Ministry of Higher Education and Scientific Research              

                        University of Technology  

        Department of Laser and Optoelectronic Engineering  

Simulation Study of Carbon Steel

Phase Transformation using Nd:YAG

laser Pulse A Thesis

Submitted to Department of Laser and Opto-Electronic Engineering of the University of Technology in Partial Fulfillment of the Requirements for the Degree of Master of Science in Laser

Engineering  

By Eng. ALAA FATHEL EDAN 

(B.sc. 2001)

Supervised by

Assistant Prof 

Dr. Kadhim A. Hubeatir 

March                          2008

 ربيع الاول 1428  

 

جمھورية العراق وزارة التعليم العالي والبحث العلمي

الجامعة التكنولوجية قسم ھندسه الليزر والبصريات الالكترونية

التحول الطوري للحديد دراسة الكاربوني باستخدام الليزر النبضي

Nd:YAG

رسالة مقدمة إلى ر والبصريات الالكترونيةقسم ھندسة الليز

في الجامعة التكنولوجية و ھي جزء من متطلبات نيل ھندسة الليزر علوم درجة الماجستير في

من قبل علاء فاضل عيدان

( )الميكانيكيةعلوم في الھندسة سبكالوريو

إشراف

كاظم عبد حبيتر. د.م.أ

ربيع الاول -ه1428 م2008 -اذار

Abstract  

In this study, a mathematical model for the hardening process by the phase

transformation , is performed on the wrought iron with nickel alloy by using

matlab program(6.5). The metal surface temperature was calculated depending on

the alloy parameters which are thermal conductivity, thermal diffusivity,

reflectivity to the used laser wavelength and original surface temperature , which is

room temperature in addition to the values of power density , which were

calculated in previous step , and the used pulse duration ..

The proper sample thickness was calculated to get the self-quenching

which is the proper condition for the complete phase transformation from austenitc

to martensite phase which is the hardening phase.

The distribution, of the residual thermal stresses at the sample surface , is

calculated by using the linear expansion coefficient and the modulus of elasticity.

The laser energy distribution inside the alloy was calculated , this helps to find out

the penetration depth of the laser radiation inside the alloy by using the alloy

surface reflectivity and energy in addition to the absorption coefficient was

plotted .The graph of the distribution of the residual surface thermal stresses by

using Matlab. Temperature distribution, at alloy surface and inside the sample

(along the thermal penetration depth of laser radiation ), was calculated and plotted

in two graphs for the two thermal distributions by using Matlab.

Introduction   

Chapter one

Introduction

1-1 GGeenneerraall Introduction

The twentieth century has witnessed many inventions and

discoveries, the mankind was unable through its long centuries to reach

these inventions and discoveries. May be one of the most important

inventions among these inventions is the laser radiation invention. Laser

radiation has great importance in many applications such as the scientific

and industrial applications. The invention of laser has led to a scientific

and technological revolution which included the conventional and

modern industries, laser helped in bring about tremendous developments

for many sciences and application fields and has become one of the

modern science achievements.

Laser is distinguished by several properties which are not available

in any other optical source, these properties are : high intensity,

coherence, monochromatic and little divergence, laser travels in a very

narrow beam for long distances . So, for all the above mentioned

properties of laser, laser technology has a big and important role in

processes connected to laser. Surface treatment of materials has been

used in industry for several years to improve some of the mechanical

properties of materials. Most failures, which may take place at material

surface as a result of lassitude , erosion and corrosion,are because of the

stresses which mostly take place at the surface and material exposure to

environment conditions. So the solution is to make the material acquire

surface properties which differ from the inner part. There are several

ways to change the surface structure for example by carburizing, nitriding

1  

Introduction   

and phase transformation hardening using flame, electrical inducation,

electron beam or plasma. Because of the advantages that laser has, it has

been used in surface thermal processes to change the microstructure of

the surface layers to improve the surface properties in comparison with

the material original properties . Using laser in surface hardening of

metals is regarded very important in several industries[1].

The advantages of laser hardening can be summarized as follows[2]:

• Selected areas can be hardened without affecting the surrounding

material.

• Minimal heat input causes little macro distortion and reduces the

need for additional machining.

.Treatment depth is accurately controlled and highly reproducible.

• Superior hardness, strength, lubrication, wear resistance and fatigue

properties can be obtained compared to conventional processes. • It can often be used without external quenching.

• No geometry specific tools such as that required for induction

hardening is necessary. • It can be integrated as an inline computer controlled process. • Time saving (no heating-up or soaking time is required). • Minimal environmental impact.

2  

Introduction   

1-2 Literature Survey

      The heat treatment operations have got a lot of attention in researches

and studies since the laser was invented. This attention has become wider

and entered the industrial field because of the multi-benefit of laser .

In the nineties witnessed a tremendous expansion in researchs and

applications to heat treatment various Ferro-alloys. In 1990

(H.J.M.Geijselaers & J.Huetink)[3] set up a finite element model to

determine temperatures, phase compositions and stresses during a thermal

hardening cycle used (steel CK45). In 1996 (K.G.Watkins, et al)[4]

studied microstructural evolution in a range of laser surface treated

aluminum alloys including laser surface melting of Al-Cu, Al-Si, Al-

Zn,Al-Fe, laser surface alloying of Al-Ni, Al-Cr & Al-Mo. The

improvements in hardness and critical pitting potential compared with

conventional alloys by CO2 laser.

At the end of years of 2001 (K.G.Watkins , et al) [6] studies the

forming of 2D sheet materials on aluminum alloys and titanium alloys by

CO2 laser. In 2003 (Geijselaers,H.J.M)[7] set up simulation of laser

hardening consists of two parts. In the first part algorithms and methods

are developed for simulating phase transformations and the stresses

which are generated by inhomogeneous temperature and phase

distributions. The second part is concerned with simulation of steady state

laser hardening (usedsteel Ck45).In same year (P .T .Mannion et al)[8]

studied the effect of damage accumulation behavior on ablation

thresholds and damage morphology in ultrafast laser micro-machining of

common metals in air the observed morphologies seem to suggest that

normal vaporization is the most probable physical mechanisms

responsible for material removal during ablation (used titanium –

3  

Introduction   

sapphire laser and stainless steel). In 2005 (Haitham El Kadiri, et al) [9]

studied the creep and tensile behaviors of Fe-Cr-Al foils and laser micro

welds at high temperature by Co2 laser and they noticed no yield point

effect and the work hardening persists at all temperatures. In the same

year (Alexander.G.P)[10] studied feasibility investigation of laser

welding aluminum alloy 7075-T6 through the use of A 300w, single-

mode,ytterbium fiber optic laser and he noticed due to their high

reflectivity and complexity in heat treatment, aluminum alloys are some

of the hardest metals to be laser welded successfully and very high laser

power is usually required . In 2005 (L .Costa.et al)[11] set up a finite

element analysis model was applied to the study of the influence of

substrate size and ideal time between the deposition of consecutive layers

on the microstructure and hardness of a ten-layer AISI 420 steel wall built

by laser powder deposition. In same year (V. Ocelik , et al) [12] studied

sliding wear resistance of SiC/Al-8Si,WC/Ti-6Al-4v & TiB2/Ti-6Al-4v

layers by Nd:YAG laser , the observed wear mechanisms are summarized

and related to detailed micro structural observations. The layers have

been found to show excellent interfacial bonding , coupled with

dramatically improved tribological properties expressed through a

relative wear resistance value ranging from 30 to 1500. In 2006

(kennth.L) [2] studied the industrial applications and practical problems

encountered with laser transformation hardening by Co2 laser and

Nd:YAG laser and he noticed the maximum hardened with near surface

hardness values of 700-150 HV . In 2004 ( Peng . C, et al) [13] studied

laser forming of complex structures by Co2 laser and used low carbon

steel AISI-1010 and they noticed that the peak temperature drop on the

unscanned surface is much larger than that of the scanned surface, and

thus the temperature gradient through the thickness direction increases

with the increasing plate thickness. In 2007 (Milton .S.F

4  

Introduction   

,Flavia.A.G,Rudimar.R,Ana.M.d)[5] studied laser surface remelting and

hardening of an automotive shaft sing a high-power fiber laser used AISI

1040 steel and they noted that during heating , the eutectoid structure of

pearlite quickly changes to austenite when the temperature rices above

Ac1.In same year (Dennis .Ket al) [14] set up model which is applied to

photo thermal measurements ,leading to depth profiles of steel hardness,

which are compared with data , destructively obtained from the same

samples using Vickers indentation techniques. In same year (David

.H.P)[15] studied the effect of the surface hardening on the microstructure

and mechanical properties as wear resistance of silicon alloyed steel and

used 55Si7 steel alloy,50CrV4 chromium steel alloy, hardened by Nd:

YAG high power laser and he noticed that ausferritic structure has an

excellent tempering resistance, and that laser hardening treatment greatly

improves wear resistance of ausferritic steels. In same year

(M.Marticorena, et al)[16] noticed that the formation of layer of Tin on

the surface and the obtained roughness, have been demonstrated to

improve bone response by using a pulsed Nd:YAG laser. (G.Labeas et

al)[17] set up the finite element code ANSYS model to calculated the

residual stresses of the paint removal process by using laser radiation

(carbon dioxide & excimer) . The material used is aluminum alloy 2024-

T3,in sheet from of 1mm thickness.

1-3 Aim of the Work

The aim of this work aims to study the surface hardening by using

laser which is known analytically by using a Matlab software to find the

used laser data and the sample . For this purpose an alloy has been

selected which is difficalt to harden by the conventional ways because it

5  

Introduction   

6  

has alow carbon ratio. The mathematical model used in this thesis put the

essential bases for the perfect choice of the laser and the material.

