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旧 2009-09-07, 12:12 PM   #1
huangyhg
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默认 anomaly in torsional stresses

anomaly in torsional stresses?
torsional stress calculation for rectangular structural stock is at the heighest at the middle of the outer fiber of the long side of stock cross section. yet, when calculating torsional stress of a rectangular weld pattern, this stress is taken at the further radial position from the center which is at the corner of the pattern. so if you butt welded two such stock along a 'v' groove having the same depth as the stock thickness,which value of that welded pattern would you give credence.
if you butt-weld two rectangular tubes end to end, you'd assume the stress distribution in the weld was the same as in the tube itself.
with a rectangular-pattern fillet weld, the assumption is that you have two rigid bodies connected by the weld; therefore, where the greatest differential motion is must be the highest stress. but this is not the same as the strctural tube design.
hi chicopee
according to my books for torsion on a box section the higher stresses occur at the inside corners of the box.
my book gives the formula's for the average shear stress on a box section under torsion at the midpoint of section as:-
s= t/(2*t(a-t)*(b-t))
where s=shear stress
t= torque
t= wall thickness of tube
a= width of tube
b= length of tube
i also concur with j.stephen in his comment
regards desertfox
desertfox- that's interesting because my reference"design and production" by kent and "design of machine elements" by mf spotts state that the max. torsional shear stress in in the middle of the longest side of rectangular structural
hi chicopee
the formula's i gave refer to hollow box section not solid
rectangular section.
i read j.stephens post and it mentioned rectangular tube.
the shear stress formula are for torsional shear stress on hollow box section.
regards
desertfox
i misread the original question to refer to rectangular tubing. however, as long as the weld has the same cross section as the parts it joins, you'd assume that stresses in the weld were the same as in those parts. this would apply to rectangular tubing or to bar stock, provided the welds were full penetration welds through the stock.
a difference i can see between the solid (monolithic) crossection and one created by welding two thick plates together (to get the same overall dim'ns) would be that the latter is going to strain more like a tube since the two plates are connected only discretely (at the ends).
i'd also challenge the op's post that ...
"torsional stress calculation for rectangular structural stock is at the heighest at the middle of the outer fiber of the long side of stock cross section." maybe i'm being pedantic, but i thought that the torsional stress on the surface (= the outer fiber) was zero and that the peak stress was at the centroid of the section (assuming a regular solid section).
rb1957,i think that you are thinking about the horizontal shearing stress being zero on the outer fiber while under bending condition.
no i'm thinking about the shear stresses on the surface of a solid section.
for the same reason as you note (under bending the surface shear stress is zero) since there is nothing outside the section to resist the section, the surface has zero stress, and the shear stress quickly builds up as you penetrate into the solid ... i was visualising the "sand pile" analogy for plastic torsional shear stresses.
i'm confused about how these are welded. is this two long pieces of flat bar slapped together and welded down the long sides? or are they welded end to end (what i'm assuming? i can't tell from the post.
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