A thin-walled copper tube is subjected to internal pressure and axial force.
1) Circumferential and longitudinal stresses are calculated.
2) Corresponding strains are determined using material properties.
3) Changes in length and diameter are computed from the strains.
The full solution is provided, calculating the stresses and strains induced in the thin-walled cylinder due to the applied internal pressure and axial force, and using these to determine the resulting changes in length and diameter.
Design of toggle jack.
This ppt includes design of basic components of jack like screw, Nut, Links etc.
Graphics will be clear all concept about design
This module contains the information about the diesel cycle,how it works and its efficency part in detail.Thiss will help you to enrich your information about diesel cycle .
The purpose of this project is to compare the Normal Stresses induced in the Knuckle-Joint due to application of Tensile Force of 12KN by manual calculations and using Ansys Workbench. Also, to find minimum and maximum stress and Deformation in the Joint. In this report, Stresses found analytically are compared with the stresses found by the Analysis Software.
Design of toggle jack.
This ppt includes design of basic components of jack like screw, Nut, Links etc.
Graphics will be clear all concept about design
This module contains the information about the diesel cycle,how it works and its efficency part in detail.Thiss will help you to enrich your information about diesel cycle .
The purpose of this project is to compare the Normal Stresses induced in the Knuckle-Joint due to application of Tensile Force of 12KN by manual calculations and using Ansys Workbench. Also, to find minimum and maximum stress and Deformation in the Joint. In this report, Stresses found analytically are compared with the stresses found by the Analysis Software.
Terms Used in Governors
Height of a governor, Equilibrium speed, Sleeve lift
Performance of Governors
Sensitiveness , Stability, Hunting, Isochronism, Governor Effort, Power
Types, Application, Function
With sketch explain construction working of
1.watt Governor
2.Porter governor
3.Proell Governor
4.Harnell Governor
Numerical Problems
Comparison of Flywheel and Governor
MCQs
Terms Used in Governors
Height of a governor, Equilibrium speed, Sleeve lift
Performance of Governors
Sensitiveness , Stability, Hunting, Isochronism, Governor Effort, Power
Types, Application, Function
With sketch explain construction working of
1.watt Governor
2.Porter governor
3.Proell Governor
4.Harnell Governor
Numerical Problems
Comparison of Flywheel and Governor
MCQs
Overview of the fundamental roles in Hydropower generation and the components involved in wider Electrical Engineering.
This paper presents the design and construction of hydroelectric dams from the hydrologist’s survey of the valley before construction, all aspects and involved disciplines, fluid dynamics, structural engineering, generation and mains frequency regulation to the very transmission of power through the network in the United Kingdom.
Author: Robbie Edward Sayers
Collaborators and co editors: Charlie Sims and Connor Healey.
(C) 2024 Robbie E. Sayers
Event Management System Vb Net Project Report.pdfKamal Acharya
In present era, the scopes of information technology growing with a very fast .We do not see any are untouched from this industry. The scope of information technology has become wider includes: Business and industry. Household Business, Communication, Education, Entertainment, Science, Medicine, Engineering, Distance Learning, Weather Forecasting. Carrier Searching and so on.
My project named “Event Management System” is software that store and maintained all events coordinated in college. It also helpful to print related reports. My project will help to record the events coordinated by faculties with their Name, Event subject, date & details in an efficient & effective ways.
In my system we have to make a system by which a user can record all events coordinated by a particular faculty. In our proposed system some more featured are added which differs it from the existing system such as security.
Courier management system project report.pdfKamal Acharya
It is now-a-days very important for the people to send or receive articles like imported furniture, electronic items, gifts, business goods and the like. People depend vastly on different transport systems which mostly use the manual way of receiving and delivering the articles. There is no way to track the articles till they are received and there is no way to let the customer know what happened in transit, once he booked some articles. In such a situation, we need a system which completely computerizes the cargo activities including time to time tracking of the articles sent. This need is fulfilled by Courier Management System software which is online software for the cargo management people that enables them to receive the goods from a source and send them to a required destination and track their status from time to time.
NO1 Uk best vashikaran specialist in delhi vashikaran baba near me online vas...Amil Baba Dawood bangali
Contact with Dawood Bhai Just call on +92322-6382012 and we'll help you. We'll solve all your problems within 12 to 24 hours and with 101% guarantee and with astrology systematic. If you want to take any personal or professional advice then also you can call us on +92322-6382012 , ONLINE LOVE PROBLEM & Other all types of Daily Life Problem's.Then CALL or WHATSAPP us on +92322-6382012 and Get all these problems solutions here by Amil Baba DAWOOD BANGALI
#vashikaranspecialist #astrologer #palmistry #amliyaat #taweez #manpasandshadi #horoscope #spiritual #lovelife #lovespell #marriagespell#aamilbabainpakistan #amilbabainkarachi #powerfullblackmagicspell #kalajadumantarspecialist #realamilbaba #AmilbabainPakistan #astrologerincanada #astrologerindubai #lovespellsmaster #kalajaduspecialist #lovespellsthatwork #aamilbabainlahore#blackmagicformarriage #aamilbaba #kalajadu #kalailam #taweez #wazifaexpert #jadumantar #vashikaranspecialist #astrologer #palmistry #amliyaat #taweez #manpasandshadi #horoscope #spiritual #lovelife #lovespell #marriagespell#aamilbabainpakistan #amilbabainkarachi #powerfullblackmagicspell #kalajadumantarspecialist #realamilbaba #AmilbabainPakistan #astrologerincanada #astrologerindubai #lovespellsmaster #kalajaduspecialist #lovespellsthatwork #aamilbabainlahore #blackmagicforlove #blackmagicformarriage #aamilbaba #kalajadu #kalailam #taweez #wazifaexpert #jadumantar #vashikaranspecialist #astrologer #palmistry #amliyaat #taweez #manpasandshadi #horoscope #spiritual #lovelife #lovespell #marriagespell#aamilbabainpakistan #amilbabainkarachi #powerfullblackmagicspell #kalajadumantarspecialist #realamilbaba #AmilbabainPakistan #astrologerincanada #astrologerindubai #lovespellsmaster #kalajaduspecialist #lovespellsthatwork #aamilbabainlahore #Amilbabainuk #amilbabainspain #amilbabaindubai #Amilbabainnorway #amilbabainkrachi #amilbabainlahore #amilbabaingujranwalan #amilbabainislamabad
Student information management system project report ii.pdfKamal Acharya
Our project explains about the student management. This project mainly explains the various actions related to student details. This project shows some ease in adding, editing and deleting the student details. It also provides a less time consuming process for viewing, adding, editing and deleting the marks of the students.
