Transcript

STATICS

EXERCISE BOOK (For BSc students in Mechanical Enigineering)

Compiled by Dr. Tamás Insperger Department of Applied Mechanincs

Budapest University of Technology and Economics

Budapest 2009

1

Problem 1 Force F is applied at point B. Determine the moment of force F about point C!

F = [ 3, 4, 0 ] [N] ; rB = [8, 3, 0 ] [m] ; rC = [1, 2, 0 ] [m]

2

Problem 2 Determine the moment of the load lifted by the trolley about point A for the lower and the upper position of the load.

3

Problem 3 The spatial force system consists of three forces. Determine the moment vector of the force system about point O!

FA = 10 kN , FB = 15 kN , FC = 20 kN , a = 1 m, b = 1 m , c = 3 m .

a

FA

FB

FC

O

b

c

x y

z

4

Problem 4 Reduce the force F to point B and to point A!

5

Problem 5 Calculate the moment of the given couple to points A and B and to the origin of the coordinate system!

F = [ 2, 1, 0 ] [N] ; rA = [4, 1, 0 ] [m] ; rB = [3, 5, 0 ] [m]

6

Problem 6 (a) Determine the resultant of the planar force system consisting of two forces by construction!

F1 = 200 N , F2 = 250 N .

F1

F2

(b) Determine the resultant of the planar force system consisting of two parallel forces by construction!

F1 = 200 N , F2 = 100 N .

F1 F2

7

Problem 7 Determine the resultant of the planar force system consisting of three forces by construction!

F1 = 200 N , F2 = 250 N , F3 = 150 N

F1

F2

F3

8

Problem 8 Determine the resultant of the given force system!

F1 = 200 N , F2 = 800 N , F3 = 600 N .

F1F2

F3

3 m

9

Problem 9 Determine the resultant of the planar force system consisting of four forces by construction!

F1 = 200 N , F2 = 250 N , F3 = 150 N, F4 = 150 N.

F1

F2

F3

F4

10

Problem 10 The rigid beam is loaded four forces. Determine the resultant of the force system by calculation!

F1 = 150 N , F2 = 600 N , F3 = 100 N , F4 = 250 N

F1

1,6 m

F2 F3 F4

1,2 m 2,0 m

11

Problem 11 Determine the resultant of the distributed force system! (Both the magnitude and the location!)

L = 2 m , p = 5 kN/m . (a)

L

p x

y

O

(b)

L

p x

y

O

12

Problem 12 Determine the location of the centroid of the given plane figures! The sizes are given in mm. (a)

100

70 20

20

20

(b)

100

30 20

60

(c)

100

70 20

20

20

13

Problem 13 Determine the location of the centroid of the bodies! a = 0,5 m . (a)

a a

a

a

a a

(b)

a a

a

a

a a

(c)

a a

a

a

a a

14

Problem 14 The structure in the drawing is in equilibrium! Determine force B and the reaction forces!

F1 = 1200 N

15

Problem 15 Determine the reaction forces for the given planar structure!

F = 600 N , p = 300 N/m , M = 300 Nm, a = 2 m . (a)

FB

a2a

A

x y

(b)

F B

a2a

A

x y

(c)

B

a2a

A

x y

p

(d)

M B

a2a

A

x y

16

Problem 16 Determine the reaction forces!

17

Problem 17 Determine the reaction forces, if G1 = 10000 N and G2= 24000 N!

18

Problem 18 Determine the reaction forces for the structure!

F = 600 N

1 m 3 m

1 mA B

C

19

Problem 19 Determine the reaction forces for the structure!

F = 1 kN

1 m 3 m

1 m A B

C

D

1,5 m

x y

20

Problem 20 Determine the reaction forces for the structure!

200 N

0,5 m

A B

C0,5 m

400 N

0,5 m

0,5 m

21

Problem 21 Determine the reaction forces for the structure! Make a separate drawing of the beam D-C and give the forces acting on it.

a = 1 m

F = 1 kN

2a a

3a

A

B C

a 2aD

G

22

Problem 22 Calculate the reaction forces for the structure and the force in the rope D-E!

F = 10 kN

A B

1 m

1 m

1 m

C

D E

x y

23

Problem 23 Determine the reaction forces! Make a separate drawing of the horizontal beam and give the forces acting on it!

a = 0,5 m , b = 0,6 m , F = 18 kN , p = 30 kN/m .

p F

b

2a

A B

C

a a

b

x y

D E

24

Problem 24 Determine the reaction forces and the forces in the bars of the given truss!

a = 1 m, F1 = 4 kN , F2 = 6 kN .

