Bruhn - Analysis And Design Of Flight Vehicles Structures Episode 2 Part 3 pptx

Al4.17, For bending about the centroidal 2 axis without twist the resultant of the internal shear flow system would obviously, due to sym- metry of section about X centroidal axis, lie

Al4.8

0.375 inches to left of point b as shown in Fig

Al4.17, For bending about the centroidal 2 axis

without twist the resultant of the internal

shear flow system would obviously, due to sym-

metry of section about X centroidal axis, lie on

the X axis, hence shear center for the given

channel section is at point O' in Fig Al4.17

The external load of 100 1b would have to be

located 0.375 inches to the left of the center-

line of the channel web if bending of the

channel without twist is to occur

A14.6 Shear Stresses for Unsymmetrical Beam Sections

In chapter Al3, which dealt with bending stresses in beams, three methods were presented

for determining the bending stresses in beams

with unsymmetrical beam sections The bending

stress equations for these three methods will be

about centroidal Z and X axes For brevity this method will be called the k method

O =~ (ky My = Ki My) > (Ka My = K, MQ) 2 (11)

BENDING SHEAR STRESSES SOUND AND OPEN SECTIONS SHEAR CENTER

Method 3 The k Method

dy = - (Ks Vy - Ki Vg) BX A ~ (Ke Vz - Ki Vy)

Zaa EXAMPLE PROBLEM USING THE THREE DIFFERENT METHODS

Fig Al4.26 shows a Zee Section subjected

to a 10,000 lb shear load acting through the shear center of the section and in the direction

as shown The problem will be to calculate the shear flow qy at two points on the beam section, namely points b and c as indicated on the figure The shear flow at these two places will be cal~ culated by all 3 methods

Since all 3 methods require the use of beam section properties and since the direction

of either the principal axes or the neutral

solution regardless of at ¬

which method is used

1s to calculate the a

section properties about centroidal X and

gives the calculations

The section has been divided into 4 portions labeled 1, 2, 3 and 4

Areal Arm | Arm ; Ị

1 1 |0.10| 1,$ö| -0 45] -.06525|.,31025|.02025 |.00838 | 000017 Kis XZ = Ix 210,34) 0.701 0 Ö | 068601 6 |.000117.02287

“Ty tf, - ifz’ ° “ly lz ~ lậu a; 0.14] -0.70| 0 0T 06880] 0 |.000117.0Z281 ¡

4 [0.10] -1.45| 0.43| -,06525| 210251 02025].00638 [000017 |

% SE ene Ty =2ZA27+2 1, = 6035 int, "

In referring back to the derivations of Iz =2ZAx*+ 2 i, = 0574

equations (5), (6) and (7) the above equations lop DA XZ = = 1205

(9), (10) and (11) can be written in terms of *z

beam external shears instead of external bending (Note: in Table 414.3 ty and tg are the mom-

its own centroidal axis)

SOLUTICN BY PRINCIPAL AXES METHOD (Method 1) Let $ be angle between Principal axes and the X and Z axes From chapter A13,

tan 2 $ =f be

Ig - ty

2 (- 0.1305) = 0.4778

* 0.0574 ~ 0.6035

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES

hence 2 $ Z 259 - 22.21 on $ = 12? - 46.1' Now substituting in equation (15)

sin $ = 0.2210 and cos 9 = 0.97527 7348

d› “ - =—=z=rz (1 X 0.1 x 1.5156) The moments of inertia about the principal axes b 0.65316 7

can now be calculated

tx =I, cos* $+ I, sin® $ ~ 2 Ix, sin } cos > - Sa (1 x 0.1} (+ 0.1184) = - 1756 + 2890

In Fig Al4.26 the external shear load is

10000 lbs acting in a direction as shown Re-

solving this shear load into Z and x components,

we obtain,

Vy = 10000 x cos 30° = 6667 1b

Vy = 10000 x sin 30° = 5000 ib

Resolving these 2 and x components further

into components along the principal axes we

Fig Al4.27 shows the position of the

principal axes as calculated The shear flow

at the free edge of the upper portion (1) is

zero For the shear flow at point (b), the

area to be used in the summations 2 zpA and

2 xXpA is the area of element (1) The arms Zp

and Xp can be calculated by simple trigonometry

Fig Al4.27 shows the value of these distances,

Calculation of shear flow at point (c)

For portion (2) area A =1.4x 0.1 = 0.14

Zp = 70 x 97527 = 0.6825 in

Xp = +70 X 2210 2 0.1547 The shear flow at point (c) equals the shear flow at point (b) plus the effect of the portion (2) between points (b) and (c)., nence

ing In Fig Al4.28,

the angle 9 is the angle between the plane of loading and the Zn prin~

cipal axis, and this

sin a = 0.9989 , cos a = 0.04742

sin* a, substituting,

In = ltp cos* a + len

Al4.10 BENDING SHEAR STRESSES,

Ty 2 0.63316 x 04742 + 02782 x S585 = 02919

The component of the given external shear load

normal to the neutral axis n-n equals

Vy, = 10000 x sin 45° ~ 29" = 7130 1b

From equation (13)

Qe ~ Fem A

Shear flow at point (b}: ~

The distance from the neutral axis to the centroid of portion (1) equal Zn "~ 0.0466 in

SOLUTION BY METHOD 3 - (The k Method)

In this method, only the section properties about the centroidal X and Z axes are needed

These properties as previously calculated in

fable Al4.3 are,

Shear flow at point (b}

For portion (1), x = ~ 0.45 in., 251.45 in

Al4.9 Beams with Constant Shear Flow Webs

Fig 414.29 shows a z

beam composed of heavy R

flange members and a curved thin web For bending about the X-X axis, the

web on the compressive,

Side of the beam absorbs very little compressive

stress, since buckling of ———

the web will take place “Flange

under low stresses, par- Zz Pig A14.29 ticularly when the curvature of the web is

small On the tension side, the wed will be

more effective, but if the flange areas are relatively large, the proportion of the total bending tensile stress carried by the web is small as compared to that carried by the tension flange Thus for beams composed of individual flange members connected by thin webs it is often assumed that the flanges develop the en~

therefore means that the shear flow 1s constant

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES over a particular web In other words in the

shear flow equation q = es zaaA, if the area

x

ot the web is neglected then q is constant be~

tween flange members

RESULTANT OF CONSTANT SHEAR FLOW FORCE SYSTEMS

Fig Al4.29 shows a beam assumed to be

carrying a downward shear load (not shown) and

to cause, bending about axis x-x without twist

Assuming the two flanges develop the entire

bending resistance, the shear flow q {s con-

stant on the wed and acts upward along the web

to balance the assumed external downward load

The resultant of this resisting shear flow

force system will give the lateral position of

the shear center for this beam section The

problem then is to find the resultant of the

shear flow system

Let q = load per inch along web (constant)

Let R = resultant of the q force system

From elementary mechancis,

R=vVðqg+ 3 q2 , where ay and qy are the x

and y components of the q forces along the web

Since q is constant, 2 qy is zero, hence,

Equation (17) states that the magnitude of

the resultant of a constant flow force system

is equal to the shear flow q times the straight

line distance between the two ends of the shear

flow system

Since & qy is zero, the direction of the

resultant 1s parallel to the straight line

joining the ends of the web

The location of the resultant force is

found by using the principle of moments, namely,

that the moment of the resultant about any

point must equal the moment of the original

force system about the same point In Fig

414.29 assume point (0) as a moment center

Then Re =qLr

but R=gqgh

In equation (18) the term Lr is equal to

twice the area (A), where area (A) is the en-

closed area formed by drawing straight Lines

from moment center (0) to the ends of the shear

flow force system Thus

The shear center thus lies at a distance e

to the left of point (0), and the external shear

load would have to act through this point if

Al4 11 EXAMPLE PROBLEM - RESULTANT OF A CONSTANT PLOW

Draw closing line between the beginning

and end points of force system (line AE) The

length h of this closing line is 20 inches

Prom eq (17) R = qh = 10 x 20 * 200 1b

The direction of the resultant is parallel

to line AE or horizontal in this problem To find the location of R take moments about any point such as (0) Draw lines from point (0) to

