Solution manual engineering mechanics statics 13th edition by r c hibbeler13e chap 06

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Determine the force in each member of the truss and state

if the members are in tension or compression Set

SOLUTION

Method of Joints: In this case, the support reactions are not required for

determining the member forces

Note: The support reactions A x and A y can be determined by analyzing Joint A

using the results obtained above

Cy = 571 lb+ c ©Fy = 0; Cy - 808.12 sin45° = 0

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Determine the force on each member of the truss and state

if the members are in tension or compression Set

SOLUTION

Method of Joints: In this case, the support reactions are not required for

determining the member forces

Note: The support reactions A x and A y can be determined by analyzing Joint A

using the results obtained above

Cy = 271.43 lb+ c ©Fy = 0; Cy - 383.86 sin 45° = 0

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Determine the force in each member of the truss, and state

if the members are in tension or compression Set

SOLUTION

Support Reactions: Applying the equations of equilibrium to the free-body diagram

of the entire truss, Fig a, we have

a

Method of Joints: We will use the above result to analyze the equilibrium of

joints C and A, and then proceed to analyze of joint B.

Joint C: From the free-body diagram in Fig b, we can write

Note: The equilibrium analysis of joint D can be used to check the accuracy of the

solution obtained above

Ay = 0.875 kN

Ay + 3.125 - 4 = 0+ c ©Fy = 0;

u = 0°

B D

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Determine the force in each member of the truss, and state

if the members are in tension or compression Set

SOLUTION

Support Reactions: From the free-body diagram of the truss, Fig a, and applying

the equations of equilibrium, we have

a

Method of Joints: We will use the above result to analyze the equilibrium of

joints C and A, and then proceed to analyze of joint B.

Joint C: From the free-body diagram in Fig b, we can write

Note: The equilibrium analysis of joint D can be used to check the accuracy of the

solution obtained above

Ay = 0.875 kN

Ay + 3.608 cos 30° - 4 = 0+ c ©Fy = 0;

u = 30°

B D

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Determine the force in each member of the truss, and state

if the members are in tension or compression

SOLUTION

Method of Joints: Here, the support reactions A and C do not need to be determined.

We will first analyze the equilibrium of joints D and B, and then proceed to analyze

FBC = 200 N (T)

FBC - 200 = 0+ c ©Fy = 0;

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Determine the force in each member of the truss, and state

if the members are in tension or compression

SOLUTION

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,

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Determine the force in each member of the Pratt truss, and

state if the members are in tension or compression

FLC = 0

R + ©Fx = 0;

FBL = 0+ c ©Fy = 0;

A

H I

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Determine the force in each member of the truss and state if

the members are in tension or compression Hint: The

horizontal force component at A must be zero Why?

5FBD - 400 = 0

FCD= 693 lb (C)+ c ©Fy = 0; FCD- 800 sin 60° = 0

FCB = 400 lb (C):+ ©Fx = 0; FCB - 800 cos 60° = 0

60

600 lb

C B

800 lb

3 ft

4 ft

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FAB = 7.5 kN (T)+ c ©Fy = 0; 45FAB- 6 = 0

Determine the force in each member of the truss and state if

the members are in tension or compression Hint: The

vertical component of force at C must equal zero Why?

B

E A

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Each member of the truss is uniform and has a mass of

Remove the external loads of 6 kN and 8 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 the member

5FAB - 157.0 = 0

B

E A

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FCF= 8.768 kN 1T2 = 8.77 kN 1T2

6.20 - FCF sin 45° = 0+ c ©Fy = 0;

FEC = 6.20 kN 1C2

23.0 - 16.33¢ 5

234≤ - 8.854¢ 1

210≤ - FEC = 0+ c ©Fy = 0;

FEA= 8.854 kN 1C2 = 8.85 kN 1C2

FEA¢ 3

210≤ - 16.33¢ 3

234≤ = 0:+ ©Fx = 0;

