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Write a JAVA program that can serve as a simple ATM (Automated Teller Machine ).
This simple ATM only provides service of withdrawals.
As ATMs in real world, a user can withdraw money from this simple ATM only when the balance of his/her account is sufficient. Moreover, withdrawals are restricted to be in thousands, with one-thousand dollar bills provided only.
You need to contruct a class named Simple_ATM_Service with implementing given interface ATM_Service.
Interface ATM_Service prepares some base function of ATM.
For our simple ATM, more specifically, checkBalance should help checking whether balance in user's account is sufficient, if not, throws an exception named ATM_Exception with type of " BALANCE_NOT_ENOUGH"; isValidAmount checks if amount of money can be divided by 1000, if not, throws an exception named ATM_Exception with type of " AMOUNT_INVALID"; withdraw first calls checkBalance and then calls isValidAmount to check if it is a valid operation.If valid, simple ATM will debit for amount of money the user specified ,and balance of user's account will also be updated. withdraw has to catch the exceptions raised by checkBalance and isValidAmount, and use getMessage defined in ATM_Exception to show the exception information.
At the end of withdraw function, it will always show updated balance in user's account in format of "updated balance : XXXupdated balance : XXX", no matter whether the user withdraws the money successfully or not.
To fulfill the whole functionality, you will needs another class named ATM_Exception ATM_Exception extending Exception.
It contains an enumerated type ExceptionTYPE which includes two kinds of exception. To record the detail of exception raised, we need a private variable exceptionCondition with the type of ExceptionTYPE we just defined and this variable would be set by constructor. For ATM to get the imformation of exception raised, use getMessage.
Account class has already been done for you, you just need to copy the code provided in Required Files section of this page.
NOTICE:
Do not write multiple try-catch blocks in withdraw. You just need to catch/print the first exception raised.
Input Format
Account account = new Account(value) create an account with specified integer valuevalue as initial balance.
Simple_ATM_Service atm = new ATM_Service() create an ATM system.
atm.checkBalance(account, value) where accountaccount is an existing account, and valuevalue is an integer.
atm.isValidAmount(value) where value is an integer.
atm.withdraw(account, value) where account is an existing account, and value is an integer.
ATM_Exception ex = new ATM_Exception(ex_type) where ex_type is an exception type defined in ATM_Exception.ExceptionTYPE.
Output Format
atm.checkBalance(account, value) returns a boolean value true if this checking process is passed successfully.
atm.isValidAmount(value) returns a boolean value true if this checking process is passed successfully.
ex.getMessage() returns a String the same as the name of exception to point out which exception happened. For more details, you can refer to sample outputsample output.
account.java:
class Account {
private int balance;
public Account(int balance) {
setBalance(balance);
}
public int getBalance() {
return balance;
}
public void setBalance(int balance) {
this.balance = balance;
}
}
ATM_Service.java:
public interface ATM_Service {
public boolean checkBalance(Account account, int money) throws ATM_Exception;
public boolean isValidAmount(int money) throws ATM_Exception;
public void withdraw(Account account, int money);
}
Sample Input
Account David = new Account(4000);
Simple_ATM_Service atm = new Simple_ATM_Service();
System.out.println("---- first withdraw ----");
atm.withdraw(David,1000);
System.out.println("---- second withdraw ----");
atm.withdraw(David,1000);
System.out.println("---- third withdraw ----");
atm.withdraw(David,1001);
System.out.println("---- fourth withdraw ----");
atm.withdraw(David,4000);
Sample Output
---- first withdraw ----
updated balance : 3000
---- second withdraw ----
updated balance : 2000
---- third withdraw ----
AMOUNT_INVALID
updated balance : 2000
---- fourth withdraw ----
BALANCE_NOT_ENOUGH
updated balance : 2000

Answer:

Flow rate = 118.8 gpm, discharge pressure= 331.5 psig, shaft power = 26.3 hp and Efficiency = 87.5%.

Explanation:

Without mincing words, let us dive straight into the solution to the question above. So, the following parameters or data are given in the question above:

gpm = 80 gpm, shaft power = 8hp, and the pressure of  150 psig.

The flow rate =[ 80 × 1500] ÷ 100g = 118.9 gpm.

Hence, the discharge pressure = 150 × [ 1500/100]² = 331.5 psig.

