AD-A159 887 DEFLECTION IN TAPERED CANTILEVER BEANS DEFLECTION (GAP li'1OPENING) IN DOUBLE.. (U) RMY RMAMENT RESEARCH ANDDEVELOPMENT CENTER NATERYLIET NY L.. B RVITZUR RUG 85

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AD-A159 887

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TECHNICAL REPORT ARLCB-TR-85027

DEFLECTION IN TAPERED CANTILEVER BEAMSDEFLECTION (GAP OPENING) IN DOUBLE CANTILEVER

TYPE FRACTURE TOUGHNESS SPECIMENS

BOAZ AVITZUR

AUGUST 1985

US ARMY ARMAMENT RESEARCH AND DEVELOPMENT CENTERLARGE CALIBER WEAPON SYSTEMS LABORATORY

BENET WEAPONS LABORATORYWATERVLIET N.V. 12189-

APPROVED FOR PUBLIC RELEASE; DISTRIBUTION UNLIMITED

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DEFLECTION IN TAPERED CANTILEVER BEAMSDEFLECTION (GAP OPENING) IN DOUBLE CANTILEVER FinalTYPE FRACTURE TOUGHNESS SPECIMENS S. PERFORMING ORO. REPORT NUMBER

* 7. AUTNO () S. CONTRACT OR GRANT NUMMER1()

*Boaz Avitzur

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111 SUPPLEMENTARY NOTES

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Gap OpeningFracture Toughness

.- Tapered Beam

1= A@STRIAC (Catffawe swrvee *sft ffnamsin, ~aitUt S b oek number)When an othervise homogeneous material under stress contains small defects(i.e., internal cracks and/or voids), the stresses at parts of the material-defect Interface significantly exceed the ones anticipated at that location in

the absence of such irregularities. Consequently, a structural member, other-wise calculated to safely sustain the applied loads, might unpredictably fail.

That branch of engineering which intends to account for such 'stress-raisers'is known as fracture mechanics. Fracture mechanics studies have found that

(CONT'D ON REVERSE)porn

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*20. ABSTRACT (CONT'D)

* "different materials (and even the same material when loaded in different-; orientations) reflect different sensitivity to such 'stress-raisers'-aamaterial property known as fracture toughness. Test samples and testing

procedur3s have been devised in order to quantify this material property. The-" relation between the applied load and its displacement (or gap opening) at the

point of crack growth is being used herein to determine (compute) materialfracture toughness.

While the equations derived for the stress field near the edge of a defect inan otherwise uniform field assume an infinite volume of material to surroundthe (relatively) very small defect, the crack to width and/or height in theselaboratory size testing samples is definitely a finite one. This report

- offers a mathematical relation between the applied load and that part of theeflection (gap opening) which is due to the cantilever-like part of the

- sample, for that class of fracture toughness test specimens which can be- described as double cantilever. A beam theory approach is used.

SECUITY CLASSIFICATION OF THIS PAGgrheM De Entea'.)

%"%

A-W

--4

-. TABLE OF CONTENTSPane

INTRODUCTION 1

PARALLEL STRESS COMPONENTS 3

Deflection Due to Bending Forces Only 8

Deflection Due to Combined Bending and Compressive Forces 12

CONVERGING STRESS COMPONENTS 18

Deflection Due to Bending Forces Only 20

Deflection Due to Combined Bending and Compressive Forces 26

SUMMARY 31

REFERENCES 32

TABLES

I. GAP OPENING (HALF) IN TAPERED DOUBLE CANTILEVER BEAMS 33

LIST OF ILLUSTRATIONS

1. Tapered double cantilever beam. 1

2. Compact fracture toughness specimen. 1

3. Definition of cartesian coordinates and major parameters. 3

4. Conversion from cartesian to cylindrical coordinate system. 18

5. Radial and cartesian components of normal stresses. 19

.

I

t.i

o-

2 . 2%

INTRODUCTION

g One type of specimen commonly used in tests designed for thedetermination of a material's fracture toughness can be looked upon as a double

cantilever beam, as shown in Figures I and 2.

-~~ 2- -- h=h x *tanoc

FO. ,cos cce -P-i , P

x

P'.sincc,,,p ./ - _. , -,

Figure 1. Tapered double cantilever beam.

hz= tz-. X'

Figure 2. Compact fracture toughness specimen.

• ; ." . " " ' L .' ' ' _ ' ' _ ._ , " .. ., ... - ... .-' -. ' , ." , , ... " ," . . .% . .- • • • .-. ., • • .. ' -/ -' ' .-. ' - .. -" ., I

The data collected in such tests to determine the material's fracture

toughness property includes the gap opening, 26', at a given load, ', and for

a given crack length, V. The stress distribution in such specimens is a

complex one and so is the resultant strain distribution and its contribution to

the gap opening. The computation of that portion of the gap opening attributed

to the elastic bending of a cantilever beam is the subject of this report.

These calculations are based on an elastic beam theory. The shortcomings of

the computed gap opening derived are as follows:

1. The fracture rarely propagates without attaining a plastically

deformed region near the tip of the crack.

2. The calculations do not account for the deformation in the remaining

portion of the specimen, namely at the areas x1 > V, beyond the crack length.

The added gap opening due to the propagation of plastic deformation should

be computed separately using the technique used by this author elsewhere (ref

1). The contribution of the deformation in the region x' > ' towards the gap

opening, can also be computed separately. (Neither one of these calculations

is included here.)

In treating each half of the specimen as a tapered cantilever, one has to

realize that its neutral axis in bending is not parallel to the specimen's

plane of symmetry (x' axis in Figure 1), but rather at an angle a to it (as

will be shown later, the effective neutral plane deviates from the bisecting

plane of each cantilever, too). As a consequence, the computation of the

lBoaz Avitzur, "Retained Deflection in Circular and Concentrically Hollowed

Beams After Local Removal," to be published.

" "' "" " ?' " °" " " "" --- " . - ' "." ,.---- - i.'- --. ..- '" ' .-. ' . .- -- --... -. , ... . -. .,2-

bending moment and of the shear forces, as well as that of the resultant

stresses thereof, will be based on the component P - P'. cos a of the appliedho

load P', where P acts normal to the axis of symmetry, x (where x - [x' - - tan2

a] cos a).

