THE DISTRIBUTION Of SWISS IN
A WOOD POLE I3ENT
A RADIAL END LOAD
February 1953
INFORMATION REVIE\NED
AND REAFFIRMED
1965
INFORMATION REVIEWED
AND REAFFIRMED
1959
No. 81933
UNITED STATES DEPARTMENT OF AGRICULTURE
FOREST SERVICE
FOREST PRODUCTS LABORATORY
Madison 5, Wisconsin
In Cooperation with the University of Wisconsin
THE DISTRIBUTION .OF STRESS IN A
WOOD POLE BENT BY A RADIAL.ENDIDAD
By
WILIITLA S. ERICKSEN, Mathematician
Forest Products Laboratory, Forest Service
U. S. Department of Agriculture
41•00110/104.0
Introduction
Wood, as an elastic material, is commonly treated as orthotropic, or
one which has three mutually perpendicular planes of elastic symmetry.
These are the plans -of the longitudinal and radial axes of the growth
rings and the planes that have these two axes as normals. A material
of this type is adequate as a model for the analysis of plywood made
from rotary-cut veneer and for the treatment of such structural members
as beams, in which the radii of curvature of the growth rings are so
large that the directions of the normals to the above planes of symmetry
may be considered fixed throughout. When a pole consisting of the entire
section of the trunk of a tree is under consideration, however, a
cylindrically aeolotropic material is a more appropriate model. For a
material of this type Hooke's law is given in terms of components of
stress and strain referred to a cylindrical coordinate system (z, r,
e), with the longitudinal axis, z, in coincidence with the axis of
symmetry of the material, the center of the pith in a pole. The elastic
symmetry of the material is expressed by the conditions that the stressstrain relations remain unchanged if the direction in which eitherz
or 8 is measured is reversed.
/It has been pointed out by Carrier (1) 2 that a material cannot be
cylindrically aeolotropic at its longitudinal axis. If the material
is continuous to this axis, a transition in properties must take place
in the vicinity of the axis, such that at the axis it is isotropic. An
analysis based on the assumption of cylindrical aeolotropy throughout
cannot, therefore, be accurate in the vicinity of the longitudinal axis
if this axis is contained in the material.
-Maintained at Addison, Wis., in cooperation with the University of
Wisconsin.
2
—Underlined numbers in parentheses refer to Literature Cited at the
end of this report.
Report No.R1933
Agriculture -Madison
The present report is concerned with finding the distribution of stress
within a hollow circular cylinder of cylindrically aeolotropic material,
bent by a terminal, radially directed force. A solution, with stress
components in closed form, satisfying the conditions of equilibrium
and compatibility is given.
The applicability of the results of this analysis to a wood pole depends,
of course, upon the extent to which the wood structure and the shape of
the pole meet the conditions assumed in analysis. Among the conditions
implied by cylindrical aeolotropy is one that the elastic coefficients
are independent of the cylindrical coordinates. No provision is therefore made for changes in these coefficients from springwood to summerwood, from heartwood to sapwood, or for changes due to any one of the
stresses exceeding its proportional limit. To satisfy the assumed
conditions of elastic symmetry, it is evident that the growth rings
should be circular cylinders. The pole itself is also taken to be of
constant inner and outer radius, and the effects of taper are therefore not investigated.
In the first section of this report approximate formulas for the stress
components are given. These formulas, which appear to be adequate for'
practical purposes exhibit the main characteristics of the stress
components and are used in this first section as a basis for discussing
the results of the analysis. Of most interest among these results, is
the manner in which the shear stress components T T
and Tzx
zr, ez,
depend upon the elastic constants of the material. The formulas show
6-,-T and,
that these components depend mainly on the parameter k = GLR
LR
in a secondary manner, upon the combination -------- . An example shows
EL
ERA
that in; the range of k for most wood species the maximum values 'Of
greatly exceed the average shear stress, the load
Tee and T
divided by the cross-sectional area, and also the maximum stress obtained
by engineering methods.
In the main body of the report the solution satisfying the equations of
equilibrium- and compatibility is given. This solution brings in effects
of cylindrical aeolotropy that are not exhibited by the approximate
solution, but which are apparently of secondary importance.
