Fx2/b-1/2qx

3/b1-2x/b+x

2/b

2

AC b07/2Fb-7/2Fx0000

CA b0-7/2Fx00

DC b-1+x/b001-2x/b+x2/b

2

01/3Xb/EJCD bx/b00x

2/b

2

BE b0-2Fb+3/2Fx0000

EB b01/2Fb+3/2Fx00

FG b01/2Fx-1/2qx2

0000

GF b0-1/2Fx+1/2qx2

00

EF b0-1/2Fb+1/2Fx0000

FE b01/2Fx00

DB b10010Xb/EJ

BD b-1001

GD √2b0Fb-√2/2Fx0000

totali29/24Fb2/EJ5/3Xb/EJ

iperstatica X=WBD-29/40Fb

Svi

lupp

i di c

alco

lo ip

erst

atic

a

Page 27: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.cmna.009PROCEDIMENTO E RISULTATI 867174 Comana Alberto

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAB = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXBA = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXDC = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXCD = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXDB = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXBD = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoAB = ∫

o

b(7/2 x/b -2 x2/b2 +1/2 x3/b3 ) Fb 1/EJ dx = [7/4 x2/b -2/3 x3/b2 +1/8 x4/b3 ]o

b Fb 1/EJ

= (7/4 b -2/3 b +1/8 b ) Fb 1/EJ = 29/24 Fb2/EJ

LXoBA = ∫

o

b(2 - x/b -1/2 x2/b2 -1/2 x3/b3 ) Fb 1/EJ dx = [2 x -1/2 x2/b -1/6 x3/b2 -1/8 x4/b3 ]o

b Fb 1/EJ

= (2 b -1/2 b -1/6 b -1/8 b ) Fb 1/EJ = 29/24 Fb2/EJ

IPERΣ01.cmna.009PROCEDIMENTO E RISULTATI 867174 Comana Alberto

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 936. mm2

Ju = 248849. mm4

Jv = 52704. mm4

yg = 20.54 mmN = -384.3 NTy = -1855. NMx = 1780800. Nmmxm = 24. mmym = 54. mmum = 6. mmvm = 33.46 mmσm = N/A-Mv/Ju = -239.9 N/mm2

xc = 18. mmyc = 40. mmvc = 19.46 mmσc = N/A-Mv/Ju = -139.7 N/mm2

τc = 2.762 N/mm2

σo = √σ2+3τ2 = 139.8 N/mm2

S* = 4446. mm3mm 0 12 24 36x

0

12

54

y

40σc,τc

σm

u

v

Page 28: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.cmna.009

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.cmna.009

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 29: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.dmns.011REAZIONI 817109 Damian Sebastiano

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1/5F

1/2F

1/5F

1/2F

A B

4/5F

1/2F

4/5F

1/2F4/5Fb

A

C

1/2F

1/2FFb

1/2F

1/2F1/2Fb

C

D

1/5F

1/2F1/5Fb

1/5F

1/2F

E

B

3/2F

1/2F1/2Fb

3/2F

1/2F

D F

3/2F

1/2F

3/2F

1/2F

F G

3/2F

1/2F

Fb

3/2F

1/2F

G

E

13/10F1/5Fb

13/10F1/5Fb

EC

IPERΣ01.dmns.011AZIONI INTERNE 817109 Damian Sebastiano

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

-1/5 -1/5

-1/2

-1/2

-1/2

3/2 3/2 3/2

-√2

-13/10

F

1/2-1/2

-4/5

1/2

-1/5

1/2 1/2-1/2

-√2/2

0

F

0 00-4

/5-1

-1/2

1/5

0

-1/20 0 0

1

0

-1/5-1/5

Fb

Page 30: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPE

RΣ0

1.dm

ns.0

11P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 8

1710

9 D

amia

n S

ebas

tiano

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

AB

C

D

E

F

G

F

W

X

X

q

q

Sch

ema

di c

alco

lo ip

erst

atic

o

00

0-1-1-1/2

0 0

-1/2

00

0

1

0

00

Mo

fless

ione

da

caric

hi a

sseg

nati

00

0-1

00

-10

00

00

0

0

11

Mx

fless

ione

da

iper

stat

ica

X=

1

IPE

RΣ0

1.dm

ns.0

11P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 8

1710

9 D

amia

n S

ebas

tiano

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

Qua

dro

cont

ribut

i PLV

per

iper

stat

ica

X=

WC

E

→M

x(x)

Mo(

x)M

xMo

MxM

x∫M

xMo/

EJd

x∫X

MxM

x/E

Jdx

AB

b0

1/2F

x-1/

2qx2

00

00

BA

b0

-1/2

Fx+

1/2q

x20

0

AC

b-x

/b-F

xF

x2 /bx2 /b

2

1/3F

b2 /EJ

1/3X

b/E

JC

A b

1-x/

bF

b-F

xF

b-2F

x+F

x2 /b1-

2x/b

+x2 /b

2

CD

b0

-Fb+

1/2F

x0

00

0D

C b

01/

2Fb+

1/2F

x0

0

EB

b-1

+x/

b0

01-

2x/b

+x2 /b

2

01/

3Xb/

EJ

BE

bx/

b0

0x2 /b

2

DF

b0

-1/2

Fb+

1/2F

x0

00

0F

D b

01/

2Fx

00

FG

b0

1/2F

x-1/

2qx2

00

00

GF

b0

-1/2

Fx+

1/2q

x20

0

GE

√2b

0F

b-√2

/2F

x0

00

0

EC

b1

00

10

Xb/

EJ

CE

b-1

00

1

tota

li1/

3Fb2 /E

J5/

3Xb/

EJ

iper

stat

ica

X=

WC

E-1

/5F

b

Svi

lupp

i di c

alco

lo ip

erst

atic

a

Page 31: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.dmns.011PROCEDIMENTO E RISULTATI 817109 Damian Sebastiano

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXCA = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXEB = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXBE = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXCE = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoAC = ∫

o

b( x2/b2 ) Fb 1/EJ dx = [1/3 x3/b2 ]o

b Fb 1/EJ

= (1/3 b ) Fb 1/EJ = 1/3 Fb2/EJ

LXoCA = ∫

o

b(1 -2 x/b + x2/b2 ) Fb 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b Fb 1/EJ

= ( b - b +1/3 b ) Fb 1/EJ = 1/3 Fb2/EJ

IPERΣ01.dmns.011PROCEDIMENTO E RISULTATI 817109 Damian Sebastiano

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 864. mm2

Ju = 251424. mm4

Jv = 62208. mm4

yg = 21. mmN = -3946. NTy = -1973. NMx = 1562400. Nmmxm = 30. mmym = 54. mmum = 6. mmvm = 33. mmσm = N/A-Mv/Ju = -209.6 N/mm2

xc = 24. mmyc = 40. mmvc = 19. mmσc = N/A-Mv/Ju = -122.6 N/mm2

τc = 2.856 N/mm2

σo = √σ2+3τ2 = 122.7 N/mm2

S* = 4368. mm3mm 0 18 30 48x

0

6

54

y

40σc,τc

σm

u

v

Page 32: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.dmns.011

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.dmns.011

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 33: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.dnsg.012REAZIONI 845411 Danesi Gabriele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

103/40F

F3Fb

63/40F

F37/40Fb

A

B

23/40F

3F3Fb

23/40F

3F

A C

23/40F23/40Fb

23/40F

D

C

F

F3/2Fb

F

F1/2Fb

B

E

FF G

F

F1/2Fb

F

E F

23/40F23/40Fb

23/40F23/40Fb

DB

FFb

F

G

D

IPERΣ01.dnsg.012AZIONI INTERNE 845411 Danesi Gabriele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

11

23/40

0

1

-1-1 -1

-23/40

√2/2

F

103/

4063

/40

-3

-23/

40

1

01 0

0

-√2/2

F

-3-3

7/40

3 0

23/4

00

-3/2

-1/2

0 0-1/2

0

-23/40-23/40

1

0

Fb

Page 34: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPE

RΣ0

1.dn

sg.0

12P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 8

4541

1 D

anes

i Gab

riele

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

A B

C

D

EF

G

F

W

X

X

q

q

-3 -3/2

30 00

-3/2 -1/2

00

-1/2

000

1

0

Mo

fless

ione

da

caric

hi a

sseg

nati

0-1

00

-10

0 0

00

001

1

0

0

Mx

fless

ione

da

iper

stat

ica

X=

1

IPE

RΣ0

1.dn

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12P

RO

CE

DIM

EN

TO

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LTA

TI 8

4541

1 D

anes

i Gab

riele

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

Quadro contributi PLV per iperstatica X=WBD

→Mx(x)Mo(x)MxMoMxMx∫MxMo/EJdx∫XMxMx/EJdx

AB b-x/b-3Fb+2Fx-1/2qx2

3Fx-2Fx2/b+1/2qx

3/bx

2/b

2

23/24Fb2/EJ1/3Xb/EJ

BA b1-x/b3/2Fb+Fx+1/2qx2

3/2Fb-1/2Fx-1/2Fx2/b-1/2qx

3/b1-2x/b+x

2/b

2

AC b03Fb-3Fx0000

CA b0-3Fx00

DC b-1+x/b001-2x/b+x2/b

2

01/3Xb/EJCD bx/b00x

2/b

2

BE b0-3/2Fb+Fx0000

EB b01/2Fb+Fx00

FG b000000

GF b0000

EF b0-1/2Fb+Fx-1/2qx2

0000

FE b01/2qx2

00

DB b10010Xb/EJ

BD b-1001

GD √2b0Fb-√2/2Fx0000

totali23/24Fb2/EJ5/3Xb/EJ

iperstatica X=WBD-23/40Fb

Svi

lupp

i di c

alco

lo ip

erst

atic

a

Page 35: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.dnsg.012PROCEDIMENTO E RISULTATI 845411 Danesi Gabriele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAB = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXBA = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXDC = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXCD = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXDB = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXBD = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoAB = ∫

o

b(3 x/b -2 x2/b2 +1/2 x3/b3 ) Fb 1/EJ dx = [3/2 x2/b -2/3 x3/b2 +1/8 x4/b3 ]o

b Fb 1/EJ

= (3/2 b -2/3 b +1/8 b ) Fb 1/EJ = 23/24 Fb2/EJ

LXoBA = ∫

o

b(3/2 -1/2 x/b -1/2 x2/b2 -1/2 x3/b3 ) Fb 1/EJ dx = [3/2 x -1/4 x2/b -1/6 x3/b2 -1/8 x4/b3 ]o

b Fb 1/EJ

= (3/2 b -1/4 b -1/6 b -1/8 b ) Fb 1/EJ = 23/24 Fb2/EJ

IPERΣ01.dnsg.012PROCEDIMENTO E RISULTATI 845411 Danesi Gabriele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 612. mm2

Ju = 149428. mm4

Jv = 27756. mm4

yg = 17.12 mmN = 276. NTy = -1440. NMx = 878400. Nmmxm = 18. mmym = 54. mmum = 3. mmvm = 36.88 mmσm = N/A-Mv/Ju = -216.4 N/mm2

xc = 15. mmyc = 38. mmvc = 20.88 mmσc = N/A-Mv/Ju = -122.3 N/mm2

τc = 4.453 N/mm2

σo = √σ2+3τ2 = 122.5 N/mm2

S* = 2773. mm3mm 0 12 18 30x

0

12

54

y

38σc,τc

σm

u

v

Page 36: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.dnsg.012

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.dnsg.012

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 37: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.dvng.013REAZIONI 774590 D’Avino Guido

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

33/40F13/40Fb

7/40F

A

B

F1/2Fb

F1/2Fb

C

D

73/40F

F2Fb

73/40F

F7/40Fb

E

C

7/40F

2F2Fb

7/40F

2F

EB

73/40F13/40Fb

73/40F13/40Fb

A C

FFb

F

F

A

FGF F

F1/2Fb

F

DG

IPERΣ01.dvng.013AZIONI INTERNE 774590 D’Avino Guido

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

00

-1-1

7/40

73/40

√2/2

-1 -1-1

F

33/4

0-7

/40

0-7

3/40

2

0

-√2/2

0-1

0

F

-13/

400

1/2

1/2

27/

40

-20

13/40 13/40

1

0

00 1/20

Fb

Page 38: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPE

RΣ0

1.dv

ng.0

13P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 7

7459

0 D

’Avi

no G

uido

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

A

B

C

D E

FG

F

W

XX

q

q

0 0

1/21/2 21/2

-200

0

1

000

1/2

0

Mo

fless

ione

da

caric

hi a

sseg

nati

1 0

00 01 00

-1-1

0

000

00

Mx

fless

ione

da

iper

stat

ica

X=

1

IPE

RΣ0

1.dv

ng.0

13P

RO

CE

DIM

EN

TO

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ISU

LTA

TI 7

7459

0 D

’Avi

no G

uido

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

Qua

dro

cont

ribut

i PLV

per

iper

stat

ica

X=

WA

C

→M

x(x)

Mo(

x)M

xMo

MxM

x∫M

xMo/

EJd

x∫X

MxM

x/E

Jdx

AB

b1-

x/b

1/2F

x-1/

2qx2

1/2F

x-F

x2 /b+

1/2q

x3 /b1-

2x/b

+x2 /b

2

1/24

Fb2 /E

J1/

3Xb/

EJ

BA

b-x

/b-1

/2F

x+1/

2qx2

1/2F

x2 /b-1

/2qx

3 /bx2 /b

2

CD

b0

1/2F

b0

00

0D

C b

0-1

/2F

b0

0

EC

bx/

b2F

b-3/

2Fx

2Fx-

3/2F

x2 /bx2 /b

2

1/2F

b2 /EJ

1/3X

b/E

JC

E b

-1+

x/b

-1/2

Fb-

3/2F

x1/

2Fb+

Fx-

3/2F

x2 /b1-

2x/b

+x2 /b

2

EB

b0

-2F

b+2F

x0

00

0B

E b

02F

x0

0

AC

b-1

00

10

Xb/

EJ

CA

b1

00

1

FA

√2b

0F

b-√2

/2F

x0

00

0

GF

b0

00

00

0F

G b

00

00

DG

b0

1/2F

b-F

x+1/

2qx2

00

00

GD

b0

-1/2

qx2

00

tota

li13

/24F

b2 /EJ

5/3X

b/E

J

iper

stat

ica

X=

WA

C-1

3/40

Fb

Svi

lupp

i di c

alco

lo ip

erst

atic

a

Page 39: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.dvng.013PROCEDIMENTO E RISULTATI 774590 D’Avino Guido

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAB = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXBA = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXCE = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXAC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXCA = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoAB = ∫

o

b(1/2 x/b - x2/b2 +1/2 x3/b3 ) Fb 1/EJ dx = [1/4 x2/b -1/3 x3/b2 +1/8 x4/b3 ]o

b Fb 1/EJ

= (1/4 b -1/3 b +1/8 b ) Fb 1/EJ = 1/24 Fb2/EJ

LXoBA = ∫

o

b(1/2 x2/b2 -1/2 x3/b3 ) Fb 1/EJ dx = [1/6 x3/b2 -1/8 x4/b3 ]o

b Fb 1/EJ

= (1/6 b -1/8 b ) Fb 1/EJ = 1/24 Fb2/EJ

LXoEC = ∫

o

b(2 x/b -3/2 x2/b2 ) Fb 1/EJ dx = [ x2/b -1/2 x3/b2 ]o

b Fb 1/EJ

= ( b -1/2 b ) Fb 1/EJ = 1/2 Fb2/EJ

LXoCE = ∫

o

b(1/2 + x/b -3/2 x2/b2 ) Fb 1/EJ dx = [1/2 x +1/2 x2/b -1/2 x3/b2 ]o

b Fb 1/EJ

= (1/2 b +1/2 b -1/2 b ) Fb 1/EJ = 1/2 Fb2/EJ

IPERΣ01.dvng.013PROCEDIMENTO E RISULTATI 774590 D’Avino Guido

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 756. mm2

Ju = 165564. mm4

Jv = 74844. mm4

yg = 15. mmN = 127.8 NTy = 1460. NMx = -963600. Nmmxm = 24. mmym = 54. mmum = 3. mmvm = 39. mmσm = N/A-Mv/Ju = 227.2 N/mm2

xc = 21. mmyc = 38. mmvc = 23. mmσc = N/A-Mv/Ju = 134. N/mm2

τc = 4.374 N/mm2

σo = √σ2+3τ2 = 134.2 N/mm2

S* = 2976. mm3mm 0 18 24 42x

0

12

54

y

38σc,τc

σm

u

v

Page 40: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.dvng.013

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.dvng.013

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 41: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.dvnl.014REAZIONI 867268 D’Avino Luca