PARALLEL STRESS COMPONENTS

As a first approximation, the gap opening 6 - 6bend + 6shear is calculated

in the cartesian coordinates system x-y-z, which is at an angle a to the

xI-y -z t coordinates and only the normal component, oxx, and the shear

component, OXY - oyx, are considered. This is followed later by correcting the

stress components to be parallel to the external surfaces at the boundaries

ho + 2x'tan a=-+±

2

h' h

P~~~~ - coc -P

P'cosx =P/P

p= P'. sine

Figure 3. Definition of cartesian coordinates and major parameters.

3

The basic dimensions in the x-y coordinates and their relation to those in

the x'-y' coordinates (see Figure 3) are as follows:

ho t

ho - = the beam's cross-sectional height at the point of load applicationCos a

ho

x - (x' - -- tan a)cos a, where x' is being measured along the top surface of2

the beam, which is the fracture toughness specimen, along the crack

surface.

hot

= (t' - -- tan a)'cos a E-distance of crack tip from point of load2

application

t - 2"tan a

P - P'cos a E component of load normal to plane of symmetry

p - P'*sin a E component of load parallel to plane of symmetry

h - ho + t*x E beam's height normal to the plane of symmetry at a distance x

from the point of load application.

Thus, the cross-sectional moment of inertia becomes

bh3 (ho+tX)3

12 12

and the local curvature p of the beam is

1 M(x) Px P x- =- . .- . 12 - - (2)SE1I(x) EI(x) Eb (ho+tx)

3

and the total deflection due to bending moment is

Yo+ eodx+ 12 f drdx (3)0 Eb 0 o (ho + tx)

3

4

/ .'."-... ...- .- -, -.,. .- , ....- .. . .* ..* - -... .-.... -. .-.. ,," -, . ..: . - ..-. .. * . .. - .- *. ..-. -.-. -..

for whiche 2P Jfo x__ . Px - + 1 ho

x

-E oo + 2h -o+tx) dx-0o+ 2 h - o + tx 2(otx2 o

p x2

00 12 -- - (4)Eb 2ho(ho+tX) 2

where 0 - angle of deflection

60 = angle of deflection at the point of load application.

However, at the tip of the crack, x = 1, OX - 0. Therefore,

p 9.2e= = e0 o - 12 -- *-------

Eb 2ho(ho+t 9)2

from which we can determine thatp 9.2

0 = 6 - --- (5)Eb ho(ho+t 9)2

By applying Eq. (5) to Eq. (3), one gets

Y W YO + 6odx + 12 !-oxj -x) dxdx

0 Eb o o (ho+tx)2

Ebho h0 t) tx d

P2 x 1 ho xYo +6-- {t------- / - -- [h°+tx - 2h° log(h°+tx) - ------- I

Ebh 0 (h0+t9.2) o t 3 ho + tx o

P t 2 1 2ho + tx ho + txY+6 --- (---x- [- tx -2h o log -o ---- (6)

Ebho (h0+tT)i t ho+ tx ho

5

Since at the tip of the crack, x - £, and the deflection y, 0, one gets

p 3 2ho + tZ ho + tiyi - 0 - Yo + 6 -- [- -- tt - 2h o log -- ]} (7)Ebho (ho+tt)2 t ho + tX ho

However, for a straight cantilever where t 0 0, one has to determine

t3 L3 I 2ho + tx ho + txyp - o y + 6- {---- ----- lim(-- [---------- tx- 2holog--------1)

Ebho (ho+tt) 2 t+0 t [ho- tx ho

whereI 2ho + tx ho + tx x

3 x 3

tO t-------- tx - 2ho log -------- lim .-------- (8)t+o t ho + tx ho t+ 3(ho+tx)

2 3ho 2

from whichp Z 3 X3 p 2 3

lim Yo - -6 ---- .--- --- -4 -- (--) = yo' (9)t+O Ebho ho

2 3ho Eb ho

which is the same as being reported in handbooks (ref 2) for a straight

cantilever of a uniform cross-section.

Another method of computing the gap opening is by invoking Castigliano's

theorem (ref 3), equating the internal deformation energy with that of the

outside work done on the structural member. This method will facilitate

computation of the gap opening due to bending moment and due to shear force

(and any other forces that impose internal deformation work).

The distribution of that component of normal stresses, 0 xx'm. which is

parallel to the axis of symmetry and which results from the bending moment, is

as follows:

2Raymond J. Roark and Warren C. Young, Formulas for Stress and Strain, FifthEdition, McGraw-Hill, NY, 1975, p. 98.

3A. C. Ugural and S. K. Fenster, Advanced Strength and Applied Elasticity,American Elsevier, NY, 1981, pp. 146-148.

6

H Paxxm = Yy (10)

where P - P' cos a, a - arctan (t/2) - arctan (Ah/2Ax), and I - bh3/12bb (ho+tx)3 .12

The component of shear stress which lies in a plane normal to the plane

of symmetry and in the direction of the axis of symmetry (or opposite to it),

ayxm and its equivalent oxy,m , is assumed to be due to gradient in that

component of the normal stress doxx,m/dx, resulting from the bending moment.

This is computed as follows:

2 Ih/2 fb/2 Alxx,m dzdy

0 o Ax

yx'm 2

where M

Oxx'm y

Thusdaxx,m d M d 12 P d x

Sd( () - P . y - -- x) 12 - y -- ( --------)dx dx I dx b(ho +tx)

3 b dx (ho +tx) 3

where

d x ho + tx - 3tx ho - 2tx 1 tx-(----------------- ------ [1 -3----

dx (hO+tx) (ho+tx) 4 (ho+tX) 4 (ho+tx) 3 ho + tx

and thusp f.(ho+tx)/2 b/2 ho - 2tx

24 -f f 'zd

b y 0 (h0+tx)4 12P (ho+tx)/2 ho - 2tx

ayx' m M f ydyb b y (ho+tX)

