188 A Textbook of Machine Design -...

188 A Textbook of Machine Design

and the theoretical stress concentration factor,

Kt = 2

1max a

b

σ ⎛ ⎞= +⎜ ⎟⎝ ⎠σWhen a/b is large, the ellipse approaches a crack transverse to the load and the value of Kt

becomes very large. When a/b is small, the ellipse approaches a longitudinal slit [as shown in Fig. 6.6(b)] and the increase in stress is small. When the hole is circular as shown in Fig. 6.6 (c), then a/b = 1and the maximum stress is three times the nominal value.

Fig. 6.6. Stress concentration due to holes.

The stress concentration in the notched tension member, asshown in Fig. 6.7, is influenced by the depth a of the notch and radiusr at the bottom of the notch. The maximum stress, which applies tomembers having notches that are small in comparison with the widthof the plate, may be obtained by the following equation,

σmax =2

1⎛ ⎞σ +⎜ ⎟⎝ ⎠

a

r

6.13 Methods of Reducing Stress ConcentrationWe have already discussed in Art 6.10 that whenever there is a

change in cross-section, such as shoulders, holes, notches or keyways and where there is an interfer-ence fit between a hub or bearing race and a shaft, then stress concentration results. The presence of

stress concentration can not be totally eliminated but it may be reduced to some extent. A device orconcept that is useful in assisting a design engineer to visualize the presence of stress concentration

Fig. 6.7. Stress concentrationdue to notches.

Crankshaft

Variable Stresses in Machine Parts 189and how it may be mitigated is that of stress flow lines, as shown in Fig. 6.8. The mitigation of stressconcentration means that the stress flow lines shall maintain their spacing as far as possible.

Fig. 6.8

In Fig. 6.8 (a) we see that stress lines tend to bunch up and cut very close to the sharp re-entrantcorner. In order to improve the situation, fillets may be provided, as shown in Fig. 6.8 (b) and (c) togive more equally spaced flow lines.

Figs. 6.9 to 6.11 show the several ways of reducing the stress concentration in shafts and othercylindrical members with shoulders, holes and threads respectively. It may be noted that it is notpracticable to use large radius fillets as in case of ball and roller bearing mountings. In such cases,notches may be cut as shown in Fig. 6.8 (d) and Fig. 6.9 (b) and (c).

Fig. 6.9. Methods of reducing stress concentration in cylindrical members with shoulders.

Fig. 6.10. Methods of reducing stress concentration in cylindrical members with holes.

Fig. 6.11. Methods of reducing stress concentration in cylindrical members with holes.

The stress concentration effects of a press fit may be reduced by making more gradual transitionfrom the rigid to the more flexible shaft. The various ways of reducing stress concentration for suchcases are shown in Fig. 6.12 (a), (b) and (c).


6.14 Factors to be Considered while Designing Machine Parts to AvoidFatigue Failure

The following factors should be considered while designing machine parts to avoid fatigue failure:1. The variation in the size of the component should be as gradual as possible.2. The holes, notches and other stress raisers should be avoided.3. The proper stress de-concentrators such as fillets and notches should be provided

wherever necessary.

Fig. 6.12. Methods of reducing stress concentration of a press fit.

4. The parts should be protected from corrosive atmosphere.

5. A smooth finish of outer surface of the component increases the fatigue life.

6. The material with high fatigue strength should be selected.

7. The residual compressive stresses over the parts surface increases its fatigue strength.

6.15 Stress Concentration Factor for Various Machine MembersThe following tables show the theoretical stress concentration factor for various types of

members.

Table 6.1. Theoretical stress concentration factor (Kt ) for a plate with hole(of diameter d ) in tension.

d

b0.05 0.1 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55

Kt 2.83 2.69 2.59 2.50 2.43 2.37 2.32 2.26 2.22 2.17 2.13

Fig. for Table 6.1 Fig. for Table 6.2

Table 6.2. Theoretical stress concentration factor (Kt ) for a shaftwith transverse hole (of diameter d ) in bending.

d

D0.02 0.04 0.08 0.10 0.12 0.16 0.20 0.24 0.28 0.30

Kt 2.70 2.52 2.33 2.26 2.20 2.11 2.03 1.96 1.92 1.90

Variable Stresses in Machine Parts 191

Table 6.3. Theoretical stress concentration factor (Kt ) for steppedshaft with a shoulder fillet (of radius r ) in tension.

Theoretical stress concentration factor (Kt)

D

dr/d

0.08 0.10 0.12 0.16 0.18 0.20 0.22 0.24 0.28 0.30

1.01 1.27 1.24 1.21 1.17 1.16 1.15 1.15 1.14 1.13 1.13

1.02 1.38 1.34 1.30 1.26 1.24 1.23 1.22 1.21 1.19 1.19

1.05 1.53 1.46 1.42 1.36 1.34 1.32 1.30 1.28 1.26 1.25

1.10 1.65 1.56 1.50 1.43 1.39 1.37 1.34 1.33 1.30 1.28

1.15 1.73 1.63 1.56 1.46 1.43 1.40 1.37 1.35 1.32 1.31

1.20 1.82 1.68 1.62 1.51 1.47 1.44 1.41 1.38 1.35 1.34

1.50 2.03 1.84 1.80 1.66 1.60 1.56 1.53 1.50 1.46 1.44

2.00 2.14 1.94 1.89 1.74 1.68 1.64 1.59 1.56 1.50 1.47

Table 6.4. Theoretical stress concentration factor (Kt ) for a steppedshaft with a shoulder fillet (of radius r ) in bending.


D

dr/d

0.02 0.04 0.08 0.10 0.12 0.16 0.20 0.24 0.28 0.30

1.01 1.85 1.61 1.42 1.36 1.32 1.24 1.20 1.17 1.15 1.14

1.02 1.97 1.72 1.50 1.44 1.40 1.32 1.27 1.23 1.21 1.20

1.05 2.20 1.88 1.60 1.53 1.48 1.40 1.34 1.30 1.27 1.25

1.10 2.36 1.99 1.66 1.58 1.53 1.44 1.38 1.33 1.28 1.27

1.20 2.52 2.10 1.72 1.62 1.56 1.46 1.39 1.34 1.29 1.28

1.50 2.75 2.20 1.78 1.68 1.60 1.50 1.42 1.36 1.31 1.29

2.00 2.86 2.32 1.87 1.74 1.64 1.53 1.43 1.37 1.32 1.30

3.00 3.00 2.45 1.95 1.80 1.69 1.56 1.46 1.38 1.34 1.32

6.00 3.04 2.58 2.04 1.87 1.76 1.60 1.49 1.41 1.35 1.33


Table 6.5. Theoretical stress concentration factor (Kt) for a stepped shaftwith a shoulder fillet (of radius r) in torsion.


D

dr/d

0.02 0.04 0.08 0.10 0.12 0.16 0.20 0.24 0.28 0.30

1.09 1.54 1.32 1.19 1.16 1.15 1.12 1.11 1.10 1.09 1.09

1.20 1.98 1.67 1.40 1.33 1.28 1.22 1.18 1.15 1.13 1.13

1.33 2.14 1.79 1.48 1.41 1.35 1.28 1.22 1.19 1.17 1.16

2.00 2.27 1.84 1.53 1.46 1.40 1.32 1.26 1.22 1.19 1.18

Table 6.6. Theoretical stress concentration factor (Kt )for a grooved shaft in tension.

Theoretical stress concentration (Kt )D

d r/d

0.02 0.04 0.08 0.10 0.12 0.16 0.20 0.24 0.28 0.30

1.01 1.98 1.71 1.47 1.42 1.38 1.33 1.28 1.25 1.23 1.22

1.02 2.30 1.94 1.66 1.59 1.54 1.45 1.40 1.36 1.33 1.31

1.03 2.60 2.14 1.77 1.69 1.63 1.53 1.46 1.41 1.37 1.36

1.05 2.85 2.36 1.94 1.81 1.73 1.61 1.54 1.47 1.43 1.41

1.10 .. 2.70 2.16 2.01 1.90 1.75 1.70 1.57 1.50 1.47

1.20 .. 2.90 2.36 2.17 2.04 1.86 1.74 1.64 1.56 1.54

1.30 .. .. 2.46 2.26 2.11 1.91 1.77 1.67 1.59 1.56

1.50 .. .. 2.54 2.33 2.16 1.94 1.79 1.69 1.61 1.57

2.00 .. .. 2.61 2.38 2.22 1.98 1.83 1.72 1.63 1.59

∞ .. .. 2.69 2.44 2.26 2.03 1.86 1.74 1.65 1.61


Table 6.7. Theoretical stress concentration factor (Kt ) ofa grooved shaft in bending.


D

dr/d

0.02 0.04 0.08 0.10 0.12 0.16 0.20 0.24 0.28 0.30

1.01 1.74 1.68 1.47 1.41 1.38 1.32 1.27 1.23 1.22 1.20

1.02 2.28 1.89 1.64 1.53 1.48 1.40 1.34 1.30 1.26 1.25

1.03 2.46 2.04 1.68 1.61 1.55 1.47 1.40 1.35 1.31 1.28

1.05 2.75 2.22 1.80 1.70 1.63 1.53 1.46 1.40 1.35 1.33

1.12 3.20 2.50 1.97 1.83 1.75 1.62 1.52 1.45 1.38 1.34

1.30 3.40 2.70 2.04 1.91 1.82 1.67 1.57 1.48 1.42 1.38

1.50 3.48 2.74 2.11 1.95 1.84 1.69 1.58 1.49 1.43 1.40

2.00 3.55 2.78 2.14 1.97 1.86 1.71 1.59 1.55 1.44 1.41

∞ 3.60 2.85 2.17 1.98 1.88 1.71 1.60 1.51 1.45 1.42

Table 6.8. Theoretical stress concentration factor (Kt ) for a groovedshaft in torsion.

Theoretical stress concentration factor (Kts)D

dr/d

0.02 0.04 0.08 0.10 0.12 0.16 0.20 0.24 0.28 0.30

1.01 1.50 1.03 1.22 1.20 1.18 1.16 1.13 1.12 1.12 1.12

1.02 1.62 1.45 1.31 1.27 1.23 1.20 1.18 1.16 1.15 1.16

1.05 1.88 1.61 1.40 1.35 1.32 1.26 1.22 1.20 1.18 1.17

1.10 2.05 1.73 1.47 1.41 1.37 1.31 1.26 1.24 1.21 1.20

1.20 2.26 1.83 1.53 1.46 1.41 1.34 1.27 1.25 1.22 1.21

1.30 2.32 1.89 1.55 1.48 1.43 1.35 1.30 1.26 — —

2.00 2.40 1.93 1.58 1.50 1.45 1.36 1.31 1.26 — —

∞ 2.50 1.96 1.60 1.51 1.46 1.38 1.32 1.27 1.24 1.23


Stepped shaft

Example 6.2. Find the maximumstress induced in the following casestaking stress concentration intoaccount:

1. A rectangular plate 60 mm ×10 mm with a hole 12 diameter asshown in Fig. 6.13 (a) and subjectedto a tensile load of 12 kN.

2. A stepped shaft as shown inFig. 6.13 (b) and carrying a tensileload of 12 kN.

Fig. 6.13

Solution. Case 1. Given : b = 60 mm ; t = 10 mm ; d = 12 mm ; W = 12 kN = 12 × 103 NWe know that cross-sectional area of the plate,

A = (b – d) t = (60 – 12) 10 = 480 mm2

∴ Nominal stress =3

212 1025 N / mm 25 MPa

480

W

A

×= = =

Ratio of diameter of hole to width of plate,12

0.260

d

b= =

From Table 6.1, we find that for d / b = 0.2, theoretical stress concentration factor,Kt = 2.5

∴ Maximum stress = Kt × Nominal stress = 2.5 × 25 = 62.5 MPa Ans.Case 2. Given : D = 50 mm ; d = 25 mm ; r = 5 mm ; W = 12 kN = 12 × 103 N

We know that cross-sectional area for the stepped shaft,

A = 2 2 2(25) 491 mm4 4

dπ π× = =

∴ Nominal stress =3

212 1024.4 N / mm 24.4 MPa

491

W

A

×= = =

Ratio of maximum diameter to minimum diameter,

D/d = 50/25 = 2

Ratio of radius of fillet to minimum diameter,

r/d = 5/25 = 0.2

From Table 6.3, we find that for D/d = 2 and r/d = 0.2, theoretical stress concentration factor,Kt = 1.64.

∴ Maximum stress = Kt × Nominal stress = 1.64 × 24.4 = 40 MPa Ans.


6.16 Fatigue Stress Concentration FactorWhen a machine member is subjected to cyclic or fatigue loading, the value of fatigue stress

concentration factor shall be applied instead of theoretical stress concentration factor. Since thedetermination of fatigue stress concentration factor is not an easy task, therefore from experimentaltests it is defined as

Fatigue stress concentration factor,

Kf =Endurance limit without stress concentration

Endurance limit with stress concentration

6.17 Notch SensitivityIn cyclic loading, the effect of the notch or the fillet is usually less than predicted by the use of

the theoretical factors as discussed before. The difference depends upon the stress gradient in theregion of the stress concentration and on the hardness of the material. The term notch sensitivity isapplied to this behaviour. It may be defined as the degree to which the theoretical effect of stressconcentration is actually reached. The stress gradient depends mainly on the radius of the notch, holeor fillet and on the grain size of the material. Since the extensive data for estimating the notch sensitivityfactor (q) is not available, therefore the curves, as shown in Fig. 6.14, may be used for determiningthe values of q for two steels.

Fig. 6.14. Notch sensitivity.

When the notch sensitivity factor q is used in cyclic loading, then fatigue stress concentrationfactor may be obtained from the following relations:

q =– 1

–1f

t

K

K

or Kf = 1 + q (Kt – 1) ...[For tensile or bending stress]

and Kfs = 1 + q (Kts – 1) ...[For shear stress]


where Kt = Theoretical stress concentration factor for axial or bendingloading, and

Kts = Theoretical stress concentration factor for torsional or shearloading.

6.18 Combined Steady andVariable Stress

The failure points from fatiguetests made with different steels andcombinations of mean and variablestresses are plotted in Fig. 6.15 asfunctions of variable stress (σv) andmean stress (σm). The most significantobservation is that, in general, thefailure point is little related to the meanstress when it is compressive but is verymuch a function of the mean stress whenit is tensile. In practice, this means thatfatigue failures are rare when the meanstress is compressive (or negative).Therefore, the greater emphasis must begiven to the combination of a variablestress and a steady (or mean) tensilestress.

Fig. 6.15. Combined mean and variable stress.

There are several ways in which problems involving this combination of stresses may be solved,but the following are important from the subject point of view :

1. Gerber method, 2. Goodman method, and 3. Soderberg method.

We shall now discuss these methods, in detail, in the following pages.

Protective colour coatings are added to make componentsit corrosion resistant. Corrosion if not taken care can magnifyother stresses.Note : This picture is given as additional information and is not a

direct example of the current chapter.


6.19 Gerber Method forCombination of Stresses

The relationship between variablestress (σv) and mean stress (σm) for axial andbending loading for ductile materials areshown in Fig. 6.15. The point σe representsthe fatigue strength corresponding to the caseof complete reversal (σm = 0) and the pointσu represents the static ultimate strengthcorresponding to σv = 0.

A parabolic curve drawn between theendurance limit (σe) and ultimate tensilestrength (σu) was proposed by Gerber in1874. Generally, the test data for ductilematerial fall closer to Gerber parabola asshown in Fig. 6.15, but because of scatter inthe test points, a straight line relationship (i.e.Goodman line and Soderberg line) is usuallypreferred in designing machine parts.

According to Gerber, variable stress,

σv = σe

21

. .. .

m

u

F SF S

⎡ ⎤σ⎛ ⎞−⎢ ⎥⎜ ⎟σ⎝ ⎠⎢ ⎥⎣ ⎦

or

21

. .. .

m v

u e

F SF S

σ σ⎛ ⎞= +⎜ ⎟σ σ⎝ ⎠...(i)

where F.S. = Factor of safety,

σm = Mean stress (tensile or compressive),

σu = Ultimate stress (tensile or compressive), and

σe = Endurance limit for reversal loading.

Considering the fatigue stressconcentration factor (Kf), the equation (i) maybe written as

21

. .. .

v fm

u e

KF S

F S

σ ×σ⎛ ⎞= +⎜ ⎟σ σ⎝ ⎠

6.20 Goodman Method forCombination of Stresses

A straight line connecting the endurancelimit (σe) and the ultimate strength (σu), asshown by line AB in Fig. 6.16, follows thesuggestion of Goodman. A Goodman line isused when the design is based on ultimatestrength and may be used for ductile or brittlematerials.

In Fig. 6.16, line AB connecting σe and

Liquid refrigerant absorbs heat as it vaporizes inside theevaporator coil of a refrigerator. The heat is releasedwhen a compressor turns the refrigerant back to liquid.

Note : This picture is given as additional information and isnot a direct example of the current chapter.

Fig. 6.16. Goodman method.

Evaporator

Gas flow

Fins radiate heat

Liquid flow

Condenser

Compressor


* Here we have assumed the same factor of safety (F.S.) for the ultimate tensile strength (σu) and endurancelimit (σe). In case the factor of safety relating to both these stresses is different, then the following relationmay be used :

1/ ( . .) / ( . .)

σ σ= −

σ σv m

F S F Se e u uwhere (F.S.)e = Factor of safety relating to endurance limit, and

(F.S.)u = Factor of safety relating to ultimate tensile strength.

σu is called Goodman's failure stress line. If a suitable factor of safety (F.S.) is applied to endurancelimit and ultimate strength, a safe stress line CD may be drawn parallel to the line AB. Let us considera design point P on the line CD.

Now from similar triangles COD and PQD,

PQ QD

CO OD=

OD OQ

OD

−= 1 –OQ

OD= ...(Q QD = OD – OQ)

∴ 1/ . . / . .

σ σ= −

σ σv m

e uF S F S

11

. . / . . . .e m m

v eu uF S F S F S

σ σ σ⎡ ⎤ ⎡ ⎤σ = − = σ −⎢ ⎥ ⎢ ⎥σ σ⎣ ⎦ ⎣ ⎦

or1

. .m v

u eF S

σ σ= +

σ σ...(i)

This expression does not include the effect of stress concentration. It may be noted that forductile materials, the stress concentration may be ignored under steady loads.

Since many machine and structural parts that are subjected to fatigue loads contain regions ofhigh stress concentration, therefore equation (i) must be altered to include this effect. In such cases,the fatigue stress concentration factor (Kf) is used to multiply the variable stress (σv). The equation (i)may now be written as

1

. .v fm

u e

K

F S

σ ×σ= +

σ σ...(ii)

where F.S. = Factor of safety,σm = Mean stress,σu = Ultimate stress,σv = Variable stress,σe = Endurance limit for reversed loading, andKf = Fatigue stress concentration factor.

Considering the load factor, surface finish factor and size factor, the equation (ii) may bewritten as

1

. .v f v fm m

u eb sur sz u e b sur sz

K K

F S K K K K K

σ × σ ×σ σ= + = +σ σ × × σ σ × × ×

...(iii)

=v fm

u e sur sz

K

K K

σ ×σ+

σ σ × × ...(Q σeb = σe × Kb and Kb = 1)

where Kb = Load factor for reversed bending load,

Ksur = Surface finish factor, and

Ksz = Size factor.

∗

Variable Stresses in Machine Parts 199Notes : 1. The equation (iii) is applicable to ductile materials subjected to reversed bending loads (tensile orcompressive). For brittle materials, the theoretical stress concentration factor (Kt) should be applied to the meanstress and fatigue stress concentration factor (Kf) to the variable stress. Thus for brittle materials, the equation(iii) may be written as

1

. .

σ ×σ ×= +σ σ × ×

v fm t

u eb sur sz

KK

F S K K...(iv)

2. When a machine component is subjected to a load other than reversed bending, then the endurancelimit for that type of loading should be taken into consideration. Thus for reversed axial loading (tensile orcompressive), the equations (iii) and (iv) may be written as

1

. .

σ ×σ= +σ σ × ×

v fm

u ea sur sz

K

F S K K...(For ductile materials)

and1

. .

σ ×σ ×= +σ σ × ×

v fm t

u ea sur sz

KK

F S K K...(For brittle materials)

Similarly, for reversed torsional or shear loading,

1

. .

τ ×τ= +τ τ × ×

v fsm

u e sur sz

K

F S K K...(For ductile materials)

and1

. .v fsm ts

u e sur sz

KK

F S K K

τ ×τ ×= +

τ τ × × ...(For brittle materials)

where suffix ‘s’denotes for shear.

For reversed torsional or shear loading, the values of ultimate shear strength (τu) and endurance shearstrength (τe) may be taken as follows:

τu = 0.8 σu; and τe = 0.8 σe

6.21 Soderberg Method for Combination of StressesA straight line connecting the endurance limit (σe) and the yield strength (σy), as shown by the

line AB in Fig. 6.17, follows the suggestion of Soderberg line. This line is used when the design isbased on yield strength.

Note : This picture is given as additional information and is not a direct example of the current chapter.

In this central heating system, a furnace burns fuel to heat water in a boiler. A pump forces the hotwater through pipes that connect to radiators in each room. Water from the boiler also heats the hotwater cylinder. Cooled water returns to the boiler.

Overflow pipeMainssupply

Hot watercylinderWater

tank Controlvalve

Radiator

Pump

Heat exchangerGas burner

Boiler Insulation

Flue

Air inlet


Proceeding in the same way as discussedin Art 6.20, the line AB connecting σe and σy, asshown in Fig. 6.17, is called Soderberg's failurestress line. If a suitable factor of safety (F.S.) isapplied to the endurance limit and yield strength,a safe stress line CD may be drawn parallel tothe line AB. Let us consider a design point P onthe line CD. Now from similar triangles CODand PQD,

PQ QD OD OQ

CO OD OD

−= =

= 1OQ

OD−

...(Q QD = OD – OQ)

∴ 1/ . . / . .

v m

e yF S F S

σ σ= −

σ σ

or1

1. . / . . . .e m m

v ey yF S F S F S

σ σ σ⎡ ⎤ ⎡ ⎤σ = − = σ −⎢ ⎥ ⎢ ⎥σ σ⎣ ⎦ ⎣ ⎦

∴1

. .m v

y eF S

σ σ= +

σ σ...(i)

For machine parts subjected to fatigue loading, the fatigue stress concentration factor (Kf)should be applied to only variable stress (σv). Thus the equations (i) may be written as

1

. .v fm

y e

K

F S

σ ×σ= +

σ σ...(ii)

Considering the load factor, surface finish factor and size factor, the equation (ii) may bewritten as

1

. .v fm

y eb sur sz

K

F S K K

σ ×σ= +

σ σ × ×...(iii)

Since σeb = σe × Kb and Kb = 1 for reversed bending load, therefore σeb = σe may be substitutedin the above equation.

Notes: 1. The Soderberg method is particularly used for ductile materials. The equation (iii) is applicable toductile materials subjected to reversed bending load (tensile or compressive).

2. When a machine component is subjected to reversed axial loading, then the equation (iii) may bewritten as

1

. .

σ ×σ= +σ σ × ×

v fm

y ea sur sz

K

F S K K

3. When a machine component is subjected to reversed shear loading, then equation (iii) may bewritten as

1

. .

τ ×τ= +τ τ × ×

v fsm

y e sur sz

K

F S K K

where K f s is the fatigue stress concentration factor for reversed shear loading. The yield strength in shear (τy)may be taken as one-half the yield strength in reversed bending (σy).

Fig. 6.17. Soderberg method.

Variable Stresses in Machine Parts 201Example 6.3. A machine component is

subjected to a flexural stress which fluctuatesbetween + 300 MN/m2 and – 150 MN/m2.Determine the value of minimum ultimate strengthaccording to 1. Gerber relation; 2. ModifiedGoodman relation; and 3. Soderberg relation.

Take yield strength = 0.55 Ultimate strength;Endurance strength = 0.5 Ultimate strength; andfactor of safety = 2.

Solution. Given : σ1 = 300 MN/m2 ;σ2 = – 150 MN/m2 ; σy = 0.55 σu ; σe = 0.5 σu ;F.S. = 2

Let σu = Minimum ultimate strength in MN/m2.

We know that the mean or average stress,

21 2 300 ( 150)75 MN/m

2 2

σ + σ + −σ = = =m

and variable stress, 21 2 300 ( 150)225 MN/m

2 2vσ − σ − −σ = = =

1. According to Gerber relationWe know that according to Gerber relation,

21

. .. .

m v

u e

F SF S

σ σ⎛ ⎞= +⎜ ⎟σ σ⎝ ⎠2

2 2

11 250 4501 75 225 11 250 4502

2 0.5 ( ) ( )

+ σ⎛ ⎞= + = + =⎜ ⎟σ σ σσ σ⎝ ⎠u

u u uu u

(σu)2 = 22 500 + 900 σu

or (σu)2 – 900 σu – 22 500 = 0

∴ σu =2900 (900) 4 1 22 500 900 948.7

2 1 2

± + × × ±=×

= 924.35 MN/m2 Ans. ...(Taking +ve sign)

2. According to modified Goodman relation

We know that according to modified Goodman relation,

1

. .m v

u eF S

σ σ= +

σ σ

or1 75 225 525

2 0.5u u u

= + =σ σ σ

∴ σu = 2 × 525 = 1050 MN/m2 Ans.

3. According to Soderberg relation

We know that according to Soderberg relation,1

. .m v

y eF S

σ σ= +

σ σ

or1 75 255 586.36

2 0.55 0.5u u u

= + =σ σ σ

∴ σu = 2 × 586.36 = 1172.72 MN/m2 Ans.

Springs often undergo variable stresses.


Example 6.4. A bar of circular cross-section is subjected to alternating tensile forces varyingfrom a minimum of 200 kN to a maximum of 500 kN. It is to be manufactured of a material with anultimate tensile strength of 900 MPa and an endurance limit of 700 MPa. Determine the diameter ofbar using safety factors of 3.5 related to ultimate tensile strength and 4 related to endurance limitand a stress concentration factor of 1.65 for fatigue load. Use Goodman straight line as basis fordesign.