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

5/2F

2/5F2Fb

3/2F

2/5F

A

B

1/2F

7/5F2Fb

1/2F

7/5F3/5Fb

A C

1/2F

3/2FFb

1/2F

3/2F1/2Fb

C D

1/2F

2/5F2/5Fb

1/2F

2/5F

EB

1/2F

1/2F1/2Fb

1/2F

1/2F

D

F

1/2F

1/2F

1/2F

1/2F

F

G

1/2F

1/2F

Fb

1/2F

1/2F

G

E

1/10F2/5Fb

1/10F2/5Fb

E

C

IPERΣ01.dvnl.014AZIONI INTERNE 867268 D’Avino Luca

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

2/5

2/5

-1/2 -1/2

-1/2

-1/2

-1/2

-1/2

0

1/10

F

-5/2

-3/2

7/5 3/2

-2/5

-1/2

-1/2

1/2

-√2/2

0

F

20

-2 -3/5 -11/2

2/50

1/2

00

01

0

-2/5

-2/5

Fb

Page 42: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPE

RΣ0

1.dv

nl.0

14P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 8

6726

8 D

’Avi

no L

uca

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

A

B

CD

EFG F

W

X X

q

q

20

-2-1

-11/

2

00

1/2000

1

0 0 0

Mo

fless

ione

da

caric

hi a

sseg

nati

00

0-1

00

-10

0000

0

0 1 1

Mx

fless

ione

da

iper

stat

ica

X=

1

IPE

RΣ0

1.dv

nl.0

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RO

CE

DIM

EN

TO

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TI 8

6726

8 D

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no L

uca

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

Qua

dro

cont

ribut

i PLV

per

iper

stat

ica

X=

WC

E

→M

x(x)

Mo(

x)M

xMo

MxM

x∫M

xMo/

EJd

x∫X

MxM

x/E

Jdx

AB

b0

2Fb-

5/2F

x+1/

2qx2

00

00

BA

b0

-3/2

Fx-

1/2q

x20

0

AC

b-x

/b-2

Fb+

Fx

2Fx-

Fx2 /b

x2 /b2

2/3F

b2 /EJ

1/3X

b/E

JC

A b

1-x/

bF

b+F

xF

b-F

x2 /b1-

2x/b

+x2 /b

2

CD

b0

-Fb+

3/2F

x0

00

0D

C b

0-1

/2F

b+3/

2Fx

00

EB

b-1

+x/

b0

01-

2x/b

+x2 /b

2

01/

3Xb/

EJ

BE

bx/

b0

0x2 /b

2

DF

b0

1/2F

b-1/

2Fx

00

00

FD

b0

-1/2

Fx

00

FG

b0

-1/2

Fx+

1/2q

x20

00

0G

F b

01/

2Fx-

1/2q

x20

0

GE

√2b

0F

b-√2

/2F

x0

00

0

EC

b1

00

10

Xb/

EJ

CE

b-1

00

1

tota

li2/

3Fb2 /E

J5/

3Xb/

EJ

iper

stat

ica

X=

WC

E-2

/5F

b

Svi

lupp

i di c

alco

lo ip

erst

atic

a

Page 43: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.dvnl.014PROCEDIMENTO E RISULTATI 867268 D’Avino Luca

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXCA = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXEB = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXBE = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXCE = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoAC = ∫

o

b(2 x/b - x2/b2 ) Fb 1/EJ dx = [ x2/b -1/3 x3/b2 ]o

b Fb 1/EJ

= ( b -1/3 b ) Fb 1/EJ = 2/3 Fb2/EJ

LXoCA = ∫

o

b(1 - x2/b2 ) Fb 1/EJ dx = [ x -1/3 x3/b2 ]o

b Fb 1/EJ

= ( b -1/3 b ) Fb 1/EJ = 2/3 Fb2/EJ

IPERΣ01.dvnl.014PROCEDIMENTO E RISULTATI 867268 D’Avino Luca

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 792. mm2

Ju = 225759. mm4

Jv = 30240. mm4

yg = 22.64 mmN = 484. NTy = -3025. NMx = 1718200. Nmmxm = 24. mmym = 54. mmum = 6. mmvm = 31.36 mmσm = N/A-Mv/Ju = -238.1 N/mm2

xc = 18. mmyc = 41. mmvc = 18.36 mmσc = N/A-Mv/Ju = -139.1 N/mm2

τc = 4.331 N/mm2

σo = √σ2+3τ2 = 139.4 N/mm2

S* = 3879. mm3mm 0 12 24 36x

0

6

54

y

41σc,τc

σm

u

v

Page 44: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.dvnl.014

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.dvnl.014

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 45: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.dcml.015REAZIONI 877976 Di Camillo Lorenzo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

F

103/40F3Fb

F

63/40F37/40Fb

A B

3F

23/40F3Fb

3F

23/40F

A

C

23/40F23/40Fb

23/40F

DC

F

F3/2Fb

F

F1/2Fb

B E

F

F

G

F

F1/2Fb

F

E

F

23/40F23/40Fb

23/40F23/40Fb

D

B

F

Fb

F

G

D

IPERΣ01.dcml.015AZIONI INTERNE 877976 Di Camillo Lorenzo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

-1 -1

-23/

40

0

-1

11

1

23/4

0

-√2/2

F

103/40 63/40

-3

-23/40

1

01

0

0

-√2/2

F

-3 -37/4030

23/400

-3/2 -1/2

00

-1/2

0-23/

40-2

3/40

1

0

Fb

Page 46: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPE

RΣ0

1.dc

ml.0

15P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 8

7797

6 D

i Cam

illo

Lore

nzo

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

AB

C

D

E

F

G

F

W

X

X

q

q

Sch

ema

di c

alco

lo ip

erst

atic

o

-3-3

/2

3 0

00

-3/2

-1/2

0 0-1/2

0

00

1

0

Mo

fless

ione

da

caric

hi a

sseg

nati

0-1

0 0

-10

00

0 00 0

11

0

0

Mx

fless

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da

iper

stat

ica

X=

1

IPE

RΣ0

1.dc

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7797

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illo

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@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

Quadro contributi PLV per iperstatica X=WBD

→Mx(x)Mo(x)MxMoMxMx∫MxMo/EJdx∫XMxMx/EJdx

AB b-x/b-3Fb+2Fx-1/2qx2

3Fx-2Fx2/b+1/2qx

3/bx

2/b

2

23/24Fb2/EJ1/3Xb/EJ

BA b1-x/b3/2Fb+Fx+1/2qx2

3/2Fb-1/2Fx-1/2Fx2/b-1/2qx

3/b1-2x/b+x

2/b

2

AC b03Fb-3Fx0000

CA b0-3Fx00

DC b-1+x/b001-2x/b+x2/b

2

01/3Xb/EJCD bx/b00x

2/b

2

BE b0-3/2Fb+Fx0000

EB b01/2Fb+Fx00

FG b000000

GF b0000

EF b0-1/2Fb+Fx-1/2qx2

0000

FE b01/2qx2

00

DB b10010Xb/EJ

BD b-1001

GD √2b0Fb-√2/2Fx0000

totali23/24Fb2/EJ5/3Xb/EJ

iperstatica X=WBD-23/40Fb

Svi

lupp

i di c

alco

lo ip

erst

atic

a

Page 47: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.dcml.015PROCEDIMENTO E RISULTATI 877976 Di Camillo Lorenzo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAB = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXBA = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXDC = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXCD = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXDB = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXBD = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoAB = ∫

o

b(3 x/b -2 x2/b2 +1/2 x3/b3 ) Fb 1/EJ dx = [3/2 x2/b -2/3 x3/b2 +1/8 x4/b3 ]o

b Fb 1/EJ

= (3/2 b -2/3 b +1/8 b ) Fb 1/EJ = 23/24 Fb2/EJ

LXoBA = ∫

o

b(3/2 -1/2 x/b -1/2 x2/b2 -1/2 x3/b3 ) Fb 1/EJ dx = [3/2 x -1/4 x2/b -1/6 x3/b2 -1/8 x4/b3 ]o

b Fb 1/EJ

= (3/2 b -1/4 b -1/6 b -1/8 b ) Fb 1/EJ = 23/24 Fb2/EJ

IPERΣ01.dcml.015PROCEDIMENTO E RISULTATI 877976 Di Camillo Lorenzo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 1080. mm2

Ju = 276955. mm4

Jv = 116640. mm4

yg = 18.6 mmN = -391. NTy = -2040. NMx = 1550400. Nmmxm = 30. mmym = 54. mmum = 6. mmvm = 35.4 mmσm = N/A-Mv/Ju = -198.5 N/mm2

xc = 24. mmyc = 39. mmvc = 20.4 mmσc = N/A-Mv/Ju = -114.6 N/mm2

τc = 3.083 N/mm2

σo = √σ2+3τ2 = 114.7 N/mm2

S* = 5022. mm3mm 0 18 30 48x

0

12

54

y

39σc,τc

σm

u

v

Page 48: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.dcml.015

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.dcml.015

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 49: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.frtg.017REAZIONI 789093 Frattini Giulietta

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1/5F

2F2Fb

1/5F

2F

A B

11/5F

F2Fb

11/5F

F1/5Fb

A

C

F

F

F1/2Fb

C

D

1/5F1/5Fb

1/5F

E

B

F

F1/2Fb

F

D FF

F G

FFb

F

G

E

6/5F1/5Fb

6/5F1/5Fb

EC

IPERΣ01.frtg.017AZIONI INTERNE 789093 Frattini Giulietta

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1/5

-1-1

-1

0

1 1 1

-√2/2

-6/5

F

2

-11/

5-1

0

1/5

1 0 0-√2

/2

0

F

-202

-1/5

0-1

/2

-1/5

0

-1/20 0 0

1

0

1/51/5

Fb

Page 50: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPE

RΣ0

1.fr

tg.0

17P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 7

8909

3 F

ratti

ni G

iulie

tta

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

AB

C

D

E

F

G

F

W

X

X

qq

Sch

ema

di c

alco

lo ip

erst

atic

o

-20

200-1/2

0 0

-1/2

00

0

1

0

00

Mo

fless

ione

da

caric

hi a

sseg

nati

00

0-1

00

-10

00

00

0

0

11

Mx

fless

ione

da

iper

stat

ica

X=

1

IPE

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tg.0

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tta

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dolfo

Zav

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si, P

olite

cnic

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Mila

no, v

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27.0

3.13

03.0

9.18

Qua

dro

cont

ribut

i PLV

per

iper

stat

ica

X=

WC

E

→M

x(x)

Mo(

x)M

xMo

MxM

x∫M

xMo/

EJd

x∫X

MxM

x/E

Jdx

AB

b0

-2F

b+2F

x0

00

0B

A b

02F

x0

0

AC

b-x

/b2F

b-2F

x-2

Fx+

2Fx2 /b

x2 /b2

-1/3

Fb2 /E

J1/

3Xb/

EJ

CA

b1-

x/b

-2F

x-2

Fx+

2Fx2 /b

1-2x

/b+

x2 /b2

CD

b0

-Fx+

1/2q

x20

00

0D

C b

01/

2Fb-

1/2q

x20

0

EB

b-1

+x/

b0

01-

2x/b

+x2 /b

2

01/

3Xb/

EJ

BE

bx/

b0

0x2 /b

2

DF

b0

-1/2

Fb+

Fx-

1/2q

x20

00

0F

D b

01/

2qx2

00

FG

b0

00

00

0G

F b

00

00

GE

√2b

0F

b-√2

/2F

x0

00

0

EC

b1

00

10

Xb/

EJ

CE

b-1

00

1

tota

li-1

/3F

b2 /EJ

5/3X

b/E

J

iper

stat

ica

X=

WC

E1/

5Fb

Svi

lupp

i di c

alco

lo ip

erst

atic

a

Page 51: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.frtg.017PROCEDIMENTO E RISULTATI 789093 Frattini Giulietta

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXCA = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXEB = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXBE = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXCE = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoAC = ∫

o

b(-2 x/b +2 x2/b2 ) Fb 1/EJ dx = [- x2/b +2/3 x3/b2 ]o

b Fb 1/EJ

= (- b +2/3 b ) Fb 1/EJ = -1/3 Fb2/EJ

LXoCA = ∫

o

b(-2 x/b +2 x2/b2 ) Fb 1/EJ dx = [- x2/b +2/3 x3/b2 ]o

b Fb 1/EJ

= (- b +2/3 b ) Fb 1/EJ = -1/3 Fb2/EJ

IPERΣ01.frtg.017PROCEDIMENTO E RISULTATI 789093 Frattini Giulietta

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 498. mm2

Ju = 141019. mm4

Jv = 31734. mm4

yg = 35.17 mmN = 102. NTy = 1020. NMx = -867000. Nmmxm = 18. mmum = -3. mmvm = -35.17 mmσm = N/A-Mv/Ju = -216.1 N/mm2

xc = 21. mmyc = 15. mmvc = -20.17 mmσc = N/A-Mv/Ju = -123.8 N/mm2

τc = 3.003 N/mm2

σo = √σ2+3τ2 = 123.9 N/mm2

S* = 2491. mm3mm 0 18 24 42x

0

48

53

y

15σc,τc

σm

u

v

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IPERΣ01.frtg.017

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.frtg.017

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 53: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.grts.018REAZIONI 792732 Gritcul Serghei

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

17/40F

1/2F17/40Fb

17/40F

1/2F

A

B

1/2F

1/2FFb

1/2F

1/2F1/2Fb

C

D

97/40F

1/2F5/2Fb

57/40F

1/2F23/40Fb

E

C

17/40F

5/2F5/2Fb

17/40F

5/2F

EB

37/40F17/40Fb

37/40F17/40Fb

A C

1/2F

1/2F

Fb

1/2F

1/2F

F

A

1/2F

1/2F

1/2F

1/2F

GF

1/2F

1/2F1/2Fb

1/2F

1/2F

DG

IPERΣ01.grts.018AZIONI INTERNE 792732 Gritcul Serghei

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

-1/2

-1/2

-1/2

-1/2

-17/40

37/40

0

-1/2-1/2 -1/2

F

17/4

0

-1/2

-97/

40-5

7/40

5/2

0

-√2/2

-1/21/2

-1/2

F

-17/

400

11/

25/

223

/40

-5/20

17/40 17/40

1

0

00 1/20

Fb

Page 54: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPE

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1.gr

ts.0

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9273

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ritcu

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Zav

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si, P

olite

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Mila

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27.0

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B

C

D E

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000

1

000

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0

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sseg

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00 01 00

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ica

X=

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IPE

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27.0

3.13

03.0

9.18

Qua

dro

cont

ribut

i PLV

per

iper

stat

ica

X=

WA

C

→M

x(x)

Mo(

x)M

xMo

MxM

x∫M

xMo/

EJd

x∫X

MxM

x/E

Jdx

AB

b1-

x/b

00

1-2x

/b+

x2 /b2

01/

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EJ

BA

b-x

/b0

0x2 /b

2

CD

b0

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x0

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C b

0-1

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b-1/

2Fx

00

EC

bx/

b5/

2Fb-

2Fx+

1/2q

x25/

2Fx-

2Fx2 /b

+1/

2qx3 /b

x2 /b2

17/2

4Fb2 /E

J1/

3Xb/

EJ

CE

b-1

+x/

b-F

b-F

x-1/

2qx2

Fb-

1/2F

x2 /b-1

/2qx

3 /b1-

2x/b

+x2 /b

2

EB

b0

-5/2

Fb+

5/2F

x0

00

0B

E b

05/

2Fx

00

AC

b-1

00

10

Xb/

EJ

CA

b1

00

1

FA

√2b

0F

b-√2

/2F

x0

00

0

GF

b0

-1/2

Fx+

1/2q

x20

00

0F

G b

01/

2Fx-

1/2q

x20

0

DG

b0

1/2F

b-1/

2Fx

00

00

GD

b0

-1/2

Fx

00

tota

li17

/24F

b2 /EJ

5/3X

b/E

J

iper

stat

ica

X=

WA

C-1

7/40

Fb

Svi

lupp

i di c

alco

lo ip

erst

atic

a

Page 55: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.grts.018PROCEDIMENTO E RISULTATI 792732 Gritcul Serghei

