Answer to Question #173399 in Mechanical Engineering for Tyrese Panue

Question #173399


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.



1
Expert's answer
2021-03-22T08:37:39-0400

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\u00d710^{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\u00d710^{42}+2.8\u00d710^{215}+2(3.61\u00d710^{85}+6.47\u00d710^{128}+1.3\u00d710^{172})]"

"W=5.6\u00d710^{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.


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