FDC= 8.40 kN 1T2

16.33¢ 3

234≤ - FDC= 0:+ ©Fx = 0;

FDE= 16.33 kN 1C2 = 16.3 kN 1C2

234≤ - 14.0 = 0+ c ©Fy = 0;

Dx = 0:+ ©Fx = 0

23.0 - 4 - 5 - Dy = 0 Dy = 14.0 kN+ c ©Fy = 0;

4162 + 5192 - Ey132 = 0 Ey = 23.0 kN+ ©MD = 0;

Determine the force in each member of the truss and state

if the members are in tension or compression

E

D

C B

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8.768 sin 45° - 6.20 = 0+ c ©Fy = 0;

FBF = 6.20 kN 1C2

FBF - 4 - 3.111 sin 45° = 0+ c ©Fy = 0;

FBA = 3.111 kN 1T2 = 3.11 kN 1T2

2.20 - FBAcos 45° = 0:+ ©Fx = 0;

Ans.

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Determine the force in each member of the truss and state if

the members are in tension or compression Set

FFE = 0FFC = 21.21 = 21.2 kN (C)

Ay = 25 kN+ c ©Fy = 0; Ay - 10 - 15 = 0

Ax = 27.5 kN:+ ©Fx = 0; Ax - 27.5 = 0

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Determine the force in each member of the truss and state

if the members are in tension or compression Set

+ c ©Fy = 0; FEC - 20 = 0

FFE = 0FFC = 28.28 = 28.3 kN (C)

Ax = 30 kN:+ ©Fx = 0; Ax - 30 = 0

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Determine the force in each member of the truss and state if

the members are in tension or compression Set

22FAB = 0

Ax = 154.0 lb:+ ©Fx = 0; Ax - 307.9 sin 30° = 0

Ay = 333.3 lb+ c ©Fy = 0; Ay - 100 - 200 - 300 + 307.9 cos 30° = 0

RD = 307.9 lb+ ©MA = 0; 200(10) + 300(20) - RDcos 30°(30) = 0

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Determine the force in each member of the truss and state if

the members are in tension or compression Set

Ay = 666.67 lb+ c ©Fy = 0; Ay - 400 - 400 + 153.96 cos 30° = 0

RD = 153.96 lb+ ©MA = 0; -400(10) + RDcos 30°(30) = 0

F

D A

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SOLUTION

Method of Joints: In this case, the support reactions are not required for

determining the member forces

Note: The support reactions and can be determinedd by analyzing Joint A

using the results obtained above

Ay

Ax

FEA = 4.62 kN 1C2

FEA + 9.238 cos 60° - 9.238 cos 60° - 4.619 = 0:+ ©Fx = 0;

Ey - 219.238 sin 60°2 = 0 Ey = 16.0 kN+ c ©Fy = 0;

FBE = 9.24 kN 1C2 FBA = 9.24 kN 1T2

F = 9.238 kN

9.238 - 2F cos 60° = 0:+ ©Fx = 0;

FCE = 9.238 kN 1C2 = 9.24 kN 1C2

FCE sin 60° - 9.238 sin 60° = 0+ c ©Fy = 0;

FDE = 4.619 kN 1C2 = 4.62 kN 1C2

FDE - 9.238 cos 60° = 0:+ ©Fx = 0;

FDC = 9.238 kN 1T2 = 9.24 kN 1T2

FDCsin 60°- 8 = 0+ c ©Fy = 0;

Determine the force in each member of the truss State

whether the members are in tension or compression Set

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If the maximum force that any member can support is 8 kN

in tension and 6 kN in compression, determine the

maximum force P that can be supported at joint D.

SOLUTION

Method of Joints: In this case, the support reactions are not required for

determining the member forces

From the above analysis, the maximum compression and tension in the truss

member is 1.1547P For this case, compression controls which requires

Ans.

