United States 

Department of 

Agriculture

Forest Service

Forest

Products

Laboratory

Madison

Wisconsin

Research

Paper

FPL 343 

May 1979 

Analysis of

Orthotropic

Beams
ABSTRACT

A plane-stress analysis of orthotropic or isotropic beams is presented. 

The loading conditions considered are: (1) a concentrated normal load 

arbitrarily located on the beam, and (2) a distributed normal load 

covering an arbitrary length of the beam. 

similarly be considered. 

exhibit close agreement with existing experimental data from Sitka 

spruce beams. 

Other loading conditions can 

Numerical results based on the present analysis 

CONTENTS

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Theoretical Analysis . . . . . . . . . . . . . . . . . . . . . . 2

Stress and Load Functions. . . . . . . . . . . . . . . . . . . . 2 

Method of Solution . . . . . . . . . . . . . . . . . . . . . . . 6

Deflection of Beam . . . . . . . . . . . . . . . . . . . . . . . 7

Numerical Results. . . . . . . . . . . . . . . . . . . . . . . . 9

Discussion and Conclusion. . . . . . . . . . . . . . . . . . . . 12 

Appendix I. Formulation of Isotropic Beam . . . . . . . . . . 28 

Appendix II. Simplified Forms of Stress Components . . . . . . 30 

Appendix III. Notation. . . . . . . . . . . . . . . . . . . . . . 34

Page
ANALYSIS OF ORTHOTROPIC BEAMS 

By

1/ J. Y. LIU 

and 2/ S. CHENG 

INTRODUCTION

The solution of the plane-stress or plane-strain problem for an infinite 

strip of orthotropic material was studied by Green (7). 3 / 

tion, both the upper and lower boundaries of the strip must be loaded, a 

situation not commonly encountered in engineering applications. Using a 

similar approach by Lamb (10) for a narrow beam of isotropic material, 

Smith and Voss (12) solved the problem of a beam of either isotropic or 

orthotropic material with a concentrated load applied at the center of 

the span. Conway (2) used polynomial stress functions in his solutions 

of an orthotropic cantilever beam loaded at the free end and a simply 

supported beam under uniformly distributed loads. He also solved the 

problem of a deep beam of orthotropic material symmetrically loaded with 

respect to the beam axis (3). 

to describe the stresses and displacements of orthotropic beams under 

distributed polynomial loads. Hashin (8) developed a method to solve 

the elasticity problem of long plane anisotropic rectangles with continuous polynomial stresses prescribed on the long sides, and force and 

moment resultants prescribed at the ends. The problems analyzed in 

(2,11) were discussed in detail for anisotropic material in (8). 

finite element technique was applied by Hooley and Hibbert (9) to solve 

the plane-stress problem of a wood beam. 

while applicable to intricate boundary conditions, is limited by the 

computer capacity; also, in areas of high stress concentrations where 

interpolation procedures must be followed for stress estimations, the 

method cannot yield accurate results. 

In his soluFunctions were derived by Silverman (11) 

The

This numerical technique, 

1 / Engineer, Forest Products Laboratory, Forest Service, U.S. Dep. 

of Agric., Madison, Wis. 

2 / Professor, Engineering Mechanics Department, University of 

Wisconsin, Madison, Wis. 

3 / Numbers in parenthesis refer to references cited near end of 

report.

----
For narrow, rectangular beams of isotropic material, solutions for 

some specified loading and support conditions can be found in (13). 

Donnell (5) applied infinite series for stresses in his solution with 

the first terms corresponding to elementary theory and the latter terms 

to increasingly minor refinements. However, his solution cannot be 

applied where discontinuities exist, and the most serious such discontinuity is that caused by a concentrated load. Boley and Tolins (1) 

used an iterative procedure to analyze the two-dimensional beam problem. 

The applied loads considered by them may consist of either normal or 

shear forces varying smoothly along the span. 

