Strength of Materials Prof S. K. Bhattacharya Department of Civil ... · Secondly, this is a...

Strength of Materials Prof S. K. Bhattacharya

Department of Civil Engineering Indian Institute of Technology, Kharagpur

Lecture -15 Application of Stress by Strain

Thin-walled Pressure Vessels - I

Welcome to the first lesson of third module which is on thin-walled pressure vessels part one which is on the application of stress and strain.

(Refer Slide Time: 00:45)

In the last two modules, we had looked into the aspects of analysis of stress and analysis of strain at a point in a stress body as to how we compute the stresses and strain? Now we will look into the applications of this stress and strains at a point in thin-walled pressure vessels. The pressure vessel has wide industrial applications. In many industries we use this kind of vessel where the pressurized air or water is used. It is necessary to evaluate the strains and the stresses at any point on the surface of pressure vessels. We will also look into how to compute the strains and stresses on the body of such pressure vessels.


Before we go into the analysis of pressure vessels let us look into aspects which we discussed in the last module. Module-2 was devoted to on analysis of strain and we had 8 lessons, so let us quickly look into those lessons which we covered in Module-2.


The module of analysis of strain includes 8 lessons. The 1st lesson was on the concept of normal strain, the strain at a point and stress strain relationship. The 2nd lesson was devoted to the aspect of normal strain in a variable section. In the first lesson we discussed the normal strain in a uniform body. If the body is having a variable section then what will be the strain in it and

correspondingly the shearing strain in it. We also discussed the concept of rigidity modulus in

the second lesson.


In the 3rd lesson, we had introduced the concept of generalized Hooke’s law and the determinate and indeterminate system were discussed. Thermal effects on strains and stresses were discussed in Module-2 and the fourth lesson. Thermal effect on compound bars and the misfit and correspondingly the pre-strain and pre-stresses was dealt with in the 5th lesson of Module-2. (Refer Slide Time: 3:15)

In the 6th lesson we discussed the transformation equation and we had discussed this aspect in the first module of stress as well wherein we say, at a particular point if you are interested to compute the values of stress and thereby the strains then we need to use the transformation equation if we orient the axes system from the rectangular axes system.


That is what was discussed in Module-2 lesson 6 in 2.6 where we had introduced the concept of transformation equation and thereby how to evaluate the principal strain. Then we saw the concept of Mohr’s circle of strain, how to compute the strain at different orientations at that particular point where we know the rectangular strain components ∈ x, ∈ y, ϒ xy how to compute the values of principal strain and strain at that particular point at different orientations. If you like to compute we can do it through the use of Mohr’s circle of strain and that is what was discussed in lesson 7.


In the 8th lesson we introduced the concept of strain gage and strain rosettes. Thereby at any point it is easier to measure the strain and it is difficult to measure the stresses when we talk about a stress body. We can measure the strains and thereby from those measures of strains we compute the stresses. But as we have seen we measure the strains in the normal direction as the normal strain. Therefore out of three strain components ∈ x, ∈ y, ϒ xy it is very difficult to measure ϒ xy (the shearing strain). Hence, we take an indirect way to measure ϒ xy by using strain gauges in three directions which we have called as strain rosettes. From the measured data of strain rosettes, we can compute the strains of ∈ x, ∈ y, ϒ xy. Using these strain components we can compute the principal strains and thereby the principal stresses.


This is what was discussed in lesson 8 Module-2. Also, we looked into the relationship between the elastic constants which were E, G and mu.


This is the summarization of the Module-2. It is expected that once this particular lesson is completed, one should be able to understand the concept of stress and strain in thin-walled cylindrical pressure vessels. Here we are concerned with two aspects, one is the vessel is subjected to internal pressure and this internal pressure is over and above the external pressure. By overcoming the external pressure on the vessel, the internal pressure is working and thereby there will be the stresses generated in the vessel and we are interested in evaluating those stresses.

Secondly, this is a thin-walled structure and this means that the thickness of the wall is very small so that the stress variation across the thickness is significantly small or we neglect that. Hence the thin-walled pressure vessel is of importance and we will look into how to compute the stresses in such vessels.


Thereby we will evaluate stresses and deformation in the thin-walled cylindrical pressure vessels subjected to internal pressure because of the content in the container.


