solution ch4

Mechanics of Aircraft structures C.T. Sun

4.1 A uniform beam of a thin-walled angle section as shown in Fig. 4.19 is

subjected to the bending (yM 0=zM ). Find the neutral axis and bending stress distribution over the cross-section.

Figure 4.19 Thin-walled angle section

Solution: (a) For finding the location of the centroid, we select the corner of the thin-walled

section as the origin of a Cartesian coordinate system with the horizontal and vertical distances between the centroid and the origin denoted by and , respectively.

cy cz

42)2/( h

hththyc ==

42)2/( h

hththzc ==

--- ANS (b) Set up a Cartesian coordinate system (y, z) in the pane of the section with the

origin at the centroid. The moments of inertia with respect to this coordinate system are (assume t


32

2 161|)

2( thytzytdyzyzdA c

c

c

c

yhyc

yh

y cA

===

(c) Using equation (4.25) in the textbook,

zIII

MIMIy

IIIMIMI

yzzy

zyzyz

yzzy

yyzzyxx 22

+=

By substituting the known values we obtain

)915(2

])8/1()24/5)(24/5[()24/5(

])8/1()24/5)(24/5[()8/1(

3

3232

yzth

M

zth

My

thM

y

yyxx

=+

=

--- ANS Maximum positive stress:

At hzhz c 43== and

4hyy c ==

23 427

)915(2 th

Myz

thM yy

xx == Maximum negative stress:

At 4hzz c == and hyhy c 4

3==

23 421

)915(2 th

Myz

thM yy

xx ==

The absolute maximum stress is 2427

thM y

xx = (d) The neutral axis is located along 0=xx

0)915(2 3

== yzth

M yxx => 0915 = yz

So the neutral plane is located at 0915 = yz in the y-z coordinate system (the centroid is the origin of this coordinate system).

--- ANS

4.1.2


4.2 Rotate the angle section of Fig. 4.19 counterclockwise for . Find the neutral axis and the maximum bending stress. Compare the load capacity with that of the original section given by Fig. 4.19.

o45

Figure 4.19 Thin-walled angle section

Solution:

Remove the primes in the coordinates

Set up a temporary Cartesian coordinate system with the origin at the corner of the thin-walled section to find the centroid. The horizontal and vertical distances from the centroid to the origin are denoted by and , respectively. cy czBecause of the symmetry, . Assuming 0yc = ht


'' 22 zIIIMIMI

yIIIMIMI

yzzy

zyzyz

yzzy

yyzzyxx

+=

and substituting the values of moments of inertia in the equation above, we obtain

zthM

12zIM

3y

y

yxx ==

--- ANS Maximum positive stress is at

22hz = , => 2

23thM y

xx =

Maximum negative stress is at

22

hz = , => 223thM y

xx =

The absolute maximum stress is 223thM y

xx = (c) The neutral axis (plane) is located along 0=xx ,

0zthM

12 3y

xx == => 0z =So the neutral axis coincides with the centroidal axis. Note that this section in this particular position is symmetric with respect to the y-z coordinate system. For symmetric sections the neutral axis always coincides with the location of the centroid.

--- ANS (d) The load capacity with the original section

For the same maximum bending stress in both beams,

2,

2,

42723

thM

thM originyrotatey

xx ==

=> 59.1212

27MM

origin,y

rotate,y ==

The load capacity of the rotated section is 1.59 times that of the original section. --- ANS

4.2.2


4.3 The stringer-web sections shown in Figs. 4.20, 4.21, and 4.22 are subjected to

the shear force , while 0zV 0=yV . Find the bending stresses in the stringers for the same bending moment . Which section is most effective in bending? yM

Figure 4.20 Stringer-web section



Solution:

The contribution of the thin sheets to bending is assumed to be negligible. Thus the

neutral axis is only depends on the cross-sectional area of the stringers. Also, assume

4.3.1


y and z are the horizontal axis and vertical axis, respectively. The origin of the system

is located at the centroid.

(a) Figure 4.20.

(1) Because of symmetry, the centroid is located at the middle of the vertical

web.

(2) Moment of inertia 222 4)2(2 AhhAzAI

iiiy ===

0)02(2 22 === AyAIi

iiz

0== i

iiiyz zyAI

(3) Bending stress. Considering 0yM and 0=zM

zAhM

zIM y

y

yxx 24

==

The stresses at the stringer are

. At , hz =AhM

zAhM yy

xx 44 2==

. At ,hz =AhM

zAhM yy

xx 44 2==

--- ANS

(b) Figure 4.21.

