Aircraft Structures 3E Episode 13 pps

In the presence of axial constraint, de/dz is no longer constant so that the longitudi- nal strain a w / a z is not zero and direct also shear stresses are induced.. At any section of th

Fig 11.23 (a) Torsion of I-section beam; (b) plan view of beam showing undistorted shape of flanges

Fig 11.24 Obviously the beam still twists along its length but the rate of twist is no longer constant and the resistance to torsion is provided by the St Venant shear stres- ses (unrestrained warping) plus the resistance of the flanges to bending The total

torque may therefore be written T = TJ + Tr, where TJ = GJ d8/dz from the uncon-

strained torsion of open sections but in which d8/dz is not constant, and Tr is obtained from a consideration of the bending of the flanges It will be instructive

to derive an expression for Tr for the I-section beam of Fig 11.25 before we turn

our attention to the case of a beam of arbitrary section

Suppose that at any section z the angle of twist of the I-beam is 8 Then the lateral displacement u of the lower flange is

11.5 Constraint of open section beams 467

tY

z

Fig 11 2 5 Torsion of I-section beam fully built-in at one end

and the bending moment MF in the plane of the flange is given by

d2u

MF = -EIF -

dz2 (see Section 9.1 for sign convention) where I, is the second moment of area of theflange cross-section about they axis It is

assumed here that displacements produced by shear are negligible so that the lateral

deflection of the flange is completely due to the self-equilibrating direct stress system

c7r set up by the bending of the flange We shall not, however, assume that the shear

stresses in the flange are negligible The shear S , in the flange is then

Similarly, there is a shear force in the top flange of the same magnitude but opposite in

direction Together they form a couple which represents the second part Tr of the

total torque, thus

and the expression for the total torque may be written

de h2 d38

T = G J - - EIF - 7

The insight into the physical aspects of the problem gained in the above will be found

helpful in the development of the general theory for the arbitrary section beam shown

in Fig 11.26

t’

Fig 11.26 Torsion of an open section beam fully built-in at one end

The theory, originally developed by Wagner and Kappus, is most generally known

as the Wagner torsion bending theory It assumes that the beam is long compared with its cross-sectional dimensions, that the cross-section remains undistorted by the loading and that the shear strain T~~ of the middle plane of the beam is negligible although the stresses producing the shear strain are not From similar assumptions is derived, in Section 9.6, an expression for the primary warping w of the beam, viz

In the presence of axial constraint, de/dz is no longer constant so that the longitudi-

nal strain a w / a z is not zero and direct (also shear) stresses are induced Thus

(11.54)

The or stress system must be self-equilibrating since the applied load is a pure torque Therefore, at any section the resultant end load is zero and

IC or,tds = 0 ( IC denotes integration around the beam section 1

or, from Eq (11.54) and observing that d29/d? is a function of z only

The limits of integration of Eq (1 1.55) present some difficulty in that AR is zero when

w is zero at an unknown value of s Let

2 4 ~ = 2 A ~ , o - 2Ak where AR.0 is the area swept out from s = 0 and A; is the value of AR!O at w = 0 (see

Fig 11.27) Then in Eq (1 1.55)

11.5 Constraint of open section beams 469

Fig 11.27 Computation of swept area AR

and

giving

( 1 1.56)

The axial constraint shear flow system, qr, is in equilibrium with the self-

equilibrating direct stress system Thus, from Eq (9.22)

The integral in this equation is evaluated by substituting p~ = (d/ds)(2A~) and integrating by parts Thus

At each open edge of the beam qr, and therefore $ 2ARtds, is zero so that the integral reduces to - Jc 4Ait ds, giving

(1 1.58) where r R = sc 4Ait ds, the torsion-bending constant, and is purely a function of the geometry of the cross-section The total torque T, which is the sum of the St Venant torque and the Wagner torsion bending torque, is then written

(Note: Compare Eq (1 1.59) with the expression derived for the I-section beam.)

tD of the beam wall so that r R , for a beam with n booms, may be generally written

In the expression for r R the thickness t is actually the direct stress carrying thickness

n

where B, is the cross-sectional area of the rth boom The calculation of r R enables the second order differential equation in dO/dz (Eq (1 1.59)) to be solved The constraint shear flows, qr, follow from Eqs (1 1.57) and (1 1.56) and the longitudinal constraint stresses from Eq (1 1.54) However, before illustrating the complete method of solution with examples we shall examine the calculation of r R

