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

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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 subỞ jected to torsional forces under operating con-

ditions, hence @ knowledge of torsional stresses and distortions of members is necessary in aircraft 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 round cylinder (2) Sections remain circular after application of torque (3) Diameters remain straight after twisting of section (4) Material 1s homogeneous, tsotropic and elastic

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

perpendicular to the axis of the shaft or

cylinder ,p

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

ABỖ under these shearing stresses and this rotation 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 moving to OBỖ The unit shearing strain in a distance L equals, BB' r6 ồ Ộ1E AB * TLỢ

Let G equal modulus of rigidity of the material and let t equal the unit shearing

stress at the extreme fiber on the cross-section

Hence, T= eẠ G@ = Ở=Ở~ -7- (1) In Fig A6.2 let t, equal the unit shearing stress on a circular strip dA ata distance p from 0 Then

ề aoe shh OG poe

2 r TLỢ L

The moment of the shearing stress on the circular strip dA about 0 the axis of the bar is equal to, 2

ay = t, po = 2 TA T and thus the total internal torsional resisting moment is,

a

x = { Goteda int u

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 = ~

ụ

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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 The resistance moment on tube AB due to side 150# = 150 x 26 = 3900 in lb to this torque is provided by the 1508 H 24 ST Al Alloy }Control Stick ly = 3,800,000 psi 8 28"

j4_Elev Control Wire Aileron Horn /Aljeron Operating TubeLWỞ+Ở 5 a x.058 Tube

Bearing 2 l4 24 St Al Al Bearing 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 Th 3900 x 28 GI ~ 5,800,000 x 0.1866 or 12 degrees, 6c = 0.2] radians TORSION

ates an aileron control

a cireular torque tube ported "reo tr Ở 0 Fo crt ee st gop ort bracket arine stress in aileron is as iso compute tween horn section Find the tụ indica 5 om p t Đ Average load on surface = 40#/0' 14 - 049 Tube(AL Alloy) 24" Fig A6.4 SOLUTION:

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) Then 3 2 9 ab a + (0.5 w}l2 = 4.16 hence w = 9,463 1b

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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 computed by using the average torque as acting on

antire length of the tube to one side of horn or a distant L = 29", nence 261 x 29 BE BSCG0G0 x C.0eS7e 97:5 * 0-86 degrees A6.4 Torsion of Members with Non-Circular Cross- Sections

The formulas derived in Art 46.2 cannot be used for non-circular shapes since the assumpỞ 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 ể max T L S "g7" Fig A6.5 ~ Fig A6.6

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 Ổhe surface and this ts

not true The theory of elasticity shows that the maximum shear stress occurs at the centerline 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 Ổts

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 Ổshape 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 measured 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 following 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 illustrated in Fig A6.9 These contour lines

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

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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 ằde- 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 A6.3 Fig A6.9

Slope Slope Slope

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 torsional 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 unit length aati, ee eee ee Tax F Gp eet tte (6) = T "Getta 3 radians -+ - (7) Values of a and $ are given in Table A6.1 TABLE A6.1 CONSTANTS ằ AND 6 b/tÍ1.09 |: 5g |1 76 ]2.00 |3.50 |3,00 | 4 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 rectangles, 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 the jun- at Fig a b can be taken as centerline length for above type of sections wa Da ey + Te ba mà bạ sty + Fig A6 10

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

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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 it an open section For the round tube Ji = 0.02474 in* w a For open tube J, = 5 x 2.14 x 035 = 0.000045 os be3 4" 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 ) = Modulus of Rigidity, (n4) From Table SECTION x SOLD ELLIITTICAL SECTION, at

Ấ Ha3 bi TMAX = (at ends

2Ừ ề=a of minor axis), abe 2a SOLD SQUARE TT T + 4 ỘMAX ồ 0 708a4 (Át a K = 0.1414! midpoint af each side) SOLID RECTANGLE, +" |Kz NI + 3.3820 - 2b rd MAX = Tas 1 8b) Sa: midpoint of longaide, Ởể SOLD TRIANGLE ~ 20T 4 = Ly eae at midpoint neo of side

For an extensive list of formulas for many shapes both solid and hollow, refer to book, "Formulas For Stress and StrainỢ 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

= and = *

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 constant The shear stress at any point tT = q/t

If the wall thickness t changes the shear stress

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AG, 6

changes but the shear force q does not change, or

Tata = ỘbÈp = 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 A T=Q2A eee errr rere (14) or _7 = as-yet (15)

where A is the enclosed area of the mean periphery 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 fe 1 + eS we 1H ds + Ậ qda ể 2 a 4 === hai 43 ỞỞỞ;

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 Ấ ảqU =Ở ặ However the shear strain 6 can te written, z2 z8 bu = S Sq tap out a sa ma TS et a hence 4U = xin 48 pa

or Uz [ae S5 the integral $ is the line integral around the periphery of the tube From Chapter a7 from CastiglianoỖs

theorem,

9U T

Ộ=Ở~ ỞỞỞỞc3 ể- - - =~ỞỞ-Ở -Ở_~ 1

9= } TRE Ư8 (17)

Since all values except t are constant, equa-

tion (17) can be written, I MB LL ee ee eee 18) 8 Ộ sên Ặ 5 (28) and since T = 2 qA, then 2150, =, ậ/$8. - AG) ea hs (198)

where 9 is angle of twist in redians per unit length of one inch of tube For a tube length

of L \

L

hàn: ể ỞỞỞỞ (20)

A6.9 Expression for Torsional Moment in Terms of Internal Shear Flow Systems for Multiple Cell Closed Sections

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 Ổhe 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

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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 Ộ2x (Ai + Aap) + 2QzÂza Ở 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,

we have (qi ~ q,) equal to the shear load per

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 integrals 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,, qẤ 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): -