Report
No., 1933
»2-
Notationa
inner radius of cylinder.
b
outer radius of cylinder.
ELYoung's Modulus of wood, longitudinal.
moduli of rigidity of wood.
GLRY GLT
GLT
k\I GLII
length of pole.
coordinates with axes oriented longitudinally, radially,
and tangentially with respect to growth rings.
applied radial end load
rectangular coordinates with axes indicated in figure 1.
cylindrical coordinates with axes indicated in figure 2.
L, R, T
P
z, x, y
z, r, A
T zz rr,
T
Tzr
TZZ,
stress components referred to cylindrical coordinates.
..
stress components referred to rectangular coordinates.
TZX
Discussion of Results
The components of stress within a wood pole subjected to a radially
directed end load as indicated in figure 1 are given approximately by the
following formulas.
TZ
4Pr
7( 134 .. a4)
(t-
z) cos 9 (R-1)
2 0-LRGuk2
1 3
4Pb2
Tir = r(b4 _ a4 ) 4).
fr\k-1) ..
( S)
EL
9 _ k2
(,\ 3+k ‘
\l - k 17 )
./
/13\ k471
-1,..3.)
- (i)2
\i
I
n a 2k (a\ 34-k-\
- )3,)
i\jD
COS
43
(R..2)
3
A number of symbols used in the main body of the report are defined
as they are introduced.
Report No. 1933
_
2 , j E
3
l
4Pb2
Tez=-
a4)
rr(b 4
b)k+1 1(a\2k
;
b
and
3+kl
(2.;.•
sin 9 (R-3)
= O.-
T
00
(R-4)
These components are transformed to those referred to rectangular
coordinates (z, x, y), figures 1 and 2, as follows:
T
4P(Q-z)x
4
4
n(b - a
zz
(R-5)
= Tzr cos e TZ
X
Tyz
and
T
=
Tir sin
8+
T
Az sin 0
T
Etz
cos 0
T
= 0,
Xy
T zx is the component called the "horizontal shearing stress" in beam
theory.
It is seen that in the isotropic case where
EL =
ES
LR
= T
, and GLA = GLT
WI7077
the preceding formulas are exact
sections 55 end 59)w The source
of error in the formulas when these relations amongtheelastic constants
do not hold, is discussed below. Formulas (R-2) and (R-3) may often be
simplified by dropping a number of powers of the ratio it.
Report No. 1933
.44.1110.
The following formulas, applicable to a solid circular beam with the
present loading, are obtained from elementary beam theory on the assume=
tion that Te z is constant along the diameter x = 0 (7, section 27).
zr
T
8z
=
4P
3n b2
,f
4P
3 nb2
-
fN''
r
cos e
(R-9)
`
sin
(R-10)
and
4P
3 Tb
X2
2
L
b2
(R-11)
These formulas are included here for purposes of comparison.
The shear stress components Tzr, Te z , and Tzx are represented
graphically for three combinations of elastic constants in figures 3 to
14. These figures mere constructed from formulas derived in the body of
the report and listed in the final section. However, the results from
formulas (R-1) to (R-8) are such that for practical purposes they may
be considered the same as those from the formulas given below. The
combinations of elastic constants used in the computations were selected
from those for species of wood for the purpose of illustrating the
characteristics of the stress components as determined by the parameter.
k=\
GLT
(R-12)
Three cases, I, k = 1; II, k > 1; and III, k < 1 are considered.
In the figures curves are given for = 0.1, and also for ,g = 0. The
former value was chosen for convenience rather than as one of typical
magnitude for the ratio of pith radius to outer radius in a wood pole,
As indicated above, the curves for a = 0 are not accurate in the vicinity
of r = 0.
Case I, k = 1
Figures 3, 4, 9, and 10 represent the shear stress components in a pole
of material for which k = 1. The elastic constants for this case are
avera g e values, for a number of specific gravities for Douglas-fir
taken from references (2) and (5). This combination of constants,
Report No. 1933
sm5-
given. in table 1, was selected for the purpose of illustrating the
characteristics of the shear, stress components for the value k = 1
rather than as one representative of Douglas-fir. As reported in
reference (2), the relation GLT) GLR exists in specimens with like
specific gravities in this species. (A combination of constants in
which the latter relation holds is used in case II below.)
