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.
