Aircraft Structures Notes 1

5/26/2009

ME4212 Aircraft Structures (formerly Mechanics of Thin-Walled Structures)

T. E. Tay, Department of Mechanical Engineering, National University of Singapore

Why study thin-walled structures?


1

5/26/2009





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5/26/2009

Aircraft Structural Analysis

Fracture Mechanics Linear Elastic Fracture Elasto-Plastic Fracture Damage Tolerance Fatigue Delamination

Mechanics of ThinWalled Structures Beams Plates Shells Stringers and Panels Multi-cell Torque Boxes Tapered Structures Buckling and Instability Vibration

Mechanics of Composite Materials Stress-Strain Relations of FiberReinforced Composites Laminate Design and Architecture Hygrothermal Effects Durability and Failure of Composites


Course Overview: Torsion Bending Idealized Beams

T.E. Tay Professor Room EA-7-17 6516-2887 6516 2887 mpetayte@nus.edu.sg

Multi-cell Sections & Tapered Beams Circular & Rectangular Plates Instability of Columns & Plates

S.L. S L Toh Associate Professor

Energy Methods in Instability


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Torsion • Non-Circular Shafts • Thin-Walled Open Sections • Thin-Walled Closed Sections • Warping of Unrestrained Sections

B di Bending • Unsymmetric Bars • Thin-Walled Open Sections • Thin-Walled Closed Sections • The Shear Center

Idealized dea ed Beams ea s – Bending e d g & Torsion oso Multi-cell Sections – Bending & Torsion Tapered Beams & Beams with Varying Moments of Area


Books: Mechanics of Materials, A.C. Ugural, McGraw-Hill, 1991 Ch 6.8, 6 8 6.9: 6 9: Torsion, Torsion Torsion of Thin-Walled Thin Walled Shafts Ch 7: Bending of Beams Ch 8.6, 8.8: Unsymmetric Bending, Shear Center

Mechanics of Materials, Roy R. Craig Jr., Wiley & Sons, 1996 Ch 4, 4 4.7, 4 7 4.8: 4 8: Torsion, Torsion Torsion of Noncircular Sections Ch 6, 6.6, 6.8-6.10, 6.12: Bending, Unsymmetric Bending, Shear Flow, Shear Center


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5/26/2009

Books (advanced): Mechanics of Elastic Structures, J.T. Oden and E.A. Ripperger, McGraw-Hill, 1981 Aircraft Structures for Engineering Students, T.H.G. Megson, Edward Arnold, 1990 (2nd Edition) Analysis of Aircraft Structures, B.K. Donaldson, McGraw-Hill, 1993 Aircraft Structures & Systems, R. Wilkinson, Addison-Wesley Longman, 1996


5

Torsion


Torsion of Circular Sections Bars of Circular Sections subjected to torsion

Radial lines on cross-section remains straight. But square elements are distorted.


1

The shear strain is defined:

γ = Δx

r

Δφ

rΔ φ Δx

And the shear stress is

τ = Gγ = lim Gr Δx→ 0

= Gr

dφ dx

Δφ Δx

= Gr θ

where G is the shear modulus.

θ=

dφ dx

is the “rate of twist”, and is a constant.


The twisting moment (i.e. torque) is

T =

∫∫ r (τ dA ) A

dF

=G

τ

r ΔA

dφ dx

= GJ

∫∫ r

2

dA

A

dφ dx

T = GJ θ

(1)

And the shear stress

τ = Gr θ =

Tr J

(2)

where A is the area of the cross-section, and J is the polar second moment of area.


2

Torsion of Non-Circular Sections Bars of Non-Circular Sections subjected to torsion




3



Consider a cross-section of arbitrary shape: A is an arbitrary point.

It is displaced to A’ under torque. z

Note: There is no change in shape. Hence

There is a point O, that has no displacement when the torque is applied, called the centre of twist.

γ yz = 0 τ yz = 0

A’

Let’s place the origin of our coordinate system at O.

O

A

y


4

z

In general, we write the displacements:

v = −θ xz w = θ xy

Horizontal displacement of A.

(3a) (3b)

− v = φz

where θ is the rate of twist. We further assume that the axial displacements are given in this form:

u = θψ ( y , z )

φ O

Vertical displacement of A.

A’

w = φy A

y

(3c)

where ψ ( y , z ) is the warping function, yet to be determined.


Axial displacements depend on the warping function

ψ ( y, z )

z Our objective is to determine u, v and w, the complete displacement field.

y

x


5

Start with the strain-displacement (compatibility) relations:

∂w ∂u ∂v , εy = , εz = , ∂z ∂x ∂y ∂u ∂v ∂v ∂w ∂u ∂w + = + , γ yz = + , γ xz = ∂y ∂x ∂z ∂y ∂z ∂x

εx = γ xy

Substituting Eqns (3) into above,

ε x = ε y = ε z = γ yz = 0

(4a)

⎛ ∂ψ ⎞ − z ⎟⎟ γ xy = θ ⎜⎜ ⎝ ∂y ⎠

(4b)

⎛ ∂ψ ⎞ + y⎟ ∂ z ⎝ ⎠

(4c)

γ xz = θ ⎜

γ yz = 0 implies cross-section does not distort in its own plane (i.e. all points are simply rotated as in rigid-body rotation).


The relevant force equilibrium equation

∂σ x ∂τ xy ∂τ xz + + =0 ∂x ∂y ∂z with

τ xy = Gγ xy , τ xz = Gγ xz , σ x = 0 and Eqns (4) yields

∂ 2ψ ∂ 2ψ + =0 ∂y 2 ∂z 2 or

∇ 2ψ = 0

(5)

This is the Laplace Equation. Clearly, the warping function ψ satisfies both compatibility and equilibrium. In order to obtain a solution, we need to specify boundary conditions.


6

Consider a small element on the boundary, which is usually traction free. Force equilibrium in the axial direction provides the equation of the boundary condition.

dx

τ xy

dz

τ xy dzdx − τ xz dydx = 0 z

(6)

y

dy

τ xz

∴τ xy dz = τ xz dy

x In general, it may be necessary to solve the Laplace Equation numerically.


Stress Functions for Torsion Prandtl assumed a stress function Φ which is twice differentiable, and has the property:

∂Φ ∂z ∂Φ τ xz = − ∂y

τ xy =

(7a) (7b)

Substituting (7) into (4b) and (4c), differentiating (4b) with respect to z, and (4c) with respect to y,

−

⎞ ∂ ⎛ ∂Φ ⎞ ∂ ⎛ ∂ψ −θ z ⎟ ⎜ ⎟ = G ⎜θ ∂z ⎝ ∂z ⎠ ∂z ⎝ ∂y ⎠

(8a)

∂ ⎛ ∂Φ ⎞ ∂ ⎛ ∂ψ ⎞ +θ y ⎟ ⎜ ⎟ = G ⎜θ ∂y ⎝ ∂y ⎠ ∂y ⎝ ∂z ⎠

(8b)


7

Ludwig Prandtl Father of Modern Fluid Mechanics (1875-1953)

Subtracting (8b) from (8a) results in

∂ 2Φ ∂ 2Φ + = −2Gθ ∂y 2 ∂z 2

(9)

or

The solution to this equation satisfies both compatibility and equilibrium. In order to obtain a solution, we need to specify boundary conditions.

∇ 2 Φ = −2Gθ

For the boundary condition, substituting (7) into (6), we obtain

∂Φ ∂Φ dz − dy = 0 ∂z ∂y or dΦ = 0

(10)

This means that at the boundary, Φ is a constant. Since the value of Φ can be arbitrary, we shall select Φ = 0 for convenience.


8

z The torque about the x-axis due to the shear stresses is given by

τ xz dA

T = ∫∫ (τ xz y − τ xy z )dA

(11)

A

y

τ xy dA

Area dA

z y

O


Introducing Eqn (7) , and noting that dA = dydz,

⎛ ∂Φ ∂Φ ⎞ T = − ∫∫ ⎜⎜ y+ z ⎟dydz ∂z ⎟⎠ ⎝ ∂y ∂Φ ∂Φ = − ∫∫ ydydz − ∫∫ zdydz ∂y ∂z

(12)

⎛ B ∂Φ ⎞ ⎛ D ∂Φ ⎞ z dz ⎟⎟dy − ∫ ⎜⎜ ∫ y dy ⎟⎟dz = − ∫ ⎜⎜ ∫ ⎝ A ∂z ⎠ ⎝ C ∂y ⎠

(

)

(Integration by parts)

(

)

= − ∫ (Φ B z B − Φ A z A ) − ∫ Φ dz dy − ∫ (Φ D y D − Φ C yC ) − ∫ Φ dy dz (13)


9

z

Φ A Φ B ΦC Φ D are values at boundary points, which we have chosen to set to zero. Hence the equation reduces to

B

T = 2 ∫∫ Φdydz

(14)

D

C

y

O

A T. E. Tay, Department of Mechanical Engineering, National University of Singapore

If we retain the expression

T = GJθ

(15)

where now we define the symbol J as the torsional constant, then the formula for J is

J=

2 Gθ

∫∫ Φ dydz

(16)

The product GJ is known as the torsional stiffness. Only for the very special case of the circular cross-section is J equal to the second polar moment of area.


10

The Membrane (Soap Bubble) Analogy Good for visualizing Φ . Equilibrium equation for small deflection w of a flat membrane subjected to internal pressure p closely resembles the torsion equation. ∂2w ∂2w p + 2 =− 2 ∂y ∂z TM

(17)

Pressure p Tension TM w z (into plane)

y


Comparing Eqns (9) and (17),

∂ 2Φ ∂ 2Φ + = −2Gθ ∂y 2 ∂z 2

(9)

∂2w ∂2w p + =− ∂y 2 ∂z 2 TM

(17)

p is analogous to 2Gθ TM w is analogous to Φ Furthermore, according to Eqn (14), the volume under the membrane is proportional to the torsion T.

T = 2 ∫∫ Φdydz

(14)


11

The slope of the membrane is proportional to the shear stresses.

∂Φ ∂w ⇔ ∂z ∂z ∂Φ ∂w ⇔− τ xz = − ∂y ∂y

τ xy =

Local slope

∂w ∂y

w z (into plane)

y Similarly for

∂w ∂z


If the section is hollow,

dΦ = 0 at both inner and outer boundaries. z y

Φ = constant

∂w =0 ∂y

Φ=0

z (into plane)

y


12

Torsion of Thin-Walled Open Sections Consider a narrow rectangular section: z

z y x Profile of w or Φ in x-z plane

∂Φ = 0 over much of y ∂y

∂Φ = 0 ∂z only at the centerline of the section (at z = 0).

x

y Profile of w or Φ in x-y plane


∂Φ = 0 , Eqn (9) reduces to Since ∂y

∂ 2Φ = −2Gθ ∂z 2

∂ 2Φ ∂ 2Φ + = −2Gθ ∂y 2 ∂z 2

(9)

(18)

Integrating twice with respect to z yields

Φ = −G θz 2 + C1z + C 2

(19)

Applying the boundary condition that Φ = 0 at z = + t/2 and – t/2, we obtain C 1 = 0 and C 2 = G θ t 2 / 4 .

⎛ t2 ⎞ ⎟ ∴ Φ = − G θ ⎜⎜ z 2 − 4 ⎟⎠ ⎝

(20)


13

The stresses are therefore

∂Φ = − 2G θz ∂z ∂Φ =− =0 ∂y

τ xy =

(21a)

τ xz

(21b)

Eqn (21a) shows that τxy varies linearly with z and is zero at the centerline of the section (at z =0).


From Eqn (16),

J =

2 Gθ

= −∫

∫∫ Φ dydz

b/2

−b / 2

∴J =

∫

t/2

−t / 2

⎛ 2 t2 ⎞ ⎜⎜ 2 z − ⎟⎟ dzdy 2 ⎠ ⎝

bt 3 3

The torsional constant for a thin rectangular section.

(22)

2T z J

(23)

The shear stress

τ xy = − 2 G θ z = −

has maximum values at z = + t/2 and – t/2, i.e.

τ xy MAX = ±

Tt J

(24)


14

Linear distribution of shear stress

τ

xy

across thickness:

z y


Now, consider the warping displacements: From Eqns (4),

⎛ ∂ψ ⎞ τ xy = G θ ⎜⎜ − z ⎟⎟ ⎝ ∂y ⎠

(25a)

From Eqn (21a), τxy = - 2Gθ, so Eqn (25a) becomes

− 2z =

∂ψ −z ∂y

⎛ ∂ψ ⎞ + y⎟ ⎝ ∂z ⎠

τ xz = G θ ⎜

(25b)

From Eqn (21b), τxz = 0, so Eqn (25b) becomes

∂ψ = −y ∂z

∂ψ = −z ∂y

Upon integration,

ψ = − yz + C 1 ( y )

Upon integration,

ψ = − yz + C 2 ( z ) Clearly, C1(y) = C2(z) = 0 and

ψ = − yz

(26)


15

The warping displacement is therefore

u = − θ yz

(27)

A hyperbolic-paraboloid surface

z y x


General Open Sections In general,

s=0

J = z y

Bt 3 3

s=B

Linear distribution of shear stress τ xy across thickness


16

b2 t2 For a section consisting of n elements,

J =

b1

1 n ∑ bi ti3 3 i =1

t1

b4

t3

t4

b3

(28)

Maximum shear stress is

τ iMAX =

Ti tiMAX Ji

t5

(29)

where the proportion of torque T acting on element i is

⎛J ⎞ Ti = ⎜ i ⎟ T ⎝ J ⎠

b5

(30)

b6

J =

t6

b1t13 b 2 t 23 b 3 t 33 b 4 t 43 b 5 t 53 b 6 t 63 + + + + + 3 3 3 3 3 3


Torsion of Thin-Walled Closed Sections Consider a thin-walled closed section: z y

Local slope

z (into plane)

∂w almost constant ∂y

y


17

Closed Sections

z y

Constant distribution of shear stress τ xy across thickness


Taking force equilibrium in the axial direction,

τ 1t1 = τ 2t2

Δx

(31)

z

τ 1 t1 Δ x 1

t1 t2

τ 2t2 Δ x

y x

2

Convenient to define shear flow

q =τ t

(32)

∴ q1 = q 2 = constant


18

The total moment then equals the torque,

T = ∫ rqds = q ∫ rds

(33)

O r

qds s=0

Moment about an arbitrary point O is therefore qrds, where r is the perpendicular distance.

Force due to shear flow on a small element of wall is qds.

Choose an arbitrary point on the wall, at s = 0.


As we move from s = 0 to s = s1, the area traced is known as the sectorial swept area, and is defined as

1 ω ( s1 ) = 2

s = s1

∫ rds

(34)

s =0

O

qds

r s=0


19

As we move from s = 0 to s = s1, the area traced is known as the sectorial swept area, and is defined as

1 ω ( s1 ) = 2

s = s1

∫ rds

s =0

O

s=0 The total area enclosed by the centerline of the wall is

Ω =

1 2

∫

and the torque, from Eqn (33) is

rds

(35)

T = 2Ω q

(36)


The section does not distort under torsion. Points on the wall experience a rotation φ = θ x about the center of twist.

η θx Center of twist O

η is the tangential displacement component in the s direction.