1­4 Outline of Thesis 

Chapter one of this thesis gives introduction to laser hardening. Laser

transformation hardening will be presented in chapter two. The

parameters effect surface hardening process, types of laser used in

industrial applications, comparison between laser hardening with other

technologies, laser-material interaction, residual stresses, wear

resistance, Nickel-alloy wrought iron, the phases of the alloy, self-

quenching condition are two as well.

Chapter three provides the detailed description and analysis of laser

transformation hardening models. Chapter four provides the results and

discussion. Finally, the conclusions taken from this work in addition to

the suggested future work are given in chapter five.

 

Theory Background  

Chapter Two Theoretical Background

 2-1 IInnttrroodduuccttiioonn

The economics of different hardening processes can only be

compared on a case-by-case basis. Comparative costs of the hardening

processes cannot be considered separately from the comparative costs of

preparing the material for hardening and the cost of subsequent

operations. A negative cost consideration for laser hardening is the need

to pre-treat the surface with paint to enhance the absorption of the beam,

and the subsequent need to remove the remnants of the paint. A positive

cost consideration is the absence of any post-hardening machining

operation to correct for distortion.

For hardening of gears, cost comparisons between laser hardening

and gas carburizing are available. In an environment where the laser is

producing two shifts per day, laser treatment was shown to be cost

effective for large gears where a limited area was to be treated.

One manufacturer that replaced gas carburizing with laser

hardening cites the reasons as follows[2]:

��Reduced hardening time ��Reduced scrap rate ��Elimination of complex quenching, plating, masking,

stripping, and cleaning steps. ��Reduced work-in-progress inventory ��Quicker turnaround, less material handling ��Reduced floor space requirements

7  

Theory Background  

��Reduced pollution by elimination of copper plating ��Reduced energy use  

2-2 The parameters effect on the surface hardening

process

The parameters effects on the surface hardening process are:

1- Laser hardening is a rather modern technique , where the surface to be

hardened is heated by high density laser light [3].

The value of power density to raise the sample temperature to get the

value proper to the surface hardening by phase transformation is given

by:

(2‐1) 

where:

E : the energy (J)

f : focal length (cm)

θ : divergence (rad)

t : pulse duration (μs)

I : power density (w/cm2)

2- Thermal conductivity values . which on this values the thermal flow

depends . Heating the material depends on the thermal flow thought.

3- The value of thermal diffusivity which indicates the speed of heat

diffusion through the material. Temperature rises a little in materials with

high thermal diffusivity with a good penetration to the materials surface a

contrary to low thermal diffusivity which suffer from a high rise in

temperature with a low thermal penetration through the  material  .               

8  

Theory Background  

where :

To : initial surface temperature (K)

I : power density (w/cm2)

R : reflectivity (%)

K : thermal conductivity (W/cm.k)

N : thermal diffusivity (cm2/s)

t : pulse duration (s)

5-The reflectivity value which is one of the important properties

determines the proper wavelength. Most materials reflect a large portion

of the laser beam so, a high laser power system is needed to affect the

metal surface. The reflectivity is also a affected by the surface condition

like the presence of greases which affects the reaction between the laser

beam and the metal surface. The reflectivity of a metal can be reduced by

coating it with a material , however, the coating is not efficient always

because of the weakness of thermal coupling between the coat and

metal[1] .

9  

Theory Background  

2-3 Types of laser used in hardening process

Laser equipments operating with high power levels, the high power

lasers, can produce highly energetic and well focusable laser beams that

are usable in hardening . [28]

2-3-1 Carbon dioxide (Co2) lasers

The most traditional high power lasers which has very high power and

power density, Moderate efficiency, reliable operation and excellent

beam quality. But, the high wavelength of 10.6µm results in a relative

low absorption of the laser beam by metals. It is usual to apply an

absorption enhancing pretreatment like graphitizing.[18]

   

2-3-2 Nd:YAG Laser

The Nd ion when doped into a solid-state host crystal produces the

strongest emission at a wavelength just beyond 1µm. The two host

materials most commonly used for this laser ion are yttrium aluminum

garnate (YAG) and glass. When doped in YAG, the Nd:YAg crystal

produces laser output primarily at 1.064 µm, when doped in glass, the

Nd:glass medium laser at wavelengths ranging from 1.054 to 1.062 µm,

depending upon the type of glass used. Nd also lases at 0.94 µm and at

1.32 µm from the same upper laser level as the 1.064 µm transition,

although these transitions have lower gain.

The Nd laser incorporates a four-level system and consequently has

a much lower pumping threshold than that of the rube laser. The upper

laser level lifetime is relatively long (230 µs for Nd:YAG and 320µs for

Nd :glass), so population can be accumulated over a relatively long

10  

Theory Background  

duration during the pumping cycle when the laser is used either in the Q-

switching mode or as an amplifier. The emission and gain line width are

45 nm for YAG and 28 nm for glass. These lasers can be pumped either

by flashlamps or by other lasers. Diode pumping is a relatively recent

technique that has led to the development of much more compact Nd

lasers, both at the fundamental wavelength of 1.06µm and at the

frequency-doubled wavelength of 0.53µm.

The Nd:YAG crystal has good optical quality and high thermal

conductivity, making it possible to provide pulse laser output at repetition

rates of up to 100Hz. The crystal size is limited to length of

approximately 0.1m and diameters of 12mm, thereby limiting the power

and energy output capabilities of this laser. Doping concentrations for

Nd:YAG crystals are typically of the order of 0.725% by weight, which

corresponds to approximately 1.4*1026 atoms per cubic meter.

For Nd:glass laser gain media, very large-size laser materials have

been produced. Rods of up to 2 m long and 0.075 m in diameter and disks

of up to 0.9 m in diameter and 0.05 m thick have been successfully

demonstrated. The large-diameter disks have been used as amplifiers to

obtain laser pulse energies of many kilojoules. The drawback of Nd:glass

laser materials is their relatively poor thermal conductivity, which

restricts these lasers to relatively low pulse repetition rates[19].

11  

Theory Background  

2-3-3 High power diode laser.

        These equipment are available at maximum 6kW power level. High

power diode laser equipment represent the newest generation of high

power lasers for materials processing. The lower wavelength (typically

0.8 and 0.94 µm) improves further the absorption characteristics of the

laser beam. Due to the very electrical/optical efficiency (30-50%), high

power diode laser equipment are remarkably smaller in size than other

lasers of the same kW level.[18]

2-4 Comparison between Co2 laser , Nd:YAG laser & high

power diode laser.

Tab.(2-2) provides a relative comparison of these three types of

lasers,.[23]

Laser type Wavelength

nm Absorption

efficiency

Initial

cost

Operating

cost

Expected

life

Co2 10,600 Low Low Moderate High

Nd:YAG 1,060 Moderate/high High High High

diode 800 High Moderate Low Low

Tab.(2-2) Relative comparison of different lasers used for heat

treatment.[23]

2-5 Laser processing

In laser processing, such as using laser energy to heat metals,

laser light is directed to the workpiece resulting in absorption

and reflection of light. Absorption by metals is highly dependent

12  

Theory Background  

on the wavelength of light, material type, angle of incidence and

surface condition. See Fig.(2-1), in general, the shorter the

wavelength, the better light is absorbed. Fig.(2-2) illustrates the

interaction of laser light with steel to produce a hardened layer.

Absorption begins at the leading edge of the spot and terminates

at the trialing edge. The time period, or exposure period,

generally is less than one second. The shape of the hardened

zone varies with the shape of the spot and the energy

distribution across the spot produced by the beam[23].

Fig.(2-1) The relationship between energy absorption and wavelength at

room temperature for iron, aluminum and copper[23].

13  

Theory Background  

Fig.(2-2) Laser hardening.[23]

2-6 Comparison between laser hardening and other

technologies

Advantages / Disadvantages    Laser is charaacterized, in comparison with other sources , which

are used in transformation hardening, by the following[1]:

1- There is no need for air evacuation of the zone where laser beam

passes through, as it is the case when using electron beam.

2- Laser may be used in hardening mini areas, and the area , which is

affected by heat, is very limited, so the mechanical properties of the other

parts aren’t affected .

14  

Theory Background  

3- Laser beam can travel long distances without attenuation.

4- Getting higher power densities in comparison with other methods.

5- Materials, which are in closed mediums , can be treated through

certain windows, also area which is hard to reach by conventional

methods can be treated.

6- Chemical cleanllness by laser treatment because there is no contact

between the sample and any other material.

7- Very high speed in performing the hardening process.

In spite of these advantages, there are several disadvantages:

1- Limitation of penetration depth and the areas treated are small.

2- Laser parameters must be controlled to get the desired effect.

3- High accuracy is required in using the optical instruments to guide and

concentrate laser radiation .

4- The training, to an efficient technical team to deal with laser system,

is necessary .

5- High cost of treatment because of the high cost of laser system.

 

2-7 laser – material interaction  

Laser beam interacts with target substances according to the

individual wavelengths. The different wavelength have several degrees of

relative absorption into the various components of hard . These

interactions include:

15  

Theory Background  

1- Reflection: a portion of the incident beam may be reflected off the

surface without penetration or interaction of light energy with the tissue.

2- Absorption: some of the light may be absorbed into a component of

the material, in this case there will be transference of the energy to the

tissue, and it is the most material interaction. When a material absorbs

laser energy, the light energy is converted into thermal energy .

Absorption of laser radiation is also determined by the chemical

composition of the component being irradiated, as well as the wavelength

(λ) of the incident laser beam[2]. The absorbed energy is determined by

following equation[24] :

            

                                     

                          

where :

E : the energy ( J )

R : reflectivity ( % )

A : absorption coefficient ( cm-1 )

Z : depth of penetration ( cm )

EZ : absorption energy (J)

Most of the incident laser radiation on the sample is absorbed at the

surface and this in turn raises the surface temperature more than

depth[24] .

