Suppose a cantilever beam (of rectangular cross-section) subjected to vertical transverse shear load at free end. If I consider any arbitrary cross-section then both bending and shear stress will be there, but now if replace this beam with another beam of same area but whose width is tending to zero and depth is very large then maximum bending stress will now be tending to zero because the section modulus will be very high and also the maximum shear stress will be same as in the previous case because Maximum shear stress = 1.5 * average shear stress and since area is not changing so maximum shear stress will not change. So why we do not use such beam practically ?
3 Answers
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The problem is that when you reduce the width but keep the area constant you are increasing one moment of area but you reduce the other.
Although there is a benefit in bending the same section with buckle (the term in structural engineering is lateral torsional buckling).
Figure: Lateral torsional buckling in a cantilever beam L. Dahmani, S. Drizi
If you try to using a sheet of metal on its side in place of a structural I Beam it will buckle under the load.
A way to maintain the stability of the column (if width/thickness cannot be increased) is to add lateral supports, but obviously that is not always convenient.
Figure: cross-section of wall loaded in compression
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$\begingroup$ as per the figure, it is a C section and since the load is not passing through the shear centre, so that will lead to torsional buckling of member. But if I consider a rectangular section of large depth, then why does it buckle? $\endgroup$– MaxNov 8, 2021 at 8:10
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$\begingroup$ @Max I tried finding an image that is a flat sheet but I couldn't find one. I was hoping that would be suffient, but apparently it isn't. I updated the answer with an image of a the cross-section of a wall under compression. Hopefully it will be obvious that if a load is applied on a thin sheet of film is similar to (not identical) having axial compression on a long and slender column, $\endgroup$– NMech ♦Nov 8, 2021 at 9:13
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$\begingroup$ @Max continuing on from the previous comment, you could think of a flat piece of A4 paper as rectangular section of large depth (if you orient it correctly). If you have that piece standing its not possible to carry much load without the paper collapsing under buckling. $\endgroup$– NMech ♦Nov 8, 2021 at 9:16
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$\begingroup$ yes the paper will collapse under buckling, so is it buckling because by making the width so small the other moment of interia is reduced significantly ? But I am not able to understand that the load which is applied is vertical so I have increased one moment of inertia significantly which has reduced bending stress, so how due to low value of other moment of inertia promotes buckling? $\endgroup$– MaxNov 8, 2021 at 10:43
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$\begingroup$ the minimum force required to cause buckling is proportional to $EI$. The formula is actually $P_{crit} = \frac{\pi}{(KL)^2} EI$. so different axis have different minimum loads. Obviously, buckling occurs when the lowest is reached. $\endgroup$– NMech ♦Nov 8, 2021 at 10:48
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You have a correct concept if there is no space limitation and side swing isn't a concern, but there is still a limit to the benefit of increasing beam depth without inducing the complexity in design.
We know that the beam theory applies to beams that possess linear-elastic behavior and the resulting plane section remains plane after loading, however, beyond a certain threshold "width-depth" ratio, and/or span-depth ratio, that usually identified as a "deep beam", the distribution of stress and strain across the depth are no longer linear, and the plane may no longer remain plane due to the increased influence of shear deformation (for the threshold, see the textbook by Timoshenko).
When you stretch the beam to a thin plate that has the same volume as the beam, do you expect them to behave the same? The answer is no. You can use beams made of cardboard to perform simple observation and verification.
ADD: As the beam gets deeper, the tendency of distortion (shape instability) increases, as the influence of shear deformation increases.
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$\begingroup$ so such a beam will fail due to shear deformation? But the same shear stress was acting before in the normal case. $\endgroup$– MaxNov 8, 2021 at 6:39
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$\begingroup$ Yes, the deep beam is prone to shear failure. The shear distribution is no longer linear and tends to warp the cross-section. The warping phenomenon is kept in check by adding intermediate stiffeners for steel deep girder, and side face reinforcing steel for deep concrete girder. Another important phenomenon is the critical shear plane is much steeper than beams in the normal size range that usually fail in the 45 degrees pattern near the support. $\endgroup$– r13Nov 8, 2021 at 16:20
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Not all the beams are prismatic.
In places where changing the section geometry is meaningful and makes sense we certainly do that.
All the appropriate steel (or any material) structures take advantage of using the most effective profile as soon as they aren't typical beam and column grid work, like churches, auditoriums.
- Bridges use tapered beams, trapezoid outriggers, and even cables if it makes sense.
- Machinery and tools use the shapes that work best even if it seems a complex geometry, and unwieldy.