CFD Simulation of By-pass Flow in a HRSG module by R&R Consult.pptxR&R Consult
CFD analysis is incredibly effective at solving mysteries and improving the performance of complex systems!
Here's a great example: At a large natural gas-fired power plant, where they use waste heat to generate steam and energy, they were puzzled that their boiler wasn't producing as much steam as expected.
R&R and Tetra Engineering Group Inc. were asked to solve the issue with reduced steam production.
An inspection had shown that a significant amount of hot flue gas was bypassing the boiler tubes, where the heat was supposed to be transferred.
R&R Consult conducted a CFD analysis, which revealed that 6.3% of the flue gas was bypassing the boiler tubes without transferring heat. The analysis also showed that the flue gas was instead being directed along the sides of the boiler and between the modules that were supposed to capture the heat. This was the cause of the reduced performance.
Based on our results, Tetra Engineering installed covering plates to reduce the bypass flow. This improved the boiler's performance and increased electricity production.
It is always satisfying when we can help solve complex challenges like this. Do your systems also need a check-up or optimization? Give us a call!
Work done in cooperation with James Malloy and David Moelling from Tetra Engineering.
More examples of our work https://www.r-r-consult.dk/en/cases-en/
Automobile Management System Project Report.pdfKamal Acharya
The proposed project is developed to manage the automobile in the automobile dealer company. The main module in this project is login, automobile management, customer management, sales, complaints and reports. The first module is the login. The automobile showroom owner should login to the project for usage. The username and password are verified and if it is correct, next form opens. If the username and password are not correct, it shows the error message.
When a customer search for a automobile, if the automobile is available, they will be taken to a page that shows the details of the automobile including automobile name, automobile ID, quantity, price etc. “Automobile Management System” is useful for maintaining automobiles, customers effectively and hence helps for establishing good relation between customer and automobile organization. It contains various customized modules for effectively maintaining automobiles and stock information accurately and safely.
When the automobile is sold to the customer, stock will be reduced automatically. When a new purchase is made, stock will be increased automatically. While selecting automobiles for sale, the proposed software will automatically check for total number of available stock of that particular item, if the total stock of that particular item is less than 5, software will notify the user to purchase the particular item.
Also when the user tries to sale items which are not in stock, the system will prompt the user that the stock is not enough. Customers of this system can search for a automobile; can purchase a automobile easily by selecting fast. On the other hand the stock of automobiles can be maintained perfectly by the automobile shop manager overcoming the drawbacks of existing system.
Welcome to WIPAC Monthly the magazine brought to you by the LinkedIn Group Water Industry Process Automation & Control.
In this month's edition, along with this month's industry news to celebrate the 13 years since the group was created we have articles including
A case study of the used of Advanced Process Control at the Wastewater Treatment works at Lleida in Spain
A look back on an article on smart wastewater networks in order to see how the industry has measured up in the interim around the adoption of Digital Transformation in the Water Industry.
TECHNICAL TRAINING MANUAL GENERAL FAMILIARIZATION COURSEDuvanRamosGarzon1
AIRCRAFT GENERAL
The Single Aisle is the most advanced family aircraft in service today, with fly-by-wire flight controls.
The A318, A319, A320 and A321 are twin-engine subsonic medium range aircraft.
The family offers a choice of engines
Forklift Classes Overview by Intella PartsIntella Parts
Discover the different forklift classes and their specific applications. Learn how to choose the right forklift for your needs to ensure safety, efficiency, and compliance in your operations.
For more technical information, visit our website https://intellaparts.com
Saudi Arabia stands as a titan in the global energy landscape, renowned for its abundant oil and gas resources. It's the largest exporter of petroleum and holds some of the world's most significant reserves. Let's delve into the top 10 oil and gas projects shaping Saudi Arabia's energy future in 2024.