Also, the required shaft power = 8 × [ 1500/100]³ = 26.3 hp.

The last part that is the efficiency can be calculated as given below:

Efficiency = [ 80 × 0.00223 × 144 × 150 ] ÷ [550 × 8 ] = 0.875 = 87.5%.

Answer:

Efficiency is defined as any performance that uses the fewest number of inputs to produce the greatest number of outputs. Simply put, you're efficient if you get more out of less.

Explanation:

After gaining access to the user's account, the attacker can steal any sensitive information they desire or perform any action that the user is authorized to perform.

In 2017, the Open Web Application Security Project (OWASP) created a list of the Top 10 Application Security Risks. The list is as follows:

InjectionBroken Authentication and Session ManagementSensitive Data ExposureXML External Entities (XXE)Broken Access ControlSecurity MisconfigurationCross-Site Scripting (XSS)Insecure DeserializationUsing Components with Known VulnerabilitiesInsufficient Logging and MonitoringNow, the given alternatives include:

a) Cross Site Request Forgery

b) Cross Site Scripting

c) Injectiond)

Broken AuthenticationFrom the above alternatives, we can say that all of the options are part of OWASP Top 10, except Cross Site Request Forgery.

Therefore, the correct answer is (a) Cross Site Request Forgery.

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Phase Field Dislocation Dynamics (PFDD) is a computational modeling technique used to simulate the behavior of dislocations in crystalline materials.

It combines the phase field method, which describes the evolution of a scalar order parameter, with the dislocation dynamics theory. PFDD allows for the study of dislocation interactions, evolution, and their effect on the material's mechanical behavior.

In the case of modeling non-Schmid behavior in body-centered cubic (BCC) metals, PFDD can be informed by atomistic simulations to incorporate more accurate and realistic dislocation properties.

Atomistic simulations, such as molecular dynamics (MD) or density function theory (DFT), provide detailed information about the atomic-scale behavior of dislocations, including dislocation core structures and their interactions with other defects.

Here is a general outline of how atomistic simulations can inform PFDD modeling for non-Schmid behavior in BCC metals:

1. Atomistic simulations: Perform atomistic simulations, such as MD or DFT, to investigate the dislocation properties and behaviors specific to the BCC metal of interest. This includes studying dislocation core structures, dislocation mobility, and dislocation interactions with other defects.

2. Extracting parameters: Extract relevant parameters from the atomistic simulations that describe the non-Schmid behavior. These parameters can include dislocation line tension, line mobility, and the Peierls barrier, which represents the energy barrier for dislocation motion.

3. Parameterization: Incorporate the extracted parameters into the PFDD model. The non-Schmid behavior can be incorporated by modifying the governing equations or adding additional terms to account for the specific dislocation properties observed in the atomistic simulations.

4. PFDD simulations: Conduct PFDD simulations using the modified model. The PFDD simulations will provide a larger-scale description of dislocation dynamics, interactions, and their effects on material behavior, accounting for the non-Schmid behavior observed in the atomistic simulations.

5. Validation and analysis: Compare the results of the PFDD simulations with experimental data or other atomistic simulations to validate the model's accuracy. Analyze the simulation results to gain insights into the dislocation behavior, material deformation mechanisms, and the effect of non-Schmid behavior on the overall material response.

By combining the strengths of atomistic simulations and PFDD, this approach allows for a multiscale understanding of dislocation behavior, capturing both the atomistic details and the larger-scale mechanical response of BCC metals. It provides a powerful tool for studying dislocation-mediated plasticity and material deformation in complex systems.

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The type of nail that has the smallest head is called C. a "brad" or "pin" nail.

Brad nails are characterized by their small, thin head, which is usually less than 1/16 of an inch in diameter. They are used for applications where the nail head needs to be concealed or where a larger head would be undesirable, such as in finish carpentry and woodworking.

Brad nails are typically driven using a brad nailer, which is a type of pneumatic or electric tool designed specifically for driving small nails like brads.

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which of the following types of nail has the smallest head?

Common nails

Finish nails:

Brad nails:

Roofing nails

Customers are vital for the success of business logistics, as they drive demand and provide feedback. Pre-transaction customer service activities include product consultation, order processing, and complaint resolution.

