We were discussing basics of shear force and bending moment diagrams and sign conventions for shear force and bending moment in our recent posts. We have also discussed the concept to draw shear force and bending moment diagrams for a cantilever beam with a point load during our previous posts.

Today we will see here the concept to draw shear force and bending moment diagrams for a cantilever beam with a uniformly distributed load with the help of this post.

Let us consider one beam AB of length L as displayed in following figure. As we can see here that one end of the beam AB is fixed at one end i.e. at A and other end is free i.e. at end B. Therefore, we can say that we have one cantilever beam here and we will have to find the method to draw shear force and bending moment diagram when cantilever beam will be loaded with uniformly distributed load.

Let us consider that cantilever beam AB is loaded with uniformly distributed load of w per unit length over the entire cantilever beam as displayed in following figure.

Let us consider one section XX at a distance x from free end i.e. from end B. Now we will have two portion of the beam AB i.e. left portion and right portion. Let us deal with the right portion here; you can also go with left portion of the beam in order to draw shear force and bending moment diagram.

### Shear force diagram

Recall the concept of different types of loads on beam, during determination of the total load, total uniformly distributed load will be converted in to point load by multiplying the rate of loading i.e. w (N/m) with the span of load distribution i.e. L and will be acting over the midpoint of the length of the uniformly load distribution.

Let us assume that FX is shear force and MX is the bending moment at section XX. Shear force at section XX will be equal to the resultant force acting to the right portion of the section.

We will recall here the sign conventions for shear force and bending moment and we can conclude here that resultant force acting to the right portion of the section will be w*x and it will be positive.

Therefore, shear force at section XX will be written as mentioned here
FX = w*x

As we can see here from above equation that shear force at section XX will follow the linear equation and on the basis of value of x we can conclude value of shear force at critical points i.e. at point A and at point B.

Shear force at free end i.e. at point B, x=0
FB = 0

Similarly, Shear force at fixed end i.e. at point A, x=L
FA = w*L
FA = W
Where, W = w*L

Now we have data for shear force at critical points i.e. at point A and at point B and as we have stated above that shear force at section XX will follow the linear equation and hence we can draw shear force diagram i.e. SFD as displayed in following figure.

### Bending moment diagram

Bending moment at section XX will be written as

MX = - (Total uniformly distributed load) x (distance between section XX and point of action of total uniformly distributed load)
MX = - (w*x) * (x/2)
MX = - w*x2/2
Above equation indicates that bending moment at section XX will follow here the parabolic equation.
Again if we will recall sign conventions for shear force and bending moment, we will conclude here that bending moment at section XX will be negative.

We can also conclude here that bending moment will be directionally proportional with the square of distance x and we can secure here the value of bending moment at critical points i.e. at point A and at point B.

Bending moment at free end i.e. at B, value of distance x = 0
MB= 0

Bending moment at fixed end i.e. at A, value of distance x = L
MB= - w.L2/2
MB= - W.L/2
Where, W = w*L

Do you have any suggestions? Please write in comment box

### Reference:

Strength of material, By R. K. Bansal