η = rθ x

(37)


20

Consider a small element of the wall

Δx

z

y

Δs

x


Δx

z

Δs Δη

y x

Displacement in the s direction is Δ η


21

Δx

z

y

Δs

x

Δu Displacement in the x direction is Δ u


Δx

z

Δs

x

Therefore, the shear strain in the wall is

∂u ∂η + ∂s ∂x ∂u = + rθ ∂s

y

γ xs =

(38)


22

Since γ xs = τ xs / G it follows that

∂u τ xs = − rθ ∂s G

(39)

Multiplying both sides by ds and integrating around the total periphery S of the tube, q ds (40) ∫ du = G ∫ t − θ ∫ rds The integral on the left-hand-side is zero, since u at s = 0 and u at s = S are the same (i.e. same point), so that the difference is zero. Substituting for Ω from Eqn (35),

ds − 2Ωθ t q ds ∴θ = 2Ω G ∫ t

0=

q G

∫

(41)


Substituting for q from Eqn (36),

θ=

T 4Ω 2G

∫

ds t

(42)

Since T = GJθ , the torsional constant J for a closed section is

J =

4Ω 2 ds ∫t

(43)

If the thickness t is constant,

J =

4Ω 2 t S

(44)


23

Hybrid Sections

J=

4Ω 2 1 n + bi ti3 ds 3 ∑ i =1 ∫t

(45)


Example 1 The round tube shown has an average radius R and a wall thickness t. Compare the torsional strength and rigidity of this tube with that of a similar tube which is slit along its entire length. Assume R/t = 20.

R

R t

Closed tube

t

Open section


24

Open section

Closed tube

4Ω 2 ds ∫t 4π 2 R 3t = = 2πR 3t 2π

bt 3 3 2πRt 3 = 3

J Closed = Torsional constant

J Open =

J Closed ⎛R⎞ = 3⎜ ⎟ J Open ⎝t ⎠

2

= 1200 For a given angle of twist, the closed section resists 1200 times the torsion of the open section.

For a given torque, the open section twists through an angle 1200 times as great as the closed section.


Open section

Closed tube

Maximum shear stress

Tt J 3T = 2πRt 2

τ Open MAX =

T 2Ω t T = 2πR 2 t

τ Closed =

τ Open ⎛R⎞ = 3⎜ ⎟ τ Closed ⎝ t ⎠ = 60 For a given torque, the shear stress in the open section is 60 times as high as in the closed tube.

If the allowable shear stresses are equal, the closed tube is 60 times as strong as the open section.


25

Example 2 A torque of T = 6000 Nmm is applied to the section shown. Determine the rate of twist and maximum shear stress. Assume shear modulus G = 80 GPa. 15 2

2

19

1 2 1 Dimensions in mm 20 T. E. Tay, Department of Mechanical Engineering, National University of Singapore

Total area enclosed:

∫

Ω = 20 × 19 = 380 mm 2

ds 20 19 20 19 = + + + = 58 .5 t 1 1 2 2

The torsional constant is

4Ω 2 1 3 + bt ds 3 ∫t 2 4(380 ) 1 3 = + (15 )(2 ) = 9.913 × 10 3 mm 4 58 .5 3

J =

Since T = GJθ , the rate of twist is

θ=

6000 9 .913 × 10 3 80 × 10 3

(

)(

)

= 7 .57 × 10 − 6 rad/mm


26

The contributions to the torsional constant are

4 (380 ) = 9 .873 × 10 3 mm 4 58 .5 1 3 = (15 )(2 ) = 40 mm 4 3 2

J Closed Part = J Fin

}

Compare !

The proportions of torque carried are

9 .873 × 6000 = 5976 Nmm 9 .913 40 = × 6000 = 24 .2 Nmm 9913

TClosed Part = TFin

The maximum stresses are

τ Closed Part MAX = τ Fin MAX =

TClosed Part 2 Ω t MIN

=

5976 = 7 .86 MPa 2 (380 )(1)

TFin t (24 .2 )(2 ) = = 1 .21 MPa J Fin 40


If the section is slit longitudinally at A as shown, what are the rate of twist and maximum shear stress? 15 2

2

A 19

1 2 1 Dimensions in mm 20 T. E. Tay, Department of Mechanical Engineering, National University of Singapore

27

The torsional constant is

J =

[

]

1 (19 )(1)3 + (20 )(1)3 + (19 )(2 )3 + (35 )(2 )3 = 157 mm 4 3

and

(6000 )(2 ) = 76 .4 MPa Tt MAX = J 157 T 6000 θ= = = 4 .78 × 10 − 4 rad/mm GJ 80 × 10 3 (157 )

τ MAX =

(

)

Note the increase in stress and rate of twist.


Warping of Unrestrained Thin-Walled Sections

Recall from Eqn (39) that

∂u τ xs = − rθ ∂s G Multiplying both sides by ds and integrating from s = 0 to some other point on the wall,

u − u0 =

1 τ xs ds − θ ∫ rds G∫

(46)

where u0 is the displacement of the point s = 0 in the x-direction.


28

Recall the sectorial swept area (Eqn (34)), defined as ω ( s ) =

1 2

1 (47) ∴ u − u 0 = ∫ τ xs ds − 2θω ( s ) G where r is measured from an arbitrary point O (called the pole).

∫ r ds

O r s=0 The equation, valid for both open and closed sections, can be rewritten as 1

G∫ u = u − u0 u=

τ xs ds − 2θω( s)


For closed sections, since τ xs = q / t , we have

u=

q ds − 2θω G∫ t

(48)

Here, u is a function of s, and is zero when the integration is completed around the closed tube.

O

u =0 s=0


29

For open sections, the contribution of the shearing strain γxs to the warping displacement u is negligible because the walls are very thin, so that

∫τ

xs

ds ≈ 0

(49)

Hence the equation reduces to

u = −2θω

(50)


Example 3 The closed box section shown is subjected to a torque T. Determine the warping displacements. tb

Rate of twist:

1

2

T ds 2 ∫ 4Ω G t 2 3 4 1 T ⎛ ds ds ds ds ⎞ = 2 2 ⎜⎜ ∫ + ∫ + ∫ + ∫ ⎟⎟ 4b h G ⎝ 1 tb 2 th 3 tb 4 t h ⎠ T ⎛ 2h 2b ⎞ = 2 2 ⎜⎜ + ⎟⎟ 4b h G ⎝ th tb ⎠

θ= th

T th

tb

3

4

h

=

T ⎛h b⎞ ⎜ + ⎟ 2b 2 h 2G ⎜⎝ t h tb ⎟⎠

b


30

Select the pole at O the center of section, and starting point at I. 1

2

For closed sections, the relative warping displacement is

q ds − 2θω G∫ t

u= O

1

I

∴ u1 =

q ds − 2θω I 1 G ∫I t h

We find that q = 3

4

T T and = 2Ω 2bh

⎛ b ⎞⎛ h ⎞ ⎝ 2 ⎠⎝ 2 ⎠

Also, 2ωI 1 = ⎜ ⎟⎜ ⎟ = Therefore, u1 =

bh 4

T ⎛h b⎞ ⎜ − ⎟ 8bhG ⎜⎝ th tb ⎟⎠

1

ds

∫t I

h

=

h 2th

Interestingly, there will be no warping if

h b = t h tb


bh bh + 4 2 3bh = 4

2ωI 2 =

Now, 1

2

2

and O

3

∫ I

I

4

1

=

u2 = =

2

ds ds ds = + t ∫I th ∫1 tb h b + 2th tb

⎛ h b ⎞ 3bh T ⎛ h b⎞ T ⎜⎜ + ⎟⎟ − 2 2 ⎜⎜ + ⎟⎟ 2bhG ⎝ 2t h tb ⎠ 2b h G ⎝ t h tb ⎠ 4 T ⎛b h⎞ ⎜ + ⎟ 8bhG ⎜⎝ tb t h ⎟⎠

∴ u 2 = −u1 Similarly, u3 = u1 , u4 = −u1 T. E. Tay, Department of Mechanical Engineering, National University of Singapore

31

Due to symmetry, we know that the axial displacement at I is zero. Since u1 = u1 −u I , we have u1 = u1.

2 1

I 3 4


Example 4 What would the warping displacements be if the section is slit longitudinally along its entire length at point I ? 1

2

Torsional constant: J =

O

I

Rate of twist: θ =

(

2 3 hth + btb3 3

)

T 3T = GJ 2G hth3 + btb3

(

)

For an open section, u = −2θω 3

4


32

1

2

O

I

3

From I to 1, u1 = −

bh 3T 3bhT ⋅ =− 4 2G hth3 + btb3 8G hth3 + btb3

From I to 2, u2 = −

9bhT = 3u1 8G hth3 + btb3

(

(

)

)

)

From I to 3, u3 = 5u1

5bh ⎞ ⎛ ⎜ 2ωI 3 = ⎟ 4 ⎠ ⎝

From I to 4, u4 = 7u1

7bh ⎞ ⎛ ⎜ 2ωI 4 = ⎟ 4 ⎠ ⎝

From I to I’, uI ' = 8u1

8bh ⎞ ⎛ ⎜ 2ωI 3 = ⎟ 4 ⎠ ⎝

4

(


2 1 Longitudinal slit

I 3 4


33

Example 5 Let us return to the closed section. What will happen if we select another point O’ as pole? 2

1

From I to 1, 2ωI 1 = 1

and

∫ I

O’

I

u1 =

3

4

bh 2

ds h = t 2th

⎛ h b ⎞ bh T ⎛ h ⎞ T ⎜⎜ ⎟⎟ − 2 2 ⎜⎜ + ⎟⎟ 2bhG ⎝ 2t h ⎠ 2b h G ⎝ t h tb ⎠ 2

=−

T ⎛b⎞ ⎜ ⎟ 4bhG ⎜⎝ tb ⎟⎠


From I to 2, 2ωI 2 = bh 2

∫

and 2

I

1

ds h b = + t 2th tb

⎛h b⎞ T ⎛ h b⎞ T ⎜ ⎜ + ⎟bh + ⎟− 2bhG ⎜⎝ 2t h tb ⎟⎠ 2b 2 h 2G ⎜⎝ t h tb ⎟⎠ T ⎛h⎞ ⎜ ⎟ =− 4bhG ⎜⎝ t h ⎟⎠

u2 = O’

I

3

4

Similarly, we can show that

u3 = −u2 , u4 = −u1

Try it yourself !


34

For illustration purposes, let

T h b = 1.0 = 30, = 10 and 2bhG th tb

Then u1 = −2.5, u2 = −7.5, u3 = 7.5, u4 = 2.5 Present Example 5:

From Example 3: 2

2

1

1

I

I 3

3 4

4


Hence changing the pole does not alter the problem (i.e. the stresses are also unaffected), but merely changes the reference plane from which the axial displacements are measured. In this example, the new reference plane is rotated about the O’I axis.


35

Bending


Bending of Bars Bars of Non-Circular Sections subjected to pure bending

Bending Moment M


1

Pure bending occurs only when the force F acts through the Shear Center E.

Built-in end

When the force F acts through any other point, there is combined bending and twisting.


x-y-z are the centroidal axes Let the loads Vz and Vy act through the shear centre E. Hence pure bending.

z

Vz

y

Centroid O

Vy

x

Shear Centre

E

t


2

x-y-z are the centroidal axes

Mz z

The shear loads give rise to the applied moments (positive right hand screw rule)

My

y

Centroid O

x

t


Contribution of Vz to bending moment

Hence

Vz

dM

y

Vz =

= V z dx ∂M y ∂x

(51a)

dx z

+ dMy E x

y (into page)


3

Contribution of Vy to bending moment Hence

Vy

dM z + V y dx = 0 ∴ Vy = −

∂M z ∂x

(51b)

dx y

E

+ dMz

x z (out of page)


Let σx denote the normal stress due to bending. Since there is no nett axial load,

∫σ z

Area dA = t ds

σx

x

dA = 0

(52a)

A

dx y x

ds

t


4

Consider a small element of the wall (in positive z)

The moment about the y-axis contributes to the normal stress σx due to bending. Consider applied moment about the y-axis

Area dA = t ds

z

σx

dx

y

ds

My

x z

t


Contribution of axial force dF due to σx to the bending moment Area dA = t ds

M y = ∫ σ x z dA

(52b)

A

(Tensile on positive z) σx dx z

z

+ dMy

x y (into page)


5

The moment about the z-axis also contributes to the normal stress σx due to bending.

Consider a small element of the wall (in positive y)

z

Consider applied moment about the z-axis

σx dx

ds

y

Mz

x

t y T. E. Tay, Department of Mechanical Engineering, National University of Singapore

Contribution of axial force dF due to σx to the bending moment Area dA = t ds

M z = − ∫ σ x y dA

(52c)

A

(Compressive on positive y) σx dx y

y

+ dMz

x z (out of page)


6

Plane sections remain plane assumption implies that the axial strain εx (and axial stress σx ) is a linear function of y and z. The cross-section rotates about a neutral axis whose orientation is dependent upon the loading and geometry of the cross-section. z

λ

y

x

Cross-section

Side view along the neutral axis T. E. Tay, Department of Mechanical Engineering, National University of Singapore

Let σ x = C1 + C 2 y + C 3 z and substituting this into Eqns (52) and solving for the constants C1, C2, C3, we have C1 = 0 and

⎛ M z I yy + M y I yz ⎞ ⎛ M I + M z I yz ⎟ y + ⎜ y zz 2 ⎟ ⎜ I I − I2 yy zz yz ⎝ I yy I zz − I yz ⎠ ⎝

σ x = − ⎜⎜

⎞ ⎟z ⎟ ⎠

(53)


7

Eqn (53) can be rewritten in the form

σx = −

M z y M yz + I zz I yy

⎛ M I ⎜ M z + y yz ⎜ I yy Mz = ⎝ ⎛ I2 ⎞ ⎜ 1 − yz ⎟ ⎜ I yy I zz ⎟⎠ ⎝ M I ⎛ ⎜⎜ M y + z yz I zz My = ⎝ 2 ⎛ ⎞ I ⎜ 1 − yz ⎟ ⎜ I yy I zz ⎟⎠ ⎝

⎞ ⎟ ⎟ ⎠

⎞ ⎟⎟ ⎠


The neutral axis can be located by setting σx = 0 in Eqn (53), and find, for angle λ,

λ

z y

tan λ =

z M z I yy + M y I yz = y M y I zz + M z I yz

(54)


8

General distribution of normal bending stress σ x

Compression z

Neutral axis

y Tension

x


If the section is symmetric about either the y- and zaxes, Iyz = 0 and Eqn (53) reduces to

σx = −

M zy M yz + I zz I yy

(55)


9

A word about sign convention for moments:

σx = − z

M zy M yz + I zz I yy Applied Moment My Positive into page

x y (into page)

This causes tensile (+) stresses for positive z

y Applied Moment Mz Positive out of page x z (out of page)

This causes compressive (-) stresses for positive y

Note that Equation (53) for normal bending stresses is valid for all general sections (whether solid or thin-walled, open or closed sections).


10

Example 6 The simply-supported beam has an unsymmetrical cross-section. Locate the neutral axis and determine the largest tensile and compressive stresses in terms of load Q and where they occur. (Assume Q acts through the shear center.) Q

Q

z 20 1000

y

1500

80

20

60 Cross-section T. E. Tay, Department of Mechanical Engineering, National University of Singapore

The maximum bending moment occurs at the point of application of load Q and is given by My = - (0.6Q)(1000) = -600 Q Nmm.

Q

Mz = 0.

1000

1500

0.6 Q

0.4 Q My

Free-body diagram: 1000

Positive into page

0.6 Q T. E. Tay, Department of Mechanical Engineering, National University of Singapore

11

First, locate the centroid C, as the equations are related to centroidal axes. Q

The total cross-sectional area: = (20)(100) + (60)(20) = 3200 mm2

z

Define the points A, B and D.

A 20

D 15 C

Therefore,

y

60 0 ⎞ 1 1⎛ ⎜ ydA 20 y dy 100 = + ∫− 20ydy ⎟⎟⎠ = 5 mm A∫ A ⎜⎝ ∫0

y=

80

measured from point B,

5

and

B

1 zdA = −15 mm A∫

z=

60

20

measured from point D.


