Single Plate Connection Problem 1

[DanteQ]

Hi, I am using RAM Connection Standalone. I am designing a Single Plate Connection (Beam to Column Flange) .

The beam is a W16x36 and the column a W10 x 33, both ASTM 992 G50. The Ultimate Shear is 39.01K = (1.2*19.44)+(1.6*9.80)

The plate is A36, 9" length, 1/4" thick, vert & horiz edge dist = 1.5" and the vertical center-to-center distance= 3". The weld size is 3/16"

I am using 3 bolts (1 column, 3 rows).

When I look at the Design Report, all design checks are OK except in the "PLATE (support side)" section,
here, I get :
Weld Capacity 64.71k Demand 39.01K OK
Reduction Factor 1
Calculation Eccentricity (in) = 2.5"
Bending at Weld Line Capacity = 7.54 k.ft Demand = 8.13 k.ft NO GOOD

I have successfully checked all the results manually except the Bending at Weld Line Capacity.

How do I calculate by hand the "Bending at weld line capacity" (7.54 k.ft) for a quick check ?

What is the procedure ?

I have tried several ways but I come up with different results, not the 7.54 k.ft

Thank you,

DanteQ

 

[kslee1000]

39.01k*2.5"/12=8.127 k-ft

 

[DanteQ]

Hi kslee1000, thanks. I knew how to get that one. My question is how do I obtain the 7.54k.ft? that is, the bending moment capacity of the weld line ?

 

[kslee1000]

try linear (trangle) stress distribution over the length of the weld line with fa (allowable shear stress of weld size)on top and bottom, zero at center, then solve for moment. see if it works.

 

[kslee1000]

Forgot to mention, remember fa=sqr(fx^2+fy^2+fx^2), you need to take out the vertical applied shear (fy)to get net fa for moment (actually produces stress normal to the weld line, fa mentioned above should be fz).

 

[COEngineeer]

It is just bending of the plate which is .9 x 36ksi = M c/ I

c and I are known. Make sure you use the h as the length of weld, b=.25 in. It sounds like you need to make the shear tab taller.

Never, but never question engineer's judgement

 

[nutte]

COEngineer, that's what I would think, but this doesn't agree with the output. t=0.25 in, d=9 in, Sx=3.375 in^3, 0.9*36 ksi*Sx = 109.35 k-in = 9.11 k-ft, not the 7.54 k-ft provided. I can't figure out how they're getting 7.54.

It's not the vector weld analysis kslee is suggesting. The 64.71k weld capacity corresponds to an instantaneous center of rotation capacity with e=2.5 in.

 

[kslee1000]

Not familar with LRFD, but see if the following makes sense:

fa = 64.71/(9*.9) = 7.99
fv = 39.01/9 = 4.33
fb = sqr(7.99^2-4.33^2) = 6.72
M = 6.72*4.5*.5*6/12 = 7.56

 

[kslee1000]

Solution above was a wild guess, the number could be jus a coincidence. I suggest follow normal design method to check the plate and weld stesses to verify the accuracy of the program output.

 

[USCeng09]

This is a beam to column flange connection? Are you welding around the plate on all three sides?

If so what is the length of your horizontal welds?

 

[USCeng09]

Took a second look at the first post...those questions are irrelevant...

 

[enginerding]

Lateral-torsional Buckling - See AISC 2005 Spec., Section F11.2

 

[enginerding]

By the way... Lb d/t^2 = 2.5" * 9" / .25^2 = 360

0.08*E/Fy = 64.4

LTB will govern...

 

[asixth]

Weld stress

Iwx=d^3
vw=M*y/Iwx

 

[USCeng09]

I'm interested in learning what the reason for that output was, but in the mean time have you considered referring to the Table 10-9a in the steel manual to get your plate and weld sizes for the single-plate connection?

 

[nutte]

enginerding: I hadn't thought of LTB. This results in Mn=1.40 My, or phiMn=12.74 k-ft. Using 0.9*Fy*Sx, you get 9.11 k-ft. Both of these are still high.

But I figured it out. They're using the von-Mises shear reduction on page 10-103.

fv = 39.01/(9*.25) = 17.34 ksi
Fy = 36 ksi
Fcr = (Fy^2-3*fv^2)^0.5 = 19.85 ksi ***This equation is actually wrong***
phi = 0.9
Zx = 5.0625 in^3

phi Mn = phi Fcr Zx = 90.44 k-in = 7.54 k-ft.



*** The equation for Fcr is actually in error in the 13th edition manual. For LRFD, the equation should be:

phi Fcr = ((phi Fy)^2-3*fv^2)^0.5 = 12.16 ksi
phi Mn = phi Fcr Zx = 61.58 k-in = 5.13 k-ft, much less.

The example problems on the CD that came with the manual confirm the equation for phi Fcr above is the correct equation. See page IIA-58. The version in the seismic design manual, that includes the effects of axial load, is also incorrect. I've derived the equation, and agree with the example problem version. That is a whole other discussion, though.

 

[DanteQ]

Nutte, the solution is the one you posted. I had found it out too but had connection problems and coudn't post it. Thanks to those who contributed to this post.

Best regards.