A material with k = 1 (Gil GLR) may, for present purposes, be considered isotropic in the sense that with an equivalent Poisson's ratio,
Ts determined by the relation
2 cr LRGIR
1+
( R-13 )
EL
formulas (R-1) to (R-4) are identical in form with those for the isotropic case. With formulas obtained in this way it is found that T
zr
and T e z have the following maximum values.
Pb 2 (3 + 2a• )(1
rz
max!
Ta3- )2
ate= 0,
=
( b4a4) (1 4. 0.)
-
max!
n (b4 - a4) +
(1
a.)2at g
2 cr (b i / I
a-)
and
IT8z maxi .
r=at
(3
2
v)
at e
+Tr
(R-14)
b
1 + 2a
Pb2(3+2a-)
ez
r
b
O
r = 0, a = O.
+
TT
-2
(R-ls)
(R-16)
Etr b2q1 a.
The maximum values ofzx,- coincide with the extreme values of T ee
For purposes of making comparisons, the value of the equivalent Poisson's
ratio in the preceding formulas may be taken as zero, because, for wood,
EL is usually large compared with GLR. With a- = 0, formula (R-16) yields
{ Taz maxi
Report No. 1933
-6-
This value is about 12 percent greater than that obtained from the
engineering method, formula (R-10).
C a s e U. 1c-> 1
for a value k >1 are
zit
given in figures 5, 6, 11, and 12. The constants used in the computaa
tions for this case, table 1, were the same as those used for the preceding case, except that the ratios
The shear stress components
T
T
GLR = 0.064 and
e z,
and
= 0.078,
which are average values of these ratios reported in reference (2), were
used in determining GLR and GLT.
With k> 1 the stress component r gz does not necessarily take on its
largest value on the inner surface of the cylinder, because this comp
ponent, formula (R-3), evaluated at r = a, vanishes with a for all
values of 0. For a pole with a small, hollow pith, the shear stresses
T
and T may therefore be small near the pith and attain their
2X
Az
greatest values within the wood of the pole. This situation is not
illustrated in figures 6 and 12 because the value t = 0.1 used in the
computation is too large.
Case III, k < 1
Figures 7, 8, 13, and 14 represent the shear stress components rzr,
T
and T in a pole of material for which k 1. For this case,
ez tzx
which is most common among wood species (2), the elastic constants used
in the computations were taken from average values for sweetgum reported
in reference (4). vith k in this range, it is seen from formulas (R-2)
= a
and (R-3) that the stress components Tzr and T ez evaluated at
take on arbitrarily large values when a is near zero and become infinite
when a = 0. This illustrates the errors that result from assuming that
a cylindrically aeolotropic material contains its longitudinal axis. The
large values attained by the stress components, near a = 0 are reflected
= 0.1. At
in the results given in figures 8 and 14 for the value
is found to be
r = a the value of maximum value of TEft out of T
*X
r
a
3.6P
This is almost three times as large as that obtained by the
b2
engineering method, formula (R-11).
Tr
Report No. 1933
-7-
Analysis
Stress-strain Relations, Equations of
Equilibrium and Compatibility
The stress-strain relations in a cylindrically aeolotropic material have
been shown by Carrier (1) to be of the forms:
Fe zz
F a ll
a12
a13
0
0
q
exr.
a 21
a22
a23
0
0
0
e99
a31
a32
a33
0
0
0
T
ere
0
0
0
a44
0
0
Tr@
0
eez
Trr
GA
TAz
a55
0
ezr
T.
ZZ
466
J
Tzr
In these relations the notation for the components of stress and strain
is that used by Sokolnikoff (6), and the matrix elements are interpreted
as follows (3):
1
all = 17L.,/
a 12
12
a21 =
a 22 - ER,
a
31
=
_crTt
crRL
ER
3
1
•LT
EL '
- _1
'44 - 2GRT'
an
RT
ART
ER ,
1
a55 = raTil
au
a
= ET
-°TR
(2)
23 = ET
1
a33 E T
1
a66 = 2GLR
we+
Here E denotes a Young's modulus, oa Poisson's ratio, and Ga modulus
of rigidity, while 1, 0 R, and T associate these quantities with directions indicated in figure 2. The following relations exist among the
elastic coefficients.