(1) Because of symmetry (when neglecting the effects of webs), the centroid is

located at the center of the section as shown in the figure.

(2) Moment of inertia 222 4)(4 AhhAzAI

iiiy ===

222 ))2

((4 AhhAyAIi

iiz === 0==

iiiiyz zyAI


zIM

zIII

MIy

IIIMI

y

y

yzzy

yz

yzzy

yyzxx =+

= 22

The stresses at the stringers are (y position is not involved)

4.3.2


. At , hz =AhM

zAhM yy

xx 44 2==

. At , hz =AhM

zAhM yy

xx 44 2==

--- ANS

(c) Figure 4.22.

(1) Again, when neglecting the effects of webs, the centroid is located at the

middle of the vertical web.


iiiy ===

222 2)(2 AhhAyAIi

iiz === 22))((2 AhhhAzyAI

iiiiyz ===


zAhM

yAhM

zAh

My

AhM

zIII

MIy

IIIMI

yy

yy

yzzy

yz

yzzy

yyzxx

22

222222

22

])2(24[2

])2(24[2

+=+=+

=

The stresses at the stringer are

At , , hz = hy =

0)(222 222

=+=+= hhAhM

zAhM

yAhM yyy

xx . At , , hz = 0=y

AhM

hAhM

zAhM

yAhM yyyy

xx 2)0(

222 222=+=+= . At , , hz = 0=y

AhM

hAhM

zAhM

yAhM yyyy

xx 2)0(

222 222==+= . At , , hz = hy =

0)(222 222

==+= hhAhM

zAhM

yAhM yyy

xx --- ANS

4.3.3


(d) Comparing the above results, sections in Figure 4.20 and Figure 4.21 are both

more effective than the section in Figure 4.22 for this particular loading.

--- ANS

4.3.4


4.4 Compare the bending capabilities of the two sections of Figs. 4.21 and 4.22 if

, . 0=yM 0zM



Solution:

The thin sheets are assumed to be negligible in bending. Thus, the location of the

centroid of the cross-section only depends on stringers. The coordinates (y, z) are set

up with the origin at the centroid with y and z designating the horizontal axis and

vertical axis, respectively.

(a) Figure 4.21.

(1) The centroid is located at the center of of the space defined by the four

stringers.


iiiy ===

4.4.1


222 ))2

((4 AhhAyAIi

iiz === 0==

iiiiyz zyAI

(3) Bending stress.

Considering and 0=yM 0zM we have

yIMz

IIIMI

yIII

MI

z

z

yzzy

zyz

yzzy

zyxx =

+= 22

The stresses in stringers 1 and 4 are:

At 2hy = , y

AhM

2z

xx = The stresses in stringers 2 and 3 are

At 2hy = , y

AhM

2z

xx =

--- ANS

(b) Figure 4.22.

(1) The centroid is located at the middle of the vertical web.


iiiy ===

222 2)(2 AhhAyAIi

iiz === 22))((2 AhhhAzyAI

iiiiyz ===

(3) Bending stress.

For and 0=yM 0zM

zAhMy

AhM

zAh

MyAh

MzIIIMI

yIII

MI

zz

zz

yzzy

zyz

yzzy

zyxx

22

222222

2

])2(24[2

])2(24[4

+=+=

+=

The stress in stringer 1 is

At , hz = hy =

4.4.2


AhMhh

AhMz

AhMy

AhM zzzz

xx 2)2(

22 222=+=+=

Stringer 2:

At , , hz = 0=yAhMh

AhMz

AhMy

AhM zzzz

xx 2)0(

22 222=+=+=

Stringer 3:

At hz = , ,0=yAhMh

AhMz

AhMy

AhM zzzz

xx 2)0(

22 222==+=

Stringer 4:

At hz = , , hy =AhMhh

AhMz

AhMy

AhM zzzz

xx 2)2(

22 222==+=

--- ANS

(c) Comparing the above results, we see that sections in Figure 4.21 and Figure 4.22

have the same bending efficiency; they both reach the same maximum bending

stress under the same moment.

--- ANS

4.4.3


4.5 Figure 4.23 shows the cross-section of a four-stringer box beam. Assume that

the thin walls are ineffective in bending and the applied bending moments are

cmNM y = 000,500 cmNM z = 000,200 .

Find the bending stresses in all stringers.

