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Schema.agsr.001 REAZIONI 877947 Agosti Riccardo @ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19 1/2F 7/4F 1/2F 7/4F 7/4Fb A B 5/4F 3/4F 5/4Fb 5/4F 3/4F B C 7/4F 1/2F 1/2Fb 7/4F 1/2F B D 3/2F 1/2F 3/2F 1/2F 3/2Fb D E 3/2F 1/2F 1/2Fb 3/2F 1/2F E F 3/2F F B 3/2F F 3/2F 1/2Fb F G 1/2F 1/2Fb 1/2F G A Schema.agsr.001 AZIONI INTERNE 877947 Agosti Riccardo @ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19 -1/2 3/4 -7/4 1/2 3/2 -3/2 3/2 3/2 0 F 7/4 -5/4 -1/2 -3/2 1/2 0 -1 0 1/2 F 0 7/4 5/4 0 1/2 0 0 -3/2 -1/2 0 0 0 0 -1/2 -1/2 0 Fb

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Page 1: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.agsr.001REAZIONI 877947 Agosti Riccardo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1/2F

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1/2F

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7/4F

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B D

3/2F

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3/2F

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D

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3/2F

1/2F

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F

B

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FG

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1/2F

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Schema.agsr.001AZIONI INTERNE 877947 Agosti Riccardo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

-1/2

3/4

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Page 2: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 3: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.agsr.001PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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

LXXBC = ∫

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

LXXCB = ∫

o

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

b 1/EJ

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

LXoAB = ∫

o

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

b Fb 1/EJ

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

LXoBA = ∫

o

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

b Fb 1/EJ

= (-1/2 b +1/2 b -1/6 b ) Fb 1/EJ = -1/6 Fb2/EJ

A = 540. mm2

Ju = 154030. mm4

Jv = 37908. mm4

yg = 36.6 mmN = 500. NTy = -1500. NMx = -885000. Nmmxm = 18. mmum = -3. mmvm = -36.6 mmσm = N/A-Mv/Ju = -209.4 N/mm2

xc = 21. mmyc = 15. mmvc = -21.6 mmσc = N/A-Mv/Ju = -123.2 N/mm2

τc = 4.251 N/mm2

σo = √σ2+3τ2 = 123.4 N/mm2

S* = 2619. mm3mm 0 18 24 42x

0

48

54

y

15σc,τc

σm

u

v

Schema.agsr.001

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 4: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.agsr.001

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.agsr.001

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 5: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.bldd.002REAZIONI 912028 Baldin Daniele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

5/4F 5/4F5/4Fb

A B

7/4F

1/4F7/4Fb

7/4F

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B

C

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Schema.bldd.002AZIONI INTERNE 912028 Baldin Daniele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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Page 6: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 7: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.bldd.002PROCEDIMENTO E RISULTATI 912028 Baldin Daniele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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

LXXBC = ∫

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

LXXCB = ∫

o

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

b 1/EJ

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

LXoAB = ∫

o

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

b Fb 1/EJ

= (1/6 b ) Fb 1/EJ = 1/6 Fb2/EJ

LXoBA = ∫

o

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

b Fb 1/EJ

= (1/2 b -1/2 b +1/6 b ) Fb 1/EJ = 1/6 Fb2/EJ

A = 756. mm2

Ju = 165564. mm4

Jv = 74844. mm4

yg = 39. mmN = -725. NTy = -1450. NMx = -928000. Nmmxm = 18. mmum = -3. mmvm = -39. mmσm = N/A-Mv/Ju = -219.6 N/mm2

xc = 21. mmyc = 17. mmvc = -22. mmσc = N/A-Mv/Ju = -124.3 N/mm2

τc = 4.541 N/mm2

σo = √σ2+3τ2 = 124.5 N/mm2

S* = 3111. mm3mm 0 18 24 42x

0

42

54

y

17σc,τc

σm

u

v

Schema.bldd.002

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 8: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.bldd.002

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.bldd.002

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 9: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.bltt.003REAZIONI 893013 Beltran Toledo Italo Jose

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1/2F

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Schema.bltt.003AZIONI INTERNE 893013 Beltran Toledo Italo Jose

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1/2

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Page 10: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 11: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.bltt.003PROCEDIMENTO E RISULTATI 893013 Beltran Toledo Italo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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

LXXBC = ∫

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

LXXCB = ∫

o

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

b 1/EJ

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

LXoAB = ∫

o

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

b Fb 1/EJ

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

LXoBA = ∫

o

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

b Fb 1/EJ

= (-1/2 b +1/2 b -1/6 b ) Fb 1/EJ = -1/6 Fb2/EJ

A = 864. mm2

Ju = 251424. mm4

Jv = 62208. mm4

yg = 33. mmN = 835. NTy = -2505. NMx = -1753500. Nmmxm = 18. mmum = -6. mmvm = -33. mmσm = N/A-Mv/Ju = -229.2 N/mm2

xc = 24. mmyc = 14. mmvc = -19. mmσc = N/A-Mv/Ju = -131.5 N/mm2

τc = 3.627 N/mm2

σo = √σ2+3τ2 = 131.7 N/mm2

S* = 4368. mm3mm 0 18 30 48x

0

48

54

y

14σc,τc

σm

u

v

Schema.bltt.003

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 12: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.bltt.003

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.bltt.003

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 13: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.bnda.004REAZIONI 888197 Bendo Alessandro

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

7/4F 7/4F7/4Fb

A B

5/4F

7/4F5/4Fb

5/4F

7/4F

B

C

5/4F

1/2F1/2Fb

5/4F

1/2F

B D

2F

1/2F

2F

1/2F2Fb

D

E

2F

1/2FFb

2F

1/2F1/2Fb

EF

F

1/2F1/2Fb

1/2F

F

B

FFG

G

A

Schema.bnda.004AZIONI INTERNE 888197 Bendo Alessandro

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0

7/4

-5/4

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0

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0

0000

Fb

Page 14: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 15: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.bnda.004PROCEDIMENTO E RISULTATI 888197 Bendo Alessandro

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoBC = ∫

o

b(-1/2 + x/b -1/2 x2/b2 ) Fb2 1/EJ dx - 1 (-1) (-1) Fb3/EJ

= [-1/2 x +1/2 x2/b -1/6 x3/b2 ]o

b Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ

= (-1/2 b +1/2 b -1/6 b ) Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ = -7/6 Fb3/EJ

LXoCB = ∫

o

b(-1/2 x2/b2 ) Fb2 1/EJ dx - 1 (-1) (-1) Fb3/EJ = [-1/6 x3/b2 ]o

b Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ

= (-1/6 b ) Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ = -7/6 Fb3/EJ

A = 1080. mm2

Ju = 276955. mm4

Jv = 116640. mm4

yg = 35.4 mmN = 625. NTy = -2500. NMx = -1875000. Nmmxm = 18. mmum = -6. mmvm = -35.4 mmσm = N/A-Mv/Ju = -239.1 N/mm2

xc = 24. mmyc = 15. mmvc = -20.4 mmσc = N/A-Mv/Ju = -137.5 N/mm2

τc = 3.778 N/mm2

σo = √σ2+3τ2 = 137.7 N/mm2

S* = 5022. mm3mm 0 18 30 48x

0

42

54

y

15σc,τc

σm

u

v

Schema.bnda.004

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 16: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.bnda.004

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.bnda.004

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 17: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.brnr.005REAZIONI 868395 Bernasconi Riccardo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

7/4F 7/4F7/4Fb

A B

5/4F

7/4F5/4Fb

5/4F

7/4F

B

C

5/4F

1/2F1/2Fb

5/4F

1/2F

B D

2F

1/2F

F

1/2F3/2Fb

D

E

F

1/2F1/2Fb

F

1/2F

EF

1/2F

F

B

FFG

G

A

Schema.brnr.005AZIONI INTERNE 868395 Bernasconi Riccardo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0

7/4

-5/4

1/2

1/21

-1/2

1

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7/4

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0

0

0

F

0 7/4

5/4

0

1/2 0

0-3

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-1/200

0

0000

Fb

Page 18: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Sch

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Page 19: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.brnr.005PROCEDIMENTO E RISULTATI 868395 Bernasconi Riccardo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b(1/2 x2/b2 ) Fb2 1/EJ dx = [1/6 x3/b2 ]o

b Fb2 1/EJ

= (1/6 b ) Fb2 1/EJ = 1/6 Fb3/EJ

LXoBA = ∫

o

b(1/2 - x/b +1/2 x2/b2 ) Fb2 1/EJ dx = [1/2 x -1/2 x2/b +1/6 x3/b2 ]o

b Fb2 1/EJ

= (1/2 b -1/2 b +1/6 b ) Fb2 1/EJ = 1/6 Fb3/EJ

A = 468. mm2

Ju = 136587. mm4

Jv = 14364. mm4

yg = 34.38 mmN = 325. NTy = -650. NMx = -789750. Nmmxm = 12. mmum = -3. mmvm = -34.38 mmσm = N/A-Mv/Ju = -198.1 N/mm2

xc = 15. mmyc = 14. mmvc = -20.38 mmσc = N/A-Mv/Ju = -117.2 N/mm2

τc = 1.824 N/mm2

σo = √σ2+3τ2 = 117.2 N/mm2

S* = 2300. mm3mm 0 12 18 30x

0

48

54

y

14σc,τc

σm

u

v

Schema.brnr.005

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 20: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.brnr.005

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.brnr.005

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 21: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.brss.006REAZIONI 887236 Berselli Samuele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

31/16F 15/16F23/16Fb

A B

23/16F

15/16F23/16Fb

23/16F

15/16F

B

C

23/16FB D

F

FFb

D

EF

EF

F

B

FFGG

A

Schema.brss.006AZIONI INTERNE 887236 Berselli Samuele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0 0

15/1

6

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0

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31/16 15/16

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Page 22: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 23: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.brss.006PROCEDIMENTO E RISULTATI 887236 Berselli Samuele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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

LXXBC = ∫

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

LXXCB = ∫

o

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

b 1/EJ

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

LXoAB = ∫

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

LXoBA = ∫

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

A = 612. mm2

Ju = 149427. mm4

Jv = 27756. mm4

yg = 36.88 mmTy = -980. NMx = -842800. Nmmxm = 12. mmum = -3. mmvm = -36.88 mmσm = -Mv/Ju = -208. N/mm2

xc = 15. mmyc = 16. mmvc = -20.88 mmσc = -Mv/Ju = -117.8 N/mm2

τc = 3.031 N/mm2

σo = √σ2+3τ2 = 117.9 N/mm2

S* = 2773. mm3mm 0 12 18 30x

0

42

54

y

16σc,τc

σm

u

v

Schema.brss.006

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 24: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.brss.006

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.brss.006

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 25: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.brmm.007REAZIONI 829837 Bormolini Matteo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

5/4F 5/4F5/4Fb

A B

7/4F

1/4F7/4Fb

7/4F

1/4F

B

C

7/4F

F1/2Fb

7/4F

B D

F

FFb

D

EF

EF

F

B

FFG

G

A

Schema.brmm.007AZIONI INTERNE 829837 Bormolini Matteo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0

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Page 26: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 27: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.brmm.007PROCEDIMENTO E RISULTATI 829837 Bormolini Matteo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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

LXXBC = ∫

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

LXXCB = ∫

o

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

b 1/EJ

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

LXoAB = ∫

o

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

b Fb 1/EJ

= (1/6 b ) Fb 1/EJ = 1/6 Fb2/EJ

LXoBA = ∫

o

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

b Fb 1/EJ

= (1/2 b -1/2 b +1/6 b ) Fb 1/EJ = 1/6 Fb2/EJ

A = 792. mm2

Ju = 225759. mm4

Jv = 30240. mm4

yg = 31.36 mmTy = -1740. NMx = -1583400. Nmmxm = 12. mmum = -6. mmvm = -31.36 mmσm = -Mv/Ju = -220. N/mm2

xc = 18. mmyc = 13. mmvc = -18.36 mmσc = -Mv/Ju = -128.8 N/mm2

τc = 2.491 N/mm2

σo = √σ2+3τ2 = 128.9 N/mm2

S* = 3879. mm3mm 0 12 24 36x

0

48

54

y

13σc,τc

σm

u

v

Schema.brmm.007

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 28: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.brmm.007

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.brmm.007

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 29: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.brsa.008REAZIONI 891317 Borsani Alessio

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

23/16F 23/16F23/16Fb

A B

15/16F

23/16F23/16Fb

31/16F

23/16F

B

C

15/16FB D

F

FFb

D

EF

EF

F

B

FFGG

A

Schema.brsa.008AZIONI INTERNE 891317 Borsani Alessio

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0

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Page 30: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Sch

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Page 31: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.brsa.008PROCEDIMENTO E RISULTATI 891317 Borsani Alessio

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b(-1/2 x2/b2 ) Fb2 1/EJ dx = [-1/6 x3/b2 ]o

b Fb2 1/EJ

= (-1/6 b ) Fb2 1/EJ = -1/6 Fb3/EJ

LXoBA = ∫

o

b(-1/2 + x/b -1/2 x2/b2 ) Fb2 1/EJ dx = [-1/2 x +1/2 x2/b -1/6 x3/b2 ]o

b Fb2 1/EJ

= (-1/2 b +1/2 b -1/6 b ) Fb2 1/EJ = -1/6 Fb3/EJ

LXoBC = ∫

o

b(-1/2 +3/2 x/b -3/2 x2/b2 +1/2 x3/b3 ) Fb2 1/EJ dx - 1 (-1) (-1) Fb3/EJ

= [-1/2 x +3/4 x2/b -1/2 x3/b2 +1/8 x4/b3 ]o

b Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ

= (-1/2 b +3/4 b -1/2 b +1/8 b ) Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ = -9/8 Fb3/EJ

LXoCB = ∫

o

b(-1/2 x3/b3 ) Fb2 1/EJ dx - 1 (-1) (-1) Fb3/EJ = [-1/8 x4/b3 ]o

b Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ

= (-1/8 b ) Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ = -9/8 Fb3/EJ

Schema.brsa.008PROCEDIMENTO E RISULTATI 891317 Borsani Alessio

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

A = 936. mm2

Ju = 248849. mm4

Jv = 52704. mm4

yg = 33.46 mmTy = -1760. NMx = -1707200. Nmmxm = 12. mmum = -6. mmvm = -33.46 mmσm = -Mv/Ju = -229.6 N/mm2

xc = 18. mmyc = 14. mmvc = -19.46 mmσc = -Mv/Ju = -133.5 N/mm2

τc = 2.62 N/mm2

σo = √σ2+3τ2 = 133.6 N/mm2

S* = 4446. mm3mm 0 12 24 36x

0

42

54

y

14σc,τc

σm

u

v

Page 32: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.brsa.008

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.brsa.008

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 33: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.brnl.009REAZIONI 889394 Brandizi Leonardo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1/2F

7/4F

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7/4F7/4Fb

A B

5/4F

3/4F5/4Fb

5/4F

3/4F

B

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7/4F

1/2F1/2Fb

7/4F

1/2F

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Schema.brnl.009AZIONI INTERNE 889394 Brandizi Leonardo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

