A simply supported beam with a span of 9.75 m is carrying full span u.d.l. of 10 kN/m. What is the magnitude and position of maximum bending moment developed?

A simply supported beam with a span of 9.75 m is carrying full span u.d.l. of 10 kN/m. What is the magnitude and position of maximum bending moment developed?

Solution

To find the support reaction,

Σ Fy = 0

Ra + Rb - (10 x 9.75) = 0

Ra + Rb = 97.5 kN ……………………………………………………………… (1)

Taking moment at A,

Σ Ma = 0

Ra x 0 + (10 x 9.75) x 9.75/2 - Rb x 9.75 = 0

475.312 = 9.75 x Rb

Rb = 475.312/9.75

Rb = 48.75 kN

Put Rb in (1) eqn

Ra + Rb = 97.5

Ra + 48.75 = 97.5

Ra = 48.75 kN

SF Calculation

S.Fal = 0

S.Far = 48.75

S.Fbl = 48.75 - (10 x 9.75)

S.Fbl = - 48.75

S.Fbr = 0

As S.F. is zero at pt. C

To find pt of contra shear

S.Fc = 0

S.Fc = 48.75 - (10 x 𝒳) = 0

10𝒳 = 48.75

𝒳 = 4.875m

BM Calculation

B.Ma = 0

B.Mb = 0

B.Mc = 48.75 x 4.875 - (10 x 4.875 x 4.875/2)

B.Mc = 118.828 kN-m

B.M at c is the max B.M. as at c

S.F. is zero