$4-37.$ Draw the shear and moment diagrams for the beam.

SOLUTION

Referring to the FBD of the left segment of the beam sectioned at an arbitrary distance of $x$ from the left support, Fig. $a,$ and setting $V=0,$

$+$uarr$F_y=0$;$,7\mathrm{.}50-\frac{1}{2}(\frac{6}{3}x)x=0,x=2\mathrm{.}739m$

$+M=0$;$,M+\frac{1}{2}{|}_{{?}_{?}}\frac{6}{3}(2\mathrm{.}739{)}_{{?}_{?}}|(2\mathrm{.}739)(\frac{2\mathrm{.}739}{3})-7\mathrm{.}50(2\mathrm{.}739)=0$

$M=13\mathrm{.}69kN*m$

$=13\mathrm{.}7kN*m$

When $x=3m,$

darr$+M=0$;$,M+\frac{1}{2}{|}_{{?}_{?}}\frac{6}{3}(3{)}_{{?}_{?}}|(3)(\frac{3}{3})-7\mathrm{.}50(3)=0,M=13\mathrm{.}5kN*m$

Write an equation as a function of $x$ for the following: $xAB{V}_{\text{m}\text{a}\text{x}}=+-7\mathrm{.}5kN$

$M_{\text{m}\text{a}\text{x}}=13\mathrm{.}7kN*m4\mathrm{.}5,do$ the cut section from $Ato$ the $3rd$ cut. Then $do$ the cut section from $Bto$ the $3rd$ cut $to$ show how the lengths differ.

Ans.

$V_{\text{m}\text{a}\text{x}}=+-7\mathrm{.}5kN$

$M_{\text{m}\text{a}\text{x}}=13\mathrm{.}7kN*m4\mathrm{.}5.$ Make sure the lengths are $inx.$

For $4\mathrm{.}5,do$ the cut section from $Ato$ the $3rd$ cut. Then $do$ the cut section from $Bto$ the $3rd$ cut $to$ show how the lengths differ.

Ans.

$V_{\text{m}\text{a}\text{x}}=+-7\mathrm{.}5kN$

$M_{\text{m}\text{a}\text{x}}=13\mathrm{.}7kN*m3$ and $4\mathrm{.}5.$ Make sure the lengths are $inx.$

For $4\mathrm{.}5,do$ the cut section from $Ato$ the $3rd$ cut. Then $do$ the cut section from $Bto$ the $3rd$ cut $to$ show how the lengths differ.

Ans.

$V_{\text{m}\text{a}\text{x}}=+-7\mathrm{.}5kN$

$M_{\text{m}\text{a}\text{x}}=13\mathrm{.}7kN*m$