Customer satisfaction is crucial for the success of any business logistics function. Customers play a vital role in driving demand, providing valuable feedback, and establishing long-term relationships. By focusing on customer needs and expectations, businesses can optimize their logistics operations to deliver products efficiently and effectively.

Effective pre-transaction-related activities of customer service are essential for building strong customer relationships. These activities involve engaging with customers before the actual transaction takes place, ensuring their needs are understood, and addressing any concerns or inquiries they may have. Examples of pre-transaction customer service activities include:

1. Product Information and Consultation: Providing detailed information about products or services, explaining their features and benefits, and offering recommendations based on customer requirements. For instance, a customer service representative might assist a customer in selecting the right size and specifications for a product they intend to purchase.

2. Order Processing and Tracking: Assisting customers in placing orders, confirming order details, and providing updates on the order status. This can involve addressing any issues related to inventory availability, payment processing, or delivery scheduling. For example, a customer service team may help a customer track their package and provide real-time updates on its location and estimated delivery time.

3. Handling Inquiries and Resolving Complaints: Promptly addressing customer inquiries, concerns, or complaints to ensure a satisfactory resolution. This can involve troubleshooting technical issues, facilitating returns or exchanges, or offering compensation for any service failures. For instance, a customer service representative may assist a customer who received a damaged product and promptly arrange for a replacement or refund.

By engaging in these pre-transaction-related customer service activities, businesses can establish trust, enhance the customer experience, and increase the likelihood of repeat purchases and positive word-of-mouth. It sets the stage for a successful transaction and lays the foundation for a long-term customer-business relationship.

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Answer:

Sole Proprietorship, General Partnership , Limited Partnership, corporation

Explanation:

Business in something that an individual or a group of people do for a living and produce products and services that benefits the society and the people. There are several entities that can be common in business. Some common form of entities are :

Sole Proprietorship : One to one

-- here there is only one owner in the business and he maintains and manages the entire business functions under his control.

Limited Partnership : many to one

-- here two or more than two partners establish business and runs it but only one or more is liable to the amount of the investments.

General Partnership : many to many

-- It is a business partnership where all the partners shares the profits, the assets, legal liabilities and financial liabilities, etc.

Corporation : many to one

It is a business entity where a group of individual or a group of companies run a single business which is generally authorized by the state.

Some of the typical types of the entities that one may think to be common in term of the business entities are about the relationships that are held within many to many and one to many.

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Answer:

Plzzzz answer fast......

Answer:

O The mix is uniform in color

the torque is required to rotate the inner cylinder at 12 r min, the outer cylinder remaining stationary is 11.0807 N-m.

we have left out some other important details, such as the time required to reach that rpm, whether the beginning state is at rest, and the axis around which the cylinder is revolving. However, we'll suppose that it will be 60 seconds (time is necessary to convert rpm to angular acceleration)

rotation around the center axis

Angular acceleration = torque + MOI

Currently, angular acceleration equals 2 rpm/(t 60).

In order to avoid getting an extremely high figure, we estimated that the diameter of the cylinder was 600 mm rather than 600 meters: angular acceleration= 1.047 rad/s² MOI for cylinder across center dia= 1/4MR²+ 1/12ML²

MOI= 10.5833 kg-m

Now, the product of these two is torque.

11.0807 N-m of torque

Various MOI equations can be used to compute for different axes of rotation.

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The complete question is:

Crude oil at 20 c fills the space between two concentric cylinders 250 mm high and with diameters of 150 mm and 156 mm. Find the torque is required to rotate the inner cylinder at 12 r min, the outer cylinder remaining stationary.

The calculated mean value is 16. The sum of all values divided by the total number of values determines the mean (also known as the arithmetic mean, which differs from the geometric mean) of a dataset.

The term "average" is frequently used to describe this measure of central tendency. By dividing the sum of the given numbers by the entire number of numbers, the mean—the average of the given numbers—is determined. Mean is equal to (Sum of All Observations/Total Observations).

Mean=12+15+17+20/4

=16

Special road signage designate 10-cent zones. At the beginning and finish of the zone, there are signs. There is one space available for as long as three or four hours each day. Only specific hours of the day are subject to the 10 cent rate. There may be a Blue Zone in locations where paid parking is not (yet) in effect.

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Consequently, it will take about 46,134.21 ft. lb. of labour to pump the entire pool's water out.

