Modelling and Analysis of Pipe Inspection Robot

International Journal of Emerging Technology and Advanced Engineering Website: www.ijetae.com (ISSN 2250-2459, ISO 9001:2008 Certified Journal, Volume 3, Issue 5, May 2013)

Modelling and Analysis of Pipe Inspection Robot

Atul Gargade ${ }^{1}$, Dhanraj Tambuskar ${ }^{2}$, Gajanan Thokal ${ }^{3}$
${ }^{1}$ M.E. Scholar, ${ }^{2,3}$ Assistant Professor, Mechanical Engineering Department, PIIT, New Panvel, Mumbai University

Abstract

Robots are used to remove human being from laborious and dangerous work. This project describes an inpipe inspection robot. This robot consist of a fore leg system, a rear leg system and a body. The fore and rear leg systems are constructed by using three worm gear system that are arranged at an angle of 120 degree with respect to each other to operate inside a pipe of different diameters. The springs are attached to each leg and the robot body to operate in pipes of 140 mm to 200 mm diameter range. Here, all major components of robot are designed. Modelling and assembly of robot components is done in Solidworks 11. Stress analysis of all major components is done in Solidworks 11 and Static stress analysis of proposed in-pipe inspection robot assembly is carried out in Ansys 13. This robot is used for offline visual inspection of gas pipelines, water pipelines and drain pipes etc. This robot also has wide applications in chemical industries as well as in gulf countries for inspection of oil and gas pipelines.

Keywords- Ansys13, In-pipe robot, mechanism, pipe inspection, Solidworks 11, worm gear.

I. INTRODUCTION

Many kinds of pipes are being utilized to construct important lifelines such as water and gas supply in our contemporary society. Also pipes are widely used in chemical industries and in gulf countries for carrying petrol, diesel, oil etc. But after some years these pipes get damaged and defects are occurring in pipe. If the defects in the pipe are caused by rust and nature calamity, it is difficult to find out the defects and the place of the defects, and also there is great amount of loss. Thus scheduled inspection must be done. If we decide to do this inspection manually then large amount of time, effort and labour is necessary to grub up the pipes that are buried in the ground. If the robot can inspect inside the pipes, fast and accurate examination will be able to be done at low cost.

There are several types of pipe inspection robots some are in-pipe inspection robot and some are out-pipe inspection robot. A wheel type in-pipe robot was proposed for the inspection of urban gas pipelines with a nominal 8in and a 4 -in diameter [2]. A mobile inspection robot was developed which can move by itself on a feeder pipe by using an inch worm mechanism [3]. The mobile inspection robot is constructed by two gripper body that can fix its body on to the pipe.

A two wheeled type in-pipe mini-robot was proposed which is based on the linkage mechanisms [4]. A sewer pipe inspection robot was developed which can move into the straight pipe without any intelligence of the controller or sensor information [5, 6]. A semi-autonomous robot that can investigates sewer pipes were proposed [7]. The robot can detect and rate defects automatically using artificial intelligence techniques. A robot that was able to move in a pipe filled with water was developed by using a caterpillar [8]. It can move only in the pipe horizontally arranged. A pipe inspection robot was proposed which is based on a helical motion of the driving body [9]. The robot uses wheeled structures on elastic suspension. With a considerable history behind the development of robotics, in-pipe robots can be classified in to seven different subcategories, based on their applications. These are named as pig type robot (figure. a), wheel type robot (figure. b), caterpillar type robot (figure. c), wall press robot (figure. d), walking type robot (figure. e), inchworm type robot (figure. f) and screw type robot.

Figure 1. Types of in-pipe inspection robot
An in-pipe inspection robot has been designed that can deal with many kinds of pipes with various diameters such as plastic pipes or metallic pipes which are in horizontal or vertical manner only. The pipe inspection robot is composed of body, fore leg system, rear leg system and springs. Three legs of each leg system are arranged at an angle of 120 degrees to each other to move inside various pipe diameters. By using spring it is able to move freely inside pipes of different diameters. A CCD camera is installed on front part of the fore leg system to do visual inspection of pipe.

International Journal of Emerging Technology and Advanced Engineering Website: www.ijetae.com (ISSN 2250-2459, ISO 9001:2008 Certified Journal, Volume 3, Issue 5, May 2013)

This paper is organised as follows. Section II describes the mechanism and construction of pipe inspection robot. Section III describes design and analysis of robot. Section IV provides result and discussion. Section V provides conclusion.

