3 December 2022

Shaeffler diagram | Dissimilar Material Welding | Joining of dissimilar metals | welding of C-Steel to austenitic SS

Shaeffler diagram | Dissimilar Material Welding

Joining of dissimilar metals

  • Dilution needs to be considered – undesirable microstructure in the weldment need to be avoided
  • Proper selection of welding consumables & process
  • Weld joint needs to at least match the properties of the metals to be joined
  • The joint needs to meet the design parameters

C-Steel to austenitic SS

  • When a weld is made using a filler wire or consumable, there is a mixture in the weld consisting of approximately 20% parent metal and  80% filler metal alloy
  • Any reduction in alloy content of 304 / 316 type austenite is likely to cause the formation of martensite on cooling.  This could lead to cracking problems and poor ductility.
  • To avoid this, over-alloyed filler metal is used – such as 309, which should still form austenite on cooling at normal dilution.
  • The Shaeffler diagram can be used to determine the microstructure of the weld when a filler metal and parent metal of differing compositions are used.

Schaeffler Diagram

  • Ni-equivalent =  %Ni + 30%C + 0.5%Mn
  • Cr-equivalent = %Cr + Mo + 1.5%Si + 0.5%Nb
  • Typical 304L
  • Ni Equiv = 11.3, Cr Equiv = 18.8
  • For a typical 309L welding consumable
  • Ni Equiv = 14.35 Cr Equiv = 24.9
  • The main disadvantage with this diagram is that it does not represent Nitrogen, which is a very strong Austenite former.

Choose what weld metal structure you wish to have, draw a line from the 50% point (B) to this desired point and then extrapolate by allowing for 25% dilution to arrive at the filler metal composition.

 

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