mil hdbk 776

MIL-HDBK-776 (AR)

15 September 1981

..

,:

MILITARY HANDBOOK

SHAFTS, ELASTIC

TORSIONAL STRESS ANALYSIS OF

●

NO DELIVERABLE DATA

8QUIRED BY THIS DOCDNSNT

..

AREA CDNC

MIL-HDBK-776(AR)15 September 1981

DEPARTMENT OF DEFENSEWASHINGTON, DC 20360

Shafts, Elastic Torsional Stress Analysis Of

MIL-HDBK-776(AR)

1. This standardization handbook is approved for use by the Armament Researchand Development Command, Department of the Army, and is available for use by alldepartments and agent’iesof the Department of Defense.

2. Beneficial comments (recommendations, additions, deletions) and any per-tinent data which may be of use in improving th:Lsdocument should be addressedto: US Army Armament Research and Development Command, ATTN: DRDAR-TST-S, Dover,NJ 07801.

I

I ii

MIL-HDBK-776(AK)15 September 1981

FOREWORO

-.

I

Design charts and tables have been developed for theelastic torsional stress analyses of free prismatic shafts,splines and sprinq bars with virtually all common industriallyencountered thick cross sections. Circular shafts with rect-angular and circular }.eyways, external. splines, and milledflats along with rectangular and X-shaped torsion bars arepresented.

A computer program (“SIIAFT”) was developed which providesa finite difference solution to the governing (pOISS~N’S)partial differential eq,lation which defines the stress functionsfor solid and hollow shafts with generalized contours. Usingthe stress function solution for the various shapes, andprandtl’s membrane analogy, dimensionless design charts (andtables) have L>een qenerated for transmitted torque and maximumshearing stress. The desiqn data have been normalized for aunit dimension of the cross section (radius or length) and areprovided for solid shapes.

The eleven solid shapes presented, along with the classicalcircular cross section solution, provides the means for analyz-ing 144 combinations of hollow shafts with various Outer andinner contours. 1!o11ow shafts’may be analyzed by usinq thecomputer program directly Or b.y using the sOlid shape. charts inthis paper and the principles .of superposition based on theconcept of parallel shafts. Stress/tOrque ratio curves arepresented as keinq more intuitively recognizable and usefulthan those of stress alone.

sample problems illustrating the use of the charts and “.,:.tables as desicm tools and the validity of the superrJositiOn” . ‘:’concept are included.

.,, . . .

iii

.-—--

:..

.:... -.

,,

_—,

... . ~.:, ,.,..-

,,,. . . .

.,.,


TABLE OF COt4TENTS

Paqe No.

The Torsion Problem

Design Charts and Tables

Accuracy of the Computerized Solution

Parallel Shaft Concept

Bibliography

Appendixes

A Mathematical Model Used in the SHAFTComputer Program

B Extension of Model to Uollow” Shafts

1

4

54

55

66

67

75


Tables

1

2

3

4

5

6

7

8

9

10

11

13

14

15

16

17

Element nomenclature

Split shaft, volume factor [v)

Split shaft, stress factor (f)

Single keyway shaft, volume factor (V)

single keyway shaft, stress factor (f)

Two keyway shaft, volume factor (V)

TWO keyway shaft, stress factor (f)

Four keyway shaft, volume

Four keyway shaft, stress

Sinqle square keyway with

Single spl

Single spl.

Two spline

Two spline

factor (V)

factor (f)

inner fillets

ne shaft, volume factor (V)

ne shaft, stress factor (f)

shaft, volume factor (V)

shaft, stress factor (f)

Four spline shaft, volume factor (V)

Four spline shaft, stress factor (f)

S[~uare keyways and external splinesvolume factor (V)

~,q{l.arckeyw. ays and external splin es,stress factor (f)

5

7

9

11

13

15

17

19

21

23

25

27

29

31

33

35

37

39

●✎✌✎✎✎

,T’””-”


19 Milled shaft,

20 Milled shaft,

volume factor (V)

stress factor (f)

21 Rectangular shaft

22 Pinned shaft, volume factor (V)

23 Pinned shaft, stress factor

24 Cross shaft, volume factor (V)

25 Cross shaft, stress factor

Figures

1

2

3

4

5

6

7

8

9

10

11

12

f)

f)

Membrane analogy

Split shsft, torque

Split stift, stress

Single keywayshsft, torque

Single keyway shaft, stress

Twokeyway shaft, torque

Two keyway shaft, stress

Four keyway shaft, torque

Four keyway shaft, stress

Single square keyway with inner fillets

Single spline shaft, torque

Single spline shaft, stress

41

43

45

47

49

51

53

3

6

8

10

12

14

16

18

20

22

24

26

,—_ VI-..

M1L-HDBK-776(AR)15 September 1981

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

l’wo spline shaft, torque

‘rwo spline shaft, stress

Four spline shaft, torque

Four spline shaft, stress

Square keyways and external splines, torque

Square keyways and external splines, stress

Milled shaft, torque

Milled shaft, stress

Rectangular shaft

Pinned shaft, torque

Pinned shaft, stress

Cross shaft, torque

Cross shaft, stress

parallel shaft concept

Milled shaft with central hole

Circular shaft with four inner splines

Superposition for two spline shaft

Superposition A for four spline shaft

Superposition B for four spline shaft

Illustrative design applications

28

30

32

34

36

38

40

42

44

46

48

50

52

56

57

58

59

60

61

63

65

●

✍✎

●

●


TORSIONAL ANALYSIS oF 5pLINED .5NILLED 51[AFT5

The elastic stress analysis of uniformly circular shaftsin torsion is a familiar and straiqhforward concept to designengineers. As the bar is twisted, plane sections remainplane, radii remain straiqht, and each section rotates aboutthe longitudinal axis. The shear stress at any point is pro-portional to the distance from the center, and the stress vec-tor lies in the plane of the circular section ancl is perpen-dicular to the radius to the point, with the maximum stresstangent. to the outer face of the bar. The torsional stiffnessis a function of material property, angle of twist, and thepolar moment of inertia of the circular cross-section. Theserelationships are expressed as:

@ = T/J.G, or T = G.O.J

and S5 = T.r/J, or S5 = G.O.r

!ihere T = twisting moment or transmitted torque, G = Modulusof Riqidity of the shaft material, O = angle of twist per unitlength of the shaft, J = polar moment of inertia of the (cir-cular) cross-section, S5 = shear stress, and r = radius to anypoint.

Rowever, if the cross-section of the bar deviates evenslightly from a circle, as in a splined shaft, the situationchanges radically and far more complex design equations arerequired. Seqtions of the bar donot remain plane, but warpinto surfaces, and radial lines through the center do not re-main straight. The distribution of shear stress on the sectionis no longer linear, and the direction of shear stress is notnormal to a radius.

