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Columns and Struts

34.1. Introduction

A structural member, subjected to an axial compressive force, is called a strut. As per defi- nition, a strut may be horizontal, inclined or even vertical. But a vertical strut, used in buildings or frames, is called a column.

34.2. Failure of a Column or Strut

It has been observed, that when a column or a strut is subjected to some compressive force, then the compressive stress induced,

σ = P A

where P = Compressive force and

A = Cross-sectional area of the column.

34 Unit-IV

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A little consideration will show, that if the force or load is gradually increased the column will reach a stage, when it will be subjected to the ultimate crushing stress. Beyond this stage, the column will fail by crushing. The load corresponding to the crushing stress, is called crushing load.

It has also been experienced that sometimes, a compression member does not fail entirely by crushing, but also by bending i.e., buckling. This happens in the case of long columns. It has also been observed that all the short columns fail due to their crushing. But, if a long column is subjected to a compressive load, it is subjected to a compressive stress. If the load is gradually increased, the col- umn will reach a stage, when it will start buckling. The load, at which the column just buckles is called buckling load, criticial load or crippling load and the column is said to have developed an elastic instability. A little consideration will show that for a long column, the value of buckling load will be less than the crushing load. Moreover, the value of buckling load is low for long columns and relatively high for short columns.

34.3. Euler’s Column Theory

The first rational attempt, to study the stability of *long columns, was made by Mr. Euler. He derived an equation, for the buckling load of long columns based on the bending stress. While deriv- ing this equation, the effect of direct stress is neglected. This may be justified with the statement that the direct stress induced in a long column is negligible as compared to the bending stress. It may be noted that the Euler’s formula cannot be used in the case of short columns, because the direct stress is considerable and hence cannot be neglected.

34.4. Assumptions in the Euler’s Column Theory

The following simplifying assumptions are made in the Euler’s column theory:

1. Initially the column is perfectly straight and the load applied is truly axial.

2. The cross-section of the column is uniform throughout its length.

3. The column material is perfectly elastic, homogeneous and isotropic and thus obeys Hooke’s law.

4. The length of column is very large as compared to its cross-sectional dimensions.

5. The shortening of column, due to direct compression (being very small) is neglected.

6. The failure of column occurs due to buckling alone.

34.5. Sign Conventions

Though there are different signs used for the bending of columns in different books, yet we shall follow the following sign conventions which are commonly used and internationally recognised.

Fig. 34.1

* As a matter of fact, mere length is not the only criterion for a column to be called long or short. But it has an important relation with the lateral dimensions of the column.

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1. A moment, which tends to bend the column with convexity towards its initial central line as shown in Fig. 34.1 (a) is taken as positive.

2. A moment, which tends to bend the column with its concavity towards its initial central line as shown in Fig. 34.1 (b) is taken as negative.

34.6. Types of End Conditions of Columns

In actual practice, there are a number of end conditions, for columns. But, we shall study the Euler’s column theory on the following four types of end conditions, which are important from the subject point of view:

1. Both ends hinged, 2. Both ends fixed,

3. One end is fixed and the other hinged, and 4. One end is fixed and the other free.

Now we shall discuss the value of critical load for all the above mentioned type of and conditions of columns one by one.

1 2 3 4

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or tan P l EI

⎛ ⎞

⎜ ⎟

⎝ ⎠ = l EIP

⎝ ⎠

A little consideration will show that the value of l EIP

⎝ ⎠in radians, has to be such that its tangent is equal to itself. We know that the only angle, the value of whose tangent is equal to itself, is about 4.5 radians.

l EIP = 4.5 or 2 P 20.25

l ×EI = or P=20.252 EI l

P

2 2

2 EI l

= π

NOTE: A little consideration will show that 20.25 is not exactly equal to 2π2, but approximately equal to 2π2. This has been done to rationalise the value of P, i.e., crippling load in various cases.

34.11. Euler’s Formula and Equivalent length of a Column

In the previous articles, we have derived the relations for the crippliing load under various end conditions. Sometimes, all these cases are represented by a general equation called Euler’s formula,

PE =

2 2 e

EI L π

where Le is the equivalent or effective length of column.

The is another way of representing the equation, for the crippling load by an equivalent length of effective length of a column. The equivalent length of a given column with given end conditions, is the length of an equivalent column of the same material and cross-section with both ends hinged and having the value of the crippling load equal to that of the given column.

The equivalent lengths (L) for the given end conditions are given below:

Table 34.1

S.No. End conditions Relation between equivalent Crippling load (P) length (Le) and actual length (l)

1. Both ends hinged Le = l P = ( )

2 2

EI l π =

2 2

EI l π

2. One end fixed and the other free Le = 2 l P =

2

(2 )2

EI l π =

2

4 2

EI l π

3. Both ends fixed Le =

2

l P =

2 2

2 EI l p Ê ˆË ¯

=

2 2

4 EI l π

4. One end fixed and the other hinged Le = 2

l P =

2 2

2 EI l p

Ê ˆ

Á ˜

Ë ¯

=

2 2

2 EI l π

NOTE.The vertical column will have two moments of inertia (viz., IXX and LYY). Since the column will tend to buckle in the direction of leas moment of inertia, therefore the least value of the two moments of inertia is to be used in tlhe relation.