-1/2

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0

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Page 34: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 35: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.brnl.009PROCEDIMENTO E RISULTATI 889394 Brandizi Leonardo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b(1/2 x2/b2 ) Fb2 1/EJ dx = [1/6 x3/b2 ]o

b Fb2 1/EJ

= (1/6 b ) Fb2 1/EJ = 1/6 Fb3/EJ

LXoBA = ∫

o

b(1/2 - x/b +1/2 x2/b2 ) Fb2 1/EJ dx = [1/2 x -1/2 x2/b +1/6 x3/b2 ]o

b Fb2 1/EJ

= (1/2 b -1/2 b +1/6 b ) Fb2 1/EJ = 1/6 Fb3/EJ

A = 936. mm2

Ju = 248849. mm4

Jv = 52704. mm4

yg = 20.54 mmN = 580. NTy = -1740. NMx = -1774800. Nmmxm = 24. mmym = 54. mmum = 6. mmvm = 33.46 mmσm = N/A-Mv/Ju = 239.3 N/mm2

xc = 18. mmyc = 40. mmvc = 19.46 mmσc = N/A-Mv/Ju = 139.4 N/mm2

τc = 2.59 N/mm2

σo = √σ2+3τ2 = 139.5 N/mm2

S* = 4446. mm3mm 0 12 24 36x

0

12

54

y

40σc,τc

σm

u

v

Schema.brnl.009

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 36: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.brnl.009

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.brnl.009

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 37: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.brga.010REAZIONI 881254 Broggi Andrea

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

5/4F 5/4F5/4Fb

A B

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C

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F

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D

E

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1/2FFb

F

1/2FFb

EF

1/2F

F

B

FFG

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A

Schema.brga.010AZIONI INTERNE 881254 Broggi Andrea

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0

1/4

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00

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Page 38: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 39: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.brga.010PROCEDIMENTO E RISULTATI 881254 Broggi Andrea

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoBC = ∫

o

b(1/2 - x/b +1/2 x2/b2 ) Fb2 1/EJ dx - 1 (-1) (-1) Fb3/EJ

= [1/2 x -1/2 x2/b +1/6 x3/b2 ]o

b Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ

= (1/2 b -1/2 b +1/6 b ) Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ = -5/6 Fb3/EJ

LXoCB = ∫

o

b(1/2 x2/b2 ) Fb2 1/EJ dx - 1 (-1) (-1) Fb3/EJ = [1/6 x3/b2 ]o

b Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ

= (1/6 b ) Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ = -5/6 Fb3/EJ

A = 468. mm2

Ju = 136587. mm4

Jv = 14364. mm4

yg = 19.62 mmN = -740. NTy = -1480. NMx = -799200. Nmmxm = 18. mmym = 54. mmum = 3. mmvm = 34.38 mmσm = N/A-Mv/Ju = 199.6 N/mm2

xc = 15. mmyc = 40. mmvc = 20.38 mmσc = N/A-Mv/Ju = 117.7 N/mm2

τc = 4.154 N/mm2

σo = √σ2+3τ2 = 117.9 N/mm2

S* = 2300. mm3mm 0 12 18 30x

0

6

54

y

40σc,τc

σm

u

v

Schema.brga.010

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 40: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.brga.010

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.brga.010

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 41: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.crlm.011REAZIONI 893348 Carlino Mauro

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1/2F

7/4F

1/2F

7/4F7/4Fb

A B

5/4F

7/4F5/4Fb

5/4F

7/4F

B

C

3/4F

1/2F1/2Fb

3/4F

1/2F

B D

3/2F

1/2F

3/2F

1/2F3/2Fb

D

E

3/2F

1/2F3/2Fb

3/2F

1/2FFb

EF

1/2F

F

B

3/2FFG

1/2F

1/2F

G

A

Schema.crlm.011AZIONI INTERNE 893348 Carlino Mauro

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1/2

7/4

-3/4

1/2

3/2

-1/2

3/200

F

7/4

-5/4

-1/2

-3/2

1/2

0

0

1/2

-1/2

F

0 7/4

5/4

0

1/2 0

0-3

/2

-3/2-1

00

0000

Fb

Page 42: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 43: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.crlm.011PROCEDIMENTO E RISULTATI 893348 Carlino Mauro

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b(-5/2 x2/b2 ) Fb2 1/EJ dx = [-5/6 x3/b2 ]o

b Fb2 1/EJ

= (-5/6 b ) Fb2 1/EJ = -5/6 Fb3/EJ

LXoBA = ∫

o

b(-5/2 +5 x/b -5/2 x2/b2 ) Fb2 1/EJ dx = [-5/2 x +5/2 x2/b -5/6 x3/b2 ]o

b Fb2 1/EJ

= (-5/2 b +5/2 b -5/6 b ) Fb2 1/EJ = -5/6 Fb3/EJ

LXoBC = ∫

o

b(-2 +4 x/b -2 x2/b2 ) Fb2 1/EJ dx + 1 (-1) (-1) Fb3/EJ

= [-2 x +2 x2/b -2/3 x3/b2 ]o

b Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ

= (-2 b +2 b -2/3 b ) Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ = 1/3 Fb3/EJ

LXoCB = ∫

o

b(-2 x2/b2 ) Fb2 1/EJ dx + 1 (-1) (-1) Fb3/EJ = [-2/3 x3/b2 ]o

b Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ

= (-2/3 b ) Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ = 1/3 Fb3/EJ

Schema.crlm.011PROCEDIMENTO E RISULTATI 893348 Carlino Mauro

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

A = 864. mm2

Ju = 251424. mm4

Jv = 62208. mm4

yg = 21. mmN = 895. NTy = -2685. NMx = -1584150. Nmmxm = 30. mmym = 54. mmum = 6. mmvm = 33. mmσm = N/A-Mv/Ju = 209. N/mm2

xc = 24. mmyc = 40. mmvc = 19. mmσc = N/A-Mv/Ju = 120.7 N/mm2

τc = 3.887 N/mm2

σo = √σ2+3τ2 = 120.9 N/mm2

S* = 4368. mm3mm 0 18 30 48x

0

6

54

y

40σc,τc

σm

u

v

Page 44: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.crlm.011

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.crlm.011

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 45: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.csrf.012REAZIONI 879105 Caserta Francesco

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

7/4F 7/4F7/4Fb

A B

5/4F

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5/4F

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B

C

5/4F

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1/2F

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Schema.csrf.012AZIONI INTERNE 879105 Caserta Francesco

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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Page 46: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 47: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.csrf.012PROCEDIMENTO E RISULTATI 879105 Caserta Francesco

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b(1/2 x2/b2 ) Fb2 1/EJ dx = [1/6 x3/b2 ]o

b Fb2 1/EJ

= (1/6 b ) Fb2 1/EJ = 1/6 Fb3/EJ

LXoBA = ∫

o

b(1/2 - x/b +1/2 x2/b2 ) Fb2 1/EJ dx = [1/2 x -1/2 x2/b +1/6 x3/b2 ]o

b Fb2 1/EJ

= (1/2 b -1/2 b +1/6 b ) Fb2 1/EJ = 1/6 Fb3/EJ

A = 612. mm2

Ju = 149428. mm4

Jv = 27756. mm4

yg = 17.12 mmN = 345. NTy = -1380. NMx = -883200. Nmmxm = 18. mmym = 54. mmum = 3. mmvm = 36.88 mmσm = N/A-Mv/Ju = 218.6 N/mm2

xc = 15. mmyc = 38. mmvc = 20.88 mmσc = N/A-Mv/Ju = 124. N/mm2

τc = 4.268 N/mm2

σo = √σ2+3τ2 = 124.2 N/mm2

S* = 2773. mm3mm 0 12 18 30x

0

12

54

y

38σc,τc

σm

u

v

Schema.csrf.012

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 48: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.csrf.012

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.csrf.012

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 49: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.cste.013REAZIONI 877793 Castiglione Ettore

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

7/4F 7/4F7/4Fb

A B

5/4F

7/4F5/4Fb

5/4F

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B

C

5/4F

1/2F1/2Fb

5/4F

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D

E

F

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F

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EF

1/2F

F

B

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A

Schema.cste.013AZIONI INTERNE 877793 Castiglione Ettore

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0

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Page 50: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Sch

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Page 51: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.cste.013PROCEDIMENTO E RISULTATI 877793 Castiglione Ettore

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b(1/2 x2/b2 ) Fb2 1/EJ dx = [1/6 x3/b2 ]o

b Fb2 1/EJ

= (1/6 b ) Fb2 1/EJ = 1/6 Fb3/EJ

LXoBA = ∫

o

b(1/2 - x/b +1/2 x2/b2 ) Fb2 1/EJ dx = [1/2 x -1/2 x2/b +1/6 x3/b2 ]o

b Fb2 1/EJ

= (1/2 b -1/2 b +1/6 b ) Fb2 1/EJ = 1/6 Fb3/EJ

A = 756. mm2

Ju = 165564. mm4

Jv = 74844. mm4

yg = 15. mmN = 460. NTy = -920. NMx = -966000. Nmmxm = 24. mmym = 54. mmum = 3. mmvm = 39. mmσm = N/A-Mv/Ju = 228.2 N/mm2

xc = 21. mmyc = 37. mmvc = 22. mmσc = N/A-Mv/Ju = 129. N/mm2

τc = 2.881 N/mm2

σo = √σ2+3τ2 = 129.1 N/mm2

S* = 3111. mm3mm 0 18 24 42x

0

12

54

y

37σc,τc

σm

u

v

Schema.cste.013

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 52: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.cste.013

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.cste.013

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 53: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.crra.014REAZIONI 914406 Corradino Andrea

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

31/16F 15/16F23/16Fb

A B

23/16F

15/16F23/16Fb

23/16F

15/16F

B

C

23/16FB D

F

FFb

D

E

FFb

FFb

EF

F

B

FFGG

A

Schema.crra.014AZIONI INTERNE 914406 Corradino Andrea

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0 0

15/1

6

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0

1

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0

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0

0

0

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00

0000

Fb

Page 54: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Sch

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Page 55: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.crra.014PROCEDIMENTO E RISULTATI 914406 Corradino Andrea

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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

LXXBC = ∫

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

LXXCB = ∫

o

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

b 1/EJ

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

LXoAB = ∫

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

LXoBA = ∫

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

A = 792. mm2

Ju = 225759. mm4

Jv = 30240. mm4

yg = 22.64 mmTy = -2300. NMx = -1725000. Nmmxm = 24. mmym = 54. mmum = 6. mmvm = 31.36 mmσm = -Mv/Ju = 239.6 N/mm2

xc = 18. mmyc = 41. mmvc = 18.36 mmσc = -Mv/Ju = 140.3 N/mm2

τc = 3.293 N/mm2

σo = √σ2+3τ2 = 140.4 N/mm2

S* = 3879. mm3mm 0 12 24 36x

0

6

54

y

41σc,τc

σm

u

v

Schema.crra.014

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 56: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.crra.014

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.crra.014

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 57: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.dlsf.015REAZIONI 772937 D’Alessio Francesca

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

5/4F 5/4F5/4Fb

A B

7/4F

1/4F7/4Fb

7/4F

1/4F

B

C

7/4F

F1/2Fb

7/4F

B D

F

FFb

D

E

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EF

F

B

FFG

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Schema.dlsf.015AZIONI INTERNE 772937 D’Alessio Francesca

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0

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0

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1 0

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0

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00

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Page 58: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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00

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00

00

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00

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00

00

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elon

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CL B

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Page 59: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.dlsf.015PROCEDIMENTO E RISULTATI 772937 D’Alessio Francesca

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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

LXXBC = ∫

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

LXXCB = ∫

o

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

b 1/EJ

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

LXoAB = ∫

o

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

b Fb 1/EJ

= (1/6 b ) Fb 1/EJ = 1/6 Fb2/EJ

LXoBA = ∫

o

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

b Fb 1/EJ

= (1/2 b -1/2 b +1/6 b ) Fb 1/EJ = 1/6 Fb2/EJ

A = 1080. mm2

Ju = 276955. mm4

Jv = 116640. mm4

yg = 18.6 mmTy = -1930. NMx = -1563300. Nmmxm = 30. mmym = 54. mmum = 6. mmvm = 35.4 mmσm = -Mv/Ju = 199.8 N/mm2

xc = 24. mmyc = 39. mmvc = 20.4 mmσc = -Mv/Ju = 115.2 N/mm2

τc = 2.916 N/mm2

σo = √σ2+3τ2 = 115.3 N/mm2

S* = 5022. mm3mm 0 18 30 48x

0

12

54

y

39σc,τc

σm

u

v

Schema.dlsf.015

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 60: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.dlsf.015

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.dlsf.015

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 61: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.dnsg.016REAZIONI 845411 Danesi Gabriele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

23/16F 23/16F23/16Fb

A B

15/16F

23/16F23/16Fb

31/16F

23/16F

B

C

15/16FB D

F

FFb

D

E

FFb

FFb

EF

F

B

FFGG

A

Schema.dnsg.016AZIONI INTERNE 845411 Danesi Gabriele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0

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Page 62: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Sch

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Page 63: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.dnsg.016PROCEDIMENTO E RISULTATI 845411 Danesi Gabriele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoBC = ∫

o

b(-1/2 x/b + x2/b2 -1/2 x3/b3 ) Fb2 1/EJ dx + 1 (-1) (-1) Fb3/EJ

= [-1/4 x2/b +1/3 x3/b2 -1/8 x4/b3 ]o

b Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ

= (-1/4 b +1/3 b -1/8 b ) Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ = 23/24 Fb3/EJ

LXoCB = ∫

o

b(-1/2 x2/b2 +1/2 x3/b3 ) Fb2 1/EJ dx + 1 (-1) (-1) Fb3/EJ

= [-1/6 x3/b2 +1/8 x4/b3 ]o

b Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ

= (-1/6 b +1/8 b ) Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ = 23/24 Fb3/EJ

A = 540. mm2

Ju = 154030. mm4

Jv = 37908. mm4

yg = 17.4 mmTy = -1020. NMx = -877200. Nmmxm = 24. mmym = 54. mmum = 3. mmvm = 36.6 mmσm = -Mv/Ju = 208.4 N/mm2

xc = 21. mmyc = 39. mmvc = 21.6 mmσc = -Mv/Ju = 123. N/mm2

τc = 2.891 N/mm2

σo = √σ2+3τ2 = 123.1 N/mm2

S* = 2619. mm3mm 0 18 24 42x

0

6

54

y

39σc,τc

σm

u

v

Schema.dnsg.016

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 64: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.dnsg.016

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.dnsg.016

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 65: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.grts.017REAZIONI 917200 Gritcul Serghei

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1/2F

5/2F

1/2F

5/2F5/2Fb

A B

2F

3/2FFb

2F

3/2FFb

B

C

5/2F

3/2F3/2Fb

5/2F

3/2F

B D

3/2F

3/2F

3/2F

3/2F3/2Fb

D

E

3/2F

3/2F3/2Fb

3/2F

3/2F

EF

5/2F

F

B

3/2F

F

3/2F1/2Fb

FG

1/2F1/2Fb

1/2F

G

A

Schema.grts.017AZIONI INTERNE 917200 Gritcul Serghei

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

-1/2

3/2

-5/2

3/2

3/2

-5/2

3/23/2

0

F

5/2

-2

-3/2

-3/2

3/2

0

-10

1/2

F

0 5/2

1-1

3/2 0

0-3

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-3/200

0

0-1/2

-1/2

0

Fb

Page 66: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Sch

ema.

grts

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27.0

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stat

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X=

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C

→M

x(x)

Mo(

x)M

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MxM

x∫M

xMo/

EJd

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MxM

x/E

Jdx

AB

b-x

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Fb2 /E

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Page 67: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.grts.017PROCEDIMENTO E RISULTATI 917200 Gritcul Serghei

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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

LXXBC = ∫

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

LXXCB = ∫

o

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

b 1/EJ

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

LXoAB = ∫

o

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

b Fb 1/EJ

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

LXoBA = ∫

o

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

b Fb 1/EJ

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

LXoBC = ∫

o

b( x/b - x2/b2 ) Fb 1/EJ dx + 1 (-1) (-1) Fb2/EJ

= [1/2 x2/b -1/3 x3/b2 ]o

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

= (1/2 b -1/3 b ) Fb 1/EJ + 1 (-1) (-1) Fb2/EJ = 7/6 Fb2/EJ

LXoCB = ∫

o

b( x/b - x2/b2 ) Fb 1/EJ dx + 1 (-1) (-1) Fb2/EJ

= [1/2 x2/b -1/3 x3/b2 ]o

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

= (1/2 b -1/3 b ) Fb 1/EJ + 1 (-1) (-1) Fb2/EJ = 7/6 Fb2/EJ

Schema.grts.017PROCEDIMENTO E RISULTATI 917200 Gritcul Serghei

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

A = 498. mm2

Ju = 141019. mm4

Jv = 31734. mm4

yg = 35.17 mmN = 975. NTy = -975. NMx = -877500. Nmmxm = 18. mmum = -3. mmvm = -35.17 mmσm = N/A-Mv/Ju = -216.9 N/mm2

xc = 21. mmyc = 15. mmvc = -20.17 mmσc = N/A-Mv/Ju = -123.6 N/mm2

τc = 2.87 N/mm2

σo = √σ2+3τ2 = 123.7 N/mm2

S* = 2491. mm3mm 0 18 24 42x

0

48

53

y

15σc,τc

σm

u

v

Page 68: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.grts.017

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.grts.017

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 69: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.dmcm.018REAZIONI 870485 D’Amico Michele Francesco

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

2F 2F2Fb

A B

5/2F

F3/2Fb

5/2F

FFb

B

C

5/2F

1/2F1/2Fb

5/2F

1/2F

B D

F

1/2F

F

1/2FFb

D

E

F

1/2FFb

F

3/2F

EF

3/2F

F

B

FFG

G

A

Schema.dmcm.018AZIONI INTERNE 870485 D’Amico Michele Francesco

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0

1

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1/2

11

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1

0

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0

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0

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dolfo

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no, v

ers.

