huangyhg

clarification of bolt strength
can someone explain to me the rationale behind this?
re: aisc's lrfd, 3rd ed. page 16.4-29, sect 5.1, design shear and tensile strengths
to compute the nominal strength of a bolt, the following is given:
rn = fa*ab, where ab is the full diameter of the shaft without considering the threads.
if one considers a 1" a325 bolt for example, the shaft area is 0.785 in2 whereas the tensile area due to the threads is only 0.606 in2, a 23% reduction in area.
table 7-14, page 7-35, shows the design tensile strength (phi*rn) of a 1" bolt to be be 53 kips (with a phi of 0.75 included). using the "nomimal area", this would relate to (53/0.785)/0.75 = 90 ksi, which is the a325 yield point.
however, under a full tensile load, it is obvious the that threads would be intersecting the plane of the load and that the actual applied stress would have be 53/0.606 = 87.4 ksi without even considering the phi factor.
so how does aisc explain the rationale or compensate for this difference? i would appreciate it if some of you with many years of structural steel experience would respond to this.

look in the bolt specification, page 16.4-31 in your manual. they say for usual bolt sizes, the ratio between the tensile area through the threads and the nominal cross-sectional area is about 75%. so, they incorporate this 0.75 factor in the nominal stress equations so that the designer can still use the nominal area for doing the strength calculation.
nutte:
thanks for pointing out that commentary, and i understand the 0.75 relationship between the full and threaded areas.
but the question still is, if they are using the 0.75 factor to allow nominal areas to be used (instead of having to look up the threaded areas), then would that not imply that another 0.75 multiplier should be put on top of that to apply the phi factor?
again, table 7-14 says design tensile strength, indicating that the phi factor has been included and that the tabulated numbers represent phi*rn. otherwise, would they not have said that the table represents the nominal tensile strength?
so, the question remains; should yet another 0.75 reduction be applied to the tabulated values, or did the phi factor simply disappear, or are we now relating reduction factors to the ultimate tensile strength instead of the yield point?

ultimate tensile stress of a325 = 120 ksi
times factor for ratio of threaded area to nominal area
nominal tensile stress = 0.75*120 = 90 ksi
times phi=0.75
design tensile stress = 0.75*90 = 67.5 ksi
table 7-14 on page 7-35 shows phi-ft for a325 to be 67.5 ksi. you don't have to take an additional 0.75 factor, and the phi hasn't been forgotten.
ok, i see where the confusion came in.

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