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F6-20. Determine the reactions at D. 4 m B 10 kN 3 m3 m3 -3 m m3 m3 m -3m- 15 kN Prob. F6-20 C D

F6-20. Determine the reactions at D. 4 m B 10 kN 3 m3 m3 -3 m m3 m3 m -3m- 15 kN Prob. F6-20 C D

International Edition---engineering Mechanics: Statics, 4th Edition
International Edition---engineering Mechanics: Statics, 4th Edition
4th Edition
ISBN:9781305501607
Author:Andrew Pytel And Jaan Kiusalaas
Publisher:Andrew Pytel And Jaan Kiusalaas
Chapter10: Virtual Work And Potential Energy
Section: Chapter Questions
Problem 10.27P
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F6-20. Determine the reactions at D. 4 m B 10 kN 3 m3 m3 -3 m m3 m3 m -3m- 15 kN Prob. F6-20 C D
Transcribed Image Text:F6-20. Determine the reactions at D. 4 m B 10 kN 3 m3 m3 -3 m m3 m3 m -3m- 15 kN Prob. F6-20 C D
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In my textbook these are the correct answer but all are positive, why is that I thought we can assume the the forces.

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This is how I solved for reaction D, is this also correct?

Problem (6-20) LOKN FAB (35) FAB 3m √ ↑ FAB (5) 2Fx= FAB (2) + (x = 0 4m Dx 3m 15KN 12Fy= FAB (4/5)-10 KN-15KN + Cy=0 (Mc = (3m 15KN) + (6m 10 KN) - (9m. FAB (45)) -0 . FAB= 14.583 KN 13.333 Dy 3m ум Cx=-(14.583 HN) (²) -- 8.749 KN (y=-14.583 (4) KN + 10 + 15 KN = 13.333 KN F37 →8.749 KN cy Ax ZMD= M-(4m 8.749 KN) = 0 ZFx=Dx+8.749 KN = 0 Fy=-13.333 + Dy = ² DX = -8.749 KN Dy = 13.333 KN M= 34.996 KN.m
Transcribed Image Text:Problem (6-20) LOKN FAB (35) FAB 3m √ ↑ FAB (5) 2Fx= FAB (2) + (x = 0 4m Dx 3m 15KN 12Fy= FAB (4/5)-10 KN-15KN + Cy=0 (Mc = (3m 15KN) + (6m 10 KN) - (9m. FAB (45)) -0 . FAB= 14.583 KN 13.333 Dy 3m ум Cx=-(14.583 HN) (²) -- 8.749 KN (y=-14.583 (4) KN + 10 + 15 KN = 13.333 KN F37 →8.749 KN cy Ax ZMD= M-(4m 8.749 KN) = 0 ZFx=Dx+8.749 KN = 0 Fy=-13.333 + Dy = ² DX = -8.749 KN Dy = 13.333 KN M= 34.996 KN.m
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