A wood strength question to ponder............

[Mark Rios]

Does a board of a given dimension and volume have the same compressive/span strentgh as a board of a different dimension but with the same volume? In other words, does a 4 foot 2 x 6 (1 1/2" x 5 1/2"), on edge, have the same strength as a 4 foot 1 x 12 (3/4" by 11")?

Anyone have any info on this?

[Greg Funk]

No, strength is not a function of volume. For a simple example take a 1x12 and compare its strength on edge or on the flat. Same volume but drastically different strength.

Greg

[Jimmy Newman]

Strength and stiffness are two different things. Strength is how much load you can put on something before it breaks. Stiffness is how much a given piece flexes under a certain load.

Stiffness is slightly complicated to calculate.

First, you have to decide how your object will be loaded. If you draw out the flexure, on one side the object elongates slightly and on one side it compresses slightly. You have to find the plane in which there is no compression or expansion (for something that is symmetric, this is just the center plane). Then, to find the stiffness, you take the integral of the square of the distance from that center plane over the cross sectional area of the object and multiply that by the elastic coefficient (which changes based on grain direction and species of wood, plus a little variation from piece to piece).

Essentially, the more cross sectional area you have further away from the no-stretch plane, the stiffer it is.

Of course, this isn't the same as strength. If you make it really, really thin but tall, then it'll be very, very stiff, but will break really easily.

[Eddie Darby]

Deflection = Load x Length^3 / Width x Depth^3 for a double supported beam.

This formula tells us that Depth plays a very important role in the resistance to the beam bending, since it is to the "power of 3".

a 2" x 6" has a number of 432
a 1" x 12" has a number of 1728

Which means a 1" x 12" is 4 times greater at resisting deflection than a 2" x 6" beam for the same span and same load, and of course the same species of wood.

[Joseph N. Myers]

Mark,

Try playing with: http://www.woodbin.com/calcs/sagulator.htm


It's a program to calculate "sagging" for shelves of different woods, width, length, thickness, etc.

HTH, Joe

[Mark Singer]

In compression without bending equal areas of cross section have equal strength. This is a short column where bending is not involved. As the column becomes proportionatley longer the "l/r" ratio comes into play and the section modulus determines the strength. There are formulas to calulate the stress based on the modulus of elasticity of the material, section modulus and colum length. When supported at the mid height the long column is 4 times as strong since it is a function squared

[Bill Wyko]

Mark,

Try playing with: http://www.woodbin.com/calcs/sagulator.htm


It's a program to calculate "sagging" for shelves of different woods, width, length, thickness, etc.

HTH, Joe
Thanks for the info. that'll come in handy.

[Mark Rios]

In compression without bending equal areas of cross section have equal strength. This is a short column where bending is not involved. As the column becomes proportionatley longer the "l/r" ratio comes into play and the section modulus determines the strength. There are formulas to calulate the stress based on the modulus of elasticity of the material, section modulus and colum length. When supported at the mid height the long column is 4 times as strong since it is a function squared

Mark, this makes sense. If I assume correctly then, this is (basically) why Truss Joists, et al, can exist with their thin web in between the top and bottom chords. Is that right?

[joe greiner]

Almost right. Stiffness is proportional to the Moment of Inertia, I, and the Modulus of Elasticity, E. Strength is proportional to the Section Modulus, S. The Moment of Inertia and the Section Modulus are geometric properties; S = I/c, where c is the distance from the neutral axis to the extreme fiber. The Modulus of Elasticity is a material property, different for different woods.

Members in compression are governed by buckling, not strength per se. The shorter the distance between stiffening locations, the greater the buckling resistance. At a minimum buckling resistance, allowable compression stress governs.

Google ["strength of materials"] for some example texts. Your public library may have such books.

DIY stress analysis is OK for furniture and such. For significant construction, a construction professional should be consulted. Building codes often have conservative prescriptive requirements which may also be relied upon. I'm a structural engineer, and I'm not trying to drum up business for my brethren; just attempting to keep you and yours alive.

