6–6.

Determine the force in each member of the truss, and stateif the members are in tension or compression.

SOLUTION

Method of Joints: We will begin by analyzing the equilibrium of joint D, and thenproceed to analyze joints C and E.

Joint D: From the free-body diagram in Fig. a,

Ans.

Ans.

Joint C: From the free-body diagram in Fig. b,

Ans.

Ans.

Joint E: From the free-body diagram in Fig. c,

Ans.

Ans.FEA = 1750 N = 1.75 kN (C)

FEA - 1000 - 900 sin 36.87° - 750 cos 73.74° = 0Q+ ©Fy¿ = 0;

FEB = 750 N (T)

- 900 cos 36.87° + FEB sin 73.74° = 0R+ ©Fx¿ = 0;

FCB = 800 N (T)

800 - FCB = 0+ c ©Fy = 0;

FCE = 900 N (C)

FCE - 900 = 0©Fx = 0;:+

FDC = 800 N (T)

1000 a45b - FDC = 0+ c ©Fy = 0;

FDE = 1000 N = 1.00 kN (C)

FDE a35b - 600 = 0©Fx = 0;:+

B

E

D

A

C

600 N

900 N

4 m

4 m

6 m

6–7.

Determine the force in each member of the Pratt truss, andstate if the members are in tension or compression.

SOLUTION

Joint A:

Joint B:

Joint L:

Joint C:

Joint K:

Joint J:

Ans.

Due to Symmetry

Ans.

Ans.

Ans.

Ans.

Ans.

Ans.FKD = FID = 7.45 kN (C)

FKJ = FIJ = 23.6 kN (C)

FCK = FEI = 10 kN (T)

FBL = FFH = FLC = FHE = 0

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

FAL = FGH = FLK = FHI = 28.3 kN (C)

FJD = 33.3 kN (T)

2 (23.57 cos 45°) - FJD = 0+ c ©Fy = 0;

FJI = 23.57 kN (L)

23.57 sin 45° - FJI sin 45° = 0©Fx = 0;:+

FKJ = 23.57 kN (C)

28.28 - 10 cos 45° + 7.454 sin (45° - 26.57°) - FKJ = 0+Q©Fy = 0;

FKD = 7.454 kN (L)

10 sin 45° - FKD cos (45° - 26.57°) = 0R+ ©Fx - 0;

FCK = 10 kN (T)

FCK - 10 = 0+ c ©Fy = 0;

FCD = 20 kN (T)

FCD - 20 = 0©Fx = 0;:+

FLK = 28.28 kN (C)

28.28 - FLK = 0+Q©Fy = 0;

FLC = 0R+ ©Fx = 0;

FBL = 0+ c ©Fy = 0;

FBC = 20 kN (T)

FBC - 20 = 0©Fx = 0;:+

FAB = 20 kN (T)

FAB - 28.28 cos 45° = 0©Fx = 0;:+FAL = 28.28 kN (C)

20 - FAL sin 45° = 0+ c ©Fy = 0;

A

B C D E FG

H

I

J

K

L

2 m

2 m

2 m 2 m

10 kN 10 kN20 kN

2 m 2 m 2 m

2 m

2 m

6–22.

SOLUTION

c

Joint F:

(1)

Joint C:

Joint A:

From Eq.(1), and Symmetry,

Ans.

Ans.

Ans.

Ans.

Ans.

Ans.

Ans.

Ans.

Ans.FCD = 0.471 P (C)

FEC = 1.41 P (T)

FBD = 1.49 P (C)

FBF = 1.41 P (T)

FAC = 1.49 P (C)

FAE = 1.67 P (T)

FAB = 0.471 P (C)

FFD = 1.67 P (T)

FFE = 0.667 P (T)

FAE =53

P = 1.67 P (T)

:+ ©Fx = 0; FAE -223

Pa 1

22b -

2253

Pa 2

25b = 0

FCD =223

P = 0.4714P = 0.471P (C)

FCA =225

3P = 1.4907P = 1.49P (C)

2

25FCA -

1

22FCD = P

:+ ©Fx = 0; FCA a 2

25b - 22Pa 1

22b - FCDa 1

22b = 0

FCA 1

25- 22P

1

22+ FCD

1

22= 0

FFD - FFE = P

1

2

FFB = = 1.41 P (T)

:+ ©Fy = 0; FFB a2b - = 0

22P

P

:+ ©Fx = 0; FFD - FFE - FFBa 1

22b = 0

+ c ©Fy = 0; Ay = P

Dy = P

+ ©MA = 0; PaL

3b + Pa2L

3b - (Dy)(L) = 0

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

= 22PSimilarly, FEC

Similarly, FFD=1.67 P (T)

ADFE

P P

B C

L/3

L/3L/3L/3

+c ©Fy = 0;

6–25.

Determine the force in each member of the truss in terms ofthe external loading and state if the members are in tensionor compression.

