Impact Pressure/Energy for indentation

[donsrno1]

I'm trying to find what pressure or energy would be required to indent copper by 1mm. To be more specific it would be for a piece of copper sheet that was about 1mm thick and resting on some dust like particles so nothing too hard underneath. It would be struck by a pin like object that had a rounded end of about 1mm diameter.

If anyone could help with what the best way to calculate this would be it would be much appreciated.

I posted this in the materials section of the forum as well but thought that I may have better luck posting here.

Thanks.

 

[BATman2]

donsrno1,

I would conduct a simple small scale quasi-static test to start. Depending on the pressure distribution, amount, and the spacing of the "particles", gross deformation might also be of concern at 1mm of copper.

To measure the depth of the indentation after test, I would apply dental impression material to the sheet.

Another approach may be to use a Rockwell hardness tester or similar.

Batman2

 

[BATman2]

donsrno1,

Consider using hardness testing formulas..

http://www.calce.umd.edu/general/Facilities/Hardness_ad_.htm

Brinell Hardness Test

Brinell hardness is determined by forcing a hard steel or carbide sphere of a specified diameter under a specified load into the surface of a material and measuring the diameter of the indentation left after the test.The Brinell hardness number, or simply the Brinell number, is obtained by dividing the load used, in kilograms, by the actual surface area of the indentation, in square millimeters.The result is a pressure measurement, but the units are rarely stated [5].

The Brinell hardness test [6] uses a desk top machine to press a 10mm diameter, hardened steel ball into the surface of the test specimen. The machine applies a load of 500 kilograms for soft metals such as copper, brass and thin stock. A 1500 kilogram load is used for aluminum castings, and a 3000 kilogram load is used for materials such as iron and steel. The load is usually applied for 10 to 15 seconds. After the impression is made, a measurement of the diameter of the resulting round impression is taken. It is measured to plus or minus .05mm using a low-magnification portable microscope. The hardness is calculated by dividing the load by the area of the curved surface of the indention, (the area of a hemispherical surface is arrived at by multiplying the square of the diameter by 3.14159 and then dividing by 2). To make it easier, a calibrated chart is provided, so with the diameter of the indentation the corresponding hardness number can be referenced. A well structured Brinell hardness number reveals the test conditions, and looks like this, "75 HB 10/500/30" which means that a Brinell Hardness of 75 was obtained using a 10mm diameter hardened steel with a 500 kilogram load applied for a period of 30 seconds. On tests of extremely hard metals a tungsten carbide ball is substituted for the steel ball. Among the three hardness tests discussed, the Brinell ball makes the deepest and widest indentation, so the test averages the hardness over a wider amount of material, which will more accurately account for multiple grain structures, and any irregularities in the uniformity of the alloy.

The Brinell hardness test was one of the most widely used hardness tests during World War II [7]. For measuring armour plate hardness the test is usually conducted by pressing a tungsten carbide sphere 10mm in diameter into the test surface for 10 seconds with a load of 3,000kg, then measuring the diameter of the resulting depression. Several BHN tests are usually carried out over an area of armour plate. On a typical plate each test would result in a slightly different number. This is due not only to minor variations in quality of the armour plate (even homogenous armour is not absolutely uniform) but also because the test relies on careful measurement of the diameter of the depression. Small errors in this measurement will lead to small variations in BHN values. As a result, BHN is usually quoted as a range of values (e.g. 210 to 245, or 210-245) rather than as a single value.

The BHN of face hardened armour uses a back slash ?\? to separate the value of the face hardened surface from the value of the rear face. For example, a BHN of 555\353-382 indicates the surface has a hardness of 555 and the rear face has a hardness of 353 to 382.

The Brinell Hardness Test described above is called ?HB 10/3000 WC? and was the type of test used by the Germans in World War II. Other types of hardness tests use different materials for the sphere and/or different loads. Softer materials deform at high BHN which is why tungsten carbide (a very hard material) is used to measure armour plate. Even so, as the BHN goes above 650 the tungsten carbide ball begins to flatten out and the BHN values indicate a greater difference in hardness than there actually is, while above 739 the ball flattens out so badly that it cannot be used.

When there are widely different values for quoted BHN then the cause may be use of a Poldi Hardness Tester instead of the Brinell Hardness Test. The Poldi Hardness Tester is less accurate but could be used in the field. The Poldi Hardness Test has the advantage that the testing unit is portable, so measurements can be carried out in the field, e.g., on captured enemy vehicles after a battle. The Poldi portable unit relies on a hammer blow impression in a standardized sample. This test is much less accurate than the Brinell Hardness Test.

ASTM E-10 is a standard test for determining the Brinell hardness of metallic materials. The load applied in this test is usually 3,000, 1,500, or 500 kgf, so that the diameter of the indentation is in the range 2.5 to 6.0 mm. The load is applied steadily without a jerk. The full test load is applied for 10 to 15 seconds. Two diameters of impression at right angles are measured, and the mean diameter is used as a basis for calculating the Brinell hardness number (BHN), which is done using the conversion table given in the standard [8].

 

[rb1957]

"nothing too hard undeneath" ... doesn't that mean that there's nothing to react the load, so alot of the applied energy will be wasted in moving the sheet, or maybe bending it ...

 

[Cockroach]

Take a look at your static textbooks, first year courses. Typically these computations are handled using Work-Energy Theorem.

Recalling that work is the applied force over a distance, then you can call work that of a resisting medium times distance of penetration. This is what you are looking for.

On the energy side, you need to understand that the TOTAL energy equation is that of potential and kinetic. Obviously the kinetic energy assumed from potential is the work content.

Finally, you can adjust the model for impact efficiency by noting elastic/inelastic collisions. In your case, it is obvious that in the absence of deformation the elastic case would apply.

So for example, a steel ball of mass falls from a specified height onto a copper plate, leaving an indent of 1.0 mm. Compute the resisting force. Clearly if that force over the surface area of the ball is greater than material yield, deformation occurs and further analysis would be by inelastic collision. Otherwise, elasticity applies and proceed to compute the velocities involved and imparted energy of the impact.

This would allow you to approximate resisting forces involved with a collision.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada