Wood Beam: Design Equations and Span Tables

12/12/2011

by hoangkybaction

Derivations of the formulas shown in the following table:

Derivation of interal shear force V at a point x from the beam’s left-end:

This is a basic physics problem:

From a Static cource, the conditions for an equilibrium are sum of all vertical foreces must be equal to zero, (no horizontal forces in this case), and all moments at point x must be equal to zero as well . Thus, at distance x from the left end of the beam, one can write:

R – w.x – V = 0      =>   V   =  R  –  w.x

With R = (w.L)/2 

V  =  (w.L)/2  –  w.x  

V  =  w(L/2 – x);    (eq.1) 

Thus V = 0 at x = L/2,  half span.

Next, sum all the moment about an axis at point x, and set it equal to zero:

((w.L)/2)x  – w.x(x/2) – M  =   0

M  =  (w.L/2)x  – (w/2)x.x

M  = (w.x/2)(L – x);  (eq.2)

Thus, at x = 0 and x = L, M = 0; and at x = L/2, M = w.L.L/8 = Maximum value .

Remember: Eq.2 will be used in the beam’s deflection equation during the double intergration process to find the deflection function y(x), and the maximum amount of deflection for a given load .

When the beam is loaded, it will bend as shown in the figure below. Imagine that the whole length L of the beam lies on the function y(x), with the beam’s left-end is located at the origin, and the beam’s right-end is also on the x=axis at the other end.  Thus, once the deflection function y(x) is determined, the maximum deflection can be calculated.

In beam design, the amount of deflection is usually small compared to the length of beam; for example, deflection may be 1 in. or less in a span of 30 feet.  Therefore, one can assume that the beam under load lies on a large circle of  radius r which has a curvature equal 1/r, and sustain an angle theta as shown in the figure.

Thus,

L  = r.theta  => theta  = L/r     (1)

Similarly,

L + dL  =  (r + c).theta  => theta  =  (L + dL)/(r + c)    (2)

equate (1) and (2):

L/r  =  (L + dL)/(r + c), or

(L + dL)/L  =  (r + c)/r

1  +  dL/L   =  1   +  c/r

but strain is defined as dL/L, thus

strain = dL/L  = c/r = c(1/r)  (3)

or,

1/r  = curvature of y(x) = strain/c   (4)

Recall that modulus of elasticity E is defined as:

E  =  stress/strain 

strain = stress/E  (5)

Substitute (5) into (4):

1/r  = curvature of y(x) = strain/c  = (stress/E)/c  = stress/E.c    (6)

From the beam’s resisting moment formula, namely M = Fb(I/c) in the preveous section, where Fb is the maximum stress in a beam’s cross section area:

Stress Fb = c.M/I ; (7)

where I is the moment of inertia of the beam’s cross-sectional area, remained undeformed under bending condition. 

Substitute (7) into (6):

1/r = curvature = stress/E.c = (cM/I)/E.c

1/r  =  curvature = M/(E.I)   (8);

where M  = (w.x/2)(L – x) = eq.2 above .

Now, from caluclus, curvature of a function y(x) is given as:

curvature  =  d(theta)/dS  =  y”/[1  (y’)^2]^3/2;  (9)

where: y” and y’ are second and first derivatives of y(x) with respect to x.

Since the amount of deflection is very small compared to beam’s length, y’ is neglectible. That is y’ = 0; Thus, (9) becomes:

1/r  =  curvature =  y”  (10)

Substitute (10) in to (8):

y”  =  M/E.I

or,

(E.I)y”   =   M  =   = (w.x/2)(L – x)

(E.I)y”   =  (wL/2)x  –   (w/2)x^2   (11)

Now, integrate (11) with respect to x:

(E.I)y’  =  (wL/2)(x^2)/2   –  (w/2)(x^3)/3    +   k1      (12)

Since the beam is bended symmetrically at the midpoint x = L/2, one can set y’ = 0 and evaulate the constant k1:

(E.I)y’  =  (wL/2)(x^2)/2   –   (w/2)(x^3/3   + k1   =  0    at x  = L/2

or,

k1   =  (1/3)(w/2)(L/2)^3   –  (1/2)(wL/2)(L/2)^2

k1  =  (wL^3)/48     –     (wL^3)/16    =  – (2wL^3)/48  

k1  =   –   (wL^3)/24   (13)

Substitute (13) into (12):

(E.I)y’  =  (wL/2)(x^2)/2   –  (w/2)(x^3)/3    –   (wL^3)/24     (14)

Now, integrate (14) to find y(x):