Figure 4.23 Thin-walled section

Solution:

(a) Set up a temporary coordinate system with stringer 1 as the origin. The location of

the centroid is

cm5.54)4124()20012002(

A

yAy

ii

iii

c =++++==

cm9.40)4124()1004501(

z

zAz

ii

iii

c =++++==

(b) The moment of inertia

4

222

i

2iiy

cm240901

)909091.40100(4)909091.4050(1)909091.40)(24(zAI

=+++==

Similarly, 4

i

2iiz cm87273yAI ==

4.5.1


4

iiiiyz cm14545zyAI ==

It is more convenient to put in the chart, for instance:

iA iy iz Stringer No. )(cm )( 2cm )(cm

2ii zA )( 4cm

2ii yA )( 4cm

iii zyA )( 4cm

1 4 -54.5 -40.9 6694 11901 89256

2 2 145.5 -40.9 3347 42314 -11901

3 1 145.5 9.1 82.6 21157 1322

4 4 -54.5 59.1 13967 11901 -12893

= 24091 87273 -14545

(c) Bending stress in the stringers.

By using the equation: zIIIMIMI

yIIIMIMI

yzzy

zyzyz

yzzy

yyzzyxx 22

+= , and

cmNM y = 000,500 cmNM z = 000,200 .

4909.24090 cmI y = 4727.87272 cmI z =

4455.14545 cmI yz = We obtain z54.21y298.1xx = Therefore the bending stresses in the stringers are:

iy Stringer No. )(cm

iz )(cm

xx )/( 2cmN

1 -54.54 -40.91 951.92

2 145.45 -40.91 692.31

3 145.45 9.09 -384.62

4 -54.54 59.09 -1201.92

--- ANS

4.5.2


4.6 Find the neutral axis in the tin-walled section of Fig. 4.23 for the loading given

in Problem 4.5.

cmNM y = 000,500 cmNM z = 000,200 .

Find the bending stresses in all stringers.

Figure 4.23 Thin-walled section

Solution:

(a) From Problem 4.5 we get the centroid position as follows.

cm5.54yc = , cm9.40zc =These are the horizontal and vertical distances, respectively, from stringer 1.

(b) Set up the coordinate system (y,z) with the origin located at the centroid. Neutral

plane is located at the position that centroid is the origin. From the bending stress

formulas we find the neutral plane by setting the bending stress to zero, i.e.,

0z538.21y298.1xx == On the cross-section, this equation represents the line passing through the centroid

with and an angle z59.16y =o45.3)

59.161(tan)

yz(tan 11 ===

--- ANS

4.6.1


4.7 Find the bending stresses in the stringers at the fixed end of the box beam loaded

as shown in Fig. 4.24. Assume that the thin sheets are negligible in bending.

Find the neutral axis.

Figure 4.24 Loaded box beam

Solution:

(a) Name the stringers from top to bottom and left to right as stringer 1, stringer 2,

and stringer 3, respectively. Relative to string 2 the centroid position is given by

cm67.2643804

A

yAy

ii

iii

c ===

cm33.1343404

A

zAz

ii

iii

c ===

(b) The bending moments at the fixed end of the box beam produced by the loads are

cmNPLM y === 200000)500)(200(22 ( is positive in positive y) yMcmNPLM z === 200000)500)(200(22 ( is positive in negative z) zM

(c) Set up the coordinate system (x,y,z) with the origin at the centroid.

Moment of inertia (see table below for details): 422

i

2ciy cm4266)67.2633.132(4zAI =+==

422

i

2ciz cm17067)33.5367.262(4yAI =+==

4.7.1


4

icciyz cm4266zyAI ==


2ii zA )( 4cm

2ii yA )( 4cm

iii zyA )( 4cm

1 4 -26.67 26.67 2844 2844 -2844

2 4 -26.67 -13.33 711 2844 1422

3 4 53.33 -13.33 711 11377 -2844

= 4266 17067 -4267 (d) Bending stress in the stringers.

Using the equation zIIIMIMI

yIIIMIMI

yzzy

zyzyz

yzzy

yyzzyxx 22

+= ,

we obtain zyxx 125.7825.31 = and the bending stresses in the stringers are:


iz )(cm

xx )/( 2cmN

1 -26.67 26.67 -1250

2 -26.67 -13.33 1875

3 53.33 -13.33 -625

--- ANS

(e) Neutral plane by angle . Neutral plane is located at the position where bending stresses vanish under this

particular loading. We have

0125.7825.31 == zyxx It is the line passing through the centroid with zy 5.2=

o8.21)5.21(tan)

yz(tan 11 ===

--- ANS

4.7.2


4.8 Find the deflection of the box beam of Fig. 4.24 using the simple beam theory.


Solution:

(a) Name the stringers from top to bottom and then left to right as stringer 1, stringer

