AUTOMATION OF LONGITUDINAL MAGNETIC COIL

INSPECTION

Submitted by

B.Naga Malleswara Rao 2011A4PS350H

Dhruv Kumar 2011A4PS283G

At

Oil Country Tubular Ltd.

A Practice School-II Station Of

Birla Institute of Technology and Science,

Pilani

July-December, 2014

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A REPORT

ON

AUTOMATION OF LONGITUDINAL MAGNETIC COIL

INSPECTION

Submitted By

B.Naga Malleswara Rao 2011A4PS350H

Dhruv Kumar 2011A4PS283G

In partial fulfillment of the course

Practice School-II

AT

OIL COUNTRY TUBULAR LIMITED,

SREEPURAM

A PRACTICE SCHOOL-II STATION OF

BIRLA INSTITUTE OF TECHNOLOGY AND SCIENCE,

PILANI

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BIRLA INSTITUTE OF TECHNOLOGY AND SCIENCE, PILANI

Practice School Division

Station: Oil Country Tubular Limited Centre: Narketpally Duration: 3 months Date of Start: 4/7/14

Date of Submission: 8 Oct 2014 Title of the Project: Automation of Longitudinal Magnetic Coil Inspection

Names of the students ID. Nos. Disciplines B. Naga Malleswara Rao 2011A4PS350H B.E. Mechanical Engineering

Dhruv Kumar 2011A4PS283G B.E. Mechanical Engineering

Name and Designation of the Experts: Mr. Satish Kumar, DGM (Operations) Mr. Anjaneyulu, In-charge (NDT), Mr. Narayana, AGM (USAI Forge) Mr. Timma Reddy, In-charge (Mechanical maintenance)

Name of the PS faculty: Mr. Sashikanth Akula Key Words: Automation, NDT, Coil Inspection

Abstract: This project aims to automate the manual inspection of pipes with the help of longitudinal fields generated by coils. The automation mechanism is useful in both UT machine and Electro Magnetic testing machine.

Signature of the Students: Signature of PS Faculty: Date: Date:

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ACKNOWLEDGEMENT

We would like to extend our heartfelt gratitude to Mr. Satish Kumar, DGM

(Operations) for having constant faith in us throughout this project and

supporting us in every possible way.

We would like to extend our heartfelt gratitude to our mentor and guide Mr.

Timma Reddy and Mr. Narayan for having constant faith in us throughout this

project and directing and supporting us in every possible way at each step.

We also express our gratitude to our PS instructor Mr. SASHIKANTH, for

constantly monitoring and helping us improve at each stage.

We extend our thanks to all the employees of OCTL for being very supportive.

Above all, we thank each and every one of those who have been

instrumental in the successful completion and presentation of this report.

We also extend our sincere thanks to Mr. Sridhar Kamineni, MD, without

whose permission we could not have carried out this project.

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TABLE OF CONTENTS

S No. Topic Page no.

1

2

3

4

5

6

Symbols and Properties

Preface

Introduction

Pneumatic Cylinder Assembly

Coil Support Assembly

Mounting Base

5

6

7

12

18

28

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Symbols and Properties

• Weight of coil = 25 – 30 kg • Outer diameter of coil = 16.5 inch – 20 inch • Outer diameter of pipes = 2.375 inch – 13.375 inch

Therefore, the radial difference = 5.5 inch

Mild Steel

• Density = 7850 kg/m3 • Modulus of Elasticity, E = 200 GPa • Yield Strength, 𝜎 = 250 MPa

Coil Support Assembly

• Volume of coil stand = 0.00375 m3 • Mass of coil stand = 29.4 kg • Volume of base = 0.0153 m3 • Mass of base = 120 kg

Formulae

• Safety Factor (SF) = 3

• Buckling force x SF = 𝜋2𝐸𝐼4𝑙2

(one end fixed, other end free)

= 4𝜋2𝐸𝐼𝑙2

(both ends fixed)

• Bending tensile stress, 𝜎𝑆𝐹

= 𝐹𝐴

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PREFACE

This project aims at designing an automated mechanism to perform the

Magnetic Particle Inspection in NDT shop. As opposed to the old method in

which the coil is dragged manually to the pipe edge, this mechanism will

replace the manpower with automated coil movement.