Q

Section properties (second moments of area):

z

( 20 )(100 ) 3 + ( 2000 )(15 ) 2 12 ( 60 )( 20 ) 3 + + (1200 )( 25 ) 2 12 = 2.907 × 10 6 mm 4

I yy =

20 15 C

y 80

5

I zz = 1 .627 × 10 6 mm 4 I yz = ( 2000 )( − 15 )( − 15 ) + (1200 )( 25 )( 25 )

20

60

= 1 .2 × 10 6 mm 4


12

Neutral axis:

tan λ =

z Compression A (-25,35)

= 36.4° C Tension

B (-5,-65)

y

∴

M y I yz + M z I yy M y I zz + M z I yz I yz I zz

=

1 .2 1 . 627

since Mz = 0.

λ = 36 . 4 °

Points A and B are furthest from the neutral axis, and therefore experience the greatest stresses. From Eqn (53), we find

(σ x )A (σ x )B

= − 0 . 0158 Q MPa = 0 . 0182 Q MPa


There is also a shear stress distribution due to bending. We will now discuss shear stresses due to bending in thin-walled members only.


13

Force equilibrium in x-direction: Shear stress distribution due to bending

z

∂σ x ∂q ⎞ ⎛ ⎞ ⎛ dx ⎟ tds − qdx + ⎜ q + ds ⎟ dx = 0 − σ x tds + ⎜ σ x + ∂x ∂s ⎠ ⎝ ⎠ ⎝ ∂q ∂σ x = −t ∴ (56) ∂s ∂x

σx

dx q y q+

σx +

ds

∂q ds x∂ s

∂σ x dx ∂x


Recall that Eqn (53) holds for normal bending stresses. Differentiating with respect to x,

∂ σ x ⎛⎜ V y I yy − V z I yz = ⎜ I I −I2 ∂x yz ⎝ yy zz

⎞ ⎛ V I + V y I yz ⎟ y + ⎜ z zz ⎟ ⎜ I I −I2 yz ⎠ ⎝ yy zz

⎞ ⎟z ⎟ ⎠

(57)

where the relations in Eqns (51) have been used. Substituting Eqn (57) into Eqn (56) and integrating with respect to dA (= t ds), we have

τ xs = where

(V y I yy − V z I yz )Q z + (V z I zz − V y I yz )Q y q =− t I yy I zz − I yz2 t

(

)

Q z = ∫ y dA = Ay

(59a)

Q y = ∫ z dA = Az

(59b)

A

(58)

( y and z are coordinates of the centroid)

A


14


⎛ V yQz VzQ y q = −⎜ + ⎜ I I yy ⎝ zz

⎞ ⎟ ⎟ ⎠

⎛ V I ⎞ ⎜ V y − z yz ⎟ ⎜ I yy ⎟⎠ Vy = ⎝ ⎛ I2 ⎞ ⎜ 1 − yz ⎟ ⎜ I yy I zz ⎟⎠ ⎝ V I ⎞ ⎛ ⎜⎜ V z − y yz ⎟⎟ I zz ⎠ Vz = ⎝ ⎛ I2 ⎞ ⎜ 1 − yz ⎟ ⎜ I yy I zz ⎟⎠ ⎝


If the section is symmetric about either the y- and zaxes, Iyz = 0 and Eqn (58) reduces to

⎛ V yQz VzQ y + q = −⎜ ⎜ I I yy ⎝ zz

⎞ ⎟ ⎟ ⎠

(60)


15

Bending of open sections

General distribution of bending shear flow q for open sections

s

z

q is a function of s and may even change direction, depending upon loading q=0

y x

q=0


∂σ ∂q = −t x We already know that ∂s ∂x . This gives the same solution for the shear flow distribution as in the open section case. But qm is not zero because the section is closed. However, qm can be found by equating to zero the sum of moments about x-axis, since no torque is applied.

Bending of closed sections

z

Vz

y

Centroid O

σx

x

Vy qm

∂q qm + ds ∂s Select an arbitrary point m on the wall for s = 0

E

m s

σx +

∂σ x dx ∂x


16

Example 7 Determine the shear flow distribution in the z-section of uniform thickness t, due to a shear load Vz applied through the shear center of the section. h/2 z

First thing to note is that due to anti-symmetry, the centroid and the shear center are the same point O. This is a special case.

h/2

Vz O

y t

2

h/2

th 3 th 3 ⎛ ht ⎞⎛ h ⎞ = I yy = 2 ⎜ ⎟⎜ ⎟ + 12 3 ⎝ 2 ⎠⎝ 2 ⎠ 3

th 3 ⎛ t ⎞⎛ h ⎞ I zz = 2 ⎜ ⎟⎜ ⎟ = 12 ⎝ 3 ⎠⎝ 2 ⎠ 3 ht h h ⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ht ⎞⎛ h ⎞⎛ h ⎞ th I yz = ⎜ ⎟⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ − ⎟⎜ − ⎟ = 8 ⎝ 2 ⎠⎝ 2 ⎠⎝ 4 ⎠ ⎝ 2 ⎠⎝ 2 ⎠⎝ 4 ⎠

h/2


Hence

⎛ V z I yz ⎞ ⎜⎜ − ⎟ I zz ⎟⎠ Vy = ⎝ = − 0 . 86 V z ⎛ I yz2 ⎞ ⎜1 − ⎟ ⎜ I yy I zz ⎟⎠ ⎝

Vz =

Vz ⎛ I yz2 ⎜1 − ⎜ I yy I zz ⎝

⎞ ⎟ ⎟ ⎠

= 2 . 28 V z

The shear flow due to bending: s ⎞ ⎛ Vy s Vz q = −⎜ ytds ztds ⎟ + ∫ ⎟ ⎜I ∫ I yy 0 ⎠ ⎝ zz 0 s ⎞ ⎛ 0 . 86 s 2 . 28 ytds ztds ⎟⎟ = V z ⎜⎜ 3 − 3 ∫ ∫ th / 3 0 ⎠ ⎝ th / 12 0 s s ⎛ ⎞ V = 3z ⎜⎜ 10 . 3 ∫ yds − 6 . 85 ∫ zds ⎟⎟ h ⎝ 0 0 ⎠


17

For the bottom flange, define a coordinate s1 such that z = - h/2 and y = - h/2 + s1, where 0 ≤ s1 ≤ h / 2

q 12 = h/2

=

z

h/2

O

h/2

t

1

s1 s1 ⎛ ⎞ ⎜ 10 . 3 ⎛⎜ s1 − h ⎞⎟ ds 1 − 6 . 85 ⎛⎜ − h ⎞⎟ ds 1 ⎟ ∫0 ⎝ 2 ⎠ ∫0 ⎝ 2 ⎠ ⎟ ⎜ ⎝ ⎠

(

Vz 5 . 15 s12 − 1 . 72 hs 1 + c h3

)

Boundary condition: At s1 = 0, q12 = 0; hence c = 0

y

s1

Vz h3

∴ q 12 =

(

Vz 5 . 15 s12 − 1 . 72 hs 1 h3

)

2 h/2


q 12 =

)

At 2, s1 = h/2, q12 = 0.43 Vz/h.

z

O

(

Vz 5 . 15 s12 − 1 . 72 hs 1 h3

y

Shear flow distribution is parabolic, with a change in sign at s1 = 0.334h. For values of s1 < 0.334h, q12 is negative and therefore in opposite direction to s1.

s1 1

2

The shear flow distribution on the top flange is similar.

0.334h


18

Along the vertical web 2-3, z = s2 – h/2, y = 0, and 0 ≤ s2 ≤ h

3

z

q 23 =

O

y

s2 1

2

Vz h3

s2

∫ (3 . 43 h − 6 . 85 s )ds 2

2

+ q2

0

where q2 is the shear flow at point 2, i.e. q2 = 0.43 Vz/h.

∴ q 23 =

(

Vz 0 . 43 h 2 + 3 . 43 hs 2 − 3 . 43 s 22 h3

)

This shear flow distribution is also parabolic, with a maximum at s2 = 0.5h.


The Shear Center There is a point where the resultant force must pass through in order that no twisting moments (torque) are developed. This point is called the shear center. Why do we wish to find the shear center? So that we will know, for a given load, how to proportion the stresses (and deformations) due to bending and torsion. How do we find the shear center ? We make use of its definition. A general procedure: 1.

Find the centroid.

2.

Find section properties.

3.

Apply Vz or Vy, in turn.

4.

The moment due to shear flows is due to application of the shear force through the shear center. The position of the shear center with respect to a convenient point can thus be obtained. T. E. Tay, Department of Mechanical Engineering, National University of Singapore

19

Example 8 Determine the position of the shear center of the section. b

Assume the shear center E exists and is located distance ye from vertical web, as shown.

z

h/2

tf O

Vz

y

Apply a shear load Vz through the shear center. E

ye

h/2

First thing to note is that since the section is symmetric, the positions of the centroid and the shear center must lie somewhere on the horizontal axis of symmetry.

Section properties:

tw

2 2 twh3 t h 3 t f bh ⎛h⎞ + 2bt f ⎜ ⎟ = w + 12 12 2 ⎝2⎠ =0 because section is symmetric.

I yy = I yz

tf

Izz is not relevant.


b

Vy = 0

∴q = −

Vz tz ds I yy ∫0

At the top flange, z = h/2, y = b – s1, where s1 defined.

z tf O

h/2

Vz = Vz , s

s1

h/2

Hence

Vz

y ye

E

∴ q s1 = −

tw

=− tf

Vz tf I yy Vz t f h 2 I yy

s1

h

∫ 2 ds

1

0

s1

At s1 = 0, q s1 = 0. (free edge).


20

Take moments about the point A, b

V z y e = h ∫ q s1 ds1

s1

b

∫ q s1 ds1

0

⎡ s2 ⎤ =− Vz ⎢ 1 ⎥ 2 I yy ⎣ 2 ⎦ 0

z

h/2

tf O

=−

Vz

y

E

ye

h/2

b

h 2t f

b

0

∴ ye = −

tw A

=−

tf

b 2 h 2t f 4 I yy

Vz

b 2 h 2t f 4 I yy

(ht

3b 2 t f w

+ 6bt f

)

Therefore, the shear center is to the left of the section. T. E. Tay, Department of Mechanical Engineering, National University of Singapore

Example 9 Determine the position of the shear center of the semi-circular section.

Assume the shear center E exists and is located distance ye from vertical web, as shown.

z r

θ

Vz

O

y ye

t

The section is symmetric, the positions of the centroid and the shear center must lie somewhere on the horizontal axis of symmetry.

Apply a shear load Vz through the shear center. E Section properties:

I yy =

πr 3 t

I yz = 0

2 because section is symmetric.



21

s

Define coordinate s. s

z θ

r

q=−

Vz

O

y

E

ye

θ

=− =−

t

Vz tz ds I yy ∫0

Vz t 2 r cos θ dθ I yy ∫0

where

z = r cosθ , ds = rdθ.

2

V z tr sin θ I yy


Take moments about the point O, S

V z y e = r ∫ q ds 0

π

= r 2 ∫ q dθ

s

0

z r

θ

=−

Vz

O

y ye

E

=−

∴ ye = − t

V z tr 4 [− cos θ ]π0 I yy 4r

π

Vz

4r

π

= −1.27 r The shear center is to the left of the section.


22

Example 10 Determine the position of the shear center of the closed section and the distribution of shear flow. Assume the shear center E exists and is located distance ye from vertical web, as shown.

150 t1

Apply a shear load Vz = 1 × 106 N through the shear center.

t1 Vz 250

E

ye

Section properties:

I yy = 25 .78 × 10 6 mm 4 I yz = 0

because section is symmetric.

t1 = 3 mm , t 2 = 6 mm .

t2


t1


Define coordinates and points as follows. s4

A s1

z

D t1

O

t1

y s3 t2

B

t1

s2

C

From A to B,

s1 = − z + 125

From B to C,

z = – 125

From C to D,

s3 = z + 125

From D to A,

z = 125

Let the shear flow at A be qm, an unknown quantity to be determined.


23

q AB = q m − qm A s1

z

s4 D t1

t1

∫ zt ds 1

1

(10 6 )(3) (125 − s1 ) ds1 25 .78 × 10 6 ∫ = q m − 0.116 125 s1 − 0.5 s12 = qm −

(

)

At B, qAB = qm

O

y s3

t1

q BC = q m −

Vz I yy

∫ zt ds 1

2

= q m − 0.116 ∫ (− 125 ) ds 2

t2 B

Vz I yy

= q m + 14 .5 s 2

C

s2

At C, qBC = 2175 + qm


qCD = q m − qm A s1

z

s4 D t1

t1

∫ zt ds 2

3

(10 6 )( 6) (s3 − 125 ) ds3 25 .78 × 10 6 ∫ = q m − 0.233 0.5 s32 − 125 s3 + 2175

= qm −

(

)

At D, qCD = qm + 2175

O

y s3

t1

s2

q DA = q m −

Vz I yy

∫ zt ds 1

4

= q m − 0.116 ∫ (125 ) ds 4

t2 B

Vz I yy

C

= q m − 14 .5 s4 + 2175


24

Rate of twist = 0, since load acts through shear center.

1 ⎛ q ⎞ dφ = ds ⎟ = 0 ⎜ dx 2Ω G ⎝ ∫ t ⎠ q q q q ∴ ∫ AB ds1 + ∫ BC ds 2 + ∫ CD ds 3 + ∫ DA ds 4 = 0 t t t t AB 1 BC 1 AB 2 AB 1 which simplifies to

675 q m + 598125 = 0 ∴ q m = −886 .1


qm = - 886.1 A

q = 1288.9 D

Therefore,

q AB = −886 .1 − 14 .5 s1 + 0.058 s12 q BC = −886 .1 + 14 .5 s 2 qCD = 1289 − 0.116 s32 + 29 s3 q DA = 1289 − 14 .5 s 4

B qm = - 886.1

C q = 1288.9


25

Take moments about point O to obtain shear center.

A

150

D

C

D

250 ∫ q BC ds 2 + 150 ∫ qCD ds 3 = V z y e B

C

150

V z y e = 250 ∫ (− 886 .1 + 14 .5 s 2 ) ds 2

Vz O ye

250

E

0

250

+ 150

∫ (1288 .9 − 0.116 s

2 3

)

+ 29 s3 ds 3

0

B

C

= 75525 + 93646250 ∴ y e = 93 .7 mm


Combined Bending and Torsion


26

P General Case: Superposition of Bending and Torsion

=

+ Pure Bending through shear center

Pure Torsion T = Pd

P d

E

E

T


Pure Torsion of Open Sections

T

Linear distribution of shear stress τ xy across thickness


27

Pure Torsion of Closed Sections

T

Constant distribution of shear stress τ xs across the thickness Shear flow q is constant anywhere in the section, but shear stress τ xs is NOT constant because of varying wall thickness T. E. Tay, Department of Mechanical Engineering, National University of Singapore

Pure Bending of Open Sections

P

E Distribution of shear flow q in walls due to bending such that nett contribution of q must be due to P, applied force. Bending stress σ x exists in walls


28

P

Pure Bending of Closed Sections

E

Distribution of shear flow q in walls is similar to that of an open section, except for a constant q0 superimposed. Bending stress σ x exists in walls


29

Idealized Beams



1



2


The section is made up of concentrated stringers of area An, joined by thin sheets of thickness tn. Let the loads Vz and Vy act through the shear centre E. Hence pure bending.

z

y Centroid O

Vz x

tn

E

Shear Centre

An

Vy zn yn

The nth stringer is located at positions yn and zn away from the origin at the centroid.


3

Taking moments as before (see Eqns (51)), we have

Vz =

∂M y

∂x M y = Vz x

(61a)

∂M z ∂x M z = −V y x

(61b)

and

Vy = −


Let σn denote the normal stress due to bending in the nth stringer. Since there is no nett axial load, N

∑σ n =1

σn

z

An = 0

(62a)

where N is the total number of stringers. An

y Centroid O

n

x


4

The moment about the y-axis contributes to the normal stress σn due to bending.