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A6.8 (3 agÂÁy + Arad + đi aÁỢ + BoyÁa^ -~ 1 Fig, A6.18 =i Zarda * 4: 2Ã -.-e da = 3 lax + aya? + | r (30) = in va an LL J=4 [Em Faiadao * Factor (31) a D D2 2S - ST TT TT an -Ở (32)

where A = Ai + Ag

A6.12 Example Problems of Torsional Stresses in Multigle-Celi-Thin-Walled Tubes

Exampie 1 - Torsional Stresses in Un-symmetrical Two- Cell-Tube

Fig A6.19 shows a typical @~-cell tubular section as formed by a conventional airfoil 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 do iS avo = See = 1075; ai = Sts os Cell areas: - A, Aa A Line integrals a Ấ 25.25 ase = DS 15.1 25.3 os * SE = 17 70ge = 1785

Solution by equating angular twist of each cell General equation 2Ge = 3ậ ds Ỷ = Clockwise flow of q is positive Call 1L Subt in general equation 1 289 = Tp [ể: x 1075 + (-q, + q,) ss | = ~ 13.23 qa + 3.165 qa TORSION Cell 2 865 di - 5.34 qa Equating (33) and (24) - 14.195 qi + 8.505 qa =O -

The summation of the external and internal re-

sisting torque must equal zera .0ạ0" 1= 25.28" = -= Ga 0407/ Seema Torque ý 4 83450" | 4: cen @ ib 050Ợ | 1 = 25, 30" 83450 ~ 2 x 105.3 qi - 2 x 387.4 da = 1 ồ

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

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ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES 1735 x 105.8 + 355 x 493.2 1738 x 105.3Ợ + đỏ5 x 495.2Ợ + 1075 x aa T 2 = ,000665 T = 000665 x 83450 = 55.6#/in qa =k 2 Brghg Đ 3u aÁ tT = 2) aeodi + iadỢ + AioAa |Ợ 1075 x _ 387.4 + 335 x 493.2 Tt 1755 x 105,82 + 225 X 495.22 + 10758 x 387.47 | 2 ỀÔ001107 T = 001107 x 83450 = 92.5#/1n Example Probiem 2

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 moment Fig, A6 21 == Ss SOLUTION: .03" Calculation of 3 terms Area of celÍS: ~ A = 100 A, = 100 A = 4, + Ag = 200 Line integrals a = ds ST = 1Ô 29 _Ấ Ẽ aro = SE To = 866.7 ~ 10 _ Bia > ORF 333.3 Ấ 20 10 _ Bao 7-5 + 73a > 916.7 Solution of Equations from Article A6.li: ~ 1 =i BagAi + Arad 2 9: * 3 |ayodi? + @iaệ + diode <1 916.7 x 100 + 355,3 x 200 U3 [318.7 x 100? + 555.5 x 2007+866.7x 1002 = 002540 T a, =i @roda + Arad T a 2 [aecdaỎ mm Tra AT 866.7 x 100 + 332.3 x 200 + 916.7 x 100Ợ + 333.3 x 2007 + 866.7 x looỢ nr T A6.9 = ,002456 T Js AagohsỢ* + Aizh* + arcdaệ @vodia + Aiadag + Aacdio ema Stak FT SicAs 4 {926.7 x 100% + 333.3 x 200% + 866.7 x 1007 866.7 x 333.3 + 333.3 x 916.7 + 916.7 x 866.7 = 89.76 T T Ế x69.76 ề01116 = (rad.) per unit e g oe - 9#Ở= 6 length of cell

A6.13 Example 3 - Three-Cell-Tube,

Fig A6.22 shows a thin-walled tubular section composed of three cells The internal shear flow pattern will be determined in re-

sisting the external torque of 100,000" as shown L 5Ợ Ở 10" ỞẨỞ tơ sả 93 + Fig AG 22 SOLUTION: Calculation of cell constants Cell areas: A, 3 39.3 Aq = 100 As = 100 Line integrals a = {= 7 ft x12 _ - 19 _ Aro = sos 629 Gia = Ge = 200 ase =O = 667 das = Os 88 = 20 10 _ 8:o Ộ+ 747 917 Equating the external torque to the

sisting torque: - internal re=

2qiA + 2Geda + 2QsÁa + T Substituting: =0 ~ (37) 78.6 q; + 200 44 + 200 qs - 100,000 writing of each Cell (1)

2z Ộ+ [asa a + (ai - aa) ass]

the expression for the angular twist

cell:

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A6, 10 Substituting: ace "sa [529 Qa + 200 aa ể 200 a] - (38) Cell (2) 289 = El = G3) aia + Qedac * (42 - Gs) ể Substituting: 2e = me [200 de - 200 q + 667 qa + 333 Gq, ~ 333 as | Cell (3) 1 2e = in Ở đa) Gast astso| Substituting: 06 = sig [505 as - S85 aa + 927 as] ~~= (40) Solving equations (37) to (40), we obtain, 145.4#/in 234.1Ậ/1n 208.8#/1n, 90.78/1n 25.38/1n qa Sa Qs Qa~- 9: 3z ~ qa 234 19/in 908 8#/in AAW + Fig A6 23 208 8#/in \ \ 208 8#/in 234 1#/in,

Fig A6.23 shows the resulting internal shear flow pattern The angle of twist, if de-~ 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 <TTT II Fig A6 24

With one unknown shear ?1ow q for sach cell, the solution by the previous equations becomes

quite labortous

The method of successive approximations provides a simple, rapid method Zor finding the shear flow in multiple ceils under pure torsion,

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 subject cell (1) to such a shear flow a, as to make GO, = 1 From equation (19) we can write, #63 ZA 5 Now assume G9 = 1, then Ge =

Since practically all cellular aircrart beams nave wall and web panels of constant thickness for each particular unit, the term ậ 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): -

Trang 11

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES 2A, 2X 89.3 128 asp 85.7 TỔ s Già = 04212 Ib,/1n, T ,0 ` ,05 + q=.109 Fig ằ