(EI)y   =   (wL/4)(x^3)/3    –   (w/6)(x^4)/4   –  (w/24)(L^3)x     +    k2

With the boundary values y(0)  =  y(L)  = 0, solve for k2:

(E.I)y(L)   =  (wL/4)(x^3)/3    –    (w/6)(x^4)/4    –    (w/24)(L^3)x    +   k2   =   0

k2    =    (w/24)(L^4)   +    (w/24)L^4    –   (wL/12)L^4    =    0

k2    =    0

Thus, the deflection function y(x) is:

y(x)   =   {(wL/12)x^3    –    (w/24)x^4    –   (w/24)(L^3)x}/(E.I)    (15)

Since the maximum deflection Dmax = y max occurs at x = L/2:

Dmax = y max  = {(wL/12)(L/2)^3    –   (w/24)(L/2)^4   –   (w/24)(L^3)(L/2)}/(E.I)

Dmax = y max  =  {(wL^4)/96    –    (wL^4)/384    –   (wL^4)/48}/(E.I)

Dmax = y max  =  (wL^4).(1/96  –  1/384   –  1/48)/(E.I)

Dmax  =   y max  =  (wL^4)(4/384  – 1/384  –  8/384)/(E.I)

Dmax  =  y max  =  (5wL^4)/(384E.I)   =   (5/384)(WL^3)/(E.I); where W  = w.L 

Thus,

Dmax =   (5/384)(w.L)(L^3)/(E.I)  =   (5/384)(WL^3)/(E.I); where W  = w.L    (16)

Done by hoangkybactien

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Longleaf pine, hardest of all pines: Modulus of Elasticity  E  =  1,445 Kg/sq.mm

E = 1,445Kg/sq.mm x 2.2lbs/Kg x 645sq.mm/sq.in.  =  2,050,455 lbs/sq.in.

E  =  2,050,455 lbs/sq.in.

Lecture 38:

Beam Stresses

***

Jack Pine: Weakest of all pines: Modulus of Elasticity

E  =  868 Kg/sq.mm, about 60% as strong as those of longleaf pines

 E = 868 Kg/sqmm x 2.2lbs/Kg  x 645sqmm/sq.in = 1,231,692 lbs/sq.in.

E  =  1,231,692 lbs/sq.in.

***

BENDING STRESSES


A formula based on bending in a beam will be derived that can be used to:

  • Select a beam that has adequate strength to carry a given load in bending.
  • Determine the bending stresses in a given beam caused by a specific loading condition.
  • Determine the greatest moment a given beam will resist (or the greatest load it will carry)


It will not be necessary to be able to derive the formula. However, it is essential to work through the derivation at least once in order to understand it and have confidence in how it is applied.

F = M/S or S = M/F  In which,

S is the Section Modulus in in^3. This beam property can either be calculated or read from tables for most beams.

M is the moment in the beam in inch-pounds (or inch-kips). In the selection of a beam it is taken from the moment diagram.

F is the bending stress (psi) (Fb is usually used to denote the allowable bending stress; fb is used to indicate the actual bending stress).



BENDING THEORY

There are a number of assumptions that were made in order to develop the Elastic Theory of Bending. These are:

  • The beam has a constant, prismatic cross-section and is constructed of a flexible, homogenous material that has the same Modulus of Elasticity in both tension and compression (shortens or elongates equally for same stress).
  • The material is lineraly elastic; the relaitonship between the stress and strain ar directly proportional.
  • The beam material is not stressed past its proportional limit.
  • A plane section within the beam before bending remains a plane after bending (see AB & CD in the image below).
  • The neutral plane of a beam is a plane whose length is unchanged by the beam’s deformation. This plane passes throught the centroid of the cross-section.
    In order to visualize this, think of a black rubber beam with three lines drawn on its side. The dashed lines in the diagram below represent this beam, and the neutral axis and lines AB and CD are drawn parallel to their respective sides. Lines AB and C D are separated by some distance that is determined, but is not of consequence for the following discussion. These two lines were parallel before bending. As the beam bends, these lines remain perpendicular to the neutral axis.delta = change between lines parallel to the neutral axis in a beam before and after bendingThus, as the beam bends, developing a curve in the neutral plane that reflects the bending, the line CD does not remain parallel to line AD. There is a distinct shortening at the top face of the beam and elongation at the bottom face. CC’ is equal to d (delta), or the shortening of the top fiber. Similarly, DD’ is equal to (delta) or the elongation of the bottom fiber (tension). If these were measured, they would be found to be equal, but opposite. Knowing that the Modulus of Elasticity (E) describes a linear relationship between the strain and the amount of stress in a material and that this material is homogeneous and has a certain value for E,we can determine the stress. The magnitude of the strain at any point along C’D’ can be found by using similar triangles. Therefore, the stress is also proportional to the distance from the neutral plane, as illustrated in the following diargram.stress prism across the cross sectionThe loads and reactions acting on this beam segment cause a tendency for clockwise rotation (a clockwise moment). An internal moment counteracts this tendency so that the beam segment remains in equilibrium. Thus, the magnitude of the internal moment is exactly equal to the moment due to the external loads and reactions; but in the opposite direction. There is a state of equilibrium:Internal Moment = External Moment

