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COSMOS: Complete Online Solutions Manual Organization System

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, DavidMazurek, 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 = 13kN 5 FAC = 2.60 kN 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 3.
Joint FBDs:

Joint B:
FAB FBC 450 lb = = 12 13 5 so FAB = 1080 lbT 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 = 725 lb 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 andDynamics, 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 TJoint C:

ΣFx = 0:

2 FAB − FBC = 0 5

FBC = 6.00 kN C

ΣFy = 0: ΣFx = 0:

6 kN −

3 FAC = 0 5 4 FAC + 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 andDynamics, 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: JointC:
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 C

Joint 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 andDynamics, 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 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:...
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