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Publicado: 24 de febrero de 2013
Chapter 6, Solution 1.
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Joint FBDs:
Joint B:
FAB 800 lb FBC = = 15 8 17 so
FAB = 1500 lb T
FBC = 1700 lb C
Joint C:
FAC Cx 1700 lb = = 8 15 17 FAC = 800 lb T
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J.Cornwell © 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 2.
Joint FBDs: Joint B:
ΣFx = 0: ΣFy = 0: 1 4 FAB − FBC = 0 5 2 1 3 FAB + FBC − 4.2 kN = 0 5 2
so
7 FBC = 4.2 kN 5
FBC = 3.00 kN C !
FAB = 3.39 kN C !
Joint C:
FAB =
12 2 kN 5
ΣFx = 0:
4 12 (3.00 kN) − FAC = 0 5 13 FAC = 13 kN 5 FAC = 2.60 kN T !
Vector Mechanicsfor Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 3.
Joint FBDs:
Joint B:
FAB FBC 450 lb = = 12 13 5 so FAB = 1080 lb T FBC = 1170 lb C
Joint C:
ΣFx = 0:
3 12 FAC −(1170 lb ) = 0 5 13 FAC = 1800 lb C
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 4.
Joint FBDs:
Joint D:
FCD FAD 500 lb = = 8.4 11.6 8 FAD = 725lb T FCD = 525 lb C
Joint C:
ΣFx = 0:
FBC − 525 lb = 0 FBC = 525 lb C
This is apparent by inspection, as is FAC = C y
ΣFx = 0:
8.4 3 (725 lb) − FAB − 375 lb = 0 11.6 5 FAB = 250lb T
Joint A:
ΣFy = 0:
FAC −
4 8 (250 lb) − (725 lb) = 0 5 11.6 FAC = 700 lb C
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E.Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 5.
FBD Truss: ΣFx = 0 :
By symmetry: C y = D y = 6 kN
Cx = 0
Joint FBDs: Joint B:
ΣFy = 0: − 3 kN + 1 FAB = 0 5
FAB = 3 5 = 6.71 kN T
Joint C:
ΣFx = 0:
2 FAB − FBC = 0 5
FBC = 6.00 kN C
ΣFy = 0: ΣFx = 0:
6 kN −
3 FAC = 0 5 4FAC + FCD = 0 5
FAC = 10.00 kN C FCD = 2.00 kN T
Joint A:
6 kN −
ΣFy = 0:
1 3 − 2 3 5 kN + 2 10 kN − 6 kN = 0 check 5 5
By symmetry:
FAE = FAB = 6.71 kN T FAD = FAC = 10.00 kN C FDE = FBC = 6.00 kN C
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J.Cornwell © 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 6.
FBD Truss:
ΣM A = 0:
(10.2 m ) C y + ( 2.4 m )(15 kN ) − ( 3.2 m )( 49.5 kN ) = 0
C y = 12.0 kN
Joint FBDs: Joint FBDs: Joint C:
FBC F 12 kN = CD = 7.4 7.4 8 FBC = 18.50 kN C FCD = 18.50 kN T 4 7 (18.5 kN) = 0 FAB − 5 7.4 FAB = 21.875 kN; ΣFy = 0: FAB = 21.9 kN CJoint B:
ΣFX = 0:
3 2.4 (21.875 kN) − 49.5 kN + (18.5 kN) + FBD = 0 5 7.4 FBD = 30.375 kN; FBD = 30.4 kN C
Joint D:
ΣFx = 0: − 4 7 FAD + (18.5 kN ) + 15 kN = 0 5 7.4 FAD = 40.6 kN T
FAD = 40.625 kN;
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 TheMcGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 7.
Joint FBDs:
Joint E:
FBE FDE 3 kN = = 5 4 3 FBE = 5.00 kN T FDE = 4.00 kN C ΣFx = 0: − FAB + 4 (5 kN) = 0 5 FAB = 4.00 kN T ΣFy = 0: FBD − 6 kN − 3 (5 kN) = 0 5 FBD = 9.00 kN C
Joint B:
Joint D:
ΣFy = 0:
3 FAD − 9 kN = 0 5 FAD = 15.00 kN T 4 (15 kN) − 4 kN = 0 5 FCD = 16.00 kN C
ΣFx = 0:...
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