Questions Int 1

7/29/2019 Questions Int 1

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Q.1 How are pipes designated under the 1) ASTM code 2) IS code (below 6 dia.)

Give examples.

1) ASTM code A106 Gr.B / A 335 P 11 / A 333 Gr.6.

2) IS code ( below 6 dia.) IS 1239.

Q.2 Define Schedule number?

Schedule number is a designation system for use in ordering of pipe (wall thickness / size

of the pipe).

Note : ( The original intent of the ASME committee was to establish a system of

schedule number for pipe size / wall thickness combinations which would have an

approximately uniform relationship equal to 1000 times the P/S expression contained

in the modified Barlow formula for piping wall thickness.The resulting numbersdeparted so far from the existing wall thickness in common use that the original

intent could not be accomplished.) The wall thickness standard / extra strong anddouble exdtra strong have been commercially used designations for many years.

Standard and Schedule 40 are same up to NPS 10 ( DN 250) inclusive, All larger

sizes of standard have 3/8 (9.53 mm) wall thicknesses. Extra strong and Schedule

80 are same up to NPS 8 ( DN 200) inclusive, All larger sizes of Extra strong have ( 12.7 mm) wall thicknesses.

Calculate the theoretical schedule number in the following cases

(a) (b) (c)

Max. design pressure 6 kg/cm2 100 kg/cm2 250 kg/cm2Stress 15000 psi 20000 psi 16000 psi.

Convert the above theoretical schedule number in the standard schedules in the case

of

(a) 1.5 dia pipe (b) 4 dia pipe (c) 8 dia pipe

&

Q.3 Give the formula for calculation of minimum wall thickness of a pipe. Calculate

the thickness in the above cases.

Wall thickness is calculated as given below.

Minimum wall thickness of a pipe = 0.875x Nominal thk. of a pipe.

T= PxD. Where P= Design Pressure, D= Nominal Dia. Of pipe & S= allowable stress.

2xS


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For example : T=6x1.5x14.22x25.4 [1Kg/cm2=14.22 Psi]

2x15000

(a1) For 1.5 dia. Case a : T= 0.11mm converting to nearest Schedule no. is

1.5 Sch.40 i.e 3.68 mm.

(a2) For 1.5 dia. Case b : T= 1.35 mm to nearest Schedule no. is 1.5 Sch.40i.e 3.68 mm.

(a3) For 1.5 dia. Case c : T= 4.23 mm to nearest Schedule no. is 1.5 Sch.80

i.e 5.08 mm.

(b1) For 4.0 dia. Case a : T= 0.29 mm to nearest Schedule no. is 4 Sch.40

i.e 6.02 mm.

(b2) For 4.0 dia. Case b : T= 3.61 mm to nearest Schedule no. is 2 Sch.40i.e 3.91 mm.

(b3) For 4.0 dia. Case c : T= 11.29 mm to nearest Schedule no. is 4 Sch.160

i.e 13.49 mm.

(c1) For 8.0 dia. Case a : T= 0.58mm to nearest Schedule no. is 8 Sch.20

i.e 6.35 mm.(c2) For 8.0 dia. Case b : T= 7.23mm to nearest Schedule no. is 8 Sch.40

i.e 8.18 mm.

(c3) For 8.0 dia. Case c : T= 22.58mm to nearest Schedule no. is 8 Sch.160

i.e 23.01 mm.

Q.4 How are Tubes designated ?

Tubes are designated with respect to the OD and wall thickness.

Q.5 Give typical material specifications for

(a) C.S.Pipe :SA 53 Typ.E Gr.A & B ERW & SA 53 Typ.S Gr.A & B SMLS / SA 106 Gr.A,B,C SMLS

C.S. Tube :SA-178 Gr.A,C,D ERW tubes / SA 179 SMLS tubes

(b) S.S Pipe :SA- 312 TP 304, TP 316, TP 317, TP 321 SMLS & Welded pipes/

SA- 358 TP 304, TP 316, TP 317, TP 321 Fusion welded pipes.

S.S Tube :SA-249 TP 304, TP 316, TP 317, TP 321 welded tubes / SA- 789

SMLS & Welded pipes / SA-213 TP 304, TP 316, TP 317, TP 321 SMLS tubes

/ SA-688 TP 304, TP 316 welded tubes.