Joe

[Chuck Saunders]

If I assume correctly then, this is (basically) why Truss Joists, et al, can exist with their thin web in between the top and bottom chords. Is that right?
That is correct Mark. If you think about a floor joist (we will exaggerate by having the joist bending like a smile) , the top of the joist is in compression while the bottom of the joist is in tension. As you progress from bottom to top you must translate from tension to compression. as you reach the neutral axis (we will make it simple and say the center of the beam) you will change from tension to compression. At this point there is little stress in the beam so there is little need for material. Less material > less cost. This is how I beams work. You must insure that the lateral forces are restrained as the trusses are rather flimsy side to side.

[Mark Singer]

This may be helpful....



Compression Members http://www.efunda.com/images/section_bar.gif
Compression members, such as columns, are mainly subjected to axial forces. The principal stress in a compression member is therefore the normal stress,

http://www.efunda.com/formulae/solid_mechanics/columns/images/NormalStress.gif
The failure of a short compression member resulting from the compression axial force looks like,

http://www.efunda.com/formulae/solid_mechanics/columns/images/CompressionMember_Short.gif
However, when a compression member becomes longer, the role of the geometry and stiffness (Young's modulus (http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/elastic_constants_E_nu.cfm)) becomes more and more important. For a long (slender) column, buckling (http://www.efunda.com/formulae/solid_mechanics/columns/theory.cfm) occurs way before the normal stress reaches the strength of the column material. For example, pushing on the ends of a business card or bookmark can easily reproduce the buckling.

http://www.efunda.com/formulae/solid_mechanics/columns/images/CompressionMember_Long.gif
For an intermediate length compression member, kneeling occurs when some areas yield before buckling, as shown in the figure below.

http://www.efunda.com/formulae/solid_mechanics/columns/images/CompressionMember_Intermediate.gif
In summary, the failure of a compression member has to do with the strength and stiffness of the material and the geometry (slenderness ratio) of the member. Whether a compression member is considered short, intermediate, or long depends on these factors. More quantitative discussion on these factors can be found in the next section.


Top of Page (http://www.efunda.com/formulae/solid_mechanics/columns/intro.cfm#PageTop)

Design Considerations http://www.efunda.com/images/section_bar.gif
In practice, for a given material, the allowable stress in a compression member depends on the slenderness ratio Leff / r and can be divided into three regions: short, intermediate, and long.
Short columns are dominated by the strength limit of the material. Intermediate columns are bounded by the inelastic limit (http://www.efunda.com/formulae/solid_mechanics/columns/inelastic.cfm) of the member. Finally, long columns are bounded by the elastic limit (i.e. Euler's formula (http://www.efunda.com/formulae/solid_mechanics/columns/columns.cfm)). These three regions are depicted on the stress/slenderness graph below,

http://www.efunda.com/formulae/solid_mechanics/columns/images/Column_Design.gif
The short/intermediate/long classification of columns depends on both the geometry (slenderness ratio) and the material properties (Young's modulus and yield strength). Some common materials used for columns are listed below:
Material Short Column
(Strength Limit) Intermediate Column
(Inelastic Stability Limit) Long Column
(Elastic Stability Limit) Slenderness Ratio ( SR = Leff / r) Structural Steel SR < 40 40 < SR < 150 SR > 150 Aluminum Alloy AA 6061 (http://www.efunda.com/materials/alloys/aluminum/show_aluminum.cfm?ID=AA_6061&prop=all&Page_Title=AA%206061) - T6 SR < 9.5 9.5 < SR < 66 SR > 66 Aluminum Alloy AA 2014 (http://www.efunda.com/materials/alloys/aluminum/show_aluminum.cfm?ID=AA_2014&prop=all&Page_Title=AA%202014) - T6 SR < 12 12 < SR < 55 SR > 55 Wood SR < 11 11 < SR < (18 ~ 30) (18 ~ 30) < SR < 50

