Answer to Question #182313 in Quantitative Methods for LF17 Gaming

Question #182313

2- The longitudinal extension of metal bar in direction of an applied force is

given by


y = LKeF×1×10−3


Where y is the longitudinal extension in m, L is the length of the bar in m which

is L= 0.15 (m), K is a constant depends on the material and is K= 1, and F is applied

force in N.

a) Find the work done if the force increases from 100 N to 500 N using:

i) An analytical integration technique

ii) A numerical integration technique (n=8 intervals)

[Note: the work done is given by the area under the curve]


b) Using a computer spreadsheet and recalculates step (ii) by increasing

the number of intervals to n=10 and compare your obtained results

with (i) and (ii)

c) Using Simpson’s rule to find the work done (n=8).

d) Analyse whether the size of numerical steps has effect on the obtained

result and explain why.


1
Expert's answer
2021-05-13T03:28:28-0400

1 STEP. Substitute all the constants that we know



"y(F)=0.15\\cdot1\\cdot10^{-3}\\cdot e^{F}\\to\\boxed{y(F)=1.5\\cdot10^{-4}\\cdot e^F}"



2 STEP. We will deal with the work done. As we know, work is equal to force multiplied by displacement, in our case angling.



"dA=F\\cdot dy=F\\cdot d\\left(1.5\\cdot 10^{-4}\\cdot e^{F}\\right)\\to\\\\[0.3cm]\ndA=1.5\\cdot 10^{-4}\\cdot\\left(F\\cdot e^{F}\\cdot dF\\right)\\to\\\\[0.3cm]\n\\boxed{A=1.5\\cdot10^{-4}\\int\\limits_{F_1}^{F_2}F\\cdot e^{F}\\cdot dF}"



3 STEP. Part (a.i)

Let's deal with integration



"\\int F\\cdot e^{F}\\cdot dF=\\int F\\cdot d\\left(e^{F}\\right)=F\\cdot e^{F}-\\int e^{F}\\cdot dF=\\\\[0.3cm]\n=F\\cdot e^{F}- e^{F}\\to\\boxed{\\int F\\cdot e^{F}\\cdot dF=\\left(F-1\\right)\\cdot e^{F}}\\\\[0.3cm]\nA=1.5\\cdot10^{-4}\\int\\limits_{100}^{500}F\\cdot e^{F}\\cdot dF=1.5\\cdot 10^{-4}\\cdot\\left.\\left(F-1\\right)\\cdot e^{F}\\right|_{100}^{500}=\\\\[0.3cm]\n=1.5\\cdot10^{-4}\\cdot\\left(499\\cdot e^{500}-99\\cdot e^{100}\\right)\\longrightarrow\\\\[0.3cm]\n\\boxed{A_{real}=1.5\\cdot10^{-4}\\cdot\\left(499\\cdot e^{500}-99\\cdot e^{100}\\right)\\approx1.05 \u00d710^{216}(N)}"

4 STEP. Part (a.ii)

1. Since no specific method of numerical integration is indicated, I will use the trapezoidal method.

2. In order not to increase the amount of work, I will attach all calculations in the form of an Excel table at the end of the solution, and I will only insert screenshots into the solution itself.

3.The integration formula by the trapezoid method has the form



"\\int\\limits_a^bf(x)dx=\\frac{\\Delta x}{2}\\cdot\\left(f(x_0)+2\\cdot\\sum\\limits_{k=1}^{n-1}f(x_k)+ f(x_n)\\right)"

More information :https://en.wikipedia.org/wiki/Trapezoidal_rule






Conclusion,



"\\boxed{A_{calc}(n=8)\\approx 2.67\\cdot 10^{217}(N)}"

5 STEP. Part (b)





Conclusion,



"\\boxed{A_{calc}(n=10)\\approx 2.1\\cdot 10^{217}(N)}"

As we can see



"\\boxed{A_{real}\\approx 1.05\\cdot 10^{216}<A_{calc}(n=10)<A_{calc}(n=8)}"

6 STEP. Part (c)


The integration formula by the Simpson's rule has the form



"\\int\\limits_a^bf(x)dx=\\frac{\\Delta x}{3}\\cdot\\left(f(x_0)+2\\cdot\\sum\\limits_{j=1}^{n\/2-1}f(x_{2j})+4\\cdot\\sum\\limits_{j=1}^{n\/2}f(x_{2j-1})+f(x_n)\\right)"


More information : https://en.wikipedia.org/wiki/Simpson%27s_rule



Conclusion,



"\\boxed{A_{calc}\\left(\\text{Simpson's rule}\\right)\\approx 1.75\\cdot 10^{217}(N)}"

7 STEP. Part (d)

Since the rule of trapeziums and Simpson's formula have different orders of accuracy:



"error\\left(\\text{Simpson's rule}\\right)=-\\frac{h^4\\cdot(b-a)\\cdot f^{(4)}\\left(\\xi\\right)}{180}\\\\[0.3cm]\nerror\\left(\\text{Trapezoidal rule}\\right)=-\\frac{h^3\\cdot n\\cdot f''\\left(\\xi\\right)}{12}\\\\[0.3cm]"

then it is advisable to compare only part(a.ii) and part(b).

We see that there is an increase in inaccuracy



"A_{calc}(n=10)\\approx 2.1\\cdot 10^{217}<2.67\\cdot 10^{217}=A_{calc}(n=8)"



But this increase is insignificant. This can be explained by the fact that the specified function "W(F)=F\\cdot e^{F}" grows too quickly and the division of the integration segment into 8 or 10 parts is too rough. Therefore, the numerical values differ by an order of magnitude from the analytical solution.


Note : calculation table


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