$(\text{'}$rod length in $m$: $\text{'}))$

$(\text{'}$force in Newtons: $\text{'}))$

$($are the units correct?$)$

# elastic modulus of steel

# the circular constant $($this line is correct$)$

# convert units

# cross$-$sectional area

# The stretch, in meters

$*1000\mathrm{.}0$ # now in milimeters

ts

vill stretch $\%6\mathrm{.}03fmm.\%$ delta$\_$mm$)]$

Run your program as often as you'd like, before submittir

input values in the first box, then click Run program and

second box$\mathrm{.}2\mathrm{.}20$ LAB: Axial stretch a uniform rod in tension

The following equation computes the stretch in a uniform steel rod subject to a tension force using the following relation:

$=\frac{lF}{AE}$

The equation above is correct for all of the quantities in base $Sl$ units. So$,$ length $/$ in $m,$ cross$-$sectional area $A$ in $m^2,$ force $F$ in $N,$ and elastic

modulus $E$ in $Pa.$ The cross$-$sectional area for a circular rod of diameter $d$ is $A=\frac{d^2}{4}.$ We will consider steel, for which $E=200{10}^{9}Pa.$

The code below attempts to compute the stretch for a rod when the user enters the diameter, length, and force. The code has one or more

Python errors. In addition, there is an incorrect number somewhere. Using hand calculations and proper unit conversions, you find that the

correct result for $d=25mm,l=1m,$ and $F=1000N$ is about $=0\mathrm{.}010mm.$ You could do the same calculations for different $d,l,$ and $F$ to

test the code.

Ex: If the input is:

$25$

$1$

$1000$

then the correct last line of output would be:

The rod will stretch $0\mathrm{.}010mm\mathrm{.}2\mathrm{.}20$ LAB: Axial stretch a uniform rod in tension

The following equation computes the stretch in a uniform steel rod subject to a tension force using the following relation:

$=\frac{lF}{AE}$

The equation above is correct for all of the quantities in base SI units. So$,$ length $/$ in $m,$ cross$-$sectional area $A$ in $m^2,$ force $F$ in $N,$ and elastic

modulus $E$ in $Pa.$ The cross$-$sectional area for a circular rod of diameter $d$ is $A=\frac{d^2}{4}.$ We will consider steel, for which $E=20010Pa.$

The code below attempts to compute the stretch for a rod when the user enters the diameter, length, and force. The code has one or more

Python errors. In addition, there is an incorrect number somewhere. Using hand calculations and proper unit conversions, you find that the

correct result for $d=25mm,l=1m,$ and $F=1000N$ is about $=0\mathrm{.}010mm.$ You could do the same calculations for different $d,l,$ and $F$ to

test the code.

Ex: If the input is:

$25$

$1$

$1000$

then the correct last line of output would be:

The rod will stretch $0\mathrm{.}010mm.$

main.py

$d=$ float$($input$(\text{'}$rod dianeter in nn: $\text{'}))$ di we will need to convert this to $n$

$1=$ float $($input$(\text{'}$rod length in $m$ : $"))$

$F=$ float $($input$(\text{'}$force in Newtons: $"))$

Q save constants $($are the units correct?$)$

$E=200**10****,$i elastic modulus of steel

fron math import pi A the circular constant $($this line is correct$)$

$d_{-}n=\frac{d}{1000\mathrm{.}0},$ in convert units

$A^{-}=p^{**}{d}_{-}{m}^{{?}^{{?}^{?}}}\frac{2}{4\mathrm{.}0},$ U cross$-$sectional area

delta $=1**\frac{F}{A}**E,"$ The stretch, in weters

delta$\_$mm $=$ delta$*1000\mathrm{.}0$ in now in wilineters

a show the results

print$("$The rod will stretch $26\mathrm{.}03f$ nm$."x$ de$1$ta$\_$nn$)$

Run your program as often as you'd like, before submitting for grading. Below, type any needed

input values in the first box, then click Run program and observe the program's output in the

second box.

Enter program input $($optional$)$

$25$

$1000$

Input $(from$ above$)$ longrightarrow Output $(shown$ below$)$