T section thickness prying calculation can be extended? 1

[Alexey881]

Hello,
2 questions:

a) I'm trying to find formulas that would permit to design angle thickness when the angle is loaded in tension and that may create prying.
In the Canadian steel manual CSA-S16 (which is very similar to AISI) (in LSD or LRFD methods) There are formulas for a T-section loaded in tension and undergoing prying. The input force in the equation is 2*P where P is the tensile load divided by a number of bolts.

The way I see it, I should be able to use this formula for a single angle (or half of T section) without any modification to the formula.
Could you please confirm my thoughts?

and if no, where could I find a formula that would apply?

b) In the explanation in the book they say that when the T-section is rigid enough then the ratio of P to Tension in the bolt is 1, hence no prying at all.
If I am to use a single angle and try to use simple moment and force summations about the furthest edge of the angle, on the anchor bolt side, I will find out that for equidistant angle sides and angle holes my P=1/2*Tension in the bolt (or Tension in the bolt gets inflated by a factor of 2 regardless of considering rigidity or not). This puzzles me.

Thanks!

 

[gwynn]

No, you cannot use this formula. With a T section the flange is a continuous member with moments in the flange at both the bolt and stem locations. I believe it treats the failure point as having plastic hinges form in the flange at both the stem and bolt lines. With an angle iron you have peak moment at the bolt line and none at the upstanding leg.

The Canadian steel manual points this out a couple of pages before the bolt prying formula you are referring to and lays out how to treat an angle iron in that section.

 

[gwynn]

b) In the explanation in the book they say that when the T-section is rigid enough then the ratio of P to Tension in the bolt is 1, hence no prying at all.
If I am to use a single angle and try to use simple moment and force summations about the furthest edge of the angle, on the anchor bolt side, I will find out that for equidistant angle sides and angle holes my P=1/2*Tension in the bolt (or Tension in the bolt gets inflated by a factor of 2 regardless of considering rigidity or not). This puzzles me.

With bolts on opposite sides of a stem in a T-section, the section is in static equilibrium when tension is applied via the stem. If the plate is infinitely rigid, there is no prying.

With an angle, there is no static equilibrium unless there is a prying force.

Draw out the free body diagrams if you cannot understand it.

 

[Alexey881]

Hey, yeah sorry, question b) was dumb. However, going back to a)

I pulled up a paper on which this formula was based (

http://www.cisc-icca.ca/files/publications/techpubs/design/bolting/highstrengthbolting.pdf

(See Ch. 6)), and you're right this is derived based on both moments. But there is this parameter Alpha that is expressing Moment 2 (that is at bolt holes) in terms of M1 (that is in the stem location). it is said that: (If α = 1.0 , then there is a plastic hinge at each of the
M1 and M2 locations, and the prying force is at maximum (It would also mean M1=M2). If α = 0 , then of course there is no prying. Hence maximum required plate thickness is at α = 0.

So, what I'm trying to see is if this equitation could be used for a single angle only when M2=M1 which happens only at α = 1.0 With this in mind I should be able to design the thinnest angle possible while having a lot of prying. [ex. for the input of 500 lb on a 2" x 2" x 2" angle with bolt holes at 1" from the edge, i'd get 750 lb tension on the bolt, and required flange thickness would be 0.15" assuming [Fy=36000psi, Phi=0.9, b=1 (used b instead of b'), bolt hole=3/8" so An=(2-.375)t and Ag=2*t]

So here, I'm getting bolt force = 1.5 Initial force whereas when done with simple moment summation it would be =2 times Initial force. The 1.5 is due to the difference in summation of forces on FBDs of the two cases and is not a real amplification. So, lets say I say my bolt feels 2 times the input force but my thinnest possible angle would be derived from the equation of the T, however I must make sure that the input force in equation for T will be such that the Moment at bolt holes in T would be equal to the moment in bolt holes of a normal angle. (and don't forget α = 1.0)

You think that might work? though I'm not too sure what I'm gaining by all this really, except the tmin. for the angle...

 

[Alexey881]

You see, In my case I have a machine (box), on each length it has 2 angles bolted to it with one bolt each, and these angles are anchored to concrete. I need to design the angle size and the connections (bolt, and anchor). My machine is overturning over the boltholes on one of the lengths. So the angle is being worked in tension on the other length.

Now, FBD of the angle in the manual is based on simply supported beam model, so the M at edges is 0. However, in my case wouldn't the fact that the angle is connected to the Unit which is rigid could change simply supported condition to something like a Wall condition on one side of the beam? And in turn change my moment diagram to the one similar with double T?