2, and stringer 3, respectively. From the solution of problem 4.7, we have the

following moments of inertia: 4

y cm4266I = 4

z cm17066I = 4

yz cm4266I = Let the origin of the coordinate system be located at the fixed end. )0x( =The bending moments produced by the forces applied at the free end are

cmNxM y = )500(400 cmNxM z = )500(400

(b) The governing equations (see p. 122 in the book) for the bidirectional bending are

)/()500(063.0 3222

cmNxIIIMIMI

dxvdE

yzzy

yyzzy == ,

)/()500(156.0 3222

cmNxIIIMIMI

dxwdE

yzzy

zyzyz ==

Integrating twice the above differential equations, we obtain

4.8.1


21

32 )

6250(063.0 CxCxxEv ++=

43

32 )

6250(156.0 CxCxxEw ++=

By applying the boundary conditions, the integration constants are solved as

0)0( ==xv , 0)0( ==xdxdv => 021 == CC

0)0( ==xw , 0)0( ==xdxdw => 043 == CC

Then the lateral (in y-direction) and vertical (in z-direction) deflections are,

respectively,

)6

250(063.0)(3

2 xxE

xv =

)6

250(156.0)(3

2 xxE

xw = In the expressions above, distance x is measured in cm, and the units of Youngs

modulus and deflection are and , respectively. 2/ cmN cm

--- ANS

As an example, consider Aluminum 2024-T3, .

The deflections in y and z directions at the free end are:

)/(107272 25 cmNGPaE ==

cmxv 36.0)6

500500250(1072

063.0)500(3

25 ===

cmxw 90.0)6

500500250(1072

156.0)500(3

25 ===

4.8.2


4.9 Find the bending stresses in the stringers of the box beam in Fig. 4.24 for the

bending moments given in Problem 4.5.

cmNM y = 000,500 cmNM z = 000,200 .


Solution:

(a) Name the stringers from top to bottom and then left to right as stringer 1, stringer

2, and stringer 3, respectively. The centroid position is given by

cm67.2643804

A

yAy

ii

iii

c ===

cm33.1343404

A

zAz

ii

iii

c ===

relative stringer 2.

(b) Moment of inertia (see the table below for details) 422

i

2ciy cm4267)67.2633.132(4zAI =+==

422

i

2ciz cm17067)333333.53666667.262(4yAI =+==

4

icciyz cm4267zyAI ==

4.9.1



2ii zA )( 4cm

2ii yA )( 4cm

iii zyA )( 4cm

1 4 -26.67 26.67 2844 2844 -2844

2 4 -26.67 -13.33 711 2844 1422

3 4 53.33 -13.33 711 11378 -2844

= 4267 17067 -4267

(c) Bending stress in the stringers.

Subsituting the moments and moments of inertia in the bending stress formula

zIIIMIMI

yIIIMIMI

yzzy

zyzyz

yzzy

yyzzyxx 22

+= ,

we obtain z62.140y44.23xx = Therefore the bending stresses in the stringers are:


iz )(cm

xx )/( 2cmN

1 -26.67 26.67 -3125

2 -26.67 -13.33 2500

3 53.33 -13.33 625

--- ANS

4.9.2


4.10 A cantilever beam of a solid rectangular cross-section is loaded as shown in Fig.

4.25. Assume that the material is isotropic. Find the deflections of the beam

using the simple beam theory and Timoshenko beam theory, respectively. Plot

the ratio of the maximum deflections of the two solutions (at the free end)

versus L/h. Use the shear correction factor 65=k .

P

L

x

t

h

Figure 4.25 Cantilever beam subjected to a shear force P

Solution:

(a) Simple beam theory

The displacement equilibrium equations for the simple beam theory is:

z40

4

y pdxwd

EI = (4.10.1)

In this particular problem, we have 3121 thI y = , 0pz = . Thus,

040

4

=dxwdEI y (4.10.2)

Integrating the equation (4.10.2) and applying boundary conditions,

PCdxwdEI y == 030

3

(shear force)

Integrating again, we obtain

120

2

CPxdxwdEI y += . (4.10.3)

At , Lx = 1202

0)( CPLLxdxwdEIM y +====

=> PLC =1From (4.10.3),

4.10.1


220

21 CPLxPx

dxdwEI y ++=

At , 0=x 0dxdw0 = => 02 =C

Finally, 323

0 21

61)( CPLxPxxwEI y ++= , and 03 =C because

at .

