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PCI 7th edition handbook issue

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JStructsteel

Structural
Aug 22, 2002
1,448
For those that have the 7th edition, on page 6-37 they are going thru a angle design. They have the formula for Vu close to the bottom. Tell me what value you get for that, and then on page 6-38 tall me how they got their value for Vu. Im not getting it to match. Thanks
 
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pg. 6-37 and 6-38 are showing the computations for two different limits on the shear.
Limit 1 is Vu based on leg flexure:
Mu = phi*Fy*z
z = b t2 /4
Mu = Vu*e

so Vu = phi*fy*b*t2/ 4*e

Limit 2 is Vu based on the anchor failure modes:
they work this out to an equation based on the tension and shear interaction equation and solve via iteration. the Vu they note under given on 6-38 is just a starting guess.

The Vu capacity is then the min[Vu,limit 1 , Vu,limit 2]

I'm making a thing: (It's no Kootware and it will probably break but it's alive!)
 
If thats the case, then the value for Vu at the end of the calc (5805lb) gives a ecc equal to 2.33". how is that correct?

Disregarding everything else, the bending capacity of the leg (vu) is Vu = phi*fy*b*t2/ 4*e correct? Thus 4.556K (Im assuming critical section is above the bolt on the vertical leg).

What am I missing?
 
JStructsteel said:
If thats the case, then the value for Vu at the end of the calc (5805lb) gives a ecc equal to 2.33". how is that correct?
how are you back calculating e?
plugging their values back into the formula for a I get 3.0046" so correct to two decimal places.

JStructsteel said:
Disregarding everything else, the bending capacity of the leg (vu) is Vu = phi*fy*b*t2/ 4*e correct? Thus 4.556K (Im assuming critical section is above the bolt on the vertical leg).
thats what I get. So limiting capacity for Vu in this case is governed by leg flexure.


I'm making a thing: (It's no Kootware and it will probably break but it's alive!)
 
If ecc is 3.0, then Vu is 4.556K correct, not 5.805K? Based on the sketch, this is the only limit state for angle bending, ecc-3", and thus Vu is 4.556K
 
yes

you are checking two limit states steel flexure and concrete anchorage
Vu,steel flexure = 4.6 k
Vu,concrete anchorage = 5.8 k

Vu, design = min[Vu,steel flexure, Vu,concrete anchorage] = 4.6 k

I'm making a thing: (It's no Kootware and it will probably break but it's alive!)
 
Thanks Celt83. Its unfortunate they dont report it that way in the calc so its clear. I just wonder where they got the 5854 value they used?
 
If you graph the function they derived you could eyeball that value being near the root. More likely they solved it previously and just used a close enough value to illustrate the iterative procedure for a solution.

I'm making a thing: (It's no Kootware and it will probably break but it's alive!)
 
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