The mechanism is capable of movement in 2 directions –

1. Horizontal direction parallel to the pipe axis to accommodate pipes of

varying length.

2. Vertical direction perpendicular to the pipe axis to accommodate the

varying pipe diameter.

The mechanism uses a pneumatic cylinder to move the coil along the pipe axis.

The stroke length of piston is given as 7ft and any required movement greater

than 7ft is accomplished by the complete mechanism movement on the rails

provided.

The vertical movement is accomplished with the help of manually operated

power screws. The vertical movement is limited to 6 inches.

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INTRODUCTION

The initial conditions for which the automation has to be done is described

further. Generally a batch contains 30-40 small diameter or 70-80 large

diameter pipes. The pipes that come for inspection are drill pipes, tubing pipes

or casing pipes. There are two machines which inspect the pipes; one is UT

machine and another is PITCO electromagnetic inspection machine. These

machines cannot do end inspection because of flux leakages. So the ends of

the pipes have to be manually inspected. We have designed the automation

for longitudinal field testing at the end of the pipes using a magnetic coil. This

coil generates a longitudinal field that extends up to a meter or 3 feet. The coil

has to be inserted 1 foot at the end of a pipe. Even 9 inches insertion is enough

to magnetize the end of a pipe. There are two sizes of coils available here in

OCTL; 16.5 inch and 20 inch OD (outer diameter) coils. In the same batch of

pipes there is a 3-4 feet variation in length. The outer diameter of the pipes

varies from 2.375 – 13.375 inches. There are different ranges of pipes

according to customer specifications or API standards. They are:

Range II (feet)

Drill pipes 27-30

Tubing pipes 28-32

Casing pipes 25-34

Range III (feet)

Drill pipes 40-45

Casing pipes 34-48

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The weight of the pipes ranges from 80-500 Kg.

The automation has to be done for the pipe inspection keeping in mind all the

constraints and conditions involved.

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The main components involved in this automation are:

• Pneumatic cylinder assembly

• Coil support assembly

• Mounting Base

Pneumatic cylinder assembly

Pneumatic cylinder assembly has individual components such as:

• Cylinder

• Piston

• Piston rod

• Cylinder stand

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Coil support assembly

The individual components in Coil support assembly are:

• Coil stand

• Power screw

• Aligning rods

• Linear bearing and the guide way

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Mounting Base

The individual components in Mounted assembly are:

• Metal sheet on which the mechanism is mounted

• Wheels

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PNEUMATIC CYLINDER ASSEMBLY

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Piston Rod

Piston rod is attached to the piston and pushes the coil support so that coil

inserts into the end of the pipe when the pressure is applied at the piston. The

dimensions of the rod are chosen such that buckling is prevented and has

enough tensile strength to withstand pull and push.

BUCKLING CALCULATIONS

𝐹 × 𝑆𝐹 = 𝜋2𝐸𝐼𝑙2

Where 𝐸 = Young’s modulus

F = Applied force

SF = factor of safety

𝐼= Moment of Inertia

𝑙 = L/2 (L is total length of the rod); (both end fixed condition)

Given E =200Gpa

F = force to attain acceleration + force to overcome friction

= (m x a) + (𝜇 x N) = (180 x 2.438) + (0.004 x 1764) = 446 N

𝐼 = 𝜋×𝐷4

64

L = 7ft = 2.1336 m

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SF = 3 (assumed)

446 x 3 = 𝜋2×200×109×𝜋𝑑4

64

�𝐿2�2

d = 0.0112 m = 0.441 inch ≈ 0.5 inch

TENSILE STRENGTH CALCULATIONS

𝜎𝑆𝐹

= 𝐹𝐴

Given, 𝜎=Yield strength of steel = 250 MPa

𝐹 = (𝜇 × 𝑁) + (𝑚 × 𝑎) = 446 N

A = 𝜋 r2

SF = 3 (assumed)

250 × 106

3=

446𝜋𝑟2

r = 1.3 mm

d = 2.6 mm = 0.103 inch

Therefore the diameter of the piston rod is taken as 0.5 inch to prevent

buckling.

