SM3225
Structural
- Mar 26, 2017
- 15
It is common knowledge that maximum practical fillet weld size is ½” (13mm) – common practice to employ another type of weld, such as groove weld, when the calculated weld is greater than ½”(13 mm).
Typically CJP welds are the most expensive weld and I feel the same should be reserved for situations in which they are the only viable option.
Coming to the original problem – this is the weld between shear lugs and base plate. In the current case, type of shear plate employed is Cross Plate Shear Lug - owing to the existence of pretty high magnitude of forces, plate thickness is worked out as 1-1/2”(40 mm) – first pass shear plate dimensions 6”(150 mm)longx8”(200 mm) deepx1-1/2” (40mm) thick.
Consideration of Fillet weld (FW) alone requires a fillet leg size of approximately 1-1/8” (28 mm) – what would be the probable combination of FW+PJP instead of a CJP one.
I am planning to employ a FW(1/2”or13 mm) + PJP(1/2” or 13 mm).
How to arrive at a calculation which possibly demonstrates design strength (fRn) of FW+PJP is more than the required strength (Ru)? I am planning to carry on as mentioned below
***********
Equate the kips/in (N/mm) of a 1-1/8” (28 mm) FW = kips/in (N/mm) of FW(1/2”or 13 mm) + PJP(1/2” or 13 mm) – more precisely setting the FW leg size to ½”(13mm) – we’d only be left with the finalization of PJP size.
Does anyone have similar experience earlier – please let me have your opinion.
Thanks & Regards.
Typically CJP welds are the most expensive weld and I feel the same should be reserved for situations in which they are the only viable option.
Coming to the original problem – this is the weld between shear lugs and base plate. In the current case, type of shear plate employed is Cross Plate Shear Lug - owing to the existence of pretty high magnitude of forces, plate thickness is worked out as 1-1/2”(40 mm) – first pass shear plate dimensions 6”(150 mm)longx8”(200 mm) deepx1-1/2” (40mm) thick.
Consideration of Fillet weld (FW) alone requires a fillet leg size of approximately 1-1/8” (28 mm) – what would be the probable combination of FW+PJP instead of a CJP one.
I am planning to employ a FW(1/2”or13 mm) + PJP(1/2” or 13 mm).
How to arrive at a calculation which possibly demonstrates design strength (fRn) of FW+PJP is more than the required strength (Ru)? I am planning to carry on as mentioned below
***********
Equate the kips/in (N/mm) of a 1-1/8” (28 mm) FW = kips/in (N/mm) of FW(1/2”or 13 mm) + PJP(1/2” or 13 mm) – more precisely setting the FW leg size to ½”(13mm) – we’d only be left with the finalization of PJP size.
Does anyone have similar experience earlier – please let me have your opinion.
Thanks & Regards.