The governing partial differential equation, from Saint-Vernant’s theory is

—

IMIL-HDBK-776(AR)15 September 1981

where @ = Saint- Venant’s torsion stress function. The prob-lem then is to find a o function which satisfies this equationand also the boundary conditions that @ = a constant along theboundary. The O function has the nature of a potential func-tion, such as voltage, hydrodynamic velocity, or gravitationalheight. Its absolute value is, therefore, not important; onlyrelative values or differences are meaningful.

The solutions to this equation required complicated math-ematics. Even simple, but commonplace, practical cross-sec-tions could not be easily reduced to manageable mathematicalformulae, and numerical approximations or intuitive methodshad to be used.

One of the most effective numerical methods to solvefor Saint-Venant’s torsion stress function is that of finite(Differences and the best intuitive method, the membrane ana-loqy, came from Prandtl. [Ie showed that the compatibilityecluation for a twisted bar was the “same” as the equation for.a mem~)ranc stretched over a hole in a flat plate, then in-flated. This concept provides a simple way to visualize thetorsional stress characteristics of shafts of any cross-section relative to those of circular shafts for which anexact analytical solution is readily obtainable. A computer pro-qrarn called SIIAPT was written and applied to produce the dimension-less desiqn charts. on the following pages.

The three-dimensional plot of @ over the cross-section isa surface and, with O set to zero (a valid constant) alonq theperiphery, the surface is a domb or O membrane. The trans-mitted torque (~) is proportional to twice the volume under themembrane and the stress (Ss) is proportional to the slope of themembrane in the direction Perpendicular to the measured slope.Neqlectinq the stress concentration of sharp re-entrant cor-ners, which are relieved with generous fillets, the maximumstress for bars with solid cross sections is at the point onthe periphery nearest the center.

‘The dcsiqn data have been normalized for a unit dimension(radius or lenqth) of the shaft cross-section and are in di-mensionless format. The! data and charts may, therefore, beused for shafts of any dimensions, materials and twist (loading) .

2

L

1

#

EM


#2 NOOES

~=:-. ~

FINITE DIFFERENCE--

/,

//

//

MEMBRANE SOLUTION ‘. :

#5a

CROSS-SECTtiPIS :

Figure 1. Membrane analogy.

3


DESICN CHARTS AND TABLES

Design charts and related data which support the elastic torsional

stress analyses conducted by Ml S. Dare shown in figures 2 through 25

and tables 2 through 25, respectively. The item nomenclature used inthe analyses is given in table 1.

These data are based on the stress function solution for various

shapes provided by the SHAFT computer program and on Prandtl’s

membrane analogy.

Since thedesign charts are dimensionless, they can be used for

shafts of any material and any dimensions.

I

I4

MIL-HDBK-776(AR)15,September 1981

Table 1. Element nomenclature

TORSIONAL PROPERTIES

OF

SOLID, NON-CIRCULAR SHAF13

T

T=e=

G.

R.

“d@f.‘=’

s~ =

T.

Ss =

Ss .

Y

TRANSMITTED TOROUE, N m (lb i.. )

ANGLE OF TWIST PER UNIT LENGTH. rad/mm (,wUin.i

MOOULUS OF RIGIDITY OR MODULUS OFELASTICITY IN SHEAR, kPa (lb/in.2)

OUTER RADIUS OF CR OSSSECTION, mm (in. )

VARIABLES FROM CHARTS (OR TABLESIRELATED TO VOLUME UNDER “SOAP FILMMEMBRANE” AND SLOPE OF “MEMBRANE’”

SHEAR STRESS. kPa (lb/in.z)

2G 0 (V)R4


I .4(

I .3!

.3C

.2!

Iv

.2C

.1!

.1(

.05

.0

I

~/~i= 0.1

i

0.3 “0.50.7

\I niTORSIONAL STIFFNESS

..-

‘N\ TRANSMITTED TORQUE,

T=2.G. e(v)R:

‘o\

I 1 I , I I.1 .2

‘-~ “3RYw “4 ““5 “6

Figure 2. Split shaft, torque.

●6

●

●

●


Table 2. Split shaft, volume factor (V)

Y/Ri Ri/llo

0.1 0.2 0.3 0.4 0.5 0.6— — — — — —

0.1 .3589 .2802 .2068 .1422 .0891 .0491

0.2 .3557 .2762 .2030 .1391 .0870 .0478

0.3 .3525 .2722 .1991 .1360 .0848 .0464

0.4 .3492 .2680 .1952 .1328 .0825 .0450

0.5 .3457 .2637 .1911 .1294 ,0801 .0436

0.6 .3423 .2593 .1869 .1260 .0777 .0421

0.7 .3387 .2548 .1824 .1223 .0750 .0405

0.8 .3350 .2499 .1776 .1183 .0722 .0387

0.9 .3312 .2447 .1725 .1139 .0689 .0367

1.0 .3269 .2389 .1665 .1087 .0649 .0340


6.5

6.0

5.5

5.0

f

4.5

4.0

3.5

3.0

2.5

2.0

SHEAR STRESS

MAXIMUM AT INNER EDGE,AT X

s~ = T (f/R:)

y/Ri = 1.0

0.9

0.7

0.5

0.3

0.1

t 1 1 1 1 1

0.1 0.2 0.3 0.4 0.5 0.6

Ri/Ro

Figure 3. Split shaft, stress

8

●

●


Table 3. Split shaft, stress factor (f)

Y/Ri Ri/Ro

0.2—

0.1 2.2140

0.2 2.2447

0.3 2.2767

0.4 2.3103

0.5 2.3461

0.6 2.3883

0.7 2.4233

0.8 2.4670

0.9 2.5142

1.0 2.5672

0.3—

2.2742

2.3162

2.3608

2.4082

2.4594

2.5142

2.5750

2.6423

2.7197

2.8140

0.4—

2.5771

2.6336

2.6942

2.7597

2.8304

2.9084

2.9955

3.0952

3.2141

3.3690

0.5—

3.2178

3.2977

3.3838

3.4771

3.5795

3.6930

3.8232

3.9744

4.1618

4.4218

0.6—

4.4650

4.5865

4.7182

4.8620

5.0233

5.2016

5.4082

5.6550

5.9691

6.4392

MIL-HDBK-776(AR)15 ‘September1981

.Bo

.7s

.70

.65

v .60

.55

.s0

.4s

.40

TORSIONAL STIFFNESS

TRANSMITTED TORQUE,

T=2.Ge(v)R4

1 I I 1 I

.1 .2 .3 .4 .5BIR

Figure 4. Single keyway shaft, torque.

10●

MTL-HDBK-776(AR)15 September 1981

●

!

