A uniform beam of a thin-walled angle section as shown in Fig. 9 is

subjected to the bending M_y (M_z =0). 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.

yc z_c

4 2

) 2 /

( h

ht h t

y_c = h⋅ ⋅ =

4 2

) 2 /

( h

ht h t

z_c = h⋅ ⋅ =

--- 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 << h)

3 2

c 3 2 c 3

y th

24 thz 5 12 ) ht z h ( 12 th

I =th + − + + = in which parallel axis theorem for

moments of inertia has been employed and the term 12 ht³

has been neglected.

3 2

c 3 2 c 3

z th

24 thy 5 12 ) ht y h ( 12 th

I =th + − + + =

3 2

1 8

1th yzdA yzdA

yzdA I

A A

yz =

∫

=

∫

+

∫

=

where, ³

2

1 16

| 1 2)

(z th

t y ztdz y

yzdA ^c

c c

c

z h

z c

z h

z c

A

=

=

= ⁻ ₋⁻

∫

−

∫

3

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

By substituting the known values we obtain

) Maximum positive stress:

At z h z_c h

Maximum negative stress:

At 4

The absolute maximum stress is ₂ 4 centroid is the origin of this coordinate system).

--- ANS

4.1.2

Address: 1650, BLVD DE MAISONNEUVE Apt. 904, Montreal, QC H3H2P3, CAN

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.

45o

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 y_c and z_c, respectively.

Because of the symmetry, y_c =0. Assuming t<< , we obtain h

2 2

h ht

2

) 2 2 / h ( t h

z_c = 2 ⋅ ⋅ =

--- ANS (a) Moment of inertia

) 12 ( 2 12 2 2

3

3 th

h I_y = × t =

) 3 ( 2 3 2 2

3

3 th

h I_z = × t =

=0

=

∫

^yzdA

I_xy (this is always true for symmetric sections)

(b) Set up a new coordinate system (y, z). Using equation (4.25) in the textbook,

'

' 2

2 z

I I I

M I M y I I I I

M I M I

yz z y

z yz y z

yz z y

y yz z y

xx −

+ −

−

= − σ

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

12M I z

M

3 y y

y

xx = =

σ

--- ANS Maximum positive stress is at

2 2

z= h , => 3 2₂ th

M_y

xx = σ

Maximum negative stress is at

2 2

z=− h , => 3 2₂ th

M_y

xx =− σ

The absolute maximum stress is 3 2₂ th

M_y

xx = σ

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

th z 12M₃^y

xx = =

σ => z=0

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

,

4 27 2

3

th M th

M_{yrotate yorigin}

xx = =

σ

=> 1.59

2 12

27 M

M

origin , y

rotate ,

y = =

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

--- ANS

4.2.2

Address: 1650, BLVD DE MAISONNEUVE Apt. 904, Montreal, QC H3H2P3, CAN

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

the shear force V_z ≠0, while V_y =0. Find the bending stresses in the stringers for the same bending moment M_y. Which section is most effective in bending?

Figure 4.20 Stringer-web section

Figure 4.21 Stringer-web section

Figure 4.22 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

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

2

The stresses at the stringer are

①. At z=h,

(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

2

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

4.3.2

Address: 1650, BLVD DE MAISONNEUVE Apt. 904, Montreal, QC H3H2P3, CAN

①. At z=h,

(1) Again, when neglecting the effects of webs, the centroid is located at the middle of the vertical web.

(2) Moment of inertia

2

The stresses at the stringer are At z=h, y=−h,

(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

Address: 1650, BLVD DE MAISONNEUVE Apt. 904, Montreal, QC H3H2P3, CAN

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

, .

=0

My M_z ≠0

Figure 4.21 Stringer-web section

Figure 4.22 Stringer-web section

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.

(2) Moment of inertia

2 2 2

4 ) (

4 A h Ah

z A I

i i i

y =

∑

= × =

2

The stresses in stringers 2 and 3 are

At 2

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

(2) Moment of inertia

2

The stress in stringer 1 is At z=h, y =−h

4.4.2

Address: 1650, BLVD DE MAISONNEUVE Apt. 904, Montreal, QC H3H2P3, CAN

Ah h M

Ah h z M Ah y M Ah

M_{z z z z}

xx ( 2 ) 2

2

2 ^{2 2}

2 + = − + =−

σ =

Stringer 2:

At z=h, y=0,

Ah h M

Ah z M Ah y M Ah

M_{z z z z}

xx (0 ) 2

2

2 ^{2 2}

2 + = + =

σ = Stringer 3:

At z=−h, y=0,

Ah h M

Ah z M Ah y M Ah

M_{z z z z}

xx (0 ) 2

2

2 ^{2 2}

2 + = − =−

σ = Stringer 4:

At z=−h, y=h,

Ah h M

Ah h z M Ah y M Ah

M_{z z z z}

xx (2 ) 2

2

2 ^{2 2}

2 + = − =

σ =

--- 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

4.5 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

cm N M_y =−500,000 ⋅

cm N M_z =200,000 ⋅ . 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

cm 5 . ) 54 4 1 2 4 (

) 200 1 200 2 ( A

y A y

i i i

i i

c =

+ + +

× +

= ×

=

∑

∑

cm 9 . ) 40 4 1 2 4 (

) 100 4 50 1 ( z

z A z

i i i

i i

c =

+ + +

× +

= ×

=

∑

∑

(b) The moment of inertia

4

2 2

2 i

2 i i y

cm 240901

) 909091 .

