Obviously, you don't have to calculate the bonding area of a butt joint. It will be the cross-section of the thinner member and that's that. But lap joints are often variable. Their length can be increased or decreased. How long should a lap joint be. The rule of thumb is to design the lap joint to be three times as long as the thickness of the thinner joint member.
A longer lap may waste brazing filler metal and use more base metal material than is really needed, without a corresponding increase in joint strength. And a shorter lap will lower the strength of the joint. For most applications, you're on safe ground with the "rule of three." More specifically, if you know the approximate tensile strengths of the base members, the lop length required for optimum joint strength in a silver brazed joint is as follows:
If you have a great man identical assemblies to braze, or if the joint strength is critical, it will help to figure the length of lap more exactly, to gain maximum strength with minimum use of brazing materials, The formulas given below will help you calculate the optimum lap length for flat and for tubular joints.
Figuring length of lap for flat joints.
| X = Length of lap |  |
| T = Tensile strength of weakest member | |
| W = Thickness of weakest member | |
| C = Joint integrity factor of .8 | |
| L = Shear strength of brazed filler metal | |
| Let’s see how this formula works, using an example. |
Problem: What length of lap do you need to join .050" annealed Monel sheet to a metal of equal or greater strength?
Solution:
| C = .8 T = 70,000 psi (annealed Monel sheet) |
| W = .050" |
| L = 25,000 psi (Typical shear strength for silver brazing filler metals) |
| X = (70,000 x .050) /(.8 x 25,000) = .18" lap length |
Problem in metric: What length of lap do you need to join 1.27 mm annealed Monel sheet to a metal of equal or greater strengths
Solution:
| C = .8 T = 482.63 MPa (annealed Monel sheet) |
| W = 1.27 mm |
| L = 172.37 MPa (Typical shear strength for silver brazing filler metals) |
| X = (482.63 x 1.27) /(.8 x 172.37) |
| X = 4.5 mm (length of lap) |
Figuring length of lap for tubular joints.
| W (D-W) T CLD |  |
| X = Length of lap area | |
| W = Wall thickness of weakest member | |
| D = Diameter of lap area | |
| T = Tensile strength of weakest member | |
| C = Joint integrity factor of .8 | |
| L = Shear strength of brazed filler metal |
Again, an example will serve to illustrate the use of this formula. Problem: What length of lap do you need to join 3/4" O.D. copper tubing (wall thickness .064") to 3/4" I.D. steel tubing?
Solution:
| W = .064" |
| D = .750" |
| C= .8 |
| T = 33,000 psi (annealed copper) |
| L = 25,000 psi (a typical value) |
| X = (.064 x (.75 – .064) x 33,000)/(.8 x .75 x 25,000) |
| X = .097" (length of lap) |
Problem in metric: What length of lap do you need to join 19.05 mm O.D. copper tubing (wall thickness 1.626 mm] to 19.05 mm I.D. steel tubing?
Solution:
| W = 1.626 mm |
| D = 19.05 mm |
| C = .8 |
| T = 227.53 MPa (annealed copper) |
| L = 172.37 MPa (a typical value) |
| X = (1.626 x l19.05 - 1.626) x 227.53)/(.8 x 19.05 x 172.37) |
| X = 2.45 mm (length of lap) |
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