Beams | Bending Moment and Shear Force Diagram


Beams are structural members which are
most commonly used in buildings in a beam transverse load is acted, which
in fact comes from the slabs to the column or walls for analysis beam can be separated out
from beam column system length of a beam is much higher than its lateral
dimensions axial strain developed in a beam is very
small compared to shear strain or strain induced due to bending So for design purposes beams, analysis of shear force and bending moment induced
are of atmost importance The interesting thing is that you can
draw a sheer force and bending moment distribution along any beam by understanding what exactly is sheer
force and bending moment first shear force. Shear force is the
resistance created in beam crosssection in order to balance transverse external
load acting on beam Consider this beam. It does not matter
from where you take a section when you add forces acting on it it, it
should be in equilibrium shear force is induced exactly for this
purpose to bring the section to equilibrium in
vertical direction It acts parallel to cross section Usual sign convention is as follows: So just by applying force balance in vertical direction on the free body
diagram, we can determine value of shear force at a particular cross-section now we can apply same concept
in different cross-section and find how shear force varies along the length of
the beam But balance of transverse forces alone does
not guarantee equilibrium of a section there is another possibility a beam rotation, if
moment acting on it is not balanced If this is the case a bending moment
will be induced in cross-sectional beam to arrest this rotation It will be induced as normal forces
acting on fiber cross-section as shown Resultant forces were be zero, but it will produce a moment to
counterbalance the external moment sign convention a bending moment is as
follows: so we can calculate moment induced at any cross-section by balancing the
external moment acting on the free body diagram with these concepts developed we can
easily calculate distribution and shear force and bending moment along the length of
the beam we will see some examples For this cantilever carryng three loads
we can start analysis from the free end so between the A&B if you take a section
the only external force acting on it is F1 so a sheer force should induce in
section to balance this force so value of here force between A and B is
F1 but force balance alone does not guarantee equilibrium of the section there is an external moment on the
section so a bending moment will be induced in Section in order to
balance the external moment since value of external moment is F1 into X bending moment will vary in the nearly Between B&C effect of F2 also comes So sheer force becomes F1 plus F2 in bending moment effect f2 also gets added similar analysis is done between section
C and D also so SFD in BMD have this problem would
look like this Now consider this problem a simply supported beam with uniformly distributed load first step here would be determination
of reaction forces since the problem is symmetrical
reaction forces will be equal and will be have have total load acting
on beam Lets start analysis from point A if you take in between points A&B it should be in
equilibrium so shear force will have equal magnitude of
reaction force so bending moment also gives a linear
variation but after point B effect a point load
and distributed load comes effect have distributed load is
something interesting take a section in BC in this section along with two point
loads there is a distributed load also this distributed load can be assumed as
a point load passing through centroid of distributed load value of this is U into X minus L by
three and it is at a distance (X-L/3)/2 from section line so shear force will have one more term
which comes from distributed load from the equation it’s clear that shear
force varies linearly you can easily predict how bad a moment varies along length from the
same force diagrams since this equation is quadratic it will
have a parabolic shape same procedure is repeated in remaining
section since this problem is symmetrical
bending moment and shear force are having a symmetrical distribution hope you got a good intern analysis beams thank you

47 thoughts on “Beams | Bending Moment and Shear Force Diagram

  1. Is shear force and bending moment diagram is unique for a given beam? If yes, why does different text books use different sign convention and diagrams? 

  2. I had always had trouble with understand what bending moment really meant. Your diagram showed the BM as something arising from the internal forces along the length of the beam; this is honestly the first time in years that I have really understood it. I've read so many lengthy explanations in books and on the internet but not one gave a succinct and visually apt explanation as yours. Thanks a million and keep up the good work!

  3. 3:57 > section "C-D" you write moment equation [F1.x + F2.(x-L1) + F3(x-L2)] but I mean it is wrong. Because instead F3(x-L2) should be F3.[x(L1-L2)].

  4. I was fascinated by how you explained car transmissions, so I subscribed and looked at your page. I was so glad you also cover structural analysis because I'm a graduating Civil Engineering student and it would be good to hear about what I'm studying and maybe I can learn more because of how you explained them!

Leave a Reply

Your email address will not be published. Required fields are marked *