Hybrid optimization of pumped hydro system and solar- Engr. Abdul-Azeez.pdffxintegritypublishin
Advancements in technology unveil a myriad of electrical and electronic breakthroughs geared towards efficiently harnessing limited resources to meet human energy demands. The optimization of hybrid solar PV panels and pumped hydro energy supply systems plays a pivotal role in utilizing natural resources effectively. This initiative not only benefits humanity but also fosters environmental sustainability. The study investigated the design optimization of these hybrid systems, focusing on understanding solar radiation patterns, identifying geographical influences on solar radiation, formulating a mathematical model for system optimization, and determining the optimal configuration of PV panels and pumped hydro storage. Through a comparative analysis approach and eight weeks of data collection, the study addressed key research questions related to solar radiation patterns and optimal system design. The findings highlighted regions with heightened solar radiation levels, showcasing substantial potential for power generation and emphasizing the system's efficiency. Optimizing system design significantly boosted power generation, promoted renewable energy utilization, and enhanced energy storage capacity. The study underscored the benefits of optimizing hybrid solar PV panels and pumped hydro energy supply systems for sustainable energy usage. Optimizing the design of solar PV panels and pumped hydro energy supply systems as examined across diverse climatic conditions in a developing country, not only enhances power generation but also improves the integration of renewable energy sources and boosts energy storage capacities, particularly beneficial for less economically prosperous regions. Additionally, the study provides valuable insights for advancing energy research in economically viable areas. Recommendations included conducting site-specific assessments, utilizing advanced modeling tools, implementing regular maintenance protocols, and enhancing communication among system components.
Hybrid optimization of pumped hydro system and solar- Engr. Abdul-Azeez.pdf
thinandthickcylinders-180511185429.pdf
1. THIN AND THICK CYLINDERS
They are,
In many engineering applications, cylinders are frequently
used for transporting or storing of liquids, gases or fluids.
Eg: Pipes, Boilers, storage tanks etc.
These cylinders are subjected to fluid pressures. When a
cylinder is subjected to a internal pressure, at any point on the
cylinder wall, three types of stresses are induced on three
mutually perpendicular planes.
INTRODUCTION:
2. 2. Longitudinal Stress (σL ) – This stress is directed along the
length of the cylinder. This is also tensile in nature and tends
to increase the length.
3. Radial pressure ( pr ) – It is compressive in nature.
Its magnitude is equal to fluid pressure on the inside wall and
zero on the outer wall if it is open to atmosphere.
1. Hoop or Circumferential Stress (σC) – This is directed along the
tangent to the circumference and tensile in nature. Thus, there
will be increase in diameter.
3. σ C σ L
1. Hoop Stress (σC) 2. Longitudinal Stress (σL) 3. Radial Stress (pr)
Element on the cylinder
wall subjected to these
three stresses
σ C
σ C
σC
p
σ L
σ L
σ L
p p
pr
σ L
σ L
σ C
σ C
pr
pr
4. INTRODUCTION:
A cylinder or spherical shell is considered to be thin when the
metal thickness is small compared to internal diameter.
i. e., when the wall thickness, ‘t’ is equal to or less than
‘d/20’, where ‘d’ is the internal diameter of the cylinder or shell,
we consider the cylinder or shell to be thin, otherwise thick.
Magnitude of radial pressure is very small compared to
other two stresses in case of thin cylinders and hence neglected.
THIN CYLINDERS
5. Longitudinal
axis
Longitudinal stress
Circumferential stress
t
The stress acting along the circumference of the cylinder is called
circumferential stresses whereas the stress acting along the length of
the cylinder (i.e., in the longitudinal direction ) is known as
longitudinal stress
6. The bursting will take place if the force due to internal (fluid)
pressure (acting vertically upwards and downwards) is more than the
resisting force due to circumferential stress set up in the material.
p
σc σc
P - internal pressure (stress)
σc –circumferential stress
7. P - internal pressure (stress)
σc – circumferential stress
dL
σc
p
t
8. EVALUATION OF CIRCUMFERENTIAL or HOOP STRESS (σC):
Consider a thin cylinder closed at both ends and subjected to internal
pressure ‘p’ as shown in the figure.
Let d=Internal diameter, t = Thickness of the wall
L = Length of the cylinder.
p d
t
σc
σc
dl
t
p
d
9. To determine the Bursting force across the diameter:
Consider a small length ‘dl’ of the cylinder and an elementary
area ‘dA’ as shown in the figure.
r
p
p dθ
dl
dA
dF ×
×
×
=
×
=
dθ
dl
dFx ×
×
×
×
= θ
cos
2
d
p
dA
σc
σc
dl
t
p
d
dθ
θ
Force on the elementary area,
Horizontal component of this force
dθ
dl ×
×
×
=
2
d
p
dθ
dl
dFy ×
×
×
×
= θ
sin
2
d
p
Vertical component of this force
10. The horizontal components cancel out
when integrated over semi-circular
portion as there will be another equal
and opposite horizontal component on
the other side of the vertical axis.
sin
2
d
p
force
bursting
l
diametrica
Total
0
∫ ×
×
×
×
=
∴
π
θ dθ
dl
dA
σc
σc
dl
t
p
θ
d
dθ
[ ]
surface.
curved
the
of
area
projected
p
d
p
cos
dl
2
d
p 0
×
=
×
×
=
−
×
×
×
= dl
π
θ
12. )
1
....(
..........
