111111111[11

orm111111FOREST PRODUCTS LABORATORY

UNITED STATES DEPARTMENT OF AGRICULTURE

FOREST SERVICE

In Cooperation with the University of WisconsinMADISON S. W15CON51N

I STRENGTH OF WOOD BEAMS OF

RECTANGULAR CROSS SECTION AS

AFFECTED 13Y SPAN-DEPTH RATIO

Information Reviewed and Reaffirmed

January 1959

No. 1910

STRENGTH OF WOOD BEAMS OF RECTANGULAR CROSS SECTION

AS AFFECTED BY SPAN-DEPTH RATIO*

BytECH .Tet-

S. C. BN*ES@Er, Engineerand

CHARLES B. NORRIS, Engineer

Forest Products Laboratory,* Forest ServiceU. S. Department of Agriculture

Introduction

The purpose of this report is to validate a theoretical equation that can beused to predict the type of failure and the loads that cause failure in woodbeams of rectangular cross section under varying span-depth ratios. The equa-tion is derived in Appendix I. It is based on the distribution of longitudinaland shear stresses in the beam. The report is divided into two parts. Thefirst part relates the longitudinal stress distribution in a beam to the stress-strain relationship in a compression specimen. In the second part, a simpli-fied assumption regarding the longitudinal stress distribution is made, thetheoretical equation is derived, and values computed with it are compared withthe results of tests.

The simplified longitudinal stress distribution was first suggested by JamesS. Mathewson of the Forest Products Laboratory in 1917. The longitudinal stressdistribution he suggested is shown in figure 1.

The longitudinal stress distribution was used by J. B. Kommers and C. B. Norrisin studies made in 1920 on the strength of wood I-beams and box beams. Themethod as used by them does not apply to beams that are likely to fail in hori-zontal shear.

In 1933, Pragerg independently arrived at the same longitudinal stress distri-bution and also applied it to wood I-beams and box beams. The equations hedeveloped are similar to those obtained by'KoMmers and Norris.

*Original report issued March 1952.

Maintained at Madison, Wis., in cooperation with the University of Wisconsin.

?Prager, Von W. Uber die Querschnittsbemessun g Zweigurtiger Holzholme. Zeit-

schrift fur Flugtechnik and Motorluftschiffahrt. October 1933.

Rept. No. 81910 -1-

In March 1949, T. Kon published a paper3 in which he developed the distri-bution of shear stress associated with this longitudinal stress distribution.His method leads to the same shear-stress distribution as the analysis givenin Appendix which was made before Kon i s paper was received at the Labor-atory.

The distribution of longitudinal and shear stresses indicates that the beammay fail in tension at the extreme fiber, in longitudinal shear at the loca-tion of the maximum shear stress, or in combined tension and shear at alocation between these two. The strength under the combined stresses isgiven in two Forest Products Laboratory reports /4 ) 5 on failure under combinedstresses.

The two stress distributions (longitudinal and shear) are combined with thecriterion of failure under combined stresses in Appendix I, and a formulafor the strength of the beam is obtained.

Part I

Relation of Stress-strain Curve Determined by Compression Tests to Load-

deflection Curve Obtained by Bending Tests

Mathematical developments in the sections that follow are based on the relation-ship of the stress-strain curve determined from compression tests to theload-deflection curve obtained from bending tests. This relationship wasdetermined by tests performed on a Sitka spruce bending specimen and com-pression specimens end-matched to it. The results from the compression testswere used in computing a load-deflection curve for the bending test, and thiscurve was compared with the load-deflection curve obtained from actual measure-ments.

Description of Tests

The set-up for the static-bending test is shown in figure 2. The test speci-men was 2 by 2 inches by 5 feet and was subjected to quarter-point loads.The loading blocks were attached to two fixed I-beams, and loads were applied

.Kon, T. On the Laws of Flexural Rupture of flood. Hokkaido University,Sapporo, Japan. March 1949.

LiNorris, C. B. and LooKinnon, P. F. Compression, Tension, and Shear Tests onYellow-Poplar Plywood Panels of Sizes That Do Not Buckle with Tests Madeat Various Angles to the Grain. Forest Products Laboratory Report No.1328. 1946.

..Norris, C. B. Strength of Orthotropic ivlaterials Subjected to CombinedStresses. Forest Products Laboratory Report No. 1816. 1950.

Rept. No. 81910 -2-

to the test beam through 2- by.2-1/2-inch, flat, wood bearing plates. Thebearing plates were built into loading yokes that had roller bearings on whichthe loading blocks rested. rtoller-bearing supports were also bolted 2 inchesfrom each end of the beam. The purpose of the roller bearings was to keep theend reactions and the application of the load in a vertical plane. The de-flection of the beam at the center line was determined over a length of 14inches. Since the bending moment over the 14-inch length was constant, thedeflection curve of the beam was circular. Two dials at the center of thebeam measured the bending deflection. Dials at the end of the beam and atthe point of application of the load measured the change in moment arm.

The compression test was performed according to American Society for TestingMaterials procedure. The test set-up is shown in figure 3. Each of the testspecimens was taken from the beam as shown in the cutting diagram of figure it.

The ends of the compression specimens were dried, so that failure would occurwithin the gage length. Curves obtained from the compression test were carriedwell beyond the maximum load in order to get a complete picture of the stress-strain relationship.

Method of Analyzing Data

Test results from the compression specimens, when plotted, produce a stress-strain curve as shown in figure 5. The stress-strain values for each of thecompression specimens tested are indicated by the dots on either side of theconstructed average-value curve. If the tension stress-strain curve isassumed to be tangent to the compression curve at the neutral axis and ofstraight-line variation, a complete stress-strain curve will result, whichresembles that of figure 5. If the wood in the beam behaves similarly tothat in the compression specimen, this stress-strain curve (or parts of it)can be used to show stress distribution in the beam.

When the complete stress-strain curve is used as a stress-distribution curvefor the beam, a load-deflection curve can be computed in the following manner:A line parallel to the stress axis is placed at random on the compressionside of the curve and extended from the y-axis to the compression curve. Thearea under this line is measured or computed. Another line is drawn parallelto the stress axis and extended from the y-axis to the tension curve at sucha distance from the stress axis that the tension area will equal the com-pression area. A scale is then chosen so that the distance on the y-axisbetween the two lines parallel to the stress axis is equal to the depth ofbeam for which the load-deflection curve is desired. The bending moment iscomputed using the compression and tension areas. The load on the beam canthen be determined from the bending moment. This process is repeated a numberof times, choosing lines parallel to the stress axis at various distances fromthe origin of the curve.

To determine the deflection of the beam, the radius of curvature must firstbe computed from the following equation:

1 E

R = d t

E + E C

Kept. No. h1910 -.3-

where R = radius of curvature

d = depth of beam

E t = strain at the bottom of the beam as determined by the line fromthe y-axis to the tension curve. (Taken as positive).

strain at the top of the beam as determined by the line fromthe y-axis to the compression curve. ( Taken as positive)•

The radius of curvature can then be substituted in the following equation todetermine the deflection of the beam:

L' 2

8 _ + 82R

where A = radius of curvature

L = increment of length over which the deflection is determined.

6 = deflection

Since the 6 2 term is small in comparison to L2, its value is of little signifi-

cance and may be omitted, resulting in the approximate equation for,deflection

L28 -87

Table 1 shows the computed loads and deflections associated with the stress-strain curve obtained from the compression tests (fig. 5). These computedload-deflection values were plotted and compared with an actual load-deflectioncurve obtained from the static-bending test (fig. 6). The computed curve issmoother than the measured curve in the upper regions, because the compressionspecimen had but one failure, while the bending specimen used for the measuredcurve had a number of small failures, spaced about 2 inches apart throughoutthe central part of the beam.

A notable feature of the computed curve is that the maximum compressive stressoccurs in the uppermost fiber of the beam at a load that is less than theapparent proportional-limit load. The explanation of this is that the pro-portional limit of the wood is first reached only in the uppermost fiber ofthe beam and has little effect on the deflection of the beam. As the load isincreased, more and more fibers are subjected to stresses equal to or greaterthan the proportional-limit stress. A point is reached where enough of theupper fibers are stressed beyond the proportional limit to affect the deflec-tion of the beam measurably. The load at this point is slightly higher thanthat associated with the maximum compressive stress. It might be possible,by use of very sensitive deflection gages, to determine the proportional limitat some lower load value than is indicated by the curve.

Comparison of the two curves of figure 6 indicates that the stress-strainrelationship in a beam is substantially the same as that in a compressionspecimen.

Rept. No. 81910 -4-

Part II

Strength of Beams

The general aim of this study is to determine the accuracy of the theoreticalequation derived in Appendix I. The equation can be used to predict the typeof failure and the loads that cause failure in beams under varying span-depthratios. The bending strength of various Sitka spruce beams was determined topermit comparison of the test values with the computed values of bendingstrength obtained with the equation.

Description of Test Material

The test material was Sitka spruce received in 1943 from the Tongass NationalForest in Alaska. The planks were in an air-dried condition when selected,having originally been kiln dried and then stored in the open. Only clear,straight-grained planks were used.

For each span-depth ratio investigated, one bending, two shear, two com-pression, and two tension specimens were tested. The specimens for eachseries of tests were matched as shown in figure 7. Each specimen was markedwith the span-depth ratio corresponding to the bending-test specimen.

Test Procedure

The span-depth ratio of the bending specimen was first varied by keeping thedepth constant and changing the span. Two complete series of tests mere run,one using center loading and one using two-point loading. In the next seriesof tests, two-point loading was used, and the desired span-depth ratio wasobtained by varying the depth of the beam and keeping the span constant. Fortwo-point loading, the span is equal to the distance between end reactionsminus the distance between loads.

The static-bending test was made in accordance with American Society forTesting Materials procedure. The test set-up for two-point loading is shownin figure 8, and the test set-up for center loading is shown in figure 9.The reason for using two-point loading was to alleviate some of the crushingeffect of the loading block on the upper fibers of the beam. , This crushingeffect was especially noticeable on the beams having the smaller span-depthratios.

The tension test was performed on a test specimen designed by the ForestProducts Laboratory. The dimensions of the specimen are shown in figure 10,and the test set-up is shown in figure 11.

Tests of shear parallel to the grain were performed in accordance with AmericanSociety for Testing Materials procedure. The test set-up is shown in figure12.

Rept. No. R1910 -5-

Tests of compression parallel to the grain were also made in accordance withAmerican Society for Testing Materials procedure. The test set-up is sham infigure 3. It was necessary to carry the stress-strain curve well beyond theultimate strength in the earlier tests, because it was not clear just howthe simplified stress-strain curve should be drawn for mathematical analysis.Later, it was determined that the simplified curve should be drawn throughthe maximum stress, and a complete curve was therefore no longer necessary.The ends of the compression specimens were oven dried before testing to assurea failure within the gage length.

All specimens tested were conditioned at 75° F. and 6L percent relativehumidity until they were of constant moisture content. Equilibrium moisturecontent of the specimens was approximately 13 percent.

Mathematical Analysis'

The derivation of the theoretical equations in Appendix I is based on thesimplified longitudinal-stress distribution • shown in figure 1. The compressionportion of the longitudinal-stress distribution is based on figure 5. Formathematical analysis, the complete stress-strain curve of figure 5 is simpli-fied by the assumptions that the proportional limit coincides with the maximumstress and that the stress is constant at greater strains. This results in thecurve as shown in figure 1 and by the dotted line in figure 5. The tensionportion is the same as that of figure 5. Using the simplified longitudinal-stress distribution, the shear-stress distribution is obtained by consideringa small differential block in the beam as shown in figure 13. When the blockis placed in longitudinal equilibrium, the variation of shear stress is foundto be parabolic in form. Figure 1 shows parabolic variation of shear stressas obtained from the engineering analysis given in Appendix I. A more accurateshear distribution resulting from a concentrated load was determined bySmith and Voss.6

The beam can be made to fail in three ways: (1) A tension failure, in whichthe bottom fibers of the beam fail first; (2) a shear failure, in which thebeam fails at some depth from the bottom of the beam; and (3) a combination ofshear and tension. In the lower part of the beam, the wood is subjected totension and shear, and failure will occur somewhere within the beam wherethese combined stresses are sufficient to satisfy the function _which isdefined by equation (8) of Appendix I. When this function reaches unity,failure will occur.5 The function 0 has various values at different positionsin the beam and will first reach 1 at the position of its maximum value.The position of its maximum value may change as the load on the beam isincreased.

Strength values obtained from the various tests performed on the beam and onthe other specimens were used in the calculation of theoretical bending-strength values. Sample problems are shown in Appendix II. Comparison of the

-Smith, C. B. and Voss, A. W. Stress Distribution in a Beam of OrthotropicMaterial Subjected to a Concentrated Load. National Advisory Committeefor Aeronautics Technical Note No. 1486. 19413.

Rept. No. R1910 -6-

theoretical values with the measured strength values substantially verifiedthe assumed stress distribution.

When equation (10) of Appendix I is used, a value of n is assumed, and thenvarious values of y are chosen to see if the combined shear and tensionfunction 0 equals 1 at any point in the beam. Figure 14 shows a typicalcurve obtained when the beam fails in tension. Figure 15 shows a typicalcurve obtained when the beam fails in combined tension and shear. Thecurve of figure 16 was obtained by proper choice of span-depth ratio. Thiscurve shows that the beam is just as likely to fail by combined tension andshear at some distance from the extreme fiber as by tension in the extremefiber. The curve also shows that the beam is close to failure at any pointbetween these two locations. It is observed from the curves that a smallchange in n results in a large change in 1 and that a decrease in n willincrease the value of 0.

Some general statements can be made in regard to the use of the equationsderived in this report. When the span-depth ratio is about 14, the failureof the beam will be in tension, with the shear stress playing a minor role.For these conditions, equation (7a) can be used to find the value of m atfailure. To find the bending strength, this value of m is multiplied by thecompressive strength. When the span-depth ratio is less than 6, the failureof the beam will be in shear, with the tension stress playing an unimportantrole. For these conditions, equation (6b) can be used to find m, and thebending strength can be determined in the same manner as for the tensionequation. Sample computations are given in Appendix As the span-depthratio decreases from 14, the results from the tension equation increase inerror. Also, as the span-depth ratio increases from 6, the results from theshear equation increase in error. Table 2 shows ratios of bending strengthcomputed by the tension and shear equations to the measured bending strengthof the beam. These values indicate that, if the bending strength is computedby both the tension and shear equations, the smaller value obtained will bevery close to the value obtained by equation (10). The greatest use of thecombined equation (10) is between the span-depth ratios of 7 and 10, wherethe error from the tension or shear equations becomes large.

Discussion of Results

Table 2 shows the various strength properties of the specimens tested and themeasured bending strength as compared to the computed bending strength. Theexperimental values confirm the theory as closely as can be expected.

The theoretical value of bending strength was, in most cases, higher than themeasured value. This was due to the use of the simplified stress-strain curve.The area under the simplified curve is slightly larger than the area underthe actual stress distribution curve shown in figure 5. This increase in areawill tend to cause computed values to be slightly higher than the measuredvalues.

Figure 17, 18, 19, and 20 show that, in general, the best beams failed nearthe locations of the concentrated loads. These locations are indicated inthe photographs by the depressions caused by the loading blocks. The theorydeveloped does not apply accurately at these locations, because the load is

Rept. No. R1910 -7-

spread over the area of contact between the loading block and the beam; thus,the total shear decreases from one edge of this area to zero at the otheredge for the two-point loading and to zero at the center of the beam for thecentral loading. The distribution of shear across the depth of the beam neara concentrated load is also somewhat different from that determined by theanalysis given in Appendix 1.6 Furthermore, the vertical stresses in the beamcaused directly by the application of the loads were neglected in the analysis.

Nevertheless, the equations developed seem to predict the strength of the beams

with reasonable accuracy.

Conclusions

1. The stress-distribution verve a mood beam of rectangular cross section

is similar at a distance from concentrated loads, to the stress-straincurve obtained from a compression test of the same material.

2. The strength values of such beams computed on the basis of this stressdistribution (slightly simplified) agree reasonably well,when the resultingshear distribution is taken into account, with the strength values deter-mined by test.

Rept. No. 81910 -8-

Appendix I

Derivation of General Equations

The longitudinal-stress distribution shown in figure 1 will be assumed. Fromthe equilibrium of longitudinal forces:

1Fe db = -f a (Fe + ao ) b

2 F0 da -

Fe ao

where Fe = the compressive strength

d = the depth of the beam

b = the width of the beam

= the tensile stress at the bottom of the beam

a = location at which proportional limit in compression is reached.

From the straight-line stress distribution in. the lower part of the beam:

a = ao - Kz

where z = a variable distance measured upward from the bottom of the beam

K = slope of the curve of this equation

a = the tensile stress at any value of z

When z = a, u = - Fc

-Fc = a - Ka

and ao + FcK =

a

Therefore, with the use of equation (1),

( °- ,z) Fc)

2

K2 Fe d

and

( act Fc)2 = t7

0

(1)

(2)2 Fe d

±ept. No. 81910 -9-

The bending moment M is obtained from figure 1 by taking moments about apoint at the bottom of the beam.

M=b Fc d2a

(FM = - b c +2 o 3

or

M=

- 6 (Fc co)b Fcd2 bat

ffith the use of equation (1),

b Fcd2 2 b Fc2d2

M = 3 Fc +0-02

or

1 2 FcM = F d -

c Fc ac

All of the values in equation (3), with the exception of ao, are independent

of the longitudinal position in the beam in the range in which the beamstarts to fail in compression; that is, in the range in which the stress dis-tribution is given by figure 1.

The shear load S is given by:

dix( 2 2 Fc d a 0S = = Fe bd ) dx

or

2 2 Fc2 d‘7- 0

S = bd (Fc a0)2 dx

Rearrangement of the last equation gives:

dc- 0 3 (Fc + a0)2

dx = 2 S

Also, S = bds, where s is the average shear stress, and therefore

dac _ 3 (Fco-0)2_ a

dx 2 d Fc2

Figure 13 shows a small differential block in the beam. Its length is Ax,its depth is z, and its width is b. The tensile force F on the left end ofthis block is given by:

(3)

(4)

Rept. No. R1910 -10-

(ZF= b a dz

With the use of equation (2),

( a0Fc)2F )12

[

z dz

o 2 Fe d

or

F = b 0-oz - ( cro + Fc)2

z2]

[

4 Fe d

Also, from longitudinal equilibrium of the small block in figure 3,

Tbax =

or

1 dF=

where T is the shear stress.

When the equation for F above is used,

d o- 0 d o0T = z ao + Fe

z2dx 2 Fe d dx

or

T = r7 _ 0-0 + Fc 2do- o

2 Fe d z i dx

With the use of equation (4),

a + Fo c 2 I 3 °o Fc) 2

2 Fez

d 2s d Fc 2

Thus, the variation of shear stress r is parabolic in form, and T = 0 whenz = 0 and also when

The maximum value of r occurs when

L. = + Fo c

Fc

3 s

( ao Fc)2 =0dz d 2 dFc'

Rept. No. 81910

T =

2Fe dz - =a.

Fc + 0-0 -

(5)

1 Fe d 13 ( cro + Fc ) 2

2a+F 2 s d Fc2c

[

d=max o + Fca

Fe 3 3 Mit 2 71773d

or when

Fcd a

The dotted curve in figure 1 shows this distribution.the shear stress 7 is:

The maximum value of

or3 cro Fc

max ' 3 F

Equations (2), (5), and (6) contain

the beam. It is more convenient toof the bending moment. This can be

the tensile stress co at

express the value of thisdone by using equation (3

(6)

the bottom of

stress in terms):

M 1 2 Fc

Fc bd2 2 3 Fe 4- a

2 Fc 1 M

3 Fe + ac 2 Fc bd2

Fc 3 3 M

Fe + ao 4 2 Fc bd2

Fc + 1

0

a o 1 1 Fc 3 3 M

4 2 Fe bd2

1 11c- 0 2 Fc T2Dd

Fc

3M2 Fe bd2

Rept. No. 81910 -12-

1 + 6 M mo o = E FC bdz

Fc

Fc bd2

.M1 + 6

o Fc bd2

Fe 3 6M Fe bd2

(7)

(2a)

Fc bd2a = Fe 6M3 - Fe bd2

For simplification, let

z _a - Y

=tFe

aoFe

6MFe bd2 m

T= q

S

where y = the distance measured upward from the bottom of the beam divided

by' the depth..of the beam.t = the tensile stress at any value of z divided by the oompressive strength

to = the tensile stress at the bottom of the beam divided by the compressivestrength

m = the bending Stress computed by means of the usual engineering formuladivided by the compressive strength

q = the shear stress at any value of z divided by the average shear stressin the beam

Equations (2), (5), (6), and (7) then become:

1t = to - 7 (to + 1) 2 y

Rept. No. R1910 -13-

3 [ 1q = y - ( to + 1 ) y21 [to + 1 2 (5a)

qmax = Oto 1)

1 + mto 3 - m

From equation (7a),

1 + mto + 1 =

3 - m

1 + m + 3 - m 4+ 1 =

3 - m 3 - m

(6a)

(7a)

with the use of this equation and equation (7a), equations(6a) become:

1 + m 8 t =

3 - m (3 - m) 2 Y

24 q-

(3 - m) 2 [Y 3

3

qmax = 3 - m

(2a), (5a), and

2 - y2

(2b)

(5b)

(6b)

If beam failure is due to shear, the modulus of rupture can be found byrearranging equation (6b) to give:

m - 3 (qmax - 1)

qmax

where q -is placed equal to qs , the shear strength divided by the average

shear stress in the beam at failure and m becomes the modulus of rupturedivided by the compressive strength.

If the beam fails due to tension in the extreme fibers, the modulus of rupturecan be found by rearranging equation (7a) to give:

3 to - 1m- to + 1

where to is placed equal to ts , the tensile strength divided by the compressive

strength, and m becomes the modulus of rupture divided by the compressivestrength.

Rept. No. 81910 -14-

In the lower part of the beam, the wood is subjected to tension and shear,and failure will occur somewhere within the beam where these combined stressesare sufficient to cause failure. The function that defines failure is:

(8)

where Ft = the tensile strength of the materialFs . the shear strength of the material

When this function reaches unity, failure will occur:.1 The function has variousvalues at different positions in the beam, and at any section of the beam, itvaries with z. It will first reach unity at the position of its maximum value.Equation (8) can be put in the form:

t \2 + 2

(6a)ts

where is = the longitudinal tensile strength divided by the compressivestrength.

qs = the shear strength divided by the average shear stress in the beam.

dith the use of equations (2b) and (5b) equation (8a) becomes:

r + m

= I1 2 I 24 2

2)

(9)1 2

is (3 m) is (38 - 111)2 i [cis - ra)2 (Y - 3 - m.J

Let n = 3 - m then

m = 3 - n

1 m = 4 - n

At failure,0 = 1, and therefore, from equation (9),

[4 - n 8

I ts n _ is n2Y

1:—"

242 2\, - Y =1

qs n (io)

Equation (10) is solved by assuming, a value of n and substituting differentvalues of z in the equation to determine if it Ban be satisfied with theassumed n. This procedure is followed until the proper value of n is found.The value of m at failure can then be determined from the relationshipm=3-n. If this value of m is multiplied by the compressive strength F0of the wood, the bending strength Fb is obtained. Refer to table 3 for

values of constants for different values of n.

Rept. No. R1910 -15-

Appendix II

Sample Computations Using Equation (10)

Equation (10) is solved by a trial and error process. The value of F b is

guessed. This results in a definite value for n, the values F c , Ft , and Fs

being known. If the proper value of Fb is chosen, substitution in equation

(10) will lead to this same value of Fb.