(c). L.T.C.SPipe :SA 333 Gr. 1,3,4,6,7,8,9,10,11 SMLS & Welded pipes /

SA- 524 Gr.I & II welded pipes.


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L.T.C.S Tube :SA- 334 Gr. 1,3,6,7,8,9,11 welded tubes.

(d) A.S Pipe :SA-335 Gr.P5, P9, P11, P22, P9, P91 SMLS pipes

A.S. Tube :SA- 199 Gr. T5, T9, T11, T22 SMLS tubes/ SA-209 Gr. T1, T1a &

T1b - C-Mo. Alloy steel SMLS tubes / SA-423 Gr.1 & 2 - low alloy steelSMLS & ERW tubes.

Q.6. Name the various types of flanges in common use. Also give the various facings.

Various types of Flanges in common use are WNRF,SORF,SWRF,BF and Threaded

Flanges.

Various facings are Flat face, Raised face For ratings 150# & 300# raised face shall

be 0.06 (1.5mm) which is included in the minimum flange thickness and for

400,600,900,1500 and 2500# - raised face shall be 0.25(6.35mm) which is additional to

the minimum flange thickness.For lapped joint flanges facings shall be raised face, Large male and female, Tongue and

Groove, Ring joint.

The Flange facings finish shall be judged by visual comparison with Ra standards and

not by instrument having stylus tracers and electronic amplification.

For Tongue & Groove and Small Male & Female Surface roughness shall not exceed

125 in.

For Ring joint Surface roughness shall not exceed 63 in.For Serrated Spiral and Serrated Concentric Surface roughness shall be from 125 in to

250 in.

Q.7 How are elbows, reducers, tees and flanges designated under the DIN code?

Q.8 How are Flanges designated under the

(1) ASA code : 3 WNRF A 105 300LB Sch 80 3.2/6.3 Ra B 16.5

(2) BS code.

Q.9 Give the equivalent of a 600# ASA in (1) DIN code (2) BS code.

Q.10 Give typical material specifications for

(a) CS flanges :A105 forgings / A 216 Gr. WCB castings / A 515 Gr.70 Plates.(b) SS flanges : A 182 Gr.F304 - forgings / A 351 Gr.CF3 - castings / A 240 Gr.304

Plates.

(c) LTCS flanges :A 350 Gr. LF3 forgings / A 352 Gr. LC3 castings / A 203

Gr.E Plates.


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(d) AS flanges :A 182 Gr.F11 CL 2 - forgings / A 217 Gr.WCB - castings / A 387

Gr.11 CL.2 Plates.

Q.11Give typical usages of eccentric reducers with suitable explanations/sketches.

Eccentric reducers are used in the pump inlets to avoid the turbulence & Gas pitting of

the impellers.

On the pipe racks when size variation are there eccentric reducers shall be used so thatlevel shall rest on the pipe support.

Q.12Calculate from first principles the centre line distance of a 6 dia. 45 deg.

Elbow ( R=1.5D).

Q. 13 Give Typical material specifications for

a) CS Tees -2x2 BW equal Tee A 234 WPB Sch 160 B 16.9b) SS Tees - 2x2 BW equal Tee A 234 WP 11 Sch 80 B 16.9

c) LTCS reducers 4x3 Conc. Reducers A 234 WPL6 Std. wt. B 16.9

d) AS Elbows 2 BW 90 El A 234 WP 5 Sch. 80 LR B 16.9

Q.14 Name the commonly used types of valves and its applications.

Commonly used types of valves are

Gate valve Used as an Isolation valves

Check valves swing chk valve, Tilting disc chk valve, Lift chk valve, stop chk valves.-Used to prevent flow reversal.

Ball valve Used as an flow with minimum turbulence and can balance or throttle easy

operation.Globe valve Regulation valve

Butterfly valve Used for high capacity with low pressure loss.

Plug valve Another type of isolation valve similar to Gate valve but with quick shut off.

Q.15 Give typical material specification for

a) CS plates :A 537 CL 1, A 515 Gr. 70

b) SS plates :A 240 TP 304 L

c) LTCS plates :A 203 Gr.E, A 516 Gr. 70

d) AS plates :A 387 Gr.11 CL 2, A 517 Gr.B

Q.16 Give a brief description of

(1) Intergranular corrosion :Corrosion which occurs at the grain boundaries due to

the carbide precipitation in SS welding is known as IGC. This can be avoided by usinglow carbon SS.

