A material test, however, requires knowing the dimensions of the cube to properly report its findings. Let’s assume the cube is one inch in length on each edge, and it takes 100 pounds of force to break the cube in half. For a force test, the 100 pound-force value is all that’s required for the force system to collect as a data value, but for a material test the system needs to translate that information to PSI. Because it is a perfect cube breaking evenly in half (without stretching), the cross section is one inch by one inch square. To determine the stress at break, the 100 pound-force is divided by the cross- sectional area of 1 square inch, which equals 100 pound-force per square inch, or 100 PSI. To help put this into context for factoring in variables when material testing, let’s think of a cube with two-inch edges, made of a different material that
A spring compression (load) test. The spring follows a linear compression and reaches a maximum load at approximately 350 pounds for one sample, and 500 for another. Pressure testing is not required for this test, and elongation is not calculated, so the stress/ strain values are not graphed.
also breaks at 100 pounds-force. Now that the cube is bigger, so is the cross-section, which is now two inches by two inches, or four-square inches. When dividing the 100 pounds-force by that area, the result is 25 PSI. It is a lower stress value – an indication of a generally weaker material. Consider that the area where the break occurs on the second cube is four times the area of the original, but still breaks at 100 pounds of force. If you had a cube of the second material that was one inch on each side like the first one, you can assume that it would break at a lower load value. You can even estimate that value: 25 PSI times 1 in2 is equal to 25 pounds. Operating with that same information, you can determine that the second cube, being two inches on each edge, would have a breaking load of 400 pounds-force; 100 PSI times 4 in2 equals 400 pounds. The example provided ignores many real-world behaviors for the sake of simplicity, but the general application is the same – material tests determine the inherent properties of what is being tested, and need to be scaled appropriately for when the samples change size so that they collect information accurately. A force test does not require that information, and can note behaviors under load without needing the material sample test data. Looking at the first test example again, the test goal could be that the sample needs to
break before exceeding 100 pounds of force to pass the test. The second material would pass as a one- inch or two-inch cube, but the first would only pass as a one-inch cube, because a two-inch cube would require four times the load to break. Force Testing Examples One of the better examples of a consistently manufactured product that must conform to specific force application and displacement standards are wire coil springs, made for either tension or compression. When testing, the springs are most often checked to see if they either extend or compress to the appropriate distance to determine their spring rates (in units of force per distance such as LBF/in, or N/mm) meet with the design
A material test progressing to sample break, determining the maximum stress ( σ max) and the strain value at that point ( ε @max). A cross-sectional area and gage length was provided for the controller to calculate the stress from the load and the strain values. The Load at Maximum Stress (L@max) and travel distance at that point (D@max) are calculated from these values.
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