shear area of semicircular lug

huangyhg

shear area of semicircular lug
for pin-connected members (square end gusset plate or lug), the following limit states apply:
1. tension on the net effective area:
   pn = (2)x(t)x(beff)x(fu) (asic 13, eq. d5-1)
2. shear on the effective area:
   pn = (0.6)x(asf)x(fu) (asic 13, eq. d5-2)
3. bearing strength
   pn = (1.8)x(fy)x(apb)  (asic 13, eq. j7-1)
4. shear in pin
   rn = (fnv)x(ap)
however, for a round-end plate (or semicircular lug), what is the asf?  (please see the next .jpg i will post to see the shorter failure plane i am referring to).
is the shear rupture (i.e., different failure plane) the only thing that changes between a square and round pin-connection plate?  thanks.
jpeg of shorter failure plane.
more specifically, what is the calculation for the new asf?  and could anyone please provide a .pdf of a document showing/supporting the calc?  thanks again.
it seems to me that asf = 2*t*r where r is the minimum radial distance from any part of the pin to the outer curve.  and the failure planes shown on your sketch should also be radial.
ba
do you have the paper by tolbert and hackett "experimental investigation of lug stresses and failures"?   
yes miecz, i have the paper by tolbert and hackett "experimental investigation of lug stresses and failures," (1974) but it does not give the area equation for the semicircular lug.
bruhn ("analysis and design of flight vehicle structures") has two approaches.
conservatively use 2*(the distance from the edge of the hole to the edge of the lug) on the cl of the hole.
less conservatively, project a radius at 40deg to the cl of the lug.  project this point (on the edge of the hole) parallel to the cl of the lug to the edge of the lug.  use 2*this distance.
quote (rb1957):
conservatively use 2*(the distance from the edge of the hole to the edge of the lug) on the cl of the hole.
this is how aisc does it for the bolt bearing checks.
rn = 1.2 lc t fu, where lc is your distance from the edge of hole to edge of plate.  this is the same as your equation 2 above, with asf = 2*lc*t, "2" being because there are two shear planes.
in your first sketch, with the square end, it seems very tedious to actually calculate the length of the dashed line you show.  i would use the lc distance and be done with it.  the extra length you'll gain isn't worth the effort.
oh, i see what your getting at.  the last line of the paper merely states that the area term changes for a semicircular end, but it doesn't say exactly what it changes to.  figure 6 does show the location and orientation of the assumed failure plane, so i think it reduces to a geometry problem.  i would simply subtract out the small area at the bottom of the dashed lines.
different strokes for different folks i guess ...
for me bearing is based on the pin diameter*thickness; edge distance enters into the bearing atrength (dependent on e/d).  ... the the op says in para (c)
shear tear-out is the lug shearing apart.  not sure why you have (a+d/2) ... i'd use "a" which is clearly defined (minimum distance from hole to edge of lug).
yes, it is shear tear-out.  aisc lumps these together as "bolt bearing," where the bearing capacity is the lesser of the tear-out or local crushing (the diameter*thickness).
i checked the equations for tension on pin-connected