                   

16  

Theory Background  

3- Scattering: This occurs when the target tissue can cause the laser

beam to spread out into a large area and this is useful in photo

polymerization of composite resin, but as the beam is scattered, its power

density eventually decreased to the point where it has no significant

effect.

4- Transmission: The energy travels directly through the material

causing no effect. Depending on the material, some lasers penetrate

deeper than others, e.g. the wavelength of argon laser is transmitted by

clear structure of the eyes, while it is absorbed by the blood vessels of the

retina.

(2-3) Laser-material itraction[20].

2-8 Residual stresses

Residual stresses are important in many of metalworking . They can

cause distortion, either directly or after subsequent heat treatment.

17  

Theory Background  

Residual stresses arise wherever there has been inhomogeneous

plastic deformation, but it is important to recognize that they are elastic

stresses due directly to differences in elastic strain and cannot exceed the

yield stress of the material [25].

Residual stresses should not be overlooked as sources of fatigue

failure. The rupture associated with fatigue failure is a brittle one, caused

by stresses acting in tension at right angles to the cleavage plane.

Therefore, residual tensile surface stresses are undesirable, because they

effectively decrease the stresses which produce failure. Residual tensile

surface stresses many exist in castings as a result of restraint of

dimensional change during cooling. They are similarly found in some

pieces rapidly cooled from hot working temperatures . Residual tensile

surface stresses are also found in welded structures when thermal

contraction is prevented by the rigidity of the structure. They are

encountered in electrodeposited metals, perhaps as a result of evolution

of hydrogen during formation of the deposit. Residual stresses are too

complicated to permit a simplified statement as to their origins[26].

The effect of temperature gradient alone. It was shown earlier, under

the effect of size and mass, that during quenching the surface is cooled

more rapidly than the inside. This results in a temperature gradient across

section of the piece or a temperature difference between the surface and

the center. Almost all solids expand as they are heated and contract as

they are cooled. This means that the surface, since it is at a much lower

temperature, should have contracted much more than the inside.

However, since the outside and inside are attached to each other, the

inside, being longer, will prevent the outside from contracting as much as

it should . it will therefore elongate the outside layers, putting them in

tension while the inside in turn will be in compression. The approximate

18  

Theory Background  

magnitude of this thermal stresses may be calculated from the following

formula[27]:

                 

(2‐5) 

where:

S : residual stresses (Mpa)

α : coefficient of linear expansion (cm/cm. Co)

e : modulus of elasticity (Mpa)

ΔT : change of temperature on the surface (Co )

Since cracks are propagated only by tensile stress, surface residual

compressive stress would be most desirable because they are opposite to

the applied load[27].

                                                        

2-9 The Transformation phases

2-9-1 Ferrite

A solid solution of one or more elements in body-centered-cubic

iron. Unless otherwise designated, the solute is generally assumed to be

carbon. The lower area is alpha ferrite, the upper, delta ferrite. If there is

no designation, alpha ferrite is assumed[27] .

Regions which have attained a temperature above Ac1 and below

Ac3 still contain ferrite. The effect is more pronounced in lower carbon

steels, since the thicker ferrite network requires a longer time above Ac3

to pass completely into solution[29] Fig.(2-5)[1] .

19  

Theory Background  

 

Fig. (2‐4). The iron‐iron carbide equilibrium diagram[1]. 

2-9-2 Cementite

         Cementite is a carbide of iron of the formula Fe3C .It may occur in

steel as free cementite or as a constituent of the eutectoid, pearlite. It

always exists in annealed steels with over 0.9 per cent of carbon and

occurs in the form of a continuous network in normalized steels with

more than 1 per cent of carbon[29].

 

2-9-3 Pearlite

Pearlite was so called on account of the iridescent colours it shows

and its resemblance to mother-of-pearl when viewed by oblique

illumination. It consists of alternate lamellae of ferrite and cementite , and

20  

Theory Background  

contains a little under 0.9 per cent of carbon. Annealed steels containing

less than 0.9 per cent of carbon (hypo-eutectoid steels) consist of pearlite

and ferrite those with more than 0.9 per cent carbon contain pearlite and

cementite. Since the whole of the carbon of a hypo-eutectoid steel

(except less than 0.01 per cent, dissolved in the ferrite) is in the

pearlite[29] .

2-9-4 Austenite

A solid solution of one or more elements in face-centered-cubic

iron. The solute is generally assumed to be carbon. An alloy steel whose

structure is normally austenitic at room temperature [27].

2-9-5 Martensite                    The name martensite honours one of the pioneers of metallurgy,

Professor A. Martens. Martensitic phase transformation is found in

numerous alloys without it necessarily being accompanied by hardening.

Cooled under equilibrium conditions (slow cooling), steel carbon atoms

diffuse out of the austenite matrix to form a pearlite structure, iron atoms

orientate themselves into a Body Centred Cubic (BCC) crystallographic

structure, and with forced cooling rates carbon atoms are trapped in the

iron matrix which results in a highly distorted Body Centered Tetragonal

crystallographic structure [2]. The volumetric difference between martensite and austenite is about

4% resulting in compressive stresses on the surface. For the formation of

martensite, the steel component first needs to heated to the Ac3

(austenite

formation) temperature. This Ac3

temperature is dependent on the

chemical composition of the steel[2] .

21  

Theory Background  

The crystallographic structure of austenite above the Ac

3 temperature

is Face Centred Cubic (FCC).The component treated then needs to be

cooled rapidly to ensure that carbon atoms are trapped in the matrix of the

iron to form martensite. The Ms

and Mf

(“Martensite start” and

“Martensite finish”) temperatures are determined by the chemical

composition of the steel [27].

2-10 Heat treating of metals

The material in aerospace applications is often chosen because of

their heat and corrosion resistance, fatigue properties or low weight.

Depending on the base material of the component is the heat treating

done at different temperature and holding time. In order to describe a

number of common heat treating processes and their purpose, are steels

used an an example. Steel is define as an alloy of iron and carbon with the

carbon content up to about 2 wt%. Other alloy elements can be up to

5wt% in a low-alloy steel and more in a high-alloy steel.Heat treatment is

general name of a large number of thermal processes where the goal is

often to obtain a satisfactory hardeness. Fig.(2-9) shows typical heating

ranges in an Iron-Carbon diagram for different heat treating processes.A1

is the eutectoid line, or the lower critical temperature for austenite

transformation and A3 is the upper critical temperature. Acm represent the

upper critical temperature for hypereutectoid steels. The most common

heat treating processes for steels are described below.

22  

Theory Background  

Fig.(2-5) Iron-Iron Carbide phase diagram showing typical temperature

ranges for different heat treatment operations[30].

      Annealing is a heat treatment process, refers to a material exposed to

an elevated temperature for an extended period of time , and thereafter

cooled down. This is primarily done in order to soften the material.

Ferrite and pearlite are the dominating phases in the material after the

annealing process. If the cooling rate is increased, then martensite will be

created, this process is called quenching. The hardness of the material is

controlled by the amount of the martensite created because of the rapid

cooling from the austenitizing (A3) or solution treating temperature, The

amount of martensite can be controlled by the selection of quench

medium. Common quench media are water, saltwater, oil, polymer

solution or some inert gas (helium, argon or nitrogen).

Tempering of steel is a process in which previously hardened or

normalized steel is heated to a temperature below the critical temperature

in order to increase ductility and toughness. The difference between

tempering and stress relife heat treating is that the aim of the tempering

operation is to create a certain microstructure. In the other case is the

primary aim to relieve stresses, but both procedures are performed in the

23  

Theory Background  

24  

same temperature interval .

The part must be heated up above the A3/Acm temperature so a

homogenous austenite phase is created, in order to be classed as a

normalizing treatment. The chosen cooling rate from the austenizing

temperature is depends of the required strength and hardeness of the

material. At higher cooling rates, more pearlite is formed and the lamellae

are finer and more closely spaced. Larger amount of pearlite and fine

lamellae gives higher strength and hardness. Observe that the cooling rate

should not be as high as for the quenching process where martesite is

created [30].

  

Description of The Model  

Chapter Three

Description of The Model 3-1 Introduction

In this chapter the steps of performing the mathematical model will be

described .                                                                                                    

              

3­2 CK45 Steel 

Material: CK 45

Standard: DIN

Country: Germany

Steel Group: Structural and constructional steels

Subgroup: DIN 1652-4 Bright steels for quenching and

Tempering

Comment: Technical delivery conditions DIN 1652-4 was

Superseded by EN 10277-5

Application: Automobile-and motor construction , mech,

engineering , Bolts and nuts, resistant to

elevated temperature up to 400 oC[31]

 

 

35  

Description of The Model  

3­3Properties (CK45 steel) 

   t<=5mm;K­Cold drawn[31] 

(Yield stress,Rpo2 MPa)                   640 

Tensile stress,Rm(MPa)                 770 

Elongation,As(%)                            4.00 

5<t<= 10mm;K old drawn  ­C

Yield stress,Rpo2(MPa)                   560 

Tensile stress,Rm(MPa)                730 

Elongation As(%)                            5.00 

 

3­4Chemical Composition(%) 

Table(3­1) The chemical composition Ck45steel.[31] 

Max. Min. Criteria 0.5000 0.4200 C 0.4000 Si 0.8000 0.5000 Mn 0.0350 P 0.0350 S 0.4000 Cr 0.1000 Mo 0.4000 Ni 0.6300 Cr + Mo

 

 

 

 

36  

Description of The Model  

3-5 Calculating the power density (I)

Laser effect on perform the surface hardening process relies on

certain parameters and these parameters are power density and pulse

duration.

Eq. (2-1) is used to evaluate the proper power density to raise the

sample surface temperature to get the value proper to the surface

hardening by phase transformation .

