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

The portion of the beam between sections 1-1 and 2-2 under the given loading are subjected to pure bending since the shear is zero in this region.. Solving for ay, For equilibrium of t

? CHAPTER A12

BENDING

A13.0 Itroduction

The bar AB in fig a is subjected to an

axial compressive load P If the compressive

stresses are such that no buckling of the bar

takes place, then bar sections such as 1-1 and

2-2 move parallel to each other as the bar

shortens under the compressive stress

Moment Dia Fig b

In Fig b the same bar is used as a simply

supported beam with two applied loads P as

shown The shear and bending moment diagrams

for the given beam Loading are also shown The

portion of the beam between sections 1-1 and 2-2

under the given loading are subjected to pure

bending since the shear is zero in this region

Experimental evidence for a beam segment

ax taken in this beam region under pure bending

shows that plane sections remain plane after

bending but that the plane sections rotate with

respect to each other as illustrated in Fig ¢,

where the dashed line represents the unstressed

beam segment and the heavy section the shape

after pure bending takes place Thus the top

fibers are shortened and subjected to compress-

ive stresses and the lower fibers are elongated

and subjected to tensile stresses Therefore at

some plane n-n on the cross-section, the fibers

suffer no deformation and thus have zero stress

This location of zero stress under pure bending

is referred to as the neutral axis

A13,1 Location of Neutral Axis

Fig ¢ shows a cantilever beam subjected to

a pure moment at its free end, and under this

applied moment the beam takes the exaggerated

deflected shape as shown

The applied bending moment vector acts parallel to the Z axis, or in other words the applied bending moment acts in a plane perpend- icular to the Z axis Consider a beam segment

of length L Fig d shows the distortion of this segment when plane sections remain plane after bending of the beam

È will be assumed that the beam section ts

homogeneous, that is, made of the same material,

and that the beam stresses are below the pro- portional limit stress of the material or in other words that Hook’s Law holds

From the geometry of similar triangles,

given a minus sign

Solving for ay,

For equilibrium of the bending stress perpend~

ticular to the beam cross-section or in the %

direction, we can write ZF, = 0, or

27-22 | yaa = 0,

however in this expression, the term 2 ts not

¢ zero, hence the term | yda must equal zero and this can only be true if the neutral axis coin- cides with the centroidal axis of the beam cross~section

A13.L

A13.2

The neutral axis does not pass through the

beam secticn centroid when the beam 1s nonhomo-

geneous that is, the modulus of elasticity is not

constant over the beam section and also when

Hook’s Law does not apply or where the stress-

strain relationship ts non-linear These beam

conditions are described later in this Chapter

A13.2 Equations for Bending Stress, Homogeneous Beams,

Stresses Below Proportional Limit Stress

In the following derivations, it will te

assumed that the clane of the external loads

contain the flexural axis of the beam and hence,

the beam is not subjected to torsional forces

which, if present, would produce bending stress~

es if free warping of the beam sections was re-

strained, as occurs at points of support The

questions of flexural axes and torsional effects

are taken up in later chapters

Fig Al3.1 represents a cross-section of a

straight cantilever beam with a constant cross-

section, subjected to external loads which ite

im a plane making an angle 9 with axis Y-Y

through the centroid 0 To simplify the figure,

the flexural axis has been assumed to coincide

with the centroidal axis, which in general is

not true

Let NN represent the neutral axis under the

given loading, and let @ be the angle between

the neutral axis and the axis X-X The problem

is to find the direction of the neutral axis and

the bending stress o at any point on the section

In the fundamental beam theory, it is as- sumed that the unit stress varies directly as

the distance from the neutral axis, within the

proportional limit of the material Thus, Fig

Al3.2 illustrates how the stress varies along a

line such as mm perpendicular to the neutral

axis N-N

Let co represent unit bending stress at any

point a distance y, from the neutral axis, Then

the stress o on da is

where k ts a constant Since the position of

the neutral axis 1s uninown, yy will be express-

ed for convenience in terms of rectangular co~

ordinates with respect to the axes X-X and Y-Y,

BEAM BENDING STRESSES

Thus, y, = (y - x tan J) cos Z n

external forces that lie on one side of the section ABCD about each of the rectangular axes

