Calculating Psi from Rectangular Box Experiment

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The experiment involved loading a rectangular box with weights until mechanical failure to assess the material's strength. Calculating psi based solely on the surface area may not yield accurate results, as the failure could depend on how the weight was applied and the box's geometry. If the load was applied symmetrically, the walls would experience uniform compression, and psi could be calculated using the total weight divided by the cross-sectional area. However, more detailed information about the dimensions, material, and failure mode is necessary for a reliable calculation. Understanding the nature of the failure is crucial for determining the appropriate testing method.
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We performed an experiment on a rectangular box of measured dimensions made from a certain material. We loaded the box with weights until mechanical failure. So, I know the dimensions of the box and the amount of force put on it until it failed. Would I be able to accurately calculate and report the psi of the material we used by simply calculating the surface area of the box and doing the math? Or, is this test inaccurate because the point of mechanical failure might be dependent on the geometry of the rectangular box?
 
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Was this experiment designed to test the strength of the component? Or was it happenstance that you applied too much weight and it failed?
 
Mech_Engineer said:
Was this experiment designed to test the strength of the component? Or was it happenstance that you applied too much weight and it failed?
It was designed to test the strength of the component.
 
Tryzxobbnym123 said:
Would I be able to accurately calculate and report the psi of the material we used by simply calculating the surface area of the box and doing the math?
The answer is almost certainly NOT based on the surface area of the top of the box. The method would depend on how the weight was applied to the box. For example the top of the box might fail due to point loads of the applied weight.

If the box was loaded symmetrically then all the walls would have been under compression. If the walls were then crushed, and so failed without buckling, the “psi” would be the total weight in pounds, divided by the cross sectional area of the walls measured in square inches.

You need to give a lot more information about dimensions, material and mode of failure before a reliable answer can be given.
 
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Are you trying to create a 3-point flexural bending test?

Three_point_flexural_test.jpg
 
Please describe the nature of the failure.
 
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