Experiments were performed by Cowan (4) to determine the horizontal 

shear stress distribution in beams of Sitka spruce. 

the vicinity of a bearing support, the shear stress distributions were 

considerably different from what the simple beam theory would predict. 

The same phenomenon was observed in (12) to exist in the vicinity of a 

load point. 

In the study presented herein the practical loading conditions not 

discussed in the cited literature will be considered, namely: (1) a 

concentrated normal load arbitrarily located on the beam, and (2) a 

distributed normal load covering an arbitrary length of the beam. 

these solutions available, a large number of practical beam problems can 

be handled using the method of superposition. 

He found that in 

With

THEORETICAL ANALYSIS 

The beam considered is assumed to be an orthotropic solid having 

its edges parallel to two perpendicular axes of elastic symmetry. 

thickness of the beam is assumed small as compared with the vertical 

depth so that the problem can be treated as one of plane stress. 

The x-axis is taken along the center line of the beam and the 

y-axis coincident with the left end of the beam as shown in figure 

For the isotropic case, the pertinent derivations are presented in 

Appendix I. 

The

1(a).

STRESS AND LOAD FUNCTIONS 

For the state of plane stress in the orthotropic beam, the stress 

function, o|, must satisfy the following equation of compatibility (12): 

2
where

Equation 1 may be satisfied by taking the function o |in the form 

In which f( η) is a function of η only.

equation 1 and using the notation m = n π / one finds the following 

equation for determining f(η ):

Substituting equation 2 into 

(3)

The general integral of this linear differential equation with constant 

coefficients is 

(4)

3

(1)

(1a)

(1b)

(1c)

(2)
where

and

The stress function is then obtained by substituting equation 4 into 

equation 2, and the corresponding stress components are 

(7)

4

(4a)

(4b)

(6)
in which the constants A

n, Bn, Cn, and Dn are to be determined from the 

loading conditions and τ represents τ

Assuming that the load function can be expressed as a Fourier series, 

xy.

the coefficients a

n in equation 8 can be written as 

Let p (x) be represented by P/2 ξ acting at x = a as shown in figure 1(b). 

Then

For a concentrated load P acting at x = a, one obtains from equation 8b 

Hence, for any specified loading condition, the load function can be 

expressed as a Fourier series following the above procedure to be consistent with the stress function defined by equation 2. 

5

(8)

(8a)
To determine A

n, Bn, Cn, and D 

four boundary conditions are needed. 

figure 1(b). 

in equations 5, 6, and 7, n

The four boundary conditions are: 

Consider the problem described in 

From these boundary conditions, one obtains 

where

6

(1)

(2)

(3)

(4)

(9)

(10)

(11)

(12)

(13)

(13a)

(13b)

(13c)

METHOD OF SOLUTION 
(13d)

(13e)

It is noted that equations 13 are applicable for a 

load function. 

lower side of the beam, all that need be done is to change the sign for 

h in equations 9-12.

The infinite series obtained by substituting equation 13 into 

equations 5, 6, and 7 can easily be solved using a digital computer. 

order to save on computer storage and time, a method of reducing the 

infinite series to the sum of a finite series and a closed form is 

described in Appendix II. 

corresponding to any n

It is also noted that if the load were applied on the 

In

DEFLECTION OF BEAM 

Deflection of a beam may indicate its performance under specified 

loading conditions. It may also serve to derive the reactions in an 

indeterminate situation. From the strain-displacement relationships and 

the generalized Hooke's law, the following equations can be obtained: 

where u and v are the displacements in the x- and y-directions,

respectively. Also, 

7

(14)

(15)
From the stress expressions in equations 5, 6, and 7, it is evendent that in 

equation 16 the following relation must hold true 

Let

Then

Likewise,

8

(16)

(17)

(18)
and

The constants KO

, K1, and K in equations 18 and 19 are to be determined 2

from the known displacements at the supports. 

Usually, only the vertical deflection of the beam center line is of 

interest in engineering applications. One then obtains from equations 5, 

6, 15, and 19, the following expression 

For a beam whose vertical displacements at the two ends are zero, it is 

seen that both K and K

o 2 in equation 20 must vanish. 

NUMERICAL RESULTS 

Numerical calculations were made for the wood beams used in Cowan's 

experiments (4) for shear stress measurements using a Univac 1100 computer. 

In his experiments, a simple beam and a two-span continuous beam were 

used, which were clear straight-grained specimens of Sitka spruce. 

Supports and load blocks also were of Sitka spruce to simulate actual 

construction conditions. 

For the simple beam, the applied load and reactions and the geometrical dimensions from (4) are shown in figure 2. The right support 

is a roller-type support, but the left support is a flat bearing block 

restrained against horizontal movement. Hence, the left support could 

be assumed to approximate a hinged support with resultant of the reaction 

forces acting closer to the load point. The reaction forces at the left 

support are therefore assumed to be distributed as shown in figure 2(b). 

The material properties taken from (4) were as follows: 

9

(20)
6

4 8 2 

E

x = 1.84 x 10 psi (1.2686 x 10 10 N/m2)

E = 6.62 x 10 psi (4.5643 x 10 N/m ) y

5 8 2 G = 1.089 x 10 psi (7.5083 x 10 N/m ) xy

= 0.26 

= 0.0094 

Figure 3 presents the orthogonal shear stress distributions in the beam 

at the left support. 

data are seen to agree closely with each other. 

of course, due to the material heterogeneity and the assumed stress 

distribution at the support. 

the beam are strongly dependent on the stress distribution at the support 

which can only be estimated. 

The vertical compressive stress distributions are shown in figure 4. 

By comparing figures 3(c) and 3(d) with figure 4, it is seen that the 

large shear stresses do not always occur where the vertical compressive 

stresses are high. 

The bending stresses plotted in figure 5 are seen to be highly nonlinear and are negative in both the upper and lower regions of the beam 

above the support. 

Figure 6 shows the applied loads and reactions and the dimensions of 

the two-span continuous beam. The middle reaction is a flat bearing 

block of wood 8 inches long. 

the resultant of the reaction forces was found to be 2 inches on the right 

of the center of the middle support. In the numerical calculations, the 

reaction forces are then assumed to be distributed as described in 

figure 6(b). 

The material properties for this beam were the same as those for the 

simple beam except the modulus of rigidity was found to be 

5 8 2 5 1.121 x 10 psi (7.7289 x 10 N/m ) rather than 1.089 x 10 psi 

(7.5083 x 10 

The shear stress distributions at the middle support are shown in 

figure 7. 

inexactness, considerable randomness exists in the experimental data. 

However, the variations of the experimental shear stress with location 

follow generally the same trend as predicted theoretically. 

The theoretical predictions and the experimental 

The differences are, 

It is to be noted that the stresses in 

However, due to material variabilities, 

8 2 N/m ) according to (4). 

Due to the material heterogeneity and the inherent experimental 

10
The vertical compressive stress distributions plotted in figure 8 

disclose their strong dependency on the assumed reactive forces at the 

support.

of the applied loads and the end supports from the middle support. 

At a given depth of the beam, the nonlinear bending stresses are 

seen to vary only slightly across the length of the support as described 

in figure 9. 

Numerical calculations were also performed for the simple beam of 

Sitka spruce loaded at the center as reported in (12). 

central load, the theoretical predictions of the shear stresses in the 

vicinity of the load point of the present study and of reference (12) 

are essentially the same. In (12), the following function was used for 

a unit load acting on the beam: 

This is, of course, partly due to the relatively large distances 

For a concentrated 

where x is measured from the center of an infinitely long beam along its 

axis and the value of is related to the length covered by the load. 

For = 0, the load is concentrated; for small values of equation 21 

was assumed suitable for representation of a load applied over a small 

area by a curved loading block. 

load was applied over a length of about 0.4 inch (10.16 mm) by a cylindrical 

block.

(0 mm) was 235 psi (1.6202 x 10 

was 116 psi (7.9978 x 10 N/m ), which was close to the measured value. 

