Evaluate Bending moment for overhanging beam:
Draw the shear force & bending moment diagrams for 15 m span overhanging beam, that overhangs on both sides. This is subjected to a u.d.l. of 5 kN/m on left side overhanging portion of length 5 m and a u.d.l. of 4 kN/m on right side overhanging portion of length 4 m. Mention the numerical values at all of salient points.
Solution
To determine Reaction at B
letting right side, the moments around A,
M_{A} =+ R _{B} × 6 - 4 × 4 × ( 6 + (4/2))
M_{A} = 6 R_{B} - 128
Letting left side, take moments around A,
M _{A} = ( 5 × 5 × (5/2)) = - 62.5
Equating these two values, we obtain
6 R_{B} - 128 = - 62.5
6 R_{B} = 65.5
R_{B} = 10.92 kN
R_{A} = (5 × 5) + (4 × 4) - RB
R_{A} = 41 - 10.92 = 30.08 kN
Shear Force (Starting from the left end C)
SF at C, F_{C} = 0
SF just left of A, F_{A} = 0 - 5 × 5 = - 25 kN
SF just right of A, F_{A} = - 25 + 30.08 = + 5.08 kN
SF just left of B, F_{B} = + 5.08 kN
SF just right of B, F_{B} = + 5.08 + 10.92 = + 16 kN
SF at D, F_{D} = + 16 - (4 × 4) = 0
Figure
Bending Moment
BM at C and D, M_{C} = M_{D} = 0
BM at A, MA = - 5 × 5 ×(5/2) = - 62.5 kN m
BM at B, MB = - 4 × 4 ×(4/2) = - 32 kN m