Solutions 10 PDF

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

600 N

D

4m

SOLUTION

900 N

E C

Method of Joints: We will begin by analyzing the equilibrium of joint D, and then proceed to analyze joints C and E. Joint D: From the free-body diagram in Fig. a, + ©Fx = 0; :

+ c ©Fy = 0;

6m

Ans.

4 1000 a b - FDC = 0 5 FDC = 800 N (T)

Ans.

Joint C: From the free-body diagram in Fig. b, + ©F = 0; : x

FCE - 900 = 0 FCE = 900 N (C)

+ c ©Fy = 0;

Ans.

800 - FCB = 0 FCB = 800 N (T)

Ans.

Joint E: From the free-body diagram in Fig. c, R + ©Fx ¿ = 0;

- 900 cos 36.87° + FEB sin 73.74° = 0 FEB = 750 N (T)

Q+ ©Fy ¿ = 0;

Ans.

FEA - 1000 - 900 sin 36.87° - 750 cos 73.74° = 0 FEA = 1750 N = 1.75 kN (C)

B

A

3 FDE a b - 600 = 0 5 FDE = 1000 N = 1.00 kN (C)

4m

Ans.

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

J 2m

K

2m

SOLUTION

I

L

H

2m A

Joint A: + c ©Fy = 0;

20 - FAL sin 45° = 0

10 kN

FAL = 28.28 kN (C) + ©F = 0; : x

FAB - 28.28 cos 45° = 0 FAB = 20 kN (T)

Joint B: + ©F = 0; : x

FBC - 20 = 0 FBC = 20 kN (T)

+ c ©Fy = 0;

FBL = 0

Joint L: R + ©Fx = 0;

FLC = 0

+Q©Fy = 0;

28.28 - FLK = 0 FLK = 28.28 kN (C)

Joint C: + ©F = 0; : x

FCD - 20 = 0 FCD = 20 kN (T)

+ c ©Fy = 0;

FCK - 10 = 0 FCK = 10 kN (T)

Joint K: R + ©Fx - 0;

10 sin 45° - FKD cos (45° - 26.57°) = 0 FKD = 7.454 kN (L)

+Q©Fy = 0;

28.28 - 10 cos 45° + 7.454 sin (45° - 26.57°) - FKJ = 0 FKJ = 23.57 kN (C)

Joint J: + ©F = 0; : x

23.57 sin 45° - FJI sin 45° = 0 FJI = 23.57 kN (L)

+ c ©Fy = 0;

G B C D E F 2m 2m 2m 2m 2m 2m

2 (23.57 cos 45°) - FJD = 0 FJD = 33.3 kN (T)

Ans.

Due to Symmetry FAL = FGH = FLK = FHI = 28.3 kN (C)

Ans.

FAB = FGF = FBC = FFE = FCD = FED = 20 kN (T)

Ans.

FBL = FFH = FLC = FHE = 0

Ans.

FCK = FEI = 10 kN (T)

Ans.

FKJ = FIJ = 23.6 kN (C)

Ans.

FKD = FID = 7.45 kN (C)

Ans.

20 kN

10 kN

6–22. B

Determine the force in each member of the double scissors truss in terms of the load P and state if the members are in tension or compression.

C

L/3

SOLUTION c + ©MA = 0;

+ c ©Fy = 0;

L/3

Joint F: FFD - FFE - FFB a

1 22

b = 0

(1)

FFD - FFE = P + ©F = 0; : y

FFB a

1 22

b - P = 0

FFB = 22P= 1.41 P (T) Similarly,

FEC = 22P

Joint C: FCA a

+ ©F = 0; : x

2 25 +c ©Fy = 0;

FCA

2 25

b - 22Pa

FCA 1 25

1 22

1 22

b - FCD a

1 22

b = 0

FCD = P

- 22P

1 22

+ FCD

1 22

=0

FCA =

225 P = 1.4907P = 1.49P (C) 3

FCD =

22 P = 0.4714P = 0.471P (C) 3

FAE -

225 22 1 2 b = 0 b Pa Pa 3 3 25 22

FAE =

5 P = 1.67 P (T) 3

Joint A: + ©F = 0; : x

Similarly, FFD=1.67 P (T) From Eq.(1), and Symmetry, FFE = 0.667 P (T)

Ans.

FFD = 1.67 P (T)

Ans.

FAB = 0.471 P (C)

Ans.

FAE = 1.67 P (T)

Ans.

FAC = 1.49 P (C)

Ans.

FBF = 1.41 P (T)

Ans.

FBD = 1.49 P (C)

Ans.

FEC = 1.41 P (T)

Ans.

FCD = 0.471 P (C)

Ans.

F L/3

P

Ay = P

+ ©F = 0; : x

E

A

2L L b - (Dy)(L) = 0 Pa b + Pa 3 3 Dy = P

D L/3

P

6–25. Determine the force in each member of the truss in terms of the external loading and state if the members are in tension or compression.

P B

L

C P u

L

L

SOLUTION A

Joint B: + c ©Fy = 0;

FBA sin 2u - P = 0 FBA = P csc 2u (C)

+ ©F = 0; : x

Ans.