Report No. 1933
a
12 = a21 ,
a 23 = a 23
a13 = a31,
(3)
The conditions of compatibility are most easily given in terms of components of strain referred to rectangular coordinates. In order to apply
them in such form the components of strain in (1) must be transformed to
a rectangular system. At any point p in the material, figure 2, the
components of strain referred to, z, r, and 9, are transformed to those
associated with the z, x, and y coordinates in the same manner in which
components referred to the rectangular L, R, and T coordinates are transformed to the z, x, and y system. The transformation is accordingly given
in the following form (6, equations 16.5).
1
0
0
0
0
0
r e zz
err
O
cos2 9 sin2 9 sin2 9
0
0
ac
eee
O
o
o
e
ere
O
0
0
e19z
O
sin2 A cos2 9 -sin2 e
-sin 28 sin2 9
cost 0
2
2
0
0
0
ezr
O
L
0
yy (4)
0
0
exY
cos 9 -sin 8
eyz
sin e
ezxj
cos 8 i
where 6 is measured from the x axis to the r axis. By equating the righthand members of (1) and (4) and then multiplying both members of the
resulting equation on the left'by the inverse of the matrix of (4), obtained by changing 9 to -9 throughout, it is found that
e zz = all Tzz a12 Tr.r a13 T99
eXX
(a21 cos26 + a31sin26) Tzz
+ (a23cos29 + a33 sin29) Te
(a21sin2e a31c°5 2
9) T zz
YY =
(a22e°a29
a32sin29) T rr
2
a44 ein Tre
(5)
(a22eine + amcos 29) Trr
+ (a23sin29 a33 c0s2e ) T ee + a44 ein29 Tr8
ems,
(a21 - a31)
2
T
(a22 a32)
zz
2
sin2) Trr
(a23 a33)
Gina) Tee + a cos29 T
3
44
re
(Equation (5) continued on next page)
Report No. 1933
-9-
eyz = e ss eese Tez
ezx = –a 55 sin
ez + a 66 cose
Let
a
21
al =
2
-
+a
T
zr
J
a21 — a31
31
2
2
a
a
a66 sine Tzr
a
22 + a 32
2
-
2
- a 52
> (6)
2
f32
a 25 + a 55
a3
22
a23 -
2
••nnn•
Then with the application of (3) the following relations are obtained
from (5),
e zz
a11 T zz + (a l +
rr + ( a l
1
e3cc = ( a l +(3 i cos28) Tzz + ( a 2 + (32 cos20)
a 5 +
3 5 cos28) T 68
(7)
0 1) Tee
rr
a44 sin2e Tre
(8)
( a 2
13 2 coe28)T ri
eyy = (a l03. cos28) T z z
a 5 --13 5 cos28)
exy = 13 1 sin2e Tzz
T
a 44 sin29 T re
132 sin28 Tyr
+ a44 cos2e T ra
eyz
= a55cose T ez
ez
=-a55 sine
Report
No, 1933
T
ez
(9)
T
GE)
-0,
(10)
a66 sine
+ a 6
T zr
cosh T zr
(U)
(1 2.)
The equations of equilibrium with stress components referred to cylindrical
coordinates are (6, equations 48.16),
aT rr
1 a7ie
aTrz
Trr
0,
az
a T..
ar
r
c)e)
2
z
aT
= 0,
Tre
+7
az
T e z
T rz 1
+
T
_ _ e
zz 1
T
0
z
r rz
••••
az
ae
and with components of strain referred to rectangular coordinates, the
conditions for compatibility are (6, equations 10.10)
2 e
xx
2
2 e
a
2 Pt e
ax2
rz ox
o-2 e
2
o.2 e
zz =
yy .4. ------
Z.-I1 z 2
ay2
., 2
2
)oc
GY
y
(16)
yz
(17)
,Y az
2
2'.
a e ZX
•
°e-xy
-
--ax
e xy
az
y ax az ay az .
