Equipment Design: Mechanical Aspects

Prof. Shabina Khanam

Department of Chemical Engineering
Indian Institute of Technology - Roorkee

Lecture 17
Design of Support

Welcome to the second lecture of week 4 and in this lecture we will discuss design of support.
Discussion on design of support we have started from the previous lecture and where we have
discussed what are supports for vertical as well as horizontal vessels, and there we have also
discussed one support that is bracket or lug support, which is used for vertical vessels of small
height. Now in this lecture we will discuss design of support and here we will discuss design of
support for horizontal vessel and that is the saddle support.

So let us start the discussion on saddle support.

(Refer Slide Time: 01:25)

So general practice of supporting horizontal cylindrical vessel is by means of saddle support and
number of support or number of saddle may be two or more. So basically this saddle support is
used for horizontal vessels and how it looks like that we can see from these images. So if you see
this image here we have the horizontal cylindrical vessel okay, different sections you can identify,
this is the shell, this is head, and it is basically supported by these supports and these supports are
called as saddle support okay.
So if you see saddle support is nothing but this structure okay and which has curvature and that
curvature fits well with the cylindrical vessel. If I want to see separately the saddle support that
you can see in this image and where this is basically the saddle and where this curvature will
depend on the diameter of vessel okay. So this is basically the saddle support. Now if you focus
on this image, here we have horizontal vessel and as far as stresses are concerned this horizontal
vessel has longitudinal stress as well as circumferential stresses okay.

So this saddle support should be designed in such a way so that it can withstand or it can hold the
vessel with all longitudinal as well as circumferential stresses. So from here onward we will
discuss the stress analysis related to the saddle and according to these stresses what should be the
condition that must be satisfied for completing the designing of saddle. So let us start the stress
analysis for saddle.
(Refer Slide Time: 03:02)

So as far as stress analysis of saddle is concerned we will first discuss the longitudinal stresses and
then we will focus on circumferential stresses. Now as far as longitudinal stresses are concerned
we will first speak on longitudinal bending movement and then that movement will be considered
to compute stresses in longitudinal direction. So let us start the discussion on longitudinal bending
movement.
And further we can see that these longitudinal stresses are available in shell. So whatever stresses
are available in shell that should be supported by saddle okay. So in this case as far as longitudinal
bending movement is concerned we focus that discussion on shell okay. So if you consider here I
am having this shell which is horizontally placed, if you see this is the shell and where it is lie on
two saddle that is this one as well as this one. Now the total weight of the shell will be supported
by these saddles and that will be taken by these saddles.

Now as far as weight is concerned what that can be that includes the weight of shell, weight of
head and whatever content is inside the vessel its weight should be added to that, and if I am having
different attachment inside this, for example, if I am having tubes or some other assembly so
weight of those assembly will also be included in total weight okay.

So here we can see the total weight and that total weight will be equally distributed to two saddles
or if I am using three saddles so that will be distributed equally to three saddles okay. So for
example, if I am having two saddles and total weight will be transferred to these two saddles
equally I can make the movement diagram okay. Now to make the movement diagram it will be
easier to consider shell as a beam which is lie on two supports okay.

Now if I am having a beam which is lying on two supports how I can make the movement diagram
that we will discuss here. So for example, this is a beam which is lie on this saddle as well as this
saddle now what happens total weight will be equally divided in these two saddles. Now what
happens when the weight will act to these supports, the supports will give the reaction in opposite
direction. So this will happen here and it will happen on second support also.

Now for drawing the movement diagram what we can consider that, for example, if this is the
beam okay and you must understand that movement diagram we prepare from left to right, when
we analyze the system from left to right. So what we have considered over here that shell is
considered as a beam which is supported on two saddles okay. So as we have discussed that when
weight is falling on the saddle, saddle will give a reaction in opposite direction.

So for example if I am having this beam or that you can consider as a shell which is we are
considering as beam. Now this beam is basically supported at two places, this is one saddle, this
is another saddle. So this is one saddle, this is second saddle. Now what happens when this saddle
reacts upward and this saddle will also react upward okay and total weight of this shell will lie at
the centre okay. So here I am having two reactions and at the centre I am having total load of the
shell.