The volume in cubic feet is obtained by dividing that number by the depth. As there are 7.5 gallons in every cubic foot, multiplying the pool's cubic feet by 7.5 will give you the pool's volume in gallons.

V_sides = πr²h

= π(12 ft)²(5 ft)

= 2,262.74 ft³

V_water = πr²h

= π(8 ft)²(24 ft)

= 6,079.63 ft³

V = V_sides - V_water

= 2,262.74 ft³ - 6,079.63 ft³

= -3,816.89 ft³

W = Fd

F = mg

m = ρV

= 0.075 lb/ft³ × 3,816.89 ft³

= 286.29 lb

F = mg

= 286.29 lb × 32.2 ft/s²

= 9,226.84 ft·lb/s² = 9,226.84 N

W = Fd

= 9,226.84 ft·lb/s² × 5 ft

= 46,134.21 ft·lb

Consequently, it will take about 46,134.21 ft. lb. of labour to pump the entire pool's water out.

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Answer:

Demonstrate the Current Divider formula for the following circuit.

Explanation:

The Current Divider Rule allows us to calculate the current flowing through each parallel resistive branch as a percentage of the total current.

Answer:

I think its A

Explanation:

Answer:

Explanation:

In summary, the XYZ Company will require:

(a) 705 processing and assembly operations,

(b) 710 workers (direct labor only), and

(c) 25,738 m2 of total floor space for the new plant.

To calculate the required processing and assembly operations, workers, and total floor space for the new factory, we can break down the problem into smaller parts and analyze each element.

(a) Processing and assembly operations:

Number of components made in the factory: 250 components * 35% = 87.5 (round up to 88 components)

Processing operations for components: 88 components * 8 processing operations = 704 processing operations

Assembly operations for final product: 1 assembly operation (as each product is assembled in one workstation)

Total operations = 704 processing operations + 1 assembly operation = 705 operations

(b) Number of workers (direct labor only):

Processing workers: 704 processing operations / 1 (one worker per work cell) = 704 workers

Assembly workers: 13 models * 1000 units/model = 13,000 units/year

Assembly time per unit: 48 min/unit = 0.8 hr/unit

Assembly time for all products: 13,000 units * 0.8 hr/unit = 10,400 hr

Assembly workers required: 10,400 hr / 2,000 hr/shift = 5.2 (round up to 6 workers)

Total workers = 704 processing workers + 6 assembly workers = 710 workers

(c) Total floor space required:

Processing floor space: 704 work cells * 25 m2/cell = 17,600 m2

Assembly floor space: 6 workstations * 25 m2/station = 150 m2

Total production area: 17,600 m2 + 150 m2 = 17,750 m2

Additional allowance (45%): 17,750 m2 * 45% = 7,987.5 m2

Total floor space = 17,750 m2 + 7,987.5 m2 = 25,737.5 m2 (round to 25,738 m2)

In summary, the XYZ Company will require:

(a) 705 processing and assembly operations,

(b) 710 workers (direct labor only), and

(c) 25,738 m2 of total floor space for the new plant.

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Answer:

The correct answer is True

Explanation:

Most business organizations especially those that are into sales, call centers, even individuals that are into marketing use CRM (Customer Relationship Management) software to drive their business forward by enabling them to interact more with customers and strive to provide better customer service. This is achievable because The software (CRM) which is a business tool stores customers' information ranging from names, contact info, sales, and other important information based on the need of the salespeople (organization) to track, meet and respond to customers needs. CRM benefits range from getting to know more about prospective or existing customers to building a better relationship with customers so as to retain them and allow better communication between customers in a bid to meet their demands.

Shunt motors are direct current motors that have parallel windings with identical voltage. The total input power of the shunt motor is 25.724 kW and the line current will be 111.843 A.

Total input power is the ratio of the output power and the efficiency, while the line current is the proportion of input power to an input voltage.

The total input power (total IP) is calculated as:

Given,

Output power (OP) = 30 hp

Efficiency = 0.87

Substituting values:

\(\begin{aligned}\text{Total IP} &= \rm \dfrac {OP}{efficiency}\\\\&= \dfrac {30 \times 746 \rm \;watts}{0.87}\\\\&= 25.724 \;\rm kW \end {aligned}\)

Line current is calculated as:

Given,

Input power (IP) = 25.724 kW

Input voltage (IV) = 230 Volt

Solving for line current:

\(\begin{aligned} \text{Line current} &= \rm \dfrac{IP}{IV}\\\\&= \dfrac{25.724 \times 10^{3}}{230}\\\\&= 111.843 \;\rm A \end{aligned}\)

Therefore, the total input power of the shunt is 25.724 kW, and the line current is 111.843 A.

