# Lecture -5 Simple Stresses In Machine Elements good day we will be having the lecture 5 for
the machine design course 1 and this lecture deals with the simple stresses in machine
elements now as you know that what is actually the stress that is being produced in the machine
element actually the stresses in the machine elements are produced due to applied load
this is something like that if you have been given too much of home assignments then you
feel stressed in the similar manner if a machine element is loaded then it is also stressed
in similar manner now in this particular case in the machine
design what one has to ensure is that the elements can sustain the induced stresses
what i mean to say is that in this particular case the all the machine elements will be
subjected due to the applied load to the different kinds of stress although we will be taking
up each and every type of stresses in short while from now but still just i am giving
an example of a lever arm attached with one lever arm you can see attached with a spring
at this particular junction and then what is happening is that you are having an hinge
then what will happen then this particular hinge pin will be transmitting the force to
this one the spring will be getting compressed so what you can see that hinge pin will be
subjected to some sort of stresses that is called as shearing stress spring will be subjected
to a pure shear stress and lever arms due to this loading it will be subjected to a
bending stress now in general if we consider that what is the stress we normally consider
is something like this suppose we are having a simple body which
is acted upon by some forces in these directions suppose it is p1 it is p2 it is p3 so and
so forth then if you make a cut at any section like this then what we will be getting? due
to its particular p1 and p2 and p3 if we are making a cut and if we just try to find out
that this portion of the cut is being removed this portion of the cut is being removed then
what we will get? we will be getting an internal forces generated
in these one now it is customary to put this internal forces one along the direction of
the plane it is a perpendicular direction to the plane one along the plane now in this
case what we will be calling is that this normal force what you will be getting is say
delta f and this particular force if we consider as delta v then we consider a stress sigma=delta
f/delta a where we consider this limit delta a tends to 0 so under these circumstances
what we will call this as a normal stress this is what we call as a normal stress now we can see that whenever you are vanishing
this area proportionately it is delta f is also coming down so in the limit we consider
this force to be a normal stress now in the similar manner if we consider this force component
then we are having another one we can call this as delta v/delta a limit delta a tending
to 0 and this is called let me write it little clearly this is called a shear stress so we
understand that any stress which is coming out perpendicular to the plane of the cut
is called the normal stress and anything which is coming as the tangential to the plane of
cut is called the shear stress now in general what we can see in the next
slide is that this what we consider is you can see this cuboid now this particular cuboid
is nothing but what we represent as a stress at a point although it is a cuboid we always
consider is a cuboid as representation for a stress at a point what we can consider is
that this cuboid what is showing to be some dimension is actually representing a point
and this actually represents a state of a stress at a point now we can see the state of a stress at a
point is defined as the normal stresses in 3 direction sigma x sigma y and sigma z and
onto the corresponding planes the shear stress is tau xy tau yz and so on and so forth you
can see tau xy means it is onto the x plane this is the x plane and what we are considering
is that in the direction of y that is a tau xy and in the similar manner it is in the
tau xz means it is in the x plane and in the direction of z the overall state of stress
at a point is defined by this particular matrix sigma x tau xy this is tau xz and similarly
you are seeing it as an tau yx sigma y tau yz tau zx tau zy and sigma z now a stress at a point also can be represented
in the form of what you call as a 2-dimensional state of stress so once we consider a two-dimensional state
of stress then we can consider just viewing from one side the cuboid if view from one
side suppose we view it from the side like this having a coordinate y and then this is
coordinate z then we understand that this one will be coming out to be sigma y this
will be coming out to be as sigma y this will be coming out to be as sigma z and this will
be coming to be sigma z now what will be the shear stresses correspondingly to this one so shear stress corresponding to this one
we can write it a shear stress in this direction as this way one in this particular manner
just this we can consider like this another one like
this another one like this and another one like this so what should be its notation it
is in the y planes so this we will consider as tau it will be yz this will be similarly
tau zy now in the similar manner we can write down these also as tau yz and this is also
tau we consider as tau zy now one of the interesting facts is that that
if we eliminate then this sigma y sigma z sigma y sigma z and we just simply retain
this one all other stresses then what we get is this that it is a state of stress for a
pure shear what we will call is that this state of stress is a pure shear so once it
is a pure shear then what we will consider is that we can see that if we in this particular
case if we take a moment about this particular point of the forces that is arising due to
the stresses on this face this face this face and this face then it will be seen that what we can see
that this particular force and this force because this area if we considered this length
and the height so what is this length this length is so that means we can write down
that force balance in the fz direction if we consider then what we can see we can see
that tau yzx this length is dzx top portion will be dx-tau yzxdzxdx=0 that means this
states that force balance=0 is maintained similarly if we consider a fact that this
was tau zy if we consider the fact that
if i considered this point about o and take moment about o then we will be finding out
that moment o=0 is satisfied and we will be finding out that tau yz=tau zy i am not doing
the entire calculations in the similar manner as earlier i have shown you can also verify
that tau yz comes out to be equals to tau zy this is a very important implication of the
stress is that this mutually always the shear stress occurs in a mutual perpendicular fashion
