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

pin so if you are loading this particular p load

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

case the loading was along the axis of the member and in this case the loading is perpendicular

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

this sort of loading along the axis we get a normal stress when there is a loading which

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

to transverse load then torsional load also due to bending in

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…

AZEEM AMIRPost authorawesome explanation….thank you..

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

md numaanPost authorthank you

thenomad014Post authorthank you very much

orume Eboka DivinePost authorthanks very much for this lectures

akshay thakurPost authorThanks'

SHOBHIT SHARANPost authorvery poor quality video, I am unable to see symbols and notations properly

mahesh babuPost authorpoor quality

aniket palPost authorgood

Ashish JaiswalPost authorfake accent

rare clown97Post authori cant hear anything

kamaljeet singhPost authorBc 1 m awaaz nhi aa rhi

kamaljeet singhPost authorRandi rona.

Wtf hrd

Nikunj gargPost authorStrange!! ek ear m awwaz nhi aa rhi

Nikunj gargPost authorwatch at 1.25x ; otherwise it is very slow

Market SentimentPost authorSir nasta ni kiye tha kya itni kam voice kyon a rhi

Dhruvil TPost authorSound and quality issue

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

Sagar ParadkarPost authorAxial loads can create shear .. if the opposing forces are displaced from the axis