Mech-Vi-Design of Machine Elements II [10me62]-Notes(1)

7/26/2019 Mech-Vi-Design of Machine Elements II [10me62]-Notes(1)

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Design of Machine Elements-II 10ME62

Department of Mechanical Engineering, SJBIT, Bengaluru. Page 1

DESIGN OF MACHINE ELEMENTS II

Subject Code :10ME62 IA Marks : 25

Hours/Week :04 Exam Hours : 03

Total Hours :52 Exam Marks : 100

PARTA

UNIT - 1

Curved Beams: Stresses in curved beams of standard cross sections used in crane hook,

punching presses & clamps, closed rings and links

Cylinders & Cylinder Heads: Review of Lames Equations; compound cylinders, stresses due

to different types of fits, cylinder heads, flats. 08 Hours

UNIT - 2

Belts Ropes and Chains: Flat belts: Length & cross section, Selection of V-belts, ropes and

chains for different applications. 05 Hours

UNIT - 3

Springs: Types of springs - stresses in Helical coil springs of circular and non-circular cross

sections. Tension and compression springs, springs under fluctuating loads, Leaf Springs:

Stresses in leaf springs. Equalized stresses, Energy stored in springs, Torsion, Belleville and

Rubber springs. 08 Hours

UNIT - 4

Spur & Helical Gears: Spur Gears: Definitions, stresses in gear tooth: Lewis equation and form

factor, Design for strength, Dynamic load and wear load. Helical Gears: Definitions, formative

number of teeth, Design based on strength, dynamic and wear loads. 07 Hours

PARTB

UNIT - 5

Bevel and Worm Gears: Bevel Gears: Definitions, formative number of teeth, Design based on

strength, dynamic and wear loads. Worm Gears: Definitions, Design based on strength, dynamic,

wear loads and efficiency of worm gear drives. 07 Hours


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UNIT - 6

Clutches & Brakes: Design of Clutches: Single plate, multi plate and cone clutches. Design of

Brakes: Block and Band brakes: Self locking of brakes: Heat generation in Brakes.05 Hours

UNIT - 7

Lubrication and Bearings: Lubricants and their properties, Mechanisms of Lubrication bearing

modulus, coefficient of friction, minimum oil film thickness, Heat Generated, Heat dissipated,

Bearing Materials, Examples of journal bearing and thrust bearing design. 07 Hours

UNIT - 8

IC Engine Parts: Design of piston, connecting rod and crank shaft. 05 Hours

DESIGN DATA HANDBOOK:

1. Design Data Hand Book , K. Lingaiah, McGraw Hill, 2nd Ed.

2. Data Hand Book, K. Mahadevan and Balaveera Reddy, CBS Publication

3. Design Data Hand Book, H.G. Patil, I. K. International Publisher, 2010.

TEXT BOOKS:

1. Mechanical Engineering Design, Joseph E Shigley and Charles R. Mischke. McGraw Hill

International edition, 6th Edition 2003.

2. Design of Machine Elements, V. B Bhandari, Tata McGraw Hill Publishing Company Ltd.,

New Delhi, 2nd Edition 2007.

REFERNCE BOOKS:

1. Machine Design, Robert L. Norton, Pearson Education Asia, 2001.

2. Design of Machine Elements, M. F. Spotts, T. E. Shoup, L. E.

Hornberger, S. R. Jayram and C. V. Venkatesh, Pearson Education, 2006.

3. Machine Design, Hall, Holowenko, Laughlin (SchaumsOutlines series) Adapted by S.K.Somani, Tata McGraw Hill Publishing Company Ltd., New Delhi, Special Indian Edition, 2008.


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CONTENTS

Sl. No. Particulars Page No.

1 Curved Beams and Cylinders 3-71

2 Design of Belts, Ropes and Chains 72-106

3 Design of Springs 107-193

4 Spur and Helical Gears 194-229

5 Bevel Gears and Worm Gears 230-242

6 Clutches and Brakes 243-294

7 Lubrication and Bearings 295-333

8 IC Engine Parts 334-367


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UNIT 1

CURVED BEAMS AND CYLINDERS

Theory of Simple Bending:Due to bending moment, tensile stress develops in one portion of

section and compressive stress in the other portion across the depth. In between these two

portions, there is a layer where stresses are zero. Such a layer is called neutral layer. Its trace on

the cross section is called neutral axis.

Assumptions:

1. The material of the beam is perfectly homogeneous and isotropic.

2. The cross section has an axis of symmetry in a plane along the length of the beam.

3. The material of the beam obeys Hookes law.

4. The transverse sections which are plane before bending remain plane after bending also.

5. Each layer of the beam is free to expand or contract, independent of the layer above or

below it.

6. Youngs modulus is same in tension & compression.

Consider a portion of beam between sections AB and CD as shown in the figure. Let e1f1be the

neutral axis and g1h

1an element at a distance y from neutral axis. Figure shows the same portion

after bending. Let r be the radius of curvature and is the angle subtended by a1b1and c1d1at

centre of radius of curvature. Since it is a neutral axis, there is no change in its length (at neutral

axis stresses are zero.)

EF = E1F1= R

Now, Strain in layer =

G1H1= (R+Y)

GH = R

Also Stress, --------- (1)

Also

OR dF = 0

there is no direct force acting on the element considered.


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but

Therefore, ---------- (2)

From (1) and (2)

CURVED BEAM

Curved beams are the parts of machine members found in C - clamps, crane hooks, frames of

presses, riveters, punches, s hears, boring machines, planers etc. In straight beams the neutral

axis of the section coincides with its centroidal axis and the stress distribution in the beam is

linear. But in the case of curved beams the neutral axis o f the section is shifted towards the

centre of curvature of the beam ca using a non-linear [hyperbolic] distribution of stress. The

neutral axis lies between the centroidal axis and the centre of curvature and will always be

present within the curved beams.

Stresses in Curved Beam

Consider a curved beam subjected to bending moment Mb as shown in the figure. The

distribution of stress in curved flexural member is determined by using the following

assumptions:


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1. The material of the beam is perfectly homogeneous [i.e., same material throughout] and

isotropic [i.e., equal elastic properties in all directions]

2. The cross section has an axis of symmetry in a plane along the length of the beam.

3. The material of the beam obeys Hooke's law.

4. The transverse sections which are plane before bending remain plane after bending also.

5. Each layer of the beam is free to expand or contract, independent of the layer above or

below it.

6. The Young's modulus is same both in tension and compression.

Derivation for stresses in curved beam

Nomenclature used in curved beam,Ci=Distance from neutral axis to inner radius of curved beam

Co=Distance from neutral axis to outer radius of curved beam

C1=Distance from centroidal axis to inner radius of curved beam

C2= Distance from centroidal axis to outer radius of curved beam

F = Applied load or Force

A = Area of cross section

L = Distance from force to centroidal axis at critical section

d= Direct stressbi= Bending stress at the inner fiber

bo= Bending stress at the outer fiberri= Combined stress at the inner fiberro= Combined stress at the outer fiber


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In the above figure the lines 'ab' and 'cd' represent two such planes before bending. i.e., when

there are no stresses induced. When a bending moment 'Mb' is applied to the beam the plane cd

rotates with respect to'ab' through an angle 'd ' to the position 'fg' and the outer fibers areshortened while the inn er fibers are elongated. The original length of a strip at a distance 'y' fro

m the neutral axis is (y + rn). It is shortened bythe amount ydand the stress in this fiber is,

= E.e

Where = stress, e = strain and E = Young's Modulus

Since the fiber is shortened, the stress induced in this fiber is compressive stress and hence

negative sign.The load on the strip having thickness dy and cross sectional area dA is'dF'

From the condition of equilibrium, the summation of forces over the whole cross-section is zero

and the summation of the moments due to these forces is equal to the applied bending moment.

Let, Mb = Applied Bending Moment

ri = Inner radius of curved beam

ro = Outer radius of curved beam


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rc= Radius of centroidal axis

rn= Radius of neutral axis

CL= Centre line of curvature

Summation of forces over the whole cross section

The neutral axis radius 'rn' can be determined from the above equation.