r
,71
e
Yz •
x
2
Pi 2 e zz
r
(18)
2
ae„
(19)
ay
L
F-)
exx
aezx
P!z
ax1
f
(1.)
fez ae
eq•
Report No. 1933
;F)x I
elexy
(20)
?jiy
aeyy
r z 12x
-11-
ae
a
(a)
Stress Resultants and Boundary Conditions
Let Q. denote the length of the cylinder, figure 1, and P the magnitude
of the radially directed force at z = R. Take the direction of 1' parallel to the x axis, as indicated in figure 1, and denote by a and b the
inner and outer radii of the cylinder, respectively. Then a distribution
of stress statically equivalent to that produced by such a force is one
for which
2ir
T
cos 8
T
r
rti
b
S
rdrde P,
(2 2)
(
(23)
2ir
ZZ
_
8z
bin;
rz
2 cos 8 drd9
=-**
- z)
+ te eo 9 rdrd8 = 0,
z
( 24)
la q
b
2
7r
f
2
zz r sin 9 drd9 = 0,
(25)
zz rdrd8 = 0,
(26)
T
a
b
I
- 21T
10 T
and
2
Telz
r2 drd9 = 0.
(27)
Since the inner and outer surfaces of the cylinder are free of stress,
the boundary conditions are
Report No. 1933
-12-
T rr
= 0 at r = a and r = bo
(28)
T
= 0 at r = a and r = bo.
(29)
r
T
zr
= 0 at r = a and r = b.
(30)
The Stress Components
The following expressions give components of stress that satisfy conditions (15) to (30), inclusive.
-
T =
zz
n
i
JA 1r + Air +Ar i (
i
r
T = 1?,r. + B-ri + B r J
rr
‘
1
Tee
r= 5Bi r +
.:
t
i . (2-
I
(31)
S/ - z) cos 9
cos
z)
(32)
0
1
(2.
+ 1)B i r i + (3 + 1)B i r i
z)
cos
e
(33)
J
TrO
+ Bi r i + Bird
=
Tez = '
1,
(2- z) sin 0
rk-1 - kF-k r-k-1 + 1-: 3H 1.
r L
L
d
2)Hi - A if ri+1 + <1 0 +
Trz.
=
t...
(34)
A
.
n 4
4
rJ+1] sin8- (55)
2)H3
2)H . - Az>
J
J+1
.4.1
- + F_O
Fkrk-1
+ H1r 2 + H-r
1 ifl + H.r
3 '
CO
4
(36)
In these expressions
i= \11
J
\I
+K
+ K -1,
4
-It is assumed in writing (30 and (36) that (n + 2) 2i , k2 o n =
(37)
(38)
1, i,
J.
Terms associated with an exponent n that violates this condition are
determined by using formulas (66) and (68) in (62) and (63).
Report No. 1933
.130,
where
a 2 + f32
a11
4-
2(
1
-0
a 3 +13 3) + 2a44.
a ll
and
-( a l "1) ( 3 a
- (a l
a 5
39)
1)2
\ Ir 66
i7--
k =
(40)
a55
The coefficients A n and B n , n = 1, i and j are determined by the
following set of equations.
(An
(n - 1) a l - (n + 1)
Bn [(n - 1) a2
+ 1) 02 +
+ 2) 1 (n-1) 0. 3 - (n+1)i33
n = 1, i, j
(41)
Bir+B.r+ B j.r J = 0, r = a and r = b
(42)
and
+ 2 4.
-I
Airj+2
Air
r
l
cast a drd6 = -P
(43)
With these determinations the quantities
2D 4
(n + 2)A " +
" a55
2) 2 k2
Report No. 1933
n = 1, i, j
(44)
where
a -ai l
))
D =A
n1
n
+13
(n ♦
a
+ a,
2)( a5
44
(45)
can be evaluated, and the coefficients F k and F_k are then found by
solving the equations
Fkrk-1 + F -kr-k-1 + H ir2 + Hiri
1
+ H ri+1.= 0, r = a and r = b (46)
To show that the distribution of stress determined by the preceding
relations satisfies the required conditions, consider first the stress
resultants and boundary conditions. Among these (23) is fulfilled by
(43), while'(22)'f011ows from (23) and (15) without reference to the
specific forms of T 7. , r r z and T.
ez The remaining conditions on the
stress resultants, 24) t o (27),inc lusive l are fulfilled because the
integral with respect to"8 vanishes' in eadh case. As for the boundary
conditions, it is seen that (28) and (29) are fulfilled by (42), while
(50) is fulfilled by (46).