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 together 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 computed for each cell =.109 Fig d = Le Cell (1), Ge = ap laze - = sy aos [22 Xa + (.212 - 109) RIF 1 25.7 = = 0.875 Cell (2) ? q6, 25h rqts= + ` ĐÁ, t - = ay 50S [: 109 x aE ~ (.212 ~ 109) | = 1 15,7 10 = 0.4375

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: ' +) - ( = a ( vJj 98H Ể) ~ da +) web 1-2 7 9 _(@) wee be web 1-2 Ộ1G (2Ạ) cen a) Q) Substituting values in equation (42) ~ - (42) 10 2 05 - 200 % = 4a |Z8.77Ở To | = age Ge * -Z57 QẤ v04 * 705

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

Trang 12

A6 12

cancel the twist effect of q, = 0.212 on web

(2-1) Therefore we can write, L ah (: x) cell (2) - a (#) wed (2-1) = 0 hence Ở9 @ web (2-1) qa = Ẽ L 3 Ậ) cell (2) Substituting in equation (43): - t- f200\ _ 4aỢ: (723) =

Thus the carry over factor from cell (1) to

cell (2) in terms of q: to make GQ, 5 1 again can be written -277 4, = 277 x 212 = 05874/in = Ể) web (2-1) _ 200 Ểt92) Gly on a 72 t C.0.F ề277

Fig @ shows the constant shear flow qi and qi that were added to make GO = 1 for each cell However these cor-

rective shear flows 4Ỗ, = 0587

were added assuming

the cells were again ||

independent of each (2) )

other or did not |

have the common wed ae

(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 &Đ call (1) - ah e) wed 1-2 = 0 Hence Fig e af = = ,0887 x 237 = 0139#%/in or n= 5 ts az (C.0.F.) 2 tol times q) 237 x ,0587 = 0129 TORSION Considering Cell (2) qa = ah (C.0.F.) = 0256 x 227 = itoe 007178/1n

Fig f shows the resulting second set cf corrective constant shear flows for each cell Since our corrective

shear flows are

rapidly getting

smaller, the con- tinuation of the process depends on the degree of accuracy we wish for the final results Suppose we add one

more set of cor-

rective constant shear flows qfand q?, Using the carry over correction factors previously found we obtain, qv 247 x 00717 = 00173/1n qỲ = 277 x 0139 = CO38S#/in Fig g shows the results

The final or re-

sulting cell shear

flows then equal the original shear flows plus all corrective

cell shear flows, or Ở== # /% = 0011 | = t a (final) 7 a+qit ai + qt de (tinal) F $a * đa * 4ã? 4z Pig h shows the final results To check the final twist of each cell the value Ge

will be computed for each cell using the q values in Fig h Cali (1) 1 : 6: = res 2584 x ng + ,0747 x1" 987

G8, = zzai l7ÊP x ST - ,07A7 x SỊ 997

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

Trang 13

orn ~ ae ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES AG 13 OPERATIONS TABLE 1 ) web 1-2 20 Celi i G.0 - t - +05 = 40 Ộ (1-2) (Ù ỘTO 296 - a (=) cell 2 Ộ5 * OT Gì Web 2~L c.0 2.25 Explanation of Table 1

Line 1 gives the carry over factors for each cell, computed as explained before Line 2 gives the necessary constant shear flow q in each cell ta give unit rate of twist to each

cell acting independently (G@=1) Line 3

gives the first set of constant corrective shear flows to add to each csll The corrective q re- farred to as the carry over q or C.0.q in the table consists of the q in the adjacent cell

times the C.0 factor of that cell

Thus 237 x 109 = 0254 is carry over from

cell (2) to cell (1) and 277 x 212 = 0587 1s carry over from ceil (1) to cell (2)

Line 4 gives the second set of corrective carry over shear flows, namely 277 x 0258 = Ấ00717 to cell (2) and 237 x 0587 = 0139 to cell (1) Line 5 repeats the corrective carry over process once more Line 6 gives the final q values which equal the original q plus all carry over q values

Example Problem 1 (2 cells)

Determine the internal shear flow system for the two cell tube in Fig 46.25 when sub~ đJected to a torque of 20,000 in lbs }_ 10" + Ở 10" Ở_,| Ƒ 08 0.10 10h |.08 CỀH 1 os) - CSH og 0 ? L 98 0.10 Fig A6.25 OPERATIONS TABLE 2 Cell i

Explanation of solution as given in Table (2): Line 1 gives the values of the carry over

factors The values are calculated as follows:

C.0 factor cell (1) to cell (2):

(2-1) "z(seaq

Line 2 gives the shear flow q in each cell, when it is assumed each cell is acting separately and is subjected to a unit rate of twist ST Q9 =1 - The calculations for the q values are as follows: - For cell (1) as ~ 24100 25 705 For cell (2) q, ềỞỞỞSAa 8 as 2X 100 , z () cei 2 +s

Ay and Ay equal cell area of cells (1) and (2)

respectively L = length of wall element and t its corresponding thickness

In order not to start out with decimal values

of q, the values above will arbitrarily be

multiplied by 100 to make q: = 25#/in and

da = 40#/in Since we want only relative values of terms this ts permissible These

values are shown in line 2 of Table (2) The corrective carry over process proper is started in line 3 of table (2) In cell (1) the amount

earrted over of q = 25 to call (2) equals the C.0 factor times 25 or 0.4 x 25 = 10 which is

written along the vertical Line under cell (2)

Likewise in cell (2) the amount of qd = 40 that

ts carried over to cell (1) equals the C.0

factor times qs or 25 x 40 = 10, which is

written along the vertical line under cell (1)

The second set of corrective carry over constant shear flows are given in line 4 of

Table (2), thus, 4 of the qi = 10 = 4.0 is

carried over to cell (2) and 25 x value of qi = 10 = 2.5 is carried over to cell {1) Line 5 repeats the process namely 0.4 x 2.5 = 1 to

cell (2) and 25 x 4 = 1.0 to cell (1)