If the moments are taken about the neutral axis,

Since represents the statical moment of area, it may be replaced by A.e., the product of

total area A and the distance 'e' from the centroidal axis to the neutral axis.

This is the general equation for the stress in a fiber at a distance 'y' from neutral axis.


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Difference between a straight beam and a curved beam

Straight Beam Curved Beam

Stresses are not proportional to the distance

of the fibres from the neutral axis and

therefore it causes nonlinear distribution of

stress usually hyperbollic in nature.

Stresses are proportional to the distance of the

fibres from the neutral axis and hence the

distribution of stress is linear.

Neutral axis of the section does not coincide

with the centroidal axis and it is shifted

towards the centre of curvature of the beam.

Neutral axis of the section coincides with its

centroidal axis.

The general Equation of bending stress in

curved beam is.

The general equation for bending stress in a

straight beam is

Stress distribution in C.B is as shown in fig Stress distribution in straight beam r is as shown fig


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Location of the neutral axis By considering a rectangular cross section


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Centroidal and Neutral Axis of Typical Section of Curved Beams


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Why stress concentration occurs at inner side or con cave side of curved beam

Consider the elements of the curved beam lying between two axial planes ab and cdseparated by angle . Let fg is the final position of the plane cd having rotated through an angle

d about neutral axis. Consider two fibers sym metrically located on ei ther side o f the neutral

axis. Deformation in both the fibers is same and equal to yd .

Since length of inner element is smaller than outer element, the strain induced and stress

developed are higher for inner element than outer element as shown.


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Thus stress concentration occur at inner side or concave side of curved beam

The actual magnitude of stress in the curved beam would be influenced by magnitude of

curvature However, for a general comparison the stress distribution for the same section and

same bending moment for the straight beam and the curved beam are shown in figure.

It is observed that the neutral axis shifts inwards for the curved beam. This results in stress to be

zero at this position, rather than at the centre of gravity.

In cases where holes and discontinuities are provided in the beam, they should be preferably

placed at the neutral axis, rather than that at the Centroidal axis. This results in a better stress

distribution.

Example:For numerical analysis, consider the depth of the section ass twice the inner radius.


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For a straight beam:

Inner most fiber:

Outer most fiber:

For curved beam: h=2ri

e = rc- rn= h0.910 h = 0.0898h, co= ro- rn= h0.9 10h = 0590h

ci= rnri = 0.910h - = 0.410h


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Comparing the stresses at the inner most fiber based on (1) and (3), we observe that the stress at

the inner most fiber in this case is:

bci= 1.522BSi

Thus the stress at the inner most fiber for this case is 1.522 times greater than that for a straightbeam.

From the stress distribution it is observed that the maximum stress in a curved beam is higher

than the straight beam.

Comparing the stresses at the outer most fiber based on (2) and (4), we observe that the stress at

the outer most fiber in this case is:

bco= 1.522BSi

Thus the stress at the inner most fiber for this case is 0.730 times that for a straight beam.

The curvatures thus introduce a non linear stress distribution.

This is due to the change in force flow lines, resulting in stress concentration on the inner side.

To achieve a better stress distribution, section where the Centroidal axis is the shifted towards

the insides must be chosen, this tends to equalize the stress variation on the inside and outside

fibers for a curved beam. Such sections are trapeziums, non symmetrical I section, and T

sections. It should be noted that these sections should always be placed in a manner such that theCentroidal axis is inwards.

Problem no.1

Plot the stress distribution about section A-B of the hook as shown in figure.

Given data:

ri= 50mm

ro= 150mm F =

22X103N

b = 20mm

h = 150-50 = 100mm

A = bh = 20X100 = 2000mm2


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e = rc- rn= 100 - 91.024 = 8.976mm

Section A-B will be subjected to a combination of direct load and bending, due to

the eccentricity of the force.

Stress due to direct load will be,

y = rnr = 91.0 24r

Mb= 22X103X1 00 = 2.2X106N-mm


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Problem no.2

Determine the value of t in the cross section of a curved beam as shown in figsuch that t he normal stress due to bending at the extreme fibers are numerically

equal.

Given data;Inner radius ri=150mm Outerradius ro=150+40 +100

=290mm

Solution;From Figure Ci+ CO= 40 + 100

= 140mm (1)

Since the normal stresses due to bending at theextreme fiber are numerically equal we have,

i.e Ci=

= 0 .51724Co (2)

Radius of neutral axis

rn=

rn=197.727 mm

ai= 40mm; bi= 100mm; b2=t;

ao= 0; bo= 0; ri= 150m m; ro= 290mm;

=

i.e., 4674.069+83.6 1t = 4000+100t;

t = 41.126mm


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Problem no.3

Determine the stresses at point A and B of the split ring shown in thefigure.

Solution:

The figure shows the cri tical section of the split ring.

Radius of centroidal axis rc= 80mm

Inner radius of curved b eam ri= 80-30

= 50mm

Outer radius of curved beam ro= 80 +30

= 110mm

Radius of neutral axisrn= 77.08mm


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Bending moment about centroidal axis Mb= Fl = 20,000x80

=16x105N-mm

Distance of neutral axis to centroidal axis

e = rc-rn= 80-77.081

=2.919mm

Distance of neutral axis to inner radiusci= rn-ri

= 77.081-50

=27.081mm

Distance of neutral axis to outer radiusco= ro-rn

= 110-77.081

=32.919mm

Direct stress d=

=7.0736N/mm2(comp.)

Bending stress at the inner fiber bi=

567/50

= 105N/mm2(compressive)

Bending stress at the outer fiberbo

=

= 58.016N/mm2

(tensile)

Combined stress at the inner fiber


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ri

=d

+bi

=7.07

= - 112.0736N/mm2(compressive)

combined stress at the outer fiber

ro=d+bo= 7.0736+58.016

= 50.9424N/mm2(tensile)

Maximum shear stressmax= 0.5xmax

=0.5x112.0736

=56.0368N/mm2, at B

The figure.


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Problem No. 4

Curved bar of rectangular section 40x60mm and a mean radius of

100mm is subjected to a bending moment of 2KN-m tending to

straighten the bar. Find the position of the Neutral axis and draw a

diagram to show the variation of stress across the section.

Solution:

Given data:

b= 40mm h= 60m m rc=100mm

Mb= 2x106N-mm

C1=C2= 30m m

ro= rc+h/2=100+30=13

0=(ri+c1+c2) ri= rc- h/2 = 100

30= 7 0mm (rc-c1) rn= 96.924mm

Distance of neutral axis to centroidal axise = rc- rn= 100-96.924 =3.075mm

Distance of neutral axis to inner radius

ci= rnri= (c1-e) = 26.925mm Distance of neutral axis to outer radius

co=c2+e= (ro-rn) = 33.075mmArea

A= bx h = 40x60 = 2400 mm2

Bending stress at the inner fiber bi=

= 104.239 N/mm2(compressive)


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Bending stress at the outer fiberbo

=4302:;540

= -68.94 N/mm2(tensile)

Bending stress at the centroidal axis =

= -8.33 N/mm2(Compressive)

The stress distribution at the inner and outer fiber is as shown in thefigure.


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Problem No. 525mm and depth of trapezium =120mm.Find the proper value of b, if the extreme

fiber stresses due to pure bending are numerically equal, if the section is subjected

to a couple which develop a maximum fiber stress of 60Mpa.Determine the

magnitude of the couple.

Solutionri= 120mm; bi= b; bo= 25mm; h = 120mm bi=bo= 60MPa

Since the extreme fibers stresses due to pure bending are numerically

equal we have,

We have,Ci/ri=co/ro

2ci=co

But h= ci+ co

120 = ci+2ci

Ci=40mm; co=80mm

The section of a crane hook is a trapezium; the inner face is b and is at adistance of 120mm from the centre line of curvature. The outer face is

ci/co=120/240

rn= ri+ ci= 120+40 =160 mm


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b=150.34mm

To find the Centroidal axis, (C2)

bo= 125.84mm; b=25mm; h=120mm

= 74.313mm.

