Evaluate the point of contraflexure:
An overhanging beam of 15 m span is carrying a consistently distributed load of 1 kN/m over the length of 10 m at a distance 5 m from the left free end and point loads 7 kN and 4 kN, at free end and at a distance 2 m from the free end, respectively. Sketch the BMD and SFD for the beam. Situate the point of contraflexure.
Solution
Taking moments around B and equating to zero,
R _{A} × 10 - (7 × 15) - (4 × 13) - (1 × 10 × (10/2) ) = 0
R_{A} = 20.7 kN
and R_{B} = 7 + 4 + (1 × 10) - 20.7 = 0.3 kN
Shear Force (beginning from the left end C)
SF just right of C, F_{C} = - 7 kN
SF just left of D, F_{D} = - 7 kN
SF just right of D, F_{D} = - 7 - 4 = - 11 kN
SF just left of A, F_{A} = - 11 kN
SF just right of A, F_{A} = - 11 + 20.7 = 9.7 kN
SF just left of B, F_{B} = + 9.7 - (1 ×10) = - 0.3 kN = Reaction at the support B.
Bending Moment
BM at B, MB = 0
BM at A, M_{A}
= (0.3 × 10) - (1 × 10 × (10/2)) = - 47 kN m
BM at D, M_{D} = - (7 × 2) = - 14 kN m
BM at C, M_{C} = 0
let a section XX at a distance x from the end B as illustrated in Figure.
Shear force at section XX,
F_{x} = - 0.3 + 1(x)
For maximum bending moment, shear force will be zero.
- 0.3 + x = 0
x = 0.3 m
We have,
M_{x} = + 0.3(x) - 1 × (x) ×(x/2) = 0.3x - x ^{2} /2
M_{max} = 0.3 × (0.3) - ((0.3)^{2} /2) = + 0.045 kN m
Figure
This is seen that the maximum negative bending moment take place at A and maximum positive bending moment take place at a distance 0.3 m from the right end B.
M_{max} (positive) = + 0.045 kN m
M_{max} (negative) = - 47 kN m
Point of Contraflexure
This is observed that the BM changes sign among A and B,
BM at any section XX,
M_{x} = 0.3 x - x ^{2} / 2
Equating this to zero,
0.3 x - x^{2}/2 = 0
0.6 x - x^{2} = 0
∴ x = 0.6 m
Thus, the point of contraflexure is at a distance of 0.6 m from the right end B.