:0

i

Table 4. Single keyway shaft, volume factor (V]

A/B B/R

0.1—

0.2

0.3

0.4

0.5 .7682

0.6 .7676

0.7 .7668

0.8 .7658

0.9 .7647

1.0 .7633

1.2 .7621

1.5 .7592

2.0 .7560

0.2—

.7379

.7341

.7290

.7262

.7224

.7190

.7162

.7125

.7079

.7012

.6945

0.3—

.6994

.6900

.6816

.6725

.6663

.6592

.6533

.6480

.6424

.6347

.6260

.6200

0.4 0.5— —

.6472 .5864

.6316 .5648

.6173 .5459

.6043 .5294

.5941 .5152

.5848 .5032

.5762 .4931

.5686 .4849

.5619 .4783

.5531 .4697

.5449 .4649

.5424

11

I

MIL-HDBK-776(AH)15 September 1981 -

1.20

1.15

1.10

1.05

f

1.00

.95

.90

0.5

0.6

0.7

0.8

0.9

1.0

1.2

1.5

AIB =04

SHEAR STRE:

MAXIMUM AT

~ 2.0 ,/

I + I 1 1 I

0.1 0.2 0.3 0.4 0.5BIR

Figure 5. Single keyway shaft, stress

12

● “iI

“

1

●

●

●

MIL-HDBK-776(AR)15 Se?tember 1981

Table 5. Single keyway shaft, stress factor (f)

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.2

1.5

2.0

0.1—

.9899

.9767

.9602

.9393

.9124

.8773

.8651

.8300

.8083

0.2—

1.1867

1.1241

1.0303

1.0077

.9746

.9466

.9334

.9131

.8993

.8829

.8752

.:,.-. . . . .

0.3—

1.2273

1.1333

1.0624

1.0387

1.0098

.9953

.9843

.9749

.9684

.9655

.9667

13

,-,.:

0.4—

1.2538

1.1642

1.1155

1.0960

+;0820

1.0737

1.0699

1.0691

1.0721

1.0774

1.0799

0.5—

1.2832

1.2234

1.1962

1.1859

1.1848

1.1885

1.1944

1.2009

1.2120

1.2198

MIL-HDBK-776(AR)

I 15 September 1981

.60

.70

.60

.50

v

.40

.,..Ju

.20

—

TORSIONAL STIFFNESS

—

.3

.4

.5

.6

.1 .2 .3 .4 .5

BIR

Figure 6. Two keyway shaft, torque,

14

●

●


Table 6. Two keyway shaft, volume factor (V)

Al B B/R

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.2

1.5

2.0

0.1—

.7524

.7511

.7496

.7477

.7454

.7426

.7404

.7346

.7283

0.2.

.6927

.6853

.6753

.6698

.6625

.6558

.6505

.6433

.6344

.6215

.6086

0.3

.6187

.6008

.5848

.5678

.5562

.5429

.5319

.5221

.5117

.4974

.4813

.4703

0.4 0.5.

.5226 .4195

.4944 .3831

.4688 .3517

.4457 .3246

.4277 .3014

.4112 .2818

.3962 .2655

.3829 .2522

.3713 .2416

.3559 .2276

.3416 .2197

.3373

15

●

.-


1.60

1.50

1.40

1.30

f

1.20

1.10

1.00

.90

SHEAR STRESS

MAXIMUM AT XAm = 0.3

~= T(f/R3)

0,50,60.70.80.9

I

1 1 , 1 1

0.1 0.2 0.3 0.4 0,5

BIR

Figure 7. TWO keyway shaft, stress.

16

P

1

‘o1

I

I

I

1

1

I

I

I

‘o

I

I

I

I

1

b

I

I

I

I

‘o

I

I

i (

MIL-HDBK-776(AR)~15.September’1981

.. . . ..Table 7.,..‘IWO ke~4a~ sh.a.ft,.stress factor (f)

,..

. . . .. . ..,., ,..!,:!.”-”’~-”i !. ”’.:

0.1 0.2 0.3 0.4— ~ 0.5 ~— — — ?—,

0.2 1.4936 1.6578

0.3 1.2487 1.3642 1..4929

1.7501 :

1.6491 -.jj5

1.6313 :

1.6555

~.7027 c.

1.7623

1.8269< @., v

1.8910

1.9502 ~1

2.0422_j ~i,,

2.1024 ~

!\

..l ,>!,

-; 01.


.80

.70

.60

.50

v .40

.30

.20

.10

TORSIONAL STIFFNESS

TRANSMITTED TORQUE,

T=2”GOOW)R4

●

I I I 1 I I.1 .2 .3 .4 .5

81R

IFigure 8. Four keyway shaft, torque.

18

●


Table 8. Four keyway shaft, volume factor (V)

A/B BIR

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.2

1.5

2.0

0.1.

.7214

.7190

.7161

.7124

.7080

.7024

.6982

.6870

.6748

0.2—

.6088

.5952

.5769

.5672

.5541

.5422

.5330

.5203

.5051

.4832

.4622

0.3

.4806

.4511

.4253

.3983

.3805

.3605

.3444

.3304

.3160

.2974

.2787

.2692

0.4 0.5.

.3361 .2114

.2965 .1705

.2624 .1384

.2333 .1140

.2119 .0962

.1935 .0842

.1783

.1662

.1572

.1482

19

I

MIL-HDBK-776(AR)’15 September.1981

2.40

2.20

2.00

I 1.80

f

~ 1.60

1.40

1.20

1.00

.80

—..

0-

.,’ , .,

....

,>. ;. (~ ;: :.,. . .

