Bruhn - Analysis And Design Of Flight Vehicles Structures Episode 1 Part 5 doc

on stick grip, determine the Shearing stress on aileron torque tube and the angle of twist between points A and B.. 3 The maxtmum slope of the deflected membrane at any point, with respe

CHAPTER A6 TORSION - STRESSES AND DEFLECTIONS

A6.1 Introduction

Problems involving torsion are common in

aircraft structures The metal covered airplane

wing and fuselage are basically thin-walled

tubular structures and are subjected to large

torsional moments in certain flight and landing

conditions The various mechanical control sys-

tems in an airplane often contain units of

vartous cross-sectional shapes which are sub—

jected to torsional forces under operating con-

ditions, hence @ knowledge of torsional stresses

and distortions of members is necessary in air-

craft structural design

A6.2 Torsion of Members with Circular Cross Sections

The following conditions are assumed in the

derivation of the equations for torsional

stresses and distortions: ~

(1) The member its a circular, solid or hollow

(5) The applied loads lie in a plane or planes

perpendicular to the axis of the shaft or

Fig Aé.1 shows a straight cylindrical bar

subjected to two equal but opposite torsional

couples The dar twists and each section 1s

subjected to a shearing stress Assuming the

lert end as stationary relative to the rest of

the bar a line AB on the surface will move to

AB’ under these shearing stresses and this ro-

tation at any section will be proportional to

the distance from the 2ixed support It is as-

A6.1

sumed that any radial line undergoes angular

displacement only, or OB remains straight when

For equilibrium, the internal resisting moment equals the external torsional moment T, and since G@/L is a constant, we can write,

T

Ge

T = Ming, = [ora = ~

ø where J = polar moment of inertia of the shaft cross section and equals twice the moment of in- ertia about a diameter a

A6.2

A6,3 Transmission of Power by a Cylindrical Shaft

The work done by a twisting scouple

moving through an angular displacement

to the product or the magnitude of the

and the angular disslacement in radians

angular displacemant is ona r

done equals 2nT IT

pounds and N is the angular velecity in revelu-

tions per minutes, then ths horsepower trens-

mitted by a rotating shaft may be written,

where 396000 represents inch pounds of work of

one horsepower for cone minute Equation (6)

may be written:

X 396000 _ 63025 H.P

EXAMPLS PRCBLEMS

Problem 1

Fig A6.3 shows a conventional control

stick-torque tube operating unit For a side

load of 150 lbs on stick grip, determine the

Shearing stress on aileron torque tube and the

angle of twist between points A and B

SOLUTION:

Torsional stick force of

Fig A6.3 aileron operating system attached to aileron

horn and the horn pull equals 3900/11 = 356 lb

The polar moment of inertia of a 13 - 0.058

round tube equals 0.1368 in*

Maximum Shearing stress = t - Tr/J

(3900 x 0.75)/0.1368 = 21400 psi

The angular twist of the tube between

points A and B equals

ates an aileron control

a cireular torque tube

The airload on the surface tends to rotate

the aileron around the terque t ; dUt move-

ment is prevented or created by a control rod attached to the torque tube over the center supporting bracket

The total load on 2 strip or aileron inch wide = 40(15 x 1/244) 18 1b

Let w equal intensity of loading per i

of aileron span at the leading edze point o

aileron surface, (see pressure diagram in Fig A6.4)

in 1b, Hence, the maximum torque, which occurs

at the center of the aileron, equals $.0 x 29 =

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES varies directly as the distance from the end of

