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Problem on Lever Loaded Safety Valve - Design of Cotter Joint Knuckle Joint Levers and Offset Links | Ekeeda | YouTubeToText
YouTube Transcript: Problem on Lever Loaded Safety Valve - Design of Cotter Joint Knuckle Joint Levers and Offset Links
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Video Transcript
Hello friends here in this video we will see
a problem on lever loaded safety valve
for that purpose here we have a
question the question is quite lengthy i will
read it so that we can understand what is
given in this part now a lever
loaded safety valve is 70 mm in diameter
and is to be designed for a boiler to blow off
at a pressure of 1 newton per a screw
gauge design a suitable mild steel lever
for rectangular cross section using the
following permissible force stresses tensile stress
70 megapascals shear stress 50 megapascals
bearing pressure 25 newtons per mm screw
the distance from the vane to the weight of the lever
is 880 mm and the distance between vane
and load is 80 mm now this is the question we
have to design the lever loaded safety
valve I will quickly draw the diagram and
valve now this is an opening
happens is that
this valve is used in case of boiler where
the steam pressure is very high so steam would be coming
from the bottom of the
boiler where this steam enters and then it will
try to lift this valve next this valve has
some pin attached to it
extended now this is the diagram of a
lever loaded safety valve once steam pressure comes it will
try to lift this
valve and this valve is kept in its
position by this value of p which is called
and this is called as the pane
and where the steam pressure acts this is called
as the load and now these are the pins that we
need to design and this is the lever Cross
section here we have this a the lever which we would be
designing so this is a lever loaded
safety valve now for simplification I will draw
problem now I'll read the
question a lever loaded safety valve is 70 mm
in diameter so the diameter of this valve is
of this
portion this I denote it by capital D
so this capital D is given to
us as it is equal to 70 mm
next and it has to be designed for boiler to
blow off at a pressure of 1 newton per millimeter
square gauge so the pressure inside this boiler this is
p this is the pressure of steam here so p is
next design suitable mild steel lever
for rector cross section the cross section
of the lever i rctor they have given using the
following permissible stresses r at the
values of total stress sig mti equal t 70
newtons per mm square Next shear stress
stress
to 50 newtons per mm square bearing pressure
P suffix B is equal to 25
newtons per mm square Next given
some distance the distance from the vane to the
weight of the lever is 880 mm So the weight of
the lever is nothing
but the effort we are applying called the weight
of the
lever so distance between vane now vane m
is at this
this so from the vane rum to the point where the
880
mm Next and the other distance is also
given in the problem is and the distance
So load is acting at this pin this second as we can see
so this is the load pin and at
this its center the distance between vane and
the load is given is 80
80
mm Now Il denote r This is the face
and face m here is
load face a denoted by f load is denoted
by d as it is very much clear from the
diagram the load is acting from down
words so the load on this pin it is from
down word direction that is d effort is
down word now I need to calculate how much is
the load w is acting so for the solution
of this what ever we have written here that is
a part of the data so it is given
data next we need to design this
entity and even the lever cross section so in
acting on the
valve will be given by weight is equal to
area pa ba fdi sk multiplied by the
so now from this I will say that for d i
pi ba
4 diameter is 70 70 sq into pressure is
one so from this I will get the value of W
and that comes out to be
be
3850 3850
Newtons now this much is the load ch e
from the downward direction
direction
now after getting double I can say there
now the reaction it can be calculated
once we have w and p values w value we
already have we have calculated just now
p we don't have so reaction result will be
effort and it can be effort minus load
depending on the value I will keep this as
on equation number one now we will go to
find p for that purpose I will say that there
for taking
moments
order now the moment we can
calculate by I can say since we want to
calculate moment at the full order r I consider
clockwise moment positive so it p in 880
Because the distance from the plank to this load is
880. So
for p in
in 880,
880,
next effort is acting downward, load is
acting upward. So here is the moment produced
by load to be anticlockwise. So it is 880 minus
0. From this p will be equal to 1. I shift
this term n to the other side n. Divide by 0. Now we have
3850. So there for the value of p it comes
out to be
be 350
350
Newtons. Now I have val and even dl. So set
there for put
dl and
p in equation number
one. So here I have rf e equal to dl. It was
dl minus p in equation number one. In equation it was
3850 minus p = 350. So from this I will
get my value of rf s 3500
know The value which is coming here
is the force now because of this reaction what will happen that since
it is a force the
face pin is acted upon by this force
and then there would be some kind of failure
of this face pin so I will say that after
reaching this stage now I will say that then we are
designing first step number
pin now when we are considering face pin face
this face
bush the length of the bush is taken approximations
of the face pin and since pin is it is made up
of metal even the bush is made up of metal so
because of metal to metal rubbing the phenomenon
it is called bearing because this pin would be
rotating inside the bush and the area which gets won
out that I can draw it to if this is a
Called
area diameter of plank pin now it will be set
for the area of all pin this would be projected
area this is
diameter of all
pin and if I have
pin so there
of all pin if we are
considering the bearing failure a set
set for
for bearing
strength it will be given by since we want
strength and strength is nothing but the load
carrying capacity at the plank the load is
nothing but reaction at all pin and it will be
equal to bearing pressure multiplied by the resisting
resisting area
now there for reaction at the plank rum that
value we have it is
3500 bearing pressure is given in the
df.com pin by
1.25 times the diameter of all pin so
from this if I calculate then go on calculating this
then at the full pin We know that the load
acting is
3500 and from this we would be getting the
it is close to 10 mm so it is I am
taking the value s12 so this is the
diameter of the alcam pin after this I can say that
for lf length of alcam pin is
1.25 in diameter of alcam pin is 12 from this
I will get l value s15
their reaction at the vane this value was
3500 newton load is
3850 so there is not much difference here so
whatever dimension we are keeping at the vane must be the same
there at the load and so after this I will say that
for taking the
taking the
for load
pin so now I will say that there
for load pin diameter I will take it as d1
and that will be equal to the vane pin
diameter which is 12 am length of load pin
I will take it as l1 its value is same
as the length of flank pin I will say that
of load
pin a1 is equal
to length
of load
pin now after getting this I can say that in our
step number one I have designed the flank
pin and even the load pin next after this what we can do is checking
flank
pin and we can say now even the load pin so it is flank
flank
shear now here we are
checking the flank pin under shear once it is shear it will
break into three parts
like this is called double shear so I will say that
d1 is the
diameter of the flank pin then there
for resisting
area it will be pi by
4 into diameter of flank
pin square into two because it is the case of
double shear after getting this resisting
area I
set the strength for
strength
of flank
pin as V So from this I will get the shear stress value as 1547 newtons per millimeter squared.
