plant layout lecture

8/7/2019 plant layout lecture

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Background:Background: Why we require piping in plants?

Because they are the conveying systems of the entire units like the

blood vessels in our body.

Why we require stress analysis of piping systems?

To establish the technical viabilities of a particular geometrical layout

of the system in relation with the determination of:

•Support/static/rotating equipment nozzle load/moments

•Stress history of the entire system

•Displacement profile of the system

•Natural frequencies of the entire system (required for any

transient loading)



What are the areas of engineering science required for this

exercise? It requires the knowledge of -

Engineering Mechanics,

Fluid Mechanics,

Mechanics of Materials,

Theory of Structures

Finite Element Methods,

Material Science,

Machine Design,

Theories of Mechanical Vibration,

Instrumentation Engineering,

Process Engineering

Background (contd.):Background (contd.):



What are the activities of piping stress analysis?

Background (contd.):Background (contd.):

Process parameters

fixation

Lay-out Dwg.

P&I Dwg.

Start Civil/Str.

Dwg.

Elect.

Dwg.

Stress Analysis

Results OK ?

Y e s

Stop

No

Downstream Engineering



Different Loading Conditions:Different Loading Conditions: Normally, a piping system can be classified as: (1) Cold Piping

And (2) Hot Piping

The usual loading conditions can be classified as:

(1) Static conditions,which includes:

(a) Self-weight of the piping

(b) Thermal loading conditions

(c) Any static load testing conditions (e.g. hydro-test, steam-out, etc.)

(2) Dynamic conditions, which includes:

(a) Seismic/Earthquake loading conditions

(b) Wind loading

(c) Safety valve blow down, or any transient loading situations



Mechanics of Static loadingConditions: Forces/Moments acting on a piping system depend upon the manner in

which it is supported spatially.

The figure shows a typical spatial

layout of a piping system with its

necessary supporting arrangements

PA

A

SH

PGPRH

PR

T K

- 3 0 3

The first thing we need to know

how to represent the support

forces and moments. For that we

must know the Degree of

Freedom at each support point.



Mechanics of Static loading Conditions:

If we carefully observe we note the equipment nozzle arrests all six

degrees of freedom. Similarly, the degrees of freedom of other supporting systems can be determined and we can draw the following

f b d of the system:

A

X

Y

Z




The governing equations of equilibrium are:

= ⇒ =

=

=

∑ ∑∑∑

ur r0 ( ) 0

( ) 0

( ) 0

x

y

z

F i F

ii F

iii F

And

( )

( )( )

= ⇒ =

==

∑ ∑

∑∑

uur r0 ( ) 0

( ) 0

( ) 0

A A x

A y

A z

M i M

ii Miii M






Hence, it is observed that the problem is a Statically Indeterminate one as the

equations of Statics are not alone to solve the problem. Deformation

characteristics of the geometry are required to be taken into account.

The most simple piping configuration which is statically determinate is:

X

Y

Z



How it is taken into account ? Let’s take an example:


For the above rod, find the support reactions.

Let us solve by starting with its FBD.

L/2 L/2

A B

X

YP

C




L/2 L/2

R A R B X

Y

Obviously, from force balance, we get:

0 (1)+ + = ⇒ + =− A B A B R R P R R P

P

Now, considering deformation characteristics:

0= = A Bu u

C

Take the segments AC and CB and draw the FBD s:




L/2

R A

C

R A

A

/2δ = − = =− AC A C A C R L

u u u AE

L/2

R A

B

R BP

L/2

R B

B

R B

CC

/2

δ = − =− = B B C B C C

R Lu u u

AE




Thus eliminating uC , we get:

(2)2 2

= ⇒ = A B A B

R L R L R R

AE AE

Hence from equations (1) and (2), we can say

that: 2

= =− A B P

R R

( )2

∴ = = ← A B P

R R

So we conclude that for statically indeterminate problems,deformation characteristics of the system need to be taken into the

consideration.




Now, we can formulate what is called as the stiffness method of formulation

of a structural mechanics problem. According to this, for an arbitrary rigid body subjected to various forces (moments), we can write:

F 1 F 2

F n

A1

A2

An

1 11 1 12 2 1

11

....

=

= + + +

= ∑

n n

n

i ii

F k u k u k u

k u

Similarly, 2 21=

= ∑n

i ii

F k u

Or, more generally,

1=

= ∑n

j ji ii

F k u




In the above expression, k ji represents the stiffness coefficient

and is defined as:

“The force (moment) at node (point) j due to a unitdeflection (rotation) at node (point) i keeping all othernodes fixed”

According to Maxwell-Betti’s reciprocity theorem, we can say:

= ji ijk k



More formally we can express the above results

in matrix format as follows:


1 11 12 1 1

2 21 22 2 2

1 2

. .

. .

. . . . . . ..

. . . . . . .

. . .

=

n

n

n n n nn n

F k k k u

F k k k u

F k k k u



Or, in compact way:


{ } [ ] { }=

F k uLet us take an example of a simple prismatic

rod:




L

Rod of area = A

Mod. of Elasticity = E

1 2

u1 u2

Obviously,11 21 12

22

, AE AE

k k k L L

AEk L

= = − =

=



Thus, we can write the stiffness matrix of the rod as:


1 1

1 1

A Ek

L

−=

−




What we do with this matrix ? We

Take this equation and solve to write:

{ } [ ] { }1−=u k F

Nodal

Displacements

(rotations)

Nodal Forces (moments)




Let us solve a simple problem:

For the structure shown,

determine the support reaction

and the displacement of thestructure.

