Angle to HSS Moment Connection 4

[SixkHz]

I'm reviewing the connection for a cantilevered L3x3x0.5" angle welded to the side of an HSS6x4x0.25". There is a load at the tip of the angle resulting in a moment at the interface of the angle and wall of the HSS (see attached sketch).

The angle strength is fine, and the weld is fine, but I'm having trouble trying to check the plastification of the HSS wall.

I tried using the AISC Design Guide 24 assuming just the vertical leg of the angle is effective (ie. acts like a transverse plate), but the plate formulas only cover axial load, not moment. I also tried using the HSS to HSS moment connection formulas (again considering only the vertical leg of the angle), but the Hb/Bb ratio is < 0.5.

Any help would be appreciated.

 

[ishvaaag]

Look as a problem for nonlinearity in FEM. If the material is properly modeled it should identify the behavior under the standing load.

 

[slickdeals]

Did you see Page 88 in DG 24, limit states being Plate yielding and HSS punching shear.

 

[BAretired]

Why are you using 1/2" angle thickness? Do you need it? If you do, then by inspection, 1/4" HSS wall thickness is not going to be adequate to develop the strength of the angle.





BA

 

[BAretired]

If you are reviewing an existing situation, the failure load can be approximated using Yield Line Analysis.

BA

 

[TLHS]

If there's any significant load there I'd probably design a thicker plate to span you back to the horizontal HSS walls. The plate and associated work is likely cheaper that you taking time trying to do an FE model.

Watch where you're loading the angle. You could have other local failure mechanisms other than full section bending. Also, make sure you're not torquing the angle.

 

[dhengr]

I would be most concerned about the tensile stress due to the bending moment, plus the shear stress in the weld (combines stress), at the upper tip of the angle. Then add to that the flexing of the HSS wall and its prying action on that weld. That’s a bad detail from the design standpoint, whichever design guide you use to justify it. And, it’s a difficult detail to weld without leaving a stress raiser right at this same high stress location. At least rotate the angle 180°, so the horiz. leg has the tension weld.

 

[SixkHz]

Thanks for the feedback, everyone.

I should have elaborated a bit more in my original post. This connection is part of a steel frame used to lift equipment (the frame connects to the equipment and a crane lifts the frame). Hundreds of these lifting frames have been manufactured already and have been used across North America for quite some time (without any noted structural issues).

I'm reviewing the lifting frame for the manufacturer to assess the rated load capacity (approximately 8,000 lbs, and there is an angle at each corner of the frame, so 2,000 lbs per angle), in accordance with ASME BTH-1-2008 "Below the Hook Lifting Devices".

I realize this is a bad detail, and I'm going to suggest modifications to make it a more robust connection, but I need to provide justification as to why it doesn't work (in the form of a design calculation package to be submitted to the client).

I've done an FEA and I'm getting very high localized stresses at the interface of the HSS wall and the tips and corner of the angle, but I want to do a hand calculation to compare to the results I'm getting from the FEA.

 

[SixkHz]

BAretired: my experience with yield line theory is very limited. Assuming a plate on an HSS with an applied moment, is my approach in the attached calculation valid?

 

[BAretired]

You have the right idea. You must assume a yield line pattern, then equate internal vs external energy.

Assuming only the vertical leg of the angle is effective (a bit conservative but a good first approximation), I don't think the yield lines will be quite as symmetrical as you have shown.

Since y₂ > y₁, Triangles A and D may be similar but not equal in size.

Trapezoids E+B and F+C cannot exist without additional yield lines such that all planes can slope as required to meet the assumed deflections.

I would have to think about the problem a little more, but can't do it right now. Let's see what others come up with.

BA

 

[ishvaaag]

What is certain is that the drawn scheme fails in identifying necessary yield lines, and the rest be planes. The trapeze edges are not in a plane, but distort. You are near it, but you need with your scheme one pyramid up and one pyramid down, i.e. 4 diagonals more in the trapezes.

 

[ishvaaag]

I see BA also saw first the issue.

 

[SixkHz]

That makes sense, thanks.