My = =

n =1 N

∑σ n =1

σn

z

N

∑M n

yn

An z n (62b)

An

y Centroid O

zn

x

Myn

yn


Contribution of axial force dF due to σn to the bending moment Area An

M yn = σ n An zn (Tensile on positive z)

σn dx z

zn

+ Myn

x y (into page)


5

The moment about the z-axis contributes to the normal stress σn due to bending.

Mz =

N

∑M n =1

zn

N

= − ∑ σ n An y n n =1

σn

z

(62c) An

y Centroid O

zn

x

Mzn yn


Contribution of axial force dF due to σn to the bending moment Area An

M zn = −σ n An yn (Compressive on positive y)

σn dx y

yn

dMzn

x z (out of page)


6

Invoke again the assumption that plane sections remain plane, which implies that the axial strain ε (and axial stress σ) is a linear function of y and z. Let σ n = C1 + C 2 y n + C 3 z n and substituting this into Eqns (62) and solving for the constants C1, C2, C3, we have C1 = 0 and

⎛ M z I yy + M y I yz ⎞ ⎛ M I + M z I yz ⎟ y n + ⎜ y zz 2 ⎟ ⎜ I I − I2 yy zz yz ⎝ I yy I zz − I yz ⎠ ⎝

σ n = − ⎜⎜

⎞ ⎟ zn ⎟ ⎠

(63)

Note the similarity with Eqn (53). The expression for the neutral axis remains the same as Eqn (54).


However, the section properties are now N

I yy = ∑ An z n2

(64a)

n =1 N

I zz = ∑ An y n2

(64b)

n =1 N

I yz = ∑ An y n z n

(64c)

n =1

They are easier to evaluate than integrals!


7

Eqn (63) can, of course, be rewritten in terms of equivalent moments:

σn = −

M z yn M y zn + I zz I yy

⎛ M I ⎜ M z + y yz ⎜ I yy Mz = ⎝ ⎛ I2 ⎞ ⎜ 1 − yz ⎟ ⎜ I yy I zz ⎟⎠ ⎝ M I ⎛ ⎜⎜ M y + z yz I zz My = ⎝ 2 ⎛ ⎞ I ⎜ 1 − yz ⎟ ⎜ I yy I zz ⎟⎠ ⎝

⎞ ⎟ ⎟ ⎠

⎞ ⎟⎟ ⎠


General distribution of normal bending stress σ n

z Tension y Neutral axis

x

Compression


8

Now, consider the shear flows in the sheets or webs.

z

Select a small length dx of a typical nth stringer with two adjacent nth and (n + 1)th sheets.

Anσ n qn

y dx x

qn+1

∂σ ⎛ ⎞ An ⎜σ n + n dx⎟ ∂x ⎠ ⎝

Force equilibrium in x-direction:

∂σ ⎞ ⎛ −σn An + ⎜σn + n dx⎟ An − qndx+ qn+1dx = 0 ∂x ⎠ ⎝ ∂σ ∴ qn+1 = − n An + qn (65) ∂x T. E. Tay, Department of Mechanical Engineering, National University of Singapore

Substituting Eqn (63) into Eqn (65) and using Eqns (61),

⎛ (V y I yy − Vz I yz )yn + (Vz I zz − V y I yz )zn ⎞ ⎟ An + qn qn +1 = −⎜ 2 ⎜ ⎟ − I I I yy zz yz ⎝ ⎠

(

)

(66)

This equation relates the shear flow of the (n + 1)th sheet to the shear flow of the adjacent nth sheet. If the section is open, starting from one free edge, the shear flows in successive sheets may be easily found. Note that the shear flow does not vary with s within each sheet or web.


9


⎛ V y yn Vz zn q n +1 = − ⎜ + ⎜ I zz I yy ⎝

⎞ ⎟⎟ An + q n ⎠

⎛ V z I yz ⎞ ⎜⎜ V y − ⎟ I yy ⎟⎠ ⎝ Vy = ⎛ I yz2 ⎞ ⎜⎜ 1 − ⎟ I yy I zz ⎟⎠ ⎝ V y I yz ⎞ ⎛ ⎜ Vz − ⎟ I zz ⎠ Vz = ⎝ ⎛ I yz2 ⎞ ⎜⎜ 1 − ⎟ I yy I zz ⎟⎠ ⎝


If the section is closed, we first analyze the problem as if one of the sheets (pick one arbitrarily) is missing.

q5 + q0 q0 q4 + q0

Replace the missing sheet with the unknown shear flow. The unknown shear flow is then superposed onto the problem and solved by taking moments of the combined shear flows about a convenient point.

q1 + q0

q3 + q0 q2 + q0


10

Example 11 Determine the shear flow distribution in the idealized channel section produced by a vertical load of 48 kN acting through its shear center. Assume the skin is effective only in resisting shear stresses, while the stringers, each of area 300 mm2, resist all the direct stresses. 2

1 z

48 kN

2000 4

I yy = ∑ Ai zi2 = 4(300)(2000)2 = 4.8×109 mm4

y

i =1

2000 3

4 2000 Dimensions in mm T. E. Tay, Department of Mechanical Engineering, National University of Singapore

q12 = − 2

6 N/mm

1

2000

q 23

− 48000 ( 300 )( 2000 ) − 6 = − 12 N/mm 4 . 8 × 10 9 V = − z A3 z 3 + q 23 I yy =

y 2000 6 N/mm 3

− 48000 ( 300 )( 2000 ) = − 6 N/mm 4 . 8 × 10 9 V = − z A2 z 2 + q12 I yy =

z 12 N/mm

Vz A1 z1 + 0 I yy

4

q 34

=

− 48000 ( 300 )( 2000 ) − 12 = − 6 N/mm 4 . 8 × 10 9

2000 Dimensions in mm

Note that we can check vertical and horizontal force equilibrium and show that the resultants are 48 kN and zero, respectively.


11

Example 12 Find the shear flows in the webs of the unsymmetrical closed beam section. 300 mm2 A

B 100 mm

2

z V z y

D 100 mm2

200 mm

Vy

100 mm

C 300 mm2

Vy = 4 kN Vz = 10 kN


4

I yy = ∑ Ai z i2 = 2(300 )(50 ) 2 + 2 (100 )(50 ) 2 = 2 .0 × 10 6 mm 4 i =1 4

I zz = ∑ Ai yi2 = 2(300 )(100 ) 2 + 2(100 )(100 ) 2 = 8.0 × 10 6 mm 4 i =1 4

I yz = ∑ Ai y i z i = (300 )( − 100 )(50 ) + (300 )(100 )( − 50 ) i =1

+ (100 )(50 )(100 ) + (100 )( −50 )( − 100 ) = −2 .0 × 10 6 mm 4 ⎛ − 2.0 ⎞ 4 00 0 − 10 0 00 ⎜ ⎟ ⎝ 2 .0 ⎠ = 1 .86 67 × 1 0 4 Vy = 2 ( − 2 .0 ) 1− ( 2 .0 )(8 .0 ) ⎛ − 2 .0 ⎞ 10 0 00 − 4 00 0 ⎜ ⎟ ⎝ 8 .0 ⎠ = 1.4 66 7 × 10 4 Vz = ( − 2.0 ) 2 1− ( 2 .0)(8 .0 ) T. E. Tay, Department of Mechanical Engineering, National University of Singapore

12

⎛ V y yn Vz zn ⎞ q n +1 = − ⎜ + ⎟ An + q n ⎜ I zz I yy ⎟⎠ ⎝ ⎛ 1.8667 y n 1.4667 z n ⎞ An + q n = −⎜ + 2 2.0 × 10 2 ⎟⎠ ⎝ 8.0 × 10 = ( − 2.33 y n − 7.33 z n ) × 10 − 3 An + q n


Let qAD be the unknown shear flow, i.e. qAD = q0. Consider stringer D: B

A

q0

D(-100,-50) 100 mm2

C 60 + q0

q DC = (− 2 .33 ( − 100 ) − 7 .33 ( − 50 ) ) × 10 − 3 (100 ) + q 0 = 60 + q 0 T. E. Tay, Department of Mechanical Engineering, National University of Singapore

13

Consider stringer C: B

A

q0

100 + q0

D

C(100,-50) 60 + q0

300 mm2

q CB = (− 2 .33 (100 ) − 7 . 33 ( − 50 ) ) × 10 − 3 (300 ) + q DC = 100 + q 0 T. E. Tay, Department of Mechanical Engineering, National University of Singapore

Consider stringer B: 40 + q0 A

100 mm2 B(100,50)

q0

100 + q0

D

C 60 + q0

q BA = (− 2 .33 (100 ) − 7 . 33 (50 ) )× 10 − 3 (100 ) + q CB = 40 + q 0 T. E. Tay, Department of Mechanical Engineering, National University of Singapore

14

Let us now take moments about, say O:

q AD (100 )(100 ) + q BA ( 200 )( 50 ) + q CB (100 )( 100 ) + q DC ( 200 )( 50 ) = 0 4 q 0 = − 200 q 0 = − 50 N/mm

40 + q0

B

A

q0

100 + q0

O

C

D

60 + q0

qAD = - 50 N/mm

qDC = 10 N/mm

qCB = 50 N/mm

qBA = - 10 N/mm


Example 13

2

I yy = ∑ Ai z i2 = 2 Ar 2

Find the shear center.

i =1

s

q=− z

r

Vz

θ O

y ye

V z z n An V =− z I yy 2r

Taking moments about O, E

πr

V z y e = r ∫ q ds 0

π

= r 2 ∫ q dθ Area of each stringer = A mm2

0

= πr 2 q π rV z =− 2 ∴ ye = −

πr 2


15

Example 14 Find the shear center. 120 B (80,40)

C (-40,40) 120 mm2 45

40 F

z O

60

y 60 mm2

D (-40,-50) 120 mm2

All dimensions in mm

60 mm2

40

A (80,-20)

First, remember to locate the centroid!

y=

1 4 2(60)(120) ∑ Ai yi = 360 = 40 A i =1

z=

1 4 (−60)(60) + (120)(−90) Ai zi = = −40 ∑ A i =1 360

measured from C

measured from C


Section properties: 4

I yy = ∑ Ai z i2 = (60 + 120 )( 40 ) 2 + (60 )( − 20 ) 2 + (120 )( −50 ) 2 = 0.612 × 10 6 mm 4 i =1 4

I zz = ∑ Ai yi2 = 2(60 )(80 ) 2 + 2(120 )( − 40 ) 2 = 1.152 × 10 6 mm 4 i =1 4

I yz = ∑ Ai y i z i = ( 60 )(80 )( − 20 ) + (60 )(80 )( 40 ) + (120 )( 40 )( − 40 ) i =1

+ (120 )( − 40 )( −50 ) = 0.144 × 10 6 mm 4


16

Let us first find the z-coordinate position of shear center E. Vy E

C ze

z Assume an arbitrary shear force Vy applied to the shear center.

y

O

F

B

A Hence

D

Vy I zz

= 0.894 × 10−6 Vy

Vz = −0.210 ×10−6 Vy I yy


qBC

C

B

z qCD

O

F

qAB

y A

D

⎛ Vy y A Vz z A ⎞ + q AB = − ⎜ ⎟ AA + 0 ⎜ I zz I yy ⎟⎠ ⎝ = − ( 0.894(80) − 0.210(−20) ) 60 × 10−6 Vy = −4.543 × 10−3Vy ⎛ Vy yB Vz zB ⎞ qBC = − ⎜ + ⎟ AB + q AB ⎜ I zz I yy ⎟⎠ ⎝ = − ( 0.894(80) − 0.210(40) ) 60 ×10−6 Vy − 4.543 × 10−3Vy = −8.33 × 10−3Vy qCD = −5.68 × 10−3Vy


17

Vy qBC

C ze qCD

B

z qAB

y

O

F

E

5

A

D Taking moments about a convenient point F,

−V y ( z e + 5) = [π (45) 2 ( − 5.68) + ( − 4.543)(60)(120) + ( − 8.330)(120)(45)] × 10 − 3 V y = − 113.83V y ∴ z e = 108.8 mm

measured from O.


Let us now find the y-coordinate position of shear center E. Vz E

C

B

z O

F

Assume an arbitrary shear force Vz applied to the shear center.

y ye A

D

Now,

Vy I zz

= −0.21×10 −6 V y

Vz = 1.68 × 10 −6 V y I yy


18

Vz E

C

B

q AB = 3.024 ×10 −3Vz

z qCD

y

O

F

qAB

qBC = 0 qCD = 7.056 ×10 −3Vz

ye A

40 D Taking moments about F,

V z ( y e + 40 ) = [π ( 45 ) 2 ( 7 . 056 ) + ( 3 .024 )( 60 )(120 )] × 10 − 3 V z = 66 . 7V z ∴ y e = 26 .7 mm measured from O.


Example 15 The idealized tube is loaded by a vertical shear force of 10 kN acting in the plane of the section. Assuming the direct stresses are carried by the stringers while the webs are effective only in carrying the shear stresses, calculate the distribution of shear flow around the section. 10 kN 2

3 z

A1 = A8 = 200 mm2 A2 = A7 = 250 mm2 A3 = A6 = 400 mm2 A4 = A5 = 100 mm2 1

4 200 100

O

y 5

60 6

120

8

7 240

240

All dimensions in mm T. E. Tay, Department of Mechanical Engineering, National University of Singapore

19

Section properties:

(

8

I yy = ∑ Ai zi2 = 2 ( 200 )(30 ) 2 + ( 250 )(100 ) 2 + ( 400 )(100 ) 2 + (100 )(50 ) 2

)

i =1

= 13 .86 × 10 6 mm 4 I yz = 0

because the section is symmetric about the horizontal axis.

Izz does not enter into calculation because Vy = 0.


Select the shear flow in web 2-3 to be the unknown shear flow, q23.

⎛V z ⎞ q34 = −⎜ z 3 ⎟ A3 + q23 ⎜ I ⎟ ⎝ yy ⎠ (10 ×103 )(100) =− (400) + q23 13.86 ×106 = −28.9 + q23 10 kN 3

q23 – 28.9 z

q23

2


4 200 100

O

y 5

60 6

120

8

7 240

240


20

⎛V z ⎞ q45 = −⎜ z 4 ⎟ A4 + q34 ⎜ I ⎟ ⎝ yy ⎠ (10 ×103 )(50) =− (100) − 28.9 + q23 13.86 ×106 = −32.5 + q23

Due to symmetry,

q56 = q34 = −28.9 + q23 q67 = q23

10 kN

q23 – 28.9 z 200 100

O

3

q23

2


4 q23 – 32.5 y 5

60 6

q23 – 28.9

q23 240

120

8

7 240


(10 × 103 )(−100) (250) + q23 13.86 × 106 = 18.1 + q23

q78 = −

(10 × 103 )(−30) (200) + 18.1 + q23 q81 = − 13.86 × 106 = 22.4 + q23

Due to symmetry,

q12 = q 78 = 18 .1 + q 23 A1 = A8 = 200 mm2 A2 = A7 = 250 mm2 A3 = A6 = 400 mm2 A4 = A5 = 100 mm2

10 kN q23 – 28.9 z 200 100

O

3

q23

2

q23 + 18.1 1

4 q23 – 32.5 y 5

60 6

q23 – 28.9 120

8

7 q23 240

q23 + 22.4

q23 + 18.1 240


21

Take moments about the point of intersection of the line of action of the shear load and the horizontal axis of symmetry.