00 =w0=x

Therefore, the deflection curve is

])(6)(2[)21

61(1)( 23230 h

xhL

hx

EtPPLxPx

EIxw

y

+=+=

--- ANS

The maximum deflection occurs at Lx = : 323

max, )(4])(6)(2[

hL

EtP

hL

hL

hL

EtPw S =+=

--- ANS

(b) Timoshenko beam theory

The displacement equilibrium equations for Timoshenko beam theory are:

0)( 022

=+ yyy dxdwkGA

dxd

EI (4.10.4)

0)( 20

2

=++ zy pdxd

dxwdkGA

(4.10.5)

and can be combined into the following equation,

2

2

40

4

dxpd

GAEI

pdxwdEI zyzy = (4.10.6)

In this particular problem, we have 3121 thI y = , and 0=zp . Hence we have

040

4

=dxwdEI y as the governing equation.

The concentrated loading at the free end produces a constant shear force along the

beam, so we have

Pforceshear)dxdw

(kGA y0 ==+ (4.10.7)

Substituting (4.10.7) in (4.10.4) yields

4.10.2


Pdxd

EI yy =22

(4.10.8)

Integrating equation (4.10.8) twice, we obtain

102

21 BxBPxEI yy ++= (4.10.9)

Using (4.10.7) and (4.10.9), we obtain

)21(1 10

20 BxBPxEIkGA

PkGAP

dxdw

yy ++==

Integrating the equation above,

212

03

0 )21

61(1)( BxBxBPx

EIx

kGAPxw

y

+++= (4.10.10)

The following boundary conditions are used to determine the arbitrary constants in

(4.10.10):

0)()( ==== Lxdxd

EILxM yy

=> PLB =0

0)0( ==xy (no rotation of the cross-section) => 01 =B 0)0(0 ==xw => 02 =B

Then the deflection equation (4.10.10) becomes

)21

61(1)( 230 PLxPxEI

xkGAPxw

y

=

With 65=k , , thA = 3

121 thI y = , and )1(2 +=

EG , we obtain

])(6)(2[)(5

)1(12)( 230 hx

hL

hx

EtP

hx

EtPxw +=

--- ANS

The maximum deflection occurs at Lx = : 3

max, )(4)(

5)1(12

hL

EtP

hL

EtPw T ++=

--- ANS

(c) The ratio of the maximum deflections of the two solutions versus L/h

Assume the material to be Aluminum 2024-T3 with GPaE 72= , 33.0= . For convenience, we let =

hL .

The maximum deflection according to the simple beam theory:

4.10.3


33max, 055556.0)(72

4 tP

tPw S ==

The maximum deflection according to the Timoshenko beam theory:

33max, 055556.0044333.0)(72

4)()72(5

)33.01(12 tP

tP

tP

tPw T +=++=

Maximum deflections vs. L/h

0

2

4

6

8

10

12

14

0 1 2 3 4 5 6 7L/h

(w

=

P/t)

SimpleTimoshenko

Define %100(%)max,

max,max, =T

ST

www

Error

Error (%) vs. L/h

0

10

20

30

40

50

60

70

80

90

100

0 1 2 3 4 5 6L/h

Erro

r (%

)

7

4.10.4


4.11 A thin-walled beam of length 2 m long with one end built into a rigid wall and

the other end is subjected to a shear force NVz 5000= . The cross-section is given by Fig. 4.21 with mh 2.0= and the wall thickness . The material is aluminum 2024-T3 with

m002.0=GPaE 70= , GPaG 27= , and the

cross-sectional area of each stringer is . Assume that thin walls carry

only shear stresses. Find the deflections at the free end using the simple beam

theory and the Timoshenko beam theory, respectively. Compare the transverse

shear stress in the vertical web obtained from the two theories.

225cm


Solution:

(a) Simple beam theory

(1) The displacement equilibrium equation for the simple beam theory is:

040

4

=dx

wdEI y (4.11.1)

Integrate the equation (4.11.1) and apply shear force boundary condition to yield,

zy VCdxwdEI == 030

3

(shear force)

Integrate again to obtain

120

2

CxVdx

wdEI zy += ,

At the free end, , Lx = 1202

0)( CLVLxdx

wdEIM zy +====

=> LVC z=14.11.1


Again by integration, we have

220

21 CLxVxV

dxdwEI zzy ++= .