In the table, Leff is the effective length (http://www.efunda.com/formulae/solid_mechanics/columns/columns.cfm#ExtendedEuler) of the column, and r is the radius of gyration (http://www.efunda.com/math/areas/RadiusOfGyration.cfm) of the cross-sectional area, defined as http://www.efunda.com/formulae/solid_mechanics/columns/images/rofGyration.gif.
Radii of gyration for standard beams (http://www.efunda.com/math/areas/IbeamIndex.cfm), common beams (http://www.efunda.com/math/areas/Common_Cross_Section_Index.cfm), and other common areas (http://www.efunda.com/math/areas/Common_Geometric_Shapes_Index.cfm) can

[Mark Singer]

Here are the beam diagrams and formulas,,,

Shear Force and Bending Moment Diagrams

Introduction
Normally a beam is analysed to obtain the maximum stress and this is compared to the material strength to determine the design safety margin. It is also normally required to calculate the deflection on the beam under the maximum expected load. The determination of the maximum stress results from producing the shear and bending moment diagrams. To facilitate this work the first stage is normally to determine all of the external loads.

Nomenclature
e = strain
σ = stress (N/m2)
E = Young's Modulus = σ /e (N/m2)
y = distance of surface from neutral surface (m).
R = Radius of neutral axis (m).
I = Moment of Inertia (m4 - more normally cm4)
Z = section modulus = I/ymax(m3 - more normally cm3)
M = Moment (Nm)
w = Distrubuted load on beam (kg/m) or (N/m as force units)
W = total load on beam (kg ) or (N as force units)
F= Concentrated force on beam (N)
S= Shear Force on Section (N)
L = length of beam (m)
x = distance along beam (m)

Calculation of external forces
To allow determination of all of the external loads a free-body diagram is construction with all of the loads and supports replaced by their equivalent forces. A typical free-body diagram is shown below.
http://www.roymech.co.uk/images11/beam_1.gif
The unknown forces (generally the support reactions) are then determined using the equations for plane static equilibrium.
http://www.roymech.co.uk/images6/force10.gif
For example considering the simple beam above the reaction R2 is determined by Summing the moments about R1 to zero
R2. L - W.a = 0 Therefore R2 = W.a / L
R1 is determined by summing the vertical forces to 0
W - R1 - R2 = 0 Therefore R1 = W - R2
Shear and Bending Moment Diagram

The shear force diagram indicates the shear force withstood by the beam section along the length of the beam.
The bending moment diagram indicates the bending moment withstood by the beam section along the length of the beam.
It is normal practice to produce a free body diagram with the shear diagram and the bending moment diagram position below

For simply supported beams the reactions are generally simple forces. When the beam is built-in the free body diagram will show the relevant support point as a reaction force and a reaction moment....

Sign Convention
The sign convention used for shear force diagrams and bending moments is only important in that it should be used consistently throughout a project. The sign convention used on this page is as below

http://www.roymech.co.uk/images11/beam_23.gif
Typical Diagrams
A shear force diagram is simply constructed by moving a section along the beam from (say)the left origin and summing the forces to the left of the section. The equilibrium condition states that the forces on either side of a section balance and therefore the resisting shear force of the section is obtained by this simple operation

The bending moment diagram is obtained in the same way except that the moment is the sum of the product of each force and its distance(x) from the section. Distributed loads are calculated buy summing the product of the total force (to the left of the section) and the distance(x) of the centroid of the distributed load.

The sketches below show simply supported beams with on concentrated force.

http://www.roymech.co.uk/images11/beam_24.gif
The sketches below show Cantilever beams with three different load combinations.

http://www.roymech.co.uk/images11/beam_25.gif
Note: The force shown if based on loads (weights) would need to be converted to force units i.e. 50kg = 50x9,81(g) = 490 N.

Shear Force Moment Relationship

Consider a short length of a beam under a distributed load separated by a distance δx.
http://www.roymech.co.uk/images11/beam_42.gif
The bending moment at section AD is M and the shear force is S. The bending moment at BC = M + δM and the shear force is S + δS.