Solution. Given : Wmin = 200 kN ; Wmax = 500 kN ; σu = 900 MPa = 900 N/mm2 ; σe = 700 MPa= 700 N/mm2 ; (F.S.)u = 3.5 ; (F.S.)e = 4 ; Kf = 1.65

Let d = Diameter of bar in mm.

∴ Area, A = 2 2 20.7854 mm4

d dπ × =

We know that mean or average force,

Wm = 3500 200350 kN 350 10 N

2 2max minW W+ += = = ×

∴ Mean stress, σm =3 3

22 2

350 10 446 10N / mm

0.7854

× ×= =mW

A d d

Variable force, Wv = 3500 200150 kN 150 10 N

2 2max minW W− −= = = ×

∴ Variable stress, σv =3 3

22 2

150 10 191 10N / mm

0.7854vW

A d d

× ×= =

We know that according to Goodman's formula,

.1 –

/ ( . .) / ( . .)m fv

e e u u

K

F S F S

σσ=

σ σ3 3

2 2

191 10 446 101.65

1700 / 4 900 / 3.5

d d

× × ×= −

Paint Manufacture : A typical gloss paint is made by first mixingnatural oils and resins called alkyds. Thinner is added to makethe mixture easier to pump through a filter that removes anysolid particles from the blended liquids. Pigment is mixed intothe binder blend in a powerful mixer called a disperser.

Pigment and paint thin-ner added

Final adjustments madeFilter tankSetting tank

Mixingtank

Thinner added

Oil and resinblendedtogether

Disperser Bead mill Holding tank



2 2

1100 28601

d d= − or 2

1100 28601

d

+ =

∴ d 2 = 3960 or d = 62.9 say 63 mm Ans.Example 6.5. Determine the thickness of a 120 mm wide uniform plate for safe continuous

operation if the plate is to be subjected to a tensile load that has a maximum value of 250 kN and aminimum value of 100 kN. The properties of the plate material are as follows:

Endurance limit stress = 225 MPa, and Yield point stress = 300 MPa.

The factor of safety based on yield point may be taken as 1.5.

Solution. Given : b = 120 mm ; Wmax = 250 kN; Wmin = 100 kN ; σe = 225 MPa = 225 N/mm2 ;σy = 300 MPa = 300 N/mm2; F.S. = 1.5

Let t = Thickness of the plate in mm.

∴ Area, A = b × t = 120 t mm2

We know that mean or average load,

Wm = 3250 100175 kN = 175 × 10 N

2 2max minW W+ += =

∴ Mean stress, σm =3

2175 10N/mm

120mW

A t

×=

Variable load, Wv = 3250 10075 kN 75 10 N

2 2max minW W− −= = = ×

∴ Variable stress, σv =3

275 10N/mm

120vW

A t

×=

According to Soderberg’s formula,1

. .m v

y eF S

σ σ= +

σ σ3 31 175 10 75 10 4.86 2.78 7.64

1.5 120 300 120 225t t t t t

× ×= + = + =× ×

∴ t = 7.64 × 1.5 = 11.46 say 11.5 mm Ans.Example 6.6. Determine the diameter of a circular rod made of ductile material with a fatigue

strength (complete stress reversal), σe = 265 MPa and a tensile yield strength of 350 MPa. Themember is subjected to a varying axial load from Wmin = – 300 × 103 N to Wmax = 700 × 103 N andhas a stress concentration factor = 1.8. Use factor of safety as 2.0.

Solution. Given : σe = 265 MPa = 265 N/mm2 ; σy = 350 MPa = 350 N/mm2 ; Wmin = – 300 × 103 N ;Wmax = 700 × 103 N ; Kf = 1.8 ; F.S. = 2

Let d = Diameter of the circular rod in mm.

∴ Area, A = 2 2 20.7854 mm4

d dπ × =

We know that the mean or average load,

Wm =3 3

3700 10 ( 300 10 )200 10 N

2 2max minW W+ × + − ×= = ×

∴ Mean stress, σm =3 3

22 2

200 10 254.6 10N/mm

0.7854mW

A d d

× ×= =


Variable load, Wv =3 3

3700 10 ( 300 10 )500 10 N

2 2max minW W− × − − ×= = ×

∴ Variable stress, σv =3 3

22 2

500 10 636.5 10N/mm

0.7854vW

A d d

× ×= =

We know that according to Soderberg's formula,

1

. .v fm

y e

K

F S

σ ×σ= +

σ σ3 3

2 2 2 2 2

1 254.6 10 636.5 10 1.8 727 4323 5050

2 350 265d d d d d

× × ×= + = + =× ×

∴ d 2 = 5050 × 2 = 10 100 or d = 100.5 mm Ans.Example 6.7. A steel rod is subjected to a reversed axial load of 180 kN. Find the diameter of

the rod for a factor of safety of 2. Neglect column action. The material has an ultimate tensilestrength of 1070 MPa and yield strength of 910 MPa. The endurance limit in reversed bendingmay be assumed to be one-half of the ultimate tensile strength. Other correction factors may betaken as follows:

For axial loading = 0.7; For machined surface = 0.8 ; For size = 0.85 ; For stressconcentration = 1.0.

Solution. Given : Wmax = 180 kN ; Wmin = – 180 kN ; F.S. = 2 ; σu = 1070 MPa = 1070N/mm2; σy = 910 MPa = 910 N/mm2 ; σe = 0.5 σu ; Ka = 0.7 ; Ksur = 0.8 ; Ksz = 0.85 ; Kf = 1

Let d = Diameter of the rod in mm.

∴ Area, A = 2 2 20.7854 mm4

d dπ × =

We know that the mean or average load,

Wm =180 ( 180)

02 2

max minW W+ + −= =

∴ Mean stress, σm = 0mW

A=

Variable load, Wv = 3180 ( 180)180 kN 180 10 N

2 2max minW W− − −= = = ×

∴Variable stress, σv =3 3

22 2

180 10 229 10N/mm

0.7854vW

A d d

× ×= =

Endurance limit in reversed axial loading,

σea = σe × Ka = 0.5 σu × 0.7 = 0.35 σu ...(Q σe = 0.5 σu)

= 0.35 × 1070 = 374.5 N/mm2

We know that according to Soderberg's formula for reversed axial loading,

1

. .v fm

y ea sur sz

K

F S K K

σ ×σ= +

σ σ × ×3

2 2

1 229 10 1 9000

2 374.5 0.8 0.85d d

× ×= + =× × ×

∴ d 2 = 900 × 2 = 1800 or d = 42.4 mm Ans.


Example 6.8. A circular bar of 500 mm length is supported freely at its two ends. It is actedupon by a central concentrated cyclic load having a minimum value of 20 kN and a maximum valueof 50 kN. Determine the diameter of bar by taking a factor of safety of 1.5, size effect of 0.85, surfacefinish factor of 0.9. The material properties of bar are given by : ultimate strength of 650 MPa, yieldstrength of 500 MPa and endurance strength of 350 MPa.

Solution. Given : l = 500 mm ; Wmin = 20 kN = 20 × 103 N ; Wmax = 50 kN = 50 × 103 N ;F.S. = 1.5 ; Ksz = 0.85 ; Ksur = 0.9 ; σu = 650 MPa = 650 N/mm2 ; σy = 500 MPa = 500 N/mm2 ;σe = 350 MPa = 350 N/mm2

Let d = Diameter of the bar in mm.

We know that the maximum bending moment,

Mmax =3

350 10 5006250 10 N-mm

4 4maxW l× × ×= = ×

and minimum bending moment,

Mmin =3

320 10 5002550 10 N -mm

4 4minW l× × ×= = ×

∴ Mean or average bending moment,

Mm =3 3

36250 10 2500 104375 10 N-mm

2 2max minM M+ × + ×= = ×

and variable bending moment,

Mv =3 3

36250 10 2500 101875 10 N -mm

2 2max minM M− × − ×= = ×

Section modulus of the bar,

Z =3 3 30.0982 mm

32d d

π × =

∴ Mean or average bending stress,

σm =3 6

23 3

4375 10 44.5 10N/mm

0.0982mM

Z d d

× ×= =

Layout of a military tank.


Shells formain gun

Loader Engine

Drivingsprocket

Main gun

Machine gun

Rubber tyres

Driver

Gunner

Commander

Sighting equipment


and variable bending stress,

σv = 3 6

23 3

1875 10 19.1 10N/mm

0.0982vM

Z d d

× ×= =

We know that according to Goodman's formula,

1

. .v fm

u e sur sz

K

F S K K

σ ×σ= +σ σ × ×

6 6

3 3

1 44.5 10 19.1 10 1

1.5 650 350 0.9 0.85

× × ×= +× × × ×d d

...(Taking Kf = 1)

=3 3 3

3 3 3

68 10 71 10 139 10

d d d

× × ×+ =

∴ d 3 = 139 × 103 × 1.5 = 209 × 103 or d = 59.3 mm

and according to Soderberg's formula,

1

. .v fm

y e sur sz

K

F S K K

σ ×σ= +

σ σ × ×

6 6

3 3

1 44.5 10 19.1 10 1

1.5 500 350 0.9 0.85

× × ×= +× × × ×d d

...(Taking Kf = 1)

=3 3 3

3 3 3

89 10 71 10 160 10

d d d

× × ×+ =

∴ d 3 = 160 × 103 × 1.5 = 240 × 103 or d = 62.1 mm

Taking larger of the two values, we have d = 62.1 mm Ans.Example 6.9. A 50 mm diameter shaft is made from carbon steel having ultimate tensile

strength of 630 MPa. It is subjected to a torque which fluctuates between 2000 N-m to – 800 N-m.Using Soderberg method, calculate the factor of safety. Assume suitable values for any other dataneeded.

Solution. Given : d = 50 mm ; σu = 630 MPa = 630 N/mm2 ; Tmax = 2000 N-m ; Tmin = – 800 N-mWe know that the mean or average torque,

Tm = 32000 ( 800)600 N-m 600 10 N-mm

2 2max minT T+ + −= = = ×

∴ Mean or average shear stress,

τm = 3

23 3

16 16 600 1024.4 N / mm

(50)

× ×= =π π

mT

d... 3

16

π⎛ ⎞= × τ ×⎜ ⎟⎝ ⎠Q T d

Variable torque,

Tv = 32000 ( 800)1400 N-m 1400 10 N-mm

2 2max minT T− − −= = = ×

∴ Variable shear stress, τv = 3

23 3

16 16 1400 1057 N/mm

(50)vT

d

× ×= =π π

Since the endurance limit in reversed bending (σe) is taken as one-half the ultimate tensilestrength (i.e. σe = 0.5 σu) and the endurance limit in shear (τe) is taken as 0.55 σe, therefore

τe = 0.55 σe = 0.55 × 0.5 σu = 0.275 σu

= 0.275 × 630 = 173.25 N/mm2

Assume the yield stress (σy) for carbon steel in reversed bending as 510 N/mm2, surface finish

factor (Ksur) as 0.87, size factor (Ksz) as 0.85 and fatigue stress concentration factor (Kfs) as 1.


Since the yield stress in shear (τy) for shear loading is taken as one-half the yield stress inreversed bending (σy), therefore

τy = 0.5 σy = 0.5 × 510 = 255 N/mm2

Let F.S. = Factor of safety.

We know that according to Soderberg's formula,

1 24.4 57 1

. . 255 173.25 0.87 0.85v fsm

y e sur sz

K

F S K K

τ ×τ ×= + = +τ τ × × × ×

= 0.096 + 0.445 = 0.541

∴ F.S. = 1 / 0.541 = 1.85 Ans.Example 6.10. A cantilever beam made of cold drawn carbon steel of circular cross-section as

shown in Fig. 6.18, is subjected to a load which variesfrom – F to 3 F. Determine the maximum load that thismember can withstand for an indefinite life using a factorof safety as 2. The theoretical stress concentration factoris 1.42 and the notch sensitivity is 0.9. Assume thefollowing values :

Ultimate stress = 550 MPa

Yield stress = 470 MPa

Endurance limit = 275 MPa

Size factor = 0.85

Surface finish factor = 0.89

Solution. Given : Wmin = – F ; Wmax = 3 F ; F.S. = 2 ; Kt = 1.42 ; q = 0.9 ; σu = 550 MPa= 550 N/mm2 ; σy = 470 MPa = 470 N/mm2 ; σe = 275 MPa = 275 N/mm2 ; Ksz = 0.85 ; Ksur = 0.89

Fig. 6.18


Army Tank

All dimensions in mm.


The beam as shown in Fig. 6.18 is subjected to a reversed bending load only. Since the point Aat the change of cross section is critical, therefore we shall find the bending moment at point A.

We know that maximum bending moment at point A,

Mmax = Wmax × 125 = 3F × 125 = 375 F N-mmand minimum bending moment at point A,

Mmin = Wmin × 125 = – F × 125 = – 125 F N-mm∴ Mean or average bending moment,

Mm =375 ( 125 )

125 N -mm2 2

max minM M F FF

+ + −= =


Mv =375 ( 125 )

250 N -mm2 2

max minM M F FF

− − −= =

Section modulus, Z =3 3 3(13) 215.7 mm

32 32d

π π× = = ...( Q d = 13 mm)

∴ Mean bending stress, σm = 21250.58 N/mm

215.7mM F

FZ

= =

and variable bending stress, σv = 22501.16 N/mm

215.7vM F

FZ

= =

Fatigue stress concentration factor,Kf = 1 + q (Kt – 1) = 1 + 0.9 (1.42 – 1) = 1.378

We know that according to Goodman’s formula

1

. .v fm

u e sur sz

K

F S K K

σ ×σ= +

σ σ × ×1 0.58 1.16 1.378

2 550 275 0.89 0.85

F F ×= +× ×

= 0.001 05 F + 0.007 68 F = 0.008 73 F

∴ F =1

57.3 N2 0.00873

=×

and according to Soderberg’s formula,

1

. .v fm

y e sur sz

K

F S K K

σ ×σ= +

σ σ × ×1 0.58 1.16 1.378

2 470 275 0.89 0.85

F F ×= +× ×

= 0.001 23 F + 0.007 68 F = 0.008 91 F

∴ F =1

56 N2 0.008 91

=×

Taking larger of the two values, we have F = 57.3 N Ans.Example 6.11. A simply supported beam has a concentrated load at the centre which fluctuates

from a value of P to 4 P. The span of the beam is 500 mm and its cross-section is circular with adiameter of 60 mm. Taking for the beam material an ultimate stress of 700 MPa, a yield stress of 500MPa, endurance limit of 330 MPa for reversed bending, and a factor of safety of 1.3, calculate themaximum value of P. Take a size factor of 0.85 and a surface finish factor of 0.9.

Solution. Given : Wmin = P ; Wmax = 4P ; L = 500 mm; d = 60 mm ; σu = 700 MPa = 700 N/mm2 ;

σy = 500 MPa = 500 N/mm2 ; σe = 330 MPa = 330 N/mm2 ; F.S. = 1.3 ; Ksz = 0.85 ; Ksur = 0.9

Variable Stresses in Machine Parts 209We know that maximum bending moment,

Mmax =4 500

500 N-mm4 4

maxW L PP

× ×= =

and minimum bending moment,

Mmin = 500

125 N-mm4 4

minW L PP

× ×= =

∴ Mean or average bending moment,

Mm =500 125

312.5 N-mm2 2

max minM M P PP

+ += =


Mv =500 125

187.5 N-mm2 2

max minM M P PP

− −= =

Section modulus, Z =3 3 3 3(60) 21.21 10 mm

32 32d

π π× = = ×

∴ Mean bending stress,

σm =2

3

312.50.0147 N/mm

21.21 10mM P

PZ

= =×


σv =2

3

187.50.0088 N/mm

21.21 10vM P

PZ

= =×

We know that according to Goodman’s formula,

1

. .v fm

u e sur sz

K

F S K K

σ ×σ= +

σ σ × ×1 0.0147 0.0088 1

1.3 700 330 0.9 0.85

P P ×= +× ×

...(Taking Kf = 1)

= 6 6 6

21 34.8 55.8

10 10 10

P P P+ =

∴ P =61 10

13 785 N 13.785 kN1.3 55.8

× = =

and according to Soderberg's formula,

1

. .v fm

y e sur sz

K

F S K K

σ ×σ= +

σ σ × ×

6 6 6

1 0.0147 0.0088 1 29.4 34.8 64.2

1.3 500 330 0.9 0.85 10 10 10

P P P P P×= + = + =× ×

∴ P =61 10

11 982 N 11.982 kN1.3 64.2

× = =

From the above, we find that maximum value of P = 13.785 kN Ans.

6.22 Combined Variable Normal Stress and Variable Shear StressWhen a machine part is subjected to both variable normal stress and a variable shear stress; then

it is designed by using the following two theories of combined stresses :

1. Maximum shear stress theory, and 2. Maximum normal stress theory.


We have discussed in Art. 6.21, that according to Soderberg's formula,

1

. .v fbm

y eb sur sz

K

F S K K

σ ×σ= +

σ σ × ×...(For reversed bending load)

Multiplying throughout by σy, we get

. .y v y fb

meb sur sz

K

F S K K

σ σ × σ ×= σ +

σ × ×The term on the right hand side of the above expression is known as equivalent normal stress

due to reversed bending.

∴ Equivalent normal stress due to reversed bending,v y fb

neb meb sur sz

K

K K

σ × σ ×σ = σ +

σ × ×...(i)

Similarly, equivalent normal stress due to reversed axial loading,v y fa

nea mea sur sz

K

K K

σ × σ ×σ = σ +

σ × ×. ..(ii)

and total equivalent normal stress,

σne = σneb + σnea = . .y

F S

σ...(iii)

We have also discussed in Art. 6.21, that for reversed torsional or shear loading,

1

. .v fsm

y e sur sz

K

F S K K

τ ×τ= +

τ τ × ×Multiplying throughout by τy, we get

. .y v y fs

me sur sz

K

F S K K

τ τ × τ ×= τ +

τ × ×The term on the right hand side of the above expression is known as equivalent shear stress.∴ Equivalent shear stress due to reversed torsional or shear loading,

v y fses m

e sur sz

K

K K

τ × τ ×τ = τ +

τ × ×...(iv)

The maximum shear stress theory is used in designing machine parts of ductile materials.According to this theory, maximum equivalent shear stress,

τes(max) =2 21

( ) 4 ( )2 . .

yne es F S

τσ + τ =

The maximum normal stress theory is used in designing machine parts of brittle materials.According to this theory, maximum equivalent normal stress,

σne(max) =2 21 1

( ) ( ) 4 ( )2 2 . .

σσ + σ + τ = y

ne ne es F SExample 6.12. A steel cantilever is 200 mm long. It is subjected to an axial load which varies

from 150 N (compression) to 450 N (tension) and also a transverse load at its free end which variesfrom 80 N up to 120 N down. The cantilever is of circular cross-section. It is of diameter 2d for thefirst 50 mm and of diameter d for the remaining length. Determine its diameter taking a factor ofsafety of 2. Assume the following values :

Yield stress = 330 MPaEndurance limit in reversed loading = 300 MPaCorrection factors = 0.7 in reversed axial loading

= 1.0 in reversed bending


Stress concentration factor = 1.44 for bending

= 1.64 for axial loading

Size effect factor = 0.85

Surface effect factor = 0.90

Notch sensitivity index = 0.90

Solution. Given : l = 200 mm; Wa(max) = 450 N; Wa(min) = – 150 N ; Wt(max) = 120 N ;Wt(min) = – 80 N; F.S. =2 ; σy = 330 MPa = 330 N/mm2 ; σe = 300 MPa = 300 N/mm2 ;

Ka = 0.7; Kb = 1; Ktb = 1.44 ; Kta = 1.64; Ksz = 0.85 ; Ksur = 0.90 ; q = 0.90

First of all, let us find the equiva-lent normal stress for point A which iscritical as shown in Fig. 6.19. It is assumedthat the equivalent normal stress at thispoint will be the algebraic sum of theequivalent normal stress due to axial load-ing and equivalent normal stress due tobending (i.e. due to transverse load act-ing at the free end).

Let us first consider the reversedaxial loading. We know that mean oraverage axial load,

Wm =( ) ( ) 450 ( 150)

150 N2 2

a max a minW W+ + −= =

and variable axial load,

Wv =( ) ( ) 450 ( 150)

300 N2 2

a max a minW W− − −= =

∴ Mean or average axial stress,

σm =2

2 2

150 4 191N/mmmW

A d d

×= =π

... 2

4

π⎛ ⎞= ×⎜ ⎟⎝ ⎠Q A d

and variable axial stress,

σv =2

2 2

300 4 382N/mmvW

A d d

×= =π

We know that fatigue stress concentration factor for reversed axial loading,

Kfa = 1 + q (Kta – 1) = 1 + 0.9 (1.64 – 1) = 1.576

and endurance limit stress for reversed axial loading,

σea = σe × Ka = 300 × 0.7 = 210 N/mm2

We know that equivalent normal stress at point A due to axial loading,

σnea = σm + 2 2

191 382 330 1.576

210 0.9 0.85

v y fa

ea sur sz

K

K K d d

σ × σ × × ×= +σ × × × × ×

= 22 2 2

191 1237 1428N/mm

d d d+ =

Fig. 6.19


Now let us consider the reversed bending due totransverse load. We know that mean or average bend-ing load,

Wm = ( ) ( )

2

+t max t minW W

120 ( 80)20 N

2

+ −= =

and variable bending load,

Wv =( ) ( )

2

−t max t minW W

120 ( 80)

100 N2

− −= =

∴ Mean bending moment at point A,

Mm = Wm (l – 50) = 20 (200 – 50) = 3000 N-mm

and variable bending moment at point A,

Mv = Wv (l – 50) = 100 (200 – 50) = 15 000 N-mm

We know that section modulus,

Z =3 3 30.0982 mm

32d d

π × =


σm =2

3 3

3000 30 550N/mm

0.0982= =mM

Z d dand variable bending stress,

σv = 23 3

15 000 152 750N/mm

0.0982= =vM

Z d dWe know that fatigue stress concentration factor for reversed bending,

Kfb = 1 + q (Ktb – 1) = 1 + 0.9 (1.44 – 1) = 1.396

Since the correction factor for reversed bending load is 1 (i.e. Kb = 1), therefore the endurancelimit for reversed bending load,

σeb = σe . Kb = σe = 300 N/mm2

We know that the equivalent normal stress at point A due to bending,

σneb = σm 3 3

30 550 152 750 330 1.396

300 0.9 0.85

v y fb

eb sur sz

K

K K d d

σ × σ × × ×+ = +σ × × × × ×

= 23 3 3

30 550 306 618 337 168N/mm

d d d+ =

∴ Total equivalent normal stress at point A,

σne = σneb + σnea 2

3 2

337 168 1428N/mm

d d= + ...(i)

Machine transporter

Variable Stresses in Machine Parts 213We know that equivalent normal stress at point A,

σne =2330

165 N/mm. . 2y

F S

σ= = ...(ii)

Equating equations (i) and (ii), we have

33 2

337 168 1428165 or 337 168 + 1428 165+ = =d d

d d∴ 236.1 + d = 0.116 d3 or d = 12.9 mm Ans. ...(By hit and trial)

Example 6.13. A hot rolled steel shaft is subjected to a torsional moment that varies from330 N-m clockwise to 110 N-m counterclockwise and an applied bending moment at a critical sectionvaries from 440 N-m to – 220 N-m. The shaft is of uniform cross-section and no keyway is present atthe critical section. Determine the required shaft diameter. The material has an ultimate strength of550 MN/m2 and a yield strength of 410 MN/m2. Take the endurance limit as half the ultimate strength,factor of safety of 2, size factor of 0.85 and a surface finish factor of 0.62.

Solution. Given : Tmax = 330 N-m (clockwise) ; Tmin = 110 N-m (counterclockwise) = – 110 N-m(clockwise) ; Mmax = 440 N-m ; Mmin = – 220 N-m ; σu = 550 MN/m2 = 550 × 106 N/m2 ;

σy = 410 MN/m2 = 410 × 106 N/m2 ; σe = 1

2σu = 275 × 106 N/m2 ; F.S. = 2 ; Ksz = 0.85 ; Ksur = 0.62

Let d = Required shaft diameter in metres.

We know that mean torque,

Tm =330 ( 110)

110 N-m2 2

max minT T+ + −= =

and variable torque, Tv =330 ( 110)

220 N-m2 2

max minT T− − −= =

∴ Mean shear stress,

τm =2

3 3 3

16 16 110 560N/m

×= =π π

mT

d d dand variable shear stress,

τv =2

3 3 3

16 16 220 1120N/mvT

d d d

×= =π π

Since the endurance limit in shear (τe) is 0.55 σe, and yield strength in shear (τy) is 0.5 σy,therefore

τe = 0.55 × 275 × 106 = 151.25 × 106 N/m2

and τy = 0.5 × 410 × 106 = 205 × 106 N/m2

We know that equivalent shear stress,

τes = τm + v y fs

e sur sz

K

K K

τ × ττ × ×

= 6

3 3 6

560 1120 205 10 1

151.25 10 0.62 0.85d d

× × ×+× × × ×

...(Taking Kfs = 1)

= 23 3 3

560 2880 3440N/m

d d d+ =

Mean or average bending moment,

Mm =440 ( 220)

110 N-m2 2

max minM M+ + −= =



Mv =– 440 ( 220)

330 N-m2 2

max minM M − −= =

Section modulus, Z = 3 3 30.0982 m32

d dπ × =


σm =2

3 3

110 1120N/m

0.0982mM

Z d d= =


σv =2

3 3

330 3360N/m

0.0982vM

Z d d= =

Since there is no reversed axial loading, therefore theequivalent normal stress due to reversed bending load,

σneb = σne = σm + v y fb

eb sur sz

K

K K

σ × σ ×σ × ×

= 6

3 3 6

1120 3360 410 10 1

275 10 0.62 0.85d d

× × ×+× × × ×

...(Taking Kfb = 1 and σeb = σe)

= 2

3 3 3

1120 9506 10626N/m

d d d+ =

We know that the maximum equivalent shear stress,

τes(max) = 2 21( ) 4 ( )

. . 2y

ne esF S

τ= σ + τ

2 26

3 3

205 10 1 10 625 34404

2 2

× ⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠d d

205 × 106 × d 3 = 6 6 3113 10 4 11.84 10 12.66 10× + × × = ×

∴ d 3 =3

6 3

12.66 10 0.0617

205 10 10

× =×

or d =0.395

0.0395 m 39.5 say 40 mm10

= = Ans.