But C1=C2

rc= ro-c2=240 - 74.313 =165.687mm

e=rc- rn= 165.687 - 160 = 5.6869 mm

Bending stress in the outer fiber,

= 1050.4mm

Mb=10.8x106N-mm


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Problem no.6

Determine the stresses at point A and B of the split ring shown infig.1.9a

Solution:

Redraw the critical section as shown in the

figure. Radius of Centroidal axis rc= 80mm

Inner radius of curved beam ri= 80/60= 50mm

Outer radius of curved beam ro= 80 +

60= 110mm

Radius of neutral axis r =

=77.081mm

,

Applied force F = 20kN = 20,000N (compressive)

Area of cross section A = 2827.433mm

Distance from centroidal axis to force -= rc= 80mm Bending

moment about centroidal axis Mb= FI = 20,000x80

=16x105Nmm

Distance of neutral axis to centroidal axis e = rc/rn

= 80/77.081

=2.919mm


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ci= rn!ri= 77.081!50 = 27.081mm

Distance of neutral axis to outer radius

co= ro!rn = 110!77.081 = 32.919mm

Direct stress 1/23,44

=!7.0736N/mm2(comp.)

Bending stress at the inner fiber

= 105N/mm2(compressive)

Bending stress at the outer fiber bo


Combined stress at the inner fiber

ri=d+bi=7.0736+105.00=112.0736N/mm2

(compressive)

Combined stress at the outer fiber

ro=d+b=7.0736+58.016

= 50.9424N/mm

2

(tensile)

Maximum shear stress

Gmax= 0.5x max= 0.5x112.0736

= 56.0368N/mm2, at B


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The figure shows the stress distribution in the critical section.


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Problem no.7

Determine the maximum tensile, compressive and shear stress induced

in a c frame of a hydraulic portable riveter shown in fig.1.6a

Solution:

R1

0

Draw the critical section as shown 50 0 9000N

in

the figure. 80

Inner radius of curved beam ri = 175 mm100mm

Outer radius of curved beam ro= 100+80

= 180mm

Radius of centroidal axis r = 100+10

h = 80mm

c2e

c1c

= 140mm

Radius of neutral axis rn=

_Criticalb

=50mm

Section

co ci r = 100mm 1io

A A nNC rc

CL

= 136.1038mm

Distance of neutral axis to centroidal axis

e = rc- rn= 140-136.1038 = 3.8962mm


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ci= rn -ri = 136.1038-100 = 36.1038mm

Distance of neutral axis to outer radius

co= ro -rn= 180-136.1038 = 43.8962mm

Distance from centroidal axis to force

-= 175+ rc= 175+140 = 315mm

Applied force F = 9000N

Area of cross section A = 50x80 = 4000mm

2

Bending moment about centroidal axis Mb= FI = 9000x315 =

2835000 N-mm

Direct stress d= 9000= 2.25N/mm2(tensile)

Bending stress at the inner fiber bi500


Bending stress at the outer fiber bo=743196/7510

=44.326N/mm2(compressive)

Combined stress at the inner fiber ri= d+ bi= 2.25+65.676

= 67.926N/mm2

(tensile)

Combined stress at the outer fiber ro= d+ bo=

= 42.112 N/mm2(compressive)


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Maximum shear stress Gmax= 0.5xmax= 0.5x67.926

= 33.963 N/mm2, at the inner fiber

The stress distribution on the critical section is as shown in the figure.

ri=67.926 N/mm2

Combined stress ro=-42.112 N/mm2

bi=65.676 N/mm2

Bending stress bo=-44.362 N/mm2

Direct stress (d) d=2.25 N/mm2

CA

NA

b = 50 mm

h =80 mm


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Problem no.8

The frame punch press is shown in fig. 1.7s. Find the stress in inner andouter surface at section A-B the frame if F = 5000N

Solution:

Draw the critical section as shown in thefigure.

Inner radius of curved beam ri= 25mm


h = 40mm F

c2e

c1

=6mm

18mm

=b

i

b

co ci ri= 25mm 100mmrc

roNA n

C FL

= 65mm

Distance of centroidal axis from inner fiber c1= 4

=

,051_/76

= 16.667mm


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Radius of centroidal axis rc= ri+ c1

=25+16.667 = 41.667 mm

Distance of neutral axis to centroidal axis e = rc-rn

= 1.667-38.8175

=2.8495mm

Distance of neutral axis to inner radius ci= rn-ri

= 38.8175-25=13.8175mm

Distance of neutral axis to outer radius co= ro-rn

= 65-38.8175=26.1825mm

Distance from centroidal axis to force -= 100+ rc= 100+41.667

= 141.667mm

Applied force F = 5000N

Area of cross section A = RS__TU Q V_= 480mm2

Bending moment about Centroidal axis

Mb= FI

= 5000x141.667

= 708335 N-mm


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Direct stress d= 10.417N/mm2(tensile)

Bending stress at the inner fiber bi= 286.232N/mm2(tensile)

Bending stress at the outer fiber bo=

= 208.606N/mm2

(compressive) Combined stress at the inner fiber ri= d+ bi

= 10.417+286.232


Combined stress at the outer fiber ro= d+ bo=

= 198.189N/mm2(compressive)

Maximum shear stress Gmax= 0.5x max= 0.5x296.649

= 148.3245 N/mm2, at the inner fiber


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The figure shows the stress distribution in the critical section.

ri=296.649 N/mm2

Combined stress ro=-198.189 N/mm2

bi=286.232 N/mm2

Bending stress bo=-208.606 N/mm2

Direct stress (d) d=10.417 N/mm2

CA

NA

b = 6 mm b = 18 mmo i

h =40 mm


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Problem no.9

The section of a crane hook is rectangular in shape whose width is

30mm and depth is 60mm. The centre of curvature of the section is at

distance of 125mm from the inside section and the load line is 100mmfrom the same point. Find the capacity of hook if the allowable stress in

tension is 75N/mm2

h=60mm

b=30mm100

1 2 5mm

Solution:

Draw the critical section as shown in thefigure.

Inner radius of curved beam ri= 125mm


= 185mm

Radius of centroidal axis rc=100+60

/

= 130mm

Radius of neutral axis rn= -H_ = -H 60I) JK8

#I*

%

J_

8%

F = ?h = 60mm

e

2 1

30mm

Load

line

b

=

100

o iri= 125mm

orn

rcl

N A CF L

= 153.045mm

Distance of neutral axis to centroidal axis e = rc- rn

= 155-153.045 = 1.955mm


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Distance of neutral axis to inner radius ci= rnri

= 153.045-125

= 28.045mm

Distance of neutral axis to outer radius co= ro- rn

= 185-153.045

= 31.955mm

Distance from centroidal axis to force l = rc-25

= 155-25

= 130mm

Area of cross section A = bh

= 30x60

= 1800mm2

Bending moment about centroidal axis Mb= Fl

= Fx130

F = 8480.4N =Capacity of the hook.


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Problem no.10

The figure shows a loaded offset bar. What is the maximum offset

distance x if the allowable stress in tension is limited to 50N/mm2

Solution:

Draw the critical section as shown in the figure.

Radius of centroidal axis rc= 100mm

Inner radius ri= 100100/2 = 50mm

Outer radius ro= 100+ 100/2 = 150mm

_

Z

2ro ri

Radius of neutral axis rn= WD

4F

F _F 2

93.3mm =

e = rc- rn= 100 - 93.3 = 6.7mm

ci= rnri= 93.350 = 43.3

mm

co= ro- rn= 150 - 93.3 =

56.7mm

A =i,x d

2=

i,x 100

2= 7853.98mm

2

Mb= FR= 5000R

Combined maximum stress at the inner fiber

(i.e., at B)


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Problem no.11

An Open S made from 25mm diameter od as shown in the figure

determine the maximum tensile, compressive and shear stress

Solution:

(I) Consider the section P-Q

Draw the critical section at P-Q as shown in the figure.


