Calculate the magnitude of maximum bending moment:
The shear force diagram for the overhanging beam is illustrated in Figure. Draw the loading diagram & bending moment diagram. Determine the magnitude of maximum bending moment & situated the point of contraflexure.
Solution
Let us analyse the shear force diagram specified in Figure .
At A
The shear force diagram suddenly reduces from 0 to - 400 N. It denotes that there is a downward point load of 400 N at point A.
Between A and B
The shear force diagram is an inclined straight line and reduces from - 400 N to - 560 N. It denote that there is a consistently distributed load of (560 - 400 = 160) ⇒ 160 × 1 = 160 kN/m among A and B.
At B
There is a sudden enhances from - 560 N to + 440 N at B. It denotes that there is a support reaction of 1000 N at B.
Between B and C
As the shear force diagram varies linearly from + 440 N to - 520 N among B & C. It denote that a u.d.l. of (440 + 520 = 960) 960 /6= 160 kN/m is acting between B and C.
At C
At C, the shear force diagram enhance suddenly from - 520 kN to 480 N. It denote that there is a support reaction of 1000 N (520 + 480) at C.
Between C and D
The shear force diagram is an inclined straight line which denote that there is a uniform ally distributed load of 480/3 = 160 kN/m from C to D.
Bending Moment
BM at A and D, M_{A} = M_{D }= 0.
BM at B, M _{B} =- (400 × 1) - (160 × 1 × (½) ) =- 480 N m
BM at C, M _{C} = - (160 × 3 × (3/2)) =- 720 N m
Maximum Bending Moment
SF at any section XX among B and C,
F_{x} = +1000 - 400 - 160x
For maximum bending moment, F_{x }must be equal to zero.
600 - 160x = 0
∴ x = 3.75 m
∴ M _{max} = (1000 × 2.75) - (400 × 3.75) - (160 × 3.75 × (3.75/2)
= + 125 N m
Maximum positive bending moment take places at a distance of 3.75 m from the end A, where SF changes sign.
Maximum negative bending moment takes place at a support C where SF changes sign.
∴ M_{max} (negative) = - 720 N m
Points of Contraflexure
Let M_{1} and M_{2} be the points of contraflexure, where BM is zero. However at any section XX among B and C at a distance x from the end A.
M _{x} = 1000 ( x - 1) - 400x - (160 × x × (x/2) )
or 1000x - 1000 - 400x - 80x^{2} = 0
80x^{2} - 600x + 1000 = 0
8x^{2} - 60x + 100 = 0
2x^{2} - 15x + 25 = 0
(2x - 5) (x - 5) = 0
Therefore, x = 2.5 m and 5 m.