Simply supported beam - Bending Moment:
Sketch the SFD and BMD for a simply supported beam of 15 m span loaded as illustrated in Figure
Figure
Solution
Taking moment about A,
R _{B} × 15 - 1.5 × 6 × (9 + (6/2)) - (7.5 × 8) + 9 - ((½) × 6 × 3 × (2/3) × 6 ) = 0
R_{B} = 13 kN
R _{A} = (( ½) × 6 × 3?)+ 7.5 + (1.5 × 6) - R_{B} = 25.5 - 13 = + 12.5 kN
Shear Force (Beginning from the Left End A)
SF at A, F_{A} = + 12.5 kN
SF at C, F _{C} =+ 12.5 - ((½) × 6 × 3 ) = + 3.5 kN
SF just left of E, F_{E} =+ 3.5 kN
SF just right of E, F_{E} =+ 3.5 - 7.5 = - 4 kN
SF at F, F_{F} =- 4 kN
SF just left of B, F_{B} =- 4 - (1.5 × 6) = - 13 kN = Reaction at B.
Bending Moment (Beginning from Right End B)
BM at B, M_{B} = 0
BM at F,
M _{F} =+ (13 × 6) - (1.5 × 6 × (6/2) =+ 51 kN-m
BM at E, M _{E} =+ (13 × 7) - 1.5 × 6 × (1 + (6 /2)) =+ 55 kN-m
BM just right of D, M _{D} =+ (13 × 8) -1.5 × 6 × (2 + (6/2)) - 7.5 =+ 51.5 kN-m
BM just left of D, M _{D} = + 51.5 + 9 = + 60.5 kN-m
BM at C, M _{C} =+ (12.5 × 6) - (½) × 3 × 6 × ((1/3) × 6 ) =+ 57 kN-m