II. Construction And Working Of Pipe Inspection ROBOT

An in-pipe inspection robot composed of body, fore leg system, rear leg system and springs. This pipe inspection robot is designed for $0.1388 \mathrm{~m} / \mathrm{s}$ speed. The robot mainly employs aluminium as structural material. Fore leg system and rear leg system are symmetric. Each leg system consists of a DC motor, worm gear driving system and three legs. DC motor is prime mover which is used to drive the robot. Three legs of each leg system are arranged at an angle of 120 degrees to each other to move inside various pipe diameters which is shown in figure 2. Worm gear system is constructed by using a worm and three wheels. Worm is mounted on motor shaft which drives three worm wheels. Each leg consists of a belt drive and two wheels. Belt drive is used to transfer a motion from worm wheel to robot wheel. Coiled spring is attached to each leg and a robot body. By using springs it is able to move freely inside pipes of 140 mm to 200 mm diameter range. A CCD camera is installed on front part of the fore leg system and wireless control system is mounted on a robot body.

Figure 2 fore and rear leg system
For inspection, robot is put into pipe. When electric supply starts, robot covers distance equal to perimeter of robot wheel and then it get stop. The ultrasonic sensor detects the cracks and damages in the pipe and it gives wireless data to computer. In this way we get size and location of crack from starting point on display screen of computer. During crack detection process, CCD camera does visual inspection of pipe and it gives an image of robot path in each crack detection cycle.

Figure 3. Pipe inspection robot
Table 1
Specification of Robot

III. DESIGN AND ANALYSIS OF PIPE INSPECTION ROBOT

A. Selection of motor

Total load on robot $=47.5 \mathrm{~N}$
$\therefore$ Power required to robot to carry weight of 47.5 N with $0.1388 \mathrm{~m} / \mathrm{s}$ speed is,
$\mathrm{P}=\mathrm{W} \times \mathrm{v}$
$=47.5 \times 0.1388$
$\therefore \mathrm{P}=6.593$ watts.
$\therefore$ In worst case if only one motor is working then it has to give total power.
$\therefore$ Power required to two DC motors to drive the robot is,
$\mathrm{P}_{\text {required }}=2 \times 6.593$
$\therefore \mathrm{P}_{\text {required }}=13.188$ watts
If a motor of $12 \mathrm{v} \& 1 \mathrm{amp}$ current is selected then power provided by two motors is,

$$\mathrm{P}_{\text {provided }}=19.2 \text { watts }$$

here, $\mathrm{P}_{\text {provided }}>\mathrm{P}_{\text {required }}$ $\qquad$ hence ok.

So, select two DC motors of $12 \mathrm{v}, 1 \mathrm{amp}$ current & 400 rpm each.

B. Design of Motor Shaft

Step I: Material selection & Calculation of motor shaft Material: $\mathrm{C}_{45}$

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We know that diameter of shaft can be calculated by using following formula,

$$\mathrm{d}^{3}=\frac{16}{\pi \tau_{\max }} \sqrt{\left(\mathrm{K}_{\mathrm{b}} \mathrm{M}_{\mathrm{b}}\right)^{2}+\left(\mathrm{K}_{\mathrm{s}} \mathrm{M}_{\mathrm{s}}\right)^{2}}$$

where,
$\left[\mathrm{M}_{\mathrm{t}}\right]=$ twisting moment
$\left[\mathrm{M}_{\mathrm{t}}\right]=0.1574 \mathrm{~N}-\mathrm{m}$
$\left[\mathrm{M}_{\mathrm{b}}\right]=$ bending moment
$\left[\mathrm{M}_{\mathrm{b}}\right]=1.8533 \mathrm{~N}-\mathrm{m}$
$\mathrm{K}_{\mathrm{t}}=$ torsional factor $=1.5 \quad \&$
$\mathrm{K}_{\mathrm{b}}=$ bending stress factor $=2$
Now for $\mathrm{C}_{45}$ shaft material,
$\sigma_{\mathrm{ut}}=$ yield stress $=353 \mathrm{~N} / \mathrm{mm}^{2}$
$\tau=0.5 \sigma_{\mathrm{ut}}$
$\tau=176.5 \mathrm{~N} / \mathrm{mm}^{2}$
Let’s take fos $=1.5$
$\therefore\left[\tau_{\max }\right]=\tau / 1.5$
$\left[\tau_{\max }\right]=176.5 / 1.5$
$\left[\tau_{\max }\right]_{\text {design }}=117.6667 \mathrm{~N} / \mathrm{mm}^{2}=117.6667 \times 10^{6} \mathrm{~N} / \mathrm{m}^{2}$
$\therefore$ Diameter of motor shaft is,
$\mathrm{d}^{3}=\frac{16}{\pi \times 117.6667 \times 10^{6}} \sqrt{(2 \times 1.8533)^{2}+(1.5 \times 0.1574)^{2}}$
$\therefore \mathrm{d}=0.005438 \mathrm{~m}=5.438 \mathrm{~mm} \approx 6 \mathrm{~mm}$
$\therefore$ Select a motor of shaft of 30 mm length and 6 mm diameter.

