Dynamics 14th edition by r c hibbeler chapter 06

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

Determine the force in each member of the truss and state

if the members are in tension or compression Set

P1 = 20 kN, P2 = 10 kN

A D

Determine the force in each member of the truss and state

if the members are in tension or compression Set

P1 = 45 kN, P2 = 30 kN

A D

FDF = 230 lb (T):+ ©Fx = 0; FDF - 140 - 35(150) = 0

FDE = 120 lb (C)+ c ©Fy = 0; FDE - 45(150)= 0

FAC = 150 lb (C)+ c ©Fy = 0; 45(FAC) - 1213 (130) = 0

Determine the force in each member of the truss State if

12 5 13

FFG = 8.078 kip = 8.08 kip (C)+R©Fx¿ = 0; FFG + 3 sin 21.80° + 4.039 sin 46.40° - 12.12 = 0

FFC = 4.039 kip = 4.04 kip (C)+Q©Fy¿ = 0; FFCcos 46.40° - 3 cos 21.80° = 0

FEF = 12.12 kip = 12.1 kip (C)+ c ©Fy = 0; 4.5 - FEFsin 21.80° = 0

:+ ©Fx = 0; FAB- 4.039 cos 21.80° = 0 FAB= 3.75 kip (T)

FAl = 4.039 kip = 4.04 kip (C)+ c ©Fy = 0; 1.5 - FAlsin 21.80° = 0

Determine the force in each member of the truss and state

FGI = 5.924 kip = 5.92 kip (C)+Q©Fy¿ = 0; FGIcos 46.40° - 3cos 21.80° - 1.40 cos 21.80° = 0

FCG = 1.40 kip (T)+ c ©Fy = 0; FCG + 0.2692 sin 21.80° - 4.039 sin 21.80° = 0

FCI = 0.2692 kip = 0.269 kip (T):+ ©Fx = 0; -FCIcos 21.80° - 4.039 cos 21.80° - 3.75 + 7.75 = 0

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

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

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 havea

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

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

FDE¢ 5

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

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;

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,

FEB = 750 N (T)

- 900cos36.87° + FEBsin73.74° = 0R+ ©Fx¿ = 0;

FCB = 800 N (T)

800 - FCB = 0+ c ©Fy = 0;

Method of Joints We will carry out the analysis of joint equilibrium according to the

sequence of joints A, D, B and C.

Determine the force in each member of the truss and state

if the members are in tension or compression Set

Method of Joints We will carry out the analysis of joint equilibrium according to the

sequence of joints A, D, B and C.

Determine the force in each member of the truss and state

if the members are in tension or compression Set P1 = 6 kN,

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

state if the members are in tension or compression

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;

FLC = 0R+ ©Fx = 0;

FBL = 0+ c ©Fy = 0;

FCE = 0+a©Fy¿ = 0; FCEcos 39.09° + 125 cos 14.04° - 500 cos 75.96° = 0

:+ ©Fx = 0; FCD - 515.39 cos 75.96° = 0 FCD = 125 lb (C)

FDE = 515.39 lb = 515 lb (C)+ c ©Fy = 0; FDEsin 75.96° - 500 = 0

Determine the force in each member of the truss and state

if the members are in tension or compression

FCE = 0+a©Fy¿ = 0; FCEcos 39.09° + 125 cos 14.04° - 500 cos 75.96° = 0

:+ ©Fx = 0; FCD - 515.39 cos 75.96° = 0 FCD = 125 lb (C)

FDE = 515.39 lb = 515 lb (C)+ c ©Fy = 0; FDEsin 75.96° - 500 = 0

Determine the force in each member of the truss and state

if the members are in tension or compression

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

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

1

— 4

Ans:

F CD = FAD = 0.687P (T)

F CB = FAB = 0.943P (C)

F DB = 1.33P (T)

Members AB and BC can each support a maximum

compressive force of 800 lb, and members AD, DC, and BD

can support a maximum tensile force of 1500 lb If ,

determine the greatest load P the truss can support.

1

— 4

Ans:

Pmax = 849 lb

Members AB and BC can each support a maximum

compressive force of 800 lb, and members AD, DC, and BD

can support a maximum tensile force of 2000 lb If a = 6 ft,

determine the greatest load P the truss can support.

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

P = 8 kN

60•• 60••

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;

60°60°

Support Reactions Not required.

Method of Joints We will perform the joint equilibrium according to the sequence

Determine the force in each member of the truss and state

if the members are in tension or compression Set

Determine the force in each member of the truss and state

if the members are in tension or compression Set P1 = 8 kN,

Support Reactions Not required.

Method of Joints We will perform the joint equilibrium according to the sequence

Determine the force in each member of the truss and state

if the members are in tension or compression Set P1 = 9 kN,

Method of Joints: By inspecting joints D and F, we notice that members DE, DC

and FA are zero force members Thus

S+ ΣF x = 0; 10.125 - FBA = 0 FBA = 10.125 kN (T) = 10.1 kN (T) Ans.