A constant value of divergence , a certain energy band between (3-7

J) a lenses focal length between (5-25 cm) and a pulse duration between

(1*10-4 to 5*10-4 s ) were used to perform this search . The procedure was

done as follows fixing energy and focal length values and using variable

values of pulse duration and within the previously mentioned range.

The calculations within all energy , pulse duration and focal length

ranges were made by taking a constant value of the divergence , a wide

range of power density was obtained under different conditions (i.e.

energy , pulse duration and focal length values) at constant divergence

value .                             

                  

3-6 Computing surface temperature (TS)

     The results were obtained from Eq. (2-1). By using Eq.(2-2) different

temperature values were obtained of the sample surface and the proper

value was chosen which was 1500Co (1773k) because it is the nearest to

the sample melting point which is (1538Co) because the phase

transformation degree is very high and near the melting point. The other

values of temperature which were obtained are neglected because they

either higher than the melting point , which is unapplicable to the phase

transformation temperature to austenite phase , or lower than the melting

37  

Description of The Model  

point (very large difference ). This is not valid with the condition that the

phase transformation temperature is very high and near the alloy melting

point .

3-7 Thermal distribution calculation

Thermal distribution calculations in the mathematical model are

divided into two parts :

hermal  distribution  inside  the  sampleCalculations  of    tPart  one  : 

Calculations of thermal distribution calculations inside the sample can be

done by using the thermal penetration depth which was calculated by

Matlab program (Appendix A) and by dividing this depth into five zones

between the surface and the thermal penetration depth, the temperature in

each zone can be calculated after that by appling the Matlab program. A

three-dimensional figure can be obtained and this figures describes the

thermal distribution inside the sample .

 ermal distribution at  the sample surfaceof    thcalculation Part  two:  

calculations of thermal distribution at sample surface are done by

using Matlab program(Appendix B). By this program the distance , at

which heat is distributed, can be calculated from the limits of the

hardened spot (thermally treated spot) which equals laser beam spot area

incident on the sample surface. After that divide this distance into five

zones and the temperature for each zone at the sample surface is

calculated. By appling Matlab program two-dimensional figures are

obtained which describe thermal distribution at sample surface .

3-8 Residual stresses calculations

38  

Description of The Model  

39  

The value of the residual thermal stresses at the sample surface is

computed by using Eq.(2-5) , at the beginning the residual thermal stress,

which is generated between the hardened spot (exposed to laser radiation

and its temperature is 1500Co) and the neighbour zone which is heated by

conductivity of heat, after that between the former zone and the

neighbour zone and soon the distribution of the compressive thermal

stresses at sample surface are computed. By using Matlab is program a

two dimensional graph obtained which clarifies the distribution of the

residual thermal stresses at the sample surface.

Results and Discussion  

Chapter Four

Results and Discussion

4-1 Calculating the power density(I)

In table(4-1) the energy value is (3J), the focal length is (5cm) , and

pulse duration between (1*10-4 to 5*10-4 s) . Values of power density

were obtained . After that a focal length value of (10 cm) was used and

the same pulse duration between (1*10-4 to 5*10-4 s) . The focal length

variation continued (15,20,25cm ) with the same data mentioned above.

The value of laser energy was changed to (4J) table(4-2) the

previous procedure continued until the calculations within all ranges of

energy , pulse duration and focal length were made by taking a constant

value of divergence also table(4-3) ,table(4-40 &table(4-5) .

1) :-The results from using Eq.(2

Table.(4-1) The results from using energy 3J.

I(w/m2) at F=0.25m

I(w/m2) at F=0.20m

I(w/m2) at F=0.15m

I(w/m2) at F=0.10m

I(w/m2) at F=0. 005m

t (s)

3.8197*1065.9683*1061.0610*1072.3873*1079.5493*1071*10-4

1.9099*1062.9842*1065.3052*1061.1937*1074.7746*1072*10-4

1.2732*1061.9894*1063.5386*1067.9577*1063.1831*1073*10-4

9.5493*1061.4921*1062.6526*1065.9683*1062.3873*1074*10-4

7.6394*1051.1937*1062.1221*1064.7746*1061.9099*1075*10-4

50  

Results and Discussion  

Table.(4-2) The results from using energy 4J.

I(w/cm2) at F=25cm

I(w/cm2) at F=20cm

I(w/cm2) at F=15cm

I(w/cm2) at F=10cm

I(w/cm2) at F=5cm

t (s)

5.0930*1067.9577*1061.4147*1073.1831*1071.2732*1081*10-4

2.5465*1063.9789*1067.0736*1061.5915*1076.3662*1072*10-4

1.6977*1062.6526*1064.7157*1061.0610*1074.2441*1073*10-4

1.2732*1061.9894*1063.5368*1067.9577*1063.1831*1074*10-4

1.0186*1061.5915*1062.8294*1066.3662*1062.5465*1075*10-4

Table.(4-3) The results from using energy 5J.

I(w/cm2) at F=25cm

I(w/cm2) at F=20cm

I(w/cm2) at F=15cm

I(w/cm2) at F=10cm

I(w/cm2) at F=5cm

t (s)

6.3662*1069.9472*1061.7684*1073.9789*1071.5915*1081*10-4

3.1831*1064.9736*1068.8419*1061.9894*1077.9577*1072*10-4

2.1221*1063.3157*1065.8946*1061.3263*1075.3052*1073*10-4

1.5915*1062.4868*1064.4210*1069.9472*1063.9789*1074*10-4

1.2732*1061.9894*1063.5368*1067.9572*1063.1831*1075*10-4

Table.(4-4) The results from using energy 6J.

I(w/cm2) at F=25cm

I(w/cm2) at F=20cm

I(w/cm2) at F=15cm

I(w/cm2) at F=10cm

I(w/cm2) at F=5cm

t (s)

2.2282*1081.1937*1072.1221*1074.7746*1071.9099*1081*10-4

1.1141*1085.9683*1061.0610*1072.3873*1079.5493*1072*10-4

7.4272*1073.9789*1067.0736*1061.5915*1076.3662*1073*10-4

5.5704*1072.9842*1065.3052*1061.1937*1074.7740*1074*10-4

4.4563*1072.3873*1064.2441*1069.5493*1063.8197*1075*10-4

51  

Results and Discussion  

Table.(4-5) The results from using energy 7J.

I(w/cm2) at F=25cm

I(w/cm2) at F=20cm

I(w/cm2) at F=15cm

I(w/cm2) at F=10cm

I(w/cm2) at F=5cm

t (s)

8.9127*1061.3926*1072.4740*1075.5704*107 2.228*108 1*10-4

4.4563*1066.9630*1061.2379*1072.7852*1071.1141*1082*10-4

2.9709*1064.6420*1068.2525*1061.8568*1077.4272*1073*10-4

2.2282*1063.4815*1066.1894*1061.3926*1075.5704*1074*10-4

1.7825*1062.7852*1064.9515*1061.1141*1074.4563*1075*10-4

4-2Calculate thermal conductivity and diffusuivity

Thermal conductivity is a propperty of materials that expressed the

power density that will flow through the material if a certain temperature

gradient exists over the material.

The thermal conductivity is usually expressed in W/m.k .the usual

formala is;

Thermal conductivity=power density/temperature

It should be noted that thermal conductivity is a property that is

describes the semi static situation, temperature gradient is assumed to be

constant. As soon as the temperature starts changing,other prameters

enter the equation.

The immediately explains why it is so very difficult to measurment

thermal conductivity. Explains why it is so very difficult to measure

thermal conductivity. Idealy this would require a steady situation. This is

far from easy because it usaualy requires a carefully planned laboratory

experimentt a lot of time to get to an equilibrium.

52  

Results and Discussion  

In thermal parameters,also the heat capacity starts playing a role. The

heat capacity is again a material property. It expresses the fact that for

changing theb temperature of a certain volume of material energy must

flow in or out. The heat capacity is usally found in the textbooks a

specific heat capacity, which must be multiplied by the density to get the

full picture.

Heat capacity=density*specific heat capacity

When dynamic processes are involved,the change of temperature

versus time,at known boundary conditions is determined by both thermal

conductivity and heat capacity.

Thermal diffusuivity= thermal conductivity/heat capacity

4-3Computing surface temperature (TS)

The energy value was selected to be (3J) and the pulse duration is

(1*10-4 s) because they are the best conditions to make the power density

with suitable value to raise the temperature of the surface to the phase

transformation temperature degree table(4-6) .                   

                                                   

When the focal length is decreased , the power density increases

according to Eq. (2-1). The relation between the surface temperature and

the power density is a direct relation this means that when the power

density is increased then the surface temperature rises according to Eq.

(2-2) . It can be realized from the previous figure that the rise in the

surface temperature is very little for the focal length values (20 to 25 cm)

that is because the resulting power density is not sufficient to raise the

53  

Results and Discussion  

surface sample temperature to a high value . For the focal length (15 cm)

the case is different, the change in temperature of the sample surface will

make the surface temperature get the required temperature value. After

the focal length value (10cm) the temperature rise will be so high because

at that point the power density will be so high that it is enough to melt

surface which will be at very high temperature. The energy value was

selected to be (3J) because it is the suitable value to get the proper power

density which raises the metal surface temperature to the phase

transformation temperature to the austenite phase also table(4-7)

,table(4-8),table(4-9),table(4-10)

.

2):-The results from using Eq.(2

Tab.(4-6) The results from using energy 3J.