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Rn = 99.6mm

Distance of neutral axis from centroidal axis e =rc- rn

=100 - 99.6 = 0.4mm

Distance of neutral axis to inner fiber ci= rnri

= 99.687.5 =12.1 mm

Radius of centroidal axis rc=100mm

Inner radius ri=100 = 87.5mm

Outer radius ro=100+12.5 = 112.5mm

Radius of neutral axis


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Distance of neutral axis to outer fiber co= ro-rn

=112.599.6 = 12.9 mm

Area of cross-section A = 490.8mm

Distance from centroidal axis I = rc= 100mm

Bending moment about centroidal axis Mb= F.l = 100 x 100

= 100000Nmm

Combined stress at the outer fiber (i.e., at Q) =Direct stress +bending

stress

=56.36N/mm2(compressive) Combined stress at inner fibre (i.e., at p)

ri=Direct stress + bending stress

F Mbci 1000

=72.466 N/mm2(tensile)

(ii)Consider the section R -S

Redraw the critical section at RS as shown in fig.

rc= 75mm

ri= 75- 25=62.5 mm

r=87.5 mm

e = rc- rn= 75 -74.4755 =0.5254 mm

ci= rnri=74.475562.5 =11.9755 mm


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co= ro- rn= 87.574.4755

= 13.0245 mm

l = rc= 75 mm

Mb= Fl = 1000 X 75 = 75000 Nmm

Combined stress at the outer fibre (at R) = Direct stress + Bending stress

= - 41.324 N/mm2(compressive)

Combined stress at the inner fiber (at S) = Direct stress + Bending stress

= 55.816 N/mm2(tensile)

Maximum tensile stress = 72.466 N/mm2at P Maximum

compressive stress = 56.36 N/mm2atQ

Maximum shear stress max=0.5

max= 0.5 X 72.466

= 36.233 N/mm2at P

Stresses in Closed Ring

Consider a thin circular ring subjected to symmetrical load F as shown in the figure.


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The ring is symmetric al and is loaded symmetrically in both the horizontal andvertical directions.

Consider the horizontal section as shown in the figure. At the two ends A and B,

the vertical forces would be F/2.

No horizontal forces would be there at A and B. this argument can be proved by

understanding that since the ring and the extern al forces are symmetrical, the

reactions too must be symmetrical.

Assume that two horizontal inward forces H, act at A and B in the upper half, asshown in the figure. In this case, the lower

half must have forces H acting outwards as shown.

This however, results in violation of symmetry and hence H

must be zero. Besides the forces, moments of equal

magnitude M0 act at A and B. It should be noted that these

moments do not violate the condition of symmetry. Thus

loads on the section can be treated as that shown in the

figure. The unknown quantity is M0. Again Considering

symmetry, We conclude that the tangents at A and B must be

vertical and must remain so after deflection or M0 does not

rotate. By Castiglianos theorem, the partial

derivative of the strain energy with respect to the load gives the displacement of

the load. In this case, this would be zero.

.(1)


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The bending moment at any point C, located at angle , as shown in

the figure.

Will be

..(2)

As per Castiglianos theorem,

From equation (2)

And, ds = Rd


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INTRODUCTION TO COMPOUND CYLINDERS

In thick walled cylinders subjected to internal pressure only, it can be seen from the

equation of the hoop stress that the maximum stresses occur at the inside radius and this

can be given by:

can be shown that for large internal pressures in thick walled cylinders the wall

thickness is required to be very large. This is shown schematically in figure 2-10.

Figure 2.10 Variation of wall thickness with the internal pressure in thick cylinder

This means that the material near the outer edge of the cylinder is not effectively used

since the stresses near the outer edge gradually reduce.

In order to make thick-walled cylinders that resist elastically large internal pressure and

make effective use of material at the outer portion of the cylinder the following methods

of pre-stressing are used:

Shrinking a hollow cylinder over the main cylinder. (Compound cylinders)

Multilayered or laminated cylinders.

Autofrettage or self hooping.

Compound cylinders

An outer cylinder (jacket) with the internal diameter slightly smaller thanthe outer diameter of the main cylinder is heated and fitted onto the main

cylinder. When the assembly cools down to room temperature, a compound

cylinder is obtained. In this process the main cylinder is subjected to an

external pressure leading to radial compressive stresses at the interface (Pc)

as shown in figure 2.11.


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Figure 2.11 contact stress Pc in a compound cylinder

The outer cylinder is subjected to an internal pressure leading to tensile circumferential

stresses at the interface (Pc) as shown in figure 2.11. Under these conditions as the internalpressure increases, the compression in the internal cylinder is first released and then onlythe cylinder begins to act in tension.

Figure. 2.11 Compound cylinders

Compound cylinders:

The tangential stress at any radius r for a cylinder open at both ends and subjected tointernal pressure (Birnies equation)


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The Radial stress at any radius r for a cylinder open at both ends and subjected to internalpressure (Birnies equation)

The tangential stress at the inner surface of the inner cylinder

The tangential stress at the outer surface of the inner cylinder

The tangential stress at the inner surface of the outer cylinder

The tangential stress at the outer surface of the outer cylinder

Total shrinkage allowance when two cylinders are made of two different materials

When both the cylinders are made of same material the pressure between the cylinders is

given by the equation


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Laminated cylinders

The laminated cylinders are made by stretching the shells in tension and then weldingalong a longitudinal seam. This is shown in figure 2.12.

Figure.2.12 Laminated cylinder

Autofrettage

Pressure vessels are now widely used in nuclear power plants for steam and powergeneration. Other pressure vessel applications may involve pressures as high as 1380

MPa and temperatures of up to 300 C, resulting in the pressure vessel material holding

immense potential energy exerted by the working fluid. The process fluid may also be a

source of hydrogen embrittlement and/or stress corrosion cracking. Such high-pressurevessels require proper understanding of the stress levels and their distributions in order

to have fail-safe designs or even to minimize the probability of disruptive failures. Past

pressure vessel catastrophic failures, arising from lack of understanding of stress levels,material properties and fluid/structure environmental interactions, particularly early in

the last century, were very expensive in terms of losses in materials and human life, and

they were the main impetus for the early studies of stresses in cylinders of various

materials.

High-pressure vessels are now of great importance in many industries and their

economic use often depends upon the occurrence of small, controlled, permanentdeformations. Before commissioning, pressure vessels are normally pressure tested at an

overstrain pressure of 1.251.5 times the design pressure in order to test for leakages.This process results in yielding of the bore and may also advantageously lead tocatastrophic failure for poorly designed or fabricated vessels. Vessels with brittle

characteristics may also fail at this stage. After overstraining, residual stresses are left in

the cylinder and the nature of these residual stresses is now widely known. However, the

residual stress levels are not documented for use in service or during de-rating after

periodic inspections. In service, the vessels are able to carry a much higher load beforere-yielding than would be the case without the leak test. Overstraining beyond the leak

test pressure is usually carried out during manufacture and this technique is called

Autofrettage or self-hooping.

In some applications of thick cylinders such as gun barrels no inelastic deformation ispermitted. But for some pressure vessel design satisfactory function can be maintained


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until the inelastic deformation that starts at inner bore spreads completely over the wall

thickness. With the increase in fluid pressure yielding would start at the inner bore and

then with further increase in fluid pressure yielding would spread outward. If now thepressure is released the outer elastic layer would regain its original size and exert a

radial compression on the inner shell and tension on the outer region.

This gives the same effect as that obtained by shrinking a hoop over an inner cylinder.This is known as Self- hooping or Autofrettage. This allows the cylinder to operate athigher fluid pressure. For a given autofrettage fluid pressure a given amount of inelasticdeformation is produced and therefore in service the same fluid pressure may be usedwithout causing any additional inelastic deformation.

2.9 Examples on compound cylinders

E1: A shrink fit assembly formed by shrinking one tube over another tube is subjectedto an internal pressure of 60 MPa. Before the fluid is admitted, the internal & externaldiameters of the assembly were 120 & 200mm and the diameter of the junction was 160mm. After shrinking, the contact pressure at the junction was 8 MPa, plot the resultant

stress distribution in each cylinder after the fluid has been admitted.

