So in this tutorial, we are going to explain you how to draw shear force and bending moment diagram in instance of cantilever beam carrying uniformly distributed load and point load.

It is carrying a uniformly distributed load ‘UDL’ of 0.8 kilo Newton/meter at a distance of 1.5 meter from the free end and also a point load of 2 kilo Newton at a distance of 1.5 meters from the free end. Now, due to the specific downward acting rule, the reaction is RA at the point A and we have to find out this reaction.

As per shear force diagram calculations, upward forces are set negative, while downward forces are set positive.

As there is force acting on the last point ‘B’, as a result shear force at point B= 0 kN Furthermore, shear force at C right = 0.8 x 1.5 = 1.2 kN (without taking into account of the point load) Shear force at C = 1.2 + 2 = 3.2 kN Shear force at A = shear force at C = 3.2 kN As a result, the reaction RA = 3.2 kN As a result, at point B, it is zero, while at point C right, it is 1.2. Therefore, there will be an inclined line.

After this, there is a point load that is a straight line of 2kN. Therefore, total load is going to be 3.2 kN. There is no additional force that is acting between C and A.

Therefore, it is going to be a horizontal line.

Furthermore, shear force at point A is 3.2 kN.

For the calculation of bending moment, the UDL is altered into point load.

Therefore, 0.8 x 1.5 is UDL load.

Bending moment calculations Bending moment at B is 0 as it is the end point.

Bending moment at C = - 0.8 x 1.5 x 0.75 = - 0.9 kN m In the same way, bending moment at A = - 0.8 x 1.5 x (0.75 + 1.5) – 2 x 1.5 = - 5.7 kN m

There will be a parabola and an inclined straight line.

The values are going to be 0.8 kN m and 5.7 kN m.

Opt from the links given below to make calculation: civilengineer.webinfolist.com For more information, go through the following tutorial.

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