QUESTION 4 a) A cylindrical shell of 1.3 m diameter is made...

QUESTION 4 a) A cylindrical shell of 1.3 m diameter is made up of 18 mm thick plates. Calculate the circumferential and longitudinal stress in the plates if the boiler is subjected to an internal pressure of 2.4 MPa. Take efficiency of the joints as 70%. [5 marks] b) A thin cylindrical shell of 400 mm diameter is to be designed for an internal pressure of 2.4 x 10^6 N/m^2. Compute the suitable thickness of the shell if the allowable circumferential stress is 50 MPa. [3 marks] c) A cylindrical shell of .055 m diameter is required to withstand an internal pressure of 4 MPa. Predict the minimum thickness of the shell if maximum tensile strength in the plate material is 400 MPa and efficiency of the joints is 65%. Take factor of safety as 5. [4 marks] d) A cylindrical thin drum 0.008 m in diameter and 450 mm long is made of 0.015 m thick plates. If the drum is subjected to an internal pressure of 2.5 MPa, identify its changes in diameter and length. Take E as 200 GPa and Poisson's ratio as 0.25. [8 marks] [Total: 20 marks]

Transcript

00:01 We are given a cylindrical cell.

00:03 So we are given here cylindrical cell and the cylindrical cell has a diameter so we can say small t which is equal to 1 .3 meter which is equal to 1300 millimeter.

00:18 So thickness we are given here t which is equal to 18 millimeter.

00:23 Now the pressure p which is equal to we are given a value is 2 .4 mega pascal which is equal to 2 .4 newton per millimeter square.

00:36 The eta we are given efficiency here is 70 here 70 percentage and the sigma c which is equal to sigma h.

00:45 Here this value we want to find also sigma l.

00:50 Sigma l here this is the longitudinal stress this is our stress this is also our stress but that is circumferential stress that is sigma c.

01:02 These two things we need to be find.

01:04 So let's solve this question and find the answer.

01:06 This is our first part of this question here.

01:09 Now let's find out these two things.

01:11 So for the formula we know the sigma c which is equal to formula is p multiplied by d divided by we can write 2t eta which is equal we can write less substitute value so we can write 2 .4 here here 1300 divided by we can write here 2 multiplied by 18 multiplied by we can write 0 .7 and now we just simplify and we can write answer is 124 mega pascal.

01:36 Now here let's talk about the sigma l.

01:40 So sigma l which is equal to p multiplied by d divided by we can write 4t and eta which is equal to answer is sigma c divided by 2 which is equal we can write 124 divided by 2 which is equal to answer is 62 mega pascal.

01:58 So correct answer.

01:59 Now let's solve our second part of this question.

02:01 So in the second part for cylindrical cell and this cell we are given diameter that is 400 millimeter the pressure we are given p which is equal to 2 .4 we can write 10 raise to 6 newton per meter square which is equal we can write 2 .4 mega pascal and we also we can write 2 .4 newton per millimeter square.

02:29 Now here we can write the sigma c the stress the allowable which is equal to value is 50 mega pascal which means we can write 50 newton per millimeter square there is another unit of the stress and the sigma c which is equal to formula we know that is p multiplied by d divided by we can write t multiplied by 2 and in this question we need to find value of thickness t.

02:57 So let's substitute value into this formula.

02:59 So here we can write value is 50 which is equal to we can write value of p here which is 2 .4 multiplied by 400 divided by we can write 2 multiplied by t.

03:09 Now simplify and we can write answer is so t which is equal to answer is 9 .6 millimeter.

03:17 Now in the third part of this question we are given a cylindrical cell and the diameter we are given here 55 millimeter also we are given in the meter so we are already convert here it's converted into the millimeter...

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