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAB = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXBA = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXCE = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXAC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXCA = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoEC = ∫

o

b(5/2 x/b -2 x2/b2 +1/2 x3/b3 ) Fb 1/EJ dx = [5/4 x2/b -2/3 x3/b2 +1/8 x4/b3 ]o

b Fb 1/EJ

= (5/4 b -2/3 b +1/8 b ) Fb 1/EJ = 17/24 Fb2/EJ

LXoCE = ∫

o

b(1 -1/2 x2/b2 -1/2 x3/b3 ) Fb 1/EJ dx = [ x -1/6 x3/b2 -1/8 x4/b3 ]o

b Fb 1/EJ

= ( b -1/6 b -1/8 b ) Fb 1/EJ = 17/24 Fb2/EJ

IPERΣ01.grts.018PROCEDIMENTO E RISULTATI 792732 Gritcul Serghei

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 714. mm2

Ju = 156210. mm4

Jv = 68670. mm4

yg = 38.15 mmN = -174.3 NTy = 1025. NMx = -922500. Nmmxm = 18. mmum = -3. mmvm = -38.15 mmσm = N/A-Mv/Ju = -225.5 N/mm2

xc = 21. mmyc = 16. mmvc = -22.15 mmσc = N/A-Mv/Ju = -131. N/mm2

τc = 3.165 N/mm2

σo = √σ2+3τ2 = 131.1 N/mm2

S* = 2894. mm3mm 0 18 24 42x

0

42

53

y

16σc,τc

σm

u

v

Page 56: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.grts.018

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.grts.018

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 57: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.lbtt.019REAZIONI 850296 Labate Tommaso

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

2F

F3/2Fb

2F

F1/2Fb

A

B

11/20F11/20Fb

11/20F

C

D

11/20F

3F5/2Fb

11/20F

2F

ED

31/20F

F5/2Fb

31/20F

F19/20Fb

E

A

FFb

F

F

C

9/20F11/20Fb

9/20F11/20Fb

C A

F

F1/2Fb

F

BG

FGF

IPERΣ01.lbtt.019AZIONI INTERNE 850296 Labate Tommaso

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1

0

-11/20-11/20

1

-√2/2

-9/20

111

F

2

-11/

20

-3-2

31/2

0

-√2/2

0

-100

F

-3/2

1/2

11/2

00 5/20

-5/2

-19/

20

1

0

-11/20 -11/20

1/2000

Fb

Page 58: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPE

RΣ0

1.lb

tt.01

9P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 8

5029

6 La

bate

Tom

mas

o

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

A

B

CDE

FG

F

X

X

q q

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0

00 1/

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fless

ione

da

caric

hi a

sseg

nati

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ione

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iper

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ica

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1

IPE

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EN

TO

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5029

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Tom

mas

o

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

Qua

dro

cont

ribut

i PLV

per

iper

stat

ica

X=

WC

D

→M

x(x)

Mo(

x)M

xMo

MxM

x∫M

xMo/

EJd

x∫X

MxM

x/E

Jdx

AB

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2Fx

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b0

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00

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00

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b-x

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2

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0F

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x0

00

0

CA

b1

00

10

Xb/

EJ

AC

b-1

00

1

BG

b0

1/2F

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x+1/

2qx2

00

00

GB

b0

-1/2

qx2

00

GF

b0

00

00

0F

G b

00

00

tota

li11

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b2 /EJ

5/3X

b/E

J

iper

stat

ica

X=

WC

D-1

1/20

Fb

Svi

lupp

i di c

alco

lo ip

erst

atic

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Page 59: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.lbtt.019PROCEDIMENTO E RISULTATI 850296 Labate Tommaso

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXCD = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXDC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEA = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXAE = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXCA = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXAC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoEA = ∫

o

b(5/2 x/b - x2/b2 ) Fb 1/EJ dx = [5/4 x2/b -1/3 x3/b2 ]o

b Fb 1/EJ

= (5/4 b -1/3 b ) Fb 1/EJ = 11/12 Fb2/EJ

LXoAE = ∫

o

b(3/2 -1/2 x/b - x2/b2 ) Fb 1/EJ dx = [3/2 x -1/4 x2/b -1/3 x3/b2 ]o

b Fb 1/EJ

= (3/2 b -1/4 b -1/3 b ) Fb 1/EJ = 11/12 Fb2/EJ

IPERΣ01.lbtt.019PROCEDIMENTO E RISULTATI 850296 Labate Tommaso

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 816. mm2

Ju = 230061. mm4

Jv = 52992. mm4

yg = 31.79 mmN = -401.5 NTy = -2190. NMx = 1733750. Nmmxm = 18. mmum = -6. mmvm = -31.79 mmσm = N/A-Mv/Ju = 239.1 N/mm2

xc = 24. mmyc = 13. mmvc = -18.79 mmσc = N/A-Mv/Ju = 141.1 N/mm2

τc = 3.13 N/mm2

σo = √σ2+3τ2 = 141.2 N/mm2

S* = 3946. mm3mm 0 18 30 48x

0

48

53

y

13σc,τc

σm

u

v

Page 60: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.lbtt.019

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.lbtt.019

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 61: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.lnge.020REAZIONI 806366 Longoni Emanuele Luca

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

4F

4/5F4Fb

4F

4/5F

A

B

F

14/5F4Fb

F

14/5F6/5Fb

A C

F

3F2Fb

F

2F1/2Fb

C D

4/5F4/5Fb

4/5F

EB

F

F1/2Fb

F

D

F

F

F

G

F

Fb

F

G

E

1/5F4/5Fb

1/5F4/5Fb

E

C

IPERΣ01.lnge.020AZIONI INTERNE 806366 Longoni Emanuele Luca

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

4/5

-1 -1 -1

0

-1-1

-1

√2/2

1/5

F

-4

14/5 3 2

-4/5

-10

0-√2/2

0

F

40

-4 -6/5 -21/2

4/50

1/2

00

01

0

-4/5

-4/5

Fb

Page 62: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPE

RΣ0

1.ln

ge.0

20P

RO

CE

DIM

EN

TO

E R

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LTA

TI 8

0636

6 Lo

ngon

i Em

anue

le

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

A

B

CD

EFG F

W

X

X

q

q

Sch

ema

di c

alco

lo ip

erst

atic

o

40

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2

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0 0 0

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fless

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caric

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nati

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01

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0000

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da

iper

stat

ica

X=

1

IPE

RΣ0

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CE

DIM

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TO

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@ A

dolfo

Zav

elan

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olite

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no, v

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27.0

3.13

03.0

9.18

Qua

dro

cont

ribut

i PLV

per

iper

stat

ica

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C

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x(x)

Mo(

x)M

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MxM

x∫M

xMo/

EJd

x∫X

MxM

x/E

Jdx

AB

b0

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4Fx

00

00

BA

b0

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x0

0

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b-4

Fb+

2Fx

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x+2F

x2 /bx2 /b

2

-4/3

Fb2 /E

J1/

3Xb/

EJ

CA

b-1

+x/

b2F

b+2F

x-2

Fb+

2Fx2 /b

1-2x

/b+

x2 /b2

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2qx2

00

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2

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qx2

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00

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li-4

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b2 /EJ

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iper

stat

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X=

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5Fb

Svi

lupp

i di c

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lo ip

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atic

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Page 63: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.lnge.020PROCEDIMENTO E RISULTATI 806366 Longoni Emanuele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXCA = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXEB = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXBE = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXCE = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoAC = ∫

o

b(-4 x/b +2 x2/b2 ) Fb 1/EJ dx = [-2 x2/b +2/3 x3/b2 ]o

b Fb 1/EJ

= (-2 b +2/3 b ) Fb 1/EJ = -4/3 Fb2/EJ

LXoCA = ∫

o

b(-2 +2 x2/b2 ) Fb 1/EJ dx = [-2 x +2/3 x3/b2 ]o

b Fb 1/EJ

= (-2 b +2/3 b ) Fb 1/EJ = -4/3 Fb2/EJ

IPERΣ01.lnge.020PROCEDIMENTO E RISULTATI 806366 Longoni Emanuele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 1032. mm2

Ju = 260495. mm4

Jv = 107424. mm4

yg = 34.56 mmN = 600. NTy = -3000. NMx = 1500000. Nmmxm = 18. mmum = -6. mmvm = -34.56 mmσm = N/A-Mv/Ju = 199.6 N/mm2

xc = 24. mmyc = 15. mmvc = -19.56 mmσc = N/A-Mv/Ju = 113.2 N/mm2

τc = 4.674 N/mm2

σo = √σ2+3τ2 = 113.5 N/mm2

S* = 4870. mm3mm 0 18 30 48x

0

42

53

y

15σc,τc

σm

u

v

Page 64: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.lnge.020

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.lnge.020

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 65: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.mnfa.021REAZIONI 853708 Manfrin Alves Fernando

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1/2F

17/40F17/40Fb

1/2F

17/40F

A B

1/2F

1/2FFb

1/2F

1/2F1/2Fb

CD

1/2F

97/40F5/2Fb

1/2F

57/40F23/40Fb

EC

5/2F

17/40F5/2Fb

5/2F

17/40F

E

B

37/40F17/40Fb

37/40F17/40Fb

A

C

1/2F

1/2F

Fb

1/2F

1/2F

F

A1/2F

1/2F

1/2F

1/2F

G

F

1/2F

1/2F1/2Fb

1/2F

1/2F

D

G

IPERΣ01.mnfa.021AZIONI INTERNE 853708 Manfrin Alves Fernando

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1/2

1/2 1/21/2

17/4

0

-37/

40

0

1/2

1/2

1/2

F

17/40

-1/2 -97/40-57/40

5/20

-√2/2

-1/2

1/2

-1/2

F

-17/400

11/2 5/223/40

-5/2

0

17/4

017

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1

0

00

1/2

0

Fb

Page 66: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPE

RΣ0

1.m

nfa.

021

PR

OC

ED

IME

NT

O E

RIS

ULT

AT

I 853

708

Man

frin

Alv

es

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

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es

@ A

dolfo

Zav

elan

i Ros

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olite

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no, v

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27.0

3.13

03.0

9.18

Qua

dro

cont

ribut

i PLV

per

iper

stat

ica

X=

WC

A

→M

x(x)

Mo(

x)M

xMo

MxM

x∫M

xMo/

EJd

x∫X

MxM

x/E

Jdx

AB

b-1

+x/

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+x2 /b

2

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BA

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Page 67: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.mnfa.021PROCEDIMENTO E RISULTATI 853708 Manfrin Alves

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAB = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXBA = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXCE = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXAC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXCA = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoEC = ∫

o

b(-5/2 x/b +2 x2/b2 -1/2 x3/b3 ) Fb 1/EJ dx = [-5/4 x2/b +2/3 x3/b2 -1/8 x4/b3 ]o

b Fb 1/EJ

= (-5/4 b +2/3 b -1/8 b ) Fb 1/EJ = -17/24 Fb2/EJ

LXoCE = ∫

o

b(-1 +1/2 x2/b2 +1/2 x3/b3 ) Fb 1/EJ dx = [- x +1/6 x3/b2 +1/8 x4/b3 ]o

b Fb 1/EJ

= (- b +1/6 b +1/8 b ) Fb 1/EJ = -17/24 Fb2/EJ

IPERΣ01.mnfa.021PROCEDIMENTO E RISULTATI 853708 Manfrin Alves

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 438. mm2

Ju = 124871. mm4

Jv = 12114. mm4

yg = 33.08 mmN = 242.3 NTy = 1425. NMx = -783750. Nmmxm = 12. mmum = -3. mmvm = -33.08 mmσm = N/A-Mv/Ju = -207. N/mm2

xc = 15. mmyc = 14. mmvc = -19.08 mmσc = N/A-Mv/Ju = -119.2 N/mm2

τc = 4.166 N/mm2

σo = √σ2+3τ2 = 119.4 N/mm2

S* = 2190. mm3mm 0 12 18 30x

0

48

53

y

14σc,τc

σm

u

v

Page 68: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.mnfa.021

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.mnfa.021

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 69: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.mrna.022REAZIONI 835477 Maranga Andrea

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

F1/2Fb

F1/2Fb

AB

1/20F1/20Fb

1/20F

C D

F

1/20F1/2Fb

1/20F

E

D

F

19/20F1/2Fb

F

19/20F9/20Fb

EA

F

Fb

F

F

C

19/20F1/20Fb

19/20F1/20Fb

C

A

F

F1/2Fb

F

B

G

F

G

F

IPERΣ01.mrna.022AZIONI INTERNE 835477 Maranga Andrea

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1

0

1/20

1/20

1

√2/2

19/2

0

-1-1

-1

F

0

-1/20

10

-19/20

-√2/2

0

10

0

F

-1/2-1/2

1/20 0

-1/2

0

1/2-9/20

1

0

-1/2

0-1

/20-1/2

00

0

Fb

Page 70: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPE

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anga

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C

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27.0

3.13

03.0

9.18

Qua

dro

cont

ribut

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per

iper

stat

ica

X=

WC

A

→M

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Mo(

x)M

xMo

MxM

x∫M

xMo/

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MxM

x/E

Jdx

AB

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00

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b

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i di c

alco

lo ip

erst

atic

a

Page 71: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.mrna.022PROCEDIMENTO E RISULTATI 835477 Maranga Andrea

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXCD = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXDC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEA = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXAE = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXCA = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXAC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoEA = ∫

o

b(1/2 x/b - x2/b2 ) Fb 1/EJ dx = [1/4 x2/b -1/3 x3/b2 ]o

b Fb 1/EJ

= (1/4 b -1/3 b ) Fb 1/EJ = -1/12 Fb2/EJ

LXoAE = ∫

o

b(-1/2 +3/2 x/b - x2/b2 ) Fb 1/EJ dx = [-1/2 x +3/4 x2/b -1/3 x3/b2 ]o

b Fb 1/EJ

= (-1/2 b +3/4 b -1/3 b ) Fb 1/EJ = -1/12 Fb2/EJ

IPERΣ01.mrna.022PROCEDIMENTO E RISULTATI 835477 Maranga Andrea

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 582. mm2

Ju = 140714. mm4

Jv = 25506. mm4

yg = 36.03 mmN = 1004. NTy = -1004. NMx = 852000. Nmmxm = 12. mmum = -3. mmvm = -36.03 mmσm = N/A-Mv/Ju = 219.9 N/mm2

xc = 15. mmyc = 15. mmvc = -21.03 mmσc = N/A-Mv/Ju = 129. N/mm2

τc = 3.053 N/mm2

σo = √σ2+3τ2 = 129.1 N/mm2

S* = 2567. mm3mm 0 12 18 30x

0

42

53

y

15σc,τc

σm

u

v

Page 72: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.mrna.022

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.mrna.022

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 73: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.mrtm.024REAZIONI 846219 Martignoni Matteo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

3/2F

1/2FFb

3/2F

1/2F1/2Fb

A

B

2/5F

1/2F2/5Fb

2/5F

1/2F

C

D

2/5F

5/2F2Fb

2/5F

3/2F

ED

7/5F

1/2F2Fb

7/5F

1/2F3/5Fb

E

A

1/2F

1/2F

Fb

1/2F

1/2F

F

C

1/10F2/5Fb

1/10F2/5Fb

C A

1/2F

1/2F1/2Fb

1/2F

1/2F

BG

1/2F

1/2F

1/2F

1/2F

GF

IPERΣ01.mrtm.024AZIONI INTERNE 846219 Martignoni Matteo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1/2

1/2

-2/5-2/5

1/2

0

-1/10

1/21/21/2

F

3/2

-2/5

-5/2-3/2

7/5

-√2/2

0

-1/2-1/21/2

F

-11/

2

2/5

0 20

-2-3

/5

1

0

-2/5 -2/5

1/2000

Fb

Page 74: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPE

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rtm

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I 846

219

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tigno

ni M

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dolfo

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si, P

olite

cnic

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Mila

no, v

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27.0

3.13

03.0

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dolfo

Zav

elan

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si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

Qua

dro

cont

ribut

i PLV

per

iper

stat

ica

X=

WC

A

→M

x(x)

Mo(

x)M

xMo

MxM

x∫M

xMo/

EJd

x∫X

MxM

x/E

Jdx

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bx/

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2

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1/2q

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00

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x20

0

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/3F

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5/3X

b/E

J

iper

stat

ica

X=

WC

A2/

5Fb

Svi

lupp

i di c

alco

lo ip

erst

atic

a

Page 75: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.mrtm.024PROCEDIMENTO E RISULTATI 846219 Martignoni Matteo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXCD = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXDC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEA = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXAE = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXCA = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXAC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoEA = ∫

o

b(-2 x/b + x2/b2 ) Fb 1/EJ dx = [- x2/b +1/3 x3/b2 ]o

b Fb 1/EJ

= (- b +1/3 b ) Fb 1/EJ = -2/3 Fb2/EJ

LXoAE = ∫

o

b(-1 + x2/b2 ) Fb 1/EJ dx = [- x +1/3 x3/b2 ]o

b Fb 1/EJ

= (- b +1/3 b ) Fb 1/EJ = -2/3 Fb2/EJ

IPERΣ01.mrtm.024PROCEDIMENTO E RISULTATI 846219 Martignoni Matteo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 900. mm2

