Svolgimento esercizi d'esame, bigliettini e formule

I

effettiva L L

≤ ≤

f f

―― ――

c_amm.trave c_amm.pilastro

200 150

sicuramente verificato

Se ho progettato per resistenza:

Ricavo la: (dalla tabella) conoscendo W,e tramite formule della freccia (da schemi noti)

sostituisco e verifico che:

I

effettiva L L

≤ ≤

f f

―― ――

c_amm.trave c_amm.pilastro

200 150

Verifico per resistenza

Se ho progettato per resistenza:

Verifico che: (vedi tabella profilato)

A ＝

area_sezione

N N

‾‾‾‾‾‾‾‾

2 2 A ＝

＝ ＝ < ＝ ⋅

σ 190

σ σ + 3 τ ―― ―― anima

amm

id 2 2

mm mm

N M T

kN ⋅ m kN

asta max max

＝ ⋅ + ＝ ＝ ＝

σ ω τ

――― ――

―――― ――― ―――

A W A

3 2

m mm

area_sezione effettiva anima

scelgo il più grande tra i due e ricavo dalla tabella sulla normativa

β ⋅ L β ⋅ L ω

＝ ＝

λx λy

―― ――

δ δ

x y

Vedi tabella schemi noti raggi di inerzia (da tabella profilato)

β ＝ δ ＝

x_y

Se ho progettato per deformabilità:

Ricavo la: (dalla tabella) conoscendo I,e Verifico come sopra

W effettiva

verifica presso/tensoflessione (punti del dominio)

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W effettiva

verifica presso/tensoflessione (punti del dominio)

1) Massimo sforzo di trazione ε ＝ ε

s sy

⎛⎝ ⎞⎠

N ＝ −f ⋅ A + A ′ ε ＝ ε

u yd s s c cu

⎛ ⎞ ⎛ ⎞

h h ε ＝ 0.186 %

＝ ⋅ − − ⋅ − ＝

M f ⋅ A δ f ⋅ A ′ δ 0

― ―

⎜ ⎟ ⎜ ⎟ su

u yd s yd s

2 2

⎝ ⎠ ⎝ ⎠

2) Massimo sforzo di compressione ε ＝ 0.35 %

se cu

A ＝ A ′

s s

⎛⎝ ⎞⎠

N ＝ f ⋅ b ⋅ H + f ⋅ A + A ′

u cd yd s s

⎛ ⎞ ⎛ ⎞

h h

＝ ⋅ − + ⋅ − ＝

M −f ⋅ A δ f ⋅ A ′ δ 0

― ―

⎜ ⎟ ⎜ ⎟

u yd s yd s

2 2

⎝ ⎠ ⎝ ⎠

3) Rottura bilanciata se

ε cu A ＝ A ′

＝ ⋅ ＝ , ‥

X d 0 65 ⋅ d

――― s s

bil ε + ε Hp ＝ ε < ε

cu sy c cu

ε ＝ ε

s sy

N ＝ b ⋅ f ⋅ 0.8 ⋅ X + f ⋅ A ′ − f ⋅ A

u cd bil yd s yd s

⎛ ⎞ ⎛ ⎞ ⎛ ⎞

h h h

＝ ⋅ − + ⋅ − + ⋅ −

M b ⋅ f ⋅ 0.8 ⋅ X 0.4 ⋅ X f ⋅ A ′ δ f ⋅ A δ

― ― ―

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

u cd bil bil yd s yd s

2 2 2

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

4) Flessione Semplice

ε su (( ))

＝ ⋅ ≤

ε ′ X − δ 1.826 %

――

s d − X 2

⎛⎝ 0.8⎞⎠ ⎛⎝ ⎞⎠ ⎛⎝ ⎞⎠

N f ⋅ b ⋅ X f ⋅ 0.8 ⋅ b ⋅ d + E ⋅ ε ⋅ A ′ + f ⋅ A X E ⋅ ε ⋅ δ ⋅ A ′ + f ⋅ d ⋅ A 0

＝ − + ＝

u cd cd s su s yd s s su s yd s

(( )) (( ))