As we keep going through we will be looking into the aspects of previous lessons or recapitulation of previous lessons which we will be doing through the question answer session. We will look into the answers for those questions which I had posed for you in the last lesson and thereby will be scanning through the aspects which we had discussed in the previous lesson.


Also, we are going to derive the formulae for evaluating stresses and strains in thin-walled cylindrical pressure vessels. In this particular lesson we will be concentrating on the cylindrical type of pressure vessels and subsequently we look into different categories of pressure vessels. We will also look through a few examples for evaluation of stresses, strains and deformations in such pressure vessels. These are the two main aspects that we will be doing in this particular lesson.


Let us look into the answers to this questions which I posed last time. The first question is how you will evaluate principal stresses at a point from measured strain using a rosette.

When we try to measure stress at a point, we cannot directly measure the stress. Hence we measure strain at that particular point and thereby we compute the values of stresses. The strain component as we have ∈x, ∈y, ϒxy under a plane strain situation we cannot measure ϒxy directly, instead we measure indirectly and from there we compute the values of principal strains. Now let us look into how that is done.



In the previous lesson we had discussed that this is the x direction and this is the y direction. We can use these kinds of strain gauges which can measure the normal strain. If we place these types of gauges in the x direction and y direction , we can measure the strain∈ x, ∈ y, but we need three

quantities∈ x, ∈ y, ϒ xy for evaluating the principal strains. Since we cannot measure ϒ xy directly what we do is that we place three strain gauges in three different orientations. Let us call this directions as a, b and c oriented at three different angles; θa, θb and θc respectively. And thereby employing the transformation equation which is:∈a = (∈x + ∈y)/2 + ((∈x’ + ∈y') cos2θ)/2 + (γxy Sin 2θa)/2

can be employed to evaluate strain along a, b and c which are ∈a, ∈b and ∈c. Keeping in mind that ∈a is acting along 'a' which is at an orientation of an angle θa with reference to x axis. Hence in the place of θ we will substitute θa , for ∈b the orientation is along θb and for ∈c the orientation is along θc.


So, if we substitute that we can get the equations in form shown above where we have three equations ∈a, ∈b and ∈c and they are oriented at an angle of θa, θb and θc. In these three equations, we know ∈a, ∈b, and ∈c since we are measuring strain along a, b and c. Also, we know θa, θb, θc as these are predetermined orientations along which gagues will be placed.


So the only unknown quantities are ∈x, ∈y, ϒxy which we can evaluate from above three equations. Once we know ∈ x, ∈ y, ϒ xy now we can compute the values of ∈ 1 and ∈2 as a function of ∈ x, ∈ y, ϒ xy.

So, we can evaluate the value of ∈1 and ∈2 from ∈x, ∈ y, ϒ xy. And as we know the direction of principal stresses and direction of principal strain coincides. So we can compute the values of principal strain from the principal stresses. So these above equations are the values of σ1 and σ2 in terma of ∈1 and ∈2.


How many elastic constants are necessary for analyzing deformation and how many are independent?

Now we have seen that when we are trying to compute the stresses and the strains, we require three constants which are E, G and μ which are the modulus of elasticity, shear modulus and the Poisson’s ratio respectively. When we were relating shear modulus with elastic modulus, we saw that out of these three elastic constants necessary for analyzing deformation in a stress body, two are independent and one is dependent on the other two. G can be evaluated once we know the value of E and μ.

G = E/2(1+μ) (Refer Slide Time: 18:06)

The third question is can you infer on the maximum value of Poisson’s ratio from the expression of bulk modulus?


Let us look into the bulk modulus. Dilatation is defined as the change in volume per unit volume.

E =∆V/Vwhere ∆V = change in the volume V = original volume

If V = 1, then E = ∆V

Now in the expression for the change in volume or the dilatation we have the Poisson’s ratio μ. There are other terms here which are 'E', elastic modulus and 'p', the hydrostatic pressure acting at that particular point in the small element. Now p is the stress which is acting in all directions in place of σx, σy and σz. These are the values of p which are called as the state of hydrostatic pressure. That means that element is subjected to a compressive stress from all sides and it is under the state of a compressive force.

So if you look into this particular expression wheree = - 3(1 - 2μ)p/EThe minus indicates that it is under compression.Now if , μ ≥ 0.5 ,then (1 - 2μ) becomes negative which means, 'e' becomes positive which indicates that ∆V is increasing i.e., the volume is increasing.