At the fixed end, , the rotation of the cross-section vanishes, i.e., 0=x00 ==

dxdw

y => 02 =C Thus, we have, after integration,

323

0 21

61)( CLxVxVxwEI zzy ++=

In which the integration constant C3 is determined by the boundary condition

00 =w at => 0=x 03 =C . The deflections curve is

)21

61(1)( 230 LxVxVEI

xw zzy

+= (4.11.2)

(2) Properties of the cross-section 442422 104)2.0)(1025(44 mAhzAI

iiiy

==== GPaE 70=

mL 2= NVz 5000=

(3) Deflections

Compute deflection curve (4.11.2):

)(107857.1109762.2

))2)(5000(21)5000(

61(

)104)(1070(1)(

2435

23490

mxx

xxxw

+=+=

Deflection at the free end:

mmm

mxw

48.010762.4

)2(107857.1)2(109762.2)2(4

24350

=

+==

--- ANS

(b) Timoshenko beam theory

(1) The displacement equilibrium equations for the Timoshenko beam theory

are:

4.11.2


0)( 022

=+ yyy dxdwGA

dxd

EI (4.11.3)

0)( 20

2

=++ zy pdxd

dxwdGA

(4.11.4)

which can be combined into the following equation,

2

2

40

4

dxpd

GAEI

pdx

wdEI zyzy = (4.11.5) In the equations above, the area A in the GA term is the effective area of the

thin-walled section that carries shear stress and should not be confused with the

stringer cross-sectional area.

Since 0=zp we have

040

4

=dx

wdEI y as the governing equation.

The concentrated shear loading at the free end produces a constant shear force

along the beam; so we have

zy VdxdwGA =+ )( 0 (4.11.6)

Substitution of (4.11.6) in (4.11.3) yields

zy

y Vdxd

EI =22

(4.11.7)

Integrating (4.11.7), we obtain

102

21 BxBxVEI zyy ++= (4.11.8)

Using (4.11.6) and (4.11.8), we have

)21(1 10

20 BxBxVEIGA

VGAV

dxdw

zy

zy

z ++==

Integrating the above equation with the result,

212

03

0 )21

61(1)( BxBxBxV

EIx

GAVxw z

y

z +++= (4.11.9)

Applying boundary conditions to equation (4.11.8) and (4.11.9), we have

0)()( ==== Lxdx

dEILxM yy

=> LVB z=0

0)0( ==xy => 01 =B

4.11.3


0)0(0 ==xw => 02 =B Then equation (4.11.9) becomes

)21

61()( 230 LxxEI

VxGAVxw

y

zz = (4.11.10)

(2) Properties of the cross-section

In the Timoshenko beam theory the area A (in the GA term) of the

thin-walled cross-section is 24108)002.0)(2.0(22 mhtAshear

=== 442422 104)2.0)(1025(44 mAhzAI

iiiy

==== GPaE 70= , GPaG 27=

mL 2= NVz 5000=

(3) Deflection

Compute the deflection curve (4.11.10) using the above properties:

)(109762.2107857.1103148.2

)107857.1109762.2()108)(1027(

5000)(

35244

2435490

mxxx

xxxxw

+==

Deflection at the free end is

mmm

mxw

94.010391.9

)2(109762.2)2(107857.1)2(103148.2)2(4

352440

=

+==

--- ANS

The difference between the two theories is

%3.49%100391.9

762.4391.9w

ww(%)Error

Tim,0

Sim,0Tim,0 ===

(c) Transverse Shear Stress

(1) Simple beam theory

From the derivation of the simple beam theory, we assume 0=xy as an approximation. As a result, the transverse shear stress can not be directly

obtained from the stress-strain relations. It is obtained usually from the

4.11.4


equilibrium equation. We have

MPamNAVshear

zxz 25.6/1025.6108

5000 264 ====

(2) Timoshenko beam theory

MPaAVGshear

zxzxz 25.6===

--- ANS

4.11.5


4.12 A 2024-T3 aluminum box beam with a thin-walled section is shown in Fig.

4.26. Assume that thin walls (thickness t = 0.3 cm) are ineffective in bending.

The cross-sectional area of each stringer is 20 cm2. Find the deflections at the

free end using the simple beam theory for shear loads and

separately. Solve the same problem using Timoshenko beam

theory. In which loading case is the simple beam theory more accurate in

predicting the deflection? Explain.

NVz 5000=NVy 5000=

Figure 4.26 Box beam with a triangular thin-walled section

Solution:

(a) First, we need to know the centroid of this section.