The last step, in which the chosen value of Fb leads to this same value, isillustrated in the following examples. —

Example 1:

Computation of bending strength of beam No. 5, table 21 `Two-point loadingwith 6-inch spacing between loads. Span-depth ratio /

IL equal to 14.d

Data from tests of beam and minors

Fe =Fb =

=FtF =

6,020 pounds per square inch -11,940 pounds per square inch17,675 pounds per square inch1,322 pounds per square inch -

- Average for two test specimens-- Assumed-- Average for two test specimens- Average for two test specimens

t !I _ 172_675 = 2.936s Fc6,020

A -

Fs - L = x ,322 x 14 = 4.65qs 11,940

Fb 11 940 - 1.016n = 3 - T = 3 6020

Then n2 = 1.032

n 4 - 1.0162.936 x 1.016

8 8 t n2 - 2.936 x 1.032

- 2.64

= 1

Rept. No. R1910 -16-

24 24 ---- n 7 4.65 x 1.032 5.00qs

2 2r7 = rzEg - 1.968

n 8 2 - 4 2[4ts 71 2Y

[2q Z AY

2 2)Y = 1 =

(1.000 - 2.64 y)2 + (5.00y - 9.84 Y2 ) 2 = 1 =

Values of and 0 for n = 1.016 are as follows (see figure 14 for typicalcurve for failure of a beam in tension):

y = 0

y = 0.05y = 0.10y = 0.15y = 0.20y = 0.25y = 0.30

= 1.000• = 0.804• = 0.704

= 0.666• 0.593

= 0.5200 = 0.425

when n = 1.016m=3 - n = 3 - 1.016 = 1.984

Therefore, the computed bending strength for = 1 is

Fb = Fein = 6,020 x 1.984 = 11,940 ig111143uare inch

Test value of Fb = 11,090/Per a:quare inch11 940 - 11,090

Error' 119090 x 100 = 7.67 percent

If the short equation (7a) is used to find the computed bending strength,then

Sts - 1 (3 x 2.936) - 1

m=

ts + 1 2 ,,936 + 1 -1.984

/11=zFb = Fel = 6,020 x 1.964 = 11,940 square inch

Therefore, the short equation gives the sum answer as equation (10).

Rept. No. 81910 -17-

= 5,760 pounds per square inch --= 17,580 pounds per square inch -= 1,329 pounds per square inch --= 10,550 pounds per square inch -

FcFtFsFb

Average for two test specimens- Average for two test specimensAverage for two test specimens

-- Assumed

From equation(6b),

3(q - 1) 3(4.65 - 1)m = s . 2.35qs 4.65

Fb = Fcm = 6,020 x 2.35 = 14,150 pounds per square inch

error = 14,150 - 11,090 11,090

x 100 = 27.6 percent11

Example 2:

Computation of bending strength of beam No. 8, table 2. Two-point loadingwith 6-inch spacing between loads. Span-depth ratio equal to 7.

From tests of beam and minors,

t = Ft = 17,580 = 3.05S 17c 00

qs 3 x TC). x

5,7

Fs a = 3 x 0

11,355029 x 7= 2.646

,

n 3 _ Fb 3 - 10,550- 1.17F 5,760

Then n2 = 1.369

4 - n 4 - 1.17t

sn - 3.05 x 1.17

= 0.793

8 8F - 1.916

t n2 - 3.05 x 1.50

24 _ 24- 6.63asn2 2.646 x 1.369

2 2= = 1.709

n 1.17

Rept. No. R1910 -18-

4 — n 8 vt777t,,. n 2 r

+ 24----f (Y Y) = 1 =4cis n

2 2 2

(0.793 - 1.916 y) 2 + (6.63 y - 11.32 y2 ) 2 = 1 = 4)

areThe following/values of and 4) for n = 1.17 (see figure 15 for typical

curve for failure of a beam in combined tension and shear):

