Solution ch 4 PDF

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Chapter 4. Structures 6-3. Determine the force in each member of the truss in terms of the load P and state if the members are in tension or compression. L B L Solution 1. Determine reaction forces at support A and D C L L L D A E L Consider structure as a rigid body L P Free-body diagram 4 B 1 3 F7...

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Chapter 4. Structures 6-3. Determine the force in each member of the truss in terms of the load P and state if the members are in tension or compression.

L

B

L

C

L

L

L

Solution 1. Determine reaction forces at support A and D

D

A E L

L

Consider structure as a rigid body

P

Free-body diagram 4

B

1

3

F7

C

5

7

Ax

D D

A

2

6

E

Ay

F6 Dy

Dy

P

Force system in equilibrium F1

( P, Ax , Ay, Dy )  0 A

Equilibrium equations  Fkx  Ax  0

Ax  0

 Fky  Ay  P  Dy  0

Dy  P / 2

 M A  PL  2DyL  0

Ay  P  Dy  P / 2

2. The Method of Joints It is assumed that all members are in tensile. Equilibrium of Joint D (note D) ( Dy , F6, F7 )  0 == >

 Fkx  F6  F7 cos 60o  0  Fky  D y  F7 sin 60o  0

Equilibrium of Joint A (note A) (Ay , F1, F2)  0 == >

 Fkx  F2  F1 cos 60o  0  Fky  Ay  F1 sin 60o  0

1

F2 Ay

Equilibrium of Joint B (note B) B

Equilibrium of Joint C (note C)

F4

F4

F1

F5 F3

F7

Check equilibrium of joint E (note E) F3

F2

C

F5

E

F6 P

2

*6-4. Each member of the truss is uniform and has a weight W . Remove the external force P and determine the approximate force in each member due to the weight of the truss. State if the members are in tension or compression. Solve the problem by assuming the weight of each member can be represented as a vertical force, half of which is applied at each end of the member.

Prob. 6-3/4 Solution 1. Determine reaction forces at support A and D Consider structure as a rigid body Free-body diagram 1.5W

1.5W 4

B

W

1

3

C

5

7

Ax

W D

A

2

6

E

Ay

2W

Dy

Force system in equilibrium ( W, 1.5 W, 1.5 W, 2 W, W, Ax, Ay, Dy)  0

Equilibrium equations  Fkx  Ax  0

Ax  0

 Fky   7W  Ay  D y  0

Dy  7W / 2

 M A   7WL  2DyL  0

Ay  7W  Dy  7 P / 2

2. The Method of Joints 3

It is assumed that all members are in tensile. F7

Equilibrium of Joint D (note D) (W, Dy , F6 , F7 )  0 == >

 Fkx  F6  F7 cos 60o  0

W

 Fky  W  Dy  F7 sin 60o  0 D F6

Equilibrium of Joint A (note A) Dy F1 W

( W, Ay , F1, F2 )  0 == >

A

 Fkx  F2  F1 cos 60o  0  Fky  W  Ay  F1 sin 60o  0

F2 Ay

Equilibrium of Joint B (note B)

Equilibrium of Joint C (note C)

1.5W

1.5W

B

F4

F4

F1

F5 F3

F7

Check equilibrium of joint E (note E)

F3

F2

C

F5

E

F6 2W

4

6-5. Determine the force in each member of the truss and state if the members are in tension or compression. Set P 1 = 20kN, P2 = 10kN.

Prob. 6-5 B

C

D

2 E

A G 1.5

F 1.5

1.5

1.5

P1

P2

Solution 1. Determine reaction forces at support A and E Consider structure as a rigid body

C

B

D

Free-body diagram 2 Ax

A

E G 1.5

F 1.5

1.5

1.5

Ay

Ey P1

Force system in equilibrium ( P1, P2, Ax, Ay, Ey)  0

Equilibrium equations  Fkx  Ax  0

Ax  0

 Fky  P1  P2  Ay  E y  0

Ey  16 (1.5P1  4.5 P2 )

 M A  6E y  1.5P1  4.5P2  0

Ay  P1  P2  Ey

5

P2

Applying the method of joints (Examine equilibrium of notes) Note A Note E Note B Note D Note F

FAB Ax

FED

A FAG

E FEF

Ay

Ey

B

D FBC

FAB

FDC

FED FDF

FBG

FCF

FBC

C

FCD

FFD

FGF

FEF F

FCG

FCF

P2

Check note C and G.

6

Method of sections B

C

D

2 Ax

A

E G 1.5

F 1.5

1.5

1.5

Ay

Ey P1

P2

7

6-7. Determine the force in each member of the truss and state if the members are in tension or compression.

