EFFECT Of ELLIPTIC OR CIRCULAR HOLES AS ORTHOTROPIC MATERIALS

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EFFECT Of ELLIPTIC OR CIRCULAR HOLES

ON THE STRESS DISTRIBUTION IN PLATES

Of WOOD OR !PLYWOOD CONSIDERED

AS ORTHOTROPIC MATERIALS

Intermatien-Pci4etued-aittl-PtatiffintitKI

-March-1-9ft

INFORMATION REVIEWED'

AND REAFFIRMED

062 t: .

No. 1510

111111111d11111111111111

MIL fib, map

FOREST PRODUCTS LABORATORY

MADISON 5, WISCONSIN

UNITED STATES DEPARTMENT OF AGRICULTURE

FOREST SERVICE la Cooperation with the University of Wisconsin

EFFECT OF ELLIPTIC OR CIRCULAR HOLES ON THE

STRESS DISTRIBUTION IN PLATES OF WOOD OR PLY-

WOOD CONSIDERED AS ORTHOTROPIC MATERIALS 1—

By

2

Forest Products Laboratory,— Forest Service

U. S. Department of Agriculture

Abstract

This is a mathematical analysis of the stress distribution existing near a hole in a wood or plywood plate subjected to tension, as, for example, near holes in the tension flanges of wood box beams. It is assumed that the strains are small and remain within the proportional limit. In this analysis a large, rectangular, orthotropic plate with a small elliptic hole at the center is subjected to a uniform tension along two opposite edges. By taking equal axes for the elliptic hole, the theoretical stress distribution in the neighborhood of a circular hole is obtained. The analysis shows that for a circular hole in a plain-sawn plate of Sitka spruce a maximum tensile stress of 5.84S is attained, S being the uniform tension applied in the direction of the grain. This maximum stress occurs on the edge of the circular hole at the ends of the diameter perpendicular to the direction of the uniform tension. Although this stress concentration may be greater than that found in isotropic materials, it is much more sharply localized and is relieved by plastic flow when the stress exceeds the proportional limit. For the reader who does not wish to follow the mathematical details of this report, the formula for computing this maximum stress for wood or plywood and an explanation of the constants involved is given in the appendix. The formula has not

1

—This progress report was originally prepared by the Forest Products

Laboratory in 1944.

2

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

Wisconsin.

Report No. 1510

-1-

Agriculture -Madison

been confirmed by an extensive series of tests, but it was found to be in reasonably satisfactory agreement with the only test made. It is also demonstrated that for the Sitka spruce plate mentioned the maximum shear stress attained is 0.71S (fig. 5). The stress function from which these formulas for the orthotropic case are obtained is shown to reduce to the well-known stress function for the isotropic case.

Introduction

An orthotropic material possesses three planes of elastic symmetry at right angles to each other. If from such a material a flat plate is cut parallel to a plane of symmetry, it will have two perpendicular axes of symmetry in the plane of the plate. In the discussion that follows, a large rectangular orthotropic plate of this type with its edges parallel to the axes of symmetry and with a small hole at its center will be considered. A uniform tension S will be taken to act on two opposite edges of the plate as shown in figure 1, and the resulting stress distribution will be obtained. Throughout the report it is assumed that the strains are small and remain within the limits of "perfect elasticity."

The solution is first derived for an elliptic hole with major and minor axes coinciding with the axes of symmetry. By taking the major and minor axes equal, the solution for a circular hole is obtained. For the orthotropic plate, the solution for the circular hole is but slightly more simple than that for the elliptic hole. This is different from the corresponding problem for the isotropic plate.2 The stress-strain distribution in the neighborhood of the circular hole is notably different in an orthotropic plate than in an isotropic plate as the concentration may be considerably higher and of a more localized character for the orthotropic than for the isotropic plate. Thus in a plate of plain-sawn Sitka spruce with a uniform tension acting in the grain direction, the greatest stress component at the edge of the hole decreases rapidly to one-half of its maximum value in a radial distance of less than one-tenth of the radius.

For the isotropic plate the corresponding stress component decreases to one-half of its maximum value in a radial distance larger than onehalf of the radius from the edge of the hole.

.11

Report No. 1510

-2-

The Effect of Elliptic or Circular Holes on the Stress Distribution in a Wood Plate

The components of stress and strain in a plate under plane stress are connected by the following relations± if the axes of x and are taken as the axes of elastic symmetry of the orthotropic plate:

xx

1

X - x y x E y e = -

YY

E

1

X + x

Y

x E y

(1) e xy

1

X y

In these equations Ex and Ey are Young's moduli in the x- and I-directions, respectively. Poisson's ratio cr is the ratio of the contraction parallel to the y-axis to the extension parallel to the x-axis associated with a tension parallel to the x-axis and similarly for cr The quantity p.3.1 is the modulus of rigidity associated with the directions of x and z.

Since the equations of equilibrium are deduced with no reference to the law connecting stresses and strains, they will be satisfied by a stress function F such that

2

8 F

Xx = 2 ,

By

Y =

Y

8

2

8x

F

2

0

2

F

8x8y

(2)

4

—March, H. W., Stress =Strain Relations in Wood and Plywood Considered as Orthotropic Materials, U. S. Forest Products Laboratory

Report No. 1503.

Report No. 1510

-3-

Substituting (2) in (1) and then making use of the compatibility equation

8

2 e xx

8y

2 a

Z e

YY ax

2 a

2 e xy axay the following differential equation, satisfied by the stress function F, obtained: is

1 8

4

F 1 crxy

84

F

E

y ax4

+ (—

E

2

x ax

ay

1 8

4

F

E x 8y

4

= 0

(3)

In deriving (3), the relation!

E x has been used.

It is convenient to rewrite equation (3) ass

8

4

F

+ 2K

8x

4

ax

8 4F

2

8/

8 4F

2 +

3.11

4 -

where

K -

2

x

E y

1

(µE

xy by making the substitution

(4)

(5)

(6)

= ey

(7)

-March, H. W. , Flat Plates of Plywood Under Uniform or Concentrated

Loads, U. S. Forest Products Laboratory Report No. 1312.

Report No. 1510

-4-

0

where

E 11

(8)

Substituting F = F(x + vu) in (5) leads to the following equation for v: v

4

+ 2Kv

2

+ 1 = 0

(9)

or

2 v

= -

K

2

"%ilk -

1

(

10)

Now let6

K

= cosh

4) then v 2 = - cosh 4 f _jcosh2

(I) -

1

= - cosh

(I) f sinh st•

2 e - e

-4)

2

= - e4) or

-e -4

-For wood or plywood,

K as defined here is probably always greater than 1.

Report No. 1510

-5-

Hence, the roots of equation (9) are

_ v l - le

4) 12 ie -4)12 v 3 = -ie4112

. -4112 v 4 = -le

Equation (5) may now be written as ax

2

+

8 2

2z a z

) (— ax2

+ 0 z2

2

)

F = o where

4)

a

The solution of equation (13) is facilitated by taking F of the form

(12)

(13)

F=R [F1 (x + + F 2 (x + ig Tl)1 + F o (x, where a = e c1)12

= K +

AiK

2

1 g = e -4/2

=

- -

-

1

(14)

,

(15) i=

I o the real part of the expression in the bracket is to be taken. That (14) is a solution of (12) is easily established by direct substitution. Equations (12) and (14) hold for the stress function associated with any elastic problem of plane stress in an orthotropic plate with x- and L- axes chosen as described here.

Report No. 1510

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n

A suitable choice of F for the problem discussed is

F=R

[

(z

1

- w i

‘ l2

+

Y i.

2 log (z 1 + wi)

(z2 w2)2

+Y2 log (z

2

+ w2)

+

ST1

2

2e

2

(16) where z 1 = x +

, z

2 = x +

, w

2

2

2 = V z

2

N2

(17)

22

cr

, y

2

2

= a

2

- /3

2 e

2 b

2

, the boundary of the hole (fig. 1) is given by

(18) x = a cos 0 ,

=ey =eh sin

,

(

1

9) and S is the uniform tension applied at the ends of the plate.

In order that the stresses may be single-valued, the square roots in (17) are to be chosen so that the following relations always hold: z 2 + w2

V2

Report No. 1510

-7-

Then

8x

[zl

1

+ w

1

1

+

B [ z

2

1

+ w

2 and

8F

811

R a

[z1

+w1

+

B

[ i z

2+w 2 s e 2

On the boundary of the hole x and ti are given by (19), so that

(20)

(21) z

1

+ w

1

= a cos 0 + ierE b sins

+ a

2

cos

2

2iae ab cos a sine -a

2 e

2 b

2 sing e - a

2

+ a

2 e

2 b

2

,

=a cos 0 +

iaeb

sin 0 +

(aeb

cos 0 + ia sin 0) ( 2 2 )

=(a+

aeb)

e i8

Similarly, on the edge of the hole z

2

+ w

2

= (a + b)ei0

(23)

(24)

Consider the boundary of the hole in figure 2, and orient the positive direction of the normal

v

and of the tangent

T relative to one another in

a way

similar to the orientation of the x- and y-axis, respectively.

Report No. 1510

-8-

Let Xv ds and Yv ds represent and x- and /-components of the force per unit thickness of the plate, acting on the element of arc ds of the boundary from the positive side indicated by the direction of positive v. Then

2

8y2

2

8 F 8 F

= — cos (x, v) -- cos (y, v) axay

-

8

2

F

8x8y cos (x,

+

8

2

F

Bx

2 cos (y, v) .

But cos (x, v) = cos

(y,

T) 511 dB

9 cos (y, v) = - cos (x,

= - ds so that

3

8

2

8y

2

F dy 8

=

2 ds exay ds ds 8y

3

8 2F dy -

8x8y ds

8x

dx

2 ds ds 8x

Report No. 1510

-9-

(25)

(26)

Equations (26) for a boundary on which no stresses act take the form d ( 8F = 0 ds 8y

8F ds 8x

)

= 0

Hence,

OF

8x and — are constant along a free boundary.

Oy

Then, on the boundary of the hole,

8FBF

Ox

1

8y

OF

811

= C 2

Substituting (23) and (24) in (20) and (21), it follows from (27) that

(27)

R e

-i0

a +

aeb

+ B e

-10 a + geb and

-

0

ale

a +

aeb

+B f3ie

a + gel)

[A

Taking the real parts

= C1

SEb sin 0

E

2

2

(28)

[ a A

+ afb

B a + f3Eb

cos 0 =c

1

Aar

+ aEb

Bp a + gib

Sb

+ — sin = —

E

These equations must be satisfied for any value of

2, hence

(

2 9)

Report No. 1510

-10-

A

a +

aeb

' B

a +

gel)

- 0 , and

Aa

a +

aeb

BI3

a

+ fieb

Therefore,

c

1 c

Z

=

0

(30)

B -

Sb (a + geb)

E ( a - 0)

By using equations (30), the stress components are found to be

a z

F

= Y =

8x2

Sb [ (a + creb)

(a -

13-) w (z + w

1 1

1) 2 2

(31) z

2 8 F

E — =

Sy

2

Seb

-a2 (a +

act)) + /3 2(a + Peb)

P) w l (z 1 w1 ) w 2 (z2 w2)

4. 5

(32)

-E

2

8 F

8x81

= X y

=

(a

Sb

+ aeb) +

+

13€1,)

13) [w

l (z 1

w 1 ) w 2 (z 2 w2]

(

3

3)

To obtain the stress components for a circular hole, it is only necessary to let b = a in equations (31), (32), and (33). Then

Report No. 1510 -11-

[

(1 wl (z

+ aE } i

+ WI) - w'

(1 + 13e)

2

(z

2

+

w '2 )

(34)

Y y

Xx =R

Sa

2

E -a

2

(1

+

oE )

cr=73 w l

1

(z + w'

1 1

13 w'

2

2

(1 ++/3Ew).