Example 6.14. A pulley is keyed to a shaft midway between two bearings. The shaft is made ofcold drawn steel for which the ultimate strength is 550 MPa and the yield strength is 400 MPa. Thebending moment at the pulley varies from – 150 N-m to + 400 N-m as the torque on the shaft variesfrom – 50 N-m to + 150 N-m. Obtain the diameter of the shaft for an indefinite life. The stressconcentration factors for the keyway at the pulley in bending and in torsion are 1.6 and 1.3 respectively.Take the following values:

Factor of safety = 1.5Load correction factors = 1.0 in bending, and 0.6 in torsionSize effect factor = 0.85

Surface effect factor = 0.88

Machine parts are often madeof alloys to improve theirmechanical properties.

Variable Stresses in Machine Parts 215Solution. Given : σu = 550 MPa = 550 N/mm2 ; σy = 400 MPa = 400 N/mm2 ;

Mmin = – 150 N-m; Mmax = 400 N-m ; Tmin = – 50 N-m ; Tmax = 150 N-m ; Kfb = 1.6 ; Kfs = 1.3 ;F.S. = 1.5 ; Kb = 1 ; Ks = 0.6 ; Ksz = 0.85 ; Ksur = 0.88

Let d = Diameter of the shaft in mm.

First of all, let us find the equivalent normal stress due to bending.

We know that the mean or average bending moment,

Mm = 3400 ( 150)125 N-m 125 10 N-mm

2 2max minM M+ + −= = = ×


Mv = 3400 ( 150)275 N-m 275 10 N-mm

2 2max minM M− − −= = = ×

Section modulus, Z =3 3 30.0982 mm

32d d

π × =


σm= 3 3

23 3

125 10 1273 10N/mm

0.0982

× ×= =mM

Z d dand variable bending stress,

σv = 3 3

23 3

275 10 2800 10N/mm

0.0982

× ×= =vM

Z d dAssuming the endurance limit in reversed bending as one-half the ultimate strength and since

the load correction factor for reversed bending is 1 (i.e. Kb = 1), therefore endurance limit in reversedbending,

σeb = σe = 2550275 N/mm

2 2uσ

= =

Since there is no reversed axial loading, therefore equivalent normal stress due to bending,

σneb = σne = σm + v y fb

eb sur sz

K

K K

σ × σ ×σ × ×

=3 3

3 3

1273 10 2800 10 400 1.6

275 0.88 0.85d d

× × × ×+× × ×

=3 3 3

23 3 3

1273 10 8712 10 9985 10N/mm

d d d

× × ×+ =

Now let us find the equivalent shear stress due to torsional moment. We know that the meantorque,

Tm = 3150 ( 50)50 N-m 50 10 N-mm

2 2max minT T+ + −= = = ×

and variable torque, Tv = 3150 ( 50)100 N-m 100 10 N-mm

2 2max minT T− − −= = = ×

∴ Mean shear stress,

τm =3 3

23 3 3

16 16 50 10 255 10N/mmmT

d d d

× × ×= =π π


and variable shear stress,

τv =3 3

23 3 3

16 16 100 10 510 10N/mmvT

d d d

× × ×= =π π

Endurance limit stress for reversed torsional or shear loading,τe = σe × Ks = 275 × 0.6 = 165 N/mm2

Assuming yield strength in shear,τy = 0.5 σy = 0.5 × 400 = 200 N/mm2

We know that equivalent shear stress,

τes = τm + v y fs

e sur sz

K

K K

τ × τ ×τ × ×

=3 3

3 3

255 10 510 10 200 1.3

165 0.88 0.85d d

× × × ×+× × ×

=3 3 3

23 3 3

255 10 1074 10 1329 10N/mm

d d d

× × ×+ =and maximum equivalent shear stress,

τes(max) = 2 21( ) 4 ( )

. . 2y

ne esF S

τ= σ + τ

2 23 3 3

3 3 3

200 1 9985 10 1329 10 5165 104

1.5 2 d d d

⎛ ⎞ ⎛ ⎞× × ×= + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∴ d 3 =35165 10 1.5

38 740 or 33.84 say 35 mm200

d× × = = Ans.

6.23 Application of Soderberg’s EquationWe have seen in Art. 6.21 that according to Soderberg's equation,

1

. .v fm

y e

K

F S

σ ×σ= +

σ σ...(i)

This equation may also be written as

1

. .m e v y f

y e

K

F S

σ × σ + σ × σ ×=

σ × σ

or F.S. =y e y

ym e v y fm f v

e

KK

σ × σ σ=

σσ × σ + σ × σ × ⎛ ⎞σ + × σ⎜ ⎟σ⎝ ⎠

...(ii)

Since the factor of safety based on yield strength is the ratio of the yield point stress to theworking or design stress, therefore from equation (ii), we may write

Working or design stress

= σm + y

f ve

Kσ⎛ ⎞

× σ⎜ ⎟σ⎝ ⎠...(iii)

Let us now consider the use of Soderberg's equation to a ductile material under the followingloading conditions.

1. Axial loadingIn case of axial loading, we know that the mean or average stress,

σm = Wm / A

Variable Stresses in Machine Parts 217and variable stress, σv = Wv / A

where Wm = Mean or average load,

Wv = Variable load, and

A = Cross-sectional area.

The equation (iii) may now be written as follows :

Working or design stress,

=

ym f v

y em vf

e

W K WW W

KA A A

σ⎛ ⎞+ ×⎜ ⎟σ σ⎛ ⎞ ⎝ ⎠+ × =⎜ ⎟σ⎝ ⎠

∴ F.S. =y

ym f v

e

A

W K W

σ ×σ⎛ ⎞

+ ×⎜ ⎟σ⎝ ⎠2. Simple bending

In case of simple bending, we know that thebending stress,

σb =.M y M

I Z= ...

⎛ ⎞=⎜ ⎟⎝ ⎠Q

IZ

y


σm = Mm / Z


σv = Mv/ Z

where Mm = Mean bending moment,

Mv = Variable bending moment, and

Z = Section modulus.

The equation (iii) may now be written asfollows :

Working or design bending stress,

σb =σ⎛ ⎞

+ ×⎜ ⎟σ⎝ ⎠

ym vf

e

M MK

Z Z

σ⎛ ⎞+ ×⎜ ⎟σ⎝ ⎠=

ym f v

e

M K M

Z

= 3

32 ⎡ σ ⎤⎛ ⎞+ ×⎢ ⎥⎜ ⎟σπ ⎢ ⎥⎝ ⎠⎣ ⎦

ym f v

e

M K Md

... 3For circular shafts,32

π⎛ ⎞= ×⎜ ⎟⎝ ⎠Q Z d

∴ F.S. =

3

32

y

ym f v

e

M K Md

σ

⎡ σ ⎤⎛ ⎞+ ×⎢ ⎥⎜ ⎟σπ ⎝ ⎠⎣ ⎦

Note : This picture is given as additional informationand is not a direct example of the current chapter.

A large disc- shaped electromagnet hangs fromjib of this scrapyard crane. Steel and iron objectsfly towards the magnet when the current isswitched on. In this way, iron and steel can beseparated for recycling.


3. Simple torsion of circular shaftsIn case of simple torsion, we know that the torque,

T = 33

16or

16

Td

d

π × τ × τ =π

∴ Mean or average shear stress,

τm = 3

16 mT

dπ

and variable shear stress, τv = 3

16 vT

dπwhere Tm = Mean or average torque,

Tv = Variable torque, and

d = Diameter of the shaft.

The equation (iii) may now be written as follows :

Working or design shear stress,

τ = 3 3 3

16 16 16τ ⎡ τ ⎤⎛ ⎞ ⎛ ⎞+ × = + ×⎢ ⎥⎜ ⎟ ⎜ ⎟τ τπ π π ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

y ym vfs m fs v

e e

T TK T K T

d d d

∴ F.S. =

3

16

y

ym fs v

e

T K Td

τ⎡ τ ⎤⎛ ⎞

+ ×⎢ ⎥⎜ ⎟τ⎝ ⎠π ⎣ ⎦where Kfs = Fatigue stress concentration factor for torsional or shear loading.

Note : For shafts made of ductile material, τy = 0.5 σy, and τe = 0.5 σe may be taken.

4. Combined bending and torsion of circular shaftsIn case of combined bending and torsion of circular shafts, the maximum shear stress theory

may be used. According to this theory, maximum shear stress,

τmax =2 21

( ) 4. . 2y

bF S

τ= σ + τ

=

2 2

3 3

1 32 164

2

⎡ ⎤ ⎡ ⎤⎧ σ ⎫ ⎧ τ ⎫⎛ ⎞ ⎛ ⎞⎪ ⎪ ⎪ ⎪+ × + + ×⎢ ⎥ ⎢ ⎥⎨ ⎬ ⎨ ⎬⎜ ⎟ ⎜ ⎟σ τπ π⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎩ ⎭ ⎩ ⎭⎣ ⎦ ⎣ ⎦

y ym f v m fs v

e e

M K M T K Td d

=

2 2

3

16 y ym f v m fs v

e e

M K M T K Td

⎡ σ ⎤ ⎡ τ ⎤⎛ ⎞ ⎛ ⎞+ × + + ×⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟σ τπ ⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

The majority of rotating shafts carry a steady torque and the loads remain fixed in space in bothdirection and magnitude. Thus during each revolution every fibre on the surface of the shaft under-goes a complete reversal of stress due to bending moment. Therefore for the usual case when Mm = 0,Mv = M, Tm = T and Tv = 0, the above equation may be written as

2

23

16

. .y y

fe

K M TF S d

τ ⎡ σ ⎤⎛ ⎞= × +⎢ ⎥⎜ ⎟σ⎝ ⎠π ⎣ ⎦

Note: The above relations apply to a solid shaft. For hollow shaft, the left hand side of the above equations mustbe multiplied by (1 – k4), where k is the ratio of inner diameter to outer diameter.

Variable Stresses in Machine Parts 219Example 6.15. A centrifugal blower rotates at 600 r.p.m. A belt drive is used to connect the

blower to a 15 kW and 1750 r.p.m. electric motor. The belt forces a torque of 250 N-m and a force of2500 N on the shaft. Fig. 6.20 shows the location of bearings, the steps in the shaft and the plane inwhich the resultant belt force and torque act. The ratio of the journal diameter to the overhung shaftdiameter is 1.2 and the radius of the fillet is 1/10th of overhung shaft diameter. Find the shaft diameter,journal diameter and radius of fillet to have a factor of safety 3. The blower shaft is to be machinedfrom hot rolled steel having the following values of stresses:

Endurance limit = 180 MPa; Yield point stress = 300 MPa; Ultimate tensile stress = 450 MPa.

Solution. Given: *NB = 600 r.p.m. ; *P = 15 kW; *NM = 1750 r.p.m. ; T = 250 N-m = 250 × 103

N-mm; F = 2500 N ; F.S. = 3; σe = 180 MPa = 180 N/mm2 ; σy = 300 MPa = 300 N/mm2 ; σu = 450MPa = 450 N/mm2

Fig. 6.20

Let D = Journal diameter,

d = Shaft diameter, and r = Fillet radius.

∴ Ratio of journal diameter to shaft diameter,

D/d = 1.2 ...(Given)

and radius of the fillet, r = 1/10 × Shaft diameter (d) = 0.1 d

∴ r/d = 0.1 ...(Given)

From Table 6.3, for D/d = 1.2 and r/d = 0.1, the theoretical stress concentration factor,

Kt = 1.62

The two points at which failure may occur are at the end of the keyway and at the shoulder fillet.The critical section will be the one with larger product of Kf × M. Since the notch sensitivity factor qis dependent upon the unknown dimensions of the notch and since the curves for notch sensitivityfactor (Fig. 6.14) are not applicable to keyways, therefore the product Kt × M shall be the basis ofcomparison for the two sections.

∴ Bending moment at the end of the keyway,

Kt × M = 1.6 × 2500 [100 – (25 + 10)] = 260 × 103 N-mm

...(Q Kt for key ways = 1.6)

and bending moment at the shoulder fillet,

Kt × M = 1.62 × 2500 (100 – 25) = 303 750 N-mm

Since Kt × M at the shoulder fillet is large, therefore considering the shoulder fillet as the criticalsection. We know that

2

23

16

. .y y

fe

K M TF S d

τ ⎡ σ ⎤⎛ ⎞= × +⎢ ⎥⎜ ⎟σπ ⎢ ⎥⎝ ⎠⎣ ⎦

* Superfluous data


23 2

3

0.5 300 16 300303750 (250 10 )

3 180

⎡× ⎛ ⎞= × + ×⎢⎜ ⎟⎝ ⎠⎣πd

... (Substituting, τy = 0.5 σy)

50 =3

33 3

16 2877 10565 10

d d

×× × =π

∴ d 3 = 2877 × 103/50 = 57 540 or d = 38.6 say 40 mm Ans.Note: Since r is known (because r/d = 0.1 or r = 0.1d = 4 mm), therefore from Fig. 6.14, the notch sensitivityfactor (q) may be obtained. For r = 4 mm, we have q = 0.93.

∴ Fatigue stress concentration factor,

Kf = 1 + q (Kt – 1) = 1 + 0.93 (1.62 – 1) = 1.58

Using this value of Kf instead of Kt, a new value of d may be calculated. We see that magnitudes of Kf andKt are very close, therefore recalculation will not give any improvement in the results already obtained.

EEEEEXEXEXEXEXERRRRRCISECISECISECISECISESSSSS

1. A rectangular plate 50 mm × 10 mm with a hole 10 mm diameter is subjected to an axial load of 10 kN.Taking stress concentration into account, find the maximum stress induced. [Ans. 50 MPa]

2. A stepped shaft has maximum diameter 45 mm and minimum diameter 30 mm. The fillet radius is 6mm. If the shaft is subjected to an axial load of 10 kN, find the maximum stress induced, taking stressconcentration into account. [Ans. 22 MPa]

3. A leaf spring in an automobile is subjected to cyclic stresses. The average stress = 150 MPa; variablestress = 500 MPa; ultimate stress = 630 MPa; yield point stress = 350 MPa and endurance limit = 150MPa. Estimate, under what factor of safety the spring is working, by Goodman and Soderberg formulae.

[Ans. 1.75, 1.3]

4. Determine the design stress for bolts in a cylinder cover where the load is fluctuating due to gaspressure. The maximum load on the bolt is 50 kN and the minimum is 30 kN. The load is unpredict-able and factor of safety is 3. The surface of the bolt is hot rolled and the surface finish factor is 0.9.

During a simple tension test and rotating beam test on ductile materials (40 C 8 steel annealed), thefollowing results were obtained :

Diameter of specimen = 12.5 mm; Yield strength = 240 MPa; Ultimate strength = 450 MPa; Endurancelimit = 180 MPa. [Ans. 65.4 MPa]

5. Determine the diameter of a tensile member of a circular cross-section. The following data is given :

Maximum tensile load = 10 kN; Maximum compressive load = 5 kN; Ultimate tensile strength = 600MPa; Yield point = 380 MPa; Endurance limit = 290 MPa; Factor of safety = 4; Stress concentrationfactor = 2.2 [Ans. 24 mm]

6. Determine the size of a piston rod subjected to a total load of having cyclic fluctuations from 15 kN incompression to 25 kN in tension. The endurance limit is 360 MPa and yield strength is 400 MPa. Takeimpact factor = 1.25, factor of safety = 1.5, surface finish factor = 0.88 and stress concentration factor= 2.25. [Ans. 35.3 mm]

7. A steel connecting rod is subjected to a completely reversed axial load of 160 kN. Suggest the suitablediameter of the rod using a factor of safety 2. The ultimate tensile strength of the material is 1100MPa, and yield strength 930 MPa. Neglect column action and the effect of stress concentration.

[Ans. 30.4 mm]

8. Find the diameter of a shaft made of 37 Mn 2 steel having the ultimate tensile strength as 600 MPa andyield stress as 440 MPa. The shaft is subjected to completely reversed axial load of 200 kN. Neglectstress concentration factor and assume surface finish factor as 0.8. The factor of safety may be takenas 1.5. [Ans. 51.7 mm]

Variable Stresses in Machine Parts 2219. Find the diameter of a shaft to transmit twisting moments varying from 800 N-m to 1600 N-m. The

ultimate tensile strength for the material is 600 MPa and yield stress is 450 MPa. Assume the stressconcentration factor = 1.2, surface finish factor = 0.8 and size factor = 0.85. [Ans. 27.7 mm]

10. A simply supported shaft between bearings carries a steady load of 10 kN at the centre. The length ofshaft between bearings is 450 mm. Neglecting the effect of stress concentration, find the minimumdiameter of shaft. Given that

Endurance limit = 600 MPa; surface finish factor = 0.87; size factor = 0.85; and factor of safety = 1.6.[Ans. 35 mm]

11. Determine the diameter of a circular rod made of ductile material with a fatigue strength (completestress reversal) σe = 280 MPa and a tensile yield strength of 350 MPa. The member is subjected to avarying axial load from 700 kN to – 300 kN. Assume Kt = 1.8 and F.S. = 2. [Ans. 80 mm]

12. A cold drawn steel rod of circular cross-section is subjected to a variable bending moment of 565 N-m to 1130 N-m as the axial load varies from 4500 N to 13 500 N. The maximum bending momentoccurs at the same instant that the axial load is maximum. Determine the required diameter of the rodfor a factor of safety 2. Neglect any stress concentration and column effect. Assume the followingvalues:

Ultimate strength = 550 MPa

Yield strength = 470 MPa

Size factor = 0.85

Surface finish factor = 0.89

Correction factors = 1.0 for bending

= 0.7 for axial load

The endurance limit in reversed bending may be taken as one-half the ultimate strength. [Ans. 41 mm]

13. A steel cantilever beam, as shown in Fig. 6.21, is subjected to a transverse load at its end that variesfrom 45 N up to 135 N down as the axial load varies from 110 N (compression) to 450 N (tension).Determine the required diameter at the change of section for infinite life using a factor of safety of 2.The strength properties are as follows:

Ultimate strength = 550 MPa

Yield strength = 470 MPa

Endurance limit = 275 MPa

Fig. 6.21

The stress concentration factors for bending and axial loads are 1.44 and 1.63 respectively, at thechange of cross-section. Take size factor = 0.85 and surface finish factor = 0.9. [Ans. 12.5 mm]

14. A steel shaft is subjected to completely reversed bending moment of 800 N-m and a cyclic twistingmoment of 500 N-m which varies over a range of ± 40%. Determine the diameter of shaft if a reductionfactor of 1.2 is applied to the variable component of bending stress and shearing stress. Assume


(a) that the maximum bending and shearing stresses are in phase;

(b) that the tensile yield point is the limiting stress for steady state component;

(c) that the maximum shear strength theory can be applied; and

(d) that the Goodman relation is valid.

Take the following material properties:

Yield strength = 500 MPa ; Ultimate strength = 800 MPa ; Endurance limit = ± 400 MPa.

[Ans. 40 mm]

15. A pulley is keyed to a shaft midway between two anti-friction bearings. The bending moment at thepulley varies from – 170 N-m to 510 N-m and the torsional moment in the shaft varies from 55 N-mto 165 N-m. The frequency of the variation of the loads is the same as the shaft speed. The shaft ismade of cold drawn steel having an ultimate strength of 540 MPa and a yield strength of 400 MPa.Determine the required diameter for an indefinite life. The stress concentration factor for the keywayin bending and torsion may be taken as 1.6 and 1.3 respectively. The factor of safety is 1.5. Take sizefactor = 0.85 and surface finish factor = 0.88. [Ans. 36.5 mm]

[Hint. Assume σe = 0.5 σu; τy = 0.5 σy; τe = 0.55 σe]

QQQQQUEUEUEUEUESTSTSTSTSTIONSIONSIONSIONSIONS

1. Explain the following terms in connection with design of machine members subjected to variableloads:

(a) Endurance limit, (b) Size factor,

(c) Surface finish factor, and (d) Notch sensitivity.

2. What is meant by endurance strength of a material? How do the size and surface condition of acomponent and type of load affect such strength?

3. Write a note on the influence of various factors of the endurance limit of a ductile material.

4. What is meant by ̀ stress concentration'? How do you take it into consideration in case of a componentsubjected to dynamic loading?

5. Illustrate how the stress concentration in a component can be reduced.

6. Explain how the factor of safety is determined under steady and varying loading by different methods.

7. Write Soderberg's equation and state its application to different type of loadings.

8. What information do you obtain from Soderberg diagram?

OBJECTOBJECTOBJECTOBJECTOBJECTIVE IVE IVE IVE IVE TTTTT YPYPYPYPYPE QE QE QE QE QUEUEUEUEUESTSTSTSTSTIONSIONSIONSIONSIONS

1. The stress which vary from a minimum value to a maximum value of the same nature (i.e. tensile orcompressive) is called

(a) repeated stress (b) yield stress(c) fluctuating stress (d) alternating stress

2. The endurance or fatigue limit is defined as the maximum value of the stress which a polishedstandard specimen can withstand without failure, for infinite number of cycles, when subjected to(a) static load (b) dynamic load(c) static as well as dynamic load (d) completely reversed load

3. Failure of a material is called fatigue when it fails(a) at the elastic limit (b) below the elastic limit(c) at the yield point (d) below the yield point

Variable Stresses in Machine Parts 2234. The resistance to fatigue of a material is measured by

(a) elastic limit (b) Young's modulus(c) ultimate tensile strength (d) endurance limit

5. The yield point in static loading is ............... as compared to fatigue loading.(a) higher (b) lower (c) same

6. Factor of safety for fatigue loading is the ratio of(a) elastic limit to the working stress(b) Young's modulus to the ultimate tensile strength(c) endurance limit to the working stress(d) elastic limit to the yield point

7. When a material is subjected to fatigue loading, the ratio of the endurance limit to the ultimatetensile strength is(a) 0.20 (b) 0.35(c) 0.50 (d) 0.65

8. The ratio of endurance limit in shear to the endurance limit in flexure is(a) 0.25 (b) 0.40(c) 0.55 (d) 0.70

9. If the size of a standard specimen for a fatigue testing machine is increased, the endurance limit for thematerial will(a) have same value as that of standard specimen (b) increase (c) decrease

10. The residential compressive stress by way of surface treatment of a machine member subjected tofatigue loading(a) improves the fatigue life (b) deteriorates the fatigue life(c) does not affect the fatigue life (d) immediately fractures the specimen

11. The surface finish factor for a mirror polished material is(a) 0.45 (b) 0.65(c) 0.85 (d) 1

12. Stress concentration factor is defined as the ratio of(a) maximum stress to the endurance limit (b) nominal stress to the endurance limit(c) maximum stress to the nominal stress (d) nominal stress to the maximum stress

13. In static loading, stress concentration is more serious in(a) brittle materials (b) ductile materials(c) brittle as well as ductile materials (d) elastic materials

14. In cyclic loading, stress concentration is more serious in(a) brittle materials (b) ductile materials(c) brittle as well as ductile materials (d) elastic materials

15. The notch sensitivity q is expressed in terms of fatigue stress concentration factor Kf and theoreticalstress concentration factor Kt, as

(a)1

1

++

f

t

K

K (b)1

1

−−

f

t

K

K

(c)1

1

++

t

f

K

K (d)1

1

−−

t

f

K

K

ANSWEANSWEANSWEANSWEANSWERRRRRSSSSS

1. (c) 2. (d) 3. (d) 4. (d) 5. (a)

6. (c) 7. (c) 8. (c) 9. (c) 10. (a)

11. (d) 12. (c) 13. (a) 14. (b) 15. (b)


Pressure Vessels

224

7CHAPTER

7.1 IntroductionThe pressure vessels (i.e. cylinders or tanks) are used

to store fluids under pressure. The fluid being stored mayundergo a change of state inside the pressure vessel as incase of steam boilers or it may combine with other reagentsas in a chemical plant. The pressure vessels are designedwith great care because rupture of a pressure vessel meansan explosion which may cause loss of life and property.The material of pressure vessels may be brittle such as castiron, or ductile such as mild steel.

7.2 Classification of Pressure VesselsThe pressure vessels may be classified as follows:

1. According to the dimensions. The pressurevessels, according to their dimensions, may be classifiedas thin shell or thick shell. If the wall thickness of the shell(t) is less than 1/10 of the diameter of the shell (d), then it iscalled a thin shell. On the other hand, if the wall thickness

1. Introduction.2. Classification of Pressure

Vessels.3. Stresses in a Thin Cylindrical

Shell due to an InternalPressure.

4. Circumferential or HoopStress.

5. Longitudinal Stress.6. Change in Dimensions of a

Thin Cylindrical Shell due toan Internal Pressure.

7. Thin Spherical ShellsSubjected to an InternalPressure.

8. Change in Dimensions of aThin Spherical Shell due toan Internal Pressure.

9. Thick Cylindrical ShellSubjected to an InternalPressure.

10. Compound CylindricalShells.

11. Stresses in CompoundCylindrical Shells.

12. Cylinder Heads and CoverPlates.

Pressure Vessels 225of the shell is greater than 1/10 of thediameter of the shell, then it is said tobe a thick shell. Thin shells are usedin boilers, tanks and pipes, whereasthick shells are used in high pressurecylinders, tanks, gun barrels etc.Note: Another criterion to classify thepressure vessels as thin shell or thick shellis the internal fluid pressure (p) and theallowable stress (σt). If the internal fluidpressure (p) is less than 1/6 of the allowablestress, then it is called a thin shell. On theother hand, if the internal fluid pressure isgreater than 1/6 of the allowable stress, thenit is said to be a thick shell.