27.0

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06.0

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per

iper

stat

ica

X=

HC

→M

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Mo(

x)M

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MxM

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Jdx

AB

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Fb3 /E

J1/

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JB

A b

-b+

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1/2F

x-1

/2F

b2 +F

bx-1

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bx+

x2

BC

bb-

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b2 +F

bxb2 -2

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JC

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-xF

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00

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b0

00

00

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00

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FG

b0

00

00

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F b

00

00

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b0

00

00

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G b

00

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elon

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one

asta

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CL B

C-F

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tota

li-5

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5/2F

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erst

atic

a

Page 71: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.dmcm.018PROCEDIMENTO E RISULTATI 870485 D’Amico Michele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b(-1/2 x2/b2 ) Fb2 1/EJ dx = [-1/6 x3/b2 ]o

b Fb2 1/EJ

= (-1/6 b ) Fb2 1/EJ = -1/6 Fb3/EJ

LXoBA = ∫

o

b(-1/2 + x/b -1/2 x2/b2 ) Fb2 1/EJ dx = [-1/2 x +1/2 x2/b -1/6 x3/b2 ]o

b Fb2 1/EJ

= (-1/2 b +1/2 b -1/6 b ) Fb2 1/EJ = -1/6 Fb3/EJ

LXoBC = ∫

o

b(-1 + x/b ) Fb2 1/EJ dx - 1 (-1) (-1) Fb3/EJ = [- x +1/2 x2/b ]o

b Fb2 1/EJ - 1 (-1) (-1) Fb3/E

= (- b +1/2 b ) Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ = -3/2 Fb3/EJ

LXoCB = ∫

o

b(- x/b ) Fb2 1/EJ dx - 1 (-1) (-1) Fb3/EJ = [-1/2 x2/b ]o

b Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ

= (-1/2 b ) Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ = -3/2 Fb3/EJ

Schema.dmcm.018PROCEDIMENTO E RISULTATI 870485 D’Amico Michele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

A = 714. mm2

Ju = 156210. mm4

Jv = 68670. mm4

yg = 38.15 mmN = 495. NTy = -990. NMx = -940500. Nmmxm = 18. mmum = -3. mmvm = -38.15 mmσm = N/A-Mv/Ju = -229. N/mm2

xc = 21. mmyc = 16. mmvc = -22.15 mmσc = N/A-Mv/Ju = -132.6 N/mm2

τc = 3.057 N/mm2

σo = √σ2+3τ2 = 132.8 N/mm2

S* = 2894. mm3mm 0 18 24 42x

0

42

53

y

16σc,τc

σm

u

v

Page 72: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.dmcm.018

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.dmcm.018

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 73: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.dcml.019REAZIONI 877976 Di Camillo Lorenzo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1/2F

5/2F

1/2F

5/2F5/2Fb

A B

2F

5/2FFb

2F

5/2FFb

B

C

3/2F

3/2F3/2Fb

3/2F

3/2F

B D

3/2F

3/2F

3/2F

3/2F3/2Fb

D

E

3/2F

3/2F3/2Fb

3/2F

3/2F

EF

3/2F

F

B

3/2FFG

1/2F

1/2F

G

A

Schema.dcml.019AZIONI INTERNE 877976 Di Camillo Lorenzo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1/2

5/2

-3/2

3/2

3/2

-3/2

3/200

F

5/2

-2

-3/2

-3/2

3/2

0

0

1/2

-1/2

F

0 5/2

1-1

3/2 0

0-3

/2

-3/200

0

0000

Fb

Page 74: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Sch

ema.

dcm

l.019

PR

OC

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ULT

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I 877

976

Di C

amill

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@ A

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si, P

olite

cnic

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27.0

3.13

06.0

9.19

A

B C

D

EF

G

W

F

X

q

Sch

ema

di c

alco

lo ip

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atic

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0

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0 0 0

00 0 0

Mo

fless

ione

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caric

hi a

sseg

nati

01

1 000

00

00 0 0

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1

Sch

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Mila

no, v

ers.

27.0

3.13

06.0

9.19

Qua

dro

cont

ribut

i PLV

per

iper

stat

ica

X=

HC

→M

x(x)

Mo(

x)M

xMo

MxM

x∫M

xMo/

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x∫X

MxM

x/E

Jdx

AB

bx

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x1/

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1/3X

b3 /EJ

BA

b-b

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bx+

1/2F

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bx+

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BC

bb-

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b-F

b2 +F

bxb2 -2

bx+

x2

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Fb3 /E

J1/

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JC

B b

-xF

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BD

b0

3/2F

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00

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b0

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Fx

00

DE

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00

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00

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0

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elon

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asta

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CL B

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tota

li-4

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2/3X

b3 /EJ

iper

stat

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erst

atic

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Page 75: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.dcml.019PROCEDIMENTO E RISULTATI 877976 Di Camillo Lorenzo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b(1/2 x2/b2 ) Fb2 1/EJ dx = [1/6 x3/b2 ]o

b Fb2 1/EJ

= (1/6 b ) Fb2 1/EJ = 1/6 Fb3/EJ

LXoBA = ∫

o

b(1/2 - x/b +1/2 x2/b2 ) Fb2 1/EJ dx = [1/2 x -1/2 x2/b +1/6 x3/b2 ]o

b Fb2 1/EJ

= (1/2 b -1/2 b +1/6 b ) Fb2 1/EJ = 1/6 Fb3/EJ

LXoBC = ∫

o

b(-1 + x/b ) Fb2 1/EJ dx - 1 (-1) (-1) Fb3/EJ = [- x +1/2 x2/b ]o

b Fb2 1/EJ - 1 (-1) (-1) Fb3/E

= (- b +1/2 b ) Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ = -3/2 Fb3/EJ

LXoCB = ∫

o

b(- x/b ) Fb2 1/EJ dx - 1 (-1) (-1) Fb3/EJ = [-1/2 x2/b ]o

b Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ

= (-1/2 b ) Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ = -3/2 Fb3/EJ

Schema.dcml.019PROCEDIMENTO E RISULTATI 877976 Di Camillo Lorenzo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

A = 816. mm2

Ju = 230061. mm4

Jv = 52992. mm4

yg = 31.79 mmN = -1740. NTy = -1740. NMx = 1740000. Nmmxm = 18. mmum = -6. mmvm = -31.79 mmσm = N/A-Mv/Ju = 238.3 N/mm2

xc = 24. mmyc = 13. mmvc = -18.79 mmσc = N/A-Mv/Ju = 140. N/mm2

τc = 2.487 N/mm2

σo = √σ2+3τ2 = 140.1 N/mm2

S* = 3946. mm3mm 0 18 30 48x

0

48

53

y

13σc,τc

σm

u

v

Page 76: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.dcml.019

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.dcml.019

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 77: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.lbtt.020REAZIONI 850296 Labate Tommaso

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

5/2F 5/2F5/2Fb

A B

2F

5/2FFb

2F

5/2FFb

B

C

2F

3/2F3/2Fb

2F

3/2F

B D

2F

3/2F

2F

3/2F2Fb

D

E

2F

3/2F2Fb

2F

3/2F1/2Fb

EF

F

3/2F1/2Fb

3/2F

F

B

FFG

G

A

Schema.lbtt.020AZIONI INTERNE 850296 Labate Tommaso

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0

5/2

-2

3/2

2

-3/2

-3/2

1

0

F

5/2

-2

-3/2

-2

3/210

0

0

F

0 5/2

1-1

3/2 0

0-2

-2-1/2

-1/2

0

0000

Fb

Page 78: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Sch

ema.

lbtt.

020

PR

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ULT

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296

Laba

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dolfo

Zav

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si, P

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cnic

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no, v

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27.0

3.13

06.0

9.19

A

B C

D

EF

G

W

F

X

q

Sch

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di c

alco

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atic

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01/

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0

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-1/20

00 0 0

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si, P

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cnic

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no, v

ers.

27.0

3.13

06.0

9.19

Qua

dro

cont

ribut

i PLV

per

iper

stat

ica

X=

HC

→M

x(x)

Mo(

x)M

xMo

MxM

x∫M

xMo/

EJd

x∫X

MxM

x/E

Jdx

AB

bx

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bx+

x2

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J1/

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JC

B b

-xF

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00

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elon

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CL B

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tota

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2/3X

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iper

stat

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X=

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Svi

lupp

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lo ip

erst

atic

a

Page 79: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.lbtt.020PROCEDIMENTO E RISULTATI 850296 Labate Tommaso

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b(1/2 x2/b2 ) Fb2 1/EJ dx = [1/6 x3/b2 ]o

b Fb2 1/EJ

= (1/6 b ) Fb2 1/EJ = 1/6 Fb3/EJ

LXoBA = ∫

o

b(1/2 - x/b +1/2 x2/b2 ) Fb2 1/EJ dx = [1/2 x -1/2 x2/b +1/6 x3/b2 ]o

b Fb2 1/EJ

= (1/2 b -1/2 b +1/6 b ) Fb2 1/EJ = 1/6 Fb3/EJ

LXoBC = ∫

o

b(-1 + x/b ) Fb2 1/EJ dx - 1 (-1) (-1) Fb3/EJ = [- x +1/2 x2/b ]o

b Fb2 1/EJ - 1 (-1) (-1) Fb3/E

= (- b +1/2 b ) Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ = -3/2 Fb3/EJ

LXoCB = ∫

o

b(- x/b ) Fb2 1/EJ dx - 1 (-1) (-1) Fb3/EJ = [-1/2 x2/b ]o

b Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ

= (-1/2 b ) Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ = -3/2 Fb3/EJ

Schema.lbtt.020PROCEDIMENTO E RISULTATI 850296 Labate Tommaso

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

A = 1032. mm2

Ju = 260495. mm4

Jv = 107424. mm4

yg = 34.56 mmN = 2145. NTy = -2860. NMx = -1515800. Nmmxm = 18. mmum = -6. mmvm = -34.56 mmσm = N/A-Mv/Ju = -199. N/mm2

xc = 24. mmyc = 14. mmvc = -20.56 mmσc = N/A-Mv/Ju = -117.5 N/mm2

τc = 4.236 N/mm2

σo = √σ2+3τ2 = 117.8 N/mm2

S* = 4630. mm3mm 0 18 30 48x

0

42

53

y

14σc,τc

σm

u

v

Page 80: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.lbtt.020

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.lbtt.020

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 81: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.lgml.021REAZIONI 835162 Ligammari Luca

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

5/2F 5/2F5/2Fb

A B

2F

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2F

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Schema.lgml.021AZIONI INTERNE 835162 Ligammari Luca

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0

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Page 82: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 83: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.lgml.021PROCEDIMENTO E RISULTATI 835162 Ligammari Luca

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoBC = ∫

o

b(3/2 -2 x/b +1/2 x2/b2 ) Fb2 1/EJ dx + 1 (-1) (-1) Fb3/EJ

= [3/2 x - x2/b +1/6 x3/b2 ]o

b Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ

= (3/2 b - b +1/6 b ) Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ = 5/3 Fb3/EJ

LXoCB = ∫

o

b( x/b +1/2 x2/b2 ) Fb2 1/EJ dx + 1 (-1) (-1) Fb3/EJ

= [1/2 x2/b +1/6 x3/b2 ]o

b Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ

= (1/2 b +1/6 b ) Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ = 5/3 Fb3/EJ

A = 438. mm2

Ju = 124871. mm4

Jv = 12114. mm4

yg = 33.08 mmN = 920. NTy = 1380. NMx = -800400. Nmmxm = 12. mmum = -3. mmvm = -33.08 mmσm = N/A-Mv/Ju = -209.9 N/mm2

xc = 15. mmyc = 14. mmvc = -19.08 mmσc = N/A-Mv/Ju = -120.2 N/mm2

τc = 4.034 N/mm2

σo = √σ2+3τ2 = 120.4 N/mm2

S* = 2190. mm3mm 0 12 18 30x

0

48

53

y

14σc,τc

σm

u

v

Schema.lgml.021

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 84: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.lgml.021

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.lgml.021

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 85: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.mnrs.022REAZIONI 878035 Manrique Silvera Gerald

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

43/16F 27/16F35/16Fb

A B

35/16F

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Schema.mnrs.022AZIONI INTERNE 878035 Manrique Silvera Gerald

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0 0

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0

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Page 86: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 87: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.mnrs.022PROCEDIMENTO E RISULTATI 878035 Manrique Silvera

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b(- x2/b2 +1/2 x3/b3 ) Fb2 1/EJ dx = [-1/3 x3/b2 +1/8 x4/b3 ]o

b Fb2 1/EJ

= (-1/3 b +1/8 b ) Fb2 1/EJ = -5/24 Fb3/EJ

LXoBA = ∫

o

b(-1/2 +1/2 x/b +1/2 x2/b2 -1/2 x3/b3 ) Fb2 1/EJ dx

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

b Fb2 1/EJ

= (-1/2 b +1/4 b +1/6 b -1/8 b ) Fb2 1/EJ = -5/24 Fb3/EJ

LXoBC = ∫

o

b(1/2 -1/2 x2/b2 ) Fb2 1/EJ dx + 1 (-1) (-1) Fb3/EJ

= [1/2 x -1/6 x3/b2 ]o

b Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ

= (1/2 b -1/6 b ) Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ = 4/3 Fb3/EJ

LXoCB = ∫

o

b( x/b -1/2 x2/b2 ) Fb2 1/EJ dx + 1 (-1) (-1) Fb3/EJ

= [1/2 x2/b -1/6 x3/b2 ]o

b Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ

= (1/2 b -1/6 b ) Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ = 4/3 Fb3/EJ