(2) Carbide precipitation :Carbon is precipitated out at the grain boundaries, of the

steel, in the temperature range of 1050 F. (565 C) to 1600 F. (870 C.). This is a

normal temperature range during the welding of stainless steel.


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This carbon combines with the chrome in the stainless steel to form chromium carbide

starving the adjacent areas of the chrome they need for corrosion protection. In the

presence of some strong corrosives an electrochemical action is initiated between thechrome rich and chrome poor areas with the areas being low in chrome becoming

attacked. The grain boundaries are then dissolved and become non existent.

Q.17 What are the steps taken to prevent the above ?

There are three ways to combat this:

Anneal the stainless after it has been heated to this sensitive range. This means

bringing it up to the proper annealing temperature and then quickly cooling it

down through the sensitive temperature range to prevent the carbides from

forming.

When possible use low carbon content stainless if you intend to do any welding

on it. A carbon content of less than 0.03% will not precipitate into a continuous

film of chrome carbide at the grain boundaries. 316L is as good example of a lowcarbon stainless steel.

Alloy the metal with a strong carbide former. The best is columbium, but

sometimes titanium is used. The carbon will now form columbium carbide rather

than going after the chrome to form chrome carbide. The material is now said tobe "stabilized" .

Q.18 What is the difference between

(1) SS 304 and SS 304 L -304 L is low Carbon C 0.03%.

(2) SS 304, SS 316 and SS 321.

SS 304 18 Cr. and 8 Ni.

Where as SS 316 contains Molybdenum 2 to 3 %.

SS 321 as the same chemical composition of SS 304 but in addition to that it contains Tiwhich is a stabilizer.

Q.19 Give the various welding positions for Butt welding along with sketches.

Plates Groove welding

1G Flat position, 2G Horizontal position, 3G Vertical position & 4G Overhead

position.

Plates Fillet welding


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1F Flat position, 2F Horizontal position, 3F Vertical position & 4F Overhead

position.

Pipes Groove welding

1G Pipe horizontal and rotated, 2G- Pipe vertical, 5G- Pipe horizontal and fixed, 6G-Pipe inclined at 45 deg.

Pipes Fillet welding

1F Pipe inclined at 45 deg. and rotated, 2F- Pipe vertical, 2FR Pipe horizontal

rotated, 4F- Pipe vertical but welding in overhead position, 5G- Pipe horizontal andfixed.

Stud welding

1S Stud vertical and welding in horizontal position, 2S- Stud horizontal, 4S- Stud

vertical but welding in overhead position.

Q.20 Explain briefly what you understand by HAZ ?

HAZ means Heat affected zone. The portion of the base metal which has not been melted

but whose mechanical properties or micro structures have been altered by the heat of the

welding or cutting.

Q.21 How is stress induced in a weld ?

During welding the base material near the weld metal and the HAZ transforms throughvarious metallurgical phases depending upon the chemistry of the metals in their areas.

Hardening occurs in various degrees, depending mainly upon the Carbon content. This is

particularly true in the HAZ adjacent to the weld metal deposit. The resultant stressesare highest due to melting and solidification. Stress due to welding is of magnitude

roughly equal to the yield strength of the base material.

Q.22 What is stress relieving ?

Stress relieving is a heat treatment process by which the fabricated equipment or thevessel or the welding is uniformily heated to a sufficiently high temperature but below thelower transformation temperature range then holding it for a sufficient time depending

upon the material thickness so that the locked up stresses are relieved and then finally

uniformily cooling it.

Q.23 What are the various types of Non destructive testing? Explain briefly

principles underlying them?


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Various types of NDT are PT, MT, RT, UT.

PT : Penetrant test is an effective means for detecting discontinuities which are open to

the surface. It works fairly well on all materials- metals, Non metals, with an exceptionon porous materials. This works on the principle of capillary action. The surface to be

examined shall be cleaned and a penetrant with a dye is applied over the surface. Aftersome dwell time the penetrant is cleaned with a cleaner and a developer is sprayed on

the surface. When the developer dries the penetrant which has got in to the surfacedefects if any will be blotted on the developed surface. Penetrating power is a very

important feature on which test sensitivity is heavily dependent.

MT :Magnetic particle is an effective means for detecting discontinuities which are on

the surface as well as sub surface (4 to 6 mm). The method is based on residual fieldinduced at the interface of discontinuity by a high intensity magnetic flux. When any

ferro magnetic material is magnetized and if fine iron powder are distributed over the

magnetized surface, then if any discontinuities exists the iron powder will get

accumulated because of the flux leakage this will be visible under the proper lightingcondition.

