Mechanical Engineering Chapter 6 Homework Start at joint C and then proceed to join D

481

SOLUTION

Method of Joints. Start at joint C and then proceed to join D.

Joint C. Fig. a

6–1.

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.

CB

A

D

1.5 m

2 m

P1

P2

Ans:

482

SOLUTION

6–2.

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.

CB

A

D

1.5 m

2 m

P1

P2

Ans:

FCB =0

483

6–3.

SOLUTION

Joint A:

Joint B:

Joint D:

+c©Fy=0; 4

5(FAC)–12

13 (130) =0

Determine the force in each member of the truss. State if

the members are in tension or compression.

3ft3ft3ft

12

5

13

130 lb

AB

C

E

D

F

4 ft 4 ft

484

*6–4.

SOLUTION

c

Joint A:

Joint B:

Joint F:

Joint H:

:

+©Fx=0; –FED –3.5 +12.12 cos 21.80° =0

+c©Fy=0; 1.5 –FAl sin 21.80° =0

+©ME=0;

Ay(40) +1.5(4) +2(12) –3(10) –3(20) =0Ay=1.5 kip

Determ

i

ne t

h

e force

i

n eac

h

mem

b

er of t

h

e truss an

d

state

if the members are in tension or compression. 2kip

1.5 kip

4ft

10 ft 10 ft 10 ft

3kip

10 ft

AB

I

H

G

F

CD

E

8ft

485

*6–4. Continued

Joint C:

J

oint G:

:

+©Fx=0; –FCI cos 21.80° –4.039 cos 21.80° –3.75 +7.75 =0

Ans:

FAl =4.04 kip (C)

FAB =3.75 kip (T)

FEF =12.1 kip (C)

FED =7.75 kip (T)

FBC =3.75 kip (T)

FDC =7.75 kip (T)

FFC =4.04 kip (C)

FFG =8.08 kip (C)

486

6–5.

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 Cand A, and then proceed to analyze of joint B.

Joint A:From the free-body diagram in Fig. c, we can write

0.875 –F

AD a3

y=0;

N

C =3.125 kN

N

C (2 +2) –4(2) –3(1.5) =0+©MA=0;

u=0°

AC

B

D

2 m

4 kN

3 kN

2 m

1.5 m

u

487

6–6.

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

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

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

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

Joint A:From the free-body diagram in Fig. c, we can write

Joint B:From the free-body diagram in Fig. d, we can write

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

solution obtained above.

u=30°

AC

B

D

2 m

4 kN

3 kN

2 m

1.5 m

u

488

6–7.

489

6–7. Continued

Ans:

FDE =16.3

kN (C)

FDC =8.40

kN (T)

FEA =8.85

FEC =6.20

490

*6–8.

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 Cand E.

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

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

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

F

CE –900 = 0©F

x=0;

:

+

B

E

D

A

C

600 N

900 N

4 m

Ans:

FDE =1.00 kN (C)

FDC =800 N (T)

FCE =900 N (C)

FCB =800 N (T)

FEB =750 N (T)

FEA =1.75 kN (C)

491

SOLUTION

Support Reactions. Referring to the FBD of the entire truss shown in Fig. a,

6(4) +3(8) – A

x=

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

sequence of joints A, D, B and C.

Joint A. Fig. b

+

c

ΣF

y

=0;

4.00 –FAE

a

=0

6–9.

Determine the force in each member of the truss and state

if the members are in tension or compression. Set

P1 = 3 kN, P2 = 6 kN.

AD

E

B C

P1P2

4 m4 m4 m

6 m

492

6–9. Continued

+

c

Σ

FDC =5.00 kN (T)

Joint B. Fig. d

+

c

Σ

Fy=0; FBE

a

3

210 b

–3=0 F

BE =2

10 kN (T)

=

3.16 kN (T) Ans.

