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 19' clearspan on a 16 foot
wide building that has 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, 19' X
10'...190 square feet. Multiply that by 50 pounds of weight on
every square foot = 9500 pounds bearing on the beam.
Fiberstress in Bending: What The Calculator Does
In order to
find the section modulus required we need to find the maximum
bending moment and divide it by the allowable fiberstress for the
species and grade you enter into the calculator. The formula for finding
the maximum bending moment is:
|Mmax = WL/8|
|Working the Example:|
|Mmax = (9500 X 19) / 8|
|Mmax = 22562.5 Ft/Lbs|
To find the section modulus required we
need to divide the maximum bending moment in inch pounds by the
|Section Required = Mmax X 12 / Fb|
|Sreq = (22562.5 X 12) / 1150 lbs/sq in|
|Sreq = 270,744 in/lbs / 1150 PSI |
|Sreq = 235.4 inches3|
We are looking for a beam with a section modulus of 235.4 in3 The
formula for determining section modulus for a rectangular beam is:
|S = bd2 / 6|
|Let's try a 6 X 8|
|S = 6 X (8 X 8) / 6 |
|S = 6 X 64 / 6|
|S = 64 in3 |
We needed a value of 235.4 or greater, we have 64...not even
close. Try again with an 8X14 it comes to 261.3...that works as
does a 10 X 12 at 240 in3
The calculator does the moment and section modulus
required calculations after you enter the load, span, size, and Fb values.
There is a link to a table of NDS values for many grades and species on the
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 pounds
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 11.25" depth.
|D =(5 * W * L3) / (384 * E * (b * d3/ 12)) |
|Working the Example:|
|D =(5 * 1066 * 192"3) / (384 * 1600000E * (1.5b * 11.25d3/ 12))|
|D = .345" Design deflection|
The calculator also checks what the maximum allowable deflection for the span entered is,
The deflection of .345" is less than the allowable deflection of .533 so the 2x12's
work in deflection.
|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 =(( 3 / 2 )* V ) / ( bf * df )|
|Shear =(( 3 / 2 )* 533 ) / ( 1.5 * 11.25 )|
|Shear = 47.37 PSI |
The calculator checks for the section needed and the section provided,
We input 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 |
Back to Beam Calculator