points A and E The enclosed area (A) equals Sx10+5x10+ 5nx5*+10x 50 = 189.3 sq in

From eq (19)

2A „2x 189.5 „

Pig Al4.30 shows the resultant of 200 lb

acting at a distance e from (0) and parallel to lime AE

Al4,10 Example Problems for Beams with Constant Shear Flows Between Flange Members

EXAMPLE PROBLEM 1 Beam Section Symmetrical

About One Axis

Fig Al4.31 shows an open beam section com— posed of 8 flange members cornected by thin

# aan ro —"

ue 40" 10t 110" 20"

_ b ws -/-| -+- NV ván Š

Fig À14.31

Al4 12

Sheet to form the weds and walls The flange

members are numbered a to h and the areas of

each are given on the figure It will be as-

sumed that the webs and walls develop no bending

resistance and thus the shear flow between ad-

jacent flange members will be constant The

problem is to determine the shear center for the

beam section

SOLUTION: -

Since the beam section is symmetrical about the X axis, the centroidal X and Z axes

are also principal axes, since the product of

inertia Ix, 1s zero

The vertical position of the beam section centroid due to symmetry is midway between the

upper and lower flanges

To find the horizontal position of the centroid, take moments of the flange areas

about the left end or line be:

0.4x158+0.2x10+0.2x5

= 5.625 tn

The moments of inertia for the section about

the centroidal x and y axes are: ~

Ty = 2A 2* = (0.8 x 5*) 2 = 40 Int

2

I, = 0.8 x 5.625 +0.2%x 0.625" +0.2Xx

2 4.375 + 0.4 x 9.375® = 64.4 int, HORIZONTAL POSITION OF SHEAR CENTER: -

The horizontal position of the shear center will coincide with the centroid of the shear

flow system due to bending about axis xx with-

out twist For simplicity, to eliminate large

decimal values for shear flow values an ex-

ternal shear load Vz = 100 1b will be assumed

and the internal resisting shear flow system

will be calculated for this external loading

From equation (8) Vv;

ay * 1 3 2A, substituting values of V„ and

edges and thus dy 1s zero In this solution,

we will start at the free edge at point (a} and

go counterclockwise around the beam section

The area of each flange member has been concen-

trated at a point coinciding with the centroid

of each flange area In solving for the q val-

ues the subscript y will be omitted, and sub-

scripts using the flange letters will be used

BENDING SHEAR STRESSES SOUND AND OPEN SECTIONS, SHEAR CENTER

in order to indicate at wnat point the shear flow { is being calculated

đạp 7 - 2.52, 2A 2 -2.5%5 xX 0.1 *

= 1.25 1b./in

The first letter of the subscript refers

to the flange member where the shear flow q is being calculated and the second ietter indicates

on which adjacent side of the particular flange member Hence q,, means the shear flow at flange (a) but on the side toward (b)

đạp = Na = - 1.25 (Since no additional flange

area is added, and thus

shear flow ts constant on sheet ab

Abe = Wa ~ 2-5 Ty ZA

=- 1.25 -2.5x%5x 0.4 = - 6.25 lb./in

đẹp * + 6-25 Ved = Ich - 2-5 UG ZA

- 6.25 - 2.5 x (- 5 x 0.4) = - 1.25

Gag = Ugg — 25 3g ZA = - 1,25 - 2.5 (- 5 x 0.1)