Ju = 233812. mm4

Jv = 48816. mm4

yg = 32.66 mmN = -492. NTy = -3075. NMx = 1722000. Nmmxm = 12. mmum = -6. mmvm = -32.66 mmσm = N/A-Mv/Ju = 240. N/mm2

xc = 18. mmyc = 14. mmvc = -18.66 mmσc = N/A-Mv/Ju = 136.9 N/mm2

τc = 4.725 N/mm2

σo = √σ2+3τ2 = 137.1 N/mm2

S* = 4311. mm3mm 0 12 24 36x

0

42

53

y

14σc,τc

σm

u

v

Page 76: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.mrtm.024

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.mrtm.024

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 77: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.msts.026REAZIONI 846325 Musto Simone

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1/2F

1/2FFb

1/2F

1/2F1/2Fb

AB

1/2F

1/5F1/5Fb

1/2F

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C D

1/2F

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1/2F

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E

D

1/2F

4/5F

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EA

1/2F

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1/2F

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13/10F1/5Fb

13/10F1/5Fb

C

A

1/2F

3/2F1/2Fb

1/2F

3/2F

B

G

1/2F

3/2F

1/2F

3/2F

G

F

IPERΣ01.msts.026AZIONI INTERNE 846325 Musto Simone

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1/2

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F

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00

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1

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00

Fb

Page 78: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

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Page 79: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.msts.026PROCEDIMENTO E RISULTATI 846325 Musto Simone

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXCD = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXDC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEA = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXAE = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXCA = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXAC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoEA = ∫

o

b(- x2/b2 ) Fb 1/EJ dx = [-1/3 x3/b2 ]o

b Fb 1/EJ

= (-1/3 b ) Fb 1/EJ = -1/3 Fb2/EJ

LXoAE = ∫

o

b(-1 +2 x/b - x2/b2 ) Fb 1/EJ dx = [- x + x2/b -1/3 x3/b2 ]o

b Fb 1/EJ

= (- b + b -1/3 b ) Fb 1/EJ = -1/3 Fb2/EJ

IPERΣ01.msts.026PROCEDIMENTO E RISULTATI 846325 Musto Simone

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 462. mm2

Ju = 129608. mm4

Jv = 14346. mm4

yg = 19.18 mmN = 1443. NTy = -721.2 NMx = 816000. Nmmxm = 18. mmym = 53. mmum = 3. mmvm = 33.82 mmσm = N/A-Mv/Ju = -209.8 N/mm2

xc = 15. mmyc = 39. mmvc = 19.82 mmσc = N/A-Mv/Ju = -121.7 N/mm2

τc = 2.09 N/mm2

σo = √σ2+3τ2 = 121.7 N/mm2

S* = 2253. mm3mm 0 12 18 30x

0

6

53

y

39σc,τc

σm

u

v

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IPERΣ01.msts.026

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.msts.026

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 81: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.nrde.027REAZIONI 835674 Nardi Edoardo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1/20F

3/2F3/2Fb

1/20F

3/2F

A B

41/20F

1/2F3/2Fb

41/20F

1/2F11/20Fb

A

C

1/2F

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1/2F

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C

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1/20F

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1/20F

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B

3/2F

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3/2F

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1/2F

G

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EC

IPERΣ01.nrde.027AZIONI INTERNE 835674 Nardi Edoardo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1/20

-1/2

-1/2

-1/2

-1/2

3/2 3/2 3/2

-√2

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00

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1

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Fb

Page 82: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

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Page 83: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.nrde.027PROCEDIMENTO E RISULTATI 835674 Nardi Edoardo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXCA = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXEB = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXBE = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXCE = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoAC = ∫

o

b(-3/2 x/b +2 x2/b2 ) Fb 1/EJ dx = [-3/4 x2/b +2/3 x3/b2 ]o

b Fb 1/EJ

= (-3/4 b +2/3 b ) Fb 1/EJ = -1/12 Fb2/EJ

LXoCA = ∫

o

b(1/2 -5/2 x/b +2 x2/b2 ) Fb 1/EJ dx = [1/2 x -5/4 x2/b +2/3 x3/b2 ]o

b Fb 1/EJ

= (1/2 b -5/4 b +2/3 b ) Fb 1/EJ = -1/12 Fb2/EJ

IPERΣ01.nrde.027PROCEDIMENTO E RISULTATI 835674 Nardi Edoardo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 852. mm2

Ju = 238569. mm4

Jv = 62064. mm4

yg = 20.54 mmN = 63. NTy = 1890. NMx = -1606500. Nmmxm = 30. mmym = 53. mmum = 6. mmvm = 32.46 mmσm = N/A-Mv/Ju = 218.6 N/mm2

xc = 24. mmyc = 39. mmvc = 18.46 mmσc = N/A-Mv/Ju = 124.4 N/mm2

τc = 2.824 N/mm2

σo = √σ2+3τ2 = 124.5 N/mm2

S* = 4277. mm3mm 0 18 30 48x

0

6

53

y

39σc,τc

σm

u

v

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IPERΣ01.nrde.027

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.nrde.027

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 85: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.omrm.028REAZIONI 743445 Omara Mohamed

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

17/8F

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A

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IPERΣ01.omrm.028AZIONI INTERNE 743445 Omara Mohamed

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1

1/8

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Page 86: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

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Page 87: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.omrm.028PROCEDIMENTO E RISULTATI 743445 Omara Mohamed

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAB = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXBA = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXDC = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXCD = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXDB = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXBD = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoAB = ∫

o

b(3 x/b -3/2 x2/b2 ) Fb 1/EJ dx = [3/2 x2/b -1/2 x3/b2 ]o

b Fb 1/EJ

= (3/2 b -1/2 b ) Fb 1/EJ = Fb2/EJ

LXoBA = ∫

o

b(3/2 -3/2 x2/b2 ) Fb 1/EJ dx = [3/2 x -1/2 x3/b2 ]o

b Fb 1/EJ

= (3/2 b -1/2 b ) Fb 1/EJ = Fb2/EJ

LXoDC = ∫

o

b(1/2 x/b - x2/b2 +1/2 x3/b3 ) Fb 1/EJ dx = [1/4 x2/b -1/3 x3/b2 +1/8 x4/b3 ]o

b Fb 1/EJ

= (1/4 b -1/3 b +1/8 b ) Fb 1/EJ = 1/24 Fb2/EJ

LXoCD = ∫

o

b(1/2 x2/b2 -1/2 x3/b3 ) Fb 1/EJ dx = [1/6 x3/b2 -1/8 x4/b3 ]o

b Fb 1/EJ

= (1/6 b -1/8 b ) Fb 1/EJ = 1/24 Fb2/EJ

IPERΣ01.omrm.028PROCEDIMENTO E RISULTATI 743445 Omara Mohamed

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 606. mm2

Ju = 141406. mm4

Jv = 27738. mm4

yg = 16.76 mmN = 41.25 NTy = -990. NMx = 891000. Nmmxm = 18. mmym = 53. mmum = 3. mmvm = 36.24 mmσm = N/A-Mv/Ju = -228.3 N/mm2

xc = 15. mmyc = 38. mmvc = 21.24 mmσc = N/A-Mv/Ju = -133.8 N/mm2

τc = 3.018 N/mm2

σo = √σ2+3τ2 = 133.9 N/mm2

S* = 2587. mm3mm 0 12 18 30x

0

12

53

y

38σc,τc

σm

u

v

Page 88: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.omrm.028

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.omrm.028

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 89: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.pdzl.030REAZIONI 871692 Peduzzi Luca

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

7/2F

13/20F7/2Fb

7/2F

13/20F

A

B

1/2F

53/20F7/2Fb

1/2F

53/20F17/20Fb

A C

1/2F

5/2F3/2Fb

1/2F

3/2F1/2Fb

C D

1/2F

13/20F13/20Fb

1/2F

13/20F

EB

1/2F

1/2F1/2Fb

1/2F

1/2F

D

F

1/2F

1/2F

1/2F

1/2F

F

G

1/2F

1/2F

Fb

1/2F

1/2F

G

E

3/20F13/20Fb

3/20F13/20Fb

E

C

IPERΣ01.pdzl.030AZIONI INTERNE 871692 Peduzzi Luca

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

13/2

0

-1/2 -1/2 -1/2

-1/2

-1/2

-1/2

-1/2

0

-3/2

0

F

-7/2

53/20 5/2 3/2

-13/20

-1/2

-1/2

1/2

-√2/2

0

F

7/2

0

-7/2 -17/20 -3/21/2

13/200

1/2

00

01

0

-13/

20-1

3/20

Fb

Page 90: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPE

RΣ0

1.pd

zl.0

30P

RO

CE

DIM

EN

TO

E R

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LTA

TI 8

7169

2 P

eduz

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uca

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dolfo

Zav

elan

i Ros

si, P

olite

cnic

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Mila

no, v

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27.0

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03.0

9.18

A

B

CD

EFG F

W

X

X

qq

Sch

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di c

alco

lo ip

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atic

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7/20

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1/2

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1

0 0 0

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caric

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sseg

nati

00

0-1

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0000

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0 1 1

Mx

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iper

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Zav

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27.0

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03.0

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Qua

dro

cont

ribut

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iper

stat

ica

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WE

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→M

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x)M

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x∫X

MxM

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AB

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J1/

3Xb/

EJ

CA

b1-

x/b

3/2F

b+2F

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1/2F

x-2F

x2 /b1-

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2

CD

b0

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Fb+

5/2F

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2qx2

00

00

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Fb+

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00

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0

GE

√2b

0F

b-√2

/2F

x0

00

0

EC

b1

00

10

Xb/

EJ

CE

b-1

00

1

tota

li13

/12F

b2 /EJ

5/3X

b/E

J

iper

stat

ica

X=

WE

B-1

3/20

Fb

Svi

lupp

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alco

lo ip

erst

atic

a

Page 91: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.pdzl.030PROCEDIMENTO E RISULTATI 871692 Peduzzi Luca

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXCA = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXEB = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXBE = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXCE = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoAC = ∫

o

b(7/2 x/b -2 x2/b2 ) Fb 1/EJ dx = [7/4 x2/b -2/3 x3/b2 ]o

b Fb 1/EJ

= (7/4 b -2/3 b ) Fb 1/EJ = 13/12 Fb2/EJ

LXoCA = ∫

o

b(3/2 +1/2 x/b -2 x2/b2 ) Fb 1/EJ dx = [3/2 x +1/4 x2/b -2/3 x3/b2 ]o

b Fb 1/EJ

= (3/2 b +1/4 b -2/3 b ) Fb 1/EJ = 13/12 Fb2/EJ

IPERΣ01.pdzl.030PROCEDIMENTO E RISULTATI 871692 Peduzzi Luca

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 780. mm2

Ju = 214152. mm4

Jv = 30096. mm4

yg = 22.16 mmN = 513.5 NTy = -2765. NMx = 1382500. Nmmxm = 24. mmym = 53. mmum = 6. mmvm = 30.84 mmσm = N/A-Mv/Ju = -198.4 N/mm2

xc = 18. mmyc = 40. mmvc = 17.84 mmσc = N/A-Mv/Ju = -114.5 N/mm2

τc = 4.085 N/mm2

σo = √σ2+3τ2 = 114.7 N/mm2

S* = 3797. mm3mm 0 12 24 36x

0

6

53

y

40σc,τc

σm

u

v

Page 92: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.pdzl.030

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.pdzl.030

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 93: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.ptrm.031REAZIONI 868523 Petronio Marco

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

F

17/8F3Fb

F

17/8F7/8Fb

A B

3F

1/8F3Fb

3F

1/8F

A

C

9/8F5/8Fb

1/8F

DC

F

F3/2Fb

F

F1/2Fb

B E

F

F

G

F

F1/2Fb

F

E

F

9/8F5/8Fb

9/8F5/8Fb

D

B

F

Fb

F

G

D

IPERΣ01.ptrm.031AZIONI INTERNE 868523 Petronio Marco

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

-1

-1/8

00

-1

11

1

9/8

-√2/2

F

17/8

-3

-9/8-1/8

1

01

0

0

-√2/2

F

-3 -7/830

5/80

-3/2 -1/2

00

-1/2

0-5/8

-5/8

1

0

Fb

Page 94: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPE

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dolfo

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cnic

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no, v

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27.0

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03.0

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C

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X

q

q

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no, v

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27.0

3.13

03.0

9.18

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dro

cont

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per

iper

stat

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x(x)

Mo(

x)M

xMo

MxM

x∫M

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EJd

x∫X

MxM

x/E

Jdx

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b/E

J

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stat

ica

X=

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b

Svi

lupp

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lo ip

erst

atic

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Page 95: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.ptrm.031PROCEDIMENTO E RISULTATI 868523 Petronio Marco

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAB = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXBA = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXDC = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXCD = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXDB = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXBD = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoAB = ∫

o

b(3 x/b -3/2 x2/b2 ) Fb 1/EJ dx = [3/2 x2/b -1/2 x3/b2 ]o

b Fb 1/EJ

= (3/2 b -1/2 b ) Fb 1/EJ = Fb2/EJ

LXoBA = ∫

o

b(3/2 -3/2 x2/b2 ) Fb 1/EJ dx = [3/2 x -1/2 x3/b2 ]o

b Fb 1/EJ

= (3/2 b -1/2 b ) Fb 1/EJ = Fb2/EJ

LXoDC = ∫

o

b(1/2 x/b - x2/b2 +1/2 x3/b3 ) Fb 1/EJ dx = [1/4 x2/b -1/3 x3/b2 +1/8 x4/b3 ]o

b Fb 1/EJ

= (1/4 b -1/3 b +1/8 b ) Fb 1/EJ = 1/24 Fb2/EJ

LXoCD = ∫

o

b(1/2 x2/b2 -1/2 x3/b3 ) Fb 1/EJ dx = [1/6 x3/b2 -1/8 x4/b3 ]o

b Fb 1/EJ

= (1/6 b -1/8 b ) Fb 1/EJ = 1/24 Fb2/EJ

IPERΣ01.ptrm.031PROCEDIMENTO E RISULTATI 868523 Petronio Marco

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 1068. mm2

Ju = 262174. mm4

Jv = 116496. mm4

yg = 18.21 mmN = -118.8 NTy = -2850. NMx = 1567500. Nmmxm = 30. mmym = 53. mmum = 6. mmvm = 34.79 mmσm = N/A-Mv/Ju = -208.1 N/mm2

xc = 24. mmyc = 38. mmvc = 19.79 mmσc = N/A-Mv/Ju = -118.4 N/mm2

τc = 4.45 N/mm2

σo = √σ2+3τ2 = 118.7 N/mm2

S* = 4913. mm3mm 0 18 30 48x

0

12

53

y

38σc,τc

σm

u

v

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IPERΣ01.ptrm.031

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.ptrm.031

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 97: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.pnzc.032REAZIONI 878201 Ponziani Camilla

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

11/40F11/40Fb

11/40F

A B

F1/2Fb

F1/2Fb

CD

F

91/40F2Fb

F

51/40F9/40Fb

EC

2F

11/40F2Fb

2F

11/40F

E

B

51/40F11/40Fb

51/40F11/40Fb

A

C

F

Fb

F

F

AF

G

F

F

F1/2Fb

F

D

G

IPERΣ01.pnzc.032AZIONI INTERNE 878201 Ponziani Camilla

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

0

1 11

11/4

0

-51/

40

-√2/2

11

1

F

11/40

0-91/40-51/40

20

-√2/2

0-1

0

F

-11/400

1/21/2 29/40

-20

11/4

011

/40

1

0

00

1/2

0

Fb

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IPE

RΣ0

1.pn

zc.0

32P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 8

7820

1 P

onzi

ani C

amill

a

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

AB

CD

E

F

G

F

W

XX

q

q Sch

ema

di c

alco

lo ip

erst

atic

o

00

1/2

1/2

21/

2

-20

0 0

1

0

00 1/20

Mo

fless

ione

da

caric

hi a

sseg

nati

-10

00

0-1

00

1 1

0

0

00 00

Mx

fless

ione

da

iper

stat

ica

X=

1

IPE

RΣ0

1.pn

zc.0

32P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 8

7820

1 P

onzi

ani C

amill

a

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

Quadro contributi PLV per iperstatica X=WAB

→Mx(x)Mo(x)MxMoMxMx∫MxMo/EJdx∫XMxMx/EJdx

AB b-1+x/b001-2x/b+x2/b

2

01/3Xb/EJBA bx/b00x

2/b

2

CD b01/2Fb0000

DC b0-1/2Fb00

EC b-x/b2Fb-2Fx+1/2qx2

-2Fx+2Fx2/b-1/2qx

3/bx

2/b

2

-11/24Fb2/EJ1/3Xb/EJ

CE b1-x/b-1/2Fb-Fx-1/2qx2

-1/2Fb-1/2Fx+1/2Fx2/b+1/2qx

3/b1-2x/b+x

2/b

2

EB b0-2Fb+2Fx0000

BE b02Fx00

AC b10010Xb/EJ

CA b-1001

FA √2b0Fb-√2/2Fx0000

GF b000000

FG b0000

DG b01/2Fb-Fx+1/2qx2

0000

GD b0-1/2qx2

00

totali-11/24Fb2/EJ5/3Xb/EJ

iperstatica X=WAB11/40Fb

Svi

lupp

i di c

alco

lo ip

erst

atic

a

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IPERΣ01.pnzc.032PROCEDIMENTO E RISULTATI 878201 Ponziani Camilla