So when I am starting drawing the movement diagram I will start from left to right. So when I am
starting from left the first reaction will occur due to first saddle and it is moving in upward direction
okay. So it will try to turn this beam towards anticlockwise direction. I hope you are getting me.
So when I am moving from left to right the first reaction will be due to the beam and as it is moving
upward it will try to revolve it or move it in anticlockwise direction.

And further the second load I am having from this is the load at the middle of the beam that is the
load of the shell okay. Now this load of the shell it is acting downward so it will try to make or it
will try to rotate this beam in clockwise direction okay. And in the similar line if I keep on moving
from here towards the end second saddle will give the reaction in upward direction and it will
revolve beam in anticlockwise direction.

So when we draw the movement diagram what happens whatever anticlockwise movement that
are denoted by minus sign and whatever clockwise movements those are denoted by plus sign that
is the normal consideration. So the movements which are coursing through saddle support these
will be denoted by minus sign and whatever will be created by weight of shell itself at the centre
of beam it will be denoted by plus sign.

So considering this if I am focusing on this movement diagram, this M2 and this M2 will be
negative, however, this M1 would be positive based on discussion we have just done. Now further
if we consider this assembly, if you consider here, what is this A, A is basically the distance at
middle of the saddle from end of the shell, from this side as well as from this side okay. And then
this H is basically the height of dome section and L is the tangent to tangent length.

Tangent to tangent length we have already discussed in previous lectures because this head is
basically making tangent to the shell and therefore it is called tangent to tangent length. If we
consider cross sectional view of this assembly, here you can see R is the inner diameter of vessel,
theta is the angle which is made by support to the centre of vessel okay. So this is minimum angle
of this support is 120 and then it keeps on moving.

And when support is attached to the shell this section we call as horn of the saddle okay. So as far
as bending movement diagram, M1 would be positive and M2 would be negative. So at mid-span
if I am considering, it means I am considering the centre part of this where M1 will play its role
𝑊1  𝐿
and M1 will have the positive sign and that would be denoted by this, which is equal to , where
4

2(𝑅 2 −𝐻 2 ) 4𝐴
W1 is the weight on each saddle and [1 + 4𝐻 − ]. So this is basically the movement at mid-
1+ 𝐿
3𝐿

span.
At support if I am considering where I am having M2 and this is basically denoted by negative
𝐴 (𝑅2 −𝐻2 )
1− +
𝐿 2𝐴𝐿
sign, why it is that we have already discussed and it is equal to −𝑊1 𝐴 [1 − 4𝐻 ]. So these
1+
3𝐿

two expressions of longitudinal movement we have considered, which will be used for
computation of longitudinal stress in shell.
(Refer Slide Time: 11:48)

So here I am having longitudinal bending stress in shell. So here we have seen that longitudinal
movements are occurring at two different places at mid-span as well as at support. So if I am
considering mid-span, mid-span will have two section, if this is the vessel mid-span would be the
at the middle of this total length and it has two section, first is at the top and second is at the bottom.
Top it means top section or top point of, top point at middle of the vessel and bottom section we
are considering at the centre or the middle point of the shell at bottom.

𝑃𝑅 𝑀
So stress at the top is given by f1, which is equal to − 𝜋 𝑅12 𝑡 okay. And at the bottom we will
2𝑡
𝑃𝑅 𝑀
calculate f1’ as a longitudinal stress and that is given by this expression 𝑓1′ = + 𝜋 𝑅12 𝑡. So here
2𝑡

you see we have at the bottom plus sign and at the top we have minus sign. So stress at the bottom
would be maximum.

Now we have longitudinal bending stress at saddle. It means wherever I am having the support,
again this saddle will include two points, first is at the top of it and second is at the bottom of it.
So at the top if I am considering it means I am considering top of shell which is just above the
saddle okay. And bottom if I am considering it means bottom of shell which is attached to the
saddle okay.

