1. Zero-Force Members EF KL - Dokuz Eylül Üniversitesi · Determine the Zero-Force Members in the...
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Transcript of 1. Zero-Force Members EF KL - Dokuz Eylül Üniversitesi · Determine the Zero-Force Members in the...
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