Based on the present work, the calculated shear for = 0 inch (0 mm) was 

234 psi (1.6134 x 10 

(1.5789 x 10 

(1.4617 x 10 N/m2).

the length covered by the load must be much larger than 0.6 inch 

(15.24 mm). 

likely to be in error and cannot be properly used to represent a load 

over a specified small length. 

that 84 percent of the area under the curve of equation 21 is over a 

length on either side of x = 0 is wrong. 

length of the percentage is 50; for a length of the percentage 

is 70; and for a length of the percentage is 84. 

In the test reported in (12), the central 

At a point close to the load, the calculated shear for = 0 inch 

6 2 N/m ) and for = 0.25 inch (6.35 mm) 

5 2

6 2 N/m ), for = 0.2 inch (5.08 mm) was 229 psi 

6 2

6

N/m ) and for = 0.3 inch (7.62 mm) was 212 psi 

To reduce the shear stress to the measured value, 

It is believed that the measured value at that point is 

In fact, the statement made in (12) 

It can be shown that for a 

11

(21)
For the isotropic case, the calculated stresses and deflections 

based on the present work agree with the best estimations in (13) for a 

simple beam loaded at the middle. 

DISCUSSION AND CONCLUSION 

The present analysis demonstrates a procedure for calculating the 

stresses and deflection of a two-dimensional orthotropic or isotropic 

beam under arbitrary loading conditions. 

infinite series has been reduced to the sum of a finite series and a 

closed form, making it possible to obtain accurate numerical results 

using a computer of relatively small capacity. 

Numerical results based on the present work indicate that they are 

in reasonable agreement with existing experimental data. For specified 

loading conditions that can be considered by existing theories, e.g., 

(12) and (13), the present method can yield the same predictions as the 

other theories. 

The method is very useful in predicting the stress distributions in 

a beam, especially in the vicinity of a support or a nominal load point. 

However, for the latter purpose, the reactive stresses at a support or 

the acting stresses at a nominal load point must be correctly estimated 

in order to obtain accurate results. 

In the solutions, each of the 

12
Figure 1.--(a) Coordinate system and geometry of beam 

(b) load function and load location on beam. 

(M 146 766) 

13
Figure 2.--Simple beam (a) applied load and reactions; (b) 

assumed compressive forces at left support (1 in. = 25.4 mm; 

1 lb. = 4.4482 N; 1 lb./in. = 175 N/m). 

(M 146 767) 

14
Figure 3.--Variations of shear stress vs. depth 

from top of simple beam (1 in. = 25.4 mm; 

1 psi = 6.8947 x l0 3 N/m2).

(M 146 768, 146 769, 146 770, 146 771) 

15
Figure 3.--Variations of shear stress vs. depth 

from top of simple beam (1 in. = 25.4 mm; 

1 psi = 6.8947 x 10 3 N/m2). (Continued)

(M 146 768, 146 769, 146 770, 146 771) 

16
Figure 3.--Variations of shear stress vs. depth 

from top of simple beam (1 in. = 25.4 mm; 

1 psi = 6.8947 x 10 3 N/m2). (Continued)

(M 146 768, 146 769, 146 770, 146 771) 

17
Figure 3.--Variations of shear stress vs. depth 

from top of simple beam (1 in. = 25.4 mm; 

1 psi = 6.8947 x 10 3 N/m2). (Continued)

(M 146 768, 146 769, 146 770, 146 771) 

18
Figure 4.--Theoretical predictions of vertical compressive 

stress vs. distance from left end of simple beam; at left 

end of support x = 4.46 inches; at right end of support 

x = 10.46 inches. (1 in. = 25.4 mm; 1 psi = 6.8947 x 10 3 N/m2).

(M 146 772) 

19
Figure 5.--Theoretical predictions of bending stress 

vs. depth from top of simple beam (1 in. = 25.4 mm; 

1 psi = 6.8947 x 10 3 N/m2).

