Q. A simply supported beam of span 3 m carries two point load of 15 kN and 10 kN are acting at distances of 1 m and 2 m from left hand support. Draw SFD and BMD. Find max S.F and max В.М and indicate where it occurs.

SOLUTION:

To find reactions
Total load= 15 + 10 = 25kN
Taking moments about A
RB × 3 = 10 × 2 + 15 × 1 = 35
RB = 35/3 = 11.67kN
RA = 25 – 11.67 = 13.33kN

(or)
Taking moments about B
RA × 3 = 15 × 2 + 10 × 1 = 40
RA = 40/3 = 13.33kN

S.F.D (SHEAR FORCE DIAGRAM )

Start drawing S.F.D from left hand support

FA = 13.33kN (and same in between AC)
FC =(just left of c )=13.33 kN
FC =(just right of c )=-1.67 kN (and is same between C & D)
FD =(just left ofD )=-1.67 kN
FD =(just right ofD )=-11.67 kN (and is same between DB)
FB = – 11.67kN
Max SF occurs at support A = 13.33kN

BMD ( SHEAR FORCE DIAGRAM)
M{A} = M{B} = 0
M{C} = 13.33 × 1 = 13.33kNm
M{D} = 11.67 * 1 = 11.67kNm
Max B.M occurs at C (i.e., where & SF changes its sign)
MC =M max =13.33 kNm