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34.12. Slenderness Ratio

We have already discussed in Art. 34.11 that the Euler’s formula for the crippling load, PE =

2 2 e

EI L

π ...(i)

We know that the buckling of a column under the crippling load will take place about the axis of least resistance. Now substituting I = Ak2 (where A is the area and k is the least radius of gyration of the section) in the above equation,

PE = 2

( )

2 2

2 2

e e

E Ak EA

L L

k

p = p

Ê ˆÁ ˜ Ë ¯

...(ii) where Le

k is known as slenderness ratio. Thus slenderness ratio is defined as ratio of equivalent (or unsupported) length of column to the least radius of gyration of the section.

Slenderness ratio does not have any units.

NOTE.It may be noted that the formula for crippling load, in the pervious articles, have been derived on the assumption the the slenderness ratio Le

k is so large, that the failure of the column occurs only due to bending, the effect of direct stress (i.e., P

A) being negligible.

34.13. Limitation of Euler’s Formula

We have discussed in Art. 32.12 that the Euler’s formula for the crippling load, PE =

2 2 e

EA L

k p Ê ˆ Á ˜ Ë ¯

∴ Euler’s crippling stress, σE =

2 2 e

P E

A L

k

= p Ê ˆÁ ˜ Ë ¯ A little consideration will show that the crippling stress will be high, when the slenderness ratio is small.

We know that the crippling stress for a column cannot be more than the crushing stress of the column mate- rial. It is thus obvious that the Euler’s formula will give the value of crippling stress of the column (equal to the crushing stress of the column material) corre- sponding to the slenderness ratio. Now consider a mild steel column. We know that the crushing stress for the mild steel is 320 MPa or 320 N/m2 and Young’s modu- lus for the mild steel is 200 GPa or 200 × 103 N/mm2.

Now equating the crippling stress to the crushing stress, 320 =

2 2 3

2 2

(200 10 )

e e

E

L L

k k

π =π × ×

⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

(6)

2

Le

k Ê ˆÁ ˜ Ë ¯ =

2 3

200 10 320 π × ×

or Le

k = 78.5 say 80

Thus, if the slenderness ratio is less than 80 the Euler’s formula for a mild steel column is not valid.

Sometimes, the columns, whose slenderness ratio is more than 80 are known as long columns and those whose slenderness ratio is less than 80 are known as short columns. It is thus obvious that the Euler’s formula holds good only for long columns.

NOTE.In the Euler’s formula, for crippling load, we have not taken into account the direct stresses induced in the material due to the load, (which increases gradually from zero to its crippling value). As a matter of fact, the combined stress, due to direct load and slight bending reaches its allowable value at a load, lower than that required for buckling ; and therefore this will be the limiting value of the safe load.

EXAMPLE 34.1. A steel rod 5 m long and of 40 mm diameter is used as a column, with on end fixed and the other free. Determine the crippling load by Euler’s formula. Take E as 200 GPa.

SOLUTION. Given : Length (l) = 5 m = 5 × 103 mm ; Diameter of column (d) = 40 mm and modulus of elasticity (E) = 200 GPa = 200 × 103 N/mm2.

We know that moment of iertia of the column section,

I = ( )4 (40)4

64π × d =64π × = 40 000 π mm4

Since the column is fixed at one end and free at the other, therefore equivalent length of the column,

Le = 2 l = 2 × (5 × 103) = 10 × 103 mm

∴ Euler’s crippling load, PE =

2 2 3

2 3 2

(200 10 ) (40000 ) (10 10 )

e

EI L

π =π × × × π

× = 2480 N

= 2.48 kN Ans.

EXAMPLE 34.2. A hollow alloy tube 4 m long with external and internal diameters of 40 mm and 25 mm respectively was found to extend 4.8 mm under a tensile load of 60 kN. Find the buckling load for the tube with both ends pinned. Also find the safe load on the tube, taking a factor of safety as 5.

SOLUTION. Given : Length l = 4 m ; External diameter of column (D) = 40 mm ; Internal diameter of column (d) = 25 mm ; Deflection (δl) = 4.8 mm ; Tensile load = 60 kN = 60 × 103 N and factor of safety = 5.

Buckling load for the tube We know that area of the tube,

A = [ 2 2] [(40)2 (25) ]2

4 D d 4

π× − =π − = 765.8 mm2

and moment of inertia of the tube,

I = [ 4 4] [(40)4 (25) ]4

64π D =d =64π − = 106 500 mm4 We also know that strain in the alloy tube,

e = 4.83 0.0012 4 10

l

δ =l =

×

(7)

and modulus of elasticity for the alloy, E =

60 103

Load

Area × Strain 765.8 0.0012

= ×× = 65 290 N/mm2

Since the column is pinned at its both ends, therefore equivalent length of the column, Le = l = 4 × 103 mm

∴ Euler’s buckling load,PE =

2 2

2 3 2

65290 106500 (4 10 )

e

EI L

π =π × ×

× = 4290 N

= 4.29 kN Ans.

Safe load for the tube

We also know that safe load for the tube

= Buckling load 4.29

Factor of safety= 5 = 0.858 kN Ans.

EXAMPLE 34.3. Compare the ratio of the strength of a solid steel column to that of a hollow of the same cross-sectional area. The internal diameter of the hollow column is 3/4 of the exter- nal diameter. Both the columns have the same length and are pinned at both ends.