So far we have referred the swept area AR, and hence r R , to the centre of twist of the beam without locating its position This may be accomplished as follows At any section of the beam the resultant of the qr shear flows is a pure torque (as is the

resultant of the St Venant shear stresses) so that in Fig 11.28

IC qr sin $ds = Sy = 0

Fig 11.28 Determination of the position of the centre of twist

11.5 Constraint of open section beams 471 Therefore, from Eq (1 1.57)

Now

dY d sin$= , - ( U R ) = P R

ds ds and the above expression may be integrated by parts, thus

The first term on the right-hand side vanishes as si 2ARt ds is zero at each open edge

of the beam, leaving

y 2 A ~ t ds = 0

Again integrating by parts

The integral in the first term on the right-hand side of the above equation may be

recognized, from Chapter 9, as being directly proportional to the shear flow produced

in a singly symmetrical open section beam supporting a shear load Sy Its value is

therefore zero at each open edge of the beam Hence

Similarly, for the horizontal component S, to be zero

(11.60)

(1 1.61)

Equations (11.60) and (11.61) hold if the centre of twist coincides with the shear

centre of the cross-section To summarize, the centre of twist of a section of an

open section beam carrying a pure torque is the shear centre of the section

We are now in a position to calculate rR This may be done by evaluating sc 4Ait ds

in which 2AR is given by Eq (1 1 S6) In general, the calculation may be lengthy unless

the section has flat sides in which case a convenient analogy shortens the work

considerably For the flat-sided section in Fig 11.29(a) we first plot the area 2AR:o

swept out from the point 1 where we choose s = 0 (Fig 11.29(b)) The swept area

AR,O increases linearly from zero at 1 to (1/2)pI2dl2 at 2 and so on Note that move-

ment along side 23 produces no increment of 2kfR,o as p23 = 0 Further, we adopt a

sign convention for p such that p is positive if movement in the positive s direction

of the foot of p along the tangent causes anticlockwise rotation about R The

increment of 2AR.0 from side 34 is therefore negative

In the derivation of Eq (1 1.56) we showed that

Suppose now that the line 1’2’3‘ .6‘ is a wire of varying density such that the weight

of each element 6s’ is tSs Thus the weight of length 1‘2’ is tdI2 etc They coordinate of the centre of gravity of the ‘wire’ is then

Comparing this expression with the previous one for 2AL, y and J are clearly analo- gous to 2AR,o and 2Ak respectively Further

Expanding and substituting

2Ak IC t ds for jc 2AR,ot ds

gives

r R = ( 2 A ~ , o ) ~ t d s - (2AL)’ tds (1 1.62) Thus, in Eq (1 1.62), r R is analogous to the moment of inertia of the ‘wire’ about an axis through its centre of gravity parallel to the s axis

An open section beam of length L has the section shown in Fig 11.30 The beam is

M ybuilt-in at one end and carries a pure torque T Derive expressions for the direct stress and shear flow distributions produced by the axial constraint (the or and qr systems) and the rate of twist of the beam

11.5 Constraint of open section beams 473

2

Fig 11.30 Section of axially constrained open section beam under torsion

The beam is loaded by a pure torque so that the axis of twist passes through the

shear centre S(R) of each section We shall take the origin for s at the point 1 and

initially plot 2AR-o against s to determine r R (see Fig 11.31) The position of the

centre of gravity, (2A’,), of the wire 1’2’3‘4’ is found by taking moments about the

s axis Thus

t(2d+h)2A’, = td (3 - + th (7) - + rd (hqd) -

from which

hd(h + d ) 2(h + 2 d ) 3A‘, =

r R follows from the moment of inertia of the ‘wire’ about an axis through its centre of gravity Hence

which simplifies to

td3h2 2 h + d

r R =- 12 (-) h + 2 d Equation (1 1.59), that is

de d3 8

T = G J - - E E ~ R T may now be solved for dO/dz Rearranging and writing p2 = G J / E r R we have

dz3 z = - p

The solution of Eq (iii) is of standard form, i.e

(ii)

(iii)

de T

dz GJ - - + Acoshpz + Bsinhpz The constants A and B are found from the boundary conditions

dO/dz = 0 at the built-in end

load Therefore, from Eq (1 1.54), d’O/d2 = 0 at the free end

(1) At the built-in end the warping w = 0 and since w = -2ARd8/dz then (2) At the free end gr = 0, as there is no constraint and no externally applied direct

twist 8, the appropriate boundary condition being 8 = 0 at the built-in end Thus

(4

1

sinh p ( L - z) - sinh p L

GJ p cosh p L p cosh pL

11.5 Constraint of open section beams 475

z z = L

Fig 11.32 Stiffening effect of axial constraint

and the angle of twist, at the free end of the beam is

e i = : E = = ( 1 - 7 ) TL tanh pL

Plotting 6 against z (Fig 1 1.32) illustrates the stiffening effect of axial constraint on

the beam

The decrease in the effect of axial constraint towards the free end of the beam is

shown by an examination of the variation of the St Venant ( T J ) and Wagner ( T r ) torques along the beam From Eq (iv)

and

d38 coshp(L - z) dz3 cosh p L

Tr = - E r R - = T

(vii)

(viii)

T j and Tr are now plotted against z as fractions of the total torque T (Fig 11.33) At

the built-in end the entire torque is carried by the Wagner stresses, but although the

constraint effect diminishes towards the free end it does not disappear entirely This is

due to the fact that the axial constraint shear flow, qr, does not vanish at z = L, for at

this section (and all other sections) d38/dz3 is not zero

Fig 11.33 Distribution of St Venant and torsion-bending torques along the length of the open section beam

shown in Fig 11.30

( a ) Fig 11.34 Distribution of axial constraint direct stress around the section

Equations (iii) to (viii) are, of course, valid for open section beams of any cross- section Their application in a particular case is governed by the value of the torsion bending constant r R and the St Venant torsion constant J [= (h + 2 d ) t 3 / 3 for this example] With this in mind we can proceed, as required by the example, to derive the direct stress and shear flow distributions The former is obtained from Eqs

(11.54) and (iv) Thus

In Eq (ix) E, G , J and r R are constants for a particular beam, T is the applied torque,

AR is a function of s and the hyperbolic term is a function of z It follows that at a

given section of the beam the direct stress is proportional to -2AR, and for the

beam of this example the direct stress distribution has, from Fig 11.31, the form

shown in Figs 11.34(a) and (b) In addition, the value of cr at a particular value of

s varies along the beam in the manner shown in Fig 11.35

I-

d &

Fig 11.35 Spanwise distribution of axial constraint direct stress

11.5 Constraint of open section beams 477

Fig 11.36 Calculation of axial constraint shear flows

Finally, the axial constraint shear flow, qr, is obtained from Eq (11.57), namely

At any section z, qr is proportional to

ring to Fig 11.36, 2AR = 2 4 0 - 2Afi so that in flange 12

2&t ds and is computed as follows Refer-

Note that in the above d38/& is negative (Eq (viii)) Also at the mid-point of

the web where s2 = h / 2 , qr = 0 The distribution on the lower flange follows from

antisymmetry and the distribution of qr around the section is of the form shown in

Fig 11.37 Distribution of axial constraint shear flows

giving

qr = “r 2ARtds from Eq (11.57)

rR 0

Hence for a given value of s, ( $ 2 A ~ t d s ) , qr is proportional to Tr (see Fig 11.33)

11.5.1 Distributed torque loading

We now consider the more general case of a beam carrying a distributed torque load- ing In Fig 11.38 an element of a beam is subjected to a distributed torque of intensity

T,(z), i.e a torque per unit length At the section z the torque comprises the St Venant torque T j plus the torque due to axial constraint Tr At the section z + Sz the torque increases to T + ST(= Tj + STj + Tr + STr) so that for equilibrium of the beam element

11.5 Constraint of open section beams 479

( 1 1.63)

(11.64)

The solution of Eq (11.64) is again of standard form in which the constants of

integration are found from the boundary conditions of the particular beam under

So far we have been concerned with open section beams subjected to torsion in which,

due to constraint effects, axial stresses are induced Since pure torsion can generate

axial stresses it is logical to suppose that certain distributions of axial stress applied

as external loads will cause twisting The problem is to determine that component

of an applied direct stress system which causes twisting

Figure 11.39 shows the profile of a thin-walled open section beam subjected to a

general system of loads which produce longitudinal, transverse and -rotational

A

SY

Fig 11.39 Cross-section of an open section beam subjected to a general system of loads

displacements of its cross-section In the analysis we assume that the cross-section of the beam is undistorted by the loading and that displacements corresponding to the shear strains are negligible In Fig 11.39 the tangential displacement ut is given by

Eq (9.27), i.e

ut = pR8 + ucos$ + usin$ (1 1.65)

Also, since shear strains are assumed to be negligible, Eq (9.26) becomes

(1 1.66) Substituting for vt in Eq (1 1.66) from Eq (1 1.65) and integrating from the origin for s

to any point s around the cross-section, we have

W, - W O = 2AR,o - -(x - X O ) - - ( y -yo) (11.67) where 2 4 0 = p~ ds The direct stress at any point in the wall of the beam is given

by

Thus, from Eq (1 1.67)

Now AR,O = A k + A R (Fig 11.39) so that Eq (1 1.68) may be rewritten

uZ = fi (z) - E - ~ A R - E - x - E - (1 1.69)

in which

The axial load P on the section is given by

where Jc denotes integration taken completely around the section From Eq (11.55)

we see that sc 2 A ~ t dr = 0 Also, if the origin of axes coincides with the centroid of the section Jc txds = Jc tyds = 0 and J tyds = 0 so that

in which A is the cross-sectional area of the material in the wall of the beam

The component of bending moment, M,, about the x axis is given by

11.5 Constraint of open section beams 481

Substituting for CT? from Eq (1 1.69) we have

M x = f i ( z ) IC tyds - 2ARtyds - E G I C txydS - .-Ic d? ty’ds

We have seen in the derivation of Eqs (11.60) and (11.61) that Jc2ARtyds = 0 Also