 

[ishvaaag]

You will need the angle between faces of a pyramid when expressing the angles rotated there as a function of the trial geometry. This link gives a straightforward way to get to the angle expression.

http://mathcentral.uregina.ca/QQ/database/QQ.09.07/h/carla5.html

Note that as BAretired hints the pyramids need not to be of equal sides (yet the method above still will apply), and you will be minimizing the inner work (as a function of the dimensions a,b,h of the pyramid) whilst meeting equilibrium with the applied moment at a concomitant rotation (also a function of a,b,h).

The solution will give the deformation in the hypothesis of plasticity. Examining if plastic behavior at the thickness is reasonable can then be done, as could be through an elastic FEM analysis.

You can also investigate this way the extent of deformation in the hypothesis of plasticity for a range of thicknesses.

 

[BAretired]

To make the problem symmetrical, I assumed a 2"x 1/2" plate centered in the 6" wall of the HSS.

Two equal and opposite forces, H are applied to the HSS wall 2" from the top and 2" from the bottom.

External Work = H(1+1) = 2H
Internal Work = 29m where m is the plastic moment of the wall per inch.

m = phi*Z*Fy = 0.9(0.25)2/4*50,000 = 703"#/"
H = 14.5m = 10,195#
M_ult = 2H = 20,390"#

The eccentricity of load P was not specified, but P*e = M_ult/SF


BA

 

[BAretired]

Correction:
m = phi*Z*Fy = 0.9(0.25)²/4*50,000 = 703"#/"

In arriving at Internal Work, I assumed yield lines at 45^o from the corners of the plate. The yield lines formed a trapezoid top and bottom and three triangles each side of the plate.

BA

 

[jeff91]

Frankly, I think everyone is looking at this detail in a more complicated way than it needs to be. I would like to offer three thoughts with regard to the detail. First, the angle is located only 1" from the end of the HSS where there is located a cap plate. The cap plate will provide substantial stiffening to the end of the member and will reduce the bending stress in the sidewall of the HSS. Second, if you want a simplified/conservative approach to determine the bending stress in the sidewall consider the wall as a simply supported beam spanning between the top and bottom of the HSSS subjected to a concentrated moment at midspan. The eccentricity of the applied load is not shown but if we assume the sketch is to scale it appears to be about 2 inches. This results in an applied moment of 2 kips x 2" = 4 kin (using the 2000 lbs per angle listed above). This results in a maximum moment in the sidewall of M/2 = 2 k in. If you assume an effective width of the sidewall equal to 5 inches then the section modulus is 5*.25^2/6 = .0520833 in^3. Subsequently this results in a bending stress of 2/.0520833 = 38.4 ksi. This is a relatively high stress but it is still less than .90*Fy = .90*46 = 41.4 ksi. Lastly, a simple check on tear-out could be determined by approximating the force in the vertical leg and dividing by the perimeter length of the portion of the angle in tension. Then compare this stress to a rupture allowable of .30Fu. The 4 k in moment results in an approximate flange force of 4/2 = 2 kips. The perimeter length of the tension side of the angle = 2.068 + .5 + 2.068 = 4.636 in. The force per unit length along the tension side of the angle = 2/4.636 = .43 kips/in. This results in a rupture stress of .43/.25 = 1.72 ksi. Which is relatively low.

 

[ishvaaag]

BAretired, please, from where do you find

Internal Work = 29m where m is the plastic moment of the wall per inch?

Is it that you have some formula for the case? Other than that we need to formulate all the rotations (both at the load applying plane and at each yield line dihedral) as a function of the deformed trial geometry, then whilst meeting equilibriums seeking between the cases we try what is the one of the minimum internal work ...

 

[BAretired]

ishvaaag,

The attached sketch shows the location of assumed yield lines. Internal Work for each segment is defined as L*m*theta where theta is the angular rotation of the segment.

Dimension 'x' was first assumed to be 2" which led to an Internal Work of 29m.

Differentiating IW with respect to x and setting the result equal to zero indicates that a lesser value of x will be more critical. So Internal Work on this revised basis turned out to be 28.7m, not much different than before.

BA

 

[ishvaaag]

I will study it ... thanks, BAretired.