So here basically we are discussing these stresses in shell only, therefore all top and bottom are
considered in shell only, whether it is lying in saddle or whether it is lying away from the saddle
okay. So when I am having longitudinal bending stress at the saddle, at top we compute that in
𝑃𝑅 𝑀2
terms of f2, which is equal to − , and at the bottom we have this
2𝑡 𝐾1  𝜋 𝑅 2 𝑡
𝑃𝑅 𝑀2
𝑓2′ = +𝐾 2𝑡
where I am having constant K2. So this K1 and K2 are design constants, which
2𝑡 2  𝜋 𝑅

will be discussed later.

So for satisfactory design all these stresses that is f1, f1’, f2, f2’, all these stresses should be less
than or equal to f J okay, where f is the allowable stress of shell. So in this way we calculate stress
in shell in longitudinal direction. Now we will discuss stress in circumferential direction. So when
I am considering circumferential stresses, it means throughout the circumference stress will form
and which we have also discussed as hoop stress okay.

So whatever stresses are forming in a shell, all these stresses will be supported by saddle only
okay. So due to shear all these stresses would be transferred to saddle only. So here we are
considering two section, first is tangential shearing stress, which is for unsupported part,
unsupported part means where saddle is not occurring or saddle is not available. And second we
will consider circumferential stresses, which is just at the support or at the saddle. So let us start
that discussion with tangential shearing stress which is related to unsupported part of vessel.
(Refer Slide Time: 15:44)

Now load is transferred from unsupported part of the shell to part over the supports by tangential
shearing stress that we have discussed already, which vary with the local stiffness of shell. What
is the meaning of that local stiffness? For example, if I am providing some structural support to
the cylinder or to the vessel the load will be transferred or the circumferential stress will also be
transferred to that structure okay.

So in that case what we can consider if I am having this much length and then I am having here
head, so head is basically attached to the shell by welding okay. So that welding will provide
structural strength to that support okay. So unsupported part in that case would be which is away
from that edge where head is connected to the shell and which is not available at saddle. So
whatever that part we have that we will consider as unsupported part.

So here I am having case 1, where A is greater than R/2. Now what is A? A if you remember we
have discussed earlier that it is the distance from edge of shell to the middle point of the nearby
saddle. So if that distance A is greater than R/2, which is basically internal diameter of shell if
that distance is greater than R/2 we can calculate maximum tangential shearing stress as give by
𝐾3 𝑊1 𝐿−2𝐴−𝐻
this expression where q is the maximum tangential shearing stress equal to ( ). So all
𝑅𝑇 𝐿+𝐻

this L, H, A and R t, all these parameters you know.

Now what is this t, that appears in f1 and f2 expression also. This t is basically the minimum
thickness of shell because we are discussing stresses in shell okay. So here we have this design
constant K3 that we will discuss further along with K1 and K2. Now second case I am having if A
is less than R/2, it means if that saddle is close to the end or close to the head, so in that case I am
having two section. So when A is less than R/2, it means the saddle is close to the end and at that
point we will consider stress in shell as well as that in head.

So here we have two sections, shell as well as head. So in shell we can calculate shearing stress
𝐾3  𝑊1
as as it is shown here, where q should be less than or equal to 0.8𝑓 okay. Now this condition
𝑅  𝑡

will be applicable for this expression also and for this expression also because in both cases we
𝐾4  𝑊1
have q. Now in the end or in the head we have stress as qe, which is equal to , where te is the
𝑅  𝑡𝑒

minimum thickness of head and K4 is again the design constant that we will discuss later on.

Now condition to be satisfied this qe is it should be less than equal to 1.15𝑓 − 𝑓𝑛 , where fn is
𝑃 𝐷𝑜 𝐶
denoted by this okay. And this C is basically the shape factor if you remember that shape
2 𝑡

factor we have discussed in design of heads. For different types of head we have different methods
to compute this shape factor, so this shape factor will be computed based on procedure whatever
we have already discussed okay. So in this way we will calculate tangential shearing stresses in
shell as well as in head.