Equations (1 1.71) and (11.72) are identical to Eqs (9.19) so that from Eqs (9.17)

E - d2u = MxIxs - M y2 I x x , E - d2v = -My Iyy + My IxJ (1 1.73) dz2 Ixx Iyy - I,, dz2 Ixx Iyy - I$

The first differential, d2$/dz2, of the rate of twist in Eq (1 1.69) may be isolated by multiplying throughout by 2ARt and integrating around the section Thus

The second two terms on the right-hand side of Eq (1 1.75) give the direct stress due to bending as predicted by elementary beam theory (see Eq (9.6)); note that the above approach provides an alternative method of derivation of Eq (9.6)

Comparing the last term on the right-hand side of Eq (1 1.75) with Eq (1 1.54), we see that

Thus, if sc az2ARt ds is interpreted in terms of the applied loads at a particular section then a boundary condition exists (for d26/d2) which determines one of the constants

in the solution of either Eq (1 1.59) or Eq (1 1.64)

The units of sc a,2ARtds are force x (distance)2 or moment x distance A simple physical representation of this expression would thus consist of two equal and opposite moments applied in parallel planes some distance apart This combination has been termed a moment couple’ or a bimoment3 and is given the symbol Mr or

B, Equation (11.75) is then written

(1 1.76)

As a simple example of the determination of Mr consider the open section beam

shown in Fig 11.40 which is subjected to a series of concentrated loads P I ,

P 2 , , Pk, , Pn parallel to its longitudinal axis The term azt ds in Jc a,2ARt ds

may be regarded as a concentrated load acting at a point in the wall of the beam Thus, Sc az2ARt ds becomes E;= Pk2Aw, and hence

11.5 Constraint of open section beams 483

The column shown in Fig 11.41(a) carries a vertical load of 100kN Calculate the

angle of twist at the top of the column and the distribution of direct stress at its

base E = 200 000 N/mm2 and G I E = 0.36

The centre of twist R of the column cross-section coincides with its shear centre at

the mid-point of the web 23 The distribution of 2AR is obtained by the method

detailed in Example 11.2 and is shown in Fig 11.42 The torsion bending constant

r R is given by Eq (ii) of Example 11.2 and has the value 2.08 x 10'omm6 The

St Venant torsion constant J = Cst3/3 = 0.17 x 105mm4 so that d m(= ,LL

in Eq (iii) of Example 11.2) = 0.54 x Since no torque is applied to the

Fig 11.42 Distribution of area 2AR in the column of Example 11.3

column the solution of Eq (iii) in Example 11.2 is

(i)

d8

- = Ccoshpz + D s i n h p

dz

At the base of the column warping of the cross-section is suppressed so that, from

Eq (9.65), dO/dz = 0 when z = 0 Substituting in Eq (i) gives C = 0 The moment

couple at the top of the column is obtained from Eq (1 1.77) and is

Mr = P ~ A R = -100 x 2.5 x IO3 = -25 x lo5 kNmm2

Therefore, from Eq (11.74) and noting that Jc uz2ARtds = M r , we have

- _ d28 - 2'5 lo5 lo3 - 0.06 1 0 - 6 / ~ ~ 2 dz2 200000 x 2.08 x 1O'O -

at z = 3000mm Substitution in the differential of Eq (i) gives D = 0.04 x

that Eq (i) becomes

At the top of the column (z = 3000mm) the angle of twist is then

8(top) = 0.08cosh0.54 x x 3000 = 0.21 rad(l2.01")

(ii)

(iii)

The axial load is applied through the centroid of the cross-section so that no bending

occurs and Eq (1 1.76) reduces to

At the base of the column

(see Eq (1 1.74)) Therefore, from Eq (ii)

The self-equilibrating shear flow distribution, qr, produced by axial constraint is

1 Argyris, J H and Dunne, P C , The general theory of cylindrical and conical tubes under

torsion and bending loads, J Roy Aero SOC., Parts I-IV, Feb 1947, Part V, Sept and Nov

1947, Part VI, May and June 1949

2 Megson, T H G., Extension of the Wagner torsion bending theory to allow for general

systems of loading, The Aeronautical Quarterly, Vol XXVI, Aug 1975

3 Vlasov, V Z., Thin-walled elastic beams, Israel Program for Scientific Translations,

Jerusalem, 196 1