the ailerons, the angle cf twist 9 can be com-

puted by using the average torque as acting on

antire length of the tube to one side of horn or

The formulas derived in Art 46.2 cannot be

used for non-circular shapes since the assump—

tions made do not nold In a circular shaft

subjected to pure torsion, the shearing stress

distribution is as indicated in Fig A6.5,

namely, The maximum shearing stress is located

at the most remote fiber from the centerline

axis of the bar and is perpendicular to the

radians to the stressed point At a given dist~

ance from the axis of rotation the shear stress

is constant in both directions as illustrated in

Fig A6.5, which means that ends of segments of

the bar as it twists remain parallel to each

other or in other words the bar sections do not

warp out of their plane when the bar twists

If the conditions of Fig A6.5 are applied

to the rectangular bar of Fig A6.6, the most

stressed fibers will be at the corners and the

stress Will be directed as shown The stress

would then have a component normal to the sur-

face as well as along ‘he surface and this ts

not true The theory of elasticity shows that

the maximum shear stress occurs at the center-

line of the long sides as illustrated in Pig

A6.6 and that the stress at the corners 18 zero

Thus when 2 rectangular bar twists, the shear

stresses are not constant at the same distances

from the axis of rotation and thus the ends of

segments cut through the bar would not remain

parallel to each other when the bar twists or in

other words, warping of the section out of ‘ts

plane takes place Fig A6.7 illustrates this

action in a twisted rectangular bar The ends

of the bar are warped or suffer distortion

normal to the original unstressed plane of the

bar ends

Further discussion and a summary of equa-

tions for determining the shear stresses and

A6,3

twists of non-circular cross-sections is given

in Art 46.6

A6.5 Elastic Membrane Analogy

The ‘shape of a warped cross-section of 4 non-circular cross-section in torsion 1s

needed in the analysis by the theory of elas~

ticity, and as a result only a few shapes such

as rectangles, 4t11pse3, triangles, etc., have been solved by the theoretical approach How- ever, a close approximation can be made ex~

perimentally for almost any shape of cross-

section by the use of the membrane analogy

It was pointed out by Prandtl that the

equation of torsion of a bar and the equation

for the deflection of a membrane subjected to uniform pressure have the same form Thus if

an elastic membrane is stretched over an open-

ing which has the same shape as the cross~

section of the bar being considered and then if

the membrane is deflected by subjecting it to 4 slight difference of pressure on the two sides, the resulting deflected shape of the membrane

provides certain quantities which can be mea- sured experimentally and then used in the

theoretical equations However, possibly the main advantage of the membrane theory is, that

it provides a method of visualizing toa considerable degree of accuracy how the stress

conditions vary over a complicated cross-section

of a bar in torsion

The membrane analogy provides the follow- ing relationships between the deflected mem-

brane and the twisted bar

(1) Lines of equal deflection on the membrane

(contour lines) correspond to shearing

stress lines of the twisted bar

(2) The tangent to a contour line at any point

on the membrane surface gives the direction

of the resultant shear stress at the corre-

sponding point on the cross-section of the bar being twisted

(3) The maxtmum slope of the deflected membrane

at any point, with respect to the edge support plane is equal in magnitude to the shear stress at the corresponding point on

the cross-section of the twisted bar

Tue applied torsion on the twisted bar is proportional to twice the volume included

between the deflected membrane and 4 plane

through the supporting edges

To illustrate, consider a bar with a

rectangular cross-section as indicated in Fig

A6.8, Over an opening of the same shape we stretch a thin membrane and deflect it normal

to the cross-section by a small uniform pres=

sure Zqual deflection contour lines for this deflected membrane will take the shape as il- lustrated in Fig A6.9 These contour lines

which correspond to direction of shearing stress in the twisted bar are nearly circular

near the center region of the bar, but tend

A6.4

to take the shape of the

boundary is approached

tion through the contour lines or the deflected

membrane along the lines 1-1, 2-2 and 3-3 of Fig

A6é.9 It is obvious that the slopes of the ¢de-

flected surface along line 1-1 will be greater

than along lines 2-2 or 3-3 From this we can

conclude that the shear stress at any point on

line 1-1 will be greater than the shear stress

for corresponding points on lines 2-2 and 3-3

The maximum slope and therefore the maximum

bar boundary as the Pig A6.Ga shows a sec-

Fig A6.9a

stress will occur at the ends of line 1-1 The

slope of the deflected membrane will be zero at

the center of the membrane and at the four

corners, and thus the shear stress at these

points will be zero

A6.6 Torsion of Open Sections Composed of Thin Plates

Members having cross-sections made up of

narrow or thin rectangular elements are some-

times used in aircraft structures to carry tor-

sional loads such as the angle, channel, and Tee

Shapes

catwo-P OF & bar of rectangular cross-section of

width b and thickness t a mathematical elasticity

analysis gives the following equations for max!-

mum shearing stress and the angle of twist per

a0 0.2 0 387 0

TORSION

From Table A6.1 it is noticed that for

large values of b/t, the values o7 the con-

stants ts 1/3, and thus for such narrow rec- tangles, equations (6) and (7) reduce to,