.-:, ,}:, ,. ..:. !

~~~ /

l;...- .!

,“’. ,. ~~.~1+ :’..:.

t I I 1 1

0.1 0.2 0.3 ~, 0.4 0.5B/R ‘i

Figure 9. Four keyway shaft, stress

●✎

1

(

●20

MILTHDBKZ7Zj,(AR):15j$epte.rn}e;1981

● Table 9. Four keyway shaft, stress factor (f)

0.1 0.2 0.3— —13.4,(...-. 0.5 ;-.. —- I

●

0.2 :..... :,..”. “ :2:7365 2;1206 2.4931,.. .,.,.. 1’ “- ,-

0.3 ‘i-.’3566-’ ‘ i.6214 ‘ 2.0014 2.5784..J

0.4 ! “’” 1.3011 1.5460 1.9971 2.8271, ..... . . ......

0.5 2“;’ i.0371 ,,,’1.2130 1.50’i6 2.0591 3.1882‘ . .’,

0.6 ~.0252 ‘1.1979-’’,.; 1..5115 2.1568 3.61393 ~+,”

0.7 ,., , ~.0102 1’1.i737 ;1:5,175 2.2660 4.0368

0.8 \ .9910 l.lsii ~ 1.5382.:,

2.3834.

0.9 1.1493”’~:.j i. “9661 :

““1,.5609 ““<.,2.4993-...

1.0 .9331 1.1398 1.58’9 9..... . . ...?.? {03 o

1.2

1.5 [,2.

2.0

<:~,

{ .9232

+.8940I{ .8796I

l..I

1<

!

.! ,:,,,I

I

I

! ,j c;. . .

1

II

..: [;::I

1’--! {.?,...

I

, .... .


.75

.70

.65

.60

I v

.55

.50

.45

.40

4

\

TRANSMiTTED TORQUE

vT=2. Gd(V)R4

SHEAR STRESS

S~=T(f/R3)

f2

R

B l--I I I I I

0.1 0.2 0.3 0.4 0.5

BIR

Figure 10. Single square keyway with inner fillets

22

1.30

1.20

1.10

f

1.00

.90

.80

.70

●

I

I

●❉

MIL-liDBK-776(AR)15 September 1981

●Table 10. Single square keyway with tight inner fillets

Stress factor(f)

Volume At keyway

!?@

At innerfactor(V) center(1) fillet(2)

0.1 .7703 .7804

0.2 .7206 .9715 .9777

0.3 .6504 .9941 1.0817

0.4 .5690 1.0735 1.1641

0.5 .4840 1.1977 1.2245

.,——.

●

23

MIL-HDBK-776(AR):15 September’.l98l

1.51.15 AIB = 2.0

//

TORSIONAL STIFFNESs ,:7., ., ,lP.:::: .-.: ‘- .,>-3,.?‘TRANSMITTED TORQUE, -

9 .‘..

T=2”G”O(V)R41.10

—. -,.. .

,

1.00

v

.95

.90

.85

.8a

if!?-+liik/ 2!/”2‘+2 . I_ ///

“’”’--//I /’”0

I 1 I I I

.1 .2 .3 .4 .5BIR

o

Figure 11. Single spline shaft, torque.

24

1

I

I

‘o

●

e

MIL-RDBK-776(AR)15 September 1981

Table 11. Single spline shaft, volume factor (V)

A/B B/R

0.1—

0.2

0.3

0.4

0.5 .7845

0.6 .7852

0.7 .7857

0.8 .7862

0.9 .7866

1.0 .7869

1.2 .7890

1.5 .7907

2.0 .7953

0.2—

.7853

.7864

.7874

.7899

.7918

.79s0

.7976

,7996

.8071

.8174

.8407

0.3—

.7853

..7870

.7903

.7933

.7993

.8059

.8113

.8202

.8253

.8456

.8754

.9420

25

0.4—

.7865

.7906

.7968

.8035

.8143

.8270

.8390

.8560

.8712

.9117

.9800

1,1404

0.5—

.7878

.7944

.8048

.8189

.8362

.8580

.8832

.9110

.9433

1.0158

1.1561


0.65

0.60

0.55

f

0.50

0.45

0.40

SHEAR STRESS

MA)(iMUM AT )(

S~= T (f/R3)

At13 =

AiB = 2.0 ‘

0.2

0.40.50.60.70.8

0.9

1.0

1.2

1.5

-tI $ 1

0.1 0.2 0.3 0.4BI’R

0.5

Figure 12. Single spline shaft, stress.

26

● ✌✎

MIL-HDBK-776(AH)15 September 1981

Table 12. Single spline shaft, stress factor (f)

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.2

1.5

2.0

0.1—

.6374

.6370

.6366

.6364

.6361

.6359

.6346

.6335

.630,7

0.2—

.6369

.6362

.6356

.6340

.6328

.6308

.6291

.6279

.6233

.6172

.6038

0.3—

.6369

.6358

.6337

.6317

.6280

.6239

.6205

.6152

.6120

.6004

.5842

.5525

0.4—

.6361

.6335

.6295

.6251

.6184

.6107

.6035

.5939

.5854

.5648

.5340

.4798

0.5—

.6352

.6309

.6241

.6152

.6047

.5920

.5781

.5638

.5483

.5173

.4104

.4331

27


1.60

1.40

1.30

1.20

~v

I 1.10

1.00

I.90

I

I .60

I.m

T=2.G%(V)R4

I

AIB = 2.o 11.5

TORSIONAL STIFFNESS

TRANSMITTED TORQUE,

I ! 1 I I

.1 .2 .3 .4 .5BIR

Figure 13. Two spline shaft, torque.

28

.9

.8 ‘

.7

.6

.5

,4.3.2

● ’


Table 13. Two spline shaft, volume factor (V)

Al B B/l?

0.2

0.3

0.4

0.5

0.6

00.7

0.8

0.9

1.0

1.2

1.5

2.0

0.1—

.7850

.7863

.7874

.7883

.7891

.7897

.7940

.7973

.8066

0.2—

.7864

.7886

.7906

.7958

.7994

.8059

.8111

.8152

.8302

.8509

.8980

0.3—

.7865

.7899

.7965

.8026

.8145

.8278

.8386

.8565

.8668

.9078

.9682

1.1045

0.4—.7889

.7970

.8095

.8229

.8446

.8701

.8945

.9288

.9595

1.0418

1.1818

1.5172

0.5—

.7914.

.8047

.8255

.8538

.8886

.9326

.9837

1.0400

1.1058

1.2547

1.5471

29

MIL-HDBK-776(AR)15 September “1981

0.65

0.60

0.55

f

0.50

0.45

0.40

0.35

4—

SHEAR STRESS

MAXIMUM AT X

S~= T (f/R3)A/B =

R

Figure 14. Two spline shaft, stress.

30

0.2

0.30.4

0.5

0.6

0.7

0.8

0.9

1.0

1.2

1.5

I I I I i

0,1 0.2 0.3 0,4 0.5I

I

● ;

LII

!I

-!

oMIL-HDBK-776(AR)15 September 19B1

Table 14. TWO spline shaft, stress factor (f)

A/B B/R

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.2

1.5

2.0

0.1—

.6371

.6363

.6357

.6351

.6346

.6342

.6315

.6295

.6240

02&

.6362

.6348

.6336

.6303

.6281

.6241

.6209

.6184

.6097

.5981

.5740

0.3—

;6362

.6340

.6298

.6259

.6187

.6108

.6043

.5944

.5887

.5678

.5402

.4903

0.4—

.6346

.6294

.6215

.6131

.6004

.5862

.5732

.5564

.5421

.5088

.4632

.3937

0.5.

.6329

.6243

.6113

.5946

.5753

.5532

.5300

.5071

.4836

.4398

.3815

3,1


1.80

1.73

1.60

1.50

140

v

130

I .m

1.lC

1.Oc

.%

.8(

TORSIONAL STIFFNESS

TRANSMITTED TORQUE,

T=2’G’e(v)R4 I1.2

.1 .2 .3 .4 .5Blfl

Figure 15. Four spline shaft, torque.

_.. .

32

●


●

II

Table 15. Four spl ine shaft, volume factor (V)

A/B B/R

0.1—

0.2

0.3

0.4

0.5 .7859

0.6 .7885

0.7 .7906

0.8 .7924

0.9 .7940

1.0 .7954

1.2 .8040

1.5 .8106

2.0 .8292

0.2—

.7887

.7932

.7971

‘.8076

.8149

.8280

.8386

.8467

.8773

.9196

1.0180

0.3—

.7888

.7957

.8090

.8213

.8452

.8723

.8944

.9310

.9519

1.0378

1.1663

1.4739

33

0.4—

.7937

.8101

.8352

.8623

.9063

.9588

1.0090

1.0808

