1. determine the zero-force members in the plane...

1. Determine the Zero-Force Members in the plane truss.

1

2. Determine the force in each member of the loaded truss. Make use of the symmetry of

the truss and of the loading. Use the Method of Joints.

2

3. Determine the force in member GM by the Method of Section.

3

4. Determine the forces in members BC and FG.

4

CutFBC

FCJ FFJ

FFG

FBC

FCJFFJ

FFG

1200 N

800 N C

CNFFM FGFGC 60002120040

Cut (Upper Side)

TNFFFF BCFGBCy 60000

600

+

EK, EF

Zero-Force Members:

5

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

25 kN

D

E

CF

GB

50 kN

25 kN25 kN

25 kN

25 kN

25 kN

25 kN

BGZero-Force Members:

6

25 kN

DE

CF

GB

50 kN

25 kN25 kN

25 kN

25 kN

25 kN

25 kN

1st Cut

1st Cut (Upper part) E

F

25 kN

D

50 kN

25 kN25 kN25 kN

25 kN

25 kN

BCFG

G

C

CG

00 CGFx+ CkNBCBCM F 1000)4()8(500

TkNFGBCFGFy 25050250100

Joint F: EF

FCF

25 kN

25 kN

FG

TkNEF

FGEF

Fy

7.45

045cos45cos2525

0

25

7

6. Calculate the forces in members DE, GJ and DG of the simple truss. State

whether they are in tension or compression.

8

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.

By=40 kN

Ax=80 kN

Ay=60 kN

From equilibrium of whole truss;

Reactions at the supports

9

Ax=80 kN

Ay=60 kN

I. Cut

GF

LM

GM

FL

I. Cut (Left Side)

10

8. If it is known that the center pin A supports one-half of the vertical loading shown,

determine the force in member BF.

11

From equilibrium of whole truss; 00 xx AF

Reactions at the supports

Center pin A supports one-half of the vertical loading.

kNAy 262

10284

Ay

Gy

Ax

Hy

Because of symmetry, kNHG yy 13

12

DE

DF

BF

AF

I. Cut

Hy=13 kN

DE

DF

BF

AF

Hy=13 kN

Ay

Gy

I. Cut (Right side)

There are four unknowns.

13

Joint AAB AF

Ay=26 kN

45o 45o

AFABAFABFx 045cos45cos0

I. Cut (Right side)

DE

DFBF

AF

Hy=13 kN

TkNBF

AFAFBF

M D

24.24

0)12(45sin)16(45cos)16()48(13)36(8)24(8)12(10

0

38.1838.18

+45o

CkNAFABAFABFy 38.1802645sin45sin0

A

D

14

9. Determine the forces acting in members DE, DI, KJ, AJ.

JK

A

L

M

DCB

I

E F

G

H

20 kN

37o

4 m 4 m 4 m

3 m

3 m

6 m

15

JK

A

L

M

DCB

I

E F

G

H

20 kN

37o

4 m 4 m 4 m

3 m

3 m

6 m

kNAATF xxx 4037cos200

From equilibrium of whole truss; kNAAF yyy 12037sin200

Reactions at the supports

+

EF, FG

Zero-Force Members:

Ax

Ay

T

kNTTM A 200)12(37sin20)6(37cos20)12(0

16

JK

A

L

M

DCB

I

E F

G

H

20 kN

37o

4 m 4 m 4 m

3 m

3 m

6 m

J

E

20 kN37o

CkNAJ

AJA

F

x

x

5

086

8

0

22

Joint A:

Ax

Ay

T

AL

Ay=12 kN

AJ

Ax=4 kN

1st Cut (Right side)

AJ

KJ

DEEI

IJ

1st Cut

+

TkNDEDE

M J

80437sin206

0

17

JK

A

L

M

DC

I

E F

G

H

20 kN

37o

4 m 4 m 4 m

3 m

6 m

E

20 kN

37o

3 mI

JK

Ax

Ay

T

2nd Cut (Right side)

AJ

KJ

DE

DI

2nd Cut

+

TkNKJAJKJDE

M I

120437sin20337cos20337cos33

0

58

KI

TkNDIDIDIAJDE

M K

5.70437sin337cos837sin20437sin6

0

58

+

18

10. Determine the forces in members DE, EI, FI, and HI of the arched roof truss.

19

Gx

GyAy

kNAAGF

kNGGM

yyyy

yyA

15001007522520

1500)36(25)30(75)20(100)10(75)4(25)40(0

From equilibrium of whole truss;

00 xx GF

Reactions at the supports

kNGA yy 150

Because of symmetry of the truss:

+

BK, HF

Zero-Force Members:

20

Ay=150 kN Gy=150 kN

1st Cut (Right side)

1st Cut

Gy=150 kN

FI

25 kN

F

EF

HI

H

G

+

I

CkNEFEFM I 48.31501615012251264

40

22

TkNHIHIEFFy 93.750251501614

14

64

40

222248.315

TkNFIHIFIEFFx 356.20501614

16

64

60

2293.75

2248.315

21

Ay=150 kN Gy=150 kN

2nd Cut (Right side)

2nd Cut

Gy=150 kN

IF

E

DE

IH

G

+

I

EI25 kN

F

75 kN

CkNDEDEDE

M I

008.29701615067512254103

106

103

3

0

2222

TkNEI

HIDEEIFy

4.26

015025751614

14

103

3

64

40

2293.75

22008.297

22

22

11. 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.

B

A

3.5 m

2 m

a

o.

.tan

7429

53

2

a

a

23

I. Cut (Left Side)

AB

CD

Ex

Ey

a a

AB

CDB

A

3.5 m

2 m

a

ABCD

CDCDABABM E

3

03sin4cos5.1sin6cos0

aaaa

o.

.tan

7429

53

2

a

a

I. Cut (Right Side)

I. Cut (Left Side)

I. Cut (Right Side)

05.1sin6cos3sin4cos6100 aaaa CDCDABABM F

+

+

CkNABABCD 78.3098.195.560 24