Although equations (8) and (S$) have been derived for a narrow rectangular shape, can be applied to an approximate analysis of

shapes made up of thin rectangular members such as {llustrated in Fig A6.10 The more generous the fillet or corner radius, the

smaller the stress concentration at these

ctions and therefore the more accuracy of these

approximate formulas Thus for 4 section made

up of a continuous plate such as {llustrated in

Fig (a) of Fig AS.10, the width b can be taken

as the total 1ghếth of the cress-section For sections such as the tee and H section in Fig

A6.10, the polar moment of inertia J can be taken as & bt°/, Thus for the tee section of Fig A&.10:

go 22 _3t g7 & 2 ots

a 1n tb > Da) ằằẮaee=a (18)

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES EXAMPLE PROBLEM SHOWING TORSIONAL STIFFNESS OF

LOSED THIN WALLED TUBE COMPARED

~~ TO OPEN OR SLOTTED TUBE

Fig A6.lla shows a 1 inch diameter tube

with 035 wall thickness, and Fig A6.11b shows

the same tube but with a cut in the wall making

Fig AG ila Flg A6.11b

Let 9; equal twist of closed tube and 9,

equal twist of open tube The twist, will then

be inversely proportional to J since 0 ay

Therefore the closed tube is Ji/J_ = 0.02474/

0.000045 = 550 times as stiff as the open tube

This result shows why open sections are not ef-

ficient torsional members relative to

torsional deflection

TABLE Aé.3 FORMULAS FOR TORSIONAL DEFLECTION AND STRESS

T

= KG = twist in radians per inch of length

= Torsional Moment (in lb )

„ Ha3 bi TMAX = (at ends

2» «=a of minor axis), abe

For an extensive list of formulas for many shapes both solid and hollow,

refer to book, "Formulas For Stress and Strain” by Roark, 1954 Edition

A6.5 A6.7 Torsion of Solid Non-Circular Shapes and Thick- Walled Tubular Shapes

Table A6.3 summarizes the formulas for torsional deflection and stress for a few

shapes These formulas are based on the as~

sumption that the cross-sections are free to

warp (no end restraints) Material is homo-

geneous and stresses are within the elastic

range

46.8 Torsion of Thin-Walled Closed Sections

The structure of aircraft wings, fuselages and control surfaces are essentially thin-walled tubes of one or more cells Flight and landing loads often produce torsional forces on these

major structural units, thus the determination

of the torsional stress and deformation of such structures plays an important part in aircraft structural analysis and design

Pig A6.12 shows a portion of a thin-

walled cylindrical tube which is under a pure

torsional moment There are no end restraints

on the tube or in other words the tube ends and

tube cross-sections are free to warp out of their plane

q constant Fig A6 13

Fig A6.12

Lat gq, be the shear force intensity at point

(3} on the cross-section and qụ that at point

b)

Now consider the segment a @ vb of the tube

wall as shown in Pig 46.12 as a free body The applied shear force intensity along the segment edges parallel to the y axis will be given the

values Say and dy as shown in Pig A6.12 For

a plate in pure shear the shearing stress at a

point in one plane equals the stress in a plane

at right angles to the first plane, hence

da = fay dw * I,

Since the tube sactions are free to warp

there can be no longitudional stresses on the tube wall Considering the equilibrium of the Segment in the Y direction,

BFy = 0 = dayl - doyl = 9, hence Yay = Uy and therefore qq = 4p or in other words the shear

force intensity around the tube wall is con- stant The shear stress at any point tT = q/t

If the wall thickness t changes the shear stress

AG, 6

changes but the shear force q does not change,

or

Tata = “bÈp = constant

The product tt is generally referred to as

the shear rlow and is given the symbol q The

name shear flow possibly came from the fact that

the equation tt = constant, resembles the equa-

tion of continuity of fluid flow qS = constant

where q is the flow velocity and S the tube

cross-sectional area

We will now take moments of the shear flow

q on the tube cross-section about some point (0)