The equations for equilibrium in 2 dimensions results in the equations.. Forces

S - w.δx = S + δS
Therefore making δx infinitely small then.. dS /dx = - w
Moments.. Taking moments about C

M + Sδx - M - δM - w(δx)2 /2 = 0
Therefore making δx infinitely small then.. dM /dx = S
Therefore putting the relationships into integral form.
http://www.roymech.co.uk/images11/beam_43.gif
The integral (Area) of the shear diagram between any limits results in the change of the shearing force between these limits and the integral of the Shear Force diagram between limits results in the change in bending moment...

[Bill White]

OH MY GOODNESS!!!! My head is gonna explode.
Bill

[Mark Rios]

OH MY GOODNESS!!!! My head is gonna explode.
Bill


ROFLOL Bill.
:D :D :D :D :D


Thanks for all of the data folks. I like that kind of stuff.

However, what I'm doing really doesn't warrant all of your great efforts although I"m very appreciative of all of your info and help.

I"ve got a bunch of clear, structural select 2x12 material that has warped a little. I am making a platform to dry some recently milled wood. The platform kinda looks like a stick-framed wall laying down. This is my second wood drying platform. I used 2x4 on my first one and supported it at 3' and 6'. I've ripped the 2x12 in half. By the time I joint and plane the 2x12 (2x6 now) rips they will be about 1" or 1 1/4" thick. So, I was wondering if a 1" x 5" (full, actual size) board would give me the same flexural strength as a 2x4 (1 1/2" x 3 1/2" actual size). From what I've read, I'm good.

So thanks to all very much for all your very helpful info.

[dan moran]

OH MY GOODNESS!!!! My head is gonna explode.
Bill

i got an "A" in that class..:p

after a day or two of perusing my statics book, i was like " OMG the maths they been trying to teach me ACTUALLY HAVE A PRACTICAL APPLICATION"

[Bob Swenson]

A year ago I took a course in sculpture at our County College the first assignment
Was to build a bridge 36 “ long and 3” wide using of a piece of ¼” foam core 30” X 24”
And hot glue. Piling 7 lb bricks in the center tested the bridges. Most student bridges failed in the 4 to 6 range. This bridge did not fail at 14 (98 pounds) But the professor would not let me put another brick on. It made a pretty scary pile. Perhaps Mark could use a little fancy math to explain why it’s so strong. I used old geezer instinct.
Bob
PS- It holds the school record

[Tom Veatch]

A year ago I took a course in sculpture at our County College the first assignment
Was to build a bridge 36 “ long and 3” wide using of a piece of ¼” foam core 30” X 24”
And hot glue. Piling 7 lb bricks in the center tested the bridges. Most student bridges failed in the 4 to 6 range. This bridge did not fail at 14 (98 pounds) But the professor would not let me put another brick on. It made a pretty scary pile. Perhaps Mark could use a little fancy math to explain why it’s so strong. I used old geezer instinct.
Bob
PS- It holds the school record

When I saw the picture, the first thing I though of was it's almost identical to a carrythru spar. That's the structure in an aircraft that ties cantilevered wings together through the fuselage. Your "sculpture" looks almost ideally designed for bending loads - top and bottom spar caps transmit the axial loads, shear web resists the shear forces and stabilizes the spar caps, and web stiffeners resist shear buckling of the web. I'd say your "old geezer instinct" was right on target. Good job!!!

[Bob Swenson]

Thanks Tom.
Maybe you guys wonder why I still go to college - silly thought

[Per Swenson]

Oh sure.

Wonder why he still goes to college?

Continuing education?

Nope at 82 he pretty much knows it all, just pretends

cranial infirmity, like people with selective hearing.

No Folks, my Daddy attends class for the view.

Are you catching a little resentful tone here?

I won't begrudge him entertainment......

But I am the one who has to say, your project will be

a little late as the shop is closed for piano lessons.

:D


Per