F2a

a

A

F1

B

a a a

1

2

3

4

5

6

7

C

D

E

x y

bar R (kN)

1

2

3

4

5

6

7

25

Problem 25 Determine the reaction forces and the forces in the bars of the given truss!

a = 1 m, F = 10 kN .

F

a

a

A

B a

1

6

5

4

3

2

7

bar +/- R (kN)

1

2

3

4

5

6

7

26

Problem 26 Determine the reaction forces and the forces in the bars!

a = 1 m , F = 1 kN . (a)

F

a

a

A B

a

(b)

F

A B

a a

a

27

Problem 27 Determine the reaction forces and the forces in the bars!

a = 0,6 m , F = 750 N .

F

a

a

A

B a

15

4

3

2 600

bar +/- R (N)

1

2

3

4

5

28

Problem 28 Determine the stress resultant diagrams for the given beams!

F = 4 kN , a = 0,5 m . (a)

F

aB

a

V

Mh

A

(b)

F

aB

a

V

Mh

A

29

Problem 29 Determine the stress resultant diagrams for the given beams!

p = 4 kN/m , a = 0,5 m . (a)

aB

a

V

Mh

A p

(b)

aB

a

V

Mh

A p

30

Problem 30 Determine the stress resultant diagrams for the given beams! (a)

90 N

a

Ba

V

Mh

A 30 N

a a=0,3 m

(b)

1 kN

a

Ba

V

Mh

A 1 kN

a a=1 m

31

Problem 31 Determine the stress resultant diagrams for the given beams! (a)

3 kN

2a

B

a

V

Mh

A 6 kN

2a a=150 mma

3 kN

(b)

aBa

V

Mh

A

p = 10 kN/m

4a a=1 m

32

Problem 32 Determine the stress resultant diagrams for the given beams! (a)

3 kN

a a

V

Mh

A

N

1 kN

1 kN a = 2 m

(b)

2a a

V

Mh

4kN/m 6kN 3kN

N

a = 2 m

33

Problem 33 Determine the stress resultant diagrams for the given structures! (a)

V

Mh

N

3 kN

1 m

0,5 m A

Mh

V

(b)

V

Mh

N

4 kN 1 m

0,5 m A

Mh

V

B

34

Problem 34 Determine the reaction forces and give the stress resultant diagrams!

a = 2 m , F1 = 4 kN , F2 = 10 kN , M = 20 kNm , p = 3 kN/m .

F1

B

a

V

Mh

A

a a a

p

M

F2

N

35

Problem 35 Determine the limit values of force F such that the body is standing in equilibrium on the frictional slope! The coefficient of friction between the slope and the block is µ.

µ = 0,2 , G = 100 N (weight).

30o

F

µG

36

Problem 36 The ladder is placed on the frictional ground (µ = 0,3) leaning against the smooth wall (with frictionless connection). How high can one climb up the latter without making the ladder to slip?

G

h

5 m

sima

2 m

érdes

smooth

frictional

37

Problem 37 Determine the limit values of force F such that the system remains in equilibrium!

G1 = 30 kN , G2 = 15 kN , µ1 = 0,3 , µ2 = 0,2 .

45o

F

B

A

µ1

µ2

1

2

38

Problem 38 The mechanical system of vertical arrangement presented in the figure is in equilibrium. Determine the limit values of force F such that the system remains in equilibrium!

G1 = 2 kN , G2 = 5 kN , µ1 = µ2 = 0,2 , µ3 = 0 .

45o

F

µ1

µ2

1

2

µ3

39

Solutions:

Problem 1: 00 Nm25

C

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

M

Problem 2: 1

00 kNm2

A

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟−⎝ ⎠

M , 2

00 kNm1, 2

A

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟−⎝ ⎠

M

Problem 3: 20

10 kN15

−⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

F , 1560 kNm

30O

⎛ ⎞⎜ ⎟= −⎜ ⎟⎜ ⎟⎝ ⎠

M

Problem 4: 00 N110

rB

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟−⎝ ⎠

F , 38,5

0 Nm0

B

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

M , 00 N110

rA

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟−⎝ ⎠

F , 38,555 Nm0

A

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

M

Problem 5: 00 Nm9

A

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

M , 00 Nm9

B

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

M , 00 Nm9

O

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

M (the couple is a free vector)