Ts(k) I(w/cm2) t=5*10-4 s

I(w/cm2) t=4*10-4 s

I(w/cm2) t=3*10-4 s

I(w/cm2) t=2*10-4 s

I(w/cm2) t=1*10-4 s

F(cm)

13575 1.9099*1072.3873*1073.1831*104.7746*1079.5193*1075 3617 4.7746*1065.9683*1067.9577*1061.1937*10 2.3873*10710 1773 2.1221*1062.6526*1063.5368*1065.3052*1061.0610*10715 1127 1.1937*1061.4921*1061.9894*1062.9842*1065.9683*10620 829 7.6394*1059.5493*1051.2732*1061.9099*1063.8197*10625

Tab.(4-7) the results from using energy 4J.

Ts(k) I(w/cm2) t=5*10-4 s

I(w/cm2) t=4*10-4 s

I(w/cm2) t=3*10-4 s

I(w/cm2) t=2*10-4 s

I(w/cm2) t=1*10-4 s

F(cm)

18001 2.5165*1073.1831*1074.2241*1076.3662*1071.2732*1085 4723 6.3662*1067.9577*1061.0610*1071.5915*1073.1831*10710 2265 2.8294*1063.5368*1064.7157*1067.0736*1061.4147*10715 1404 1.5915*1061.9894*1062.6526*1063.9789*1067.9577*10620 1006 1.0186*1061.2732*1061.6977*1062.5465*1065.0930*10625

 

54  

Results and Discussion  

55  

Tab.(4-8) The results from using energy 5J.

Ts(k) I(w/cm2) t=5*10-4 s

I(w/cm2) t=4*10-4 s

I(w/cm2) t=3*10-4 s

I(w/cm2) t=2*10-4 s

I(w/cm2) t=1*10-4 s

F(cm)

22427 3.1831*1073.9789*1075.3052*1077.9577*1071.5915*1085 5830 7.9572*1069.9472*1061.3263*1071.9894*1073.9789*10710 2756 3.5368*1064.4210*1065.8946*1068.8419*1061.7684*10715 1681 1.9894*1062.4868*1063.3157*1064.9736*1069.9472*10720 1183 1.2732*1061.5915*1062.1221*1063.1831*1066.3662*10625

Tab.(4-9) The results from using energy 6J.

Ts(k) I(w/cm2) t=5*10-4 s

I(w/cm2) t=4*10-4 s

I(w/cm2) t=3*10-4 s

I(w/cm2) t=2*10-4 s

I(w/cm2) t=1*10-4 s

F(cm)

26853 3.8197*1074.7740*1076.3662*1079.5493*1071.9099*1085 6936 9.5493*1061.1937*1071.5915*1072.3873*1074.7764*10710 3248 4.2441*1065.3052*1067.0736*1071.0610*1072.1221*10715 1957 2.3873*1062.9842*1063.9789*1065.9683*1061.1937*10720 1360 1.5279*1061.9099*1062.5465*1063.8197*1067.6394*10625

Tab.(4-10) The results from using energy 7J.

Ts(k) I(w/cm2) t=5*10-4 s

I(w/cm2) t=4*10-4 s

I(w/cm2) t=3*10-4 s

I(w/cm2) t=2*10-4 s

I(w/cm2) t=1*10-4 s

F(cm)

31278 4.4563*1075.5704*1077.4272*1071.1141*108 2.2282*108 5 8043 1.1141*1071.3926*1071.8568*1072.7852*107 5.5704*107 10 3740 4.9515*1066.1894*1068.2525*1061.2379*107 2.4740*107 15 2234 2.7852*1063.4815*1064.6420*1066.9630*106 1.3926*107 20 1537 1.7825*1062.2282*1062.9709*1064.4563*1068.9127*106 25

Results and Discussion  

4-3 Energy distribution and depth of heat penetration

calculations

The energy distribution is calculated by using Eq.(2-4) . When the

sample thickness, which was obtained by Eq.(2-7), is used and laser pulse

duration is (1*10-4 s) and by substituting it in the equation , the absorbed

energy (Ez) at the thickness equals zero . When the thickness is made

half, the absorbed energy equals zero too. This means that there is no

absorbed energy at this depth and when the depth of 0.002795 cm was

reashed , there was an absorbed energy value and this is a proof that it is

the thermal penetration depth because when laser rays penetrate the

material , energy absorption happens. And by appling this equation to

various locations between the above mentioned depth and the surface , it

can be noticed that most energy is absorbed at the surface, as a result the

surface temperature will rise to average high value which is the phase

transformation temperature table(4-11) .

 )4‐2q.(E ingus The results        

Ez(J) Z(cm) 00.0111000 00.0055500 00.0027960 

1.5692*10‐990.0027950 9.8896*10‐800.0018236 6.2325*10‐600.0013677 3.9278*10‐400.0009118 2.4753*10‐200.0004559 

1.560

Tab.(4-11)

56  

Results and Discussion  

4-4 Thermal distribution calculation

Thermal distribution calculations in the mathematical model are

divided into two parts :

ermal distribution inside the sampleth  calculations of Part one : 

The fig.(4-1) shows the thermal distribution in the selected areas of

the depth . It can be noticed at the surface exposed to radiation area is a

small are compared to the others because it represents the laser radiation

incidence area at the sample surface, the temperature of this area rises to

about 1500Co and the size of this area depends on the laser spot size and

this area is the hardened area . Also it can be noticed that when moving

inside the sample deeper temperature decreases and this is because that

(most thermal energy is absorbed at the surface and the rest is distributed

on the depth ) and the zone area increases because of the thermal

diffusivity which diffuses the thermal energy. The cause behind energy

penetration from the surface to the depth is the thermal conductivity it can

be noticed also the big difference in temperature between the surface and

the area near to it and this decreases with the increase in depth . This is

because of the short time of heating (laser pulse duration ) this prevents

radiation from penetrating to big depth and because of the surface

reflectivity which reflects part of the radiation table(4-12) .

        

57  

Results and Discussion  

The results    

Tab.(4-12)

Temperature(k) Depth(cm) 177305380.0004559 3440.0009118 3090.0013677 3010.0018236 2980.0027950 

Fig.(4‐1) The temperature distribution in the depth. 

                                                              

 surface  ermal distribution at the sampleth of calculationPart two:  

The fig.(4-2) explains thermal distribution in the selected areas of the

sample surface . It can be noticed at the middle of the figure a small red

58  

Results and Discussion  

zone which represents laser radiation falling area on the sample surface

whose temperature reaches (1500Co) which is the hardening area (zone)

which is a small zone because the laser spot is small. A big difference in

temperature between the hardening zone and the adjacent one is because

surface sample exposure time to laser radiation is short and not enough

for large heat transfer to the adjacent zones table(4-13) .

The results

Tab.(4-13)

Temperature(K) Distance on the surface(cm) 177305930.00160366 3710.00320732 3220.00481098 3090.00641464 2980.00801830 

59  

Results and Discussion  

 Fig.(4‐2)The temperature distribution in the sample surface. 

4-5 Residual stresses calculations

The fig.(4-3) clarifies the generated thermal stresses after the

hardening process. The small spot in the middle of the figure represents

the hardening zone which is the zone of the sample surface where the

laser radiation falls, because of the rapid rise in its temperature which

reaches about (1500Co). This zone expands, for the rest of the sample

surface which is not influenced by the temperature rise because the

thermal exposure is short (short laser pulse duration), this leads to oppose

the expansion and this in turn causes compressive stresses directed

towards the hardening zone. These stresses are useful because they act in

the opposite direction to the load which is applied on the sample increases

as shown in Eq.(2-5) & table (4-14) .           

                           

 

 

60  

Results and Discussion  

61  

  )5‐2q.(E ingsu The results

Tab.(4-14)

S (psi) ∆T (K)F (cm) Ts (K) 2589015 132775 13575 647205 331910 3617 287205 147515 1773 161655 82920 1127 103545 53125 829 

Fig.(4‐3)The thermal stresses distribution in the sample surface. 

Conclusion and Suggestion for Future work  

Chapter Five

Conclusion and Suggestions for Future Work

 

5-1 Conclusions

The main conclusions from the present project are :

1- Not every laser system can be used in hardening, but a certain

condition must be available so as to use the laser system and alloy in

hardening.

A certain condition in the laser system:

I- A suitable pulse duration was found is equal (1*10-4s).

II-A suitable focal length of lenses was found is equal (15cm).

III- A suitable energy was found is equal (3J).

IV- A suitable power density to get the phase transformation temperature

(1500Co) of the surface alloy (1.061*107W/cm2).

The best conditions in the alloy transformation hardening on the surface:

I- A suitable reflectivity of the surface alloy for this laser wavelength

(1.064µm) is (48%) .

II- A suitable thickness of alloy was found to be equal (0.111mm)

because the remaining parts of the sample from a heat sink sufficient to

cool the heat surface at a cooling rate to form the martensite at the alloy

surface and there is no need for an exterior cooling medium and this

know self –quenching .

2- In the thermal distribution inside the sample noticed the big difference

in temperature between the surface and the area near to it and this

72  

Conclusion and Suggestion for Future work  

73  

decreases with the increase in depth. This is because of the short time

heating (laser pulse duration) this prevents radiation from pentrating to

big depth and because of the surface reflectivity which reflects part of the

radiation.

3- In the thermal distribution at the sample surface a big difference in

temperature between the hardening zone and adjacent one is because

surface sample exposure time to laser radiation is short and not enough

for large heat transfer to the adjacent zones.

4- The residual stresses are compressive stresses directed towards the

hardening zone. These stresses are useful because they act in the opposite

direction to the load which is applied on the sample.

5-2 Suggestions for future work

1- Performing the practical application and comparing the results with the

mathematical model.

2- Set up a mathematical model to study the suitable parameters to make

the reflectivity match the laser wavelength and this is done by coating the

surface with a suitable paint and studying the conditions that lead to the

occurance of the thermal coupling between the paint and the surface.

 3- Set up a mathematical model to study the effects of the residual

thermal stresses on the surface by using a pulse laser and a continuous

laser and comparing between them .