This process of carrying over values of q

is continued until the values are small or negligible Line 8 gives the final q values in

each cell as the summation of the assumed q

value plus all carry over values of q Thus in

cell (1} q = 38.85#/in and for cell (2)

q = 55.5 Line 9 gives the <orque that these values of cell shear flow can produce,

Trang 14

TORSION

AG.14

For cell (1) T= 2x 100 x 38.85 = 7770 in ib For cell (2) T= 2x 100 x 55, 8 5 = 1100 in, lb Line 10 gives the sum of the above two yalues

which equals 18870 in lb

The original requirement of the problem was the shear flow system for a torque of 20,000"#

Therefore the required q values follow by direct proportion, whence _ 20000 - 3: Ộ T8g;o * 38.85 = 41.28/1n _ 80090 Ấ | 4a = To X 55.5 = õ8.98/1n,

These values are shown in line 11 of Table

2 Check on twist of cells under final q values

The relative total strain around each cell boundary is given by the term = 5 at for the cell Thus for cell (1) For cell (2), 1 bil 10 30 c3 3$ Ộ18g [ (8.8 - 4.2) Ze + 53.9 x (& |

Thus both cells have the same twist In the above calculations q, and qq act

clockwise in each cell, hence the shear flow on

the interior common web is the difference of the

two q values

Example Problem 2 Three cells

Explanation of solution as given in Table 3: ~

cell (1) cell (2) cell (3) Ay = 144 A, = 96 As = 56.5

3z E= 1140 t z+= 1040 t oe 3# = 1085 Shear flow q for

dependently: ~ G@ = 1 for each cell acting in- Ạ Cell (1) ast - St Ấ228 Ấ Tag 7 265 t = 242 = 122 = (1645 Cell (2) as thig = -2648 + - 8Á, - 112 _ Ấ Call (3) qa TL" O55 7 107 t

To avoid small numbers these values of q are multiplied by 100 and entered on line 2 of

the table Calculation of carry factors as

Trang 15

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES

Cell (3) to (2)

The balance of the solution or procedure in Table 3 is the same as explained for Problem 1 in Table (2) It should be noticed that cell

(2) being between two cells receives carry over q values from both adjacent cells and these two values are added together before being distribu- ted or carried over again to adjacent cells For example consider iine 3 in Table (3) and esll (2) The-q value 5.82 representing 23 x 25.3 is brought over from cell (1) and the q

value 3.C8 = 2388 x 10.70 from cell (3) These

two values are added together or 5.82 + 3.08 ề 8.90 The carry over q to cell (1) is then 21

x 6.90 = 1.87 and to cell (3) is 284 x 8.90 =

2.535

In line 8, the final q in cell (2) equals the original q of 18.45 plus all carry over q values from each adjacent cell

Line 10 in Table (3) shows the total terque developed by the resultant internal shear flow

is 17435"# Since the problem was to find the

shear flow system for a torque of - 100,000 tn ib., the values of q in line 8 must be multipli-

ed by the factor 100,000/17435

Line 11 shows the final q values,

Example Problem 3 Four cells

Determine the internal shear flow system for the four cell structure in fig 46.27 when subjected to a torsional moment of - 100,000 tn ib (500) (625) Celli Cell 2 Can 3 Celt 4 Fig A6 27 OPERATIONS TABLE 4 Cell 2 q A 65.3 +5, A

In Fig 46.27 the values in the rectangles

represent the cell areas The values in ( } represent the L/t values for the particular wall

or web, After studying example problems 1 and 2

one should have no trouble checking the values

AG 15 as given in Table 4 Line 10 shows the correction of q values to develop a resisting

torque of 100,000ồ# The multiplying factor ts

100 ,000/80630

A6,16 Torsion of Thin-Wailed Cylinder Having Closed Type Stiffeners

The airplane thin-walled structure usually

contains lontitudinal stiffeners spaced around

the outer walls as illustrated in Figs A6é.2e and A6.29, Ở xe Fig Aậ.28 Fig A6 29 ~Ở Ở

Open Type Stiffener Closed Type Stiffener For the open type stiffener as illustrated in Fig A6.28, the torsional rigidity of the in-

dividual stiffeners as compared to the torsional

rigidity of the thin-walled cell is so small to be negligible However a closed type

stiffener is essentially a small sized tube and

its stiffness is much greater than an open section of similar size Thus a cell with closed type stiffeners attached to its outer

walls could be handled as a multiple cell structure, with each st_ffener acting as a cell with

a@ common wall with the outside surrounding cell

Since in general the stiffness provided by the stiffeners 1s comparatively small compared to the over-all cell, the approximate simplified procedure as given in NACA T.N 542 by Kuhn can be used to usually give sufficient accuracy In this approximate method, the thin-walled tube

and closed stiffeners are converted or trans-

formed into a single thin-walled tube by modi- fying the closed stiffeners by either one of the following precedures: -

(1) Replace each closed stiffener by a doubler

plate having an effective thickness

tg=tgps/d, and calculate J ds/t with these

doubler plates in place The enclosed area

of the torsion tube still remains (A) or

Trang 16

A6 18

(2) Replace the skin over each stiffener by a "liner" in the stiffener having a thickness

te = tgx d/s (See Fig A6.31.) The en~ closed area (A) of the cell now equals the

original area less that area cross-latched in Fig A6.31

Procedure (1) slightly overestimates and

procedure (2) slightly underestimates the stiffness effect of the stiffeners,

The corner members of a stiffened cell are usually open or solid sections and thus their torsional resistance can be simply added to the torsional stiffness of the thin-walled over-all cell