..........
t
2
d
p
σ
stress,
ntial
Circumfere c
×
×
=
∴
Force due to fluid pressure = p × area on which p is acting = p ×(d ×L)
(bursting force)
Force due to circumferential stress = σc × area on which σc is acting
(resisting force) = σc × ( L × t + L ×t ) = σc × 2 L × t
Under equilibrium bursting force = resisting force
p ×(d ×L) = σc × 2 L × t
Assumed as rectangular
13. LONGITUDINAL STRESS (σL):
p
σL
The force, due to pressure of the fluid, acting at the ends of the
thin cylinder, tends to burst the cylinder as shown in figure
P
A
B
The bursting of the cylinder takes
place along the section AB
14. EVALUATION OF LONGITUDINAL STRESS (σL):
d
4
π
p
cylinder)
of
end
(on the
force
bursting
al
Longitudin 2
×
×
=
p
t
σL
t
d
π
σ
force
Resisting L ×
×
×
=
∴
t
d
π
force
this
resisting
section
cross
of
Area ×
×
=
cylinder.
the
of
material
the
of
stress
al
Longitudin
σ
Let L =
17. EVALUATION OF STRAINS
A point on the surface of thin cylinder is subjected to biaxial
stress system, (Hoop stress and Longitudinal stress) mutually
perpendicular to each other, as shown in the figure. The strains due
to these stresses i.e., circumferential and longitudinal are obtained
by applying Hooke’s law and Poisson’s theory for elastic
materials.
σ C=(pd)/(2t)
σ C=(pd)/(2t)
σL=(pd)/(4t)
σ L=(pd)/(4t)
20. final volume = final area of cross section × final length
[ ] [ ]
[ ] [ ]
[ ]
L
d
d
L
d
L
d
d
d
L
L
d
L
d
L
L
d
d
d
d
L
L
d
d
δ
δ
δ
δ
δ
δ
δ
π
δ
δ
δ
π
δ
δ
π
2
)
(
2
)
(
4
2
)
(
4
4
2
2
2
2
2
2
2
+
+
+
+
+
=
+
×
+
+
=
+
×
+
=
[ ]
L
d
d
d
L
L
d
volume
Final
L
d
d
and
L
d
L
d
as
such
quantities
smaller
the
neglecting
δ
δ
π
δ
δ
δ
δ
δ
2
2
2
2
2
4
2
)
(
,
)
(
+
+
=
[ ] [ ]
[ ]
L
d
d
d
L
V
L
d
L
d
d
d
L
L
d
V
volume
in
change
δ
δ
π
δ
π
δ
δ
π
δ
2
2
2
2
2
4
4
2
4
+
=
−
+
+
=
21. [ ]
L
d
4
π
2
4
π
V
dv
2
2
×
×
+
=
d
L
d
L
d δ
δ
= εL + 2 × εC
V
dV
)
5
.......(
..........
μ)
4
5
(
E
t
4
d
p
V
dv
i.e., ×
−
×
×
×
=
μ)
2
(
E
t
4
d
p
2
μ)
2
1
(
E
t
4
d
p
−
×
×
×
×
+
×
−
×
×
×
=
d
d
2
L
L δ
δ
×
+
=
24. PROBLEM 1:
A thin cylindrical shell is 3m long and 1m in internal diameter. It is
subjected to internal pressure of 1.2 MPa. If the thickness of the sheet is
12mm, find the circumferential stress, longitudinal stress, changes in
diameter, length and volume . Take E=200 GPa and μ= 0.3.
1. Circumferential stress, σC:
σC= (p×d) / (2×t)
= (1.2×1000) / (2× 12)
= 50 N/mm2
= 50 MPa (Tensile).
SOLUTION:
2. Longitudinal stress, σL:
σL = (p×d) / (4×t)
= σC/2 = 50/2
= 25 N/mm2
= 25 MPa (Tensile).
ILLUSTRATIVE PROBLEMS
27. A copper tube having 45mm internal diameter and 1.5mm wall
thickness is closed at its ends by plugs which are at 450mm apart. The
tube is subjected to internal pressure of 3 MPa and at the same time
pulled in axial direction with a force of 3 kN. Compute: i) the change
in length between the plugs ii) the change in internal diameter of the
tube. Take ECU = 100 GPa, and μCU = 0.3.
A] Due to Fluid pressure of 3 MPa:
Longitudinal stress, σL = (p×d) / (4×t)
= (3×45) / (4× 1.5) = 22.50 N/mm2
= 22.50 MPa.
SOLUTION:
Change in length, δL= εL × L = 9 × 10-5
×450 = +0.0405 mm (increase)
E
)
μ
2
1
(
t
4
d)
(p
ε
strain,
Long. L
×
−
×
×
×
=
5
3
10
9
10
100
)
3
.
0
2
1
(
5
.
22 −
×
=
×
×
−
×
=
28. Change in diameter, δd= εc × d = 3.825 × 10-4
×45
= + 0.0172 mm (increase)
B] Due to Pull of 3 kN (P=3kN):
Area of cross section of copper tube, Ac = π × d × t
= π × 45 × 1.5 = 212.06 mm2
Longitudinal strain, ε L = direct stress/E = σ/E = P/(Ac × E)
= 3 × 103
/(212.06 × 100 × 103
)
= 1.415 × 10-4
Change in length, δL=εL× L= 1.415 × 10-4
×450= +0.0637mm (increase)
E
)
μ
2
(
t)
(4
d)
(p
ε
strain
ntial
Circumfere C
−
×
×
×
=
Pd/4t = 22.5
4
3
10
825
.
3
10
100
)
3
.
0
2
(
5
.
22 −
×
=
×
−
×
=
29. Lateral strain, εlat= -μ × Longitudinal strain = -μ × εL
= - 0.3× 1.415 × 10-4
= -4.245 × 10-5
Change in diameter, δd = εlat × d = -4.245 × 10-5
×45
= - 1.91 × 10-3
mm (decrease)
C) Changes due to combined effects:
Change in length = 0.0405 + 0.0637 = + 0.1042 mm (increase)
Change in diameter = 0.01721 - 1.91 × 10-3
= + 0.0153 mm (increase)
30. PROBLEM 3:
A cylindrical boiler is 800mm in diameter and 1m length. It is
required to withstand a pressure of 100m of water. If the permissible
tensile stress is 20N/mm2
, permissible shear stress is 8N/mm2
and
permissible change in diameter is 0.2mm, find the minimum thickness
of the metal required. Take E = 200GPa, and μ = 0.3.
