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Thanks

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- Thread starter ttlg
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Thanks

- #2

FredGarvin

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http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html#pax

- #3

Q_Goest

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I think what ttlg is asking is if the two beams are simply sitting on top of each other such that there can be no transverse shear between the two (like a leaf spring on a car). The two beams, stacked one on top of the other, are then simply supported at the ends and loaded in the center with a verticle force. In that case, since there's no transverse shear between the two (ie: the beams act independantly) the parallel axis theorem doesn't apply to the beams as a set. In this case, the load acting on the beams can simply be split 50/50 between the two beams (each beam supports 1/2 the load).

- #4

FredGarvin

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The loads can be split the way you mention, but to find the area MOI of the assembly, i.e. the composite section, one needs to use the parallel axis theorem. I am using the interface between the two beams as the neutral axis (with no shear between the two as you mentioned). From there take the two individual beams' respective area MOIs and use the parallel axis theorem to calculate the overall area MOI.

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ok, thanks for your help

- #6

FredGarvin

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The loads can be split the way you mention, but to find the area MOI of the assembly, i.e. the composite section, one needs to use the parallel axis theorem. I am using the interface between the two beams as the neutral axis (with no shear between the two as you mentioned). From there take the two individual beams' respective area MOIs and use the parallel axis theorem to calculate the overall area MOI.

QGoest said:since there's no transverse shear between the two (ie: the beams act independantly) the parallel axis theorem doesn't apply to the beams as a set.

Looking at Q's comments about the shear stress...he's right on that. The MOI is simply two times the individual MOIs.

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