A500 steel tubing beam deflection

[coyote50]

TL;DR

I'm looking to construct a simple rectangular 13'x17' hollow steel frame over my patio, but I'm getting conflicting beam deflection calculations.
I need to determine my beam size for my long spans of 17'. My plan was to use 3.5"OD A500 structural steel. 11 gauge (.120 in) wall thickness, 58 ksi, 5.64 lbs/ft. Beams will be sitting on and bolted (might weld) to large steel posts. So technically the ends of the beam will be free.

Hello,
I'm looking to construct a simple rectangular 13'x17' hollow steel frame over my patio, but I'm getting conflicting beam deflection calculations.

I need to determine my beam size for my long spans of 17'. My plan was to use 3.5"OD A500 structural steel. 11 gauge (.120 in) wall thickness, 58 ksi, 5.64 lbs/ft. Beams will be sitting on and bolted (might weld) to large steel posts. So technically the ends of the beam will be free.

Online calculations show deflections from .75" all the way past 6". Clearly something isn't right.

I'd like these two beams to be strong enough to handle a horizontal sunshade, and maybe a few hanging plants. I calculated a 150 lb downward force in the middle per beam just to be safe.

I guess I'm left scratching my head because they sell 16' span shade pergolas with these tiny cheap aluminum beams with no center support. Here at work I have 3"square, thin walled steel racking beams spanning 12' that can hold thousands of pounds with no real deflection. I would've thought my 3.5" structural steel beam would be ok.

Thoughts?

Thanks, Matt.

 

[jrmichler]

Note that a 150 lb load in the center of a beam is equivalent to a 300 lb load distributed evenly over the entire length of the beam. Have you allowed for snow loads on the sunshade?

The trouble with online calculators is difficulty knowing exactly what assumptions they incorporate, which is why I prefer to use the beam deflection equations directly. You have two separate loadings to calculate - the weight of the beam, and the weight of the stuff supported by the beam. Search beam deflection equations and area moment to find the equations.

Note that deflection from a point load is proportional to the cube of the length, so a 17 foot beam will have almost 3 times the deflection of a 12 foot beam under a point load It will have 4 times the deflection under its own weight (weight is proportional to length).

Beam weight = $5.64 * 17 = 96 lbs$.
Beam length = $12 * 17 = 204 inches$.
E = $30E6 lb/inch^2$
I = $1.82 in^4$

Beam deflection due to its own weight = $96 * 204^3 * 5 / 384EI = 0.194 in$
Beam deflection due to 150 lb load in center = $150 * 204^3 / 48EI = 0.486 in$
Total beam deflection = $0.194 + 0.486 = 0.68 inches$

 

[coyote50]

Note that a 150 lb load in the center of a beam is equivalent to a 300 lb load distributed evenly over the entire length of the beam. Have you allowed for snow loads on the sunshade?

The trouble with online calculators is difficulty knowing exactly what assumptions they incorporate, which is why I prefer to use the beam deflection equations directly. You have two separate loadings to calculate - the weight of the beam, and the weight of the stuff supported by the beam. Search beam deflection equations and area moment to find the equations.

Note that deflection from a point load is proportional to the cube of the length, so a 17 foot beam will have almost 3 times the deflection of a 12 foot beam under a point load It will have 4 times the deflection under its own weight (weight is proportional to length).

Beam weight = $5.64 * 17 = 96 lbs$.
Beam length = $12 * 17 = 204 inches$.
E = $30E6 lb/inch^2$
I = $1.82 in^4$

Beam deflection due to its own weight = $96 * 204^3 * 5 / 384EI = 0.194 in$
Beam deflection due to 150 lb load in center = $150 * 204^3 / 48EI = 0.486 in$
Total beam deflection = $0.194 + 0.486 = 0.68 inches$

Thank you for the reply!
No snow load as I will be removing shade during the winter.
Your .68” deflection is much more in line with what I was expecting. Looking like I’ll be ok with my beam selection. Thank you for the equations, very helpful!