Analysis of overhanging beam with a uniformly distributed load

Overhanging beam

It’s a simply supported beam which overhangs ( extends in the form of a cantilever) from its support.

For the purpose of shear force and bending moment diagrams, the overhanging beam may overhang on one side only or on both sides of the support.

Point of Contraflexure

Point of contraflexure occurs in overhanging beam. Over hanging beam is analyzed as a combination of simply supported beam and a cantilever. In cantilever the bending moment is negative, whereas that in simply supported beam is positive. It’s thus obvious that in an overhanging beam, there will be a point , where the bending moment will change sign from negative to positive or vice versa .Such a point , where the bending moment changes sign, is known as point of contraflexure.

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An overhanging beam ABC is loaded as shown above .Draw the shear force and bending moment diagrams and find the point of contraflexure, if any.

Solution

Given span =4m, uniformly distributed load, w= 4.5KN/M

First of all, let us find the reactions RA and RB.

Taking moment about A, EMA=0

-RB(L) + ql*l/2=0

-3RB + 4.5*4*4/2=0

-3RB + 36=0

RB=12KN

Taking moment about B, EMB=0

RA(L) – ql*l/2 + ql*l/2=0

3RA – 4.5*3*3/2 + 4.5*1*1/2=0

3RA-20.25 + 2.25=0

RA= 6KN

Computation of shear force

Shear force at A= 6KN

Shear force at B= 6-(4.5*3)+12= 4.5KN

Shear force at C= 6-(4.5*3)+12-4.5*1=0

Computation of bending moment

MA=0

MB= -(4.5*1*1/2)=-2.25KNM

MC=0

We know that maximum bending moment will occur at Y, where the shear force changes sign . using the principle of similarity of triangle

X/6=3-X/7.5, cross multiply

You will have 7.5x=18-6x

Collect like terms

13.5x=18

X=1.33m

MY= 6*1.33-4.5*1.33*1.33/2=4KN

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Point of contraflexure

Let P be the point of contraflexure at a distance Y from the support A. We know that bending moment at P.

Mp= 6*y – 4.5*y*y/2=0

2.25y^2-6y=0

Y= 6/2.25= 2.67m