(z

2

2

+w''

2

(35) where

X = R fSa.

2

a -

(3

-ai (1 + a€) w'

(z + w' )

-4-

WI

1

2 p i pz + +t),

)1

1

2 2

(36) wiz = z z

2 2

13

2

E

2

) .

(37) wi = V

2 z i a2

a

2

E

2

) ,

Expressions (31), (32), (33) are complicated and do not permit easy computation of the numerical values of the stresses. The most important information they contain is, of course, in the neighborhood of the hole.

Stresses at the Edge of the Hole

From equations (22), (23), and (24), it is found that over the edge of the elliptic hole

Sbe

-i0

1

Y y = 1 e

(a -

13) Lae b cos 0 + la sin 0

1

PE b cos 0 + is sin 0 '

(38)

_ t

S be ie

- a

2

X x

= R -i c p aEb cos 0 + ia sin 0 i_

+

13 k n2 b cos 0+iasin0

+

S ,

(39)

Report No. 1510

-12-

X =

She

-ai

-

p)

[aE b cos

0 + is sin 0

13i

13E1) cos 0 + is sin 0

(40)

Since, by (15) af3 = 1, the real parts are given by

Y = y

T

+ b

2 cos

2

0

b 2 cos

2 0

(41)

X x

= 2

TT

2 b

2 cos 2 0 a2 sin2 0

(42) where

-T

X -

2 . 2 b

2 cos 2 0

ab sin 0 cos 0

(43)

TT

S(a

2

sin

2

+ b

2

cos

2

0)

[

-E

2 b

2 cos

2

0 + a(aEb + pEb + a) sin

2

0]

(a2e2b2 cos 0 + a sin Axe

E

2 b2 cos 0 + a2 sin 0)

(44)

It is of interest to refer these stresses at the edge of the hole to a pair of axes at right angles to each other having the directions of the normal v and the tangent

T, respectively (fig. 2). In this coordinate system, the shearing stress and the tension (or compression) in the v-direction must vanish, as this is required by the boundary conditions. It is readily shown that

T T of equation (44) is the tension (or compression) in the tangential direction. The component of direct stress in the tangential direction is given by

X cos 2 (x,

T)

Y y

(y, 7) +

2X

y cos (x,

T) cos (y,

T)

Report No. 1510 -13-

From (19) and (25), this becomes

Xx a

2

sin

2

0 +

Yy b

2 cos

2

0 - 2X ab cos 0 sin

8 a

2

sin g 0 + b

2 cos t

0

Substituting from (41), (42), and (43), the tangential stress is found to be identical with T

T of equation (44). If E 1, it follows that at the end of the axis perpendicular to the direction of the tension, that is for 0 = 90°

( fig. 1), T attains its maximum value, while at the end of the

axis

parallel to the tension, that is 0 = 0°, T T attains its minimum value.

Then

T

T

(max.)

= a

(ae

b +

b + a)

(45a)

T

S

T (min.)

=

7

(46)

For an isotropic material, it follows from (6), (8), (11), and (15) that

a =

=

e =

1. Introducing these values, (45a) and (46) become S(— + 1) a and -S, respectively, as in the isotropic case.2

Equation (45a) gives the tension at C (fig. 1) the highest stress occurring in the plate. Numerical values of this stress will be computed in the two b a 50 elastic moduli are given in table 1. The grain direction is taken parallel to the line 0 = 0°.

b 1

When — = — a 50 the tension at C (fig. 1) is 1.097 times the mean tension.

Report No. 1510

-14-

The ellipse when a-

=

50 would appear as a straight crack perpendicular to the grain which might be approximated by a break in the fibers, such as a compression failure. A very small tension applied to the plate across the crack would set up a tension at the ends of very high intensity.

If this high tension at the ends caused the crack to tear, the increase in length would exaggerate the stress still further, and the crack would continue to spread.

When — = — the ellipse would appear as a straight crack running in the a 50 direction of the grain or of the tension. Such a crack does not produce great local stress, a conclusion which is almost self-evident.

If the hole is circular (11 = 1), equations (41), (42), (43), and (44) reduce to a

T' cos

T

2

0 X = T' sin g 0 ,

X

T

X = - T1 sin 0 cos 0 , where

T I -

T

S [- e

2 cos

2

+

(cre + 13E +

1) sin z

(a

2 e

2 cos

2

0 + sin

2

0X/3

2 e

2

cos

2

0 + sin 0)

The stress at the end of the diameter perpendicular to the direction of the tension for a circular hole is given by

(max.)

(ae + Pe +

1)

For an isotropic plate

(a =

13 = e = 1) this reduces to 3S, as is well as 5.84S (fig. 8).