=0

ded = dag = 0 dep =O - 2.52, 2A50-2.5 (-5 x 0.1) = 1.25

dre F der * 1.25

Qgg = 1.2 ~ 2.5 Bp 2A

(+5 x 0.2) = 3.75 Ger = Gpg = 3-75

or in the xz plane It is only necessary to determine this sense at the beginning point, that

is in sheet panel ab The surest way to deter- mine this sense is to draw a simple free body sketch of flange member (a) as illustrated in Pig Al4.31 The shear flow on the cut face is Qy(ab) = - 1.25 and

this value is shown

on the free body By + simple rule given at a the end of Art Al4.6,

the shear flow in the plane of the cross~

Fig Al4.31

ANALYSIS AND DESIGN OF section is also directed toward the common

boundary line and thus 4x(ap) has a sense as

shown in Fig Ald.31 The sense of the shear

flow on the cross-section will now continue in

this direction until the sign changes in the

origional calculation, which means therefore the

shear flow sense will reverse Fig Al4.22

shows a plot of the shear flow pattern with the

sense indicated by the arrow heads

static equilibrium exists relative to 2 Fy and

The shear flow force system in Fig Al4.32

causes the section to bend about axis xx with-

out twist The resultant of this system is 100

lb acting down in the 2 direction The posi-

tion of this resultant will thus locate the

lateral position of the shear center

Equating the moments of the shear flow

system about some point such as (c) to the mo~

ment of the resultant about the same point we

obtain:

100 a = 10 x 3.75 x 15 - 1.25 x0.5x2x5-

1.985 xO.5x 2x 10 + 1.25xX0.5X2x 15

hence e # 562.5/100 = 5.625 lnches

Thus the shear center lies on a vertical line

5.625 inches to right of line oc

CALCULATION OF VERTICAL POSITION OF SHEAR

CENTER

For convenience as before, we will assume

a shear load Vy, = 100 lb and compute the re-

sisting shear flow system to resist this load

in vending about axis zz without twist The

resultant of this shear flow system will give

the vertical location of the shear center The

shear flow equation is,

We will again start at the free edge ad- jacent to flange (a) where dy = 9,

Qde = 7.087 ~ 1.55 (- 0.625 x 0.1) = 7.184 đạa 2 7.184

dạ; = 7.184 ~ 1.55 x 4.375 x 0.1 = 6.504

Sraạ = 6.504 ote = 6.504 = 1.55 x 9.375 x 0.2 = 3,589

dge = 5.689

Gen * 3.589 - 1.55 x 9.375 x 0.2 = 0.674 Fig Al4.33 shows the plotted shear flow resuits The signs of the calculated shear flows are for shear flows in the y direction

Simple consideration of a free body of flange member (a) will give the sign or sense of the shear flow in the plane of the beam section

Thus in Fig Al4.34 qy must act as shown when

Al4.14 BENDING SHEAR STRESSES

Checking to see if 2 Fz =O and 2 Fy = 0: ~

toward the left To find the location of the

resultant take moments about a point 0.5 inch

location for the given beam section

EXAMPLE PROBLEM 2 Unsymmetrical Beam Section

Fig Al4.36 shows a four flange beam sec- tion The areas of each flange are shown ad-

jacent to flange The external shear load V

equals 141.14 lb and acts in a direction as

shown The problem is to find the line of

action of V so that section will bend without

To locate centroidal x and z axes: -

zg.Ă5AX _1xX12+0.5x8_ Bese RE SS in,

z= BAS = {0:5 + 0:5) 16 5 5.333 in

SOUND _AND OPEN SECTIONS, SHEAR CENTER

Calculation of Ix, I„ and lyz —

l„ “3 Á 27 = 5 (5.3334 + 2.6679) + 1 , (5.535° + 6.667*%) = 90,667

lạ =ZA X® = 1 (10.667%) + 2 (5.333%) = 170.667 Tyg = 2A x2 = 1 xX 6.667 (- 5,333) + 1 (- 5.333) (+ 5.333) + 0.5 x 10.667 x 2.667 + 0.5 x 10.667

ANALYSIS AND DESIGN OF

deg = 6-249 - 0.7487 x 0.5 x 10.667 - 1.276 x

0.5 (= 5.333) = 5.680 dao = Igg = 5-880

ga = 5-660 ~ 0.7457 x 5 x 10,667 - 1.278 x 0.5

X 2.667 = 5.680 - 3.977 + 1.704 =

0 (checks free edge at d where qy must be zero.)

