• Author / Uploaded
• Howard Johann Gomez V

• Categories
• Braguero
• Diseño arquitectonico
• Ingeniería
• Elementos arquitectónicos
• Edificios y estructuras de transporte

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1. Determine the Zero-Force Members in the plane truss. Also find the forces in members EF, KL and GL for the Fink truss shown by the Method of Joints.

2. Determine the force in member BE by the Method of Section.

3. Determine the forces in members BC and FG.

FCJ

Cut FBC

FFJ FG

4. Determine the forces in members FG, CG, BC, and EF for the loaded crane truss.

5. If it is known that the center pin A supports one-half of the vertical loading shown, determine the force in member BF.

Joint A

AB

DE

AF

DF

BF AF

Ay=26 kN

I. Cut

Hy=13 kN

DE DF BF AF

Hy=13 kN

6. a) By inspection, identify the zero-force members in the truss. b) Find the forces in members GI, GJ and GH.

H 0.125 m

J

F

0.125 m

L

D

G

B

0.8 m

M

E

0.125 m

I

0.125 m

K

C

0.25 m

450 N N

A 0.6 m

0.3 m

0.3 m

0.3 m

0.3 m

0.6 m

I. Cut H 0.125 m

HJ J

F

0.125 m

L

D

GJ G

B

0.8 m

M

GI

E

0.125 m

I

0.125 m

K

C

0.25 m

900 N

A 0.6 m

Ay=450 N

0.3 m

0.3 m

0.3 m

N 0.3 m

0.6 m

Ny=450 N

7. The truss shown consists of 45° triangles. The cross members in the two center panels that do not touch each other are slender bars which are incapable of carrying compressive loads. Determine the forces in members GM and FL.

I. Cut

Ax

Ay

By

8. The hinged frames ACE and DFB are connected by two hinged bars, AB and CD, which cross without being connected. Compute the force in AB.

9. Determine the force acting in member JI. 20 kN

4 kN

D F

3 kN

3 kN

G

H

3m 5 kN

E

C L

K

J

3m

I B

A M 5 kN 4m

P

N 4m

10 kN 4m

4m

3m

10. Determine the force acting in member DK.

Ux Uy

Vy

I. Cut

Uy=15 kN

III. CutII. Cut

Vy=20 kN

11. Determine the forces in members ME, NE and QG.

I. Cut

II. Cut

FDE

FEK FME

FMB FLB

FFQ FFG

12. In the truss system shown determine the forces in members EK, LF, FK and CN, state whether they work in tension (T) or compression (C). Crossed members do not

touch each other and are slender bars that can only support tensile loads. 4 kN 10 kN

6 kN

F

E

H G

2m

B

C

D

J 2m

N K

L

2m A 3m

20 kN

M

P

3m

4m

4m

4m

4m

Radii of pulleys H, F and K 400 mm

4 kN

By

10 kN

6 kN

F

E

10 kN

10 kN

H

G 2m

B

C

D

10 kN

Bx

10 kN

2m

20 kN

N K

L

2m

Ax

J

A 3m

10 kN

M

P 3m

4m

4m

4m

4m

Radii of pulleys H, F and K 400 mm

1st cut 4 kN

By 2m

B

10 kN

6 kN

E

C

10 kN

10 kN

H

G

D

2m

10 kN

FEK

N L

2m A 3m

F

FFL

Bx

Ax

FEF

FKL K

10 kN

J

20 kN

10 kN

M

P 3m

4m

4m

4m

4m

Radii of pulleys H, F and K 400 mm

2nd cut 4 kN

By 2m

B

10 kN

6 kN

C

E

FEF

20 kN

FJK K

L A 3m

J

10 kN

N

2m

H

10 kN

FFK

2m

10 kN

10 kN G

FFL

D

Bx

Ax

F

10 kN

M

P 3m

4m

4m

4m

4m

Radii of pulleys H, F and K 400 mm

4 kN

By

3rd cut

6 kN

10 kN

F

E

10 kN

10 kN

H

G 2m

B

C FCD

Bx

D

10 kN 10 kN

FDN

2m

J

20 kN

N

FMN

2m

Ax

A 3m

P FPM 3m

K

L

10 kN

M 4m

4m

4m

4m

Radii of pulleys H, F and K 400 mm

4 kN

By 2m

6 kN

10 kN

F

E

10 kN

4th cut B

C FCD

2m

FCN FPN

2m A 3m

P FPM 3m

H

G D

10 kN

Bx

Ax

10 kN

10 kN

J

20 kN

N K

L

10 kN

M 4m

4m

4m

4m

Radii of pulleys H, F and K 400 mm

13. Determine the forces in members ON, NL and DL.

Ax

Ay

Iy

From equilibrium of whole truss;