Step II: Checking for bending
Design shear stress is given by formulae,
$\left[\sigma_{\mathrm{d}}\right]=\frac{32\left[\mathrm{M}_{\mathrm{b}}\right]}{\pi \mathrm{d}^{2}}$
Where,
$\mathrm{d}=$ diameter of shaft $(\mathrm{mm})$
$\mathrm{M}_{\mathrm{b}}=$ bending moment $(\mathrm{N}-\mathrm{mm})$
$\left[\sigma_{\mathrm{d}}\right]=\frac{32 \times 1.8533 \times 10^{6}}{\pi \times 6^{2}}$
$\left[\sigma_{\mathrm{d}}\right]=87.4404 \mathrm{~N} / \mathrm{mm}^{2}<117.6667 \mathrm{~N} / \mathrm{mm}^{2} \ldots \ldots .$. hence design is safe.
C. Design of Worm and worm wheel

Step I: System selection and Strength calculation
To control abrasive wear and to provide continuous lubrication select close system.
Tooth profile: Involve profile for both
Pressure angle $(\alpha): 20^{\circ}$, full depth system.
Quality of gear: To control seizure and pitting high quality precision gear is selected.

Correction of gear: No correction.
i.e. $\mathrm{X}_{1}=\mathrm{X}_{2}=0$

Step II: Material selection:
Worm: steel ( $40 \mathrm{Ni} 2 \mathrm{Cr} 1 \mathrm{Mo} 28$ ) or $\mathrm{C}_{45}$
Worm Wheel: Bronze
Design stress $\left(\varphi_{\mathrm{b}}\right)$ is,
For $\mathrm{C}_{45}$,
$\left[\sigma_{\mathrm{b}}\right]_{\mathrm{w}}=1400 \mathrm{kgf} / \mathrm{cm}^{2}$
For bronze and sand casting & rotation in one direction only,
$\left[\sigma_{\mathrm{b}}\right]_{\text {ww }}=500 \mathrm{kgf} / \mathrm{cm}^{2}$
$\therefore\left[\sigma_{\mathrm{b}}\right]_{\mathrm{w}} \times \mathrm{Yv}_{\mathrm{w}}=1400 \times 0.4549=636.86 \mathrm{kgf} / \mathrm{cm}^{2} \&$
$\left[\sigma_{\mathrm{b}}\right]_{\text {ww }} \times \mathrm{Yv}_{\text {ww }}=500 \times 0.2854=142.7 \mathrm{kgf} / \mathrm{cm}^{2}$
Here,
$\left[\sigma_{\mathrm{b}}\right]_{\text {ww }} \times \mathrm{Yv}_{\text {ww }}<\left[\sigma_{\mathrm{b}}\right]_{\mathrm{w}} \times \mathrm{Yv}_{\mathrm{w}}$ i.e. Worm Wheel is weaker element.
$\therefore$ Worm Wheel should be designed.
Step III: Checking for bending
Bending stress $\left(\sigma_{\mathrm{b}}\right)$ is,
$\sigma_{\mathrm{b}}=\frac{1.9\left[\sigma_{\mathrm{b}}\right]}{\mathrm{mg} \times \mathrm{q} \times 2 \times \mathrm{Te}}$
$\therefore$ For addendum modification coefficient $\mathrm{X}=0 \& \mathrm{z}_{\mathrm{v}}=$ $14.4571 \approx 14$
$\mathrm{Y}_{\mathrm{v}}=0.33$
$\sigma_{\mathrm{b}}=\frac{1.9 \times 4.0315}{0.128^{2} \times 11 \times 12 \times 0.93}$
$\sigma_{\mathrm{b}}=90.03313 \mathrm{kgf} / \mathrm{cm}^{2}<\quad\left[\sigma_{\mathrm{b}}\right]=\quad 500$
$\mathrm{kgf} / \mathrm{cm}^{2} \ldots \ldots \ldots$. hence design is safe.
$\sigma_{\mathrm{b}}=9.0033 \mathrm{~N} / \mathrm{mm}^{2}$
Step IV: Check for $\mathrm{F}_{\mathrm{d}}, \mathrm{F}_{\mathrm{s}} \& \mathrm{~F}_{\mathrm{w}}$
Dynamic load $\left(\mathrm{F}_{\mathrm{d}}\right)$ :
$\mathrm{F}_{\mathrm{d}}=\mathrm{F}_{\mathrm{i}} \times \mathrm{C}_{\mathrm{v}}$
where,
$\mathrm{F}_{\mathrm{i}}=$ tangential force $(\mathrm{kgf})$
$\mathrm{C}_{\mathrm{v}}=$ Barth velocity factor
$\therefore \mathrm{C}_{\mathrm{v}}=\frac{6+\mathrm{V}_{\mathrm{mg}}}{6}$
$\mathrm{V}_{\mathrm{mg}}=\frac{\pi \mathrm{d}_{2} \mathrm{~N}_{4}}{60}$
$\mathrm{d}_{2}=\mathrm{m}_{\mathrm{s}} \ell$
$=1.25 \times 12$
$\mathrm{d}_{2}=15 \mathrm{~mm}=0.015 \mathrm{~m}$
$\therefore \mathrm{V}_{\mathrm{mg}}=\frac{5.14 \times 0.015 \times 132.64}{60}$
$\therefore \mathrm{V}_{\mathrm{mg}}=0.1041 \mathrm{~m} / \mathrm{s}$
Now,

