me3255_group5

Final Project (Fahim & Yasser) #Problem 1 Using analytical techniques to solve for the shape of the beam, the function shape_simple_support.m returns the displacement w(x) given q, the transverse distributed load and x, the location on the beam.

a.) The plot of q vs the maximum deflection of the beam is provided in the shape_simple_support_plot.m file in the repository and here as well:

b.) The Monte Carlo Method is used to determine the mean and standard deviation for the maximum deflection if the height and the width of the beam are normally distributed random variables with 0.1% standard deviations at q=50N/m.

#Problem 2 Central difference approximation was used to set up a system of equations of the beam when P=0, w=0 to determine the value of the endpoints. d. The plot of the numerical results of q vs dx is provided in the repository as problem2_plot.m and here as well:

e. For our graph for the analysis portion, there was linear decay which made sense because as the distributed load increases, the maximum deflection goes more negative. It is kind of the same as the numerical approach except the y values are positive for the numerical approach so that the maximum deflection increases as distributed load increases. Both mean the same thing but using different techniques to get the same result.

#Problem 3 The central difference method was used again except with P>0 for 6,10, and 20 segments.

d) The plots for problem 3 are done out and are located in the repository for when the load is P=0,100,200, and 300 for each segment. As loading increases, so does the deflection for each case.

#Problem 4 A function called eigenvalue.m was used to solve for the natural frequencies of the simply supported beam.

Frequency values for six segments:

First Natural Frequency	Second Natural Frequency	Third Natural Frequency
141.7854	529.1503	1.0583e+03

Frequency values for ten segments:

First Natural Frequency	Second Natural Frequency	Third Natural Frequency
143.8803	561.4373	1.2118e+03

Frequency values for twenty segments:

First Natural Frequency	Second Natural Frequency	Third Natural Frequency
144.7715	575.5213	1.2816e+03

4e. The plots for each segment and their natural frequencies are below by inputting N=6,10,20 and P=0:

For 6 segments:

For 10 segments:

For 20 Segments:

fmi13001/me3255_group5

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