In the physical state, a particular element is subjected to hydrostatic pressure from all directions. Thereby σ x = σ y= σ z= - p and now this is trying to compress the body.


So μ ≥0.5 shows that the body is expanding which is contradictory to the physical state.So the maximum value of μ = 0.5


Pressure vessels are the containers in which the gas or liquid is put under pressurized condition and thereby it exerts pressure in the internal surface of the container and if that internal pressure is higher than the external pressure the body will be experiencing stresses and basically there will be tensile stress in the body. Our objective here is to evaluate those stresses in such pressurized containers. We use compressed air in most applications and another example of this type of pressure vessel is the water flowing through a thin-walled pipe.


What is stated here is that the pressure vessels are closed structures containing liquid or gas under pressure. And we have qualified the terms pressure vessels with thin-walled because the

thickness of the wall or the thickness of the container wall is sufficiently small. It is so small that the variation of the stress across the thickness is insignificant and thereby we consider that there is a state of stress which is on the surface in two directions and the third direction stress is 0.

For those vessels for which the thickness is so small that in comparison to the other dimension that the ratio of r/t ≥ 10 where 'r' is the radius of the cylindrical tank and 't' is the thickness.So if r/t ≥ 10 , then we call those kinds of vessels as thin-walled vessel.The term ten or number 10 here actually is considered because it has been observed if r/t ratio exceeds 10, then the error in the stress level is the minimum and thus it is considered the limiting value for thin-walled vessels.

Also, the internal pressure must exceed the external pressure otherwise there will be other kinds of problems.


Basically in pressure vessels, we try to go for a simplified analysis which is based on the effect that the internal pressure exerts pressure or the forces on the body and thereby the body is subjected to stress. That is why we try to analyze how much stress the body undergoes when the container is subjected to pressure but we neglect other effects such as effects of the external loads, the weights of content and the weight of the container.

Also, when we try to compute the elastic deformation of such a container this needs the application of generalized Hooke’s law. We have seen this where we compute the strain in terms of stresses in a generalized Hooke’s law form where

∈x = σx/E - μ.σy/E - μ.σz/EThese are the generalized Hooke’s law application and we take the applications of these for analyzing strains and thereby the deformation in the pressure vessels.


We can have different forms of pressure vessels but in this particular lesson we are concentrating on pressure vessels which are of cylindrical form. If we have a cylindrical pressure vessel which is exerted by pressure inside which is more than the external pressure then what will be the state of stress that will be acting on such containers?

Let us look into one such configuration that this is the cylinder, if we cut this cylinder by two planes AA and BB and also we are cutting this cylinder along this center half then we get a configuration as shown in the figure below.Here, t is the thickness and r is the radius of this cylindrical and r/t ≥ 10 and we call this kind of cylindrical form as thin-walled cylinder.

Now, here internal pressure will be exerted by the content on the walls. Here,t = thickness of the cylinderri = internal radiusro = external radiusσ1 = stresses acting in the circumferential directionσ2 = stresses acting in the longitudinal directionP = pressure exerted by the content on cylinder wallsP = σ x areaIf we take the distance between the two cuts AA and BB as 'L' which is this length of the segment and if we are computing stress for this particular length, then

P = σ1 . t . L

We can compute the values of σ1 in two ways.One is, if we take this thin wall part and we say that this is subjected to the pressure, p from inside. Now if we consider a small element which is at an angle of θ making an angle dθ, then the force which will be exerted here is, = p.r.dθ.L which is acting over the length L.If we take the component of this force in the horizontal direction which is, p.r.L.dθ cosθ acting over the whole of this segment.So, 2P = 2 ∫ p.r.L. cosθ dθ = p.r.L

Circumferemtial stress is acting along the circumference of the cylinder or sometimes we call this as Hoops stress.

0

π/2

σ1.t.L = p.r.L

σ1 = pr/t -- circumferential stress

P = σ1 . t . L

We can compute this directly as well. Now here again,

Also, P = projected length over which this force is acting x internal pressureP = 2r.L.p

So,

2σ1.t.L = 2 p.r.L

σ1 = pr/t -- circumferential stress/ hoop stress

Though initially I stated two values of r where one is ri and ro, but since the thickness of the vessel is very small we consider that there is little variation between the internal radius and the external radius and we consider the radius(r) as internal radius and compute the stresses .