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAB = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXBA = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXCE = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXAC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXCA = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoEC = ∫

o

b(-2 x/b +2 x2/b2 -1/2 x3/b3 ) Fb 1/EJ dx = [- x2/b +2/3 x3/b2 -1/8 x4/b3 ]o

b Fb 1/EJ

= (- b +2/3 b -1/8 b ) Fb 1/EJ = -11/24 Fb2/EJ

LXoCE = ∫

o

b(-1/2 -1/2 x/b +1/2 x2/b2 +1/2 x3/b3 ) Fb 1/EJ dx

= [-1/2 x -1/4 x2/b +1/6 x3/b2 +1/8 x4/b3 ]o

b Fb 1/EJ

= (-1/2 b -1/4 b +1/6 b +1/8 b ) Fb 1/EJ = -11/24 Fb2/EJ

IPERΣ01.pnzc.032PROCEDIMENTO E RISULTATI 878201 Ponziani Camilla

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 534. mm2

Ju = 146122. mm4

Jv = 37890. mm4

yg = 16.99 mmN = 203.5 NTy = 1480. NMx = -888000. Nmmxm = 24. mmym = 53. mmum = 3. mmvm = 36.01 mmσm = N/A-Mv/Ju = 219.2 N/mm2

xc = 21. mmyc = 38. mmvc = 21.01 mmσc = N/A-Mv/Ju = 128. N/mm2

τc = 4.331 N/mm2

σo = √σ2+3τ2 = 128.3 N/mm2

S* = 2566. mm3mm 0 18 24 42x

0

6

53

y

38σc,τc

σm

u

v

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IPERΣ01.pnzc.032

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.pnzc.032

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 101: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.rclf.033REAZIONI 843694 Recalcati Francesca

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

19/40F

FFb

19/40F

F

A B

61/40F

FFb

61/40F

F21/40Fb

A

C

F1/2Fb

F1/2Fb

C

D

21/40F1/40Fb

19/40F

E

B

F

F1/2Fb

F

D F

FF G

FFb

F

G

E

61/40F1/40Fb

61/40F1/40Fb

EC

IPERΣ01.rclf.033AZIONI INTERNE 843694 Recalcati Francesca

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

-19/40

-1-1

00

1 1 1

-√2/2

-61/40

F

1

-61/

400

21/4

0-1

9/40

1 0 0

-√2/2

0

F

-101

-21/

40-1

/2-1

/2

-1/4

00

-1/20 0 0

1

0

1/401/40

Fb

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IPE

RΣ0

1.rc

lf.03

3P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 8

4369

4 R

ecal

cati

Fra

nces

ca

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

AB

C

D

EF

G

F

W

X

X

q

q

Sch

ema

di c

alco

lo ip

erst

atic

o

-10

1-1/2-1/2-1/2

0 0

-1/2

00

0

1

0

00

Mo

fless

ione

da

caric

hi a

sseg

nati

00

0100

1 0

00

00

0

0

-1-1

Mx

fless

ione

da

iper

stat

ica

X=

1

IPE

RΣ0

1.rc

lf.03

3P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 8

4369

4 R

ecal

cati

Fra

nces

ca

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

Qua

dro

cont

ribut

i PLV

per

iper

stat

ica

X=

WE

C

→M

x(x)

Mo(

x)M

xMo

MxM

x∫M

xMo/

EJd

x∫X

MxM

x/E

Jdx

AB

b0

-Fb+

Fx

00

00

BA

b0

Fx

00

AC

bx/

bF

b-3/

2Fx

Fx-

3/2F

x2 /bx2 /b

2

01/

3Xb/

EJ

CA

b-1

+x/

b1/

2Fb-

3/2F

x-1

/2F

b+2F

x-3/

2Fx2 /b

1-2x

/b+

x2 /b2

CD

b0

-1/2

Fb

00

00

DC

b0

1/2F

b0

0

EB

b1-

x/b

1/2F

x-1/

2qx2

1/2F

x-F

x2 /b+

1/2q

x3 /b1-

2x/b

+x2 /b

2

1/24

Fb2 /E

J1/

3Xb/

EJ

BE

b-x

/b-1

/2F

x+1/

2qx2

1/2F

x2 /b-1

/2qx

3 /bx2 /b

2

DF

b0

-1/2

Fb+

Fx-

1/2q

x20

00

0F

D b

01/

2qx2

00

FG

b0

00

00

0G

F b

00

00

GE

√2b

0F

b-√2

/2F

x0

00

0

EC

b-1

00

10

Xb/

EJ

CE

b1

00

1

tota

li1/

24F

b2 /EJ

5/3X

b/E

J

iper

stat

ica

X=

WE

C-1

/40F

b

Svi

lupp

i di c

alco

lo ip

erst

atic

a

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IPERΣ01.rclf.033PROCEDIMENTO E RISULTATI 843694 Recalcati Francesca

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXCA = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXEB = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXBE = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXCE = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoAC = ∫

o

b( x/b -3/2 x2/b2 ) Fb 1/EJ dx = [1/2 x2/b -1/2 x3/b2 ]o

b Fb 1/EJ

= (1/2 b -1/2 b ) Fb 1/EJ = 0

LXoCA = ∫

o

b(-1/2 +2 x/b -3/2 x2/b2 ) Fb 1/EJ dx = [-1/2 x + x2/b -1/2 x3/b2 ]o

b Fb 1/EJ

= (-1/2 b + b -1/2 b ) Fb 1/EJ = 0

LXoEB = ∫

o

b(1/2 x/b - x2/b2 +1/2 x3/b3 ) Fb 1/EJ dx = [1/4 x2/b -1/3 x3/b2 +1/8 x4/b3 ]o

b Fb 1/EJ

= (1/4 b -1/3 b +1/8 b ) Fb 1/EJ = 1/24 Fb2/EJ

LXoBE = ∫

o

b(1/2 x2/b2 -1/2 x3/b3 ) Fb 1/EJ dx = [1/6 x3/b2 -1/8 x4/b3 ]o

b Fb 1/EJ

= (1/6 b -1/8 b ) Fb 1/EJ = 1/24 Fb2/EJ

IPERΣ01.rclf.033PROCEDIMENTO E RISULTATI 843694 Recalcati Francesca

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 582. mm2

Ju = 166519. mm4

Jv = 44082. mm4

yg = 37.89 mmN = -712.5 NTy = 1500. NMx = -1005000. Nmmxm = 18. mmum = -3. mmvm = -37.89 mmσm = N/A-Mv/Ju = -229.9 N/mm2

xc = 21. mmyc = 16. mmvc = -21.89 mmσc = N/A-Mv/Ju = -133.3 N/mm2

τc = 4.308 N/mm2

σo = √σ2+3τ2 = 133.6 N/mm2

S* = 2870. mm3mm 0 18 24 42x

0

48

55

y

16σc,τc

σm

u

v

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IPERΣ01.rclf.033

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.rclf.033

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 105: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.rzzm.034REAZIONI 833111 Rizzo Michele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

2/5F

1/2F2/5Fb

2/5F

1/2F

A

B

1/2F

1/2FFb

1/2F

1/2F1/2Fb

C

D

7/5F

1/2F2Fb

7/5F

1/2F3/5Fb

E

C

2/5F

3/2F2Fb

2/5F

5/2F

EB

9/10F2/5Fb

9/10F2/5Fb

A C

1/2F

1/2F

Fb

1/2F

1/2F

F

A

1/2F

1/2F

1/2F

1/2F

GF

1/2F

1/2F1/2Fb

1/2F

1/2F

DG

IPERΣ01.rzzm.034AZIONI INTERNE 833111 Rizzo Michele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

-1/2

-1/2

-1/2

-2/5-2/5

9/10

0

-1/2-1/2 -1/2

F

2/5

-1/2

-7/5

3/25/2

0

-√2/2

-1/21/2

-1/2

F

-2/5

0

11/

22

3/5

-20

2/5 2/5

1

0

00 1/20

Fb

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IPE

RΣ0

1.rz

zm.0

34P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 8

3311

1 R

izzo

Mic

hele

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

A

B

C

D E

FG

F

W

XX

q

q

0 0

11/2 21

-200

0

1

000

1/2

0

Mo

fless

ione

da

caric

hi a

sseg

nati

-10

00 0-1

001

1

0

000

00

Mx

fless

ione

da

iper

stat

ica

X=

1

IPE

RΣ0

1.rz

zm.0

34P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 8

3311

1 R

izzo

Mic

hele

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

Qua

dro

cont

ribut

i PLV

per

iper

stat

ica

X=

WA

B

→M

x(x)

Mo(

x)M

xMo

MxM

x∫M

xMo/

EJd

x∫X

MxM

x/E

Jdx

AB

b-1

+x/

b0

01-

2x/b

+x2 /b

2

01/

3Xb/

EJ

BA

bx/

b0

0x2 /b

2

CD

b0

Fb-

1/2F

x0

00

0D

C b

0-1

/2F

b-1/

2Fx

00

EC

b-x

/b2F

b-F

x-2

Fx+

Fx2 /b

x2 /b2

-2/3

Fb2 /E

J1/

3Xb/

EJ

CE

b1-

x/b

-Fb-

Fx

-Fb+

Fx2 /b

1-2x

/b+

x2 /b2

EB

b0

-2F

b+3/

2Fx+

1/2q

x20

00

0B

E b

05/

2Fx-

1/2q

x20

0

AC

b1

00

10

Xb/

EJ

CA

b-1

00

1

FA

√2b

0F

b-√2

/2F

x0

00

0

GF

b0

-1/2

Fx+

1/2q

x20

00

0F

G b

01/

2Fx-

1/2q

x20

0

DG

b0

1/2F

b-1/

2Fx

00

00

GD

b0

-1/2

Fx

00

tota

li-2

/3F

b2 /EJ

5/3X

b/E

J

iper

stat

ica

X=

WA

B2/

5Fb

Svi

lupp

i di c

alco

lo ip

erst

atic

a

Page 107: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.rzzm.034PROCEDIMENTO E RISULTATI 833111 Rizzo Michele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAB = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXBA = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXCE = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXAC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXCA = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoEC = ∫

o

b(-2 x/b + x2/b2 ) Fb 1/EJ dx = [- x2/b +1/3 x3/b2 ]o

b Fb 1/EJ

= (- b +1/3 b ) Fb 1/EJ = -2/3 Fb2/EJ

LXoCE = ∫

o

b(-1 + x2/b2 ) Fb 1/EJ dx = [- x +1/3 x3/b2 ]o

b Fb 1/EJ

= (- b +1/3 b ) Fb 1/EJ = -2/3 Fb2/EJ

IPERΣ01.rzzm.034PROCEDIMENTO E RISULTATI 833111 Rizzo Michele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 798. mm2

Ju = 175127. mm4

Jv = 81018. mm4

yg = 39.82 mmN = -292. NTy = 1095. NMx = -1051200. Nmmxm = 18. mmum = -3. mmvm = -39.82 mmσm = N/A-Mv/Ju = -239.4 N/mm2

xc = 21. mmyc = 17. mmvc = -22.82 mmσc = N/A-Mv/Ju = -137.3 N/mm2

τc = 3.329 N/mm2

σo = √σ2+3τ2 = 137.4 N/mm2

S* = 3194. mm3mm 0 18 24 42x

0

42

55

y

17σc,τc

σm

u

v

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IPERΣ01.rzzm.034

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.rzzm.034

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 109: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.rssl.035REAZIONI 866706 Rossi Luisa

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

2F

F3/2Fb

2F

F1/2Fb

A

B

23/40F23/40Fb

23/40F

C

D

23/40F

3F3Fb

23/40F

3F

ED

103/40F

F3Fb

63/40F

F37/40Fb

E

A

FFb

F

F

C

17/40F23/40Fb

17/40F23/40Fb

C A

F

F1/2Fb

F

BG

FGF

IPERΣ01.rssl.035AZIONI INTERNE 866706 Rossi Luisa

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1

0

-23/40

11

-√2/2

-17/40

111

F

2

-23/

40

-3

103/

4063

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-√2/2

0

-100

F

-3/2

1/2

23/4

00 30

-3-3

7/40

1

0

-23/40 -23/40

1/2000

Fb

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IPE

RΣ0

1.rs

sl.0

35P

RO

CE

DIM

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TO

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6670

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27.0

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olite

cnic

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no, v

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27.0

3.13

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9.18

Quadro contributi PLV per iperstatica X=WAC

→Mx(x)Mo(x)MxMoMxMx∫MxMo/EJdx∫XMxMx/EJdx

AB b0-3/2Fb+2Fx0000

BA b0-1/2Fb+2Fx00

CD b-1+x/b001-2x/b+x2/b

2

01/3Xb/EJDC bx/b00x

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2

ED b03Fb-3Fx0000

DE b0-3Fx00

EA b-x/b-3Fb+2Fx-1/2qx2

3Fx-2Fx2/b+1/2qx

3/bx

2/b

2

23/24Fb2/EJ1/3Xb/EJ

AE b1-x/b3/2Fb+Fx+1/2qx2

3/2Fb-1/2Fx-1/2Fx2/b-1/2qx

3/b1-2x/b+x

2/b

2

FC √2b0Fb-√2/2Fx0000

CA b10010Xb/EJ

AC b-1001

BG b01/2Fb-Fx+1/2qx2

0000

GB b0-1/2qx2

00

GF b000000

FG b0000

totali23/24Fb2/EJ5/3Xb/EJ

iperstatica X=WAC-23/40Fb

Svi

lupp

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IPERΣ01.rssl.035PROCEDIMENTO E RISULTATI 866706 Rossi Luisa

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXCD = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXDC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEA = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXAE = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXCA = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXAC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoEA = ∫

o

b(3 x/b -2 x2/b2 +1/2 x3/b3 ) Fb 1/EJ dx = [3/2 x2/b -2/3 x3/b2 +1/8 x4/b3 ]o

b Fb 1/EJ

= (3/2 b -2/3 b +1/8 b ) Fb 1/EJ = 23/24 Fb2/EJ

LXoAE = ∫

o

b(3/2 -1/2 x/b -1/2 x2/b2 -1/2 x3/b3 ) Fb 1/EJ dx = [3/2 x -1/4 x2/b -1/6 x3/b2 -1/8 x4/b3 ]o

b Fb 1/EJ

= (3/2 b -1/4 b -1/6 b -1/8 b ) Fb 1/EJ = 23/24 Fb2/EJ

IPERΣ01.rssl.035PROCEDIMENTO E RISULTATI 866706 Rossi Luisa

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 912. mm2

Ju = 272448. mm4

Jv = 71424. mm4

yg = 34.13 mmN = -391. NTy = -2040. NMx = 1591200. Nmmxm = 18. mmum = -6. mmvm = -34.13 mmσm = N/A-Mv/Ju = 198.9 N/mm2

xc = 24. mmyc = 14. mmvc = -20.13 mmσc = N/A-Mv/Ju = 117.1 N/mm2

τc = 2.844 N/mm2

σo = √σ2+3τ2 = 117.3 N/mm2

S* = 4558. mm3mm 0 18 30 48x

0

48

55

y

14σc,τc

σm

u

v

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IPERΣ01.rssl.035

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.rssl.035

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 113: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.rsss.036REAZIONI 877254 Russo Stefano

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

3F

1/8F3Fb

3F

1/8F

A

B

F

17/8F3Fb

F

17/8F7/8Fb

A C

F

2F3/2Fb

F

2F1/2Fb

C D

9/8F5/8Fb

1/8F

EB

F

F1/2Fb

F

D

F

F

F

G

F

Fb

F

G

E

1/8F5/8Fb

1/8F5/8Fb

E

C

IPERΣ01.rsss.036AZIONI INTERNE 877254 Russo Stefano

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1/8

-1 -1

00

-1-1

-1

√2/2

-1/8

F

-3

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0

F

30

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5/80

1/2

00

01

0

-5/8

-5/8

Fb

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IPE

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usso

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27.0

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27.0

3.13

03.0

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ribut

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per

iper

stat

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X=

WE

B

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x(x)