Now this value of shear stress is less than the given value in the question
shear stress 50 and we are getting it as 1547 so this value is less than the given value in the question
getting it as
1547 so this value is less than the given
safe under
shear. Now in this step number one we have
designed both the face pin and the load pin. Next
I will say that we will design the lever cross
two it
of the
lever cross section. Now for the lever cross section
section design
design
diagram I'm just taking this section to judge
a part of this lever Now the lever looks something like
position this is called a
boss
and here we have an internal hole and inside
this we are inserting the pin So first I will say
that for the lever cross section when I am
designing I will say that since the
pin the pin is at both the ends this is the
face which is a and this is the
load and so at that end we have designed the pin
So the diameter of the pin is equal to 12 so we have got
the value 12 now this pin is
inserted in a bush and after the bush it is
inserted in this lever section So here we need to
calculate how much is the hole in this lever
2mm
So there for the
diameter of the hole in the lever
hole in the lever
the Diameter of hole inside this
lever it will be the diameter of the pin which is 12
plus since I am using 2 am thickness of the
bush so there for this diameter of hole in the
lever I will denote it by d suffix a so there
for this value will be 16
am it means inside this lever section we
have to make a drill of 16 aa now this
external position is called boss so I will say that there for
for
diameter of
boss it is equal to 2 inside diameter of hole
that is 16 so diameter of boss comes out to be a
denote it as capital D it comes out
to be 32 am
now as it is given in the problem that the
cross section of the lever is a rectangle so I will say that there
section
thickness and here I have
inside of the lever cross section aa se So let the
relation be equal to four times of t so I'm asking now I
'm considering lever cross
section I will design this
lever re close to this flexion section that is here I will
okay I will
take a section here that is close to the load
load
section so now here
the area which is there which is the cross
section area it is rectangle in
section so this rectangle section we have to
design and effort is acting so here I need to know
the l the distance so a say that
taking moments close to this section f I
call it section
vava so a set after this deer for
for taking
moments at
section vava no no taking the moment to will be equal
equal
to p is the effort effort multiplied by this
distance and this distance I can take it
as 80 my 80 from this I will subcut the
radius of the hole so I will say that for
this m is equal to p in the The distance is
880 minus 0 from that I will subcut the radius
of the hole by the diameter of the hole. The radius of the hole is divided by 1. So here I am
taking the moment section 880
minus 0 minus the radius of this hole will give
me this
distance. So now I'll put the
values p is given to me.
This was 350. Next, this
is 800 minus 8. So finally, it comes out to be
27700 Newtons. So this is the value of
bending moment f I'm getting. After this, I set the value for the
model. It will be equal to t.
For the section model, this is the section model.
equal to t. So finally, the bending stress,
if I look into this
problem, the bending stress is not given in the
data, but we can take the tunnel stress equal to the
bending stress. Because both are principal
stresses, now sigma is 70. It
70. It was
27700. If it was tb2 /
6, so here I have 70 =
6. I will go into the numerator
t. Here, I can say that it is equal to
40. Because I have added 6 since tb is
4t. So from this, here I will get
this value
as this becomes t. I will shift it to the
277. 20000 * 6 divided by this 70 will come
here 4s * 16. That is, so finally I will
get the cube root of t and that is
is
11.4 and we can get
12mm. Now once we have got the thickness, I will say that even
with this, it can be calculated. So
So
for tb, it is equal to
4t. So it is 4 in 12 so it comes out to be 48
48 mm
mm
now this is the lever cross section so we
have this lever here the thickness of the lever we found
out the thickness is 12 and this width of the
lever that is equal
to 48 mm so we have designed this lever
cross section we have designed both the vane pin
and the load pin
so in this video we have seen how to
design the lever loaded safety valve by
considering the pressure of steam converting
that into load and then designing the vane pin
load pin and finally the lever press I
hope what ever is explained here that is understood
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