Prob. 6-7 Solution 1. Determine reaction forces at support E and F Consider structure as a rigid body

Fy

Free-body diagram 0.3

0.3

A

13

12

0.3

B

D

0.4

0.4

F

Fx

5 P=130 N E C

Force system Ey

(P, Ey , Fx , Fy )  0

Equilibrium Equations

Fy

2. Applying method of Joints (Equilibrium of Notes) B

Note A, B, F, E and C (check note D)A

D

F Fx

13

12

0.4

FBC

5 P=130 N E C

Remark: FBC = 0, (based on property of joint B)

8

Ey

*6-8. Determine the force in each member of the truss and state if the members arc in tension or compression. Hint: The resultant force at the pin E acts along member ED. Why? Solution Applying method of joints (notes) P1

P2

3 A

3 B

C P2

P1 4 D

B FAB

FCB FDB

FDC

E FAD

FDB

FDC

D FED

Equilibrium of notes : C, B, D.

9

FCB

C

6-9. Each member of the truss is uniform and has a mass of 8 kg/m . Remove the external loads of 3 kN and 2 kN and determine the approximate force in each member due to the weight of the truss. State if the members are in tension or compression. Solve the problem by assuming the weight of each member can be represented as a vertical force, half of which is applied at each end of th e member. P1

P2

3 A

3 B

4 D

Prob. 6-8/9

E

10

C

6-14. Determine the force in each member of the truss and state if the members are in tension or compression. The load has a mass of 40 kg. T (mg) D

6 FDE 3.5 FDC

G (mg)

FAG 2.5 FAB A

Ax

Probs. 6-14/15 Solution Let us consider pulley at joint D. (G, T, FDC , FDE )  0

Equations of equilibrium T  FDC cos 45o  FDE cos   0 G  FDC sin 45o  FDE sin   0

2

with

cos   6 / 6  3.5

2

sin   3.5 / 62  3.52

T  G  mg

It is found FDC  G 2  mg 2,

FDE  0

Or it can be seen that the resultant force of G and T lies on AD, so that FDE  0 . Remark: if three members form a truss joint for which two of the members are collinear, the third member is a zero-force member provided no external force or support reaction is applied to the joint (2D structures). Consider note E: we have FEC  0, FEF  FDE  0 Consider note C: because FEC  0 , we have FCF  0 and FCB  FCD  mg 2 Consider note F, a nd then note B… FAB  FBC  mg 2

Finally, consider note A: ( Ax , FAB, FAG )  0 =>

Ax  FAB cos 45o  0 o

FAG  FAB sin 45  0

FAG  FAB sin 45o  mg

FEC  FCF  FFB  FBG  0, FDE  F EF  F FG  0

Answer:

FDC  FCB  FBA   mg 2, FAG  mg

11

6-15. Determine the largest mass m of the suspended block so that the force in any member does not exceed 30 kN (T) or 25 kN (C).

Probs. 6-14/15 Remark: if three members form a truss joint for which two of the members are collinear, the third member is a zero-force member provided no external force or support reaction is applied to the joint (2D structures).

12

*6-16. Determine the force in each member of the truss in terms of the load P and state if the members are in tension or compression. Ey E

Ex d

D A

Ax

C P

Prob. 6-16/17

d/2 d/2

B

Solution

d

d

1. Determine reaction forces at supports A and E (By hardening all structure) rigidization we have: ( P, Ax, Ex, Ey)  0 Equation of equilibrium  Fx  ..  0  Fy  ..  0  mE  ..  0

Solving for unknowns Ax , Ex , Ey

2. Consider joints A , B, C and E (check joint D) FBD FAD

Ax

FBC FDE

FAB

FAB

B

D

FCE

FDC Ey

FDA FDB

FCD E Ex

C FBC

FED

13

FCE

6-17. The maximum allowable tensile force in the members of the truss is (F t )max = 5kN, and the maximum allowable compressive force is (F c )max = 3kN. Determine the maximum magnitude of the load P that can be applied to the truss. Take d = 2 m.

Prob. 6-16/17

14

6-31. The Howe bridge truss is subjected to the loading shown. Determine the force in members DE, EH, and HG , and state if the members are in tension or compression.

Prob. 6- 31 P4

30 kN 20 kN

P1

20 kN P3

P2 C

B

40 kN

D

E

F

4m Ax

A

G J Ay

I

H Gy

16 m = 4@4

1. Make some comments from system structure From Joints B, I, F we can find out in members BC, DI, EF and FG have no forces. 2. Determine reaction forces at supports A and G. ( P1, P2, P3, P4, Ax, Ay, Gy )  0 == > G y

3. Determine the force in members DE, EH, and HG :

P4

Method of section ( P4, Gy , FGH , FEH , FED)  0

E

Eqs of Equilibrium

F

FDE

 Fx   FDE  FHG  0  Fy  Gy  P4  FHE  0

FHE G

 mE ()  4Gy  4 FHG  0 FHG

[4. Method of Joints To determine forces in all members

Gy

Joint B, A, C, J Joint G, E, H]

15

6-34. The Pratt bridge truss is subjected to the loading shown. Determine the force in members LD, LK, CD, and KD, and stale if the members are in tension or compression.

