How Accurate Is the Analytical Method for Calculating Truss Forces?

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Stacyg
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Hi. The question is attached as well as my attempt at the answer, which I'm pretty sure is at least partially wrong. I have'ne included my calculations at point B as I have uploaded the limit of 3 files but I got BC=18.30kN (compression)
Any help would be great thanks.
 
Attachments
  • Hand-drawn truss framework with supports at A and B, joint D, E and C, a 3 m span marked between D and E, members at 45 degrees, a 2 tonne load at D and a 3 tonne load at E.
    Question.png
    12.6 KB · Views: 478
  • Handwritten truss diagram with joints A, B, C, D, E, showing loads of 2 tonne at D and 3 tonne at E, reactions 22.07 at A and 26.98 at B, and member forces including 22.09, 31.22, 2.88 kN and 44.13; below, the moment equation sum of M_A = 0 and the vertical force balance R_A = 19.62 + 29.43 - 26.98 = 22.07 kN.
    Answer Pt 1.png
    18 KB · Views: 489
  • Handwritten truss force calculations: force-triangle sketches at joints with equilibrium sums such as ΣV = 0 and ΣH = 0, solving for member forces including AD = -31.22 kN, AC = 2.10 kN, DE = -44.17 kN, BE 0.707 + BE = 44.17 kN, and CE = 25.88 kN (tension).
    Answer Pt 2.png
    22.1 KB · Views: 468

Discussion

I'm having some trouble with legibility of the files; so I only checked joint A. Although I cannot really read it, it appears everything at joint A is correct, except on your free-body diagram you assumed the force in member AD is compression. Then when that force variable turned out to be negative, you claimed the force is compression. That is incorrect. Negative means the force is the opposite of the direction you assumed for the unknown vector in your free-body diagram. Try it again.
 

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