4- Set up a mathematical model to study the hardening process by

surface melting by laser .

 5- Set up a mathematical model to study the hardening process by

surface alloying by laser .

Contents 

No.p Subject I II

  

IIIVI I

VIIX  1 3 5 6  7 8 

10 10 10 12 12  12 14  15 17 18 20 22 23 25 26 26  27 27 

 Dedication     Acknowledgements  Abstract  Contents List of Figures   List of Symbols                    

ctioion

                                            Chapter one ‐ Introdu

t 

n1‐1General  Introduc1‐2Literature survey1‐3 Aim of the work1‐4 Outline of thesisChapter two – Theortical Background2‐1 Introduction2‐2 The parameters effect on the surface hardening process2‐3 Types of laser used in hardening process2‐3‐1 Carbon dioxide (Co2) laser2‐3‐2 Nd:YAG laser2‐3‐3 High power laser 2‐4 Comparison between Co2laser,Nd:YAG laser & high power diode laser 2‐5 Laser processing2‐6 Comparison between laser hardening with other Technologies2‐7 Laser‐ material interaction2‐8 Residual Stresses2‐9 The Transformation phases2‐9‐1 Ferrite2‐9‐2 Cementite2‐9‐3 Pearlite2‐9‐4 Austenite2‐9‐5 Martensite 2‐10 Heat treating of metals  Chapter Three – Description of  The Model

VI 

3‐1 Introduction

VI 

27  28  35 35 36 36 38  50 52 53  57 60   for Future Work  72 73  

3‐2 CK45 Steel

3‐3Properties (CK45 steel)

3‐4Chemical Composition(%)

3‐5 Calculating the power density(I)3‐6 Computing surface temperature(Ts)3‐7 Thermal Distribution Calculation3‐8 Residual stresses Calculations

 Chapter Four‐Discussion and Results4‐1 Introduction4‐2 Computing surface temperature(Ts)4‐3 Energy distribution and depth of heat penetration calculation4‐4 Thermal Distribution Calculation4‐5 Residual stresses Calculation

  Chapter Five‐Conclusions and Suggestions 

5‐1 Conclusion5‐2 Suggestions for Future Work

  AppendixesReferences

 

 

References  

 74 

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31-Key to metal steel.

O=2*10^‐3; 

E=3; 

t=1*10^‐4; 

F=5; 

I=E/(pi*F^2*O^2*t( 

I=9.5493*10^7 

t=2*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=4.7746*10^7 

t=3*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=3.1831*10^7 

t=4*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=2.3873*10^7; 

t=5*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.9099*10^7 

F=10; 

t=1*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=2.3873*10^7 

t=2*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.1937*10^7 

t=3*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=7.9577*10^6 

t=4*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=5.9683*10^6 

t=5*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=4.7746*10^6 

F=15 ;% 

t=1*10^‐4;  

I=E/(pi*F^2*O^2*t ( 

I=1.0610*10^7  

t=2*10^‐4 ; 

I=E/(pi*F^2*O^2*t ( 

I=5.3052*10^6  

t=3*10^‐4 ; 

I=E/(pi*F^2*O^2*t( 

I=3.5368*10^6 

t=4*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=2.6526*10^6 

t=5*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=2.1221*10^6 % 

F=20; 

t=1*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=5.9683*10^6 

t=2*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=2.9842*10^6 

t=3*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.9894*10^6 

t=4*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.4921*10^6 

t=5*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.1937*10^6 

F=25; 

t=1*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=3.8197*10^6 

t=2*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.9099*10^6 

t=3*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.2732*10^6 

t=4*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=9.5493*10^5 

t=5*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=7.6394*10^5 

E=4; 

F=5; 

t=1*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.2732*10^8 

t=2*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=6.3662*10^7 

t=3*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=4.2441*1067 

t=4*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=3.1831*1067 

t=5*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=2.5465*10^7 

F=10; 

t=1*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=3.1831*10^7 

t=2*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.5915*10^7 

t=3*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.0610*10^7 

t=4*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=7.9577*10^6 

t=5*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=6.3662*10^6 

F=15; 

t=1*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.4147*10^7; 

t=2*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=7.0736*10^6; 

t=3*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=4.7157*10^6 

t=4*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=3.5368*10^6 

t=5*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=2.8294*10^6 

F=20; 

t=1*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=7.9577*10^6 

t=2*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=3.9789*10^6 

t=3*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=2.6526*10^6 

t=4*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.9894*10^6 

t=5*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.5915*10^6 

F=25 

t=1*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=5.0930*10^6 

t=2*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=2.5465*10^6 

t=3*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.6977*10^6 

t=4*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.2732*10^6 

t=5*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.0186*10^6 

E=5; 

F=5; 

t=1*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.5915*10^8 

t=2*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=7.9577*10^7 

t=3*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=5.3052*10^7 

t=4*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=3.9789*10^7 

t=5*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=3.1831*10^7 

F=10; 

t=1*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=3.9789*10^7 

t=2*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.9894*10^7 

t=3*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.3263*10^7 

t=4*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=9.9472*10^6 

t=5*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=7.9577*10^6 

F=15; 

t=1*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.7684*10^7 

t=2*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=8.8419*10^6 

t=3*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=5.8946*10^6 

t=4*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=4.4210*10^6 

t=5*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=3.5368*10^6 

F=20; 

t=1*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=9.9472*10^6 

t=2*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=4.9736*10^6 

t=3*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=3.3157*10^6 

t=4*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=2.4868*10^6 

t=5*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.9894*10^6; 

F=25; 

t=1*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=6.3662*10^6 

t=2*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=3.1831*10^6 

t=3*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=2.1221*10^6 

t=4*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.5915*10^6 

t=5*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.2732*10^6 

E=6; 

F=5; 

t=1*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.9099*10^8 

t=2*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=9.5493*10^7 

t=3*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=6.3662*10^7 

t=4*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=4.7746*10^7 

t=5*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=3.8197*10^7 

F=10; 

t=1*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=4.7746*10^7 

t=2*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=2.3873*10^7 

t=3*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.5915*10^7 

t=4*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.1937*10^7 

t=5*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=9.5493*10^6; 

F=15; 

t=1*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=2.1221*10^7 

t=2*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.0610*10^7 

t=3*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=7.0736*10^6 

t=4*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=5.3052*10^6 

t=5*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=4.244*10^6 

F=20; 

t=1*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.1937*10^7 

t=2*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=5.9683*10^6 

t=3*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=3.9789*10^6 

t=4*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=2.9842*10^6 

t=5*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=2.3873*10^6 

F=25; 

t=1*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.1937*10^7 

t=2*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=5.9683*10^6 

t=3*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=3.9789*10^6 

t=4*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=2.9842*10^6 

t=5*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=2.3873*10^6 

F=25; 

t=1*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=7.6394*10^6 

t=2*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=3.8197*10^6 

t=3*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=2.5465*10^6 

t=4*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.9099*10^6 

t=5*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.5279*10^6 

E=7; 

F=5; 

t=1*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=2.2282*10^8 

t=2*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.1141*10^8 

t=3*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=7.4272*10^7 

t=4*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=5.5704*10^7 

t=5*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=4.4563*10^7 

F=10; 

t=1*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=5.5704*10^7 

t=2*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=2.7852*10^7 

t=3*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.8568*10^7 

t=4*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.3926*10^7 

t=5*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.1141*10^7 

F=15; 

t=1*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=2.4757*10^7 

t=2*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.2379*10^7 

t=3*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=8.2525*10^6 

t=4*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=6.1894*10^6 

t=5*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=4.9515*10^6 

F=20; 

t=1*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.3926*10^7 

t=2*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=6.9630*10^6 

t=3*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=4.6420*10^6 

t=4*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=3.4815*10^6 

t=5*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=2.7852*10^6 

F=25; 

t=1*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=8.9127*10^6 

t=2*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=4.4563*10^6 

t=3*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=2.9709*10^6 

t=4*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=2.2282*10^6 

t=5*10^‐4; 

I=E/(pi*F^2*O^2*t( 

I=1.7825*10^6 

IF  E=3;   F=5; 

K=0.1800; 

To=298; 

N=0.3064; 

R=0.4800; 

t=1*10^‐4; 

I=9.5493*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=13575 

t=2*10^‐4; 

I=4.7746*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=13575 

t=3*10^‐4; 

I=3.1831*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=13575 

t=4*10^‐4; 

I=2.3873*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=13575 

t=5*10^‐4; 

I=1.9099810^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=13575 

IF E=3;  F=10; 

t=1*10^‐4; 

I=2.3873*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=3617 

t=2*10^‐4; 

I=1.1937*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=3617 

t=3*10^‐4; 

I=7.9577*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=3617 

t=4*10^‐4; 

I=5.9683*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=3617 

t=5*10^‐4; 

I=4.7746*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=3617 

IF E=3;  F=15 ; % 

t=1*10^‐4; 

I=1.0610*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1773 

t=2*10^‐4; 

I=5.3052*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1773 

t=3*10^‐4; 

I=3.5368*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1773 

t=4*10^‐4; 

I=2.6526*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1773 

t=5*10^‐4; 

I=2.1221*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1773 % 

IF E=3;   F=20; 

t=1*10^‐4; 

I=5.9683*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1127 

t=2*10^‐4; 

I=2.9842*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1127 

t=3*10^‐4; 

I=1.9894*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1127 

t=4*10^‐4; 

I=1.4921*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1127 

t=5*10^‐4; 

I=1.1937*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1127 

IF E=3;  F=25; 

t=1*10^‐4; 

I=3.8197*10^6 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=829 

t=2*10^‐4; 

I=1.9099*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=829 

t=3*10^‐4; 

I=1.2732*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=829 

t=4*10^‐4; 