10 kN

z 200 100

4 q45 O

q23

3

q34

2

q12 1

y

5

A

60

6

8

7

120

q81

240

240



(50 q 45 )(120 ) + (120 q34 + 240 q 23 + 240 q12 )(100 ) + ( 70 q12 )( 240 ) + (30 q81 )( 480 ) = 0 70q12

10 kN 120q34

3

z 200 100

4 50q45 O

240q23

5

A

60

6 120

30q81 1

240q12

50q34 y

2

8

7 240

240


22

Substituting for q23 and solving, we obtain

q23 = −5.4 N/mm q34 = −34.3 N/mm q45 = −37.9 N/mm q56 = −34.3 N/mm q67 = −5.4 N/mm q78 = 12.7 N/mm q81 = 17.0 N/mm

10 kN 34.3

3

5.4

2

q12 = 12.7 N/mm 12.7 1

4 37.9

17.0 5

6

8

7 12.7

34.3

5.4


Question to ponder: How do we determine the separate proportions of these shear flows due to bending and torsion?

10 kN 34.3

3

5.4

2

12.7 1

4 37.9

17.0 5 34.3

6

8

7 12.7 5.4


23

Example 15 Determine the position of the shear center. (Let the thickness of each web = 1 mm.)

2

3 z


4 200 100

O

y 5

60 6

8

7

120

240

240 All dimensions in mm



(50q45 )(120) + (120q34 + 240q23 + 240q12 )(100) + (70q12 )(240) + (30q81 )(480) = (10 ×103 ) ye 70q12 120q34

3

z 200 100

4 50q45 O

240q23

50q34 y

5

A 6

120

E

2

30q81 1

240q12 10 kN

ye

60 8

7 240

240


24

Substituting for the unknown shear flow,

6000(q23 − 32.5) + 12000q23 + 24000q23 + 24000q23 + 16800q23 + 304080 − 346800 + 434400 + 14400q23 + 322560 = (10 × 103 ) ye ∴ 97.2q23 + 519.24 = ye The additional equation is supplied by the rate of twist equation.



dφ 1 ⎛ q ⎞ ds ⎟ = 0 = ⎜ dx 2ΩG ⎝ ∫ t ⎠ q q q q q ∴ 2 ∫ 12 ds12 + 2 ∫ 23 ds 23 + 2 ∫ 34 ds34 + ∫ 45 ds 45 + ∫ 81 ds81 = 0 t t t t t 1− 2 12 2 − 3 23 3− 4 34 4 − 5 45 8 −1 81 3

2

z

1

4 O

y 5

A 6

7

8


25


2 ∫ (18 .1 + q23 ) ds12 + 2 ∫ q 23 ds 23 + 2 ∫ ( q 23 − 28 .9) ds34 1− 2

+

∫ (q

2 −3

23

− 32 .5) ds 45 +

4 −5

∫ (22.4 + q

3− 4

23

) ds81 = 0

8 −1

2(18 .1 + q 23 )( 250 ) + 2 q 23 ( 240 ) + 2( q 23 − 28 .9)(130 ) + ( q 23 − 32 .5)(100 ) + ( 22 .4 + q 23 )(60 ) = 0 1400 q 23 = 370 ∴ q 23 = 0.264 Substituting this into the moment equation, we have

ye = 544.9 mm


26

5/26/2009

Multicell Thin-walled Sections The multicell tube in pure torsion is a statically indeterminate problem. The condition of consistent twist must be used for each cell of the tube in order to obtain the solution. Consider the n-celled tube of general shape subjected to a torque T. Clearly, the shear flow in the web between cell i and cell j is qj-qi. Similarly, in the web between cell j and cell k, the shear flow is qj-qk. H Hence, th the rate t off twist t i t off cell ll j becomes b

1 ⎛ dφ ⎞ ⎜ ⎟ = Ωj dx 2 G ⎝ ⎠j where Ωj is the area enclosed by cell j.

⎛ ⎞ ⎜ q ds − q ds − q ds ⎟ i ∫ k ∫ ⎜ j s∫ t ⎟ t t s ji s jk ⎝ j ⎠

(1)

j 1

,

2

,…

qi

qj

qk

…n

k i

The first integral is evaluated over the entire cell parameter sj, the second integral over the web between cell i and cell j (i.e. sji), and the third integral over the web between cell j and cell k (i.e. sjk). 1


Although cross-sections warp, they do not distort in their own plane. This means that the entire cross-section and each cell rotate at the same rate of twist dφ . Thus dx

dφ ⎛ dφ ⎞ ⎛ dφ ⎞ ⎛ dφ ⎞ ⎛ dφ ⎞ = ⎜ ⎟ = ⎜ ⎟ = ... = ⎜ ⎟ = ... = ⎜ ⎟ dx ⎝ dx ⎠1 ⎝ dx ⎠ 2 ⎝ dx ⎠ j ⎝ dx ⎠ n

(2)

Therefore, for a general case where cell j is bounded by m cells,

dφ 1 ⎡ ds m ⎛ ds ⎞⎤ ⎢ q j ∫ − ∑ ⎜ qr ∫ ⎟ ⎥ = ⎜ dx 2GΩ j ⎢ s j t r =1 t ⎟ ⎝ s jr ⎠⎥⎦ ⎣

(3)

For n cells, there are n such equations.


2

1

5/26/2009

However, the number of unknowns involved is (n+1), i.e. q1, q2, q3, …, qj, …, qn and dφ . The additional equation required is the moment equilibrium equation, i.e. dx n

T = 2∑ qi Ω i

(4)

i =1

where qi is the shear flow in the

ith

cell.

The foregoing procedure can also be used for multicell thin-walled tubes loaded by ttransverse a sve se forces, o ces, in w which c case tthee moments o e ts induced duced by the t e transverse t a sve se forces o ces should be taken into account in the moment equilibrium equation, and the shear flow expressions for bending should be used.

3


Example 1: The section shown has constant thickness t throughout. It is subjected to a constant torque T. Determine the rate of twist. 2a

a q1

q2

1

a

2

Consider cell 1:

Consider cell 2:

4a a dφ 1 ⎡ ds1 ds ⎤ = − q2 ∫ 12 ⎥ q 2 ⎢ 1∫ dx 2Ga ⎣ 0 t t ⎦ 0 1 (4q1 − q2 ) = 2tGa

dφ 1 = dx 4Ga 2

(1)

=

a ⎡ 6 a ds d 2 d ⎤ ds − q1 ∫ 12 ⎥ ⎢ q2 ∫ t t ⎦ 0 0 ⎣

1 (6q2 − q1 ) 4tGa


(2)

4

2

5/26/2009

Moment equilibrium: 2

T = 2∑ qi Ω i i =1

= 2q1Ω1 + 2q2 Ω 2 = 2a 2 (q1 + 2q2 )

(3)

From (1) and (2) we have:

q1 =

Substituting these into (3),

16 dφ aGt 23 dx

T=

andd

q2 =

dφ 23 T = dx 104 Gta 3

∴

18 dφ aGt 23 dx

104 3 dφ a Gt 23 dx


5

Example 2. Calculate the shear stress distribution in the walls of the three-cell wing section shown when it is subjected to an anticlockwise torque of 11.3kNm. G (N/mm2)

Thickness (mm)

Cell Area (mm2)

Wall

Length (mm)

12o

1650

1.22

24200

AI = 258000

12i

508

2 03 2.03

27600

AII = 355000 AIII = 161000

13, 24

775

1.22

24200

34

380

1.63

27600

35, 46

508

0.92

20700

56

254

0.92

20700

Note: The superscript symbols o and i are used to distinguish between outer and inner walls connecting the same points. 11300Nm 3

1

III

II

I 2

5

4

6


6

3

5/26/2009

∗ Let Gref = 27.6 GPa and t =

G ×t Gref

Hence

24.2 × 1.22 = 1.07 27.6 ∗ t13∗ = t24 = 1.07

t12∗ o =

∫

ds12o

∫

ds12i

12 o

12

∗ ∗ ∗ t35 = t46 = t56 = 0.69

i

∗ 12 o

t

t12∗ i

= =

1650 = 1542 1.07

508 = 250 2.03

ds13 ds = ∫ ∗24 = 725 ∗ t t 13 13 24 24

∫

ds34 = 233 t∗ 34 34

∫

ds35 ds = ∫ ∗46 = 736 ∗ t t 46 46 35 35

∫

ds56 = 368 t∗ 56 56

∫


For cell I,

⎛ ds12o ds i ds i 1 dφ ⎜ qI = + qI ∫ ∗12 − qII ∫ ∗12 dx 2Gref (258000 ) ⎜⎝ 12∫o t12∗ o i t i i t i 12 12 12 12 =

7

⎞ ⎟ ⎟ ⎠

1 (qI (1542 + 250) − 250qII ) 2Gref (258000 )

For cell II,

dφ 1 = (− 250qI + (250 + 725 + 233 + 725)qII − 233qIII ) dx 2Gref (355000 )

For cell III,

dφ 1 = (− 233qII + (736 + 736 + 233 + 368)qIII ) dx 2Gref (161000 )

3

1

II

I

qI

2

qII

5

III 4

qIII

6


8

4

5/26/2009

From the moment equilibrium equation

2(258000q I + 355000q II + 161000q III ) = 11.3 ×10 6 These equations are solved to give qI = 7.1 N/mm q II = 8.9 N/mm qIII = 4.2 N/mm

8.9 1

7.1

4.2

3

1.8

5

4.7 2

8.9

4

4.2 6

4.2


9

The Method of Successive Approximations for Pure Torque Consider a two-cells section free to warp and subjected to torque T.

I

II qI

qII

If the cells were separated, we see that ⎛⎜ dφ ⎞⎟ ≠ ⎛⎜ dφ ⎞⎟ and this violates the constant ⎝ dx ⎠ I ⎝ dx ⎠ II rate of twist condition.

II

I qI

qII


10

5

5/26/2009

⎛ dφ ⎞ ⎛ dφ ⎞ ⎟ = G⎜ ⎟ = 1, then ⎝ dx ⎠ I ⎝ dx ⎠ II

If we arbitrarily let G⎜

For cell I,

For cell II,

q ⎛ ds ⎞ ⎛ dφ ⎞ G⎜ ⎟ = II ⎜ ∫ ⎟ = 1 ⎝ dx ⎠ II 2Ω II ⎝ t ⎠ II qII δ II =1 2Ω II 2Ω

q ⎛ ds ⎞ ⎛ dφ ⎞ G⎜ ⎟ = I ⎜ ∫ ⎟ = 1 ⎝ dx ⎠ I 2Ω I ⎝ t ⎠ I qI δ I =1 2Ω I 2Ω


11

Considering the separated cell I, we see that at the wall shared with cell II, the shear flow must be corrected for the shear flow due to cell II.

I

qII qI


12

6

5/26/2009

The contribution of qII to the rate of twist is represented by an equivalent shear flow qI′ over the perimeter of cell I.

⎛ ds ⎞ ⎛ ds ⎞ q′I ⎜ ∫ ⎟ = qII ⎜ ∫ ⎟ ⎝ t ⎠I ⎝ t ⎠ I , II

I

qII

q′I

I

13


Or,

q′I = qII

where

δ I , II δI

δ I , II = ∫

ds for the wall common to cells I and II t

The “correction carry-over factor” is then defined as

C II , I =

δ I , II δI

∴ q′I = qII C II , I

q′I

I qI


14

7

5/26/2009

Since the correction carry-over factor is less than unity, if the procedure is repeated, the correction become successively smaller until they are finally negligible. The number of iterations depends on the accuracy required.

(qI )Final = qI + q′I + q′I′ + q′I′′ + K where

q′I′ = q′II C II , I q′I′′ = q′II′ C II , I


15

Similarly for cell II,

C I , II =

δ I , II δ II

qI

II

qII

∴ q′II = qI C I , II

q′II

II qII


16

8

5/26/2009

Example 3: Let us solve the problem of the previous example using the method of successive approximations.

⎛ ds ⎞ ⎟ ⎝ t ⎠ ds12o

δI = ⎜ ∫ =

∫

12

3

2

+

∫ 12

i

ds12i t12∗ i

δ II = 250 + 725 + 233 + 725

5

= 1933

III

II

I

t12∗ o

= 1542 + 250 = 1792

11300Nm 1

o

4

δ III = 736 + 233 + 736 + 368

6

= 2073

δ I , II = 250 δ II , III = 233 17


The carry-over factors are:

δ I , II 250 = = 0.129 δ II 1933 δ 250 = I , II = = 0.140 δ I 1792

C I , II = C II , I

233 = 0.112 2073 233 = = 0.121 1933

The initial assumed values of shear flows:

2

qI = qII = qIII =

=

2Ω II

=

2Ω III

=

δ II

δ III

4

6

2(258000) = 287.9 1792

2Ω I

δI

5

III

II

I

C II , III = C II , III

3

1

2(355000) = 367.3 1933

2(161000 ) = 155.3 2073


18

9

5/26/2009

Cell I

Cell II

CI,II = 0.129

CII,I = 0.140

Cell III CII,III = 0.112

CIII,II = 0.121

A Assumed dq

287 9 287.9

367 3 367.3

155 3 155.3

q′

(0.140)(367.3)=51.4

(0.129)(287.9)=37.14

(0.121)(155.3)=18.8

(0.112)(367.3)=41.1

q′′

(0.140)(37.14)=5.2

(0.129)(51.4)=6.63

(0.121)(41.1)=4.97

(0.112)(18.8)=2.10

q′′′

0.93

0.67

0.25

0.56

q′′′′

0.09

0.12

0.07

0.03

Final q

345 5 345.5

435 9 435.9

199 1 199.1


19

We still need the torque equilibrium equation to arrive at the correct solution. From the torque equilibrium equation,

T = 2qI Ω I + 2qII Ω II + 2qIII Ω III

= 2(345.5)(258000) + 2(435.9 )(355000) + 2(199.1)(161000) = 5.52 ×108

But the actual applied torque is 0.113x108 So the values of q must be scaled down by a factor of 0.113/5.52=0.0205

Cell I

Cell II

Cell III

Final q

345.5

435.9

199.1

Scaled q

7.07

8.92

4.1


20

10

5/26/2009

Bending Shear Stresses in Multi-cell Beams Consider a three-cells section carrying a shear load through the shear centre (no twist):

VZ

I

II

III


21

If each cell were “cut”, the open-section shear flow would be given by

qb = =

(I

yz

Qz − I zz Q y ) Vz I yy I zz − I yz2

(I ∫ yt ds − I ∫ zt ds ) V yz

zz

I yy I zz − I

z

2 yz

Distribution of qb:

I

II

III


22

11

5/26/2009

If we now close the cells, the constant shear flows qI, qII and qIII must be superposed onto the shear flow qb in the respective cells:

qb

qI

qII

qIII


23

Considering each cell separately, we now have to correct for the shear flows in the shared walls. Consider, for example, cell II, where qI acts on the left wall, and qIII on the right wall:

qII qI

qIII

qb


24

12

5/26/2009

Since the twist is zero, 1 ⎛ ds ⎞ ⎛ dφ ⎞ G⎜ ⎟ = ⎜∫q ⎟ = 0 t ⎠ II ⎝ dx ⎠ II 2Ω II ⎝

⎛ ds ⎞ ⎜∫q ⎟ = 0 t ⎠ II ⎝

or

From the figure, taking positive torque anti-clockwise, ds ⎞ ⎛ ⎛ ds ⎞ ⎛ ds ⎞ ⎛ ds ⎞ =0 ⎜ ∫ qb ⎟ + q II ⎜ ∫ ⎟ − q I ⎜ ∫ ⎟ − q III ⎜ ∫ ⎟ t ⎠ II ⎝ ⎝ t ⎠ II ⎝ t ⎠ I , II ⎝ t ⎠ II , III

The other two cells have similar equations, i.e.: ds ⎞ ⎛ ⎛ ds ⎞ ⎛ ds ⎞ ⎛ ds ⎞ =0 ⎜ ∫ qb ⎟ + qI ⎜ ∫ ⎟ − qII ⎜ ∫ ⎟ − qIII ⎜ ∫ ⎟ t ⎠I ⎝ ⎝ t ⎠I ⎝ t ⎠ II , I ⎝ t ⎠ I , III

and

ds ⎞ ⎛ ⎛ ds ⎞ ⎛ ds ⎞ ⎛ ds ⎞ − qI ⎜ ∫ ⎟ =0 ⎜ ∫ qb ⎟ + q III ⎜ ∫ ⎟ − qII ⎜ ∫ ⎟ t ⎠ III ⎝ ⎝ t ⎠ III ⎝ t ⎠ II , III ⎝ t ⎠ I , III

These three equations can be solved to obtain qI , qII and qIII The above equations are also applicable to idealized multicell beams.