Now next point I am having is the circumferential stresses and these stresses are falling in shell,
which is just above the saddle okay. So this is basically we can call as supported part.
(Refer Slide Time: 20:16)
So at the support at lowest point of cross section we can calculate stress as f3, which is equal to
𝐾5  𝑊1
and that should be equal to 0.5 into yield point stress. And at the horn of the saddle, what
𝑡(𝐵+10 𝑡)

is that at horn of the saddle, where edge of horn is appearing in shell, so that is basically the end
−𝑊1
of support in shell. So there we have circumferential stress as f4, which is equal to −
4𝑡(𝐵+10 𝑡)
3  𝐾6  𝑊1
and this expression we will use if L/R is greater than 8.
2 𝑡 2

And in the similar line if L/R is less than 8 we will calculate f4 from this expression. Now what is
this +10 𝑡, if you see this image here this 𝐵 + 10 𝑡 is appearing, B is the total width of the saddle
and t is the thickness of shell, so 𝐵 + 10 𝑡 is this section. Now what is this section, it happens that
when we weld the saddle to the shell wherever that welding occur it makes the shell weaker at that
particular point, so for that purpose we provide a reinforcement strip of width 𝐵 + 10 𝑡 okay.

So sometimes we provide it if thickness of the shell is not significant and if it is significant we do

not provide that reinforced strip okay. So that 𝐵 + 10 𝑡 is nothing but the width of reinforced strip
and all these circumferential stresses should be satisfied or should be falling less than or equal to
1.25f. So in this way we will consider different stresses and their conditions to be satisfied. If
saddle is designed very well, all these conditions must be satisfied.
(Refer Slide Time: 22:36)
So till now we have discussed stresses which are falling in shell due to shell. Now we will discuss
design of saddle because whatever stresses are occurring in shell or what as or whatever forces are
apply in metal sheet of vessel or the weight of vessel itself that should be withstand by the support.
So support should also be strong enough to bear all these loads. So for that purpose we have to
satisfy the design equation which is related to saddle. So let us discuss that.

So saddle should be strong enough to withstand the forces imposed by vessel okay and for that
purpose we have design equation and that is basically horizontal component of all radial load
which are falling on saddle that should be F which is equal to K9W1. Now if you focus on this
image here we have this saddle support with reinforcement plate and here we have this cross
sectional view.

Now if you consider this force it means the horizontal component of radial forces that would be F
which is equal to K9W1. So when I am having a saddle okay and when failure will occur it will
fall either in one direction like this or like this okay. So in that sense or in that way we should
consider horizontal component of all radial forces together and then we will apply the stress
condition.

So total horizontal component of all forces will be given as F and that would be 𝐾9  𝑊1 as it is
𝐹
shown here and that should be F and (𝑅⁄3)𝐵that should be less than or equal to 2⁄3 𝑓. Now this f
is basically horizontal component and this is the acting area, now why it is acting area. If you
consider this image, here we have this distance or this length R/3, R is basically inner diameter of
vessel and its one-third part would be this height into B okay. So what is B, B is basically the
width of saddle.

𝐹
So this is basically the acting area for F, so (𝑅⁄3)𝐵 and that should be less than or equal to 2⁄3 𝑓,

where this f is the allowable stress of saddle material. So in this way we consider all stresses in
shell, which is due to internal pressure or design pressure and due to saddle and further we have
checked the strength of the saddle. Now whatever design equation I have considered in this lecture
it includes some design constants like K1, K2 like that up to K9. So what are the values of these
constants that we can see from this slide. So let us discuss that.
(Refer Slide Time: 25:45)

Here we have values of constant for saddle support. So here we have contact angle that is 120 is
the minimum angle and 122 and we have long table which I will show further. And here we have
K1, K2 for different conditions like A is less than R/2 or other condition and then K3 when shell is
stiffened by end of the vessel and we have other conditions also. And similarly K5, K6, K7, K8 and
K9. So here you see values are available at 120 and 122.
(Refer Slide Time: 26:20)
And further we have the table from 122 to 158, we have different values for these constants. (Refer
Slide Time: 26:30)