AG.17 Effect of End Restraint on Members Carrying Torsion

The equations derived in the previous part of this chapter assumed that cross-sections throughout the length of the torsion members

were free to warp out of their plane and thus there could be no stresses normal to the cross- sections in actual practical structures re-

straint against this free warping of sections is however often present For example, the airplane cantilever wing from its attachment to a rather rigid fuselage structure is restrained

against warping at the wing-fuselage attachment

point Another example of restraint ts a heavy wing balkhead such ag those carrying a landing gear or power plant reaction The flanges of these heavy balkheads often possess considerable lateral bending Stiffness, hence they tend to prevent warping of the wing cross-section Since only torsional forces are being considered here as being applied to the member, the stress~

es produced normal to the cross-section of the

member namely, tension and compression must add up Zero for equilibrium Thus the applied torque is carried by pure torsion action of the member and part by the longitudional stresses

normal to the member cross-sections The per-

centage of the total torque carried by each action depends on the dimensions and shape of the cross-section and the length of the member,

Fig A6.32 111ustrates the distortion of an

open Section, namely, a channel Section subject~

ed to a pure torsional force T at its free end

and fixed at the other or Supporting end Near the fixed end the applied torque is practically all resisted by the lateral bending of the top and bottom legs of the channel acting as short cantilever beams, thus forming the couple with 8 forces as {llustrated in the Figure Near the free end of the member, these top and bottom

legs are now very long cantilever Deams and thus

their bending rigidity 1s small and thus the pure torsional rigidity of the section in this region is greater than the bending rigidity of

the channel legs

TORSION

Partly lateral shear and partly torsionai shear

Torsional shear,

Fig A6 32

A6.13 Example Problem lustrating Effect of End, Restraint on a Member in Torsion

Fig A6.33 shows an I-beay Subjected to a

torsional moment T at its free end The preb-

lem will be to determine what proportion of the torque T is taken by Ổhe flanges in bending and what proportion by pure Shear, at two different sections, namely 10 inches and 40 inches from

the fixed end of the i-beam

Fig AG 33 Fig A6 34 Fig A6, 35 Fig A6.34 shows the torque dividing into two Parts, namely the couple force F-F formed by bending of the flanges cf the i-deam and the

Pure Shearine stress system on Ộhe cross-

Section Fig A6.35 shows the twisting of the Section through a distance 5

The solution will consist tn computing the

angle of twist 9 under the two stress con-

ditions and equatinz them

Trang 17

ANALYSIS AND DESIGN OF

Note: Tne deflection of a cantilever beam with a load F at its and equals FLồ/3E1, and I the

moment of tnertia of a rectangle about its cen-

ter axis = toồ/12

hence h 9p = Bot ay 3

Now let Tt be the portion of the total torque carried by the aember in pure torsion The ap~

proximate solution for open sections composed of

rectangular elements as given in Art A6.6,

equation (12) will de used, Saat 8 Ộ GEY (TS 7 bạ) Equating 6p to Sy , we can write Tạ _ 3 En? bề == Te 8b?at* (2b + by) Substituting values when L = 10 inches 2 2 Ta - 3x 10.5 x 10% x 3.4.x 1.75 Te 8x 102 x 3.7 x 10ồ x O.14(2 x 1.75 + 3.5) = 9.65

FLIGHT VEHICLE STRUCTURES A6 L7

onstant speed The dif~

on two sides of a pulley

Calculate the maxim 1-3/4 cac

The shaft rotates at gerence in belt wuld 9 are shovm.on the figure torsional shearing stress in

shaft between pulleys (1) and (2) 4

(2) ane (3)

(2) A 1/2 HP motor operating at 10C0 RP

rotates a 3/4~.035 aliminum alloy torque tube

30 inches long wnich drives t:

for operating a wing flap Decaraine the mum torsicnal stress Ín the torque shazt

full rower and RPM ind the an

of shaft in the 3O inch Length Polar inertia of tube = 0l in Modulus of ẹ G 3ặ00,000 psi tre l (3) In the cellular s determine the tersiona = mare 3

0.5 percent | che axvernal torque 22 Ạ0CCO !n,15,

wall thickness are given on the f

the tude 1s 100 In long and finc

it we consider the section 40 inches from deflection Material {s alumima 4 the fixed end, then L = 40 inches Thus if 40% Beco, 000 pst.)

is placec in the above sudstitution instead of (4) In Pig A6.36 remove che tntartor 035

107 the results for Ts/Tz would be 0.602 and web and compute torsional shear flow and de-

the percent of tne torque carried by the flenges | elections

in bending would be 36 as compared to 90.5 per-

cent at L = 10 inches tron support Thus in T = 100,000 in tb general the eZfact of the end restraint decreas:

Tapidiy with increasing value of L

ne effect of end restraint on thin wallec tubes with Longitudional stiffeners is 2 more

invelved prob and cannot be nandled in such a simplified manner This problem is censica in other Chapters A6.18 Problems + 12Ợ 5 + Fig A6 37 fe Lo te grt

3208 (5) In the 3-cell structure of Fig A6.57

Fig A6.36 deternine the tnt ermal resisting shear 710w dua to external torque of 100,000 in lb Fora {1} In Fig A6.36 pulley (1) ts the driving | length of 100 inches calculate twist of cellular

pulley and (2) and (3) are the driven pulleys, structure 12 G 1s assumed 3,800,000 psi

Trang 18

AG 18 TORSION

(6) Remove the 05 intertor web of Fig EỞ 14814

A6.37 and calculate shear flow and twist 1ã 08 ĐT

(04 L05 04 031 joe (a)

(7) Remove both intertcr webs of Ftg |

A6.37 and calculate shear flow and twist 03 -03 ji

POM ee Bt 10" Wale 10"

(8) Each of the cellular structures in 03 | 03 | 03 08

Fig A6.38 is subjected to a torsional moment 10" i

of 120,000 in.lbs Using the method of 04 704 035/088 Lũ3 04 03] (b)

successive approximation calculate resisting iL 03 | 03 | 03 03 Shear flow pattern

187 Ở 10" te BB et eg

| lo

All interior webs = 05 thickness Top skin 084" thickness Bottom skin 064" thickness Pig A6 38 "havcoms wy 2 The big helicopters of the future will be used itn many important industrial and military opera-