Determine the force in each member of the truss and state

if the members are in tension or compression Set P1 =

30 kN, P2 = 15 kN

Solution

Support Reactions

a+ΣMA = 0; NC(6) - 15(3) - 30(4) = 0 NC = 27.5 kN

Method of Joints: By inspecting joints D and F, we notice that members DE, DC

and FA are zero force members Thus

FFB = = 1.41 P (T)

:+ ©Fy = 0; FFB a

2b - = 022PP

:+ ©Fx = 0; FFD - FFE - FFBa 1

22b = 0

+ c ©Fy = 0; Ay = P

Dy = P+ ©MA = 0; PaL3 b + Pa2L3 b - (Dy)(L) = 0

Determine the force in each member of the double scissors

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

Determine the force in each member of the truss in terms

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

FDB = 0.3333P = 0.333P (T)+ c ©Fy = 0; 2(0.3727P sin 26.57°) - FDB= 0

FDC = 0.3727P = 0.373P (C):+ ©Fx = 0; 0.3727P cos 26.57° - FDCcos 26.57° = 0

FAB= FAD = F = 0.373P (C):+ ©Fx = 0; 0.6667P - 2F cos 26.57° = 0 F = 0.3727P

+ c ©Fy = 0; FAB sin 26.57° - FAD sin 26.57° = 0 FAB= FAD = F

+ c ©Fy = 0; P - FED - 1.202P cos 33.69° = 0 FED = 0

FEC = 1.202P = 1.20P (T):+ ©Fx = 0; FECsin 33.69° - 0.6667P = 0

The maximum allowable tensile force in the members of the

truss is and the maximum allowable

maximum magnitude of the load P that can be applied to

the truss Take d = 2 m

1Fc2max = 3 kN

1Ft2max = 5 kN,

SOLUTION

Maximum tension is in member EC.

Maximum compression is in members AB, AD, DC, and BC.

Thus, the maximum allowable load is

FDB = 0.3333P = 0.333P (T)+ c ©Fy = 0; 2(0.3727P sin 26.57°) - FDB = 0

FDC = 0.3727P = 0.373P (C):+ ©Fx = 0; 0.3727P cos 26.57° - FDCcos 26.57° = 0

FAB = FAD = F = 0.373P (C):+ ©Fx = 0; 0.6667P - 2F cos 26.57° = 0 F = 0.3727P

+ c ©Fy = 0; FAB sin 26.57° - FAD sin 26.57° = 0 FAB= FAD = F

+ c ©Fy = 0; P - FED - 1.202P cos 33.69° = 0 FED = 0

FEC = 1.202P = 1.20P (T):+ ©Fx = 0; FECsin 33.69° - 0.6667P = 0

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

the external loading and state if the members are in tension

Support Reactions Not required.

Method of Joints: We will start the joint equilibrium analysis at joint C, followed by

joint D and finally joint A.

+cΣF y = 0; FDA cos 30° - F DB sin 30° - 2 = 0 (2)

Solving Eqs (1) and (2)

The maximum allowable tensile force in the members of the

truss is (Ft)max = 5 kN, and the maximum allowable

compressive force is (F c)max = 3 kN Determine the

maximum magnitude P of the two loads that can be applied

Support Reactions Not required.

Method of Joints: We will start the joint equilibrium analysis at joint C, followed by

joint D and finally joint A.

+cΣF y = 0; FDA cos 30° - FDB sin 30° - P = 0 (2)

Solving Eqs (1) and (2),

F DA = 2.3094P (C) FDB = 2.00P (T)

Joint C Fig c

S+ ΣFx = 0; 2.3094 P sin 30° - FAB = 0 FAB = 1.1547P (C)

+cΣF y = 0; NA - 2.3094 p cos 30° = 0 NA = 2.00P

By observation, members DA and DB are subjected to maximum compression and

tension, respectively Thus, they will reach the limit first

F DA = (FC)max; 2.3094P = 3 P = 1.299 kN = 1.30 kN (Control!) Ans.

F DB = (F C)max; 2.00P = 5 P = 2.50 kN

Ans:

P = 1.30 kN

Determine the force in members DC, HC, and HI of the

truss, and state if the members are in tension or

compression

SOLUTION

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

the free - body diagram of the truss, Fig a,

FHI = 42.5 kN (T)+ ©MC = 0; 70(3) - 32.5(2) - 40(1.5) - FHI(2) = 0

+ c ©Fy = 0; 57.5 - 40 - 50 + Ay = 0; Ay = 32.5 kN

Fy = 57.5 kN+ ©MA= 0; 40(1.5) + 30(3) + 40(2) - Fy(4) = 0

A

C G

H F

SOLUTION

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

the free - body diagram of the truss, Fig a,

FGH = 76.7 kN (T)+ ©ME = 0; -57.5(2) + FGH(1.5) = 0

+ c ©Fy = 0; 57.5 - 40 - 50 + Ay = 0; Ay = 32.5 kN

Fy = 57.5 kN+ ©MA= 0; 40(1.5) + 30(3) + 40(2) - Fy(4) = 0

Determine the force in members ED, EH, and GH of the

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

A

C G

H F

Ans:

F GH = 76.7 kN (T)

F ED = 100 kN (C)

F EH = 29.2 kN (T)

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;

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;

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

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

H I

J K

L

F E

D C

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

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

H I

J K

L

F E

D C

FGF = 12.5 kN (C)+ ©MD = 0; - 45FGF(1.5) - 2(2) + 9.5(2) = 0

+ ©MA = 0; Ey(8) - 2(8) - 5(6) - 5(4) - 5(2) = 0 Ey = 9.5kN

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

The Howe truss is subjected to the loading shown.

Determine the force in members GH, BC, and BG of the

truss and state if the members are in tension or compression

FBG = 6.01 kN (T)+ ©MA = 0; -5 (2) + FBGsin 56.31°(2)= 0

FGH = 12.5 kN (C)+ ©MB = 0; -7.5(2) + FGHsin 36.87°(2) = 0

Determine the force in members EF, CF, and BC, and state

if the members are in tension or compression

Support Reactions Not required.

Method of Sections FBC and FEF can be determined directly by writing the moment

equations of equilibrium about points F and C, respectively, by referring to the FBD

of the upper portion of the truss section through a–a shown in Fig a.