FBC =3.00 kN (T)

a

3

210 b

Ans:

FAE =5.66 kN (C)

FAB =4.00 kN (T)

493

SOLUTION

Support Reactions. Referring to the FBD of the entire truss shown in Fig. a,

a+

ΣMA=0;

ND(12) –6(4) –9(8) =0

ND=8.00 kN

9(4) +6(8) – A

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

sequence of joints A, D, B and C.

Joint A. Fig. a

+

c

ΣF

2

Joint D. Fig. c

+

c

ΣF

y=

0;

8.00 –FDE

=0 F

DE =

8

2

2 kN (C)

=

11.3 kN (C) Ans.

6–10.

Determine the force in each member of the truss and state

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

P2 = 9 kN.

AD

E

B C

P1P2

4 m4 m4 m

6 m

494

Joint B. Fig. d

+

c

ΣF

y

=0;

FBE

a

3

210 b

–6=0 F

BE =

2

10 kN (T)

=

6.32 kN (T) Ans.

Joint C. Fig. e

+

c

Σ

Fy=0; FCE

210 b

–9=0 F

CE =

3

2

10 kN

=

9.49 kN (T) Ans.

6–10. Continued

Ans:

495

6–11.

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

state if the members are in tension or compression.

SOLUTION

Joint A:

Joint L:

Joint C:

Joint K:

Joint J:

Due to Symmetry

Ans.

F

AL =F

GH =F

LK =F

HI =28.3 kN (C)

23.57 sin 45° –F

JI sin 45° =0©F

x=0;

:

+

10 sin 45° –F

KD cos (45° –26.57°) =0R+©F

x–0;

F

+

F

LC =0R+©F

x=0;

F

AL =28.28 kN (C)

20 –F

AL sin 45° =0+c©F

y=0;

A

BCDEF

G

H

I

J

K

L

2 m

2 m 2 m

2 m 2 m 2 m

2 m

496

*6–12.

SOLUTION

Joint D:

Joint C:

Joint E:

Ans.

+Q©Fy¿=0; FEB cos u=0FEB =0

+Q©Fx¿=0; FCB –500 sin 75.96° –125 sin 14.04° =0

+c©Fy=0; FDE sin 75.96° –500 =0

Determine the force in each member of the truss and state

if the members are in tension or compression.

500 lb

3 ft

500 lb

C

B

E

D

6 ft

3 ft 3 ft

Ans:

FDE =515 lb (C)

FCD =125 lb (C)

FCB =515 lb (C)

497

6–13.

SOLUTION

Joint A:

Joint D:

:

+©Fx=0; 4

214 (FAD)–1

22FAB =0

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

the load Pand state if the members are in tension or

compression.

B

D

P

A

C

a a

a

3

—

4

1

—

4

Ans:

FCD =FAD =0.687P (T)

FCB =FAB =0.943P (C)

FDB =1.33P (T)

498

6–14.

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 Pthe truss can support.

a=10 ft

SOLUTION

Joint A:

Joint D:

¢

1

217

≤

+FDB =0

:

+©Fx=0; –800

¢

1

22

≤

+FAD

¢

4

217

≤

=0

B

D

P

A

C

a

3

—

4

1

—

4

6–15.

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.

B

D

A

C

a a

a

3

—

4

1

—

4

SOLUTION

Joint A:

S

+

ΣF

x=

0; –800

a1

22b

+FAD

a4

217 b

=0

+

c

Joint D:

+

c

ΣF

y

=0;

–848.5297 –583.0952(2)

+FDB =0

500

*6–16.

SOLUTION

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

determining the member forces.

Joint D:

Joint B:

Thus,

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

using the results obtained above.

A

y

A

x

F

BE

sin 60° –F

BA

sin 60° =0+c©F

y

=0;

F

CE

y

=0;

F

DE

–9.238 cos 60° =0:

+©F

x

=0;

F

DC

y

=0;

Determine the force in each member of the truss. State

whether the members are in tension or compression. Set

P=8 kN.

60••

4m 4m

B

ED

C

A

4m

P

Ans: