It's standard practice to get first year mech-eng students to do this (at least at my uni!), just because it seems strange the first time around.
As mjolnirman has said, it's hard to type out the formulae, so I'm going to have to hope that everything comes out right and I don't misplace anything.
An actual bayonet gets rather complicated since you have dimensional variations in all axes (cross-section varies as you go up and down the blade), but the same principle applies, so let's use I beams.
For a rectangle, the second area moment is Ix = (b*h^3)/12. Note that this means I will change depending on which way the force is acting. Iy = (h*b^3)/12. Let's take a solid beam that is 10 mm by 10 mm. It's a square, so I in either direction will be 833.3 mm^4.
For composite objects (like an I beam), we need to find out the 2nd area moment, like before, and also the neutral axis of this composite. Fortunately for us, due to symmetry the neutral axis is identical to that of the simple beam. Let's take the same 10 x 10 beam, but now machine two grooves in the sides (read: remove material). Symmetrical in both x and y directions, but obviously not to scale:
Sounds good?
Since we're discussing bayonets, if you draw the cross-section as I've described it above, we're interested in forces going from left to right, or vice versa.
To find the 2nd moment for this composite beam, we need to break it down and use the parallel axis theorem, I(around an axis) = I(element) + A*d^2 where A is the cross-sectional area of the element, and d is the distance between the axis of the overall object and the axis of the individual element.
I of the second element (6x4) is easy to compute, using Iy = (h*b^3)/12 = (4*6^3)/12 = 72 mm^4.
Thanks to symmetry we can double I for either of the two other elements. For one element:
Area of one element = 3*10 = 30 mm^2.
Distance from neutral axis to centroid of object's moment = 3.5 mm (a quick drawing will show this quite easily).
I(around neutral axis) = (3*10^3)/12 + (3*10)*3.5^2 = 617.5 mm^4.
That's for one of the side elements.
Now we just sum it all together to get Iy = 72 + 2(617.5) =1307 mm^4.
Compare this with the solid beam's value for I of 833.3 mm^4.
That's the basic stuff out of the way.
So what happens when we put this under stress? If we exert a force at one end, we introduce a moment into the beam, this moment acts on each tiny little element of the beam to stretch or compress it. The formula is sigma = My/I where y is the distance from the neutral axis to the element in question. The maximum stress will thus be found at the point the furthest from the neutral axis. If the stress exceeds a certain limit, the element will permanently deform. Those of you who are interested can look up yield strengths, but I'll leave those for another day, all that matters here is that we want to keep sigma as low as possible.
In both cases, the max distance will be 5 mm. Let's take a moment of 10,000 N.mm, because I'm uncreative. For the solid beam:
sigma = 10,000*5/833.3 = 60 N/mm^2.
For the I beam:
sigma = 10,000*5/1307 = 38.26 N/mm^2.
If the yield strength of the material in question is less than 38.26 N/mm^2, nothing happens to either beam. If it's greater than 60 N/mm^2, both will bend. Permanently. If it's between these two numbers, then the solid beam will bend, while the I-beam won't. Observe that due to this, a similar case can be made for elastic (non-permanent) movement, for a given moment, the I-beam will deflect less.
I trust that this helps everyone understand how material removal can help strengthen a beam (or conversely, material addition can weaken a beam
).
Brought to you by a second-year mech-eng student, the letter sigma, and the number 10,000.
On a side note and for my own personal amusement, observe how in the above, if we remove the middle component of the I-beam, and are thus left with two supporting beams, I will still be greater than that of the standard beam (1235 compared with 833.3 mm^4). I leave this as a problem to the reader.
As mjolnirman has said, it's hard to type out the formulae, so I'm going to have to hope that everything comes out right and I don't misplace anything.
An actual bayonet gets rather complicated since you have dimensional variations in all axes (cross-section varies as you go up and down the blade), but the same principle applies, so let's use I beams.
For a rectangle, the second area moment is Ix = (b*h^3)/12. Note that this means I will change depending on which way the force is acting. Iy = (h*b^3)/12. Let's take a solid beam that is 10 mm by 10 mm. It's a square, so I in either direction will be 833.3 mm^4.
For composite objects (like an I beam), we need to find out the 2nd area moment, like before, and also the neutral axis of this composite. Fortunately for us, due to symmetry the neutral axis is identical to that of the simple beam. Let's take the same 10 x 10 beam, but now machine two grooves in the sides (read: remove material). Symmetrical in both x and y directions, but obviously not to scale:
Sounds good?
Since we're discussing bayonets, if you draw the cross-section as I've described it above, we're interested in forces going from left to right, or vice versa.
To find the 2nd moment for this composite beam, we need to break it down and use the parallel axis theorem, I(around an axis) = I(element) + A*d^2 where A is the cross-sectional area of the element, and d is the distance between the axis of the overall object and the axis of the individual element.
I of the second element (6x4) is easy to compute, using Iy = (h*b^3)/12 = (4*6^3)/12 = 72 mm^4.
Thanks to symmetry we can double I for either of the two other elements. For one element:
Area of one element = 3*10 = 30 mm^2.
Distance from neutral axis to centroid of object's moment = 3.5 mm (a quick drawing will show this quite easily).
I(around neutral axis) = (3*10^3)/12 + (3*10)*3.5^2 = 617.5 mm^4.
That's for one of the side elements.
Now we just sum it all together to get Iy = 72 + 2(617.5) =1307 mm^4.
Compare this with the solid beam's value for I of 833.3 mm^4.
That's the basic stuff out of the way.
So what happens when we put this under stress? If we exert a force at one end, we introduce a moment into the beam, this moment acts on each tiny little element of the beam to stretch or compress it. The formula is sigma = My/I where y is the distance from the neutral axis to the element in question. The maximum stress will thus be found at the point the furthest from the neutral axis. If the stress exceeds a certain limit, the element will permanently deform. Those of you who are interested can look up yield strengths, but I'll leave those for another day, all that matters here is that we want to keep sigma as low as possible.
In both cases, the max distance will be 5 mm. Let's take a moment of 10,000 N.mm, because I'm uncreative. For the solid beam:
sigma = 10,000*5/833.3 = 60 N/mm^2.
For the I beam:
sigma = 10,000*5/1307 = 38.26 N/mm^2.
If the yield strength of the material in question is less than 38.26 N/mm^2, nothing happens to either beam. If it's greater than 60 N/mm^2, both will bend. Permanently. If it's between these two numbers, then the solid beam will bend, while the I-beam won't. Observe that due to this, a similar case can be made for elastic (non-permanent) movement, for a given moment, the I-beam will deflect less.
I trust that this helps everyone understand how material removal can help strengthen a beam (or conversely, material addition can weaken a beam
).Brought to you by a second-year mech-eng student, the letter sigma, and the number 10,000.
On a side note and for my own personal amusement, observe how in the above, if we remove the middle component of the I-beam, and are thus left with two supporting beams, I will still be greater than that of the standard beam (1235 compared with 833.3 mm^4). I leave this as a problem to the reader.
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