(45b)

Report No. 1510

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Experimental Verification of Maximum Stress

This maximum stress was experimentally verified by compressing a quarter-sawn yellow-poplar board containing a circular hole. The dimensions of the board and hole and the position of the 1/8-inch metalectric strain gages at P on the circumference of the hole and at Q at some distance from the 1tole are shown in figure 3. The gage measuring the maximum strain at P was placed inside the hole as shown in figure 3, since the maximum strain occurs just on the edge of the hole and dies off rapidly in the radial direction. The gage at Q is placed as far from the hole as is feasible for the specimen used. The gage length was assumed short enough to give a close approximation to the actual strain at the end of the diameter. Table 2 gives the experimental and the theoretical values of the strains at points P and Q of figure 3. The theoretical strains were computed in terms of S from the stresses given by formulas (35) and (45b) using the elastic moduli of the yellow-poplar plate. These moduli were obtained from coupons cut from the specimen and are given in table 1.

The experimental value of the ratio of the strain at P to the strain at Q

(fig. 3) is about 16 percent less than the theoretical one. This difference is not considered serious as the mathematical analysis was based on the assumption that the dimensions of the plate were large as compared to the dimensions of the hole. Also the impossibility of measuring strain at a point acted to decrease the apparent ratio. It is important to note that the experimental ratio is considerably higher than the corresponding ratio existing for a circular hole in an isotropic plate.

Shear Stress at the Edge of the Hole

The expression (43) for the shear component at the edge of the hole is highly important when considering holes in wood plates. When the uniform tension is parallel to the grain direction, this shear acts in the same direction, and its distribution may indicate where failure caused by shear will take place if the uniform tension is sufficiently high. This b

= 1, as this type of faila ure occurred in wood box beams with small circular holes in their tension flanges tested at the U. S. Forest Products Laboratory (see fig. 4).

The failure occurred in the vicinity of the point of maximum shear, which occurs theoretically, as shown in figure 5, at about 0 = 78°. The shear stress at the edge of the hole was evaluated numerically when 7, = 1 for

Report No. 1510

-16-

table 1. The results are given in table 3 for values of 0 between 0* and

90° when the line 0 = 0° is taken to be parallel to the grain of the wood.

The shear at other points on the circle may be easily filled in from the given values. Figure 5 illustrates the variation of IC, over the edge of the hole for spruce and an isotropic material.

Stress Components at Points

Along the x- and_y -Axes

Simple expressions will now be obtained for the stress components at points along the x- and i-axes.

Writing equation (31) as

Y

Sb

(a + aEb)(z i - w1)

E(a P) [wl (z 1 w1 )(z1 wl)

(a + f3E13)(z 2 - w2) w

2

(z

2

+ w

2

)(z

2

- w

2

) and performing similar operations on equation (32) and (33), it is found that

=

Sb

-

0)

1

[a - aEb z1 wl

1

- 1)

+ a - f3Eb z2 w2

14

Xx

=

R

SEb

- 0)

-a2 a - aeb (w 1 - 1)

+ a

R

2

Ed) 'w2

1 11

+

5 , (47)

Xy

=

Sb

-

0)

[ -ai.

a -

aEb p

/ fi t

2 a - Peb w2

1

1

These equations can be reduced considerably when x

=

0 or

/ =

0.

The resulting equations then readily show how the stress concentration dies off with increasing distance from the hole.

Report No. 1510

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For x

=

0, y a it follows that

Y

Y

Sb

e ( fr -

13) a - aEb cre y

a

2

E y

2

+

.y

1

2 a -

1 xx

SE b

(ci _13)

-a

2 a - ac

ae

-Va2

e 2 y 2

+ y 2

1

1)

+

a -

i3E Y b

\

42 e

2 y

2 .y22

0eY

2 2 2 e Y

V

2

2

+5 (48)

X y = 0 and for y

0 , x a

Y y

c(a

Scb

1

3)

-

a -

acb

2 x

.Y1

2

X

x

-

S

(a - 13) x

Z

-

2 a -

1 b x

Y

2

2

R2 a -b

)x2 x

Y

2

2

+ 5 (49)

X = 0 .

When -a- =1 the distances from the center of the hole may be expressed as multiples of the radius by writing x' = a , y' = a in equations (48) and (49). Hence, the stress components at points along the positive y-axis for a circular hole are given by

Report No. 1510

-18-

Y

r

c(a

,

-

X x

(a

SE

at y'

a

2 e

2

(y

12

-

1) +1

trey'

\-Vce2e2

(y'

2

- 1)+ 1

12

-

OE

2 2

(y'

Z

1)+ 1

X y

= 0 and at points along the positive x-axis r

S

[

I E(

11x

12

-

1 + a

2

E

2

-

)

I -

1 ( x'

-

1

Vx

12

-

1 A- 13 2

E

2

X -

SE x

( a

P)

-a2

1 -

at I

X'

-

I + a2ea

+

1 j3-2JSE

V x

,2 xt

4. 02E2

(50)

X y

= 0 .

Numerical values of the stress components

Xx

and Y at points along the x- and b a

L-

1 are given in tables 4 and 5 for a plain-sawn plate of Sitka spruce.

The variation of these components is illustrated in figures 6, 7, 8, and 9, and the variation of the corresponding stress components for an isotropic material is also shown for comparison. It is evident from these figures that considerable differences may exist between the stress distribution developed in the orthotropic plate as compared to the isotropic plate. In particular, figure 8 shows that the maximum stress developed in the spruce plate is nearly twice the maximum stress developed in the isotropic plate.

Report No. 1510

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Displacements in the Neighborhood of a Circular Hole

The expressions for the displacements in the neighborhood of a circular

hole (1

3

- = 1) will now be obtained.

Substituting (34) and (35) in the first two equations of (1), it is found that e

X x au ax

= R

L

Sa

2

E

-aZ (1

+

)

(a -

13)E

[w' (z + v.r! )

x

1

1

1

0

2

( 1

+

) wt(z z

+ w

2

1 )

Sa

2 o- yx e(a - (3)E

y

[ (1 +

(le)

(1 + E3E )

w' (z 1 + w') w'

(z 2 +

1 1

1

w'2)

(52) ev e =— =

YY ay

Sa t

-

[3)E

Y

(1 + aE ) w'

(z 1

+

Iv!)