1.1455

1.3239

1.6438

0.5—

.7989

.8254

.8674

.9250

.9962

1.0877

1.1950

1.3158

1.4601

1.8021


0.65

0.60

0.55

f

0.50

0.45

0.40

......

R

SHEAR STRESS

MAXIMUM AT X

.S~= T (f/f33)

AIB =

V15

~

12

L+’

A

L B

0.2

0.30.4

0.5

0.6

0.7

0.8

0.91.0

+—U.1 0.2 0.3 0.4 0.5

131R

Figure 16. Four spline shaft, stress

34

● ’,1

4

1

● ✎

MIL-HDBK-776(AK)15 September 1981

Table 16. Four spline shaft, stress factor (f)

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.2

1.5

2.0

0.1—

.6369

.6358

.6349

.6341

.6334

.6328

.6293

.6265

.6192

0.2—

.6356

.6336

.6318

.6273

.6240

.6187

.6144

.6109

.5998

.5860

.5630

0.3.

.6356

.6323

.6263

.6206

.6106

.6001

.5913

.5794

.5720

.5508

.5279

0.4—

.6332

.6256

.6142

.6019

.5848

.5670

.5516

.5344

.5199

.4989

0.5—

.6305

.6176

.5986

.5756

.5510

.5257

.5028

.4842

.47L4

35

MIL-HDBK-776(AK)

15September1981

I

●’


●

●

Table 17. Square keyways and external splines, volume factor

BjR One keyway TWO keyways— Four keyways

0.1 .7633 .7426 .7024

0.2 .7125 .6433 .5203

0.3 .6424 .5117 .3160

0.4 .5619 .3713 .1572

0.5 .4783 .2416

0/R One spline Two splines—

0.1 .7869 .7897

0.2 .7996 .8152

0.3 .8253 .8668

0.4 .8712 .9595

0.5 .9433 1.1058

37

●

Four splines

.7954

.8467

.9519

1.1455

1.4601

(v)

. .. . ,,,. =,, .7. / .-. .

FLIL-tUJDh-l IO(AK)

15 September 1981

2.40

2.20

2.00

1.80f

KEYWAYS

1.60

1.40

1.20

1.00

.80

4 KEYWAYS /

SHEAR STRESS

MAXIMUM AT X

S~= T (f/R3) /

SPLINES

R

KEYWAYS3

4 SPLINES

1 KEYWAY

I I I 1 1

0.1 0.2 0.3 0.4 0.5

BIR

0.65

0.60

0.55f

SPLINES

0.50

0.45

0.40

0.35

Figure 18. Square keyways and external splines, stress

38

●

●

●


Table 18. Square keyways & e%ternal splines, stress factor(f)

B/R—

0.1

0.2

0.3

0.4

0.5

B/R—

0.1

0.2

0.3

0.4

0.5

one keyway

.8773

.9131

.9749

1.0691

1.2009

One spline

.6359

.6279

.6120

.58S4

.5483

39

Two keyways

.8978

.9916

1.1625

1.4532

1.9502

Two splines

.6342

.6184

.5887

.5421

.4836

Four keyways

.9331

1.1398

1.5899

2.6030

Four splines

.6328

.6109

.5720

.5199

.4714

MIL-HDBK-776(AR)

15September1981

40

l-vi?—

o

0.1

0.20.29289

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

Table 19

One flat

.7813

.7617

.7018

.6291

.s510

.4717

.3951

.3228

.2568

.1980

.1460


Milled shaft, volume factor (V)

Two flats Four flats

.7811 .7811

.7149 .6520

.5998 .4501.2777

.4667

.3349

.2168

.1225

.0559

.0173

41

MIL-HDBK-776(AR)

15September1981

\

\

x1-aL0;1-a@

’.“

7:

\\

t’----I

u

00

00

00

kI.ri

+ei

w--

m

42

0!.40(

LJm.-L


Table 20. Milled shaft, stress factor (f)

H/ii—

0.1

0.2

0.29289

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

One flat

.7749

.8571

.9485

1.0593

1.1977

1.3725

1.5987

1.8975

2.3049

2.8935

Two flats Four flats

.8199 .8743

.9776 1.1848

1.7004

1.2045

1.5520

2.1237

3.1455

5.3129

11.5433

43

MIL-HDBK-776(AR)15 Sept

1.1[

1.OC

.90

.80

.70

v

.60

.50

.40

.30

.20

.10

]er“1961

\

El=+T

L-d_f-TORSIONAL STIFFNESS

TRANSMITTED TORQUE /

/

T=2G0(v)A4

SHEAR STR&S

(MAXIMUM AT

=1 I 1 1 I I0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

B/A

Figure 2T. Rectangular shaft.

44

5.0

4.5

4.0

3.5

f

3.0

2.5

2.0

1.5

1.0

E!Z!?0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0


Table 21. Rectangular shaft

volumefactor(v)

.05635

.1248

.2250

.3559

.5146

.6971

.8991

1.1167

Stressfactor(f)

5.2697

3.0928

2.0587

1.4805

1.1230

.8862

.7212

.6015

45

MIL-HDBK-776(AR)

15 September 1981

v

TRANSMITTED TORQUE,

T=2. G.0(WR4 \

I 1 1 1 1.1 .2 .3 A/R .4 .5

Figure 22. Pinned shaft, torque.

.46

IIIIII

●

o


Table 22. Pinned shaft, volume factor (V)

AIR One groove TWO grooves Four grooves—

0.1 ,7700 .7558 .7280

0.2 .7316 .6803 .5855

0.3 .6760 .5738 .4062

0.4 .6087 .4521 .2374

0 0.5 .5349 ,3300 1118


4.0

3.5

3.0

f

2.5

2.0

1.5

1.0

SHEAR STRESS

MAXIMUM AT BOTTOM

OF CIRCULAR CUSPS, AT X

5~=T (f/R3)

1 1 I I 1

0.1 0.2 0.3 0.4 0.5

AIR

Figure 23. Pinned shaft, stress.

48

e


Table 23. Pinned shaft, stress factor(f)

A/R One groove Two grooves Four— grooves

0.1 1.1197 1.1374 1.1674

0.2 1.1804 1.2520 1.3800

0.3 1.2286 1.3939 1.7281

0.4 1.2894 1.6015 2.3912

0.5 1.3822 1.9211 3.8744

49

MIL-HDBK-776(AR)

15September1981

II1-Ii;

_

-ul-

+-

50

●

)(/s—

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8


Table 24. Cross shaft, volume factor (V)

Shape P

.00741

.05219

.1642

.3538

.5947

.8302

1.0058

1.0981

51

Shape M

.09907

.2120

.3767

.5714

.7639

.9247

1.0368

1.0981


8

7

6

Ij

f

4

3

2

1

+s-’+

a “ ‘~SHAPE - “P”

(P)a

\+

1; x

a a

\

III SHEAR STRESS

MAXIMUMS AT a & b

S~= T (f/s3)

‘“dL-+4)4.,(M)I,

I I I I I I I

“0.1 0.2 0.3 0.4 0.5 0.6 0.7 0:8Xls

Fiqure 25. Cross shaft, stress

52

●

I

●✍

●MIL-HDBK-776(AH)15 September 1981

Table 25. Cross shaft, stress factor(f)

x/s—

Shape PAt a At b

Shape MAt a At b

0.1 26.8826 .8.7805 4.0676 .7564

0.2 7.2946 2.7669 2.3702 .9225

0.3 3.5252 1.4172 1.5818 .8651

0.4 2.1192 .9709 1.1844 .7806

0.5 .7849 .9210 .7109

0.6 .6903 .7576 .6606

0.7 .6366 .6059 .6275

0.8 .6090 .4763 .6090

●

53


ACCURACY o~ THE COUPUTERIZED SOLUTION

To compare the SHAFT (computer) analysis of the torsionof a solid circular shaft with the exact, classical textbooksolution, one quadrant of a unit-radius shaft was run withtwo finite-different qrid spacinqs and the results of theequations were compared, as-follows:

