huangyhg

block shear vs. shear rupture at connections?
i have only been working as an engineer for (1) year so please forgive me if this question comes across as "stupid".
here is the question:
the new 13th edition manual (section j4.2 and j4.3 pp16.1-112) has equation j4-4 which gives the shear rupture capacity of a member or connection element at the connection. anv is listed as the net area subject to shear. if you have a double angle shear connection with (4)bolts going through the beam flange, the shear rupture anv would not include the area above the top bolt for the angles). is that an accurate statement? if that is an inaccurate statement please explain how? if it is an accurate statement, then why would block shear have to be checked since anv + ant will always be greater than just anv for shear rupture.
for the connection i mention above, the shear rupture would cause a shear rupture below the bolts leaving the vertical edge distance above the top bolt intact. the block shear failure would fail the same shear area plus a tension failure for the horizontal edge distance to the right (or left depending how you are viewing the connection) of the top bolt.
in this instance, wouldn't the shear rupture capacity always be less than the block shear capacity?
i know the block shear now has a shear yielding component in it, but this seems very minor compared to the relatively large tensile area that is added in the block shear calc.

hmmm... i haven't done it in a while, and i don't have the book in front of me, but i think shear rupture is more associated with a partial moment connection where the rotation of the connecting elements is limited. like a shear plate connected to a braced column. it's more of a "tearing" of the steel at the bolt location, and is horizontal. your moment is the end reaction multiplied by the distance from the face of the column, to the center of the bolts. the moment is transferred to the bolts, top bolt pulling away from the column, and the bottom bolt pushing into the column, and resolved into components, such that the total resultant shear in the bolts doesn't exceed the allowable shear, and the connecting material (beam web, shear plate) doesn't tear.
definitely double check me on this. i haven't done connection design in 6 years.
chipb-
that isn't the way it is presented in the 13th edition steel manual. if you have one at the office, check out pp16.1-112 (j4.2 and j4.3).
"if you have a double angle shear connection with (4)bolts going through the beam flange, the shear rupture anv would not include the area above the top bolt for the angles). is that an accurate statement? "
no it is not an accurate statement. the rupture area would include the full height of the angle minus the holes. this represents one possible plain of failure, the others being bearing and tearout of the bolts and block shear. bearing and tearout can control with a thin web where the bolts just rip out through the web. the shear rupture state can control over block shear when there is a large width of material adjacent to the bolts but the top edge distance is small (and therefore the weaker plane is to just rip straight through the angle rather than overcoming the tensile component of the block shear equation).
i'm having a monday. i need a picture so i can understand. any suggestions?
i am not necessarily disagreeing, i am just failing to "see" it. the shear is coming into the angles through the bolts bearing on the angles. i don't see how the shear is getting above the top bolt. the anv for the block shear calc would not include this area above the top bolt (for the angle). they should be more clear that anv for shear rupture and anv for block shear are not equal.
i understand this represents only one failure possibility but i am only interested in this one at the moment.
just to talk about it - if you have a double angle with (4) bolts through the beam web and the geometry of the bolts are as follows: (1) vertical row of (4) bolts with a vertical spacing of 3", the vertical edge distance from the top bolt to the edge of angle is infinite and the horizontal edge distance is also infinite (the vertical edge distance form the bottom bolt to edge of angle is 3"). now, you wouldn't consider the shear rupture capacity as infinite just because technically speaking the anv is infinite, would you? i would think you would take (12-3.5(db+1/16))*(thickness of angle)*fu*0.6 all the while neglecting the area above the top bolt. the 12-3.5(db+1/16) represents the length of the angle subject to shear rupture, not the entire angle length.
where am i going astray?
i will be glad to post a pic, can someone tell me how to do it!! lol
i think what willisv is stating is correct - that there are multiple possible failure mechanisms and you need to check the various possibilities.
look at the commentary (page 16.1-351). there is the possible pure shear failure (all v) and then there are potential block shear failures where shear and tension failures must both occur. this must be checked as well.
also, i found this q and a on aisc:
block shear
question
02/01/2007
i noticed that the tension yield component of the block shear check is no longer included in the 2005 specification, eq. j4-5. the new procedure is obviously simpler without the need for dealing with the tension yield component. could you explain the change?


answer(s)
yes, this change was intentional. when the asd model (shear rupture-tension rupture) and lrfd model (shear yield-tension rupture, shear rupture-tension yield) were combined in the 2005 aisc specification, we re-evaluated the data. we noticed that the shear component was always dominant and that true block shear rupture did not depend upon the tension yielding. rather, the shear mechanism would have to form in yield or rupture, and then the tension plane would fail in rupture. said another way, tension yielding occurred before the shear mechanism was reached, making it unnecessary to consider.
the new approach is easier. we tried to use the even simpler asd model, but we couldn’t assume that rupture-rupture would always be the controlling mechanism.
charlie carter, s.e., p.e.

for your (odd) example, yes you would consider the shear rupture (and the block shear rupture) capacities as infinite. if the shear strength of the bolts were adequate, bolt bearing and tearout would control this example.
willisv-
i appreciate your patience with me. the failure mechanism i am envisioning is the (4) bolts tearing down through the angle. i was thinking of this as a shear rupture. is this what you are calling tear-out? can you point me in the right direction to find this in the spec?
yes that mode is different then pure rupture and is called a bearing/tear-out failure. see section j3.10. the terms to the left side in equations j3-6a thru c represent a tearout failure (where the length of material between the bolts, lc, actually rips out as you are envisioning) and the terms to the right side represent a bearing type failure in which the material "squishes" and the hole elongates/deforms somewhat.

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