1 w' (z

2

(1 +

RE)

2

+ w 1 )

2

Sa

2 co.xy

(a - 13) Ex

- a2 (1

+

ac)

wl (z

l

wl )

02(1

+ w12 (z 2

) w12)

ST

E x

Integration of equations (52), gives

Report No. 1510

-20-

u

V I'm

R

( a

2

+

E x cr yx)

(1

+

EE

Y a€) z + w'

1

( t p

E x

E yx

(1

+

f3E

IZ 2+w'

2

Sx

Ex

(53)

1 1

E

E

2 a x-y

E

(1 z

1

+

+

w'

1

)

1

X

1

E c

E x xy (1 + 13e ) z 2 + w'

2

S a-

E x

These are the actual displacements except for the addition of a linear function of y and a linear function of x to the expressions for u and v, respectively. This is easily shown by deriving the expression for xy av y ax

1 tive functions may be made to vanish by fixing the origd eliminating rotation about the origin.

xy in an

Next the real and imaginary parts of (z

1

+ wi)

-1

and of a similar expression for

z

2

and w

2

will be found. To do this, it will be useful to write z

1

+ w

1

- 2

1 a (1 - wi

2 2

E

Let

1 (cos + i sin T

1

) (54) where w'

1

R

1

=

V

[x2 - a

2

- a

2 e

2

(y

2

- a

2

)1

2

+ 4a

2

E

2 x

2 y

2

(55)

Report No. 1510

-21-

-1 t

ae

xy

Then

1 z + w'

1 x - -VIC cos 11/1

+ gaey -

-Vti. sin 1,2) a

2

(1 -

a

2

E

2

)

Similarly

1

+ w12 -

x - R 2

2

+

i(13ey -

a

2

(1 - 02E

2 ) where

(56)

(57)

(58)

(59)

2

= tan

-1

213E xy

R + x2 - a2 -

0

2

e 2 (y2 - a 2 )

To keep the displacements single -valued, Ili and 1, 2 must always be taken so that the following relations hold: —

(60)

1.

2

Report No. 1510 a -\11 - a

2 e

2 a

J

1- 02E2

-22-

Taking the real parts, equations (53) become

S u -

( a -

1

3 )

Ex

Z

(Ea +

(r xy (x - -

E

VR i

1 - aE cos WO

,

-

(Ea

cr.

+ xy,

e

)

(x - -VR

2

1 - f3e

cos 11

2

) o

x

+ E— x

(61) v -

S

(a ...

(3) Ey

_ (

1 4.

Ea

2 cr )

YX

[( crey - -Ft].

sin Ilii)

ae

(1 -

ae )

+(

E

+

EP

2 cr

Yx

)

(3Ey - f3E (1 sin x li,)

L

- 0E )

5

6 yx

Y

EY

.

(62)

These equations for the displacements would be used to check data given by strain gages of finite length, since the strains computed from the gage readings are average strains.

Isotropic Stress Function as the Limiting Case of the Orthotropic Stress Function

It is of interest to obtain from (16) the stress function for the b correspond-

Substituting from (15) and (30), (16) becomes

Report No. 1510

-23-

F =R

2E (e 442 -

1 a - e

./ eb 2

(z

1

y

2 i log (z1

w1)

1

/2 a - e Pi eb

1

2 (z 2 - w2 )

2

2E 2

(63)

For an isotropic material becomes equal to 0, and

a,

0,

and

E each becomes equal to unity as shown by equations (6), (8), (11), and (15). For

0

0

Evaluating by differentiating both numerator and denominator with respect to (remembering that z 1 , z 2 , and y2 all contain ±), it follows that

Report No. 1510

-24-

•:7

N

0

N r q

,.0

I

0rd

I r\i,

N ra

w

N

N

r0 ra s

N

.....

ga)

...„

-0-0 al

N

0

,

±

....„..i

...,

3-1

N

—I

......0

ell

1

N

......

-e-

I

0

• r•I

I

....

.---.

N

3

I rq

N

N

..,..._:________,.

N t \I

..0

w

--

8

4)

±

N

...,

-0-

1

4)

•r4

-I-

N

rg

-0 uo

-6-

0

N

(N.)

NI

..,

N

.-4

+

N

...,.

-e-

I a)

1

RI

I

F-

N

.0

r%)

.....„...„....

I n

43

-$

-

0

-

N

+

--1

N N r

I

+

NI

4.—

P"

V-4

.--

N

,--I

IN3

......

.--1

N

.0

r.) i24 a)

• ri

N

,

—1

+

-1

N '

v..I

N

v--

1-

(U

+

+

+

...,

e

-

a)

N

.....

-e-

P'`

4)

N

.--.

Fl

N

-

P

N

.......

a)

--I

N

>-

O

+

N (.•1 x

N

+ r`l

..n

,-1 -9-

II

Report No. 1510

-25-

tv

N

W rf) N

N

N

N N

Let = 0 and e = 1. Then (64) takes the form

F=R

[

1 a -b

(z w) ( iyw - iyz - b2" a + w iy

+ b

2 z + w

- b log (z

+ w)

Y2 b

(z w)

2

2(a - 1)2

(65) where w Vz

2

- a + b2

On setting b = a, parts of equation (65) assume the indeterminate form

On evaluating (65), it is found that for the circular hole, o

0

F=R

[a3 a

3

2z

Zaiy z

- a

3

- a log 2z + a

(66)

= R

- a

4

2

2.

