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Queren Suriano
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Homework Statement
Calcular the force P when a bar starts yield, when 2 bars start yield, and when the 3 bars start yield
Diameter = 25 mm (for every bar), Yield Stress of Steel = 250 MPa.
Homework Equations
The Attempt at a Solution
[/B]
[F][/AD] + [F][/BE] + [F][/CF] = P
2[Δ[/AD] - 3[Δ][/BE] + [Δ][/AC]= 0
I calculate first what happened when the 3 bars were in elastic zone, with the Hooke law I obtain:
[F][/AD]= 0.14P
[F][/BE]= 0.29 P
[F][/CF] = 0.57P
So the firs bar in reach yield stress is CF.
0.57P = [σ][/y] * A => P=215, 295.54 NWhen the 2 bars (BE and CF) are with the yield stress, [F][/BE]=[F][/CF]
From equations of equilibrium, sumatory of moments in C:
0.4P = 1.2 [F][/AD] + 0.8 [F][/BE]
and I know from the eq (1) [F][/AB] + [F][/BE] + [F][/CF] = P, so 2[F][/BE] + [F][/AD]= P. Solving this 2 eq. I obtain P=243, 605.30 N and [F][/AD]= -1831.62 N ...So I don't know if this answer is correct, because it says that the bar AD now is in compresion, when before it was in tension
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