Solution:

Hoop stresses due to shrink fit in both the cylinders

The tangential stress at the inner surface of the inner cylinder

The tangential stress at the outer surface of the inner cylinder

The tangential stress at the inner surface of the outer cylinder

The tangential stress at the outer surface of the outer cylinder


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Variation of tangential stresses due to shrink fit alone

Stress distribution only due to internal pressure alone

Lames equation for internal pressure:

At r= 60 mm, t = +127.5 MPa (tensile)At r= 80 mm t = +86.4 MPa (tensile)At r=100 mm t = + 67.5 MPa (tensile)

Resultant stress distribution

Resultant Stress at the inner surface of the inner cylinder= -36.57+127.5 = + 90.3 MPa

Resultant Stress at the outer surface of the inner cylinder= -28.57+86.41 = + 57.19 MPa

Resultant Stress at the inner surface of the outer cylinder=36.44+86.48

= 122.92 MPa

Resultant Stress at the outer surface of the outer cylinder= 28.44+67.5

= 95.94 MPa


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Resultant stress distribution

2.10 Introduction to press and shrink fits

Press fits, or interference fits, are similar to pressurized cylinders in that the placementof an oversized shaft in an undersized hub results in a radial pressure at the interface.In

a press fit, the shaft is compressed and the hub is expanded. There are equal and

opposite pressures at the mating surfaces. The relative amount of compression and

expansion depends on the stiffness (elasticity and geometry) of the two pieces. The sum

of the compression of the shaft and the expansion of the hub equals the interference

introduced.


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Commonly used Interference Fits

H7/p6 Locational Interference: Fit for parts requiring rigidity and alignment with prime

accuracy of location, but without special bore pressure requirements.

H7/r6 Medium Drive fit for ferrous parts and light drive fit for non ferrous parts that can be

dismantled.

H7/s6 Permanent or semi -permanent assemblies of steel and cast iron. Fit for ordinary steel

parts or shrink fits on light sections, the tightest fit usable with cast iron. For light alloys this

gives a press fit.

H7/u6 Force fit: Fit suitable for parts which can be highly stressed or for shrink fits where

the heavy pressing forces required are impractical.

Analysis of Press Fits:

Start by finding the interface pressure.

Where B is the shrinkage allowance for hub and shaft of the same material with

modulus of elasticity E.

If the shaft is solid, di= 0 and

If the shaft and hub are of different materials

Once we have the pressure, we can use the cylinder equations to compute the hoop

stresses at the interface. The hoop stress at the ID of the hub istensile.


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The hoop stress at the OD of the shaft is compressive.

Strain Analysis of Press Fits

The press fit has no axial pressure and it is a biaxial stress condition. The circumferentialstrain which equals the radial strain (because C = 2pr):

c

c

r

E E

Because the radial change, we get the increase in Inner diameter of the outer member (hub):

The decrease in Outer diameter of the inner member (shaft):

The total axial force required to assemble a force fit (approximately): Fa dLfpc

The torque transmitted:

Where,

f= coefficient of friction


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SHRINK FITS

The temperature to which a piece to be shrunk must be heated for assembling:

t2 is the final temperature to which hub must be heated

t1 is the temperature of the shaft .

is the thermal expansion coefficient of material of the hub.

Examples on Press and shrinkfit:

E1. A cast steel crank is to be shrunk on a 250 mm steel shaft. The outside diameter of thecrank

hub is 444.5 mm. The maximum tangential stress in the hub is limited to 150 MPa. Coefficient

of friction between the hub and the shaft is 0.15. Determine:

7. The required bore of the crank.8. Probable value of the normal pressure between the hub and the shaft.9. The torque that may be transmitted without using a key, if the hub length is 250 mm.

Solution:

The hoop stress is maximum at the ID of the hub and is given by:

Limiting value of hoop stress is given to be 150 MPa

do=444.5mm, dc=250mm. Substitute and determine Pc.

Pc= 78 MPa.

Determine B, the total shrinkage allowance for hub and shaft of the same material, withmodulus of elasticity, E.If the shaft is solid, di= 0.

E=200x103MPa

B = total shrinkage allowance = 0.28mm


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Assuming that the assembly is done using selective assembly technique,

The required bore of the crank = basic size- shrinkage allowance

=250-0.28=249.72 mm

The total axial force required to assemble a force fit (approximately): Fa dLfpc

f = 0.15, Pc=78 MPa, L=75mm, d=250mm

Substitute and determine the force required to assemble

Fa= 2.29x 106N

If this force calculated cannot be obtained with the regular presses,it becomes obvious that

we should reduce the interference at the time of assembly. This can be achieved by heating

the hub and expanding its internal diameter by a calculated amount and then slipping it over

the shaft. This produces the required shrink fit once assembly cools and reaches the room

temperature.

F d d Lfpc

T e torque transm tte , T a2 2

Fa= 2.29x106N, d= 250 mm. Substitute and determine T.

T= 287x106N-mm.

The torque calculated is based on shrink fit only (without the presence of the key).

E2. A steel hub of 440 mm outside diameter, 250mm inside diameter & 300 mm length has an

interference fit with a shaft of 250 mm diameter. The torque to be transmitted is 30x104N-m.

The permissible stress for the material of the shaft & hub is 120 MPa. The coefficient of friction

is 0.18. Determine:

The contact pressure

The interference required

The tangential stress at the inner and outer surface of the hub.

Force required to assemble.

Radial stress at the outer & inner diameter of the hub.

The contact pressure can be determined based on the torque transmitted.F d d Lfp

cThe torque transmitted, T a

2 2


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T= 30x104N-m= 30x10

7N-mm. d = 250mm,

L= 300 mm, f=0.18. Substituting the above values,

Pc= 56.6 MPa.

The hoop stress at the ID of the hub is tensile and is given by:

Substitute do= 440mm, dc=250mm and obtain the value of hoop stress at the ID of the hub.

tc 110.56 MPa < 120 MPa

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

di =0, Pc = 56.6 MPa and determine B. B=0.21 mm.

The interference required is 0.21 mm.

The total axial force required to assemble a force fit (approximately):

Radial Stress at the inner surface of the hub= -Pc= -56 .6 MPa.

Radial Stress at the outer surface of the hub = 0


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2.1 Introduction to pressure vessels

Pressure vessels in the form of cylinders and tanks are used for storing variety of liquids

and gasses at different temperatures and pressures. Some of the substances stored can

be lethal to human beings if exposed and some can be highly explosive. Bursting of

pressure vessels due to improper design can prove fatal to human life and property. It isimperative that a designer should have a comprehensive understanding of the principles

of designing such pressure vessels based on national and international standards.

2.2 Stresses in thin cylinders

If the wall thickness is less than about 7% of the inner diameter then the cylinder

may be treated as a thin one. Thin walled cylinders are used as boiler shells, pressure

tanks, pipes and in other low pressure processing equipments. A thin cylinder is also

defined as one in which the thickness of the metal is less than 1/20 of the diameter of the

cylinder. In thin cylinders, it can be assumed that the variation of stress within the metalis negligible, and that the mean diameter, dm is approximately equal to the internal

diameter, di.

In general three types of stresses are developed in pressure cylinders viz. circumferentialor hoop stress, longitudinal stress in closed end cylinders and radial stresses. These

stresses are demonstrated in figure 2.1.

Figure 2.1

Radial stress in thin cylindrical shells can be neglected as the radial pressure is not generallyhigh and that the radial pressure acts on a larger area.

The internal pressure, p tends to increase the diameter of the cylinder and this produces ahoop or circumferential stress (tensile). If the stress becomes excessive, failure in the formof a longitudinal burst would occur.


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Consider the half cylinder shown. Force due to internal pressure, piis balanced by the force

due to hoop stress, ti.e. hoop stress x area = pressure x projected area

t x 2 L t = p x L x di

t = (p di) / 2 t

Longitudinal stress in a cylinder:

Figure 1.2

The internal pressure p also produces a tensile stress in the longitudinal direction as

shown in figure 1.2.

The force P acting on an area (d2i/ 4) is balanced by longitudinal stress tacting overan approximate area dit.

tx dit = p (d2

i/ 4)

t= pdi / 4t

Since hoop stress is twice longitudinal stress, the cylinder would fail by tearing along a

line parallel to the axis, rather than on a section perpendicular to the axis. The equation

for hoop stress is therefore used to determine the cylinder thickness.

Pressure vessels are generally manufactured from curved sheets joined by welding.

