1. Determine the forces in members CG and GH of the symmetrically loaded

truss. Indicate whether the members work in tension {T} or compression {C}.

(4/32)

FCD

FCG

FHG

Fy

FBD of whole system

Ay

Ax=0

0 AM LFFLFL

LL yyy 5.1,1015,0)10()10(2

)7()3(

Cut, right side 0 CM

CG and GH

0)7(5.1)7(2

)4()3( LL

LFGH TLFGH

+

+

0 yF

0

05.1)7(2

)53

(

CG

CG

F

LL

LF

Cut

2. Determine the force in member DG of the loaded truss. (4/37)

TLF

FF

F

DG

DGGH

y

004.14sin

0

FDC

FBD of whole system

0 BM+

CutAy

Bx=0

By

LAALALLLLL yyy 3,2060,0)20()20()16()12()8()4(

Cut, left side 0 DM+

LF

LLLF

GH

GH

12.4

0)8(3)8()4()3(04.14cos

FGH

FFG

FDG

Joint G

FDH

FGH

y

xq

04.14,41

tan qq

3. Determine the forces in members BC and FG.(4/41)

BC and FG

CutFBC

FCJ FFJ

FFG

TNF

F

M

BC

BC

F

600

0)2(1200)4(

0

Cut, upper side

+

CNF

F

M

BC

FG

C

600

0)2(1200)4(

0

+

4. 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. (4/47)

AB and CD cross without being connected. AB=?

ABCD

CDAB

CDCDABAB

FF

FF

FFFF

99.2

99.195.5

0)3(sin)4(cos)5.1(sin)6(cos

0 EM+

I. Cut, left side

7.29,5.3

2tan

FAB

I.Cut

FCD

Ex

Ey

TkNF

CkNF

FF

FF

FF

FF

CD

AB

ABAB

AB

F

CD

ABAB

CDCD

AB

33.11

79.3

6099.179.17

99.16095.5

0)6(10)3(sin)4(cos

)5.1(sin)6(cos

99.2

0 FM+

II. Cut, right side

D

C

3.5 m

2 m

FAB

FCD

Fy

Fx

II.Cut

5. Find the force in member JQ for the Baltimore truss where all angles are

30°, 60°, 90° or 120°. Indicate whether the member works in tension {T} or

compression {C}. (4/55)

JQ = ?, all angles are 30°, 60°, 90° or 120°

NN=125 kNNA=75 kN

a

1.73a

1.73a

3.46a

JQ=? All angles are 30°, 60°, 90° or 120°

0 GM+

CkNFaaaaF WXWX 130,0)6(125)2(100)(100)46.3(

I. Cut, right side

FWX

NN=125 kN

FQX

NA=75 kN

1.73a

1.73aFGQ

FGH

FWX

FJQ

FHJ

0 GM+

CkNF

aaaFaF

JQ

JQWX

85.57

0)6(125)2(100)2(60sin)46.3(

II. Cut, right side

I.Cut

II.Cut

6. For the planar truss shown, determine the forces in members

CD, DE, CE and DG. State whether they work in tension {T} or

compression {C}.

A

9810 N

B

C D

E F

G

HI

J4 m 4 m

4 m

4 m

4 m

2 mK

r=0.4 m

r/2

CD, DE, CE and DG

A

9810 N

B

C D

E F

G

HI

J4 m 4 m

4 m

4 m

4 m

2 mK

r=0.4 m

r/2

4 m

A B

D

C

HGF

E

K J IL

NM P

4 m 4 m4 m

3 m

3 m

3 m

20 kN

3 kN

5 kN

10 kN5 kN

4 kN

3 kN

7. Determine the force acting in member JI for the truss shown. Indicate

whether the member works in tension {T} or compression {C}.

4 m

A B

D

C

HGF

E

K J IL

NM P

4 m 4 m4 m

3 m

3 m

3 m

20 kN

3 kN

5 kN

10 kN5 kN

4 kN

3 kN

JI=?

8. 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 left side

CkNF

FFAM

ON

ONON

kN

yC

014.9

0)3(64

62

64

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

2222

4

Ax

Ay

Iy

+

Joint M

4 kN

FMLFMN

)(605.3064

4240

064

6

64

60

22

2222

CkNFFFF

FFFFF

MLMNMNy

MLMNMLMNx

Ax

Ay

Iy

II.cut

FMN

FNL

FDL

FDE

)(0054

64

420

5.4

0)4(464

63

64

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

22

22

605.3

22

605.34

memberforceZeroFFFAF

CkNF

FFFAM

DLDLMNyy

NL

NLMNMN

kN

yD

II.cut

Ax

Ay

Iy

+

20 kN

9. Determine the forces in members HG and IG.

20 kN

I.cut

II.cut

20 kN

20 kN20 kN

20 kN

20 kN

20 kN

forces in members HG and IG

20 kN

I.cut

II.cut

FCD

20 kN

20 kN20 kN

20 kN

20 kN

20 kN

FHG

FGIFGJ

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

FHIFBA

forces in members HG and IG

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

C

B

A

D E F G

HO

L K J

I

N

1 kN

2 kN 2 kN2 kN 5 kN

2 kN 2 kN2 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

HO

L K J

I

N

1 kN

2 kN 2 kN2 kN 3 kN

4 kN

2 kN 2 kN2 kN

4 m

4 m

3 m 3 m 3 m 3 m

I. Cut

Top Part

Ay Iy

M

Ax

FHI

FHOFMOFMNFBN

FBA

forces in members EF, NK and LK

II. Cut

MM=0

FEF and FMF are determined

C

B

A

D E F G

HO

L K J

I

N

1 kN

2 kN 2 kN2 kN 3 kN

4 kN

2 kN 2 kN2 kN

4 m

4 m

3 m 3 m 3 m 3 m

II. Cut

Top Part M

FEF

FMF

FMOFMNFBN

FBA

forces in members EF, NK and LK

III. Cut

MN=0

FLK and FNK are determined

C

B

A

D E F G

HO

L K J

I

N

1 kN

2 kN 2 kN2 kN 3 kN

4 kN

2 kN 2 kN2 kN

4 m

4 m

3 m 3 m 3 m 3 m

III. Cut

Left Side

MFMO

FLK

FNK

FMF

FEF

forces in members EF, NK and LK

11. Determine the forces in members KN and FC.

kN

kN

kN

kN

kN

1 m

1 m

1 m

2 m

2 m1 m1 m2 m

A B

C D

O

E

G

P F

NM

I

JK

L

H

225

210

220

210 210

kN

kN

kN

kN

kN

1 m

1 m

1 m

2 m

2 m1 m1 m2 m

A B

C D

O

E

G

P F

NM

I

JK

L

H

225

210

220

210 210

I. Cut

II. Cut

III. Cut

ByAy

Ax

forces in members KN and FC