In Fig A6.13 the force dF on the wall element

ds = qds Its arm from the assumed moment cen~

ter (0) ts h Thus the moment of dF about (0)

is qdsh However, ds times h is twice the area

of the shaded triangle in Fig A6.13

Hence the torsional moment dT of the force

on the element ds equals,

a? = qhds = 2qdA and thus for the total torque for the entire

shear flow around the tube wall equals,

T= | 2qdA and since q is constant

where A is the enclosed area of the mean periph-

ery of the tube wall

The shear stress + at any point on the tube wall ts equal to q, the shear force per inch of

wall divided by the area of this one inch length

or lx tor

TUBE TWIST Consider a small element cut from the tube

wall and treated as a free body in Fig A6.14,

with ds in the plane of the tube cross-section

and a unit length parallel to the tube axis

Under the shearing strains the plate element

Fig A6 14 Fig A6, 15

deforms as illustrated in Fig 46.15, that ts,

the face a~a moves with respect to face 2-2 a

distance 6, The force on edge a-a equals q ds

and it moves through a distance 6

TORSION

Tre elastic strain energy dU stored in tnis

element therefore equals,

ó8 „ äqU =— £ However the shear strain 6 can te written,

or Uz [ae S5 the integral $

is the line integral around the periphery of the tube From Chapter a7 from Castigliano’s

theorem,

Since all values except t are constant, equa-

tion (17) can be written,

Fig A&é.16 shows the internal shear flow pattern for a 2-cell thin-walled tube, when the tube is subjected to an external torque

di, da and a, represent the shear load per inch

on the three different portions of the cell walls

For equilibrium of shear forces at ‘he junction point of the interior web with the out-

side wall, we know that

Choose any moment axis such as point (o)

Referring back to Fig A6.13, we found that the

moment of a constant shear force q acting along

a wall length ds about a point (o} was equal in

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES magnitude to twice the area of the geometrical

Shape formed by radii from the moment center to

the ends of the wall element ds times the shear

flow q

Let T, ~ moment of shear flow about point

(o) Then from Fig A6.16,

Ta “2x (Ai + Aap) + 2QzÂza — 2dsAay

= 2q,A + 2q.Aay + 2dedag ~ 2daday ~ - - - (22)

But from equation (21), qs = di - qa

Substituting the value of qs in (22)

Ta # 2Q:Á: + 2QxÂap + 2QuÀaa ~ 2qiAap + 2dahay

But A, = Aag + Aah

Hence, T, = 24,4, + 24,4, -

where A = area of cell (1) and Az = area of

cell (2) Therefore, the moment of the internal

Shear system of a multiple cell tube carrying

pure torsional shear stresses 1s equal to the

Sum of twice the fuctosed area of cach cell

times the shear léad per inch which exists in

the outside wall of that cell (Note: The web

mn is referred to as an inside wall of either

cell)

A6,10 Distribution of Torsional Shear Stresses in a

Multiple-Cell Thin-Walled Closed Section

Angle of Twist

Fig A6.17

Fig A6.17 shows in general the internal

Shear flow pattern on a 3-ceil tube produced by

a@ pure torque load on the tube The cells are

numbered (1), (2) and (3), and the area outside

the tube is designated as cell (0) Thus, to

designate the outside wall of cell (1), we re~

fer to it as lying between cells 1-0; for the

outside wall of cell 2, as 2-43; and for the web

between cells (1) and (2) as 1-2, etc

di = Shear load per inch = t,t, in the out-

Side wall of cell (1), where t, equals the unit

Stress and t, = wall thickness Likewise,

Ga * Tata and qs = Tats = Shear load per inch in

outside walls of cells (2) and (3) respectively

For equilibrium of shear forces at the dJunetien

points of Intearior webs with the outside walls,

A6.7

inch in the web 1-2 and (aq, ~ 4,) for web 2-3, For equilibrium, the torsional moment o

the internal shear system must equal the ex-

ternal torque on the tube at this particular section Thus, from the conclusions of article

A6.9, we can write:

T = 2qiA + 2qaA, + 2q3A, - ~ - For elastic continuity, the twist of each cell

must be equal, or 0 = 9, = Os

From equation (19), the angular twist of a cell is

(25)