Schema.mnrs.022PROCEDIMENTO E RISULTATI 878035 Manrique Silvera

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

A = 582. mm2

Ju = 140714. mm4

Jv = 25506. mm4

yg = 36.03 mmN = 1370. NTy = -1370. NMx = -863100. Nmmxm = 12. mmum = -3. mmvm = -36.03 mmσm = N/A-Mv/Ju = -218.6 N/mm2

xc = 15. mmyc = 15. mmvc = -21.03 mmσc = N/A-Mv/Ju = -126.6 N/mm2

τc = 4.166 N/mm2

σo = √σ2+3τ2 = 126.8 N/mm2

S* = 2567. mm3mm 0 12 18 30x

0

42

53

y

15σc,τc

σm

u

v

Page 88: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.mnrs.022

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.mnrs.022

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 89: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.mrna.023REAZIONI 835477 Maranga Andrea

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

2F 2F2Fb

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Schema.mrna.023AZIONI INTERNE 835477 Maranga Andrea

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0

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Page 90: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 91: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.mrna.023PROCEDIMENTO E RISULTATI 835477 Maranga Andrea

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b(- x2/b2 ) Fb2 1/EJ dx = [-1/3 x3/b2 ]o

b Fb2 1/EJ

= (-1/3 b ) Fb2 1/EJ = -1/3 Fb3/EJ

LXoBA = ∫

o

b(-1 +2 x/b - x2/b2 ) Fb2 1/EJ dx = [- x + x2/b -1/3 x3/b2 ]o

b Fb2 1/EJ

= (- b + b -1/3 b ) Fb2 1/EJ = -1/3 Fb3/EJ

LXoBC = ∫

o

b(-1/2 +2 x/b -3/2 x2/b2 ) Fb2 1/EJ dx + 1 (-1) (-1) Fb3/EJ

= [-1/2 x + x2/b -1/2 x3/b2 ]o

b Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ

= (-1/2 b + b -1/2 b ) Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ = Fb3/EJ

LXoCB = ∫

o

b( x/b -3/2 x2/b2 ) Fb2 1/EJ dx + 1 (-1) (-1) Fb3/EJ

= [1/2 x2/b -1/2 x3/b2 ]o

b Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ

= (1/2 b -1/2 b ) Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ = Fb3/EJ

Schema.mrna.023PROCEDIMENTO E RISULTATI 835477 Maranga Andrea

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

A = 756. mm2

Ju = 207276. mm4

Jv = 26352. mm4

yg = 30.31 mmN = 2340. NTy = -2340. NMx = -1591200. Nmmxm = 12. mmum = -6. mmvm = -30.31 mmσm = N/A-Mv/Ju = -229.6 N/mm2

xc = 18. mmyc = 47. mmvc = 16.69 mmσc = N/A-Mv/Ju = 131.2 N/mm2

τc = 3.613 N/mm2

σo = √σ2+3τ2 = 131.4 N/mm2

S* = 3841. mm3mm 0 12 24 36x

0

48

53

y

47σc,τc

σm

u

v

Page 92: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.mrna.023

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.mrna.023

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 93: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.mrtm.024REAZIONI 846219 Martignoni Matteo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

35/16F 35/16F35/16Fb

A B

27/16F

35/16F19/16Fb

43/16F

35/16FFb

B

C

27/16F

FFb

27/16F

F

B D

F

F

F

FFb

D

E

F

FFb

F

F

EF

F

F

B

FFG

G

A

Schema.mrtm.024AZIONI INTERNE 846219 Martignoni Matteo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0

35/1

635

/16

-27/16

1

1

-1

1

0

F

35/16

-27/

16-4

3/16

-1

-1

1

0

0

0

F

0 35/16

19/1

6-1

1 0

0-1

-100

0

0000

Fb

Page 94: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Sch

ema.

mrt

m.0

24P

RO

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E R

ISU

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4621

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noni

Mat

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@ A

dolfo

Zav

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si, P

olite

cnic

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Mila

no, v

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27.0

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A

B

C

DEF

G

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X

q

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atic

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00 0 0

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fless

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Zav

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27.0

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9.19

Qua

dro

cont

ribut

i PLV

per

iper

stat

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X=

HD

→M

x(x)

Mo(

x)M

xMo

MxM

x∫M

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x/E

Jdx

AB

b-x

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bx+

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+x

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3/2F

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x2 -1/2

qx3

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+x2

-1/2

4Fb3 /E

J1/

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JC

B b

xF

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00

00

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b0

00

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00

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00

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F b

00

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b0

00

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00

00

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elon

gazi

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asta

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CF

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lo ip

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atic

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Page 95: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.mrtm.024PROCEDIMENTO E RISULTATI 846219 Martignoni Matteo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b(-3/2 x2/b2 ) Fb2 1/EJ dx = [-1/2 x3/b2 ]o

b Fb2 1/EJ

= (-1/2 b ) Fb2 1/EJ = -1/2 Fb3/EJ

LXoBA = ∫

o

b(-3/2 +3 x/b -3/2 x2/b2 ) Fb2 1/EJ dx = [-3/2 x +3/2 x2/b -1/2 x3/b2 ]o

b Fb2 1/EJ

= (-3/2 b +3/2 b -1/2 b ) Fb2 1/EJ = -1/2 Fb3/EJ

LXoBC = ∫

o

b(-1/2 +3/2 x/b -1/2 x2/b2 -1/2 x3/b3 ) Fb2 1/EJ dx + 1 (-1) (-1) Fb3/EJ

= [-1/2 x +3/4 x2/b -1/6 x3/b2 -1/8 x4/b3 ]o

b Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ

= (-1/2 b +3/4 b -1/6 b -1/8 b ) Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ = 23/24 Fb3/EJ

LXoCB = ∫

o

b( x/b -2 x2/b2 +1/2 x3/b3 ) Fb2 1/EJ dx + 1 (-1) (-1) Fb3/EJ

= [1/2 x2/b -2/3 x3/b2 +1/8 x4/b3 ]o

b Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ

= (1/2 b -2/3 b +1/8 b ) Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ = 23/24 Fb3/EJ

Schema.mrtm.024PROCEDIMENTO E RISULTATI 846219 Martignoni Matteo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

A = 900. mm2

Ju = 233812. mm4

Jv = 48816. mm4

yg = 32.66 mmN = 2340. NTy = -2340. NMx = -1731600. Nmmxm = 12. mmum = -6. mmvm = -32.66 mmσm = N/A-Mv/Ju = -239.3 N/mm2

xc = 18. mmyc = 14. mmvc = -18.66 mmσc = N/A-Mv/Ju = -135.6 N/mm2

τc = 3.595 N/mm2

σo = √σ2+3τ2 = 135.7 N/mm2

S* = 4311. mm3mm 0 12 24 36x

0

42

53

y

14σc,τc

σm

u

v

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Schema.mrtm.024

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.mrtm.024

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 97: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.mccg.025REAZIONI 867110 Mecacci Giorgia

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1/2F

FFb

1/2F

F2Fb

A B

1/2F1/2Fb

1/2F

B

C

F

3/2F3/2Fb

F

3/2F

B D

3/2F

3/2F

3/2F

3/2F3/2Fb

D

E

3/2F

3/2F3/2Fb

3/2F

3/2F

EF

5/2F

F

B

3/2F

F

3/2F1/2Fb

FG

1/2F1/2Fb

1/2F

G

A

Schema.mccg.025AZIONI INTERNE 867110 Mecacci Giorgia

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

-1/2

0

-1

3/2

3/2

-5/2

3/23/2

0

F

1

-1/2

-3/2

-3/2

3/2

0

-10

1/2

F

1 2

1/2

0

3/2 0

0-3

/2

-3/200

0

0-1/2

-1/2

0

Fb

Page 98: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Sch

ema.

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g.02

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27.0

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W

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27.0

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9.19

Qua

dro

cont

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iper

stat

ica

X=

VA

→M

x(x)

Mo(

x)M

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MxM

x∫M

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bb-

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Fbx

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Fx2

b2 -2bx

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Fb3 /E

J1/

3Xb3 /E

JC

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01/

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BC

elon

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asta

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CL B

C-F

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tota

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atic

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Page 99: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.mccg.025PROCEDIMENTO E RISULTATI 867110 Mecacci Giorgia

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b( x/b ) Fb2 1/EJ dx = [1/2 x2/b ]o

b Fb2 1/EJ

= (1/2 b ) Fb2 1/EJ = 1/2 Fb3/EJ

LXoBA = ∫

o

b(1 - x/b ) Fb2 1/EJ dx = [ x -1/2 x2/b ]o

b Fb2 1/EJ

= ( b -1/2 b ) Fb2 1/EJ = 1/2 Fb3/EJ

LXoBC = ∫

o

b(-1/2 + x/b -1/2 x2/b2 ) Fb2 1/EJ dx - 1 (-1) (-1) Fb3/EJ

= [-1/2 x +1/2 x2/b -1/6 x3/b2 ]o

b Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ

= (-1/2 b +1/2 b -1/6 b ) Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ = -7/6 Fb3/EJ

LXoCB = ∫

o

b(-1/2 x2/b2 ) Fb2 1/EJ dx - 1 (-1) (-1) Fb3/EJ = [-1/6 x3/b2 ]o

b Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ

= (-1/6 b ) Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ = -7/6 Fb3/EJ

Schema.mccg.025PROCEDIMENTO E RISULTATI 867110 Mecacci Giorgia

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

A = 924. mm2

Ju = 235641. mm4

Jv = 52560. mm4

yg = 20.11 mmN = 1785. NTy = -1785. NMx = -1410150. Nmmxm = 24. mmym = 53. mmum = 6. mmvm = 32.89 mmσm = N/A-Mv/Ju = 198.8 N/mm2

xc = 18. mmyc = 39. mmvc = 18.89 mmσc = N/A-Mv/Ju = 115. N/mm2

τc = 2.746 N/mm2

σo = √σ2+3τ2 = 115.1 N/mm2

S* = 4349. mm3mm 0 12 24 36x

0

12

53

y

39σc,τc

σm

u

v

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Schema.mccg.025

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.mccg.025

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 101: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.mssf.026REAZIONI 889516 Messa Federico

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1/2FFb

1/2F3/2Fb

A B

F

1/2FFb

F

1/2F

B

C

F

1/2F1/2Fb

F

1/2F

B D

F

1/2F

F

1/2FFb

D

E

F

1/2FFb

F

3/2F

EF

3/2F

F

B

FFG

G

A

Schema.mssf.026AZIONI INTERNE 889516 Messa Federico

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0

-1/2

-1

1/2

11

-3/2

1

0

F

1/2

-1

-1/2

-1

1/23/2

0

0

0

F

1 3/2

10

1/2 0

0-1

-100

0

0000

Fb

Page 102: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Sch

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27.0

3.13

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Qua

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cont

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x(x)

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x)M

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bxb2 -2

bx+

x2

BC

bb-

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2Fb-

1/2F

x1/

2Fb2 -F

bx+

1/2F

x2b2 -2

bx+

x2

1/6F

b3 /EJ

1/3X

b3 /EJ

CB

b-x

-1/2

Fx

1/2F

x2x2

BD

b0

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2Fx

00

00

DB

b0

-1/2

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00

DE

b0

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00

00

ED

b0

Fb-

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00

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b0

-Fb+

1/2F

x+1/

2qx2

00

00

FE

b0

3/2F

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2qx2

00

FB

b0

00

00

0B

F b

00

00

FG

b0

00

00

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00

00

GA

b0

00

00

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G b

00

00

BC

elon

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one

asta

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Cε B

CL B

C-F

b3 /EJ

tota

li-1

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ica

X=

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Svi

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erst

atic

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Page 103: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.mssf.026PROCEDIMENTO E RISULTATI 889516 Messa Federico

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b( x/b ) Fb2 1/EJ dx = [1/2 x2/b ]o

b Fb2 1/EJ

= (1/2 b ) Fb2 1/EJ = 1/2 Fb3/EJ

LXoBA = ∫

o

b(1 - x/b ) Fb2 1/EJ dx = [ x -1/2 x2/b ]o

b Fb2 1/EJ

= ( b -1/2 b ) Fb2 1/EJ = 1/2 Fb3/EJ

LXoBC = ∫

o

b(1/2 - x/b +1/2 x2/b2 ) Fb2 1/EJ dx - 1 (-1) (-1) Fb3/EJ

= [1/2 x -1/2 x2/b +1/6 x3/b2 ]o

b Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ

= (1/2 b -1/2 b +1/6 b ) Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ = -5/6 Fb3/EJ

LXoCB = ∫

o

b(1/2 x2/b2 ) Fb2 1/EJ dx - 1 (-1) (-1) Fb3/EJ = [1/6 x3/b2 ]o

b Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ

= (1/6 b ) Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ = -5/6 Fb3/EJ

Schema.mssf.026PROCEDIMENTO E RISULTATI 889516 Messa Federico

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

A = 462. mm2

Ju = 129608. mm4

Jv = 14346. mm4

yg = 19.18 mmN = 475. NTy = -950. NMx = -798000. Nmmxm = 18. mmym = 53. mmum = 3. mmvm = 33.82 mmσm = N/A-Mv/Ju = 209.3 N/mm2

xc = 15. mmyc = 39. mmvc = 19.82 mmσc = N/A-Mv/Ju = 123.1 N/mm2

τc = 2.753 N/mm2

σo = √σ2+3τ2 = 123.2 N/mm2

S* = 2253. mm3mm 0 12 18 30x

0

6

53

y

39σc,τc

σm

u

v

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Schema.mssf.026

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.mssf.026

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 105: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.mnta.027REAZIONI 887897 Montagner Alessio

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1/2F

FFb

1/2F

F2Fb

A B

1/2F

F1/2Fb

1/2F

F

B

C

3/2F3/2Fb

3/2F

B D

3/2F

3/2F

3/2F

3/2F3/2Fb

D

E

3/2F

3/2F3/2Fb

3/2F

3/2F

EF

3/2F

F

B

3/2FFG

1/2F

1/2F

G

A

Schema.mnta.027AZIONI INTERNE 887897 Montagner Alessio

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1/2

1

0

3/2

3/2

-3/2

3/200

F

1

-1/2

-3/2

-3/2

3/2

0

0

1/2

-1/2

F

1 2

1/2

0

3/2 0

0-3

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-3/200

0

0000

Fb

Page 106: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Sch

ema.

mnt

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27.0

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Qua

dro

cont

ribut

i PLV

per

iper

stat

ica

X=

HC

→M

x(x)

Mo(

x)M

xMo

MxM

x∫M

xMo/

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x∫X

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x/E

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AB

bx

Fb+

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bx+

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b3 /EJ

BA

b-b

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BC

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00

00

AG

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1/2q

x20

0

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elon

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one

asta

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Cε B

CL B

C-F

b3 /EJ

tota

li-1

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b3 /EJ

2/3X

b3 /EJ

iper

stat

ica

X=

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1/2F

Svi

lupp

i di c

alco

lo ip

erst

atic

a

Page 107: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.mnta.027PROCEDIMENTO E RISULTATI 887897 Montagner Alessio

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b( x/b +1/2 x2/b2 ) Fb2 1/EJ dx = [1/2 x2/b +1/6 x3/b2 ]o

b Fb2 1/EJ

= (1/2 b +1/6 b ) Fb2 1/EJ = 2/3 Fb3/EJ

LXoBA = ∫

o

b(3/2 -2 x/b +1/2 x2/b2 ) Fb2 1/EJ dx = [3/2 x - x2/b +1/6 x3/b2 ]o

b Fb2 1/EJ

= (3/2 b - b +1/6 b ) Fb2 1/EJ = 2/3 Fb3/EJ

A = 852. mm2

Ju = 238569. mm4

Jv = 62064. mm4

yg = 20.54 mmN = 1770. NTy = -1770. NMx = -1593000. Nmmxm = 30. mmym = 53. mmum = 6. mmvm = 32.46 mmσm = N/A-Mv/Ju = 218.8 N/mm2

xc = 24. mmyc = 39. mmvc = 18.46 mmσc = N/A-Mv/Ju = 125.3 N/mm2

τc = 2.644 N/mm2

σo = √σ2+3τ2 = 125.4 N/mm2

S* = 4277. mm3mm 0 18 30 48x

0

6

53

y

39σc,τc

σm

u

v

Schema.mnta.027

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 108: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.mnta.027

