PLF Load based on 16" spacing

[mgilbertson]

I am trying to design a beam based on 810# end reaction of the supported products. The products are spaced 16" apart. I know to take that end reaction divide by 2 to get a plf load on that beam for 24" spacing. But do not recall how to get for 16" spacing.

 

[Gr8blu]

PLF = PSF x spacing in feet
To calculate PLF (Permanent Load Factor) use the formula above. FOr distances in inches, convert the distance to feet by dividing the inches by 12.
Example 1: 24 inch spacing. PLF = PSF * 24/12 = PSF * 2
Example 2: 16 inch spacing. PLF = PSF * 16/12 = PSF * 1.3333

A uniform load is a continuous load along the entire length of a member and is expressed in PLF.

 

[XR250]

(Permanent Load Factor)

Huh?

Honestly, this is pretty basic stuff. I am not sure I would trust the OP to do a proper analysis.

 

[Ron247]

I assume you are saying you have multiple point loads of 810# on a beam and want to convert 810# at a 16" spacing to a uniform linear load expressed in pounds a lineal foot. Assuming that is your question, it is (810x12)/16 or 810#/1.33 or 607.5 plf. The "2" in your original end reaction division stands for 24 inches expressed as feet, not 2 as in "divide into 2 equal pieces".

With that said, I agree with XR250, this is very basic stuff and in fact, does not get much more basic. The complexity of ANY beam design is FAR greater than this calc. So, if this is a real beam and not a homework assignment problem, get some help designing it.

Also, considering terminology, we would tend to say, "I have a point load of 810#", not a "product".