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$$\begin{aligned} & \mathrm{C}_{\mathrm{v}}=\frac{6+0.1841}{\&} \\ & \mathrm{C}_{\mathrm{v}}=1.0173 \end{aligned}$$

Now we know that,
$\left[m_{t}\right]=F_{t} \times\left(d_{2} / 2\right)$
$4.0315=\mathrm{F}_{\mathrm{t}} \times(1.5 / 2)$
$\therefore \mathrm{F}_{\mathrm{t}}=5.3753 \mathrm{kgf}$
$\mathrm{F}_{\mathrm{d}}=5.3753 \times 1.0277$
$\mathrm{F}_{\mathrm{d}}=5.4683 \mathrm{kgf}$
$\mathrm{F}_{\mathrm{d}}=54.683 \mathrm{~N}$
Static load $\left(\mathrm{F}_{\mathrm{s}}\right)$ :
$\mathrm{F}_{\mathrm{s}}=\left[\sigma_{\mathrm{ts}}\right] \times \mathrm{Yv}_{\mathrm{ww}} \times \mathrm{b} \times \mathrm{m}_{\mathrm{s}}$
For $\mathrm{Z}=4$
$\mathrm{b}=0.67 \mathrm{~d}_{1}$
but, $\mathrm{d}_{1}=\mathrm{q} \mathrm{m}_{\mathrm{s}}$

