Hello, All these timber sizing questions..... I know this is a lot but for you all who are adept at this help would be greatly appreciated.

I am having a little trouble with how to proceed on this formula to find allowable bending force.

S=M/F

I have not derived S but know its formula to be bd^2/6. A transverse section such as (8")(12")^2/6= 192in^3.

An example: (Longleaf Pine) I want to size a simply supported beam. The tributary area is 220 ft^2 (10'X 22').

DL= 174 lb/linear foot
LL= 1000 lb/linear foot
TL= 1174 lb/linear foot

W= 22' x 1174= 25828 lbs

V (Shear)= W/2= 12914

M (Internal Moment) = WL^2/8
= (1174 lb/ft)(22')^2/8
= 71027 lb/ft

I think I now have M and S, so I can calculate F

f =actual bending stress

S=M/F
f=M/S
f= 71072lb/ft / 192in^3
f= 370lb/ft/in3

Now, first of all how many errors are in this? Is there a table to cross reference f with? And lastly is it best to solve for f this way or use a table (if there is one) to find S?

Mo, one thing I notice is your units for the final calculation of the imposed bending stress are mixed. In the last steps, M is expressed in lb × ft, but divided by S which is given in in³. To play it safe it’s best to convert all units to the same denomination from the get-go. Units are treated like any other algebraic terms in a formula and may be multiplied and cancelled; a good way to check your work. A brief summary of our simple rectangular beam:

Section Modulus = S = bd²/6 = 8 in × (12 in)² ÷ 6 = 192 in³
Total Load = L = 1174 lbs ÷ 12 in = 97.83333 lb/in
L = 22 ft × 12 in/ft = 264 in
Bending Moment = M = wL² ÷ 8 = 97.83333 lb/in × (264 in)² ÷ 8 = 852324 lb × in
Bending Stress = F = M ÷ S = 852324 lb × in ÷ 192 in³ = 4439.1875 lb/in² … as is customary, the stress is expressed in terms of pounds per square inch.

The units of your original force calc would have to be written as:
F = M ÷ S = 71027 lb × ft ÷ 192 in³ = 370 lb × ft / in³ … a bit awkward to work with. But technically correct and easy to fix, since there are 12 in/ft:
F = M ÷ S = (71027 lb × ft × 12 in/ft) ÷ 192 in³ = 4439.1875 lb/in² or psi.

The permitted or allowable stresses found in tables are determined by empirical testing of specimens. I generally work with Eastern White Pine. Looking in my Log Span Tables the value of the allowable bending stress, Fb, is given as 875 psi. For the beam to pass, the actual stresses or forces imposed upon it must be less than or equal to the allowed values. Here we have a large discrepancy between the actual and allowed bending stress. Rather than assuming the beam dimensions it’s best to solve the section modulus based on the allowed bending stress:

S = M/F = (852324 lb × in) ÷ (875 lb/in²) = 974.08457 in³

Assuming you want your width to remain 8 in:
bd²/6 = 974.08457 in³
d² = 6 × 974.08457 in³ ÷ 8 in = 730.56343 in²
d = √730.56343 in² = 27.02894 in

In theory we can make the width whatever we wish so there are an infinite number of solutions here. Only in theory because the beam also has to satisfy the allowed shear and deflection.

Checking the area needed to accommodate the shear stress imposed on this beam:
Reaction Force at ends = Shear Load = W ÷ 2 = (1174 lbs/ft × 22 ft) ÷ 2 = 12914 lbs
From the Log Span Tables the allowable shear parallel to the grain for Eastern White Pine is given as Fv = 65 psi.

Required area to pass in shear:
A = bd = (3 × V) ÷ (2 × Fv)
A = (3 × 12914 lbs) ÷ (2 × 65 lbs/in²) = 298.01538 in²

But our area based on the bending stress calc above is:
8 in × 27.02894 in = 216.23150 in² … hmmmm … that falls a bit short. This could take a long time and we haven’t even considered the deflection yet.

So let’s try working the beam section backwards from the allowed stresses and see if we can economise on the math.