Now we have values of all these constants from 160 to 180. So here you can see that contact angle
of the saddle to the shell vary from 120 to 180.
(Refer Slide Time: 26:53)
And now we will discuss different dimensions related to saddle and that can be better understood
by these diagrams, which are shown over here. If you see this image, here we have the vessel
diameter of 1.2 m or lesser than that. Now what would be the standard dimension, dimension
includes this J, E, C, G, Y, tbp that is the thickness of bearing plate. All these parameters have
standard dimensions.
(Refer Slide Time: 27:25)

In the similar line when vessel diameter exceeds 1.2 m, in that case all these parameters whatever
are shown in this image. These are summarized over here, like here we have vessel diameter and
then maximum operating weight V, Y, C. All these dimension which are available in previous
images are summarized or shown in this table, and similarly other parameters also. So these are
standard dimension used for design of saddle.
(Refer Slide Time: 27:47)

Now we will see few examples for support design. So here I am having example 1 and in this
example the cylindrical vessel of 3 m outside diameter and 8 m length is supported by 4 lugs, so
this example is basically related to bracket or lug support, which we have discussed in last lecture.
So the vessel is filled with the liquid or fluid having density 870 kg/m3 up to 5 m height and is
being operated at 1.5 MN/m2 in gauge.

It is covered with flat head from both end having 12 kN of each head, corrosion allowance is 2
mm and it is class 2 vessel with double welded butt joint with full penetration and vessel is made
with IS: 2041-1962 20Mo55 material and its design temperature is 450. Wind pressure on the
vessel is 0.9 kN/m2, height of the vessel from foundation is 1. So total height of the vessel is 8
and height from the foundation is 1. So from foundation total height of vessel should be 9 m.

So height of bracket from foundation is 5 m, so this is basically length of the column, end diameter
of anchor bolt 3.1 m and steel channels are used to support the lugs. Now here we are given
allowable stresses of horizontal base and gusset plate that is 100 MN/m2, allowable stress of
channel 150 MN/m2, density of shell material and µ are given as this and this respectively.
What I need to find is standard thickness of the shell, then I need to compute the minimum
thickness of horizontal and gusset plates for this dimension and we have to find out the area of
cross section of channel column, for section modulus of column as 200 cm3 okay. Also we need
to suggest a suitable column.
(Refer Slide Time: 30:00)

So here I am having different parameters where operating pressure and all other parameters are
given, which we have discussed in the problem itself and let us start the solution of that.
(Refer Slide Time: 30:12)

The first part is I need to find out standard thickness of shell okay. Now if you consider this
problem, it has operating pressure as 1.0 MN/m2 and liquid is available up to 5 m height, it means
it has static load of liquid also okay. So when this type of condition is there when I am having
operating pressure as well as static head of the liquid I have to compute the design pressure in
slightly different way. For that first of all I have to find static load of the liquid that is ρgh of the
liquid and then I have to compare that with condition.

Now condition is when that load is greater than 5% of operating pressure I will add static head
with the operating pressure to compute design pressure. Otherwise if it is lesser than 5% I will
consider 5% extra as design pressure as we usually do. So static head would be 𝜌𝑔ℎ and that can
be found as 42673.5 N/m2 and which is equal to this much meganewton and when we consider 5%
of 1.5 MN/m2 it comes out as this and as this value is lesser than this I will consider design pressure
as 5% extra than operating pressure in gauge.

So here we have this is the design pressure and further calculation is similar as we have discussed
in previous lectures for design of shell, so this is the minimum thickness of shell by standard
expression, adding corrosion it gives 28.24 mm thickness and then standard is available at 32 in
table B1. So standard thickness of shell is 32 mm.
(Refer Slide Time: 32:26)

Now I need to find out thickness of horizontal as well as gusset plates. For that case if you consider
this is the expression to find out thickness of horizontal plate okay. And to compute this I need to
calculate Pav first and for calculation of Pav I need to find out total load on each lug okay and that
is given by this particular expression.
In this expression we need to find out 𝑃𝑤 (𝐻 − 𝐻𝑐 ) is the total height of the vessel, which is
basically 8 m, n is the number of lugs and c is the anchor diameter, Wmax is the maximum weight
of shell and accessories. So first of all I need to compute Pw and Wmax, so that I can find out value
of P. So Pw is coming as 17.01, which can be computed as 0.7 * 1 that is K1, K2, if you remember.