X-X and Y-Y must be equal and opposite, respect- ively, te the sum of the moments of the internal

stresses on the section about the same axes

Let M represent the bending moment in the plane of the loads; then the moment about axis

X-X and Y-Y¥ is M, = Mcos 6 and M, «= M sin 9

The moment of the stresses on the’ beam section about axis X-X is fo day Hence, taking

moments about axis X-X, we obtain for equil- 1briưn,

Mecos 9= /ở đa y

-/ (cos 9 y*da - sin Ø xyda)

=k cos @ / y*da - k sin Ø / xyda ()

In similar manner, taking moments about

the Y-¥ axis

Msine=/adax whence

M sin 9 = ~ x sin Ø /x”da + x cos Ø / xyda(4a)

A13.3 Method 1l, Stresses for Moments About the Principal Axes,

In equation (4), the term / y“da ts the

moment of inertia of the cross-sectional area about axis X-X, which we wili denote by Ix, and

the term / xyda represents the product of in-

ertia about axes X-X and Y-Y We know, however, that the product of inertia with respect to the principal axes is zero Therefore, if we se- lect XX and YY in such a way as to make them

coincide with the principal axes, we can write

equation (4):

Mcos @ =k cos 9 ly

In like manner, from equation (4a) Msin 9“-ksinØ 1

Tp

To find the unit stress ở at any point on the

cross-section, we solve equation (5) for cos Ø

and equation (6) for sin g, and then substitut-

ing these values in (3), we obtain the follow-

ing expressing, giving o the subscript b to represent bending stress: -

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES

Let the resolved bending moment M cos @ and

M sin @ about the principal axes be given the

symbols Mp and M, Then we can write

The minus signs have been placed before each

term in order to give a negative value for dy

when we have a positive bending moment, or Mẹp

is the moment of a couple acting about Xp~Xp›

positive when it produces compression in the

upper right hand quadrant My» is the moment

about the Yp-Yp axis, and is also positive when

it produces compression in the upper risht hand

quadrant

BENDING STRESS EQUATION FOR SYMMETRICAL

BEAM SECTICNS

Since symmetrical axes are principal axes

(teim / xyda = 0), the bending stress equation

for bending about the symmetrical XX and YY axes

is obviously,

oy = - Mxy _ Myx we ee ee ee ee eee (74) Ix ly

Al13.4 Method 2 Stresses by use o{ Neutral Axis for

Given Plane of Loading

The direction of the neutral axis NN, mea-

sured from the Xy principal axis is given ty

dividing equation (6) by (5)

Tan 9 = - Ixy

The negative sign arises from the fact that

@ is measurad from one principal axis and 2 is

measured in the same direction from the other

principal axis

Since equation (8) gives us the location of

the neutral axis for a particular plane of load-

ing,the stress at any point can be found by re-

solving the external moment into a plane perpen-

dicular to the neutral axis N-N and using the

A13.3 Moment of inertia about the neutral axis, hence

oy = Lites (6 - Ø)]ÿn - tmờn TT _ (3) Tn tn

Ip can be determined from the relationship ex- pressed in Chapter A5, namely,

In = Ix,008 8 + ly, sin"ố eee eee eee (10)

Al3.5 Method 3 Stresses from Moments, Section Prop- erties and Distances Referred to any Pair of

Rectangular Axes through the Centroid of the Section

The fiber stresses can be found without

resort to principal axes or to the neutral

axis, Equation (4) can be written:

Me =k cos @Iy -kK sin J lyy -+ - - (11)

Where ly = / y*da and Ixy > / xyda, and My =

Substituting these values in Equation (13): -

Oy = - (KaMy - KiMy) X¬ (XaWy ~ KiMy)y - (14)

In Method 2, equation (8) was used to find

the position of the neutral axis for a given Plane of loading The location of the neutral axis can also be found relative to any pair of rectangular centroidal axes X and Y as follows:

Since the stress at any point on the neutral axis must be zero, we can write trem equation (14) that: -

(Kstly ~ KiMy)x = - (Kelly - KaMy)y for all

points located on the neutral axis.” From Fig

Alg tan p=,

Thus tan g = - (Katty = Katt) meee tee {15)

at > Kitty)

