Static Loads and Design Considerations – Design Against Static Loads – Machine Design I

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from Ekeeda Hello friends welcome to the subject of machine design wa so far we
have finished with 2 modules we’re in the first model we have started with the
basics we went through different aspects of design so in the second model we went
through the design and this a numerical of curved beams and thick cylinders in
the current model we are going to start with the real-life problems where we
will be little closer to the engineering problems so the topic itself says that
it’s the design against these static loads people there are two types of
which keeps on changing with time is called your fluctuating you in next
model will be dealing with the components which experience fluctuating
aspects and design procedures for the components some of the components not
all some of the components which experience static loops so let us begin
applied thing you need to understand that if
it’s just an assumptions that load are applied very gradually what happens when
effects that take place or the elongation for example that take place
inside the material inside the part happens gradually and hence we can use
certain set of formally to define them that is that is why we consider them
that is gradually applied loads fine then what are the modes of failure when
elongation like stress like strain all these effects are gradual and this
effects may lead the object may lead the part to failure failure happens or
failure may not happen these are two different aspects so we have to design
the component considering that this component is going to fail under this
load so that if that load is applied with the design aspect the component
will not fail for example if there is a chair that I need to design I’ll
consider that the chair will fail at 150 Newton if I apply the load of 150 Newton
so let me design it to certain higher extent I’ll design it for 200 Newton so
I’ll design in such a manner that it will be sufficient to carry 200 Newton
so if at all I am applying the load of say 150 Newton which is smaller than 200
the chair will not fail so I have to consider the extreme case where the
chair may fail that is 200 Newton similarly I have to design different
products for their braking or for their failure loads so what other types of
failure that we need to understand so there are basically three types of
failure the first is elastic deflection
deflection may happen in different directions and that depends upon the
applied load for example if the load is applied in axial manner the deflection
takes place in axial manner the load applied is in tangential manner a
deflection can happen Schulman of the product then general
yielding yes we know that when the load is applied on the material depending
upon the strength of material or the elasticity of the material plasticity of
the material it yields yielding simply means change in dimension it may gain
the dimension it may reduce the dimension so based on the given or based
on the applied loads this yielding may happen now if the yielding is beyond the
specified limit or if the yielding is beyond the tolerable limit then I’ll say
that the product has failed for example if I am applying some load on say one
member one one metallic member of length 100 mm after that load is applied that
member will elongate know if that elongation if someone has already
specified can be tolerated up to 1 mm and somehow due to that load the
material goes beyond 102 mm that means its elongation is its new length minus
initial length new length is 1 1 0 to initial in this 100 so if I take the
subtraction it comes out to be 2 mm elongation somehow the specified limit
is 1 mm so in that case I’ll say the mattel has failed the failure actually
doesn’t mean the breaking the deflecting or distorting of the object failure can
is always a subjective thing for some people for some define problems it can
be just the crossing the limit for some problem it can be the actual physical
failure I hope you are getting my pond so
general building can also lead to the failure and the third is the fracture
the fracture is the thing that we are not considering at all right now we’ll
consider in the subsequent modules but fracture actually means the sudden
failure of the product due to the cyclic loading okay so these are the three main
modes of failure which someone can observe in case of machine parts let us
begin with it let’s go ahead with very important concept that is called factor
of safety of course we have learned this concept before this let us quickly
revise it factor of safety is the margin through which the product is kept safe
like I said if I have to design a chair to sustain 150 Newton but if I design it
for 200 Newton this margin of 200 minus 150 that is 50 Newton is is giving me
the safety range or safe range so that even after the application of floor
greater than 150 the chair will be in safe position so this fraction of 50
Newton or the margin of 50 Newton is expressed in terms of factor of safety
the very generalized formula is failure stress divided by allowable stress that
as we know that stress is directly proportional to the applied load so we
can also say that failure load divided by allowable load we know that failure
load failure stress is always greater and allowable stress is always smaller
so in that case I will get the factor of safety but I must take care that the
factor of safety is greater than one then only I will be able to pursue it
and hence it is actually in words of it so it makes it the lore
which I have allowed / the lower that is actually working inside the material
in simple manner allowed / actual we are supposed to keep the actual load lower
than the allowed load there is the simple mean like 200 divided by 150 that
kind of thing so there are two ways we can define factor of safety factor of
safety for ductile materials and factor of safety for brittle materials we know
that different parts are made up of different materials we are basically
considering ductile and brittle so when you consider the pile material if you
remember stress-strain curve for ductile materials its the yielding point that we
consider the prime point because after yielding point your plastic yielding
actually starts and that’s why we have to consider the extreme point as the
yielding burn and hence in case of ductile materials factor of safety will
be given by yield stress divided by working stress or yield force divided by
working force Yale forces the force at which the
material will actually start yielding and hence for ductile its yield about
working but in case of brittle metal we know that there is no specific yield
point right that’s why in case of brittle materials we consider the basic
point as the ultimate point and hence it is the ultimate point divided by working
stress for the brittle material or it is the ultimate force divided by working