P = 5.20 kN1.1547P = 6

FBE sin 60° - FBA sin 60° = 0 FBE = FBA = F+ c ©Fy = 0;

211.1547P cos 60°2 - FCB= 0 FCB = 1.1547P 1T2:+ ©Fx = 0;

FCE = 1.1547P 1C2

FCE sin 60° - 1.1547P sin 60° = 0+ c ©Fy = 0;

FDE - 1.1547P cos 60° = 0 FDE = 0.57735P 1C2:+ ©Fx = 0;

FDCsin 60° - P = 0 FDC = 1.1547P 1T2+ c ©Fy = 0;

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Determine the force in each member of the truss and state if

the members are in tension or compression Hint: The

resultant force at the pin E acts along member ED Why?

E

3 m

4 m

3 m

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Each member of the truss is uniform and has a mass of

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 the member

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Determine the force in each member of the truss in terms of

the load P, and indicate whether the members are in tension

+ c ©Fy = 0; 2.404P¢ 1

23.25≤ + FBD¢ 1

21.25≤ - FBF¢ 1

21.25≤ = 01.00P - 0.4472FBF - 0.4472FBD = 0

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From the above analysis, the maximum compression and tension in the truss members

are 2.404P and 2.00P, respectively For this case, compression controls which requires

P = 1.25 kN2.404P= 3

If the maximum force that any member can support is 4 kN

in tension and 3 kN in compression, determine the maximum

force P that can be applied at joint B Take d = 1 m

A

B

C D

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FFB = = 1.41 P (T)

:+ ©Fy = 0; FFB a

2b - = 022PP

Determine the force in each member of the double scissors

truss in terms of the load P and state if the members are in

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FCE = 0.577 P (T)+ c ©Fy = 0; 0.577 P sin 60° - FCE sin 60° = 0

+ c ©Fy = 0; P

2 - P + Ay = 0

Dy = P2+ ©MA = 0; -P(L) + Dy (2 L) = 0

Determine the force in each member of the truss in terms of

the load P and state if the members are in tension or

D

L

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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

FCE = 1.1547W = 1.15 W (T)

+ c ©Fy = 0; 2.887W sin 60° - 3

2W - FCE sin 60° = 0

FDE = 1.44 W (T):+ ©Fx = 0; 2.887W cos 60° - FDE = 0

FCD = 2.887W = 2.89 W (C)+ c ©Fy = 0; 72W - W - FCD sin 60° = 0

Dy = 7

2W+ ©MA = 0; -32WaL2b - 2 W(L) - 32Wa32Lb - W(2 L) + Dy(2 L) = 0

D

L

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Determine the force in each member of the truss in terms of

the external loading and state if the members are in tension

FCA = cot 2u + 1cos u - sin ucot 2uP+ c ©Fy = 0; FCDsin 2u - FCAsin u = 0

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

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

FBA = P csc2u (C)+ c ©Fy = 0; FBA sin 2u - P = 0

A L

u

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The maximum allowable tensile force in the members of the

maximum magnitude P of the two loads that can be applied

to the truss Take and L= 2 m u = 30°

FCA = 2.732 P (T)

FCD = a tan30° + 1

23 cos 30° - cos 60°b P = 1.577 P (C):+ ©Fx = 0; P tan30° + P + FCDcos 60° - FCA cos 30° = 0

FCA = FCDasin 60°

sin 30°b = 1.732 FCD+ c ©Fy = 0; -FCA sin 30° + FCD sin 60° = 0

FBC = P tan30° = 0.57735 P (C):+ ©Fx = 0; FAB sin 30° - FBC= 0

FBA= cos 30°P = 1.1547 P (C)+ c ©Fy = 0; FBA cos 30° - P = 0

u

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Determine the force in members HG, HE, and DE of the

truss, and state if the members are in tension or compression

SOLUTION

Method of Sections: The forces in members HG, HE, and DE are exposed by

cutting the truss into two portions through section a–a and using the upper portion

of the free-body diagram, Fig a From this free-body diagram, and can be

obtained by writing the moment equations of equilibrium about points E and H,

respectively can be obtained by writing the force equation of equilibrium along

FDE = 3375 lb (C)

