1. determine the force in members gm and - ki??isel...
TRANSCRIPT
1. Determine the force in members GM and MN.
2. Determine the forces in members BC and FG.
CutFBC
FCJ FFJFFG
FBCFCJ FFJ
FFG
1200N
800N C
CNFFM FGFGC 60002120040
Cut (UpperSide)
TNFFFF BCFGBCy 60000
600
+
EK,EF
Zero‐ForceMembers:
3. If it is known that the center pin A supports one-half of the vertical
loading shown, determine the force in member BF.
Fromequilibriumofwholetruss; 00 xx AF
Reactionsatthesupports
Center pin A supports one-half of the vertical loading. kNAy 26
210284
Ay
GyAx
Hy
Because of symmetry, kNHG yy 13
DE
DF
BF
AF
I.Cut
Hy=13kN
DE
DF
BF
AF
Hy=13kN
Ay
Gy
I.Cut (Rightside)Therearefourunknowns.
Joint AAB AF
Ay=26kN
45o 45o
AFABAFABFx 045cos45cos0
I.Cut (Rightside)
DE
DF
BF
AF
Hy=13kN
TkNBF
AFAFBFM D
24.24
0)12(45sin)16(45cos)16()48(13)36(8)24(8)12(100
38.1838.18
+
45o
CkNAFABAFABFy 38.1802645sin45sin0
4. Determine the forces in members FG, CG, BC, and EF for the loaded crane truss.
25kN
DE
CF
G B
50kN
25kN25kN25kN
25kN
25kN
25kN
BGZero‐ForceMembers:
25kN
DE
CF
G B
50kN
25kN25kN25kN
25kN
25kN
25kN1st Cut
1st Cut(Upperpart) E
F
25kND
50kN
25kN25kN25kN
25kN
25kN
BCFGG
CCG
00 CGFx+ CkNBCBCM F 1000)4()8(500
TkNFGBCFGFy 25050250100
JointF: EF
FCF
25kN
25kN
FG
TkNEF
FGEF
Fy
7.45
045cos45cos2525
0
25
5. Determine the forces in members DE, EI, FI, and HI of the arched roof truss.
GxGyAy
kNAAGF
kNGGM
yyyy
yyA
15001007522520
1500)36(25)30(75)20(100)10(75)4(25)40(0
Fromequilibriumofwholetruss;
00 xx GF
Reactionsatthesupports
kNGA yy 150
Becauseofsymmetryofthetruss:
+
BK,HF
Zero‐ForceMembers:
Ay=150kN Gy=150kN
1st Cut(Rightside)
1st Cut
Gy=150kN
FI25kN
F
EF
HIH
G
+
I
CkNEFEFM I 48.31501615012251264
4022
TkNHIHIEFFy 93.750251501614
1464
402222
48.315
TkNFIHIFIEFFx 356.20501614
1664
6022
93.7522
48.315
Ay=150kN Gy=150kN
2nd Cut(Rightside)
2nd Cut
Gy=150kN
IF
25kN
EDE
IH
G
+
IEI
25kN
F
75kN
CkNDEDEDE
M I
008.29701615067512254103
106103
30
2222
TkNEI
HIDEEIFy
4.26
015025751614
14103
364
4022
93.7522
008.29722
6. Determine the forces acting in members DE, DI, KJ, AJ.
JK
A
L
M
DCB
I
E F
G
H20 kN
37o
4 m 4 m 4 m
3 m
3 m
6 m
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
Fromequilibriumofwholetruss; kNAAF yyy 12037sin200
Reactionsatthesupports
+
EF,FG
Zero‐ForceMembers:
Ax
Ay
T
kNTTM A 200)12(37sin20)6(37cos20)12(0
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
80
22
JointA:
Ax
Ay
TAL
Ay=12kN
AJ
Ax=4kN
1st Cut(Rightside)
AJ
KJ
DEEI
IJ
1st Cut
+
TkNDEDEM J
80437sin2060
JK
A
L
M
DC
I
E F
G
H
20 kN
37o
4 m 4 m 4 m
3 m
6 mJ
E
20 kN37o
3 mI
JK
Ax
Ay
T2nd Cut(Rightside)
AJ
KJ
DE
DI
2nd Cut
+
TkNKJAJKJDEM I
120437sin20337cos20337cos330
58
KI
TkNDIDIDIAJDEM K
5.70437sin337cos837sin20437sin60
58
+
7. 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.
I.Cut (Left Side)
AB
CD
Ex
Ey
ABCDB
A
3.5m
2m
ABCD
CDCDABABM E
303sin4cos5.1sin6cos0
o..
tan
742953
2
I.Cut (Right Side)
I.Cut (Left Side)
I.Cut (Right Side)
05.1sin6cos3sin4cos6100 CDCDABABM F
+
+
CkNABABCD 78.3098.195.560