We were discussing meaning and importance of Shear force and bending moment and also discussed shear force and bending moment diagrams for a simply supported beam with a point load acting at midpoint of the loaded beam during our previous posts.
We have also seen shear force and bending moment diagrams for a simply supported beam with an eccentric point load in our recent post.
Today we will see here the concept to draw shear force and bending moment diagrams for a simply supported beam with uniform distributed load 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 distributed load of w per unit length over the entire length of beam AB 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 distribution loading and we can refer the post different types of load acting on the beam in order to understand the uniformly distribution load.
In case of uniformly distributed load, total uniform 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.
Total uniform distributed load = w. L
ƩFX =0, ƩFY =0, ƩM =0,
RA + RB – w. L =0
RA + RB = w. L
RA * L – w. L*(L/2) = 0
RA = w L/2
RB = w L/2
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 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; you can also go with right portion of the beam in order to draw shear force and bending moment diagram.
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 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.
FX = RA – w. x
FX = w. L/2 – w. x
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 at all critical points i.e.
Shear force at point A, x= 0
FA= + w. L/2
Shear force at point B, x= L
FB = - w. L/2
Let us think that point C is the midpoint of the uniformly distributed load AB and let us also find shear force value at x= L/2,
FC = w. L/2 – w. L/2
FC = 0
As we can see that shear force is following linear 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 at a distance x from the end A between A and B and hence bending moment at section XX i.e. MX will be determined as mentioned here.
MX = RA. x – w. x. x/2
MX = (w. L/2). x – w. x2/2
MX = w. x. L/2 – w. x2/2
Bending moment at point A, x=0
MA = 0
Bending moment at point C, x=L/2
MC = w.L2/4 – w. L2/8
MC = + w. L2/8
Bending moment at point B, x=L
MB = w. L. L/2 – w. L2/2
MB = w. L2/2 – w. L2/2
MB = 0
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 parabolic 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
Strength of material, By R. K. Bansal
Image Courtesy: Google
We will see another important topic i.e. shear force and bending moment diagrams for a simply supported beam with uniform varying load in the category of strength of material.