,
Among the equations of equilibrium, (13) and (14) are found to be fulfilled by direct substitution. The remaining condition, (10, will be
dealt with later.
Now consider the first three equations of compatibility, (16), (17), and
(18). By observing the forms of the stress components (31) to (36), it
is seen that among the strain components, (7) to (12), ems , eyy, and
exy are linear in z while eyz and e zx are independent of z. The component e zz is made linear in x = r cos 8 and independent of y = r sin 0
by imposing the conditions
A na l). + Bn
<
+ ( n + 2)( a l
+
- f3 ) =
n
and n = j
(47)
These relations are shown below to be equivalentto (41) for n = i and
n =
respectively, and therefore hold by previous assumptions. Equations
(16), (17), and (18) are then satisfied because the components of strain
are such that all of their terms vanish.
With the substitutions of (51) to (34), inclusive, into (8) and (10)
and the application of the operators
,
--,
0 _ sin 0 a
--57" 0 cos 8 Ar
a9
r
' '
1
(48)
a
, in 9
ay .-
Report No, 1933
a .,
_. 2)r
Cos et. a
r
29
1
J
it is found that
exx
Cn rn-1
Tx
2•
2, sin4
(491
n=1,i1j
with
- 1)0.
+B
n
[(n - 1)
1
- (n
Q2 -
3) 0l
(n + 5) 0 2 + (n+2)-(n-1)a3-( n+3)03?
(so)
Similarly, from (9); (10), and (31) to (34), using (48),
22xx
r X
aeXY
Y
n
n . 1 1
cosie + ( ►+1)D nz)rn-1 (51)
j-
where, as previously given,
D n - An (a1
-P i )
Bri
<0. 2
-
0 2 (n +
2)(0.3 -,33)
a44‘fr
(45)
J
With a second
ay
application of (48), (49) and (5]) yield
-
'11-77r
(n +
GC
1)1
10
. n +
ey
(n 1)1):1 p(12- z)rn-2 cos9
(52)
n = 1, 1, j
Equation (19) can therefore be satisfied by taking
Cn + (n - 1) Dn . 0, 1 for n 1,
Ta
j
(53)
-The right member of (52) indicates the possibility of including terms
with the exponent n. -1 in the expressions for the stress components
T
T
and Tre. This possibility is not considered for the
zzl re
reason that terms with this exponent lead to multiple-valued displacement components.
Report No. 19S3 -16-
1
or, from (50) and (45),
1.-
1
An < (n -
1)
a- (n + 1)13
;.>
1j
1
+ B 1(n-1)
n ,
.
+ (n+2))‘ (n-1) a 3 ... (ri+1) '3 5 "2a44 = o
a 2 - (n+ 1) 0
2
L
L
-I
j
n = 1, i,
j
....(41)
For n =
j these conditions must be imposed simultaneously with (47).
This is made possible by the choice of i and j as the roots of the
determinant of the system or of
(54)
n2 + 2n K = 0
where K is given by (39).
With the substitutions of (49) and(51) the last two equations of
compatibility, (20) and (21), yield, respectively,
I es e zx
---.
