Angle Support Design for Uniformly Distributed Dead Load

[jsu0512]

Hello Everyone,

I need some help with analyzing the strength of the angle leg when the uniformly distributed load is applied in perpendicular to the angle leg. (See the attachment)

As shown in the attached drawing, the angle size is 5"x5"x1/4" thickness with 49 inch length and applied with the uniformly distributed load at 105mm offset from the concrete curb. To check the strength of the angle leg, the thickness (1/4") and length (49") of the angle leg was used to calculate the section modulus, before calculating the moment resistance of angle leg . (Mr=0.9*S*Fy)

And for the Factored Moment, I calculated the point load by multiplying the uniformly distributed load(kN/m) by the angle length (1244mm), and then calculated the factored moment accordingly. (Mf=1.25*w*b*r)

Please let me know if I am on the right track.

 

[WARose]

For the leg strength (since that is a unfactored load and is continuous), I'd put the leg strength at 0.75Fy.

For the developed moment, your span would be between the screws (with a overhang from the last screw). As far as capacity goes, AISC has a section for angles. To be on the safe side you may want to take the full length as your unbraced length. (Doubt it would change things though.)

 

[rb1957]

there are many issues ... for a start ...
1) shear into the supporting fasteners,
2) beam bending of the angle, as a beam on many supports (or simplify to a SS beam between fasteners),
3) flange bending of the loaded flange,
4) bending of the supporting flange ... the out-of-plane reaction from the wall, reacting the offset load.

another day in paradise, or is paradise one day closer ?

 

[jayrod12]

Let's keep it in perspective, he asked about the angle leg in bending. He is on the right track. Provided he compares apples to apples, i.e. factored loads versus LRFD capacities.

The rest of the design obviously needs to be fleshed out as rb1957 indicates, but regarding his specific question, he's going down the right path.

 

[mike20793]

I'll add that there is a special case for equal leg angles that simplifies the calculations.

 

[wannabeSE]

Switching back and forth between metric and US customary units is a disaster waiting to happen. I have a hard time following the Original Post. I became lost in the factored moment paragraph. Where is the point load. What code is being used? I don't understand what is Mf = 1.25*w*b*r. Some of my confusion is probably because I am not accustomed to the symbols. For instance in AISC, Mr stands for required flexural strength (demand) rather than resisting moment (capacity). what is w,b, and r.

To check the bending in the angle leg, I would consider a 1 foot long piece (if you are using the metric, use a convient unit, meters?) rather than the entire length. I think you are on the right track, but I couldn't follow how the demand is calculated. Also, is S the elastic section modulus or the plastic section modulus (it is eleastic in the US, but I think is is the plastic section modulus elsewhere).

 

[dik]

I usually use the plastic section bd^2/4 and you have to be careful with the high alkali of concrete and aluminum.

[added] Also, usually have the moment arm closer to the support.

[added^2] should have added galvanic corrosion, too.

Dik

 

[Settingsun]

I'm with Jayrod12. The calcs were easy enough to understand but I'm a native speaker of metric and limit states. It looks like you've been given a problem whose geometry is imperial but you prefer to work in metric so have converted all your inputs. Take on board the feedback in this thread if you have to present your calcs to imperial/allowable people.