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.

Cut FBC

FCJ 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.

DE

DF

BF

AF

I. Cut

Joint A AB AF

Ay=26 kN

DE

DF

BF

AF

Hy=13 kN

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.

0.6 m 0.3 m 0.3 m 0.3 m 0.3 m 0.6 m

A

B

C

E

G

H

F

0.8 m

0.125 m

0.125 m

D

J

I

L

K

M

N

0.125 m

0.125 m

0.25 m 450 N

0.6 m 0.3 m 0.3 m 0.3 m 0.3 m 0.6 m

A

B

C

E

G

H

F

0.8 m

0.125 m

0.125 m

D

J

I

L

K

M

N

0.125 m

0.125 m

0.25 m

I. Cut

Ay=450 N Ny=450 N

900 N

HJ

GJ

GI

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.

Ax

Ay By

I. Cut

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.

4 m

A B

D

C

H G F

E

K J I L

N M P

4 m 4 m 4 m

3 m

3 m

3 m

20 kN

3 kN

5 kN

10 kN 5 kN

4 kN

3 kN

10. Determine the force acting in member DK.

Ux

Uy

Vy

Uy=15 kN Vy=20 kN

I. Cut II. Cut III. Cut

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

I. Cut II. Cut FDE

FME

FMB

FLB

FEK

FFQ

FFG

2 m

2 m

2 m

4 m 3 m 3 m 4 m 4 m 4 m

A

B C D

E F

G

N

M

L K

J

H

P

10 kN 6 kN

Radii of pulleys H, F and K 400 mm

4 kN

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.

20 kN

2 m

2 m

2 m

4 m 3 m 3 m 4 m 4 m 4 m

A

B C D

E F

G

N

M

L K

J

H

P

10 kN 6 kN

20 kN

4 kN

10 kN 10 kN

10 kN

10 kN

10 kN

Ax

By

Bx

Radii of pulleys H, F and K 400 mm

2 m

2 m

2 m

4 m 3 m 3 m 4 m 4 m 4 m

A

B C D

E F

G

N

M

L K

J

H

P

10 kN 6 kN

20 kN

4 kN

10 kN 10 kN

10 kN

10 kN

10 kN

Ax

By

Bx

1st cut

Radii of pulleys H, F and K 400 mm

FEF

FFL

FEK

FKL

2 m

2 m

2 m

4 m 3 m 3 m 4 m 4 m 4 m

A

B C D

E F

G

N

M

L K

J

H

P

10 kN 6 kN

20 kN

4 kN

10 kN 10 kN

10 kN

10 kN

10 kN

Ax

By

Bx

2nd cut

Radii of pulleys H, F and K 400 mm

FEF

FFL

FFK

FJK

2 m

2 m

2 m

4 m 3 m 3 m 4 m 4 m 4 m

A

B C D

E F

G

N

M

L K

J

H

P

10 kN 6 kN

20 kN

4 kN

10 kN 10 kN

10 kN

10 kN

10 kN

Ax

By

Bx

3rd cut

Radii of pulleys H, F and K 400 mm

FCD

FDN

FMN

FPM

2 m

2 m

2 m

4 m 3 m 3 m 4 m 4 m 4 m

A

B C D

E F

G

N

M

L K

J

H

P

10 kN 6 kN

20 kN

4 kN

10 kN 10 kN

10 kN

10 kN

10 kN

Ax

By

Bx

4th cut

Radii of pulleys H, F and K 400 mm

FCD

FCN

FPN

FPM

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

Ax

Ay Iy

kNIIAF

kNAAM

kNAF

yyyy

yyA

xx

60100

40)3(2)6(2)9(4)15(2)2(6)18(0

60

From equilibrium of whole truss;

FON

FOC

FBC

I.cut

I.cut

)(014.9

0)3(64

62

64

4)3(2)2(6)6(0

22224

nCompressiokNF

FFAM

ON

ONON

kN

yC

Joint M

4 kN

FML FMN

)(605.30

64

4240

064

6

64

60

22

2222

CkNFFFF

FFFFF

MLMNMNy

MLMNMLMNx

II.cut

FMN

FNL

FDL

FDE

)(005

4

64

420

)(5.4

0)4(464

63

64

4)2(6)6(2)9(0

22

22605.3

22605.34

memberforceZeroFFFAF

CkNF

FFFAM

DLDLMNyy

NL

NLMNMN

kN

yD

II.cut

20 kN

14. Determine the forces in members HG and IG.

20 kN

I.cut

II.cut

20 kN

20 kN 20 kN

20 kN

20 kN

20 kN

20 kN

I.cut

II.cut

FCD

20 kN

20 kN 20 kN

20 kN

20 kN

20 kN

FHG

FGI FGJ

I.cut MG=0 FCD=54.14 kN (T)

II.cut MA=0 FHG=81.21 kN (C)

I.cut Fx=0 FGI=18.29 kN (T)

FCD

FHG

FHI FBA

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

C

B

A

D E F G

H O

L K J

I

N

1 kN

2 kN 2 kN 2 kN 5 kN

2 kN 2 kN 2 kN

4 m

4 m

3 m 3 m 3 m 3 m

M

3

4

From the equilibrium of whole truss

Ax, Ay and Iy

are determined.

I. Cut

MH=0

FAB is determined

C

B

A

D E F G

H O

L K J

I

N

1 kN

2 kN 2 kN 2 kN 3 kN

4 kN

2 kN 2 kN 2 kN

4 m

4 m

3 m 3 m 3 m 3 m

I. Cut

Top Part

Ay Iy

M

Ax

FHI

FHO FMO FMN FBN

FBA

II. Cut

MM=0

FEF and FMF are determined

C

B

A

D E F G

H O

L K J

I

N

1 kN

2 kN 2 kN 2 kN 3 kN

4 kN

2 kN 2 kN 2 kN

4 m

4 m

3 m 3 m 3 m 3 m

II. Cut

Top Part M

FEF

FMF

FMO FMN FBN

FBA

III. Cut

MN=0

FLK and FNK are determined

C

B

A

D E F G

H O

L K J

I

N

1 kN

2 kN 2 kN 2 kN 3 kN

4 kN

2 kN 2 kN 2 kN

4 m

4 m

3 m 3 m 3 m 3 m

III. Cut

Left Side

M FMO

FLK

FNK

FMF

FEF

16. Determine the forces in members KN and FC.

kN

kN

kN

kN

kN

1 m

1 m

1 m

2 m

2 m 1 m 1 m 2 m

A B

C D

O

E

G

P F

N M

I

J K

L

H

225

210

220

210 210

kN

kN

kN

kN

kN

1 m

1 m

1 m

2 m

2 m 1 m 1 m 2 m

A B

C D

O

E

G

P F

N M

I

J K

L

H

225

210

220

210 210

I. Cut

II. Cut

III. Cut

By Ay

Ax