Marine Pipelines Nm966 Course Work Write Up

8/10/2019 Marine Pipelines Nm966 Course Work Write Up

http://slidepdf.com/reader/full/marine-pipelines-nm966-course-work-write-up 1/6

OGBIGHELE AKPOBOGHENE 201156617

MSC SUBSEA ENGINEERING

COURSE TITLE MARINE PIPELINES NM 966

COURSE WORK 1

Question 1

For X60 steel pipe, plot pc versus R/t for various f o=( f o=1,2,3%)(pc - pecr )(pc

2- py

2 )=2pecr py pc f oR/t ……………………………………….(1)

py = y t ; f o= (Dmax-Dmin)

R 2

pecr= E t 3

4(1-v2) R

Where f o=1, 2, 3%; =X60 =413.69MPa; E=210GPa

By expanding equation (1) above, we will get the cubic equation below

Pc3-pc

2 pecr -pc

2 pecr +pecr py

2-2pecr py pc f o R/t=0………………………………………(2)

Let’s consider two equations

i.) x3+ax

2+bx+c =0 and (ii.)

Let pc=x on the cubic equation and remove the coefficient of x, you get

a= -pecr; b=-2pecr py pc f o R/t; c=pecr py 2

Any cubic equation of the form above (ii) can be reduced to one of the form of the second

equation by substituting:

We get;

By expanding the above and simplifying (noting that t2 cancels out), we find:

By reducing the Cubic we get

………………………………………………………..………………………………………………………(3)

…………………………………………………………..……………………………………..(4)

To solve the reduced equation above we remember our earlier equation

……………………………………….(i)






COURSE WORK 1

Since P and q are not zero we consider, that, for two numbers u and v:

(u-v)3- 3uv(u-v) = u

3-v

3

This is the cubic equivalent of completing the square in quadratics.

If we substitute (u-v)3

- 3uv (u-v) = u3-v

3with the equation above we get

p=3uv, and -q=u3-v

3 ……………………………………………………..(5)

Therefore and

To get:

Multiplying throughout by u3:

…………………………………………………………………….[6]

Now, this equation is a quadratic in u3, so we can now solve the cubic

Or, simplified:

………………………………………………………………………….[7]

From equation 5, we can find v3:

……………………………………………………………………………..[8]

Quite clearly, the discriminant, D or Δ, is the bit in the square root.

……………………………………………………………………………………………...……[9]

Substituting Δ in 7 and 8, we have:

……………………………………………………………………………………………… [10]

And:

………………………………………………………………………………………………… [11]

If Δ=0, then from 10 and 11, taking cube roots, noting √(Δ)=0:

Notice that v1=-u1, when Δ=0.






COURSE WORK 1

x=u-v-a/3

Also, x1=u1-v1-a/3

=2u1-a/3 (because v1=-u1)

..……………………………………………………………………………………… *12]

We can find the other x's by multiplying by the cube roots of one. (The complex numbers

must be conjugate because 3uv=p, which is real):

……………………………………………………………………………………..[13]

As v1=-u1

…………………………………………………………………………………….[14]

X2,3= U2,3 - V2,3 -a/3

So we add the corresponding parts

……………………………………………………………………………………………… [15]

That is both of these roots are the same (and real).

………………………………………………………………………………………[16]

Or

……………………………………………………………………………………………….[17]

One Real and Two Complex Roots

Here, Δ is not negative, so the square root is a real number. We continue from Equation

7 and reproduce the equations below, as a reminder:

[7]

[8]

The discriminant is:

http://www.trans4mind.com/personal_development/mathematics/polynomials/cubicAlgebra.htm#7_u3










COURSE WORK 1

Substituting Δ in Equations 7, 8:

Taking cube roots, we find:

……………………………….………………………………………………………..[18]

……………………………………………………………………………………………..[19]

………………………………………………………………………………………………...[20]

We can find the other u's using the cube roots of unity

…………………………………………………………………………………….[21]

Then v2 has the conjugate complex number:

…………………………………………………………………………………….[22]

Again we use the formula: x=u-v-a/3:

So, by simplifying, we get:

……………………………………….....[23]

And, similarly:

………………………………………….[24]

………………………………………………........[25]

To get the discriminant complex number we use the equations above

So, in this case, Δ is negative, so √(Δ) is an imaginary number. If Δ <0, then √(Δ)=√(-Δ)i.

Therefore:

……………………………………………………………….[26]






COURSE WORK 1

…………………………………………………………………………………………[27]

Please note that the complex number is of the form a+bi. So we convert to trigonometric

form, absolute value of the complex number, r=√(a2+b2). The argument, φ is the slope of r.

The complex number, (a+bi) can be represented as r(cos (φ) +i sin (φ) ), where cos (φ)=a/r,

and sin (φ)=b/4. Therefore we get

…………………………..[28]

First, it is noticeable that the "r" for both u3 and v

3 is the same. Also note that

cos(φv)=-cos( iφu), and sin(φv)=sin( iφu),

Because the real parts of u and v have an opposite sign and the imaginary bits have the

same sign.

Trigonometric Representation of u3

We find r2 using the normal formula for the absolute value of a complex number:

………………………………………..………………….. [29]

Expanding Δ (it becomes minus):

………….………………………………………..[30]

The q's cancel out leaving the p term:

…………………………………………….……………………………… [31]

Finally, take the square root:

...……………………………………………………………………..[32]

Now we know r, and we can find cos(φ):

…………………………………………………….…[33]

We need φ for our calculations, and now we will go on to find the u's and v's and therefore

the x's:

Finding u and v

Similarly, noting that for v, the r value is the same, the cosine negative, and the sign the

same as the u value, we can just write down the v:






COURSE WORK 1

Our first x, calling it x1=u1-v1-a/3

Because we know or can compute the value of the cosine, (getting phi from the inverse

cosine, and dividing it by 3), we have only to deal with the remaining x's. We simply add

multiples of 2π to the old φ values. So:

Therefore after calculating X1,X2,X3 we use X3 = pc. since it is the less positive figure, then we

plot the graph below for fo= 1,2,3%.

Marine Pipelines Nm966 Course Work Write Up

Documents

Transcript of Marine Pipelines Nm966 Course Work Write Up