and they are equal in magnitude now this is what we can consider about the stress at a
point now if we try to see another aspect is that while we are considering the stress
onto the machine element the machine element also undergoes certain amount of deformation now that particular amount of deformation
what we are considering is being known to all of us what we call as a strain suppose
we write down the word ex then this strain in the direction of x axis can be written
as what we consider as a sigma x/e what is this e this we call elastic modulus or it
can be called as modulus of elasticity or simply young s modulus similarly we can have
the idea that means the same way we can write down as epsilon y in the similar manner as
epsilon y/e and like that we can also write but if we consider a body acted upon a body
which is something like this that we consider a body against suppose a cuboid acted upon
by stresses like sigma x sigma y sigma z then what we can say that in that case we can write
down in the expressions for all the forces in the similar manner as a 3-dimensional constant
is epsilon x we write down as epsilon x/e-muxepsilon y/e-muxepsilon z/e what this implies? this implies that once
you are having this particular idea then if a body is acted upon by the load as shown
in this figure then a body can have a tendency of deforming in this manner suppose you are
considering this as an x direction so we can see it will then increment in the x direction
and it will be an increment in the x direction but there will be a substantial amount of
what we call a deformation or a lateral contraction that takes place and this situation is taken
care of by this idea what is called a poisson ratio so in this particular case if we write down
the expressions for epsilon x similarly we can write down this expression for epsilon
y to be sigma y/e-mu sigma x/e-mu sigma z/e and epsilon z can be written as sigma z/e-mu
sigma x/e-mu sigma y/e so this way we can consider the strain of a machine member by
this expressions as it has been shown here now in the similar manner whenever we have
considered this particular element just earlier case that it was a shear stress coming in
this direction shear stress coming in this direction shear stress coming in this direction
and shear stress coming in this direction then what we can see that this effect of this
particular one this was y this is z then effect of this situation comes out something like
this as we can see it will be taking up some shape like this because it is getting stretched
like this in this manner now what we consider that this particular
angle this angle is actually what we consider as notations as gamma/2 and this is gamma/2
now due to such deformation due to the pure shear of an element this sort of distortion
gives rise to this angular deformation and this gamma is called shearing strain so in
this particular case of a pure element under shear we get an idea of the shearing strength
as such now in the similar manner the way we have
discussed about the case what we are considering as the normal stress in the similar manner
when we are considering the shear stress the shearing strain is also we can consider this
shearing strain and the shearing strain is defined by the relationship tau=gxgamma this
is another important relationship where we understand that if we considered this idea then this tau is the shear stress this is
shearing strain and what we consider this g is called shear modulus of elasticity so
this one what we call as a shear modulus of elasticity now once we know that these are
the relationship between the stress and the strain or the deformation pattern we can think
of something more in the later part what i was trying to show you is that we classify
some type of stresses depending upon the different type of loading one thing one should remember that we are
having only 2 type of stresses what are those 2 type of stresses just we have discussed
about one is the normal stress another is a shear stress now depending upon the type
of loading one can have different type of stresses that is created onto an machine element so if we consider that this particular idea
that what are the type of stresses that comes into picture that means we can categorise one as say normal
stress another as the shear stress in the normal stress if you are having a member like
this suppose a prismatic bar acted upon by the load p and load p here then what will
happen that this at any section if you cut then this will create a normal stress sigma
over here similarly if you are having a plate supposed placed onto a base and some amount
of it is projecting out and suppose a load is being applied to this one then what will
happen? this member will just cut cross this line this member will just simply cut across
this line and it will come down over here and this one just after deformation will look
like this this is an example where you can see a sheet metal when it is being cut you
give a load like this and it is getting sheared off so we call this word as sheared off that
means what is happening by the application of load a shear stress comes into picture
and the machine element could not take up that shear stress and thereby it fails in
shear so in contrast to the same thing in this particular
to the axis of member this is the thing and this was the perpendicular to the axis of
the member so this gives you an idea of the shear stress this is the normal stress so
what is the relationship of this stress so we will be writing this sigma as the stress
created and this stress created is p/a where a stands for what this area let me draw it
here again this is the area this area of the cross section is that a and here also see that tau we will be writing
as also p/a see the relationship between these 2 are the same but only thing is that one
is creating a shear mode and other is creating a normal stress mode so the relationship between
these 2 will be coming out to be the same thing and here if we consider an area suppose
in this direction a cut then what will happen if we take a free body diagram there will
be load like this one along the normal direction and one along this tangential direction so
this tangential direction will create our tau stress so here it will be getting a tau stress and
here it will be getting a sigma stress so what we understand that whenever there is
is acting perpendicular to the axis of the member but tangential to the cross-section
then we are getting a shear stress what are the other forms where we can take the normal
stress and shear stress respectively let us consider one situation like this what is that situation we are having just
simply a bar and this bar is acted upon by a moment now once it is acted upon by a moment
m this is simply being deemed a bar then what we get we understand this particular bar will
take a shape of this particular one now if we consider here on this top most fibres then