Mo(

x)M

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MxM

x∫M

xMo/

EJd

x∫X

MxM

x/E

Jdx

AB

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x2 /b2

Fb2 /E

J1/

3Xb/

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CA

b1-

x/b

3/2F

b+3/

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3/2F

b-3/

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b-1

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Fb2 /E

J1/

3Xb/

EJ

BE

bx/

b1/

2Fx-

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x21/

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qx3 /b

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b0

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GE

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CE

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li25

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b

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IPERΣ01.rsss.036PROCEDIMENTO E RISULTATI 877254 Russo Stefano

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXCA = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXEB = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXBE = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXCE = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoAC = ∫

o

b(3 x/b -3/2 x2/b2 ) Fb 1/EJ dx = [3/2 x2/b -1/2 x3/b2 ]o

b Fb 1/EJ

= (3/2 b -1/2 b ) Fb 1/EJ = Fb2/EJ

LXoCA = ∫

o

b(3/2 -3/2 x2/b2 ) Fb 1/EJ dx = [3/2 x -1/2 x3/b2 ]o

b Fb 1/EJ

= (3/2 b -1/2 b ) Fb 1/EJ = Fb2/EJ

LXoEB = ∫

o

b(1/2 x/b - x2/b2 +1/2 x3/b3 ) Fb 1/EJ dx = [1/4 x2/b -1/3 x3/b2 +1/8 x4/b3 ]o

b Fb 1/EJ

= (1/4 b -1/3 b +1/8 b ) Fb 1/EJ = 1/24 Fb2/EJ

LXoBE = ∫

o

b(1/2 x2/b2 -1/2 x3/b3 ) Fb 1/EJ dx = [1/6 x3/b2 -1/8 x4/b3 ]o

b Fb 1/EJ

= (1/6 b -1/8 b ) Fb 1/EJ = 1/24 Fb2/EJ

IPERΣ01.rsss.036PROCEDIMENTO E RISULTATI 877254 Russo Stefano

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 1128. mm2

Ju = 293725. mm4

Jv = 125856. mm4

yg = 36.21 mmN = 85. NTy = -2040. NMx = 1693200. Nmmxm = 18. mmum = -6. mmvm = -36.21 mmσm = N/A-Mv/Ju = 208.8 N/mm2

xc = 24. mmyc = 15. mmvc = -21.21 mmσc = N/A-Mv/Ju = 122.4 N/mm2

τc = 2.991 N/mm2

σo = √σ2+3τ2 = 122.5 N/mm2

S* = 5168. mm3mm 0 18 30 48x

0

42

55

y

15σc,τc

σm

u

v

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IPERΣ01.rsss.036

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.rsss.036

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

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IPERΣ01.slvg.037REAZIONI 834781 Salvatori Gabriele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1/2F

2/5F2/5Fb

1/2F

2/5F

A B

1/2F

1/2FFb

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CD

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B

9/10F2/5Fb

9/10F2/5Fb

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C

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1/2F

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1/2F

1/2F

F

A

1/2F

1/2F

1/2F

1/2F

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1/2F

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D

G

IPERΣ01.slvg.037AZIONI INTERNE 834781 Salvatori Gabriele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1/2

1/2 1/2

2/5

2/5

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0

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1/2

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0

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IPE

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stat

ica

X=

WC

A2/

5Fb

Svi

lupp

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alco

lo ip

erst

atic

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Page 119: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.slvg.037PROCEDIMENTO E RISULTATI 834781 Salvatori Gabriele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAB = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXBA = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXCE = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXAC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXCA = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoEC = ∫

o

b(-2 x/b + x2/b2 ) Fb 1/EJ dx = [- x2/b +1/3 x3/b2 ]o

b Fb 1/EJ

= (- b +1/3 b ) Fb 1/EJ = -2/3 Fb2/EJ

LXoCE = ∫

o

b(-1 + x2/b2 ) Fb 1/EJ dx = [- x +1/3 x3/b2 ]o

b Fb 1/EJ

= (- b +1/3 b ) Fb 1/EJ = -2/3 Fb2/EJ

IPERΣ01.slvg.037PROCEDIMENTO E RISULTATI 834781 Salvatori Gabriele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 498. mm2

Ju = 147997. mm4

Jv = 16614. mm4

yg = 35.6 mmN = 208. NTy = 780. NMx = -915200. Nmmxm = 12. mmum = -3. mmvm = -35.6 mmσm = N/A-Mv/Ju = -219.7 N/mm2

xc = 15. mmyc = 15. mmvc = -20.6 mmσc = N/A-Mv/Ju = -126.9 N/mm2

τc = 2.221 N/mm2

σo = √σ2+3τ2 = 127. N/mm2

S* = 2529. mm3mm 0 12 18 30x

0

48

55

y

15σc,τc

σm

u

v

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IPERΣ01.slvg.037

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.slvg.037

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 121: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.sblm.038REAZIONI 877022 Sblendido Maria Angela

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

F1/2Fb

F1/2Fb

AB

1/40F1/40Fb

1/40F

C D

F

1/40FFb

F

1/40F

E

D

F

79/40FFb

F

39/40F19/40Fb

EA

F

Fb

F

F

C

39/40F1/40Fb

39/40F1/40Fb

C

A

F

F1/2Fb

F

B

G

F

G

F

IPERΣ01.sblm.038AZIONI INTERNE 877022 Sblendido Maria Angela

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1

0

1/40

11

√2/2

39/4

0

-1-1

-1

F

0

-1/40

1

-79/40-39/40

-√2/2

0

10

0

F

-1/2-1/2

1/40 0

-10

1-19/40

1

0

-1/4

0-1

/40-1/2

00

0

Fb

Page 122: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPE

RΣ0

1.sb

lm.0

38P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 8

7702

2 S

blen

dido

Mar

ia

@ A

dolfo

Zav

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i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

AB

C

D

E

F

G

F

W

X

X

q

-1/2

-1/2

00

-10

1-1

/2

1

0

00

-1/20 0 0

Mo

fless

ione

da

caric

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sseg

nati

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10

0 0

01

0

0

-1-1

0 0 0 0

Mx

fless

ione

da

iper

stat

ica

X=

1

IPE

RΣ0

1.sb

lm.0

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TO

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7702

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Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

Quadro contributi PLV per iperstatica X=WCA

→Mx(x)Mo(x)MxMoMxMx∫MxMo/EJdx∫XMxMx/EJdx

AB b0-1/2Fb0000

BA b01/2Fb00

CD b1-x/b001-2x/b+x2/b

2

01/3Xb/EJDC b-x/b00x

2/b

2

ED b0-Fb+Fx0000

DE b0Fx00

EA bx/bFb-2Fx+1/2qx2

Fx-2Fx2/b+1/2qx

3/bx

2/b

2

-1/24Fb2/EJ1/3Xb/EJ

AE b-1+x/b1/2Fb-Fx-1/2qx2

-1/2Fb+3/2Fx-1/2Fx2/b-1/2qx

3/b1-2x/b+x

2/b

2

FC √2b0Fb-√2/2Fx0000

CA b-10010Xb/EJ

AC b1001

BG b0-1/2Fb+Fx-1/2qx2

0000

GB b01/2qx2

00

GF b000000

FG b0000

totali-1/24Fb2/EJ5/3Xb/EJ

iperstatica X=WCA1/40Fb

Svi

lupp

i di c

alco

lo ip

erst

atic

a

Page 123: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.sblm.038PROCEDIMENTO E RISULTATI 877022 Sblendido Maria

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXCD = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXDC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEA = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXAE = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXCA = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXAC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoEA = ∫

o

b( x/b -2 x2/b2 +1/2 x3/b3 ) Fb 1/EJ dx = [1/2 x2/b -2/3 x3/b2 +1/8 x4/b3 ]o

b Fb 1/EJ

= (1/2 b -2/3 b +1/8 b ) Fb 1/EJ = -1/24 Fb2/EJ

LXoAE = ∫

o

b(-1/2 +3/2 x/b -1/2 x2/b2 -1/2 x3/b3 ) Fb 1/EJ dx

= [-1/2 x +3/4 x2/b -1/6 x3/b2 -1/8 x4/b3 ]o

b Fb 1/EJ

= (-1/2 b +3/4 b -1/6 b -1/8 b ) Fb 1/EJ = -1/24 Fb2/EJ

IPERΣ01.sblm.038PROCEDIMENTO E RISULTATI 877022 Sblendido Maria

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 642. mm2

Ju = 158306. mm4

Jv = 30006. mm4

yg = 37.71 mmN = 25.75 NTy = 1030. NMx = -957900. Nmmxm = 12. mmum = -3. mmvm = -37.71 mmσm = N/A-Mv/Ju = -228.1 N/mm2

xc = 15. mmyc = 16. mmvc = -21.71 mmσc = N/A-Mv/Ju = -131.3 N/mm2

τc = 3.092 N/mm2

σo = √σ2+3τ2 = 131.4 N/mm2

S* = 2852. mm3mm 0 12 18 30x

0

42

55

y

16σc,τc

σm

u

v

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IPERΣ01.sblm.038

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.sblm.038

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 125: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.scll.039REAZIONI 835340 Scalia Luca

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

59/20F

1/2F9/2Fb

59/20F

1/2F31/20Fb

A

B

19/20F

9/2F9/2Fb

19/20F

9/2F

A C

19/20F

1/2F19/20Fb

19/20F

1/2F

D

C

5/2F

1/2F5/2Fb

3/2F

1/2F1/2Fb

B

E

3/2F

1/2F

3/2F

1/2F

F G

3/2F

1/2F1/2Fb

3/2F

1/2F

E F

9/20F19/20Fb

9/20F19/20Fb

DB

3/2F

1/2F

Fb

3/2F

1/2F

G

D

IPERΣ01.scll.039AZIONI INTERNE 835340 Scalia Luca

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1/2

19/20

1/2

1/2

1/2

-3/2 -3/2-3/2

-9/20

√2

F

59/2

0

-9/2

-19/

20

5/2

3/2

1/2-1/2

1/2

0

-√2/2

F

-9/2

-31/

20

9/2 0

19/2

00

-5/2

-1/2

0 0-1/2

0

-19/20-19/20

1

0

Fb

Page 126: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPE

RΣ0

1.sc

ll.03

9P

RO

CE

DIM

EN

TO

E R

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LTA

TI 8

3534

0 S

calia

Luc

a

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dolfo

Zav

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si, P

olite

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Mila

no, v

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27.0

3.13

03.0

9.18

A B

C

D

EF

G

F

W

X

X

q

q

Sch

ema

di c

alco

lo ip

erst

atic

o

-9/2 -5/2

9/2

0 00

-5/2 -1/2

00

-1/2

000

1

0

Mo

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ione

da

caric

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sseg

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0-1

00

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0

Mx

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IPE

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no, v

ers.

27.0

3.13

03.0

9.18

Qua

dro

cont

ribut

i PLV

per

iper

stat

ica

X=

WB

D

→M

x(x)

Mo(

x)M

xMo

MxM

x∫M

xMo/

EJd

x∫X

MxM

x/E

Jdx

AB

b-x

/b-9

/2F

b+2F

x9/

2Fx-

2Fx2 /b

x2 /b2

19/1

2Fb2 /E

J1/

3Xb/

EJ

BA

b1-

x/b

5/2F

b+2F

x5/

2Fb-

1/2F

x-2F

x2 /b1-

2x/b

+x2 /b

2

AC

b0

9/2F

b-9/

2Fx

00

00

CA

b0

-9/2

Fx

00

DC

b-1

+x/

b0

01-

2x/b

+x2 /b

2

01/

3Xb/

EJ

CD

bx/

b0

0x2 /b

2

BE

b0

-5/2

Fb+

5/2F

x-1/

2qx2

00

00

EB

b0

1/2F

b+3/

2Fx+

1/2q

x20

0

FG

b0

1/2F

x-1/

2qx2

00

00

GF

b0

-1/2

Fx+

1/2q

x20

0

EF

b0

-1/2

Fb+

1/2F

x0

00

0F

E b

01/

2Fx

00

DB

b1

00

10

Xb/

EJ

BD

b-1

00

1

GD

√2b

0F

b-√2

/2F

x0

00

0

tota

li19

/12F

b2 /EJ

5/3X

b/E

J

iper

stat

ica

X=

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D-1

9/20

Fb

Svi

lupp

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lo ip

erst

atic

a

Page 127: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.scll.039PROCEDIMENTO E RISULTATI 835340 Scalia Luca

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAB = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXBA = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXDC = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXCD = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXDB = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXBD = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoAB = ∫

o

b(9/2 x/b -2 x2/b2 ) Fb 1/EJ dx = [9/4 x2/b -2/3 x3/b2 ]o

b Fb 1/EJ

= (9/4 b -2/3 b ) Fb 1/EJ = 19/12 Fb2/EJ

LXoBA = ∫

o

b(5/2 -1/2 x/b -2 x2/b2 ) Fb 1/EJ dx = [5/2 x -1/4 x2/b -2/3 x3/b2 ]o

b Fb 1/EJ

= (5/2 b -1/4 b -2/3 b ) Fb 1/EJ = 19/12 Fb2/EJ

IPERΣ01.scll.039PROCEDIMENTO E RISULTATI 835340 Scalia Luca

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 828. mm2

Ju = 244195. mm4

Jv = 34128. mm4

yg = 32.37 mmN = 380. NTy = -1800. NMx = 1764000. Nmmxm = 12. mmum = -6. mmvm = -32.37 mmσm = N/A-Mv/Ju = 234.3 N/mm2

xc = 18. mmyc = 14. mmvc = -18.37 mmσc = N/A-Mv/Ju = 133.2 N/mm2

τc = 2.618 N/mm2

σo = √σ2+3τ2 = 133.2 N/mm2

S* = 4262. mm3mm 0 12 24 36x

0

48

55

y

14σc,τc

σm

u

v

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IPERΣ01.scll.039

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.scll.039

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

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IPERΣ01.shhe.040REAZIONI 809828 Shehu Elton

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

3/2F

1/2FFb

3/2F

1/2F1/2Fb

A

B

17/40F

1/2F17/40Fb

17/40F

1/2F

C

D

17/40F

5/2F5/2Fb

17/40F

5/2F

ED

97/40F

1/2F5/2Fb

57/40F

1/2F23/40Fb

E

A

1/2F

1/2F

Fb

1/2F

1/2F

F

C

3/40F17/40Fb

3/40F17/40Fb

C A

1/2F

1/2F1/2Fb

1/2F

1/2F

BG

1/2F

1/2F

1/2F

1/2F

GF

IPERΣ01.shhe.040AZIONI INTERNE 809828 Shehu Elton

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1/2

1/2

-17/40

1/2

1/2

0

-3/40

1/21/21/2

F

3/2

-17/

40

-5/2

97/4

057

/40

-√2/2

0

-1/2-1/21/2

F

-11/

2

17/4

00 5/20

-5/2

-23/

40

1

0

-17/40 -17/40

1/2000

Fb

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IPE

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X

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1

0

00 1/

20

00

Mo

fless

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da

caric

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sseg

nati

0 0

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00

0-1

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IPE

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ISU

LTA

TI 8

0982

8 S

hehu

Elto

n

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

Qua

dro

cont

ribut

i PLV

per

iper

stat

ica

X=

WA

C

→M

x(x)

Mo(

x)M

xMo

MxM

x∫M

xMo/

EJd

x∫X

MxM

x/E

Jdx

AB

b0

-Fb+

3/2F

x0

00

0B

A b

0-1

/2F

b+3/

2Fx

00

CD

b-1

+x/

b0

01-

2x/b

+x2 /b

2

01/

3Xb/

EJ

DC

bx/

b0

0x2 /b

2

ED

b0

5/2F

b-5/

2Fx

00

00

DE

b0

-5/2

Fx

00

EA

b-x

/b-5

/2F

b+2F

x-1/

2qx2

5/2F

x-2F

x2 /b+

1/2q

x3 /bx2 /b

2

17/2

4Fb2 /E

J1/

3Xb/

EJ

AE

b1-

x/b

Fb+

Fx+

1/2q

x2F

b-1/

2Fx2 /b

-1/2

qx3 /b

1-2x

/b+

x2 /b2

FC

√2b

0F

b-√2

/2F

x0

00

0

CA

b1

00

10

Xb/

EJ

AC

b-1

00

1

BG

b0

1/2F

b-1/

2Fx

00

00

GB

b0

-1/2

Fx

00

GF

b0

-1/2

Fx+

1/2q

x20

00

0F

G b

01/

2Fx-

1/2q

x20

0

tota

li17

/24F

b2 /EJ

5/3X

b/E

J

iper

stat

ica

X=

WA

C-1

7/40

Fb

Svi

lupp

i di c

alco

lo ip

erst

atic

a

Page 131: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.shhe.040PROCEDIMENTO E RISULTATI 809828 Shehu Elton