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.mnta.027

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 109: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.mntm.028REAZIONI 844334 Montemarano Mauro

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

FFb

F2Fb

A B

1/2F

F1/2Fb

1/2F

F

B

C

1/2F

3/2F3/2Fb

1/2F

3/2F

B D

2F

3/2F

2F

3/2F2Fb

D

E

2F

3/2F2Fb

2F

3/2F1/2Fb

EF

F

3/2F1/2Fb

3/2F

F

B

FFG

G

A

Schema.mntm.028AZIONI INTERNE 844334 Montemarano Mauro

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0

1

-1/2

3/2

2

-3/2

-3/2

1

0

F

1

-1/2

-3/2

-2

3/210

0

0

F

1 2

1/2

0

3/2 0

0-2

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0

0000

Fb

Page 110: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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CF

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Page 111: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.mntm.028PROCEDIMENTO E RISULTATI 844334 Montemarano Mauro

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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

LXXBC = ∫

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

LXXCB = ∫

o

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

b 1/EJ

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

LXoAB = ∫

o

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

b Fb 1/EJ

= (-1/2 b -1/6 b ) Fb 1/EJ = -2/3 Fb2/EJ

LXoBA = ∫

o

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

b Fb 1/EJ

= (-3/2 b + b -1/6 b ) Fb 1/EJ = -2/3 Fb2/EJ

A = 606. mm2

Ju = 141406. mm4

Jv = 27738. mm4

yg = 16.76 mmN = 690. NTy = -920. NMx = -874000. Nmmxm = 18. mmym = 53. mmum = 3. mmvm = 36.24 mmσm = N/A-Mv/Ju = 225.1 N/mm2

xc = 15. mmyc = 38. mmvc = 21.24 mmσc = N/A-Mv/Ju = 132.4 N/mm2

τc = 2.805 N/mm2

σo = √σ2+3τ2 = 132.5 N/mm2

S* = 2587. mm3mm 0 12 18 30x

0

12

53

y

38σc,τc

σm

u

v

Schema.mntm.028

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 112: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.mntm.028

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.mntm.028

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 113: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.mttv.029REAZIONI 853842 Motta Valentina

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

FFb

F2Fb

A B

1/2F

F1/2Fb

1/2F

F

B

C

1/2F

3/2F3/2Fb

1/2F

3/2F

B D

2F

3/2F

F

3/2F3/2Fb

D

E

F

3/2F3/2Fb

F

3/2F

EF

3/2F

F

B

FFG

G

A

Schema.mttv.029AZIONI INTERNE 853842 Motta Valentina

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0

1

-1/2

3/2

3/21

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1

0

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0

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0

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Page 114: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 115: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.mttv.029PROCEDIMENTO E RISULTATI 853842 Motta Valentina

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b( x/b +1/2 x2/b2 ) Fb2 1/EJ dx = [1/2 x2/b +1/6 x3/b2 ]o

b Fb2 1/EJ

= (1/2 b +1/6 b ) Fb2 1/EJ = 2/3 Fb3/EJ

LXoBA = ∫

o

b(3/2 -2 x/b +1/2 x2/b2 ) Fb2 1/EJ dx = [3/2 x - x2/b +1/6 x3/b2 ]o

b Fb2 1/EJ

= (3/2 b - b +1/6 b ) Fb2 1/EJ = 2/3 Fb3/EJ

A = 750. mm2

Ju = 156599. mm4

Jv = 74826. mm4

yg = 14.69 mmN = 650. NTy = 975. NMx = -975000. Nmmxm = 24. mmym = 53. mmum = 3. mmvm = 38.31 mmσm = N/A-Mv/Ju = 239.4 N/mm2

xc = 21. mmyc = 37. mmvc = 22.31 mmσc = N/A-Mv/Ju = 139.8 N/mm2

τc = 3.019 N/mm2

σo = √σ2+3τ2 = 139.9 N/mm2

S* = 2910. mm3mm 0 18 24 42x

0

12

53

y

37σc,τc

σm

u

v

Schema.mttv.029

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 116: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.mttv.029

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.mttv.029

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 117: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.nzza.030REAZIONI 881418 Nuzzo Arianna

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

19/16FFb

3/16F27/16Fb

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Schema.nzza.030AZIONI INTERNE 881418 Nuzzo Arianna

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0 0

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Page 118: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Sch

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Page 119: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.nzza.030PROCEDIMENTO E RISULTATI 881418 Nuzzo Arianna

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b( x/b -1/2 x3/b3 ) Fb2 1/EJ dx = [1/2 x2/b -1/8 x4/b3 ]o

b Fb2 1/EJ

= (1/2 b -1/8 b ) Fb2 1/EJ = 3/8 Fb3/EJ

LXoBA = ∫

o

b(1/2 +1/2 x/b -3/2 x2/b2 +1/2 x3/b3 ) Fb2 1/EJ dx

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

b Fb2 1/EJ

= (1/2 b +1/4 b -1/2 b +1/8 b ) Fb2 1/EJ = 3/8 Fb3/EJ

LXoBC = ∫

o

b(-1/2 + x/b -1/2 x2/b2 ) Fb2 1/EJ dx - 1 (-1) (-1) Fb3/EJ

= [-1/2 x +1/2 x2/b -1/6 x3/b2 ]o

b Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ

= (-1/2 b +1/2 b -1/6 b ) Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ = -7/6 Fb3/EJ

LXoCB = ∫

o

b(-1/2 x2/b2 ) Fb2 1/EJ dx - 1 (-1) (-1) Fb3/EJ = [-1/6 x3/b2 ]o

b Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ

= (-1/6 b ) Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ = -7/6 Fb3/EJ

Schema.nzza.030PROCEDIMENTO E RISULTATI 881418 Nuzzo Arianna

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

A = 780. mm2

Ju = 214152. mm4

Jv = 30096. mm4

yg = 22.16 mmN = 2570. NTy = -2570. NMx = -1362100. Nmmxm = 24. mmym = 53. mmum = 6. mmvm = 30.84 mmσm = N/A-Mv/Ju = 199.4 N/mm2

xc = 18. mmyc = 40. mmvc = 17.84 mmσc = N/A-Mv/Ju = 116.8 N/mm2

τc = 3.797 N/mm2

σo = √σ2+3τ2 = 116.9 N/mm2

S* = 3797. mm3mm 0 12 24 36x

0

6

53

y

40σc,τc

σm

u

v

Page 120: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.nzza.030

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.nzza.030

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 121: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.ptre.031REAZIONI 888074 Pietrobelli Eleonora

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1/2FFb

1/2F3/2Fb

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B

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D

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A

Schema.ptre.031AZIONI INTERNE 888074 Pietrobelli Eleonora

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0

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Page 122: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 123: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.ptre.031PROCEDIMENTO E RISULTATI 888074 Pietrobelli Eleonora

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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

LXXBC = ∫

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

LXXCB = ∫

o

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

b 1/EJ

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

LXoAB = ∫

o

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

b Fb 1/EJ

= (-1/2 b +1/6 b ) Fb 1/EJ = -1/3 Fb2/EJ

LXoBA = ∫

o

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

b Fb 1/EJ

= (-1/2 b +1/6 b ) Fb 1/EJ = -1/3 Fb2/EJ

A = 1068. mm2

Ju = 262174. mm4

Jv = 116496. mm4

yg = 18.21 mmN = 2690. NTy = -2690. NMx = -1560200. Nmmxm = 30. mmym = 53. mmum = 6. mmvm = 34.79 mmσm = N/A-Mv/Ju = 209.6 N/mm2

xc = 24. mmyc = 38. mmvc = 19.79 mmσc = N/A-Mv/Ju = 120.3 N/mm2

τc = 4.2 N/mm2

σo = √σ2+3τ2 = 120.5 N/mm2

S* = 4913. mm3mm 0 18 30 48x

0

12

53

y

38σc,τc

σm

u

v

Schema.ptre.031

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 124: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.ptre.031

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.ptre.031

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 125: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.ppvy.032REAZIONI 849430 Popova Yulia

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

11/16FFb

11/16F27/16Fb

A B

3/16F

11/16F11/16Fb

19/16F

11/16F

B

C

3/16F

FFb

3/16F

F

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F

F

F

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D

E

F

FFb

F

F

EF

F

F

B

FFG

G

A

Schema.ppvy.032AZIONI INTERNE 849430 Popova Yulia

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0

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0000

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Page 126: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 127: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.ppvy.032PROCEDIMENTO E RISULTATI 849430 Popova Yulia

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b( x/b ) Fb2 1/EJ dx = [1/2 x2/b ]o

b Fb2 1/EJ

= (1/2 b ) Fb2 1/EJ = 1/2 Fb3/EJ

LXoBA = ∫

o

b(1 - x/b ) Fb2 1/EJ dx = [ x -1/2 x2/b ]o

b Fb2 1/EJ

= ( b -1/2 b ) Fb2 1/EJ = 1/2 Fb3/EJ

LXoBC = ∫

o

b(1/2 x/b - x2/b2 +1/2 x3/b3 ) Fb2 1/EJ dx - 1 (-1) (-1) Fb3/EJ

= [1/4 x2/b -1/3 x3/b2 +1/8 x4/b3 ]o

b Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ

= (1/4 b -1/3 b +1/8 b ) Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ = -23/24 Fb3/EJ

LXoCB = ∫

o

b(1/2 x2/b2 -1/2 x3/b3 ) Fb2 1/EJ dx - 1 (-1) (-1) Fb3/EJ

= [1/6 x3/b2 -1/8 x4/b3 ]o

b Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ

= (1/6 b -1/8 b ) Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ = -23/24 Fb3/EJ

Schema.ppvy.032PROCEDIMENTO E RISULTATI 849430 Popova Yulia

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

A = 534. mm2

Ju = 146122. mm4

Jv = 37890. mm4

yg = 16.99 mmN = 1400. NTy = -1400. NMx = -882000. Nmmxm = 24. mmym = 53. mmum = 3. mmvm = 36.01 mmσm = N/A-Mv/Ju = 220. N/mm2

xc = 21. mmyc = 38. mmvc = 21.01 mmσc = N/A-Mv/Ju = 129.4 N/mm2

τc = 4.097 N/mm2

σo = √σ2+3τ2 = 129.6 N/mm2

S* = 2566. mm3mm 0 18 24 42x

0

6

53

y

38σc,τc

σm

u

v

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Schema.ppvy.032

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.ppvy.032

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 129: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.pzzm.033REAZIONI 891221 Pozzoni Marco

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1/2F

7/4F

1/2F

7/4F7/4Fb

A B

5/4F

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5/4F

3/4F

B

C

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7/4F

3/2FFb

B D

3/2F

3/2F

3/2F

3/2F3/2Fb

D

E

3/2F

3/2F3/2Fb

3/2F

3/2F

EF

5/2F

F

B

3/2F

F

3/2F1/2Fb

FG

1/2F1/2Fb

1/2F

G

A

Schema.pzzm.033AZIONI INTERNE 891221 Pozzoni Marco

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

-1/2

3/4

-7/4

3/2

3/2

-5/2

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0

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7/4

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F

0 7/4

5/4

0

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0-3

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0

0-1/2

-1/2

0

Fb

Page 130: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 131: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.pzzm.033PROCEDIMENTO E RISULTATI 891221 Pozzoni Marco

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b(-3/2 x2/b2 ) Fb2 1/EJ dx = [-1/2 x3/b2 ]o

b Fb2 1/EJ

= (-1/2 b ) Fb2 1/EJ = -1/2 Fb3/EJ

LXoBA = ∫

o

b(-3/2 +3 x/b -3/2 x2/b2 ) Fb2 1/EJ dx = [-3/2 x +3/2 x2/b -1/2 x3/b2 ]o

b Fb2 1/EJ

= (-3/2 b +3/2 b -1/2 b ) Fb2 1/EJ = -1/2 Fb3/EJ

LXoBC = ∫

o

b(-1 +2 x/b - x2/b2 ) Fb2 1/EJ dx + 1 (-1) (-1) Fb3/EJ

= [- x + x2/b -1/3 x3/b2 ]o

b Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ

= (- b + b -1/3 b ) Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ = 2/3 Fb3/EJ

LXoCB = ∫

o

b(- x2/b2 ) Fb2 1/EJ dx + 1 (-1) (-1) Fb3/EJ = [-1/3 x3/b2 ]o

b Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ

= (-1/3 b ) Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ = 2/3 Fb3/EJ

Schema.pzzm.033PROCEDIMENTO E RISULTATI 891221 Pozzoni Marco

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

A = 582. mm2

Ju = 166519. mm4

Jv = 44082. mm4

yg = 37.89 mmN = 1425. NTy = -1425. NMx = -1011750. Nmmxm = 18. mmum = -3. mmvm = -37.89 mmσm = N/A-Mv/Ju = -227.8 N/mm2

xc = 21. mmyc = 16. mmvc = -21.89 mmσc = N/A-Mv/Ju = -130.6 N/mm2

τc = 4.093 N/mm2

σo = √σ2+3τ2 = 130.8 N/mm2

S* = 2870. mm3mm 0 18 24 42x

0

48

55

y

16σc,τc

σm

u

v

Page 132: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.pzzm.033

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.pzzm.033

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 133: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.prva.034REAZIONI 853194 Previtali Alessia

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

5/4F 5/4F5/4Fb

A B

7/4F

1/4F7/4Fb

7/4F

1/4F

B

C

7/4F

1/2F1/2Fb

7/4F

1/2FFb

B D

F

1/2F

F

1/2FFb

D

E

F

1/2FFb

F

3/2F

EF

3/2F

F

B

FFG

G

A

Schema.prva.034AZIONI INTERNE 853194 Previtali Alessia

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0

1/4

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1

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0

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0-1

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0

0000

Fb

Page 134: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 135: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.prva.034PROCEDIMENTO E RISULTATI 853194 Previtali Alessia

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b(-1/2 x2/b2 ) Fb2 1/EJ dx = [-1/6 x3/b2 ]o

b Fb2 1/EJ

= (-1/6 b ) Fb2 1/EJ = -1/6 Fb3/EJ

LXoBA = ∫

o

b(-1/2 + x/b -1/2 x2/b2 ) Fb2 1/EJ dx = [-1/2 x +1/2 x2/b -1/6 x3/b2 ]o

b Fb2 1/EJ

= (-1/2 b +1/2 b -1/6 b ) Fb2 1/EJ = -1/6 Fb3/EJ

LXoBC = ∫

o

b(-1 +2 x/b - x2/b2 ) Fb2 1/EJ dx + 1 (-1) (-1) Fb3/EJ

= [- x + x2/b -1/3 x3/b2 ]o

b Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ

= (- b + b -1/3 b ) Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ = 2/3 Fb3/EJ

