- #1
vesu
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Homework Statement
A beam (note: part of a truss, I chose to use the beam in the truss with the most force and length for my calculations, which I assume is the correct thing to do) of length [itex]6m[/itex] with a force of [itex]9N[/itex] is being constructed out of a material with a Young's Modulus of [itex]E = 70 \times 10^9 N/m^2[/itex] and it's ultimate stress is [itex]σ_{u} = 110 \times 10^6 N/m^2[/itex]. What is the minimum dimension of the square beam (i.e. cross-section) so that the structure will not fail?
Homework Equations
Spring constant of beam: [itex]k_{beam} = \frac{A \times E}{L}[/itex]
Hooke's Law: [itex] F = k \times ΔL[/itex]
[itex]E = \frac{σ}{ε} = \frac{F/A}{ΔL/L}[/itex]
The Attempt at a Solution
I'm pretty lost on what to do here, we barely covered this stuff in lectures. My first thought was simply letting [itex]σ = F/A[/itex] but that doesn't make much sense because surely length is a factor and that method doesn't use the young's modulus at all. My second thought was to substitute [itex]k_{beam}[/itex] into the hooke's law equation, giving [itex] F = \frac{A \times E}{L} \times ΔL[/itex] => [itex]9 = \frac{A \times 70 \times 10^9}{6}[/itex] but I'm not sure where this gets me.
P.S. Sorry if I didn't explain the question very well, I paraphrased a bit because the original question is part of a series of questions and doesn't make a lot of sense on it's own.