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 (D**istance 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

## 32 comments

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## Abhishek shukla

February 12, 2012 at 1:55 am (UTC 5.5) Link to this comment

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

## Anonymous

February 13, 2012 at 10:34 pm (UTC 5.5) Link to this comment

Thank you

## Dilini

March 13, 2012 at 7:58 am (UTC 5.5) Link to this comment

Thank you sir.

## E915hern

April 18, 2012 at 3:31 am (UTC 5.5) Link to this comment

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?

## Norazmah1963

April 28, 2012 at 6:28 am (UTC 5.5) Link to this comment

How to make BMD for u shape concrete drain and culvert

## MechGuru

April 28, 2012 at 8:33 pm (UTC 5.5) Link to this comment

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.

## MechGuru

April 28, 2012 at 8:43 pm (UTC 5.5) Link to this comment

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.

## Skahmad

June 2, 2012 at 4:41 pm (UTC 5.5) Link to this comment

Visit http://civilengineer.webinfolist.com/mech/bmcalc.htm for Online calculation of Bending moment and shear force due to different loading cases.

## Chazz

June 5, 2012 at 1:37 am (UTC 5.5) Link to this comment

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

## MechGuru

June 5, 2012 at 7:39 pm (UTC 5.5) Link to this comment

Chazz,

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

Thank you

shibashis

## Mattyreevz

June 7, 2012 at 5:47 am (UTC 5.5) Link to this comment

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

## Mattyreevz

June 7, 2012 at 5:47 am (UTC 5.5) Link to this comment

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

## Mattyreevz

June 7, 2012 at 5:47 am (UTC 5.5) Link to this comment

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

## MechGuru

June 7, 2012 at 7:57 pm (UTC 5.5) Link to this comment

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.

## Pratik kale

August 20, 2012 at 10:06 pm (UTC 5.5) Link to this comment

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

## Gshaland

August 31, 2012 at 8:38 pm (UTC 5.5) Link to this comment

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

## Gshaland

August 31, 2012 at 8:38 pm (UTC 5.5) Link to this comment

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

## Pratik kale

August 31, 2012 at 9:28 pm (UTC 5.5) Link to this comment

Thank you sir!

## Dhruvspatel1829

October 23, 2012 at 5:38 pm (UTC 5.5) Link to this comment

thank u very much SIR….

## Civilwebindex

October 29, 2012 at 8:09 pm (UTC 5.5) Link to this comment

another very useful tool for bending moment and shear force calculations is availble at http://civilengineer.webinfolist.com/mech/bmcalc.htm

## Viraj Baraskar

November 9, 2012 at 1:19 pm (UTC 5.5) Link to this comment

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

## Asanka_4286

December 26, 2012 at 4:13 pm (UTC 5.5) Link to this comment

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

## CivilEng.

January 24, 2013 at 7:07 pm (UTC 5.5) Link to this comment

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

## Vladimir Bovt

January 28, 2013 at 12:55 am (UTC 5.5) Link to this comment

One simple step – visit to http://www.soromat.ho.ua/en

– for any beam

– bending moment

– shear force

– size of cross-sections

– deflections

## pritesh

February 26, 2013 at 3:05 pm (UTC 5.5) Link to this comment

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

## MechGuru

February 26, 2013 at 9:50 pm (UTC 5.5) Link to this comment

Good question. By plotting the SFD and BMD you came to know what is the max. stress and moment value and where it is located. These values (especially max bending moment) are used as input for deciding the minimum cross sectional area required to withstand the loads (which you used for plotting the SFD and BMD). You can refer this article for better clarity: http://blog.mechguru.com/machine-design/why-bending-stress-is-more-important-than-shear-stress-in-beam-design/

## q

March 16, 2013 at 9:26 pm (UTC 5.5) Link to this comment

The bending moment is wrong!!!

## pinky

July 10, 2013 at 10:57 am (UTC 5.5) Link to this comment

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?

Plz reply

52 minutes ago

– 4 days left to answer.

Additional Details

20kN 30kN

……………!………………..!..…

3m 3m 3m

udl of 5kN/m all through the span. Design a beam.

Can u plz help in arriving design moment?

## abhimanyu

October 3, 2013 at 10:03 pm (UTC 5.5) Link to this comment

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

## shrishti srivastava

December 5, 2013 at 8:55 pm (UTC 5.5) Link to this comment

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

Plzz reply asap

## pawan

July 26, 2014 at 10:57 am (UTC 5.5) Link to this comment

i want to learn basics of beams and applied loads concepts….plese provide me some link to learn all of these ….

## Joel Monteiro

August 31, 2014 at 12:30 pm (UTC 5.5) Link to this comment

Isn’t it usually positive for anticlockwise moment and negative for clockwise moment? So why is the downward force of 20 kn producing positive moment?