Mostly Vbutt welded joints are used. The riveted joints may also be used. Since theplates are weakened at the joint due to the rivet holes, the plate thickness should be

enhanced by taking into account the joint efficiency. Allowance is made for this by

dividing the thickness obtained in hoop stress equation by efficiency (i.e. tearing andshearing efficiency) of the joint. A typical welded construction of a pressure vessel is

shown in figure 2.3 and riveted construction is shown in figure 2.4.


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Figure 2.3. Welded construction of a pressure vessel

Figure 2.4. Rivetted construction of a pressure vessel

2.3 Stress induced in a spherical shell:

A sphere is the most favorably stressed shaped for a vessel requiring minimum wall thickness. Itis used for extremely high pressure options. It is used in space vehicles and missiles for thestorage of liquefied gasses at lower pressures but with light weight thin walls. Spheres also havethe greatest buckling resistance. Spherical vessels are used as pressure carrying structures and asliving space in most deep submerged vehicles for oceanography. The stress induced in aspherical vessel is as shown in figure 2.5 and is given by:

t = pdi/ 4t

Figure 2.5 Stress in a spherical pressure vessel


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2.4 EXAMPLE ON THIN CYLINDERS

E1. An air receiver consisting of a cylinder closed by hemispherical ends is shown in Figure

below. It has a storage capacity of 0.25 m3and an operating internal pressure of 5 MPa. It is

made of plain carbon steel 10C4 with an ultimate tensile strength of 340 MPa. Factor ofsafety to be used is 4. Neglecting the effect of welded joints, determine the dimensions of

the receiver.

Volume of the vessel V= d2iL/4 + d

3i/6

Substituting L=2diand simplifying;

di = (3V/2)

= 0.492 m or 500 mm

L=2di = 500 X 2 =1000 mm.

Allowable stress= 340/4 =85 MPa.

Thickness of the cylinder = (neglecting the effect of welded joint)

t= pdi/2t

t= 14.7 mm or 15 mm.

Thickness of the hemispherical Ends:

t= pdi/4t

t=7.35 mm or 7.5 mm.

Problems for practice:

EP1. A seamless pipe 800 mm in diameter contains air at a pressure of 2 MPa. If the permissible stress of the pipe material is 100 MPa, find the minimum thickness of the pipe.

EP2. A cylindrical air receiver for a compressor is 2m in internal diameter and made of

plate15 mm thick. If the hoop stress is not to exceed 90 MPa and longitudinal stress is notto exceed 60 MPa find the safe air pressure.

EP3. A cylindrical shell of 2.2m internal diameter is constructed of mild steel plate. Theshell is subjected to an internal pressure of 0.8 MPa. Determine the thickness of the shellplate by adopting a factor of safety of 6. The ultimate tensile strength of steel is 470 MPa.The efficiency of the longitudinal joint may be taken as 78%.


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2.5 THICK CYLINDERS

If the wall thickness is more than about 7% of the inner diameter then the cylinder may be

treated as a Thick Cylinder.

Difference in Treatment between Thin and Thick Cylinders

In thin cylinders the hope stress is assumed to be constant across the thickness of thecylinder wall. In thin cylinders there is no pressure gradient across the wall. In thincylinders, radial stress is neglected (while it is of significant magnitude in thick cylinders).In thick cylinders none of these assumptions can be used and the variation of hoop andradial stress will be as shown in figure 2.6.

Figure 2.6 Variation of stresses in thin and thick cylinders

A thick cylinder subjected to both internal and external pressure is shown in figure 2.7.

Figure 2.7 Thick cylinder subjected to both internal and external pressure

Figure 2.8 Radial and Tangential stress (hoop stress) distribution in a thick cylinder

subjected to internal pressure only.


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Figure 2.9 Radial and Tangential stress (hoop stress) distribution in a thick cylinder

subjected to external pressure only

2.6 REVIEW OF LAMES EQUATION FOR THICK CYLINDERS:

When the material of the cylinder is brittle, such as Cast iron or Cast steel, Lames equation

is used to determine the wall thickness. It is based on the Maximum Principal stress theory,where maximum principal stress is equated to permissible stress of the material.

The three principal stresses at the inner surface of the cylinder are:

Tangential or hoop stress tLongitudinal stress ltRadial stress r

Thick Cylinders

Lames Equations:

The tangential stress in the cylinder wall at radius r

The radial stress in the cylinder wall at radius r


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Lames equation for internal pressure:

The cylinder wall thickness for brittle materials based on the maximum normal stress theory

The cylinder wall thickness for ductile materials based on the maximum shear theory

Clavarinos equation is applicable to cylinders with closed ends and made of ductile

materials. Clavarinos equation is based on maximum strain theory.

The thickness of a thick cylinrer based on Clavarinos equation is given by

Birnies equation for open ended cylinders made of ductile materials is given by


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2.7 EXAMPLE ON THICK CYLINDERS

E1: A pipe of 400 mm internal diameter and 100 mm thickness contains a fluid at 8 MPa.Find the maximum and minimum hoop stress across the section. Also sketch the hoopstress and radial stress distribution across the section.

Solution:

t= a + (b/r2) and t= a - (b/r

2)

At r = 200 mm, r= -8 Mpa.

At r= 300 mm, r= 0

Determine the values of a & b.

a= 6.4 b=576000

t= 6.4 + 576000/ r2

r= 6.4576000/ r2

Calculate the values of Hoop stress and Radial stress at 200,225,250,275 & 300 mmradius. Plot the variation on a graph sheet. Variation shown in the figure is not to scale.

r(mm) t(MPa) r(MPa)

200 20.8 -8.0

225 17.77 - 4.97

250 - 2.8115.61

275 - 1.2114.0

300 012.8

Problems for practice:

EP1:The piston rod of a hydraulic cylinder exerts an operating force of 10kN. The

frictiondue to piston packings and stuffing box is equivalent to 10% of the operating

force. The pressure in the cylinder is 10 MPa. The cylinder is made of cast iron FG

200 and the factor of safety is 5. Determine the diameter and thickness of the cylinder.

EP2: The inner diameter of a cylindrical tank for storing a liquefied gas is 250mm.The

gaspressure is limited to 15 MPa. The tank is made of plain carbon steel 10C4 (Sut=

340 MPa & = 0.27) and the factor of safety is 5. Calculate the thickness of the

cylinder.


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E3. A steel hub of 50 mm outside diameter, 28mm inside diameter & 75 mm length is tohave an interference fit with a shaft of 28 mm diameter employing a heavy press fit notusing selective assembly. Determine:

The tolerances and interference of mating parts.The maximum radial contact pressureThe actual maximum and minimum tangential stress at the contact surface of the hub.

The maximum and minimum tangential stress based on maximum strain theory.Force required to assemble the hub onto the shaft. The coefficient of friction is 0.18.The maximum torque that can be transmitted by this assembly.

Choose 28 H7/s6 as the fit for the assembly.

28 s6= Upper limit=28+0.048

=28.048 mm

Lower limit=28+0.035

=28.035 mm

28 H7 = upper limit=28+0.021

=28.021 mm

Lower limit=28+0.000

=28.000 mm.

Maximum Interference= Min.bore size- Max. shaft size= 28.000-28.048= - 0.048mm

Minimum Interference= Max. bore size- Min. shaft size= 28.021-28.035= - 0.014 mm

The interference in the assemblies varies between the limits of a maximum of 0.048mm and

a minimum of 0.014mm.

B max=0.048mm Bmin=0.014mm

The contact stress will be maximum when interference is maximum.BmaxE (dc

2di

2)(d0

2dc

2)

Pc max 2dc3(do2di2)

dc=28mm, do=50mm, E=210x103MPa, Bmax= 0.048

mm & determine Pc max.

Pc max= 123.55 MPa.

BminE (dc2di

2)(d0

2dc

2)

Pc min 2dc3(do2di2)

Pc min= 36.04 MPa.

The contact stress varies between the limits 123 MPa to 36 MPa.