RT : Radiographic testing method is based on recording the varying degree of

absorption of penetrating radiation by an object such as weld. In conventionalradiography this varying degree of absorption produces a latent image of the object

being examined on a film. The film is chemically processed, transforming the latent

image in to a permanent shadow image of the internal and external condition of theobject. The processed film is called the radiograph. The radiograph can be interpreted

and the integrity of the object can be evaluated.

UT :

Ultrasonic testing is based on the phenomenon of acoustic wave reflection uponencountering obstacles to propagation within a material. If the obstacle lies normally tothe ultrasonic incident beam the wave is reflected back to its generating source. By

observing the reflected signals the defects can be analysed.

Q.24 Name the various types of isotopes giving their half-life periods. Define half-

life?

Half-life is the time taken for the number of radioactive atoms to be reduced to half of the

original number.

Various types of Radioactive isotopes are Ir-192 Iridium 192 74.5 days half-lifeperiod, Co-60 Cobalt 60 5.3 years half-life period, Cs-137 Caesium 137,

Q.25 Explain what is meant by Stereography ?

Stereographic imaging methods, he says, uses information from two 2-D projections to

calculate the 3-D location of features in an object. Two images can be acquired

simultaneously using the two-source/detector pairs in the facility and appropriate


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software controls for the two CCD cameras. If the point of interest is identifiable in both

images, the 3-D coordinate can be established.

Hence, stereographic imaging can produce a 3-D map of a feature of interest as a

function of time.

Q.26 Calculate the minimum load that a 20 cm long 8 mm fillet weld can take. ( f =

1025 kg/cm2).

W=P/A, where W=Load on the fillet weld; P=Allowable load; A=length of the weld.

w=W/f, where w = Fillet weld leg size cm; W=load on the fillet weld in Kg/cm;

f=allowable load on weld in Kg/cm2.

A=20 cm, w = 8 mm, f=1025 kg/cm2. Therefore W=wxf = 8/10x1025=820 Kg/cm.

P=WxA = 820x20 = 16400 Kg = 16.4 Tons.

Q.27 Give the material specification for high temperature stud/nuts ?

Stud bolts SA 193 Gr. B7 & Nuts SA-194 Gr.2H.

Q.28 What do you understand by a 6013 electrode ?

60- Indicates the min. UTS of undiluted weld metal in Ksi ie. 60 Ksi = 60000 psi.

1 Indicates welding position All position except vertical down.( if it is 2 then it is Flat

& Horizontal fillet, 3 then Flat, 4 then vertical down).

3 Indicates the usability and covering of the electrodes Light penetration and

Titania-sodium covering. ( if it is 0 means then Deep penetration and Cellulose-sodium

covering, 8 means medium penetration and Low Hydrogen-iron powder covering).

Q.29 What do you understand by Hydrogen Induced Cracking?

Microcracking in steel that is caused by the absorption of hydrogen, usually due tocorrosion or welding. Hydrogen atoms produced, for example, by the corrosion reaction

of iron, usually combine to form hydrogen gas molecules. However, in the presence of

sulphide or cyanide, the hydrogen recombination reaction is poisoned so that the nascent

hydrogen atoms diffuse into the steel rather than recombining on the metal surface toform hydrogen gas. Hydrogen atoms that enter the pipeline can cause embrittlement and

failure. Usually HIC failures occur within a few weeks of the pipeline being put inservice.

Q.30 Define K value for an insulation material ?

K is the Thermal conductivity of an insulation material.


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Q.31 Explain why mineral wool is preferred to glass wool for insulation?

Mineral wool density can ber maintained whereas glass wool is more fibrous and

disintegrates easily which will cause more pollution. Nowadays mineral wools arereadily available and glass wool is not available.

Q.32 Explain why mineral wool and thermocole cannot be used for cold and hot

insulation respectively?

Because of the Thermal conductivity this cannot be used.

Q.33 Explain how coating and wrapping is done for U/G piping ?

Coating and wrapping for U/G piping is carried out by two methods, one by Hot applied

Coal Tar Enamel protective coating complete with inner and outer reinforcing wrapsand another by Cold applied self adhesive PVC backed tape or liquid Polyurethane Tar

for field joints, valves, spools and fittings etc. The installation design temperature forpiping shall be assumed to be 43C.