I Shear stress (max)

Equation Comparison SHAFT

Torque 2Ge(v)R’2(V)Rh2(V)R*2V

~8(d~,R

ds

(d~)ds

SHAFT

Torque (h=O.125 1.5546(h=O.0625) 1.5669

Shear stress, (h=O.125) 1.0000(h=O.0625) 1.0000

Area* (h=O.125) 3.13316(h=O.0625) 3.13984

*

% >

Exact

1.57081.5708

1.01.0

3.141593.14159

Exact

GOJJ(n/2)R*(n/2)

GeR

1.

Deviation (%)

1.030.25

0.0.

0.2680.056

‘used for internal program checking.

The mathematical model used in the .?~IAFTcomputer programgeneration of this handbook is described in appendix A.

●II

I 54

MIL-HDBK-776(AK)15.September 1981

PARALLEL SHAFT CONCEPT

The torsional rigidity of a uniform circular shaft, i.e. , the torque

required to produce unit (one radian) displacement, is:

C = _i_/6J = G-J

In the terminology of the membrane analogy, the torsional rigidity

of non-circular shafts is defined as:

C = T/e = 2. G+)(V) f(R)/13

The overall torsional rigidity of a system consisting of a-numberof shafts in parallel (fig. 26) is simply the sum of the torsional

rigidities of the individual component shafts.

N

z ci=c1+c2+c3+-. +cNi=l

N N

z Tiei = ez Ti=e(Tl +T1+T~+.. +TN)i=l i=l

The torsional rigidity of hollow shafts can be determined by re-

garding the configuration as a parallel shaft arrangement. The over-

all torsional rigidity can be obtained by subtracting the torsional

rigidity of a shaft having the dimensions of the tmre (or inner contour)

from that of a shaft having the dimensions of the outer contour. The

advantages of being able to apply the principles of superposition

(fig. 27-31) to combinations of concentric (inner and outer] shaft con-

tours are obvious. If, for example, design charts have been prepared

for 20 different shaft shapes, then 400 different solutions to all possiblecombinations of inner and outer shaft contours (20 inner x 20 outer)

are available.

55

MIL-RDBK-776(AR)

15September1981

$0-1azaUI

zaKmz:

56

MIL-HDBK-776(AR)

15September1981

e +I

I

211I

0/’

“b\

“F!B

um

m0II

0II.

“d

(n

e

—Om

o1

!II

II>1-1-

57

MIL-RDBK-776(AR)

15September1981

1-

-UY

of/Jn+

CC

L

Wa

n=

3~(n.

W

/11m

..

-..,,...

...

,’,

5%

..;F

,.,’,,

,,.,

:G

ma)c.-

Zms.x3&Na$m.-U.

I

●

—

●●

MIL-HDBK-776(AR)

15September1981

D-4-

-/“

K

z

F>I

*.o_!,

z

Kiiim

L--

%<-+

bt~-[

.,,.‘-A

-““-

Klx

Nw

.II

IIII

GU

iN

0’ .i.~-i-_t

----A-

“.’

aj,..‘

1

w’‘K>

0>

59

.

MIL-HDBK-776(AR)

15September19811.

m,

@ a+.-

L.-,-

-.a

T

uo&,,cmo

-i4-”””-

a-

m0m60

1-a

a

%.,,

,,*II

IIII

omalL3m.-U

e0I

●

I

MIL-HDBK-776(M)

15September1981

-0 a\‘--“”+

“-

eInmf-

0>mIn

w0>

m

—,!

II

IIII

mm

00ZZJ00

61


I1,LU!;’I’RA’I’lVEDISIGN APPI,lCA’rT.ON

Find tl~c m~jxj.mum torque that may be transmitt,:d by the

,:ijc(]].ar:;lIaft with {he intc]-i.or splints (shown in Fig. 32)

2) Maxjl[,urnShear StrcsS S~ not to excccd 15,000 kPa (psi)

TOIIIUC T =1 T = E 2GG(V)R4 = 2G6 E (V)R4

Z (V)R4 = o.7854- (O.1O58-O.O491)

= 0.7854(1” circle)-O.0567(8 tooth spline)

= 0.7287

Condition 1:

(3= 2x(n/180)xl/18) = 0.001939(rad/in)

T = 2 (12x106)(0.001939)(0. 7287) = 33,~of)(in-lb)

Confliti.on 2:

S /T will be maximum at the outer contour, which is acomple?c cil-cle (for which d$/ds = 1.0).

.S~/T = GO(d$/CIS)R/2GOZ(VR4))

= (dO/dS)R/2Z(VR4))

J (1.0)(1.0)/2(0.7287)) = 0.6862

T = Ss/0.6862 = 15,000/0.6862 = 21,860(in-lb)

Use T of 21,860(in lb) as maximum design Torque

62 ● 1


.O”R

-—. .(J.5”’r-i I

VR4=(7V4)R4

=(7774) 14 =0.7854

—

(From Table 15) VR4=(7T14)R4

‘2 V(R)4=2(0.8468) (0.5)4 =(7T/4)(o.5)4

=0.1058 =0.0491

Figure 32. Illustrative design application,,. .

,., ... .. ... ,,..

.,:., .

~


(,mor]lm) IllUStratiVe DESIGN APPLICATION

Find the maximum shear stress and angular twist per unit length produced byan imposed torque of 20,000 lb–in. The double milled steel shaft (G=12xl(36Psi)

has a 4-spline inner hole as shown (Figure 33) .

ouTER rx)N’rouR: Double Milled Shaft

Ii/l{=0.1V (T+LbLQ 19) = O.7149 (This is really V(R)4 where R=L.0)f (’l’able20) = 0.8199

INNER CONTOUR : 4–Spl ine Shaft

A/B = 1.0, B/R = 0,1/0.5 = 0.2

V (Table 1.5)= 0.8467 (Really V(R)& for a 1.O”R shaft)

‘r= 2.C.0LV(R)4

~l,ere );V(R)4 = 0.7149(1)4 - 0.8467(0.5)4

= 0.6620

20,000 = 2 (12xI.O)0(0.6620)and angular twist,El= 0.001259 radians/inch

F,,r ;]SUJ id double-milled shaft, the maximum shear stress is at the midpointof the flats:

Ss = T(f/R3) = 20,000 (0.8199)/(1)3=,16,398 psi

fkcz.use of the inner hole, however, the V(R)4 term is reduced from O.7149 to

0.6620. ‘l’liemaximum shear stress is still at the midpoint of the milled flats:

adjusted f1.. . ~.. —..-_

/

Ss = ‘r(f’/R3) = 20,000 (0.8199x 0.7149 ) 30.6620 (1)

= 17,708 psi

‘he effect of the inner hole is to increase the maximum shear stress by7.99%.

64

ttI

NilL-HDBK-776(AR)

15

September

1981

L1

?11

1%m


BIBLIOGRAPHY

I F, S, Sha W, ,,The TOr~ion Of Solid and Hollow Prisms in the Elastic

and Plastic Range by Relaxation Methods, “ Austral ian Council for

Aeronautics, Report ACA-11, November 1944.

2. F. S, Shaw, An Introduction to Relaxation Methods, Dover Publ ica -

tions, Inc. , Melbourne, Australia, 1953.

3. D. N. deG. Al Ien, Relaxation Methods, McGraw-Hill Book Company,

Inc. New York, 1954.

4. D. H. Plettaand F. J Maher, “The Torsional Properties of Round-

Edged Flat Bars, II Bulletin of the Virginia Polytechnic tnStitllte,

Vol XXXV, No. 7, Engineering Experiment Station Series No. 50,