Oe
Cnr
Fiy
n- 1
sin28
(ss)
n=113.0
and
3e
y ) ra
y
cax j
=C cosZe
+ (n + 1)D(56)
n
n=1,1,3 L.,
By the use of (48) these are transformed to the two following equivalent relations,
aezx
ae
+ n+1)Dn rn- 1 sine
__,ZE =
y
ZNx
j
n=11/0
(57)
and
Ey
T/F9
8ezx
ay
1 Drn-lcos8
nj
6eyz
,- 71x
i,j
Report No. 1933
.17—
(58)
From these equations it is concluded, with the use of (53), that
ae
= 2 /
zx
-17i-
D
nrn sine + C
(59)
n=1,i,j
where C is independent of r and 8. It is evident from the expression
on the left that C represents the mean local twist over a section (6,
page 220). In order that (27) be satisfied, it is specified that
C=0
p ith
(60)
expression (31) substituted for T zz equation (15) is written
a
aT rz aT rz 1
r
car
elr
T
8z
ae
=
An rn cose
( 61)
n=1,i,j
This equation is identically satisfied by the functions
ao n=1,i
rz r 3r
A
n + 2 r ncos9
(62)
and
T
=
•
(63)
9z Zr
With e and e zx obtained from (11) and (12), respectively, by the
yz
substitution: of these 'expressions, it is found by applying (48) that
e zx
aY
(--)eyz
f :.,..)29/
= a55
1
41 +
a 66 ")295
4
L ?) r
'
A,
r2 a62
rn sin e
//
a66 n+2
n=1,i,j
(6 4)
By equating the right member of this equation with that of (59) and
introducing the notation
a66
k =\ a55
Report No. 1933
(40)
—18.
the following equation in 0 is obtained.
k 2 p1 20
r2 -2.,(a 2
2 r
/
<1kk2AA nn 4, 2D n
n=1,i,j
rn
sin e
(65)
a55 In+2
A solution of this equation that is suitable for present purposes is
+
0 '
-k'
+
Ord
n=1,i,j
D
.2_11
where2An
ass L
2
rn+2 if ( n+2 ) 2
On - ; (n 2)2_ k2
and
1
-
(66)
sine
2D
k 2A n
k2
(67)
n
(n+2) log r-1>r - 2
4(n+2) 2n + 2 a55
if (n+2) 2 = k2
(68)
2
A situation in which (n+2) = k 2 arises in the isotropic case where
k = 1 and one of the values of n is -3. However, for this value of n
the coefficients A n and Bn vanish and On, expression (68), is identically
zero. The possibility that (n+2) 2 = k2 , exactly, in an aeolotropic
material is considered remote, and formulas for the shear stress components Trz and T re based on (68) are therefore not given.
with 0 given by (66) and (67), formulas (62) and (63) yield expressions
The preceding analysis
(35) and (36) for T and T rz,
ez
shows that these stress components are so determined that equations (15),
(20), and (21) are satisfied and all of the prescribed conditions (13)
to (30), inclusive, have now been shown to be fulfilled.
Application of Results. Simplifications
Formulas (35) to (45), inclusive, have been used in computing the shear
Stress components T zr and T ez for the three combinations of elastic
constants listed in table 1. The results of these computations, given
in figures 3 to 14, have been described in the section entitled
Discussion of Results. In order to show by examples that a number of
Report No. 1933
the terms introduced in the preceding section can be neglected, the
complete formulas, after determining all of the constants, are listed
as follows for the three combinations.
Case I: k = 1, i = 2.18, j =-3.18
TT
rz a
P
22
Tr (b -a
[1.47 - 0.0145
-8 b()5.18
5 x 10
r
-P
n(b2-a2)
Tez
(T) 2 - 1.45 (E) 2 - 0.01 (E) 5.18
cos e
1.47 + 0.0145
(12)2
(19 - 0.02 (
- 0.388 1 2
- 9x108 (b)3.18 sin e
Case II: k = 1.104, i = 2.18, j = -338
T a
rz
-
T
6z
r1.49 ( .)
i
n (b2...a2)
0.01
(r)3.18
0.104 _
04093:3
-7x10-a ( 1 3.18
2)
cos
(1,2)2.104..
1
.47 (i)2
e
r 0.104
1.65(-)
+ 0.0101(7b )2.104 - 0.47(11
13
b'
Tr (b2-a2) [
•
•n•n•••••n••nn•11101n
•
5.18 10-7(b15.18
.
002(I)
sin 6
Case III: k = 0.827, i = 0.741, j =-2.74
rz
(32-22) [1.450)
r
0.173
+ 0.03(1)1.74++ 2x10 -6 q.)
Report No, 1933
- 0.03230.) 1.83 1.45( E) 2
-
cos 0
3.18
h 0 173 . 0.0268C;b 1.83
1.18(.7.)
-
T
ez
r2
. 0.51(-)
,T (b2 -a 2
-6 b 1.74] sin
.
