# Create Bending Moment Diagrams in Four Simple Steps

This article will clear your concepts of bending moment diagrams (BMD). You will know the important five steps required for how to calculate and draw bending moment diagram.

For the ease of understanding we will discuss the same example what we have discussed in the shear force diagram tutorial

The shear force diagram of the above example looks like below:

Now, for creating the bending moment diagram of the simply supported beam you have to follow the five steps:

Step-1: Get the Reaction Forces: While creating shear force diagram of the beam you already have calculated the vertical reaction forces at different points as below:

Rc = 60 KN
Re = -20 KN

These reaction forces will be useful for calculating the bending moments at different points on the beam.

Step-2: Sign Convention: The bending moment, which will cause “sagging” to the beam, will be considered positive and the bending moment, which cause “hogging” to the beam, will be considered negative. We will consider this sign convention for the rest of this article.

Step-3: Calculate the Bending Moments: You need to calculate the bending moments at the different points on the beam. For calculating that you need to start from the extreme left (point A) and gradually you have to approach toward right hand side support (point A). You will use the following formula :

Bending moment (M) = (Force) X (Distance between the point of application of the force and the point at which you need to calculate the bending moment (BM))

BM @ A:

Ma       = -20 * 0

= 0

BM @ B:

Mb       = -20 * 1

= – 20 KN-M

BM @ C:

Mc       = bending moment due to the 20KN force + bending moment due to the 10KN\M    UDL

= – 20 * (1+1) – (10*1*0.5)

= – 45 KN-M

BM @ D:

Md       = BM due to the 20KN force + BM due to the 10KN \ M UDL + BM due to the reaction force Rc

= – 20 * (1+1+1) – (10*2X1) + (60*1)

= -20 KN-M

BM @ E:

Me       = BM due to the 20KN force + BM due to the 10KN \ M UDL + BM due to the reaction force Rc

= – 20 * (1+1+1+1) – (10*2X2) + (60*2)

= 0

Step-4: Plot the Bending Moments: Just now you have calculated the BM values at different points of the beam, now plot the values and you will get the bending moment diagram like below:

In the above BMD you might have observed for the point load bending moment diagram is a straight line and for UDL bending moment is a curve. But how to decide whether the curve should be a convex or concave one? We will discuss in the next article.

So, as of now you are ready to calculate bending moments and create BM diagram of a simply supported beam. Creating the BMD  is the important first step toward design of structural member

## 33 thoughts on “Create Bending Moment Diagrams in Four Simple Steps”

1. Abhishek shukla says:

sir u hav xplained the steps of making SFD &BMD very clearly

2. Anonymous says:

Thank you

3. Dilini says:

Thank you sir.

4. E915hern says:

how to solve the diagrams with a beam having three loads, plus the weight of the beam?
is the weight of the beam treated as a distributed load?

5. Norazmah1963 says:

How to make BMD for u shape concrete drain and culvert

6. MechGuru says:

The weight of the beam can either to be applied at the CG of the beam or, as you rightly said, as a distributed load.

7. MechGuru says:

To be very simple,the value of bending Moment (BM) does not depends upon the cross section, the value of bending stress does. So, you can calculate the BM values as discussed in the article and draw the bending moment diagram. After you got the BM values at different locations throughout the length, you can use the equation-1 of the article:

http://blog.mechguru.com/machine-design/why-a-hollow-shaft-stronger-than-a-same-weight-solid-shaft-in-bending-2/

To calculate bending stress values.

To get more accurate bending stress values you can always use the FEA tools.

8. Chazz says:

Thank you, could you possibly add a moment to this tutorial, would be helpful. I know how the moment works in the BM diagram, but an illustration would be nice regardless

9. MechGuru says:

Chazz,
Thank you for stopping by. That’s a good suggestion, i will definitely incorporate that.

Thank you
shibashis

10. Mattyreevz says:

Where does the “0.5” come from at point C? Wouldn’t it just be -(20*1+1) – (10*1) = -50KNm?

11. Mattyreevz says:

Where does the “0.5” come from at point C? Wouldn’t it just be -(20*1+1) – (10*1) = -50KNm?

12. Mattyreevz says:

Where does the “0.5” come from at point C? Wouldn’t it just be -(20*1+1) – (10*1) = -50KNm?

13. MechGuru says:

The “0.5” is because of the fact that the “resultant of the distributed loads”, which is, as you know, will be at the centre of the spread of the loads.

14. Pratik kale says:

sir what are 2nd degree and 3rd degree parabolic curves? Why they are used for only for UDL and UVL in BMD and SFD?

15. Gshaland says:

Thank you MechGuru. Did you ever answer Mettyreev’s comment about the determination of the 0.5 at point C?

16. Gshaland says:

Thank you MechGuru. Did you ever answer Mettyreev’s comment about the determination of the 0.5 at point C?

17. Pratik kale says:

Thank you sir!

18. Dhruvspatel1829 says:

thank u very much SIR….

thanx a lot sir….had plenty of doubts regarding BMD…all solved in just one example…thanx

20. Asanka_4286 says:

thank you very very much, I got it………….

21. CivilEng. says:

or you can use a computer programs http://beamsbending.com/Applet.html

One simple step – visit to http://www.soromat.ho.ua/en
– for any beam
– bending moment
– shear force
– size of cross-sections
– deflections

23. pritesh says:

what is the importance of calculate sfd bmd and moment and shear force

24. q says:

The bending moment is wrong!!!

25. pinky says:

Design a fixed beam with 5kNm UDL and point loads 20kN and 30kN acting L/3 distance. span is 9m.?
For what moment shud we design? and how to arrive at that moment in fixed beam?
52 minutes ago
– 4 days left to answer.
20kN 30kN
……………!………………..!..…
3m 3m 3m

udl of 5kN/m all through the span. Design a beam.
Can u plz help in arriving design moment?

26. abhimanyu says:

how do we get 0.5 for C,1 for D & 2 for E i very confused sir please reply

27. shrishti srivastava says:

Wanted to ask that when we find out shear forces , in left hand rule we take all the forces
But one force at right side get left
So do I should apply right hand rule for that left force ?
Or I should leave it