Problems 6-9: Construction Problem 10: 600NeF = , 3,133mex = Problem 11: (a) 10kNeF = , 1mex = (b) 5kNeF = , 4 / 3mex = Problem 12: (a) 28,53mmSx = , 58,24mmSy = (b) 16,21mmSx = , 53,10mmSy = (c) 36,25mmSx = , 50mmSy = Problem 13: (a) 0, 4643mSx = , 0, 4643mSy = , 0, 4643mSz = (b) 0,375mSx = , 0,375mSy = , 0,375mSz = (c) 0, 4167mSx = , 0,5mSy = , 0,5mSz = Problem 14: 1200NxA = , 500NyA = , 500NB = Problem 15: (a) 0NxA = , 300NyA = − , 900NB = (b) 0NxA = , 200NyA = , 400NB = (c) 0NxA = , 400NyA = , 800NB = (d) 0NxA = , 50NyA = , 50NB = − (Couple can only be balanced by couple!)

40

Problem 16: 770NxA = , 0NyA = , 770NxB = , 350NyB = Problem 17: 58kNxA = − , 34kNyA = , 58kNxB = Problem 18: 450NxA = , 450NyA = , 450NxB = − , 150NyB = Problem 19: 1125NxA = , 1125NyA = , 1125NxB = − , 125NyB = Problem 20: 150NxA = , 250NyA = , 250NxB = , 50NyB = − Problem 21: 333,3NxA = , 333,3NyA = , 333,3NxB = − , 666,7NyB = 0NxD = , 166,7NyD = − , 333,3NxG = , 500NyG = Problem 22: 10kNxA = − , 5kNyA = − , 5kNyB = , 5kND = Problem 23: 35kNxA = , 42kNyA = , 35kNxB = − , 42kNyB = , 36kNyC = − 35kNxD = , 42kNyD = , 35kNxE = − , 6kNyE = Problem 24: 1 2,83kNR = − , 2 6kNR = , 3 2,83kNR = , 4 4kNR = − , 5 5,66kNR = 6 4kNR = , 7 5,66kNR = − Problem 25: 1 14,14kNR = − , 2 10kNR = , 3 10kNR = , 4 10kNR = − , 5 14,14kNR = − 6 20kNR = , 7 10kNR = Problem 26: (a) 1kNxA = , 0,5kNyA = , 0,5kNyB = − , 1 0,5kNR = − , 2 1kNR = − , 3 1,118kNR = (b) 0kNxA = , 0,5kNyA = , 0,5kNyB = , 1 0,707kNR = − , 2 0,5kNR = , 3 0,707kNR = − Problem 27: 1 918,6NR = − , 2 1024,5NR = , 3 918,6NR = , 4 1299NR = − , 5 0NR = Problem 28: (a) (b)

V

Mh

2

[kN]

-2

[kNm]

-1

V

Mh

4

[kN]

0

[kNm]

2

0

41

Problem 29: (a) (b)

V

Mh

2

[kN]

-2

[kNm]

-0,5

V

Mh

0,5

[kN]

-1,5

[kNm]

2

0,25 0,2813 Problem 30: (a) (b)

V

Mh

[N]

[Nm]

50

-40 -10

-15

-3

V

Mh

[kN]

[kNm]

1

0

-1

-1 Problem 31: (a) (b)

V

Mh

[kN]

[kNm]

2,25

-0,3375

-6,75

3

-0,1123

0,9

-0,75

V

Mh

[kN]

[kNm]

-10

20

-20

10

5

-15

5

Problem 32: (a) (b)

V

Mh

N

[kN]

[kN]

[kNm]

3

6

2 1

V

Mh

N

[kN]

[kN]

[kNm]

-6

11

-5

12

-3,125

-3 0

42

Problem 33: (a)

V

Mh

N

3 kN

1 m

0,5 m A

Mh

V

[kN]

[kN]

[kNm]

0

+3

0

+3

+1,5

+1,5

0

Problem 33: (b)

V

Mh

N

Mh

V

[kN]

[kN]

[kNm]

-4

+8

0

+8

+4

+4

0

4 kN 1 m

0,5 m A

B

Problem 34:

V

Mh

N

10

[kN]

[kN]

[kNm]

-2

6 2

-4

10 -14

6 4 6 3,33 3,33

4

43

Problem 35: min 33,83NF = , max 87,88NF = Problem 36: 3,75mh = Problem 37: max 15,25NF = , min 18,375NF = − Problem 38: max 4, 4kNF = , min 0,0667kNF = −

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