we can see the top most fibres are getting compressed and the bottom fibres are getting
extended so this extension of the fibre and this is a compression of the fibre so what we are getting that as along the fibre
it is getting compressed or extended respectively then these also the stress here we consider
as an stress called sigma c and here we consider the stress as sigma t what is the t t stands
for a tensile type of stress and this stands for a compressive type of stress so in this
case also the stress created will be of a nature sigma that is normal stress but it
will have 2 situations one is sigma compressive another is sigma tensile so somewhere in between
we expect that there will be a zone where the fibres are not to be stressed at all by
the application of this movements external moments so here also we find out one case where the
normal stress arise in a similar manner if we consider another example over here that
we are having a bar simple cylindrical bar if you just considers a cylindrical bar like
this and if you apply torsion t and another reverse torsion t over there then what will
happen the bar will get twisted that means this bar will have a something like that original
line was there so with respect to this it gets twisted like this so in this case if we considered one section
over here then this portion of the section with respect to this section rotates in this
direction so thereby a similar situation occurs that as if this block is moving out with respect
to this block in this direction so hereby we can say that at this juncture that is at
this juncture what we can get we can get a stress of tau that is called the shear stress so we understand that if we consider or categorise
the type of loads that comes onto that machine element then we can say that what is happening
is this that normal stress are created by what a) direct load what we call normally
as axial load this creates stress of 2 types one is tensile that means both the loads are
acting on to the element like this another is compressive when both the loads are acting
like this number b is the bending similarly we are having shear stress we considered again
if we write direct load i hope that you understand direct load but
this will be something like as in transverse direction then you get and another one that
is due to torsion we will be discussing later on that we can have also shear stress developed
due to the action of bending onto a machine member so let us consider that how we the
situations or the expressions for this type of normal stress due to direct load normal
stress arising due to bending load and the last one i am not explaining here by any figure
but i am just telling it is a shear stress due to bending what is this one we will be
discussing later on that is also we get a shear stress due to bending so if we see that what is this one as the
figure comes out to be then tensile stress we have already seen the figure and this we
have seen this particular expression also we have written the compressive stress it
is just same axial loading but in the reverse direction normal stress perpendicular to the
section is always that one you understand so if there was a load i also have shown this
one that it will give rise to a sigma normal stress in this direction and what we get that
shear stress in these line is parallel to this section so this is what that gives as
a idea of the tensile and compressive stresses and if we think of a shear stress developed
at a plane parallel to the applied force now let us consider this idea you can see this
particular 2 plates these are the 2 plates and that these 3 plates these 2 plates are attached by a bolt or it
is just fixed by a bolt and here also you can see 2 and this is the third plate because
all the three plates are just kept in position by one bolt now if we look what happens to
this particular figure then let us try to explain this one this is the first situation this is one plate
and this is another plate so we are putting a bolt something like this
then what we get you can see that okay this is the bolt so with the 2 nuts we have fastened
it like that now we apply a load p although this p and this p are not in the same line
it creates a little bit of offset but ignoring the factor more or less what is happening
this p will create that bolt to be dragged on this site and this p will be creating a
situation to drag this part of the bolt in this side so if the bolt fails under the load
p then it takes a shape of this something like this so what we can see that there is a failure
how the load acted the load acted along the plane along the cross-section okay that means
transverse to the axis so thereby we call this as a shear failure so this is an example
what we are trying to show for the shear stress due to the transverse loading so this is a
shear failure how many plain it has failed it has failed in one lane so the shear failure
is normally termed as an single shear failure and what will be the mathematical expression
for the stress developed the tau=the load p/a where a stands for the
area of the bolt to be more precise this is the cross sectional are of the bolt now let us take another situation where we
find that a bolt or this situation what we have just seen 3 plates are being pulled by
this force p this is total p so let us give it as p/2 and something like that p/2 now
again similarly we are having a bolt attached with the nut over here now we can see that
at this particular position what is happening this portion of the bolt and this portion
of the bolt will have a tendency to move in this direction and this portion also in this
direction and this portion will also go like this direction so that means if the bolt fails then it will
take up a shape like this so what was the load on to this one the load onto this one
was p so that stress developed is again p divided by but in this case we are having
twice area because this is one area and this is another area so what we call it is failure
in double shear so this is also another fact that it is then called the failure in double
shear so we can find out this expressions like that in single shear it is in the double
shear now next thing we can find out that in finding
the normal or shear stress in design problems the critical sections must be considered what
it is meant by this it is meant by something like this that if we consider a plate and
there is a hole in the plate and you are acted this plate is acted upon by a load p in this
way this one suppose having a dimension d this is having a dimension suppose small d
then you can see the load is acting like this so it will be at this particular plane it is normal to the load so what type of stress
we get a sigma stress so one is p/dx suppose t where this stands for the thickness and
another case we get sigma=p/suppose is this section so it is sigma 1 and that is sigma
