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Hi! I hope you are having a happy new year!
How can I calculate how much force (in newtons) a 1/2 in and 1 in thick steel (preferably stainless steel) can take?
Thanks for your help!
Question Date: 2008-12-31
Answer 1:

Ooh - mechanically, this is out of my league, but intuitively, I think you want a force per unit area rather than a pure force, as well as be more specific about your question. You need to think of the difference between tensile strength and compression, as well as the geometry of the situation since almost any force you could apply is going to cause both tension and compression. In any case, the force is distributed over a small enough area, it will punch through the steel even if it would not were it distributed over a wider area. Think, for example, of getting stabbed with a sword or stabbed with a baseball bat - the force behind the thrust is the same, but the pressure of the sword point will be so much greater because the area of contact is so much smaller, and as a result the sword will pierce while the bat's blunt head will not.

You also need the constants for the tensile strength of steel. I don't know what these are or how to get those for compressive stress, but here is something you could calculate for the tensile stress: The guns of the Japanese WWII battleship /Yamato/ could fire shells massing approx. 1500 kg a distance of about 40 km (specifically, 45,000 yards). The angle that the guns could fire that could give them the maximum range will have them with equal parts vertical and horizontal velocity, so you can calculate the muzzle velocity from these numbers. At point blank range, those shells could penetrate a layer of steel 60 cm (2 ft.) thick. The cross-sectional shape of the shells is circular, and they had a diameter of 46 cm - but remember also that the shell is bullet-shaped and first contacts its target on the point of the bullet, so the actual force is distributed over a much smaller area than the cross-section of the shell. Anyhow, a thinner layer of steel would stop proportionately less of an impulse - a less massive shell, or one not moving as fast. Also, remember that this "force" is an IMPULSE, i.e. force integrated over time (it has units of N/s), hardly a static situation!

It should also be noted that the U.S. battleship /Iowa/ can fire shells that mass only about 1250 kg only 42,000 yards and have roughly equal penetrating power, so shape of the shell does matter!

Happy New Year by the way - it's nice to see you asking good, difficult questions again!

Answer 2:

This is a bit tricky as there are several possible meanings of "How much force can it take?", and furthermore, steel comes in a variety of grades, with greatly varying tensile and Youngs modulus factors.Finally, the question depends on how the plate is supported. A plate that is supported at its edges is elastically deformed (but will comeback to its initial position) by a small force away from the supports. A much larger force is required if the force is applied close to the support. Once you have reached the yield point, the material deforms unelastically -- so does not return to its original shape and position.Unfortunately, the yield point is strongly dependent on the alloys in the steel, i.e. how much carbon, tungsten, vanadium, molybdenum... are present in the iron as well as the morphology - i.e. how the steel was made. As a case in point, suitable heat treatment of a steel screw can change its yield point by a factor of 4-5x. Making the issue more complicated, the yield strength can be (and often is) highly directional in the final product -- witness the utility of 'forging'. Despite these complexities, there are classical strength tables for mechanical design which are based on averages for given steel types in texts like the "Machinery's Handbook" or ASTM tables. They allow calculation of deformation of a plate of a given thickness given shape,force and apsect ratio. One can also estimate the deformation by using an estimated Young's modulus and solving the stress PDE.

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