Today we will see here the concept to draw shear force and bending moment diagrams for a simply supported beam with uniform varying load from zero at one end to w per unit length at the other end with the help of this post.

Let us see the following figure, we have one beam AB of length L and beam is resting or supported freely on the supports at its both ends. And therefore, displayed beam indicates the simply supported beam AB and of length L.

Let us consider that simply supported beam AB is loaded with uniformly varying load zero at one end to w per unit length at the other end as displayed in following figure.

First of all we will remind here the important points for drawing shear force and bending moment diagram. Now always remember that we will have to first determine the reaction forces at each support.

Let us consider that RA and RB are the reaction forces at end support A and end support B respectively and we will use the concept of equilibrium in order to determine the value of these reaction forces.

Before going ahead, we must have to understand here the uniformly varying loading and we can refer the post different types of load acting on the beam in order to understand the uniformly varying load.
In case of uniformly varying load, total uniformly varying load will be converted in to point load by determining the area of the triangle and will be acting through the C.G of the triangle.

Total uniform varying load = Area of the triangle ABO
Total uniform varying load = (AB * BO)/2
Total uniform varying load = w*L/2

ƩFX =0, ƩFY =0, ƩM =0,
RA + RB – w. L/2 =0
RA + RB = w. L/2
ƩMB =0
RA * L – w. L/2*(L/3) = 0
RA * L = w. L2/6
RA = w L/6
RB = w L/3

Now we have values of reaction forces at end A and end B and it is as mentioned above. Let us determine now the value of shear force and bending moment at all critical points.

Let us consider one section XX between A and B at a distance x from end A. Now we will have two portion of the beam AB i.e. left portion and right portion. Let us deal with the left portion here. Let us assume that FX is shear force at section XX and bending moment is MX at section XX.

### Shear force diagram

As we have assumed here section XX between A and B at a distance x from end A and therefore loaded beam AB will be divided in two portion and let us consider the left portion of the beam. Shear force at section XX will be equivalent to the resultant of forces acting on beam to the left side of the section.

Let us first determine the rate of loading at section XX, because rate of loading at section XX will be used during calculation of shear force.

XC/OB = x/ L
XC/w = x/ L
XC= w. x/L

Let us determine the load of uniformly varying load up to section XX and we will have
Load of uniformly varying load up to section XX = Area of triangle AXC
Load of uniformly varying load up to section XX = (x. w. x/L)/2

Load of uniformly varying load up to section XX will act through a point which will be at a distance of x/3 from section XX.

FX = RA – w.x2/2L
FX = w. L/6 – w.x2/2L

Force acting to the left of the section and in upward direction will be considered as positive and force acting to the left of the section but in downward direction will considered as negative and we can refer the post sign conventions for shear force and bending moment in order to understand the sign of shear force which is determined here.

Now we have information here in respect of shear force equation and therefore we will determine shear force at all critical points i.e.

Shear force at point A, x= 0
FA= + w. L/6

Shear force at point B, x= L
F= w. L/6 – w.L2/2L = w. L/6 – w. L/2
FB= - w. L/3

Let us also determine the value of x for zero value of shear force, we can determine as mentioned here
0 = w. L/6 – w.x2/2L
L2/3 = x2
Let us consider that point C is at a distance of
Therefore shear force will be zero at point C.

Again we can refer the post sign conventions for shear force and bending moment in order to understand the sign of shear force which is determined here.
As we can see that shear force is following parabolic equation and therefore we can now draw here shear force diagram and it is displayed here in above figure.

### Bending moment diagram

As we have considered above one section XX between A and B at a distance x from the end A and hence bending moment at section XX i.e. MX will be determined as mentioned here.
M= RA. x – (w.x2/2L) x/3
M= (w. L/6).x – w. x3/6L
M= w. x. L/6 – w. x3/6L
As we know that bending moment will be maximum at a point where shear force will be zero. And as we have shown above that shear force will be zero at point C and hence bending moment will be maximum at point C.

Bending moment at point A, M= 0
Bending moment at point B, M= 0
Please refer the post Basics of shear force and bending moment diagrams , in case of simply supported beam, bending moment will be zero at end supports.

Therefore we have bending moment for all critical points and we have also information about the equation followed by bending moment i.e. bending moment equation stated above is following the cubic equation and therefore we can draw here the BMD i.e. bending moment diagram and we have drawn it as displayed in above figure.

Do you have any suggestions? Please write in comment box

### Reference:

Strength of material, By R. K. Bansal