2. According to the endconstruction. The pressure vessels,according to the end construction, may be classified as open end or closed end. A simple cylinderwith a piston, such as cylinder of a press is an example of an open end vessel, whereas a tank is anexample of a closed end vessel. In case of vessels having open ends, the circumferential or hoopstresses are induced by the fluid pressure, whereas in case of closed ends, longitudinal stresses inaddition to circumferential stresses are induced.

7.3 Stresses in a Thin Cylindrical Shell due to an Internal PressureThe analysis of stresses induced in a thin cylindrical shell are made on the following

assumptions:

1. The effect of curvature of the cylinder wall is neglected.

2. The tensile stresses are uniformly distributed over the section of the walls.

3. The effect of the restraining action of the heads at the end of the pressure vessel is neglected.

Fig. 7.1. Failure of a cylindrical shell.

When a thin cylindrical shell is subjected to an internal pressure, it is likely to fail in the followingtwo ways:

1. It may fail along the longitudinal section (i.e. circumferentially) splitting the cylinder intotwo troughs, as shown in Fig. 7.1 (a).

2. It may fail across the transverse section (i.e. longitudinally) splitting the cylinder into twocylindrical shells, as shown in Fig. 7.1 (b).

Pressure vessels.


* A section cut from a cylinder by a plane that contains the axis is called longitudinal section.

Thus the wall of a cylindrical shell subjected to an internal pressure has to withstand tensilestresses of the following two types:

(a) Circumferential or hoop stress, and (b) Longitudinal stress.

These stresses are discussed, in detail, in the following articles.

7.4 Circumferential or Hoop StressConsider a thin cylindrical shell subjected to an internal pressure as shown in Fig. 7.2 (a) and

(b). A tensile stress acting in a direction tangential to the circumference is called circumferential orhoop stress. In other words, it is a tensile stress on *longitudinal section (or on the cylindrical walls).

Fig. 7.2. Circumferential or hoop stress.

Let p = Intensity of internal pressure,d = Internal diameter of the cylindrical shell,l = Length of the cylindrical shell,t = Thickness of the cylindrical shell, and

σt1 = Circumferential or hoop stress for the material of thecylindrical shell.

We know that the total force acting on a longitudinal section (i.e. along the diameter X-X) of theshell

= Intensity of pressure × Projected area = p × d × l ...(i)and the total resisting force acting on the cylinder walls

= σt1 × 2t × l ...(Q of two sections) ...(ii)From equations (i) and (ii), we have

σt1 × 2t × l = p × d × l or 1 2tp d

t

×σ = or 12 t

p dt

×=σ

...(iii)

The following points may be noted:1. In the design of engine cylinders, a value of 6 mm to 12 mm is added in equation (iii) to

permit reboring after wear has taken place. Therefore

t =1

6 to 12 mm2 t

p d× +σ

2. In constructing large pressure vessels like steam boilers, riveted joints or welded joints areused in joining together the ends of steel plates. In case of riveted joints, the wall thicknessof the cylinder,

t =12 t l

p d×σ × η

where ηl = Efficiency of the longitudinal riveted joint.

Pressure Vessels 227

* A section cut from a cylinder by a plane at right angles to the axis of the cylinder is called transversesection.

3. In case of cylinders of ductile material, the value of circumferential stress (σt1) may be taken0.8 times the yield point stress (σy) and for brittle materials, σt1 may be taken as 0.125 timesthe ultimate tensile stress (σu).

4. In designing steam boilers, the wall thickness calculated by the above equation may becompared with the minimum plate thickness as provided in boiler code as given in thefollowing table.

Table 7.1. Minimum plate thickness for steam boilers.

Boiler diameter Minimum plate thickness (t)

0.9 m or less 6 mm

Above 0.9 m and upto 1.35 m 7.5 mm

Above 1.35 m and upto 1.8 m 9 mm

Over 1.8 m 12 mm

Note: If the calculated value of t is less than the code requirement, then the latter should be taken, otherwise thecalculated value may be used.

The boiler code also provides that the factor of safety shall be at least 5 and the steel of theplates and rivets shall have as a minimum the following ultimate stresses.

Tensile stress, σt = 385 MPa

Compressive stress, σc = 665 MPa

Shear stress, τ = 308 MPa

7.5 Longitudinal StressConsider a closed thin cylindrical shell subjected to an internal pressure as shown in Fig. 7.3 (a)

and (b). A tensile stress acting in the direction of the axis is called longitudinal stress. In other words,it is a tensile stress acting on the *transverse or circumferential section Y-Y (or on the ends of thevessel).

Fig. 7.3. Longitudinal stress.

Let σt2 = Longitudinal stress.In this case, the total force acting on the transverse section (i.e. along Y-Y)

= Intensity of pressure × Cross-sectional area

= 2( )4

p dπ× ...(i)

and total resisting force = σt2 × π d.t ...(ii)


From equations (i) and (ii), we have

σt2 × π d.t = 2( )4

p dπ×

∴ σt2 =4

p d

t

× or24 t

p dt

×=σ

If ηc is the efficiency of the circumferential joint, then

t =24 t c

p d×σ × η

From above we see that the longitudinal stress is half of the circumferential or hoop stress.Therefore, the design of a pressure vessel must be based on the maximum stress i.e. hoop stress.

Example 7.1. A thin cylindrical pressure vessel of 1.2 m diameter generates steam at apressure of 1.75 N/mm2. Find the minimum wall thickness, if (a) the longitudinal stress does notexceed 28 MPa; and (b) the circumferential stress does not exceed 42 MPa.

Solution. Given : d = 1.2 m = 1200 mm ; p = 1.75 N/mm2 ; σt2 = 28 MPa = 28 N/mm2 ;σt1 = 42 MPa = 42 N/mm2

(a) When longitudinal stress (σσσσσt2) does not exceed 28 MPaWe know that minimum wall thickness,

t =2

. 1.75 1200

4 4 28t

p d ×=σ × = 18.75 say 20 mm Ans.

(b) When circumferential stress (σσσσσt1) does not exceed 42 MPaWe know that minimum wall thickness,

t =1

. 1.75 120025 mm

2 2 42t

p d ×= =σ ×

Ans.

Example 7.2. A thin cylindrical pressure vessel of 500 mm diameter is subjected to an internalpressure of 2 N/mm2. If the thickness of the vessel is 20 mm, find the hoop stress, longitudinal stressand the maximum shear stress.

Solution. Given : d = 500 mm ; p = 2 N/mm2 ; t = 20 mm

Cylinders and tanks are used to store fluids under pressure.

Pressure Vessels 229Hoop stress

We know that hoop stress,

σt1 =. 2 500

2 2 20

p d

t

×=×

= 25 N/mm2 = 25 MPa Ans.

Longitudinal stressWe know that longitudinal stress,

σt2 =. 2 500

4 4 20

p d

t

×=×

= 12.5 N/mm2 = 12.5 MPa Ans.

Maximum shear stressWe know that according to maximum shear stress theory, the maximum shear stress is one-half

the algebraic difference of the maximum and minimum principal stress. Since the maximum principalstress is the hoop stress (σt1) and minimum principal stress is the longitudinal stress (σt2), thereforemaximum shear stress,

τmax = 1 2– 25 – 12.5

2 2t tσ σ

= = 6.25 N/mm2 = 6.25 MPa Ans.

Example 7.3. An hydraulic control for a straight line motion, as shown in Fig. 7.4, utilises aspherical pressure tank ‘A’ connected to a working cylinder B. The pump maintains a pressure of3 N/mm2 in the tank.

1. If the diameter of pressure tank is 800 mm, determine its thickness for 100% efficiency ofthe joint. Assume the allowable tensile stress as 50 MPa.

Fig. 7.4

2. Determine the diameter of a cast iron cylinder and its thickness to produce an operatingforce F = 25 kN. Assume (i) an allowance of 10 per cent of operating force F for friction in thecylinder and packing, and (ii) a pressure drop of 0.2 N/mm2 between the tank and cylinder. Take safestress for cast iron as 30 MPa.

3. Determine the power output of the cylinder, if the stroke of the piston is 450 mm and the timerequired for the working stroke is 5 seconds.

4. Find the power of the motor, if the working cycle repeats after every 30 seconds and theefficiency of the hydraulic control is 80 percent and that of pump 60 percent.

Solution. Given : p = 3 N/mm2 ; d = 800 mm ; η = 100% = 1 ; σt1 = 50 MPa = 50 N/mm2 ;F = 25 kN = 25 × 103 N ; σtc = 30 MPa = 30 N/mm2 : ηH = 80% = 0.8 ; ηP = 60% = 0.6


1. Thickness of pressure tank

We know that thickness of pressure tank,

t =1

. 3 80024 mm

2 . 2 50 1t

p d ×= =σ η × ×

Ans.

2. Diameter and thickness of cylinder

Let D = Diameter of cylinder, and

t1 = Thickness of cylinder.

Since an allowance of 10 per cent of operating force F is provided for friction in the cylinderand packing, therefore total force to be produced by friction,

F1 = F + 10

100 F = 1.1 F = 1.1 × 25 × 103 = 27 500 N

We know that there is a pressure drop of 0.2 N/mm2 between the tank and cylinder, thereforepressure in the cylinder,

p1 = Pressure in tank – Pressure drop = 3 – 0.2 = 2.8 N/mm2

and total force produced by friction (F1),

27 500 =4

π × D2 × p1 = 0.7854 × D2 × 2.8 = 2.2 D2

∴ D2 = 27 500 / 2.2 = 12 500 or D = 112 mm Ans.We know that thickness of cylinder,

t1 = 1 . 2.8 1125.2 mm

2 2 30tc

p D ×= =σ ×

Ans.

3. Power output of the cylinder

We know that stroke of the piston= 450 mm = 0.45 m ...(Given)

and time required for working stroke= 5 s ...(Given)

∴Distance moved by the piston per second

=0.45

0.09 m5

=

Jacketed pressure vessel.

Pressure Vessels 231We know that work done per second

= Force × Distance moved per second

= 27 500 × 0.09 = 2475 N-m

∴ Power output of the cylinder

= 2475 W = 2.475 kW Ans. ...(Q 1 N-m/s = 1 W)

4. Power of the motorSince the working cycle repeats after every 30 seconds, therefore the power which is to be

produced by the cylinder in 5 seconds is to be provided by the motor in 30 seconds.

∴ Power of the motor

=H P

Power of the cylinder 5 2.475 50.86 kW

30 0.8 0.6 30× = × =

η × η × Ans.

7.6 Change in Dimensions of a Thin Cylindrical Shell due to an InternalPressure

When a thin cylindrical shell is subjected to an internal pressure, there will be an increase in thediameter as well as the length of the shell.

Let l = Length of the cylindrical shell,

d = Diameter of the cylindrical shell,

t = Thickness of the cylindrical shell,

p = Intensity of internal pressure,

E = Young’s modulus for the material of the cylindrical shell, and

μ = Poisson’s ratio.The increase in diameter of the shell due to an internal pressure is given by,

δd =2.

1 –2 . 2

p d

t E

μ⎛ ⎞⎜ ⎟⎝ ⎠

The increase in length of the shell due to an internal pressure is given by,

δl =. . 1

–2 . 2

p d l

t E⎛ ⎞μ⎜ ⎟⎝ ⎠

It may be noted that the increase in diameter and length of the shell will also increase its volume.The increase in volume of the shell due to an internal pressure is given by

δV = Final volume – Original volume = 4

π (d + δd)2 (l + δl) –

4

π × d 2 .l

=4

π (d 2.δl + 2 d.l.δd ) ...(Neglecting small quantities)

Example 7.4. Find the thickness for a tube of internal diameter 100 mm subjected to an internalpressure which is 5/8 of the value of the maximum permissible circumferential stress. Also find theincrease in internal diameter of such a tube when the internal pressure is 90 N/mm2.Take E = 205 kN/mm2 and μ = 0.29. Neglect longitudinal strain.

Solution. Given : p = 5/8 × σt1 = 0.625 σt1 ; d = 100 mm ; p1 = 90 N/mm2 ; E = 205 kN/mm2

= 205 × 103 N/mm2 ; μ = 0.29

Thickness of a tubeWe know that thickness of a tube,

t = 1

1 1

0.625 100.31.25 mm

2 2t

t t

p d σ ×= =

σ σ Ans.


Increase in diameter of a tubeWe know that increase in diameter of a tube,

δd =2 2

13

90 (100) 0.291 – 1 – mm

2 . 2 22 31.25 205 10

p d

t E

μ⎛ ⎞ ⎡ ⎤=⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦× × ×

= 0.07 (1 – 0.145) = 0.06 mm Ans.

7.7 Thin Spherical Shells Subjected to an Internal PressureConsider a thin spherical shell subjected to an internal pressure as shown in Fig. 7.5.

Let V = Storage capacity of the shell,


d = Diameter of the shell,

t = Thickness of the shell,

σt = Permissible tensile stress for theshell material.

In designing thin spherical shells, we have to determine

1. Diameter of the shell, and 2. Thickness of the shell.

1. Diameter of the shellWe know that the storage capacity of the shell,

V =4

3 × π r3 =

6

π × d 3 or

1/36 V

d⎛ ⎞= ⎜ ⎟π⎝ ⎠

2. Thickness of the shellAs a result of the internal pressure, the shell is likely to rupture along the centre of the sphere.

Therefore force tending to rupture the shell along the centre of the sphere or bursting force,

= Pressure × Area = p × 4

π × d 2 ...(i)

and resisting force of the shell

= Stress × Resisting area = σt × π d.t ...(ii)Equating equations (i) and (ii), we have

p × 4

π × d2 = σt × π d.t

or .

4 t

p dt =

σIf η is the efficiency of the circumferential

joints of the spherical shell, then

.

4 .t

p dt =

σ ηExample 7.5. A spherical vessel 3 metre

diameter is subjected to an internal pressure of1.5 N/mm2. Find the thickness of the vessel requiredif the maximum stress is not to exceed 90 MPa. Takeefficiency of the joint as 75%.

Solution. Given: d = 3 m = 3000 mm ;p = 1.5 N/mm2 ; σt = 90 MPa = 90 N/mm2 ; η = 75% = 0.75

Fig. 7.5. Thin spherical shell.

The Trans-Alaska Pipeline carries crude oil 1, 284kilometres through Alaska. The pipeline is 1.2metres in diameter and can transport 318 millionlitres of crude oil a day.

Pressure Vessels 233We know that thickness of the vessel,

t = . 1.5 3000

4 . 4 90 0.75t

p d ×=σ η × ×

= 16.7 say 18 mm Ans.

7.8 Change in Dimensions of a Thin Spherical Shell due to an InternalPressure

Consider a thin spherical shell subjected to an internal pressure as shown in Fig. 7.5.

Let d = Diameter of the spherical shell, t = Thickness of the spherical shell,

p = Intensity of internal pressure, E = Young’s modulus for the material of the spherical shell, and

μ = Poisson’s ratio.Increase in diameter of the spherical shell due to an internal pressure is given by,

δd =2.

4 .

p d

t E (1 – μ) ...(i)

and increase in volume of the spherical shell due to an internal pressure is given by,

δV = Final volume – Original volume = 6

π (d + δd)3 –

6

π × d3

=6

π ( 3 d 2 × δ d ) ...(Neglecting higher terms)

Substituting the value of δd from equation (i), we have2 2 43 .

(1 – ) (1 – )6 4 . 8 .

d p d p dV

t E t E

⎡ ⎤π πδ = μ = μ⎢ ⎥⎣ ⎦

Example 7.6. A seamless spherical shell, 900 mm in diameter and 10 mm thick is being filledwith a fluid under pressure until its volume increases by 150 × 103 mm3. Calculate the pressureexerted by the fluid on the shell, taking modulus of elasticity for the material of the shell as200 kN/mm2 and Poisson’s ratio as 0.3.

Solution. Given : d = 900 mm ; t = 10 mm ; δV = 150 × 103 mm3 ; E = 200 kN/mm2

= 200 × 103 N/mm2 ; μ = 0.3

Let p = Pressure exerted by the fluid on the shell.

We know that the increase in volume of the spherical shell (δV),

150 × 103 =4

8

p d

t E

π (1 – μ) =

4

3

(900)

8 10 200 10

π× × ×

p (1 – 0.3) = 90 190 p

∴ p = 150 × 103/90 190 = 1.66 N/mm2 Ans.

7.9 Thick Cylindrical Shells Subjected to an Internal PressureWhen a cylindrical shell of a pressure vessel, hydraulic cylinder, gunbarrel and a pipe is subjected

to a very high internal fluid pressure, then the walls of the cylinder must be made extremely heavy orthick.

In thin cylindrical shells, we have assumed that the tensile stresses are uniformly distributedover the section of the walls. But in the case of thick wall cylinders as shown in Fig. 7.6 (a), the stressover the section of the walls cannot be assumed to be uniformly distributed. They develop bothtangential and radial stresses with values which are dependent upon the radius of the element underconsideration. The distribution of stress in a thick cylindrical shell is shown in Fig. 7.6 (b) and (c). Wesee that the tangential stress is maximum at the inner surface and minimum at the outer surface of theshell. The radial stress is maximum at the inner surface and zero at the outer surface of the shell.


In the design of thick cylindrical shells, the following equations are mostly used:

1. Lame’s equation; 2. Birnie’s equation; 3. Clavarino’s equation; and 4. Barlow’s equation.

The use of these equations depends upon the type of material used and the end construction.

Fig. 7.6. Stress distribution in thick cylindrical shells subjected to internal pressure.

Let ro = Outer radius of cylindrical shell,

ri = Inner radius of cylindrical shell,

t = Thickness of cylindrical shell = ro – ri,


μ = Poisson’s ratio,

σt = Tangential stress, and

σr = Radial stress.

All the above mentioned equations are now discussed, in detail, as below:

1. Lame’s equation. Assuming that the longitudinal fibres of the cylindrical shell are equallystrained, Lame has shown that the tangential stress at any radius x is,

2 2 2 2

2 2 2 2 2

( ) – ( ) ( ) ( ) –

( ) – ( ) ( ) – ( )i i o o i o i o

to i o i

p r p r r r p p

r r x r r

⎡ ⎤σ = + ⎢ ⎥⎣ ⎦

While designing a tanker, the pressure added by movement of the vehicle also should beconsidered.

Pressure Vessels 235and radial stress at any radius x,

2 2 2 2

2 2 2 2 2

( ) – ( ) ( ) ( ) ––

( ) – ( ) ( ) – ( )i i o o i o i o

ro i o i

p r p r r r p p

r r x r r

⎡ ⎤σ = ⎢ ⎥⎣ ⎦

Since we are concerned with the internal pressure ( pi = p) only, therefore substituting the valueof external pressure, po = 0.

∴ Tangential stress at any radius x,

2 2

2 2 2

( ) ( )1

( ) – ( )i o

to i

p r r

r r x

⎡ ⎤σ = +⎢ ⎥

⎣ ⎦...(i)

and radial stress at any radius x,

2 2

2 2 2

( ) ( )1

( ) – ( )

⎡ ⎤σ = −⎢ ⎥

⎣ ⎦i o

ro i

p r r

r r x...(ii)

We see that the tangential stress is always a tensile stress whereas the radial stress is a compressivestress. We know that the tangential stress is maximum at the inner surface of the shell (i.e. whenx = r i) and it is minimum at the outer surface of the shell (i.e. when x = ro). Substituting the value ofx = ri and x = ro in equation (i), we find that the *maximum tangential stress at the inner surface of theshell,

2 2

( ) 2 2

[( ) ( ) ]

( ) – ( )o i

t maxo i

p r r

r r

+σ =

and minimum tangential stress at the outer surface of the shell,

2

( ) 2 2

2 ( )

( ) – ( )i

t mino i

p r

r rσ =

We also know that the radial stress is maximum at the inner surface of the shell and zero at theouter surface of the shell. Substituting the value of x = ri and x = ro in equation (ii), we find thatmaximum radial stress at the inner surface of the shell,

σr(max) = – p (compressive)

and minimum radial stress at the outer surface of the shell,

σr(min) = 0

In designing a thick cylindrical shell of brittle material (e.g. cast iron, hard steel and castaluminium) with closed or open ends and in accordance with the maximum normal stress theoryfailure, the tangential stress induced in the cylinder wall,

σt = σt(max) = 2 2

2 2

[( ) ( ) ]

( ) – ( )o i

o i

p r r

r r

+

Since ro = ri + t, therefore substituting this value of ro in the above expression, we get2 2

2 2

[( ) ( ) ]

( ) – ( )i i

ti i

p r t r

r t r

+ +σ =

+σt (ri + t)2 – σt (ri)

2 = p (ri + t)2 + p (ri)2

(ri + t)2 (σt – p) = (ri)2 (σt + p)

2

2

( )

–( )i t

ti

r t p

pr

+ σ +=

σ

* The maximum tangential stress is always greater than the internal pressure acting on the shell.


–i t

i t

r t p

r p

+ σ +=

σ or 1–

t

i t

pt

r p

σ ++ =

σ

∴ – 1–

t

i t

pt

r p

σ +=

σor – 1

–t

it

pt r

p

⎡ ⎤σ += ⎢ ⎥σ⎢ ⎥⎣ ⎦

...(iii)

The value of σt for brittle materials may be taken as 0.125 times the ultimate tensilestrength (σu).

We have discussed above the design of a thick cylindrical shell of brittle materials. In case ofcylinders made of ductile material, Lame’s equation is modified according to maximum shear stresstheory.

According to this theory, the maximum shear stress at any point in a strained body is equal toone-half the algebraic difference of the maximum and minimum principal stresses at that point. Weknow that for a thick cylindrical shell,

Maximum principal stress at the inner surface,

σt (max) =2 2

2 2

[( ) ( ) ]

( ) – ( )o i

o i

p r r

r r

+

and minimum principal stress at the outer surface,

σt(min) = – p∴ Maximum shear stress,

2 2

2 2( ) ( )

[( ) ( ) ]– (– )

– ( ) – ( )

2 2

o i

t max t min o imax

p r rp

r r

+σ σ

τ = τ = =2 2 2 2 2

2 2 2 2

[( ) ( ) ] [( ) – ( ) ] 2 ( )

2[( ) – ( ) ] 2[( ) – ( ) ]o i o i o

o i o i

p r r p r r p r

r r r r

+ += =

2

2 2

( )

( ) – ( )i

i i

p r t

r t r

+=

+ ... (Q ro = ri + t)

or τ ( r i + t)2 – τ(ri)2 = p(ri + t)2

(ri + t)2 (τ – p) = τ(ri)2

2

2

( )

–( )i

i

r t

pr

+ τ=τ

–i

i

r t

r p

+ τ=τ or 1

–i

t

r p

τ+ =τ

∴ – 1–i

t

r p

τ=τ

or – 1–it r

p

⎡ ⎤τ= ⎢ ⎥τ⎣ ⎦...(iv)

The value of shear stress (τ) is usually taken as one-half the tensile stress (σt). Therefore theabove expression may be written as

– 1– 2

ti

t

t rp

⎡ ⎤σ= ⎢ ⎥σ⎢ ⎥⎣ ⎦

...(v)

From the above expression, we see that if the internal pressure ( p ) is equal to or greater thanthe allowable working stress (σt or τ), then no thickness of the cylinder wall will prevent failure.Thus, it is impossible to design a cylinder to withstand fluid pressure greater than the allowableworking stress for a given material. This difficulty is overcome by using compound cylinders (SeeArt. 7.10).

Pressure Vessels 2372. Birnie’s equation. In

case of open-end cylinders (suchas pump cylinders, rams, gunbarrels etc.) made of ductilematerial (i.e. low carbon steel,brass, bronze, and aluminiumalloys), the allowable stressescannot be determined by means ofmaximum-stress theory of failure.In such cases, the maximum-straintheory is used. According to thistheory, the failure occurs when thestrain reaches a limiting value andBirnie’s equation for the wallthickness of a cylinder is

(1 – )

– 1– (1 )

ti

t

pt r

p

⎡ ⎤σ + μ= ⎢ ⎥σ + μ⎢ ⎥⎣ ⎦The value of σt may be taken

as 0.8 times the yield point stress(σy).

3. Clavarino’s equation.This equation is also based on themaximum-strain theory of failure,but it is applied to closed-end cyl-inders (or cylinders fitted withheads) made of ductile material.According to this equation, thethickness of a cylinder,

(1 – 2 )– 1

– (1 )t

it

pt r

p

⎡ ⎤σ + μ= ⎢ ⎥σ + μ⎢ ⎥⎣ ⎦

In this case also, the value of σt may be taken as 0.8 σy.

4. Barlow’s equation. This equation is generally used for high pressure oil and gas pipes.According to this equation, the thickness of a cylinder,

t = p.ro / σt

For ductile materials, σt = 0.8 σy and for brittle materials σt = 0.125 σu, where σu is the ultimatestress.

Example 7.7. A cast iron cylinder of internal diameter 200 mm and thickness 50 mm issubjected to a pressure of 5 N/mm2. Calculate the tangential and radial stresses at the inner, middle(radius = 125 mm) and outer surfaces.

Solution. Given : di = 200 mm or ri = 100 mm ; t = 50 mm ; p = 5 N/mm2

We know that outer radius of the cylinder,

ro = ri + t = 100 + 50 = 150 mm

Oil is frequently transported by ships called tankers. The larger tank-ers, such as this Acrco Alaska oil transporter, are known as super-tankers. They can be hundreds of metres long.


