Path A,B-J-P-S,T Net Area
Since thickness is not constant along the path, it is easier to look at this problem in different manner than you would if you had a constant thickness along the path.
The logical approach is to look at things in terms of AREA. In this approach you add and subtract area from the member gross area.
Net Area = Gross area - Cross Sectional Area of Holes + Additional Area Due to Stagger
For this problem, there are holes on the web and the flange. The area reduction for a hole on the web equals the effective diameter of the hole times the thickness of the element with the hole.
EFFECTIVE HOLE DIAMETER = bolt dia. + 1/16" for std hole + 1/16" damage allowance
HOLE AREA = (effective hole diameter) * t
There are four flange holes and two web holes:
WEB HOLE AREA = = (.875" + .125") * (0.415") = 0.415 in2
FLANGE HOLE AREA = (.875" + .125") * (0.720") = 0.720 in2
There are two identical staggers on the web:
STAGGER AREA = 0.415" * (3")2/(4*3.25") = 0.287 in2
This gives a net area of:
Net Area = Ag - 4 flange hole areas - 2 web hole area + 2 stagger areas
Net Area = 20.0 in2 - 4*(0.720 in2) - 2*(.415 in2) + 2*(0.287 in2) = 16.86 in2