force for the brittle material so I’ll say force ultimate divided by force
working in the same manner for all material it is the force yield /
force who working so in short the load which is which is causing yield / the
load which is allowable for working so this factor is very important and this
factor of safety let’s move ahead with some of the
parameters which we need to consider as far as the design against static loads
are concerned of course we have studied those concepts already in different
manner but today we are going to look at them in the light of design so very
first thing is strength strain we know that it’s the ratio of
change in length divided by original length no matter which direction it is
it is change in length divided by original length and hence it’s a
unitless term we need to consider this strain that is actually causing incident
that is actually generating inside a product if the design specification is
in terms of strength for example if the design is to be done for the limited
strain of a product I must find out the strain that is causing inside the
product I have to compare this actual strain which is being caused with the
limiting strength in that case I must keep the actual strain caused inside the
material smaller than the limiting but how I will do that for how a person will
do that the person has to design its dimensions person has to design its
material in such a manner that the strain that is generated is smaller than
that so there are two things in a hand we can design the material and we can
design the dimensions when I say design the material I have to use an
appropriate material the material which is suitable for that condition may be
the condition may be in tension may be in compression or may be in shear so
depending upon the given condition I will use it appropriate material based
on its material property for example if there is a product which has to
undergo compression so we know that brittle materials are better in
compression than the ductile materials so for that particular design
I will prima facie consider the brittle material for design if that material is
found safe I will go for the manufacturing with the same material of
course the cost and other considerations are there but as of now we are not
considering those considerations we are considering only material only
dimensions and only given loads right so in that sense we will design it
for material that means we build is we will select an appropriate material or
we will design its dimensions basically selecting material and designing
dimension is what we are going to do throughout this module so that was about
the strength next thing is elongation it goes in line with the elongation again
we know that elongation is the change in length rate or the children’s which has
changed after application of load for certain product let us say they have
specified the allowable change in length or the allowable length will be only 1 0
2 mm whereas its original length is 100 mm but after application of load if it
is going beyond 1 0 3 mm it is not within the range so its elongation is
getting out of the range out of the range which is allowed so in that case I
will say that the material is failing for its elongation
so that’s the second part third part is direct stress we all know that direct
stress is nothing but the ratio of the force divided by area
but which force and which area if there is a product of this kind for
some part of the product of this kind in which there is a force acting on its
cross-section directly so this trace which is actually generated in this
direction due to this force when it was acting on this area will be given by
stress is equal to force upon area that that is called that it stays in majority
of the cases direct stress is considered for the design nowaday our friends see
you need to understand that when a product is actually brought into its
working condition it’s not required that it will always experience the direct
stress it is not necessary that it will always experience the shear stress it
can be a combination of this tube it can be only one of them it can be multiple
stresses right so depending upon the problem statement we need to see that
the product is safe in both the stress shear stress as well as direct stress or
it should be safe for bending stress and added stress and so on so different
combinations are possible right so of course when we are going to solve this
problem we are going to look at these possibilities anyway
so that was about direct stress next come this shear stress all of us know
that shear stress is nothing but again force upon area of cross-section
which force and which area of cross-section let us say there is a
product which is failing due to this kind of load which is tangential to
certain area so that kind of failure is called shear
failure and if this is happening in that case shear stress that is the resistance
to shear is given by the applied force P which is tangential to the area a
no doubt the formula looks like similar but the way the forces are acting on the
bodies are different it was found that and of course we are
going to prove this in the latter part of this topic this subject that shear
stress the maximum shear stress that occurs in
any body rather a ductile body is exactly half is the yield strength
let me repeat the shear strength of any ductile body is exactly half its
yield strength and hence if I mentioned the shear stress by tau the Tau is
actually equal to 0.5 times Sigma Y now since we are discussing this
particular term for only ductile material I will not use it for brittle
materials so it has to be only Sigma y naught Sigma ultimate right so this is
how the relation is established between shear stress and direct stress for yield
stress the next thing is bending stress bending stress come into picture when
the loads act in transverse manner the example is being
in such cases the stresses which are inducing inside body will be bending
stresses next thing is talk talk is something which tries to change the area
of the cross-section in the circular manner
and hence the talk is also an important factor now you’ll ask why why I am
discussing all these points you need to understand that these static loads which
we are going to consider include all of them or in other language the static
loads which we are going to consider will cause all these effects on the body
so in short in this particular chapter we are going to design those parts or
those machine parts which are going to experience static loads which makes
variants all this kind of effects and we have to design them for sustaining all
this kind of effects so that is the introduction part of static loads in the
start with a very simple example and then we’ll go topic by topic
different parts we will go on designing thank you so much for watching this
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