4a 3Y

.4, 4

2

- Za z

2 log z -

2 i n

(67)

Since the constant

2a

2

log 2 will not appear in the stresses obtained from (67), it may be omitted. The real part of (67) is in polar form:

F =

4

4 az cos 20 - 4a

2

sin

2

0 + 2r

2

sin g

0 - Za

2

log

(68)

= S [ r2 - 2a

2

log r -

(r

2 r

2

Report No. 1510

-26cos 20

which is the well-known form for the stress function in the corresponding problem for the isotropic plate.2

The Effect of Elliptic or Circular Holes on the Stress Distribution in a Plywood Plate

A plywood panel having the grain direction of each ply perpendicular to that of the adjacent plies and having a symmetrical construction may be considered as an orthotropic plate. The strains in such plywood may be obtained from the average stresses (to be defined later), or the average stresses from the strains,± by exactly the same relations as exist between the strains and stresses for an orthotropic plate if the apparent elastic moduli of the plywood are used as the elastic moduli of the orthotropic plate. The panel is assumed to remain flat under the application of the loading considered.

The x-axis will be taken parallel to the grain direction of the face plies.

The subscript 1 will be used to denote quantities associated with plies having grain direction parallel to the x-axis, and the subscript 2 will be used to denote quantities associated with plies having grain direction parallel to the y-axis.

Let h1 denote the total thickness of all plies having grain direction paral-

lel to the x-axis.

Let h2 denote the total thickness of all plies having grain direction parallel to the y-axis.

For plies having grain direction parallel to the x-axis, the following relations hold

(exx)i

1

-r-

11

(Xx)i

(cr xY) 1 (YY) 1 1

(69)

(e

YY

)

1

1

( E )

Y

1

[

-

(o- yx

)

Report No. 1510

-27-

and for plies having grain direction parallel to the Y-axis corresponding equations hold if the subscripts are changed.

Since the panel remains flat, it will be assumed that (e ) = (e ) ,

XX XX

2

(e )

= (e ) , (e ) = (e ) , and it is no longer necessary to write

YY

1

YY 2 (e x (ex 2 the strain components with subscripts.

Solving equations (69) for the stress components gives

[e ) e

Yx 1 YY1

(70) where

(Y) =

Y

1

( E )

Y

1

X

[e

YY

+ (crx__)i y e xx

X =1 - (o- xY

) (o- )

1

Yx

1

Similar relations hold in the remaining plies if the subscripts are changed throughout.

Assuming the plies are all rotary cut and from the choice of axes, the elastic constants have the following values in the respective plies:

Report No. 1510

-28-

(Ex)1 = EL

( E ) r

1

= E

T

(°xy

)

=

Cr

LT

(E x

)

2

= E

T

(E) = ET

Y

2

(0xy

)

2

=

TL

(a.

yx

) i

= (rTL

(Cr

yx ) 2 = Cr LT

After making these changes, equations (70) become

EL

(Xx)i = X

(Xx)2

E

T x.

[ e

YY

+ cr

LT e xd

E

T x

[e,

+ cr

e

LT YY1

(Y

Y

)

2

=

E

L

[e YY + cr

TL exx] where

= 1 -

TL cr

LT

Report No. 1510

-29-

(71)

It is now important to define the mean stress components X

x

and Y by the following equations:

X x

=

(X x

) i

+ h.

2

(Xx)

2

Y

h

1

(Y y

)

1

+ h

2

(Y y

)

2 hi. E

L

+ h

2

ET hX

exx cr TL EL x

YY

(72) h

1

E

T

+ h

2

E

L hX.

e +

YY

TL

X

EL

exx

1 2 ,

CT

LT

E

Cr

TL

E

T has been used.

Equations (72) may be rewritten as follows:

Ea

X

X m. X e +

0" yx eYY1

(73)

E b

= -

[e +

Y X e

YY xY xx where

E a

= h

1

E

L

+ h

2

E T

E b

= h

1

E

T

+ h

2

E

L

(74)

Cr

TL

E

L

E b

Report No. 1510

7

=

Yx

-30cr

TL EL

E a

Solving equations (73) for the strain components gives e XX

(1 -

Tr" 7 )

xy yx

1

E a

1

(1 — Tr

sTr" )

xy yx

E

b

Y 7 .5Z

y yx x or -

1 e xx E

[ x xy

T y

YY E

1

[V

-

c7 3-E y yx x] where?

T x

-

E a

(75)

(76)

It is easily established that

7

E x

7

—Since o, xy xy , imately equal to E a and Eb , respectively.

Report No. 1510

-31-

In plies with grain direction parallel to the x-axis, the conditions of equilibrium require

8(X x

)

1

8x

8(X__)

Y

1

=0

8y

8(x„)

Y

1

8x

a( y )

Y 1

8y

- o and similarly in the remaining plies

a(x)

8x

2

+

8( Xy ) 2

=0

By a(x)

8x

2

a( y )

Y 2

ay

Then h

8(Xx )

1 a(X )

Y l

8x

+ h1 ay

= 0

(77) h 1

8(x)

1

ex a(yy),

a

Y

=

0

(78) and

Report No. 1510

-32-

h

2 8x

2

+ h2

a(x )

ay

2

= 0

(79) a(x )

2

+ h 2

8(Y)

2 ay ax

= 0

Adding equations (77) and (79) and dividing by h, it is found that arc x

Bx

+ aR.

By = 0 where, since in all plies the relation

(80)

(81)

XY = ILLT exy holds, the mean stress component ponent X .