0()
r
+ 0.03(-)1.74
b
e
The terms in these formulas are given in the order in which they appear
in formulas (35) and (56)0
The fourth and fifth terms in each of the preceding formulas are
associated, respectively, with the exponents i + 1 and .j + 1. These
terms are so small throughout the range of4 that they may be neglected
for practical purposes. The corresponding terms, those with exponents
i and j, in formula (31) for T zz are likewise found to be negligible,
and the possibility of simplifying the results by neglecting these
terms in the analysis is thus indicated. Formulas (R-1) to (R-4),
inclusive, are based on such a simplification. They are derived as
follows.
Take A = A = 0. Then from (43),
i
A
and
T
zz
4P
ii.(b4.a4)
l
takes the form (R-1). Assume ti n = 0 for n = 1, 1, and j, so
that
2
H1 =
G
°- LR LT
1
5A]
.2, + 24 A 5 EL
D
9 - k2
1
0
-LT
GLit
I
as obtained from (44), (45), (6), and (2); while H i = H 3 = 0, and
= 0. With F k and F_ k evaluated in terms of H1 from
rr = T 89 = TDB
(46), T zr and T ez take the forms given by (R-2) and (R-3), respectively.
T
It is seen that among conditions (13) to (30), inclusive, (19) is the
only one that is not in general satisfied by the stress components
determined in this way. When
cr LT -TM
Pi
2E
=0 •
as it is in the isotropic case, condition (19) is also satisfied and the
simplified solution is exact.
Report No. 1933
n21-
The stress components (R. ' S) to (R-8) are obtained from those given
by (R.1) to (11.4) by the application of (4) with the strain components
replaced by corresponding stress components. Formula R-6 was used in
plotting figures 11 to 14.
Literature Cited
(1)
CARRIER, G. F. Stress Distributions in Cylindrically Aeolotropic
Plates. Journal of Applied Mechanics, Sept. 1943, vol. 10, No. 3.
(2)
DOYLE, D. V., McBURNEY, R. S., DR0 1 4, J. T. Elastic Properties of
Wood. The Moduli of Rigidity of Douglas-fir at about 11 Percent
Moisture Content. U. S. Forest Products Laboratory Report No.
1528-E, Nov. 1946.
(3)
MARCH, H. W. Stress-strain Relations in Wood and Plywood Considered
as Orthotropic Materials. U. S. Forest Products Laboratory Report
No. 1503. 1944.
(4)
McBURNEY, R. S., DOYLE, D. V., DHOW, J. T. Elastic Properties
of Wood. Young's Moduli, Poisson's Ratios, and Moduli of
Rigidity of Sweetgum at Approximately 11 Percent Moisture Content.
U. S. Forest Products Laboratory Report No, 1528.F. Nov. 1946.
(5)
DROW, J. T. Elastic Properties of Wood. Young's
Moduli and Poisson's Ratios of Douglas-fir and Their Relations to
Moisture Colthent. U. S. Forest Products Laboratory Report No.
1528-D * Nov. 19480
(6)
SOKOLNTXOFF, I. S. Mathematical Theory of Elasticity. McGrawHill Book Co., N. Y. 1946.
(7)
TIMOSHENKO, S. Strength of Materials, Part I. D. Van Nostrand
Co., Inc., N. Y. 1940.
Report No. 1933
Table I..► .Elastic constants used in com utations
Case: EL : ER : EET :
T LR :• T LT
•
• 1,000 :1,000 :1,000 •
1-1 : 2,280 : 154
II-2 . 2 280 •• 154
III 1,800 : 195
: 113
: 113
: 85
RT
::Case: GLR
s:
: :
GLT
GRT
-1 000 :11,000 -11000
*--
sp.s•i.spes•i.:p•s•i.
:0.292 :0.449 :0.390 ::I-‘ : 4
-128 : 4128 v4-12.7
: .292 ; .449 : .390 ::II=0 : h46 : =178 x412.7
.7
: .325 : .403 : .682 ::III=: 168 : 115 : 38.8
1
-Reference (5).
--Reference (4).
3
-Reference (2).
4
-Obtained from averaged values.
5
-Obtained from average ratios to E L .
Report No.1933
T