2 s stands 1 section first section and second section so we consider this as pxd-dxt now
obviously we understand that area of this particular one is this section 2 is smaller
than what we get in the area in the section 1 so what we are getting we will be always
getting a less stress at this particular section this is 1s and this is 2s so 1s will be a
less normal stress compared to what you are getting at the 2s so obviously what we will
have in the failure if at all of the plate is taking place it will take place at 2s so
this 2s stress what we are calling can be talked about as something like this c what
c stands for c stands for critical section so what we want to know is that in case of
finding the normal stress in the design either for the normal case or normal loading or the
shear loading always one should have an idea of the critical section so what we have gathered today what we have
gathered today is that what we can find is this situation that we know the definition
one is the stress we could know stress at a point it is the 3d view/2d view
we know what is strain that is in the normal and shear strain we have found out that
critical section means what and before that it should not be
the last but before that we have found out that sigma is created by normal force what
is called axial then bending moment and
we have found out that number six is that tau or the shear is due
the next class we will continue with a discussion of the simple stresses in machine elements
where we will be finding out that how we can develop the expressions for bending stresses
torsional stresses shear stresses due to bending and something more on to this subject of simple
stresses on machine element thank you so today we continue the lecture on simple
stresses in machine elements and this week is the lecture 6 professor – student conversation starts (()) (48:31)
professor – student conversation ends now in the bending stress we consider a beam
subjected to pure bending which you can see by this figure so this is a beam which has
got a prismatic cross-section and it is under a bending moment m please note that this bending
moment is equal in both the sides which signifies that the beam is under a pure bending what
we call is a pure bending means there is no change in bending moment from this section
to this section now under the action of such pure bending onto a beam element what we can
see that the top layer will be under compression the bottom layer will be under tension so obviously somewhere in between there will
be a layer where neither there will be any tension or compression now this particular
axis is called neutral axis some other things what we can also consider that in this case
of pure bending the cross-section of the beam has got an axial symmetry and again the assumption
is that after such bending the plane section or it means that the cross-section remains
plane even after the beam has undergone a bending now i will show you with the help
of one small rubber model that how a bending takes place now this is one simple rubber which you can
consider to be a beam segment and my 2 hands one on this side and another on this side
is approximately applying a uniform bending moment equal bending moment rather to be more
precise at the 2 ends so if i apply a bending like that you can see that what we have just
discussed earlier that the top fibre is under compression so that the length of the beam
segment this particular segment is getting a compression and the beam segment at the bottom side is
getting extended that means this is under tension so obviously what here has been picturized
by neutral axis na is actually what we call as a neutral axis that means at the line along
that neutral axis there is neither any tension nor any compression so this is precisely the
model of a beam under a pure bending next we go back to the text once again so what we have seen is that a beam which
is coming out to be under bending means this particular beam what we have considered in
the earlier figure was something like this this what we call as a neutral axis this we
call as a neutral axis and this is acted upon by a pure bending that means bending moment
at both the sections are coming out to be the same now under the action of this bending moment
as we have just seen that earlier that we can have some idea is something like this
once again we draw it in a bent shape mode let me redraw this once again in a little
clarified manner this is the beam this is the line segment now this is the line after
it has come across a bending now this was the section got bent in the similar manner
and this is what we call as a neutral axis now we understand that here if we considered
the original one then this saves the original one and which got twisted to this line as
it is shown in the figure so obviously if from this one at any distance y please note
that we consider this as a positive direction of y at any direction y if this was the length
of the original fibre and that we considered as dx then this is the increment what have
obtained at this condition now when it has come to this bent shape let
us consider that this particular line this particular line is having a radius of curvature
we considered that as to be rho and this angle we considered to be say a small angle d theta
so obviously what is the length of this particular elongation so if we consider suppose this
was a now it has come down to aa dash then the length of this particular segment we as
increment aa dash and what is the original length the original length was the thing what we
considered as a line this length a suppose a double prime so aa double prime was the
original length and this what we can consider as a strain in x direction so this we can
consider as an epsilon x so if the strain of this particular fibre at a distance can
be even written what is this one with respect to this angle this is yd theta/this we can
consider as an dx again one thing we can see that this yd theta is an increment and what
is the relationship between d theta and dx once again we can see that this particular
one when we considered this length dx then immediately we can see that this row whatever
a considering the distance this rho d theta was nothing but dx because this was original
length so once again we can consider this one as rho d theta=this is dx so obviously
we substitute then we can see that this d theta/dx if we substitute it comes out to
be equals to y/rho now this gives us a simple idea that epsilon x or the strain of the fibre
is coming out to be equals to y/rho now let us consider again that epsilon x we
found out to be equals to be y/rho so what is a stress sigma=exepsilon x=ey/rho so we
can see the stress is varying for a given bending moment e and y constants so the stress
varies with the distance y from the neutral axis now let us consider…

