Calculating Forces in Truss Members using Method of Section

AI Thread Summary
The discussion focuses on calculating forces in truss members using the Method of Sections. The participant successfully determined the reaction forces at points A and B, identifying the vertical reaction at A as 4000N upwards and at B as 6000N upwards. They calculated forces in members AC, AF, BK, and BE, confirming their values as correct. Further calculations for members FC and FG were discussed, with the participant arriving at FC = 5000N and FG = -3000N, indicating compression. The conversation emphasizes the importance of understanding the signs of force values to determine tension and compression in truss members.
Mattmiles
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Homework Statement



a) Find the reactions provided by the foundations at A and B.
b) Use the method of Resolution at the Joints to calculate the forces in all the members.
c) Which member will fail first and why ?

Homework Equations



ƩFx = 0
ƩFy = 0
ƩM = 0

The Attempt at a Solution



I am really just concerned with whether I have got the following correct. If I have, then I have grasped the concept if not then if someone could point me in the right direction and advise me on where to go next of what I am doing wrong I would be very grateful.

Horizontal Reaction Forces in the X direction at point A = 0
Vertical Reaction Forces in the Y direction at A = 4000N Upwards
Reaction Forces about the moment A at point B = 6000 Newtons Upwards

Forces in Member AC = 0N (0 + AC + AF = 0)
Forces in Member AF = 4000N in Compression (4000 + AC + AF = 0)
Forces in Member BK = 6000N in Compression (6000 + FBK.Sin(90) + FBE = 0)
Forces in Member BE = 0N (-FBE + FBK = 0)

Thanks
 

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This is all good, you got them all correct:smile::approve::cool:
 
Thanks very much! One last thing. I am now finding the forces in Members FC and FG. Am I right in saying:

Forces in direction:
-(-4000) - FC x sin(53.1) = 0 and then I just rearrange for FC = 5000 Newtons

and now I know FC I can do the following:

Force in direction:
0 + (FC x cos(53.1)) + FG = 0 and again rearrange for FG = -3002 Newtons
 
Mattmiles said:
Thanks very much! One last thing. I am now finding the forces in Members FC and FG. Am I right in saying:

Forces in direction:
-(-4000) - FC x sin(53.1) = 0 and then I just rearrange for FC = 5000 Newtons
yes , compression or tension?
and now I know FC I can do the following:

Force in direction:
0 + (FC x cos(53.1)) + FG = 0 and again rearrange for FG = -3002 Newtons
You have a round off error, these are 3-4-5 right triangles, FG = -3000 N. What does the minus sign mean?
 
Sorry a negative number means it is compression and a positive number means it is in tension.
 
Mattmiles said:
Sorry a negative number means it is compression and a positive number means it is in tension.
OK, you are again correct.
 
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