$$=11 \times 1.25$$

$\therefore \mathrm{d}_{1}=13.75 \mathrm{~mm}$
$\therefore \mathrm{b}=0.67 \times 13.75$
$\mathrm{b}=9.2125 \mathrm{~mm}=0.92125 \mathrm{~cm} \&$
$\mathrm{m}_{\mathrm{s}}=\mathrm{m}_{\mathrm{s}} \cos \gamma$
$=1.25 \times \cos 19.9830^{0}$
$=1.1747 \mathrm{~mm}$
$\therefore \mathrm{m}_{\mathrm{s}}=0.1174 \mathrm{~cm}$
$\therefore \mathrm{F}_{\mathrm{s}}=500 \times 0.2854 \times 0.9212 \times 0.1174$
$\therefore \mathrm{F}_{\mathrm{s}}=15.4328 \mathrm{kgf}$
$\therefore \mathrm{F}_{\mathrm{s}}=154.328 \mathrm{~N}>\mathrm{F}_{\mathrm{d}}=54.683 \mathrm{~N} \quad \ldots \ldots \ldots$ hence design is safe.
Wear load $\left(\mathrm{F}_{\mathrm{w}}\right)$ :
$\therefore$ Wear load $\mathrm{F}_{\mathrm{w}}$ is,
$\mathrm{F}_{\mathrm{w}}=\mathrm{D}_{\mathrm{g}} \mathrm{b} \mathrm{K}_{\mathrm{w}}$
where,
$\mathrm{D}_{\mathrm{g}}=$ pitch diameter of gear (mm)
$\mathrm{K}_{\mathrm{w}}=$ wear factor $\left(\mathrm{kgf} / \mathrm{cm}^{2}\right)$
For hardened steel & Bronze combination & for full depth pressure angle $=20^{\circ}$
$\mathrm{K}_{\mathrm{w}}=5.6 \mathrm{kgf} / \mathrm{cm}^{2}$
$\mathrm{F}_{\mathrm{w}}=1.5 \times 0.9212 \times 5.6$
$\mathrm{F}_{\mathrm{w}}=7.7380 \mathrm{kgf}$
$\mathrm{F}_{\mathrm{w}}=77.38 \mathrm{~N}>\mathrm{F}_{\mathrm{d}}=54.683 \mathrm{~N} \quad \ldots \ldots \ldots$ hence design is safe.
Step V: Gear proportions
Worm:
PCD of worm, $\mathrm{d}_{1}=\mathrm{q} \mathrm{m}_{\mathrm{s}}$

$$\begin{aligned} & =11 \times 1.25 \\ & =13.75 \mathrm{~mm} \end{aligned}$$

$\therefore$ Tip diameter (addendum), $\mathrm{da}_{1}=\mathrm{d}_{1}+2 \mathrm{~m}_{\mathrm{s}}$

$$\begin{aligned} & =13.75+(2 \times 1.25) \\ & =16.25 \mathrm{~mm} \end{aligned}$$

$\therefore$ Root diameter, $\mathrm{df}_{1}=\mathrm{d}_{1}-2 \mathrm{f}_{\mathrm{o}} \mathrm{m}_{\mathrm{s}}-2 \mathrm{c}$

$$\begin{aligned} & =\mathrm{d}_{1}-2 \mathrm{~m}_{\mathrm{s}}-2(0.2) \mathrm{m}_{\mathrm{s}} \\ & =\mathrm{d}_{1}-2 \mathrm{~m}_{\mathrm{s}}-0.4 \mathrm{~m}_{\mathrm{s}} \\ & =\mathrm{d}_{1}-2.4 \mathrm{~m}_{\mathrm{s}} \\ & =13.75-(2.4 \times 1.25) \\ & =10.75 \mathrm{~mm} \end{aligned}$$

$\therefore$ For $\mathrm{X}=0 \& \mathrm{Z}=4$
Length of worm, $\mathrm{L} \geqq(12.5+0.09 \mathrm{z}) \mathrm{m}_{\mathrm{x}}$

$$\begin{aligned} & =[12.5+(0.09 \times 12)] \times 1.25 \\ & =16.975 \mathrm{~mm} \approx 17 \mathrm{~mm} \end{aligned}$$

$\therefore$ For ground worm,
For $\mathrm{m}_{\mathrm{x}}<10 \mathrm{~mm}$
$\mathrm{L}=\mathrm{L}+25$
$=17+25$
$=42 \mathrm{~mm}$
Worm Wheel:
PCD of worm wheel, $\mathrm{d}_{2}=\mathrm{m}_{\mathrm{x}} \mathrm{z}$

$$\begin{aligned} & =1.25 \times 12 \\ & =15 \mathrm{~mm} \end{aligned}$$

$\therefore$ Tip diameter (addendum), $\mathrm{da}_{2}=\mathrm{d}_{2}+2 \mathrm{~m}_{\mathrm{s}}$

$$\begin{aligned} & =15+(2 \times 1.25) \\ & =17.5 \mathrm{~mm} \end{aligned}$$

$\therefore$ Root diameter, $\mathrm{df}_{2}=\mathrm{d}_{2}-2 \mathrm{f}_{\mathrm{o}} \mathrm{m}_{\mathrm{x}}-2 \mathrm{c}$

$$\begin{aligned} & =\mathrm{d}_{2}-2 \mathrm{~m}_{\mathrm{s}}-2(0.2) \mathrm{m}_{\mathrm{x}} \quad \vee \mathrm{f}_{\mathrm{o}}=1 \\ & =\mathrm{d}_{2}-2 \mathrm{~m}_{\mathrm{s}}-0.4 \mathrm{~m}_{\mathrm{x}} \\ & =\mathrm{d}_{2}-2.4 \mathrm{~m}_{\mathrm{x}} \\ & =15-(2.4 \times 1.25) \\ & =12 \mathrm{~mm} \end{aligned}$$