25

The Method of Successive Approximations for Bending Shear Flows Since the twist is zero,

If shear force acts through shear centre

1 ⎛ ds ⎞ ⎛ dφ ⎞ G⎜ ⎟ = ⎜∫q ⎟ = 0 t ⎠ II ⎝ dx ⎠ II 2Ω II ⎝

or

d ⎞ ⎛ ds ⎜∫q ⎟ = 0 t ⎠ II ⎝

Hence, from the figure, taking positive torque anti-clockwise, for cell II, ds ⎞ ⎛ ⎛ ds ⎞ ⎛ ds ⎞ ⎛ ds ⎞ =0 ⎜ ∫ qb ⎟ + q II ⎜ ∫ ⎟ − q I ⎜ ∫ ⎟ − q III ⎜ ∫ ⎟ t ⎠ II ⎝ ⎝ t ⎠ II ⎝ t ⎠ I , II ⎝ t ⎠ II , III

qb

ds ⎞ ⎛ ⎜ ∫ qb ⎟ + q II δ II − q I δ I , II − q III δ II , III = 0 t ⎠ II ⎝

Rearranging,

qII = −

ds ⎞ ⎛ ⎜ ∫ qb ⎟ t ⎠ II ⎝

δ II

δ δ + q I I , II + qIII II , III δ II δ II

qI

qII

qIII

= q′II + C I , II q I + C III , II qIII T. E. Tay, Department of Mechanical Engineering, National University of Singapore

26

13

5/26/2009

To begin, we first neglect the contribution of qI and qIII, so that

q II = q′II = −

ds ⎞ ⎛ ⎜ ∫ qb ⎟ t ⎠ II ⎝

δ II

q I = q′I q III = q′III

Subsequently, q II = q′II + C I , II q′I + C III , II q′III = q′II + q′II′

etc.


27

Example 4. Determine the shear flow distribution in the single symmetrical three-cell wing section below when it carries a shear load of 100 kN applied through its shear centre and hence find the distance of the shear centre from the spar web 34. Assume all direct stresses are resisted by the stringers while the skin is effective only in shear. The shear modulus G is constant throughout. Stringer areas: B1 = B6 = 2500 mm2, B2 = B5 = 3800 mm2, B3 = B4 = 3200 mm2 C ll areas: AI = 265000 mm2, AII = 580000 mm2, AIII = 410000 mm2 Cell Wall

12, 56

23, 45

34o

16

25

34i

Length (mm)

1025

1275

2200

330

450

400

Thickness (mm)

1.25

1.65

2.25

1.65

2.65

2.65

100kN

2

3

1

E

I

4

ξs 1270mm

II

III

5

6 1020mm


28

14

5/26/2009

The carry-over factors:

δ I = 1128.7 δ II = 1870.0 δ III = 2013.6

δ I , II 150.9 = = 0.081 1870 δ II δ 150.9 = I , II = = 0.134 δ I 1128.7

C I , II = C II , I

δ I , II = 150.9

C II , III = 0.086

δ II , III = 173.6

C III , II = 0.093

29


For an open idealized section symmetric about the horizontal axis,

qbi = −

∴

Vz Ai zi I yy

100 ×103 (3200)(− 200) = 80.6 N/mm 7.94 ×108 0 = 110.1 N/mm

qb 43 = − qb 52

3

qb 61 = 52.0 N/mm Also,

0

0

2

80.6

0

4

52.0

110.1

0

(80.6)(400) = −12166.0 N/mm ⎛ ds ⎞ −⎜∫q ⎟ = − t ⎠I 2.65 ⎝ (110.1)(460) − 12166.0 = −6945.7 N/mm ⎛ ds ⎞ −⎜∫q ⎟ = − t ⎠ II 2.65 ⎝

0 1

0

5

0

6

⎛ ds ⎞ − ⎜ ∫ q ⎟ = 6218.9 N/mm t ⎠ III ⎝


30

15

5/26/2009

4.4

3

11.4

The solution proceeds in the tabulated form shown. Taking moments about the centre of spar web 34,

2.7

2

4

54.7

III

II

I

1

103.0

73.6

5

4.4

2.7

6

100 ×103 ξ s = 2Ω I qI + 2Ω II qII + 2Ω III q III + (110.1)(460)(1270) + (52)(330)(2290)

∴

ξ s = 946.8 mm Cell I

Cell II

Cell III

1128.7

1870.0

2013.6

-12166.0

-6945.7

6218.9

δ −∫

qb ds t

CI,II=0.08 ⎛ q ds ⎞ q′ = − ⎜ ∫ b ⎟ δ ⎝ t ⎠ q′′

CII,I=0.134

-10.78 10.78

CII,III=0.086

CIII,II=0.093

-3.71 3.71

(0.134)(-3.71) =-0.50

3.09

(0.08)(-10.78)=-0.86

(0.093)(3.09)=0.29

(0.086)(-3.71)=-0.32

q′′′

(0.134)(-0.86)=-0.115

(0.08)(-0.50)=-0.04

(0.093)(0.32)=-0.03

(0.086)(0.29)=0.02

q′′′′

(0.134)(-0.04)=-0.005

(0.08)(-0.115)=-0.01

(0.093)(0.02)=0

(0.086)(-0.03)=0

Final q

-11.4

-4.36

2.7


31

Tapered Beams Beams in aerospace vehicle structures are often tapered in order to achieve maximum structural efficiency with minimum weight. The simple beam shown consists of two stringers of equal areas, joined by a vertical web that does not carry bending stresses. x x0

b

α1

h1 h0

α2

h2

Vz T. E. Tay, Department of Mechanical Engineering, National University of Singapore

32

16

5/26/2009

Consider forces acting on a section of the beam at a distance b from the free end. The shear force in the web is Vw = qh

(1)

The total shear force in the stringers is equal to the sum of the vertical forces in the stringers:

P tanα1

V f = P (tan α1 + tan α 2 ) ⎛h h ⎞ = P⎜ 1 + 2 ⎟ ⎝x x⎠ Ph = x

P q

VW (2)

h

P tanα2 P

33


Now, the presence of P is caused by the bending moment due to the external load Vz , so that th t Ph = VZ b ∴ Vf =

Vz b x

(3) from (2)

Vz

P

(4) h

P b


34

17

5/26/2009

Also, considering vertical forces,

Vz = V f + Vw

(5)

∴ Vw = Vz (1 − b x ) = Vz x0 x

from (3)

(6)

From similar triangles, x0 x = h0 h

(7)

so that

Vw = Vz h0 h

(8)

and

V f = Vz (h − h0 ) h

(9)


35

Hence the shear flow in the web at a section of depth h is q=− =−

Vw I yy

∑Az

i i

Vw

(Ah 2) 2

*Note that we have used Vw instead of Vz

Ah 2

= − Vw h

(10)

h0 h2

(11)

∴ q = −Vz

in terms of applied shear load Vz and geometry.


36

18

5/26/2009

Let us now seek a more general analysis of the bending and twisting of tapered idealized beams. Consider the section of an idealized tapered multicell tube subjected to shear loads Vy and Vz , acting through an arbitrary point. The reference axis passes through the centroid of the section. Pn is the force in the nth stringer, and is in general composed p of three components p in the x,, y and z directions. Let these components p be denoted Pxn, Pyn and Pzn, respectively. Pn z

Vz

x

y

Vy

37


The axial component is given by

Pxn = σ xn An

where σxn is calculated by the flexure formula and An is the cross-sectional area of the nth stringer.

Pxn

We see that

x

Pxn Δx = Pzn Δz n Δz ∴ Pzn = n Pxn Δx Pyn Pzn

=

z

Pzn

(13) Pyn

Δzn

Δyn Δz n

Δy Δy ∴ Pyn = n Pzn = n Pxn Δz n Δx

Δzn Δx

L t us isolate i l t stringer ti id forces f i the th yz Let n andd consider in and xz planes.

Also,

z

Pzn (12)

Δy n (14)


y 38

19

5/26/2009

Now consider moment equilibrium about some convenient point C. Let the horizontal and vertical distances from stringer n to the point C be ξn and ηn, respectively. Also let the corresponding distances of the point of application of the shear loads Vy and Vz to the point C be ξ0 and η0. The total number of stringers is N and the total number of sheets is M. Pzn Pyn Vz

ηn

Vy

η0

C

ξn

ξ0


39

Taking moments about C, N

N

n =1

n =1

M

∴ Vzξ 0 − V yη 0 = ∑ Pznξ n − ∑ Pynη n + ∑ ∫ ri qi

Due to forces in stringers

i =1 i

ds ti

(15)

Due to shear flow in webs

Vertical and horizontal equilibrium of forces yield N

V y = ∑ Pyn + V yw

(16)

n =1

N

and

Vz = ∑ Pzn + Vzw

(17)

n =1

Here Vyw and Vzw are resultants of the shear forces carried by the webs. webs These values should be used in the calculation of shear flows in the webs. The solution procedure is similar to that for multicell boxes, except that the modified moment equilibrium equation (i.e. Eqn 15) should be used.


40

20

5/26/2009

Example 1: A tapered two-cell tube with symmetrical cross-sections 1.2m apart has the dimensions shown. The tube supports load which produce a bending moment My =-1.65 kNm and a shear force Vz =10 kN in the plane of the internal spar web at the larger cross-section. The shear modulus G is constant throughout. Determine the forces in the stringers and the shear flow distribution at this cross-section. Stringer areas: A1 = A3 = A4 = A6 = 600 mm2 , A2 = A5 = 900 mm2 Vz =10 kN 150mm

100mm

0.8mm

2

3

0.8mm

I 1.0mm

1.0mm

4

II

5

0.8mm

80 mm 180mm

1.0mm

0.8mm

200mm

1

6

400mm


41

At the larger cross-section, I yy = 4(600)(90) + 2(900)(90) = 34 ×106 mm 4 2

2

The direct stresses are given by σ xn = M y zn

Hence

Pxn =

e.g.

Px1 = −

I yy

M y zn I yy

An

(1600)(90)(600)(103 ) = −2620 N 34 ×10 6

The forces in the other stringers are similarly calculated. Note that Δx = 1200mm. e.g. For stringer 1, as x increases, z decreases; ∴ Δz = −50 mm

and

Δz = −0.042 Δx


42

21

5/26/2009

ξn and ηn are measured from the midpoint of the internal web 2-5. Stringer

Pxn

Δzn/Δx

Pzn

Δ y n / Δx

Pyn

Pznξn

Pynηn

1 2 3 4 5 6

-2620 -3930 -2620 2620 3930 2620

-0.042 -0.042 -0.042 0.042 0 042 0.042 0.042

110 165 110 110 165 110

0 0.083 0.125 0.125 0 083 0.083 0

0 -326 -328 328 326 0

44000 0 -22000 -22000 0 44000

0 29340 29520 29520 29340 0

Note:

Pzn =

Δz n Pxn Δx

6

Therefore,

∑P ξ n =1

zn n

∑P η

yn n

∑P

zn

6

∑P n =1

Hence, the shear forces carried by the webs are V yw = V y − ∑ Pyn = 0

= 117720 Nmm

n =1 6

6

n =1

= 44000 Nmm

Δy n Pxn Δx

6

6

n =1

Pyn =

and

yn

Vzw = Vz − ∑ Pzn = 10000 − 770 = 9230 N

= 770 N

n =1

=0


43

Now let shear flows q12 and q23 be unknowns. 1

2

3

II

I

4

5

6

Force equilibrium of stringers: Stringer 2,

Stringer 1,

V z A q25 + q23 = − zw 2 2 + q12 I yy = −22 + q12

V zA q12 = − zw 1 1 + q61 I yy = −14.7 + q61

Stringer 3, q34 = −

Vzw z3 A3 + q23 I yy

= −14.7 + q23

Also, q45 = q23 and q56 = q12


44

22

5/26/2009

Rate of twist, Consider cell I:

Consider cell II:

1 dφ ⎛ 2(200 )q23 180q23 − (14.7 )(180 ) 180q25 ⎞ = + − ⎜ ⎟ 1.0 1.0 ⎠ dx 2G (200 )(180 ) ⎝ 0.8 1 (860q23 − 180q12 + 1314) = 72000G

dφ 1 (− 180q23 + 1360q12 − 1314) = dx 72000G

(1)

(2)

Taking moments about the midpoint of web 2-5, we have (1.47)(180)(400) – (14.7)(180)(200) + 72000q23 + 144000q12 + 44000 + 117720 = 0

(3)

E Equating i (1) andd (2), (2) andd solving l i with i h (3), (3) one obtains b i q23 = -4.4 N/mm

and

q12 = -2.6 N/mm


45

Effect of Varying Moments of Area The box beam shown is subjected to a pure bending moment about the y axis. The stringer areas change from section A to section B over a distance d along the beam. The cross-sections of the beam are symmetric about the y axis. z Pi d

y Pi+ΔPi x

Section A

Section B


46

23

5/26/2009

Let us consider equilibrium of forces in the axial direction. ΔPi + (qi +1 − qi )d = 0 ∴ qi +1 = ΔPi d + qi

(18)

Now, the direction of ΔPi shown here produces a positive moment My. Thus ΔPi produces a change in the bending moment from sections A to B. Pi

ΔPi =

ΔM y zi Ai

∴ qi +1 =

qi+1

(19)

I yy

qi

ΔM y zi Ai I yy d

+ qi

(20)

Pi+ΔPi


47

Example 2: The single cell box beam is loaded by a transverse force of 1.0 kN. Between points B and C, the beam has constant stringer section as shown in section B-B. Between points B and D, two stringers taper uniformly, with the stringer areas at point A as indicated in section A-A. Calculate the shear flow at section A-A. Area of stringers: A1=100 mm2 A2=300 mm2 1.0 kN

A2

A1

A1

100

A

Section A-A

B D

C

A

A2 A1

B

50 C

A1

50 A1

A1 A1

100

*All dimension in mm.

Section B-B

A1 A1 50

50

A1


48

24

5/26/2009

Consider section B-B: 3B

2B

1B

Consider section A-A: 3A

I yy = ∑ An z n2

2A

1A

I yy = ∑ An z n2

= 6(100)(50 )

= 4(100 )(50 ) + 2(300 )(50 )

2

2

= 1.5 × 106 mm 4 4B

5B

= 2.5 × 106 mm 4

6B

4A

Stress in stringers due to bending σb = −

2

5A

6A

Stress in stringers due to bending

M yz

σb = −

I yy

(300)(1000)(50) =

M yz I yy

(500)(1000)(50) = 2.5 ×106 = 10 MPa

1.5 ×106 = 10 MPa

∴ Axial force in each stringer = (100)(10) = 1000 N

∴ Axial force in each corner stringer = 1000 N, and axial force in each of stringers 2A and 5A is = 3000 N

49


∴ ΔP 1 = ΔP 3 = ΔP 6 = ΔP 6 = 0 N but ΔP2 = ΔP5 = 2000 N

3 2 1

2000 N

4

A-A 5

2000 N

6

B-B


50

25

5/26/2009

Assume q61 unknown. Then q61 = q12. Vertical equilibrium of forces gives 2q61 (100) = 1000 ∴ q61 = 5 N/mm

and

q12 = 5 N/mm

Check for the middle stringer, q12 =

5 N/mm

ΔP2 2000 + q23 = + q23 d 200

5 N/mm

∴ q23 = 5 − 10 = −5 N/mm

5 N/mm

which is correct, because q23 = -q12 due to symmetry.