Now for σ2 acting in the longitudinal direction, let us take the section here shown in the figure below. Here you see that we have internal pressure is acting on the cylinderical surface and the thin wall which is subjected to the longitunal forces as represented resisting this internal one.

P' = p πri2

The pressure which is exerted by the content inside the pressure vessel is,

So, from the whole of the cylindrical surface with external radius ro if we subtract the internal part with the ri, then the force given by the walls is,

P' = σ2 (πro2 - πri

2)

This force is going to resist this internal pressure is exerted by the content. So.

p πri2 = σ2 (πro

2 - πri2)

p πri2 = σ2 π(ro - ri)(ro + ri)

p πri2 = σ2 π t (ro + ri)

Also approximately, ro = ri = r, so, p πr2 = 2 σ2 π t r

σ2 = pr/2 t -- Longitudnal Stress


Now we have obtained the longitudinal stress is in fact that half of this circumferential stress. So if we know the internal pressure, if we know the radius of the cylindrical vessel, if we know the thickness of the wall, we can compute the stresses; the circumferential stress and the longitudinal stress. The circumferential stress or the Hoop stress and the longitudinal stress are basically tensile in nature.


Having looked into the concept of the stresses in the cylindrical vessels, let us look into some of the problems. Here are some problems with reference to the stress system or with reference to the Mohr circle of strain and the transformation equation. Let us also look into the examples of the pressure vessels.


Now let’s us look into how you compute the values of the principal strains such as the maximum shear strains and the strains which are at orientation which is oriented at an angle of 75 degrees with reference to the x-axis.


Here is the first example. What is stated here is that an element of material in a plane strain

condition is subjected to

∈x = 480 x 10-6

∈y = 70 x 10-6

γxy = 420 x 10-6

Now lets us look into how you compute the values of the principal strains such as the

maximum shear strains and the strains which are at orientation which is oriented at an angle of

75° with reference to the x axis.

(Refer Slide Time: 41:00 - 51:57)

Now, let me draw the Mohr circle corresponding to the state of stress which is given

∈x = 480 x 10-6

∈y = 70 x 10-6

γxy = 420 x 10-6

If we try to plot the Mohrs circle based on these values of strains then we have, let us say the x-

axis is ϵ axis and y-axis is +ve γxy/2. Now we can plot the points, (∈x, γxy/2) and (∈y, γxy/2).

If we join these two points together, the line (AA’) crosses the ∈ axis at point O, which we call

as the centre of Mohr’s circle and considering OA’ or OA as the radius, we can draw the Mohr

circle as shown in the figure above.

Here point X indicates the value of maximum strain which is the principal strain, ∈1. Point Y is

the minimum normal strain which we call as ∈2, and as you have seen the distance of the

center(O) from the origin(O’), OO’ is (∈x + ∈y)/2 and the distance B and B’ is (∈x - ∈y)/2.

OO’ = (∈x + ∈y)/2 = 480 + 70 = 550/2 = 275 x 10-6

OB = OB’ = (∈x - ∈y)/2 = 410/2 = 205 x 10-6.

AB = 210 x 10-6.

Thereby from this particular triangular configuration, OAB,

AB = 210 x 10-6 and OB = 205 x 10-6 Therefore, OA = Radius (R) = 210

2052 2 x 10-6 = 293.5 x 10-6.

Hence, the principal strains, ∈1 and ∈2,

Maximum principal strain, ∈1 = OO’ + R = (∈x + ∈y)/2 + R

∈1 = (275 x 10-6) + (293.5 x 10-6) = 568.5 x 10-6

Minimum principal strain, ∈2 = O’O – R

∈2 = 275 x 10-6 – 293.5 x 10-6 = – 18.5 x 10–6 (negative sign is there because this value

is to the left of origin in the mohr circle.)

Maximum shearing strain, γmax

Maximum value of the R = γmax/2

293.5 = γmax/2

So, γmax = 2R = 587 x 10-6

Now let us look into the angle. Since X is the point of the maximum principal strain the angle

from the reference point is this which is 2Θp

Now, tan2Θp = 210/205

2Θp = 45.70

Θp = 22.850.

This is the orientation along which the principal strain acts. Now we will have to

compute the value of strain with reference to axis.