M ＝ f ⋅ b ⋅ 0.8 ⋅ X d − 0.4 X + E ⋅ ε ′ ⋅ d − s ⋅ A ′ ＝ N ⋅ mm

u cd s s s

5) rottura contemporanea

ε cu

＝ ⋅ ＝ , ‥

X d 0 26 ⋅ d ＝

―――

bil ε + ε

cu su

N ＝ b ⋅ f ⋅ 0.8 ⋅ X + f ⋅ A ′ − f ⋅ A

u cd yd s yd s

⎛ ⎞ ⎛ ⎞ ⎛ ⎞

h h h

＝ ⋅ − + ⋅ − + ⋅ −

M b ⋅ f ⋅ 0.8 ⋅ X 0.4 ⋅ X f ⋅ A ′ δ f ⋅ A δ

― ― ―

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

u cd yd s yd s

2 2 2

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

6) Transizione (cls) ε su

N ＝ b ⋅ f ⋅ 0.8 ⋅ H + f ⋅ A − E ⋅ ε ′ ⋅ A ′ ＝ ε < ε ε ＝ ε ＝ ⋅ <

ε ′ δ 0.186 %

――

u cd yd s s s s s su c cu s M

⎛ ⎞ ⎛ ⎞ ⎛ ⎞

h h h

＝ ⋅ − + ⋅ − − ⋅ − ＝

M b ⋅ f ⋅ 0.8 ⋅ H 0.4 ⋅ H f ⋅ A δ E ⋅ ε ′ ⋅ A δ N ⋅ mm

― ― ―

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

u cd yd s s s s

2 2 2

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

7) Transizione (acciaio) ε su

＝ ⋅ <

ε ′ δ 0.186 %

ε ′ < ε ε ＝ ε ――

s

s sy s su d

N ＝ −E ⋅ ε ′ ⋅ A ′ − f ⋅ A ＝

u s s s yd s

⎛ ⎞ ⎛ ⎞

h h

＝ ⋅ − + ⋅ − ＝

M −E ⋅ ε ′ ⋅ A ′ δ f ⋅ A δ

― ―

⎜ ⎟ ⎜ ⎟

u s s s yd s

2 2

⎝ ⎠ ⎝ ⎠

Se ho solo da calcolare altri 2 punti in più

A ≠ A ′

s s

Creato con PTC Mathcad Express. Per ulteriori informazioni, vedere www.mathcad.com.

⎛ ⎞ ⎛ ⎞

h h

＝ ⋅ − + ⋅ − ＝

M −E ⋅ ε ′ ⋅ A ′ δ f ⋅ A δ

― ―

⎜ ⎟ ⎜ ⎟

u s s s yd s

2 2

⎝ ⎠ ⎝ ⎠

Se ho solo da calcolare altri 2 punti in più

A ≠ A ′

s s

punto 1 esplicito

⎛ ⎞ ⎛ ⎞

h h σ

＝ ⋅ − − ⋅ − ＝

0 f ⋅ A ′ δ σ ⋅ A δ

― ―

⎜ ⎟ ⎜ ⎟ s

yd s s s

2 2

⎝ ⎠ ⎝ ⎠

N ＝ b ⋅ f ⋅ H + f ⋅ A ′ + σ ⋅ A ＝

u cd yd s s s

punto 2 esplicito

⎛ ⎞ ⎛ ⎞

h h σ

＝ ⋅ − − ⋅ − ＝

0 f ⋅ A δ σ ⋅ A ′ δ

― ―

⎜ ⎟ ⎜ ⎟ s

yd s s s

2 2

⎝ ⎠ ⎝ ⎠

N ＝ −f ⋅ A ′ − σ ⋅ A ＝

u yd s s s

Diagramma momento curvatura

Dati: As ＝ As

′ ＝ H ＝ δ ＝ b ＝

f ＝ 25 Mpa R ＝ 30 MPa

ck ck

1) Punto di 1° fessurazione (primo stadio) Sn ＝ 0

f , ff ⋅ I ‥

ctk n KN ⋅ m

＝ ＝

M N ⋅ mm

――――

cr H − X

f , ff ‥ H

1

ctk −SE ⋅ A ＝ A ′ ＝

X

＝ ＝

χ ―

――

―――― s s

cr (( )) 2

mm

E ⋅ H − X

c N N

3 ‾‾‾

＝ ＝ ⋅

f 0.3 f 2.56 f ＝ 0.7 ⋅ f 1.79

＝ ⋅ ―― ――

ctm ck ctk ctm

2 2

mm mm

N N

, ＝ ⋅ ‥ ＝ ⋅

f ff ＝ 1.2 ⋅ f 2.51 f ＝ f + 8 ⋅ MPa 33

―― ――

ctk ctk cm ck

2 2

mm mm

0.3

⎛ ⎞

f N

cm

E ＝ 22000 ⋅ ＝ 31475.8 ――

――

⎜ ⎟

c 10 2

⎝ ⎠ mm

3

3 (( ))

b H − X

b ⋅ X 2 2 4

(( )) (( ))

