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1. Determine the Zero-Force Members in the plane truss.
1
2. Determine the forces in members FG, CG, BC, and EF for the loaded crane truss.
Use the Method of Joints.
3. Determine the forces in members CG and GH of the symmetrically loaded truss. Indicate whether the members work in tension or compression.
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. Calculate the forces in members DE, GJ and DG of the simple truss. State
whether they are in tension or compression.
6
6. 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
7
Ax=80 kN
Ay=60 kN
I. Cut
GF
LM
GM
FL
I. Cut (Left Side)
8
7. 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
9
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
10
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
11
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
+
12
8. Determine the force acting in member JI.
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
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
From equilibrium of whole truss;
kNAAF xxx 80350
Reactions at the supports:
By
Ax
Ay
kNABAF yyyy 25.23010532040
+ kNBBM yyA 75.180651631210864563)16(0
I. Cut
2. Cut
Joint D
DE
20 kN
a
DCDEFx 0
CkNDCDCFy 48.2802083
320
54.8
22
D
aDC
Joint C
3 kN
C 5 kN
DC
a
CHBC
)(68.21054.8
850
48.28
TkNCHDCCHFx
CkNBCDCBCFy 13054.8
330
48.28
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
B
D
C
HG
J I
N P
4 m 4 m
3 m
3 m
3 kN
5 kN
10 kN 18.75 kN=By
I. Cut (Right Side) DC
TkNPN
PNDC
M H
83.3
0)6()4(54.8
3)4(75.18)4(3
0
48.28
+GH
JN
PN
a
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
B
J I
P
4 m
3 m
10 kN 18.75 kN=By
2. Cut (Right Side)
BC
CkNHIHIPNBC
M J
904.00)4(32
3)3()2(10)6()6(75.18
0
2283.313
+
JP
PN
JI
HI
TkNJPJPHIBC
Fy
01.601032
3
32
375.18
0
2222904.013
CkN.JI
JPHIJIPNF...
x
6656
032
2
32
20
22016
229040833
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 kN5 kN
10 kN5 kN
4 kN
3 kN
9. 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
18
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 19
10. If it is known that the center pin A supports one-half of the vertical loading
shown, determine the force in member BF.
20
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
21
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.
22
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
23
11. Determine the forces in members DE, EI, FI, and HI of the arched roof truss.
24
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:
25
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
26
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
27
12. 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
FMLFMN
)(605.3064
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
13. 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
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
14. 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