Similarly equation (78) and (80) yield s identical with the stress corna3E

8x aV ayY

= 0

(82)

Equations (81) and (82) will be satisfied by a stress function F, such that

2

= 8

x

8y

F

1

=F ax

'

XY

8

2

F

exey

(83)

Substituting (83) in (75) and then making use of the compatibility equation

8

2 e

By

XX

2

+

8

2 e yy

8 x

2

_

2 e

8x8y

Report No. 1510 -33-

the following differential equation satisfied by the stress function F is found:

1 8

4

F

E y

8x4

[ 1

1-LLT

27 xy

Tx

8x

2 ay

2

1

8

4

F

ry

3y

4

As in the treatment of equation (3), it will be convenient to rewrite this equation as

84

8x 4

+ 2K

8

4

F

8 4 F

2

871

4

= 0

(84) where

K =

--\/Txfy

1

2

-

217 xy

T x

)

(85) by making the substitution

=ey

(86) where

(87)

The differential equations (84) and (5) are of the same form mathematically, and any solution of (5) is a solution of (84) if the appropriate constants are used. It is then possible by using equations (31), (32), and

(33) to obtain the stress distribution in the neighborhood of an elliptic hole situated in the middle of a plywood plate subjected to a uniform tension, provided that the axes of the ellipse are parallel and perpendicular to the grain of the face plies and that the tension is applied parallel or perpendicular to the axes of the ellipse.

Report No. 1510

-34-

Hence, if the

K and E calculated from the apparent elastic moduli of plywood are used, all of the preceding discussion for orthotropic plates containing elliptic holes holds for plywood panels.

Numerical values for the stresses at various points along the x- and yb axes in the case — a

= 1 for rotary-cut Sitka spruce plywood having half the material with grain direction parallel to the x-axis are given in tables

6 and 7. Figures 10, 11, 12, and 13 illustrate how these stresses vary in the neighborhood of the hole.

The foregoing analysis for plywood is based on the assumption that all plies are rotary cut and of the same species of wood. When these assumptions are not fulfilled, it is easy to make suitable modifications in the analysis (see appendix).

Report No. 1510

-35-

Appendix

A plain-sawn or a quarter-sawn wood plate or a plywood plate may be considered to have two perpendicular axes of elastic symmetry in the plane of the plate. The discussion given in this appendix applies to a rectangular plate of this type with its edges parallel to the axes of symmetry and with a small circular hole at the center. The x- and y-axes are taken as the axes of symmetry. A uniform tension S directed parallel to the x-axis is supposed to act on the two opposite edges of the plate, and the tensile stress on the edge of the hole at the end of the diameter perpendicular to the uniform tension is given by

S (ae + OE +1) .

For wood, the constants that appear in this expression are defined as follows: x y a

=

e4)/2

13

= e

-1)/2 e

= natural logarithmic base

E x Young's modulus in the x-direction

K

, -1

= cosh -1

E x

E y

2

1

1-1

xy

2a-

E xy ,

x

/ p. = xy modulus of rigidity in the xy-plane

Report No. 1510

-

36-

cr xy

= the Poisson's ratio given by the ratio of the contraction

parallel to the y-axis to the extension parallel to the xaxis associated with a tension parallel to the x-axis.

For plywood, equations similar to those already given for wood define

E , a, j3, and

K if the remaining constants are as subsequently defined.

The plies will be considered to be numbered consecutively from the upper to the lower face. The elastic moduli in the ith ply will be denoted by

That of the plate is h.

Then approximately h

Ey = Eb =

E (Ey ) i hi

Fixy

(1-1.

xy

) hi h xy

=

Z h i

(E )

( 0- ) y xy h E b

The shear parallel or perpendicular to direction of the uniform tension and on the edge of the circular hole is given in the mathematical part of the report by equation (43) if b is set equal to a. By substituting the constants defined previously, this shear stress may be plotted as a function of the polar coordinate 0 (fig. 5), and its maximum value read from the curve.

Report No. 1510 -37-

.4-52

Table 1. --Elastic moduli of Sitka spruce and yellow-poplar

Species

Sitka spruce

: E :

E

T :

E R :

µ

1_,T : I'l.,R : cr

LT

: 1,679: 76 :

1,000 lb. per sq. in.

(2) :

112 : (2) : 0.464

Yellow-poplar : 1,775 :

(2) :

130 : (2) :

113.6:

(2)

:

: cr-

LR

(2)

:0.337

-The elastic moduli of Sitka spruce are average values obtained from a large number of tests made at the U. S. Forest Products Laboratory. The elastic moduli of yellow-poplar are values obtained from the one specimen used in the experimental verification of this report.

— Not used.

Table 2. --Observed and theoretical values of the strains at the points P and Q of figure 3

Type

: Strain at P : Strain at

Q

Observed

: 12.3 x

10

-4

Inch per inch

:

2.75 x

10

-4

Theoretical

(computed from : equations (35) and (45b))

: 3.22 x 10

-6

'

S :

0.603 x 10

-6

S :

:

Ratio of strain at

: P to strain at Q

4.47

5.34

Report No. 1510

4)

:64

Id

H

F-r rd

U

• r-I

U a) a)

0 a) a)

U

;-1

0

(13

X

>

1 a)

0

0

0.1

0

U rd

4) rd a)

4-1

0

Kr) a) rd

/-1

O

.+4 a)

0

7c1 o tfl

0

O

444

•• ••

0

O

M

0

M q•

0

0

to

0

O eN3

0

to

• . .

.0

to

0

0

• .

• •

.43

to

M

O

O

O

O

•et4 to

M

O

0

0'

to

O

0

O

0

- •

co ret

0

O

. .

1

n

1 to

O

0

• •

O

0 tO

CO r•I

0

0

••

0

0

I

0

›ti cc

Report No. 1510

0

Cr,

O

•• ••

(NI

00 t11

• .

to

CO

1

0

0

00

:

•• •• •• ••

CO

03 n

00

.0

0

.• co

N

1 rN.1

r-

0

0 s.0

0 a

• • —

Cre

CT,

0

0 co

0

cb I t.

