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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}.

(4/34)

Fy

FBD of whole system

Ay

Ax=0

0 AM LFFLFL

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

)7()3( +

LALLAFAF yyyxx 5.1035.10,00

FCD

FCG

FHG

Fy Ay

Ax=0

Cut, right side 0 CM

CG and GH

0)7(5.1)7(2

)4()3( LL

LFGH TLFGH +

0 yF

0

05.12

)53

(

CG

CG

F

LL

LF

Cut

CG and GH = ?

2. Determine the force in member DG of the loaded truss.

TLF

FF

F

DG

DGGH

y

004.14sin

0

FDC

FBD of whole system

0 BM+

Cut Ay

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

x q

04.14,41

tan qq

DG = ?

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

BC and FG

Cut FBC

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. In the planar truss shown, the crossed members in the center panel which cross without touching are slender tie rods incapable of supporting compression. Determine whether member DH or GE works in tension. Also determine the forces in members JI, JE, and EK. State whether they work in tension {T} or compression {C}.

A

B C D E F

G H

I

J

4 m 4 m 4 m

5 m

4 m

5 m

K

4 m

L

12 kN

8 kN

10 kN 20 kN

Determine whether member DH or GE works in tension. Also determine the forces in members JI, JE, and EK.

A

B C D E F

G H

I

J

4 m 4 m 4 m

5 m

4 m

5 m

K

4 m

L

12 kN

8 kN

10 kN 20 kN

6. 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 kN NA=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

1.73a FGQ

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

7. Determine the forces acting in members GE, HI and DI in the truss loaded as shown. State whether they work in tension {T} or compression {C}.

.

3

4 5 kN

C

D

J I

4 kN

R=0.4 m

4 m

4 m

3

4 5 kN

C

D

J I

4 kN

R=0.4 m

4 m

4 m

GE, HI, DI = ?

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

8. 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

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

JI=?

9. 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

FML FMN

)(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

10. 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

forces in members HG and IG

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 FIG=18.29 kN (T)

FCD

FHG

FHI FBA

forces in members HG and IG

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