Fig A14.37 shows the resulting shear flow re-

sisting pattern The sense of the shear flow in

Fig Al4.37

the plane of the cross-section is determined in

web at flange member (a) by the simple free

body diagram of stringer (a) in Fig Al4.37a

Check 3 Fy and 3 Fz to see if each equals

100

E Py = - 6.249 x 16 = - 99.99 (checks Vy = 100)

RF, =- 12x 4.544 ~ 8 x 5.68 = - 99.94 (checks

Vz = 100)

The resultant of the internal resisting

shear flow system equals V¥ 1002 + 1007 = 141.14

1b

To locate this resultant we use the prin-

ciple of moments Taking point (b) as a moment

center,

141.14 e = 8 x 5.68 x 16 - 4.544 x đề n

212 hence e “TT 12 = 1.50 inch

Therefore external load must act at a distance

@ = 1.50" from (pb) as shown in Fig Al4.37 The

load so located will pass thru shear center of

section To obtain the shear center location,

another loading on the beam can be assumed, and

where the line of action of the resultant of the

resisting shear flow system intersects the re-

sultant as found above would locate the shear

center as a Single point If the shear center

location is desired it is convenient to assume

a unit V, and Vy acting separately and find the

horizontal and vertical locations of the shear

center from the 2 separate shear flow force

systems

Al4.i11 Shear Center Location By Using Neutral Axis

Method

In a beam subjected to bending there is a definite neutral axis position for each differ- ent external plane of loading on the beam The shear flow equation with respect to the neutral axis is,

QW Fo In Š Znà TT xxx TT (20)

where, Vy = Shear resolved normal to neutrai axis

In = Moment of inertia about neutral axis

Zy = Distance to neutral axis

In finding the shear center location of an unsymmetrical section, it 1s convenient to as- sume that the Z and X axes are neutral axis and find the shear flow system for bending about each axis by equation (20) The resultant of each of these shear flow force systems will pass through the shear center, thus the intersection

of these two resultant forces will locate the shear center

Example Problem The same beam section as used in the previous article (see Fig Al4.36) wiil be used

to illustrate the neutral axis method

Fig Al4.38 shows the section with the centroidal axis drawn in The X axis will now

Fig Al4 38 Fig Al4.39

be assumed as the neutral axis for an external plane of loading as yet unknown We will further assume that when this unknown external loading

is resolved normal to the X neutral axis, that

it will give a value of 100 lb., or Vz = 100

From the previous article Iy = 90.667

Since the X axis has been assumed as the neutral axis, equation (20) can be written

=: 100 = đục “ — 7-55 = gocggy X (~6.3ã5)1 2 -1.47

= 100 < c

Gog = ~1:47 - ap gay (-5-33) 0,5 Z 1.47

Fig Al4.39 shows the resulting shear flow

values

100 R= 103 hence 9 = 13° - 16'

Let e = distance from 33.52

resultant R to point b

Equating moments of resultant about (b) to that of shear flow

must obtain another resultant force which

passes through the shear center before we can

definitely locate the shear center Therefore

we will now assume that the 2 centroidal axis

is a neutral axis and that a resolution of the

external load system gives a shear Vx 2 100 1b,

100 170.66

ded * 6.25 ~ (0.5) (10.667) = 3,13 Fig Al4.41 shows the-shear flow results

Fig Al4.40 shows the position of this re- sultant force Where it intersects the previous resultant force gives the shear center location Al4.12 Problems