Trang 19

ng CHAPTER A7 DEFLECTIONS OF STRUCTURES ALFRED F, SCHMITTỢ AT.1 Introduction

Calculations of structural deflections are important for two reasons: -

(1} A knowledge of the load-deformation characteristics of the airplane is of primary

tmportance in studies of the influence of structural flexibility upon airplane perform- ance

(2) Calculations of deflections are neces-

sary in solving for internai load distri-

putions of complex redundant structures the The elastic deflection of a structure under load is the cumulati.> result of the

strain deformation of the individual elements composing the structure As such, one method

of solution for the tctal deflection might in-

volve a vectorial addition of tnese individual contrioution! @ involved geometry of most practical structures makes such an approach

proniditively difficult

For complex structures the more popular

techniques 2r2 analytical rather than vectorial They deal directly with quantities which are not

themselves deflections tut from which deflec-

tions may be obtained by suitable Operations Ộhe methods employed nerein for deflection cal-

culations are analytical in nature

AT.2 Work and Strain Energy

Work as defined in mechanics is the product of fcrce times distance I? the force varies over the distance then the work is com-

puted by the integral calculus Thus the work

done by a varying force P in deforming a body an amount 6 is

Work = L Pd6 and is represented

by the area under the load deformation (P-5)

curve as shown in Fig A7.1

IZ the deformed oedy is ?ectly elastic

the energy stored in the body may Se completely

recovered, the dody unloading along the same P-6 curve followed for increasi load 18

energy is called the elastic s

deformation (hereafter the str:

orevity) and is denoted by the symbol U Thus

u = ff Pdô, Should the body Ổbe linearly

elastic (as are most dodies dealt with in

structural analysis) then the load-deformation (1) Design Specialist, Convair-Astronautics Ai n work? > Pi A Oj it in the limit as 46.0 work {op đồ Flg AT.L

curve is a straight line whose equation ts

P = kô and the strain energy ts readily com~ puted to be

K6* P2

Us =U, equally, U = a AT.3 Strain Energy Expressions for Various Loadings,

STRAIN ENERGY OF TENSION

load S acting at the end of 4 A tensile

length L, cross sectional area uniform bar of

A and elastic modulus ặ causes a deformation 6= SỢ⁄AE, Hence 8 = > ô and

us sag = 32 & eee ee ee ee (1)

a L 2

Alternately,

aan ole

Use 7 TTT (2)

Squations (1) and (2) are equivalent expressions for the strain energy in a uniform bar,

the former expressing U in terms of the de-

formation and the latter expressing U in terns cf toad, The second form of expression is more

ặ neral usage and succeeding

e energ)

: a tensile bar having non-uniform prop-

erties (varying AE), or for wnicn the axial lcad

Trang 20

AT.2 bar is obtained oy of the bar Example Problem 1 A linea:

y tapered aluminum Dar is under

an axial load of 15CO lbs as showr Ổn Fig, AT.2 Find U Ao A(x) = A,(1-3) sọ ` 80 Ags 2 int t 1s00# Fig AT.2 Solution: From statics, the internal load ậ = -1500" at every section L v =f s Stax _ Ở= AE x ồ 4(1 - 39/20 X 102 ( eax/so introduced

- ark for convene

= 2(-1800)"f" ne Ce a7 2 4.5 in 2 inch Ibs ieneed Note that although P was a negative (com~ pressive) load the strain energy remained positive Example Problem 2 Find U in a uniform bar under a running load re @ = do cos 5 (Fig 47.3) root | ! bw 40" Ổ 2 4 Ấ k.2in i L~40 B= 10x10ồ pet q(x) | đạ= 25 1b/ta | Fig A7.3 Solution: The equation for S (x) was found by statics, - fl fl 2 T% Six) =f qdx = Ge S08 ar ax

= do 2 sin Ỏ] = 2h (1 -sin Se m7 Sin or Son DI,

Substitution into eq (3) gave DEFLECTIONS OF STRUCTURES A uniform beam of a pure couple un proportional to the couple tary strength of materials ể- >

Where the constant Z1, the product oặ

modulus by section moment of tnertia,

the "bending stiffness"

Since the rotation @ butlds up linearly with M the strain energy stored ts

elastic

is called

For a Jeam having non-unirorm properties (varying EI) and/or for which M varies along

the beam, the strain energy is comouted oy the calculus, From eq (4) the strain energy in a beam element of differential length is

idx ar

au = 5

Hence summing over the complete beam to get the

total strain energy one has

= - 1 ;Mồ*dx

U=/4U=s/`'%Ƒ TT tr rt tte (8)

Example Problem 3

For, the Deam of Fig A7.4 derive the strain

energy expression as a function of Mo

Me

# El = constant

Ởx:Ở Fig A7.4

Solution:

Trang 21

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES Then L a ề L :( -#) v= df wli-t) & = Mab 6EI Example Problem 4

Determine the strain energy of flexure of the beam of Fig A7.5 Neglect shear strain energy : Po = = Constant Ị he +% L + Ở~ 2 3 Flg AT Solution:

The bending moment diagram was drawn Zirst (Fig A7.5a) and analytic expressions were written fcr M : L M(X)? PgX 0<Xayt [fx Fig AT 5a

Inspection of the diagram revealed that

the energy of flexure in the right nalf of the

beam must be identical with that of the left half Hence a u ta |- | Ộ2t, x)*4x L ồ ta t1 1 = Poth? 24 st