4

3p ho - 2tx 3p ho - 2tx y 2

oyx'm - - ------- [(h+tx)2 - 4y 2 ] - 2-----2 [1 - (2-ho-tX

(ho 2b (hootX) h[+tx

3P ho- t

2b (ho +tx)2 ho+tx

7

Thus

Ubend3 -(12 +- t2)t5 bE o (h0.itx)3~d

whereX2 1 ho + tx 2ho + 3tx

dx -- [lo - -------- - txo (ho+tx) 3 d 3 ho 2(ho+tx) 2

Thus

12 9 1 p2 ho + tx 2ho + 3txUbend3 (-I + - -) - [lo9 -- 2--- -3--------- tx] = P6bend3

t 5 t bE ho 2(h0+tx)

from which12 9 1 P ho + tx 2ho + 3tx

6bend3 f (- 5 + - -) -- [o10 ------- --------- tx] (26)

t 5 t bE ho 2(h0+tx)2

and as before

12 9 1 p ho + tx 2ho + 3tx

bend3 + bE ho 2(ho+tX)2 txcos2

At the limit, as t + 0, for a beam of a constant cross-section,

P 12 9 1 ho + tx 2ho + 3tx

lim 6bend3 -- z ((- +- JI9---------2txt+O bE t+O 3 5 t ho 2(ho+tx) 2 tx]

whered ho + tx 2ho + 3tx-- [log ---------------- tx]

ho + tx 2ho + 3tx dt ho 2(ho+tx)

2

lim [log tx] nlim -------------------t+O t ho 2(ho+tx) t*0 d

-- (t 3)db

x [(2ho+3tx)x + 3tx2](ho+tX)2 - 2(2ho+3tx)•tx"(ho+tX)"x

ho + tx 2(ho+tx)4= lim 2- - - - - - - -- - - -t+O 3t2

21

.-.. :,...

.... Z . ........ ..

Since fxxm (or arr,m) is tensile in the range 0 4 8 < a (or 0 4 y 4

(ho+tx)/2) and compressive In the range -a 4 6 4 0 (or -(ho+tx)/2 < y 4 0),

ayy,m , Oxx,mtan e maintains its upwards component and these do not cancel

each other (Figure 5). Nevertheless, no discontinuity prevails at 6 - 0 (or

y - 0), since ayy,m = 0 on that (internal) surface. The contribution of the

stress component, ayy,m to the deflection is accounted for by the energy

balance in its totality.

Deflection Due to Bending Forces Only

Since

P xyUxx,m = 1217xxM 12b (ho +tX) 3

p t 2y2 xyOrr,m = 12 - +-- -------

b (ho+tx)2 (ho+tx) 3

from which the local strain energy, due to bending moment, ubend3 becomes

yrr,m 2 p 2 t2y 2 x2y2

ubend3 °rr,m"crr,m 144 (h+t2 (h;+tx2 6

Hence, one can compute the total strain energy due to the bending moment as

follows:p2 f(ho+tX)/2 b/2 x~y

2 x2y 2

Ubend3 f Ubend3"dv - 576 -. r P [ - -- -Idz'dy'dx

b2E o o 10 (he+tx) 2 (he+tx)6

p2 x(ho+tx)/2 3 2Y2l288-f f I1 + --- - - ' dydx

bE o o (h+x)2 (h0+tx)rwhere

(ho+tx)/2 t2y2 x2y2 X2 y3 t 2 y5 (hei-tx)/2f ~[ + - - I....... ........a(h +tx)2 (h0+t) 6 (h ttx)6 3 (h+x25o

x 2 1 t 2

(h0-4tx)3 24 160

20

% **...*% . --*..- - - . % .- -. .....- . ". . . .. -. . . -* . -*",-* "• -. - . - ... o. ,- .".. -. ...- .. ...-.-. . . ...-. . .. . .. . ..".-.

of any cross-sectionl at a distance x from the line of load application is

L a L'+x - + x from the origin of these cylindrical coordinates (wheret

t - 2-tan ai), and the distance I of any point (x,y) on a plane x and at a

distance y from the ais of symmetry is:

I -7 + y'- ( )2 + y 2 - -(h 0+tx) 2

t t

from whichOrr = '(ho+t) + t'Y' -

ox L ho+ tx ho+tx

or

Orr 2 [+ -~y 2(ho+tx) 1 Ox

Figure S. Radial and cartesian components of normal stresses.

19

CONVERGING STRESS COMPONENTS

By assuming that the normal stresses, oxx, are parallel to the plane of

symmetry, one violates the boundary condition at y - ± (ho+tx)/2, in a tapered

beam. Therefore, it will be assumed here that the normal stresses, Orr- at the

surfaces of y - ± (ho+tx)/2 (or e - ± a) (see Figure 4) are parallel to these

surfaces and that the radial stresses gradually align themselves with the axis

of symmetry (6 - 0). Or, in other words, these are radial stresses, as their

subscript suggests, in a cylindrical coordinate with the origin at L'

(ho/2)*tan a - ho/t distance from the point where x - 0. Thus, the distance L

h '

Figure 4. Conversion from cartesian to cylindrical coordinate systems.

18

"'' " '' . / " " "' " '""""" "" "', ' " "".. . . . . . . . . . . . . ."" ".-- - - - -" -".'- '.' '- - - ,- - -,,- - °'.-.-'.

Hence HnvP 12 1 ho + tx 2ho + 5tx P ho + txsheer2 " - (-- (41og 3 tx] + -- og -- ) (24)

E bho 2(ho+tx) 2 2 -

and1+v P' 12 1 ho + tx 2ho + 5tx t 3 ho + tx-- --{--. [4-log --- - 3 tx] + - log

-.--- cos2 ashear2 b 5 t ho 2(ho+tx) 2 24 (4

(24a)

Combining Eqs. (22) and (24), one gets

1+v P 12 1 ho + t5tx t3 ho + tx

--- { - [4.log -------- 3 - - tx] +--log--

6shear2 E b 5 t ho 2(ho+tx)2 24 ho

------------ ----------------------------------- ------------------------6bend2 P 12 ho+tx 2ho + 3tx t ho

+ tx- {2 [log -- -------------- xl+ --- ------bE t 3 ho 2(ho+tx) 2 4 (25)

As before (in computing lim 6bend),t+o

ho + txlia {_ log ---- I - 0t*0 24 ho

Therefore12 P 1+v x

lim 6shear2 - - - (--)c+0 5 b E ho

as in Eq. (16), and

6shear2 3 holim ( 1+V) (-_) 2

t+O 6bend2 5 x

as in Eq. (17).