We have the reference axis x and y. So, now we have to find out the state of strain at an

orientation of the axis which is 750 in an anticlockwise direction with reference to x. We will

have to find out the strain along X1 and Y1, now when we take 750 in the reference plane, it

becomes 2Θ in the Mohr plane = 1500.

From this particular reference point, OA if we go to 1500 anticlockwise, the point will give us

the state of stress which is oriented at an angle of 750 with reference to X axis. From OA to

reference point = 1500

2Θp = 45.70

Reference Point to OB = 900

So, if we call the angle as α, then α = 150 – 45.7 – 90 = 14.30

So, α = 14.30

Now from here if we drop a perpendicular from the reference point, then the perpendicular

distance from reference point to the Y-axis will give us ∈75 and perpendicular distance from P to

X-axis is going to give us the value of γ/2.

If we take a point diametrically opposite to the reference point, we will get ∈y.

If we know that the angle α = 14.30, and this is R then we can compute the horizontal distance ‘a’

and the vertical distance ‘b’ as shown in figure.

Now let us calculate the value of a and b.

So a = R sin(14.3) = 293.5 x 10-6 x sin 14.30 = 72.5 x 10-6

b = 293.5 x 10-6 x cos14.50 = 284.4 x 10-6.

Once we know a and b then the

∈x1 = OO’ – a = 275 – 72.5 x 10-6

∈y1 = OO’ – a = 275 + 72.5 x 10-6

γx1 y1/2 = R’ = b

γx1 y1 = 2b = 2 x 284.4 x 10-6 = 568.8 x 10-6.

So these are the values of strain ɛx1, ɛy1 and

x1y1 with reference to the axis which is oriented at

an angle of 750 with reference to the reference x and y plane.

So those are the values of the strains and the principal strains and the maximum shearing

stress we have seen how to compute it.


Now this is the orientation/ the final form of the reference plane.

When we have the orientation for the principal plane which is Θp = 14.30.

Now ∈2 was -ve and that is why it is compressed in one direction and elongated in the other.

Therefore this is the form of the principal strain.

Now let us look into the second problem as shown in the figure above.


Now if we plot the values of the Mohr’s circle or if we plot the Mohr’s circle based on the

values given here, note that here we have the values of

∈x = 121 x10–6

∈y = - 450 x 10–6

γxy = - 360 x 10-6

Now if we plot the Mohr Circle in a similar manner as the previous problem, we find out that

∈1 = 172 x 10-6

∈2 = – 502 x 10-6

γmax = 674 x10-6


Here is the third problem which is related to the example as discussed today.

Let us look into the values of the stresses as we compute.


Circumferential stress/ hoop stress, σ1 = pr/t

Longitudinal stress, σ2 = pr/2t

P = 1MPa, r = 1500 mm and t = 10 mm.

Hence σ1 = 1 x 1500 / 10 = 150MPa

σ2 = pr/2t = σ1/2 = 750MPa.

The σ2 acts in the longitudinal direction of the vessel which is longitudinal stress and σ1 acts in

the circumferential direction.

Now if we like to find out the increase in the diameter then it means you must know how much

strain it is undergoing in the circumferential direction. So we are interested to evaluate the strain

in the circumferential direction which we call as ∈1


Now ∈1 = σ1/E – µ σ2/E = 0.6375 x 10-3

So this is the value of strain it is undergoing. Now we can write this strain in the

circumferential direction as

∈1 = (2π(r + Δ) – 2πr)/2πr = Δ /r.

Δ = ∈. r

Hence the increase in the diameter, 2Δ = 2 x 0.6375 x 1500 = 1.912 mm.


(Refer Slide Time: from 58:50 - 59:09)

Here is another problem set for you which as shown in the above figure. (Refer

Slide Time: 59:09 - 59:11)

(Refer Slide Time: from 59:12 - 59:14)


Summary:

We have discussed the concept of stresses and strain in thin walled cylindrical pressure vessels.


We have also discussed Hoop stress and longitudinal stresses, evaluation of

deformation using generalized Hooke’s law and examples to evaluate stresses, strains and

deformation in thin walled cylindrical pressure vessels.

(Refer Slide Time: 59: 30 - 59:48)

The question set for you is:

What Hoop stress is and how is it related to longitudinal stress in cylindrical pressure

vessels?

What is the value of in plane shear stress and are the stresses same in spherical vessels as

well as we seen in cylindrical vessels?

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