Tangential stresses at the inner, middle and outer surfacesWe know that the tangential stress at any radius x,

σt =2 2

2 2 2

( ) ( )1

( ) – ( )i o

o i

p r r

r r x

⎡ ⎤+⎢ ⎥

⎣ ⎦∴ Tangential stress at the inner surface (i.e. when x = ri = 100 mm),

σt(inner) =2 2 2 2

2 2 2 2

[( ) ( ) ] 5 [(150) (100) ]

( ) – ( ) (150) – (100)o i

o i

p r r

r r

+ += = 13 N/mm2 = 13 MPa Ans.

Tangential stress at the middle surface (i.e. when x = 125 mm),

σt(middle) =2 2

2 2 2

5 (100) (150)1

(150) – (100) (125)

⎡ ⎤+⎢ ⎥

⎣ ⎦ = 9.76 N/mm2 = 9.76 MPa Ans.

and tangential stress at the outer surface (i.e. when x = ro = 150 mm),

σt(outer) =2 2

2 2 2 2

2 ( ) 2 5 (100)

( ) – ( ) (150) – (100)i

o i

p r

r r

×= = 8 N/mm2 = 8 MPa Ans.

Radial stresses at the inner, middle and outer surfaces

We know that the radial stress at any radius x,

σr =2 2

2 2 2

( ) ( )1 –

( ) – ( )i o

o i

p r r

r r x

⎡ ⎤⎢ ⎥⎣ ⎦

∴Radial stress at the inner surface (i.e. when x = ri = 100 mm),

σr(inner) = – p = – 5 N/mm2 = 5 MPa (compressive) Ans.

Radial stress at the middle surface (i.e. when x = 125 mm)

σr(middle) =2 2

2 2 2

5 (100) (150)1 –

(150) – (100) (125)

⎡ ⎤⎢ ⎥⎣ ⎦

= – 1.76 N/mm2 = – 1.76 MPa

= 1.76 MPa (compressive) Ans.

and radial stress at the outer surface (i.e. when x = ro = 150 mm),

σr(outer) = 0 Ans.

Example 7.8. A hydraulic press has a maximumcapacity of 1000 kN. The piston diameter is 250 mm.Calculate the wall thickness if the cylinder is made ofmaterial for which the permissible strength may betaken as 80 MPa. This material may be assumed as abrittle material.

Solution. Given : W = 1000 kN = 1000 × 103 N ;d = 250 mm ; σt = 80 MPa = 80 N/mm2

First of all, let us find the pressure inside thecylinder (p). We know that load on the hydraulic press(W),

1000 × 103 =4

π × d 2 × p =

4

π (250)2 p = 49.1 × 103p

∴ p = 1000 × 103/49.1 × 103 = 20.37 N/mm2

Let ri = Inside radius of the cylinder = d / 2 = 125 mm

Hydraulic Press

Pressure Vessels 239We know that wall thickness of the cylinder,

80 20.37– 1 125 – 1 mm

– 80 – 20.37t

it

pt r

p

⎡ ⎤ ⎡ ⎤σ + += =⎢ ⎥ ⎢ ⎥σ⎢ ⎥ ⎣ ⎦⎣ ⎦= 125 (1.297 – 1) = 37 mm Ans.

Example 7.9. A closed-ended cast iron cylinder of 200 mm inside diameter is to carry aninternal pressure of 10 N/mm2 with a permissible stress of 18 MPa. Determine the wall thickness bymeans of Lame’s and the maximum shear stress equations. What result would you use? Give reasonfor your conclusion.

Solution. Given : di = 200 mm or ri = 100 mm ; p = 10 N/mm2 ; σt = 18 MPa = 18 N/mm2

According to Lame’s equation, wall thickness of a cylinder,

80 10– 1 100 – 1 87 mm

– 80 – 10t

it

pt r

p

⎡ ⎤ ⎡ ⎤σ + += = =⎢ ⎥ ⎢ ⎥σ⎢ ⎥ ⎣ ⎦⎣ ⎦

According to maximum shear stress equation, wall thickness of a cylinder,

t = – 1–ir p

⎡ ⎤τ⎢ ⎥τ⎣ ⎦

We have discussed in Art. 7.9 [equation (iv)], that the shear stress (τ) is usually taken one-halfthe tensile stress (σt). In the present case, τ = σt / 2 = 18/2 = 9 N/mm2. Since τ is less than the internalpressure ( p = 10 N/mm2), therefore the expression under the square root will be negative. Thus nothickness can prevent failure of the cylinder. Hence it is impossible to design a cylinder to withstandfluid pressure greater than the allowable working stress for the given material. This difficulty isovercome by using compound cylinders as discussed in Art. 7.10.

Thus, we shall use a cylinder of wall thickness, t = 87 mm Ans.Example 7.10. The cylinder of a portable hydraulic riveter is 220 mm in diameter. The pressure

of the fluid is 14 N/mm2 by gauge. Determine suitable thickness of the cylinder wall assuming thatthe maximum permissible tensile stress is not to exceed 105 MPa.


Since the pressure of the fluid is high, therefore thick cylinder equation is used.

Assuming the material of the cylinder as steel, the thickness of the cylinder wall (t) may beobtained by using Birnie’s equation. We know that

t =(1 – )

– 1– (1 )

ti

t

pr

p

⎡ ⎤σ + μ⎢ ⎥σ + μ⎢ ⎥⎣ ⎦

105 (1 – 0.3) 14110 – 1 16.5 mm

105 – (1 0.3) 14

⎡ ⎤+= =⎢ ⎥+⎣ ⎦ Ans.

...(Taking Poisson’s ratio for steel, μ = 0.3)

Example 7.11. The hydraulic cylinder 400 mm bore operates at a maximum pressure of5 N/mm2. The piston rod is connected to the load and the cylinder to the frame through hinged joints.Design: 1. cylinder, 2. piston rod, 3. hinge pin, and 4. flat end cover.

The allowable tensile stress for cast steel cylinder and end cover is 80 MPa and for piston rodis 60 MPa.

Draw the hydraulic cylinder with piston, piston rod, end cover and O-ring.

Solution. Given : di = 400 mm or ri = 200 mm ; p = 5 N/mm2 ; σt = 80 MPa = 80 N/mm2 ;

σtp = 60 MPa = 60 N/mm2


1. Design of cylinderLet do = Outer diameter of the cylinder.We know that thickness of cylinder,

80 5– 1 200 – 1 mm

– 80 – 5t

it

pt r

p

⎡ ⎤ ⎡ ⎤σ + += =⎢ ⎥ ⎢ ⎥σ⎢ ⎥ ⎣ ⎦⎣ ⎦= 200 (1.06 – 1) = 12 mm Ans.

∴ Outer diameter of the cylinder,do = di + 2t = 400 + 2 × 12 = 424 mm Ans.

2. Design of piston rodLet dp = Diameter of the piston rod.We know that the force acting on the piston rod,

F =4

π (di)

2 p = 4

π (400)2 5 = 628 400 N ...(i)

We also know that the force acting on the piston rod,

F =4

π (di)

2 σtp = 4

π (dp)2 60 = 47.13 (dp)

2 N ...(ii)

From equations (i) and (ii), we have(dp)2 = 628 400/47.13 = 13 333.33 or dp = 115.5 say 116 mm Ans.

3. Design of the hinge pinLet dh = Diameter of the hinge pin of the piston rod.Since the load on the pin is equal to the force acting on the piston rod, and the hinge pin is in

double shear, therefore

F = 2 × 4

π (dh)

2 τ

628 400 = 2 × 4

π (dh)

2 45 = 70.7 (dh)2 ...(Taking τ = 45 N/mm2)

∴ (dh)2 = 628 400 / 70.7 = 8888.3 or dh = 94.3 say 95 mm Ans.

When the cover is hinged to the cylinder, we can use two hinge pins only diametrically oppositeto each other. Thus the diameter of the hinge pins for cover,

dhc =95

2 2hd

= = 47.5 mm Ans.

Fig. 7.7

Pressure Vessels 2414. Design of the flat end cover

Let tc = Thickness of the end cover.

We know that force on the end cover,

F = di × tc × σt

628 400 = 400 × tc × 80 = 32 × 103 tc∴ tc = 628 400 / 32 × 103 = 19.64 say 20 mm Ans.The hydraulic cylinder with piston, piston rod, end cover and O-ring is shown in Fig. 7.7.

7.10 Compound Cylindrical ShellsAccording to Lame’s equation, the thickness of a cylindrical shell is given by

t = – 1–

ti

t

pr

p

⎛ ⎞σ +⎜ ⎟⎜ ⎟σ⎝ ⎠

From this equation, we see that if the internal pressure(p) acting on the shell is equal to or greater than the allowableworking stress (σt) for the material of the shell, then no thicknessof the shell will prevent failure. Thus it is impossible to designa cylinder to withstand internal pressure equal to or greaterthan the allowable working stress.

This difficulty is overcome by inducing an initialcompressive stress on the wall of the cylindrical shell. Thismay be done by the following two methods:

1. By using compound cylindrical shells, and

2. By using the theory of plasticity.

In a compound cylindrical shell, as shown in Fig. 7.8,the outer cylinder (having inside diameter smaller than theoutside diameter of the inner cylinder) is shrunk fit over the inner cylinder by heating and cooling. Oncooling, the contact pressure is developed at the junction of the two cylinders, which inducescompressive tangential stress in the material of the inner cylinder and tensile tangential stress in thematerial of the outer cylinder. When the cylinder is loaded, the compressive stresses are first relievedand then tensile stresses are induced. Thus, a compound cylinder is effective in resisting higherinternal pressure than a single cylinder with the same overall dimensions. The principle of compoundcylinder is used in the design of gun tubes.

In the theory of plasticity, a temporary high internal pressure is applied till the plastic stage isreached near the inside of the cylinder wall. This results in a residual compressive stress upon theremoval of the internal pressure, thereby making the cylinder more effective to withstand a higherinternal pressure.

7.11 Stresses in Compound Cylindrical ShellsFig. 7.9 (a) shows a compound cylindrical shell assembled with a shrink fit. We have discussed

in the previous article that when the outer cylinder is shrunk fit over the inner cylinder, a contactpressure ( p ) is developed at junction of the two cylinders (i.e. at radius r2 ) as shown in Fig. 7.9 (b)and (c). The stresses resulting from this pressure may be easily determined by using Lame’s equation.

According to this equation (See Art. 7.9), the tangential stress at any radius x is

σt =2 2 2 2

2 2 2 2 2

( ) – ( ) ( ) ( ) –

( ) – ( ) ( ) – ( )i i o o i o i o

o i o i

p r p r r r p p

r r x r r

⎡ ⎤+ ⎢ ⎥⎣ ⎦

...(i)

Fig. 7.8. Compound cylindrical shell.



σr = 2 2 2 2

2 2 2 2 2

( ) – ( ) ( ) ( ) ––

( ) – ( ) ( ) – ( )i i o o i o i o

o i o i

p r p r r r p p

r r x r r

⎡ ⎤⎢ ⎥⎣ ⎦

...(ii)

Considering the external pressure only,

σt =2 2

2 2 2

– ( ) ( )1

( ) – ( )o o i

o i

p r r

r r x

⎡ ⎤+⎢ ⎥

⎣ ⎦...(iii)

...[Substituting pi = 0 in equation (i)]

and σr =2 2

2 2 2

– ( ) ( )1 –

( ) – ( )o o i

o i

p r r

r r x

⎡ ⎤⎢ ⎥⎣ ⎦

...(iv)

Fig. 7.9. Stresses in compound cylindrical shells.

Considering the internal pressure only,

σt =2 2

2 2 2

( ) ( )1

( ) – ( )

⎡ ⎤+⎢ ⎥

⎣ ⎦i i o

o i

p r r

r r x...(v)

...[Substituting po = 0 in equation (i)]


and σr =2 2

2 2 2

( ) ( )1 –

( ) – ( )i i o

o i

p r r

r r x

⎡ ⎤⎢ ⎥⎣ ⎦

...(vi)

Since the inner cylinder is subjected to an external pressure (p) caused by the shrink fit and theouter cylinder is subjected to internal pressure (p), therefore from equation (iii), we find that thetangential stress at the inner surface of the inner cylinder,

σt1 =2 2 2

2 1 22 2 2 2 2

2 1 1 2 1

– ( ) ( ) – 2 ( )1

( ) – ( ) ( ) ( ) – ( )

p r r p r

r r r r r

⎡ ⎤+ =⎢ ⎥

⎢ ⎥⎣ ⎦ (compressive) ...(vii)

... [Substituting po = p, x = r1, ro = r2 and ri = r1]

This stress is compressive and is shown by ab in Fig. 7.9 (b).

Radial stress at the inner surface of the inner cylinder,

σr1 =2 2

2 12 2 2

2 1 1

– ( ) ( )1 – 0

( ) – ( ) ( )

p r r

r r r

⎡ ⎤=⎢ ⎥

⎢ ⎥⎣ ⎦...[From equation (iv)]

Similarly from equation (iii), we find that tangential stress at the outer surface of the innercylinder,

σt2 =2 2 2 2

2 1 2 12 2 2 2 2

2 1 2 2 1

– ( ) ( ) – [( ) ( ) ]1

( ) – ( ) ( ) ( ) – ( )

p r r p r r

r r r r r

⎡ ⎤ ++ =⎢ ⎥⎢ ⎥⎣ ⎦

(compressive) ...(viii)

...[Substituting po = p, x = r2, ro = r2 and ri = r1]

This stress is compressive and is shown by cd in Fig. 7.9 (b).

Radial stress at the outer surface of the inner cylinder,

σr2 =2 2

2 12 2 2

2 1 2

– ( ) ( )1 – –

( ) – ( ) ( )

p r rp

r r r

⎡ ⎤=⎢ ⎥

⎢ ⎥⎣ ⎦

Air outSubmarine

Ballast tanks

Air pumpAir in

Water out

Water in

Submarines consist of an airtight compartment surrounded by ballast tanks. The submarine dives byfilling these tanks with water or air. Its neutral buoyancy ensures that it neither floats nor sinks.



Now let us consider the outer cylinder subjected to internal pressure ( p). From equation (v),we find that the tangential stress at the inner surface of the outer cylinder,

σt3 =2 2 22

3 3 222 2 2 2 2

3 2 2 3 2

( ) [( ) ( ) ]( )1

( ) – ( ) ( ) ( ) – ( )

r p r rp r

r r r r r

⎡ ⎤ ++ =⎢ ⎥

⎢ ⎥⎣ ⎦ (tensile) ...(ix)

...[Substituting pi = p, x = r2, ro = r3 and ri = r2]This stress is tensile and is shown by ce in Fig. 7.9 (c).Radial stress at the inner surface of the outer cylinder,

σr3 =22

322 2 2

3 2 2

( )( )1 – –

( ) – ( ) ( )

rp rp

r r r

⎡ ⎤=⎢ ⎥

⎢ ⎥⎣ ⎦...[From equation (vi)]

Similarly from equation (v), we find that the tangential stress at the outer surface of the outercylinder,

σt4 =22 2

32 22 2 2 2 2

3 2 3 3 2

( )( ) 2 ( )1

( ) – ( ) ( ) ( ) – ( )

rp r p r

r r r r r

⎡ ⎤+ =⎢ ⎥

⎢ ⎥⎣ ⎦ (tensile) ...(x)

...[Substituting pi = p, x = r3, ro = r3 and ri = r2]

This stress is tensile and is shown by fg in Fig. 7.9 (c).

Radial stress at the outer surface of the outer cylinder,

σr4 =22

322 2 2

3 2 3

( )( )1 – 0

( ) – ( ) ( )

rp r

r r r

⎡ ⎤=⎢ ⎥

⎢ ⎥⎣ ⎦

The equations (vii) to (x) cannot be solved until the contact pressure ( p) is known. In obtaininga shrink fit, the outsidediameter of the inner cylinderis made larger than the insidediameter of the outer cylinder.This difference in diameters iscalled the interference and isthe deformation which the twocylinders must experience.Since the diameters of thecylinders are usually known,therefore the deformationshould be calculated to findthe contact pressure.

Let δo = Increase in inner radius of the outer cylinder,δi = Decrease in outer radius of the inner cylinder,

Eo = Young’s modulus for the material of the outer cylinder,Ei = Young’s modulus for the material of the inner cylinder, andμ = Poisson’s ratio.

We know that the tangential strain in the outer cylinder at the inner radius (r2),

εto = 2 2

2 2

2 ( ) – 2Change in circumference

Original circumference 2o or r

r r

π + δ π δ= =

π...(xi)

Also the tangential strain in the outer cylinder at the inner radius (r2),

εto =.

–to ro

o oE E

σ μ σ...(xii)

Submarine is akin a to pressure vessel. CAD and CAM were used todesign and manufacture this French submarine.

Pressure Vessels 245We have discussed above that the tangential stress at the inner surface of the outer cylinder (or

at the contact surfaces),

σto = σt3 = 2 2

3 22 2

3 2

[( ) ( ) ]

( ) – ( )

p r r

r r

+...[From equation (ix)]

and radial stress at the inner surface of the outer cylinder (or at the contact surfaces),

σro = σr3 = – p

Substituting the value of σto and σro in equation (xii), we get

εto =2 2 2 2

3 2 3 22 2 2 2

3 2 3 2

[( ) ( ) ] ( ) ( ).

[( ) – ( ) ] ( ) – ( )o oo

p r r r rp p

E EE r r r r

⎡ ⎤+ +μ+ = + μ⎢ ⎥⎢ ⎥⎣ ⎦

...(xiii)

From equations (xi) and (xiii),

δo =2 2

3 222 2

3 2

( ) ( ).

( ) – ( )o

r rp r

E r r

⎡ ⎤++ μ⎢ ⎥

⎢ ⎥⎣ ⎦...(xiv)

Similarly, we may find that the decrease in the outer radius of the inner cylinder,

δi =2 2

2 2 12 2

2 1

– . ( ) ( )–

( ) – ( )i

p r r r

E r r

⎡ ⎤+μ⎢ ⎥

⎢ ⎥⎣ ⎦...(xv)

∴Difference in radius,

δr = δo – δi = 2 2 2 2

3 22 2 2 12 2 2 2

3 2 2 1

( ) ( ). . ( ) ( )–

( ) – ( ) ( ) – ( )o i

r rp r p r r r

E Er r r r

⎡ ⎤ ⎡ ⎤+ ++ μ + μ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦

If both the cylinders are of the same material, then Eo = Ei = E. Thus the above expression maybe written as

2 2 2 23 22 2 1

2 2 2 23 2 2 1

( ) ( ). ( ) ( )

( ) – ( ) ( ) – ( )r

r rp r r r

E r r r r

⎡ ⎤+ +δ = +⎢ ⎥

⎢ ⎥⎣ ⎦2 2 2 2 2 2 2 2

3 2 2 1 2 1 3 222 2 2 2

3 2 2 1

[( ) ( ) ] [( ) – ( ) ] [( ) ( ) ] [( ) – ( ) ].

[( ) – ( ) ] [( ) – ( ) ]

r r r r r r r rp r

E r r r r

⎡ ⎤+ + += ⎢ ⎥

⎢ ⎥⎣ ⎦2 2 2

2 3 122 2 2 2

3 2 2 1

2( ) [( ) – ( ) ].

[( ) – ( ) ] [( ) – ( ) ]

r r rp r

E r r r r

⎡ ⎤= ⎢ ⎥

⎢ ⎥⎣ ⎦

or

2 2 2 23 2 2 1

2 2 22 2 3 1

[( ) – ( ) ] [( ) – ( ) ].

2( ) [( ) – ( ) ]r r r r rE

pr r r r

⎡ ⎤δ= ⎢ ⎥

⎢ ⎥⎣ ⎦Substituting this value of p in equations (vii) to (x), we may obtain the tangential stresses at the

various surfaces of the compound cylinder.Now let us consider the compound cylinder subjected to an internal fluid pressure ( pi). We

have discussed above that when the compound cylinder is subjected to internal pressure ( pi ), thenthe tangential stress at any radius (x) is given by

σt =2 2

2 2 2

( ) ( )1

( ) – ( )i i o

o i

p r r

r r x

⎡ ⎤+⎢ ⎥

⎣ ⎦∴Tangential stress at the inner surface of the inner cylinder,

σt5 =2 2 2 2

1 3 3 12 2 2 2 2

3 1 1 3 1

( ) ( ) [( ) ( ) ]1

( ) – ( ) ( ) ( ) – ( )i ip r r p r r

r r r r r

⎡ ⎤ ++ =⎢ ⎥

⎢ ⎥⎣ ⎦ (tensile)

... [Substituting x = r1, ro = r3 and ri = r1]


This stress is tensile and is shown by ab' in Fig. 7.9 (d).

Tangential stress at the outer surface of the inner cylinder or inner surface of the outer cylinder,

σt6 =2 2 2 2 2

1 3 1 3 22 2 2 2 2 2

3 1 2 2 3 1

( ) ( ) ( ) ( ) ( )1

( ) – ( ) ( ) ( ) ( ) – ( )i ip r r p r r r

r r r r r r

⎡ ⎤ ⎡ ⎤++ =⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ (tensile)

... [Substituting x = r2, ro = r3 and ri = r1]

This stress is tensile and is shown by ce' in Fig. 7.9 (d),

and tangential stress at the outer surface of the outer cylinder,

σt7 =2 2 2

1 3 12 2 2 2 2

3 1 3 3 1

( ) ( ) 2 ( )1

( ) – ( ) ( ) ( ) – ( )i ip r r p r

r r r r r

⎡ ⎤+ =⎢ ⎥

⎢ ⎥⎣ ⎦ (tensile)

...[Substituting x = r3, ro = r3 and ri = r1]

This stress is tensile and is shown by fg' in Fig. 7.9 (d).

Now the resultant stress at the inner surface of the compound cylinder,

σti = σt1 + σt5 or ab' – ab

This stress is tensile and is shown by ab'' in Fig. 7.9 (e).

Resultant stress at the outer surface of the inner cylinder

= σt2 + σt6 or ce' – cd or cc'

Resultant stress at the inner surface of the outer cylinder

= σt3 + σt6 or ce + ce' or c'e''

∴Total resultant stress at the mating or contact surface,

σtm = σt2 + σt6 + σt3 + σt6

This stress is tensile and is shown by ce'' in Fig. 7.9 (e),

and resultant stress at the outer surface of the outer cylinder,

σto = σt4 + σt7 or fg + fg'

This stress is tensile and is shown by fg'' in Fig. 7.9 (e).

Example 7.12. The hydraulic press, having a working pressure of water as 16 N/mm2 andexerting a force of 80 kN is required to press materials upto a maximum size of 800 mm × 800 mmand 800 mm high, the stroke length is 80 mm. Design and draw the following parts of the press :1. Design of ram; 2. Cylinder; 3. Pillars; and 4. Gland.

Solution. Given: p = 16 N/mm2 ; F = 80 kN = 80 × 103 N

The hydraulic press is shown in Fig. 7.10.

1. Design of ram

Let dr = Diameter of ram.We know that the maximum force to be exerted by the ram (F),

80 × 103 =4

π (dr)

2 p = 4

π (dr)

2 16 = 12.57 (dr)2

∴ (dr)2 = 80 × 103/12.57 = 6364 or dr = 79.8 say 80 mm Ans.

In case the ram is made hollow in order to reduce its weight, then it can be designed as a thickcylinder subjected to external pressure. We have already discussed in Art. 7.11 that according toLame’s equation, maximum tangential stress (considering external pressure only) is


σt(max) =2 2 2 2

2 2 2 2 2

– ( ) ( ) ( ) ( )1 –

( ) – ( ) ( ) ( ) – ( )o ro ri ro ri

oro ri ro ro ri

p d d d dp

d d d d d

⎡ ⎤ ⎡ ⎤++ =⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ (compressive)

and maximum radial stress,

σr(max) = – po (compressive)

where dro = Outer diameter of ram = dr = 80 mm

dri = Inner diameter of ram, and

po = External pressure = p = 16 N/mm2 ...(Given)

Now according to maximum shear stress theory for ductile materials, maximum shear stress is

τmax =

2 2

2 2( ) ( )

( ) ( )– – (– )

– ( ) – ( )

2 2

ro rio o

t max r max ro ri

d dp p

d d

⎡ ⎤+⎢ ⎥σ σ ⎢ ⎥⎣ ⎦=

=2

2 2

( )–

( ) – ( )ri

oro ri

dp

d d

⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦

Fig. 7.10. Hydraulic press.

Since the maximum shear stress is one-half the maximum principal stress (which is compressive),therefore

σc = 2 τmax = 2 po

2

2 2

( )

( ) – ( )ri

ro ri

d

d d

⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦

The ram is usually made of mild steel for which the compressive stress may be taken as75 N/mm2. Substituting this value of stress in the above expression, we get


75 = 2 × 16

2 2

2 2 2

( ) 32 ( )

(80) – ( ) 6400 – ( )

⎡ ⎤=⎢ ⎥

⎢ ⎥⎣ ⎦

ri ri

ri ri

d d

d d

or

2

2

( ) 752.34

326400 – ( )ri

ri

d

d= =

(dri)2 = 2.34 [6400 – (dri)

2] = 14 976 – 2.34 (dri)2

3.34 (dri)2 = 14 976 or (dri)

2 = 14 976/3.34 = 4484

∴ dri = 67 mm Ans.and dro = dr = 80 mm Ans.2. Design of cylinder

Let dci = Inner diameter of cylinder, and

dco = Outer diameter of cylinder.

Assuming a clearance of 15 mm between the ram and the cylinder bore, therefore innerdiameter of the cylinder,

dci = dro + Clearance = 80 + 15 = 95 mm Ans.The cylinder is usually made of cast iron for which the tensile stress may be taken as 30 N/mm2.

According to Lame’s equation, we know that wall thickness of a cylinder,

t =95 30 16

– 1 – 1 mm2 – 2 30 – 16ci t

t

d p

p

⎡ ⎤ ⎡ ⎤σ + +=⎢ ⎥ ⎢ ⎥σ⎢ ⎥ ⎣ ⎦⎣ ⎦

= 47.5 (1.81 – 1) = 38.5 say 40 mm

In accordance with Bernoulli’s principle, the fast flow of air creates low pressure above the painttube, sucking paint upwards into the air steam.