The hoop stress at the ID of the hub is maximum and this also varies as the interference in

the assemblies varies and the actual maximum and minimum at the ID is given by:


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The hoop stress at the ID of the hub is maximum and this also varies as the interference in

the assemblies varies and the equivalent maximum and minimum at the ID based on Strain

theory is given by:

Substituting poisons ratio=0.3 for steel and maximum and minimum radial contact stress

Pc, determine tmax& tminbased on Max. strain theory.

tmax273MPa

tmin +79.7 MPa

Maximum torque can be transmitted when interference is maximum.When interference is maximum, Pc is maximum.The maximum torque can be calculated by:

F d d Lfpcmaxmax

a2 2

T max= 1.37x106N-mm.

The maximum axial force required to assemble a force fit (approximately):

TmaxFa

max

d

2

F a max= 97.8x103N

UNFIRED PRESSURE VESSELS

An unfired pressure vessel is defined as a vessel or a pipe line for carrying, storing orreceiving steam, gasses or liquids under pressure. I.S.2825 -1969 code for pressure vessels

gives the design procedure for welded pressure vessels of ferrous materials, subjected to aninternal pressure of 0.1 MPa to 20 MPa. This code does not cover steam boilers, nuclearpressure vessels or hot water storage tanks.Classification of pressure vessels:Pressure vessels are classified into 3 groups:

Class 1 Class 2 Class 3


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Class 1 pressure vessels are used to contain lethal and toxic substances. Substances likeHydrocyanic acid, carbonyl chloride or mustard gas which are dangerous to human life arecontained by class 1 pressure vessels. Class 1 pressure vessels are used when the operating

temperature is less than -200 C. Safety is of paramount importance in the design of these

pressure vessels. Two types of welding are employed in this case: double welded butt joint with full penetration Single welded butt joint with backing strip.

Welded joints of class 1 pressure vessels are fully radiographed.

Class 3 pressure vessels are used for relatively light duties. They are not recommended for

service when the operating temperature is less than 00C or more than 250

0C. The maximum

pressure is limited to 1.75 MPa. The maximum shell thickness is limited to 16 mm. They areusually made from carbon and low alloy steels. Welded joints are not radiographed.

Class 2pressure vessels are those which do not come under class 1 or class 3. The maximumshell thickness is limited to 38mm. Welded joints used are similar to class 1 but the joints areonly spot radiographed.

THICKNESS OF SHELLS SUBJECTED TO INTERNAL PRESSURE

The minimum thickness of shell plates exclusive of corrosion allowance in mm

a) For cylindrical shells: t pDi pDo

2fJp 2fJ p

b) For spherical shells: tpDi pDo

4fJp 4fJ p

Where J is the joint efficiency, f is the allowable tensile stress & p is the internal pressure,Di & Do are the inside and outside diameter of the shell respectively.

Corrosion in the pressure vessels can be due to:

= Chemical attack where the metal is dissolved by a chemical reagent.= Rusting due to air and moisture.

= Erosion due to flow of reagents over the surfaces of thewalls at high velocities

= Scaling or oxidation at high temperatures.

Corrosion allowance (CA) is additional metal thickness over and above that required towithstand internal pressure. A minimum corrosion allowance of 1.5 mm is recommendedunless a protective coating is employed.

END CLOSURESFormed heads used as end closures for cylindrical pressure vessels are:

Doomed heads Conical heads.

Doomed heads can further be classified into three groups: Hemispherical heads Semi ellipsoidal heads Torispherical heads


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The section below gives various formulae connected with the calculations of the thicknessof end closures.

THICKNESS OF SPHERICALLY DISHED COVERS

The thickness of spherically dished ends secured to the shell to a flange connection by

means of bolts shall be calculated by the formula (provided that the inside crown radius Rof the dished cover does not exceed 1.3 times the shell inside diameter Di and the value of

100t/R is not greater than 10)

Where J = joint factor (table 8.2)

UNSTAYED FLAT HEADS AND COVERS

Thickness of flat heads and covers:

The thickness of flat unstayed circular heads and covers shall be calculated by the formulawhen the head or cover is attached by bolts causing an edge movement

t CD p /f

For C and D refer table 8.1

The thickness of flat unstayed non-circular heads and covers shall be calculated by the

formulat CZa p /f

Where Z= the factor for non-circular heads depending upon the ratio of short span to longspan a/b


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Unstayed flat plates with uniformly distributed loads

Integral flat head

Circular plate uniformly loaded:

The thickness of the plate with a diameter D supported at the circumference and subjected to auniformly distributed pressurepover the total area.

t C1D p /f

The maximum deflection

yc

2D

4

p

Et3

Where c1and c2are coefficients from table 8.5

Circular plate loaded centrally:

The thickness of a flat cast iron plate supported freely at the circumference with a diameter D andsubjected to a load F distributed uniformly over an area D

2o/4

t 1.2 (10.670D /D)Fo f

The deflection in mm, y0.12D

2F

3Et

Grashofs formula for the thickness of a plate rigidly fixed around the circumference with the

above given type loading,

F 1/ 2t 0.65 loge(D/Do)

The deflection in mm, y0.055D F

Et 3


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Rectangular plates:

The thickness of a rectangular plate subjected to uniform load (Grashof & Bach)

The thickness of a rectangular plate on which a concentrated load F acts at the intersection ofdiagonals

Where a = length of plate, mm b =breadth of plate, mm

c3, c4= coefficients from table 8.5

Examples on Cylinder heads

E1. A cast iron cylinder of inside diameter 160mm is subjected to a pressure of 15 N/mm2. The

permissible working stress may be taken as 25MPa for cast iron. If the cylinder is closed by means ofa flat head cast integral with cylinder walls, find the thickness of the cylinder wall and the head.

Solution: Since the cylinder is made of cast iron, use the normal stress theory to calculate thewallthickness.

The cylinder wall thickness for brittle materials based on maximum normal stress theory

di= 160 mm, t=25 MPa, Pi= 15MPa Substitute and find out the value of the thickness of the cylinder

wall.

t= 80mm

Thickness of cylinder headCircular plate uniformly loaded:

The thickness of the plate with a diameter D supported at the circumference and subjected to auniformly distributed pressurepover the total area.

t C1D p /f

Since the head is cast integral with the wall, c1= 0.44 (from table 8.5)

D=160 mm, p=15 MPa, f=25 MPa,

Substitute and determine the value of tthickness of the head. t= 54.5mm or 55mm.E2. The steam chest of a steam engine is covered by a rectangular plate 240x380 mm in size. The


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plate is made of cast iron and is subjected to a steam pressure of 1.2 N/mm2and the plate is assumed

to be uniformly loaded and freely supported at the edges. Find the thickness of the plate if theallowable stress of the cast iron plate is 35 MPa.

Solution:

Rectangular plates:

The thickness of a rectangular plate subjected to uniform load (Grashof &

Bach)

Where a = length of plate, mmb = breadth of plate, mm

c3, c4= coefficients from table 8.5

a= 380mm, b= 240 mm, f= 35 MPa, p= 1.2 MPa,c3= 0.75 for a rectangular plate with the edges freely supportedDetermine the thickness t.t= 28.2 mm or 30 mm.

Problems for practice

EP1. A steel cylinder is 160 mm ID and 320 mm OD. If it is subject to an internal pressure of150

MPa, determine the radial and tangential stress distributions and show the results on a plot (using a

spreadsheet). Determine the maximum shear stress in the cylinder. Assume it has closed ends.

( Ans: t= 250 to 100 MPa, r= 0 to 150 MPa, max shear stress = 200 MPa.)

EP2. A cylinder is 150 mm ID and 450 mm OD. The internal pressure is 160 MPa and the external

pressure is 80 MPa. Find the maximum radial and tangential stresses and the maximum shear stress.

The ends are closed.

(Ans: t= 20 to 60 MPa, r= 80 to 160 MPa, max shear stress = 90 MPa.)

EP3. A 400 mm OD steel cylinder with a nominal ID of 240 mm is shrunk onto another steel cylinder

of 240 mm OD and 140 mm ID. The radial interference is 0.3 mm. Use Young's Modulus E = 200 GPa

and Poisson's Ratio = 0.3. Find the interface pressure pcand plot the radial and tangential stresses in

both cylinders. Then find the maximum internal pressure which may be applied to the assembly if the

maximum tangential stress in the inside cylinder is to be no more than 140 MPa.( Ans : pc= 120.3 MPa. inner cylinder : t= -365 to -244 MPa, r= 0 to -120.3 MPa. outer cylinder :t= 256 to 135 MPa,r= -120.3 to 0 MPa. Maximum internal pressure = 395 MPa.)