$\therefore$ Face width, $\mathrm{b}=0.67 \mathrm{~d}_{1}$

$$\begin{aligned} & =0.67 \times 13.75 \\ & =9.2125 \mathrm{~mm} \end{aligned}$$

D. Design of pulley and belt
i) Design of Pulley

Step I: Material Selection & Calculation of pulley diameter
Material- Grey Cast Iron
Type- Flat belt pulley
Diameter of small pulley is given by formula (d) :

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Step II: Checking for Bending Allowable stress in arm $\left(\sigma_{\mathrm{d}}\right)$ :
$\mathrm{a}=2.73 \frac{\sqrt{\frac{\mathrm{~F}_{\mathrm{t}}}{\mathrm{a}} \times \overline{\mathrm{D}}}}{\sqrt{1+\sigma_{\mathrm{d}}}}$
where,
$\mathrm{F}_{\mathrm{t}}=$ net belt pull $(\mathrm{N})$
$\therefore 5.7548=2.73 \frac{\sqrt{138.7923 \times 12}}{4 \times \sigma_{\mathrm{d}}}$
$\sigma_{\mathrm{d}}=44.6229 \mathrm{~N} / \mathrm{mm}^{2}<\left[\sigma_{\text {permissible }}\right]=95 \mathrm{~N} / \mathrm{mm}^{2} \ldots \ldots$. safe
ii) Design of Belt

Step I: Material Selection & Calculation of belt tensions
Material: Rubber
Type: Flat belt
Tensions in belt are,
$\mathrm{T}_{1}=138.7923 \mathrm{~N}$
$\mathrm{T}_{2}=59.6725 \mathrm{~N}$
Step II: Checking for Tensile stress
Allowable stress or design stress in belt $\left(\sigma_{\mathrm{d}}\right)$ :
$\left[\sigma_{\mathrm{d}}\right]=\frac{\mathrm{T}_{\max }}{\left(\mathrm{b} \times \tau_{1}\right.}$
$=\frac{138.7923}{\left(\mathrm{~B} \times 1.6667\right)}$
$\left[\sigma_{\mathrm{d}}\right]=16.6550 \mathrm{~N} / \mathrm{mm}^{2}$
$\left[\sigma_{\mathrm{d}}\right]=16655000 \mathrm{~N} / \mathrm{m}^{2}$
For soft fabric and warp
$\left[\sigma_{\text {permissible }}\right]=24.5195 \mathrm{~N} / \mathrm{mm}^{2}$
$\left[\sigma_{\mathrm{d}}\right]<\left[\sigma_{\text {permissible }}\right] \ldots \ldots \ldots \ldots$. hence design is safe.