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Piston

The piston is contained inside a foot mounting pneumatic cylinder. The piston

is connected to piston rod which moves the coil forward and backward. The

pressurized air of approximately 4 bar pushes the piston. The area of piston is

calculated as follows:

𝐹 = 𝑃 × 𝐴

F is calculated as,

𝐹 = (𝜇 × 𝑁) + (𝑚 × 𝑎)

Where µ = coefficient of friction

N = normal force on linear bearing

M = mass of the coil support plus coil

𝑎 = pseudo acceleration

Given µ=0.004,

𝑁 = 𝑚 × 𝑔 = 1764 N,

P= 4 bar = 4 x 105 Pa

𝑚 × 𝑎 = 180 x 2.438 = 438.8 N

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Then 𝐹 = 446 N

A = area of piston – area of piston rod

= 𝜋𝐷2

4− 𝜋𝑑2

4, where d = 0.5 inch = 0.0127 m

A = 𝐹𝑃

𝜋𝐷2

4− 𝜋𝑑2

4 =

4464×105

D = 39.77 mm = 1.565 inch

The diameter of piston is calculated as 39.77 mm or 1.565 inch.

We take 50mm standard cylinder for our purpose.

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Cylinder Wall Thickness

The wall thickness of the cylinder is calculated as follows

𝐹 = 𝑃 × 𝐴 = 𝜎 × 𝐴′

𝐹 = 4 × 105 × (𝐷𝑎𝑣𝑔 × 𝑙) = 250 × 106 × (2 × 𝑙 × 𝑡)

Where t = cylinder wall thickness

Davg = average diameter of cylinder = Inner diameter + t = 0.05 + t

𝑙 = length of cylinder

F = 0.05 + 𝑡 = 1250 × 𝑡

𝑡 = 4.0 × 10−2𝑚𝑚

The calculated wall thickness is very small.

Cylinder Stand

The cylinder stand beneath the pneumatic cylinder is placed to provide the

required height to connect the piston rod to the center of coil support base.

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COIL SUPPORT ASSEMBLY

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Calculations for Stroke Time and Acceleration

Assuming max velocity be 4ft/s, a total stroke of 7ft out of which first feet be

acceleration phase and last feet be deceleration phase.

For acceleration

v2=u2+2as

16=2 x a x 1

a = 8ft/s2 = 2.438m/s2

v=u+at

4=8 × t

t1 = 0.5s

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For constant velocity

t=s/u

t=5/4=1.25s

t2 = 1.25s

For deceleration

a = -8ft/s2

t3 = 0.5s

Total time, T = t1+t2+t3 = 2.25s

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Support Rod

The support rod consists of 2 aligning rods and one power screw rod. The

aligning rods constrain the motion of the coil support in vertical. With the help

of power screw, the coil support is centralized for different pipe diameters. The

support rods are welded to the coil support.

