Checking a uniformly loaded simple beam in bending:
Determining Beam Load
This is a rectangular wood beam supported on both ends and loaded evenly along
its length. A rafter, joist, girder or header is often a beam of this type. We
need to know what the load is on the beam. Use as an example a door header with
a 9' clearspan on a 16 foot wide building with 2 foot overhangs.
The combined Live Load and Dead Load on the building's roof is 50 pounds per
square foot. The area in gray shows the portion of the roof bearing on the beam,
9' X 10'...90 square feet. Multiply that by 50 pounds of weight on every
square foot = 4500 pounds bearing on the beam.
Fiberstress in Bending: What The Calculator
In order to find the section modulus required we need to find the maximum
bending moment and divide it by the allowable fiberstress (Fb) for the species and
grade you enter into the calculator.
The formula for finding the maximum bending
||Mmax = WL/8|
||Working the Example:|
||Mmax = (4500 X 9) / 8|
||Mmax = 5062.5 Ft/Lbs|
To find the section modulus required we need to divide the maximum bending
moment in inch pounds by the allowable fiberstress in bending for that particular wood:
The NDS and major grading agencies publish design values for different species, grades and sizes of wood.
For the example, assume I've found some #2 Dense, Southern Yellow Pine 2x10's. They have a base Fb design value of 1200 psi in bending.
Because we are using several members closer than 24" to each other we are allowed
to adjust the strength value upwards by 10%
The maximum load on the beam is a snow load. Wood can take heavy loads of short duration very well.
For situations where the snow load is the maximum load we are allowed an additional 15% allowable fiberstress in bending.
1200 x 1.1 x 1.15 = 1518 psi adjusted Fb, allowable fiber stress in bending.
||Section Required = Mmax X 12 / Fb|
||Sreq = (5062.5 X 12) / 1518 lbs/sq in|
||Sreq = 60750 in/lbs / 1518 PSI |
||Sreq = 40
We are looking for a beam with a section modulus of 40 in3 The
formula for determining section modulus for a rectangular beam is:
||S = bd2 / 6|
||Let's try a 2 ply 2x10 header|
||S = 3 X (9.25 X 9.25) / 6 |
||S = 3 X 85.56 / 6|
||S = 42.78 in3 |
We needed a value of 40 or greater, we have 42.78...The header passes in bending
The calculator does the moment and required section modulus
calculations after you enter the load, span, size, and Fb values. There is a
link to a table of NDS values for some grades and species of heavy timber on the calculator page.
A joist is basically a small beam, normally considered to be uniformly loaded.
Around here they are typically designed for a 40 or 50 pounds per square foot
loading. A joist is normally limited by deflection, it gets too bouncy or sags
enough to crack plaster before it is in fiberstress trouble, always work all the
checks to be sure though.
Going back to our same example, lets run 16' floor joists across the right
side of the building on 16" centers. Each joist will support an area halfway to
each adjoining joist. 8" along one side and 8" along the other side, 16" wide by
16' long. The joists are in blue, the area of bearing on one is in green.
(192" X 16") / 144 = 21.33 square feet
21.33 SF X 50 PSF = 1066
I like yellow pine for joists, I'm guessing it will take a pretty decent 2X12 so
will start with a no.2. Don't forget to plug in actual dimensions, in this case
I'll plug in 1.5 x 11.25. Run it and it comes up larger than needed in
fiberstress and fine on deflection (I used loblolly)...try to economize go to a
2 x 10, deflection fails, too bouncy. As a rafter with the same loading the 2 x
10 would be fine, more deflection is allowed there, Fb typically limits rafters.
Shear is fine. I have this in tables...that would be the correct way to spec
this, My CABO book says that 2 x 12 is good out to 19' 1" and a 2 x 10 is only
good to 15' 8". CWC book says 18'10" and 16'1" respectively. Now with wall
framing the span will work by code at 2 x 10, its near the limit, judgement
call, I would bump up if at all possible. Code is limiting to 1/360 span
deflection (a 15' span would deflect 1/2" at design load). Many floors are now
specced at 1/480, consumers are wanting stiffer floors nowadays. Rafter? Same
way, just use horizontal clearspan not the angled run length, deflection can be
up to 1/180 span. These are minimums, you can always go bigger.
Deflection: What the Calculator Does
The formula for deflection is:
You Have entered a
1066 lb load, 192" span, Modulus of Elasticity of 1.6 million, 1.5" width and
||D =(5 * W * L3) / (384 * E * (b * d3/ 12))
||Working the Example:|
||D =(5 * 1066 * 192"3) / (384 * 1600000E * (1.5b *
||D = .345" Design deflection|
The calculator also checks what the maximum allowable
deflection for the span entered is,
of .345" is less than the allowable deflection of .533 so the 2x12's work in
||D = L/360 |
||D = 192"/ 360|
||D = .533" Allowable deflection |
Shear: What The Calculator Does
The Calculator halves the
load of 1066 lbs to give V a value of 533 lbs. Yellow pine has an allowable
horizontal shear of 90 PSI, the 47.37 PSI design shear passes.
||Shear =(( 3 / 2 )* V ) / ( bf * df )|
||Shear =(( 3 / 2 )* 533 ) / ( 1.5 * 11.25 )|
||Shear = 47.37 PSI |
calculator checks for the section needed and the section provided,
a section of 16.875 sq" so are fine in shear.
||Section Required =(v *(bf * df))/ Fv|
||Section Required =(47.37 *(1.5 * 11.25))/ 90|
||Section Required =8.88"2 |