LXoCB = ∫

o

b(- x2/b2 ) Fb2 1/EJ dx + 1 (-1) (-1) Fb3/EJ = [-1/3 x3/b2 ]o

b Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ

= (-1/3 b ) Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ = 2/3 Fb3/EJ

Schema.prva.034PROCEDIMENTO E RISULTATI 853194 Previtali Alessia

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

A = 798. mm2

Ju = 175127. mm4

Jv = 81018. mm4

yg = 39.82 mmN = 685. NTy = -1370. NMx = -1054900. Nmmxm = 18. mmum = -3. mmvm = -39.82 mmσm = N/A-Mv/Ju = -239. N/mm2

xc = 21. mmyc = 17. mmvc = -22.82 mmσc = N/A-Mv/Ju = -136.6 N/mm2

τc = 4.165 N/mm2

σo = √σ2+3τ2 = 136.8 N/mm2

S* = 3194. mm3mm 0 18 24 42x

0

42

55

y

17σc,τc

σm

u

v

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Schema.prva.034

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.prva.034

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 137: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.prft.035REAZIONI 867435 Prifti Tomas

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1/2F

7/4F

1/2F

7/4F7/4Fb

A B

5/4F

7/4F5/4Fb

5/4F

7/4F

B

C

3/4F

3/2F1/2Fb

3/4F

3/2FFb

B D

3/2F

3/2F

3/2F

3/2F3/2Fb

D

E

3/2F

3/2F3/2Fb

3/2F

3/2F

EF

3/2F

F

B

3/2FFG

1/2F

1/2F

G

A

Schema.prft.035AZIONI INTERNE 867435 Prifti Tomas

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1/2

7/4

-3/4

3/2

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3/200

F

7/4

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0

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5/4

0

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0

0000

Fb

Page 138: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Sch

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CF

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Page 139: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.prft.035PROCEDIMENTO E RISULTATI 867435 Prifti Tomas

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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

LXXBC = ∫

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

LXXCB = ∫

o

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

b 1/EJ

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

LXoAB = ∫

o

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

b Fb 1/EJ

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

LXoBA = ∫

o

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

b Fb 1/EJ

= (-1/2 b +1/2 b -1/6 b ) Fb 1/EJ = -1/6 Fb2/EJ

A = 912. mm2

Ju = 272448. mm4

Jv = 71424. mm4

yg = 34.13 mmN = 1965. NTy = -1965. NMx = -1611300. Nmmxm = 18. mmum = -6. mmvm = -34.13 mmσm = N/A-Mv/Ju = -199.7 N/mm2

xc = 24. mmyc = 14. mmvc = -20.13 mmσc = N/A-Mv/Ju = -116.9 N/mm2

τc = 2.74 N/mm2

σo = √σ2+3τ2 = 117. N/mm2

S* = 4558. mm3mm 0 18 30 48x

0

48

55

y

14σc,τc

σm

u

v

Schema.prft.035

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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Schema.prft.035

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.prft.035

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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Schema.rttl.036REAZIONI 892357 Rattaggi Lorenza

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

7/4F 7/4F7/4Fb

A B

5/4F

7/4F5/4Fb

5/4F

7/4F

B

C

5/4F

3/2F1/2Fb

5/4F

3/2FFb

B D

2F

3/2F

2F

3/2F2Fb

D

E

2F

3/2F2Fb

2F

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EF

F

3/2F1/2Fb

3/2F

F

B

FFG

G

A

Schema.rttl.036AZIONI INTERNE 892357 Rattaggi Lorenza

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0

7/4

-5/4

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3/210

0

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0

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0

0000

Fb

Page 142: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Sch

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Page 143: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.rttl.036PROCEDIMENTO E RISULTATI 892357 Rattaggi Lorenza

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoBC = ∫

o

b(-1/2 + x/b -1/2 x2/b2 ) Fb2 1/EJ dx - 1 (-1) (-1) Fb3/EJ

= [-1/2 x +1/2 x2/b -1/6 x3/b2 ]o

b Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ

= (-1/2 b +1/2 b -1/6 b ) Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ = -7/6 Fb3/EJ

LXoCB = ∫

o

b(-1/2 x2/b2 ) Fb2 1/EJ dx - 1 (-1) (-1) Fb3/EJ = [-1/6 x3/b2 ]o

b Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ

= (-1/6 b ) Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ = -7/6 Fb3/EJ

A = 1128. mm2

Ju = 293725. mm4

Jv = 125856. mm4

yg = 36.21 mmN = 1455. NTy = -1940. NMx = -1707200. Nmmxm = 18. mmum = -6. mmvm = -36.21 mmσm = N/A-Mv/Ju = -209.2 N/mm2

xc = 24. mmyc = 15. mmvc = -21.21 mmσc = N/A-Mv/Ju = -122. N/mm2

τc = 2.845 N/mm2

σo = √σ2+3τ2 = 122.1 N/mm2

S* = 5168. mm3mm 0 18 30 48x

0

42

55

y

15σc,τc

σm

u

v

Schema.rttl.036

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 144: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.rttl.036

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.rttl.036

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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Schema.rsse.037REAZIONI 917955 Rossi Edoardo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

7/4F 7/4F7/4Fb

A B

5/4F

7/4F5/4Fb

5/4F

7/4F

B

C

5/4F

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Schema.rsse.037AZIONI INTERNE 917955 Rossi Edoardo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0

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3/21

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1

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0

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0

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Page 146: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 147: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.rsse.037PROCEDIMENTO E RISULTATI 917955 Rossi Edoardo

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b(1/2 x2/b2 ) Fb2 1/EJ dx = [1/6 x3/b2 ]o

b Fb2 1/EJ

= (1/6 b ) Fb2 1/EJ = 1/6 Fb3/EJ

LXoBA = ∫

o

b(1/2 - x/b +1/2 x2/b2 ) Fb2 1/EJ dx = [1/2 x -1/2 x2/b +1/6 x3/b2 ]o

b Fb2 1/EJ

= (1/2 b -1/2 b +1/6 b ) Fb2 1/EJ = 1/6 Fb3/EJ

A = 498. mm2

Ju = 147997. mm4

Jv = 16614. mm4

yg = 35.6 mmN = 650. NTy = 975. NMx = -906750. Nmmxm = 12. mmum = -3. mmvm = -35.6 mmσm = N/A-Mv/Ju = -216.8 N/mm2

xc = 15. mmyc = 15. mmvc = -20.6 mmσc = N/A-Mv/Ju = -124.9 N/mm2

τc = 2.776 N/mm2

σo = √σ2+3τ2 = 125. N/mm2

S* = 2529. mm3mm 0 12 18 30x

0

48

55

y

15σc,τc

σm

u

v

Schema.rsse.037

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 148: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.rsse.037

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.rsse.037

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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Schema.slrm.038REAZIONI 896150 Salari Mehdi

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

31/16F 15/16F23/16Fb

A B

23/16F

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A

Schema.slrm.038AZIONI INTERNE 896150 Salari Mehdi

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0 0

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Page 150: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 151: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.slrm.038PROCEDIMENTO E RISULTATI 896150 Salari Mehdi

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b(1/2 x2/b2 -1/2 x3/b3 ) Fb2 1/EJ dx = [1/6 x3/b2 -1/8 x4/b3 ]o

b Fb2 1/EJ

= (1/6 b -1/8 b ) Fb2 1/EJ = 1/24 Fb3/EJ

LXoBA = ∫

o

b(1/2 x/b - x2/b2 +1/2 x3/b3 ) Fb2 1/EJ dx = [1/4 x2/b -1/3 x3/b2 +1/8 x4/b3 ]o

b Fb2 1/EJ

= (1/4 b -1/3 b +1/8 b ) Fb2 1/EJ = 1/24 Fb3/EJ

A = 642. mm2

Ju = 158306. mm4

Jv = 30006. mm4

yg = 37.71 mmN = -1380. NTy = -960. NMx = -950400. Nmmxm = 12. mmum = -3. mmvm = -37.71 mmσm = N/A-Mv/Ju = -228.5 N/mm2

xc = 15. mmyc = 16. mmvc = -21.71 mmσc = N/A-Mv/Ju = -132.5 N/mm2

τc = 2.882 N/mm2

σo = √σ2+3τ2 = 132.6 N/mm2

S* = 2852. mm3mm 0 12 18 30x

0

42

55

y

16σc,τc

σm

u

v

Schema.slrm.038

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 152: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.slrm.038

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.slrm.038

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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Schema.slvg.039REAZIONI 834781 Salvatori Gabriele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

5/4F 5/4F5/4Fb

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Schema.slvg.039AZIONI INTERNE 834781 Salvatori Gabriele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0

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Page 154: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 155: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.slvg.039PROCEDIMENTO E RISULTATI 834781 Salvatori Gabriele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b(- x2/b2 ) Fb2 1/EJ dx = [-1/3 x3/b2 ]o

b Fb2 1/EJ

= (-1/3 b ) Fb2 1/EJ = -1/3 Fb3/EJ

LXoBA = ∫

o

b(-1 +2 x/b - x2/b2 ) Fb2 1/EJ dx = [- x + x2/b -1/3 x3/b2 ]o

b Fb2 1/EJ

= (- b + b -1/3 b ) Fb2 1/EJ = -1/3 Fb3/EJ

LXoBC = ∫

o

b(-3/2 +3 x/b -3/2 x2/b2 ) Fb2 1/EJ dx + 1 (-1) (-1) Fb3/EJ

= [-3/2 x +3/2 x2/b -1/2 x3/b2 ]o

b Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ

= (-3/2 b +3/2 b -1/2 b ) Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ = 1/2 Fb3/EJ

LXoCB = ∫

o

b(-3/2 x2/b2 ) Fb2 1/EJ dx + 1 (-1) (-1) Fb3/EJ = [-1/2 x3/b2 ]o

b Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ

= (-1/2 b ) Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ = 1/2 Fb3/EJ

Schema.slvg.039PROCEDIMENTO E RISULTATI 834781 Salvatori Gabriele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

A = 828. mm2

Ju = 244195. mm4

Jv = 34128. mm4

yg = 32.37 mmN = -2993. NTy = -1710. NMx = -1778400. Nmmxm = 12. mmum = -6. mmvm = -32.37 mmσm = N/A-Mv/Ju = -239.4 N/mm2

xc = 18. mmyc = 14. mmvc = -18.37 mmσc = N/A-Mv/Ju = -137.4 N/mm2

τc = 2.487 N/mm2

σo = √σ2+3τ2 = 137.5 N/mm2

S* = 4262. mm3mm 0 12 24 36x

0

48

55

y

14σc,τc

σm

u

v

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Schema.slvg.039

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.slvg.039

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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Schema.sccm.040REAZIONI 853403 Scoccimarro Martina

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

23/16F 23/16F23/16Fb

A B

15/16F

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Schema.sccm.040AZIONI INTERNE 853403 Scoccimarro Martina

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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Schema.sccm.040PROCEDIMENTO E RISULTATI 853403 Scoccimarro Martina

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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

LXXBC = ∫

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

LXXCB = ∫

o

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

b 1/EJ

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

LXoBC = ∫

o

b(-1/2 x/b + x2/b2 -1/2 x3/b3 ) Fb 1/EJ dx + 1 (-1) (-1) Fb2/EJ

= [-1/4 x2/b +1/3 x3/b2 -1/8 x4/b3 ]o

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

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

LXoCB = ∫

o

b(-1/2 x2/b2 +1/2 x3/b3 ) Fb 1/EJ dx + 1 (-1) (-1) Fb2/EJ

= [-1/6 x3/b2 +1/8 x4/b3 ]o

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

= (-1/6 b +1/8 b ) Fb 1/EJ + 1 (-1) (-1) Fb2/EJ = 23/24 Fb2/EJ

A = 972. mm2

Ju = 264196. mm4

Jv = 56592. mm4

yg = 34.24 mmN = -2588. NTy = -2760. NMx = -1518000. Nmmxm = 12. mmum = -6. mmvm = -34.24 mmσm = N/A-Mv/Ju = -199.4 N/mm2

xc = 18. mmyc = 15. mmvc = -19.24 mmσc = N/A-Mv/Ju = -113.2 N/mm2

τc = 4.19 N/mm2

σo = √σ2+3τ2 = 113.4 N/mm2

S* = 4813. mm3mm 0 12 24 36x

0

42

55

y

15σc,τc

σm

u

v

Schema.sccm.040

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 160: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.sccm.040

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.sccm.040

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 161: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.shhe.041REAZIONI 809828 Shehu Elton

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1/2F

7/4F

1/2F

7/4F7/4Fb

A B

5/4F

3/4F5/4Fb

5/4F

3/4F

B

C

7/4F

3/2F3/2Fb

7/4F

3/2F

B D

3/2F

3/2F

3/2F

3/2F3/2Fb

D

E

3/2F

3/2F3/2Fb

3/2F

3/2F

EF

5/2F

F

B

3/2F

F

3/2F1/2Fb

FG

1/2F1/2Fb

1/2F

G

A

Schema.shhe.041AZIONI INTERNE 809828 Shehu Elton

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

-1/2

3/4

-7/4

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F

0 7/4

5/4

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0

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0

Fb

Page 162: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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X=

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5/4F

Svi

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lo ip

erst

atic

a

Page 163: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.shhe.041PROCEDIMENTO E RISULTATI 809828 Shehu Elton

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b(1/2 x2/b2 ) Fb2 1/EJ dx = [1/6 x3/b2 ]o

b Fb2 1/EJ

= (1/6 b ) Fb2 1/EJ = 1/6 Fb3/EJ

LXoBA = ∫

o

b(1/2 - x/b +1/2 x2/b2 ) Fb2 1/EJ dx = [1/2 x -1/2 x2/b +1/6 x3/b2 ]o

b Fb2 1/EJ

= (1/2 b -1/2 b +1/6 b ) Fb2 1/EJ = 1/6 Fb3/EJ

A = 948. mm2

Ju = 262515. mm4

Jv = 52848. mm4

yg = 20.97 mmN = -3098. NTy = -2655. NMx = 1593000. Nmmxm = 24. mmym = 55. mmum = 6. mmvm = 34.03 mmσm = N/A-Mv/Ju = -209.8 N/mm2

xc = 18. mmyc = 40. mmvc = 19.03 mmσc = N/A-Mv/Ju = -118.8 N/mm2

τc = 4.025 N/mm2

σo = √σ2+3τ2 = 119. N/mm2

S* = 4776. mm3mm 0 12 24 36x

0

12

55

y

40σc,τc

σm

u

v

Schema.shhe.041

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 164: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.shhe.041

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.shhe.041

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 165: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.smbm.042REAZIONI 886588 Simbula Marco

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

5/4F 5/4F5/4Fb

A B

7/4F

1/4F7/4Fb

7/4F

1/4F

B

C

7/4F

1/2F1/2Fb

7/4F

1/2F

B D

F

1/2F

F

1/2FFb

D

E

F

1/2FFb

F

3/2F

EF

3/2F

F

B

FFG

G

A

Schema.smbm.042AZIONI INTERNE 886588 Simbula Marco

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0

1/4

-7/4

1/2

11

-3/2

1

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5/4

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-1

1/23/2

0

0

0

F

0 5/4

7/4

0

1/2 0

0-1

-100

0

0000

Fb

Page 166: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Sch

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Page 167: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.smbm.042PROCEDIMENTO E RISULTATI 886588 Simbula Marco

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b(-1/2 x2/b2 ) Fb2 1/EJ dx = [-1/6 x3/b2 ]o

b Fb2 1/EJ

= (-1/6 b ) Fb2 1/EJ = -1/6 Fb3/EJ

LXoBA = ∫

o

b(-1/2 + x/b -1/2 x2/b2 ) Fb2 1/EJ dx = [-1/2 x +1/2 x2/b -1/6 x3/b2 ]o

b Fb2 1/EJ

= (-1/2 b +1/2 b -1/6 b ) Fb2 1/EJ = -1/6 Fb3/EJ

LXoBC = ∫

o

b(-1 +2 x/b - x2/b2 ) Fb2 1/EJ dx + 1 (-1) (-1) Fb3/EJ

= [- x + x2/b -1/3 x3/b2 ]o

b Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ

= (- b + b -1/3 b ) Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ = 2/3 Fb3/EJ

