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Up Topic Welding Industry / General Welding Discussion / Calculating Weld Stress
- - By Zeek (**) Date 02-03-2010 15:55
Is there a general guideline or specification that calls out the correct way to calculate residual stresses in weldments?
Parent - - By 803056 (*****) Date 02-03-2010 18:20
Not that I have used. In general, the magnitude of the residual stress is on par with the yield strength of the base metal. The contributing area is any base metal that experienced a delta T of about 230 degrees F.

The area of base metals contributing to the distortion as a result of the residual stress is much larger than that of the weld. Weld shrinkage, occurring when the weld solidifies, is minimal because the weld can not exert a large tensile force while it is at high temperature, instead, it yields. As the weld cools and contracts it is capable of exerting more tensile stress on the surrounding base metal. However, at temperatures above 800 degrees the maximum force is about 1/2 the yield strength of the weld deposit. As it cools to room temperature it is capable of exerting more tensile force, on par with the yield strength of the weld.

The base metal is a different story. As the metal heats a localized area, it exerts a compressive force on the surrounding base metal due to expansion. As it reaches a temperature that is about 220 degrees above the surrounding metal the compressive force is about equal to the yield point of the base metal, thus localized yielding takes place as the temperature increases further yet. Upon cooling the compressive force becomes a tensile force as the yielded base metal cools and contracts. Clearly, the area that experiences the greatest delta T is going to experience the greatest yielding/plastic deformation, but the maximum force it can exert is equal to the yield strength. Any greater loading will cause localized yielding by plastic deformation. A joint that has some freedom to move will experience distortion as a result of the residual stresses.

The gradient of delta T can be reduced by preheating the part before welding. The higher the preheat, the less thermal gradient from one area of the welded joint to another.

There are some computer modeling software that is being used to predict the amount of distortion that will result under given conditions. I have never used the programs, but it was discussed when I was taking courses in the mid 1980's. There is an article in last month’s Welding Journal that talks about distortion and some of the computer models being used.

Best regards - Al
Parent - - By Tommyjoking (****) Date 02-04-2010 00:48 Edited 02-04-2010 00:51
Well my 1.5 yr old daughter decided my post needed to go to never never land.....I will try again.

I am SURE I am misunderstanding my much esteemed colleague AL in this case, I will just blurt my stupidity and sit and listen.

"in General" you can consider the tension in a joint "in general joint configuration"  to be at or below the yield strength of the materials in question...base metals + filler metals.  Not bringing process or any type of stress relief into the picture,,,,,,,,,,,,,,,So If I can consider that my material and my filler are let's say both rated to 80KSI in a welding situation......then I am to "assume" the tension in the weld joint could be as high or just below the yield strength, depending on the distortion involved?     So if a given joint may be loaded with this much "as welded" tension PRE existing before any other mechanical loads applied .....I could be only counting on say "5KSI"   actual load strength  be it tensile, compressive, shear or what have you loads?  I know that's not the actuality, that's just how I am interpreting Al's post.  If possible please clarify for brain density issues like mine.

Still yet, it's an engineering question purely.  To answer it fully as opened ended as it is ......would require a lot of charts and information to establish definitive answers.  I am interested on the availability of resources like this...it could prove extremely usefull.   Neat post...cannot wait to see where it ends up.
Best regards folks
TOmmy
Parent - - By 803056 (*****) Date 02-04-2010 05:47 Edited 02-04-2010 19:07
Ultimate tensile strength is the force required cause fracture under tensile loads.

Yield strength is the force required to cause plastic deformation.

Residual stress is the level of stress in the material as a result of forming, thermal cutting, welding, etc.

When you bend a piece of plate you have to exceed the yield strength of the material or the material will spring back into it's original shape. Even when you do exceed the yield strength of the material there is some spring back that has to be accounted for. The spring back is due to the deformation that occurs when the load is within the elastic range of the base metal.

To put it a different way. When you apply a small force on the plate being bent, it will spring back to its original position. The load is said to be within the elastic range for the material. Looking at a stress strain diagram for steel, the elastic range is the straight portion of the graph toward the left vertical axis. The slope of the straight line is what we refer to as the modulus of elasticity (E) which for steel is equal to 29 x 10 to the sixth power (29 million). E is a measure of stiffness for a material.

Once the applied load exceeds the yield strength of the material permanent deformation occurs. If you are bending a piece of plate, you have to exceed the yield strength of the material in order for the plate to stay bent. You usually have to bend beyond the required angle to account for the spring back (that portion of the load that is below the yield point and within the elastic range.

Once localized yielding occurs, the material is strain hardened if it is a ductile material (unhardened steel). The strain hardening (work hardening) strengthens the steel so that it can sustain additional load, but the rate of strain (stretching) increases. In other words the material stretches faster with increased loads.

Consider our ever faithful ASTM A36 carbon steel plate. It has a yield strength of 36 ksi and a tensile strength of 58 ksi. It begins to plastically  deform at 36 000 psi. Still it doesn't fracture. We can increase the stress up to 58 000 psi before the steel is in danger of fracturing.

Our designs usually limit the maximum stress to some fraction of the yield strength. For structures it is usually about 2/3's of the yield strength, well within the elastic range. We are talking about maximum stresses on the order of 21 to 27 ksi. There is plenty of reserve strength to accommodate the occasional overload and the uncertainty in our design calculations. That probably accounts for why building don't collapse when the erector or fabricator doesn't pay close attention to details and let some "quality" slip.

The residual stress act similar to pretensioned bolts in a bolted connection. The residual stress is greater than the applied stress. Plastic deformation has already occurred due to the residual stresses introduced by welding and the allowable stress is with in the elastic range. Limiting our discussion to CJP grooved joints, as long as the weld doesn't cause the connection to fail, there is little danger that a static tensile stress that is within the range permitted by design parameters is going to cause failure.

Additional complexity enters the picture due to the fact that the residual stresses are not uniformly distributed and they are both transverse to the weld and longitudinal to the axis of the weld. They act as vectors that can be added there by increasing the resultant. This is a real problem when you have two perpendicular intersecting welds or worse yet three intersecting welds (think of a web welded to a flange of a built-up girder with a stiffener welded continuously to the web and the flange, yikes!).

The problem of intersecting welds is more critical for the higher strength steels where the ratio of yield strength to tensile strength approach 1. Good old A36 has a YS/TS ratio of about 0.6. There is plenty of reserve strength and rarely do intersecting welds crack at the intersect. The material is ductile, as is the weld, so the resultant residual stress (by vector addition) is 51 ksi which is less than the UTS of 58 ksi.  The resultant simply exceeds YS so plastic deformation occurs and no crack is formed. The problem is a concern with steels like A992 which have higher YS (50 ksi) and TS (65 ksi). The ratio of YS/TS is 0.77. When two welds intersect the resultant vector force is 71 ksi which exceeds the TS. What happens when the applied force exceeds the UTS? That's right, you have a high probability of developing a crack at the point of intersection. When you have three intersecting welds in either A36 or A992, there is a   high probability of developing a crack at the point of intersection.

Interesting stuff.

Best regards - Al
Parent - By ssbn727 (*****) Date 02-04-2010 13:20
Excellent explanation Al... As usual!!! :) :) :)

Respectfully,
Henry
Parent - By Tommyjoking (****) Date 02-05-2010 01:37
Thx Al  the definitions cleared it up for me.

Best regards
TOmmy
Up Topic Welding Industry / General Welding Discussion / Calculating Weld Stress

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