Knobs to adjustflow of air andpaint

Paint supplytube

Compressedair supply

Paintreservoir

Air supply tubeNozzle

Cutaway ofair supplytube

Reducedpressure

Cutaway ofpaint supplytube

Paint is drawn up thetube and broken intotiny droplets

Pressure Vessels 249and outside diameter of the cylinder,

dco = dci + 2 t = 95 + 2 × 40 = 175 mm Ans.

3. Design of pillars

Let dp = Diameter of the pillar.

The function of the pillars is to support the top plate and to guide the sliding plate. When thematerial is being pressed, the pillars will be under direct tension. Let there are four pillars and the loadis equally shared by these pillars.

∴ Load on each pillar

= 80 × 103/4 = 20 × 103 N ...(i)

We know that load on each pillar

=4

π (dp)

2 σt = 4

π (dp)

2 75 = 58.9 (dp)2 ...(ii)

From equations (i) and (ii),

(dp)2 = 20 × 103/58.9 = 340 or dp = 18.4 mm

From fine series of metric threads, let us adopt the threads on pillars as M 20 × 1.5 having majordiameter as 20 mm and core diameter as 18.16 mm. Ans.4. Design of gland

The gland is shown in Fig 7.11. The width (w) of the

U-leather packing for a ram is given empirically as 2 rd

to 2.5 rd , where dr is the diameter (outer) of the ram in mm.

Let us take width of the packing as 2.2 rd .

∴ Width of packing,

w = 2.2 80 = 19.7 say 20 mm Ans.and outer diameter of gland,

DG = dr + 2 w = 80 + 2 × 20 = 120 mm Ans.We know that total upward load on the gland

= Area of gland exposed to fluid pressure × Fluid pressure

= π (dr + w) w.p = π (80 + 20) 20 × 16 = 100 544 N

Let us assume that 8 studs equally spaced on the pitch circle of the gland flange are used forholding down the gland.

∴ Load on each stud = 100 544 / 8 = 12 568 N

If dc is the core diameter of the stud and σt is the permissible tensile stress for the stud material,then

Load on each stud,

12 568 =4

π (dc)

2 σt = 4

π (dc)

2 75 = 58.9 (dc)2 ... (Taking σt = 75 N/mm2)

∴ (dc)2 = 12 568 / 58.9 = 213.4 or dc = 14.6 mm

From fine series of metric threads, let us adopt the studs of size M 18 × 1.5 having majordiameter as 18 mm and core diameter (dc) as 16.16 mm. Ans.

The other dimensions for the gland are taken as follows:

Fig. 7.11


Pitch circle diameter of the gland flange,

P.C.D. = DG + 3 dc = 120 + 3 × 16.16 = 168.48 or 168.5 mm Ans.

Outer diameter of the gland flange,

DF = DG + 6 dc = 120 + 6 × 16.16 = 216.96 or 217 mm Ans.

and thickness of the gland flange = 1.5 dc = 1.5 × 16.16 = 24.24 or 24.5 mm Ans.

Example 7.13. A steel tube 240 mm external diameter is shrunk on another steel tube of 80 mminternal diameter. After shrinking, the diameter at the junction is 160 mm. Before shrinking, thedifference of diameters at the junction was 0.08 mm. If the Young’s modulus for steel is 200 GPa,find: 1. tangential stress at the outer surface of the inner tube; 2. tangential stress at the innersurface of the outer tube ; and 3. radial stress at the junction.

Solution. Given: d3 = 240 mm or r3 = 120 mm ; d1 = 80 mm or r1 = 40 mm ; d2 = 160 mm orr2 = 80 mm ; δd = 0.08 mm or δr = 0.04 mm ; E = 200 GPa = 200 kN/mm2 = 200 × 103 N/mm2

First of all, let us find the pressure developed at the junction. We know that the pressuredeveloped at the junction,

p =2 2 2 2

3 2 2 12 2 2

2 2 3 1

[( ) – ( ) ] [( ) – ( ) ].

2( ) [( ) – ( ) ]r r r r rE

r r r r

⎡ ⎤δ⎢ ⎥⎢ ⎥⎣ ⎦

=3 2 2 2 2

2 2 2

200 10 0.04 [(120) – (80) ] [(80) – (40) ]

80 2 (80) [(120) – (40) ]

⎡ ⎤× ×⎢ ⎥

×⎣ ⎦

= 100 × 0.234 = 23.4 N/mm2 Ans.1. Tangential stress at the outer surface of the inner tube

We know that the tangential stress at the outer surface of the inner tube,

σti =2 2 2 2

22 12 2 2 2

2 1

– [( ) ( ) ] – 23.4 [(80) (40) ]– 39 N/mm

( ) – ( ) (80) – (40)

p r r

r r

+ += =

= 39 MPa (compressive) Ans.

A long oil tank.

Pressure Vessels 2512. Tangential stress at the inner surface of the outer tube

We know that the tangential stress at the inner surface of the outer tube,

σto =2 2 2 2

23 22 2 2 2

3 2

– [( ) ( ) ] 23.4 [(120) (80) ]60.84 N/mm

( ) – ( ) (120) – (80)

p r r

r r

+ += =

= 60.84 MPa Ans.3. Radial stress at the junction

We know that the radial stress at the junction, (i.e. at the inner radius of the outer tube),

σro = – p = – 23.4 N/mm2 = 23.4 MPa (compressive) Ans.Example 7.14. A shrink fit assembly, formed by shrinking one tube over another, is subjected to

an internal pressure of 60 N/mm2. Before the fluid is admitted, the internal and the external diametersof the assembly are 120 mm and 200 mm and the diameter at the junction is 160 mm. If after shrinkingon, the contact pressure at the junction is 8 N/mm2, determine using Lame’s equations, the stresses atthe inner, mating and outer surfaces of the assembly after the fluid has been admitted.

Solution. Given : pi = 60 N/mm2 ; d1 = 120 mm or r1 = 60 mm ; d3 = 200 mm or r3 = 100 mm ;d2 = 160 mm or r2 = 80 mm ; p = 8 N/mm2

First of all, let us find out the stresses induced in the assembly due to contact pressure at thejunction ( p).

We know that the tangential stress at the inner surface of the inner tube,

σt1 =2 2

22 2 2 2

2 1

– 2 ( ) – 2 8 (80)

( ) – ( ) (80) – (60)

p r

r r

×= = – 36.6 N/mm2

= 36.6 MPa (compressive)

Tangential stress at the outer surface of the inner tube,

σt2 =2 2 2 2

2 12 2 2 2

2 1

– [( ) ( ) ] – 8 [(80) (60) ]

( ) – ( ) (80) – (60)

p r r

r r

+ += = – 28.6 N/mm2

= 28.6 MPa (compressive)

Tangential stress at the inner surface of the outer tube,

σt3 =2 2 2 2

3 22 2 2 2

3 2

[( ) ( ) ] 8 [(100) (80) ]

( ) – ( ) (100) – (80)

p r r

r r

+ += = 36.4 N/mm2

= 36.4 MPa (tensile)

and tangential stress at the outer surface of the outer tube,

σt4 =2 2

22 2 2 2

3 2

2 ( ) 2 8 (80)

( ) – ( ) (100) – (80)

p r

r r

×= = 28.4 N/mm2


Now let us find out the stresses induced in the assembly due to internal fluid pressure ( pi ).

We know that the tangential stress at the inner surface of the inner tube,

σt5 =2 2 2 2

3 12 2 2 2

3 1

[( ) ( ) ] 60 [(100) (60) ]

( ) – ( ) (100) – (60)ip r r

r r

+ += = 127.5 N/mm2



Tangential stress at the outer surface of the inner tube or inner surface of the outer tube (i.e.,mating surface),

σt6 =2 2 2 2 2 2

1 3 22 2 2 2 2 2

2 3 1

( ) ( ) ( ) 60 (60) (100) (80)

( ) ( ) – ( ) (80) (100) – (60)ip r r r

r r r

⎡ ⎤ ⎡ ⎤+ +=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦

= 86.5 N/mm2


and tangential stress at the outer surface of the outer tube,

σt7 =2 2

12 2 2 2

3 1

2 ( ) 2 60 (60)

( ) – ( ) (100) – (60)ip r

r r

×= = 67.5 N/mm2 = 67.5 MPa (tensile)

We know that resultant stress at the inner surface of the assembly

σti = σt1 + σt5 = – 36.6 + 127.5 = 90.9 N/mm2 =90.9 MPa (tensile) Ans.Resultant stress at the outer surface of the inner tube

= σt2 + σt6 = – 28.6 + 86.5 = 57.9 N/mm2 = 57.9 MPa (tensile)

Resultant stress at the inner surface of the outer tube

= σt3 + σt6 = 36.4 + 86.5 = 122.9 N/mm2 = 122.9 MPa (tensile)

∴Total resultant stress at the mating surface of the assembly,

σtm = 57.9 + 122.9 = 180.8 N/mm2 = 180.8 MPa (tensile) Ans.and resultant stress at the outer surface of the assembly,

σto = σt4 + σt7 = 28.4 + 67.5 = 95.9 N/mm2 = 95.9 MPa (tensile) Ans.

7.12 Cylinder Heads and Cover PlatesThe heads of cylindrical pressure vessels and the sides of rectangular or square tanks may have

flat plates or slightly dished plates. The platesmay either be cast integrally with the cylinderwalls or fixed by means of bolts, rivets orwelds. The design of flat plates forming theheads depend upon the following two factors:

(a) Type of connection between thehead and the cylindrical wall, (i.e.freely supported or rigidly fixed);and

(b) Nature of loading (i.e. uniformlydistributed or concentrated).

Since the stress distribution in thecylinder heads and cover plates are ofcomplex nature, therefore empirical relationsbased on the work of Grashof and Bach areused in the design of flat plates. Let usconsider the following cases:

1. Circular flat plate with uniformly distributed load. The thickness (t1 ) of a plate with adiameter (d ) supported at the circumference and subjected to a pressure ( p) uniformly distributedover the area is given by

t1 = k1.d t

p

σwhere σt = Allowable design stress.

This 2500-ton hydraulic press is used to forge machineparts a high temperature.Note : This picture is given as additional information and is

not a direct example of the current chapter.

Pressure Vessels 253The coefficient k1 depends upon the material of the plate and the method of holding the edges.

The values of k1 for the cast iron and mild steel are given in Table 7.2.

2. Circular flat plate loaded centrally. The thickness (t1) of a flat cast iron plate supportedfreely at the circumference with a diameter (d) and subjected to a load (F) distributed uniformly over

an area 4

π (d0)

2, is given by

t1 = 00.673 1 –

t

d F

d⎛ ⎞⎜ ⎟ σ⎝ ⎠

If the plate with the above given type of loading is fixed rigidly around the circumference, then

t1 =0

1.65 loget

F d

d⎛ ⎞⎜ ⎟σ ⎝ ⎠

3. Rectangular flat plate with uniformly distributed load. The thickness (t1) of a rectangularplate subjected to a pressure (p) uniformly distributed over the total area is given by

t1 = a.b.k2 2 2( )t

p

a bσ +where a = Length of the plate; and

b = Width of the plate.

The values of the coefficient k2 are given in Table 7.2.

Table 7.2. Values of coefficients k1, k2, k3 and k4.

Material of the Type of Circular plate Rectangular plate Elliptical

cover plate connection plate

k1 k2 k3 k4

Cast iron Freely supported 0.54 0.75 4.3 1.5

Fixed 0.44 0.62 4.0 1.2

Mild Steel Freely supported 0.42 0.60 3.45 1.2

Fixed 0.35 0.49 3.0 0.9

4. Rectangular flat plate with concentrated load. The thickness (t1) of a rectangular platesubjected to a load (F) at the intersection of the diagonals is given by

t1 = 3 2 2

. .

( )t

a b Fk

a bσ +

The values of coefficient k3 are given in Table 7.2.

5. Elliptical plate with uniformly distributed load. The thickness (t1) of an elliptical platesubjected to a pressure ( p) uniformly distributed over the total area, is given by

t1 = a.b.k4 2 2( )t

p

a bσ +

where a and b = Major and minor axes respectively.

The values of coefficient k4 are given in Table 7.2.


6. Dished head with uniformly distributed load. Let us consider the following cases of dishedhead:

(a) Riveted or welded dished head. When the cylinder head has a dished plate, then thethickness of such a plate that is riveted or welded as shown in Fig. 7.12 (a), is given by

t1 = 4.16 .

u

p R

σwhere p = Pressure inside the cylinder,

R = Inside radius of curvature of the plate, and

σu = Ultimate strength for the material of the plate.

When there is an opening or manhole in the head, then the thickness of the dished plate is given by

t1 =4.8 .

u

p R

σIt may be noted that the inside radius of curvature of the dished plate (R) should not be greater

than the inside diameter of the cylinder (d).

Fig. 7.12. Dished plate with uniformly distributed load.

(b) Integral or welded dished head. When the dished plate is fixed integrally or welded to thecylinder as shown in Fig. 7.12 (b), then the thickness of the dished plate is given by

t1 =2 2( 4 )

16 t

p d c

c

+σ ×

where c = Camber or radius of the dished plate.

Mostly the cylindrical shells are provided with hemispherical heads. Thus for hemispherical

heads, c = 2

d. Substituting the value of c in the above expression, we find that the thickness of the

hemispherical head (fixed integrally or welded),

t1 =

22 4

.4416

2t

t

dp d

p dd

⎛ ⎞+ ×⎜ ⎟

⎝ ⎠ =σσ ×

...(Same as for thin spherical shells)

Pressure Vessels 2557. Unstayed flat plate with uniformly distributed load. The minimum thickness (t1) of an

unstayed steel flat head or cover plate is given by

t1 =.

t

k pd

σ

Fig. 7.13. Types of unstayed flat head and covers.

The following table shows the value of the empirical coefficient (k) for the various types ofplate (or head) connection as shown in Fig. 7.13.

Table 7.3. Values of an empirical coefficient (k).

S.No. Particulars of plate connection Value of ‘k’

1. Plate riveted or bolted rigidly to the shell flange, as shown in 0.162

Fig. 7.13 (a).

2. Integral flat head as shown in Fig. 7.13 (b), d ≤ 600 mm, 0.162

t1 ≥ 0.05 d.

3. Flanged plate attached to the shell by a lap joint as shown in 0.30

Fig. 7.13 (c), r ≥ 3t1.

4. Plate butt welded as shown in Fig. 7.13 (d), r ≥ 3 t2 0.25

5. Integral forged plate as shown in Fig. 7.13 (e), r ≥ 3 t2 0.25

6. Plate fusion welded with fillet weld as shown in Fig. 7.13 ( f ) , 0.50

t2 ≥ 1.25 t3.

7. Bolts tend to dish the plate as shown in Fig. 7.13 (g) and (h). 0.3 + G1.04 .

.

W h

H d,

W = Total bolt load, and

H = Total load onarea bounded bythe outside diameterof the gasket.


Steam engine.

Example 7.15. A cast iron cylinder of inside diameter 160 mm is subjected to a pressure of15 N/mm2. The permissible working stress for the cast iron may be taken as 25 MPa. If the cylinderis closed by a flat head cast integral with the cylinder walls, find the thickness of the cylinder walland the flat head.


Thickness of the cylinder wallWe know that the thickness of the cylinder wall,

t =25 15

– 1 80 – 1 80 mm– 25 – 15

ti

t

pr

p

⎡ ⎤ ⎡ ⎤σ + += =⎢ ⎥ ⎢ ⎥σ⎢ ⎥ ⎣ ⎦⎣ ⎦ Ans.

Thickness of the flat headSince the head is cast integral with the cylinder walls, therefore from Table 7.2, we find that

k1 = 0.44.

∴ Thickness of the flat head,

t1 = k1.d t

p

σ = 0.44 × 160 15

25 = 54.5 say 60 mm Ans.

Example 7.16. The steam chest of asteam engine is covered by a flat rectangularplate of size 240 mm by 380 mm. The plate ismade of cast iron and is subjected to a steampressure of 1.2 N/mm2. If the plate is assumedto be uniformly loaded and freely supportedat the edges, find the thickness of plate foran allowable stress of 35 N/mm2.

Solution. Given: b = 240 m ; a = 380mm ; p = 1.2 N/mm2 ; σt = 35 N/mm2

From Table 7.2, we find that for arectangular plate freely supported, thecoefficient k2 = 0.75.

We know that the thickness of a rectangular plate,

t1 = a.b.k2 2 2( )t

p

a bσ + = 380 × 240 × 0.75 2 2

1.2

35 [(380) (240) ]+

= 68 400 × 0.412 × 10–3 = 28.2 say 30 mm Ans.Example 7.17. Determine the wall thickness and the head thickness required for a 500 mm

fusion-welded steel drum that is to contain ammonia at 6 N/mm2 pressure. The radius of curvature ofthe head is to be 450 mm.

Solution. Given: d = 500 mm ; p = 6 N/mm2 ; R = 450 mmWall thickness for a steel drum

For the chemical pressure vessels, steel Fe 360 is used. The ultimate tensile strength (σu) of thesteel is 360 N/mm2. Assuming a factor of safety (F.S.) as 6, the allowable tensile strength,

σt = 236060 N/mm

. . 6u

F S

σ= =

We know that the wall thickness,

t =. 6 500

25 mm2 2 60t

p d ×= =σ ×

Ans.

Pressure Vessels 257Head thickness for a steel drum

We know that the head thickness,

t1 =4.16 . 4.16 6 450

360u

p R × ×=σ = 31.2 say 32 mm Ans.

Example 7.18. A pressure vessel consists of a cylinder of 1 metre inside diameter and is closedby hemispherical ends. The pressure intensity of the fluid inside the vessel is not to exceed 2 N/mm2.The material of the vessel is steel, whose ultimate strength in tension is 420 MPa. Calculate therequired wall thickness of the cylinder and the thickness of the hemispherical ends, considering afactor of safety of 6. Neglect localised effects at the junction of the cylinder and the hemisphere.

Solution. Given: d = 1 m = 1000 mm ; p = 2 N/mm2 ; σu = 420 MPa = 420 N/mm2 ; F.S. = 6

We know that allowable tensile stress,

σt = 242070 N/mm

. . 6u

F S

σ= =

Wall thickness of the cylinderWe know that wall thickness of the cylinder,

t =. 2 1000

2 2 70t

p d ×=σ ×

= 14.3 say 15 mm Ans.

Thickness of hemispherical endsWe know that the thickness of hemispherical ends,

t1 =. 2 1000

4 4 70t

p d ×=σ ×

= 7.15 say 8 mm Ans.

Example 7.19. A cast steel cylinder of 350 mm insidediameter is to contain liquid at a pressure of 13.5 N/mm2. It isclosed at both ends by flat cover plates which are made of alloysteel and are attached by bolts.

1. Determine the wall thickness of the cylinder if themaximum hoop stress in the material is limited to 55 MPa.

2. Calculate the minimum thickness necessary of the coverplates if the working stress is not to exceed 65 MPa.

Solution. Given : di = 350 mm or ri = 175 mm ; p = 13.5N/mm2 ; σt = 55 MPa = 55 N/mm2 ; σt1 = 65 MPa = 65 N/mm2

1. Wall thickness of the cylinderWe know that the wall thickness of the cylinder,

t =55 13.5

– 1 175 – 1– 55 – 13.5

⎡ ⎤ ⎡ ⎤σ + +=⎢ ⎥ ⎢ ⎥σ⎢ ⎥ ⎣ ⎦⎣ ⎦

ti

t

pr

p= 49.8 say 50 mm Ans.

2. Minimum thickness of the cover platesFrom Table 7.3, we find that for a flat cover plate bolted to the shell flange, the value of coeffi-

cient k = 0.162. Therefore, minimum thickness of the cover plates

t1 = di1

. 0.162 13.5350

64t

k p ×=σ = 64.2 say 65 mm Ans.

Steel drums.



1. A steel cylinder of 1 metre diameter is carrying a fluid under a pressure of 10 N/mm2. Calculate thenecessary wall thickness, if the tensile stress is not to exceed 100 MPa. [Ans. 50 mm]

2. A steam boiler, 1.2 metre in diameter, generates steam at a gauge pressure of 0.7 N/mm2. Assumingthe efficiency of the riveted joints as 75% , find the thickness of the shell. Given that ultimate tensilestress = 385 MPa and factor of safety = 5. [Ans. 7.3 mm]

3. Find the thickness of a cast iron cylinder 250 mm in diameter to carry a pressure of 0.7 N/mm2. Takemaximum tensile stress for cast iron as 14 MPa. [Ans. 6.25 mm]

4. A pressure vessel has an internal diameter of 1 m and is to be subjected to an internal pressure of 2.75N/mm2 above the atmospheric pressure. Considering it as a thin cylinder and assuming the efficiencyof its riveted joint to be 79%, calculate the plate thickness if the tensile stress in the material is not toexceed 88 MPa. [Ans. 20 mm]

5. A spherical shell of 800 mm diameter is subjected to an internal pressure of 2 N/mm2. Find thethickness required for the shell if the safe stress is not to exceed 100 MPa. [Ans. 4 mm]

6. A bronze spherical shell of thickness 15 mm is installed in a chemical plant. The shell is subjected toan internal pressure of 1 N/mm2. Find the diameter of the shell, if the permissible stress for the bronzeis 55 MPa. The efficiency may be taken as 80%. [Ans. 2.64 m]

7. The pressure within the cylinder of a hydraulic press is 8.4 N/mm2. The inside diameter of the cylinderis 25.4 mm. Determine the thickness of the cylinder wall, if the allowable tensile stress is 17.5 MPa.

[Ans. 8.7 mm]8. A thick cylindrical shell of internal diameter 150 mm has to withstand an internal fluid pressure of 50

N/mm2. Determine its thickness so that the maximum stress in the section does not exceed 150 MPa.[Ans. 31 mm]

9. A steel tank for shipping gas is to have an inside diameter of 30 mm and a length of 1.2 metres. Thegas pressure is 15 N/mm2. The permissible stress is to be 57.5 MPa. [Ans. 4.5 mm]

10. The ram of a hydraulic press 200 mm internal diameter is subjected to an internal pressure of10 N/mm2. If the maximum stress in the material of the wall is not to exceed 28 MPa, find the externaldiameter. [Ans. 265 mm]

11. The maximum force exerted by a small hydraulic press is 500 kN. The working pressure of the fluidis 20 N/mm2. Determine the diameter of the plunger, operating the table. Also suggest the suitablethickness for the cast steel cylinder in which the plunger operates, if the permissible stress for caststeel is 100 MPa. [Ans. 180 mm ; 20 mm]

12. Find the thickness of the flat end cover plates for a 1 N/mm2 boiler that has a diameter of 600 mm. Thelimiting tensile stress in the boiler shell is 40 MPa. [Ans. 38 mm]

This vessel holds oil at high pressure.



1. What is the pressure vessel ?

2. Make out a systematic classification of pressure vessels and discuss the role of statutory regulations.

3. How do you distinguish between a thick and thin cylinder?

4. What are the important points to be considered while designing a pressure vessel ?

5. Distinguish between circumferential stress and longitudinal stress in a cylindrical shell, when sub-jected to an internal pressure.

6. Show that in case of a thin cylindrical shell subjected to an internal fluid pressure, the tendency toburst lengthwise is twice as great as at a transverse section.

7. When a thin cylinder is subjected to an internal pressure p, the tangential stress should be the criterionfor determining the cylinder wall thickness. Explain.

8. Derive a formula for the thickness of a thin spherical tank subjected to an internal fluid pressure.

9. Compare the stress distribution in a thin and thick walled pressure vessels.

10. When the wall thickness of a pressure vessel is relatively large, the usual assumptions valid in thincylinders do not hold good for its analysis. Enumerate the important violations. List any two theoriessuggested for the analysis of thick cylinders.

11. Discuss the design procedure for pressure vessels subjected to higher external pressure.

12. Explain the various types of ends used for pressure vessel giving practical applications of each.

OBJECTOBJECTOBJECTOBJECTOBJECTIVE IVE IVE IVE IVE TTTTTYPYPYPYPYPE QE QE QE QE QUEUEUEUEUESTSTSTSTSTIONSIONSIONSIONSIONS

1. A pressure vessel is said to be a thin cylindrical shell, if the ratio of the wall thickness of the shell toits diameter is

(a) equal to 1/10 (b) less than 1/10

(c) more than 1/10 (d) none of these

2. In case of pressure vessels having open ends, the fluid pressure induces

(a) longitudinal stress (b) circumferential stress

(c) shear stress (d) none of these

3. The longitudinal stress is ...... of the circumferential stress.

(a) one-half (b) two-third

(c) three-fourth

4. The design of the pressure vessel is based on

(a) longitudinal stress (b) hoop stress

(c) longitudinal and hoop stress (d) none of these

5. A thin spherical shell of internal diameter d is subjected to an internal pressure p. If σt is the tensilestress for the shell material, then thickness of the shell (t) is equal to

(a).

σt

p d(b)

.

2 σt

p d

(c).

3 σt

p d(d)

.

4 σt

p d


6. In case of thick cylinders, the tangential stress across the thickness of cylinder is

(a) maximum at the outer surface and minimum at the inner surface

(b) maximum at the inner surface and minimum at the outer surface

(c) maximum at the inner surface and zero at the outer surface

(d) maximum at the outer surface and zero at the inner surface

7. According to Lame’s equation, the thickness of a cylinder is equal to

(a)(1 – 2 )

– 1– (1 – 2 )

⎡ ⎤σ + μ⎢ ⎥σ μ⎢ ⎥⎣ ⎦

ti

t

pr

p (b)(1 – )

– 1– (1 – )

⎡ ⎤σ + μ⎢ ⎥σ μ⎢ ⎥⎣ ⎦

ti

t

pr

p

(c) – 1–

⎡ ⎤σ +⎢ ⎥σ⎢ ⎥⎣ ⎦

ti

t

pr

p (d) – 1– 2

⎡ ⎤σ⎢ ⎥σ⎢ ⎥⎣ ⎦

ti

tr

p

where

ri = Internal radius of the cylinder,

σt = Allowable tensile stress,

p = Internal fluid pressure, and

μ = Poisson’s ratio.