LXoCB = ∫

o

b(- x2/b2 ) Fb2 1/EJ dx + 1 (-1) (-1) Fb3/EJ = [-1/3 x3/b2 ]o

b Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ

= (-1/3 b ) Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ = 2/3 Fb3/EJ

Schema.smbm.042PROCEDIMENTO E RISULTATI 886588 Simbula Marco

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

A = 474. mm2

Ju = 143796. mm4

Jv = 14382. mm4

yg = 20.06 mmN = 680. NTy = -1360. NMx = -897600. Nmmxm = 18. mmym = 55. mmum = 3. mmvm = 34.94 mmσm = N/A-Mv/Ju = 219.6 N/mm2

xc = 15. mmyc = 40. mmvc = 19.94 mmσc = N/A-Mv/Ju = 125.9 N/mm2

τc = 3.893 N/mm2

σo = √σ2+3τ2 = 126.1 N/mm2

S* = 2470. mm3mm 0 12 18 30x

0

6

55

y

40σc,τc

σm

u

v

Page 168: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.smbm.042

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.smbm.042

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 169: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.stcm.043REAZIONI 886709 Steccanella Marco

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1/2F

7/4F

1/2F

7/4F7/4Fb

A B

5/4F

7/4F5/4Fb

5/4F

7/4F

B

C

3/4F

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3/4F

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3/2F

3/2F

3/2F

3/2F3/2Fb

D

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3/2F

3/2F3/2Fb

3/2F

3/2F

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3/2F

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1/2F

1/2F

G

A

Schema.stcm.043AZIONI INTERNE 886709 Steccanella Marco

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1/2

7/4

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3/2

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3/200

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Page 171: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.stcm.043PROCEDIMENTO E RISULTATI 886709 Steccanella Marco

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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

LXXBC = ∫

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

LXXCB = ∫

o

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

b 1/EJ

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

LXoBC = ∫

o

b(-1/2 + x/b -1/2 x2/b2 ) Fb 1/EJ dx - 1 (-1) (-1) Fb2/EJ

= [-1/2 x +1/2 x2/b -1/6 x3/b2 ]o

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

= (-1/2 b +1/2 b -1/6 b ) Fb 1/EJ - 1 (-1) (-1) Fb2/EJ = -7/6 Fb2/EJ

LXoCB = ∫

o

b(-1/2 x2/b2 ) Fb 1/EJ dx - 1 (-1) (-1) Fb2/EJ = [-1/6 x3/b2 ]o

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

= (-1/6 b ) Fb 1/EJ - 1 (-1) (-1) Fb2/EJ = -7/6 Fb2/EJ

A = 876. mm2

Ju = 264708. mm4

Jv = 62352. mm4

yg = 21.46 mmN = 2520. NTy = -2520. NMx = -1789200. Nmmxm = 30. mmym = 55. mmum = 6. mmvm = 33.54 mmσm = N/A-Mv/Ju = 229.6 N/mm2

xc = 24. mmyc = 41. mmvc = 19.54 mmσc = N/A-Mv/Ju = 135. N/mm2

τc = 3.537 N/mm2

σo = √σ2+3τ2 = 135.1 N/mm2

S* = 4459. mm3mm 0 18 30 48x

0

6

55

y

41σc,τc

σm

u

v

Schema.stcm.043

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 172: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.stcm.043

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.stcm.043

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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Schema.srzp.044REAZIONI 919311 Suarez Pena Bryam Abraham

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

7/4F 7/4F7/4Fb

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Schema.srzp.044AZIONI INTERNE 919311 Suarez Pena Bryam Abraham

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Page 174: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 175: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.srzp.044PROCEDIMENTO E RISULTATI 919311 Suarez Pena Bryam

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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

LXXBC = ∫

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

LXXCB = ∫

o

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

b 1/EJ

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

LXoAB = ∫

o

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

b Fb 1/EJ

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

LXoBA = ∫

o

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

b Fb 1/EJ

= (-1/2 b +1/2 b -1/6 b ) Fb 1/EJ = -1/6 Fb2/EJ

A = 618. mm2

Ju = 157731. mm4

Jv = 27774. mm4

yg = 17.48 mmN = 975. NTy = -1300. NMx = -1001000. Nmmxm = 18. mmym = 55. mmum = 3. mmvm = 37.52 mmσm = N/A-Mv/Ju = 239.7 N/mm2

xc = 15. mmyc = 39. mmvc = 21.52 mmσc = N/A-Mv/Ju = 138.1 N/mm2

τc = 3.893 N/mm2

σo = √σ2+3τ2 = 138.3 N/mm2

S* = 2834. mm3mm 0 12 18 30x

0

12

55

y

39σc,τc

σm

u

v

Schema.srzp.044

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 176: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.srzp.044

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.srzp.044

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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Schema.thid.045REAZIONI 868406 Taha Iad

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

7/4F 7/4F7/4Fb

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Schema.thid.045AZIONI INTERNE 868406 Taha Iad

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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Page 178: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 179: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.thid.045PROCEDIMENTO E RISULTATI 868406 Taha Iad

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoBC = ∫

o

b(1/2 - x/b +1/2 x2/b2 ) Fb2 1/EJ dx + 1 (-1) (-1) Fb3/EJ

= [1/2 x -1/2 x2/b +1/6 x3/b2 ]o

b Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ

= (1/2 b -1/2 b +1/6 b ) Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ = 7/6 Fb3/EJ

LXoCB = ∫

o

b(1/2 x2/b2 ) Fb2 1/EJ dx + 1 (-1) (-1) Fb3/EJ = [1/6 x3/b2 ]o

b Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ

= (1/6 b ) Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ = 7/6 Fb3/EJ

A = 762. mm2

Ju = 174852. mm4

Jv = 74862. mm4

yg = 15.31 mmN = -887.5 NTy = -1065. NMx = 873300. Nmmxm = 24. mmym = 55. mmum = 3. mmvm = 39.69 mmσm = N/A-Mv/Ju = -199.4 N/mm2

xc = 21. mmyc = 38. mmvc = 22.69 mmσc = N/A-Mv/Ju = -114.5 N/mm2

τc = 3.229 N/mm2

σo = √σ2+3τ2 = 114.6 N/mm2

S* = 3181. mm3mm 0 18 24 42x

0

12

55

y

38σc,τc

σm

u

v

Schema.thid.045

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 180: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.thid.045

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.thid.045

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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Schema.tscr.046REAZIONI 829283 Tascau Radu Alexandru

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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Page 183: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.tscr.046PROCEDIMENTO E RISULTATI 829283 Tascau Radu

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b(1/2 x2/b2 -1/2 x3/b3 ) Fb2 1/EJ dx = [1/6 x3/b2 -1/8 x4/b3 ]o

b Fb2 1/EJ

= (1/6 b -1/8 b ) Fb2 1/EJ = 1/24 Fb3/EJ

LXoBA = ∫

o

b(1/2 x/b - x2/b2 +1/2 x3/b3 ) Fb2 1/EJ dx = [1/4 x2/b -1/3 x3/b2 +1/8 x4/b3 ]o

b Fb2 1/EJ

= (1/4 b -1/3 b +1/8 b ) Fb2 1/EJ = 1/24 Fb3/EJ

A = 804. mm2

Ju = 237762. mm4

Jv = 30384. mm4

yg = 23.11 mmN = -2516. NTy = -1750. NMx = 1540000. Nmmxm = 24. mmym = 55. mmum = 6. mmvm = 31.89 mmσm = N/A-Mv/Ju = -209.7 N/mm2

xc = 18. mmyc = 41. mmvc = 17.89 mmσc = N/A-Mv/Ju = -119. N/mm2

τc = 2.565 N/mm2

σo = √σ2+3τ2 = 119.1 N/mm2

S* = 4181. mm3mm 0 12 24 36x

0

6

55

y

41σc,τc

σm

u

v

Schema.tscr.046

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 184: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.tscr.046

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.tscr.046

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 185: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.ttrm.047REAZIONI 876617 Totaro Mario

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

5/4F 5/4F5/4Fb

A B

7/4F

1/4F7/4Fb

7/4F

1/4F

B

C

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B D

F

F

F

FFb

D

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F

F

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F

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B

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A

Schema.ttrm.047AZIONI INTERNE 876617 Totaro Mario

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0

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1

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0-1

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1

0

0

0

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0 5/4

7/4

0

1/2 0

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0

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Fb

Page 186: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 187: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.ttrm.047PROCEDIMENTO E RISULTATI 876617 Totaro Mario

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b(- x2/b2 ) Fb2 1/EJ dx = [-1/3 x3/b2 ]o

b Fb2 1/EJ

= (-1/3 b ) Fb2 1/EJ = -1/3 Fb3/EJ

LXoBA = ∫

o

b(-1 +2 x/b - x2/b2 ) Fb2 1/EJ dx = [- x + x2/b -1/3 x3/b2 ]o

b Fb2 1/EJ

= (- b + b -1/3 b ) Fb2 1/EJ = -1/3 Fb3/EJ

LXoBC = ∫

o

b(-3/2 +3 x/b -3/2 x2/b2 ) Fb2 1/EJ dx + 1 (-1) (-1) Fb3/EJ

= [-3/2 x +3/2 x2/b -1/2 x3/b2 ]o

b Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ

= (-3/2 b +3/2 b -1/2 b ) Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ = 1/2 Fb3/EJ

LXoCB = ∫

o

b(-3/2 x2/b2 ) Fb2 1/EJ dx + 1 (-1) (-1) Fb3/EJ = [-1/2 x3/b2 ]o

b Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ

= (-1/2 b ) Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ = 1/2 Fb3/EJ

Schema.ttrm.047PROCEDIMENTO E RISULTATI 876617 Totaro Mario

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

A = 1092. mm2

Ju = 292252. mm4

Jv = 116784. mm4

yg = 18.99 mmN = 1900. NTy = -1900. NMx = -1767000. Nmmxm = 30. mmym = 55. mmum = 6. mmvm = 36.01 mmσm = N/A-Mv/Ju = 219.4 N/mm2

xc = 24. mmyc = 40. mmvc = 21.01 mmσc = N/A-Mv/Ju = 128.7 N/mm2

τc = 2.78 N/mm2

σo = √σ2+3τ2 = 128.8 N/mm2

S* = 5131. mm3mm 0 18 30 48x

0

12

55

y

40σc,τc

σm

u

v

Page 188: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.ttrm.047

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.ttrm.047

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 189: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.vnnv.048REAZIONI 877057 Vennettilli Vittorio

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

23/16F 23/16F23/16Fb

A B

15/16F

23/16F23/16Fb

31/16F

23/16F

B

C

15/16F

FFb

15/16F

F

B D

F

F

F

FFb

D

E

F

FFb

F

F

EF

F

F

B

FFG

G

A

Schema.vnnv.048AZIONI INTERNE 877057 Vennettilli Vittorio

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

0

23/1

623

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-15/16

1

1

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1

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1/16

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0

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Page 190: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Sch

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Page 191: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.vnnv.048PROCEDIMENTO E RISULTATI 877057 Vennettilli Vittorio

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoBC = ∫

o

b(1/2 x/b - x2/b2 +1/2 x3/b3 ) Fb2 1/EJ dx - 1 (-1) (-1) Fb3/EJ

= [1/4 x2/b -1/3 x3/b2 +1/8 x4/b3 ]o

b Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ

= (1/4 b -1/3 b +1/8 b ) Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ = -23/24 Fb3/EJ

LXoCB = ∫

o

b(1/2 x2/b2 -1/2 x3/b3 ) Fb2 1/EJ dx - 1 (-1) (-1) Fb3/EJ

= [1/6 x3/b2 -1/8 x4/b3 ]o

b Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ

= (1/6 b -1/8 b ) Fb2 1/EJ - 1 (-1) (-1) Fb3/EJ = -23/24 Fb3/EJ

A = 546. mm2

Ju = 162198. mm4

Jv = 37926. mm4

yg = 17.81 mmN = 1000. NTy = -1000. NMx = -990000. Nmmxm = 24. mmym = 55. mmum = 3. mmvm = 37.19 mmσm = N/A-Mv/Ju = 228.8 N/mm2

xc = 21. mmyc = 39. mmvc = 21.19 mmσc = N/A-Mv/Ju = 131.2 N/mm2

τc = 2.88 N/mm2

σo = √σ2+3τ2 = 131.3 N/mm2

S* = 2802. mm3mm 0 18 24 42x

0

6

55

y

39σc,τc

σm

u

v

Schema.vnnv.048

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 192: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.vnnv.048

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.vnnv.048

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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Schema.vgnd.049REAZIONI 890834 Vignati Davide

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1/2F

3/4F

1/2F

3/4F3/4Fb

A B

9/4F

5/4F9/4Fb

9/4F

5/4F

B

C

7/4F

3/2F3/2Fb

7/4F

3/2F

B D

1/2F

3/2F

1/2F

3/2F1/2Fb

D

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1/2F

1/2F1/2Fb

1/2F

1/2F

EF

1/2F

F

B

1/2F

F

1/2F1/2Fb

FG

1/2F1/2Fb

1/2F

G

A

Schema.vgnd.049AZIONI INTERNE 890834 Vignati Davide

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1/2

-5/4

-7/4

-3/2

-1/2

-1/2

-1/2-1/2

0

F

3/4

-9/4

3/2

1/2

-1/2

0

-10

-1/2

F

0 3/4

9/4

0

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01/

21/2000

0-1/2

1/2

0

Fb

Page 194: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 195: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.vgnd.049PROCEDIMENTO E RISULTATI 890834 Vignati Davide

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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

LXXBC = ∫

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

LXXCB = ∫

o

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

b 1/EJ

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

LXoAB = ∫

o

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

b Fb 1/EJ

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

LXoBA = ∫

o

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

b Fb 1/EJ

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

A = 456. mm2

Ju = 127247. mm4

Jv = 25560. mm4

yg = 33.58 mmN = -1068. NTy = 915. NMx = -896700. Nmmxm = 18. mmum = -3. mmvm = -33.58 mmσm = N/A-Mv/Ju = -239. N/mm2

xc = 21. mmyc = 14. mmvc = -19.58 mmσc = N/A-Mv/Ju = -140.3 N/mm2

τc = 2.676 N/mm2

σo = √σ2+3τ2 = 140.4 N/mm2

S* = 2233. mm3mm 0 18 24 42x

0

48

52

y

14σc,τc

σm

u

v

Schema.vgnd.049

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 196: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.vgnd.049

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.vgnd.049

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 197: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.vrtd.050REAZIONI 898022 Vretenar Daniele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

F

1/4F

F

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Schema.vrtd.050AZIONI INTERNE 898022 Vretenar Daniele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1

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Page 198: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 199: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.vrtd.050PROCEDIMENTO E RISULTATI 898022 Vretenar Daniele

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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

LXXBC = ∫

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

LXXCB = ∫

o

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

b 1/EJ

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

LXoAB = ∫

o

b(5/2 x2/b2 ) Fb 1/EJ dx = [5/6 x3/b2 ]o

b Fb 1/EJ

= (5/6 b ) Fb 1/EJ = 5/6 Fb2/EJ

LXoBA = ∫

o

b(5/2 -5 x/b +5/2 x2/b2 ) Fb 1/EJ dx = [5/2 x -5/2 x2/b +5/6 x3/b2 ]o

b Fb 1/EJ

= (5/2 b -5/2 b +5/6 b ) Fb 1/EJ = 5/6 Fb2/EJ

A = 672. mm2

Ju = 147014. mm4

Jv = 62496. mm4

yg = 37.25 mmN = -1050. NTy = 1500. NMx = -780000. Nmmxm = 18. mmum = -3. mmvm = -37.25 mmσm = N/A-Mv/Ju = -199.2 N/mm2

xc = 21. mmyc = 16. mmvc = -21.25 mmσc = N/A-Mv/Ju = -114.3 N/mm2

τc = 4.775 N/mm2

σo = √σ2+3τ2 = 114.6 N/mm2

S* = 2808. mm3mm 0 18 24 42x

0

42

52

y

16σc,τc

σm

u

v

Schema.vrtd.050

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 200: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.vrtd.050