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXCD = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXDC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEA = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXAE = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXCA = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXAC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoEA = ∫

o

b(5/2 x/b -2 x2/b2 +1/2 x3/b3 ) Fb 1/EJ dx = [5/4 x2/b -2/3 x3/b2 +1/8 x4/b3 ]o

b Fb 1/EJ

= (5/4 b -2/3 b +1/8 b ) Fb 1/EJ = 17/24 Fb2/EJ

LXoAE = ∫

o

b(1 -1/2 x2/b2 -1/2 x3/b3 ) Fb 1/EJ dx = [ x -1/6 x3/b2 -1/8 x4/b3 ]o

b Fb 1/EJ

= ( b -1/6 b -1/8 b ) Fb 1/EJ = 17/24 Fb2/EJ

IPERΣ01.shhe.040PROCEDIMENTO E RISULTATI 809828 Shehu Elton

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 972. mm2

Ju = 264196. mm4

Jv = 56592. mm4

yg = 34.24 mmN = -505.8 NTy = -2975. NMx = 1547000. Nmmxm = 12. mmum = -6. mmvm = -34.24 mmσm = N/A-Mv/Ju = 200. N/mm2

xc = 18. mmyc = 14. mmvc = -20.24 mmσc = N/A-Mv/Ju = 118. N/mm2

τc = 4.294 N/mm2

σo = √σ2+3τ2 = 118.2 N/mm2

S* = 4576. mm3mm 0 12 24 36x

0

42

55

y

14σc,τc

σm

u

v

Page 132: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.shhe.040

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.shhe.040

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 133: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.smnd.041REAZIONI 847315 Simonini Davide

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1/2F

59/20F9/2Fb

1/2F

59/20F31/20Fb

A B

9/2F

19/20F9/2Fb

9/2F

19/20F

A

C

1/2F

19/20F19/20Fb

1/2F

19/20F

DC

1/2F

5/2F5/2Fb

1/2F

3/2F1/2Fb

B E

1/2F

3/2F

1/2F

3/2F

F

G

1/2F

3/2F1/2Fb

1/2F

3/2F

E

F

9/20F19/20Fb

9/20F19/20Fb

D

B

1/2F

3/2F

Fb

1/2F

3/2F

G

D

IPERΣ01.smnd.041AZIONI INTERNE 847315 Simonini Davide

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

-1/2

-19/

20

-1/2

-1/2 -1/2

3/2

3/2

3/2

9/20

-√2

F

59/20

-9/2

-19/20

5/2 3/2

1/2

-1/2

1/20

-√2/2

F

-9/2 -31/20

9/2

0

19/200

-5/2 -1/2

00

-1/2

0-19/

20-1

9/20

1

0

Fb

Page 134: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPE

RΣ0

1.sm

nd.0

41P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 8

4731

5 S

imon

ini D

avid

e

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

AB

C

D

E

F

G

F

W

XX

q

q

Sch

ema

di c

alco

lo ip

erst

atic

o

-9/2

-5/2

9/2 0

00

-5/2

-1/2

0 0-1/2

0

00

1

0

Mo

fless

ione

da

caric

hi a

sseg

nati

0-1

0 0

-10

00

0 00 0

11

0

0

Mx

fless

ione

da

iper

stat

ica

X=

1

IPE

RΣ0

1.sm

nd.0

41P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 8

4731

5 S

imon

ini D

avid

e

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

Qua

dro

cont

ribut

i PLV

per

iper

stat

ica

X=

WD

C

→M

x(x)

Mo(

x)M

xMo

MxM

x∫M

xMo/

EJd

x∫X

MxM

x/E

Jdx

AB

b-x

/b-9

/2F

b+2F

x9/

2Fx-

2Fx2 /b

x2 /b2

19/1

2Fb2 /E

J1/

3Xb/

EJ

BA

b1-

x/b

5/2F

b+2F

x5/

2Fb-

1/2F

x-2F

x2 /b1-

2x/b

+x2 /b

2

AC

b0

9/2F

b-9/

2Fx

00

00

CA

b0

-9/2

Fx

00

DC

b-1

+x/

b0

01-

2x/b

+x2 /b

2

01/

3Xb/

EJ

CD

bx/

b0

0x2 /b

2

BE

b0

-5/2

Fb+

5/2F

x-1/

2qx2

00

00

EB

b0

1/2F

b+3/

2Fx+

1/2q

x20

0

FG

b0

1/2F

x-1/

2qx2

00

00

GF

b0

-1/2

Fx+

1/2q

x20

0

EF

b0

-1/2

Fb+

1/2F

x0

00

0F

E b

01/

2Fx

00

DB

b1

00

10

Xb/

EJ

BD

b-1

00

1

GD

√2b

0F

b-√2

/2F

x0

00

0

tota

li19

/12F

b2 /EJ

5/3X

b/E

J

iper

stat

ica

X=

WD

C-1

9/20

Fb

Svi

lupp

i di c

alco

lo ip

erst

atic

a

Page 135: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.smnd.041PROCEDIMENTO E RISULTATI 847315 Simonini Davide

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAB = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXBA = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXDC = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXCD = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXDB = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXBD = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoAB = ∫

o

b(9/2 x/b -2 x2/b2 ) Fb 1/EJ dx = [9/4 x2/b -2/3 x3/b2 ]o

b Fb 1/EJ

= (9/4 b -2/3 b ) Fb 1/EJ = 19/12 Fb2/EJ

LXoBA = ∫

o

b(5/2 -1/2 x/b -2 x2/b2 ) Fb 1/EJ dx = [5/2 x -1/4 x2/b -2/3 x3/b2 ]o

b Fb 1/EJ

= (5/2 b -1/4 b -2/3 b ) Fb 1/EJ = 19/12 Fb2/EJ

IPERΣ01.smnd.041PROCEDIMENTO E RISULTATI 847315 Simonini Davide

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 948. mm2

Ju = 262515. mm4

Jv = 52848. mm4

yg = 20.97 mmN = -589. NTy = -2790. NMx = 1590300. Nmmxm = 24. mmym = 55. mmum = 6. mmvm = 34.03 mmσm = N/A-Mv/Ju = -206.8 N/mm2

xc = 18. mmyc = 41. mmvc = 20.03 mmσc = N/A-Mv/Ju = -122. N/mm2

τc = 4.022 N/mm2

σo = √σ2+3τ2 = 122.2 N/mm2

S* = 4541. mm3mm 0 12 24 36x

0

12

55

y

41σc,τc

σm

u

v

Page 136: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.smnd.041

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.smnd.041

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 137: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.sldl.042REAZIONI 866259 Soldavini Luca Giovanni

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1/2F

1/2FFb

1/2F

1/2F1/2Fb

AB

1/2F

7/40F7/40Fb

1/2F

7/40F

C D

1/2F

7/40F1/2Fb

1/2F

7/40F

E

D

1/2F

73/40F1/2Fb

1/2F

33/40F33/40Fb

EA

1/2F

3/2F

Fb

1/2F

3/2F

F

C

53/40F7/40Fb

53/40F7/40Fb

C

A

1/2F

3/2F1/2Fb

1/2F

3/2F

B

G

1/2F

3/2F

1/2F

3/2F

G

F

IPERΣ01.sldl.042AZIONI INTERNE 866259 Soldavini Luca Giovanni

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1/2

1/2

7/40

1/21/2

√2

53/4

0

-3/2

-3/2

-3/2

F

1/2

-7/40

1/2

-73/40-33/40

-√2/2

01/2

1/2

-1/2

F

-1-1/2

7/40 0

-1/2

0

1/2-33/40

1

0

-7/4

0-7

/40-1/2

00

0

Fb

Page 138: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPE

RΣ0

1.sl

dl.0

42P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 8

6625

9 S

olda

vini

Luc

a

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

AB

C

D

E

F

G

F

W

X

X

q

q

-1-1

/2

00

-1/20

1/2

-1

1

0

00

-1/20 0 0

Mo

fless

ione

da

caric

hi a

sseg

nati

00

-10

0 0

0-1

0

0

11

0 0 0 0

Mx

fless

ione

da

iper

stat

ica

X=

1

IPE

RΣ0

1.sl

dl.0

42P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 8

6625

9 S

olda

vini

Luc

a

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

Quadro contributi PLV per iperstatica X=WAC

→Mx(x)Mo(x)MxMoMxMx∫MxMo/EJdx∫XMxMx/EJdx

AB b0-Fb+1/2Fx0000

BA b01/2Fb+1/2Fx00

CD b-1+x/b001-2x/b+x2/b

2

01/3Xb/EJDC bx/b00x

2/b

2

ED b0-1/2Fb+1/2Fx0000

DE b01/2Fx00

EA b-x/b1/2Fb-2Fx+1/2qx2

-1/2Fx+2Fx2/b-1/2qx

3/bx

2/b

2

7/24Fb2/EJ1/3Xb/EJ

AE b1-x/bFb-Fx-1/2qx2

Fb-2Fx+1/2Fx2/b+1/2qx

3/b1-2x/b+x

2/b

2

FC √2b0Fb-√2/2Fx0000

CA b10010Xb/EJ

AC b-1001

BG b0-1/2Fb+1/2Fx0000

GB b01/2Fx00

GF b01/2Fx-1/2qx2

0000

FG b0-1/2Fx+1/2qx2

00

totali7/24Fb2/EJ5/3Xb/EJ

iperstatica X=WAC-7/40Fb

Svi

lupp

i di c

alco

lo ip

erst

atic

a

Page 139: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.sldl.042PROCEDIMENTO E RISULTATI 866259 Soldavini Luca

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXCD = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXDC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEA = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXAE = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXCA = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXAC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoEA = ∫

o

b(-1/2 x/b +2 x2/b2 -1/2 x3/b3 ) Fb 1/EJ dx = [-1/4 x2/b +2/3 x3/b2 -1/8 x4/b3 ]o

b Fb 1/EJ

= (-1/4 b +2/3 b -1/8 b ) Fb 1/EJ = 7/24 Fb2/EJ

LXoAE = ∫

o

b(1 -2 x/b +1/2 x2/b2 +1/2 x3/b3 ) Fb 1/EJ dx = [ x - x2/b +1/6 x3/b2 +1/8 x4/b3 ]o

b Fb 1/EJ

= ( b - b +1/6 b +1/8 b ) Fb 1/EJ = 7/24 Fb2/EJ

IPERΣ01.sldl.042PROCEDIMENTO E RISULTATI 866259 Soldavini Luca

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 474. mm2

Ju = 143796. mm4

Jv = 14382. mm4

yg = 20.06 mmN = 2093. NTy = -1047. NMx = 917600. Nmmxm = 18. mmym = 55. mmum = 3. mmvm = 34.94 mmσm = N/A-Mv/Ju = -218.6 N/mm2

xc = 15. mmyc = 40. mmvc = 19.94 mmσc = N/A-Mv/Ju = -122.8 N/mm2

τc = 2.996 N/mm2

σo = √σ2+3τ2 = 123. N/mm2

S* = 2470. mm3mm 0 12 18 30x

0

6

55

y

40σc,τc

σm

u

v

Page 140: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.sldl.042

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.sldl.042

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 141: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.spgs.043REAZIONI 833411 Spagnolo Silvia

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

5/8F

1/2F1/2Fb

5/8F

1/2F

A B

11/8F

1/2F1/2Fb

11/8F

1/2F7/8Fb

A

C

1/2F

1/2FFb

1/2F

1/2F1/2Fb

C

D

3/8F

1/2F1/8Fb

5/8F

1/2F

E

B

3/2F

1/2F1/2Fb

3/2F

1/2F

D F

3/2F

1/2F

3/2F

1/2F

F G

3/2F

1/2F

Fb

3/2F

1/2F

G

E

15/8F1/8Fb

15/8F1/8Fb

EC

IPERΣ01.spgs.043AZIONI INTERNE 833411 Spagnolo Silvia

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

-5/8

-1/2

-1/2

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F

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-1/2

1/8

0

-1/20 0 0

1

0

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Fb

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IPERΣ01.spgs.043PROCEDIMENTO E RISULTATI 833411 Spagnolo Silvia

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXCA = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXEB = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXBE = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXCE = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoAC = ∫

o

b(-1/2 x/b +3/2 x2/b2 ) Fb 1/EJ dx = [-1/4 x2/b +1/2 x3/b2 ]o

b Fb 1/EJ

= (-1/4 b +1/2 b ) Fb 1/EJ = 1/4 Fb2/EJ

LXoCA = ∫

o

b(1 -5/2 x/b +3/2 x2/b2 ) Fb 1/EJ dx = [ x -5/4 x2/b +1/2 x3/b2 ]o

b Fb 1/EJ

= ( b -5/4 b +1/2 b ) Fb 1/EJ = 1/4 Fb2/EJ

LXoEB = ∫

o

b(-1/2 x/b + x2/b2 -1/2 x3/b3 ) Fb 1/EJ dx = [-1/4 x2/b +1/3 x3/b2 -1/8 x4/b3 ]o

b Fb 1/EJ

= (-1/4 b +1/3 b -1/8 b ) Fb 1/EJ = -1/24 Fb2/EJ

LXoBE = ∫

o

b(-1/2 x2/b2 +1/2 x3/b3 ) Fb 1/EJ dx = [-1/6 x3/b2 +1/8 x4/b3 ]o

b Fb 1/EJ

= (-1/6 b +1/8 b ) Fb 1/EJ = -1/24 Fb2/EJ

IPERΣ01.spgs.043PROCEDIMENTO E RISULTATI 833411 Spagnolo Silvia

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 876. mm2

Ju = 264708. mm4

Jv = 62352. mm4

yg = 21.46 mmN = -3748. NTy = -1874. NMx = 1775500. Nmmxm = 30. mmym = 55. mmum = 6. mmvm = 33.54 mmσm = N/A-Mv/Ju = -229.3 N/mm2

xc = 24. mmyc = 41. mmvc = 19.54 mmσc = N/A-Mv/Ju = -135.3 N/mm2

τc = 2.63 N/mm2

σo = √σ2+3τ2 = 135.4 N/mm2

S* = 4459. mm3mm 0 18 30 48x

0

6

55

y

41σc,τc

σm

u

v

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IPERΣ01.spgs.043

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.spgs.043

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 145: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.sbsa.044REAZIONI 870941 Subasingha Arachchige Fernando Sh

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

14/5F

F4Fb

14/5F

F6/5Fb

A

B

4/5F

4F4Fb

4/5F

4F

A C

4/5F4/5Fb

4/5F

D

C

2F

F2Fb

F

F1/2Fb

B

E

FF G

F

F1/2Fb

F

E F

4/5F4/5Fb

4/5F4/5Fb

DB

FFb

F

G

D

IPERΣ01.sbsa.044AZIONI INTERNE 870941 Subasingha Arachchige Fernando

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1

4/5

0

11

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-4/5

√2/2

F

14/5

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-4/5

21

01 0

0

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4 0

4/5

0

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0 0-1/2

0

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1

0

Fb

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Page 147: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.sbsa.044PROCEDIMENTO E RISULTATI 870941 Subasingha

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAB = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXBA = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXDC = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXCD = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXDB = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXBD = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoAB = ∫

o

b(4 x/b -2 x2/b2 ) Fb 1/EJ dx = [2 x2/b -2/3 x3/b2 ]o

b Fb 1/EJ

= (2 b -2/3 b ) Fb 1/EJ = 4/3 Fb2/EJ

LXoBA = ∫

o

b(2 -2 x2/b2 ) Fb 1/EJ dx = [2 x -2/3 x3/b2 ]o

b Fb 1/EJ

= (2 b -2/3 b ) Fb 1/EJ = 4/3 Fb2/EJ

IPERΣ01.sbsa.044PROCEDIMENTO E RISULTATI 870941 Subasingha

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 618. mm2

Ju = 157731. mm4

Jv = 27774. mm4

yg = 17.48 mmN = 280. NTy = -1400. NMx = 1008000. Nmmxm = 18. mmym = 55. mmum = 3. mmvm = 37.52 mmσm = N/A-Mv/Ju = -239.3 N/mm2

xc = 15. mmyc = 39. mmvc = 21.52 mmσc = N/A-Mv/Ju = -137.1 N/mm2

τc = 4.192 N/mm2

σo = √σ2+3τ2 = 137.3 N/mm2

S* = 2834. mm3mm 0 12 18 30x

0

12

55

y

39σc,τc

σm

u

v

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IPERΣ01.sbsa.044

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.sbsa.044

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

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IPERΣ01.trns.045REAZIONI 843775 Tarantola Stefano