17

* - --- - ~ - ----. 7 7 7'7'- -,.7" - ' _ _'. . :- I C -.-- ' _.

Thus, the total shear strain energy due to the combined bend and

compressive forces and over the entire volume of the beam Ushear2 becomes

f(ho+tX)/2 l+v p 2 1 hO+tx)/2___ar- 2b_ ser*yd 9(ho 24hotx+4t 2x2).Ushear2 02b 0 E b 00

I 8y2 16y 4

- ---- ------------(0 tx)' (h0+tx)

6 (h0+tx)8

t4 1 41yI 4y2

+--_ [ -------- ----- --------- ]dydx4 (ho +tX)

2 (ho+tx)3 ' (ho+tx)

As in the derivation of Eq. (4)

(ho+tx)/2 1 8y2 16y 2 4 1f - - ----- ---- ----- ------ dy -. ----S(h+tx) 4 (ho+tx)

6 (ho+tx) 15 (ho+tX)

and

(ho+tx)/2 1 41yl 4y 2 1 1 1 1 1f -- - ---------- dy[- -- +--------- ---------

0 (h0+tx)2 (h0 tx)

3 (h0+tx)4 2 2 6 ho + tx 6(ho+tx)

Thusl+v p2 12 ho2 - 4hotx + 4t 2x 2 t4 1

Ushear2 . - -- . f {-----------------}dxE b o 5 (ho+tx) 3 24 ho + tx

As in the derivation of Eq. (14)

ho 2 - 4hotx + 4t2x2 1 ho + tx 2ho + 5tx(ho+tX)3 - dx - - {4"log --- - 3 ------- tx}(h+x3 t ho3 2(ho-9-X)

2

whereas

x dx 1 ho + tx

f - - logo ho+tx t

Thus

l+v P 2 12 1 ho + tx 2ho + 5tx t3 ho + tx

Ushear2 7 -- [4 "log ------ - 3 -h-)... tx] + -- log --5tho 2(h 0 +tx)2 24 h

6shear2

16

an in Eq. (9), since

lim -0 og -+ t 0t*O 4 ho

By adding the two components of shear stresses due to the bend forces, P'

* Cos a, and due to the compressive forces, P' sin a, from Eqs. (11) and (18),

respectively, one gets

P ho -2tx yt 2 lyla YXMCrY' + a {3 *1 - (2 Ty-)21-----1- - 2----11

2b 0 tx h07x 2(h 0 +tx) h0 7t

Thus, the local combined shear strain energy, ushear, due to the bending and

compressive forces is

1+IV 1+v P2 ho - 2txC 2 a 2 --_ --- { - ---- [ 2 Y-,ushear2 OY -X 2 3TX E 2b2 (h0+tx) - ho,+tx

t2 lyl(ho+tx) ho+ tx

- +V p2 {9h 02

-4hoti + 4t2-x2 (2 y--2(2

E 2b2 (h0-Itx)4- -[ h0+tx ho-Itx

ho- 2tx -( - --- t 2 + -t'h07t).l ( ho.itx ho-Itx 4(h0+tx)

2

a s4 -4 l y- - + -e(

ho + tx (h+tx)3

where, due to symmetry around the -axis, the term

h o 2tx3 " 1 . .- (2 (Y--)2 ]*[1 - 2 tt(h-+tx) ho+tx ho + t(

cancels itself on both sides of the axis of symmetry.

15

r.~~i - % 2+.2 o - t

- SS.. ........ - 2 ...---. .-~~ ~ - 5 .. , 2(ho*tX h o-... + tx *-.-< '.

,z(h.+tx)/2 P2ho+tx) 2 z2y2 +t2(' +./ f. Odeo(7

2b e7d-------- + t2 )

Ubend2 Jo o 2bE o o (ho+tx)4

0 1

dyodx(ho+tx) 2

where, as in the derivation of Eq. (12)

f (ho+tx)/2 x2y2 1 x2o hIX dy =

0 (h0+tx)6 24 (ho+tx)3

." In addition,

(ho+tx)/2 t 2 t2

fo (ho+tx) 2 dy 2(ho+tX)

Thus

p2 X2 t2

Ubend2 - {24 + dx2bE o (ho+tX) 3 2(ho+tx)

p2 24 2h0 ho2 t x

{ -- ( log(ho+tx) + -- - - -1+ - log(h0 +tx) }ho t tx 2(ho+tx) 2 0

P2 12 ho + tx 2ho + 3tx t ho + tx

- -I- [log ho -- ho--- tx] + - log -- } - P6bend2 (21)bE t3 ho 2(h0+tx)

2 4 h

from which P 12 ho + tx 2ho + 3tx t ho + tx6bend2 {3 (log h+ - log -- (22)

b' ho 2(ho+tx) 2 4

and as before

P' 12 ho + tx 2ho + 3tx t ho + tx

6'en2 -{- (og - -- -- - x]+ - log -- coo2 a (22a)bEbend2 { [ h0 2(h+tx) ho

which, as was previously shown (following the derivation of Eq. (12))

*.. P x

lim 6bend - -4 - (--)3

t+O bE ho

14

*° ..." .

*, *'*.- %i' *-* ", .** % "e , *"'. ."" .',;' - 7''''..-,..,:-,- , '- '7' ' ' '_'.'_' , '. :.' ..-.'.-.' •.

Considering the symmetry of the beam, one arrives at

-Pt2 lyl -P't lylOy~ p- bhot (1 - 2----] - [----- - 2h -+ ui tx1a- bh o + tx 4b(ho+tx) [- 2 tx

By adding the two components of normal stresses due to the bend forces,

P''cos a, and due to the compressive forces, P' sin a, one gets

P xy P t P xy 1Oxx axx,m + -xx,p 12 -------- -- 24---- - t)

b (ho+tx)3 2b ho + tx 2b (ho+tX)2 ho + tx

(19)The term

{24 ....- - ti(ho+tx) 2

in Eq. (19) is not symmetric in respect to the plane of symmetry (y - 0)

(except for beams of constant cross-section, where t - 0). Thus the beam's

geometrical plane of symmetry is not necessarily the plane of symmetry for the

stress distribution.