Fluid pressure, p = 100m of water = 100×9.81×103
N/m2
= 0.981N/mm2
.
SOLUTION:
1. Thickness from Hoop Stress consideration: (Hoop stress is critical
than long. Stress)
σC = (p×d)/(2×t)
20 = (0.981×800)/(2×t)
t = 19.62 mm
31. 2. Thickness from Shear Stress consideration:
3. Thickness from permissible change in diameter consideration
(δd=0.2mm):
E
μ)
(2
t)
(4
d)
(p
d
δd −
×
×
×
=
Therefore, required thickness, t = 19.62 mm.
t)
(8
d)
(p
τmax
×
×
=
12.26mm.
t
t)
(8
800)
(0.981
8
=
∴
×
×
=
6.67mm
t
10
200
)
3
.
0
2
(
)
t
4
(
)
800
981
.
0
(
800
2
.
0
3
=
×
−
×
×
×
=
32. PROBLEM 4:
A cylindrical boiler has 450mm in internal diameter, 12mm thick and
0.9m long. It is initially filled with water at atmospheric pressure.
Determine the pressure at which an additional water of 0.187 liters
may be pumped into the cylinder by considering water to be
incompressible. Take E = 200 GPa, and μ = 0.3.
Additional volume of water, δV = 0.187 liters = 0.187×10-3
m3
= 187×103
mm3
SOLUTION:
3
6
3
2
mm
10
143.14
)
10
9
.
0
(
450
4
π
V ×
=
×
×
×
=
μ)
4
5
(
E
t
4
d
p
V
dV
×
−
×
×
×
=
Solving, p=7.33 N/mm2
)
33
.
0
4
5
(
10
200
12
4
450
p
10
143.14
10
187
3
6
3
×
−
×
×
×
×
=
×
×
33. JOINT EFFICIENCY
Longitudinal
rivets
Circumferential
rivets
Steel plates of only particular lengths and width are available. Hence
whenever larger size cylinders (like boilers) are required, a number
of plates are to be connected. This is achieved by using riveting in
circumferential and longitudinal directions as shown in figure. Due
to the holes for rivets, the net area of cross section decreases and
hence the stresses increase.
34. JOINT EFFICIENCY
The cylindrical shells like boilers are having two types of joints
namely Longitudinal and Circumferential joints. Due to the holes for
rivets, the net area of cross section decreases and hence the stresses
increase. If the efficiencies of these joints are known, the stresses
can be calculated as follows.
Let η L= Efficiency of Longitudinal joint
and η C = Efficiency of Circumferential joint.
...(1)
..........
η
t
2
d
p
σ
L
C
×
×
×
=
Circumferential stress is given by,
36. If A is the gross area and Aeff is the effective resisting area then,
Efficiency = Aeff/A
Bursting force = p L d
Resisting force = σc ×Aeff = σc ×ηL ×A = σc ×ηL ×2 t L
Where η L=Efficiency of Longitudinal joint
Bursting force = Resisting force
p L d = σc ×ηL × 2 t L
...(1)
..........
η
t
2
d
p
σ
L
C
×
×
×
=
37. If η c=Efficiency of circumferential joint
Efficiency = Aeff/A
Bursting force = (π d2
/4)p
Resisting force = σL ×A′eff =σL ×ηc ×A′ = σL ×ηc ×π d t
Where η L=Efficiency of circumferential joint
Bursting force = Resisting force
...(2)
..........
η
t
4
d
p
σ
C
L
×
×
×
=
38. A cylindrical tank of 750mm internal diameter, 12mm thickness and
1.5m length is completely filled with an oil of specific weight
7.85 kN/m3
at atmospheric pressure. If the efficiency of longitudinal
joints is 75% and that of circumferential joints is 45%, find the
pressure head of oil in the tank. Also calculate the change in volume.
Take permissible tensile stress of tank plate as 120 MPa and E = 200
GPa, and μ = 0.3.
Let p = max permissible pressure in the tank.
Then we have, σL= (p×d)/(4×t) η C
120 = (p×750)/(4×12) 0.45
p = 3.456 MPa.
SOLUTION:
Also, σ C= (p×d)/(2×t) η L
120 = (p×750)/(2×12) 0.75
p = 2.88 MPa.
39. Max permissible pressure in the tank, p = 2.88 MPa.
μ)
4
5
(
E)
t
(4
d)
(p
V
dv
Strain,
Vol. ×
−
×
×
×
×
=
litres.
0.567
m
10
0.567
.
mm
10
567
.
0
1500
750
4
π
10
8.55
V
10
8.55
dv
10
8.55
0.3)
4
-
(5
)
10
200
12
(4
750)
(2.88
3
3
-
3
6
2
4
-
4
-
4
-
3
=
×
=
×
=
×
×
×
×
=
×
×
=
×
=
×
×
×
×
×
×
=
40. A boiler shell is to be made of 15mm thick plate having a limiting
tensile stress of 120 N/mm2
. If the efficiencies of the longitudinal and
circumferential joints are 70% and 30% respectively determine;
i) The maximum permissible diameter of the shell for an
internal pressure of 2 N/mm2
.