A13.4

It frequently happens that tre plane of the

bending moment coincides with 2ither the X-X or

the Y-Y axis, thus making either My, or My equal

to zero In this case, equation (15) can be

simplified For example, if My =O

and 1f My = O

Al13,6 Advantages and Disadvantages of the Three

Methods

Method 2 (bending about the neutral axis

for a given plane of loading) no doubt gives a

better picture of the true action of the beam

relative to its bending as a wnole The point

of maximum fiber stress is easily determined by

placing a scale perpendicular to the neutral

axis and moving it along the neutral axis to

find the point on the beam section farthese away

from the neutral axis In airplane design,

there are many design conditions, which change

the direction of the plane of loading, thus,

Saveral neutral axes must be computed for each

beam section, which is a disadvantage as com-

pared to the other two methods

In determining the shears and moments on

airplane structures, it is common practice to

resolve air and landing forces parallel to the

airplane XYZ axes and these results can be used

directly in method 3, whereas method 1 requires

a further resolution with respect to the prin-

cipal axes Methods 1 and 3 are more widely

used than method 2

Since bending moments about one principal

axis produces no bending about the other prin-

cipal axis, the principal axes are convenient

axes to use when calculating internal shear flow

distribution

Al3.7 Deflections,

The deflection can be Zound by using the

beam section properties about the neutral axis

for the given plane of loading and the bending

moment resolved in a plane normal to the neutral

axis The deflection can also be found by re-

solving the bending moment into the two prin-

cipal planes and then using the properties about

the principal axes The resultant deflection is

the vector sum of the deflections in the direc-

BEAM BENDING STRESSES

A13.8 Dlustrative Problems

Fig 4135.3 shows

a unsymmetrical one

cell dox beam with

four corner flange members a, 5, c and d

Let it be required to determine the bending axial stress in the

four corner members

due to the loads Py

and Py acting 50"

from the section

abed

In this example

solution the sneet

connecting the corner

members will be con~

The first step common to all three methods

is the calculation of the moments of inertia about the centroidal X and Y axes Table AlZ.1

gives the detailed calculations The proper-

ties are first calculated about the reference

axes x'x’ and y'y' and then transferred to the parallel centroidal axes

>, DAx' _ + 28.3 _ aann x= Se = ayy 7 7 10-867

za fay! 16 _ Ysa 7 a 5,926 ®

ly = 176 - 2.7 x 5.926" = 81.18

Ty = 460.8 = 2.7 x 10,667” = 183.88

L xy = 7 192 - (2.7 x - 10.867 x 5.926) = ~ 21.33

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES Solution by Method 1 (bending about principal axes)

The angle @ between the x axis and the

principal axes is given by the equation,

Ixy = Ty cos* g + ly sin? g-+-2 lay sin g cos @

Fig Al3.3a shows the Location of the principal

axes, as well as the perpendiuclar distances

from each corner member to each of the principal

axes These distances are readily determined

from elementary trigonometry:

The external bending moment about the X and

Ÿ axes equal,

6000 x SO = 300,000"#

M

My = 1600 x 50 = - 30,000"#

These moments will be resolved into bending

moments about the Xp and yp principal axes tr = 300,000 x cos 15° ~ 15' - 30000 x

= 25400 ~ 5520 = 21880 and similarly for stringer d, ơớp = 21900,

Solution by Method 2 (Neutral Axis Method)

“Let 6° = angle between yy axis and plane

of loading

hence

@" = 14° ~ 56" and @ = 14° - 56" + 15° ~ 15' =

= 30° - 11 Let a =

the Ấp axis, angle between neutral axis NN and

Ixy tan @ -¬ 75.38 (- 0.5816)

tana - “Ty ss = 275

hence a = 15° - 24' (see Fig A1l3.3b)

Since the angle between the X axis and the

neutral axis is only 9', we can say

Iy = Iy = 91.18

Resolving the external bending moments

normal to neutral axis, we obtain

Solution by Method 3 (Method Using Properties

BEAM BENDING STRESSES

hence oy) = 8 xX - 5.33 - 3697 x 6.074 = ~ 22450 Stringer b

X = 10.667, y = 2.074

Op = 8 x 10.667 - (3697) 2.074 = 85-7660 = ~ 7575 Stringer ¢

X= + 5,333, y= - 5.926

% = 3x - 5.333 - (3697 x - 5.926) =

42 + 21800 = 21862 Stringer d

x = 10.667, y = - 5.926

Oy = & x 10.667 - (3697 x ~ 5,826)