(1) Pig Al4.42 fan artird

1000 1b load acting through the shear center

as shown, Find the shear stress

at sections a-a and b-b by three different methods

(3) Determine the shear flow diagram and the

shear center location for bending about hori- zontal centroidal axis for the beam sections

as given in Figs Al4.44 to Al4.46

.10n," 00! op" fi 10" | Fig A144

(4) Determine the shear center location for the beam sections in Figs Al4.47 and Al4.48 Ase

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES sume flange members develop entire bending

{5) Determine the shear center for the beam

section of Fig Al4.49 Assume only the 8

stringers as being effective in bending Area

of stringers (a) and (b) = 2 sq in seach

All other stringers 1 sq in each

Fig Al4 50

(6) Determine the shear center for the un-

symmetrical beam section of Fig Al4.50 As-

sume sheet connecting the four stringers as

ineffective Areas of stringers shown on Fig

À14.17

c d

Fao 50" Fig Al4 52

Fig, Al4, 51

(7) In Pig Al4.51, the shell structure {s sub-

jected to a torsional moment M = 50,000 in lb

The shell skin shown dashed is cut out, thus the torsional moment is resisted by the constant

shear flow on the two curved sheet elements ac,

and bd Determine the value of the shear flow

(8) Determine the moment of the constant flow force system in Fig, Al4,52 about point (0)

Also find the resultant of this force system

(10) For the wing cell beam section in Fig

Al4.54, determine the location of the shear

center Assume webs and walls ineffective in bending

Boeing Tôi: Jet jet Stratoliner

Fuselage being Assembled |

CHAPTER A 15

SHEAR FLOW IN CLOSED THIN - WALLED SECTIONS

A151 Introduction The wing, fuselage and em=

pennage structure of modern aircraft is essen-

tially a single or multiple cellular beam with

thin webs and walls The design of such

structures involves the consideration of the

distribution of the internal resisting shear

stresses This chapter introduces the student

to the general problems of shear flow distri-

bution Chapter Al4 should be covered before

taking up this chapter

Al5.2 Single Cell Beam Symmetrical About One Axts

All Material Effective in Resisting Bending

Stresses,

Fig Al5.1 shows a single cell rectangular

beam carrying the load of 100 1b, as shown

The problem is to find the internal resisting

shear flow pattern at section abcd

Due to symmetry of material the X cen-

troidal axis lies at the mid-height of the

beam The shear flow equation requires the

value of ly, the moment of inertia of the

section about the X axis

ly = aex.15x10°+ 2 [20 x.06 x 6 * ] = 62,5 in.t

From Chapter Al4, the equation for shear

flow is,

ay = _ ZÁ j j {Tên nh (1)

This equation gives the change in shear

flow between the limits of the summation In

open sections we could start the summation ata free surface where q would be zero, thus the summation to any other point would give the true shear flow qy Ina closed cell there is no free end, therefore the value of Qy is unknown for any point

Equation (1) gives the shear distribution for bending about the X axis githout twist

The general procedure is to ải i a value of

the shear flow dy at some point and then find the shear flow pattern for bending without twist under the given external load The cen~

troid of this internal shear flow system will

be the location where the external shear load _ should act for bending without twist Since the given external shear would have a moment about this centroid, this unbalanced moment must be made zero by adding a constant shear flow system to the cell

To tllustrate we will assume dy to be zero

at point 0 on the web ad

= -2

God = -2-1.8 25 ZA #-2~1.6(-2.5)5x0.150

Fig Al5.2 shows a plot of the shear flow results, On the vertical web the increase in shear is parabolic since the area varies directly with distance z

The intensity of qy and qg in the plane of the cross-section is equal to the values of q found above which are in the y direction The sense of dy and qd, is determined as explained

in detail in Art Al4.6 of Chapter A14,

Al5.1