Python for ENG

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naab112

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I attached the following assignment. I need it to done by your own work and write a memo.

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Huron QHECH 630 kg/year- 3820 kolye Lake Erie QEOCE Figu springs ski reso CH CE Ontario Bros. Michigan QMHCM 810 kg/year См The EPA is also considering a bypass stream that would go directly to Lake Michigan to Lake Ontario with a flow rate of 20 km reduce the concentration of PCBs in Lake Michigan. This bypass do not change any of the existing flow rates, it would just be an additional flow out of Lake Michigan and into Lake Ontario. Report the potentia year in orde impact of the bypass. 6.2 A new type of chair for a ski lift has been developed, and the manufactun. Upon has designed a simplified model (Figure 6.4) of the chair's behavior loading a group of individuals. Recall that the basic spring equation is W = k·X, where W is the weight or force applied to the spring, k the spring constant , and x the displacement (or stretch) of the spring. You have been hired to calculate the total displacement of the system of springs and weights shown in Figure 6.4. The properties of the system are as follows: Parameter Value ki k₂ ka k ks w W 10,000.0 N.m 5,000.0 N.m 8,000.0 N.m 3,500.0 N.m 4,500.0 N·m 500.0 N 1,000.0 N 1,000.0 N Iyear year year yeas year Problems 109 Figure 6.4 A system of weights and linear springs that model a new chair lift design for kg/year ski resorts. ki ario acolo X1 W K2 Ka rectly from in order to W2 ypass does additional potential ka W w ks X3 nufacture ior upon W₃ Eon is anstant, The first step is to derive a force balance on each weight. For the first weight, W, you need to consider every spring touching the weight, including the displacement and direction of force: tem of em are W = k1.*1 - kz . (x2 - x) and for the second weight: W₂ = K2. x2 + K₃ . (x2 - x) - K4 (x3 - X 2) – kg (x3 - x2). After deriving the force balance for the third weight, the system of equations can be written as a linear matrix problem: [ki + K₃ -kz 0.0 41 500.0 1000.0 -K₃ k₂ + kz + k + ks -k4 - kg (6.16) ? ? ? 1000.0 3-L If the linear matrix has been derived correctly, the terms along the main diagonal will all be positive, off-diagonal terms will be negative, and the matrix will be symmetric. Solve the linear matrix problem and determine the displacement of each weight (i.e., each skier) in m. EGEN 102 Introduction to Engineering Computations and Applications Fall 2019 Homework #5 Due: Tuesday, February 26, 2019 at 12:00 to Brightspace folder Problem 6.2 in the textbook. I ı addition to the basic set of parameters given in the problem statement, the manuIQUÍMICI has also asked you to include a plot of the displacement of each weight (i.e., each skier) as a function of W3. The plot should have a range of values for W, on the x-axis (e.g., W3 = 500N, 1000N, and 2000N), and then the displacement (on the y-axis) of the weight(s). You may generate the plot using matplotlib or Excel. You are free to use any linear system solver from chapter 6, including the solvers that are part of the scipy and/or numpy packages (e.g., numpy.linalg.solve()).
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Explanation & Answer

Hi, Here are the answers dude. Please feel free to let me know if you need any other help regarding this. I am attaching the python code in the word document. Also, you can see them in the pdf as well. I am also uploading a concepts files including the the concepts and some reference sites for the ease of your understanding.

Concepts
1. Free Body Diagram (FBD)1, 2
The FBD or Free Body Diagram looks in to the bodies at equilibrium,
skier in this case to look in the forces acting upon them by different
elements.
For example, for the first skier the downward forces are W1 and
downward pool by spring 3 (with spring constant k3), which has a
downward stretch of x2(that of the Skier 2) less the downward stretch of
the skier 1 (x1).
The Upward forces are the pull in the spring 1 (with spring constant k1).
Since, the skier is at equilibrium, i.e., there is no acceleration acting upon
it, hence the downward forces and upward forces equate each other.

 W1 = k1 . x1 – k3.(x2 – x1)
2. Hooke’s Law for Spring3
The hook’s law for spring suggest that any force acting on a spring and
the displacement (stretch or shrinkage) are proportional. They can be
equated with a constant depending on different properties of the spring,
known as the spring constant.

F = K . x
Where,

1

F = Force acting on the Spring
x = displacement
K = The spring constant

https://en.wikipedia.org/wiki/Free_body_diagram

2

https://www.kpu.ca/sites/default/files/Faculty%20of%20Science%20%26%20Horticulture/Physics/Ch5%201%20-%20Static%20Equilibrium_0.pdf
3

https://en.wikipedia.org/wiki/Hooke%27s_law

3. Symmetric Matrix4
The Symmetric matrix is a square which remains unchanged when
transposed. Some properties of a symmetric matrix are –
i.
The transposed matrix and the original matrix are the same.
ii.
The elements on the main diagonal of the matrix remains
unchanged while transposing.
iii.
The elements are equal with following equation

aij = aji
4. Solution of Linear equations using numpy.linalg.solve(C, K)5, 6
The numpy linear equation solver takes two matrices as the input and
output the solution vector. For the solution of the following system of
the linear equation, the inputs are: first input – the coefficient matrix
and the second input – the constant matrix.
𝑐11
(𝑐21
𝑐31

𝑐12
𝑐22
𝑐32

𝑐13 𝑥1
𝑘1
𝑐23) (𝑥2) = (𝑘2)
𝑐33 𝑥3
𝑘3

 C.X = K
𝑐11
Where, C = (𝑐21
𝑐31

𝑐12
𝑐22
𝑐32

𝑐13
𝑥1
𝑘1
𝑐23), K = (𝑘2) and X = (𝑥2)
𝑐33
𝑥3
𝑘3

5. Matplotlib Plot Functions7
The matplotlib pyplot functions are used to create the plots of the displacements as
functions of the 3rd Weight.

4

https://en.wikipedia.org/wiki/Symmetric_matrix

5

https://docs.scipy.org/doc/numpy-1.13.0/reference/routines.linalg.html

6

https://docs.scipy.org/doc/numpy-1.13.0/reference/generated/numpy.linalg.solve.html#numpy.linalg.solve

7

https://matplotlib.org/api/pyplot_api.html


The Ski Lift Chair Problem
With FBD (Free Body Diagram) of the balances of the W1 and W2 and Hooke’s
law applied on the Springs k1, k2 and k3 the following equations of equillibrium
are given –
For First Weight, Eqn. (1):

W1 = k1 . x1 – k3.(x2 – x1)
For Second Weight, Eqn. (2):

W2 = k2.x2 + k3.(x2 – x1) – k4. (x3 – x2) – k5. (x3 – x2)
Hence, using the same FBD and Hook’s law on the third weight.

K4.(x3-x2)

K5.(x3-x2)

W3

 For Third Weight, Eqn. (3) :

W3 = k4. (x3 – x2) + k5. (x3 – x2)
Now, rearranging...


Anonymous
Very useful material for studying!

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