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Plastic capacity of beam

Plastic capacity of beam

Plastic capacity of beam

(OP)
A cantilever beam develops a fully plastic cross section at its base due to high bending moment. The cross section is rectangular, so the upper half will be in tension, the lower in compression with stresses equal to yield.

What happens if I then apply an axial load (e.g. tension)?

My guess:
Upper part (in tension) will just elongate due to perfect-plastic behavior.

Lower part (in compression) will experience a decrease in stress and strain, and thus the cross section as a whole will be able to handle this additional axial load.

Is this correct?
Thanks.

RE: Plastic capacity of beam

The rectangular stress block stays... but the area of compression/tension adjusts to suit the new axial loading.

A cantilever is not really a good application for plastic design... there is little redistribution <G>.

Dik

RE: Plastic capacity of beam

fully plastic behaviour assumes the tension side is worked up to ftu.

are you applying the tension at the geometrical centroid, or at the elastic-plastic NA ?

i guess you're asking about the interaction of tension endload and plastic bending.  i think you start with zero tension endload, calc up the full plastic bending moment.  then apply perhaps 90% of this, get a lower stress, which'll give you some room to apply an axial load.  and again at 70% (i don't think it'll be linear

RE: Plastic capacity of beam

petrabb, I don't think your model is quite correct.  When a rectangular section fails in pure bending, the stress distribution at failure is tensile yield stress above the neutral axis and compressive yield stress below it.  The neutral axis is located at mid-height.

When a beam fails in combined tension and bending, the stress distribution is similar to the above but the neutral axis moves down below mid-height.

If a rectangular section of width 'b' and depth 'd' fails in combined tension and bending, then if stressed in compression to Fy for a distance of 'a' from the bottom, it failed at a moment of b*a*Fy(d-a) and a tensile force of b(d-2a)Fy.

If that is not clear, draw a diagram and check it out.

BA

RE: Plastic capacity of beam

Fully plastic moment stress distribtution diagram is the answer, and it is a bit dull, since it just consists of two rectangles, width =yield stress, one positive, one negative. Call the height of the top on n and the bottom one m

beam width is b

The axial force is b*Sy*(n-m)

The moment is b*Sy*(m*(m/2+n)-n*n/2)

 

Cheers

Greg Locock


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RE: Plastic capacity of beam

sorry, but fully plastic bending (at least in my business) means stresses above yield.

but, whatever the case, if the section's full capacity is used up in bending (however you limit it) you can't add anything to it.  you can't say that the tension load will seek out the compression stresses and relieve them (which seems to be the suggestion).

if you want the section to react bending and tension you can't apply the full plastic moment, you can only apply a fraction.

If you want tension yield = bending stress + axial stress.  I know you're thinking about the compression side of the bending stress field, but you've got to think about the tension side at the same time.  

RE: Plastic capacity of beam


1)

If it is concrete section, it is a typical bending-axial problem, Just calculate the interaction diagram by sectional analysis using equilibrium equation

(Sum Force = Axial load, Max compressive Concrete strain = 0.0035)

For tension side of the interactiondiagram, the bending capacity of the section will decrease with the increasing tension up to no resistance at all !

2) This mean that because the capacity decrease, the curvature will increase up to the rupture of your rebars (Note : Moment capacity will increase a little due to strain hardening of the bar)

RE: Plastic capacity of beam

engipaul,
I don't think your post is relevant to the question.

Quote (petrabb)

A cantilever beam develops a fully plastic cross section at its base due to high bending moment. The cross section is rectangular, so the upper half will be in tension, the lower in compression with stresses equal to yield.

What happens if I then apply an axial load (e.g. tension)?

My guess:
Upper part (in tension) will just elongate due to perfect-plastic behavior.

Lower part (in compression) will experience a decrease in stress and strain, and thus the cross section as a whole will be able to handle this additional axial load.

Is this correct?
Thanks.

The answer is no, it is not correct. The member will fail in bending.

BA

RE: Plastic capacity of beam

BA... or a combination thereof...

Dik

RE: Plastic capacity of beam

Let us know what your professor proposes as the correct answer.

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