Soluzioni Esame minutolo

�� .� ��.� C�� .. ��.. ��. ��. ��.

�

�= 2 ��

� �

�� =60 �� =60 ��. �� = 2 �� 1= 1 ��. �� =60 �� τ�� τ�� .

�1= 1 ��

� C A

B �

15/09/2015

� �

�

��. 2 �� : ��: 1. �� 2. ��

� �� / �� C��

1

� C A

�

B

�

�

�

��. 2 �� : 1. �� 2. ��

�

�� 3��=3�2*2=�1 � �� =�1 �=0 �=1 �� B � �� M � �� :

ϕ

B

=0 M

A

C

B

=

ϕ

�

B �

ϕ

�

F B

� �

+

ϕ

�

M

B

��

�� , �� C �� (�� ) �� A � � � ��

B

C

� �

A �

�� A �� . ��. ��

�

 −R 3L + F 2L = 0 ⇒ R = 2F D  D 3  F RA =  3 R = 0  B  2FL MC = 2LRA = 3

�

��

()

v z = v 0 − ϕ0z −

()

(

)

v z = vC − ϕC z − 2L −

M0

2EI

MC

2EI

(

z2 −

z − 2L

T 0

z3 6EI 2 T C

)

−

z < 2L

(

z − 2L 6EI

3

)

2L < z < 3L

C��

 v d = −ϕ z − F z 3  v = 0 0 18EI z =2L  C  0 d s   v = v 2 z ⇒ ϕC = ϕ0 + ⇒ ϕC = ϕ0 + ⇒  d  ϕC = ϕC s  6EI z =2L 3EI    2 3 F 2FL2  2FL 2F v (3L ) = 0  3    ϕ ϕ v z L L z L z L z L = − 2 − 8 − + − 2 − − 2 + −  ) 6EI ( ) 18EI ( ) (  ( ) 0  0  EI EI 18 3     v 0 = 0   4FL3 d vC = −2ϕ0 L −  9EI 2 ⇒  d 2 2 2  ϕ = ϕ + 2FL 2 F FL FL FL F F 2 0  C ϕ =ϕ + ⇒ ϕB = − + =− z 3EI �� B 0 2  6 9 6 18EI EI EI EI FL 2 z =L  ϕ =− 0  9EI

�� M �� , �� B �� (�� ∆M = −M ); �� A � � � ��

M B

�

A �

�

�� A �� . ��. ��

�

 −R 3L + M = 0 ⇒ R = M D  D 3L  M RA = −  3L R = 0  B  M Bs = −RAL =

M

3

M Bd = −RAL − M = −

2M 3

��

()

v z = v 0 − ϕ0z −

()

(

)

v z = vB − ϕB z − 2L −

M0

2EI

MB

2EI

(

z2 −

z − 2L

T 0

z3 6EI 2 T B

)

−

z
(

z − 2L 6EI

3

)

L < z < 3L

C��

 d v = −ϕ z − M z 3  v = 0 0  B 18LEI z =L  0   v d = v s B B d ϕ ϕ + z 3 ⇒ ϕB d = ϕ0 + ⇒  d s ⇒  B = 0  ϕB = ϕB  18LEI z =L 6EI   2 v (3L ) = 0 v z = −ϕ L − M L − ϕ + M L  z − L − M z − L 2 + M z − L 3   0 ) 3EI ( ) 18EI ( ) ( 0  ( ) EI EI 18 6     v 0 = 0  2  M L d vB = −ϕ0 L −  18EI ⇒  d 2  ϕ = ϕ + M L M z ML ML ML 0 M F  B 6EI ϕ ϕ + ϕ = = ⇒ + = �� B B 0  6EI L z =L 6EI 6EI 3EI M L  ϕ = 0  3EI

�� , �� : ϕ

F B

−

M

+ ϕ B = 0 F L2

+

18 EI

M L

3EI

= 0⇒

M

=

FL

6

�� M =

FL

6

A

�

B �

M Bs =

18

C �

�

A �

FL

B

, M Bs = −

�

FL

9

�

, M C s = −

B

FL

MB =

18

M =

�

�

�

FL

3

; M C =

FL

6

�

�

C

A

�

M B s =

FL

18

M B s = −

�

M C s = −

+

FL

9

FL

18

2FL 3 7FL , 3 18 2FL FL + = , 3 9 2FL 11FL + = 3 18 FL

=

� A ��

B C

�= 2 ��

�

�

� �

�� =60 �� =60 ��. �� = 2 �� 1= 1 ��. �� =60 �� τ�� τ�� .

�1= 1 ��

�� . �� ,� �� . �� . �� . �� , �� dx G = 0

 2 s (h − s )(h + s )  bs 1  = + dy G =   2 bs + s1 (h − s )  2     60 ⋅ 22 1(60 − 2)(60 + 2) 1 = +  = 10.7753cm   60 ⋅ 2 + 1 ⋅ 58  2 2 1

��

I x =

b ⋅ dy G 3

−

3 = 51490.3cm 4

(b − s ) ⋅ (dy 1

G

3

3

− s)

+

(

3

)

s1 h − dyG

3

=

�� τ �� τ � �� . �� τ. �� : T n I n

b  s  S n A = − s dyG −  = −586.517cm 3  

( )

2 

( )

( )

( )

( )

(A) = −

τ

(B ) = −

( )

τ

− (C ) =

= −1211.54cm 3

τ

2

Sn B = Sn A

zx

( )

S n C = Sn A + Sn B = 2Sn A = −1173.03cm 3

( ) ( ) (E ) = 0

S n D = Sn C − Sn

zx

(

s1 dyG − s

2

zx

2

)

zx

( )

Ty S x A

τ

I x s

( )

Ty S x B

I x s

( )

Ty S x C

I x s

zx

60000 = 1.16527Ncm −4 51490.3

=

1.16527 ⋅ 586.517 = 341.724Ncm −2 (uscente da A) 2

=

1.16527 ⋅ 586.517 = 341.724Ncm −2 2

=

1.16527 ⋅ 1173.03 = 1366.9Ncm −2 1

T S (A) (D ) = − I s = 1.16527 1⋅ 1211.54 = 1411.76Ncm (E ) = 0Ncm y

x

x

τ

=

−2

−2

Soluzioni Esame minutolo

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