I=9.5493*10^5; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=829 

t=5*10^‐4; 

I=7.6394*10^5; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=829 

IF E=4;  F=5; 

t=1*10^‐4; 

I=1.2732*10^8; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=18001 

t=2*10^‐4; 

I=6.3662*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=18001 

t=3*10^‐4; 

I=4.2441*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=18001 

t=4*10^‐4; 

I=3.1831*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=18001 

t=5*10^‐4; 

I=2.5465*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=18001 

IF E=4;  F=10; 

t=1*10^‐4; 

I=3.1831*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=4723 

t=2*10^‐4; 

I=1.5915*1067; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=4723 

t=3*10^‐4; 

I=1.0610*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=4723 

t=4*10^‐4; 

I=7.9577*10^6 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=4723 

t=5*10^‐4; 

I=6.3662*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=4723 

IF E=4;  F=15; 

t=1*10^‐4; 

I=1.4147*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=2265 

t=2*10^‐4; 

I=7.0736*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=2265 

t=3*10^‐4; 

I=4.7157*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=2265 

t=4*10^‐4; 

I=3.5368*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=2265 

t=5*10^‐4; 

I=2.8294*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=2265 

IF E=4;  F=20; 

t=1*10^‐4; 

I=7.9577*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1404 

t=2*10^‐4; 

I=3.9789*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1404 

t=3*10^‐4; 

I=2.6526*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1404 

t=4*10^‐4; 

I=1.9894*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1404 

t=5*10^‐4; 

I=1.5915*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1404 

IF E=4;  F=25; 

t=1*10^‐4; 

I=5.0930*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1006 

t=2*10^‐4; 

I=2.5465*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1006 

t=3*10^‐4; 

I=1.6977*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1006 

t=4*10^‐4; 

I=1.2732*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1006 

t=5*10^‐4; 

I=1.0186*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1006 

IF E=5;  F=5; 

t=1*10^‐4; 

I=1.5915*10^8; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=22427 

t=2*10^‐4; 

I=7.9577*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=22427 

t=3*10^‐4; 

I=5.3052*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=22427 

t=4*10^‐4; 

I=3.9789*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=22427 

t=5*10^‐4; 

I=3.1831*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=22427 

IF E=5;  F=10; 

t=1*10^‐4; 

I=3.9789*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=5830 

t=2*10^‐4; 

I=1.9894*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=5830 

t=3*10^‐4; 

I=1.3263*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=5830 

t=4*10^‐4; 

I=9.9472*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=5830 

t=5*10^‐4; 

I=7.9577*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=5830 

IF E=5;  F=15; 

t=1*10^‐4; 

I=1.7684*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=2756 

t=2*10^‐4; 

I=8.8419*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=2756 

t=3*10^‐4; 

I=5.8946*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=2756 

t=4*10^‐4; 

I=4.4210*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=2756 

t=5*10^‐4; 

I=3.5368*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=2756 

IF E=5;  F=20; 

t=1*10^‐4; 

I=9.9472*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1681 

t=2*10^‐4; 

I=4.9736*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1681 

t=3*10^‐4; 

I=3.3157*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1681 

t=4*10^‐4; 

I=2.4868*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1681 

t=5*10^‐4; 

I=1.9894*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1681 

IF E=5;  F=25; 

t=1*10^‐4; 

I=6.3662*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1183 

t=2*10^‐4; 

I=3.1831*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1183 

t=3*10^‐4; 

I=2.1221*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1183 

t=4*10^‐4; 

I=1.5915*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1183 

t=5*10^‐4; 

I=1.2732*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1183 

IF E=6;  F=5; 

t=1*10^‐4; 

I=1.9099*10^8; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=26853 

t=2*10^‐4; 

I=9.5493*1067; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=26853 

t=3*10^‐4; 

I=6.3662*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=26853 

t=4*10^‐4; 

I=4.7740*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=26853 

t=5*10^‐4; 

I=3.8197*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=26853 

IF E=6;  F=10; 

t=1*10^‐4; 

I=4.7746*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=6936 

t=2*10^‐4; 

I=2.3873*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=6936 

t=3*10^‐4; 

I=1.5915*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=6936 

t=4*10^‐4; 

I=1.1937*10^7 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=6936 

t=5*10^‐4; 

I=9.5493*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=6936 

IF E=6;  F=15; 

t=1*10^‐4; 

I=2.1221*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=3248 

t=2*10^‐4; 

I=1.0610*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=3248 

t=3*10^‐4; 

I=7.0736*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=3248 

t=4*10^‐4; 

I=5.3052*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=3248 

t=5*10^‐4; 

I=4.2441*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=3248 

IF E=6;  F=20; 

t=1*10^‐4; 

I=1.1937*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1957 

t=2*10^‐4; 

I=5.9683*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1957 

t=3*10^‐4; 

I=3.9789*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1957 

t=4*10^‐4; 

I=2.9842*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1957 

t=5*10^‐4; 

I=2.3873*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1957 

IF E=6;  F=25; 

t=1*10^‐4; 

I=7.6394*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1360 

t=2*10^‐4; 

I=3.8197*10^6 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1360 

t=3*10^‐4; 

I=2.5465*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1360 

t=4*10^‐4; 

I=1.9099*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1360 

t=5*10^‐4; 

I=1.5279*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1360 

IF E=7;  F=5; 

t=1*10^‐4; 

I=2.2282*10^8; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=31278 

t=2*10^‐4; 

I=1.1141*10^8; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=31278 

t=3*10^‐4; 

I=7.4272*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=31278 

t=4*10^‐4; 

I=5.5704*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=31278 

t=5*10^‐4; 

I=4.4563*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=31278 

IF E=7;  F=10; 

t=1*10^‐4; 

I=5.5704*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=8043 

t=2*10^‐4; 

I=2.7852*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=8043 

t=3*10^‐4; 

I=1.8568*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=8043 

t=4*10^‐4; 

I=1.3926*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=8043 

t=5*10^‐4; 

I=1.1141*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=8043 

IF E=7;  F=15; 

t=1*10^‐4; 

I=2.4740*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=3740 

t=2*10^‐4; 

I=1.2379*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=3740 

t=3*10^‐4; 

I=8.2525*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=3740 

t=4*10^‐4; 

I=6.1894*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=3740 

t=5*10^‐4; 

I=4.9515*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=3740 

IF E=7;  F=20; 

t=1*10^‐4; 

I=1.39226*10^7; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=2234 

t=2*10^‐4; 

I=6.9630*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=2234 

t=3*10^‐4; 

I=4.6420*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=2234 

t=4*10^‐4; 

I=3.4815*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=2234 

t=5*10^‐4; 

I=2.7852*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=2234 

IF E 7;  F=25; 

t=1*10^‐4; 

I=8.9127*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1537 

t=2*10^‐4; 

I=4.4563*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1537 

t=3*10^‐4; 

I=2.9709*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1537 

t=4*10^‐4; 

I=2.2282*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1537 

t=5*10^‐4; 

I=1.7825*10^6; 

Ts=To+(2*I*(1‐R)*(N*t)^1/2)/K*pi^1/2 

Ts=1537 

IF  E=3;   F=15; 

N=0.3064; 

X=0.2775; 

T=(x^2)/(4*N( 

T=6*10^‐2 

t=1*10^‐4; 

r=t/T 

r=1*10^‐3; 

X=0.0555; 

T=(x^2)/(4*N( 

T=2*10^‐3; 

t=1*10^‐4; 

r=t/T 

r=5*10^‐3 

X=0.0111 ;% 

T=(x^2)/(4*N( 

T=1*10^‐4; 

t=1*10^‐4; 

r=t/T 

r=1 % 

X=0.0022 

T=(x^2)/(4*N( 

T=3*10^‐6; 

t=1*10^‐4; 

r=t/T 

r=33.33 

X=0.0004; 

T=(x^2)/(4*N( 

T=1*10^‐7; 

t=1*10^‐4; 

r=t/T 

r=1000 

X=0.3925; 

T=(x^2)/(4*N( 

T=1*10^‐3; 

t=2*10^‐4; 

r=t/T 

r=2*10^‐3 

X=0.0785; 

T=(x^2)/(4*N( 

T=5*10^‐3; 

t=2*10^‐4; 

r=t/T 

r=4*10^‐1 

X=0.0157 ;% 

T=(x^2)/(4*N( 

T=2*10^‐4; 

t=2*10^‐4; 

r=t/T 

r=1 % 

X=0.0031; 

T=(x^2)/(4*N( 

T=7*10^‐6; 

t=2*10^‐4; 

r=t/T 

r=2.85 

X=0.0006; 

T=(x^2)/(4*N( 

T=2*10^‐7; 

t=2*10^‐4; 

r=t/T 

r=100 

X=0.4800; 

T=(x^2)/(4*N( 

T=1*10^‐1; 

t=3*10^‐4; 

r=t/T 

r=3*10^‐3 

X=0.0960; 

T=(x^2)/(4*N( 

T=7*10^‐3; 

t=3*10^‐4; 

r=t/T 

r=4*10^‐3 

X=0.0192 ;% 

T=(x^2)/(4*N( 

T=3*10^‐4; 

t=3*10^‐4; 

r=t/T 

r=1 % 

X=0.0038; 

T=(x^2)/(4*N( 

T=1*10^‐5; 

t=3*10^‐4; 

r=t/T 

r=3 

X=0.0007; 

T=(x^2)/(4*N( 

T=3*10^‐7; 

t=3*10^‐4; 

r=t/T 

r=100 

X=0.5525; 

T=(x^2)/(4*N( 

T=2*10^‐1; 

t=4*10^‐4; 

r=t/T 

r=2*10^‐3 

X=0.1105; 

