Bucklig: deflection at Euler load

  • #1
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Hello

I was trying to calculate the horizontal deflection of the free end of vertical clamped beam. The beam would be loaded at the free end with a horizontal force [itex]H[/itex] and a vertical force [itex]P[/itex]. My idea was to calculate an initial deflection due to the force [itex]H[/itex]. Then calculate the additional deflection due to the previous deflection and the force [itex]P[/itex]. I'd expect that when the force [itex]P[/itex] is bigger than the Euler buckling load that the deflection would diverge but that's not happening when I try to calculate this.

I tried it on a beam with a length [itex]L[/itex] of 6 m, stiffness [itex]EI = 1476600 Nm^2[/itex] ([itex]E=69 GPa[/itex], [itex]I=2140 *10^4 mm^4[/itex], [itex]H=1 kN[/itex]. I calculated that the Euler buckling load [itex]P_{cr}=(Pi)^2EI/(2L)^2=101,20 kN[/itex] and used a way bigger [itex]P=5000 kN[/itex].

For the initial deflection I used [itex]v_0=(1/3EI)HL^3=4,87607 *10^{-5} m[/itex]. For the next iteration steps I used [itex]v_i=v_0 + (1/3EI)L^2P*v_{i-1}[/itex] which eventually converges to [itex]v=5,08259 * 10^{-5} m[/itex] instead of diverging.

I know 5000 kN isn't a realistic value and that the beam would probably yield with such a high load but shouldn't this diverge?

Also for loads smaller than the Euler buckling load, is this the right way to calculate the deflection? If not what would be a good way then?

Thanks in advance
 

Answers and Replies

  • #2
SteamKing
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Hello

I was trying to calculate the horizontal deflection of the free end of vertical clamped beam. The beam would be loaded at the free end with a horizontal force [itex]H[/itex] and a vertical force [itex]P[/itex]. My idea was to calculate an initial deflection due to the force [itex]H[/itex]. Then calculate the additional deflection due to the previous deflection and the force [itex]P[/itex]. I'd expect that when the force [itex]P[/itex] is bigger than the Euler buckling load that the deflection would diverge but that's not happening when I try to calculate this.

I tried it on a beam with a length [itex]L[/itex] of 6 m, stiffness [itex]EI = 1476600 Nm^2[/itex] ([itex]E=69 GPa[/itex], [itex]I=2140 *10^4 mm^4[/itex], [itex]H=1 kN[/itex]. I calculated that the Euler buckling load [itex]P_{cr}=(Pi)^2EI/(2L)^2=101,20 kN[/itex] and used a way bigger [itex]P=5000 kN[/itex].

For the initial deflection I used [itex]v_0=(1/3EI)HL^3=4,87607 *10^{-5} m[/itex]. For the next iteration steps I used [itex]v_i=v_0 + (1/3EI)L^2P*v_{i-1}[/itex] which eventually converges to [itex]v=5,08259 * 10^{-5} m[/itex] instead of diverging.

I know 5000 kN isn't a realistic value and that the beam would probably yield with such a high load but shouldn't this diverge?

Also for loads smaller than the Euler buckling load, is this the right way to calculate the deflection? If not what would be a good way then?

Thanks in advance

I'm just taking a quick glance at the situation you are trying to analyze here.

The usual formulas for the Euler buckling load are derived assuming that the only force applied is applied in the axial direction. I think the situation you are describing here is for what is called the buckling of a beam-column, since lateral and axial loads are being applied simultaneously to the tip of the cantilever. The Euler critical load must be modified in this case over that calculated for a simple column with no loads applied in the lateral direction.

There are methods for analyzing such beam-columns, but it will take a little research to confirm what is required.
 
  • #3
SteamKing
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Further to the discussion in Post #2 above, this article discusses the effect on the deflection of a laterally loaded member caused by an axial load. See pp. 5-6 for a discussion of the effect:
 

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  • #4
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I didn't realise I couldn't use Euler buckling for my situation. Thank you!
 

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