  • The illustration above shows the stress prisms developed due to the straining of the material. These prisms (compression and tension) can each be resolved into a single force that acts through the centroid of each prism. These two resulting forces (one is a compressive force (C) and the other a tensile force (T) ) are found to have equal but opposite magnitudes. They are separated by a distance that is approximatly 2/3 of the depth of the beam (because of their triangular geometry) and create an internal couple(!) which causes the internal moment resistance of the beam. The exact location of the resultant forces of the stress prisms depends upon the locaton of the centroid of the entire stress prism. The location of the centroid is in turn dependent upon the exact distribution of the stresses across the section which this is determined by the geometry of the cross-section.


    GENERAL CASE
    If one accepts the assumptions given in the discussion of the bending theory above, the formula for determining the bending stress at any given fiber in a section can be derived as follows.

    stress prism for a beam

    Given the beam as shown above loaded so that it has an internal bending moment. The stress at the extreme face (the distance C in this case) from the Neutral Axis (NA) is equal to Sigma or Fb. The stress at any distance “y”; from the NA can be found using similar triangles. Thus,

    Fy = (Fb/c)(y)

    If “a” is a very small area at a distance “y” from the NA, then magnitude of the stress on area “a” equals (Fb/c)(y). Remembering that stress is equal to a force distributed over an area, the force on the area “a” is equal to the stress times that area, or (Fb/c)(y)(a). And, the moment of the force on area “a” about the NA is equal to (Fb/c)(y)(a)(y).

    Now, if all of the moments of all of the forces on all of the tiny “a” areas across the face of the beam are added together we arrive at an expression of the internal moment resistance of a beam;

    M =Sum (F/c) a y^2 or M = (F/c) Sum a y^2

    In this case, the value of (F/c) is a constant for any beam and Sum ay^2 is the second moment of the area, better known as the Moment of Inertia. Therefore, substituting items that we know, the equation becomes:

    M = (Fb/c) I

    or rearranging

    M = (I/c) Fb
    or
    M = S*fb


  • (I/c) is known as the Section Modulus (S) and is also tabulated.


    BENDING STRESSES REVIEW


  • The flexure formula:M = (F/c) I or M = (I/c) F

    M is the internal moment in a beam created by the loads and reactions. It is numerically equal to the external moment at any section. Its value can be taken directly from the moments diagram. M is in inch-pounds (or inch-kips) .

    F is the stress at any distance “c” from the NA. If “c” is taken as the distance from the NA to the most remote fiber, then the stress “F” is the greatest stress. c is in inches; F is psi.

    I is the Moment of inertia of the beam. I is in inches^4.

    S is the Section Modulus. It is (I/c.) and has the units of inches^3.

The flexure formula is useful in the following ways:

  • M = (Fb/c) I to find the resisting moment of a beam
  • Fb = M c /I to find the stress in a beam
  • I / c = Fb to select a beam
  • Sreq = M / Fb to select a beam


The last formula is the most useful.
The value of I/c, or S, is constant for any given beam and is utilized so often that its value is listed in any table of beam properties.

http://pages.uoregon.edu/struct/courseware/461/461_lectures/461_lecture38/461_lecture38.html

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From: http://www.engineersedge.com/beam_bending/beam_bending1.htm

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Physical Properties of Common Woods

http://www.csudh.edu/oliver/chemdata/woods.htm

***

References:

Provide good basic theory

http://www.awc.org/pdf/wsdd/c2a.pdf

***

Very good basic theory

http://www.freestudy.co.uk/engineering%20science%20h1/outcome%201%20t3.pdf

***

Mechanical Property of Wood  –  Elasticity Data of  Various kinds of Wood for calculating deflection and span

http://www.conradfp.com/pdfs/ch4-Mechanical-Properties-of-Wood.pdf

***

American Forest and Paper Association – American Wood Council

Beam Design Formulas

http://www.awc.org/pdf/DA6-BeamFormulas.pdf

***

Lumbers spantables for Joists and Rafters

http://www.awc.org/pdf/wsdd/C2B.pdf

***

Lumber Spantables for Joists and Rafters

http://www.msrlumber.org/spantables.pdf

***

Wood Studies

http://www.awc.org/pdf/wsdd/C1.pdf

***

Values of density marked * are for air dry samples.