(ii) Permissible intensity of internal pressure when the shell
diameter is 1.5m.
(i) To find the maximum permissible diameter of the shell for an
internal pressure of 2 N/mm2
:
SOLUTION:
η
t
2
d
p
σ
e.,
i.
L
c
×
×
×
=
a) Let limiting tensile stress = Circumferential stress = σ c =
120N/mm2
.
d = 1260 mm
7
.
0
5
1
2
d
2
120
×
×
×
=
41. η
t
4
d
p
σ
e.,
i.
C
L
×
×
×
=
b) Let limiting tensile stress = Longitudinal stress = σ L = 120N/mm2
.
The maximum diameter of the cylinder in order to satisfy both the
conditions = 1080 mm.
d = 1080 mm
.
3
.
0
5
1
4
d
2
120
×
×
×
=
42. The maximum permissible pressure = 1.44 N/mm2
.
(ii) To find the permissible pressure for an internal diameter of 1.5m:
(d=1.5m=1500mm)
a) Let limiting tensile stress = Circumferential stress = σ c =
120N/mm2
.
η
t
2
d
p
σ
e.,
i.
L
c
×
×
×
=
b) Let limiting tensile stress = Longitudinal stress = σ L = 120N/mm2
.
η
t
4
d
p
σ
e.,
i.
C
L
×
×
×
=
.
N/mm
1.68
p
7
.
0
5
1
2
500
1
p
120
2
=
×
×
×
=
.
N/mm
1.44
p
3
.
0
5
1
4
500
1
p
120
2
=
×
×
×
=
43. PROBLEM 1:
Calculate the circumferential and longitudinal strains for a boiler of
1000mm diameter when it is subjected to an internal pressure of
1MPa. The wall thickness is such that the safe maximum tensile stress
in the boiler material is 35 MPa. Take E=200GPa and μ= 0.25.
(Ans: ε C=0.0001531, ε L=0.00004375)
PROBLEM 2:
A water main 1m in diameter contains water at a pressure head of
120m. Find the thickness of the metal if the working stress in the pipe
metal is 30 MPa. Take unit weight of water = 10 kN/m3
.
(Ans: t=20mm)
PROBLEMS FOR PRACTICE
44. THIN AND THICK
CYLINDERS -33
PROBLEM 3:
A gravity main 2m in diameter and 15mm in thickness. It is subjected
to an internal fluid pressure of 1.5 MPa. Calculate the hoop and
longitudinal stresses induced in the pipe material. If a factor of safety
4 was used in the design, what is the ultimate tensile stress in the pipe
material?
(Ans: σC=100 MPa, σL=50 MPa, σU=400 MPa)
PROBLEM 4:
At a point in a thin cylinder subjected to internal fluid pressure, the
value of hoop strain is 600×10-4
(tensile). Compute hoop and
longitudinal stresses. How much is the percentage change in the
volume of the cylinder? Take E=200GPa and μ= 0.2857.
(Ans: σC=140 MPa, σL=70 MPa, %age change=0.135%.)
45. THIN AND THICK
CYLINDERS -34
PROBLEM 5:
A cylindrical tank of 750mm internal diameter and 1.5m long is to be
filled with an oil of specific weight 7.85 kN/m3 under a pressure head
of 365 m. If the longitudinal joint efficiency is 75% and
circumferential joint efficiency is 40%, find the thickness of the tank
required. Also calculate the error of calculation in the quantity of oil
in the tank if the volumetric strain of the tank is neglected. Take
permissible tensile stress as 120 MPa, E=200GPa and μ= 0.3 for the
tank material. (Ans: t=12 mm, error=0.085%.)
47. INTRODUCTION:
The thickness of the cylinder is large compared to that of thin
cylinder.
i. e., in case of thick cylinders, the metal thickness ‘t’ is more
than ‘d/20’, where ‘d’ is the internal diameter of the cylinder.
Magnitude of radial stress (pr) is large and hence it cannot be
neglected. The circumferential stress is also not uniform across the
cylinder wall. The radial stress is compressive in nature and
circumferential and longitudinal stresses are tensile in nature.
Radial stress and circumferential stresses are computed by using
‘Lame’s equations’.
48. LAME’S EQUATIONS (Theory) :
4. The material is homogeneous, isotropic and obeys Hooke’s law. (The
stresses are within proportionality limit).
1. Plane sections of the cylinder normal to its axis remain plane and
normal even under pressure.
2. Longitudinal stress (σL) and longitudinal strain (εL) remain constant
throughout the thickness of the wall.
3. Since longitudinal stress (σL) and longitudinal strain (εL) are constant,
it follows that the difference in the magnitude of hoop stress and
radial stress (pr) at any point on the cylinder wall is a constant.
ASSUMPTIONS:
49. LAME’S EQUATIONS FOR RADIAL PRESSURE AND
CIRCUMFERENTIAL STRESS
Consider a thick cylinder of external radius r1 and internal radius
r2, containing a fluid under pressure ‘p’ as shown in the fig.
Let ‘L’ be the length of the cylinder.
p
r2
r1
p
50. Consider an elemental ring of radius ‘r’ and thickness ‘δr’ as shown
in the above figures. Let pr and (pr+ δpr) be the intensities of radial
pressures at inner and outer faces of the ring.
pr
pr+δpr
r2
r1
r
pr
pr+δpr
r2
r1
r
σc σc
r δr
Pr
pr+δpr
External
pressure
51. Consider the longitudinal
section XX of the ring as
shown in the fig.