= 86 + 21900 = 21985 NOTE: The stresses op by the three methods were calculated by 10" slide rule, uence ths small discrepancy between the results for che

X› * Ixy/(lxly - lấy) ° ST TR TY TE3.Sồ — 21.58”

= 21.33 _ Bae > Bends About X and Y Centroidal Axes Due to My,

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES Stringer b - cont’d

In like manner for stringers © and d

Øp, = 19150 and ơp, = 27440 @ e

Comparing these results with the previous

results it is noticed that considerable error

exists Under these eroneous stresses the in-

ternal resisting moment does not equal the ex-

ternal bending moments about the X and Y axis

Example Problem 2

Fig 413.4 shows a portion of a cantilever

2-cell stressed skin wing box Deam in this ex-

ample, the beam section is considered constant,

and the section is identical to that used in

14260¢

Fig Al3.4

Fig AG.13 of Chapter AS for which the section

properties were computed and are as follows: =

The resultant air load on the wing outboard

of section ABC is 14260% acting up in the Y di-

rection and 760# acting forward in the X direc-

tion, and the location of these resultant loads

is 5O from section ABC (Fig Al3.4)

The bending stress intensity at the cen-

troids of stringers number (1), (9) and (12)

Will be calculated using all three methods

Soiution by Method 1 (Bending about Principal Axes)

The bending moment at section ABC about the

My, = = 71300 x sin @ - 38000 x cos Ø =-140000”#

From equation (7), the general formula for oy is:

Solution by Method 2 (Neutral Axis Method)

in Fig Al3.5 let 6' be the angle between

the Y-Y¥ axis and the plane of the resuitant

bending moment Resultant bending moment,

M = V 7140002 + 580007 = 714600" 1b

tan o! = 738000 _ 712000 0523, hence 9 nh e 9' =~ 59 3 - 1! 3 Let @ equal the angle between the plane of the resultant moment and the Y, axis

À13.8

Plane of et x Resultant \_g Ƒ Moment —=

the neutral axis, principal axes, and plane of

loading

The component of the external resultant

moment about the neutral axis N equals: -

My = 714060 x sin 93° - 26' = 709350 in ib

Iy = Ixp co8* a + Ip, sin? a

NOTE: In the three solutions, the distance

from the axis tn question to the stringers 1,

9, and 12 have been taken to the centroid of each stringer unit Thus, the stresses ob-

tained are average axial stresses on the

stringers If the maximum stress ts desired, the arm should refer to the most remote part of

the stringer or the skin surzace

Approximate Method

It is sometimes erroneously assumed that the external bending moments My and My produce

bending about the X and Y axes as though they

were neutral axes To show the error of this assumption, the stresses will be computed for

stringers 1 and 9,

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES

dy = 7718000 x 4.39 _ (+ $8000 x = 17.41) Fig A13.8 Fig A13.9

“ ——“Te.46 — T317 — 106 „46 stress intensity at bottom edge of portion A is 2 :

The previous example problems were solved

by substituting in the bending stress equations

The student should solve vending stre:s prob-

lems by equating the internal resistiig moment

at a beam section to the external berding mom-

ent at the same section To illustrite Fig

Al3.6 shows a simply supported loaded beam The

shear and bending moment diagram for the given

beam loading is also shown Fig A13.7 shows

the beam section which is constant along the

span

The maximum bending moment occurs over the

left support and equals - 24000 in lbs Due to

symmetry of the beam cross-section the centroidal

horizontal axis is the center line of the beam

and thus the neutral axis is at the midpoint of

the beam

Fig Al3.& shows a free body of that por-

{on of the beam from the left end to a section

over the left support where the bending moment

is maximum The triangular bending stress in-

tensity diagram is shown acting on the cut sec-

tion, with a value of o, at the most remo.e

fiver The forces Ty, Tg etc represent the

total load on the beam cross-sectional areas

The

labeled A and B respectively in Fig, Al2.3,

= (mae) (1.75 x 0.75) = 1.1605

The distance from the neutral axis to the

centroid of the trapizoidal stress loading on portion A is 2.64 inches In like manner,

Tg = Cg’ = C = 3 x 0.25 = 0.3780p

The arm to Tp is 0.667 x 3 = 2 inches

For equilibrium of the beam free body, take moments about point (0) and equate to zero

For Portions A and A!