The following table presents values for the properties of common woods.

This table is in the public domain and may be copied without limit. The user is encouraged to download it for private use and public distribution in any form including that of making it available on other Web servers.

*

*

*

*

*

*

*

*

*

*

*

*

source: http://www.csudh.edu/oliver/chemdata/woods.htm

***

Beam Deflection theory

http://courses.washington.edu/me354a/chap3.pdf

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Dec 5, 2011

Lecture 40:

Beam Deflection

A loaded beam deflects by an amount that depends on several factors including:

  • the magnitude and type of loading
  • the span of the beam
  • the material properties of the beam (Modulus of Elasticity)
  • the properties of the shape of the beam (Moment of Inertia)
  • the beam type (simple, cantilever, overhanging, continuous)

Deflection may or may not be critical. Excessive deflection may result in cracked plaster, objectionable appearance, or sags in flat roofs which then “pond”. It may transfer loads to non- bearing members under the beam such as partitions, doors, windows, etc., which in turn may cause partitions to crack or doors and windows to stick. A floor beam which deflects excessively is apt to be “springy”, creating an undesirable walking surface, even if it is in no danger of failing. A springy floor is especially unsatisfactory for a room housing sensitive instruments.

A general rule for limiting deflection of simple spans for floor construction, or for plastered ceilings, is that the deflection should not exceed the span (in inches) divided by 360 (max D= L/360). The deflection for exposed ceiling beams at the roof is often allowed to be 50% to 100% greater (l/240 or l/180). Codes usually specify that these deflections are based on live load only, but experience shows that this is sometimes excessive. A conservative approach is to limit the deflection to these values for total load in lightweight construction (such as wood and steel). These guidelines are general and apply in most cases, but certainly not all. For example, the springiness of a floor is influenced more by the mass of the floor than by the total deflection; more dead load could cause more deflection, but probably less springiness.

Formulas given in tables will be used to compute deflection for some loading conditions; these will be expanded to approximate deflections for other conditions. The deflection formulas will not be derived. In deflection formulas, “w” refers to pounds per inch of length of loading (not pounds per foot). “W” refers to the total distributed load. The beam length is in inches. Most mistakes in computing deflections are caused by using the length in feet instead of inches and/or using “w” to mean pounds per foot instead of pounds per inch. Also, convert any distributed load “w” to “W”.
There are four parts to deflection formulas:

1. COEFFICIENT, which takes into account:

  • Type of beam
    simple, cantilever, overhanging, etc.
  • Loading condition

2. LOAD FACTOR

  • Distributed loads “W” (total weight)
  • Distributed loads “w” (weight per inch of length)
  • Concentrated loads “P” (weight of one concentrated load)

3. LENGTH FACTOR

  • L^3 (inches) usually for load factor “P” and “W”
  • L^4 (inches) usually for load factor “w”
  • La, Lb (inches) used to locate some types of loads

(There is some confusion about the symbol “L” and “l”;. Lower case “l” is often used to indicate length in inches and upper case “L” for length in feet. However, lower case “l” can be confused with the number “1” which causes more serious problems. Therefore, in this class, “L” refers to inches in these deflection formulas.)

4. STIFFNESS FACTOR

1/EI

  • E = material stiffness (modulus of elasticity)
  • I = stiffness of the section based on the geometry of the section (moment of inertia).

CAUTION: The most common mistake in computing deflection is caused by using “w” as load per foot instead of load per inch. The derivation of the deflection formulas uses the unit of inches for all the factors in the formula. Uniformly distributed beam loads “w” are normally described as load per foot which must be converted to the proper load per inch value before insertion into the deflection formula. If “W” is used for the total distributed load instead of the load per unit of length (“w”) this conversion is not necessary.
Note that W = wL; will simplify many of the formulas.

The following are equations for finding the deflection of the more common beam types and their associated loading. Note how the co-efficient reflects the stiffness of the system and the loading.

equations for calculating maximum deflection

Source: http://pages.uoregon.edu/struct/courseware/461/461_lectures/461_lecture40/461_lecture40.html

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