T=(x^2)/(4*N( 

T=9*10^‐3; 

t=4*10^‐4; 

r=t/T 

r=4*10^‐4 

X=0.0221 ;% 

T=(x^2)/(4*N( 

T=4*10^‐4; 

t=4*10^‐4; 

r=t/T 

r=1 % 

X=0.0044; 

T=(x^2)/(4*N( 

T=1*10^‐5; 

t=4*10^‐4; 

r=t/T 

r=4 

X=0.0008; 

T=(x^2)/(4*N( 

T=5*10^‐7; 

t=4*10^‐4; 

r=t/T 

r=80 

X=0.6200; 

T=(x^2)/(4*N( 

T=3*10^‐1; 

t=5*10^‐4; 

r=t/T 

r=1*10^‐3 

X=0.1240; 

T=(x^2)/(4*N( 

T=1*10^‐2; 

t=5*10^‐4; 

r=t/T 

r=5*10^‐3 

X=0.0248 ;% 

T=(x^2)/(4*N( 

T=5*10^‐4; 

t=5*10^‐4; 

r=t/T 

r=1 % 

X=0.0049; 

T=(x^2)/(4*N( 

T=1*10^‐5; 

t=5*10^‐4; 

r=t/T 

r=5 

X=0.0009; 

T=(x^2)/(4*N( 

T=6*10^‐7; 

t=5*10^‐4; 

r=t/T 

r=83.33 

IF  E=3; 

a=6.5*10^‐6; 

e=30*10^6; 

Ti=298; 

Ts=13575; 

IF F=5; 

)Ts‐Ti)=13277 

S=a*e*(Ts‐Ti( 

S=2589015 

Ts=3617; 

IF F=10; 

)Ts‐Ti)=3319 

S=a*e*(Ts‐Ti( 

S=647205 

Ts=1773 ;% 

IF F=15; 

)Ts‐Ti)=1475 

S=a*e*(Ts‐Ti( 

S=287625 % 

Ts=1127; 

IF F=20; 

)Ts‐Ti)=829 

S=a*e*(Ts‐Ti( 

S=161655 

Ts=829; 

IF F=25; 

)Ts‐Ti)=531 

S=a*e*(Ts‐Ti( 

S=103545 

IF 

E=3; 

R=0.48; 

A=10^5; 

Z=0.011100; 

Ez=(E*(1‐R))*(exp(‐A*Z( 

Ez=0 

Z=0.005550; 

Ez=(E*(1‐R))*(exp(‐A*Z( 

Ez=0 

Z=0.002796; 

Ez=(E*(1‐R))*(exp(‐A*Z( 

Ez=0 

Z=0.002795; 

Ez=(E*(1‐R))*(exp(‐A*Z( 

Ez=1.5692*10^‐99 

Z=0.0018236; 

Ez=(E*(1‐R))*(exp(‐A*Z( 

Ez=9.8896*10^‐80 

Z=0.0013677; 

Ez=(E*(1‐R))*(exp(‐A*Z( 

Ez=6.2325*10^‐40 

Z=0.0004559; 

Ez=(E*(1‐R))*(exp(‐A*Z( 

Ez=2.4753*10^‐20 

Z=0; 

Ez=(E*(1‐R))*(exp(‐A*Z( 

Ez=1.56 

IF  d1=0;          T1=1773; 

IF  d2=0.0004559;  T2=; 

IF  d6=0.002795;   T6=298; 

0.0004559-0=0.0004559

0.002795-0=0.002795

1773-298=1475

)1475/0.002795*(0.0004559=240

T2=240+298 

T2=538 

IF  d2=0.00045559;  T2=538; 

IF  d3=0.0009118;   T3=; 

IF  d6=0.002795;    T6=298; 

0.0009118-0.0004559=0.0004559

0.002795-0.0004559=0.0023391

538-298=240

)240/0.0023391*(0.0004559=46

T3=298+46 

T3=344 

IF  d3=0.0009118;   T3=344; 

IF  d4=0.0013677;   T4=; 

IF  d6=0.002795;    T6=298; 

0.0013677-0.0009118=0.0004559

0.002795-0.0009118=0.0018832

344-298=46

)46/0.0018832*(0.0004559=11

T4=298+11 

T4=309 

IF  d4=0.0013677;    T4=309; 

IF  d5=0.0018236;    T5=; 

IF  d6=0.002795;     T6=298; 

0.0018236-0.0013677=0.0004559

0.002795-0.0013677=0.0014273

309-298=11

)11/0.0014273*(0.0004559=3

T5=298+3=301 

IF 

I=2.1221*10^6; 

t=5*10^‐4; 

spot area=E/(I*t( 

spot area=0.00282738 

IF 

density=0.87; 

V=1/density 

V=1.14 

M=density*V 

M=0.9918 

C=0.064; 

Ts=1773; 

Ti=298; 

Q=M*C(Ts‐Ti( 

Q=94.19 

area=0.0028273; 

K=0.18; 

depth=(k*area*(Ts‐Ti))/Q 

depth=0.0080183 

area=0.00282738; 

)radius1)^2=area/pi 

radius1=0.02999971 

radius2=radius1+depth 

radius2=0.03801801 

diameter=2*radius2 

diameter=0.07603602 

IF  D1=0;              T01=1773; 

IF  D2=0.00160366;     T02=; 

IF  D6=0.0080183;      T06=298; 

0.00160366-0=0.00160366

0.00801830-0=0.0080183

1773-298=1475

)1475/0.0080183*(0.00160366=295

T02=298+295 

T02=593 

IF  D2=0.00160366;     T02=593; 

IF  D3=0.00320732;     T03=; 

IF  D6=0.00801830;     T06=298; 

0.00320732-0.00160366=0.00160366

0.00801830-0.00160366=0.00641464

593-298=295

)295/0.00641464*(0.00160366=73

T03=298+73 

T03=371 

IF  D3=0.00320732;     T03=371; 

IF  D4=0.00481098;     T04=; 

IF  D6=0.00801830;     T06=298; 

0.00481098-0.00320732=0.00160366

0.00801830-0.00320732=0.00481098

371-298=73

)73/0.00481098*(0.00160366=24

T04=298+24 

T04=322 

IF  D4=0.00481098;     T04=322 

IF  D5=0.00641464;     T05=; 

IF  D6=0.00801830;     T06=298; 

0.00641464-0.00481098=0.00160366

0.00801830-0.00481098=0.00320740

322-298=24

)24/0.00320740*(0.00160366=11

T05=298+11 

T05=309 

List of symbols

symbol The energy of the laser (J)E Focal length of lens (cm)f The divergence of laser (rad)θ Pulse duration (µs)t Power density (w/cm2)I Initial surface temperature (k)To 

Reflectivity (%)R Thermal conductivity(cal/sec/cm/Co)K Thermal diffusivity (cm2/s)N Additional energy is provided by the oxygenPch 

Process energy (laser energy available for heating , melting or viporisation

Pproc 

Energy loss due to heat conduction into the work piecePc 

Energy loss due to reflectionPR Energy loss due to convectionPcon 

Absorption coefficient ( cm‐1 ) A Depth of penetration ( cm ) Z Residual stresses (psi) S Coefficient of linear expansion (cm/cm. Co) α Modulus of elasticity (psi) e Change of temperature on the surface (Co ) ΔT Time cycle (s) Interlamellar spacing (cm) ℓ Diffusion coefficient, about 10

‐9 cm

2.s

‐1at elevated 

temperatures

D 

Thermal time constant ( s/cm ) T Thickness of the alloy ( cm ) X 

X  

 

 الخلاصة

باستخدام برنامج الماتلاب ( ألطوريفي ھذه الدراسة تم عمل موديل رياضي لعملية التصليد بالتحول الكربونالتي يصعب تصليدھا بالطرق الاعتيادية وذلك لكون نسبة (لسبيكة الحديد المطاوع مع النيكل ) 6.5

.ياك النبضي - باستخدام ليزر الندميوم) فيھا منخفضة

, حساب درجة حرارة سطح المعدن بالاعتماد على مواصفات السبيكة التي ھي التوصيلية الحرارية تم لليزر المستخدم ودرجة الحرارة الابتدائية للسطح التي ھي ألموجيالانعكاسية للطول , الانتشارية الحرارية

ابق وزمن النبضة المستخدمة قيم كثافة القدرة التي تم حسابھا في الس إلى بالإضافةدرجة حرارة الغرفة ألطوريدرجة انصھار السبيكة والتي بلغت لان درجة حرارة التحول إلى الأقربواخترنا درجة الحرارة

طور الاوستنايت بالكامل ھي قريبة جدا من درجة الانصھار وبذلك نختار مقدار كثافة القدرة التي ترفع إلى .ي والطاقة التي تؤدي للحصول على كثافة القدرة ھذهوالبعد البؤر إلىدرجة حرارة السطح

ألطوريالذاتي والذي ھو الظرف المناسب للتحول الإخمادتم حساب سمك العينة المناسب لحصول البعد , الكامل من طور الاوستنايت الى طور المارتنسايت الذي ھو طور التصليد فعند ما تكون الطاقة

.البؤري وزمن نبضة الليزر

ومعامل  تم حساب توزيع الاجھادات الحرارية المتخلفة على سطح العينة باستخدام معامل التمدد الطولي داخل السبيكة باستخدام انعكاسية سطح السبيكة والطاقة الأشعةنتمكن من معرفة عمق نفاذ المرونةالمتخلفة بواسطة الى معامل الامتصاص ورسم مخطط توزيع الاجھادات الحرارية السطحية بالإضافة

.برنامج الماتلاب

على طول عمق النفاذ الحراري (تم حساب توزيع درجات الحرارة على سطح السبيكة وداخل العينة .ورسم مخطط للتوزيعين الحراريين بواسطة برنامج الماتلاب ) الليزر داخل العينة لأشعة