The bursting force is
evaluated by considering
the projected area,
‘2×r×L’ for the inner face
and ‘2×(r+δr)×L’ for the
outer face .
The net bursting force, P = pr×2×r×L - (pr+δpr)×2×(r+δr)×L
=( -pr× δr - r×δpr - δpr × δr) 2L
Bursting force is resisted by the hoop tensile force developing at the
level of the strip i.e.,
Fr=σc×2 ×δr×L
L
r
pr
pr+δpr
r+δr
X X
52. Thus, for equilibrium, P = Fr
(-pr× δr - r×δpr- δpr × δr) 2L = σ c×2×δr×L
-pr× δr - r×δpr- δpr × δr = σ c×δr
Neglecting products of small quantities, (i.e., δpr × δr)
σ c = - pr – (r× δpr )/ δr ...…………….(1)
Longitudinal strain is constant. Hence we have,
since σL, E and μ are constants (σc – Pr) should be constant . Let it be
equal to 2a. Thus
constant
E
p
μ
E
σ
μ
E
σ r
C
L
=
×
+
×
− Since Pr is compressive
ε L =
constant
)
p
σ
(
μ
E
σ
r
C
L
=
−
−
E
ε L =
53. σ c- pr = 2a,
i.e., σc = pr + 2a, ………………(2)
From (1), pr+ 2a = - pr – (r× δpr ) / δr
r
r
r
p
-
a)
p
(
2
δ
δ
×
=
+ r
)
3
.(
..........
a)
p
(
p
2
r
r
r
+
=
×
−
δ
δ
r
i. e.,
Integrating, (-2 ×loge r) + c = loge (pr + a)
Where c is constant of integration. Let it be taken as loge b, where
‘b’ is another constant.
Thus, loge (pr+a) = -2 ×loge r + loge b = - loge r2
+ loge b = loge
2
r
b
54. .....(4)
..........
a
r
b
p
stress,
radial
or,
r
b
a
p
i.e., 2
r
2
r −
=
=
+
The equations (4) & (5) are known as “Lame’s Equations” for radial
pressure and hoop stress at any specified point on the cylinder wall.
Thus, r1≤r ≤r2.
a
2
a
b
a
2
p
σ
stress,
Hoop 2
r
c +
−
=
+
=
r
.......(5)
..........
..........
a
r
b
σ
i.e., 2
c +
=
Substituting it in equation 2, we get
55. ANALYSIS FOR LONGITUDINAL STRESS
Consider a transverse section near the end wall as shown in the fig.
Bursting force, P =π×r2
2
×p
Resisting force is due to longitudinal stress ‘σ L’.
i.e., FL= σ L× π ×(r1
2
-r2
2
)
For equilibrium, FL= P
σ L× π ×(r1
2
-r2
2
)= π ×r2
2
×p
Therefore, longitudinal stress,
(Tensile)
)
r
(r
r
p
σ 2
2
2
1
2
2
L
−
×
=
p
p
r2
r1
σL
σ L
σL
σL
L
56. NOTE:
1. Variations of Hoop stress and Radial stress are parabolic across
the cylinder wall.
2. At the inner edge, the stresses are maximum.
3. The value of ‘Permissible or Maximum Hoop Stress’ is to be
considered on the inner edge.
4. The maximum shear stress (σ max) and Hoop, Longitudinal and
radial strains (εc, εL, εr) are calculated as in thin cylinder but
separately for inner and outer edges.
57. ILLUSTRATIVE PROBLEMS
PROBLEM 1:
A thick cylindrical pipe of external diameter 300mm and internal
diameter 200mm is subjected to an internal fluid pressure of 20N/mm2
and external pressure of 5 N/mm2
. Determine the maximum hoop
stress developed and draw the variation of hoop stress and radial
stress across the thickness. Show at least four points for each case.
SOLUTION:
External diameter = 300mm. External radius, r1=150mm.
Internal diameter = 200mm. Internal radius, r2=100mm.
Lame’s equations:
For Hoop stress, .........(1)
For radial stress, .........(2)
a
r
b
σ 2
c +
=
a
r
b
p 2
r −
=
58. Boundary conditions:
At r =100mm (on the inner face), radial pressure = 20N/mm2
i.e.,
Similarly, at r =150mm (on the outer face), radial pressure = 5N/mm2
i.e.,
)
........(3
..........
a
100
b
20 2
−
=
)
........(4
..........
a
150
b
5 2
−
=
..(5)
..........
7
r
2,70,000
σ 2
c +
=
..(6)
..........
7
r
2,70,000
p 2
r −
=
Solving equations (3) & (4), we get a = 7, b = 2,70,000.
Lame’s equations are, for Hoop stress,
For radial stress,
59. To draw variations of Hoop stress & Radial stress :
At r =100mm (on the inner face),
(Comp)
MPa
20
7
100
2,70,000
p
stress,
Radial
(Tensile)
MPa
34
7
100
2,70,000
σ
stress,
Hoop
2
r
2
c
=
−
=
=
+
=
At r =120mm,
(Comp)
MPa
11.75
7
120
2,70,000
p
stress,
Radial
(Tensile)
MPa
25.75
7
120
2,70,000
σ
stress,
Hoop
2
r
2
c
=
−
=
=
+
=
At r =135mm,
(Comp)
MPa
7.81
7
135
2,70,000
p
stress,
Radial
(Tensile)
MPa
21.81
7
135
2,70,000
σ
stress,
Hoop
2
r
2
c
=
−
=
=
+
=
61. PROBLEM 2:
Find the thickness of the metal required for a thick cylindrical shell of
internal diameter 160mm to withstand an internal pressure of 8 N/mm2
.