8. In a thick cylindrical shell, the maximum radial stress at the outer surfaces of the shell is

(a) zero (b) p

(c) – p (d) 2p

9. For high pressure oil and gas cylinders, the thickness of the cylinder is determined by

(a) Lame’s equation (b) Clavarino’s equation

(c) Barlow’s equation (d) Birnie’s equation

10. The thickness of a dished head that is riveted or welded to the cylindrical wall is

(a)4.16 .

σu

p R(b)

5.36 .

σu

p R

(c)6.72 .

σu

p R(d)

8.33 .

σu

p R

where

p = Internal pressure,

R = Inside radius of curvature of the dished plate, and

σu = Ultimate strength for the material of the plate.


1. (b) 2. (b) 3. (a) 4. (b) 5. (b)

6. (b) 7. (c) 8. (a) 9. (c) 10. (a)

Pipes and Pipe Joints 261

Pipes and Pipe Joints

261

1. Introduction.2. Stresses in Pipes.3. Design of Pipes.4. Pipe Joints.5. Standard Pipe Flanges for

Steam.6. Hydraulic Pipe Joint for

High Pressures.7. Design of Circular Flanged

Pipe Joint.8. Design of Oval Flanged Pipe

Joint.9. Design of Square Flanged

Pipe Joint.

8CHAPTER

8.1 IntroductionThe pipes are used for transporting various fluids like

water, steam, different types of gases, oil and other chemicalswith or without pressure from one place to another. Cast iron,wrought iron, steel and brass are the materials generally usedfor pipes in engineering practice. The use of cast iron pipesis limited to pressures of about 0.7 N/mm2 because of itslow resistance to shocks which may be created due to theaction of water hammer. These pipes are best suited for waterand sewage systems. The wrought iron and steel pipes areused chiefly for conveying steam, air and oil. Brass pipes, insmall sizes, finds use in pressure lubrication systems on primemovers. These are made up and threaded to the samestandards as wrought iron and steel pipes. Brass pipe is notliable to corrosion. The pipes used in petroleum industry aregenerally seamless pipes made of heat-resistant chrome-molybdenum alloy steel. Such type of pipes can resistpressures more than 4 N/mm2 and temperatures greater than440°C.


8.2 Stresses in PipesThe stresses in pipes due to the internal fluid pressure are determined by Lame's equation as discussed

in the previous chapter (Art. 7.9). According to Lame's equation, tangential stress at any radius x,

σt =2 2

2 2 2

( ) ( )1

( ) ( )i o

o i

p r r

r r x

⎡ ⎤+⎢ ⎥

− ⎣ ⎦...(i)


σr =2 2

2 2 2

( ) ( )1

( ) ( )i o

o i

p r r

r r x

⎡ ⎤−⎢ ⎥

− ⎣ ⎦...(ii)

where p = Internal fluid pressure in the pipe,ri = Inner radius of the pipe, andro = Outer radius of the pipe.

The tangential stress is maximum at the inner surface (when x = ri) of the pipe and minimum atthe outer surface (when x = ro) of the pipe.

Substituting the values of x = ri and x = ro inequation (i), we find that the maximum tangentialstress at the inner surface of the pipe,

σt(max) =2 2

2 2

[( ) ( ) ]

( ) ( )o i

o i

p r r

r r

+−

and minimum tangential stress at the outer surfaceof the pipe,

σt(min) =2

2 2

2 ( )

( ) ( )i

o i

p r

r r−The radial stress is maximum at the inner

surface of the pipe and zero at the outer surface ofthe pipe. Substituting the values of x = ri and x = roin equation (ii), we find that maximum radial stressat the inner surface,

σr(max) =– p (compressive)

and minimum radial stress at the outer surface of the pipe,

σr(min) =0

The thick cylindrical formula may be applied when

(a) the variation of stress across the thickness of the pipe is taken into account,

(b) the internal diameter of the pipe (D) is less than twenty times its wall thickness ( t ) , i.e.D/t < 20, and

(c) the allowable stress (σt) is less than six times the pressure inside the pipe ( p ) i.e.σt / p < 6.

According to thick cylindrical formula (Lame's equation), wall thickness of pipe,

t = R 1t

t

p

p

⎡ ⎤σ +−⎢ ⎥σ −⎢ ⎥⎣ ⎦

where R = Internal radius of the pipe.

The following table shows the values of allowable tensile stress (σ t) to be used in the aboverelations:

Cast iron pipes.


Table 8.1. Values of allowable tensile stress for pipes of different materials.

S.No. Pipes Allowable tensile stress (σt ) in MPa or N/mm2

1. Cast iron steam or water pipes 14

2. Cast iron steam engine cylinders 12.5

3. Lap welded wrought iron tubes 60

4. Solid drawn steel tubes 140

5. Copper steam pipes 25

6. Lead pipes 1.6

Example 8.1. A cast iron pipe of internal diameter 200 mm and thickness 50 mm carries waterunder a pressure of 5 N/mm2. Calculate the tangential and radial stresses at radius (r) = 100 mm ;110 mm ; 120 mm ; 130 mm ; 140 mm and 150 mm. Sketch the stress distribution curves.

Solution. Given : di = 200 mm or ri = 100 mm ; t = 50 mm ; p = 5 N/mm2

We know that outer radius of the pipe,

ro =ri + t = 100 + 50 = 150 mm

Tangential stresses at radius 100 mm, 110 mm, 120 mm, 130 mm, 140 mm and 150 mmWe know that tangential stress at any radius x,

σt = 2 2 22

2 2 2 2 2 2

( ) ( ) ( )5 (100)1 1

( ) ( ) (150) (100)i o o

o i

p r r r

r r x x

⎡ ⎤ ⎡ ⎤+ = +⎢ ⎥ ⎢ ⎥

− −⎣ ⎦ ⎣ ⎦

=2

22

( )4 1 N/mm or MPaor

x

⎡ ⎤+⎢ ⎥

⎣ ⎦∴ Tangential stress at radius 100 mm (i.e. when x = 100 mm),

σt1 =2

2

(150)4 1 4 3.25 13 MPa

(100)

⎡ ⎤+ = × =⎢ ⎥

⎣ ⎦ Ans.

Tangential stress at radius 110 mm (i.e. when x = 110 mm),

σt2 =2

2

(150)4 1 4 2.86 11.44 MPa

(110)

⎡ ⎤+ = × =⎢ ⎥

⎣ ⎦ Ans.


σt3 =2

2

(150)4 1 4 2.56 10.24 MPa

(120)

⎡ ⎤+ = × =⎢ ⎥

⎣ ⎦ Ans.


σt4 =2

2

(150)4 1 4 2.33 9.32 MPa

(130)

⎡ ⎤+ = × =⎢ ⎥

⎣ ⎦ Ans.


σt5 =2

2

(150)4 1 4 2.15 8.6 MPa

(140)

⎡ ⎤+ = × =⎢ ⎥

⎣ ⎦ Ans.

and tangential stress at radius 150 mm (i.e. when x = 150 mm),

σt6 =2

2

(150)4 1 4 2 8 MPa

(150)

⎡ ⎤+ = × =⎢ ⎥

⎣ ⎦ Ans.


Fig. 8.1

Radial stresses at radius 100 mm, 110 mm, 120 mm, 130 mm, 140 mm and 150 mmWe know that radial stress at any radius x,

σr =2 2 22

2 2 2 2 2 2

( ) ( ) ( )5 (100)1 1

( ) ( ) (150) (100)i o o

o i

p r r r

r r x x

⎡ ⎤ ⎡ ⎤− = −⎢ ⎥ ⎢ ⎥

− −⎣ ⎦ ⎣ ⎦

=2

22

( )4 1 N/mm or MPaor

x

⎡ ⎤−⎢ ⎥

⎣ ⎦∴ Radial stress at radius 100 mm (i.e. when x = 100 mm),

σr1 =2

2

(150)4 1 4 1.25 5 MPa

(100)

⎡ ⎤− = × − = −⎢ ⎥

⎣ ⎦ Ans.

Radial stress at radius 110 mm (i.e., when x = 110 mm),

σr2 =2

2

(150)4 1 4 0.86 3.44 MPa

(110)

⎡ ⎤− = × − = −⎢ ⎥

⎣ ⎦ Ans.

Radial stress at radius 120 mm (i.e. when x = 120 mm),

σr3 =2

2

(150)4 1 4 0.56 2.24 MPa

(120)

⎡ ⎤− = × − = −⎢ ⎥

⎣ ⎦ Ans.


σr4 =2

2

(150)4 1 4 0.33 1.32 MPa

(130)

⎡ ⎤− = × − = −⎢ ⎥

⎣ ⎦ Ans.


σr5 =2

2

(150)4 1 4 0.15 0.6 MPa

(140)

⎡ ⎤− = × − = −⎢ ⎥

⎣ ⎦ Ans.


σr6 =2

2

(150)4 1 0

(150)

⎡ ⎤− =⎢ ⎥

⎣ ⎦ Ans.

The stress distribution curves for tangential and radial stresses are shown in Fig. 8.1.


8.3 Design of PipesThe design of a pipe involves the determination of inside diameter of the pipe and its wall

thickness as discussed below:

1. Inside diameter of the pipe. The inside diameter of the pipe depends upon the quantity offluid to be delivered.

Let D = Inside diameter of the pipe,

v = Velocity of fluid flowing per minute, and

Q = Quantity of fluid carried per minute.

We know that the quantity of fluid flowing per minute,

Q = Area × Velocity = 2

4D v

π × ×

∴ D =4

1.13Q Q

v v× =

π2. Wall thickness of the pipe. After deciding upon

the inside diameter of the pipe, the thickness of the wall(t) in order to withstand the internal fluid pressure ( p)may be obtained by using thin cylindrical or thickcylindrical formula.

The thin cylindrical formula may be applied when

(a) the stress across the section of the pipe isuniform,

(b) the internal diameter of the pipe (D) is morethan twenty times its wall thickness (t), i.e.D/t > 20, and

(c) the allowable stress (σt) is more than sixtimes the pressure inside the pipe (p),i.e. σt /p > 6.

According to thin cylindrical formula, wall thickness of pipe,

t =. .

or2 2σ σ ηt t l

p D p D

where ηl = Efficiency of longitudinal joint.

A little consideration will show that the thickness of wall as obtained by the above relation istoo small. Therefore for the design of pipes, a certain constant is added to the above relation. Now therelation may be written as

t =.

2 t

p DC+

σThe value of constant ‘C’, according to Weisback, are given in the following table.

Table 8.2. Values of constant ‘C’.

Material Cast iron Mild steel Zinc and LeadCopper

Constant (C) in mm 9 3 4 5

Pipe Joint


Example 8.2. A seamless pipe carries 2400 m3 of steam per hour at a pressure of 1.4 N/mm2.The velocity of flow is 30 m/s. Assuming the tensile stress as 40 MPa, find the inside diameter of thepipe and its wall thickness.

Solution. Given : Q = 2400 m3/h = 40 m3/min ; p = 1.4 N/mm2; v = 30 m/s = 1800 m/min ;σt = 40 MPa = 40 N/mm2

Inside diameter of the pipeWe know that inside diameter of the pipe,

D =40

1.13 1.13 0.17 m 170 mm1800

Q

v= = = Ans.

Wall thickness of the pipeFrom Table 8.2, we find that for a steel pipe, C = 3 mm. Therefore wall thickness of the pipe,

t =. 1.4 170

3 6 mm2 2 40t

p DC

×+ = + =

σ × Ans.

8.4 Pipe JointsThe pipes are usually connected to vessels from which they transport the fluid. Since the length

of pipes available are limited, therefore various lengths of pipes have to be joined to suit any particularinstallation. There are various forms of pipe joints used in practice, but most common of them arediscussed below.

1. Socket or a coupler joint. The mostcommon method of joining pipes is by means of asocket or a coupler as shown in Fig. 8.2. A socket isa small piece of pipe threaded inside. It is screwedon half way on the threaded end of one pipe and theother pipe is then screwed in the remaining half ofsocket. In order to prevent leakage, jute or hemp iswound around the threads at the end of each pipe.This type of joint is mostly used for pipes carryingwater at low pressure and where the overall smallnessof size is most essential.

2. Nipple joint. In this type of joint, a nipple which is a small piece of pipe threaded outside isscrewed in the internally threaded end of each pipe, as shown in Fig. 8.3. The disadvantage of thisjoint is that it reduces the area of flow.

Fig. 8.3. Nipple joint. Fig. 8.4. Union joint.

3. Union joint. In order to disengage pipes joined by a socket, it is necessary to unscrew pipefrom one end. This is sometimes inconvenient when pipes are long.

The union joint, as shown in Fig. 8.4, provide the facility of disengaging the pipes by simplyunscrewing a coupler nut.

Fig. 8.2. Socket or coupler joint.

Pipes and Pipe Joints 2674. Spigot and socket joint. A spigot and socket joint as shown in Fig. 8.5, is chiefly used for

pipes which are buried in the earth. Some pipe lines are laid straight as far as possible. One of theimportant features of this joint is its flexibility as it adopts itselfto small changes in level due to settlement of earth which takesplace due to climate and other conditions.

In this type of joint, the spigot end of one pipe fits into thesocket end of the other pipe. The remaining space between thetwo is filled with a jute rope and a ring of lead. When the leadsolidifies, it is caulked-in tightly.

5. Expansion joint. The pipes carrying steam at highpressures are usually joined by means of expansion joint. Thisjoint is used in steam pipes to take up expansion and contractionof pipe line due to change of temperature.

In order to allow for change in length, steam pipes are not rigidly clamped but supported onrollers. The rollers may be arranged on wall bracket, hangers or floor stands. The expansion bends, asshown in Fig. 8.6 (a) and (b), are useful in a long pipe line. These pipe bends will spring in eitherdirection and readily accommodate themselves to small movements of the actual pipe ends to whichthey are attached.

Fig. 8.6. Expansion bends.

Fig. 8.7. Expansion joints.

The copper corrugated expansion joint, as shown in Fig. 8.7 (a), is used on short lines and issatisfactory for limited service. An expansion joint as shown in Fig. 8.7 (b) (also known as gland andstuffing box arrangement), is the most satisfactory when the pipes are well supported and cannot sag.

Fig. 8.5. Spigot and socket joint.


6. Flanged joint. It is one of the most widely used pipe joint. A flanged joint may be made withflanges cast integral with the pipes or loose flanges welded or screwed. Fig. 8.8 shows two cast ironpipes with integral flanges at their ends. The flanges are connected by means of bolts. The flanges

have seen standardised for pressures upto2 N/mm2. The flange faces are machinedto ensure correct alignment of the pipes.The joint may be made leakproof byplacing a gasket of soft material, rubberor convass between the flanges. Theflanges are made thicker than the pipewalls, for strength. The pipes may bestrengthened for high pressure duty byincreasing the thickness of pipe for a shortlength from the flange, as shown in Fig. 8.9.

For even high pressure and for largediameters, the flanges are further strengthened by ribs or stiffners as shown in Fig. 8.10 (a). The ribsare placed between the bolt holes.

Fig. 8.10

For larger size pipes, separate loose flanges screwed on the pipes as shown in Fig. 8.10 (b) areused instead of integral flanges.

Fig. 8.8. Flanged joint.

Fig. 8.9. Flanged joint.

Pipes and Pipe Joints 2697. Hydraulic pipe joint. This type of joint has oval flanges and are fastened by means of two bolts,

as shown in Fig. 8.11. The oval flanges are usually used for small pipes, upto 175 mm diameter. Theflanges are generally cast integral with the pipe ends. Such joints are used to carry fluid pressure varyingfrom 5 to 14 N/mm2. Such a high pressure is found in hydraulic applications like riveting, pressing, liftsetc. The hydraulic machines used in these installations are pumps, accumulators, intensifiers etc.

Fig. 8.11. Hydraulic pipe joint.

8.5 Standard Pipe Flanges for SteamThe Indian boiler regulations (I.B.R.) 1950 (revised 1961) have standardised all dimensions of

pipe and flanges based upon steam pressure. They have been divided into five classes as follows:

Class I : For steam pressures up to 0.35 N/mm2 and water pressures up to 1.4 N/mm2. This isnot suitable for feed pipes and shocks.

Class II : For steam pressures over 0.35 N/mm2 but not exceeding 0.7 N/mm2.

Class III : For steam pressures over 0.7 N/mm2 but not exceeding 1.05 N/mm2.

Class IV : For steam pressures over1.05 N/mm2 but not exceeding 1.75 N/mm2.

Class V : For steam pressures from1.75 N/mm2 to 2.45 N/mm2.

According to I.B.R., it is desirable thatfor classes II, III, IV and V, the diameter offlanges, diameter of bolt circles and numberof bolts should be identical and thatdifference should consist in variations of thethickness of flanges and diameter of boltsonly. The I.B.R. also recommends that allnuts should be chamfered on the side bearingon the flange and that the bearing surfacesof the flanges, heads and nuts should be true.The number of bolts in all cases should be amultiple of four. The I.B.R. recommends that for 12.5 mm and 15 mm bolts, the bolt holes should be1.5 mm larger and for higher sizes of bolts, the bolt holes should be 3 mm larger. All dimensions forpipe flanges having internal diameters 1.25 mm to 600 mm are standardised for the above mentionedclasses (I to V). The flanged tees, bends are also standardised.

The Trans-Alaska Pipeline was built to carry oil across thefrozen sub-Arctic landscape of North America.


Note: As soon as the size of pipe is determined, the rest of the dimensions for the flanges, bolts, bolt holes,thickness of pipe may be fixed from standard tables. In practice, dimensions are not calculated on a rationalbasis. The standards are evolved on the basis of long practical experience, suitability and interchangeability. Thecalculated dimensions as discussed in the previous articles do not agree with the standards. It is of academicinterest only that the students should know how to use fundamental principles in determining various dimen-sions e.g. wall thickness of pipe, size and number of bolts, flange thickness. The rest of the dimensions may beobtained from standard tables or by empirical relations.

8.6 Hydraulic Pipe Joint for High PressuresThe pipes and pipe joints for high fluid pressure are classified as follows:

1. For hydraulic pressures up to 8.4 N/mm2 and pipe bore from 50 mm to 175 mm, the flangesand pipes are cast integrally from remelted castiron. The flanges are made elliptical and securedby two bolts. The proportions of these pipe jointshave been standardised from 50 mm to 175 mm,the bore increasing by 25 mm. This category isfurther split up into two classes:

(a) Class A: For fluid pressures from5 to 6.3 N/mm2, and

(b) Class B: For fluid pressures from6.3 to 8.4 N/mm2.

The flanges in each of the above classesmay be of two types. Type I is suitable for pipesof 50 to 100 mm bore in class A, and for 50 to175 mm bore in class B. The flanges of type IIare stronger than those of Type I and are usuallyset well back on the pipe.

2. For pressures above 8.4 N/mm2 withbores of 50 mm or below, the piping is of wroughtsteel, solid drawn, seamless or rolled. The flangesmay be of cast iron, steel mixture or forged steel. These are screwed or welded on to the pipe and aresquare in elevation secured by four bolts. These joints are made for pipe bores 12.5 mm to 50 mmrising in increment of 3 mm from 12.5 to 17.5 mm and by 6 mm from 17.5 to 50 mm. The flanges andpipes in this category are strong enough for service under pressures ranging up to 47.5 N/mm2.

In all the above classes, the joint is of the spigot and socket type made with a jointing ring ofgutta-percha.

Notes: The hydraulic pipe joints for high pressures differ from those used for low or medium pressure in thefollowing ways:

1. The flanges used for high pressure hydraulic pipe joints are heavy oval or square in form, They use two orfour bolts which is a great advantage while assembling and disassembling the joint especially in narrow space.

2. The bolt holes are made square with sufficient clearance to accomodate square bolt heads and to allowfor small movements due to setting of the joint.

3. The surfaces forming the joint make contact only through a gutta-percha ring on the small area providedby the spigot and recess. The tightening up of the bolts squeezes the ring into a triangular shape and makes aperfectly tight joint capable of withstanding pressure up to 47.5 N/mm2.

4. In case of oval and square flanged pipe joints, the condition of bending is very clearly defined due tothe flanges being set back on the pipe and thickness of the flange may be accurately determined to withstand thebending action due to tightening of bolts.

Hydraulic pipe joints use two or four bolts which isa great advantage while assembling the jointespecially in narrow space.


8.7 Design of Circular Flanged Pipe JointConsider a circular flanged pipe joint as shown in Fig. 8.8. In designing such joints, it is assumed

that the fluid pressure acts in between the flanges and tends to separate them with a pressure existing atthe point of leaking. The bolts are required to take up tensile stress in order to keep the flanges together.

The effective diameter on which the fluid pressure acts, just at the point of leaking, is thediameter of a circle touching the bolt holes. Let this diameter be D1. If d1 is the diameter of bolt holeand Dp is the pitch circle diameter, then

D1 = Dp – d1

∴ Force trying to separate the two flanges,

F = 21( )

4D p

π...(i)

Let n = Number of bolts,

dc = Core diameter of the bolts, and

σt = Permissible stress for the material of the bolts.

∴ Resistance to tearing of bolts

= 2( )4 c td nπ σ × ...(ii)

Assuming the value of dc, the value of n may be obtained from equations (i) and (ii). Thenumber of bolts should be even because of the symmetry of the section.

The circumferential pitch of the bolts is given by

pc =pD

n

π

In order to make the joint leakproof, the value of pc should be between 20 1d to 30 1d ,where d1 is the diameter of the bolt hole. Also a bolt of less than 16 mm diameter should never be usedto make the joint leakproof.

The thickness of the flange is obtained by considering a segment of the flange as shown in Fig. 8.8 (b).

In this it is assumed that each of the bolt supports one segment. The effect of joining of thesesegments on the stresses induced is neglected. The bending moment is taken about the section X-X,which is tangential to the outside of the pipe. Let the width of this segment is x and the distance of thissection from the centre of the bolt is y.

∴ Bending moment on each bolt due to the force F

=F

yn

× ...(iii)

and resisting moment on the flange

= σb × Z ...(iv)

where σb = Bending or tensile stress for the flange material, and

Z = Section modulus of the cross-section of the flange = 21

( )6 fx t×

Equating equations (iii) and (iv), the value of tf may be obtained.

The dimensions of the flange may be fixed as follows:

Nominal diameter of bolts, d = 0.75 t + 10 mm

Number of bolts, n = 0.0275 D + 1.6 ...(D is in mm)


Thickness of flange, tf = 1.5 t + 3 mm

Width of flange, B = 2.3 d

Outside diameter of flange,

Do = D + 2t + 2B

Pitch circle diameter of bolts,

Dp = D + 2t + 2d + 12 mm

The pipes may be strengthened by providing greater thickness near the flanges equal to2

ft t+⎛ ⎞⎜ ⎟⎝ ⎠

as shown in Fig. 8.9. The flanges may be strengthened by providing ribs equal to thickness of 2

ft t+ ,

as shown in Fig. 8.10 (a).

Example 8.3. Find out the dimensions of a flanged joint for a cast iron pipe 250 mm diameterto carry a pressure of 0.7 N/mm2.

Solution. Given: D = 250 mm ; p = 0.7 N/mm2

From Table 8.1, we find that for cast iron, allowable tensile stress, σt = 14 N/mm2 and fromTable 8.2, C = 9 mm. Therefore thickness of the pipe,

t =. 0.7 250

9 15.3 say 16 mm2 2 14t

p DC

×+ = + =σ ×

Ans.

Other dimensions of a flanged joint for a cast iron pipe may be fixed as follows:

Nominal diameter of the bolts,

d = 0.75 t + 10 mm = 0.75 × 16 + 10 = 22 mm Ans.Number of bolts, n = 0.0275 D + 1.6 = 0.0275 × 250 + 1.6 = 8.475 say 10 Ans.Thickness of the flanges, tf = 1.5 t + 3 mm = 1.5 × 16 + 3 = 27 mm Ans.Width of the flange, B = 2.3 d = 2.3 × 22 = 50.6 say 52 mm Ans.Outside diameter of the flange,

Do = D + 2t + 2B = 250 + 2 × 16 + 2 × 52 = 386 mm Ans.Pitch circle diameter of the bolts,

Dp = D + 2t + 2d + 12 mm = 250 + 2 × 16 + 2 × 22 + 12 mm

= 338 mm Ans.Circumferential pitch of the bolts,

pc =338

106.2 mm10

pD

n

π × π ×= = Ans.

In order to make the joint leak proof, the value of pc should be between 20 1d to 30 1dwhere d1 is the diameter of bolt hole.

Let us take d1 = d + 3 mm = 22 + 3 = 25 mm

∴ 120 20 25 100 mmd = =

and 130 30 25 150 mmd = =

Since the circumferential pitch as obtained above (i.e. 106.2 mm) is within 1 120 to 30d d ,therefore the design is satisfactory.

Pipes and Pipe Joints 273Example 8.4. A flanged pipe with internal diameter as 200 mm is subjected to a fluid pressure

of 0.35 N/mm2. The elevation of the flange is shown in Fig. 8.12. The flange is connected by meansof eight M 16 bolts. The pitch circle diameter of the bolts is 290 mm. If the thickness of the flange is20 mm, find the working stress in the flange.

Solution. Given : D = 200 mm ; p = 0.35 N/mm2 ; n = 8 ;* d = 16 mm ; Dp = 290 mm ; tf = 20 mm

First of all, let us find the thickness of the pipe. Assuming the pipe to be of cast iron, we find fromTable 8.1 that the allowable tensile stress for cast iron, σt = 14 N/mm2 and from Table 8.2, C = 9 mm.