Blacksburg, Virginia, March 1942.

5. W. Ker Wilson, Practical Solution of Torsional Vibration Problems,

Volume 1, 3rd Edition, John Wiley & Sons, Inc. , New York, 1956.

6. R. 1. Isakower and R. E. Barnas, “The Book of CLYDE - With aTOrque-ing Chapter, II u .S. Army ARRADCOM Users Manual MISD

UM 77-3, Dover, NJ, October 1977.

7. R. 1. Isakower and R. E. Barnas, ,lTorsional Stresses in SlOttecf

Shafts, “ Machine Design, Volume 49, No. 21, Penton/l PC,

Cleveland, Ohio, September 1977.

8. R.I. Isakower, “Design Charts for TorsionalProperties ~f Non-Circular Shafts” , U.S. ArmyTechnical Report ARMID-TR-78001, Dover, NJ,November 1978.

,.. -66

● ’


●APPENDIX A

i4ATHE)mTICAL MODEL USED IN TllE S1lAFT cOMpuTER PROGRAM

As the term implies, boundary value problems are thosefor which conditions are known at the boundaries. These con-ditions may be the value of the problem variable itself, thenormal gradient or variable slope, or higher derivatives ofthe problem variable. For some problems, mixed boundarYconditions may have to be specified: different conditionsatdifferent parts of the boundary. The SHAFT computer codesolves those problems for which the problem variable itself(the stress function) is known at the boundary.

Given sets of equally spaced arguments and correspondingtables of function values, the finite difference analyst canemploy forward, central, and backward difference operators.SHAFT is based upon central difference operators which approxi-mate each differential operator in the partial differentialequation (PDE) .

The problem domain is overlayed with an appropriately

● selected grid. There are mariy shapes (and sizes) of over-laying Cartesian and polar coordinate grids:

rectangularsquareequilateral-triangularequilangular-hexagonaloblique

Throughout the area of the problem, SIIAFT uses a con-stant-size-square grid for which the percentage errors are ofthe order of the grid size squared (hz). This grid (or net)consists of parallel vertical lines spaced h units apart, andparallel horizontal lines, also spaced h units apart, whichblanket the problem area from left-to-right and bottom-to-top .

The intersection of the grid lines with the boundariesof the domain are called boundary nodes. The intersectionsof the grid lines with each other within the problem domainare called inner domain nodes. It is at these inner domainnodes that the finite difference approximations are applied.The approximation of the partial differential equation withthe proper finite difference operators replaces the PDE witha set of subsidiary linear algebraic equations, one at each

●inner domain node. In practical applications, the method

67


must be capable of solving problems whose boundaries may becurved. In such cases, boundary nodes are not all exactlyh units away from an inner node, as is the case between ad-jacent inner nodes. The finite difference approximation ofthe harmonic operator at each inner node involves not onlythe variable value of that node and at the four surroundingnodes (above, below, left, and right) , but also the distancebetween these four surrounding nodes and the inner node. Atthe boundaries, these distances vary unpredictably. Compensa-tion for the variation must be included in the finitedifference solution. SIIAFT represents the problem variableby a second-degree polynomial in two variables, and employsa generalized irregular “star” in all directions for eachinner node. In practice, one should avoid a grid so coarsethat more than two arms of the star are irregular (or.lessthan h units in length). The generalized star permits, andautomatically compensates for, a variation in length.of anyof the Cour arms radiating from a node. For no vzl.iation inany arm, the algorithm reduces exactly to the stzzldardharmonic “computation stencil”.

At each inner domain node, a finite difference approxima-tion to the qoverning partial differential equation (PDE) isqeneratcd by SllArT. The resulting set of linear algebraicequations is solved simultaneously by the program for the un-known problem variable (stress function) at each node in theoverlaying finite difference grid. A graphics version ofthe proqram also qenerates, and displays on the CRT screen,iso-value contour maps for any desired values of the variable.This way, a niore meaningful picture of the solution in theform of stress concentration contour lines of differentvalues is made available to the engineer.

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MIL-HDBK-776.(AR)15 September 1981

BOUNDARY NODESOverlaying finitedifferertce GRID

tfiiYvariable isknown at the boundary)

— .-—

v

the

f thev=MIe tO be ~hxi

.INNER DOMAIN NODES(value 0found by finite differencesolution)

Figure A-1 . Finite difference grid,

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If the problem geometry is symmetric,does not have to display and work with theof the problem, he need only work with the

r the designerentire picture“repeating sec-

tion” . In essence, the graphics user may examine the problemsolution at will and redesign the problem (contour, boundaryconditions, equation coefficients, etc.) at the screen re-solving the “new design” problem.

Consider the general

Where A,B,D are arbitrary

Vaf=A;~+B;~

expression:

constants.

=D

II

.

lJsing central differences, the finite difference approxi-mations to the partial differential operators of function fat representative node O are:

e

af.+(f,-f,)#$=2h

