Single Plate Connection Problem
Single Plate Connection Problem
(OP)
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
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






RE: Single Plate Connection Problem
RE: Single Plate Connection Problem
RE: Single Plate Connection Problem
RE: Single Plate Connection Problem
RE: Single Plate Connection Problem
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
RE: Single Plate Connection Problem
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.
RE: Single Plate Connection Problem
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
RE: Single Plate Connection Problem
RE: Single Plate Connection Problem
If so what is the length of your horizontal welds?
RE: Single Plate Connection Problem
RE: Single Plate Connection Problem
RE: Single Plate Connection Problem
0.08*E/Fy = 64.4
LTB will govern...
RE: Single Plate Connection Problem
Iwx=d^3
vw=M*y/Iwx
RE: Single Plate Connection Problem
RE: Single Plate Connection Problem
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.
RE: Single Plate Connection Problem
Best regards.