TENSILE TEST

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Mass of coil stand = density x volume = 7850kg/m3 x 0.00375m3 = 29.4kg

Mass of coil = 30kg

Pseudo force acting during acceleration = M x a = (29.4+30)*2.438 = 144.81 N

Centre of mass of coil stand, y1 = 0.273 m from bottom

Centre of mass of coil, y2 = 0.4826 m inch from bottom

Equivalent Centre of Mass of coil and coil stand assembly,

y = 𝑚1𝑦1+𝑚2𝑦2𝑚1+𝑚2

= 29⋅4×0.27+30×0.48

29⋅4+30 = 0.376 m

Bending stress in the support rod, 𝜎 = 𝑀𝑦𝐼

Here, M = y ×𝐹3

= 0.376 x 144.81

3 = 18.317 N-m

y = radius = d/2

𝐼 = 𝜋64𝑑4

Yield strength of mild steel = 250MPa

Assuming SF = 3

250×106

3 = 18.317×𝑑

2𝜋64𝑑

4

d = 0.0131 m = 0.515inch ≈ 0.6inch

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BUCKLING TEST

F = 𝜋2𝐸𝐼𝑙2

Here, E (mild steel) = 200 x 109 Pa

𝐼 = 𝜋64𝑑4, d = 0.6 inch = 0.01524 mm

𝑙 = 2 x rod length (one end fixed, other end free condition) =2 x

0.1778m

F = weight of coil and coil stand = (30+29.4) x 9.8 = 582.12 N

Calculated permissible force, F = 41334.81 N, which is more than the applied

force.

Therefore the support rod will not buckle.

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Power Screw

The square power screw supports most of the load for the coil support. The

power screw is designed in such a way that it supports the coil and locks the

position of coil so that it centralizes with the diameter of the pipe. The head of

the screw is aligned to the coil support. The thread spacing and thread angle

are calculated. The standard power screw which can take the given load is

taken from standard tables.

Screw Specifications:

Type = self-locking, square threaded

Pitch = 0.157 inch = 4 mm

Minor diameter, di = 0.6 inch = 15.24 mm

Major Diameter, do = di + p = 0.757 inch = 19.24 mm

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The condition for self-locking is

𝜋𝑓𝑑𝑚 > 𝑙

Where dm = mean diameter

𝑙 = lead

f = coefficient of friction of thread

𝜋 × 0.08 × 17.24 = 4.33

For single threaded screw, 𝑙 = 1 × 𝑝 = 4

Since 4.33 > 4

Therefore under the load, the screw self-locks to the nut.

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Coil Stand

The coil stand is where the coil is placed in position. It is made of metal sheets

welded together. The dimensions are given in figure.

The given dimensions are for the largest coil of outer diameter 20 inch. For the

coils smaller than that, a wedge has to be inserted from either side to keep the

coil in position. The dimensions of the wedge is given in figure

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Linear Guideways

The linear bearing gives a smooth way for the push and pull of the coil support.

The wheels in the bearing provide line contact with the surface to prevent and

minimize surface friction. This is attached to the coil support assembly.

The linear guideway used belongs to the EG series EGH15SA.

Permissible values –

MR = 0.08 kN-m

MP = 0.04 kN-m

MY = 0.04 kN-m

Static load rating CO = 9.4 kN

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Applied Load –

MR = 0

MY = 0

MP = r x F

Where r = height of Centre of Mass from bottom

F = pseudo force due to acceleration

= m*a = 180*2.438 = 438.84 N

To find r –

r = 𝑚1𝑦1+𝑚2𝑦2+𝑚3𝑦3

𝑚1+𝑚2+𝑚3

= 120×0.095+29⋅4×0.46+30×0.67

120+29⋅4+30

= 0.251 m

MP = r x F = 110.13 N-m, which is less than the permissible value.

Static load, CO = mg = 180*9.8 = 1.764 kN, which is less than the permissible

value.

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Therefore the guideway is safe.

Mounting base

The complete mechanism is mounted on a plate of thickness 0.5 inch and area

190×25 inch. The whole mechanism is fixed for a batch and usually adjusted

with the changing length of the pipes. This is enabled by wheels which are

welded at the bottom of the base plate. A stopper fixes the position for a

particular batch.

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Wheels

The whole mechanism on top of the base plate can be moved by 3 pairs of

cylindrical rollers on a guideway. The guideway is welded on an I-beam which

gives a rigid support and strength.

The dimensions are as specified in the following figure -