Take stringer 2 as the origin of a coordinate system. Then the centroid is located at

cmA

yAy

ii

iii

c 202036020 =

==

cmA

zAz

ii

iii

c 20203)2040(20 =

+==

The moments of inertia with respect to the coordinate system with the origin at the

centroid are 4222 16000))20(20(20 cmzAI

iciy =+==

4222 48000))20(240(20 cmyAIi

ciz =+== 4.12.1


0)]20()20(20)20(040[4 =++== i

cciyz zyAI (This should be

obvious because the section is symmetric with respect to y-axis)

For 2024-T3, , 25 /107272 cmNGPaE == 33.0= => 25 /10068.27 cmNG =

(b) Simple beam theory

The displacement equilibrium equations for the simple beam theory are:

040

4

=dxwdEI y , for loading (4.12.1) zV

040

4

=dxvdEI z , for loading (4.12.2) yV

Integrating the above equations, we get

zy VdxwdEI =30

3

(4.12.3)

yz VdxvdEI =30

3

(4.12.4)

Thus,

)/1(103403.4160001072

5000 2853

03

cmEIV

dxwd

y

z ===

)/1(104468.1480001072

5000 2853

03

cmEIV

dxvd

z

y ===

Integrating the above equations, we have

322

139

0 2110234.7)( CxCxCxxw +++=

652

439

0 2110411.2)( CxCxCxxv +++=

The arbitrary constants are determined by the boundary conditions,

For )(0 xw

0)0(0 ==xw , => 03 =C

0)0(0 ==xdxdw

=> 02 =C

4.12.2


0)()(20

2

==== LxMLxdxwdEI y

=> 0)160001072()200(5000)200( 15

20

2

=+== CcmxdxwdEI y

=> 61 10681.8=C

So, (4.12.5) 26390 10340.410234.7)( xxxw +=

For )(0 xv

0)0(0 ==xv , => 06 =C

0)0(0 ==xdxdv

=> 05 =C

0)()(20

2

==== LxMLxdxvdEI z

=> 0)480001072()200(5000)200( 45

20

2

=+== CcmxdxvdEI z

=> 64 10894.2=C

So, (4.12.6) 26390 10447.110411.2)( xxxv +=

---

Therefore deflections at the free end can be obtained from (4.12.5) and (4.12.6) by

setting : cmx 200=

cmcmxw

116.0)200(10340.4)200(10234.7)200( 26390

=+==

cmcmxv

039.0)200(10447.1)200(10411.2)200( 26390

=+==

--- ANS

(c) Timoshenko beam theory

The displacement equilibrium equations for Timoshenko beam theory for

loading are:

zV

0)( 022

=+ yzyy dxdwGA

dxd

EI (4.12.7)

0)( 20

2

=++ zyz pdxd

dxwd

GA

(4.12.8)

4.12.3


which can be combined into the following equation,

2

2

40

4

dxpd

GAEI

pdxwdEI z

z

yzy = (4.12.9)

Note that is the projection of the cross-sectional area of the thin sheets onto

z-axis. In this case, .

zA2243.0402 cmAz ==

In this particular problem, we have 0=zp . Hence

040

4

=dxwdEI y

is the governing equation.

The concentrated shear loading at the free end produces a constant shear force

along the beam, so we have

zyz VdxdwGA =+ )( 0 (4.12.10)

Substituting the above in equation (4.12.7) yields

zy

y Vdxd

EI =22

(4.12.11)

Integrating equation (4.12.11), we obtain

102

21 BxBxVEI zyy ++= (4.12.12)

Using equation (4.11.10) and (4.11.12), we have

)21(1 10

20 BxBxVEIGA

VGAV

dxdw

zyz

zy

z

z ++==

Integrating the above equation,

212

03

0 )21

61(1)( BxBxBxV

EIx

GAVxw z

yz

z +++= (4.12.13)

Applying boundary conditions to equation (4.12.13)

0)()( ==== Lxdxd

EILxM yy

=> LVB z=0

0)0( ==xy => 01 =B 0)0(0 ==xw => 02 =B

Then equation (4.12.13) becomes

)21

61()( 230 LxxEI

VxGAVxw

y

z

z

z = (4.12.14)

4.12.4


Similarly, the deflection in y-direction due to is yV

)21

61()( 230 LxxEI

Vx

GAV

xvz

y

y

y = (4.12.15)

where is the projection of the cross-sectional area of the thin sheets onto y-axis.

We have .

yA

2363.0602 cmAy ==

The deflection due to is zV

)10340.410234.7(10697.7

))200(21

61(

)16000)(1072(5000

)24)(10068.27(5000)(

26395

23550

xxx

xxxxw

==

And for : yV

)10447.110411.2(10131.5

))200(21

61(

)48000)(1072(5000

)36)(10068.27(5000)(

26395

23550

xxx

xxxxv

===

---

At the free end the respective deflection can be obtained from (4.12.5) and

(4.12.6) by substituting in cmx 200=

cmcmxw

131.0])200(10340.4)200(10234.7[)200(10697.7)200( 263950

===

cmcmxv

049.0])200(10447.1)200(10411.2[)200(10131.5)200( 263950

===

--- ANS

(d) Summary

(1) Deflections at the free end

Simple Beam Theory Timoshenko

Beam Theory Error (%)

(1) NVz 5000= 0.116 cm 0.131 cm 11.5 (2) NVy 5000= 0.039 cm 0.049 cm 20.4

Timoshenko

SimpleTimoshenko

ddd

Error=(%) , where 00 vorwd =

4.12.5


(2) Case 1 ( ) of the above results is more accurate. It is mainly

because of the reason that is smaller than , and, as a result, the bending

behavior for z-direction is more likely to resemble a slender beam than that

for y-direction.