SOLUTIONJoint B:

Ans.

Ans.

Joint C:

Ans.

Ans.

Joint D:

Ans.FDA = (cot 2 u + 1)(cos 2 u) (P) (C)

:+ ©Fx = 0; FDA - (cot 2 u + 1)(cos 2 u)P = 0

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

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

FCA =cot 2 u + 1

cos u - sin ucot 2 uP

+ c ©Fy = 0; FCD sin 2 u - FCA sin u = 0

:+ ©Fx = 0; P cot 2 u + P + FCD cos 2 u - FCA cos u = 0

FBC = P cot 2 u (C)

:+ ©Fx = 0; P csc 2u(cos 2u) - FBC = 0

FBA = P csc 2u (C)

+ c ©Fy = 0; FBA sin 2u - P = 0

A

L

B

P

PL C

L

DL

u

6–26.

The maximum allowable tensile force in the members of thetruss is and the maximum allowablecompressive force is Determine themaximum magnitude P of the two loads that can be appliedto the truss.Take and u = 30°.L = 2 m

1Fc2max = 1.2 kN.1Ft2max = 2 kN,

SOLUTION

Joint B:

Joint C:

Joint D:

1) Assume

(O.K.!)

Thus, Ans.Pmax = 732 N

FCD = 1.577(732.05) = 1154.7 N 6 (Fc)max = 1200 N

P = 732.05 N

FCA = 2 kN = 2.732 P

FDA = 0.7887 P (C)

:+ ©Fx = 0; FDA - 1.577 P sin 30° = 0

FCA = 2.732 P (T)

FCD = a tan 30° + 1

23 cos 30° - cos 60°b P = 1.577 P (C)

:+ ©Fx = 0; P tan 30° + P + FCD cos 60° - FCA cos 30° = 0

FCA = FCD a sin 60°sin 30°

b = 1.732 FCD

+ c ©Fy = 0; -FCA sin 30° + FCD sin 60° = 0

FBC = P tan 30° = 0.57735 P (C)

:+ ©Fx = 0; FAB sin 30° - FBC = 0

FBA =P

cos 30°= 1.1547 P (C)

+ c ©Fy = 0; FBA cos 30° - P = 0

(FC)max = 1.2 kN

(Tt)max = 2 kN A

L

B

P

PL C

L

DL

u

6–31.

Determine the force in members CD, CJ, KJ, and DJ of thetruss which serves to support the deck of a bridge. State ifthese members are in tension or compression.

SOLUTION

a

Ans.

a

Ans.

Ans.

Joint D: Ans.FDJ = 0

FCJ = 3125 lb = 3.12 kip (C)

:+ ©Fx = 0; -9375 + 11 250 -35

FCJ = 0

FCD = 9375 lb = 9.38 kip (C)

+ ©MJ = 0; -9500(27) + 4000(18) + 8000(9) + FCD(12) = 0

FKJ = 11 250 lb = 11.2 kip (T)

+ ©MC = 0; -9500(18) + 4000(9) + FKJ(12) = 0

A G

HIJKL

FEDCB

4000 lb8000 lb

5000 lb

9 ft 9 ft 9 ft 9 ft 9 ft 9 ft

12 ft

*6–32.

Determine the force in members EI and JI of the trusswhich serves to support the deck of a bridge. State if thesemembers are in tension or compression.

SOLUTION

a

Ans.

Ans.FEI = 2500 lb = 2.50 kip (C)

+ c ©Fy = 0; 7500 - 5000 - FEI = 0

FJI = 7500 lb = 7.50 kip (T)

+ ©ME = 0; -5000(9) + 7500(18) - FJI(12) = 0

A G

HIJKL

FEDCB

4000 lb8000 lb

5000 lb

9 ft 9 ft 9 ft 9 ft 9 ft 9 ft

12 ft

6–46.

SOLUTIONSupport Reactions:

a

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

Method of Sections:

a

Ans.

a

Ans.FCM = 2.00 kN T

FCM 142 - 2142 = 0+ ©MA = 0;

FCD = 5.625 kN 1T2FCD142 - 5.625142 = 0+ ©MM = 0;

Ax = 0:+ ©Fx = 0;

Ay = 5.625 kN

21122 + 5182 + 3162 + 2142 - Ay 1162 = 0+ ©MI = 0;

Determine the force in members CD and CM of theBaltimore bridge truss and state if the members are intension or compression. Also, indicate all zero-forcemembers.

16 m, 8 @ 2 m

A

B C D E F G H

J

KLM

N O P2 m

2 m

2 kN5 kN

2 kN3 kN

I

6–47.

Determine the force in members EF, EP, and LK of theBaltimore bridge truss and state if the members are intension or compression. Also, indicate all zero-forcemembers.

SOLUTIONSupport Reactions:

a

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

Method of Sections:

a

Ans.

a

Ans.