L

Area = 2 A L

P

Area = A

1

2

3



Mechanics of Static loading Conditions:We have identified different nodal points on the structure. We write our equations for the elements:

1

1 1

1 1

A Ek

L

− =

− and:

2

1 1 2 22

1 1 2 2

A E A Ek

L L

− = = −




Now writing the stiffness for the entire structure:

1

1 1 0

1 1 0

0 0 0

A EK

L

− = −

and:

2

0 0 0

0 2 20 2 2

A EK L

= − −

∴ combing

1 2K K K = +

1 1 0

1 3 20 2 2

A E

K L

−

= − − −




Note the final stiffness matrix is still symmetric.symmetric. ThusThus

the final equation is:the final equation is:

1

2

3

1

3

1 1 0

1 3 2

0

0

2 2

A E

uL

F u

uF

− − −

−

=

Noting F 1 = P and u3 = 0 we get:




− = − −

−

1

2

3

1 1 0

1 3 20

0 2 2 0

uP A E

uL

F

Solving….

( )1 2

1 2

3 2

3 0

2

AEu u P

Lu u

F u

− =

− + =

= −

1

2

3

1.5

0.5

Lu P

AE

Lu P

AE

F P

=

=

= −




Let us consider the same problem with little modification:

Area = 2 A

L

L

P

Area = A

1

2

3

In this case springs witheffective stiffness K are addedto the system. The equationsare modified as:

1

2

3

1

3

1 1 0

1 3 2

0

0

2 2

A Eu

L

F u

uF

− − −

−

=

with3 3=− F Ku




Effectively, the governing equations are written as:

( )

( )

1 2

1 2

3 2 3 3

3 0

2 2

AEu u P

L

u u

AEF u u KuL

− =

− + =

= − + = −




Or the equations are rewritten as:

( )1 2

1 2

2 3

3 0

2 2 0

AEu u P

L

u u

AE AE

u K uL L

− =

− + =

− + + =




Hence the equation of equilibrium is :

11

2

3

0

3 2

2 2

0

0

0

A E A E

L L

A E A E A EL L L

A E A E

K L L

u

u

u

F

−

− −

− +

=

stiffnlocally

modified




Let us now take example of a 2-d beam:

X

Y R

1 R2

M 1M 2

Equilibrium equations can be written as:

+ =+ + =

1 2

1 2 2

0

0

R RM M R L




X

Y

R1

M 1

= − +1 1 x M M R x

= = − +2

1 12 x d y E I M M R x dx

M x

V x

On integrating and substituting the boundary conditions,we get the following results:




2

6EI

L

312EIL

3

12EI

L

2

6EI

L

4EI

L

2EI

L

26EI

L2

6EIL




3

12EI

L3

12EI

L

2

6EI

L

2

6EI

L

4EI

L

2EI

L

2

6EI

L2

6EI

L




−

−

− − −

= 2

2

3

2

2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

L L

L L L L

L L

L L L

K

L

E I

L

From these results, we can construct the stiffness matrixof a 2-d beam as:




Let us now evaluate the matrix of a 3-d beam with all

degrees of freedom:



−

−

−

−

−

−

− − −

−

=

3 2 3 2

2 2

3 2 3 2

2 2

12 6 12 6

6

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . .

4 6 2

12 6 12 6

6 2

. .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

6 4

A E A E

L L

A E A E

L L

G J G J

L L

G J G J

L L

EI EI EI EI

L L L L

EI EI EI EI

L LL LEI EI EI EI

L L L L

EI EI EI EI

L LL L

K

−

−

− − −

−

3 2 3 2

2 2

3 2 3 2

2 2

. . . . . . . .

. . . . . . . .

. . . . . . . .

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

EI EI EI EI

L L L L

EI EI EI EIL LL L

EI EI EI EI

L L L L

EI EI EI EI

L LL L



Mechanics of Static loading Conditions:Having developed the stiffness matrix, we need to orientthem for Global set of co-ordinates, as different elementsare oriented locally.

This is done as follows:

X

Y

Z

x

y

z

The transformation of co-ordinates take place

according to the equation:

[ ]

=

x X

y T Y z Z




Basically we find stiffness matrix for locally oriented

elements.Then the stiffness matrix can be transformed as:

{ } [ ] { }

[ ] { } [ ] [ ] { }

x y z x y z x y

X Y Z x y z

F K u

T F K

=

⇒ =



{ } [ ] [ ] [ ]

[ ]

{ }1−⇒ =

1 4 4 2 4 43

XYZ

XYZ xyz XYZ

K

F T k T u


Pre-multiplying both sides by we obtain:[ ]

1−T

Thus: [ ] [ ] [ ] [ ]1−=

XYZ xyz K T K T

Note: [ ] [ ]1−

=T

T T




Thus we finally conclude that the analysis

proceeds as:1. Building of geometry of the structure.

2. Defining material database and other properties.

3. Defining load vector and different boundary conditions.

1. Building of stiffness of the different elements of the structure in local

co-ordinates.

2. Transforming the stiffness matrices in Global system.

3. Assembling to get the whole stiffness of the structure.

4. Solve the equation:

Stress analysis proceeds as:

{ } [ ] { }1−=u k F




Finally, what we get as output?

We have obtained:

Nodal Displacements and rotations for the entire structure.

Forces and Moments at the different boundary locations

Challenges still left:

To determine the stress history of the

structure.



End of Part IEnd of Part I

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plant layout lecture