Mechanical Engineering Exam > Mechanical Engineering Videos > SFD and BMD for Simply Supported Beam with Concentrated Load SFD and BMD Tutorial 4, Strength of Materials
SFD and BMD for Simply Supported Beam with Concentrated Load SFD and BMD Tutorial 4, Strength of Materials Video Lecture - Mechanical Engineering
FAQs on SFD and BMD for Simply Supported Beam with Concentrated Load SFD and BMD Tutorial 4, Strength of Materials Video Lecture - Mechanical Engineering
| 1. What is SFD and BMD in strength of materials? |  |
Ans. SFD stands for Shear Force Diagram and BMD stands for Bending Moment Diagram. In strength of materials, SFD and BMD are graphical representations of the variation of shear force and bending moment along the length of a simply supported beam subjected to various loads.
| 2. How do you determine the SFD and BMD for a simply supported beam with a concentrated load? | |
Ans. To determine the SFD and BMD for a simply supported beam with a concentrated load, follow these steps:
1. Calculate the reactions at the supports using equilibrium equations.
2. Draw the shear force diagram by starting from one end of the beam and considering the sign convention for shear forces.
3. Plot the bending moment diagram by integrating the shear force diagram.
| 3. What does a positive shear force indicate in an SFD for a simply supported beam? | |
Ans. A positive shear force in an SFD for a simply supported beam indicates that the forces on the left side of the beam are greater than the forces on the right side. It implies that the beam is experiencing upward forces on the left side and downward forces on the right side.
| 4. How can we determine the maximum bending moment in a BMD for a simply supported beam? | |
Ans. The maximum bending moment in a BMD for a simply supported beam can be determined by finding the point where the shear force changes its sign. This point indicates the location of maximum bending moment. The magnitude of the maximum bending moment can be calculated by evaluating the bending moment equation at that point.
| 5. What does a negative bending moment indicate in a BMD for a simply supported beam? | |
Ans. A negative bending moment in a BMD for a simply supported beam indicates that the beam is experiencing compression on the top side and tension on the bottom side. It implies that the beam is bending in a concave upward shape.
Text Transcript from Video
hello in this video I am going to share
with you how to solve question for
simply supported beam well you can see
here I have a typical question of simply
supported beam and in this question I am
going to share with you how you can do
the same well let me get the question
that's why you can do or you can repeat
the same steps
this is a question is simply separate
beam well we come to know is a question
of simply accepted beam that's it of
length 12 meters so let me draw a beam
so it is simply supported and its length
is 12 meter and carried point load
Carey's point load 4 kilo Newton and 6
Newton at the distance of 3 meter and 6
meter so from here to here 3 meter and
Lodi is 4 kilo Newton and from here to
here is 6 kilo Newton and the distance
is 6 so this is 3 and total from here to
here is 6 1 so to draw shear force and
bending wind diagram let me exchange
these all lines that's why it will
become easier for you as well as for me
to make you understand how you can do s
FD shear force diagram I will draw here
and I am taking this line as a reference
line so first of all this case off
simply separate beam so first of all let
me find out how much total load is
acting on this beam beam is a B so are a
plus RB is equal to 6 plus force is
equal to 10 kilo Newton so total but
maybe you're thinking why I have written
RA and RB and what are these arianna p
ra is reaction force acting at E and
this is reinforced acting as B that
means if load is acting from this side
tan that means there is a reaction force
acting from this side that is also 10 so
10 kilo Newton is acting from this side
that is why there is supporting load 10
counter
this site from here from these two
points a and B so now we have to find
out the value of our a and our B so this
is our late arrays fixed first of all
let me assume our a is fixed now if R is
fixed that means B RB into 12 is equal
to 6 into this is 6 plus 4 into 3 so 36
plus 12 is equal to 48 so RB is equal to
42 12 is equal to 4 so this is 4 kilo
Newton and if this is for Kingdom that
means the meaning is 6 Clinton is here
so at the point R B there is 4 kilo
Newton DX and force is acting and well
in so here what we have
we deed RB is equal to 4 X M force is
acting from this side is 40 Newton and
from this side is 6 kilo Newton now we
need to find out shear force Tyga
so here we have to draw this is acting
this side that's why this is negative so
this is 4 kilo Newton here and here we
do not have any load that is why it
would be state line and this line is
Pinotti into 4 kilo Newton now this is
also for Compton but again here positive
this is negative this positive minus 4
plus 6 is 2 so this is here 2 kilo
Newton now this is again straight line
because there is no road acting from
this spin to this spin so that's why I
can say straight line you know I am
adding years for because it is positive
so 4 plus 2 is 6 kilo Newton so this is
600 Newton and here this is 3 so this is
3 this is 2
Arisa 6 because 6 is total from here to
here so 6 minus 3 is equal to 3 and this
is 6 is from from here to here 6 or
remaining 6 so this is 6 now I need to
draw the bending moment diagram so this
is super simple my dear yes as I share
with you in a previous video
we use in the case of simply supported
beam if you have fixed a point then we
need to start drawing bmd diagram
bending moment diagram from this point
only so let me start drawing this
diagram well to draw this diagram we
have to start from here so the area of
this this this rectangle how much 6 into
3 18 because this is straight straight
line that's why it will become this is
become what is this inclined but it's