I cm

＝ + + + ＝

n ⋅ A ′ X − δ n ⋅ A d − X

⋅ ⋅

―― ――――

n s s

3

3

Se l'armatura è asimmetrica mi tocca calcolare l'asse neutro

⎛ ⎞

2

b ⋅ H

⎛⎝ ⎛⎝ ⎞⎠⎞⎠ ⎛⎝ ⎞⎠

− + ＝

b ⋅ H + n ⋅ A + A ′ X n A ⋅ d + A ′ ⋅ δ 0

――

⎜ ⎟

s s s s

2

⎝ ⎠

2

b ⋅ X ⎛⎝ ⎞⎠

+ n ⋅ A ⋅ d + A ′ ⋅ δ

―― s s

3 H

X cm

＝ ＝ >

―――――――― ―

⎛ ⎞

Diagramma momento curvatura

b ⋅ H + n ⋅ A + A ′ 2

⎝ ⎠

s

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2

b ⋅ X ⎛⎝ ⎞⎠

+ n ⋅ A ⋅ d + A ′ ⋅ δ

―― s s

3 H

X cm

＝ ＝ >

―――――――― ―

⎛⎝ ⎞⎠

b ⋅ H + n ⋅ A + A ′ 2

s

Diagramma momento curvatura

Analizzo il salto dovuto alla riduzione dell'area di sezione

M ＝ cost

cr ε 1

c

＝ ＝

χ ′ ――

―

cr X 2

mm -No

+Si

‾‾‾‾‾‾

2

b ⋅ X + −

−b b − 4 ac

(( )) (( ))

+ − ＝

n ⋅ A ′ ⋅ X − δ n ⋅ A ⋅ d − X 0 ＝

X

―― ――――――

s s

2 2 a

M N

cr

＝ ⋅ ＝

σ X ――

――

c I 2

mm

n 3

b X

⋅ 2 2 4

(( )) (( ))

＝ + + ＝

I cm

n ⋅ A ′ ⋅ X − δ n ⋅ A ′ ⋅ d − X

――

n s s

3

σ

c

＝ ＝

ε adimesionale

―

c E c

Punto di 1° snervamento (2° stadio)

=cls comportamento elastico lineare

Hp f

yd

＝ ⋅ ＝

M I N ⋅ mm KNm

―――

y n

(( ))

n d − X

−3

ε 10

⋅ 1

sy

χ ＝ ＝ ――

―――

y d − X mm

ε < 0.2 % ε ＝ ε ε ＝ 0.186 %

c s sy sy

X ＝

I ＝

n

Verifico Hp

ε sy

＝ ⋅ ＝ ‥

ε X . < 0.2 %

――

c d − X

Diagramma momento curvatura

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Diagramma momento curvatura

Punto di rottura (3° stadio)

(( )) (( ))

M ＝ b ⋅ f ⋅ 0.8 ⋅ X ⋅ d − 0.4 ⋅ X + E ⋅ ε ′ ⋅ A ⋅ d − δ ＝ N ⋅ mm KNm

u cd s s s

−3

ε ⋅ 10 1

su

χ ＝ ＝ ――

―――

u d − X mm

Hp: ε ＝ ε ε ＝ 1 %

s su su

ε ′ < ε ε ＝ 0.186 %

s sy sy

0.83 ⋅ 0.85 ⋅ R N

ck

＝ ＝

f 14.11 ――

―――――

cd 1.5 2

mm

(( )) (( )) (( ))

f ⋅ b ⋅ 0.8 ⋅ X ⋅ d − X + E ⋅ ε ⋅ X − δ ⋅ A ′ − f ⋅ d − X ⋅ A ＝ 0

cd s su s yd s

‾‾‾‾‾‾

+ −

−b b − 4 ac

X ＝ ――――――

2 a

Verifico Hp se è maggiore cambio ipotesi

ε su (( ))

＝ ⋅ ＝ ‥

ε ′ X − δ . < 1.86 %