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1/4F1/4Fb

1/4F

A

B

F1/2Fb

F1/2Fb

C

D

5/4F

F3/2Fb

5/4F

F1/4Fb

E

C

1/4F

F3/2Fb

1/4F

2F

EB

5/4F1/4Fb

5/4F1/4Fb

A C

FFb

F

F

A

FGF

F

F1/2Fb

F

DG

IPERΣ01.trns.045AZIONI INTERNE 843775 Tarantola Stefano

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

0

-1-1

-1/4-1/4

5/4

√2/2

-1 -1-1

F

1/4

0-5

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12

0

-√2/2

0-1

0

F

-1/4

0

1/2

1/2

3/2

1/4

-3/20

1/4 1/4

1

0

00 1/20

Fb

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a

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IPERΣ01.trns.045PROCEDIMENTO E RISULTATI 843775 Tarantola Stefano

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAB = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXBA = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXCE = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXAC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXCA = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoEC = ∫

o

b(3/2 x/b - x2/b2 ) Fb 1/EJ dx = [3/4 x2/b -1/3 x3/b2 ]o

b Fb 1/EJ

= (3/4 b -1/3 b ) Fb 1/EJ = 5/12 Fb2/EJ

LXoCE = ∫

o

b(1/2 +1/2 x/b - x2/b2 ) Fb 1/EJ dx = [1/2 x +1/4 x2/b -1/3 x3/b2 ]o

b Fb 1/EJ

= (1/2 b +1/4 b -1/3 b ) Fb 1/EJ = 5/12 Fb2/EJ

IPERΣ01.trns.045PROCEDIMENTO E RISULTATI 843775 Tarantola Stefano

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 762. mm2

Ju = 174852. mm4

Jv = 74862. mm4

yg = 15.31 mmN = -187.5 NTy = 750. NMx = -877500. Nmmxm = 24. mmym = 55. mmum = 3. mmvm = 39.69 mmσm = N/A-Mv/Ju = 198.9 N/mm2

xc = 21. mmyc = 38. mmvc = 22.69 mmσc = N/A-Mv/Ju = 113.6 N/mm2

τc = 2.274 N/mm2

σo = √σ2+3τ2 = 113.7 N/mm2

S* = 3181. mm3mm 0 18 24 42x

0

12

55

y

38σc,τc

σm

u

v

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IPERΣ01.trns.045

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.trns.045

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 153: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.vnls.048REAZIONI 878465 Vanoli Simone

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1/4F1/4Fb

1/4F

A B

F1/2Fb

F1/2Fb

CD

F

5/4F3/2Fb

F

5/4F1/4Fb

EC

F

1/4F3/2Fb

2F

1/4F

E

B

5/4F1/4Fb

5/4F1/4Fb

A

C

F

Fb

F

F

A

F

G

F

F

F1/2Fb

F

D

G

IPERΣ01.vnls.048AZIONI INTERNE 878465 Vanoli Simone

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

0

1 1

1/4

1/4

-5/4

-√2/2

11

1

F

1/4

0-5/4

12

0

-√2/2

0-1

0

F

-1/40

1/21/2 3/21/4

-3/2

0

1/4

1/4

1

0

00

1/2

0

Fb

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IPE

RΣ0

1.vn

ls.0

48P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 8

7846

5 V

anol

i Sim

one

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

AB

CD

E

F

G

F

W

XX

q

q Sch

ema

di c

alco

lo ip

erst

atic

o

00

1/2

1/2

3/2

1/2

-3/20

0 0

1

0

00 1/20

Mo

fless

ione

da

caric

hi a

sseg

nati

-10

00

0-1

00

1 1

0

0

00 00

Mx

fless

ione

da

iper

stat

ica

X=

1

IPE

RΣ0

1.vn

ls.0

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RO

CE

DIM

EN

TO

E R

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7846

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anol

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one

@ A

dolfo

Zav

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i Ros

si, P

olite

cnic

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no, v

ers.

27.0

3.13

03.0

9.18

Qua

dro

cont

ribut

i PLV

per

iper

stat

ica

X=

WA

B

→M

x(x)

Mo(

x)M

xMo

MxM

x∫M

xMo/

EJd

x∫X

MxM

x/E

Jdx

AB

b-1

+x/

b0

01-

2x/b

+x2 /b

2

01/

3Xb/

EJ

BA

bx/

b0

0x2 /b

2

CD

b0

1/2F

b0

00

0D

C b

0-1

/2F

b0

0

EC

b-x

/b3/

2Fb-

Fx

-3/2

Fx+

Fx2 /b

x2 /b2

-5/1

2Fb2 /E

J1/

3Xb/

EJ

CE

b1-

x/b

-1/2

Fb-

Fx

-1/2

Fb-

1/2F

x+F

x2 /b1-

2x/b

+x2 /b

2

EB

b0

-3/2

Fb+

Fx+

1/2q

x20

00

0B

E b

02F

x-1/

2qx2

00

AC

b1

00

10

Xb/

EJ

CA

b-1

00

1

FA

√2b

0F

b-√2

/2F

x0

00

0

GF

b0

00

00

0F

G b

00

00

DG

b0

1/2F

b-F

x+1/

2qx2

00

00

GD

b0

-1/2

qx2

00

tota

li-5

/12F

b2 /EJ

5/3X

b/E

J

iper

stat

ica

X=

WA

B1/

4Fb

Svi

lupp

i di c

alco

lo ip

erst

atic

a

Page 155: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.vnls.048PROCEDIMENTO E RISULTATI 878465 Vanoli Simone

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAB = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXBA = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXCE = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXAC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXCA = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoEC = ∫

o

b(-3/2 x/b + x2/b2 ) Fb 1/EJ dx = [-3/4 x2/b +1/3 x3/b2 ]o

b Fb 1/EJ

= (-3/4 b +1/3 b ) Fb 1/EJ = -5/12 Fb2/EJ

LXoCE = ∫

o

b(-1/2 -1/2 x/b + x2/b2 ) Fb 1/EJ dx = [-1/2 x -1/4 x2/b +1/3 x3/b2 ]o

b Fb 1/EJ

= (-1/2 b -1/4 b +1/3 b ) Fb 1/EJ = -5/12 Fb2/EJ

IPERΣ01.vnls.048PROCEDIMENTO E RISULTATI 878465 Vanoli Simone

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 546. mm2

Ju = 162198. mm4

Jv = 37926. mm4

yg = 17.81 mmN = 177.5 NTy = 710. NMx = -990450. Nmmxm = 24. mmym = 55. mmum = 3. mmvm = 37.19 mmσm = N/A-Mv/Ju = 227.4 N/mm2

xc = 21. mmyc = 39. mmvc = 21.19 mmσc = N/A-Mv/Ju = 129.7 N/mm2

τc = 2.045 N/mm2

σo = √σ2+3τ2 = 129.8 N/mm2

S* = 2802. mm3mm 0 18 24 42x

0

6

55

y

39σc,τc

σm

u

v

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IPERΣ01.vnls.048

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.vnls.048

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

Page 157: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.vrga.049REAZIONI 843591 Virga Alessandro

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

1/40F

FFb

1/40F

F

A B

79/40F

FFb

39/40F

F19/40Fb

A

C

F1/2Fb

F1/2Fb

C

D

1/40F1/40Fb

1/40F

E

B

F

F1/2Fb

F

D F

FF G

FFb

F

G

E

39/40F1/40Fb

39/40F1/40Fb

EC

IPERΣ01.vrga.049AZIONI INTERNE 843591 Virga Alessandro

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

-1/40

-1-1

-1

0

1 1 1

-√2/2

-39/40

F

1

-79/

40-3

9/40

0

-1/4

0

1 0 0

-√2/2

0

F

-101

-19/

40-1

/2-1

/2

1/40

0

-1/20 0 0

1

0

-1/40-1/40

Fb

Page 158: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPE

RΣ0

1.vr

ga.0

49P

RO

CE

DIM

EN

TO

E R

ISU

LTA

TI 8

4359

1 V

irga

Ale

ssan

dro

@ A

dolfo

Zav

elan

i Ros

si, P

olite

cnic

o di

Mila

no, v

ers.

27.0

3.13

03.0

9.18

AB

C

D

E

F

G

F

W

X

X

q

q

Sch

ema

di c

alco

lo ip

erst

atic

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-10

1-1/2-1/2-1/2

0 0

-1/2

00

0

1

0

00

Mo

fless

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caric

hi a

sseg

nati

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0-1

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Mx

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IPE

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Zav

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i Ros

si, P

olite

cnic

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Mila

no, v

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27.0

3.13

03.0

9.18

Quadro contributi PLV per iperstatica X=WCE

→Mx(x)Mo(x)MxMoMxMx∫MxMo/EJdx∫XMxMx/EJdx

AB b0-Fb+Fx0000

BA b0Fx00

AC b-x/bFb-2Fx+1/2qx2

-Fx+2Fx2/b-1/2qx

3/bx

2/b

2

1/24Fb2/EJ1/3Xb/EJ

CA b1-x/b1/2Fb-Fx-1/2qx2

1/2Fb-3/2Fx+1/2Fx2/b+1/2qx

3/b1-2x/b+x

2/b

2

CD b0-1/2Fb0000

DC b01/2Fb00

EB b-1+x/b001-2x/b+x2/b

2

01/3Xb/EJBE bx/b00x

2/b

2

DF b0-1/2Fb+Fx-1/2qx2

0000

FD b01/2qx2

00

FG b000000

GF b0000

GE √2b0Fb-√2/2Fx0000

EC b10010Xb/EJ

CE b-1001

totali1/24Fb2/EJ5/3Xb/EJ

iperstatica X=WCE-1/40Fb

Svi

lupp

i di c

alco

lo ip

erst

atic

a

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IPERΣ01.vrga.049PROCEDIMENTO E RISULTATI 843591 Virga Alessandro

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXCA = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXEB = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXBE = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXCE = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoAC = ∫

o

b(- x/b +2 x2/b2 -1/2 x3/b3 ) Fb 1/EJ dx = [-1/2 x2/b +2/3 x3/b2 -1/8 x4/b3 ]o

b Fb 1/EJ

= (-1/2 b +2/3 b -1/8 b ) Fb 1/EJ = 1/24 Fb2/EJ

LXoCA = ∫

o

b(1/2 -3/2 x/b +1/2 x2/b2 +1/2 x3/b3 ) Fb 1/EJ dx

= [1/2 x -3/4 x2/b +1/6 x3/b2 +1/8 x4/b3 ]o

b Fb 1/EJ

= (1/2 b -3/4 b +1/6 b +1/8 b ) Fb 1/EJ = 1/24 Fb2/EJ

IPERΣ01.vrga.049PROCEDIMENTO E RISULTATI 843591 Virga Alessandro

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 456. mm2

Ju = 127247. mm4

Jv = 25560. mm4

yg = 33.58 mmN = -24.25 NTy = 970. NMx = -902100. Nmmxm = 18. mmum = -3. mmvm = -33.58 mmσm = N/A-Mv/Ju = -238.1 N/mm2

xc = 21. mmyc = 14. mmvc = -19.58 mmσc = N/A-Mv/Ju = -138.9 N/mm2

τc = 2.837 N/mm2

σo = √σ2+3τ2 = 138.9 N/mm2

S* = 2233. mm3mm 0 18 24 42x

0

48

52

y

14σc,τc

σm

u

v

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IPERΣ01.vrga.049

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.vrga.049

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

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IPERΣ01.vtlm.050REAZIONI 843551 Vitali Martina

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

13/20F

1/2F13/20Fb

13/20F

1/2F

A

B

3/2F

1/2F3/2Fb

1/2F

1/2F1/2Fb

C

D

53/20F

1/2F7/2Fb

53/20F

1/2F17/20Fb

E

C

13/20F

7/2F7/2Fb

13/20F

7/2F

EB

23/20F13/20Fb

23/20F13/20Fb

A C

1/2F

1/2F

Fb

1/2F

1/2F

F

A

1/2F

1/2F

1/2F

1/2F

GF

1/2F

1/2F1/2Fb

1/2F

1/2F

DG

IPERΣ01.vtlm.050AZIONI INTERNE 843551 Vitali Martina

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

-1/2

-1/2

-1/2

-1/2

-13/20

23/20

0

-1/2-1/2 -1/2

F

13/2

0

-3/2

-1/2

-53/

20

7/2

0

-√2/2

-1/21/2

-1/2

F

-13/

200

3/2

1/2

7/2

17/2

0

-7/20

13/20 13/20

1

0

00 1/20

Fb

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IPE

RΣ0

1.vt

lm.0

50P

RO

CE

DIM

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dolfo

Zav

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si, P

olite

cnic

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no, v

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27.0

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03.0

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B

C

D E

FG

F

W

X

X

q

q

0 0

3/21/2 7/23/2

-7/2

000

1

000

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0

Mo

fless

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caric

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sseg

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00 0-1

001

1

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IPE

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27.0

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Qua

dro

cont

ribut

i PLV

per

iper

stat

ica

X=

WC

A

→M

x(x)

Mo(

x)M

xMo

MxM

x∫M

xMo/

EJd

x∫X

MxM

x/E

Jdx

AB

b-1

+x/

b0

01-

2x/b

+x2 /b

2

01/

3Xb/

EJ

BA

bx/

b0

0x2 /b

2

CD

b0

3/2F

b-3/

2Fx+

1/2q

x20

00

0D

C b

0-1

/2F

b-1/

2Fx-

1/2q

x20

0

EC

b-x

/b7/

2Fb-

2Fx

-7/2

Fx+

2Fx2 /b

x2 /b2

-13/

12F

b2 /EJ

1/3X

b/E

JC

E b

1-x/

b-3

/2F

b-2F

x-3

/2F

b-1/

2Fx+

2Fx2 /b

1-2x

/b+

x2 /b2

EB

b0

-7/2

Fb+

7/2F

x0

00

0B

E b

07/

2Fx

00

AC

b1

00

10

Xb/

EJ

CA

b-1

00

1

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b-√2

/2F

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00

0

GF

b0

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Fx+

1/2q

x20

00

0F

G b

01/

2Fx-

1/2q

x20

0

DG

b0

1/2F

b-1/

2Fx

00

00

GD

b0

-1/2

Fx

00

tota

li-1

3/12

Fb2 /E

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3Xb/

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Page 163: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.vtlm.050PROCEDIMENTO E RISULTATI 843551 Vitali Martina

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

LXXAB = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXBA = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXEC = ∫

o

b( x2/b2 ) 1/EJ dx = [1/3 x3/b2 ]o

b 1/EJ

= (1/3 b ) 1/EJ = 1/3 b/EJ

LXXCE = ∫

o

b(1 -2 x/b + x2/b2 ) 1/EJ dx = [ x - x2/b +1/3 x3/b2 ]o

b 1/EJ

= ( b - b +1/3 b ) 1/EJ = 1/3 b/EJ

LXXAC = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXXCA = ∫

o

b(1 ) 1/EJ dx = [ x ]o

b 1/EJ

= ( b ) 1/EJ = b/EJ

LXoEC = ∫

o

b(-7/2 x/b +2 x2/b2 ) Fb 1/EJ dx = [-7/4 x2/b +2/3 x3/b2 ]o

b Fb 1/EJ

= (-7/4 b +2/3 b ) Fb 1/EJ = -13/12 Fb2/EJ

LXoCE = ∫

o

b(-3/2 -1/2 x/b +2 x2/b2 ) Fb 1/EJ dx = [-3/2 x -1/4 x2/b +2/3 x3/b2 ]o

b Fb 1/EJ

= (-3/2 b -1/4 b +2/3 b ) Fb 1/EJ = -13/12 Fb2/EJ

IPERΣ01.vtlm.050PROCEDIMENTO E RISULTATI 843551 Vitali Martina

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

A = 672. mm2

Ju = 147014. mm4

Jv = 62496. mm4

yg = 37.25 mmN = -292.5 NTy = 1575. NMx = -771750. Nmmxm = 18. mmum = -3. mmvm = -37.25 mmσm = N/A-Mv/Ju = -196. N/mm2

xc = 21. mmyc = 16. mmvc = -21.25 mmσc = N/A-Mv/Ju = -112. N/mm2

τc = 5.014 N/mm2

σo = √σ2+3τ2 = 112.3 N/mm2

S* = 2808. mm3mm 0 18 24 42x

0

42

52

y

16σc,τc

σm

u

v

Page 164: REAZIONI 781715 Bao Hong Da IPERΣ01 ... - intranet dica: Homeintranet.dica.polimi.it/uploads/media/IPER01-soluti_01.pdf · 01.bnst.003 PROCEDIMENTO E RISULTATI 843782 Benassai Tommaso

IPERΣ01.vtlm.050

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18

IPERΣ01.vtlm.050

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 03.09.18