SOLUTION. Give : Area of solid steel colum AS = AH (where AH = Area of hollow column) ; Internal diameter of hollow column (d) = 3 D/4 (where D = External diameter) and length of solid column (lS) = lH (where lH = Length of hollow column).

Let D1 = Diameter of the solid column,

kH = Radius of gyration for hollow column and kS = Radius of gyration for solid column.

Since both the columns are pinned at their both ends, therefore equivalent length of the solid column,

LS = lS = LH = lH = L We know that Euler’s crippling load for the solid column,

PS =

2 2

2

2 2

· S· S

H

E A k EI

L L

π = π ...(i)

Similarly Euler’s crippling load for the hollow column PH =

2 2

2

2 2

· H· H

H

E A k EI

L L

π =π ...(ii)

Dividing equation (ii) by (i),

H S

P

P =

2 2 2

2 2 2 2

2 2 2

1 1 1

3 16 4

16

H S

D d D D

k D d

k D D D

⎛ ⎞

+ =⎜ ⎟

⎛ ⎞ = = + = ⎝ ⎠

⎜ ⎟

⎝ ⎠

=

2 2 1

25 16

D

D ...(iii)

Since the cross-sectional areas of the both the columns is equal, therefore

2

4 D1

π × =

2 2

2 2 2 3 7

( )

4 4 4 4 16

D D

D dD

π − =π⎢⎢⎣ −⎛⎜⎝ ⎞⎟⎠ ⎥⎥⎦= ×π

(8)

D12 = 7 2

16 D

Now substituting the value of 2

D1 in equation (iii),

H S

P

P =

2 2

25 25

7 7 16 16

D D =

×

Ans.

EXAMPLE 34.4. An I section joist 400 mm × 200 mm × 20 mm and 6 m long is used as a strut with both ends fixed. What is Euler’s cripp;ing load for the column? Take Young’s modulus for the joist as 200 GPa.

SOLUTION. Given : Outer depth (D) = 400 mm ; Outer width (B) = 200 mm ; Length (l) = 6 m = 6 × 103 mm and modulus of elasticity (E) = 200 GPa = 200 × 103 N/mm2.

From the geometry of the figure, we find that inner depth, d = 400 – (2 × 20) = 360 mm and inner width, b = 200 – 20 = 180 mm

We know that moment of inertia of the joist section about X-X axis, IXX = 1 [ 2 3]

12 BDba

= 1 [200 (400)3 180 (360) ]mm3 4

12 × − ×

= 366.8 × 106 mm4 ...(i) Similarly, IYY =

3 3

2 (200) 360 (20) 4

2 mm

12 12

⎡ × × ⎤+ ×

⎢ ⎥

⎢ ⎥

⎣ ⎦

= 2.91 × 106 mm4 ...(ii)

Since IYY is less than IXX, therefore the joist will tend to buckle in Y-Y direction. Thus, we shall take the value of I as IYY = 2.91 × 106 mm4. Moreover, as the column is fixed at its both ends, therefore equivalent length of the column,

Le =

3 3

(6 10 )

3 10 mm

2 2

l = × = ×

∴ Euler’s crippling load for the column, PE =

2 2 3 6

3

2 3 2

(200 10 ) (2.91 10 )

638.2 10 N (3 10 )

e

EI L

π =π × × × × = ×

×

= 638.2 kN Ans.

EXAMPLE 34.5. A T-section 150 mm × 120 mm × 20 mm is used as a strut of 4 m long with hinged at its both ends. Calculate the crippling load, if Young’s modulus for the material be 200 GPa.

SOLUTION. Given : Size of T-section = 150 mm × 120 mm × 20 mm ; Length (l) = 4 m = 4 × 103 mm and Young’s modulus (E) = 200 GPa = 200 × 103 N/mm2.

First of all, let us find the centre of the T-section; Let bottom of the web be the axis of reference.

Fig. 34.6

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34.15. Rankine’s Formulae for Columns

We have already discussed that the Euler’s formula gives correct results only for very long columns. Though this formula is applicable for columns, ranging from very long to short ones, yet

it does not give reliable results. Prof. Rankine, after a number of experiments, gave the following empirical formula for columns.

1

PR = 1 1

CS E

P +P ...(i)

where PR = Crippling load by Rankine’s Formula

PCS = σCS · A = Ultimate crushing load for the column and PE =

2 2 e

EI L

π = Crippling load obtained by Euler’s formula.

A little consideration will show that the value of PCS will remain constant irrespective of the fact whether the column is a long one or short one. Now, we shall study the effect of PE in short as well as long columns one by one.

1. Short columns. In case of short columns, the value of PE will be very high, therefore the value of 1

PE will be quite negligible as compared to 1

PCS. It is thus obvious that the Rankine’s formula will give the value of its crippling load (i.e., P) approximately equal to the ultimate crushing load (i.e., ).

2. Long columns. In case of long columns, the value of PE will be very small, therefore the value of 1

PE will be quite considerable as compared to 1

PCS. It is thus obvious that the Rankine’s formula will give the value of its crippling load (i.e., P) approximately equal to the crippling load by Euler’s formula (i.e., PE). Thus, we see that the Rankine’s formula gives a fairly correct result for all cases of columns, ranging from short to long columns.

