**Determine reactions at supports of loaded beam:**

**Determine reactions at supports ***A *and *B *of loaded beam as shown in the figure given below.

**Sol.: **First consider the *free body diagram *of the figure give to us. In which ?*CEA*, *AED *and

*FH**G *shows point load and rectangle *FHDB *shows point load.

Point load of ?*CEA *= 1/2 × *AC *× *AE *= 1/2 × 1 × 10 = 5KN, act at a distance 1/3 of *AC *(that is, 0.333m) from point A

Point load of ?*AED *= 1/2 × *AD *× *AE *= 1/2 × 2 × 10 = 10KN

act at a distance 1/3 of *AD *(that is, 0.666m )from point A

Now divided the diagram *DBGF *in to two parts A ?*FHG *and a rectangle *FHDB*. Point load of Triangle *FHG *= 1/2 × *FH *× *HG *= 1/2 × 3 × (20 - 10) = 15KN

act at distance 1/3 of FH (that is, 1.0m )from point *H*

Point load of Rectangle *FHDB *= *DB *× *BH *= 3 × 10 = 30KN

act at a distance 1/2 of *DB *(that is, 1.5m )from point *D*

At Point *A *roller support that is, only vertical reaction (*R**A**V*), and point *B *hinged support that is, a horizontal reaction (*R**B**H*) and a vertical reaction (*R**B**V*). All point load are shown in the given figure

*∑H *= 0;

*R*_{B}_{H} = 0

*R*_{B}_{H} **= 0 .......ANS**

*∑V *= 0;

*R*_{A}_{V} + *R*_{B}_{V} - 5 - 10 - 30 - 15 = 0

*R*_{A}_{V} + *R*_{B}_{V} = 60KN ...(*i*)

Now taking moment about '*A*'

- 5 × 1/3 + 10 × 0.66 + 30 × 3.5 + 15 × 4 - *R**B**V *× 5 = 0

*R*_{B}_{V} **= 34 KN .......ANS**

Putting the value of RDV in the equation (1)

*R*_{A}_{V} **= 26KN .......ANS**