Q.34 Explain with suitable sketches the terms roll,set and travel as applied

to piping?

Q.35 Give the advantages and disadvantages of a Horton sphere when compared to

a conventional storage tank?

Horton sphere is used to store liquid/gas above the atmospheric pr. Whereasconventional storage tanks shall be used for storing liquids at atmospheric pr. only.

Q.36 What is the function of a rupture disc? What is the major difference when

compared to a safety/relief valve ?

Rupture disc is used for releasing excess pr. by rupturing and once ruptured we have to

replace the disc whereas in safety/relief valve after releasing the valve can be reset.

Q.37 Differentiate between a safety valve and relief valve. What types of a safety

device would you use for steam service?

Safety valve is used for only in steam generating equipments whereas relief valve is usedfor any any pressure service.

Q.38 Explain what is flash steam?

When condensate leaves the steam traps - flash steam is generated.


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Q.39 Explain why it is important to remove condensate from steam lines?

Condenste shall be removed from the steam line other wise it will cause water

hammering when the steam flows which may results in to the failure of the pipe line.

Q.40 What is the function of a steam trap?

Function of the Steam trap is to remove the condensate.

Q.41 Explain why (1) Stainless steel (2) cast iron cannot be cut by a conventional

oxy-acetylene flame?

Because of the carbide formation the materials become very hard and difficult to melt.

Q.42 Explain briefly what is cathodic protection?

Cathodic protection is a technique applied to steel where metals anodic to iron (eg, zinc,aluminum, magnesium) are applied to the surface on the steel workpiece to provide a

corrosion resistant surface. The process relies on the fact that where a cell exists

between two metals with an electrolyte, one of the metals will corrode and in the processof corroding protect the other metal.The more active metal will serve as an anode andwill be corroded instead of the less active metal. The anode is then called a sacrificial

anode.

Q.43. Explain when a centrifugal pump is used and when a reciprocating pump is

used ?

Centrifugal pumps and rotary pumps are best suited for low to medium pressure

applications whereas Reciprocating pumps are usually specified for high pressure

service, with capabilities exceeding 100,000 psi.

The main difference between kinetic and positive displacement pumps lies in the method

of fluid transfer.

Centrifugal pumps are known as Kinetic pumps. A kinetic pump imparts velocity energy

to the fluid, which is converted to pressure energy upon exiting the pump casing.

Reciprocating pump is used for positive displacement.A positive displacement pumpmoves a fixed volume of fluid within the pump casing by applying a force to moveable

boundaries containing the fluid volume.


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Q.44. What are the 3 precautions that has to be taken during the welding of Orifice

flanges ?

1) Welding nearer to the flanges shall be flush ground to avoid any turbulence.

2) The orientation of the tapping holes shall be maintained.

3) No bends shall be available nearer to the flange it shall be straight for a distance ofatleast 5 times the Diameter of the pipe on the Up stream.

Q.45. What are Killed and Semi killed Steels ?

Killed Steel

Steel deoxidized with a strong deoxidizing agent, such as silicon or aluminum, to reduce

the oxygen content to such a level that no reaction occurs between carbon and

oxygen during solidification.

The term killed indicates that the steel has been sufficiently deoxidized to quiet the

molten metal when poured into the ingot mold. The general practice is to use aluminumferrosilicon or manganese as deoxidizing agents. A properly killed steel is more uniform

as to analysis and is comparatively free from aging. However, for the same carbon and

manganese content Killed Steel is harder than Rimmed Steel. In general all steels above

0.25% carbon are killed, also all forging grades, structural steels from 0.15% to 0.25%carbon and some special steels in the low carbon range. Most steels below 0.15% carbon

are rimmed steel.

Semikilled Steel

Steel that is incompletely deoxidized and contains sufficient dissolved oxygen to reactwith the carbon to form carbon monoxide and thus offset solidification shrinkage.

Steel incompletely deoxidized, to permit evolution of sufficient carbon monoxide to offsetsolidification shrinkage.

Q.45. What are Rimmed Steels ?

Low-carbon steel in which incomplete deoxidation permits the metal to remain liquid at

the top of the ingot, resulting in the formation of a bottom and side rim of considerablethickness. The rim is of somewhat purer composition than the original metal poured. Ifthe rimming action is stopped shortly after pouring of the ingot is completed, the metal is

known as capped steel. Most steels below 0.15% carbon are rimmed steels. For the same

carbon and manganese content rimmed steel is softer than killed steel.