From equation (i), we know that 1

PR = 1 1

·

E CS

CS E CS E

P P

P P P P

+ = +

PR = ·

1

CS E CS

CS E CS

E

P P P

P

P P

P + = +

(10)

Now substituting the values of PCS and PE is the above equation

PR = 2 2

2 2 2

· ·

1 · 1

CS CS

e CS e

CS

A A

L AL

A

E E Ak

σ = σ

+ σ × + σ ×

π π

...(ä I = Ak2)

or PR = · 2 2

1 1

CS CS

e e

A P

L L

a a

k k

σ =

⎛ ⎞ ⎛ ⎞

+ ⎜ ⎟⎝ ⎠ + ⎜ ⎟⎝ ⎠

where PCS = Crushing load of the column material σCS = Crushing stress of the column material,

A = Cross-sectional area of the column, a = Rankine’s constant equal to 2C

E

⎛ σ ⎞

⎜ ⎟

⎜ π ⎟

⎝ ⎠

Le = Equivalent length of the column, and k = Least radius of gyration.

The folowing table gives the values of crushing stress (σC) and Rankine’s constant (a) for vari- ous materiabls:

Table 34.2

S.No. Material σC in MPa a = 2C

E σ π

1. Mild Steel 320 1

7500

2. Cast Iron 550 1

1600

3. Wrought Iron 250 1

9000

4. Timber 40 1

750

Note : The above values are only for a column with both ends hinged. For other end conditions, the equivalent length should be used.

EXAMPLE 34.6. Find the Euler’s crippling load for a hollow cylindrical steel column of 38 mm external diameter and 2.5 mm thick. Take length ofthe column as 2.3 m and hinged at its both ends. Take E = 205 GPa.

Also determine crippling load by Rankine’s formula using constants as 335 MPa and 1 7500 SOLUTION. Give : External diameter (D) = 38 mm ; Thickness = 2.5 mm or inner diameter (d) = 38 – (2 × 2.5) = 33 mm ; Length of the column (l) = 2.3 m = 2.3 × 103 mm ; Yield stress (σC) = 335 MPa = 335 N/mm2 and Rankine’s constant (a) = 1

7500. Euler’s crippling load

We know that moment of inertia of the column section,

I = ( 4 4) [(38)4 (33) ] 14.05 104 3 mm4

64π Dd =64π − = × π

Since the column is hinged at its both ends, therefore effective length of the column, Le = l = 2.3 × 103 mm

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∴ Euler’s crippling load,PE =

2 2 3 3

2 3 2

(205 10 ) (14.05 10 ) (2.3 10 )

e

EI L

π =π × × × π

× = 16 880 N

= 16.88 kN Ans.

Rankine’s crippling load

We know that area of the column section,

A = ( 2 2) [(38)2 (33) ]2 88.75 mm2

4 D d 4

π − = π − = π

and least radius of gyration, k =

14.05 103

88.75 I

A

× π

= π = 12.6 mm

∴ Rankine’s crippling load,

PR = 2 2

3

· 335 88.75

2.3 10

1 1 1

7500 12.6

CS e

A a L

k

s = ¥ p

Ê ˆ Ê ¥ ˆ

+ Á ˜Ë ¯ + ÁË ˜¯

= 17 160 N

= 17.16 kN Ans.

EXAMPLE 34.7. Figure 34.8 shows a built-up column consisting of 150 mm × 100 mm R.S.J.

with 120 mm × 12 mm plate riveted to each flange.

Fig. 34.8

Calculate the safe load, the column can carry, if it is 4 m long having one end fixed and the other hinged with a factor of safety 3.5. Take the properties of the joist as Area = 2167 mm2, IXX

= 8.391 × 106 mm4, IYY = 0.948 × 106 mm4. Assume the yield stress as 315 MPa and Rankine’s constant (a) = 1

7500.

SOLUTION. Given : Length of the column (l) = 4 m = 4 × 103 mm ; Factor of safety = 3.5 ; Yield stress (σC) = 315 MPa = 315 N/mm2 ; Area of joist = 2167 mm2 ; Moment of inertia, about X-X axis (IXX) = 8.391 × 106 mm4 ; Moment of inertia about Y-Y axis (IYY) = 0.948 × 106 mm4 and Rankine’s constant (a) = 1

7500.

From the geometry of the figure, we find that area of the column section, A = 2167 + (2 × 120 × 12) = 5047 mm2 and moment of inertia of the column section bout X-X axis,

IXX = (83.91 × 106) +

3 2

120 (12)

2 120 12 (81)

12

⎡ × + × × ⎤

⎢ ⎥

⎢ ⎥

⎣ ⎦mm4

= (8.391) × 106) + (18.93 × 106) = 27.32 × 106 mm4

(12)

Similarly, IYY = (0.948 × 106) +

12 (120)3

2 12

⎡ × ⎤

⎢ ⎥

⎢ ⎥

⎣ ⎦ mm4

= (0.948 × 106) + (3.456 × 106) = 4.404 × 106 mm4

Since IYY is less than IXX, therefore the column will tend to buckle in Y-Y direction. Thus we shall take I equal to IYY = 4.404 × 106 mm4 (i.e., least of two). Moreover as the column is fixed at one end and hinged at the other, therefore equivalent length of the column.

