Shear Force and Bending Moment as Structural Basics

by Civil-Guy on March 29, 2010 in Civil Subjects

Basic structural learning begins with an analyzing of a simply supported beam. A beam is a structural member (horizontal) that is design to support the applied load (vertical). It resists the applied loading by a combination of internal transverse shear force and bending moment. An accurate analysis required in order to make sure the beam is construct without any excessive loads which affect its strength.

Types Of Load and Support

The Men-at-Work in construction...

Two types of typical loadings:

  • Concentrated load is one which can be considered to act at a point although of course in practice it must be distributed over a small area (normally vertical or incline loads). (Unit in kN)
  • Distributed load is one which is spread in some manner over the length or a significant length of the beam. It is usually quoted at a weight per unit length of beam and it may either be uniform or varying loading from point to point. (Unit in kN/m)

Three types of support:

  • Namely as Pinned support, Roller support and Fixed or Built-in support.
  • Covered in previous post together the reactions explained in diagrams.

The Sign Conventions

The general sign convention of a beam.

The sign convention depends on the direction of the stress resultant with respect to the material against which it acts. It is used for both shear force and bending moments in analyzing the directions. Positive (+ve) shear forces always deform right hand face downward with respect to the left hand face and negative (-ve) would be the other way round. Positive (+ve) bending moments always elongate the lower section of the beam and negative (-ve) would elongate the mid-section upward of the beam.

Shear Force and Bending Moment in Simply Supported Beam
For a simply supported beam, the reactions are generally simple forces. When the beam is built-in, the free body diagram will show the relevant support point as a reaction force and a reaction moment. It is normal practice to produce a free body diagram with the shear force diagram and bending moment diagram position.

A simply supported beam with brief details.

The unknown forces (generally the support reactions) are then determined using the Equation of Equilibrium:

  • Horizontal reaction respect to x-axis; ΣFx = 0
  • Vertical reaction respect to y-axis; ΣFy = 0
  • Bending reaction (clockwise or anti-clockwise), ΣM = 0

Shear Force and Bending Moment Diagrams
The shear force diagram indicates the shear force withstood by the beam section along the length of the beam. The bending moment diagram indicates the bending moment withstood by the beam section along the length of the beam.

shear force

Shear Force and Bending Moment Diagrams in simply supported beam with two different loads.

Simple calculation of determine the reactions:
* Take example the 1st one with point load, W…
ΣMc = 0 (taking clockwise as +ve at point C)
(Ra x L) – (W x L/2) = 0
So, Ra = W/2 [in kN]

ΣFy = 0 (taking vertical ↑ as +ve)
Ra + Rc – W = 0
W/2 + Rc – W = 0
So, Rc = W/2 [in kN]

The shear force diagram (SFD) is simply constructed by moving a section along the beam from the left origin and summing the forces to the left of the section. The equilibrium condition states that the forces on either side of a section balance and therefore the resisting shear force of the section is obtained by this simple operation.

On the other hand, bending moment diagram (BMD) is obtained in the same way except that the moment is the sum of the product of each force and its distance of x from the section either left or right. Distributed loads are calculated buy summing the product of the total force (to the left of the section) and the x distance of the distributed load centroid.

bending moment

Shear Force and Bending Moment Diagrams in cantilever beam with two different loads.

The basic procedure for determining the shear and moment is to determine the values of V and M at various sections along the beam and plotting the results from point to point. By doing so, we will be able to determine critical sections within the beam where a critical or maximum stress might occurs:

  • Section of Maximum Shear – Since the shear, V, at any transverse section of the beam is the algebraic sum of the transverse forces to the left of the section, the shear, in most cases can be estimated at a glance.
  • Section of Maximum Moment – It can be obtained mathematically, that when the shear force is zero or changes sign; the bending moment, M will be either a maximum or relative maximum.

In conclusion, we could obtained the reactions value at the supports by using the equation of equilibrium. The shear force and bending moment diagram used for determine both of force, V and moment, M maximum values. Both of this reactions are the fundamental forces which becomes available in mechanics of structures.