a; ~J- (f,-f,)

Y I

azf I~ (fla~ = ~x -2fo+f3)

II

I

IEor a square grid h =h = h and the harmonic operator Vzfbecomes:

x Y

h2V2f0 = [A (fl +f,) +B (fz +f,) - (A+B) 2fO]=h2D

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MIL-HJ)BK-776(AR)

15 September 1981

hX-h

r

COMPUTATION STENCIL

2 AT NODE O

+

B

T A ‘ -2(A+B)0 A ‘

FOR

h, =h.

~,””, v’f=A~2+B:$C)

t“

●

●Figure A-2. Harmonic operator for square star in X-Y grid.

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MIL-HDBK-776 (AR).15 September 1981

This finite difference equation at node zero involves the unknown

variable at node zero (fO) plus the unknown value of the variable at the

four surrounding nodes (fl , ft, f, , f,) , plus the grid spacing (h) The

five nodes involved form a four-arm star with node zero at the center.

This algebraic (or difference) equation could be conveniently visualized,,blloon~,i connected in aas a four-arm computation stencil made up of five

four-arm star pattern and overlayed on the grid nodes. The value within

each balloon is the coefficient by which the variable (f) at that node is

multiplied to make up the algebraic approximation equation.

The numerical treatment of an irregular star (h, + hz+ h,# h,) re-

presents the function f near the representative node O by a second-degree

polynomial in X and Y:

f(X, Y) =fO +al X +a2Y+a~X2 +a4Y2 +asXY

Evaluating this polynomial at the neighboring nodes (1, 2, 3, 4)

produces the following set of equations:

fl =fO +alhl +azhlz

fz =fo+azhz +a, hzz

f,= fo-alh~+a, h~z

f, =fo,-ath, +a, h,z

which are tnen solved for a$ and a, which are necessary to satisfy the

harmonic operator V2f, since:

E= azf

axa, + 2a,X + a$Y, a~ . 2a,

gJ azf

ay=az + 2a4Y+a~X, ~

ay

and

Vzf =A (2a3) + B (2a, )

= 2a4

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Performing the necessary algebraic operations, substituting results,collecting terms, and using the following ratios.

b, . ‘# b2 . ‘$

b, . ‘f b, . ‘:

The harmonic operator becomes:

[

. .

h2’v2f0 = 2A f, + 2Bbl (bl+b~) bz (b2+b4)

f, +

2Af~ +

20’+ b~ (bl+b, ) b, (bi +b, )

f, +

2A- (—

+ _2B

b, bz bz b, 1)f, = ~1~

-1

,., ,

.,>.;,, . ... ! : .,1, ;1- : ., :

73

I


IRREGULAR STAR AT NOOE Oa

NEIGHBORING NoDES(I,2,3,4. )

f@

W%-W / “/ h =h,,h

] COMPUTATION STENCiL .

0AT tiOOE O

FOR

20 Vzf.A~ +B~,. Ob. (b, ,b. ) i

\

\.

3

Figure A-3. Harmontc operator for irregular star in X-Y grid.

74

MIL-HDBK-77 6(m)

‘o 15 September ‘1981

APPENDIX .B

EXTENSION OF-MODEL TO HOLLOW SHAFTS

This would appear to be a slmPle matter of solving thegoverning PDE over a multiply-connected boundary, were itnot for the uncertainty conqernlng boundary conditions. Theactual value of the problem-variable at the boundary was notimportant in the torsion application, only the differencein the problem variable at variOus., points mattered. Theproblem variable at the boundary could be assumed to haveany value, as long as there was only one boundary. With

two or more boundaries the solution calls for a differentapproach.

The stress function is obtained as the superposition’of two solutions, one of which is adjusted by a factor (k).This is the programmed solution to shafts with a hole.The hole may be of any shape, size, and location. The twosolutions, to he combined, are shown in figure B-1: equa-tions and boundary conditions. Once the contour integralsare taken around the inner boundary of area A , the onlyunknown, k, may be readily obtained. PThe con our integral,which need not be evaluated around the’actual boundary,may be taken around any contour that encloses that boundary,and includes none other, (for example, see shaded area AB)in figure B-1.

..,.i.

,. ,,

.’, 1

‘F.S. Shaw, The Torsion of Solid and Hollow Prisms in theElastic and~ic Range by Relaxation Methods, Austral-ian Council for Aeronautics, Report ACA-11, November 1944,pp 8,11,23.

,., ., .,, ..~.,

75

MIL-HDBK-776(AR), 15 September 1981

.

.,,.:

V2*O=L2. ‘, ,.. .“-,””~’2w;1= ~

,., ,,’. ., :..,’,. ,.. . .-. .:~ ;:, ..., ,,

,. ,,

,.

I I I 1

m:’:.:.flI I I I I

-....-

. .,:

Figure B-1 . Mathematical approach to hollow shaft problem

76

Custodian:Army - AR

User activity:Navy - OS

Review activitiesArmy - AV, ME

NIL-HJ)BK-776(AR)15 September 1981

Preparing activity:

Army - AR

Project Number: CDNC-0012

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a

STANDARDIZATION Docuh4Er4T IMPROVEMENT pR0p05ALI

INSTRUCTIONS: This form is provided tosolicit beneficed comment iwhichmay improve this dcm.metenhance its use. DoD contractors, government activities, man. fact. rers, vendors, or other prospective .s<the document are invited tos.bmit comments to the government, FoldonIineson reverse side, staplt- inand send to preparing activity, Attach any pertinent data which may be of use i“ improving this doc. nwthere are additional papers, attacb to form and place both inan envelope addressed to preparing activity,response will be provided to the submitter, when name and address is provided, within 30 days indicatingtbe1426was received and whe”a”y appropriate actio”on itwill recompleted.NOTE: l%if, form shall not be used tosubmit req.ests forwaivers, deviations orclarification ofspecXicarequirements on current contracls, Cmmmmts submitted on this form do not constitute or imply authorito waive any portion of the k. fer.mced document(s)cm tomne”d contractual requirements.

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