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An agrarian and handicraft economy was replaced by one that was dominated by industry and machine manufacturing during the Industrial Revolution in modern history. The world as a whole was affected by this process after it started in Britain in the 18th century.

These first three industrial revolutions shaped our contemporary society. The world we live in underwent a fundamental transformation with the advent of each of these three innovations: the steam engine, the age of science and mass production, and the rise of digital technology. And right now, it's taking place once more, for the fourth time.

The term is frequently used to describe emerging technologies, or those expected to become available in the next five to ten years. It is typically reserved for technologies that are having, or are anticipated to have, significant social or economic effects.

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Atomic radius of  metal that has the face-centered cubic crystal structure is 0.137 nm, Then volume of its unit cell is V = 3.42×10^(-29) m³

The volume is a measurement of the cell's interior space. The surface area/volume ratio is calculated as follows. This demonstrates the amount of surface area that is available in relation to the size of the cell. A small surface area to volume ratio indicates a large cell.

Volume of unit cell

The atomic radius of aluminum =0.121 nm= 0.121×10⁻⁹ m

The volume of the unit cell in  l^3 and  signifies the length of the side of the unit cell.

Generally  equation for the Volume is mathematically given as:-

V = (4r/√2)³

Therefore

V = (4(0.121×10⁻⁹ m)/√2)³

V = 3.42×10^(-29) m³

so, volume of its unit cell is V = 3.42×10^(-29) m³.

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Approximately 170700 cubic meters of fill can be constructed from 191000 m of borrowed material.

The volume of fill that can be constructed from a given amount of borrowed material can be calculated using the formula

Vf = Vb (e1-e2)/(e2-e3)

where Vf is the volume of fill, Vb is the volume of borrowed material, e1 is the void ratio of the borrowed material, e2 is the void ratio of the fill and e3 is the void ratio of the solids.

In this case, Vb is 191000 m, e1 is 1.2, e2 is 0.7 and e3 is 0. Thus, Vf = 191000 (1.2-0.7)/(0.7-0) = 170700 m. Therefore, approximately 170700 cubic meters of fill can be constructed from 191000 m of borrowed material.

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Answer:

5

Explanation:

If he plants in 36 feet half the bushes, then for 10 feet, 5 bushes he would plant. You could also use the rule of three.

a) The employed statistical technique employed in this situation is known as "sampling",

The agency dedicatedly selected an exemplary sample of 75 industries out of a complete population of 500, to gain cognizance into the yearly sales distribution of the entire group.

This procured data was afterwards consolidated into a tabular form with a proclivity for representing it visually through a graph; an expanding practice habitually utilized for displaying figures.

b) The analysis conducted here relies upon a calculative method called "estimation".

By calculating the average annul turnover of the specifically pinpointed seventy-five industries, the office created an assessment of the per annum sales of the full store of 500 cottage industries.

This implementation would be referred to as "statistical inference"; it involves using data from a segment to make determinations or prophecies concerning a larger populous. The exactness of the judgement depends on how well the sample exemplifies the merchandise and its respective variability within the figures.

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The decomposition of Liquid A follows a first-order kinetics. It means that the rate of reaction is proportional to the concentration of A present at any given time.

The rate constant for the reaction is k. The formula for the rate of a first-order reaction is given as follows:r = k[A]where, r is the rate of reaction, and [A] is the concentration of A at any given time.The time taken for the conversion of 50% of A is given as 5 minutes. The concentration of A remaining is 50% of the initial concentration. The rate of reaction at this point is:r = k[0.5 A0]where, A0 is the initial concentration of A.