(M 146 773) 

20
Figure 6.--Continuous beam (a) applied loads and reactions; 

(b) assumed compressive forces at middle support 

(1 in. = 25.4 mm; 1 lb. = 4.4482 N; 1 lb./in. = 175 N/m). 

(M 146 774) 

21
from left end of continuous beam (1 in. = 25.4 mm; 

1 psi = 6.8947 x 10 3 N/m2).

(M 146 775, 146 776, 146 777, 146 778) 

22

Figure 7.--Variations of shear stress vs. distance 
Figure 7.--Variations of shear stress vs. distance 

from left end of continuous beam (1 in. = 25.4 mm; 

1 psi = 6.8947 x 10 3 N/m2). (Continued)

(M 146 775, 146 776, 146 777, 146 778) 

23
Figure 7.--Variations of shear stress vs. distance 

from left end of continuous beam (1 in. = 25.4 mm; 

1 psi = 6.8947 x 10 3 N/m2). Continued) 

(M 146 775, 146 776, 146 777, 146 778) 

24
Figure 7.--Variations of shear stress vs. distance 

from left end of continuous beam (1 in. = 25.4 mm; 

1 psi = 6.8947 x 10 3 N/m2). (Continued)

(M 146 775, 146 776, 146 777, 146 778) 

25
Figure 8.--Theoretical predictions of vertical compressive 

stress vs. distance from left end of continuous beam; at 

left end of middle support x = 77.5 in.; at right end of 

middle support x = 85.5 in. (1 in. = 25.4 mm; 1 psi = 

6.8947 x 103 N/m2).

(M 146 779) 

26
Figure 9.--Theoretical predictions of bending stress 

vs. depth from top of continuous beam (1 in. = 25.4 mm; 

1 psi = 6.8947 x 10 3 N/m2).

(M 146 780) 

27
APPENDIX I.--FORMULATION OF ISOTROPIC BEAM 

For an isotropic beam, one cannot obtain a solution by simply setting 

α = β = ε = 1 in equations 13 and then substitute them into equations 5, 

6, and 7, since this will cause the constants An, Bn, Cn, and Dn

in these 

equations to become indeterminate. To effect the conversion, one must 

set ε = 1, β = 1 and then differentiate D, Da, Db, Dc, and Dd in equations 13 

α 

with respect to α twice before setting α = 1. 

manipulations, the following results are obtained: 

After some tedious algebraic 

where

28

(I-1)

(I-2)

(I-3)

(I-4)
(I-5)

(I-7)

(I-8)

in which an is expressed by equation 8c. 

It is to be noted that if the stress function for the isotropic 

case as described in reference (13) were used, one would have obtained 

the same results for the same loading condition as shown in figure lb. 

This also serves to confirm that the derivations for the orthotropic 

beam in the present study are correct. 

The vertical deflection of the center line of the isotropic beam 

can be shown to take the following form: 

(I-9) 

which corresponds to equation 20 for the orthotropic beam. 

29

(I-6)
APPENDIX II.--SIMPLIFIED FORMS OF STRESS COMPONENTS 

It is possible to reduce the infinite series for each stress component 

to the sum of a finite series and a closed form. To do this, one can 

first express the hyperbolic functions in terms of exponential functions 

and then, after some simplifications, reduce the infinite series comprising 

exponential functions and trigonometrical functions to closed form. 

A. ORTHOTROPIC BEAM 

It can be shown that equation 13 can be reduced to 

(II-2)

(II-3)

(II-4)

when e is negligibly small compared with unity. 

and 7 can now be expressed in the following form when a 

is used: 

Equations 5, 6, 

n in equation 8c 

(II-5)

30

(II-1)
(II-6)

(II-7)

For other expressions of a 

the procedures of derivation remain the same. 

From reference (6), the following identities can be derived: 

the above equations need to be modified but n,

where

(II-8)

(II-9)

31
It can be seen that the series in equations II-5, II-6, and II-7 take the 

same form as either equation II-8 or equation II-9. Therefore, the stress 

components can all be reduced to closed form, which, though complicated, 

can easily be programed for a computer. 