FDE(4) - 1500(6) - 1500(3) = 0 + ©MH = 0;

FHG = 1125 lb (T)

FHG(4) - 1500(3) = 0 + ©ME = 0;

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Determine the force in members CD, HI, and CJ of the truss,

and state if the members are in tension or compression

SOLUTION

Method of Sections: The forces in members HI, CH, and CD are exposed by cutting

the truss into two portions through section b –b on the right portion of the free-body

diagram, Fig a From this free-body diagram, and can be obtained by writing

the moment equations of equilibrium about points H and C, respectively. can be

obtained by writing the force equation of equilibrium along the y axis.

FHI = 6750 lb (T)

FHI(4) - 1500(3) - 1500(6) - 1500(9) = 0 + ©Mc = 0;

FCD = 3375 lb (C)

FCD(4) - 1500(6) - 1500(3) = 0 + ©MH = 0;

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Determine the force developed in members GB and GF of

the bridge truss and state if these members are in tension or

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Determine the force in members EC, EF, and FC of the

bridge truss and state if these members are in tension or

compression

SOLUTION

Support Reactions: Applying the moment equation of equilibrium about point A by

referring to the FBD of the entire truss shown in Fig a,

Method of Sections: Consider the FBD of the right portion of the truss cut through

sec a–a, Fig b, we notice that and can be obtained directly by writing

moment equation of equilibrium about joint C and force equation of equilibrium

along y-axis, respectively.

FEF = 728.57 lb (C) = 729 lb (C)

728.57(10) - FEF(10) = 0+ ©MC = 0;

FFC

FEF

ND (28) - 600(10) - 800(18) = 0 N D = 728.57 lb+ ©MA = 0;

600 lb

B

D A

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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

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

H I

J K

L

F E

D C

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Determine the force in members EI and JI of the truss

which serves to support the deck of a bridge State if these

members are in tension or compression

H I

J K

L

F E

D C

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Determine the force in member GJ of the truss and state if

this member is in tension or compression

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Determine the force in member GC of the truss and state if

this member is in tension or compression

+ c ©Fy = 0; -1000 + 2(2000 cos 60°) - FGC = 0

FGC = 1.00 kip (T):+ ©Fx = 0; FHG = 2000 lb

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Determine the force in members BC, HC, and HG After

the truss is sectioned use a single equation of equilibrium

for the calculation of each force State if these members are

Ay = 8.25 kN+ ©ME = 0; -Ay(20) + 2(20) + 4(15) + 4(10) + 5(5) = 0

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Determine the force in members CD, CF, and CG and state

if these members are in tension or compression

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FFB = 692.82 lb 1T2 = 693 lb 1T2

FFB sin 60°1102 - 800110 cos230°2 = 0+ ©MA = 0;

FGF = 1800 lb 1C2 = 1.80 kip 1C2

FGF sin 30°1102 + 800110 - 10 cos230°2 - 11001102 = 0+ ©MB = 0;

Ax = 0:+ ©Fx = 0;

2Ay - 800 - 600 - 800 = 0 Ay = 1100 lb+ c ©Fy = 0;

Determine the force in members GF, FB, and BC of the

Fink truss and state if the members are in tension or

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Determine the force in members FE and EC of the Fink truss

and state if the members are in tension or compression

FFE = 1.80 kip (C)

1100(10) - 800(10 - 7.5) - (FFE sin30°)(10) = 0+ ©MC = 0;

2By - 800 - 600 - 800 = 0; By = 1100 lb+ c ©Fy = 0;

800 lb

600 lb

30 60

30 60

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Determine the force in members IC and CG of the truss

and state if these members are in tension or compression

Also, indicate all zero-force members

SOLUTION

By inspection of joints B, D, H and I,

AB, BC, CD, DE, HI, and GI are all zero-force members. Ans

FIC = 5.625 = 5.62 kN (C)+ ©MG = 0; -4.5(3) + FICa35b(4) = 0

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