P csc 2u(cos 2u) - FBC = 0 FBC = P cot 2 u (C)

Ans.

Joint C: + ©F = 0; : x

P cot 2 u + P + FCD cos 2 u - FCA cos u = 0

+ c ©Fy = 0;

FCD sin 2 u - FCA sin u = 0 FCA =

cot 2 u + 1 P cos u - sin u cot 2 u

FCA = (cot u cos u - sin u + 2 cos u) P (T)

Ans.

FCD = (cot 2 u + 1) P (C)

Ans.

Joint D: + ©F = 0; : x

D L

FDA - (cot 2 u + 1)(cos 2 u) P = 0 FDA = (cot 2 u + 1)(cos 2 u) (P)

(C)

Ans.

6–26. T he maximum allowa ble tensile force in t he members of the truss is 1Ft2max = 2 kN, and the maximum allowable compressive force is 1Fc2max = 1.2 kN. Determine the maximum magnitude P of the two loads that can be applied to the truss.Take L = 2 m and u = 30°.

P B

L

L

SOLUTION A

Joint B: FBA cos 30° - P = 0 FBA = + ©F = 0; : x

P = 1.1547 P (C) cos 30°

FAB sin 30° - FBC = 0 FBC = P tan 30° = 0.57735 P (C)

Joint C: + c ©Fy = 0;

-FCA sin 30° + FCD sin 60° = 0 FCA = FCD a

+ ©F = 0; : x

sin 60° b = 1.732 FCD sin 30°

P tan 30° + P + FCD cos 60° - FCA cos 30° = 0 FCD = a

tan 30° + 1 23 cos 30° - cos 60°

b P = 1.577 P (C)

FCA = 2.732 P (T) Joint D: + ©F = 0; : x

FDA - 1.577 P sin 30° = 0 FDA = 0.7887 P (C)

1) Assume FCA = 2 kN = 2.732 P P = 732.05 N FCD = 1.577(732.05) = 1154.7 N 6 (Fc)max = 1200 N Thus,

D L

(FC)max = 1.2 kN

+ c ©Fy = 0;

Pmax = 732 N

P u

L

(Tt)max = 2 kN

C

(O.K.!) Ans.

6–31. Determine the force in members CD, CJ, KJ, and DJ of the truss which serves to support the deck of a bridge. State if these members are in tension or compression.

8000 lb

5000 lb

4000 lb B

A

C

E

D

F

G 12 ft

SOLUTION

L 9 ft

a + ©MC = 0;

-9500(18) + 4000(9) + FKJ(12) = 0 FKJ = 11 250 lb = 11.2 kip (T)

a + ©MJ = 0;

Joint D:

Ans.

-9500(27) + 4000(18) + 8000(9) + FCD(12) = 0 FCD = 9375 lb = 9.38 kip (C)

+ ©F = 0; : x

K 9 ft

-9375 + 11 250 -

Ans.

3 F = 0 5 CJ

FCJ = 3125 lb = 3.12 kip (C)

Ans.

FDJ = 0

Ans.

J 9 ft

I 9 ft

H 9 ft

9 ft

*6–32. Determine t he force in mem bers EI and JI of the truss which serves to support the deck of a bridge. State if these members are in tension or compression.

8000 lb

5000 lb

4000 lb B

C

D

E

F

A

G 12 ft

SOLUTION

L 9 ft

a + ©ME = 0;

-5000(9) + 7500(18) - FJI(12) = 0 FJI = 7500 lb = 7.50 kip (T)

+ c ©Fy = 0;

K 9 ft

Ans.

7500 - 5000 - FEI = 0 FEI = 2500 lb = 2.50 kip (C)

Ans.

J 9 ft

I 9 ft

H 9 ft

9 ft

6–46. Determine the force in members CD and CM of the Baltimore bridge truss and state if the members are in tension or compression. Also, indicate all zero-force members.

M N

O

SOLUTION

C

21122 + 5182 + 3162 + 2142 - Ay 1162 = 0 Ay = 5.625 kN Ax = 0

Method of Joints: By inspection, members BN, NC, DO, OC, HJ LE and JG are zero force members.

Ans.

Method of Sections: a+ ©MM = 0;

a+ ©MA = 0;

D

E

F

5 kN 3 kN 16 m, 8 @ 2 m

Support Reactions:

FCD142 - 5.625142 = 0

FCD = 5.625 kN 1T2

Ans.

FCM = 2.00 kN T

Ans.

FCM 142 - 2142 = 0

2m

J

P

I

2 kN

+ ©F = 0; : x

K

A B

a + ©MI = 0;

L

G 2 kN

H

2m

6–47. Determine the force in members EF, EP, and LK of the Baltimore bridge truss and state if the members are in tension or compression. Also, indicate all zero-force members.

M N

O

SOLUTION

C

Iy 1162 - 21122 - 31102 - 5182 - 2142 = 0 Iy = 6.375 kN

Method of Joints: By inspection, members BN, NC, DO, OC, HJ LE and JG are zero force members.

