SOLVE FOR #$3$ USE THE BEAM AT THE TOP OF THE PAGE Another classic: the uniformly$-$loaded constant$)$ simply$-$supported beam. Using our

steps above, do the following: a$)$ Solve for $w(x)$; b$)$ show that $d\frac{w}{d}x(\frac{L}{2})=0$ to assure that the

maximum displacement occurs at $\frac{L}{2}$;$c)$ find the maximum displacement at $x=\frac{L}{2}$;$d)$ find the

value of $V(0)$ and notice it is the same force we would have obtained by statics. $($Answers: $w(x)=$

$(\frac{q_o}{E}l)(L\frac{x^3}{12}-\frac{x^4}{24}-L^3\frac{x}{24}),$ max value is $5q_o\frac{L^4}{384}El,$ and shear force value is $q\frac{L}{2}.)$ Note

that as always for our problems, El$=$constant, length $=L.$

Consider problem $1$ above EXCEPT now put a roller at the right end so that it cannot move up or

down. The point moment is still there. This problem is very famous and we can use its results to

help us solve other problems. Do the following: a$)$ solve the $4-$th order ODE to find $w(x)($Answer:

$\{$:$\frac{M^{**}\frac{x^3}{4}L-M^{**}\frac{x^2}{4}}{E}l)$; solve for the end reactions of the beam $($M$(0),$V$(0),$ and $\{$:$V(L))$ to show

that the moment at the fixed support is exactly half the applied point load value going in the

same physical direction. This is something called "Moment Carry$-$Over" and is an especially

useful concept. $($The shear values are $3\frac{M^{**}}{2}L$ going up at the left and the opposite on the right$).$

$BC$ hints at bottom of page.