Determine the point of contraflexure:
Draw the shear force & bending moment diagram for 10 m span overhanging beam along overhanging part of 4 m subjected to a system of loads as illustrated in Figure . Determine the maximum bending moment and also situated the point of contraflexure.
Solution
Taking moments around A and equating it to zero,
R_{ B} × 6 - (8 × 10) - 3 × 4 × ( 6 + (4/2) - (20 × 4) - (10 × 2) = 0
R_{B }= 46 kN
R_{A} = 10 + 20 + (3 × 4) + 8 - R_{B} = 50 - 46 = 4 kN
Figure
Shear Force (beginning from the end A)
SF at A, F_{A }= + 4 kN
SF just left of D, F_{D} = + 4 kN
SF just right of D, F_{D} = + 4 - 10 = - 6 kN
SF just left of E, F_{E} = - 6 kN
SF just right of E, F_{E} = - 6 - 20 = - 26 kN
SF just left of B, F_{B} = - 26 kN
SF just right of B, F_{B} = - 26 + 46 = + 20 kN
SF just left of C, F_{C} = + 20 kN = Load at C
Bending Moment
BM at C, M_{C} = 0
BM at B, M _{A}= (0.3 × 10) - (1 × 10 × (10/2) ) = - 47 kN m
BM at E, M_{E }= + (4 × 4) - (10 × 2) = - 4 kN m
BM at D, M_{D} = + (4 × 2) = + 8 kN m
Maximum Bending Moment
As SF changes sign at D & B, the maximum positive bending moment shall occur at D and maximum -ve bending moment shall take place at B.
M_{max} (positive) = + 8 kN m
M_{max} (negative) = - 56 kN m
Point of Contraflexure
BM changes sign among D & E. Thus, consider a section XX at distance x from the end A.
BM at section XX,
M_{x} = + 4 x - 10 (x - 2)
Equating this to zero, we obtain
4 x - 10 (x - 2) = 0
4 x - 10x + 20 = 0
- 6x + 20 = 0
∴ x = 3.333 m
The point of contraflexure is at any distance of 3.333 m from the left end of A.