STRAIN ENERGY OF TORSION

A uniform circular shaft carrying a torque

T experiences a total twist in a length L pro-

portional to the torque Thus from elementary

strength of materiais

L

O=G5T See ener (6)

where the constant GJ, the product of the elastic modulus in shear by the cross section polar mom-

ent of inertia, is called the Ộtorsional stifr-

nessỢ,

Since the twist > builds up linearly with

T the strain energy stored is

For a shaft of non-uniform properties and

varying loading one has

In passing it is worth remarking that one often encounters the group symbol "Gu" in use

for the torsional stiffness of a non-circular

shaft or beam such as an aircraft wing In

such a case the torsional stiffness nas not been computed literally as G x J, but rather a$ de-

fined by oq (6), viz the ratio of torque to

rate of twist

example Problem 5

For a Certain flight condition the torque on an airplane wing due to aerodynamic loading

is given as shown graphically in Fig A7.6 The

variation of torsional stiffness GJ is given in like manner Find the strain energy stored 1.04 = + Fig A7.8 0 + Y:o Y*L Root Tip Solution:

A numerical integration of eq (8) was

performed using SimpsonỖs rule The work is

shown in tabular form in table A7.1,

Values of T and GJ for selected wing stations were taken from the graphs provided

and were entered in the table For convenience

in handling the numerical work all the variables

were treated in non-dimensional form, eq (8)

being changed as follows

_ 1 fire =1 1R?( `):

Ves) av re {8 at

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AT.4 DEFLECTIONS OF STRUCTURES

where - strain energy ?n a beam under the action of 2

T7 *ẾT/Ta, t+=y/L transverse shear loading V(ibs)* For this purpose ầV=7x dy x t and t x dy is called a,

GJ = Gd/GIR the Seam cross sectional area H

The subscript R denotes "root" value (y=0) aU = Vidx nd

- AG 9 &

The coefficients for SimpsonỖs rule appearing in 2A

colum (6) were taken from the expression for

odd n therefore the elastic strain energy of shear for

At tne entire beam is given by L2H (tf + 4f, + 2, +4f,+ Vaax 5 us Ặ -2 - ee ee eee ee (19) ~ -+; 4 n-2 3t nì +22) 41 2AG , Coeff Be Coeff Ấ0887 0667 Therefore the strain energy was a u = LR GJR x ề994

STRAIN ENERGY OF SHEAR

A rectangular slement "dx" by "dy" of thickness "t" into the page under the action of uniform shear stress t (psi) is shown in Fig A7.6a

From elementary strength of materials the angle of

shear strainỖ ts pro-

portional to the shear stress tT as

Ts = where G is the material elastic modulus

in shear For the dis- Placements as shown in

the sketch only the down

load on the left hand side dees any work (In general all four sides move, out if the motion

is referred to axes lying along two adjacent

sides of the element, as was done here, the de-

rivation is simplified) This load is equal to

wx dy x t and moves an amount3x dx Then au = Ọ xẾb dx dy

Fig AT 6a

Eq (9) may be used to compute the shear

Example Problem 6

Determine the strain energy of shear in the beam of Example Problem 4, Solution: The shear load diagram was drawn first ` L + P b= Fae Po Po Et ỘSAG ề CONSTANT v (9 COC renee Then, Ly 2 = + pa U= BAG x2x Py* dx 9 - Pe?L 2AG

Eq (9) may be used to commite the shear strain energy in a thin sheet The element ax x dy is visualized as but one of many in the

sheet and the total energy 1s obtained by

Summing Thus

=Ư[[1t4X3 2 lll lll )

uss 1J S - - (11)

Here a double integration is required, the Summation being carried out in both directions over the sheet In dealing with thin sheet the

use of the shear flow q = tt is often convenient

so that eq (11) rewrites

1 24x dy

1= = - (11a)

A very important special case occurs when a

homogenous sheet of constamt thickness is analyzed assuming q is constant éverywhere In this case one has

where S = [{ dxdy 1s the surface area of the

sheet

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Example Prod

ĩ ar strain energy stored in @ cantilever Dox beam of uniform rectangular cross section under the action of torque T (Fig A7.8) 4 L ẹ + aa Tp Ở4 ⁄ Flg AT.8 Solution:

The shear flow was assumed given by

BredtỖs formula (Ref Chapter A-6) et Tồ oa * Boe , congtant around the perimeter of any section Then c5 2T U=Ọ Bat * 8Bb o7 aL(b+c ) _ TAL (b+c) 4Gp 707 THE TOTAL ELASTIC STRAIN ENERGY OF A STRUCTURE

The strain energy by its definition is always a positive quantity It also isa scalar quantity (one having magnitude but not

direction) and uence the total energy of a

composite structure, having a variety of ele-

ments under various loadings, is readily found as 4 simple sun (Stax, (Medex ,Ấ ( Tax sop = | RE 2E1 28 +) Ee If ae

The integral symbols and common use of "x"

an an index of integration should not be taken

too literally It is probably best to read

these terms as "the sum of so and so over the

structureỢ rasher than "the integral of", for

quite often the terms are formed as simple sums without resort to the calculus The calculus is only used as an aid in some applications

It is seldom that ali the terms of eq

(13) need de employed in a calculation Many

of the loadings, if actually present, may be of a localized or of a secondary nature and their

energy contribution may be neglected

AT.5 AT.4 The Theorems of Virtual Worx and Minimum

Potential Energy

An important relationship between load and deformation stems directly from the definitions of work and strain energy Consider Fig 47.9(a) 4U=PA6 Limit A6 + O, dU = Pdõ ôm Fig AT.9a Thus au ep - lee ee eee eee eee ee 4 để 7P (14) In words,

"The rate of change of strain energy with respect t to deflection is equal to the associated

ToadỢ Bq (14) and the above quotation are stata-

ments of the Theorem of Virtual Nork The

reader may find this theorem stated quite dif- ferentiy in the literature on rigid body me-

chanics but should be able to satisfy himself

that the expressions are nevertheless compatible

A useful restatement of the above theorem

is obtained by rewriting eq (14) as

du - Pdd = 0

It is next argued that if the change in dis-

placement d6 is sufficiently small the load P

remains sensibly constant anc hence

au - d(P6) =

a(U-P6) =O + -7- (15)

