ME-3263-Honors-Project

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This project contains the tools to solve for the maximum allowable load that can be applied to a steel, aluminum, or wood column without failure. It first asks for the function to be defined and the arguments to be inputted before running the program. This is where the type of material you are trying to solve for is determined and what the given values should entail. Then, it will ask for the user input on what the type of cross section the column is. Once entered, the program will ask for the necessary variables that need to be input to start running calculations. The final result will tell you the maximum allowable load that specific type of column can endure.
This program draws upon Euler's Critical Load Formula. With the previously given value of the Modulus of Elsasticity (in MPa) and the user input values of base, length, and radius (in terms of millimeters), the area of the cross section is calculated as well as the Moment of Intertia in the x and y-axis. The minimum value of the moment of inertias is taken and placed into the radius of gyration formula. The result is then used to divide the length of the column to produce the slenderness ratio. These values are inserted into the critical load formla. This formula allows for us to implement the slenderness ratio and the radius of gyrtation to formulate a reuduced equation to solve for the maximum allowable load. These calculations are of basic and generalized steel, aluminum and wood. While there are vast different grades of each metals, this serves as a simple proof of concept.

Example Problem:

The goal of this sample question is to calculate the allowable load that can be applied on the steel column. To do this, the user first calls the function for the specifc type of material and the arguments required for the column. In this case the user types:

findPressureSteel(5000,200000,250)

where the length of the column (5 m) is 5000 mm, the Modulus of Elasticity (200 GPa) is 200000 MPa, and Yielding Stress is 250 MPa. Note all argument and variables are in terms of mm and MPa.
The user runs the program and it has them input the type of cross section of the column. The example column's cross section is a "hollow rectanlge". The program will then ask for the variables required to run for the specific cross section. A hollow rectangle requries the thickness of the tube t, the base length b, and the height/width is h. The example problem shows a thickness of 8 mm, a base length of 127 mm, and a height/width length of 178 mm. The program only requires that the numerical value be input. Once these variables are plugged in, the program runs multiple equations based on the specific type of material, cross section, lengths and more. The final result will return the maximum allowable pressure load that can applied to the column in terms of kN. The maximum alowable load in this example is 414 kN.

This image represents the view that the user would see when the program runs and the input and reutrns of the variables and the final answers.

Function findPressureSteel (Function 1) and findPressureAluminum (Function 2) run similar equations but ask for different arguements. Function 1 asks for the length, Modulus of Elasticity, and the Yield Stress. Function 2 only asks for the length arguemnt. On the other hand, Function findPressureWood (Function 3) runs different equations but also asks for similar arguments. It asks for the length, Modulus of Elasticity, and the Compressive stress of the column. Each function and cross section will run similarily to the example shown above. The differences are only the arguements, the input, and the equations used to solve for the maximum allowable load.

Equations:

Moment of Inertia with respect to the x-axis


Moment of Inertia with respect to the y-axis


critical stress


radius of gyration


slenderness ratio

Steel:

steel constant


critical stress when slenderness ratio is greater than the steel constant


critical stress when slenderness ratio is less than the steel constant


allowable stress


allowable maximum load

Aluminum:

allowable stress when slenderness ratio is less than 66


allowable stress when slenderness ratio is greater than 66

Wood:

allowable stress


column stability factor

As shown in the figure above, the maximum load is applied axially on the columns. This creates a compressive stress. This applies to all materials and all cross sections that this program involes.

A point of thought to think of is the height, length, and width of a column. A long slender column fails by lateral buckling. A wider pier/squat column fails by crushing. The shorter a column is, the more load it can carry. There is also the idea of uniaxially loaded columns. Whne a column is subjuected to axial compression and a moment or a off axis load, is it uniaxially loaded. This can cause for the column to bend and flex due to the tension acting on one half and a compressive force acting on the other half of the column.

Work Cited:
Eurler's Critical Load Formula: http://www.engineeringcorecourses.com/solidmechanics2/C5-buckling/C5.1-eulers-buckling-formula/theory/
Eurler's Critical Load Formula Wood Columns: https://www.wood.tcaup.umich.edu/lectures/WoodColumns544_20.pdf
Moment of Inertia: https://wp.optics.arizona.edu/optomech/wp-content/uploads/sites/53/2016/10/OPTI_222_W61.pdf Uniaxially Loaded Columns : https://structurepoint.org/publication/pdf/Designing-Columns-for-Biaxial-Bending-using-Manual-Design-Procedure.pdf