Thus, for sach cell of a multiple cell struc-

ture an expression 3 | s can be written and

equated to the constant value 200 Let Ara,

das + for cell wall

1-0, and aia, Aaa; @ay and ay, the line in- tegrals is for the other outside wall and interior web portions of the 3-cell tube Let

clockwise direction of wall shear stresses in any cell be positive in sign Now, substituting

in Equation (25), we have:

represent a line integral

cell (1) x [s: Aro + (Qi - da) ass] = 260 (26)

+ (da = G:)4ia + Gedao + (da - Gs)aag

Aa

cell (3) ny [te ~ da)aas + as80] = 260 - (28)

Equations (24, 26, 27 and 28) are sufficient to determine the true values of Q:, Ge; ds and 6, Thus, to determine the torsional stress istribution in a multiple cell structure, we

write equation (25) for each cell anc these

equations together with the general torque

equation, Similar to equation (24), provides

suffictent conditions for the solution of the

Shear stresses and the angle of twist

A6.11 Stress Distribution and Angle of Twist for 2-Cell Thin-Wall Closed Section

For a two cell tude, the equations can be

Simplified to give the values of q,, q„ and 9

directly For tubes with more than two cells, the equations become too complicated, and thus the equations should be solved simultaneously

Equations for two-cell tube (Fig A6.18): -

A6.8

(3 agÂÁy + Arad + đi aÁ” + BoyÁa^

where A = Ai + Ag

A6.12 Example Problems of Torsional Stresses in

shape, and having one interior web An external

applied torque T of 83450 in lb is assumed

acting as shown, The internal shear resisting

pattern is required

Calculation of Cell Constants

105.8 sq in

387.4 sq 1n, 493.2 sq in

„ 25.25 ase = DS 15.1 25.3

os * SE = 17 70ge = 1785

Solution by equating angular twist of each cell

General equation 2Ge = 3§ ds Ỷ = Clockwise flow of q is positive

Call 1L Subt in general equation

The summation of the external and internal re-

sisting torque must equal zera

Solving equations (35) and (36), q, = 55.6#/in

and dq = 92.5#/in Since results come out

positive, the assumed direction o? counter-

clockwise was correct for qy and q, or true

signs are q, = - 55.6 and qa = - 82.5

Gia = ~ 55.6 + 92.5.2 36.9#/in (as viewed from Call 1)

Fig A6.20 shows the resulting shear ?at- tern The angular twist of the cemplete cell

can be found by substituting values of q and

Ga in either equations (33) or (34), since twist

of each cell must be the same and equal to

twist of tube as a whole

36 9#/in

92, 5#/in,

SOLUTION BY SUBSTITUTING IN EQUATIONS (29) & (30)

al 8guaày † 3; 2À -

+: “3 [Sim + aaah t9 |1

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES

Determine the torsional shear stresses in

the symmetrical 2 cell section of Fig A6.21

when subjected to a torque T Neglect any re-

sistance of stringers in resisting torsional

„ 20 10 _ Bao 7-5 + 73a > 916.7

Solution of Equations from Article A6.li: ~

sisting the external torque of 100,000" as

Calculation of cell constants Cell areas:

A, 3 39.3 Aq = 100 As = 100 Line integrals a = {= 7

Aro = sos 629 Gia = Ge = 200

ase =O = 667 das = Os 88

= 20 10 _ 8:o “+ 747 917 Equating the external torque to the

2qiA + 2Geda + 2QsÁa + T

2z “+ [asa a + (ai - aa) ass]

the expression for the angular twist

cell:

25.38/1n

qa

Sa

Qs Qa~- 9:

sired, can be found by substituting values of

shear flows in any of the equations (38} to (40)

TORSION

A6.14 Torsional Shear Flow in Muitiple Cell Beams by Method of * Successive Corrections

The trend in airplane wing structural de-

sign particularly in high speed airplanes is

toward multiple cell arrangement as illustrated

in Fig A6.24, namely a wing cross-section

made up of a relatively large number of cells

EXPLANATION OF SUCSESSIVE CORRECTION METHCD

Consider a two celi tube as shown in Fig

a To begin with assume each cell as acting independently, and

§ ss for simplicity will be written ats where

L equals the length of a wall or web panel and

t its thickness Thus we can write,

Therefore assuming G@, = 1 for cell (1) of

Fig a, we can write from equation (41): -

* Based on Paper, "Numerical Transformation Procedures for Shear Flow Calculations" by S, U Benscoter, Journal, Institute of Aeronautical Sciences, Aug 1946

