A planar $($i$.$e$.$ two$-$dimensional$)$ hanging structure is shown in Figure $1.$ The coordinates $($in mm$)$ and connectivity are:

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coor$=1000*[-1\mathrm{.}5863542487700591\mathrm{.}132433911165869$ ;

$-0\mathrm{.}9328770356499280\mathrm{.}375487610721078$ ;

$00\mathrm{.}015292535957528$ ;

$0\mathrm{.}9328770356499280\mathrm{.}375487610721078$ ;

$1\mathrm{.}5863542487700591\mathrm{.}132433911165869$ ;$]$;

conn$=[12$; $23$; $34$; $45$; $]$;

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$($You can Screen$-$copy, CTRL$+$C $+$ CTRL$-$V$)$ the numbers from the pdf to your m$-$file, do not re$-$type them by hand!$)$

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There are no initial elongations. Joints $1$ and $5$ are foundation joints. The structure is made up of $4$ pin$-$jointed bars with EA$=2000$kNmm$2.$

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$1$a$)$ How many states of self$-$stress are there and for each state of self$-$stress, describe the distribution of forces in the members.

$($Screen copy the numerical values to your Word file$).$

$1$b$)$ How many mechanisms are there?

$($Screen copy the numerical values of the mechanisms to your Word file & do a $$save$$ on the file now$).$

$1$C$)$ The mechanisms in Q$1$b can be better viewed by preventing Joint $3$ from vertical movements, and then, again, separately, from horizontal movements. Show the mechanism that results each time, both screen copying the numerical values and describe what it looks like $($e$.$g$.$ sketch on paper$).$

$1$e$)$Three equal vertical download loads of $5$N $($i$.$e$.$ in negative y$-$direction$)$ are now applied at Joints $2,3$ and $4.$ Use your linear$-$program to calculate the displacements and bar forces. $($Screen copy the numerical values to your Word file.$)$

$1$f$)$Apply the same loading case of Q$1$d to your $2$D nonlinear program. $($It is suggested you use the fopt command, and $$ones$$ as initial guess for all displacements and tensions in fsolve.$)$

the code provided to us is clear

$\%$ start with a memory clear

coor$=1000*[-(1+$sqrt$(3))/2-(3+$sqrt$(3))/6$; $\%$ x$-$ and y$-$coordinates, size nJ by $2$

$0(3+$sqrt$(3))/3$;

$(1+$sqrt$(3))/2-(3+$sqrt$(3))/6$;

$-0\mathrm{.}5-$sqrt$(3)/6$;

$0$ sqrt$(3)/3$;

$0\mathrm{.}5-$sqrt$(3)/6$;$]$;

conn$=[14$ ; $\%$ connectivity, size nB by $2$

$25$ ;

$36$ ;

$45$ ;

$56$ ;

$46$ ;$]$;

nJ$=$size$($coor$,1)$; $\%$ finding num of joints from largest number in conn

nB$=$size$($conn$,1)$; $\%$ finding numb bars from numb of rows in conn

$\%$ plot the structure with joint numbers

clf; $\%$ start with a "clear figure", ie wipe clean any current plot

hold on;

for i$=1$:nB$,$

j$1=$conn$($i$,1)$; j$2=$conn$($i$,2)$;

plot$([$coor$($j$1,1)$ coor$($j$2,1)],[$coor$($j$1,2)$ coor$($j$2,2)],\text{'}$r$-\text{'})$;

$\%$ if plotting $3$D structure, then use plot$3$

$\%$plot$3([$coor$($j$1,1)$ coor$($j$2,1)],[$coor$($j$1,2)$ coor$($j$2,2)],...$

$\%[$coor$($j$1,3)$ coor$($j$2,3)],\text{'}$r$-\text{'})$;

end;

for i$=1$:nJ$,$ text$($coor$($i$,1),$coor$($i$,2),$ num$2$str$($i$))$; end;

axis$(\text{'}$square$\text{'})$; grid;

EA$=200000*[505050101010]$;

$\%$ setting up the matrices

H$=$zeros$(2*$nJ$,$nB$)$; F$=$zeros$($nB$,$nB$)$; L$=$zeros$($nB$,1)$;

for i$=1$:nB$,\%$ looping through all the bars

j$1=$conn$($i$,1)$; j$2=$conn$($i$,2)$; $\%$ making j$1$ and j$2$ the $1$st and $2$nd joints of bar $1$

r$1=($j$1-1)*2$; r$2=($j$2-1)*2$; $\%$ r$1$ and r$2=$ how many rows before the rows in H for j$1$ and j$2$

x$1=$coor$($j$1,1)$; x$2=$coor$($j$2,1)$; $\%$ x$-$coor of j$1$ and j$2,$ x$-$coor in column $1$

y$1=$coor$($j$1,2)$; y$2=$coor$($j$2,2)$; $\%$ y$-$coor of j$1$ and j$2,$ y$-$coor in column $2$

xdiff$=$x$1-$x$2$; ydiff$=$y$1-$y$2$; $\%$ diff in the x$-$ccords, and the y$-$coords

L$=$sqrt$($xdiff$^2+$ydiff$^2)$; $\%$ calc length of bar i by Pythagoras

$\%$ putting the h of the bar straight onto the H of structure, note: a new col for each bar

H$($r$1+1,$i$)=$xdiff$/$L; H$($r$1+2,$i$)=$ydiff$/$L; $\%$ note: Xdiff$/$L is the cos of the inclination angle

H$($r$2+1,$i$)=-$xdiff$/$L; H$($r$2+2,$i$)=-$ydiff$/$L;

F$($i$,$i$)=$L$/($EA$($i$))$; $\%$ putting f of bar $($ie l$/$EA$)$ into Flex matrix of structure

end;

H$=$H$([7$:$12],$:$)$; $\%$ row reduction in H because of supports. Note, only rows, not cols as well

e$0=$zeros$($nB$,1)$; e$0(6)=-0\mathrm{.}048$; $\%$ e$0$ is vector of initial bar length imperfection

P$=[0-70000-70]\text{'}$; $\%$ the load vector,

$\%$ the analysis routine

S$=$null$($H$)$; M$=$null$($H$\text{'})$;

tH$=$pinv$($H$)*$P;

alpha$=$inv$($S$\text{'}*$F$*$S$)*(-$S$\text{'}*($e$0+$F$*$tH$))$;

t$=$tH$+$S$*$alpha;

e$=$e$0+$F$*$t;

d$=$pinv$($H$\text{'})*$e;

P$\text{'}*$M $\%$ testing to see if load is exciting mech

S $\%$ the states of selfstress $($if any$)$

M $\%$ the mechanisms $($if any$)$

t$\text{'}\%$ bar tensions

d$\text{'}\%$ nodal displacements

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