Shear force - Maximum at 45 degrees...
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Shear force - Maximum at 45 degrees...
Hi all,
Quick question. Why is it that the maximum strain induced by strain guauges is when the strain guages are mounted at +/- 45 degrees to the nuertal axis?
My step-son has a piece of college work on shear strain guages, where one strain gauge is in compression and the other in tension.
Please be fairly basic in your replies...this REALLY isn't my field
Thanks in advance for your help.
Kelly
Quick question. Why is it that the maximum strain induced by strain guauges is when the strain guages are mounted at +/- 45 degrees to the nuertal axis?
My step-son has a piece of college work on shear strain guages, where one strain gauge is in compression and the other in tension.
Please be fairly basic in your replies...this REALLY isn't my field
Thanks in advance for your help.
Kelly
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Apparently the strain gauge system is wired up in a Wheatstone Bridge configuration, where two resistors are compressed and the other two are stretched.
how sensors work - load cells
He's been told to look at the The Shear Beam Load Cell section in the above link for information - but apparently it's not explaining why the strain is largest at 45 degrees.
No lab results I'm afraid, he's been asked to briefly explain why the shear strain is largest at 45 degrees to the 'neutral axis'
Kelly
how sensors work - load cells
He's been told to look at the The Shear Beam Load Cell section in the above link for information - but apparently it's not explaining why the strain is largest at 45 degrees.
No lab results I'm afraid, he's been asked to briefly explain why the shear strain is largest at 45 degrees to the 'neutral axis'
Kelly
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The fact that the shear stress is a maximum at 45 degrees to the loading axis is shown by an equilibrium diagram of a small element of material under load. The best source of these can be found in derivations and descriptions of Mohr's Circle, such as this:
Mohr's Circle
This shows that the shear stress is a maximum when two-theta (sorry I don't know how to write Greek letters on this thing...) is 90 degrees, so theta is 45 degrees.
When a sample of ductile metal such as steel is pulled in a testing machine until it fails, the failure always exhibits a 45 degree 'lip'. In a rod, this failure has a 'cup and cone' appearance because the lip is circular. The 45 degree planes are where the maximum slippage has occurred between the crystals in the material, which is the mode of failure, and of course the slippage occurred along the plane of maximum shear stress. When you are examining wreckage from an aircraft accident, the bits that failed in tension are immediately obvious from these 45 degree features. They tell you the tensile loading direction too!
I hope this helps. Mohr's Circle is the key; hopefully it will be somewhere a few pages ahead in your stepson's textbook, but if not the internet has many more references than this one.
By the way, strain gauges work very well in compression.
Mohr's Circle
This shows that the shear stress is a maximum when two-theta (sorry I don't know how to write Greek letters on this thing...) is 90 degrees, so theta is 45 degrees.
When a sample of ductile metal such as steel is pulled in a testing machine until it fails, the failure always exhibits a 45 degree 'lip'. In a rod, this failure has a 'cup and cone' appearance because the lip is circular. The 45 degree planes are where the maximum slippage has occurred between the crystals in the material, which is the mode of failure, and of course the slippage occurred along the plane of maximum shear stress. When you are examining wreckage from an aircraft accident, the bits that failed in tension are immediately obvious from these 45 degree features. They tell you the tensile loading direction too!
I hope this helps. Mohr's Circle is the key; hopefully it will be somewhere a few pages ahead in your stepson's textbook, but if not the internet has many more references than this one.
By the way, strain gauges work very well in compression.
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Confused ! or am I to understand that the 45 degree lip occurs inwardly from the surface and thus can act as a pointer along the direction of the breaking strain?
Wish I had known about this when I examined the Valiant prototype No2 main spar failure at Boscombe Down in the mid 50s. Fortunately the flight load at max AUW was redistributed and we managed to get back to base with the broken wing.
Anyone for some diagrams?
Wish I had known about this when I examined the Valiant prototype No2 main spar failure at Boscombe Down in the mid 50s. Fortunately the flight load at max AUW was redistributed and we managed to get back to base with the broken wing.
Anyone for some diagrams?
I refrained from trying to explain Mohr's which shows where the maximum stress in a solid under shear is. It is non-intuitive; unless you want to do a fair amount of preliminary study the explanation won't make much sense, and PPRune is not the ideal place to write a Stress & Strain course.