It is to be noted that the first terms in equations 5, 6, and 7 must be 

retained until n is so large that e becomes negligibly small when 

compared with unity, and that the same number of terms must be deducted 

from equations II-5, II-6, and II-7 before the closed form expressions are 

added to the finite series. All these can easily be handled in a computer 

program.

B. ISOTROPIC BEAM 

For the isotropic beam discussed in Appendix I, the following 

results are obtained: 

when e is small compared with unity. 

equation 8c, the stress components can be written as 

Again, with a represented by n

32

(II-10)

(II-11)

(II-12)

(II-13)

(II-14)
(II-15)

(II-16)

To reduce the infinite series in these equations to closed form, in 

addition to equations II-8 and II-9, the following identities, also 

derived from (6), will be needed: 

(II-17)

(II-18)

where

The numerical procedures are the same as in the orthotropic case. 

33
APPENDIX III.--NOTATION

The following symbols are used in this paper: 

A

n, Bn, Cn, Dn = constants of integration; 

a = x-coordinate of a load point; 

n

a = Fourier coefficients; 

Ex, E = moduli of elasticity in x- and y-directions, y

respectively;

f(h), r(y), s(x) = functions; 

xy

G = modulus of rigidity associated with 

h = half depth of beam; 

K

o, K 1, K 2 = constants; 

= length of beam; 

m = 

n = integer; 

P = applied load; 

p(x) = load function; 

u, v = displacements in x-and y-directions, respectively; 

x, y = Cartesian coordinates 

34
| o= stress function; 

= Poisson's ratio associated with 

stress in x-direction and strains in xand y-directions;

x, y 

σ σ = normal stresses in x- and y-directions, respectively; 

τ = orthogonal shear stress. 

35
REFERENCES

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1956. On the stresses and deflections of rectangular beams, 

Journal of Applied Mechanics, ASME trans., Vol. 23, No. 3, 

p. 339-342, Sept. 

2. Conway, H. D. 

1952. Bending of orthotropic beams, Journal of Applied Mechanics 

ASME trans, Vol. 19, No. 2, p. 227, June. 

3. Conway, H. D. 

1953. Some problems of orthotropic plane stress, Journal of 

Applied Mechanics, ASME trans., Vol 20, No. 1, p. 72-76,

Mar.

4. Cowan, W. C. 

1962. Shear stress in two wood beams over wood block supports, 

U.S. For. Prod. Lab., FPL Rep. No. 2249, Aug. 

5. Donnell, L. H. 

1952. Bending of rectangular beams, Journal of Applied Mechanics, 

ASME trans., Vol. 19, No. 1, p. 123, Mar. 

6. Edwards, J. 

1922. Treatise on integral calculus, Vol. II, Chelsea Publishing 

Co., New York. 

7. Green, A. E. 

1939. Stress systems in aeolotropic plates, part 2, Proceedings, 

Royal Society of London, Vol. 173, p. 173-192.

8. Hashin, Z. 

1967. Plane anisotropic beams, Journal of Applied Mechanics, 

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9. Hooley, R. F., and P. D. Hibbert. 

1967. Stress concentration in timber beams, Journal of Structural 

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April.

10. Lamb, H. 

1909. Flexure of a narrow beam, Atti IV Cong. Intern. Matemat. 

(Rome), Vol. 3, p. 12-32.

36 2.5·5/79
11. Silverman, I. K. 

1964. Orthotropic beams under polynomial loads, Journal of 

Engineering Mechanics Division, Proceedings of ASCE, Vol. 90, 

No. EM5, p. 293-319, Oct. 

12. Smith, C. B. and A. W. Voss. 

1948. Stress Distribution in a beam of orthotropic material 

subjected to a concentrated load, Nat. Adv. Comm. Aeron. 

Tech. Note 1486, Mar. 

13. Timoshenko, S., and J. N. Goodier. 

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