# Apostila de Resistência dos Materiais (52 Páginas)

(Parte 3 de 4)

FC1 D = 4 – 2 = 2t

E = 2t

FC2 D = 4 - 2 = 2t

E = -2t

FC3 D = – 2 - 2 = -4t

E = - 4 – 4 = -8t

FCB D = – 8 + 8 = 0

11 EXERCÍCIOS DE FIXAÇÃO 1

1)

Fig. 10.1

a) ∑ FV = 0 b) ∑ MB = 0

RA + RB = 2 + 4 + 2 RA x 6 – 2 x 5 – 4 x 3 – 2 x 1 = 0

RA + RB = 8t 6 RA = 10 + 12 + 2

RB = 4t RA = 24 = 4t

6

2)

Fig. 10.2

a) ∑ FV = 0 b) ∑ MB = 0

RA + RB = 2 + 4 + 2 RA x 5 – 2 x 4 – 4 x 3 – 2 x 1 = 0

RA + RB = 8t 5 RA = 8 + 12 + 2

RB = 3,60t 5RA = 22

RA = 22 = 4,40t

5

3)

Fig. 10.3

a) ∑ FV = 0 b) ∑ MB = 0

RA + RB = 2 + 2 + 2 RA x 6 – 2 x 5 – 2 x 3 + 2 x 1 = 0

RA + RB = 6t 6RA – 10 – 6 - 2 = 0

RB = 3,66t 6RA = 14

RA = 14 = 2,33t

6

4)

Fig. 10.4

a) ∑ FV = 0 b) ∑ FH = 0

RA + RB = 4 + 4 + 2 HA = 4t

RA + RB = 10t

c) ∑ MB = 0

RA x 5 + HA x 5 – 2 x 3 – 4 x 5 – 4 x 3 = 0

5RA + 20 – 6 – 20 - 12 = 0

5RA = 18

RA = 18 = 3,60t // RB = 6,40t

5

5)

Fig. 10.5

a) ∑ FV = 0 b) ∑ FH = 0

RA + RB = 4 + 6 + 2 HA + 2 – 4 – 2 + 3,48 = 0

RA + RB = 12t HA – 0,52 = 0

HA = 0,52

c) ∑ MB = 0

RA x 5 + 2 x 2 – 4 x 5 – 6 x 2 – 3,48 x 2 + 0,52 x 3 + 2 x 2 = 0

5RA + 4 – 20 – 12 – 6,96 + 1,56 + 4 = 0

5RA = 29

RA = 29 = 5,88t // RB = 6,12t

5

6)

Fig. 10.6

a) ∑ FV = 0 b) ∑ FH = 0

RA + RB = 2 + 4 + 6 HA + 2 – 4 = 0

RA + RB = 12t HA = 8t

RB = 5,20t

c) ∑ MB = 0

RA x 5 + 2 x 2 – 2 x 5 – 4 x 3 – 4 x 4 = 0

5RA + 4 – 10 – 12 – 16 = 0

5RA = 36

RA = 36 = 6,80t

5

7)

Fig. 10.6

a) ∑ FV = 0 b) ∑ FH = 0

RA + RB = 4 + 2 + 2 HA + 2 – 4 = 0

RA + RB = 8t HA = 2t

c) ∑ MB = 0

RA x 4 + 4 x 6 – 4 x 4 – 2 x 2 – 2 x 3 = 0

4RA + 24 – 16 – 4 – 6 = 0

4RA = 2

RA = 2 = 0,50t // RB = 7,50t

4

12 EXERCÍCIOS DE FIXAÇÃO 2

* Traçam os D.M.F e D.E.C

1)

Fig. 11.1

a) ∑ FV = 0

RA + RB = 6 + 4 + 2 + 6 + 2

RA + RB = 20t

RB = 11,67t

b) ∑ MB = 0

RA x 6 – 6 x 5,5 – 4 x 3 – 2 x 2 - 6 x 0,5 + 2 x 1 = 0

6RA - 33 – 12 – 4 – 3 + 2 = 0

6RA = 50

RA = 50 = 8,33t

6

MF1 = MF5 = 0

MFA = - 2 x 0,5 = - 1tm

MF2 = - 6 x 1,5 + 8,33 x 2 = 7,66tm

MF3 = 8,33 x 3 – 6 x 2,5 = 9,99tm

MF4 = - 2 x 3 + 1,67 x 2 – 6 x 1,5 = 8,34tm

MFB = - 2 x 1 – 2 x 0,5 = - 3tm

E = 0

FC1 D = 0

E = 0 – 2 = -2t

FCA D = – 2 + 8,33 = 6,33t

E = 6,33 – 4 = 2,33t

FC2 D = 2,33t

E = 2,33t

FC3 D = 2,33 - 4 = -1,67t

E = - 1,67t

FC4 D = – 1,67 - 2 = -3,67t

E = 4 – 2 = 2t

FC5 D = 2 - 2 = 0t

Diagrama do momento fletor:

Fig. 11.2

a) ∑ FV = 0

RA + RB = 2 + 6 + 2 + 2 + 6 + 2

RA + RB = 20t

RB = 10,33t

b) ∑ MB = 0

- 2 x 7 + RA x 6 – 6 x 5,5 – 2 x 4 – 2 x 1 - 6 x 0,5 + 2 x 1 = 0

- 14 + 6RA - 33 – 8 – 2 – 3 + 2 = 0

RA = 4,67t

MF1 = MF5 = 0

MFB = MFA = - 2 x 0,5 – 2 x 1= - 3tm

MF2 = - 6 x 1,5 + RA x 2 – 2 x 3 = - 9 + 9,67 x 2 – 6 – 4,34 tm

MF3 = 9,67 x 4 – 6 x 3,5 – 2 x 2 – 2 x 5 = 38,68 – 21 – 4 – 2 x 5= 3,68 tm

MF4 = - 2 x 6 + 9,67 x 5 – 6 x 4,5 – 2 x 3 – 2 x 0,5 = - 12 + 48,35 – 27 – 6 – 1 = 2,37 tm

E = 0

FC1 D = 0 – 2t

E = - 2 – 2 = -4t

FCA D = – 4 + 9,67 = 5,67t

E = 5,67 – 4 = 1,67t

FC2 D = 1,67 – 2 = - 0,33t

E = - 0,33t

FC3 D = - 0,33t

E = 0,33 – 2 = - 2,33t

FC4 D = – 2,33 - 2 = -4,33t

E = - 4,33 – 2 = -6,33t

FCB D = - 6,33 + 10,33 = 4t

E = 4 – 2 = 2t

FC5 D = 2 – 2 = 0

Diagrama do momento fletor:

Fig. 11.3

a) ∑ FV = 0

RA + RB + 2 = 2 + 6 + 9 + 2 + 6 + 2

RA + RB = 25t

RB = 13,07t

b) ∑ MB = 0

- 2 x 8 + RA x 7 – 6 x 6,5 + 2 x 5 – 9 x 3,5 - 2 x 3 – 6 x 0,5 + 2 x 1 = 0

- 16 + 7RA – 39 + 10 – 31,5 – 6 – 3 + 2 = 0

RA = 11,93t

c) MF1 = MF5 = 0

MFB = MFA = - 2 x 0,5 – 2 x 1= - 3tm

MF2 = - 2 x 3 + – 6 x 1,5 + 11,93 x 2 = – 6 – 9 + 23,86 = 8,86 tm

MF3 = - 2 x 5 – 6 x 3,5 + RA x 4 + 2 x 2 – 6 x 1= 14,72 tm

MF4 = - 2 x 6 + 11,93 x 5 – 6 x 4,5 + 2 x 3 – 9 x 1,5 – 2 x 1 = - 12 + 59,65 – 27 + 6 – 13,50 - 2 = 11,15 tm

MF4 = - 6 + 26,14 – 4 = 11,14 tm

E = 0

FC1 D = 0 – 2 = -2t

E = - 2 – 2 = -4t

FCA D = – 4 + 11,93 = 7,93t

E = 7,93 – 4 = 3,93t

FC2 D = 3,93 + 2 = 5,93t

E = 5,93 – 6 = - 0,07t

FC3 D = - 0,07 – 2 = -2,07t

E = - 2,07 – 3 = - 5,07t

FC4 D = – 5,07t

E = - 5,07 – 4 = - 9,07t

FCB D = - 9,07 + 13,07 = 4t

E = 4 – 2 = 2t

FC5 D = 2 – 2 = 0

Diagrama do Momento Fletor:

* Traçar os D.M.F e D.E.C das estruturas:

Sen30° = 0,50

Cos30° = 0,87

(Parte 3 de 4)