However: in a horizontal I beam (which is an adequate description of a mainspar) with a vertical force applied to one end, the lower boom is in tension, the upper in compression, and the material in between in shear, trying to stop the upper boom disappearing inwards and the lower departing outwards. The maximum shear stress in the material is at 45 degrees to the neutral axis, and simple shear failure will usually occur at the point along the beam of maximum material stress (obviously). A spar is usually tapered to try to achieve failure at all points simultaneously, and tests prove that no spar is ever perfect, as they usually fail at a single point!
If you can follow the argument in D120's link above, then this is obviously true. If you can't, then the reason is 'because it is'.
However: in a horizontal I beam (which is an adequate description of a mainspar) with a vertical force applied to one end, the lower boom is in tension, the upper in compression, and the material in between in shear, trying to stop the upper boom disappearing inwards and the lower departing outwards. The maximum shear stress in the material is at 45 degrees to the neutral axis, and simple shear failure will usually occur at the point along the beam of maximum material stress (obviously). A spar is usually tapered to try to achieve failure at all points simultaneously, and tests prove that no spar is ever perfect, as they usually fail at a single point!
If you can follow the argument in D120's link above, then this is obviously true. If you can't, then the reason is 'because it is'.
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The key word is shear.
Imagine an elemental square at the surface of the specimen. Draw it aligned with X and Y axes.
Draw a shear vector vertically up on the right side.
To be in equilibrium there must be an equal and opposite vector on the left side - downward.
That is not the whole story - it just covers forces in the vertical direction.
To stop it twisting in the anti-clockwise direction, it needs balanced forces on the top and bottom edges.
For equilibrium you need a vector to the right on the top edge and to the left on the bottom.
The result is a tendency to pull the square up and right and down and left. It would make the square lozenge shaped.
This gives the concept of principal stresses at 45 deg. to the X Y axes.
Now do it for an elemental cube!
HTH - usually used visual aids when I did it for a living.
Imagine an elemental square at the surface of the specimen. Draw it aligned with X and Y axes.
Draw a shear vector vertically up on the right side.
To be in equilibrium there must be an equal and opposite vector on the left side - downward.
That is not the whole story - it just covers forces in the vertical direction.
To stop it twisting in the anti-clockwise direction, it needs balanced forces on the top and bottom edges.
For equilibrium you need a vector to the right on the top edge and to the left on the bottom.
The result is a tendency to pull the square up and right and down and left. It would make the square lozenge shaped.
This gives the concept of principal stresses at 45 deg. to the X Y axes.
Now do it for an elemental cube!
HTH - usually used visual aids when I did it for a living.
I know. To keep awake you have to play it on the flute.
---tough class
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A slightly different explanation that may help...
When a beam bends, e.g. a wing spar, the top surface is in compression and the bottom surface is in tension. In between there is a theoretical "neutral surface", where there is no axial stress, only bending. Because the top surface and bottom surface are under axial stress, there is also strain, i.e. the top of the wing spar gets a bit shorter, the bottom gets a bit longer.
Now imagine this with respect to the "neutral surface", which is not under strain so it has not changed length. Any material between the neutral surface and the top or bottom surface will be under shear stress as well as axial stress, because of the variation of axial stress and strain throughout the thickness of the beam.
The shear force on an element of the beam above or below the neutral surface is parallel to the beam top and bottom surfaces (due to strain) and also normal to the surfaces. If the element is a square, the top and bottom shear forces are in oposite directions, as are the vertical forces either side of the element. Do the vector sum and the resultant shear force is at 45 degrees to the axial stress.
A good reference article is here: Bending - Wikipedia, the free encyclopedia
Regards,
WF
When a beam bends, e.g. a wing spar, the top surface is in compression and the bottom surface is in tension. In between there is a theoretical "neutral surface", where there is no axial stress, only bending. Because the top surface and bottom surface are under axial stress, there is also strain, i.e. the top of the wing spar gets a bit shorter, the bottom gets a bit longer.
Now imagine this with respect to the "neutral surface", which is not under strain so it has not changed length. Any material between the neutral surface and the top or bottom surface will be under shear stress as well as axial stress, because of the variation of axial stress and strain throughout the thickness of the beam.
The shear force on an element of the beam above or below the neutral surface is parallel to the beam top and bottom surfaces (due to strain) and also normal to the surfaces. If the element is a square, the top and bottom shear forces are in oposite directions, as are the vertical forces either side of the element. Do the vector sum and the resultant shear force is at 45 degrees to the axial stress.
A good reference article is here: Bending - Wikipedia, the free encyclopedia
Regards,
WF