E. Design of spring

Step I: Material Selection & Calculation of spring stiffness Material: stainless steel wire for normal corrosion resistance
Type: Ground end
$\therefore$ Maximum elongation of spring is given as,
$\delta \mathrm{L}=\mathrm{L}_{2}-\mathrm{L}$
$=98.7368-75.4718$
$\therefore \delta \mathrm{L}=23.265 \mathrm{~mm}$
Spring force calculation:
In vertical case, total load acting on robot is additional sum of weight of robot and frictional force i.e. 47.5 N
$\therefore$ we have to design a spring which will hold the load of 47.5 N
$\therefore$ Design the spring for 40 N force.
Calculation of spring stiffness ( K ):
Spring stiffness $=\frac{\text { spring force }}{\text { maximum elongation of spring }}$
$\mathrm{K}=\frac{40}{23.265}$
$\mathrm{K}=1.7193 \mathrm{~N} / \mathrm{mm}$
Step II: Spring Proportions
Calculation of spring wire diameter (d):
here, spring wire diameter and mean diameter are unknown.
$\therefore$ for (2.10-4.50) mm wire diameter and average service
Design stress is,
$\tau=48.5 \mathrm{kgf} / \mathrm{mm}^{2}$
$\tau=485 \mathrm{~N} / \mathrm{mm}^{2}$
but, shear stress in spring is given by formulae,
$\tau=\frac{\mathrm{SPUN}_{\text {scal }}}{\pi \mathrm{d}^{2}}$
$485=\frac{8 \times 40 \times 6 \times 1.2525}{\pi \mathrm{~d}^{2}}$
$\therefore \mathrm{d}=1.2566 \mathrm{~mm}$
But, above calculated ’ d ’ is not fitting in (2.10-4.50) mm diameter range.
$\therefore \mathrm{d}=1.2566 \mathrm{~mm}$ is coming in first range i.e. upto 2.10 mm
$\therefore$ shear stress upto 2.10 mm diameter is,
$\tau=52.7 \mathrm{kgf} / \mathrm{mm}^{2}$
$\tau=527 \mathrm{~N} / \mathrm{mm}^{2}$
$\tau=527000000 \mathrm{~N} / \mathrm{m}^{2}$
we know that,
$\therefore \tau=\frac{\mathrm{SPUN}_{\text {scal }}}{\pi \mathrm{d}^{2}}$
$\therefore 527=\frac{8 \times 40 \times 6 \times 1.2525}{\pi \mathrm{~d}^{2}}$
$\mathrm{d}=1.4532 \mathrm{~mm}<2.10 \mathrm{~mm} \ldots \ldots .$. hence safe.
Pitch or mean diameter of spring:
$\therefore$ From spring index C,
$6=\frac{\mathrm{D}}{1.4532}$
$\mathrm{D}=8.7194 \mathrm{~mm}$
Calculation of no. of turns and angular deflection:
Spring force ’ F ’ is given by eq’,
$\mathrm{F}=\frac{\mathrm{yG} \mathrm{d}^{\prime}}{\mathrm{sD} / \mathrm{t}}$
where,
$y=$ axil deflection of spring (mm)
$\mathrm{G}=$ Modulus of rigidity $\left(\mathrm{N} / \mathrm{mm}^{2}\right)$

EXPLORING RESEARCH AND INNOVATIONS

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$\mathrm{G}=81370 \mathrm{~N} / \mathrm{mm}^{2}$
$40=\frac{23.265 \times 81370 \times(1.4532)^{4}}{8 \times(8.7194)^{2} \times 1}$
$\therefore \mathrm{i}=39.7999=40$ turns
$\therefore$ for ground end spring type,
Solid length $=\mathrm{i} \times \mathrm{d}$
Free length $=\mathrm{i} \times \mathrm{p}$
where,
$\mathrm{p}=$ pitch $(\mathrm{mm})$
Solid length $=40 \times 1.4532$

$$=57.8341 \mathrm{~mm}$$

$75.4758=40 \times \mathrm{p}$
$\therefore \mathrm{p}=1.8868 \mathrm{~mm}$
Angular deflection $(\theta)$ :
$\theta=\frac{16 \mathrm{PUL}}{\pi \mathrm{Gd}^{4}}$
$=\frac{16 \times 40 \times 8.7194 \times 75.4758}{3.14 \times 81370 \times(1.4532)^{4}}$
$\therefore \theta=0.3696$ degree
Step III: Checking for Shearing
Design shear stress is,
$\left[\tau_{\mathrm{d}}\right]=527 \mathrm{~N} / \mathrm{mm}^{2}$
$\therefore$ For stainless steel wire for normal corrosion resistance & wire diameter $\mathrm{d}=(1.30-2.00) \mathrm{mm}$
$\tau=170 \mathrm{kgf} / \mathrm{mm}^{2}$
$\tau=1700 \mathrm{~N} / \mathrm{mm}^{2}$
let’s take fos $=2$
$\therefore\left[\tau_{\text {permissible }}\right]=850 \mathrm{~N} / \mathrm{mm}^{2}$
$\therefore\left[\tau_{\mathrm{d}}\right]<\left[\tau_{\text {permissible }}\right] \ldots \ldots .$. hence design is safe.
Stress analysis of robot assembly

Figure 4. Stress analysis of robot assembly

IV. RESULT AND DISCUSSION

In order to confirm the effectiveness of the mechanism of the pipe inspection robot, movement of robot for two different diameters is conducted in horizontal and vertical pipe in Solidworks 11.