A low-carbon steel containing sufficient iron oxide to give a continuous evolution ofcarbon monoxide while the ingot is solidifying, resulting in a case or rim of metal


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virtually free of voids. Sheet and strip products made from the ingot have very good

surface quality.

Q.46. What is Carbon Equivalent ?

Referring to the rating of weld-ability, this is a value that takes into account theequivalent additive effects of carbon and other alloying elements on a particular

characteristic of a steel. For rating of weld-ability, a formula commonly used is:

CE = C + (Mn/6) + [(Cr + Mo + V)/5] + [(Ni + Cu)/15].

The most common carbon equivalent (CE) formulas to evaluate weldability depends ifthe metal is an alloy steel or modern carbon steel. Here are both common CE equations;

for low alloy steel, elements are expressed in weight percent amounts;

CIIw=C+Mn/6+(Cr+Mo+V)/5+(Ni+Cu)/15

for modern low carbon steels or microalloy steels, elements are expressed in weightpercent amounts ;

Ceq=C+Si/25+(Mn+Cr)/16+(Cr+Ni+Mo)/20+V/15

Also from the Welding Journal for low carbon, microalloyed steels, the Ito-Besseyocarbon equivalent

Ceq=C+Si/30+(Mn+Cu+Cr)/20+Ni/60+Mo/15+V/10+5*B

Expressed in weight percent amounts

For CE%


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ARC CRACKS

Definition: A depression left at the termination of the weld where the weld pool is leftunfilled.

Cause: Improper weld termination techniques

Repair: If no cracks exist, simply fill in the crater. Generally welding from beyond the

crater back into the crater.

Longitudinal Crack

Definition: A crack running in the direction of the weld axis. May be found in the weld orbase metal.

Cause: Preheat or fast cooling problem. Also caused by shrinkage stresses in high

constraint areas.

Prevention: Weld toward areas of less constraint. Also preheat to even out the coolingrates.

Repair: Remove and reweld

Transverse Crack

Definition: A crack running into or inside a weld, transverse to the weld axis direction.

Cause: Weld metal hardness problem

Crater Crack

Definition: A crack, generally in the shape of an X which is found in a crater. Cratercracks are hot cracks.

Cause: The center of the weld pool becomes solid before the outside of the weld pool,

pulling the center apart during cooling

Prevention: Use crater fill, fill the crater at weld termination and/or preheat to even out

the cooling of the puddle

Throat Crack

Definition: A longitudinal crack located in the weld throat area.

Cause: Transverse Stresses, probably from shrinkage. Indicates inadequate filler metal

selection or welding procedure. May be due to crater crack propagation.


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Prevention: Correct initial cause. Increasing preheat may prevent it. be sure not to leave

a crater. Use a more ductile filler material.

Repair: Remove and reweld using appropriate procedure. Be sure to correct initial

problem first.

Toe Crack

Definition: A crack in the base metal beginning at the toe of the weld

Cause: Transverse shrinkage stresses. Indicates a HAZ brittleness problem.

Prevention: Increase preheat if possible, or use a more ductile filler material.

Root Crack

Definition: A crack in the weld at the weld root.

Cause: Transverse shrinkage stresses. Same as a throat crack.

Prevention: Same as a throat crack

Underbead Crack

Definition: A crack in the unmelted parent metal of the HAZ.

Cause: Hydrogen embrittlement

Prevention: Use LOW HYDROGEN electrodes and/or preheat

Repair: (only found using NDT). Remove and reweld.

Hot Crack

Definition: A crack in the weld that occurs during solidification.

Cause: Micro stresses from weld metal shrinkage pulling apart weld metal as it coolsfrom liquid to solid temp.

Prevention: Preheat or use a low tensile filler material.

Repair: Remove and reweld, correct problem first, preheat may be necessary, increase

weld size.


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Cold Crack

Definition: A crack that occurs after the metal has completely solidified

Cause: Shrinkage, Highly restrained welds, Discontinuities

Prevention: Preheat, weld toward areas of less constraint, use a more ductile weld metal

Repair: Remove and reweld, correct problem first, preheat may be necessary.

Repairs to Cracks

Determine the cause

Correct the problem

Take precautions to prevent reoccurrence

Generally required to repair using a smaller electrode

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