As in the derivation of Eq. (12), the local strain energy due to normal

stresses, ubend2, is

dxx 2 p 2 {5 2 Y 1Ubend 2 - Oxxcxx - -- (76 - 8(ot)2t+ t2 (ho+tX) 2

E 4bE (ho+tx)' (h0+tx)2 ( 0 t)

(20)

However, in integrating over the entire volume and due to the symmetry

around the x-axis, the form

48 xy

(ho+tX)2

cancels itself on both sides of this axis. Thus, the total strain energy due

to the normal stresses and over the entire volume, Ubend, is

13

.. ..... .. -.. - ..... -...*. .*.* .. , -..?. ,., -. .. ?.,*, ,. *?.? . .......... ,.......''. "-.* -,*'.-.. , o''.- . '."-" . .. . ,. ., .

Thus12P1+v x 12 P 1+v x

S 6shearl - - - -(4-3)(;) - () (16)t.o 5 6 R 5 b E (;0,

and together with Eq. (9) (where yo - 6bend)

12 P 1+v x

6shearl 5 b E ho 3 ho.rn - M - _ M - (,+V) (_-) 2 (17)

t+O 6bendl P x 3 5 x4 b- (ho)bE ho

Deflection Due to Combined Bending and Compressive Forces

If one incorporates the effect of the compressive stresses

PS P sin a Pt-xx-p . sin a-b(ho+tx) b(ho+tx) cos a 2b(ho+tx)

then the change in compressive stresses due to change in cross-sectional area

becomes" '. d oxx 2 P1' tX X S P~i' t - - i

dx 2b (ho+tx) 2 b (ho tx) 2

from which the associated shear stresses will be

(ho+tx)/2 b/2 doxx,p Pt 2 (ho+tx)/2 /22f f - dz'dy -2 ..... dz*dyy 0 dx 2b(ho+tx)2 y o- - -- ----------- -

-p b b

--Ft2 ho + tx _Ft2 y

Ibh+t) 2 - YJ w Ill 2 ---- Ix2b(ho+tx)2 2 ho + tx

-P't y- [1-2 -----.sin a (18)

4b(ho+tx) ho + tx

12

i-.-;'..'.'.'..'.-"....'..'v,. ,-....--..-.,.. ..-..--..-,.--.---..--..,-,-.....,-..-.-",-,----,...-,.,.--.,,.----.'--,---'--'...,-. . . . ... ''.,-

and as before

12 P' 1+v 1 ho + tx 2h o + Stx

8shearl 5 b 4 log 2 txco a (14a)

which, together with Eqs. (12) and (12a), yields:

12 P 1+v I ho + tx 2ho + 5tx

-- -{4 log - -3 - tx)ashearl 5 b E t ho 2(h0+tx)

2 61 shear.

6bendl P 1 ho + tx 2ho + 3tx 6'bendl

12-- log tx}bE t 3 ho 2(ho+tx)

2

or++v ho+tx 2ho + 5tx

--- (4 log . . 3 ------- x6shearl 5 ho 2(ho+tX) 2 6S shearl

.... ...-- (15)6bendl 1 ho + tx 2ho + 3tx

6 bendl

2 (log ho 2(ho+tx) }

However, at the limit as t+O, which is equivalent to a straight cantilever

of uniform cross-section, one getsho + tx

4 log

12 P 1+v ho 2ho + 5txlim 6shearl - - - lira {-.. . .. 3 x}t+O 5 b E t+O t 2(hoytx)

2

whereho + tx d ho

+ txlog ---- - log---

ho dt ho x x

Irn (- -} lie { .- .- .. } lie -tO t t+O d t+O ho + tx ho•--(t)

dtand

2ho + 5tx x1a -- x ---t+0 2(ho+tx)2 ho

p1

.1

|o

** *, w.* ,.**-..tt~g r'.." " " " " " '" " " ". . '-/" ." ' " ' . . ' " . - ', .: ''... .''' " " , ." " " " '. " -.-. ' '''- ...'''' -.- ' '. -2 .

Thus 9 ht tx

Usheari l dxb 15 E o (h4,+tx) 3

for which

dx ho2 1 h0 2 . 1 2h + txho 2 .--- , / - tIo(h0+tx)3 2 (h0+tx)2 it (h04-tx) t2h~x

andx xdx 1 1ho x 1 4t~,c2

-4hot f --- ~-4hot -. 1 [ ------- 72-J -0(h0+tx)3 t ho+tx 2(h0+t)a t 2(h0+tx) 2

andx x2dx 4 2h0 h0 2 x

4t 2 J f -- I log(h0+tx) ------ ----------- ]o (h0+tx) 3 t ho+tx 2(h0+tx) 2 0

4 ho,+tx 2h0 + 3tx

t holo - 2(h0+tx)2 tx

ThusSh0

2 - 4hotx +4t 2 E2 1 h + txI ----- dx { 4 log --0 (h 0 +tx) 3 t h

2hotx + ~ 2 - 4t~x - 8hotx - 12 t 2 X2

2(ho+tx)2

1 ho +tx 2h + 5tx-- (4 log - --- 3 ---- tx}

t ho2(h 0 +tx) 2

from whichP712p 2 1+v 1 ho +tx 2h + 5tx

U5 heart - - - - (4 log - - 3 -t)Pser5 b E t ho 2(ho+tx) 2 - her

and thus

12p 2 1+V 1 ho +tx 2h 0 + tx6shearl - {4E log 3-- 2h+x txl (14)

10

If, however, in the derivation of Eq. (12) one uses P''cos a for P and is

to determine the displacement 6' in the x'-y' coordinate system, one will get

the following:

P' 1 ho + tx 2ho + 3txbendl - 12 - 7 {l g h ( ° t ) x }-co 2 ( 2 )

--- (h,;tx)

The local strain energy due to that component of shear stress resulting

from the gradient in normal bend stresses and acting on planes normal to the

plane of symmetry and in the axial direction, ayxm (or its equivalent acting

on planes parallel to the plane of symmetry and in the direction normal to the

plane of symmetry, axym) Ushearl, is

l+v 9P2 l+v (ho-2tx) 2

S= - (--- - --------- -(2_ --- 2] 2Ushearl -yxmCyx,m 2 - 2 ..4b2 E (h0+tX) 2 2

9P 2 1+V 1 8y2 16y 4

S2b -E- (h°2-4hotx+4t 2x2)[( -h--)-- (h+tX) 6 -(h+ 8 (13)

for which the total shear strain energy, Ushearl, over the entire volume

becomes:

9p 2 l+v , (ho+tx)/2Ushearl - T b-2 f (ho 2 -4hotx+4t 2x 2)•a00

1 8y2 16y 4

(ho+tX) 4 (ho+tx) 6 (ho+tx)8

for which(ho+tx)/2 1 8y 2 16y 4

(ho+tx)4 (h+tx)6 h h;tx)e

]dy

y 8y3 16y 5 (ho+tx)/2 4 1

(Zho tX) 3(ho+tX)6 5(ho+tx)8o "15 (ho8t

6 9

Deflection Due to Bending Forces Only

The local strain energy due to normal stresses, ubendl, is

O2 P2 x 2y 2 p2 x2y2--- 144 - _144 cos 2 aUbendi -xxm " xm - E b2 (h+t) 6 b 2E (ho+tX) 6

Thus, the total bending strain energy, Ubendi, over the entire volume of the

(deformed) beam becomes:

,°t .(ho+tx)/2 p2 0 ho+tX)/2 x2y2

Ubend - 2bf f Ubenddyddx 288 -bE - -- dy'dx0 o o o (ho+tx)

where(hO+tx)/2 x2y2 1 2(h+tx)/2 1 x 2

oa (ho+tX) 6 dy 3 (h+tX)6 y3 (htx) 3 (ho o 24 (ho+tX)3

for which

x x 2 1 ho + tx 2ho + 3txo (ho+tx)3 "dx [log -x]th o 2(ho+tX)

or288 p2 1 ho + tx 2h o + 3tx

Ubendi -- - {log ---- - ------- tx}24 bE t 3 { ho 2(h°+tx) 2

p2 1 ho + tx 2ho + 3tx- 12 - - {log -x} = P

bE t 3 h 2(h+tx) 6bendl

from which

p 1 ho + tx 2ho + 3tx6bendl - 12 - -- {log --- -- (12)bE t ho 2(ho+tx) 2 t}(

which is the same as Yo arrived at in Eq. (7) through the double integration of

the bending moment method, and as such

• P x

lim 6bendl -4 -(;;)t+O bEh

as in Eq. (9).

8

-*.i:-:. .... *.~*

x 2h. 2% + 8h~tX2 + 60 X3

-4h.tx2 - 6t 2x 3

ho + tx 2(h 0+tx) 3

t*O 3t2

*2h 02x + 4hotX2 +- 2t 2-x3 -2h 0~x - 8hotx

2 - 6t 2x3 + 4hotX2 + 6t 2x3

liii 2(h-ttx) 3

t+o 3t2

*2t 2x3

2(h 0 +tx)3 3 1 x 3---------------------------- - i-----------= ()t+o 3t2 t+0 3(h0+tx) 3 3 ho

to which one may add thatho+ tx

log---- --

1 ho +tx 2h + 3tx ho2h + 3txlirn{- [log ------- ------ tI lirn- - lim- - x

t+o t ho 2(h0+tx) 2 t-+J t t+ 2(h04-tx)2

* whereho+ tx

t+O t ho +tx ho

and

2h + 3tx

t0 2(h0+tx) ho

Thus1 o h+tx ZhQ+ 3tx

urn ( [log 0- -t+O t ho 2(h0+tx) 2tx

ThereforeP 1 x P x

lim 6 bend3 n2 -(-) - 4 -(;)t+o bE 3ho bE h

22

which, except for the sign, is the same as that arrived at in Eq. (9), and

which concurs with the value given in handbooks (ref 2) for a straight

cantilever of a uniform cross-section.

Even though the shear stress O approaches zero at the specimen's

boundaries, y - (ho+tx)/2, and thus it is not in violation on the boundaries,

it is assumed here that, as with the normal stresses, Oyx is the cartesian

component of a radial shear stress, a0 r, which converges at (or radiates from)

the same cylindrical coordinate origin at L = -(ho/t + x) as before.

• : Oe ,= A + ( ty___2

aer,m h-tr- ho+tX

or22 ]2

c12 r,m + h+--------21 yx,m

Thus, since3P ho -2tx Y 2

,. o,m- 2b 4tx)2 ho+tx

3p ho -2tx )12. / y 2erm -2b (hotx ' ho+tX h+tx

Thus, the local shear strain energy'due to the bending forces, Ushear3 is

1+v 9p2 1+V

Ushear3 - Oer,mcer,m =2 - 2 r,m - -- (h°2-4h tx+4t2i 2).

1 8y2 _ ___4 t2y2

- +tX) +i +

2Raymond J. Roark and Warren C. Young, Formulas for Stress and Strain, FifthEdition, McGraw-Hill, NY, 1975, p. 98.

23

r"P.

°r c

*7%

from which the total shear energy due to bending forces only over the entire

" beam's volume

Ushear3 4 f ushear3"dz"dy'dx

9p2 1+V x (ho+tx)/2 1 8y2 16y 49P2 l~~~v fx (ho~2-4hotx+4t2x2).[-h;) h~X) -°;8

2b E o o (ho (h0+tx)6 (h0 Itx)

"2y2

[1 + ho )2dydx(h0+tx)2

where

(ho+tx)/2 1 8y2 16y4 t2y2

(ho2-4hotx+4t2 2)[------- -------- - -------9dy0"f (ho+tX)4 (ho+tx)6 (ho+tx)8 (h o + tx)2j~

y (8-t 2 )y 3 8(2-t 2 )y 5 16t 2y 7 (ho+tx)/2

(ho+tX) 4 3(ho+tx) 6 ' 5(ho+tx) 8 ' 7(h0+tx)l 0]

h 2 - 4hotx + 4t2x 2 1 8-t2 2-t2 t 2

(hoItx) 3 [2 24 20 +56

""ho2 -4hotx + 4t2x 2 1 1 1 1 1 1-- - + - ) + (----- + -)t 2]

(h0+tx)3 L%2 3 10 24 20 56

4 t 2 ho - 4hotx + 4t2 2

15 105 (h0+tx)3

-"1 t 2 ho - 4hotx + 4t 2x 2

Thus,3 t 2 p 1+v , h 4hotx

+ 4t2x 2

Ushear3 - (4 + -) - - - - - dx10 7 b E o (h0+tX)