Prob. 6-34/35

50 kN P1

50 kN P2

N

50 kN P3

L

K

H

4m Ax

A

G B

C

Ay

D

18 m = 6@3 m

Gy

Solution 1. Determine reaction force at supports A and G. ( G y )  mA (.)  ..  0



Gy 

2. Method of section Consider the section including joints: G, H, … , K, D We find out the force in members LD, LK, CD 3. Method of joint Consider the joint K We find out the force in member KD

6-35. The Pratt bridge truss is subjected to the loading shown. Determine th e force in members JI, JE, and DE, and state if the members are in tension or compression. 16

50 kN P1

50 kN P2

N

50 kN P3

L

K

J

I

H

4m Ax

A

G B

C

Ay

D

E Gy

18 m = 6@3 m

Solution 1. Determine reaction force at supports A and G. ( G y )  mA (.)  ..  0



Gy 

2. Method of section Consider the section including joints: G, H, I, E, F We find out the force in members JI , JE , and DE Eqs of Equilibrium FJE

 Fx   FDE  FJI  0  Fy  Gy  FJE  0

I

J

H

FJI

 mJ (.)  6Gy  4 FED  0

4m G FED

E 6m

Gy

*6-40. Determine the force developed in members BC and CH of the roof truss and state if the members are in tension or compression.

17

6-41. Determine the force in members CD and GF of the truss and state if the members arc in tension or compression. Also indicate all zero -force members.

Prob. 6-40/41

6-42. The Howe truss is subjected to the loading shown. Determine the force in members GF, CD, and GC, and state if the members are in tension or compression. 6-43. The Howe truss is subjected to the loading shown. Determine the force in members GH, BC, and BG of the truss and stale if the members are in tension or compression.

Probs. 6-42/43

18

6-49. The tower truss is subjected to the loads shown. Determine the force in members BC, BF, and FG, and state if the members are in tension or compression. The left side, ABCD, stands vertical. Solution Using method of section

6-50. The tower truss is subjected to the loads sho wn. Determine the force in members BG and CF, and state if the members are in tension or compression. The left side ABCD stands vertical. FFE FFC F FFG

FBC FBF FBG

B FBA

Probs. 6-49/50 Solution Consider Note F and B Note F: FFG and FBF are known (Prob. 6-49) == > Note B: FBC and FBF are known (Prob. 6-49) == > 19

FBF

6-71. Determine the horizontal and vertical components of force that pins A and C exert on the frame.

Prob. 6-71 Solution Free-Body Diagram

Six unknowns and six equilibrium equations OR Free-Body Diagram

Three unknowns and three equilibrium equations for member AB.

6-75. The compound beam is pin supported at B and supported by rockers at A and C. There is a hinge (pin) at D. Determine the reactions at the supports.

20

Prob. 6-75 6-79. Determine the horizontal and vertical components of force that pin A and C exert on the two-member arch. // arch /ɑ:t∫/ khung hinh cung

Prob. 6-79 6-92. Determine the reactions at the fixed support E and the smooth support A. The pin, attached to member BD, passes through a smooth slot at D. // slot [s l ɔ t]: ranh, x C

F=600N 0.4 m B

D

0.4 m A E

Prob. 6-92 0.3 m

21

0.3 m

0.3 m

0.3 m

6-98. Determine the horizontal and vertical components of force at pins A and C of the two-member frame.

Prob. 6-98 6-106. The two disks each have a mass of 20 kg and are attached at their centers by an elastic cord that has a stiffness of k = 2 kN/m. Determine the stretch of the cord when the system is in equilibrium and the angle  of the cord. // cord [kɔ:d] day thung nho

l A 5

3



B

4

Prob. 6-106

6-111. Determine the force created in the hydraulic cylinders EF and AD in order to hold the shovel in equilibrium. The shovel load has a mass of 1.25 Mg and a center of gravity at G. All joints arc pin connected. // shovel ['∫ʌvl] = gau xuc cua may xuc

22

Prob. 6-111 6-121. The level mechanism for a machine press serves as a toggle which develops a large force at E when a small force is applied at the handle H. To show that this is the case, determine the force at E if someone applies a vertical force of 80 N a t H. The smooth head at D is able to slide freely dow nw ard. All members are pin connected. // toggle ['tɔgl] (ky  thuâ  t) đo n khuy  u (như) toggle-joint

Prob. 6-121 6-123. Determine the couple moment M that must be applied to member DC for equilibrium of the quick-return mechanism. Express the result in terms of the angles  and  , dimension L , and the applied horizontal load P . The block at C is confined to slide within the slot of member AB. // confine: [kən'fain] giam giư; giam ham;

Prob. 6-123

23

6-127. The structure is subjected to the loading shown. Member AB is supported by a ball-and-socket at A and smooth collar at B. Member CD is supported by a pin at C. Determine the x, y, z components of reaction at A and C. //collar ['kɔlə] (ky thuât) vo ng đai; vo ng đêm, (ky thuât) đong đai, vong dai

Prob. 6-127

24...