This is the expression K1, K2, K1 0.7 for cylindrical vessel, K2 will be 1 and P1 which is the wind
pressure, it is given as 0.9 and then H1 is the total height of vessel from foundation and that is 9,
Do outer diameter of vessel that is 3. Considering all these values we can find Pw like this.
Standard thickness of shell is given as 32 mm and then we need to find out maximum weight of
vessel.

Now how I can find out maximum weight of vessel, it includes weight of shell, weight of head and
weight of internal attachment. In this case internal attachments are not given, but liquid is given
up to 5 m height. However, for maximum weight we consider maximum possible situation and
that could be when vessel is completely with the liquid okay.

So here we will consider weight of shell, weight of head plus weight of liquid when it is completely
filled in the vessel. So to compute this, first of all we have to find out weight of shell. How I can
find out that? That is given by this expression 𝜋𝐷𝑜 − 𝑡𝑠 that is the standard thickness of shell.
This is basically for average diameter, so that is periphery into ts is the area into total length, that
is 8 m and then this basically gives the volume into density of material of shell and then g, it will
give value 177.87.

Weight of head 12 because one head includes 12 kN, so it comes accordingly. Weight of liquid,
how I can find weight of liquid that would be the internal volume of the vessel that is
𝜋
(𝐷𝑜2 − 𝐷𝑖2 ) Lρ𝑔. So considering those parameters you can calculate weight of liquid, which
4

comes out as 462.02. And similarly I can find out maximum weight.

Once I am having maximum weight I can find P as this. Now once I am having P, I can calculate
Pav and then thp that is the thickness of horizontal plate and which comes out as 27.10 m. I am not
adding corrosion to this because it is not coming in contact with the liquid. So directly next value
available in table B1 that I can choose as horizontal plate thickness.

And similarly for gusset plate we have this expression, where I need to calculate a which is
basically 0.05 and that you can find out from this expression okay and then cosθ, θ is 60o which is
given and then you can find out this t g value, which is 0.02798 m okay. Now what is this f. This
f here we have considered allowable stress of horizontal plate and here it is allowable stress of
gusset plate. In this case these both values are equal, otherwise we need to consider accordingly.
(Refer Slide Time: 36:58)

Next step I am having is to determine the area of cross section of the channel column for sectional
modulus of 200 cm3 okay. And we need to suggest the suitable column for the bracket okay. So
this is the expression to calculate the cross sectional area of column, where X is the cross sectional
area of column. All these parameters we have already discussed in the last lecture okay.
So considering all these parameters where f is the allowable stress of column okay. So that you
need to consider and solving this equation equating to 1 because that would be the extreme
condition and accordingly I can find X as 47.29 cm2. So as cross sectional area of the column
comes as 47.29 cm2, I have to choose the column which is higher than this area okay.
(Refer Slide Time: 38:03)

So for that purpose I have to see the table of channel, which is basically table C3 in B. C.
Bhattacharya book. So here you see 47.29 was the minimum area, so here what I can suggest, I
can suggest this one or I can suggest this one, which is greater than 47 cm2. So either ISLC 350
can be considered as suitable column or ISMC 350 can be considered as suitable column, so that
depends on us that which column I should use, but both column I can use easily.
(Refer Slide Time: 39:04)
Now here I am having second example, which is on design of saddle and in this example I need to
design a saddle support for horizontal drum, which is designed for internal pressure 2.1 MN/m2,
internal radius 1.2 m is given, tangent to tangent length is 15, depth of the dish of head that is H
as 0.4 m is given, allowable stress yield point of the material are given as this.