S = M ÷ Fb
bd²/6 = M/Fb
A = bd = (6 × M) ÷ (Fb × d) … the area needed to allow the theoretical bending stress

The shear limit is already written in terms of the area and since the two areas must be equal:
(6 × M) ÷ (Fb × d) = (3 × V) ÷ (2 × Fv)
Solving for d:
d = (6 × M × 2 × Fv) ÷ (3 × V × Fb) = (4 × M × Fv) ÷ (V × Fb)
Substituting for the variables:
d = (4 × 852324 lb × in × 65 lbs/in²) ÷ (12914 lbs × 875 lbs/in²) = 19.611429 in

Since S = bd²/6 = 974.08457 in³
b = 6 × 974.08457 in³ ÷ 19.611429 in² = 15.19601 in

We have a beam with a section modulus = 974.08457 in³, which meets the imposed bending stress requirement, and an area of 298.01538 in² that can handle the shear load. Now to check the deflection, let’s assume that:

Allowed Deflection = L/360 = 264 in ÷ 360 = .73333 in

For a simple beam:
Actual Deflection = (5 × w × L^4) × (384 × E × I)

E = Modulus of Elasticity = 1100000 psi (for Eastern White Pine, from the Log Span Tables)
I = Second Moment of Area = bd³/12 = 15.19601 in × 19.611429 in³ ÷ 12 = 9551.59500 in^4

Substituting in the deflection equation:
Actual Deflection = (5 × 97.83333 lb/in × (264 in)^4) × (384 × 1100000 lbs/in² × 9551.59500 in^4) = .58593 in … less than the permitted deflection so our beam is good. Or let’s look at it this way:

Length of Beam in inches/Number = .58593 in
Number = Length of Beam in inches ÷ .58593 in = 264 in ÷ .58593 in = 450.56577

So the beam deflection is L/450.56577, better than the L/360 specified.

Sorry about the lengthy post, hope your not too board …

And I hope the "Edit" function is working because there are probably a LOT of typos that have to be fixed.

Well, there's no sense in taking the time to proofread that last message because there's no way to edit the damn thing anyway.

Joe, is it possible you are using an older span table?
My 2001 edition of the NDS book shows the Fv value for eastern white pine as 125, not 65.
These values corrected when it was discovered that a value was double deducted, as I remember from out engineering workshop at a conference a while ago...

Jim, it's not only possible ... I am using an older version of the Log Span Tables published by the ILBA. I don't have it with me to check the exact date of issue but I'm fairly certain I acquired my copy before 2000 (A.D., that is ... :D).

I should have thought to mention I was using an older table. The solutions posted were only meant to be a generic guide or template, the numbers will vary anyway with species, beam length, etc, etc.

Joe, Firstly thank you for your time on the post. Definetly not board.
This is all new to me but I am trying to figure it out. Even just plugging variables can be difficult! I have followed you up to:

S = M ÷ Fb
bd²/6 = M/Fb
A = bd = (6 × M) ÷ (Fb × d)

Can you help me with the algebraic transition from the second line to the third?

Mo, multiply both sides of the equation by 6/d. Since both sides of the equation are multiplied by the same term the value of the equation does not change. Cancel equal terms on the left hand side, this leaves b × d or the area on the left hand side and (6 × M)/(Fb × d) on the right side.

bd²/6 = M/Fb
b × d × d/6 × 6/d = M/Fb × 6/d ... cancel equal terms on the left side
bd = (M × 6)/(Fb × d) ... collect multiplications and divisions on the right side

I played a few more games with the algebra ... whenever you are ready for them here are a some more formulas for a simply supported beam, rectangular in cross section. Since the "Edit" function isn't enabled I'm linking to an image file; this way if there are any boo-boos I can fix them.

The units are as I prefer to use them but there are other options ... kips, kN, lb × ft, etc. I find it's easiest to kill two birds with one stone and find a value for S which satisfies both bending and shear, then check the deflection.

Solve for M and V, Fb and Fv are found in tables.

Test drive them, have fun ...