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Thank You and Best Regards,
Author: Civil-Guy | Blog: Civil Craft Structures | Subscribe: RSS or Email

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28 comments

{ 28 comments… read them below or add one }

Anthony Fasano March 31, 2010 at 8:57 am

Excellent post, especially for those who are studying for their P.E., this is very important information to know for the exam!

Reply

Mr. Vine April 1, 2010 at 12:06 am

Hi Anthony, thank you so much for your sincere compliment. Wow, even the pros also in need of clear understanding of this fundamental concept. Awesome and thanks for sharing from your point of view. May GOD bless…

Reply

Wijaya Basnayake April 25, 2010 at 12:09 pm

Your expalnations are very simple, and can be easily understood. Your works are highly commended.

Reply

Mr. Vine April 25, 2010 at 7:18 pm

Thanks for your comment Wijaya… Thanks for the support as well…

Reply

Civil Engineer May 6, 2010 at 1:41 pm

Wow…!!! so interesting and illustrated article

Reply

Mr. Vine May 6, 2010 at 7:37 pm

Hi there, thanks for the complement and visiting my blog…

Reply

macdonald June 18, 2010 at 7:30 pm

this is good.

Reply

sasi September 20, 2010 at 12:19 pm

nice information. thanks a lot.

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rahul verma September 24, 2010 at 7:02 pm

you have created very interesting way of learning sfd and bmd

Reply

GOUTHAMI.GOWDA September 25, 2010 at 3:20 pm

DEAR SIR, I got cleared my doudts which i did’n get i class thanx a lot………..

Reply

eye candy December 2, 2010 at 9:24 pm

hi i m a student of civil engineering!!!!! its a v v v cool web site containing v v best knowledge regarding to engineering stuff !!!! Allah bless u all

Reply

Civil-Guy January 10, 2011 at 1:23 am

Thank you very-very much upon your nice comment. May God bless you too…

Reply

SRINI December 16, 2010 at 4:21 am

AFTER SO MANY DAYS I GET CLEAR INFORMATION .THANKS TO U

Reply

john February 6, 2011 at 3:10 pm

This is a good article, I stumbled across this calculator which draws shear force and bending moment diagrams for beams (learntoengineer.com/beam, also in the website of this post) it’s worth a look for anyone learning this stuff so that you can see if your calculations are correct.

Reply

Civil-Guy February 7, 2011 at 10:46 pm

Thank you very much for sharing dear friend…

Reply

intehab hussain rizvi February 16, 2011 at 1:45 pm

really after a long journey of books for sfd bmd ,i got a clear cut concept.may god bless u……………..

Reply

intekhab hussain February 16, 2011 at 1:49 pm

really after a long time i got a clear cut concept for sfd bmd.thanks

Reply

vijay February 16, 2011 at 10:29 pm

it is always nice when something improves our basics,thanks for doing the same for me

Reply

Ishak March 31, 2011 at 8:19 am

I was studying this back in 1992-1998, but it was in my country Indonesia.

Its been a while I tried to find the term in English, and I found your web, so usefull

Thank you.

Reply

Callum April 6, 2011 at 1:37 am

I found this a very useful site that really helped when revising for exams. Illustrations are extremely useful. Very good job!!!

Reply

ainmardiana April 16, 2011 at 8:23 pm

really good..keep it up sir….

Reply

Malik Fahad April 20, 2011 at 10:32 pm

its really interesting and informative! i am a DAE studnt n itz use in RCC design

Reply

Naka sai Naka April 21, 2011 at 1:21 am

Thank for creating this side I found this very easy and simple calculation
Thank God Bless you

Reply

omega May 21, 2011 at 4:54 pm

THANKS A LOT. IT IS SO INTERESTING AND CLEAR. ALL THE BLESSINGS FROM OUR ALMIGHT god

Reply

aida June 8, 2011 at 9:50 am

Mr. Vine, thanks for this information. I am a teacher and my students will be thankful too tonight at class.

Reply

loyd June 12, 2011 at 1:53 pm

Its a nice website giving d basic concepts.i got to know d basics.thank u so much keep going.

Reply

adithya gs June 16, 2011 at 2:21 am

really a superb article.. like the lecturer type explanation… so easy to understand..

Reply

Boni Amin June 20, 2011 at 12:03 pm

This is the ABC for structural engineers…very nice explaination

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