Fx  0 

Ax  6 kN

M A  0 

Ay (18)  6(2)  2(15)  4(9)  2(6)  2(3)  0  Ay  4 kN

Fy  0 

Ay  I y  10  0 

I y  6 kN

I.cut M C  0 

A (6)  6(2)  2(3)  FON y 4 kN

 FON  9.014 kN

I.cut FON

FOC

FBC

4 4 6

(Compression)

2

2

2  FON

6 4 6 2

2

(3)  0

Joint M 4 kN

FMN

FML

6

Fx  0 

 FMN

Fy  0 

4  2 FMN

 FML

42  62 4

4 6 2

2

6 42  62

0 

0



FMN  FML

FMN  FML  3.605 kN (C )

II.cut M D  0 

A (9)  2(6)  6(2)  FMN  y

3.605

4 kN

 FNL  4.5 kN Fy  0 

4 4 6 2

2

3  F MN

3.605

6 4 6 2

2

4  FNL (4)  0

(C )

Ay  2  FMN

4 4 6 2

2

 FDL

4  0 5

II.cut FMN FNL

FDL FDE

FDL  0 ( Zero  force member )

20 kN

14. Determine the forces in members HG and IG.

II.cut I.cut

20 kN 20 kN 20 kN

20 kN

20 kN 20 kN

20 kN

II.cut I.cut

FCD

FCD FHG

FHG FGI FBA

FHI

20 kN 20 kN FGJ

20 kN

20 kN

20 kN 20 kN

20 kN I.cut MG=0

FCD=54.14 kN (T)

MA=0

FHG=81.21 kN (C)

II.cut

I.cut Fx=0

FGI=18.29 kN (T)

C

2 kN

2 kN

2 kN

D

E

F

5 kN G

3 4

1 kN 4m O

N

H

B M

4m A

I L 2 kN 3m

J

K

2 kN

2 kN 3m

3m

3m

15. Determine the forces in members EF, NK and LK.

C

2 kN

2 kN

2 kN

3 kN

D

E

F

G

4 kN

From the equilibrium of whole truss

1 kN 4m

I. Cut N

Top Part

O

M B

FMN

FBN

FHO

FMO

FHI A

I L

Ay

2 kN 3m

J

K

2 kN

2 kN 3m

3m

are determined.

H

FBA Ax

Ax, Ay and Iy

3m

Iy

4m

I. Cut MH=0

FAB is determined

C

2 kN

2 kN

D

E

3 kN

2 kN

FEF

G

F

4 kN

1 kN

FMF

II. Cut Top Part

4m

II. Cut

N

O

M

B

FMN

FBN

MM=0

H

FMO

FBA

4m A

I L 2 kN 3m

J

K

2 kN

2 kN 3m

3m

3m

FEF and FMF are determined

C

2 kN

2 kN

D

E

3 kN

2 kN

FEF

G

F

4 kN

1 kN

FMF N

FMO

M

B

4m

O

H

III. Cut 4m

FNK A

I L 2 kN 3m

K

J

2 kN

2 kN

FLK 3m

III. Cut Left Side

3m

3m

MN=0 FLK and FNK are determined

10 2

G

kN

1m H

I

F

P

E

10 2 kN

10 2 kN

1m N

M

O

1m

L

J

K

D

C

25 2 kN

20 2 B

A

2m

kN

1m

1m

2m

16. Determine the forces in members KN and FC.

2m

kN

10 2

G

III. Cut

I. Cut

H

I

F

P

1m E

10 2 kN

10 2 kN

1m N

M J

K

O

1m D

C L

II. Cut 25 2 kN

20 2

Ax

B

A

2m

1m

Ay

2m

1m

By

kN

2m