A 3 hinged parabolic arch of span 20m and rise 4m carries a uniformly distributed load of 20 kN/m on the left half of the span. What is the maximum bending moment for the arc?

 A 3 hinged parabolic arch of span 20m and rise 4m carries a uniformly distributed load of 20 kN/m on the left half of the span. What is the maximum bending moment for the arc?

Solution;

 Given condition;

Span of the parabolic arch L=20m

Rise of the arch R=4m

UDL(uniformly distributed load) =20kN/m

Max. BM=?

Firstly we will draw line diagram of the said arch

Now taking Moments about hinge1

We have ;

V2 × 20 =20 ×10×10/2

V2×20=20×10×5

V2=50KN

Now we have ;

V1 +V2=20×10

V1+V2=200

Or V1+50=200

~V1=150KN

Again taking Moment about hinge3 forces on the right side side of hinge3

We have;

H×4=50×10

~H=125KN

This horizontal thrust will be same on both sides as shown in dia;

Now at any section distant X from hinge 1or 2

We have ;

Y= 4hX(L-X) /L^2~~~~~formula

Y= 4×4×X(20-X) /20×20

Y=1×X(20-X) /25

Maximum Bending Moment in 1 3

At any section distant X from hinge1,

Mx= 150X-20X×X/2–125×1×X×(20-X) /25

Mx=50X-5X^2….. . . . . eq1

Or Mx=5X(10-X) . ….eq2

B ut for the condition of max. Bending Moment

We have;

dMx/dx = 0

~dMx/dx=50–10X ( by differentiating eq. 1)

~50–10X=0

Or X=5m

Putting X=5 in eq 2

Mmax=5×5(10–5)

Mmax=+125kNm

Hence maximum Bending Moment will be 125kNm at a distance of 5m from hinge1