## 18 comments on “Lecture -5 Simple Stresses In Machine Elements”

1. AZEEM AMIR Post author

awesome explanation….thank you..

the person who commented this lecture as "useless" is indeed USELESS FELLOW

2. md numaan Post author

thank you

3. thenomad014 Post author

thank you very much

4. orume Eboka Divine Post author

thanks very much for this lectures

5. akshay thakur Post author

Thanks'

6. SHOBHIT SHARAN Post author

very poor quality video, I am unable to see symbols and notations properly

7. mahesh babu Post author

poor quality

8. aniket pal Post author

good

9. Ashish Jaiswal Post author

fake accent

10. rare clown97 Post author

i cant hear anything

11. kamaljeet singh Post author

Bc 1 m awaaz nhi aa rhi

12. kamaljeet singh Post author

Randi rona.
Wtf hrd

13. Nikunj garg Post author

Strange!! ek ear m awwaz nhi aa rhi

14. Nikunj garg Post author

watch at 1.25x ; otherwise it is very slow

15. Market Sentiment Post author

Sir nasta ni kiye tha kya itni kam voice kyon a rhi

16. Dhruvil T Post author

Sound and quality issue

17. MechanicaLEi Post author

Sir, loved the video. However, the volume is a bit low. Please add subtitles so that I can understand the explanation too. Thanks for uploading!

18. Sagar Paradkar Post author

Axial loads can create shear .. if the opposing forces are displaced from the axis