3

r--

24

where, as in the derivation of Eq. (14)

h 2 - 4hotx + 4t2x2 1 ho+tx 2ho + 5tx

fo (ho-t------ dx - -( 4 "log L - 3 --- txI0 (h 0 +tx)

3 t ho2(h 0 +tx)2

Thus3 4 t p 2 1+v ho + tx 2ho + 5txU 4-1og -- ... 3x

Ushear3 0 0 ( 7 b E ho 2(ho+tx)2 shear3

Hence3 4 t P 1+v ho

+ tx 2ho + 5tx6shear3 - (- + -) - {4og------------------ tx} (27)

10 t 7 b E ho 2(ho+tX)2

and as before,

3 4 t P 1+v ho + tx 2ho + 5tx

6 shear3 -(- + -4log ------- 3 -------- 2tx )cos2 a (27a)

10 t 7 b E ho 2(ho+tx)2

Sinceho + tx 2ho + 5tx

lim {t[4"log- - 3 tx]} 0

t0 ho 2(h+tx)

as in the derivation of Eq. (16) above

12 P 1+v xlim 6shear3 = (.--)t+O 5 b E h0

as it should be. Also,

3 4 t P 1+v ho + tx 2ho + 5tx

- + {4.og -3 - -tx6shear3 10 t 7 b E ho

2(ho+tX)2

6bend3 12 9 1 P ho + tx 2ho + 3tx

- lo g -txlt 5 t bE ho 2(ho+tx)

2

or

3 4 t ho + tx 2ho + 5tx

- -+ -)(1+v)({4'log - -3 - -- tx}6shear3 10 t 7 h 2(ho+tx)2 tx }.. . = - - - - - - (28)6bend3 12 9 1 ho + tx 2ho + 3tx(I+ -) (os tx)

t 5 t ho 2(ho+tx) 2

25

-. . .i".. .--.-. - .-. = .."-. .--. - -"-. -'-,.., '< '-''-, '.-. ,".'.. " -.'''''- -" . . .,. '" ,., -", -%

Deflection Due to Combined Bending and Compressive Forces

The same process of considering the converging stresses used in the above

calculations of the gap-opening due to the bending forces can be used for the

combined bending and compressive forces.

As in Eq. (19) before, the Cartesian component of the combined forces

oaxx - Oxx,m + Oxx,p - b {24 (h-) t 1

2b (h0 .tx) ho + tx

* Thus, as before, the full normal stress will be

P ty xy0r~ i 0;+t) 24 --- x-- t}--------

b(h0+t)2 ho + tx

• - and the local strain energy due to normal stresses, ubend4, will be

02rr p 2 t 2 y 2

Ubend4 0rr~rr - --- ---E 4b2 (h0+tx)

x 2 y2 txy 2{576----48 -- -;-+----

(ho+tx)6 (ho+tx) (ho+tx) 2}

However, due to the beam's symmetry, when integrating throughout the entire

beam's volume, the term

48 txy

(ho+tX)4

from both sides of the plane of symmetry cancels itself, and one gets for the

total bending strain energy, Ubend4, the following:

x hX +tx/2 b/2 p 2 0 (h°+tx)/2 t2y2

Ubend4 f f f ubend4"dz-dy-dx - 2bE [ + (h0+X)2 ]

{56 x2y2 t 2

{576 -t)-- - -- }dydx

(h-+tx) 6 (ho+tx)2

26

a.

a.' " """': , " " , - -,,: ,-,, ..- ,-,- -- "-""" - ."'''' 2'. ----------. i';" . " -- '- :" - ---- -,-.

'.pi' ,'. : - , ti:2. -_. ,,: ao .... ,,,, ,. •.: -, . .,, ,

|.1

' where

f (ho+tx)/2 t2y2 x2y2 t2[1+t 21 {576 + zb d

fo(h+t) h+tx) 6 (h+tx) 2d

2_ X2 _4 y 53 t~s . (ho+tx)/2- ---- + (576 __ _+ Ox 3 576 0 tx -o(ho+tx)2 (h0+tx)

6 ( 70 t'-j) 5htx8

18 X2 1 t2 t 2

5 (ho+x) 24 2 ho + tx

Thusp 2 x 18 x2 1 t 2 t 2

Ubend4 [(24 + -- t 2) ---- + )--------]dx2bE o 5 (h0+tx) 3 24 2 ho+ ]tx

where

x x 2 1 ho + tx 2ho + 3txf - tx) dx;-3[log ------ ----------

o (ho+tX)3 d ho 2(ho+tx) 2

and

x dx 1 ho + txf -- log

Sho+ tx t ho

as has been shown before. Thus,

p2 24 18 1 ho + tx 2ho + 3tx 1 t3 ho + txObend4 -{(- +- )og + (2 t + 2)log- . }

ibE {t 5 2(h~tho

= P6bend4

ThusP 24 18 1 ho + tx 2ho + 3tx

6bend4 -- {- +--)lg- tX]b bE ((t + -5 t)[1o8 ho 2(ho+tx)3

1 t 3 ho + tx+ (- t +_ -)log - "(29)

24 2 h -(9

27*5. . * - - - . . . * ' , . * - . . . . . . * , 5 ..i- ' . - .* . . . . . . . - . . .. . * - 5 . . . .?