.

. .

M en

0

0

Table 4. --Values of the stress components at points along the x-axis exterior to a circle of radius a and with center at the origin for a plain-sawn plate of Sitka spruce x

X x

S

1.0a

1, la

1.2a

1.3a

1.4a

1.5a

1.6a

1.7a

5.0a

6.

Oa

7.0a

8. Oa

:

:

•.

:

:

0

-0.0157

.0216

.0198

.0118

+ .0010

.0175

.0368

1.8a

3.0a

4.0a

: .0582

1.

9a

.0811

2.0a

.1050

.3448

.5280

.6530

.7379

.7967

.8384

9. Oa

10. Oa

.8689

.8918

Y

_Y.

S

-0.2128

-,,. 1762

.1458

.1205

.0993

.0817

.0669

.0554

.0440

.0352

.0278

+ .0052

.0106

.0102

.0087

.0073

.0060

.0050

.0042

Report No. 1510

4.0a

5.0a

6 .

0 a

7.0a

8. Oa

9 .

0 a

10.0a

1.0a

1. la

1.2a

1.3a

1.4a

1.5a

1.6a

1.7a

1.8a

1.9a

2. Oa

3.0a

Table 5. --Values of the stress components at points along the y-axis exterior to a circle of radius a and with center at the origin for a plain-sawn plate of Sitka spruce

X x

S

5.8439

2.7262

2.0069

1.6871

1.5089

1.3968

1.3205

1.2657

1.2248

1.1932

1.1682

1.0629

1.0335

1.0209

1.0143

1.0104

1.0079

1.0063

1.0050

S

.

0

0.1566

. 1522

. 1345

.1171

. 1021

.0895

. 0790

. 0701

. 0627

. 0564

.0245

. 0137

. 0087

.0060

. 0044

. 0034

.0027

. 0022

Report No. 1510

Table 6. --Values of the stress components at points along the x-axis exterior to a circle of radius a and with center at the origin for a rotary-cut plywood plate of Sitka spruce

x

Xx

1.00a

1.05a

1. 10a

1.15a

1.20a

1.25a

1.50a

2.00a

3.00a

4.00a

5.00a

:

:

0

-0.0223

.0195

.0076

+ .0085

+ .0268

+ .1275

+ .3130

+ .5682

+ .7143

+ .8005

-1.0000

- .4324

- .2467

- .1934

- .0630

+ .0062

+ .0229

+ .0191

+ .0146

Table 7. --Values of the stress components at points along the y-axis exterior to a circle of radius a and with center at the origin for a rotary-cut plywood plate of Sitka spruce

Xx

1.0a

1.2a

1.3a

1.5a

2.0a

3.0a

4.0a

5.0a

Report No. 1510

4.1335

2.0108

1.7417

1.4557

1.2018

1.0733

1.0375

1.0227

Yy

S

0

0.2377

.2547

.2583

.2212

.1451

.0973

.0683

b

Y

Figure 1.--Elliptic hole in orthotropic plate.

5

Figure 2.--Orientation of positive normal and

- tangent on the boundary of the elliptic hole.

Z M 54248 F.

_5

TRAIN GA

G

E f(

5TRA/N GAGE

IN5I D E HOLE)

Figure 3.--Circular hole in a quarter-sawn yellowpoplar board subjected to a uniform compression.

Z 14 54249 F

Figure 4.--The tension flange of a loaded wood box beam showing shear failure caused by vertical holes. The beam is subjected to a lower load in A than in B. The small cracks occurred in the stress-coat, which was applied to the beam before testing. The small round objects paralleling the edge of the flange are Whittemore strain gage points.

Z M 54250 F

oZ

t0.2

)N.

Lk.1

Lu

I

Q '24

L k.

ki k5

(1)

#0./

0

-0./

-0.2

-0.3

-0 4

-0 .5

-0.6

-0.7

-0.8

A

1 1

Pli

w-,

,,

.

n k

U

.1NLA

A

miirl

1 i

El

-0.9

19•

10°

20* 30' 40- o (DEGREES)

SO'

t0 • 70'

80°

90'

Figure 5.--Variation of the shear stress component X, around the edge of a circular hole for a plain-sawn plate of Sitka spruce and for an isotropic plate.

1.0

• K

L.-,

a

7 k

0.6

4.1

0.5

k

cr 0.4

k k

lu03

1:21

)c

02

O.

1

0

-0. /

„„npic

MI

=Vali

,

2a 3a.

40.

7a

8a 9a ioa

Figure 6.--Variation of the normal stress component

X x at points along the x-axis exterior to a circular hole of radius a and with center at the origin for a plain-sawn plate of Sitka spruce and for an isotropic plate.

Z M 54 251 F

t t

1 ii

HY

In e,

Q

0 v

't

.--

In

0 N

O

1

0

.

'Zi

^'1

0

I

41

Is, CO

0 0 CS CS i

r

I I I

IZ?

A1INZ.VA// AI CSIZY1S 2111 S. 01

C/XV-X

71/1 9/1/07V

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6a. 7a

Figure 8.--Variation of the normal stress component X x at points along the y-axis exterior to a circular hole of radius a and with center at the origin for a plain-sawn plate of

Sitka spruce and for an isotropic plate.

Z M 54253 F

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4a

Figure 12.--Variation of the normal stress Component X x at points along the y-axis exterior to a circular hole of radius a and with center at the origin for a rotary-cut plywood plate of Sitka spruce.

5a

z

0.28

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Figure 13.--Variation of the normal stress component Y v at points along the v-axis exterior to a circular hole of radius a and with center at the origin for a rntarv-cut p l ywood plate of Sitka spruce.

Z M 54256 F

5a

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