Le =

4 103

2 2

l = × = 2.83 × 103 mm

We know that least radius of gyration, k =

4.404 106

5047 I

A

= × = 29.5 mm

∴ Rankine’s crippling load on the column

PR = 2 2

3

· 315 5047

2.83 10

1 1 1

7500 29.5

C e

A a L

k

s = ¥

Ê ˆ Ê ¥ ˆ

+ Á ˜Ë ¯ + ÁË ˜¯

= 714 × 103 N = 714 kN and safe load on the column

= Crippling load 714

Factor of safety= 3.5 = 204 kN Ans.

EXAMPLE 34.8. A column is made up of two channels. ISJC 200 and two 250 mm × 10 mm flange plates as shown in Fig. 34.9.

Determine by Rankine’s formula the safe load, the column of 6 m length, with both ends fixed, can carry with a factor of safety 4. The properties of one channel are Area = 1777 mm2, IXX = 11.612 × 106 mm4 and IYY = 0.842 × 106 mm4. Distance of centroid from back to web = 19.7 mm.

Take σC = 320 MPa and Rankine’s constant = 1 7500

Fig. 34.9

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Thin Cylindrical and Spherical Shells

31.1. Introduction

In engineering field, we daily come across vessels of cylindrical and spherical shapes containing fluids such as tanks, boilers, compressed air receivers etc. Generally, the walls of such vessels are very thin as compared to their diameters. These vessels, when empty, are subjected to atmospheric pressure internally as well as externally. In such a case, the resultant pressure on the walls of the shell is zero. But whenever a vessel is subjected to internal pressure (due to steam, compressed air etc.) its walls are subjected to tensile stresses.

In general, if the thickness of the wall of a shell is less than 1/10th to 1/15th of its diameter, it is known as a thin shell.

31

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31.2. Failure of a Thin Cylindrical Shell due to an Internal Pressure

We have already discussed in the last article that whenever a cylindrical shell is subjected to an internal pressure, its walls are subjected to tensile stresses.

It will be interesting to know that if these stresses exceed the permissible limit, the cylinder is likely to fail in any one of the following two ways as shown in Fig. 31.1 (a) and (b).

(a) Split into two troughs. (b) Split into two cylinders.

Fig. 31.1 1. It may split up into two troughs and

2. It may split up into two cylinders.

31.3. Stresses in a Thin Cylindrical Shell

We have already discussed that whenever a cylindrical shell is subjected to an internal pressure, its walls are subjected to tensile stresses. A little consideration will show that the walls of the cylindri- cal shell will be subjected to the following two types of tensile stresses:

1. Circumferential stress and 2. Longitudinal stress.

In case of thin shells, the stresses are assumed to be uniformly distributed throughout the wall thickness. However, in case of thick shells, the stresses are no longer uniformly distributed and the problem becomes complex. In this chapter, we shall discuss the stress in thin shells only.

NOTE: The above theory also holds good, when the shell is subjected to compressive stress.

31.4. Circumferential Stress

Fig. 31.2

Consider a thin cylindrical shell subjected to an internal pressure as shown in Fig. 31.2(a) and (b). We know that as a result of the internal pressure, the cylinder has a tendency to split up into two troughs as shown in the figure.

Let l = Length of the shell,

d = Diameter of the shell,

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t = Thickness of the shell and p = Intensity of internal pressure.

Total pressure along the diameter (say X-X axis) of the shell,

P = Intensity of internal pressure × Area = p × d × l and circumferential stress in the shell,

σc = Total pressure Resisting section =

2 2

pdl pd

tl = t ...(ä of two sections) This is a tensile stress across the X-X. It is also known as hoop stress.

NOTE. If η is the efficiency of the riveted joints of the shell, then stress, σc =

2 pd

tη

31.5. Longitudinal Stress

Consider the same cylindrical shell, subjected to the same internal pressure as shown in Fig. 31.3 (a) and (b). We know that as a result of the internal pressure, the cylinder also has a tendency to split into two pieces as shown in the figure.

Let p = Intensity of internal pressure,

l = Length of the shell, d = Diameter of the shell and

t = Thickness of the shell.

Fig. 31.3. Longitudinal stress.

Total pressure along its length (say Y-Y axis) of the shell

P = Intensity of internal pressure × Area

= ( )2

p×π4 d and longitudinal stress in the shell,

σl =

( )2

Total pressure 4

Resisting section 4

p d pd

dt t

×π

= =

π

This is also a tensile stress across the section Y-Y. It may be noted that the longitudinal stress is half of the circumferential or hoop stress.

NOTE. If η is the efficiency of the riveted joints of the shell, then the stress, σl =

4 pd

(16)

EXAMPLE 31.1. A stream boiler of 800 mm diameter is made up of 10 mm thick plates. If the boiler is subjected to an internal pressure of 2.5 MPa, find the circumferential and longitudinal stresses induced in the boiler plates.

SOLUTION. Given : Diameter of boiler (d) = 800 mm ; Thickness of plates (t) = 10 mm and internal pressure (p) = 2.5 MPa = 2.5 N/mm2.

Circumferential stress induced in the boiler plates

We know that circumferential stress induced in the boiler plates, σc = 2.5 800

2 2 10

pd t

= ×× = 100 N/mm2 = 100 MPa Ans.

Longitudinal stress induced in the boiler plates

We also know that longitudinal stress induced in the boiler plates, σl = 2.5 800

4 4 10

pd t

= ×

× = 50 N/mm2 = 50 MPa Ans.

