This photo is balanced Symmetrically: (what is on the left is basically on the right)
Slayt 1 - DEUkisi.deu.edu.tr/emine.cinar/B16 Statics_Trusses - Problems.pdf · Determine the forces...
Transcript of Slayt 1 - DEUkisi.deu.edu.tr/emine.cinar/B16 Statics_Trusses - Problems.pdf · Determine the forces...
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