EXAMPLE $1\mathrm{.}1$

Determine the resultant internal loadings acting on the cross section at $C$ of the cantilevered beam shown in Fig. $1-4$a$.$

SOLUTION

Support Reactions. The support reactions at $A$ do not have to be determined if segment $CB$ is considered.

Free$-$Body Diagram. The free$-$body diagram of segment $CB$ is shown in Fig. $1-4$b$.$ It is important to keep the distributed loading on the segment until after the section is made. Only then should this loading be replaced by a single resultant force. Notice that the intensity of the distributed loading at $C$ is found by proportion, i$.$e$.,$ from Fig. $1-4$a$,\frac{w}{6}m=\frac{270\frac{N}{m}}{9}m,w=180\frac{N}{m}.$ The magnitude of the resultant of the distributed load is equal to the area under the loading curve $($triangle$)$ and acts through the centroid of this area. Thus, $F=\frac{1}{2}(180\frac{N}{m})(6m)=540N,$ which acts $\frac{1}{3}(6m)=2m$ from $C$ as shown in Fig. $1-4$b$.$

Equations of Equilibrium. Applying the equations of equilibrium we have

$+F_x=0$;$,-N_C=0$

$N_C=0$

$+$uarr$F_y=0$;$,V_C-540N=0$

$,V_C=540N$

$+M_C=0$;$,-M_C-540N(2m)=0$

$,M_C=-1080N*m$

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NOTE: The negative sign indicates that $M_C$ acts in the opposite direction to that shown on the free$-$body diagram. Try solving this problem using segment $AC,$ by first obtaining the support reactions at $A,$ which are given in Fig. $1-4$c$.$

PLEASE SOLVE FOR A$,$ im having trouble getting the MA and FA