EXAMPLE 31.2. A cylindrical shell of 1.3 m diameter is made up of 18 mm thick plates. Find the circumferential and longitudinal stress in the plates, if the boiler is subjected to an internal pressure of 2.4 MPa. Take efficiency of the joints as 70%.

SOLUTION. Given: Diameter of shell (d) = 1.3 m = 1.3 × 103 mm ; Thickness of plates (t) = 18 mm; Internal pressure (p) = 2.4 MPa = 2.4 N/mm2 and efficiency (η) = 70% = 0.7.

Circumferential stress

We know that circumferential stress, σc =

2.4 (1.3 10 )3

2 2 18 0.7

pd t

× ×

η= × × = 124 N/mm2 = 124 MPa Ans.

Longitudinal stress

We also know that longitudinal stress, σl =

2.4 (1.3 10 )3

4 4 18 0.7

pd t

× ×

η= × × = 62 N/mm2 = 62 MPa Ans.

EXAMPLE 31.3. A gas cylinder of internal diameter 40 mm is 5 mm thick. If the tensile stress in the material is not to exceed 30 MPa, find the maximum pressure which can be allowed in the cylinder.

SOLUTION. Given: Diameter of cylinder (d) = 40 mm ; Thickness of plates (t) = 5 mm and tensile stress (σc) = 30 MPa = 30 N/mm2.

Let p = Maximum pressure which can be allowed in the cylinder.

We know that circumferential stress (σc),

30 = 40

2 2 5

p pd

t

= ×× = 4p

p = 30

4 = 7.5 N/mm2 = 7.5 MPa Ans.

NOTE:1. Since the circumferential stress (σc) is double the longitudinal stress (σl), therefore in order to find the maximum pressure the given stress should be taken as circumferential stress.

2. If however, we take the given tensile stress of 30 N/mm2 as the longitudinal stress, then

30 = 40

4 4 5

p pd

t

= ×

× = 2p

(17)

p = 30

4 = 15 N/mm2 = 15 MPa

Now we shall provide a pressure of 7.5 MPa i.e. (Lesser of the two values) obtained by using the tensile stress as circumferential stress and longitudinal stress.

31.6. Design of Thin Cylindrical Shells

Designing of thin cylindrical shell involves calculating the thickness (t) of a cylindrical shell for the given length (l), diameter (d), intensity of maximum internal pressure (p) and circumferential stress (σc). The required thickness of the shell is calculated from the relation.

t = 2 c

pd

σ ... (See Article 31.4)

If the thickness so obtained, is not a round figure, then next higher value is provided.

NOTE: The thickness obtained from the longitudinal stress will be half of the thickness obtained from circum- ferential stress. Thus, it should not be accepted.

EXAMPLE 31.4. A thin cylindrical shell of 400 mm diameter is to be designed for an internal pressure of 2.4 MPa. Find the suitable thickness of the shell, if the allowable circumferential stress is 50 MPa.

SOLUTION. Given: Diameter of shell (d) = 400 mm ; Internal pressure (p) = 2.4 MPa = 2.4 N/mm2 and circumferential stress (σc) = 50 MPa = 50 N/mm2.

We know that thickness of the shell,

t = 2 2.42 50400

c

pdσ = ×× = 9.6 mm say 10 mm Ans.

EXAMPLE 31.5. A cylindrical shell of 500 mm diameter is required to withstand an internal pressure of 4 MPa. Find the minimum thickness of the shell, if maximum tensile strength in the plate material is 400 MPa and efficiency of the joints is 65%. Take factor of safety as 5.

SOLUTION. Given: Diameter of shell (d) = 500 mm ; Internal pressure (p) = 4 MPa = 4 N/mm2; Tensile strength = 400 MPa = 400 N/mm2 ; Efficiency (η) = 65% = 0.65 and factor of safety = 5.

We know that allowable tensile stress (i.e., circumferential stress), σc = Tensile strength 400

Factor of safety = 5 = 80 N/mm2 and minimum thickness of shell,

t = 4 500

2 c 2 80 0.65

pd = ×

σ η × × = 19.2 mm say 20 mm Ans.

31.7. Change in Dimensions of a Thin Cylindrical Shell due to an Internal Pressure

We have already discussed in the chapter on Elastic Constants that lateral strain is always accom- panied by a linear strain. It is thus obvious that in a thin cylindrical shell subjected to an internal pressure, its walls will also be subjected to lateral strain. The effect of the lateral strains is to cause some change in the dimensions (i.e., length and diameter) of the shell. Now consider a thin cylindrical shell subjected to an internal pressure.

Let l = Length of the shell,

d = Diameter of the shell, t = Thickness of the shell and p = Intensity of the internal pressure.

(18)

We know that the circumferential stress, σc =

2 pd

t

and longitudinal stress, σl = 4 pd

t

Now let δd = Change in diameter of the shell, δl = Change in the length of the shell and 1

m = Poisson’s ratio.

Now changes in diameter and length may be found out from the above equations, as usual (i.e., by multiplying the strain and the corresponding linear dimension).