Ans.

Method of Sections:

a+ ©ME = 0;

+ c ©Fy = 0;

D

E

F

5 kN 3 kN 16 m, 8 @ 2 m

S upport Reactions:

3122 + 6.375142 - FEF142 = 0

FEF = 7.875 = 7.88 kN 1T2

Ans.

FLK = 9.25 kN 1C2

Ans.

6.375182 - 2142 - 3122 - FLK 142 = 0

6.375 - 3 - 2 - FED sin 45° = 0 FED = 1.94 kN T

Ans.

2m

J

P

I

2 kN

a+ ©MK = 0;

K

A B

a + ©MA = 0;

L

G 2 kN

H

2m

6–71. Determine the support reactions at A, C, and E on the compound beam which is pin connected at B and D.

10 kN

9 kN

10 kN m B C

E D

SOLUTION

A

Equations of Equilibrium: First, we will consider the free-body diagram of segment DE in Fig. c.

1.5 m 1.5 m 1.5 m 1.5 m 1.5 m 1.5 m

+ ©MD = 0;

NE(3) - 10(1.5) = 0 NE = 5 kN

+ ©ME = 0;

Ans.

10(1.5) - Dy(3) = 0 Dy = 5 kN

+ ©F = 0; : x

Dx = 0

Ans.

Subsequently, the free-body diagram of segment BD in Fig. b will be considered using the results of Dx and Dy obtained above. + ©MB = 0;

NC(1.5) - 5(3) - 10 = 0 NC = 16.67 kN = 16.7 kN

+ ©MC = 0;

Ans.

By(1.5) - 5(1.5) - 10 = 0 By = 11.67 kN

+ ©F = 0; : x

By = 0

Finally, the free-body diagram of segment AB in Fig. a will be considered using the results of Bx and By obtained above. + ©F = 0; : x

Ax = 0

+ c ©Fy = 0;

11.67 - 9 - A y = 0 A y = 2.67 kN

+ ©MA = 0;

Ans.

Ans.

11.67(3) - 9(1.5) - MA = 0 MA = 21.5 kN # m

Ans.

6–102. The tractor boom supports the uniform mass of 500 kg in the bucket which has a center of mass at G. Determine the force in each hydraulic cylinder AB and CD and the resultant force at pins E and F. The load is supported equally on each side of the tractor by a similar mechanism.

B

A

G

0.25 m E

C 1.5 m 0.3 m

0.1 m

SOLUTION a + ©ME = 0;

2452.510.12 - F AB10.252 = 0 FAB = 981 N -Ex + 981 = 0;

Ex = 981 N

+ c ©Fy = 0;

Ey - 2452.5 = 0;

Ey = 2452.5 N

FE = 2198122 + 12452.522 = 2.64 kN

0.6 m D 0.4 m 0.3 m

Ans.

2452.512.802 - FCD1cos 12.2°210.72 + FCD1sin 12.2°211.252 = 0 FCD = 16 349 N = 16.3 kN

+ ©F = 0; : x

F

Ans.

+ ©F = 0; : x

a + ©MF = 0;

Ans.

Fx - 16 349 sin 12.2° = 0 Fx = 3455 N

+ c ©Fy = 0;

1.25 m

0.2 m

-Fy - 2452.5 + 16 349 cos 12.2° = 0 Fy = 13 527 N

FF = 21345522 + 113 52722 = 14.0 kN

Ans.

6–109. The symmetric coil tong supports the coil which has a mass of 800 kg and center of mass at G. Determine the horizontal and v ertical component s of force the linka g e e xerts on plate DEIJH at points D and E. The coil exerts only vertical reactions at K and L.

H 300 mm

D

J E

I

400 mm

SOLUTION

B 100 mm

Free-Body Diagram: The solution for this problem will be simplified if one realizes that links BD and CF are two-force members. Equations of Equilibrium : From FBD (a), 78481x2 - FK12x2 = 0

a + ©ML = 0;

FK = 3924 N

F BD cos 45°11002 + FBD sin 45°11002 - 39241502 = 0 FBD = 1387.34 N

+ ©F = 0; : x

A x - 1387.34 cos 45° = 0

+ c ©Fy = 0;

A y - 3924 - 1387.34 sin 45° = 0

A x = 981 N

A y = 4905 N From FBD (c), a+ ©ME = 0;

4905 sin 45°17002 - 981 sin 45°17002 - FCF cos 15°13002 = 0 FCF = 6702.66 N

+ ©F = 0; : x

Ex - 981 - 6702.66 cos 30° = 0 Ex = 6785.67 N = 6.79 kN

+ c ©Fy = 0;

Ans.

Ey + 6702.66 sin 30° - 4905 = 0 Ey = 1553.67 N = 1.55 kN

Ans.

Dx = FBD cos 45° = 1387.34 cos 45° = 981 N

Ans.

Dy = FBD sin 45° = 1387.34 sin 45° = 981 N

Ans.

At point D,

30° F

50 mm

100 mm

K

From FBD (b), a + ©MA = 0;

C 45° 30° A 45°

G

L...