I S+gzX 176 (6° - £55) = 12,26 int, For Web Portions b, b",

1 zi x 0.25 x 6% = 4,50 in*

Inotay = 22.76 in*

Fig Als.lO shows a loaded beam and Fig

Al3.11 shows the cross-section of the Seam at

section a-a’ Determine the magnitude of the

maximum bending stress at section a-a' under the

given beam loading The beam section is unsym~

metrical about the horizontal centroidal axis

Simple calculations Locate the neutral axis as

shown in Fig Al3.1l

3/4

—#z' —

Fig Al3.12 shows a free body of the por-

tion of the beam to the right of section aa

The bending stress intensity diagram is shown by

the horizontal arrows acting on the beam face at

section a~a Fig Al3.13 snows the cross-sec~

tion of the beam at section a-a

The general procedure will be to determine the total bending stress load on each portion,

1 to 6, of the cross-section and then the mom-

ent of each of these loads about the neutral

axis, the summation of which must equal the ex-

ternal bending moment

the bending stress formula, however, the student

should obtain a better understanding of the in-

BEAM BENDING STRESSES

ternal force action from this metned of solu-

sion In solving nonhomogeneous deams 4 beams stressed above the elastic limit stress, this method of solution often proves necessary

or advantageous because no simple beam bending stress formula can be derived

In Fig Al3.12, let oy te the intensity on the most remote fider, or 3.09" above the neu-

tral axis Table A shows the calculations of

the total stress on each of the pertions of the

cross-section and their moment about the neutral

axis, all in terms of the unknown stress, a

nd

TABLE A

Portion | Area | Average Total y = arm | Resisting

A Stress Load to N.A | moment

4 1.09 „350 „ 387 -1.44 -0 557

5 - 5' |0.36 „578 ,323 ~-1.93 -0.622

6 1,50 | ,82lỢp |1.23T0p | -2.57 | -3.180 op Totals | 6.19 0, 0000p, -8.82 o4

Explanation of Table A

Column 3 gives the average stress on each

of the 6 portions For example, the stress on portion (1) varies from zero at the neutral

2.09 Oy 3,08 “b = * .674 op at the upper edge of b +874 dp + 0 2

Thus, tne average stress ~

Then, the average stress = Oy = 337 đụ, Column 5 is the distance from the neutral axis to the centroid of the load on

each of the blocks For portion (1) the stress

pattern is triangular, and y = 2/3 x 2.09 = 1.349", On portions (2), (3), (5) and (6) the

bending stress distribution is trapezoidal, and the arm is to the centroid of this trapezoid