The quantity U - P6 is called the total potential or the system and eq (18), resembling

as it does the mathematical condition for the minimum value of a functicn, is said to be 2

statement of t Theorem inimum Potential lear that the is a restatement of

in Structural analysis the most important these theorems are made in problems con-

buckling tnstability and other non- tes No applications will be made at

this point

A?.5 The Theorem of Complementary Energy and Castigliano's Theorem

Again in the case of an elastic body, ex- amination of the area above the load-deformation curve snows that increments in this area (called

Trang 24

AT.8 DEFLECTIONS 0 the complementary energy,U*), are related to the load and deformation by (see Fig A7.S(b) + AP au* 4a gp 7Oe - elle (16}

This is the Theorem of Complementary Energy

Now for the Tinéarly élastic body a very important theorem follows since (Fig A7.9c) aU = dU" so that P ặ 4 o lá au 2g A 6 Fig A7.9ằ q ap7O -+ - ee lll (17) In words, "The rate of chan respect to load i deflectionỢ ~

ặq (17) and the above Quotation are state-

ments of CastiglianoỖs Theorem

For a body under the simultaneous action of Several loads the theorem is written so as to apply individually to each load and its associ-

ated deflection, thus*

ge of strain energy with

equal to the associated

aw oP, = 6,

The partial derivative sign in eq indicates that the increment in strain e

due to 4 small change in the Particular

all other loads held constant (17a) nergy is load Py Note that by "load" and "deflection" may be meant: Associated LOBd~-Ở-ỞỞỞỞỞesction ỞỞỞỞ_Ở _ B_

ỘThe proof of the theorem for

generally formulated more rigorously, appeal to a simple diagram such as Fig A?.9c being less effective See, for example, ỘTheory of Elasticity" by S Timoshenko,

the case of multiple loads is F STRUCTURES Translation (inches) Rotation (radians } Rotation (radians) Volume (in*) area (in?) Force (ibs )} Moment (in lbs.) Torque (in lbs.) Pressure (lbs/in?) Shear Flow (lbds/in)

Any generalizations of the meanings of "force" and "deflection" are possible only so

long as the units are such that their product yields the units of strain energy (in ibs)

Once again for emphasis it is repeated that while the complementary nature or eqs (14) and

(17) are clearly evident, the use of eq (17) (CastiglianoỖs Theorem) {s restricted to linear2 elastic structures, A orief example will serve for illustration of the Possible pitfalls

The strain energy stored in an initially straight uniform colum under an axial load P when deflected into a hal? sine wave is , y y ề P2ÊL? ~ EI 1x Y= Yo sin 55; M= PY Consider Flexural Energy Only | u = (Meas = P*YgL = Ì=+ ~ PĂL 3 x Flg, A7.10 WET

where 6 is the end shortenin, due to bowing, Because the deflections grow rapidly as P ap- proaches the critical (buckling) load the problem is non-linear, The details of the - calculation of U are given with Fig 47.10,

Now according to the Theorem of Virtual Work (eq 14) ap but qu _ PRL? d6 ~ nỎgI Therefore P12 mar *P or L (Zuler formula for uniform column) The correct result,

Application of CastigiianoỖs Theorem eq (17), leads to the erroneous result: qu a7 đU _ 2P6L^ _ ap rửET ỞỞỞỞỞỞỞ_Ở _

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a

Pe (incorrect }

Moral: Do not use CastiglianoỖs theorem for

fon-linear problems

Fortunately the above restriction upon che use of CastiglianoỖs theorem 1s not a very severe one, the majority of every day structural problems being linear CastiglianoỖs theorem

is quite useful in performing deflection calcu

lations and a variety of applications will be

made in the following sections,

AT.6 Calculations of Structural Deflections by Use of Castigliano's Theorem

As the examples of Art A7.3 have 111U5~ trated,the strain energy of a structure can be expressed as a function of the external loadings provided the internal load or stress distribution is calculable Having the strain energy so expressed the deflections at points of load application can be determined with the aid of eq

{17), CastiglianoỖs Theorem

In the examples to follow the deflections of a variety of statically determinate struc-

tures are computed Methods of handling redun-

dant structures are considered in subsequent

articles

Example Problem 8

Find the vertical deflection at the point of load application of the crane of Fig A7.11

Cross sectional areas are given on each

member The stranded cables have effective

modulii of 13.5 x 10ồ psi ặ = 29,000,000 for

other members

Fig A7.11

Solution:

The strain energy considered here was that due to axial lozding in each of the four mem-

bers The load distribution was obtained from statics and the energy calculation was made in ATT Then us 2U = 43.26 x 107 P* LB FT = 43.26 x 10ồ P ft,

Note that in this problem the only use made of the calculus was in the differentiation A

simple sum was used to form U, as per eq (2)

Example Problem 9

Derive BredtỖs formula for the rate of

twist of a hollow, closed, thin-walled tube: =~ { # Ộqnng} + - Ref chap A6 9 The strain energy 1s stored in sheat according to us 1 iJ q?dxdy 2 ỘGE +

This is the only energy stored, secondary ef- fects neglected Over a given cross section q 18 a constant and is given by BredtỖs equation 4= x Ừ Where A is the enclosed area of the tube

(Ref Chap A.5)

Solution:

Since the twist per unit length was desired the strain energy per unit length only was

written Thus, assuming y in the axtal direction, no integration was made with respect to y The

integration in the remaining direction was to be carried out around the perimeter of the tube and so the index was changed from "x" to the more ap=

propriate "s" Hence (compare with Example Prob 7)

oe Ư*Ẽề T5 Ật# ồ t 8A0 t

The symbol ậ means the integration is carried out

around the tube perimeter Therefore

_dữ_ 7 {4s

o=Readts -

USE OF FICTITIOUS LOADS

In the following example the desired deặlec~

tion is at the free end of the bar where no load is applied A fictitious load will be added for

The rate of change of strain energy with respect to this fictitious

found after which the load will be This technique gives the de~

in as much as the deflection is

rate of change of strain energy to the load and such a rate exists

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