- - . - ' -. . -...- . ..- ..

andP 24 18 1 ho + tx 2h o + 3tx

S6 t bend4 - - {t + -- )[log - - -- X tx]2ed bEF t 5 5th 0 2h+

1+ (3 t + -)log - --"}cos 2 a (29a)24 ho

Since1 t3 ho

+ txin {(-- t + -)log - - 0t+O 24 2ho

as before' P X

lim 6bend4 - 4 ( -)t+o bE ho

Incorporating the bending and the compressive components of the applied

force into the computation of the resultant radial shear stresses, aer, yields

the following:P hv-_2tx - t2

aer - 00r,m + Gerp {3 3- -[1(2---2b (h0+tx)2 ho-Itx 2(h0+tx)

[1 . A ty -

ho + tx ho+tx

Thus the local shear strain energy resulting from the radial shear stresses due

to the combined bending and compressive forces is

1+ VUshear4 - o08rc(r = 2 - 2 Or

Ep2 +v h0 2 -4hotx + 4t 2x 2 2 4•{9 [1 -2(2 -+-- (2 Y-- -

2b 2 E (ho+tX) 4 ho+tx ho+tx

ho - 2tx ll2 -Y) 3

-3" t 2 lyl - ( )-- '-) + ( -- 1,(ho+tx i ho + tx h+tx h+tx

t4lyl 2 2 ~ y

4(hh+tx)2 ho+tx ho+tx (ho+tx)2

28

However, when integrating over the entire volume one can omit the term

ho - 2tx lyl y 2 3y3h t 2 [1 - 2 - (2 ) + (2 -)J

(ho+tx) 3 ho + tx ho+tx ho+tx

since it cancels itself on both sides of the plane of symmetry. Therefore

Ushear4 4 f (hO,+tx)/2 /2 ushear4dzdydx0 0 0

p2 1+v _x (ho+tx)/2 ho2 - 4hOtx + 4t2x2 'y2f f ---- --- [1 -2(2----) + (2----) ]

b E o o (ho+tX)4 ho+tx ho+tx

t4 yly_2 t2y2tlyl y

+ ---- 2 [1 - 4 ---- + (2 -- ]1[ + ---- dydx4(ho+tx) ho+tx ho+tx (ho+tx)

where

(ho+tx)/2 ho2 - 4hotx + 4t2x2 y_2 4h09 --.----- [1 - 2(2 + (2 ]

o(ho+tX) t x ho+tx

+ 4"'o"x''1 - 4 ---t + (2 ho tx ]'1 + 'I+x' dy

h12 - 4hotx + 4t2x 2 8-t2 y3 16 8t2 y5 16t2 y7-9 ----- ... [' o ; + -.-; '- + -o- '- --

42 2 24 16 20 (otx

t4 4 y2 4+t2 y 4t2 y4 4t2 y5_}h~X/. . .. .y .. .. . . ...+-4(ho+tX [Y ho + tx 2 '(ho+tX)

2 3 (ho+tX) 3 4 (hot4 5 o

ho 2 -4hotx + 4t:2x2 1 - 2 2-t2 t2 t4

-9 (ho+tx)3 L[2 24 -20 56 4(ho+tX)

1 1 4+t2 t2 t2

[i -i + "- -- - -; + -1

29

; .-. -. ' .. _.._ , V '- - .... _-.;.. - . .,.. • .. - .-... ' ,.. '..'. .....--..

3 t2 h. 4htX + 4t2x 2 + 7 t 2 ) t

-3 (4 + j_ J#- + (+5 7 (h0+t)

3 6 240 4(ho+tx)

for hich ho - 4hotx + 4t2x2 1 ho + tx 2ho + 5tx

hot- -- dx (4 -log o 3 -tX) 2 txl

0 (ho+tx)3 t h (o

anddx 1 ho + tx

---- - log----o ho + tx t ho

Thus P 2(1+v) 3 4 t ho + tx 2ho + 5tx

Ushear4- ---------- (- + -)'[41log ------- 3 -------- tx]bE 5 tho

2(ho+tx)2

t 3 7 ho + tx

+ (6- + i- t 5 ).log - } - P 6shear4

from which

P 1+v 3 4 t ho + tx 2ho + 5tx

6shear4 - - - {-(- + -)°[4"log ho 3 --------- tx]b E 5 t 7 ho2(h 0-Itx)

2

t 3 7 ho + tx+ (--+ - t5)-_log ... --(30)6 240 ho

and similarly

P' 1+v 3 4 t ho + tx 2ho + 5tx

shear4 {( + -)[41-g 3 -x]a b E 5 tho 2(o

tX) 2

t3 7 ho + tx

+ (-- + -o t5).log ---- }cos 2 a (30a)6 240

since

t3 7 ho + txlira (- + --- t 5) log ___-- _ 0

t+0 6 240 ho

as before3 h

lir 6shear4 - 1 V)(_) 2t+O 5 x

30

, . o, . . ..,. . "~ ~~~. . .. . . . ......, % . % o . . o . .. -. ... . * • • . • ~ . Q . o.. .. ~

and6shear4

6bend4

3 4 t ho + tx 2ho + 5tx 1 7 ho + tx(1+v){-.(- + -)'[4log - - 3 - - , txj + ( + -- t

2)t3.log

5 t 7 h 2(h+tx) 6 240 ho

1 ((!4 18 1 ho + tx 2ho + 3tx 1 t2 ho + tx+ ---) -[lo +tx 2 t ] + ( 4+ -)t~log -h

2 t 5 t ho3 2(h0+x 2 (31)

SUMMARY

'riis report provides equations for the determination of that part of the

gap opening in fracture toughness specimens, which is due to the deflection of

the cantilever portion of the specimen. The method used is based on beam

theory and it covers elastic deformation only. Four different field

combinations were considered.

1. Where only the stress components which are parallel (or normal) to the

cantilever beam's plane of symmetry are being considered.

2. Where the stresses are assumed to be parallel to the beam's outer

surfaces, at these surfaces, and where they gradually rotate in between.

Each of the above stress fields was considered as the result of

a. the bending component of the applied force only, and

b. the combined bending and compressive components of the applied force.

In all of the above cases, the contributions of bending stresses and shear

stresses were computed. Table I is a computer printout for the computed half

gap opening for varying angles of specimen taper and for varying crack length

to beam's cross-sectional height ratios. The gap openings to be anticipated by

each of the above assumed stress fields are compared.

31

"i -. . . . - ..L". . .. ..-... . .. . .. . : ... .. > . . . .: . . - .-.-..... :. .... .. . ,... . : . - . .

REFERENCES

1. Boaz Avitzur, "Retained Deflection in Circular and Concentrically Hollowed

Beams After Local Removal," to be published.

2. Raymond J. Roark and Warren C. Young, Formulas for Stress and Strain, Fifth

Edition, McGraw-Hill, NY, 1975, p. 98.

3. A. C. Ugural and S. K. Fenster, Advanced Strength and Applied Elasticity,

American Elsevier, NY, 1981, pp. 146-148.

32

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