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES

In a similar manner assume cell (2) sub-

jected to a shear flow qe to make GO = 1, Then

Ag 2x 39.3

dae 5h IO, 15.7 = G8 = 0.1068/1n

$ 205 203

Fig c shows the results

Now assume the two cells are joined to-

gether with the interior web (1-2) as a common

part of both cells 7

See Fig d The in~

terior web is now

subjected to a re-

sultant shear flow

of di = da = (.212 -

.109) = 103#/i1n

Obviously this change

of shear flow on the

interior web will cause

the cell twist to be different for each cell in-

svead of the same when the cells were considered

acting separately To verify this conclusion

the twist measured oy the term GO will be com-

puted for each cell

Since @, must equal 9, tf

fs not to distort from

evident that the above

true ones wren the two

unit

Now consider cell {1) in Fig d In bring-

ing up and attaching cell (2) the common web

the cross-section

its original shape, it ts

shear flows are not the

calls act together as a

A6.11

(1-2) ts subjected to a shear flow q, = - 1O9#

/in (counterclockwise with respect to cell (1)

and therefore negative), in addition to the

shear flow q, = 212 of cell (1) The negative shear flow q, = - 109 on web 1-2 decreases the

twist of cell (1) as calculated above with the

resulting value for G61 = 0.975 instead of 1.0

as started with

Thus in order to make G6, = 1 again, we

will have to add a constant shear flow qi to cell (1) which will cancel the negative twist due to q, acting on web (1-2) Since we are

considering only cell (1) we can compare ceil wall strains instead of cell twist since in

equation (41), the term 2A is constant

Thus adding a constant shear flow q, to

cancel influences of qa on web 1-2, we can write:

Thus to make GQ, equal to 1 we must correct

the shear flow in cell (1) by adding a constant shear flow equal to 237 times the shear flow qz

in cell (2) which equals 237 x 109 = ,025e%/in

Since this shear flow is in terms of the shear flow qa of the adjacent ceil it will be referred

to as a correction carry over shear flow, and

will consist of a carry over correction factor

times qa

Thus the carry over factor from cell (2) to cell (1) may de written as

KG) wed (1-2) C.0.F , =

(2 to 1) = cell (1) which equals

-227 as found above in substitution in equation (42)

Now consider cell (2) in Fig d In bringing up and attaching cell (1), the common wed ( } is subjected to a shear flow of q, =

~ 0,212%/in (counterclockwise as viewed from

cell 2 and therefore negative) This additional

ear flow changes G6, twist cf cell (2) toa relative value of 0.4375 instead of 1.0 (see

previous G@, calculations) Therefore to make

G8, equal to 1.0 again, a corrective constant shear flow q, must

be added to cell (2) to

A6 12

cancel the twist effect of q, = 0.212 on web

(2-1) Therefore we can write,

t- f200\ _

4a”: (723) =

Thus the carry over factor from cell (1) to

cell (2) in terms of q: to make GQ, 5 1 again can

Fig @ shows the constant shear flow qi and

qi that were added to make GO = 1 for each cell

However these cor-

were added assuming

the cells were again ||

(1-2), Thus in

bringing the celis

together again the interior web is subjected to

be resultant shear flow of qi ~ qi In other

words if we were to add the shear flows of Fig e

to those of Fig d, we would not have GQ, and Ge,

equal to 1 The resulting values would be closer

to 1.0 than were found for the shear flow system

of Fig d

Considering Fig e, we will now add 4 second

set of corrective shear flows q{ and q} to cells

(1) and (2) respectively to make GO, and GO, =1

for cells acting independently

Considering cell (1), and proceeding with

same reasoning as before,

at &Ð call (1) - ah e) wed 1-2 = 0

Suppose we add one

more set of cor-

rective constant shear flows qfand q?,

The final or re-

sulting cell shear

flows then equal the original shear flows plus all corrective

cell shear flows, or

will be computed for each cell using the

G8, = zzai l7ÊP x ST - ,07A7 x SỊ 997

A6.15 Use of Operations Table to Organize Solution by Successive Corrections

Operations Table 1 arranges the calculations

so that the steps dealing with the corrective Shear flows can be carried cut rapidly and with

3 minimumm of thought