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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

loads that actually are generated one kind of flow is your static load and

second is your fluctuating load in simple language the load which is

constant or which is applied very gradually is your static load the load

which keeps on changing with time is called your fluctuating you in next

model will be dealing with the components which experience fluctuating

loads and the design in the current model we are going to study the design

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

with it so as you can see static loads are the loads which are gradually

applied thing you need to understand that if

it’s just an assumptions that load are applied very gradually what happens when

we apply the loads gradually mainly when the loads are applied gradually the

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

I was discussing about gradually applied loads they have certain effects like

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

next part we are going to start with the design procedure of static load we will

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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