∴ δd = ε1 · d =

1 2 1

1 1

2 2 2 2

pd pd

tE m d tE m

⎛ − ⎞× = ⎛ − ⎞

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

and δl = ε2 · l = 1 1 1 1

2 2 2 2

pd pdl

tE m l tE m

⎛ − ⎞× = ⎛ − ⎞

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

EXAMPLE 31.6. A cylindrical thin drum 800 mm in diameter and 4 m long is made of 10 mm thick plates. If the drum is subjected to an internal pressure of 2.5 MPa, determine its changes in diameter and length. Take E as 200 GPa and Poisson’s ratio as 0.25.

SOLUTION. Given: Diameter of drum (d) = 800 mm ; Length of drum (l) = 4 m = 4 × 103 mm ; Thickness of plates (t) = 10 mm ; Internal pressure (p) = 2.5 MPa = 2.5 N/mm2 ; Modulus of elasticity (E) = 200 GPa = 200 × 103 N/mm2 and poisson’s ratio 1

m

⎛ ⎞⎜ ⎟

⎝ ⎠ = 0.25.

Change in diameter

We know that change in diameter, δd =

2 2

3

2.5 (800)

1 0.25

1 1 mm

2 2 2 10 (200 10 ) 2

pd

tE m

⎛ − ⎞= × ⎛ − ⎞

⎜ ⎟ ⎜ ⎟

⎝ ⎠ × × × ⎝ ⎠

= 0.35 mm Ans.

Change in length

We also know that change in length, δl =

3 3

2.5 800 (4 10 )

1 1 1

0.25 mm

2 2 2 10 (200 10 ) 2

pdl

tE m

× × ×

⎛ − ⎞= ⎛ − ⎞

⎜ ⎟ ⎜ ⎟

⎝ ⎠ × × × ⎝ ⎠

= 0.5 mm Ans.

31.8. Change in Volume of a Thin Cylindrical Shell due to an Internal Pressure

We have already discussed in the last article, that there is always an increase in the length and diameter of a thin cylindrical shell due to an internal pressure. A little consideration will show that increase in the length and diameter of the shell will also increase its volume. Now consider a thin cylindrical shell subjected to an internal pressure.

Let l = Original length,

d = Original diameter,

δl = Change in length due to pressure and δd = Change in diameter due to pressure.

(19)

We know that original volume,

V = 2 ( )2 ( ) 2

4 d l 4 d d l l 4 d l

π× × =⎡⎢⎣π + δ × × δ ⎤⎥⎦− ×π ×

= ( 2 2 )

4 d l dl d

π ⋅ δ + ⋅ δ ...(Neglecting small quantities)

V

V δ =

2

2

( 2 ) 2

4 4

d l dl d l d

l d

d l

π ⋅ δ + ⋅ δ =δ + δ

π × × = εl + 2εc

or δV = V (εl + 2εc)

where εc = Circumferential strain and

εl = Longitudinal strain.

EXAMPLE 31.7. A cylindrical vessel 2 m long and 500 mm in diameter with 10 mm thick plates is subjected to an internal pressure of 3 MPa. Calculate the change in volume of the vessel.

Take E = 200 GPa and Poisson’s ratio = 0.3 for the vessel material.

SOLUTION. Given: Length of vessel (l) = 2 m = 2 × 103 mm ; Diameter of vessel (d) = 500 mm

; Thickness of plates (t) = 10 mm ; Internal pressure (p) = 3 MPa = 3 N/mm2 ; Modulus of elasticity (E) = 200 GPa = 200 × 103 N/mm2 and poisson’s ratio ⎛ ⎞⎜ ⎟⎝ ⎠m1 = 0.3.

We know that circumferential strain,

εc = 1 1 3 500 3 1 0.3

2 2 2 10 (200 10 ) 2

pd

tE m

⎛ − ⎞= × ⎛ − ⎞

⎜ ⎟ ⎜ ⎟

⎝ ⎠ × × × ⎝ ⎠ = 0.32 × 10–3 ...(i)

and logitudinal strain, εl = 1 1 3 500 3 1 0.3

2 2 2 10 (200 10 ) 2

pd

tE m

⎛ − ⎞= × ⎛ − ⎞

⎜ ⎟ ⎜ ⎟

⎝ ⎠ × × × ⎝ ⎠ = 0.075 × 10–3 ...(ii)

We also know that original volume of the vessel,

V = ( )2 (500)2 (2 10 )3

4 d l 4

π × =π × × = 392.7 × 106 mm3

∴ Change in volume,

δV = V (εc + 2εl) = 392.7 × 106 [0.32 × 10–3 + (2 × 0.075 × 10–3)] mm3

= 185 ××××× 103 mm3 Ans.

EXERCISE 31.1

1. A cylindrical shell 2 m long and 1 m internal diameter is made up of 20 mm thick plates. Find the circumferential and longitudinal stresses in the shell material, if it is subjected to an internal

pressure of 5 MPa. (Ans. 125 MPa ; 62.5 MPa)

2. A steam boiler of 1.25 m in diameter is subjected to an internal pressure of 1.6 MPa. If the steam boiler is made up of 20 mm thick plates, calculate the circumferential and longitudinal stresses. Take efficiency of the circumferential and longitudinal joints as 75% and 60%

respectively.