The total internal resisting moment of

~ 8.82 o) from Table A equals the external bend-

ing moment of Z6670"$,

Thus,

~ 26870

Bp = 736870 5 _ 4160#/in.? 3.82 ca] œ

For equilibrium, the total compressive

stresses on the cross-section of the beam must

be equal to the totai tensile stresses, or =H

must equal zero Column 4 of Table A gives the

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES

total load on each portion of the cross-section

and the total of this column ts zero

The bending stress on the lower fiber of

the cross-section is directly proportional to

the distance from the neutral axis or, blower *

Gy (upper fiber) = = $6670 27.2 X 8-08 2 _ greg psi

which checks the above solutlon

Al3,9 Bending Stresses in Beams with Non-Homogeneous

Sections, Stresses within the Elastic Ranges

In general, beams are usually made of one

material, but special cases often arise where

two different materials are used to form a beam

section For example, a commerical airplane in

its lifetime often undergoes a number of changes

or modifications such as the addition of ad-

ditional installations, fixed equipment, larger

engines, etc, This increase in airplane weight

increases the structural stresses and it often

becomes necessary to strengthen various struct-

ural members at the critical stress points

Since space limitations are usually critical, it

often necessitates that the reinforcing material

be stiffer than the original material Years

ago, when spruce wood was a common material for

wing beams, it was normal practice to use rein-

Zoreements at critical stress points of stiffer

wood material such as maple in order to cut down

the size of the reinforcements In aluminum

alloy beams, the use of heat-treated alloy steel

reinforcements are often used because steel is

29000,000/10500,000 or 2.76 times stiffer than

alumimm alloy

SOLUTION BY MEANS OF TRANSFORMED

SEAM SECTION

To illustrate how the stresses in a com-

posite beam can be determined, two example prob-

lems will be presented

Example Probiem 3

Fig Al3.14 shows the original beam section

which is entirely spruce wood Fig Al3.15

shows the reinforced or modified beam section

Two ears of maple wood have 2een added te the

upper part as shown and a stesl strap nas deen

added to the bottom face of the beam The prob—-

lem is to find the stress at the top and bottom

points on the beam section when the >eam section

is subjected to an external vending moment of

60000 in lbs about a horizontal axis which

causes compression in the upper beam portion

The modulus of elasticity (Z) for the 5 differ-

into Equivalent Spruce

ent materials is: -

A13.15 into an equivalent beam section composed

of the same material throughout This is possi- ble, because the modulus of elasticity of each Material gives us the measure of stiffness for

that material in this solution the reinforced beam will be transformed into a spruce beam

section as illustrated in Fig A13.16

in Fig Alg.15 into spruce we increase the

width of each strip to 1.23 x 0.5 = 0.615 inches

as shown in Fig Al3.16 Likewise, to trans- form the steel reinforcing strip into spruce, we make the width equal to 22.3 x 1.5 = 33.4 inches

as shown in Fig 413.16

This transformed equivalent section Is now

handled like any homogeneous beam section which

is stressed within the elastic iimit of the materials The usual calculation would locate the neutral axis as shown and the moment of in~

ertia of the transformed section about ‘the

neutral axis would dive a value of 51.30 in.*

Bending stress at upper edge of beam section: -

A13.12

the stress in the spruce section Since the re-

inforcing strips are maple, the stress at the

top edge of these maple strips would be 1.23

times (- 3900) = ~ 4880 psi

The bending stress at the lower edge of the

transformed beam section of Fig Al13.16 would be:

60000 x ( - 2.73) _

Sp, The stress in the steel reinforcing strap

thus equals 22.3 times 3200 = 71500 psi

Since all these stresses are below the

elastic limit stress for the 3 materials the

beam bending stress formula as used is

applicable

Example Problem 6

Fig Al3.17 shows an unsymmetrical beam

section composed of four stringers, a,b,c and

ad of equal area each and connected by a thin

wed The web will be neglected in this example

problem Zach of the stringers is made from

different material as indicated on Fig Al3.17

The beam section is subjected to the bending

moment My and My as indicated Let it be re-

quired to determine the stress and total load on

each stringer in resisting these applied extern-

al bending moments

k——eb" — Steel ip

Magnesium Area of Each

cả i Aloy 1g Stringer = 1 sq in

án — Alum Alloy

Fig A13 17

SOLUTION;

Since the 4 stringers are made of different

materials we will transform all the materials

into an equivalent beam section with all 4

stringers being magnesium alloy

Emag 6.5

Using the ratio of stiffness values as indicated

above gives the transformed beam section of Fig

A13.18 where all material is now magnesium

BEAM BENDING STRESSES

alloy The original area of 0.1 sq in each have been multiplied dy these stiffness ratio

Alloy

{ 0.10 c 0,431

The solution for the beam section of Fig

Al3.18 is the same as for any other unsym- metrical homogeneous beam section

The first step is to locate the centroid

of this section and determine the moments of inertia of this section about centroidal X and

ly = 0.546 x 2/388

a 1.635 = 6.34 in.*, Iyy = 0.446 (- 2.265)(4.65) 2 - 4,90 0.1615 x 3.635 x 4.65 = 2.73 0.10 (- 2.365)(~ 5.35) = 1.27 0.431 x 1.635 (- 5.35) = ~ 3.77

TOTAL - - 4,67 in.“ = Iyy

The bending stresses will be calculated

by using method 3 of Art A13.5S

From Equation (14) Art Al3.5

os ~ (KeMy - Kit )x - (Kel - KaMy ly

- 4,67 28.27 x 6.34 - 4.67