Answer to Question #182313 in Quantitative Methods for LF17 Gaming

Question

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.

Answer 1

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.

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

3 STEP. Part (a.i)

Let's deal with integration

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

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:

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

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