NVz 5000=yI zI

--- ANS

4.12.6


4.13 Consider the structure with a cutout as shown in Fig. 4.17. Find the axial force

distribution in stringers 3-4 and 5-6. Assume that both stringers and webs have

the same material properties of GPaE 70= and GPaG 27= . Also assume that , the thickness of the web mmb 200= mmt 2= , and the cross-sectional area of the stringer . Hint: The zero-stress condition in the web at

the cutout cannot be enforced because of the simplified assumption that shear

stress and strain are uniform across the width of the web. Use the known

condition that the force in the side stringers is at the cutout.

264mmA =

P5.1

1 2

3 4 5 6

P

P

P

L1 L2

Figure 4.17 Cutout in a stringer sheet panel

Solution:

(a) First, we consider the part left hand side of the cutout.

3 4

1.5P

1.5P

L1

F1

F2

F1

b

x

The balance of forces in the x-direction yields

PFF 32 21 =+ (4.13.1) Also we have the differential equation

4.13.1


dxdF

t11= (4.13.2)

)()2/( 2

2

1

1

AF

AF

bEG

dxd = (4.13.3)

Combining equations (4.13.2) and (4.13.3), we have

)(6)()2/(

11

2

2

1

12

12

PFEAbG

AF

AF

bEG

dxFd

t==

=> )(6 121

2

PFEAbGt

dxFd = , let

EAbGt62 =

=> PFdxFd 2

12

21

2

= (4.13.4) The general solution of this second-order differential equation is

PxCxCxF ++= sinhcosh)( 211

(where 2

coshxx eex

+= and 2

sinhxx eex

= ) Applying the boundary conditions,

At (fixed end) => 0=x 0= , that is 0)0(1 ==xdxdF

=> 0)0( 21 === CxdxdF => 02 =C

At => 1Lx = PLxF 5.1)( 11 == => PPLCLxF 5.1cosh)( 1111 =+== =>

11 cosh2 L

PC =

Therefore the solution of the differential equation is

)1cosh2cosh()(

11 += L

xPxF (4.13.5)

--- The axial force distribution in stringers 3-4 can be obtained from (4.13.1) and (4.13.5),

that is

)coshcosh1()(23)(

112 L

xPxFPxF ==

with 016.19)2.0)(1064)(1070(

)102)(1027(6669

39

===

EAbGt )1( m

4.13.2


--- ANS

(b) Next, we consider the part to the right of the cutout:

5

F1

F2

F1x

1.5P

1.5P

From the balance of forces in the x-direction we have

PFF 32 21 =+ (4.13.6) Also we have the differential equation

dxdF

t11= (4.13.7)

)()2/( 2

2

1

1

AF

AF

bEG

dxd = (4.13.8)

Combining equations (4.13.7) and (4.13.8), we have

)(6)()2/(

11

2

2

1

12

12

PFEAbG

AF

AF

bEG

dxFd

t==

=> )(6 121

2

PFEAbGt

dxFd = , let

EAbGt62 =

=> PFdxFd 2

12

21

2

= (4.13.9) The general solution of this second-order differential equation is

PxCxCxF ++= sinhcosh)( 211 Applying the boundary conditions,

at => 0=x PxF 5.1)0(1 == => PPCxF 5.1)0( 11 =+== => PC 5.01 =

at => 2Lx = PLxF == )( 21 => PPLCLPxF =++= 2221 sinhcosh5.0)(

4.13.3


=> 2

2 tanh2 LPC =

The solution of the differential equation is

)2tanhsinh(cosh

2)(

21 += L

xxPxF (4.13.10)

--- The axial force distribution in stringer 5-6 can be obtained from (4.13.6) and (4.13.10),

that is

)tanhsinhcosh1()(23)(

212 L

xxPxFPxF +==

where 016.19)2.0)(1064)(1070(

)102)(1027(6669

39

===

EAbGt )1( m

--- ANS

4.13.4

solution ch4

Documents

walled section

angle section of

original section

maximum bending stress

rotated section

absolute maximum stress

maximum negative stress

ans maximum positive