EP4. A cylinder with closed ends has outer diameter D and a wall thickness t = 0.1D. Determine the

%age error involved in using thin wall cylinder theory to calculate the maximum value of tangential

stress and the maximum shear stress in the cylinder.


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( Ans: tangential stress 9.75% : max. shear stress 11.1%)

EP5. A gun barrel has an ID of 150 mm. It is made by shrink fitting an outer sleeve of ID 190 mm and

OD of 210 mm over an inner sleeve of ID 150.0 mm and OD of 190.125 mm. The maximum pressure

seen in the barrel is 50 MPa. The inner sleeve is made from steel with yield strength 320 MPa and theinner sleeve is made from steel with yield strength 670 MPa.

Determine the factor of safety using the Von Mises yield criterion.

Determine the minimum temperature difference needed between the sleeves to permit shrink

fitting to take place.


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UNIT 2

DESIGN OF BELTS, ROPES AND CHAINS

INTRODUCTIONPower is transmitted from the prime mover to a machine by means of intermediate

mechanism called drives. This intermediate mechanism known as drives may be belt or

chain or gears. Belt is used to transmit motion from one shaft to another shaft with the hel

p of pulleys, preferably if the centre distance is long. It is not positive drive since there is

slip in belt drive.

Three types of belt drives are commonly used. They are:

Flat belt drive

V-belt drive

Rope or circular belt drive

FLAT BELT DRIVE

When the distance between two pulleys is around 10 meters and moderate power is

required then flat belt drive is preferred. This may be arranged in two ways

Open belt drive

Cross belt drive

When the direction of rotation of both the pulleys are required in the same direction ,

then we can use open belt drive; if direction of rotation of pulleys are required in opposite

direction then cross belt is used. The pulleys which drives the belt is known as driver and

the pulley which follows driver is known as driven or follower.

MERITS AND DEMERITS OF FEAT BELT DRIVE

Merits:1. Simplicity, low cost, smoothness of operation, ability to absorb shocks, flexibilityand efficiency at high speeds.

2. Protect the driven mechanism against breakage in case of sudden overloadsowing to belt slipping.

3. Simplicity of care, low maintenance and service.

4. Possibility to transmit power over a moderately long distance.


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Open belt drive

Cross belt drive

Demerits:=It is not a positive drive.

=Comparatively large size.

=Stretching of belt calling for resewing when the centre distance is constant.

=Not suitable for short centre distance.

=Belt joints reduce the life of the belt.

=High bearing loads and belt stresses.

=Less efficiency due to slip and creep.

Creep in Belts

Consider an open belt drive rotating in clockwise direction as shown in figure. Theportion of the belt leaving the driven and entering the driver is known as tight side andportion of belt leaving the driver and entering the driven is known as slack side.

During rotation there is an expansion of belt on tight side and contraction of belt on theslack side. Due to this uneven expansion and contraction of the belt over the pulleys,there will be a relative movement of the belt over the pulleys, this phenomenon isknown as creep in belts.


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Flat belt drive system

Serpentine belt system

Flat belt drive system

3D arrangement of belt drive

Velocity Ratio

The ratio of angular velocity of the driver pulley to the angular velocity of the drivenpulley is known as velocity ratio or speed ratio or transmission ratio.

Let,

d1 = Speed of driver pulley

d2 = Speed of driver pulley

n1 = Speed of driver pulley

n2 = Speed of driver pulleyNeglecting slip and thickness ofbelt

Linear speed of belt on driver = Linear speed of belt on driven.


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Considering the thickness of belt

Slip in Belts

Consider an open belt drive rotating in clockwise direction, this rotation of belt over the

pulleys is assumed to be due to firm frictional grip between the belt and pulleys.

When this frictional grip becomes in sufficient, there is a possibility of forward motion of

driver without carrying belt with it and there is also possibility of belt rotating without

carrying the driver pulley with it, this is known as slip in belt.

Therefore slip may be defined as the relative motion between the pulley and the belt in it.This reduces velocity ratio and usually expressed as a percentage.

Effect of Slip on Velocity Ratio

Let s1 = Percentage of slip between driver pulley rim and the belt.

s2 = Percentage of slip between the belt and the driven pulley rim.

Linear speed of driver =d1 n1

Material Used for Belt

Belts used for power transmission must be strong, flexible, and durable and must have acoefficient of friction. The most common belt materials are leather, fabric, rubber, balata,camels hair and woven cotton.Length of Open Belt

Consider an open belt drive as shown in Figure.


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Let, D = diameter of larger pulley d = diameter of smaller pulley

C = distance between centers of pulley L = length of belt

Where L and s are in radians.

For equal diameter pulleys L = s = radians.

For unequal diameters pulleys, since slip will occur first on the smaller diameter pulley,

it is necessary to consider s while designing the belt.

Length of Cross Belt

Consider a cross-belt drive as shown in Figure Let, D = diameter of larger pulley

d = diameter of smaller pulley L = Length of belt


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Ratio of Belt Tensions

Consider a driven pulley rotating in clockwise direction as shown in

Figure. Let,

T1 = Tension on tight side

T2 = Tension on slack side = Angle of lap RN = Normal Reaction

F = Frictional force = RN

Now consider a small elemental portion of the belt PQ subtending an angle at the centre.

The portion of the belt PQ is in equilibrium under the action of the following forces, (i)

Tension T at P (ii) Tension T + T at Q (iii) Normal reaction RN (iv) Frictional force F =

RN


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Centrifugal Tension

Consider a driver pulley rotating in clockwise direction, because of rotation of pulley

there will be centrifugal force which acts away from the pulley. The tensions created

because of this centrifugal force both on tight and slack side are known as centrifugal

tension.

Let,

m = Mass of belt per meter length v = Velocity in

m/sec TC = Centrifugal tension in N r = Radius of

pulley

FC = Centrifugal force

Consider a small elemental portion of the belt PQ subtending an angle shown in

Figure. Now the mass of belt PQ = M = Mass per unit length x Arc length PQ =

mrd


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Effect of Centrifugal Tension on Ratio of Tensions

Ratio of belt tension considering the effect of centrifugal tension is

T1TCT2TC

Power transmitted by belt drive


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Initial Tension

The motion of the belt with the pulleys is assumed to be due to firm frictional grip

between the belt and pulleys surface. To increase this grip the belt is mounted on the

pulleys with some tension when the pulleys are stationary.The tension provided in the belt while mounting on the pulley is Initial tension

and is represented by T0. Since in actual practice the belt is not perfectly elastic,

C.G.Barth has given the relation as

Design Procedure for Flat Belt Drive


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Centrifugal stress

Capacity

Calculate e L Land e s s and take the smaller value as the capacity. If

the coefficient of friction is same for both pulleys then find only e s since itis smaller than e L


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Example:

A belt is required to transmit 18.5 kW from a pulley of 1.2 m diameter running at 250rpm to

another pulley which runs at 500 rpm.The distance between the centers of pulleys is 2.7 m. The

following data refer to an open belt drive, = 0.25. Safe working stress for leather is 1.75 N/mm2.

Thickness of belt = 10mm. Determine the width and length of belt taking centrifugal tension

into account. Also find the initial tension in the belt and absolute power that can be transmitted

by this belt and the speed at which this can be transmitted.


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Area of cross section of belt

Total given

power 18.5 1503.18

Power transmitted per mm2 mm

2

area 0.01231Also A = b x t

1503.18

= b x 10

Therefore, b = 150.318 mm

Standard width b = 152 mm


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V- BELT DRIVE

Introduction

When the distance between the shafts is less, then V-belts are preferred. These areendless and of trapezoidal cross section as shown in Figure. It consists of central layer of

fabric and moulded in rubber or rubber like compound.

Documents

Mech-Vi-Design of Machine Elements II [10me62]-Notes(1)