Since the reaction follows a first-order kinetics, the rate constant k will remain constant throughout the reaction.To calculate the time taken for the conversion of 75% of A, we can use the following equation:ln ([A]t/[A]0) = -ktwhere, [A]t is the concentration of A remaining after time t, and [A]0 is the initial concentration of A. We know that [A]t = 0.25[A]0.Substituting these values, we get:ln (0.25) = -k(t2 - t1)where, t1 = 5 minutes (time taken for 50% conversion), and t2 is the time taken for 75% conversion.Solving for t2, we get:t2 = t1 + (1/k) ln(0.25)Substituting the value of k from the rate equation, we get:t2 = 5 + (1/k) ln(0.25 [A]0)Therefore, we need to know the value of the rate constant k to calculate the time taken for 75% conversion.

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Answer:

545454cm

Explanation:

The blue print drawing of the house shows 666 centimeters, but the real picture of the house is 333 meters. So let the number of cm on the blueprint that represent the distance of 272727 meters be x. Firstly convert the meters to centimeters 666cm = 333m, x=272727m ; then cross multiply, 666cm=33300cm x=27272700cm ; x =(666cm×272727cm)/33300cm =545454cm.

Answer:

communities of practice

Explanation:

Knowledge management systems often include communities of practice groups of employees who work together and learn from each other.

The question is incomplete. The complete question is :

A compound cylinder is formed by shrinking a tube of 250 mm internal diameter and 25 mm wall thickness onto another tube of 250 mm external diameter and 25 mm wall thickness, both tubes being made of the same material. The stress set up at the junction owing to shrinkage is 10 MN/m2. The compound tube is then subjected to an internal pressure of 80 MN/m2. Compare the hoop stress distribution now obtained with that of a single cylinder of 300 mm external diameter and 50 mm thickness subjected to the same internal pressure.

Solution :

Internal pressure = \($80 \ MN/m^2 $\)

Stress set up at the junction owing to shrinkage = \($10 \ MN/m^2 $\)

Therefore shrinkage at the outer tube at r = 0.15, \($\sigma_{r} = 0$\) and r = 0.125, \($\sigma_{r} = -10 \ MN/m^2$\)

∴ \($0=A-\frac{B}{(0.15)^2}=A-44.5B$\)  .....................(i)

  \($-10=A-\frac{B}{(0.125)^2}=A-64B$\)  .................(ii)

By solving the above equations we get A = 22.85 and B = 0.514

Now, hoop stress at radius = 0.15 m  :

A + 44.5 B = 22.85+44.5 (0.514)

                  = 45.7 MPa

Hoop stress at radius 0.125 m :

A + 64 B = 22.85 + 64 (0.514)

              = 55.74 MPa

Now shrinkage in the inner tubes

At r = 0.10, \($\sigma_{r} = 0$\) and r = 0.125, \($\sigma_{r} = -10 \ MN/m^2$\)

\($0=A-\frac{B}{(0.1)^2}=A-100B$\)  ....................(iii)

\($-10=A-\frac{B}{(0.125)^2}=A-64B$\) ...............(iv)

By solving the above equations,

A = -27.8  and B  = -0.278

Now hoop stress at 0.125 m radius :

A + 64 B = -45.6 MPa

Hoop stress at 0.10 m radius:

A + 100 B = -55.6 MPa

Considering the internal pressure only on complete cylinder at r = 0.15, \($\sigma_{r} = 0$\) and r = 0.10 ,  \($\sigma_{r} = -80$\)

\($0=A-\frac{B}{(0.15)^2}=A-44.5B$\) .............(v)

\($-80=A-\frac{B}{(0.1)^2}=A-100B$\) ...........(vi)

∴ A = 64.2 and B = 1.44

At r = 0.15 m,

\($\sigma_{H} = A+44.5B$\)

      \($=128.4 \ MN/m^2$\)

At r = 0.125 m,

\($\sigma_{H} = A+64B$\)

     \($=156.4 \ MN/m^2$\)

At r = 0.125 m,

\($\sigma_{H} = A+100B$\)

     \($=208.2 \ MN/m^2$\)

The stress for the combined shrinkage and the internal pressure are :

Outer tube

r = 015

\($\sigma_{H} = 128.4 + 45.1$\)

     \($=174.1 \ MN/m^2$\)

r = 0.125

\($\sigma_{H} = 156.4 + 55.75$\)

     \($=212.15 \ MN/m^2$\)

For inner tubes,

r = 0.125

\($\sigma_{H} = 156.4 - 45.6 = 110.8 \ MN/m^2$\)

r = 0.1

\($\sigma_{H} = 208.2 - 55.6 = 152.6 \ MN/m^2$\)