I don't understand why my #1 should be compressive strength; I am talking about toughness. Toughness is the energy required to break a sample, and there are many ways to break something. You can break it by stretching, or you can break it by compression, or you can break it by propagating fractures, etc. Each way of breaking the material can cause a different type of failure, which in turn may use up a different amount of energy. I specificaly meant compressive toughness. Similarly, meant tensile toughness as well.
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I think you are saying that tensile-strength is the integral of the stress-strain curve? If so, this is incorrect. Toughness is the integral of the stress-strain curve, because toughness is energy which is basically work = force * distance, where force is basically stress, and distance is basically strain. So to compute the energy required to break a sample, we can integrate the stress-strain curve until the sample breaks. This can be done for compressive toughness and also can be done for tensile toughness. Notice that this is a different type and a different test of toughness than the Charpy impact test.
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I don't know a lot about impacts, and am not an engineer. But as a physics major, I know that as the duration of the impact goes to zero, the force can go to infinity...
Dang, I typed a very convoluted reply and lost it ALL just now
But I'll try again.
First, the silly part, regarding force and time. A 2.5g penny fired at 1000 m/s has 500,000 Joules of kinetic energy. On impact with a human skull, it experiences a change in acceleration of <0.001 seconds, the equivalent of >100,000 G's, giving it a virtual weight of 2500 kg (2.75
tons)... that's a LOT of force!!! But it cannot penetrate a human skull, not even impacting on edge, a very small surface area.
Why not? Perhaps it has to do with
impulse, that the penny is carrying only 2.5 N*s momentum in its 2.5g mass, the 2500 Newton force cannot be transferred to the skull? I don't really get it, but what I am trying to demonstrate is that the math behind force reaching infinity as time is reduced to zero might be missing something...
ANYway, the material properties of tensile and yield strength are both measured in force, as in amount of force required to permanently bend or break the sample. The total applied force of our hypothetical sample in the suggested instance where full deceleration takes 0.0015 seconds (VERY quick deceleration) is only 3332 Newtons (where that all goes, who knows) against the steel's yield strength of 70,000 N/cc -
there isn't enough kinetic energy to generate the force required to break that knife at any section more than 0.05 cc in volume. Please understand how small that is. To reach that 70,000 N/cc, the kinetic energy of the thrown knife must be exerted as force back onto a focused volume within 68
microseconds, >20X faster than the suggested instance. Is it understood how FAST that is? In a real-world situation, this is the wrong path to tread.
You are absolutely correct about my confusion of the curve-integral = toughness vs. the curve itself = strength. I cannot even figure out which units of measurement to use!
So by way of apology, I took the time to gather and compute a lot more material-science data.
The 'toughness' we are discussing, energy required to induce fracture, equates to the "
modulus of resilience" in situations where we limit the discussion to yield strength as we did above (that 70,000 N/cc value). The
modulus of elasticity for 1095 steel (at any Rc apparently...) is 2x10^11 N/m^2; yield strength is converted values is 700x10^6 N/m^2, so the
Modulus of Resilience, i.e. the amount of energy able to be absorbed prior to fracture, equates to
1.2 J/cc of 1095 steel @ 60 Rc. Note that here the units are per volume. That drives me crazy. Shouldn't it be cross-sectional? Feel free to double-check my math.
But that's pretty low, right? The total kinetic energy of our thrown knife is 33.32 J ... but I still do not understand how we focus that into a unit
volume vs unit
area of cross-section
And perhaps yield strength & modulus of resilience isn't the way to go, after all we have the Charpy value. I think that, as you seem to surmise, the impact scenario attempts to model a collision where duration of impact (time through which force is applied,
impulse?) is minimized to 0. But impulse equates to change in momentum, which again is conserved. If we take the suggested 0.01 m penetration and 0.00143 seconds, 3332 Newtons of force is exerted but only 4.76 N*s impulse. What does that value even mean?
The Charpy value tells us how much energy is transferred into fracturing the sample upon impact, precisely what you were asking for - the force able to be applied over a realistic minimum duration of time.
1095 @ 60 Rc absorbs 20.4 J/cm^2 - that's much higher than the modulus value, and this value is used to define the 'toughness' of steel, not the modulus...
With 33.32 J of total energy to be distributed on impact, but resistance to fracture as high as 20.4 J/cc, if 40% of the kinetic energy is dissipated by other means on impact (into the target, as sound/vibration, as kinetic energy in an alternate direction), then fracture will not occur unless the structure of the material was below standard values, i.e. severely compromised.
P.S. Out of curiosity, about how fast is a knife tumbling when you throw it? I'm curious about the rotational energy versus the translational energy etc. Also, what is the approximate size, weight, and moment-of-intertia for a throwing knife? If you don't know the moment of inertia that's fine; we can just estimate it from the size and weight (approximate the knife as a rectangular bar, etc.).
Alright, I took the time to estimate:
Moment of inertia = mass*(a^2 + b^2 assumed rectangle)/12 = 0.34 kg * ((0.3m)^2 + (0.04m)^2)/12 = 0.0026 kg*m^2 for our maintained hypothetical sample
Using the excellent
website previously linked as my data source, the average 14 m/s throw includes 1 rotation every 2.16 translational meters = 1 revolution every 0.15 seconds.
Angular velocity = 2*pi*(revolutions/second) => 6.28/0.15 = 41.87 rads/s
Rotational energy = 1/2*(moment of inertia)*(angular velocity)^2 = 0.5*0.0026*(41.87)^2
= 2.28 Joules
SO shall we say that total kinetic energy, rotational + translation, is 35.6 Joules?
One of the complications is the difference between average force during the impact, versus peak-force. Average force can be estimated with some kind of analysis similar to the one you showed. As for peak-force, I don't know how to estimate it without doing complicated computer simulations. But there should be some way to make a reasonable guess at peak-forces? Maybe a mechanical engineer wants to chime in?
