E-Boogie
Mechanical
- Nov 26, 2018
- 13
Good morning all,
I have done a bunch of searching on the forum and have not found a thread that directly aids in solving my issue. I did find this thread that was really close, and I have based my understanding of the correct methodology on it and the ASME interpretation included.
I am designing a lifting lug per ASME BTH-1. I have analyzed my structure using STAADPro with AISC 360 and ASCE 7. Our standard lifting lugs project upwards through beam flanges but every once in a while, we have skid dimensions that are supplied by the customer or a building on top of our skid that does not allow for vertical lug projection. In these cases, we use a horizontal projection from the beam web and often have to offset the pin hole so that we have shackle clearance in lifting. When our lugs are horizontally projected, I request a 90° sling angle, BUT I also want to consider a 5% load in the lateral direction.
For my vertical lugs I implement 3-2.4 for stress combination due to major and minor axis bending as well as axial loading, this makes perfect sense to me. When I have a horizontal lug, I believe that I should be using 3-2.5 for combined normal and shear stresses. I am struggling to fully comprehend my inputs to this equation (3-37), though through my struggles yesterday I believe I am now understanding.
The stress plane to be evaluated is the vertical XY-plane through point B, which is directly below the beam flange face as the lug is supported by the welds to the beam. Correct? OR should it be evaluated at point A which is where the lug and beam web meet?
Evaluating on XY-plane through point B
Converting bending and torsional moments to force couples:
Bending Moments
Myy = Fx * Lz2
fx = (Myy*c)/Iyy ; c = t/2 ; Iyy = ((t^3)*h)/12
Mxx = Fy * Lz2
fy = (Mxx*c)/Ixx ; c = h/2 ; Ixx = (t*(h^3))/12
Torsional Moment
T = Fx * Ly
fvz = T / Q ; a=t/2 ; b=h/2; Q = (8*(a^2)*(b^2))/(3*a + 1.8*b)
fv (computed shear stress)
fvy = Fy / (b*t)
fvx = Fx / (b*t)
Sum all fv terms:
fv = fvz + fvy + fvx
Use 3-2.5 to combine normal and shear stresses
fcr = √(fx^2 - fx*fy+fy^2+3*fv^2) ≤ Fcr = Fy/Nd
Is my evaluation plane correct? Is there anything that I am missing?
I have done a bunch of searching on the forum and have not found a thread that directly aids in solving my issue. I did find this thread that was really close, and I have based my understanding of the correct methodology on it and the ASME interpretation included.
I am designing a lifting lug per ASME BTH-1. I have analyzed my structure using STAADPro with AISC 360 and ASCE 7. Our standard lifting lugs project upwards through beam flanges but every once in a while, we have skid dimensions that are supplied by the customer or a building on top of our skid that does not allow for vertical lug projection. In these cases, we use a horizontal projection from the beam web and often have to offset the pin hole so that we have shackle clearance in lifting. When our lugs are horizontally projected, I request a 90° sling angle, BUT I also want to consider a 5% load in the lateral direction.
For my vertical lugs I implement 3-2.4 for stress combination due to major and minor axis bending as well as axial loading, this makes perfect sense to me. When I have a horizontal lug, I believe that I should be using 3-2.5 for combined normal and shear stresses. I am struggling to fully comprehend my inputs to this equation (3-37), though through my struggles yesterday I believe I am now understanding.
The stress plane to be evaluated is the vertical XY-plane through point B, which is directly below the beam flange face as the lug is supported by the welds to the beam. Correct? OR should it be evaluated at point A which is where the lug and beam web meet?
Evaluating on XY-plane through point B
Converting bending and torsional moments to force couples:
Bending Moments
Myy = Fx * Lz2
fx = (Myy*c)/Iyy ; c = t/2 ; Iyy = ((t^3)*h)/12
Mxx = Fy * Lz2
fy = (Mxx*c)/Ixx ; c = h/2 ; Ixx = (t*(h^3))/12
Torsional Moment
T = Fx * Ly
fvz = T / Q ; a=t/2 ; b=h/2; Q = (8*(a^2)*(b^2))/(3*a + 1.8*b)
fv (computed shear stress)
fvy = Fy / (b*t)
fvx = Fx / (b*t)
Sum all fv terms:
fv = fvz + fvy + fvx
Use 3-2.5 to combine normal and shear stresses
fcr = √(fx^2 - fx*fy+fy^2+3*fv^2) ≤ Fcr = Fy/Nd
Is my evaluation plane correct? Is there anything that I am missing?