A. Horizontal movement

Using the pipe inspection robot, motion of robot in forwards and backwards direction inside the pipe that was laid horizontally is done in solidworks 11. So, the horizontal movement of the robot is achieved as shown in figure 5.

Figure 5. Horizontal movement of robot
B. Vertical movement

The vertically upward and downward motions of robot are conducted in Solidworks 11. In this way a vertical movement of robot is achieved which is shown in figure 6.

Figure 6. Vertical movement of robot
The static stress analysis result of robot assembly is shown in figure and results are satisfactory.

V. CONCLUSION AND FUTURE SCOPE

An in-pipe inspection robot has been designed which is able to move freely inside pipes with a different diameter. Mechanical design of all robot components is safely done. Modelling and assembly of robot component is done in Solidworks 11. Stress analysis of major components of robot is separately carried out in Solidworks 11. Stress analysis result is matching with analytical result and both values are less than permissible values.

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Also static stress analysis of robot assembly is done in Ansys 13 and analysis results are found to be satisfactory. In future we can fabricate this robot. After doing proper configuration of mechanical and electronic parts we can use this robot for inspection. Above pipe inspection robot is designed for 750 mm distance and its diameter range is $140 \mathrm{~mm}-200 \mathrm{~mm}$. In future we can increase strength and size of robot component to run for long distance.

REFERENCES

[1] Hun-ok Lim and Taku Ohki, "Development Of Pipe Inspection Robot"ICROS-SICE International Joint Conference 2009,
[2] H.R. Choi and S.M Ryew , “Robotics system with active steering capability for internal inspection of urban gas pipelines”, Mechatronics, vol.26, no.1,pp.105-112,2002.
[3] C. Choi, S.Jung and S. Kim, “Feeder Pipe Inspection Robot with an Inch-Worm Mechanism using Pneumatic Actuators,” International Journal of Control, Automation, and Systems, vol.4, no. 1, pp. 87$95,2006$.
[4] O. Tatar, D. Mandru, and I. Ardelean, “Development of mobile minirobots For in pipe inspection tasks,” ISSN 1392-120, MECHANIK, Nr.6(68) pp. 60-64, 2007.
[5] Amir A. F. Nassiraei, Y. Kawamural, A. Ahrary, Y. Mikuriya and K. Ishii, “Concept and Design of a Fully Autonomous Sewer Pipe Inspection Mobile Robot ‘KANTARO’,” IEEE international Conference on Robotics and Automation, pp. 136-143, 2007.
[6] Amir A. F. Nassiraei, A. Ghaderi and K. Ishii, “A Novel Approach in Correcting the Tilt Angle of a Special 2D-Pipe Inspection Robot Platform using Material Morphology,” Proceedings of the 2008 JSME Conference on Robotics and Mechatronics, 2P1-C17 (1)-(4), June 5-7, 2008.
[7] G. Cambell, K.Rogers and J.Gilbert, “PIRAT-A system for quantitative sewer assessment,” 2th International No-Dig Conference (No-Dig’95), Hamburg, pp. 455-462, 1995.
[8] R. Bradbeer, “The Pearl Rover underwater inspection robot,” In Mechatronics and Machine vision, J.Billingsley (ed), Research studies press, pp.255-262, 2000.
[9] M.Horodinca, A. Preumont, I. Burda and E. Mignon, “The heli-pipe inspection robots architecture for curved pipes,” de masini Bul. Inst. Polit. Iasi, t. XLIX (LIII), pp.155-160, 2003.
[10] S. G. Roh, S. M. Ryew, J. H. Yang, H. R Choi, “Actively steerable In-pipe inspection robots for underground urban gas pipelines,” IEEE International conference on Robotics & Automation, 2001.
[11] Yun-Jong Kim, Kyung-Hyun Yoon, Young-Woo Park, “Development of In-pipe robot for various sizes,” ASME International Conference on Advanced Intelligent Mechatronics Suntec Convention and Exhibition Center Singapore,2009 IEEE.
[12] Mohd Zafri Baharuddin, Juniza Md Saad et.al, “Robot for Boiler Header Inspection “LS-01,”” International Symposium on Robotics and Intelligent Sensors, 2012.