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.vrtd.050

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 201: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.wxng.051REAZIONI 832033 Wu Xiangyu

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

3/2F

3/4F

3/2F

3/4F3/4Fb

A B

9/4F

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9/4F

1/4F

B

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3/4F

3/2F

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A

Schema.wxng.051AZIONI INTERNE 832033 Wu Xiangyu

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

3/2

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Fb

Page 202: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

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Page 203: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.wxng.051PROCEDIMENTO E RISULTATI 832033 Wu Xiangyu

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b(-3/2 x2/b2 ) Fb2 1/EJ dx = [-1/2 x3/b2 ]o

b Fb2 1/EJ

= (-1/2 b ) Fb2 1/EJ = -1/2 Fb3/EJ

LXoBA = ∫

o

b(-3/2 +3 x/b -3/2 x2/b2 ) Fb2 1/EJ dx = [-3/2 x +3/2 x2/b -1/2 x3/b2 ]o

b Fb2 1/EJ

= (-3/2 b +3/2 b -1/2 b ) Fb2 1/EJ = -1/2 Fb3/EJ

A = 768. mm2

Ju = 208192. mm4

Jv = 43776. mm4

yg = 30.5 mmN = -1245. NTy = 2490. NMx = -1419300. Nmmxm = 18. mmum = -6. mmvm = -30.5 mmσm = N/A-Mv/Ju = -209.5 N/mm2

xc = 24. mmyc = 13. mmvc = -17.5 mmσc = N/A-Mv/Ju = -120.9 N/mm2

τc = 3.732 N/mm2

σo = √σ2+3τ2 = 121.1 N/mm2

S* = 3744. mm3mm 0 18 30 48x

0

48

52

y

13σc,τc

σm

u

v

Schema.wxng.051

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 204: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.wxng.051

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.wxng.051

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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Schema.zhny.052REAZIONI 846641 Zhan Yuchen

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

F

3/4F

F

3/4F3/4Fb

A B

9/4F

1/4F9/4Fb

9/4F

1/4F

B

C

5/4F

3/2F3/2Fb

5/4F

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D

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EF

F

1/2F1/2Fb

1/2F

F

B

FFG

FFb

F

G

A

Schema.zhny.052AZIONI INTERNE 846641 Zhan Yuchen

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1

-1/4

-5/4

-3/2

0

1/2

1/2

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0

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Page 206: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Sch

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CF

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Page 207: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.zhny.052PROCEDIMENTO E RISULTATI 846641 Zhan Yuchen

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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

LXXBC = ∫

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

LXXCB = ∫

o

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

b 1/EJ

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

LXoAB = ∫

o

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

b Fb 1/EJ

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

LXoBA = ∫

o

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

b Fb 1/EJ

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

A = 984. mm2

Ju = 244285. mm4

Jv = 98208. mm4

yg = 33.68 mmN = -2113. NTy = 2535. NMx = -1571700. Nmmxm = 18. mmum = -6. mmvm = -33.68 mmσm = N/A-Mv/Ju = -218.9 N/mm2

xc = 24. mmyc = 14. mmvc = -19.68 mmσc = N/A-Mv/Ju = -128.8 N/mm2

τc = 3.877 N/mm2

σo = √σ2+3τ2 = 129. N/mm2

S* = 4483. mm3mm 0 18 30 48x

0

42

52

y

14σc,τc

σm

u

v

Schema.zhny.052

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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Schema.zhny.052

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.zhny.052

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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Schema.zlln.053REAZIONI 878043 Zullo Nicola

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

F

3/4F

F

3/4F3/4Fb

A B

9/4F

1/4F9/4Fb

9/4F

1/4F

B

C

5/4F

3/2F3/2Fb

5/4F

3/2F

B D

3/2F

F

3/2F1/2Fb

D

E

F

1/2F1/2Fb

F

1/2F

EF

1/2F

F

B

FFG

FFb

F

G

A

Schema.zlln.053AZIONI INTERNE 878043 Zullo Nicola

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1

-1/4

-5/4

-3/2

-3/2

-1

1/2

-1

0

F

3/4

-9/4

3/2

01

-1/2

0

0

-1

F

0 3/4

9/4

0

-3/20

01/

21/2000

0010

Fb

Page 210: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Sch

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Page 211: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.zlln.053PROCEDIMENTO E RISULTATI 878043 Zullo Nicola

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

LXXAB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBA = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXBC = ∫

o

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

b b2 1/EJ

= ( b - b +1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXXCB = ∫

o

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

b b2 1/EJ

= (1/3 b ) b2 1/EJ = 1/3 b3/EJ

LXoAB = ∫

o

b(- x2/b2 ) Fb2 1/EJ dx = [-1/3 x3/b2 ]o

b Fb2 1/EJ

= (-1/3 b ) Fb2 1/EJ = -1/3 Fb3/EJ

LXoBA = ∫

o

b(-1 +2 x/b - x2/b2 ) Fb2 1/EJ dx = [- x + x2/b -1/3 x3/b2 ]o

b Fb2 1/EJ

= (- b + b -1/3 b ) Fb2 1/EJ = -1/3 Fb3/EJ

LXoBC = ∫

o

b(-5/2 +5 x/b -5/2 x2/b2 ) Fb2 1/EJ dx + 1 (-1) (-1) Fb3/EJ

= [-5/2 x +5/2 x2/b -5/6 x3/b2 ]o

b Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ

= (-5/2 b +5/2 b -5/6 b ) Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ = 1/6 Fb3/EJ

LXoCB = ∫

o

b(-5/2 x2/b2 ) Fb2 1/EJ dx + 1 (-1) (-1) Fb3/EJ = [-5/6 x3/b2 ]o

b Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ

= (-5/6 b ) Fb2 1/EJ + 1 (-1) (-1) Fb3/EJ = 1/6 Fb3/EJ

Schema.zlln.053PROCEDIMENTO E RISULTATI 878043 Zullo Nicola

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

A = 408. mm2

Ju = 112717. mm4

Jv = 9864. mm4

yg = 31.65 mmN = -1000. NTy = 1200. NMx = -804000. Nmmxm = 12. mmum = -3. mmvm = -31.65 mmσm = N/A-Mv/Ju = -228.2 N/mm2

xc = 15. mmyc = 13. mmvc = -18.65 mmσc = N/A-Mv/Ju = -135.5 N/mm2

τc = 3.48 N/mm2

σo = √σ2+3τ2 = 135.6 N/mm2

S* = 1961. mm3mm 0 12 18 30x

0

48

52

y

13σc,τc

σm

u

v

Page 212: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.zlln.053

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.zlln.053

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Page 213: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.xxxx.054REAZIONI Nome:

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

F

15/16F

F

1/16F7/16Fb

A B

39/16F

17/16F39/16Fb

39/16F

17/16F

B

C

23/16F

2F2Fb

23/16F

2F

B D

F

2F

F

2FFb

D

E

F

FFb

F

F

EF

F

F

B

FFG

FFb

F

G

A

Schema.xxxx.054AZIONI INTERNE Nome:

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1 1

-17/

16

-23/16

-2

-1

1

-1

0

F

15/16-1/16

-39/

16

2

1

-1

0

0

-1

F

0 7/16

39/1

60

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01100

0

0010

Fb

Page 214: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Sch

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iper

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WB

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→M

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x)M

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Page 215: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.xxxx.054PROCEDIMENTO E RISULTATI Nome:

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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

LXXBC = ∫

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

LXXCB = ∫

o

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

b 1/EJ

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

LXoAB = ∫

o

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

b Fb 1/EJ

= (1/2 b +1/8 b ) Fb 1/EJ = 5/8 Fb2/EJ

LXoBA = ∫

o

b(2 -9/2 x/b +3 x2/b2 -1/2 x3/b3 ) Fb 1/EJ dx = [2 x -9/4 x2/b + x3/b2 -1/8 x4/b3 ]o

b Fb 1/EJ

= (2 b -9/4 b + b -1/8 b ) Fb 1/EJ = 5/8 Fb2/EJ

A = 552. mm2

Ju = 132127. mm4

Jv = 23256. mm4

yg = 35.13 mmN = -891.3 NTy = 1240. NMx = -892800. Nmmxm = 12. mmum = -3. mmvm = -35.13 mmσm = N/A-Mv/Ju = -239. N/mm2

xc = 15. mmyc = 15. mmvc = -20.13 mmσc = N/A-Mv/Ju = -137.6 N/mm2

τc = 3.89 N/mm2

σo = √σ2+3τ2 = 137.8 N/mm2

S* = 2487. mm3mm 0 12 18 30x

0

42

52

y

15σc,τc

σm

u

v

Schema.xxxx.054

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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Schema.xxxx.054

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.xxxx.054

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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Schema.xxxx.055REAZIONI Nome:

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

F

1/4F

F

1/4F1/4Fb

A B

11/4F

7/4F11/4Fb

11/4F

7/4F

B

C

7/4F

3F5/2Fb

7/4F

2F

B D

F

2F

F

2FFb

D

E

F

FFb

F

F

EF

F

F

B

FFG

FFb

F

G

A

Schema.xxxx.055AZIONI INTERNE Nome:

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1

-7/4

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3 2

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Page 219: Schema.agsr - intranet dica: Homeintranet.dica.polimi.it/uploads/media/Schema-soluti.pdf · Schema.agsr.001 PROCEDIMENTO E RISULTATI 877947 Agosti Riccardo A @ Adolfo Zavelani Rossi,

Schema.xxxx.055PROCEDIMENTO E RISULTATI Nome:

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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

LXXBC = ∫

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

LXXCB = ∫

o

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

b 1/EJ

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

LXoAB = ∫

o

b(5/2 x2/b2 ) Fb 1/EJ dx = [5/6 x3/b2 ]o

b Fb 1/EJ

= (5/6 b ) Fb 1/EJ = 5/6 Fb2/EJ

LXoBA = ∫

o

b(5/2 -5 x/b +5/2 x2/b2 ) Fb 1/EJ dx = [5/2 x -5/2 x2/b +5/6 x3/b2 ]o

b Fb 1/EJ

= (5/2 b -5/2 b +5/6 b ) Fb 1/EJ = 5/6 Fb2/EJ

A = 720. mm2

Ju = 188659. mm4

Jv = 22464. mm4

yg = 29.2 mmN = -1138. NTy = 1950. NMx = -1267500. Nmmxm = 12. mmum = -6. mmvm = -29.2 mmσm = N/A-Mv/Ju = -197.8 N/mm2

xc = 18. mmyc = 12. mmvc = -17.2 mmσc = N/A-Mv/Ju = -117.1 N/mm2

τc = 2.878 N/mm2

σo = √σ2+3τ2 = 117.2 N/mm2

S* = 3341. mm3mm 0 12 24 36x

0

48

52

y

12σc,τc

σm

u

v

Schema.xxxx.055

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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Schema.xxxx.055

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.xxxx.055

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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Schema.xxxx.056REAZIONI Nome:

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

F

7/16F

F

7/16F7/16Fb

A B

31/16F

9/16F39/16Fb

47/16F

9/16F

B

C

15/16F

2F2Fb

15/16F

2F

B D

F

2F

F

2FFb

D

E

F

FFb

F

F

EF

F

F

B

FFG

FFb

F

G

A

Schema.xxxx.056AZIONI INTERNE Nome:

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1

-9/1

6-9

/16

-15/16

-2

-1

1

-1

0

F

7/16

-31/

16-4

7/16

2

1

-1

0

0

-1

F

0 7/16

39/1

60

-20

01100

0

0010

Fb

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Sch

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Schema.xxxx.056PROCEDIMENTO E RISULTATI Nome:

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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

LXXBC = ∫

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

LXXCB = ∫

o

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

b 1/EJ

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

LXoAB = ∫

o

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

b Fb 1/EJ

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

LXoBA = ∫

o

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

b Fb 1/EJ

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

LXoBC = ∫

o

b(-1/2 x/b + x2/b2 -1/2 x3/b3 ) Fb 1/EJ dx + 1 (-1) (-1) Fb2/EJ

= [-1/4 x2/b +1/3 x3/b2 -1/8 x4/b3 ]o

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

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

LXoCB = ∫

o

b(-1/2 x2/b2 +1/2 x3/b3 ) Fb 1/EJ dx + 1 (-1) (-1) Fb2/EJ

= [-1/6 x3/b2 +1/8 x4/b3 ]o

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

= (-1/6 b +1/8 b ) Fb 1/EJ + 1 (-1) (-1) Fb2/EJ = 23/24 Fb2/EJ

Schema.xxxx.056PROCEDIMENTO E RISULTATI Nome:

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

A = 864. mm2

Ju = 219048. mm4

Jv = 44928. mm4

yg = 31.83 mmN = -806.3 NTy = 1720. NMx = -1427600. Nmmxm = 12. mmum = -6. mmvm = -31.83 mmσm = N/A-Mv/Ju = -208.4 N/mm2

xc = 18. mmyc = 14. mmvc = -17.83 mmσc = N/A-Mv/Ju = -117.2 N/mm2

τc = 2.73 N/mm2

σo = √σ2+3τ2 = 117.3 N/mm2

S* = 4172. mm3mm 0 12 24 36x

0

42

52

y

14σc,τc

σm

u

v

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Schema.xxxx.056

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.xxxx.056

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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Schema.xxxx.057REAZIONI Nome:

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

1/2F

3/4F

1/2F

3/4F3/4Fb

A B

9/4F

5/4F9/4Fb

9/4F

5/4F

B

C

11/4F

3/2F3/2Fb

11/4F

3/2F

B D

1/2F

3/2F

1/2F

3/2F1/2Fb

D

E

1/2F

1/2F1/2Fb

1/2F

1/2FFb

EF

1/2F

F

B

1/2F

F

1/2F1/2Fb

FG

1/2F1/2Fb

1/2F

G

A

Schema.xxxx.057AZIONI INTERNE Nome:

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

-1/2

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9/4

0

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0-1/2

-1/2

0

Fb

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Schema.xxxx.057PROCEDIMENTO E RISULTATI Nome:

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

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

LXXBC = ∫

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

LXXCB = ∫

o

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

b 1/EJ

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

LXoAB = ∫

o

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

b Fb 1/EJ

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

LXoBA = ∫

o

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

b Fb 1/EJ

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

A = 912. mm2

Ju = 222885. mm4

Jv = 52416. mm4

yg = 19.68 mmN = -3190. NTy = 1740. NMx = -1531200. Nmmxm = 24. mmym = 52. mmum = 6. mmvm = 32.32 mmσm = N/A-Mv/Ju = 218.5 N/mm2

xc = 18. mmyc = 39. mmvc = 19.32 mmσc = N/A-Mv/Ju = 129.2 N/mm2

τc = 2.62 N/mm2

σo = √σ2+3τ2 = 129.3 N/mm2

S* = 4027. mm3mm 0 12 24 36x

0

12

52

y

39σc,τc

σm

u

v

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Schema.xxxx.057

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19

Schema.xxxx.057

@ Adolfo Zavelani Rossi, Politecnico di Milano, vers.27.03.13 06.09.19