Ans.FED = 1.94 kN T

6.375 - 3 - 2 - FED sin 45° = 0+ c ©Fy = 0;

FLK = 9.25 kN 1C26.375182 - 2142 - 3122 - FLK 142 = 0+ ©ME = 0;

FEF = 7.875 = 7.88 kN 1T23122 + 6.375142 - FEF142 = 0+ ©MK = 0;

Iy = 6.375 kN

Iy 1162 - 21122 - 31102 - 5182 - 2142 = 0+ ©MA = 0;

16 m, 8 @ 2 m

A

B C D E F G H

J

KLM

N O P2 m

2 m

2 kN5 kN

2 kN3 kN

I

6–71.

Determine the support reactions at A, C, and E on thecompound beam which is pin connected at B and D.

SOLUTION

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

Ans.

Ans.

Subsequently, the free-body diagram of segment BD in Fig. b will be consideredusing the results of and obtained above.

Ans.

Finally, the free-body diagram of segment AB in Fig. a will be considered using theresults of and obtained above.

Ans.

Ans.

Ans.MA = 21.5 kN # m

11.67(3) - 9(1.5) - MA = 0+ ©MA = 0;

Ay = 2.67 kN

11.67 - 9 - Ay = 0+ c ©Fy = 0;

Ax = 0©Fx = 0;:+ByBx

By = 0©Fx = 0;:+By = 11.67 kN

By(1.5) - 5(1.5) - 10 = 0+ ©MC = 0;

NC = 16.67 kN = 16.7 kN

NC(1.5) - 5(3) - 10 = 0+ ©MB = 0;

DyDx

Dx = 0©Fx = 0;:+Dy = 5 kN

10(1.5) - Dy(3) = 0+ ©ME = 0;

NE = 5 kN

NE(3) - 10(1.5) = 0+ ©MD = 0;

A B C DE

1.5 m

9 kN

10 kN m

10 kN

1.5 m1.5 m 1.5 m 1.5 m 1.5 m

6–102.

1.5 m

0.25 m

0.3 m

0.1 m0.2 m 1.25 m

0.6 m

0.4 m 0.3 m

A B

C

F

D

G

E

SOLUTION

a

Ans.

Ans.

a

Ans.

Ans.FF = 21345522 + 113 52722 = 14.0 kN

Fy = 13 527 N

-Fy - 2452.5 + 16 349 cos 12.2° = 0+ c ©Fy = 0;

Fx = 3455 N

Fx - 16 349 sin 12.2° = 0:+ ©Fx = 0;

FCD = 16 349 N = 16.3 kN

2452.512.802 - FCD1cos 12.2°210.72 + FCD1sin 12.2°211.252 = 0+ ©MF = 0;

FE = 2198122 + 12452.522 = 2.64 kN

Ey = 2452.5 NEy - 2452.5 = 0;+ c ©Fy = 0;

Ex = 981 N-Ex + 981 = 0;:+ ©Fx = 0;

FAB = 981 N

2452.510.12 - FAB10.252 = 0+ ©ME = 0;

The tractor boom supports the uniform mass of 500 kg inthe bucket which has a center of mass at G. Determine theforce in each hydraulic cylinder AB and CD and theresultant force at pins E and F. The load is supportedequally on each side of the tractor by a similar mechanism.

6–109.

SOLUTIONFree-Body Diagram: The solution for this problem will be simplified if one realizesthat links BD and CF are two-force members.

Equations of Equilibrium : From FBD (a),

a

From FBD (b),

a

From FBD (c),

a

Ans.

Ans.

At point D,

Ans.

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

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

Ey = 1553.67 N = 1.55 kN

Ey + 6702.66 sin 30° - 4905 = 0+ c ©Fy = 0;

Ex = 6785.67 N = 6.79 kN

Ex - 981 - 6702.66 cos 30° = 0:+ ©Fx = 0;

FCF = 6702.66 N

- FCF cos 15°13002 = 0

4905 sin 45°17002 - 981 sin 45°17002+ ©ME = 0;

Ay = 4905 N

Ay - 3924 - 1387.34 sin 45° = 0+ c ©Fy = 0;

Ax - 1387.34 cos 45° = 0 Ax = 981 N:+ ©Fx = 0;

FBD = 1387.34 N

FBD cos 45°11002 + FBD sin 45°11002 - 39241502 = 0+ ©MA = 0;

78481x2 - FK12x2 = 0 FK = 3924 N+ ©ML = 0;

The symmetric coil tong supports the coil which has a massof 800 kg and center of mass at G. Determine the horizontaland vertical components of force the linkage exerts on plate DEIJH at points D and E. The coil exerts only vertical reactions at K and L.

400 mm

45°

300 mm

45°30° 30°

100 mm

100 mm50 mm

AB

C

D J

E I

H

G LK

F