40 100 ( 4 ) 909091 .

40 50 ( 1 ) 909091 .

40 )(

2 4 ( z A I

=

−

× +

−

× + +

=

=

∑

Similarly,

4 i

2 i i

z Ay 87273cm

I =

∑

=

4.5.1

Address: 1650, BLVD DE MAISONNEUVE Apt. 904, Montreal, QC H3H2P3, CAN

4 i

i i i

yz Ayz 14545cm

I =

∑

=−

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

Ai y_i z_i

Stringer

No. (cm²) (cm) ^(cm)

2 i iz A

) (cm⁴

2 i iy A

) (cm⁴

i i iy z A

) (cm⁴

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: z

I I I

M I M y I I I I

M I M I

yz z y

z yz y z

yz z y

y yz z y

xx 2 − 2

+ −

−

= −

σ , and

cm N M_y =−500,000 ⋅

cm N M_z =200,000 ⋅ .

909 4

. 24090 cm I_y =

727 4

. 87272 cm I_z =

455 4

. 14545 cm I_yz =−

We obtain σ_xx =−1.298y−21.54z

Therefore the bending stresses in the stringers are:

yi

Stringer

No. (cm)

zi

) (cm

σxx

) / (N cm²

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

4.6 Find the neutral axis in the tin-walled section of Fig. 4.23 for the loading given in Problem 4.5.

cm N M_y =−500,000 ⋅

cm N M_z =200,000 ⋅ . 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.

cm 5 . 54

y_c = , z_c =40.9cm

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.,

0 z 538 . 21 y 298 .

xx =−1 − =

σ

On the cross-section, this equation represents the line passing through the centroid with y=−16.59z and an angle

45o

. 3 59) . 16 ( 1 tan y)

( z

tan ¹ − = ¹ =

= ^{− −}

α

--- ANS

4.6.1

Address: 1650, BLVD DE MAISONNEUVE Apt. 904, Montreal, QC H3H2P3, CAN

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

cm 67 . 4 26 3

80 4 A

y A y

i i i

i i

c =

×

= ×

=

∑

∑

cm 33 . 4 13 3

40 4 A

z A z

i i i

i i

c =

×

= ×

=

∑

∑

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

N PL

M_y =−2 =−2(200)(500)=−200000 ⋅ (M_y is positive in positive y) cm

N PL

M_z =−2 =−2(200)(500)=−200000 ⋅ (M_z is positive in negative z) (c) Set up the coordinate system (x,y,z) with the origin at the centroid.

Moment of inertia (see table below for details):

4 2

2 i

2 c i

y Az 4(2 13.33 26.67 ) 4266cm

I =

∑

= × + =

4 2

2 i

2 c i

z Ay 4(2 26.67 53.33 ) 17067cm

I =

∑

= × + =

4 i

c c i

yz Ay z 4266cm

I =

∑

=−

Ai y_i z_i

Stringer

No. (cm²) (cm) ^(cm)

2 i iz A

) (cm⁴

2 i iy A

) (cm⁴

i i iy z A

) (cm⁴

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 z

I I I

M I M y I I I I

M I M I

yz z y

z yz y z

yz z y

y yz z y

xx 2 − 2

+ −

−

= −

σ ,

we obtain σ_xx =−31.25y−78.125z

and the bending stresses in the stringers are:

yi

Stringer

No. (cm)

zi

) (cm

σxx

) / (N cm²

1 -26.67 26.67 -1250 2 -26.67 -13.33 1875 3 53.33 -13.33 -625

--- ANS (e) Neutral plane by angle α.

Neutral plane is located at the position where bending stresses vanish under this particular loading. We have

0 125 . 78 25 .

31 − =

−

= y z

σxx

It is the line passing through the centroid with y =−2.5z 8o

. 21 5) . 2 ( 1 tan y)

( z

tan ¹ − = ¹ =

= ^{− −}

α

--- ANS

4.7.2

Address: 1650, BLVD DE MAISONNEUVE Apt. 904, Montreal, QC H3H2P3, CAN

In document C T Sun Mechanics of Aircraft Structures Solution (Page 100-116)