Corrosion allowance is 0, joint deficiency factor 1, modulus of elasticity is given and total load on
each saddle, which is given over here as 1.4 MN, contact angle is 120. So what I need to find is
check if the following assumed saddle design data that is A/R equal to 0.6 and B equal to 0.5 m,
satisfy all design stress conditions. If not, find the new values of A/R and B, which satisfy all
design stress conditions.
(Refer Slide Time: 40:08)
So let us start the solution of this. Here basically in saddle I need to compute different stresses
and that should be satisfied with the permissible values okay. So let us start that. So here I am
having all parameters, which we have just discussed and A/R is given as 0.6, so from that we can
compute A, which is basically 0.72 and it is greater than R/2 because this condition I need to use.
So here A is greater than R/2, B is given as 0.5 m, which is the width of the saddle okay. So all
parameters I guess you understand.

W1 that is weight on each saddle is given as 1.4 MN, M1 and M2 I can find out directly through
the expression and M1 comes out as 4.119 MN-m and M2 as -0.0236 MN-m.
(Refer Slide Time: 40:59)
So considering these movements we will calculate all stresses, but these stresses will also require
thickness of shell and therefore we can compute minimum thickness of shell using usual
expression and here if you remember 1.2 m R is given, which is internal radius, so internal diameter
would be 2.4 okay. So that we have used over here and therefore this minus sign appears and
minimum thickness comes as 16.92 mm.

Considering this value we can find out f1 as 20.656, f1’ as 128.28 okay, f2 as 71.58 considering K1
as 0.107 and f2’comes as 76.07 considering K2 as 0.192. Now from where this K1 and K2 comes.
(Refer Slide Time: 42:04)

If you see here I am having this K1 where this condition will be applicable, a is greater than r/2
and this K2 is given where A is greater than R/2. So K1 0.107 corresponding to 120o angle and K2
come as 0.192. So considering these value we can find f2 and f2’. Now you see this should be
satisfied with f J, so that should be less than or equal to 150 MN/m2. So all conditions are well
satisfied.
(Refer Slide Time: 42:42)
And further we have to find out tangential shearing stress and that case comes as when A is greater
than R/2, so we will use this expression, where K3 is 1.171 and that is given by this expression,
where K3 I am considering.
(Refer Slide Time: 42:58)

Here I have different condition I have taken this condition that is shell unstiffened by rings, where
no rings is provided. If rings will be provided it will be given in the problem. So corresponding
to 120 K3 value comes as 1.171.
(Refer Slide Time: 43:20)
So considering K3 as this we can find out q, which is coming out as this and this should be lower
than 0.8 * f, so it is well satisfying the condition. Next we have the f3 value and f3 will be given
𝐾5  𝑊1
as and K5 you can see as 0.76 and here you can see K5 as 0.76 from this table
𝑡(𝐵+10 𝑡)

corresponding to 120o, and considering these points K3 comes as 94.108, which is well satisfying
the condition that is it is less than 120 MN/m2.
(Refer Slide Time: 43:44)

And further we will calculate f4, for this we need K6 and K6 is given as 0.18. So further you see
this table, where K6 we can see figure or table 10.5 from book okay and book is B. C. Bhattacharya
book.
(Refer Slide Time: 44:27)

When A/R is 0.6 okay and θ is 120, so corresponding value of K6 is coming as 0.018. This is the
table from book okay. So considering this, you can calculate f4, which is coming as -163.305.
(Refer Slide Time: 44:48)

Now we have the horizontal component of all radial forces because till now we have discussed,
we have computed stresses due to internal pressure as well as due to saddle in shell. Now we need
to check for saddle strength and for this we have to calculate f4, for which I need K9 and K9 is
given as 0.204 in this table okay.
(Refer Slide Time: 45:10)
Considering that value we can find f as 0.2856 and then this condition we need to satisfy which is
this expression comes as 1.428 and this expression comes as 100 MN/m2, so it is well below the
permissible limit. So in this way we can design the bracket as well as saddle support.
(Refer Slide Time: 45:41)

(Refer Slide Time: 45:43)

And here we have some of the references to study the topic and here we have summary of the
video and this summary of lecture 1 as well as lecture 2 which is on design of support and it goes
as supports for vertical and horizontal pressure vessels are discussed, design of bracket or lug
support for vertical pressure vessel is described, design of saddle support for horizontal pressure
vessel is discussed and then few worked example we have discussed with detailed solution. And
that is all for now, thank you.