The maximum hoop stress in the section is not to exceed 35 N/mm2
.
SOLUTION:
Internal radius, r2=80mm.
......(2)
..........
a
r
b
p
stress,
Radial
for
(1)
..........
..........
a
r
b
σ
Stress,
Hoop
for
are,
equations
s
Lame'
2
r
2
c
−
=
+
=
64. PROBLEM 3:
A thick cylindrical pipe of outside diameter 300mm and internal
diameter 200mm is subjected to an internal fluid pressure of 14 N/mm2
.
Determine the maximum hoop stress developed in the cross section.
What is the percentage error if the maximum hoop stress is calculated
by the equations for thin cylinder?
SOLUTION:
Internal radius, r2=100mm. External radius, r1=150mm
.
Lame’s equations:
For Hoop stress, .........(1)
For radial pressure, .........(2)
a
r
b
σ 2
c +
=
a
r
b
p 2
r −
=
65. Boundary conditions:
At x =100mm Pr = 14N/mm2
i.e.,
Similarly, at x =150mm Pr = 0
i.e.,
)
........(1
..........
a
100
b
14 2
−
=
)
........(2
..........
a
150
b
0 2
−
=
..(3)
..........
2
.
11
r
22,500
σ
stress,
Hoop
for
equation
s
Lame' 2
r +
=
∴
Solving, equations (1) & (2), we get a =11.2, b = 2,52,000.
66. MPa.
36.4
2
.
11
100
252000
σ 2
max =
+
=
Max hoop stress on the inner face (where x=100mm):
.
23.08%
100
)
36.4
28
-
36.4
(
error
Percentage =
×
=
14MPa.
p
and
50mm
t
200mm,
D
e
wher
t
2
d
p
σ
formula,
cylinder
By thin max
=
=
=
×
×
=
MPa.
28
50
2
200
14
σmax =
×
×
=
∴
67. PROBLEM 4:
The principal stresses at the inner edge of a cylindrical shell are
81.88 MPa (T) and 40MPa (C). The internal diameter of the
cylinder is 180mm and the length is 1.5m. The longitudinal
stress is 21.93 MPa (T). Find,
(i) Max shear stress at the inner edge.
(ii) Change in internal diameter.
(iii)Change in length.
(iv) Change in volume.
Take E=200 GPa and μ=0.3.
SOLUTION:
2
(-40)
-
81.88
2
p
-
σ
τ
:
face
inner
on the
stress
shear
Max
i)
r
C
max =
=
= 60.94 MPa
70. PROBLEM 5:
Find the max internal pressure that can be allowed into a thick pipe of
outer diameter of 300mm and inner diameter of 200mm so that tensile
tress in the metal does not exceed 16 MPa if, (i) there is no external
luid pressure, (ii) there is a fluid pressure of 4.2 MPa.
SOLUTION:
External radius, r1=150mm.
Internal radius, r2=100mm.
Boundary conditions:
At r=100mm , σc = 16N/mm2
At r=150mm , Pr = 0
Case (i) – When there is no external fluid pressure:
72. Boundary conditions:
At r=100mm , σc= 16 N/mm2
At r=150mm , pr= 4.2 MPa.
Case (ii) – When there is an external fluid pressure of 4.2 MPa:
)
........(2
..........
a
150
b
4.2
)
........(1
..........
a
100
b
16
i.e.,
2
2
−
=
+
=
Solving we get, a = 2.01 & b=139.85×103
)
........(4
..........
01
.
2
10
139.85
p
)
........(3
..........
01
.
2
10
139.85
σ
that
so
2
3
r
2
3
r
−
×
=
+
×
=
r
r
74. PROBLEM 1:
A pipe of 150mm internal diameter with the metal thickness of 50mm
transmits water under a pressure of 6 MPa. Calculate the maximum
and minimum intensities of circumferential stresses induced.
(Ans: 12.75 MPa, 6.75 MPa)
PROBLEM 2:
Determine maximum and minimum hoop stresses across the section
of a pipe of 400mm internal diameter and 100mm thick when a fluid
under a pressure of 8N/mm2
is admitted. Sketch also the radial
pressure and hoop stress distributions across the thickness.
(Ans: σmax=20.8 N/mm2
,σmin=12.8 N/mm2
)
PROBLEM 3:
A thick cylinder with external diameter 240mm and internal diameter
‘D’ is subjected to an external pressure of 50 MPa. Determine the
diameter ‘D’ if the maximum hoop stress in the cylinder is not to
exceed 200 MPa. (Ans: 169.7 mm)
PROBLEMS FOR PRACTICE
75. THIN AND THICK
CYLINDERS -63
PROBLEM 4:
A thick cylinder of 1m inside diameter and 7m long is subjected to an
internal fluid pressure of 40 MPa. Determine the thickness of the
cylinder if the maximum shear stress in the cylinder is not to exceed
65 MPa. What will be the increase in the volume of the cylinder?
E=200 GPa, μ=0.3. (Ans: t=306.2mm, δv=5.47×10-3
m3
)
PROBLEM 5:
A thick cylinder is subjected to both internal and external pressure.
The internal diameter of the cylinder is 150mm and the external
diameter is 200mm. If the maximum permissible stress in the
cylinder is 20 N/mm2
and external radial pressure is 4 N/mm2
,
determine the intensity of internal radial pressure. (Ans:
10.72 N/mm2
)