straight so this is 18 kilonewton meter
the unit or bending madhi's kilo Newton
meter again I have to find out area of
this this is 2 height and 3 is what 3 is
length so 2 into 3 is 6 6 plus 18 is 24
this would be 24 kilo Newton meter so
don't mix with this with let it take
little as yes let me use another pen yes
now 2014 ton a so now again we to find
out one thing this is 6 and this is 4 so
24 minus 4 is because this is minus and
this is positive that's why I have added
and this is negative that's why I will
do - so 24 minus 24 is 0 so this is the
way how you can draw bending moment
diagram for simply supported being in
the case of point load my dear believe
me if you follow this concept then you
can solve any question in this topic I
hope you are mes to see this way to
solve this type of question well thanks
for watching hope to see you again in my
next video ya know ya
another more thing if you are getting
problem to solve any question in this
topic let me know I will try to solve
you and I will give you the proper
solution and definitely I will upload
that video to my youtube channel
well thanks for watching hope to see you
again in my next video bye bye
with you how to solve question for
simply supported beam well you can see
here I have a typical question of simply
supported beam and in this question I am
going to share with you how you can do
the same well let me get the question
that's why you can do or you can repeat
the same steps
this is a question is simply separate
beam well we come to know is a question
of simply accepted beam that's it of
length 12 meters so let me draw a beam
so it is simply supported and its length
is 12 meter and carried point load
Carey's point load 4 kilo Newton and 6
Newton at the distance of 3 meter and 6
meter so from here to here 3 meter and
Lodi is 4 kilo Newton and from here to
here is 6 kilo Newton and the distance
is 6 so this is 3 and total from here to
here is 6 1 so to draw shear force and
bending wind diagram let me exchange
these all lines that's why it will
become easier for you as well as for me
to make you understand how you can do s
FD shear force diagram I will draw here
and I am taking this line as a reference
line so first of all this case off
simply separate beam so first of all let
me find out how much total load is
acting on this beam beam is a B so are a
plus RB is equal to 6 plus force is
equal to 10 kilo Newton so total but
maybe you're thinking why I have written
RA and RB and what are these arianna p
ra is reaction force acting at E and
this is reinforced acting as B that
means if load is acting from this side
tan that means there is a reaction force
acting from this side that is also 10 so
10 kilo Newton is acting from this side
that is why there is supporting load 10
counter
this site from here from these two
points a and B so now we have to find
out the value of our a and our B so this
is our late arrays fixed first of all
let me assume our a is fixed now if R is
fixed that means B RB into 12 is equal
to 6 into this is 6 plus 4 into 3 so 36
plus 12 is equal to 48 so RB is equal to
42 12 is equal to 4 so this is 4 kilo
Newton and if this is for Kingdom that
means the meaning is 6 Clinton is here
so at the point R B there is 4 kilo
Newton DX and force is acting and well
in so here what we have
we deed RB is equal to 4 X M force is
acting from this side is 40 Newton and
from this side is 6 kilo Newton now we
need to find out shear force Tyga
so here we have to draw this is acting
this side that's why this is negative so
this is 4 kilo Newton here and here we
do not have any load that is why it
would be state line and this line is
Pinotti into 4 kilo Newton now this is
also for Compton but again here positive
this is negative this positive minus 4
plus 6 is 2 so this is here 2 kilo
Newton now this is again straight line
because there is no road acting from
this spin to this spin so that's why I
can say straight line you know I am
adding years for because it is positive
so 4 plus 2 is 6 kilo Newton so this is
600 Newton and here this is 3 so this is
3 this is 2
Arisa 6 because 6 is total from here to
here so 6 minus 3 is equal to 3 and this
is 6 is from from here to here 6 or
remaining 6 so this is 6 now I need to
draw the bending moment diagram so this
is super simple my dear yes as I share
with you in a previous video
we use in the case of simply supported
beam if you have fixed a point then we
need to start drawing bmd diagram
bending moment diagram from this point
only so let me start drawing this
diagram well to draw this diagram we
have to start from here so the area of
this this this rectangle how much 6 into
3 18 because this is straight straight
line that's why it will become this is
become what is this inclined but it's
straight so this is 18 kilonewton meter
the unit or bending madhi's kilo Newton
meter again I have to find out area of
this this is 2 height and 3 is what 3 is
length so 2 into 3 is 6 6 plus 18 is 24
this would be 24 kilo Newton meter so
don't mix with this with let it take
little as yes let me use another pen yes
now 2014 ton a so now again we to find
out one thing this is 6 and this is 4 so
24 minus 4 is because this is minus and
this is positive that's why I have added
and this is negative that's why I will
do - so 24 minus 24 is 0 so this is the
way how you can draw bending moment
diagram for simply supported being in
the case of point load my dear believe
me if you follow this concept then you
can solve any question in this topic I
hope you are mes to see this way to
solve this type of question well thanks
for watching hope to see you again in my
next video ya know ya
another more thing if you are getting
problem to solve any question in this
topic let me know I will try to solve
you and I will give you the proper
solution and definitely I will upload
that video to my youtube channel
well thanks for watching hope to see you
again in my next video bye bye
About this Video
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Sep 30, 2025
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