2012_04_30_1T_Resolução

�� C�� L�� E�� C�� A�� L�� 2011�2012 �� 2012.04.30 � 1� TESTE � SEMESTRE DE VER�O � ��

�� . �� − B�� C25/30 �� α�� = 1.0 � �� σ�ε ��; − A�� A500 NR �� σ�ε �� ; − S�� ; E�� STR. − D�� : NP EN1992�1�1, 2010

�. (�0.� ��.) C�� F�� . q g A

B

C

7.00

0 6 . 0

7.00 0.30

g k = 20 KN/m

(pp incluido)

q k = 15 KN/m

(ψ 0 = 0.7; ψ 1 = 0.5; ψ 2 = 0.3 )

d = = 0.55

�) E�� , �� , �� (�� ): 1) M�� (0.�) (0.�) 2) M�� (0.�) 3) E�� (0.�) 4) R��

P��. 1/7


�) R�� , �� . T�� V E� �� = R � � �� .3) ��, �� : VE� = 3/8 � 27 � 7.00 + 7/16 � 22.5 � 7.00 = 70.9 + 68.9 = 139.8 �N O �� M E� �� .2) P�� , �� , R � = 3/8 � (27 + 22.5) � 7.00 = 70.9 + 68.9 = 129.9 �N

ME� = 129.9 � 7 � 49.5 � 7 � 7/2 = � 303.5 �N.�

�) V�� EL �� . S� �� ), �� = 300 KN�. P�� , �� NP EN 1992�1�1. As

Fs

�

Fc 3.5�

P��. 2/7

� ��


F� = 0.8095 � � � �� x MR� = F� � � = 0.8095 � � � � �� x � (� � λ x) = 303.5 0.8095 � 0.30 � 16700 � x � (0.55 � 0.416 x) = 303.5 1687.128 � x2 � 2230.577 � x + 303.5 = 0 x1 = 0.154 � x2 = 1.168 � F� = 0.8095 � 0.3 � 16700 � 0.154 = 624.6 �N = F � A� = 624.6 / (435 � 10 3) = 0.00144 � 2 = 14.4 ��2 (5φ20 = 15.7 ��2) A�� ε� = 0.0035 � (0.55 � 0.154) / 0.154 = 0.0091 > 0.002175 A�� = 0.04 � A � = 0.04 � 0.30 � 0.60 � 10 4 = 72 ��2 A�� = �� 0.26 � � � � � (� �� / � ��) ; 0.0013 � � � �� A�� = �� 0.26 � 0.30 � 0.55 � (2.6 / 500) ; 0.0013 � 0.30 � 0.55� A�� = �� 0.000223 ; 0.000215� = 0.000223 � 2 = 2.23 �� 2 V�� (5 φ20 = 15.7 ��2) �� = �� 1 � φ ; �� + �2 ; 10 �� T�� φ = 20 �� = 20 �� = �� 1 � 20 ; 20 + 5 ; 10 �� ⇒ �� = �� 20 �� ; 25 �� ; 10 �� = 25 ��

0 5

4 3

E��

F�� 2 � 50 = 100 �� 2 � 0.5 φ20 2 � 0.5 � 20 = 20 �� 3 � φ20 3 � 20 = 60 �� 100 + 20 + 60 = 180 �� 4 � = 300 � 180 = 120 �� ⇒ � = 120 / 4 = 30 �� > 25 �� ( A �� ) P��. 3/7


�) V�� φ6//0.10 (2 ��) � �� 30� �� . S� �� ), �� = 140 KN. VR�,�� = α�� ν1 � � �� / (�� θ + �� θ) α�� = 1.0 ; � = 0.9 � � = 0.9 � 0.55 = 0.495; ν1 = 0.6 � (1 � �� / 250) = 0.54 VR�,�� = 1.0 � 0.30 � 0.495 � 0.54 � 16700 / �� (30 °) + �� (30 °)� = 461.8 �N > 139.8 �N E�� φ6//0.10 (2 ��) (A �� / �) = 2 � 0.28 / 0.10 = 5.6 �� 2 / � (A�� / �)�� = ρ�� α ρ�� = 0.08 � � ��0.5 / � �� = 0.08 � 25 0.5 / 500 = 0.0008 (A�� / �)�� = 0.0008 � 0.30 � 104 = 2.4 ��2 / � < 5.6 A�� SL,�� = 0.75 � � � (1 + �� α) = 0.75 � 0.55 � (1 + �� 90 °) = 0.41 � > 0.10 A� �� . VR� = VR�,� = 0.00056 � 0.9 � 0.55 � 435000 � �� 30 ° = 208.8 �N > 139.8 �N �) D�� φ6//0.20. E�� φ6//0.20 (2 ��) �� (A �� / �) = 2 � 0.28 / 0.20 = 2.8 �� 2 / � �� φ6//0.10 (2 ��) �� . E�� VR�,� (φ6//0.20) = 0.00028 � 0.9 � 0.55 � 435000 � �� 30 ° = 104.4 �N Q�� 104.4 �N �� φ6//0.20. A �� 104.4 �N �� : � = (139.8 � 104.4) / 49.5 = 0.715 �

�. (�.� ��.) S�� XC3. �) D�� 50 ��. �� = �� + ∆�� P��. 4/7


∆�� = 10 �� = �� ,� ; ��,�� + ∆��,γ � ∆��,�� ∆��,�� ; 10 �� = �� ,� ; ��,�� + ∆��,γ � ∆��,�� ∆��,�� ; 10 ��

A�� ( φ = 20 ��) ��,� = 20 �� ,�� = 25 �� (C�� S4, Q�� 4.4N) ∆��,γ = ∆��,�� = ∆��,�� = 0 �� = �� 20 �� ; 25 �� ; 10 �� = 25 �� A�� ( φ = 6 ��) ��,� = 6 �� ,�� = 25 �� (C�� S4, Q�� 4.4N) ∆��,γ = ∆��,�� = ∆��,�� = 0 �� = �� 6 �� ; 25 �� ; 10 �� = 25 �� = �� + ∆�� = 25 + 10 = 35 ��

�) C�� , �� , � , �� . A�� , �� φ6 �� φ20 ��, �� : � = � � � �� φ�� φ�� = 0.60 � 0.035 � 0.006 � 0.02 / 2 = 0.549 � ≈ 0.55, �� , �� .

�. (�.� ��.) C�� F�� . 3.5 0 00 4Ø16

5 0 . 0

2.75 0 00

(-) (-)

0 9 . 0 5 0 . 0

4Ø20 0.40

(+) 1.0 0 00

1.0 0 00

(1)

(2)

P��. 5/7


�) J�� . O �� (1) �� , �� , � �� ε� = � 3.5 ‰. O �� (2) �� , �� (�� , �� ), � �� ε� = � 2.0 ‰ �� 3/7 � �� .

�) D�� (�� , �� ) �� (1).

x 0 9

. 0

�� : x = 0.90 � 3.� / 4.� = 0.70 � ε��

= 0.003� � (0.�� 0.7) / 0.7 = 0.0007�

σ�� = ε�� E� = 0.0007� � �00 � �0 � = ��0 000 �� / � � F�� = ��0 000 � 4 � 3.�4 � �0 �4 = ��.4 �� (�)

ε��

= 0.003� � (0.7 � 0.0�) / 0.7 = 0.003��

σ�� = 43� 000 �� / � � F�� = 43� 000 � 4 � �.0� � �0 �4 = 349.7 �� (C)

F� = 0.�09� � 0.40 � 0.70 � ��700 = 37��.� �� (C) �� = ��.4 (�) � 349.7 (C) � 37��.� (C) = 394�.� �� (C) �� = F�� (� / � � 0.0�) � F �� (� / � � 0.0�) � F � � (� / � � 0.4�� x) �� = ��.4 � (0.4� � 0.0�) � 349.7 � (0.4� � 0.0�) � 37��.� � (0.4� � 0.4�� 0.7) �� = 7�.4 � �39.9 � �0�.� = ��.4 ��.� P��. 6/7

�� C�� E�� C�� A�� 0��0�� 0��.04.30 � �� ES�E � SE�ES��E DE �E��

�. (�.0 ��.) �� , �� , �� , �� 0�. C�� 30�. V Rd , s =

Asw s

z f ywd (cotgθ + cotgα )senα

θ = 30° α = �0 ° �� 90°

r =

V Rd , s (90 º ) V Rd , s ( 60 º )

Asw

= s Asw s

z f ywd (cotg30º +cotg90º ) sen90º

= z f ywd (cotg30º + cotg60º ) sen 60º

(1.732 + 0) × 1 (1.732 + 0.577) × 0.866

= 0.866

�� , V Rd , s (90 º ) = 0.866 × V Rd , s ( 60 º )

�� , � �� (1 � 0.866) = 0.134 ≈ 13.4 %

�� p L

3 pL 8

p

L

10 pL 8

L

3 pL 8

7 pL 16

��. 7/7

L

10 pL 16

1 pL 16

2012_04_30_1T_Resolução

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