Answer to Question #173399 in Mechanical Engineering for Tyrese Panue

Question

3 The extension, , of a material with an applied force, , is given by .

a)     Calculate the work done if the force increases from 200N to 600N using:

i)    An analytical integration technique

ii)   A numerical integration technique

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

b)    Compare the two answers

c)     Using a computer spreadsheet increase the number of values used for your numerical method

d)    Analyse any affect the size of numerical step has on the result.

Answer 1

If we get F, is given by y = e^F x 1 x 10^-3. (e is to the power of F x 1 x 10; 10 is to the power of -3 in that equation)

A.i) $y=10^{-3}e^{F}$

$dy=10^{-3}e^FdF$

$W=\int_{100}^{500}F.dy$

$W=\int_{100}^{500}F.10^{-3}e^FdF$

$W=10^{-3}\int_{100}^{500}F.e^FdF$

$W= 10^{-3} [(F-1).e^F]_{100}^{500}$

$W=10^{-3}[499e^{500}-99e^{100}]$

$W=7×10^{216}J$

ii)Solving by Trapezoidal Rule

Let $n=4$

$h=(500-100)/4$

$x_o=100$

$x_1=200 ....x_4=500$

$y_o=10^{-3}100.e^{100}=e^{100}/10$

$y_1=10^{-3}200.e^{200}=e^{200}/20$

$y_2=10^{-3}300.e^{300}=e^{300}/30$

$y_3=10^{-3}400.e^{400}=e^{400}/40$

$y_4=10^{-3}500.e^{500}=e^{500}/50$

According to Trapezoidal rule

$W=(h/2)[y_o+y_4+2(y_1+y_2+y_3)]$

$W=2[e^{100}/10+e^{500}/50+2(e^{200}/20+e^{300}/30+e^{400}/40)]$

$W=2[2.69×10^{42}+2.8×10^{215}+2(3.61×10^{85}+6.47×10^{128}+1.3×10^{172})]$

$W=5.6×10^{215}J$

B)Answer obtained in part(ii) is lesser than that obtained in part(i).

C)If we increase the values of n to n=5,n=6,n=7...The obtained Answer is closer to the actual value.

D)As n becomes larger and larger,the more accurate the answer is.