y=0 = 0.629

y = .0.05

(I) = 0.578y = 0.10

= 0.663y = 0.15 = 0,803

y = 0.20

4) = 0.931y = 0.25 = 1.000

y = 0.30 = 0.992

When n = 1.17

m = 3 - n = 3 - 1.17 = 1.83

Therefore, the computed bending strength forci5= 1 is

poundsFb = Fcm = 5,760 x 1.83 = 10,550/per square inch

poundsTest value of Fb = 10,500/Per square inch

Error = 10

,550 — 10

'500

x 100 = 0.48 percent10,500

If the short equation (6b) is used to find the computed bending strength, then

3 (qs — 1) 3 (2.646 — 1) m = = 1866

qs 2.646

poundsFb = Fcm = 5,760 x 1.866 = 10,760/per square inch

The value obtained from the short equation is slightly higher than that ob-tained from equation (10). This could be expected, since the tension partof the equation is not considered in computing the value of m.

Rept. No. R1910 -19-

Table 1.- -Load-deflection values for a wood beam computed from the stress-strain curve for compression specimens,

Strain on compres-: Strain on tension: Computed load on : Computed deflectionsion side of beam : side of beam beam • of beam

Inch Inch

0.001570 :0.00157

.00171 : .001710 :

.00185 : .001848 •.

. :

.00199 .001987 :

.00211 .002104 :

.00225 .002243

.00240 .002392

.00255 ...002540

.00269 .002678

.00285 .002832

.00303 •. .003006

.00331 •. .003268

.00349 .003429

.00370 .003598

.00400 .003810

.00430 .004001

.o0460 .004180

.00520 .004518

.0058o •. .004829

.00640 •. .005120

.00700 : .005395

.00760 : .005651

.00820 : .005900

Pounds Inch

571 0.0385

622 .0419

672 .0453

722 .0487

765 : .0516

815 : .0550

869 .0587

922 :..

.0624

971 .o658

1,026 .0696

1,0e7 .0739

1,178 .o8o6

1,230 : .0847

1,278 .0894

1,329 .o957

1,368 : .1017

1,40o •. .1076

1,456. : .1190

1,503 : .1320

: 1,544 .1411

: 1,579 .1518

: 1,609 .1624

: 1,636 .1726

Rept. No. R1910

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n2 : 4 n 8 24 2n 2 n2n

o.90 0.810 3.444 9.877 29.631 . 2.222.91 .828 • 3.396 : 9•662 28.986 . 2.198

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1.03 : 1.061 2.883 7.540 22.620 1.9421.04 : 1.082 2.846 7.394 22.182 1.9231.05 - 1.102 2.810 . 7.260 21.780 1.9051.061.07 :

1.1241.145

2.7742.738 •.

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21.35120.961

1.8871.869

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6.7346.612

20,20219,836 1.818

1.111.12

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2.6042.571

6.496.380

19.48219.140 :

1.8021,,786

1.13 1.277 2.540 : 6.265 18.795 1,770

1.14 1.300 2.509 6.154 18.462 1.754

1.15 : 1.322 2.478 : 6.051 18.153 1.739

1.16 1.346 2.448 : 5.944 17.8321.359 2.419 5.844 17.532

1179:1,7091.17

1.18:: 1.392 2.390 : 5.747 : 17.241

1.19 1.416 2.361 : 5.650 16 950 : 1.6811.20 1.440 2.333 : 5.556 : 16.668 1,667

Rept. No. R1910

Figure 2.--Set-up for test of static bending with quarter-point loading. Thetwo indicators at the left of the center of the span measure the change inmoment arm, and the indicators at the center of the span measure the deflec-tion. Distance between the loads is 28 inches.zx 83484 F

Figure 3.--Set-up for test of compression parallel to grain,showing the spring-suspended, spherical bearing block, aIambs' roller extensometer of 6-inch gage length attached to2- by 2- by 8-inch specimen, the projection lamp with crosshairs, and the rotating drum with data sheet on which load-deflection readings are plotted.

Z}I 27176 F

Figure 11.--Set-up for test of tension parallel to grain, usingthe Forest Products Laboratory test specimen.

zu 66832 IT

Figure 12.--Set-up for test of shear parallel to grain, using thestandard shear specimen.

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