∴ Thickness of the pipe,

t =. 0.35 200

9 11.5 say 12mm2 2 14t

p DC

×+ = + =σ ×

Since the diameter of the bolt holes (d1) is taken larger than the nominal diameter of the bolts(d), therefore let us take diameter of the bolt holes,

d1 = d + 2 mm = 16 + 2 = 18 mm

Fig. 8.12

and diameter of the circle on the inside of the bolt holes,

D1 = Dp – d1 = 290 – 18 = 272 mm

∴ Force trying to separate the flanges i.e. force on 8 bolts,

F = 2 21( ) (272) 0.35 20 340 N

4 4D p

π π= =

Now let us find the bending moment about the section X-X which is tangential to the outside ofthe pipe. The width of the segment is obtained by measuring the distance from the drawing. Onmeasuring, we get

x = 90 mm

and distance of the section X-X from the centre of the bolt,

y =290 200

12 33 mm2 2 2 2

pD Dt

⎛ ⎞ ⎛ ⎞− + = − + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Let σb = Working stress in the flange.

We know that bending moment on each bolt due to force F

=20340

33 83 900 N-mm8

Fy

n× = × = ...(i)

* M16 bolt means that the nominal diameter of the bolt (d) is 16 mm.


and resisting moment on the flange

=21

( )6b b fZ x tσ × = σ × ×

=21

90 (20) 6000 N-mm6b bσ × × = σ ...(ii)

From equations (i) and (ii), we have

σb = 83 900 / 6000

= 13.98 N/mm2 = 13.98 MPa Ans.

8.8 Design of Oval Flanged Pipe JointConsider an oval flanged pipe joint as shown

in Fig. 8.11. A spigot and socket is provided forlocating the pipe bore in a straight line. A packing oftrapezoidal section is used to make the joint leakproof. The thickness of the pipe is obtained asdiscussed previously.

The force trying to separate the two flanges hasto be resisted by the stress produced in the bolts. If alength of pipe, having its ends closed somewherealong its length, be considered, then the forceseparating the two flanges due to fluid pressure isgiven by

F1 = 2

4D p

π × ×

where D = Internal diameter of the pipe.The packing has also to be compressed to make the joint leakproof. The intensity of pressure

should be greater than the pressure of the fluid inside the pipe. For the purposes of calculations, it isassumed that the packing material is compressed to the same pressure as that of inside the pipe.Therefore the force tending to separate the flanges due to pressure in the packing is given by

F2 = 2 21( ) ( )

4D D p

π ⎡ ⎤× −⎣ ⎦

where D1 = Outside diameter of the packing.∴ Total force trying to separate the two flanges,

F = F1 + F2

2 2 2 21 1( ) ( ) ( )

4 4 4D p D D p D p

π π π⎡ ⎤= × × + − =⎣ ⎦

Since an oval flange is fastened by means of two bolts, therefore load taken up by each bolt isFb = F /2 . If dc is the core diameter of the bolts, then

Fb = 2( )4 c tbdπ σ

where σtb is the allowable tensile stress for the bolt material. The value of σtb is usually kept low toallow for initial tightening stress in the bolts. After the core diameter is obtained, then the nominaldiameter of the bolts is chosen from *tables. It may be noted that bolts of less than 12 mm diameter

* In the absence of tables, nominal diameter = Core diameter

0.84

Oval flanged pipe joint.

Pipes and Pipe Joints 275should never be used for hydraulic pipes, because very heavy initial tightening stresses may be inducedin smaller bolts. The bolt centres should be as near the centre of the pipe as possible to avoid bendingof the flange. But sufficient clearance between the bolt head and pipe surface must be provided forthe tightening of the bolts without damaging the pipe material.

The thickness of the flange is obtained by considering the flange to be under bending stressesdue to the forces acting in one bolt. The maximum bending stress will be induced at the section X-X.The bending moment at this section is given by

Mxx =2bF

F e e× = ×

and section modulus, Z = 21

( )6 fb t×

where b = Width of the flange at the section X-X, andtf = Thickness of the flange.

Using the bending equation, we have

Mxx = σb.Z

or Fb × e =21

( )6b fb tσ × ×

where σb =Permissible bending stress for the flange material.

From the above expression, the value of tf may be obtained, if b is known. The width of theflange is estimated from the lay out of the flange. The hydraulic joints with oval flanges are known asArmstrong's pipe joints. The various dimensions for a hydraulic joint may be obtained by using thefollowing empirical relations:

Nominal diameter of bolts, d =0.75 t + 10 mm

Thickness of the flange, tf =1.5 t + 3 mm

Outer diameter of the flange,

Do =D + 2t + 4.6 d

Pitch circle diameter, Dp =Do – (3 t + 20 mm)

Example 8.5. Design and draw an oval flanged pipe joint for a pipe having 50 mm bore. It issubjected to an internal fluid pressure of 7 N/mm2. The maximum tensile stress in the pipe material isnot to exceed 20 MPa and in the bolts 60 MPa.

Solution. Given: D = 50 mm or R = 25 mm ; p = 7 N/mm2 ; σt = 20 MPa = 20 N/mm2 ;σtb = 60 MPa = 60 N/mm2

First of all let us find the thickness of the pipe (t). According to Lame's equation, we know thatthickness of the pipe,

t = 20 7

1 25 – 120 7

t

t

pR

p

⎡ ⎤ ⎡ ⎤σ + +− =⎢ ⎥ ⎢ ⎥σ − −⎢ ⎥ ⎣ ⎦⎣ ⎦= 11.03 say 12 mm Ans.

Assuming the width of packing as 10 mm, therefore outside diameter of the packing,

D1 =D + 2 × Width of packing = 50 + 2 × 10 = 70 mm

∴ Force trying to separate the flanges,

F = 2 21( ) (70) 7 26 943 N

4 4D p

π π= =

Since the flange is secured by means of two bolts, therefore load on each bolt,

Fb =F / 2 = 26 943 /2 = 13 471.5 N

Let dc =Core diameter of bolts.


We know that load on each bolt (Fb),

13 471.5 = 2 2 2( ) ( ) 60 47.2 ( )4 4c tb c cd d dπ πσ = =

∴ (dc)2 =13 471.5 / 47.2 = 285.4 or dc = 16.9 say 17 mm

and nominal diameter of bolts,

d =17

20.2 say 22 mm0.84 0.84

cd= = Ans.

Outer diameter of the flange,

Do =D + 2t + 4.6 d = 50 + 2 × 12 + 4.6 × 22

=175.2 say 180 mm Ans.and pitch circle diameter of the bolts,

Dp =Do – (3t + 20 mm) = 180 – (3 × 12 + 20) = 124 mm

The elevation of the flange as shown in Fig. 8.13 (which is an ellipse) may now be drawn bytaking major axis as Do (i.e. 180 mm) and minor axis as (Dp – d) i.e. 124 – 22 = 102 mm. In order tofind thickness of the flange (tf), consider the section X-X. By measurement, we find that the width ofthe flange at the section X-X,

b = 89 mm

and the distance of the section X-X from the centre line of the bolt,

e = 33 mm

∴ Bending moment at the section X-X,

Mxx = Fb × e = 13 471.5 × 33 N-mm

= 444 560 N-mm

and section modulus, Z =2 21 1

( ) 89 ( )6 6f fb t t= ×

= 14.83 (tf)2

We know that Mxx = σb × Z

444 560 = 20 × 14.83 (tf)2 = 296.6 ( t f)

2

∴ (tf)2 = 444 560 / 2 96.6 = 1500

or tf = 38.7 say 40 mm Ans.

8.9 Design of Square Flanged Pipe Joint

The design of a square flanged pipe joint, as shown in Fig. 8.14, is similar to that of an ovalflanged pipe joint except that the load has to be divided into four bolts. The thickness of the flangemay be obtained by considering the bending of the flange about one of the sections A-A, B-B, or C-C.

A little consideration will show that the flange is weakest in bending about section A-A.Therefore the thickness of the flange is calculated by considering the bending of the flange, aboutsection A-A.

Fig. 8.13


Fig. 8.14. Square flanged pipe joint.

Example 8.6. Design a square flanged pipe joint for pipes of internal diameter 50 mm subjectedto an internal fluid pressure of 7 N/mm2. The maximum tensile stress in the pipe material is not toexceed 21 MPa and in the bolts 28 MPa.

Solution. Given: D = 50 mm or R = 25 mm ; p = 7 N/mm2 ; σt = 21 MPa = 21 N/mm2 ;σtb = 28 MPa = 28 N/mm2

First of all, let us find the thickness of the pipe. According to Lame's equation, we know thatthickness of the pipe,

t =21 7

1 25 1 10.35 say 12 mm21 7

t

t

pR

p

⎡ ⎤ ⎡ ⎤σ + +− = − =⎢ ⎥ ⎢ ⎥σ − −⎢ ⎥ ⎣ ⎦⎣ ⎦Assuming the width of packing as 10 mm, therefore outside diameter of the packing,

D1 = 50 + 2 × Width of packing = 50 + 2 × 10 = 70 mm

∴ Force trying to separate the flanges,

F = 2 21( ) (70) 7 26 943 N

4 4D p

π π= =

Since this force is to be resisted by four bolts, therefore force on each bolt,

Fb = F / 4 = 26 943 / 4 = 6735.8 N

Let dc = Core diameter of the bolts.

We know that force on each bolt (Fb),

6735.8 = 2 2 2( ) ( ) 28 22 ( )4 4c tb c cd d dπ πσ = =

∴ (dc)2 = 6735.8/22 = 306 or dc = 17.5 mm

and nominal diameter of the bolts,

d =17.5

20.9 say 22 mm0.84 0.84

cd= = Ans.

The axes of the bolts are arranged at the corners of a square of such size that the corners of thenut clear the outside of the pipe.

∴ Minimum length of a diagonal for this square,

L = Outside diameter of pipe + 2 × Dia. of bolt = D + 2t + 2d

= 50 + (2 × 12) + (2 × 22) = 118 mm


and side of this square,

L1 = 118

83.5 mm2 2

L = =

The sides of the flange must be of sufficient length to accommodate the nuts and bolt headswithout overhang. Therefore the length L2 may be kept as (L1 + 2d ) i.e.

L2 = L1 + 2d = 83.5 + 2 × 22 = 127.5 mm

The elevation of the flange is shown in Fig. 8.15. In order to find the thickness of the flange,consider the bending of the flange about sectionA-A. The bending about section A-A will takeplace due to the force in two bolts.

Fig. 8.15

∴ Bending moment due to the force in two bolts (i.e. due to 2Fb),

M1 = 1 83.52 2 6735.8 562 440 N-mm

2 2bL

F × = × × =

Water pressure acting on half the flange

= 2 Fb = 2 × 6735.8 = 13 472 N

The flanges are screwed with pipe having metric threads of 4.4 threads in 10 mm (i.e. pitch ofthe threads is 10/4.4 = 2.28 mm).

Nominal or major diameter of the threads

= Outside diameter of the pipe = D + 2t = 50 + 2 × 12 = 74 mm

∴ Nominal radius of the threads

= 74 / 2 = 37 mm

Depth of the threads = 0.64 × Pitch of threads = 0.64 × 2.28 = 1.46 mm

∴ Core or minor radius of the threads

= 37 – 1.46 = 35.54 mm

∴ Mean radius of the arc from A-A over which the load due to fluid pressure may be taken tobe concentrated

= 1

2 (37 + 35.54) = 36.27 mm

The centroid of this arc from A-A= 0.6366 × Mean radius = 0.6366 × 36.27 = 23.1 mm

∴ Bending moment due to the water pressure,

Square flanged pipe joint.

Pipes and Pipe Joints 279M2 = 2 Fb × 23.1 = 2 × 6735.8 × 23.1 = 311 194 N-mm

Since the bending moments M1 and M2 are in opposite directions, thereforeNet resultant bending moment on the flange about section A-A,

M = M1 – M2 = 562 440 – 311 194 = 251 246 N-mmWidth of the flange at the section A-A,

b = L2 – Outside diameter of pipe = 127.5 – 74 = 53.5 mmLet tf = Thickness of the flange in mm.∴ Section modulus,

Z = 2 2 2 31 1( ) 53.5 ( ) 8.9 ( ) mm

6 6f f fb t t t× = × =

We know that net resultant bending moment (M),

251 246 = σb.Z = 21 × 8.9 (tf)2 = 187 (tf)

2

∴ (tf)2 = 251 246 / 187 = 1344 or tf = 36.6 say 38 mm Ans.


1. A cast iron pipe of internal diameter 200 mm and thickness 50 mm carries water under a pressure of5 N/mm2. Calculate the tangential and radial stresses at the inner, middle (radius = 125 mm) and outersurfaces. [Ans. 13 MPa, 9.76 MPa, 8 MPa ; – 5 MPa, – 1.76 MPa, 0]

2. A cast iron pipe is to carry 60 m3 of compressed air per minute at a pressure of 1 N/mm2. The velocityof air in the pipe is limited to 10 m/s and the permissible tensile stress for the material of the pipe is14 MPa. Find the diameter of the pipe and its wall thickness. [Ans. 360 mm ; 22 mm]

3. A seamless steel pipe carries 2000 m3 of steam per hour at a pressure of 1.2 N/mm2. The velocity offlow is 28 m/s. Assuming the tensile stress as 40 MPa, find the inside diameter of the pipe and its wallthickness. [Ans. 160 mm ; 5.4 mm]

4. Compute the dimensions of a flanged cast iron pipe 200 mm in diameter to carry a pressure of0.7 N/mm2.

[Ans. t = 20 mm ; d = 16 mm ; n = 8 ; tf = 33 mm ; B = 37 mm ; Do = 314 mm ; Dp = 284 mm]

5. Design an oval flanged pipe joint for pipes of internal diameter 50 mm subjected to a fluid pressure of7 N/mm2. The maximum tensile stress in the pipe material is not to exceed 21 MPa and in the bolts 28MPa. [Ans. t = 12 mm ; d = 30 mm ; tf = 38 mm]


1. Discuss how the pipes are designed.

2. Describe with sketches, the various types of pipe joints commonly used in engineering practice.

3. Explain the procedure for design of a circular flanged pipe point.

4. Describe the procedure for designing an oval flanged pipe joint.

OBJECTOBJECTOBJECTOBJECTOBJECTIVE IVE IVE IVE IVE TTTTTYPYPYPYPYPE QE QE QE QE QUEUEUEUEUESTSTSTSTSTIONSIONSIONSIONSIONS

1. Cast iron pipes are mainly used

(a) for conveying steam

(b) in water and sewage systems

(c) in pressure lubrication systems on prime movers


(d) all of the above

2. The diameter of a pipe carrying steam Q m3/min at a velocity v m/min is

(a)Q

v(b)

Q

v

(c)4

π Q

v(d) 1.13

Q

v

3. When the internal diameter of the pipe exceeds twenty times its wall thickness, then .............. cylin-drical shell formula may be applied.

(a) thin (b) thick

4. Which of the following joint is commonly used for joining pipes carrying water at low pressure?

(a) union joint (b) spigot and socket joint

(c) socket or a coupler joint (d) nipple joint

5. The pipes which are burried in the earth should be joined with

(a) union joint (b) spigot and socket joint

(c) coupler joint (d) nipple joint

6. The expansion joint is mostly used for pipes which carry steam at .............. pressures.

(a) low (b) high

7. The pipes carrying fluid pressure varying from 5 to 14 N/mm2 should have

(a) square flanged joint (b) circular flanged joint

(c) oval flanged joint (d) spigot and socket joint

8. An oval type flange is fastened by means of

(a) two bolts (b) four bolts

(c) six bolts (d) eight bolts

9. A flanged pipe joint will be a leakproof, if the circumferential pitch of the bolts is

(a) less then 20 d (b) greater than 30 d

(c) between 20 and 30d d (d) equal to one-third of inside diameter of pipe

where d = Diameter of the bolt hole.

10. The flanges in a circular flanged pipe joint are strengthened by providing ribs between the bolt holes.The thickness of such ribs is taken as

(a) t (b) tf

(c)2

− ft t(d)

2

+ ft t

where t =Thickness of pipe, and

tf = Thickness of flange.


1. (b) 2. (d) 3. (a) 4. (c) 5. (b)

6. (a) 7. (c) 8. (a) 9. (c) 10. (d)

Riveted Joints

281

1. Introduction.2. Methods of Riveting.3. Material of Rivets.4. Essential Qualit ies of a

Rivet.5. Manufacture of Rivets.6. Types of Rivet Heads.7. Types of Riveted Joints.8. Lap Joint.9. Butt Joint.

10. Important Terms Used inRiveted Joints.

11. Caulking and Fullering.12. Failures of a Riveted Joint.13. Strength of a Riveted Joint.14. Efficiency of a Riveted Joint.15. Design of Boiler Joints.16. Assumptions in Designing

Boiler Joints.17. Design of Longitudinal Butt

Joint for a Boiler.18. Design of Circumferential

Lap Joint for a Boiler.19. Recommended Joints for

Pressure Vessels.20. Riveted Joint for Structural

Use–Joints of UniformStrength (Lozenge Joint).

21. Eccentric Loaded RivetedJoint.

9CHAPTER

9.1 IntroductionA rivet is a short cylindrical bar with a head integral

to it. The cylindrical portion ofthe rivet is called shank or bodyand lower portion of shank isknown as tail, as shown in Fig.9.1. The rivets are used to makepermanent fastening between theplates such as in structural work,ship building, bridges, tanks andboiler shells. The riveted jointsare widely used for joining lightmetals.

The fastenings (i.e. joints)may be classified into the following two groups :

1. Permanent fastenings, and2. Temporary or detachable fastenings.

Head

ShankorBody

Tail

Fig. 9.1. Rivet parts.


The permanent fastenings are those fastenings which can not be disassembled withoutdestroying the connecting components. The examples of permanent fastenings in order of strengthare soldered, brazed, welded and riveted joints.

The temporary or detachable fastenings are those fastenings which can be disassembledwithout destroying the connecting components. The examples of temporary fastenings are screwed,keys, cotters, pins and splined joints.

9.2 Methods of RivetingThe function of rivets in a joint is to make a connection that has strength and tightness. The

strength is necessary to prevent failure of the joint. The tightness is necessary in order to contribute tostrength and to prevent leakage as in a boiler or in a ship hull.

When two plates are to be fastened together by a rivet as shown in Fig. 9.2 (a), the holes in theplates are punched and reamed or drilled. Punching is the cheapest method and is used for relativelythin plates and in structural work. Since punching injures the material around the hole, thereforedrilling is used in most pressure-vessel work. In structural and pressure vessel riveting, the diameterof the rivet hole is usually 1.5 mm larger than the nominal diameter of the rivet.

Original head

Backing up bar

Tail

Die

( ) Initial position.a ( ) Final position.b

Point

Fig. 9.2. Methods of riveting.

The plates are drilled together and then separated to remove any burrs or chips so as to have atight flush joint between the plates. A cold rivet or a red hot rivet is introduced into the plates and thepoint (i.e. second head) is then formed. When a cold rivet is used, the process is known as coldriveting and when a hot rivet is used, the process is known as hot riveting. The cold riveting processis used for structural joints while hot riveting is used to make leak proof joints.

A ship’s body is a combination of riveted, screwed and welded joints.

Riveted Joints 283The riveting may be done by hand or by a riveting machine. In hand riveting, the original rivet

head is backed up by a hammer or heavy bar and then the die or set, as shown in Fig. 9.2 (a), is placedagainst the end to be headed and the blows are applied by a hammer. This causes the shank to expandthus filling the hole and the tail is converted into a point as shown in Fig. 9.2 (b). As the rivet cools,it tends to contract. The lateral contraction will be slight, but there will be a longitudinal tensionintroduced in the rivet which holds the plates firmly together.

In machine riveting, the die is a part of the hammer which is operated by air, hydraulic or steampressure.Notes : 1. For steel rivets upto 12 mm diameter, the cold riveting process may be used while for larger diameterrivets, hot riveting process is used.

2. In case of long rivets, only the tail is heated and not the whole shank.

9.3 Material of RivetsThe material of the rivets must be tough and ductile. They are usually made of steel (low carbon

steel or nickel steel), brass, aluminium or copper, but when strength and a fluid tight joint is the mainconsideration, then the steel rivets are used.

The rivets for general purposes shall be manufactured from steel conforming to the followingIndian Standards :

(a) IS : 1148–1982 (Reaffirmed 1992) – Specification for hot rolled rivet bars (up to 40 mmdiameter) for structural purposes; or

(b) IS : 1149–1982 (Reaffirmed 1992) – Specification for high tensile steel rivet bars forstructural purposes.

The rivets for boiler work shall be manufactured from material conforming to IS : 1990 – 1973(Reaffirmed 1992) – Specification for steel rivets and stay bars for boilers.Note : The steel for boiler construction should conform to IS : 2100 – 1970 (Reaffirmed 1992) – Specifica-tion for steel billets, bars and sections for boilers.

9.4 Essential Qualities of a RivetAccording to Indian standard, IS : 2998 – 1982 (Reaffirmed 1992), the material of a rivet must

have a tensile strength not less than 40 N/mm2 and elongation not less than 26 percent. The materialmust be of such quality that when in cold condition, the shank shall be bent on itself through 180°without cracking and after being heated to 650°C and quenched, it must pass the same test. The rivetwhen hot must flatten without cracking to a diameter 2.5 times the diameter of shank.

9.5 Manufacture of RivetsAccording to Indian standard specifications, the rivets may be made either by cold heading or

by hot forging. If rivets are made by the cold heading process, they shall subsequently be adequatelyheat treated so that the stresses set up in the cold heading process are eliminated. If they are made byhot forging process, care shall be taken to see that the finished rivets cool gradually.

9.6 Types of Rivet HeadsAccording to Indian standard specifications, the rivet heads are classified into the following

three types :

1. Rivet heads for general purposes (below 12 mm diameter) as shown in Fig. 9.3, according toIS : 2155 – 1982 (Reaffirmed 1996).


Fig. 9.3. Rivet heads for general purposes (below 12 mm diameter).

2. Rivet heads for general purposes (From 12 mm to 48 mm diameter) as shown in Fig. 9.4,according to IS : 1929 – 1982 (Reaffirmed 1996).

1.6 d

1.5 d

1.5 d2 d

1.5 dR

0.5d

0.5d 0.25 d

1.6 d 1.6 d

0.7 d 0.7 d 0.7 d

0.5 d

Len

gth

Len

gth

Len

gth

Len

gth

Len

gth

Len

gth

( ) Snap head.a

( ) Round counter

sunk head 60º.

d ( ) Flat counter

sunk head 60º.

e ( )Flat head.f

( ) Pan head.b ( ) Pan head with tapered neck.c

d

d d d

d

d

dd

15º

60º 60º

Fig. 9.4. Rivet heads for general purposes (from 12 mm to 48 mm diameter)

Riveted Joints 2853. Rivet heads for boiler work (from 12 mm to 48 mm diameter, as shown in Fig. 9.5, according to

IS : 1928 – 1961 (Reaffirmed 1996).

Fig. 9.5. Rivet heads for boiler work.

The snap heads are usually employed for structural work and machine riveting. The countersunk heads are mainly used for ship building where flush surfaces are necessary. The conical heads(also known as conoidal heads) are mainly used in case of hand hammering. The pan heads havemaximum strength, but these are difficult to shape.

9.7 Types of Riveted JointsFollowing are the two types of riveted joints, depending upon the way in which the plates are

connected.

1. Lap joint, and 2. Butt joint.


9.8 Lap JointA lap joint is that in which one plate overlaps the other

and the two plates are then riveted together.

9.9 Butt JointA butt joint is that in which the main plates are kept in

alignment butting (i.e. touching) each other and a cover plate(i.e. strap) is placed either on one side or on both sides of themain plates. The cover plate is then riveted together with themain plates. Butt joints are of the following two types :

1. Single strap butt joint, and 2. Double strap buttjoint.

In a single strap butt joint, the edges of the main platesbutt against each other and only one cover plate is placed onone side of the main plates and then riveted together.

In a double strap butt joint, the edges of the main platesbutt against each other and two cover plates are placed onboth sides of the main plates and then riveted together.

In addition to the above, following are the types of riv-eted joints depending upon the number of rows of the rivets.

1. Single riveted joint, and 2. Double riveted joint.

A single riveted joint is that in which there is a single row of rivets in a lap joint as shown inFig. 9.6 (a) and there is a single row of rivets on each side in a butt joint as shown in Fig. 9.8.

A double riveted joint is that in which there are two rows of rivets in a lap joint as shown inFig. 9.6 (b) and (c) and there are two rows of rivets on each side in a butt joint as shown in Fig. 9.9.

X

X XX

Y Yp

pb pd

m

( ) Single riveted lap joint.a ( ) Double riveted lap joint

(Chain riveting).

b ( ) Double riveted lap

joint (Zig-zag riveting).

c

Fig. 9.6. Single and double riveted lap joints.

Similarly the joints may be triple riveted or quadruple riveted.Notes : 1. When the rivets in the various rows are opposite to each other, as shown in Fig. 9.6 (b), then the jointis said to be chain riveted. On the other hand, if the rivets in the adjacent rows are staggered in such a way that

Riveted Joints 287every rivet is in the middle of the two rivets of the opposite row as shown in Fig. 9.6 (c), then the joint is said tobe zig-zag riveted.

2. Since the plates overlap in lap joints, therefore the force P, P acting on the plates [See Fig. 9.15 (a)] arenot in the same straight line but they are at a distance equal to the thickness of the plate. These forces will forma couple which may bend the joint. Hence the lap joints may be used only where small loads are to be transmit-ted. On the other hand, the forces P, P in a butt joint [See Fig. 9.15 (b)] act in the same straight line, thereforethere will be no couple. Hence the butt joints are used where heavy loads are to be transmitted.

( ) Chain riveting.a ( ) Zig-zag riveting.b

XX

Y Y

m mm

pd

Fig. 9.7. Triple riveted lap joint.

Fig. 9.8. Single riveted double strap butt joint.

188 A Textbook of Machine Design -...

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