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solution: \rightarrowTo find supporf reaction: \rightarrow\rightarrow The given beam is of simplysupported overhang type beam\rightarrow given beam is determinate beam\rightarrow Hence we can use equilibrium conditions only\rightarrow As the loading is vertical,\text { Henie } \begin{array}{c}\sum f_{y}=0 \\\sum M_{O A}=0\end{array}\Rightarrow \sum \sum_{y}=0 \Rightarrow algebric sum of all vertical forces is zero.\begin{array}{l}\Rightarrow \quad F_{y}=0(+\downarrow) \\\Rightarrow \quad R_{A}+R_{B}-(6 \times 1.5)-(9 \times 3)-(6 \times 1.5)=0 \\\Rightarrow \quad R_{A}+R_{B}-9-27-9=0 \\\Rightarrow \quad R_{A}+R_{B}-45=0 \\\Rightarrow \quad R_{A}+R_{B}=45\end{array}\Rightarrow take summation of moment about point A is equal to zero\Rightarrow \sum M M_{A}=0,((t, S) \text { as clockwise moment = positive }anticlockwise moment negativeScanned withCamscanner\begin{aligned}& \text { EMAA }=0 \\\Rightarrow \quad & -(6 \times 1.5) \times\left(\frac{1.5}{2}\right)+(9 \times 3) \times(1.5)-R_{B} \times 3+(6 \times 1.5) \times 3.75=0 \\\Rightarrow \quad & -(9 \times 0.75)+(27 \times 1.5)-R_{B} \times 3+9 \times 3.75=0 \\\Rightarrow \quad & -6.75+40.5-R_{B} \times 3+33.75=0 \\\Rightarrow \quad & -R_{B} \times 3+67.5=0 \\\Rightarrow \quad & R_{B \times 3}=67.5 \\\Rightarrow \quad & R_{B}=\frac{67.5}{3}\end{aligned}\Rightarrow \quad R_{B}=22.5 \mathrm{kN} (4) As Answer is positive, so our Assummed direction of force is correct.put R_{B}=2.2 .5 \mathrm{kN} in egh (1)\begin{array}{l}R_{A}+f_{B}=45 \\R_{A}+22.5=45 \\R_{A}=45-22.5 \\R_{A}=22.5 \mathrm{kN} \\R A=22.5 \mathrm{kN}(\$)\end{array}SummaryReaction at A=R_{A}=22.5 \mathrm{kN}Reaction At B=R_{B}=22.5 k_{N}Note: \rightarrow Both reactions at A \& & are same due to IS scanned symmetic loading on beamsummary : reaction at A =Ra=22.5 kn                    Reaction at B= Rb=22.5 kn Note : Both reaction at support are equal due to symmetrical loading acting on beam span.  ...