(Ans. 67 MPa ; 42 MPa) 3. A pipe of 100 mm diameter is carrying a fluid under a pressure of 4 MPa. What should be the minimum thickness of the pipe, if maximum circumferential stress in the pipe material is 12.5

MPa. (Ans. 16 mm)

(20)

4. A cylindrical shell 3 m long has 1 m internal diameter and 15 mm metal thickness. Calculate the circumferential and longitudinal stresses, if the shell is subjected to an internal pressure of 1.5 MPa. Also calculate the changes in dimensions of the shell. Take E = 200 GPa and Poisson’s ratio = 0.3. (Ans. 50 MPa ; 25 MPa ; δd = 0.21 mm ; δl = 0.15 mm) 5. A cylindrical vessel 1.8 m long 800 mm in diameter is made up of 8 mm thick plates. Find the hoop and longitudinal stresses in the vessel, when it contains fluid under a pressure of 2.5 MPa.

Also find the changes in length, diameter and volume of the vessel. Take E = 200 GPa and 1/m = 0.3. (Ans. 125 MPa ; 62.5 MPa ; 0.42 mm ; 0.23 mm ; 1074 mm3)

31.9. Thin Spherical Shells

Consider a thin spherical shell subjected to an internal pressure as shown in Fig. 31.4.

Let p = Intensity of internal pressure,

d = Diameter of the shell and t = Thickness of the shell, As a result of this internal pressure, the shell is likely to be torn away along the centre of the sphere. Therefore, total pressure acting along the centre of the sphere,

P = Intensity of internal pressure × Area

= p× ×4π d2 and stress in the shell material,

σ =

2

Total pressure 4

Resisting section 4

p d pd

d t t

× ×π

= =

π × Note. If η is the efficiency of the riveted joints of the spherical shell, then stress,

σ = 4

pd tη

EXAMPLE 31.8. A spherical gas vessel of 1.2 m diameter is subjected to a pressure of 1.8 MPa. Determine the stress induced in the vessel plate, if its thickness is 5 mm.

SOLUTION. Given: Diameter of vessel (d) = 1.2 m = 1.2 × 103 mm ; Internal pressure (p) = 1.8 MPa = 1.8 N/mm2 and thickness of plates (t) = 5 mm.

We know that stress in the vessel plates, σ =

1.8 (1.2 10 )3

4 4 5

pd t

× ×

= × = 108 N/mm2 = 108 MPa Ans.

EXAMPLE 31.9. A spherical vessel of 2 m diameter is subjected to an internal pressure of 2 MPa. Find the minimum thickness of the plates required, if the maximum stress is not to exceed 100 MPa. Take efficiency of the joint as 80%.

SOLUTION. Given: Diameter of vessel (d) = 2 m = 2 × 103 mm ; Internal pressure (p) = 2 MPa = 2 N/mm2 ; Maximum stress (σ) = 100 MPa = 100 N/mm2 and efficiency of joint (η) = 80% = 0.8.

Let t = Minimum thickness of the plates in mm.

We know that stress in the plates (σ), 100 =

2 (2 10 )3 1250

4 4 0.8

pd

t t t

= × × =

η × ×

t = 1250

100 = 12.5 mm Ans.

Fig. 31.4. Spherical shell

(21)

31.10. Change in Diameter and Volume of a Thin Spherical Shell due to an Internal Pressure

Consider a thin spherical shell subjected to an internal pressure.

Let d = Diameter of the shell,

p = Intensity of internal pressure and t = Thickness of the shell.

We have already discussed in the last article that the stress in a spherical shell, σ =

4 pd and strain in any one direction, t

ε =

E mE

σ − σ ...(äσ1 = σ2 = σ)

= 1 1

4 4 4

pd pd pd

tE tEm tE m

⎛ ⎞

− = ⎜⎝ − ⎟⎠

∴ Change in diameter,

δd = ε · d =

1 2 1

1 1

4 4

pd pd

tE m d tE m

⎛ − ⎞× = ⎛ − ⎞

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

We also know that original volume of the sphere,

V = ( )3

6 d

π × and final volume due to pressure,

V + δV = ( )3

6 d d

π × + δ

where (d + δd) = Final diameter of the shell.

∴ Volumetric strain, V V δ =

3 3

3

( )

( ) 6 6

6

d d d

V V V

V d

π + δ − ×π + δ − =

π ×

=

3 2 3

3

(3 )

d d d d

d + ⋅δ −

...(Ignoring second and higher power of δd)

= 3 d

d

⋅δ = 3ε

and δV = V · 3ε =

3 1 4 1

( ) 3 1 1

6 4 8

pd pd

d tE m tE m

π × × ⎛⎜⎝ − ⎞⎟⎠=π ⎛⎜⎝ − ⎞⎟⎠

EXAMPLE 31.10. A spherical shell of 2 m diameter is made up of 10 mm thick plates.

Calculate the change in diameter and volume of the shell, when it is subjected to an internal pressure of 1.6 MPa. Take E = 200 GPa and 1/m = 0.3.

SOLUTION. Given: Diameter of shell (d) = 2 m = 2 × 103 mm ; Thickness of plates (t) = 10 mm ; Internal pressure (p) = 1.6 MPa = 1.6 N/mm2 ; Modulus of elasticity (E) = 200 GPa = 200 × 103 N/mm2 and Poisson’s ratio (1/m) = 0.3.

Change in diameter

We know that change in diameter, δd =

3 2 2

3

1.6 (2 10 )

1 1 (1 0.3)

4 4 10 (200 10 )

pd

tE m

⎛ − ⎞= × × −

⎜ ⎟

⎝ ⎠ × × ×

= 0.56 mm Ans.

References

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