Note that if we had used qi +1 = −

Vz zi Ai + qi we would have obtained, I yy

for section B-B,

q61 = −

51

for section A-A,

(1000)(50)(100) + q

1.5 ×10 6 = −3.33 + q12

12

q12 = 8.33 N/mm

q61 = −

(1000)(50)(100) + q

2.5 ×10 6 = −3.33 + q12

12

q12 = 7 N/mm

These results would have been wrong.


52

26

5/26/2009


53

27

THE GREEK ALPHABET A

a

Bfl

r r

A 8 E C

z? 6

K K A h Mr

z e long, as in scene th as in thin

MU

o

ET

v P P

Z ~,fimI s

T T Yu

*+

Zeta

Nu

Nu 8 0

a

Eta Theta Iota Kappa Lambda

Htl Q 0 1

Alpha Beta Gamma Delta Epsilon

X X ./ P # Q W

Xi Omicron Pi Rho . Sigma Tau Uhilon P~ Chi Psi Omega

b g hard, as in begin'

d e short, as in met

i

k 1 m a X

o shmt, as in lot

P

r se t U

ph ch hard, as in lock PS

o long, as in throne

Except that before a, x or another y it is nasal-ng, as in anchor. Sharp as in this, but flat before or p, as in asbestos, dismal.

ME4212 Mechanics of Thin-Walled Structures Tutorial Sheet - Torsion of Thin-Walled Members 1. A hollow aluminium tube of rectangular cross section as shown in Fig. 1 is subjected to a torque 56.6 kNm along its longitudinal axis. Determine the shear stresses and the rate of twist. Assume G = 28 GPa. 500

250

6 10 12

6 Fig. 1

Ans: τ1 = 18.9 MPa; τ2 = 37.7 MPa; τ3 = 22.6 MPa; θ = 0.00687 rad/m 2. The aluminium shafts (G = 28 GPa) of cross-sections shown in Figs. 2(a) to 2(c) are each subjected to a torque of 2 kNm. Neglecting the effect of stress concentrations, calculate (i) the maximum shear stress and (ii) the angle of twist in a 2-m length. A 3 4 A R100 B R75 6

60o B

3 2 D

2 C

R100

150

4 C

(a)

(b)

B 3

4

All dimensions in mm 2

60o

A (c) Ans:

C 120

Fig. 2 (a) τAC = 10.04 MPa, τAB = 6.70 MPa; 0.31° (using exact dimensions) ME4212 Mechanics of Thin-Walled Structures T.E.Tay, Department of Mechanical Engineering, NUS 1

(b) τAB = τBC = 8.05 MPa, τAC = 5.37 MPa; 0.31° (using exact dimensions) (c) τAC = 80.18 MPa, τAB = 40.09 MPa, τBC = 53.46 MPa; 6.84°

3. The thin-walled closed section beam shown in Fig.3 is subjected to a torque of 4500 Nm. The section is constrained to twist about O, the centre of the semi-circular arc 512. For the curved wall 512, the thickness is 2 mm and the shear modulus is 22 GPa. For the plane walls 23, 34 and 45, the thickness is 1.6 mm and the shear modulus is 27.5 GPa. Determine the warping displacements of points 2, 3, 4 and 5.

1.6 mm 5

4

1.6 mm

O

1

100 mm R 50 mm

2.0 mm

2

3 1.6 mm 200 mm

Fig. 3 Ans: u 2 = −u 5 = 0.053 mm, u 3 = −u 4 = 0.187 mm.

4. a) The closed cross-section of a thin-walled tube shown in Fig. 4 is subjected to a torque of 6000 Nmm. Determine the magnitude and location of the maximum shear stress experienced by the tube, and the torsional constant. (All dimensions in mm.) ME4212 Mechanics of Thin-Walled Structures T.E.Tay, Department of Mechanical Engineering, NUS 2

b) If a longitudinal slit is cut through the wall of the section at Point A, determine the magnitude and location of the maximum shear stress when a torque of 6000 Nmm is applied. Ans: 2.89 MPa, 77412 mm4, 11.96 MPa.

10 2 3

4

30o

30o

60o

A

60o 5

1

20

6 20

30

Fig. 4

ME4212 Mechanics of Thin-Walled Structures T.E.Tay, Department of Mechanical Engineering, NUS 3

ME4212 Mechanics of Thin-Walled Structures Tutorial Sheet - Bending of Thin-Walled Members

1.

A beam 250 mm wide by 300 mm deep by 4.2 m long is used as a simply-supported beam on a span of 4 m. It is subjected to a concentrated load P at the midsection of the span. The plane of the loads makes an angle φ = - 4π/9 with the horizontal y-axis. The beam is made of material with a yield stress 25.0 MPa. If the beam has been designed with a factor of safety 2.50 against initiation of yielding, determine the magnitude of P and orientation of the neutral axis. Ans:

2.

31.4 kN, 14.25o

In Fig. 1 let b = 300 mm, h = 300 mm, t = 25.0 mm, L = 2.50 m, and P = 16.0 kN. Calculate the maximum tensile and compressive stresses in the beam, and determine the orientation of the neural axis. Ans:

89.81 MPa, - 66.83 MPa, 30.71°

60 mm 10 mm

10 mm 10 mm

30 mm t

L

70 mm

P 10 mm

h

70 mm t

b

Fig. 2

Fig. 1

3.

An extruded bar of aluminium alloy has the cross section shown in Fig. 2. A 2.10 m length of this bar is used as a simple beam on a span of 2.00 m. A concentrated load P = 5.0 kN is applied at mid-length of the span and makes an angle of φ = 4.54 rad with the y-axis. Determine the maximum tensile and compressive stresses in the beam. Ans: 80.93 MPa, - 75.3 MPa

1

ME4212 Mechanics of Thin-Walled Structures T.E. Tay, Dept of Mechanical Engineering, NUS

4.

A cantilever beam 1.6 m long has the section shown in Fig. 3. It carries a load P = 2.5 kN in the –z direction at the free end. Find the magnitude and location of the maximum tensile flexural stress. (All dimensions in mm.) Ans: 184.1 MPa, at y = 55, z = 70 z

z

10

A

B

70 y 10 70

y

O

100

10 D

55

C 42

55 Fig. 3

Fig. 4

5. A column has the I-section shown in Fig. 4. A compressive force P parallel to the axis of the column is applied at the corner A. Find the axial stresses in terms of P at the remaining three corners B, C and D. (All dimensions in mm.) I yy = 1.105 × 10 6 mm 4 I zz = 4.94 × 10 4 mm 4 A = 680 mm 2

Ans: 0.005195P, 0.009715P and -0.008135P

6. A uniform thin-walled beam has the open cross-section shown in Fig. 5. The wall thickness is constant. Determine the position of the neutral axis and the maximum direct stress for a bending moment of My = 3.5 kNmm. Take r = 5 mm and t = 0.64 mm. Ans: -51.67°, 100 MPa t

z r y

r

Fig. 5 2


7. The cross-section of a beam is shown in Fig. 6. The beam is subjected to a bending moment about the y-axis of 6000 Nmm. Determine the bending stresses at A, B, C and D. All dimensions in mm. (All dimensions in mm.) Ans: 0.206 MPa, -0.411 MPa, 0.411 MPa, -0.206 MPa. z 50 D

C 5

50

O

y 5

50

5 A B 50

Fig. 6 8. The closed two-cell cross-section of a fuselage shown in Fig. 7 is subjected to a bending moment My = 5000 kNm. a) Derive an expression for the position of the centroid in terms of r, tw and tf. b) Hence determine the maximum tensile and compressive bending stresses in the crosssection, if r = 1500 mm, tw = 5 mm and tf = 8 mm. Ans: −

rt f 9 19 ⎛⎜ 10 ⎜⎝ 10πt w + 9t f

⎞ ⎟ from O, 143.7 MPa, - 109.1 MPa. ⎟ ⎠ tw r O

tf

9r/5

Fig. 7 3


ME4212 Mechanics of Thin-Walled Structures Tutorial Sheet – The Shear Center 1. Fig.1 shows the cross-section of an open thin-walled beam of uniform thickness t. a) Determine the position of the shear center. b) Derive expressions for the torsional constants of the open cross-section, and the cross section if it were closed at point A. If the allowable shear stress is 30 MPa, a = 25 mm and t = 2 mm, determine the maximum torques that can be applied to the open and closed sections. B

A

a t D

C a

Fig. 1 Ans: (a)

a 4 2

from C. (b) 4 Nm, 75 Nm.

2. Find the shear centers of the following open thin-walled beam sections (Fig.2): z

z t

t

y

y r

Fig. 2 (a)

Fig. 2 (b) z

z 60° y

2c

y h

t

60° t b

Fig. 2 (d)

Fig. 2 (c) Ans: (a) -2r (b) -2r (c) -0.577c (d) 3b2/(h+3b)


3. Find the shear center if the thickness of the web tw is small compared with the thickness of the flanges tf (Fig. 3). z c y O tf tw b

Fig. 3

Ans: - b/2

4. A thin-walled beam has the cross-section shown in Fig. 4. The thickness of each flange varies linearly from t1 at the tip to t2 at the junction with the web. The web itself has a constant thickness t3. Calculate the position of the shear center from the web. t2

t1

h t3

d

Fig. 4

Ans:

(2t1 + t 2 )d 2 3d (t1 + t 2 ) + ht 3

5. The thin-walled open tube (Fig. 5) of constant thickness t has a narrow longitudinal slit at the corner 1-5. Calculate and sketch the shear flow distribution due to a vertical shear force V acting through the shear center E, and note the values at points 2, 3 and l 4. Hence show that the distance ye is . 2(1 + a / b ) 4 ye 5

3

h E

1 a

b 2 l

Fig. 5 ME4212 Mechanics of Thin-Walled Structures T.E.Tay, Department of Mechanical Engineering, NUS 2

6. Determine the positions of the shear centers of the following thin-walled sections:

t

t 20°

r

20° t

t t

r

50 mm

Fig. 6 (b)

Fig. 6 (a)

t

t

45° 45° r t

r

O

O t

t

Fig. 6 (d)

Fig. 6 (c)

Ans: (a) – 1.47r (b) 37.3 mm (c) – 1.35r (d) – 0.53r


ME4212 Mechanics of Thin-Walled Structures Tutorial Sheet – Idealized Beams

1. Find the shear centers of the following idealized thin-walled beam sections (Fig.1) (All dimensions in mm): A E

A

1.5

10

C

40

B

B

1.5

40

20

C

1.5

40

F D

10

70

Fig. 1 (a)

2

1.5

D

1.5

Areas of stringers A, B, C and D are 120 mm2 each, Areas of stringers E and F are 60 mm2 each. Thickness of each sheet is 2 mm.

A

E F

20 Area of each stringer is 25 mm2.

Fig. 1 (b)

B z 60

2 y

E 60

2 D

C 60

80

2

Areas of stringers A, B, C and D are 120 mm2 each, Area of stringer E is 80 mm2.

Fig. 1 (c)

Ans: (a) -21.7 to the left of BC (b) 3.33 to the right of CD (c) 40 to the left of E.


2. The idealized symmetrical cross-section of an aircraft fuselage (Fig. 2) is subjected to a bending moment of 100 kNm about the horizontal axis. If all the direct (normal) stresses are carried by the stringers, determine the average stress in each stringer. The stringer areas are: A1 = 640 mm2, A2 = 600 mm2, A3 = 600 mm2, A4 = 600 mm2, A5 = 620 mm2, A6 = 640 mm2, A7 = 640 mm2, A8 = 850 mm2, A9 = 640 mm2. Suppose the bending moment arises from the application of a shear force Vz = -100 kN, determine the shear flows in the webs.

1 2

2

3

3

4

4 5

1200 1140

5 960 768

6

6

565 336

7

7

144 8

8 9

38

All dimensions in mm.

Fig. 2 Ans (selected): σx1 = 35.6 MPa, σx2 = 32.3 MPa, σx6 = -11.0 MPa, σx8 = -27.0 MPa, q21 = -11.82 N/mm, q32 = -31.24 N/mm, q54 = -52.21 N/mm, q76 = -46.01 N/mm.

3. Find the stresses in the stringers and the shear flow in the webs of the 1.2 m-long beam shown in Fig. 3. The area of each stringer is A mm2. 1.4 kN

t t

100 mm

45º

30º t

Fig. 3 Ans: 12246/A, 21286/A, -33522.5/A, 0 N/mm, -10.28 N/mm, 17.81 N/mm.


4. The single cell cross-section of a thin-walled tube (Fig. 4) is symmetric about the horizontal axis. The normal stresses are carried by the stringers 1 to 4, while the webs are effective only in carrying shear stresses. Calculate the shear stresses in each web. Cell area = 135000 mm2. Stringer areas: A1 = A4 = 450 mm2; A2 = A3 = 550 mm2. Web

Length (mm)

Thickness (mm)

12, 34 23

500 580

0.8 1.0

41

200

1.2 1.0 kN

2

1 100

3

4

100

500

Fig. 4

Ans: τ12 = 2.16 MPa, τ23 = -1.02 MPa, τ34 = 2.16 MPa, τ41 = 3.32 MPa.


ME4212 Aircraft Structures Tutorial Sheet – Multicell Sections

1. The cross-section of an idealized three-cell torque box is shown in Figure 1. The thickness of each web is t and the cross sectional area of each stringer is As. The shear modulus of the web material is G and the Young’s modulus of the stringer material is E. If the section is subjected to a constant torque T and allowed to warp freely, determine the rate of twist and the maximum shear flow. A

a

B

C

2a

a

a

D a

E

a 2a

F

a

a 3a

G

H

Figure 1 Ans:

.

,

0.0867

2. The two cell thin-walled section is fres to warp, and is subjected to a load of 10 kN as shown in Figure 2. Determine the rate of twist. R = 50 mm z t1 = 2 mm t2 = 2 mm t1 G = 20 GPa Vz=10 kN R y

2R

O

t2

t1

t1 2R Figure 2 Ans:

3.86

10 rad/mm ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 1

3. Use the method of successive approximations to find the position of the shear centre of the idealized thin-walled section shown in Figure 3. Each stringer has an area A and each web has a thickness t. Assume all direct bending loads are carried by the stringers and the webs carry only shear loads. 5

4

a

a a

a

3 a

a

1

2

Figure 3 Ans: ye = 0.025a =

to the right of 15

4. The cross-section of a three-cell torque box is shown in Figure 4. It is constructed of seven stringer of area 400 mm2 each and nine webs. Aa vertical force of 13.5 kN is applied to stringer 2. Assume that the webs carry only shear stresses and that the section is free to warp. Determine the shear flow distributions in the webs. Shear modulus of the web material G = 60 GPa. Young’s modulus of the stringer material E = 200 GPa. 13.5 kN

(All dimension in mm) 0.5

0.5

3

2

0.4

1

0.5

0.5

0.4

7

0.5

6 0.5

100

4

150

5 0.5

200

200

Figure 4 Ans: q23=+43.56 N/mm, q34=+34.135 N/mm, q12=+42.59 N/mm, q72=+30.96 N/mm, q63=+20.58 N/mm, q45=+4.14 N/mm.

ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 2

ME4212 Aircraft Structures Tutorial Sheet – Tapered Beams & Varying Moments of Area

1. A uniformly tapered thin-walled beam consisting of four sheets and four stringers is shown in Figure 1. The thickness of each sheet is 2 mm and the area of each stringer is 200 mm2. The distance between the larger and smaller corss-sections is 800 mm. A vertical load of 2 kN us applied at stringer 2 in the plane of the smaller cross-section. Calculate the forces in the stringers and shear flow distribution at the larger crosssection. (All dimension in mm)

2 kN

1

2 120 100

3

4 100

180 Figure 1 Ans: q31=4.63 N/mm, q12=-2.31 N/mm, q43=-2.31 N/mm, q24=-9.26 N/mm.

2. Figure 2 shows a uniformly tapered thin-walled beam consisting of four sheets and four stringers. The thickness of each sheet is 2 mm and the cross-sectional area of each stringer is 200 mm2. The distance between the larger and smaller cross-sections is 800 mm. A vertical load of 3 kN is applied to stringer 4 in the plane of the smaller cross-section. Assuming that the sheets carry only shear stresses and the stringers carry only normal stresses, determine the shear flow distributions in the sheets at the larger cross-section.


120 30

40 1 z

2

O

y

3

100

60

4

40

3 kN

Figure 2 Ans: q31=3.5 N/mm, q12=12.5 N/mm, q43=12.5 N/mm, q24=21.5 N/mm. 3. The tapered thin-walled beam shown in Figure 3 has the cross-section 600 mm apart and is made of six stringers and six webs. The webs have a thickness of 2 mm each. A vertical load of 2 kN is applied to stringer 1 in the plane of the smaller cross section. Assuming that the webs carry only sehar stresses and the stringers carry normal stresses, determine the shear flow distributions in the webs ant the larger crosssection. Shear modulus of the web material G = 40 GPa. The areas of the stringers: A1 = A3 = A4 = A6 = 500 mm2 A2 = A5 = 300 mm2 2 kN

2

2

2

1

z

3 2

2 y

200 100

4

100 5

2

150

100

2

6

150 Figure 3

Ans: q12= q54=3.91 N/mm, q23= q65=2.756 N/mm, q36=0.8333 N/mm, q41=5.83 N/mm.


4. The cantilever box beam of length 2000 mm shown in Figure 4 consists of six sheets and six stringers. It is built-in at one end and loaded by a vertical force of 800 N at the other end. Stringers 2 and 5 have constant cross-sectional areas throughout the length of the beam. Stringers 1, 3, 4 and 6 have uniformly varying cross-sectional areas. Determine the shear flows in the sheets at a distance 600 mm away from the loaded end. Assume that the sheets carry only shear and the stringers carry only normal stresses. 800 N B

B 600 2000 Figure 4 (All dimension in mm) 1

3

2

800 N

1

3

2

120 4

5

6

120 5

4

120

120

Section B-B

Loaded end

Areas of stringers: A1 = A3 = A4 = A6 = 150 mm2 A2 = A5 = 50 mm2

6

Areas of stringers: A1 = A2 = A3 = A4 = A5 = A6 = 50 mm2

Ans: q41= q63=3.33 N/mm, q12= 0.473 N/mm, q65=-0.473 N/mm.


5. The box beam shown in Figure 5 is loaded by a vertical force of 800 N. Stringers 1, 3 and 5 have uniformly tapered areas, while stringers 2, 4 and 6 have constant areas along the span of the beam. Determine the shear stresses in the webs at section A-A, a distance of 540 mm from the loaded end. (All dimension in mm)

800 N A

A

540

2000 Figure 5

100 1

3

2

5

1

3

2

90

3 4

800 N

100

90

3

6

4

Section A-A

5

6

Loaded end

Areas of stringers: A1 = A3 = A5 = 120 mm2 A2 = A4 = A6 = 80 mm2

Areas of stringers: A1 = A2 = A3 = 80 mm2 A4 = A5 = A6 = 80 mm2

Ans: q41= q63=4.444 N/mm, q12= 1.111 N/mm, q65=-1.905 N/mm.


ME4212 Mechanics of Thin-Walled Structures Solutions to Tutorials Torsion of Thin-Walled Sections

z, = 4

Hence

-

226400

tI

=

18.87 MPa

0.0 12

z2 = 226400 = 37.73 MPa

0.006

Z3

=

226400 0.01

=

22.64 MPa

To calculate rate of mist. use

6

=

6.87~10-' radlm.

j-

4 ds 2 G 6 =R t

Using

.r

=

2nt

~ B I I

=

ZA(

=

2000 2 x 4 9 . 7 8 3 ~lo-' x 2 x lo-'

=

10.04 MPa

AH

=

TC,/I

=

2000 2 x 49.783 x lo-' x 3 x 1 o

=

6.696 MPa

Angle of twist,

-~

T L 4n2G

4

= ---

=

0.005376391 rad

tan 30'

= 97/x

x

=

168

97 sin 30' = -3 AB = 194 AB

rB(.

=

r~~

Z~~

=

~ U I=

=

Angle of twist,

2000 = 8.05 MPa 2 x 3 1 . 0 6 8 ~ 1 0 -x~4 x 1 0 - ~ 2000 = 5.36 MPa 2 x 3 1 . 0 6 8 ~lo-; x 6 x 1 0 - ~

4

=

2 x lo-; 2 4 x 28x lo9 x ( 3 1 . 0 6 8 ~ 1 0 - ~ ) ~

+ --4

B

1 (120) (103.92) = 6.235 x lo-; m2 2 T From r = 2Qt 2000 ' , 4 ~ = = 40.09 MPa 2xRx4x10-~ Q

=

-

Z,K

=

2000 2 x R~ 2 x lo-'

rB(,

=

2000 = 53.46 MPa 2 x a x3 10-3

Angle of twist,

'=

=

80.19 MPa

4 =

2 A

120

2000 x 2 4~28xl0~~(6.235~10~)~

C

120 cos 30 103.92 mm

Fr&,Locate A m A =

z

@

0 '

rads 2 80.2' I n 9 0

Mere

I

(I- )

I -- 06+60l =

as173

119 cirtulw ,

- IYY, + 6

c~*r&)6205*55)~ = 5.5X~dowm+

3T as2m s ,ts

%at3 tb* Jop-l3b t 3 = 3 OP

@

d

p uLb c & _-&

o

7r

t-r

w 'dosed

'-

'

ia~t

.a.

-'.

Iy45

-- + x b

N&

12

Tyy= fLz[a 5 +b) .

f4w

8 d

s.tavte8 e ws

% ,, ly

ibt

Lr*=O.

C

o

-tb web

walyci+i

%$H. Tf

lroea d 1 2 ,%,> be m'Geu6bGseie

%--&~3,

I = rcor (608) = r(cosw0ctm + ~ i ~ r tsine) 50

srCoSC3

ds, = rde

sa=z+r

r't

Y€

ME4212 Aircraft Structures Tutorial Solutions – Multicell Sections

Question 1. A

a a

B

C

2a

q1

a

q2

a

D a

E

2a

F

q3

a

a

3a

G

H

Thickness of all webs = t Area of all stringers = As Shear modulus of web = G Determine the rate of twist.

Cell ABED: 1

2 1 2 1 2

4

1

4

Cell BCFE: 1 2 2 1 4 1 4

6 6

2 2

2


Cell DEFHG: 1 3

2 1

8

6

3

2

Moment equilibrium: Ω

2

2

2

2

2

3

4

3

From 1 4

2

1a

From 2

6

2

4

2a

From 3

2

8

6

3a

4

2a

1a

3a

2a

5a

9 10

23

8 4a

18

10

2

15.8

1.253

9

19.8

1.202

4a 5a 6a

1.11

Substitute into □ 4 , 2

1.11

2 1.253

14.44

3 1.202

0.0692

0.0867


Question 2. Define the coordinates and points as shown. q1T and q2T are constant shear flows due to torsion, to be superposed with the shear flow due to bending. s1

D

C

q2T

θ

q1T E

O

s3 A

B

s2

cos

sin

27.62 sin

1 2

0.017

1 2

2 50

55.2

2

on AB

0.55

cos on DEA

0.552

on BC

At B,

on CD and at C,

55.2 ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 3

1 2

1 2

0.011

50

Rate of twist for the ith cell: 1 2 Ω

For cell ADE, 2

2

2

2 2

2

2 3

2 2

4 3

2

1

For cell ABCD, 1 4

2

2 4

2

2

3

1

2

4

1

2

Substituting values into □ 1 and □ 2 , we have 2

0.03273

2

0.005

0.01273 0.01833

0.2345 and 0.4604

Solving for q1T and q2T, 115.84 140.708

2.921 25.914

Here, 2

14 3

2 3

1 3

1.81

10 mm


Consider now moment equilibrium about O: 2

2

0

which reduces to 8

4 3

10

0

On substitution for the values, the equation becomes 20000

7853.98

828729.3

0

Substituting for q1T and q2T into the above, we have 7.72

3.86

10 10 rad/mm


Question 3. Ω

2 2

0.3927

8

Ω √3 2 2

Ω

√3 4

0.433

1

1

2 4 3

2.571

2

,

,

Carry over factors: 1 4

,

,

4

2

,

,

1 3

2 1 3

4

1 4

,

,

,

,

2

0.389

Let open section: 5

4

0 0

3

1

2 2 2

2

4

2

2 ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 6

2

2

2

0

2

2

2

Cell I

δ

Cell II 4

2.571

1 4

0.389

0.195

2 1 3

1 4 0.167

0

0

0.0487

0.042

0

0.019

0

0

0.014

0

0.005

0.214

Final q

3

0

2 C

Cell III

0.0035 0.1

0 0.181

Check for vertical equilibrium: 0.214

0.386

0.581

0.181

0.1

5

4

0.214

0.181

3

0.581

0.386

0.181

1

2

0.1


Take moments about midpoint of 15, 0.214

2

0.1 2

2

0.181 2

2 cos 60°

0.581 0.181 2

2

sin 60°

0.025

40

to the right of 15.


Question 4. 13.5 kN 0.5

0.5

0.5

200 mm

2

200 mm

3

4 z

0.4

0.5

1 I 0.5

II

150 mm

III

0.4

125 mm

5

6

7 0.5

1

y

0.5

But

0 400 75

13.5 2

6

10 mm

13500 75 400 13.5 10 30

3

30

4

30

By symmetry, q56= q34 , q67= q23 6

30 30

7

30 30

Assume unknowns q12, q23 and q34. 1 2 Ω


For cell I, 1 50 150

2

1 7500 1 7500

125 0.5

2

500

375

875

375

150 0.4 30 11250 _____ i

For cell II, 1

2

150 0.4

150 200 1 30000

200 0.5

2

375

30

375

150 0.4 800

30

1 375 30000

1550

375

_____ ii

For cell III, 1

2

2

150 200 1 30000 1 30000

800

200 0.5

150 0.5

300

30

1475

2250

375

150 0.4 375

30

_____ iii

Taking moments about the point 1, 150 100

13500 100

200 75

2 13500 13500

150 150

150

200 75 750

4500 300

600

150

2

450

22500 54000

150 300

450 300

600

150 500 0

300

300

450

0

13500

750

0

0 _____ iv

Combining (ii) & (iii), 375

1550

375

375

1925

1850

1475 2250

2250

375

0 _____ v

Combining (i) & (ii), 300 75

875

375

11250

375

1550

375

3500

1500

45000

375

1550

3875

3050

375

45000

0 _____ vi

375


From iv ,

360

4

Substitute into v , 137250 Substitute into vi , From vii ,

4 3425

1350000

9.786

350

18550

0 _____ vii 15875

0 _____ viii

392.143

Substitute into (viii), 1350000 173902.75

18550

15875 9.786

392.143

0

7575270.125 43.56 N/mm 34.135 N/mm 42.59 N/mm 30.96 N/mm 20.58 N/mm 4.14 N/mm

(Check vertical equilibrium)


ME4212 Aircraft Structures Tutorial Solutions – Tapered Beams & Varying Moments of Area

Question 1. At the larger cross‐section,

4 200 60 2.88

2000 800 60 2.88 10

Bending stresses

200

∆

800 mm

10 mm

33.33

33.33 MPa

6.67 kN

Stringer no.

Δz

Δy

Pz

Py

Px

1

-20

+80

+166.75

-667.0

-6670

2

-20

0

+166.75

0

-6670

3

0

+80

0

+667.0

+6670

4

0

0

0

0

+6670

333.5

Δ Δ Δ 6.67 800

Δ Δ

0

Δ Δ

10

∴ Shear force carried by webs: 2000

333.5

1666.5 N


1

2 z

6.94

y

120 mm

180 mm 3

4

6.94 13.89

Take moments about 2: 180 120

180 180 120

166.75 120

666.7 180

166.75 180

667 120

0

0 2.315

4.63 N/mm

2.31 N/mm

2.31 N/mm

9.26 N/mm

Check vertical equilibrium: 120

1666.8 N


Question 2. 4 200 50

At larger cross‐section, 2

10 mm

10 mm

3000 800 50 2.88 10

Bending stresses

2

60 MPa

12000 N

∆

800 mm

Δ Δ

and

Δ Δ

Stringer no.

Δz

Δy

Pz

Py

1

-30

+40

-450

+600

2

-30

-40

-450

-600

3

+10

+40

-150

-600

4

+10

-40

-150

+600

1200 N

0

∴ Shear force carried by webs: 3000 1

1200

1800

2

9 3

4

18

Take moments about 4: 100 120

120 100 600 100

450 120 600 100

150 120 3000 40

16 2

7

3.5 N/mm

12.5 N/mm

12.5 N/mm

21.5 N/mm ME4212 Aircraft Structures T.E.Tay, Department of Mechanical Engineering, NUS 3

Question 3. At larger cross‐section,

26

2000 600 100 26 10

Bending stresses

10 mm

4.615 MPa

Pzn

z Δ Δ

Pxn

Δz

Δ Δ

Δx x Pzn

z

Δ Δ Δ Δ

Pyn

Δz Δy

y

Stringer no.

Pxn

Δz

Δy

Pzn

Pyn

1

-2307.5

-50

100

192.3

-384.58

2

-1384.5

-50

50

115.375

-115.375

3

-2307.5

-50

0

192.3

0

4

2307.5

50

100

192.3

384.58

5

1384.5

50

50

115.375

115.375

6

2307.5

50

0

192.3

0

ΣPzn=1000

ΣPyn=0

Shear forces carried by webs: 0 1000 N


1

4

2

3

5

6

1.154 1.154

3.077

1.154

_23

Σ moments about 4 = 0 150 200

150 200

200 300

115.375 200

115.375 150

115.375 150

192.3 300

2

384.58 200 192.3 300 0

8.332

Substitute for q23 & q36 and solving for q12, 2

1.154

2

2 3.077 4

8.332 15.64 3.91 N/mm 2.756 N/mm 0.833 N/mm 5.83 N/mm


Question 4. At section B‐B:

4 150 60

Forces in streingers:

600 800 60 2.52 10

1714.3 N

571.43 N

1

2

26

10 mm

Top bending stress

2 50 60

11.43 MPa

1714.3 N 571.43 N 3

Vertical equilibrium: 2

120

800 N

1 3 N/mm 3 Δ 4

6

5

1714.3 3.33 600 1714.3 3.33 600

0.473 N/mm

1

2

0.473 N/mm 0.473 N/mm

0.473 N/mm

3

3.33 N/mm

3.33 N/mm

4 0.473 N/mm

5 0.473 N/mm

6


Question 5. At section A-A: 3 80

120

1

600 mm 2

1

1 90 80 600

+ C 1.21

48

42 mm

42

240

80

48

120

160

10 mm

1.2096

10 mm

6

5

Forces in stringers:

800 540 42 1.2096 10

Top, 1800

1371.44 2

17.143

2057.16 3

15 1800

1200

800 540 48 1.2096 10

Bottom,

1

120

3 42

4

80

1371.44

Vertical equilibrium: 2

90

800 4.444

4

6

5 1800 540

1.111 N/mm

1

4.444

2

1.111 N/mm

3

1.111 N/mm 1371.44 540

4.444 N/mm

4.444

4.444 N/mm

1.905 N/mm 4

1.905 N/mm 5

1.905 N/mm

6


Aircraft Structures Notes 1

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