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cantilever beam large deflection 3
- Thread starter EcoMan
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1
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From Mechanical Springs 2nd Edition A. M. Wahl page 179 it is clear that up to deflection/beam_length=0.25 to 0.3 the behavior of the force vs deflection is linear. There is a graph there for larger deflections too. The linearity/non-linearity doesn't depend on deflection of a cantilever beam relative to the beam’s thickness but only to the beam's length. However, a long beam is a beam where the thickness is small relative to the beam length.
GregLocock
Automotive
I find that hard to believe. The shear deflection is crudely a function of the beam depth, not its length.
Cheers
Greg Locock
Cheers
Greg Locock
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- #4
Greg,
When I went to university, BOTH bending deflection AND shear deflection were functions of the cross-section geometry AND the span. As far as I know, this is still true today.
However, shear deflection is only one component of non-linearity in beam deflection. Non-linear behaviour due to span shortening or generation of catenary (membrane) stresses are other significant sources of non-linearity. I would not be prepared to offer a rule of thumb as to when non-linear behaviour generally becomes an issue, because there are just too many factors involved. (Are there any in-plane stresses such as axial tension or compression; are the ends of the span anchored, such that catenary stresses can be generated; etc.)
With respect to shear deflection only, some classic cases include:
Simply supported beam, total uniformly distributed load W:
Bending Deflection = 5.W.L^3/(384.E.I)
Shear Deflection = F.W.L/(8.A.G)
(F is a modifier for the shape of the cross section; equal to 6/5 for a rectangular section)
Simply supported beam, central point load P
Bending Deflection = P.L^3/(48.E.I)
Shear Deflection = F.P.L/(4.A.G)
Cantilver, end load P:
Bending Deflection = P.L^3/(3.E.I)
Shear Deflection = F.P.L/(A.G)
In each case:
Total Deflection = Bending Deflection + Shear Deflection
It is common practice to ignore shear deflection, except for "short" spans. As a rule of thumb, you need to get a Span-to-Depth Ratio of less than about 10:1 for simply supported beams, or about 5:1 for cantilevers, before the error in ignoring shear deflection effects is greater than about 3%. The error is typically less than 1% if your span-to-depth ratio is greater than about 20:1 for simply supported spans, or about 10:1 for cantilevers.
To see whether shear deflections are significant in your case, try calculating both deflection components (using approximate loads and geometry, if necessary), and compare the relative magnitudes of the bending and shear deflections. If the error is excessive, you will need to allow for shear deflection behaviour.
Hope this helps.
When I went to university, BOTH bending deflection AND shear deflection were functions of the cross-section geometry AND the span. As far as I know, this is still true today.
However, shear deflection is only one component of non-linearity in beam deflection. Non-linear behaviour due to span shortening or generation of catenary (membrane) stresses are other significant sources of non-linearity. I would not be prepared to offer a rule of thumb as to when non-linear behaviour generally becomes an issue, because there are just too many factors involved. (Are there any in-plane stresses such as axial tension or compression; are the ends of the span anchored, such that catenary stresses can be generated; etc.)
With respect to shear deflection only, some classic cases include:
Simply supported beam, total uniformly distributed load W:
Bending Deflection = 5.W.L^3/(384.E.I)
Shear Deflection = F.W.L/(8.A.G)
(F is a modifier for the shape of the cross section; equal to 6/5 for a rectangular section)
Simply supported beam, central point load P
Bending Deflection = P.L^3/(48.E.I)
Shear Deflection = F.P.L/(4.A.G)
Cantilver, end load P:
Bending Deflection = P.L^3/(3.E.I)
Shear Deflection = F.P.L/(A.G)
In each case:
Total Deflection = Bending Deflection + Shear Deflection
It is common practice to ignore shear deflection, except for "short" spans. As a rule of thumb, you need to get a Span-to-Depth Ratio of less than about 10:1 for simply supported beams, or about 5:1 for cantilevers, before the error in ignoring shear deflection effects is greater than about 3%. The error is typically less than 1% if your span-to-depth ratio is greater than about 20:1 for simply supported spans, or about 10:1 for cantilevers.
To see whether shear deflections are significant in your case, try calculating both deflection components (using approximate loads and geometry, if necessary), and compare the relative magnitudes of the bending and shear deflections. If the error is excessive, you will need to allow for shear deflection behaviour.
Hope this helps.
Deflection is most definitely a function of beam depth. Take the instance of a cantilever beam of square cross section (see eg. Roark Table 1, case 1a). The deflection due to an end load is proportional to the cube of the length and inversely proportional to the cube of the depth of the beam. However, if the CSA/width/depth remains constant and the only dimensional change is the beam's length, deflection obviously only depends on the length (and vice-versa). Deflection, therefore, is and is not purely dependent on the deflection depending on how you look at it. One would assume that the equation given above (deflection/beam_length=0.25 to 0.3) assumes the area properties remain constant(?)
-- drej --
-- drej --
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- #6
Shear deflection isn't a non-linearity but is considered negligible in most cases and so is ignored. The questioner is referring to the force displacement curve and the effect of the geometry change in the beam due to large displacements. The typical beam expressions assume small deflections. In that respect, given a known section, the non-linearity is a function of beam length.
corus
corus
just to add more smoke to the topic.....
the max bending stress in a cantilever beam as a function of tip displacement X is....
stress = 1.5 * X * h * E / (L^2)
where h = deam depth
X = tip displacement
if the question is....when does the tip deflection (as a function of depth) go plastic (exceed yield and become non-linear), it is when....
X * h = (yield stress) * (L^2) / (1.5 *E )
hope i got the algebra right....you should check it.
daveleo
the max bending stress in a cantilever beam as a function of tip displacement X is....
stress = 1.5 * X * h * E / (L^2)
where h = deam depth
X = tip displacement
if the question is....when does the tip deflection (as a function of depth) go plastic (exceed yield and become non-linear), it is when....
X * h = (yield stress) * (L^2) / (1.5 *E )
hope i got the algebra right....you should check it.
daveleo
- Thread starter
- #8
Thank you, all.
My question concerns a cantilever beam of given cross-sectional area. Assume negligible shear deflection and no residual stresses.
mentions an analytical solution for this problem and references Gere and Timoshenko.
confirms the posts by israelkk and corus.
My question concerns a cantilever beam of given cross-sectional area. Assume negligible shear deflection and no residual stresses.
mentions an analytical solution for this problem and references Gere and Timoshenko.
confirms the posts by israelkk and corus.
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