Show that there is no warping in a bar of circular cross-section

Solution:

(a) Saint-Venant assumed that as the shaft twists the plane cross-sections are warped but the projections on the x-y plane rotate as a rigid body, then,

zy u=−θ

zx

v=θ (3.1.1)

) , (x y w=θψ

where ψ(x,y) is some function of x and y, called warping function, and θ is the angle of twist per unit length of the shaft and is assumed to be very small.

(b) From the displacement field above, it is easy to obtain that

=0

=

=

= _{yy zz xy}

xx ε ε γ

ε

So from the stress-strain relationship, we have

=0

=

=

= _{yy zz xy}

xx σ σ τ

σ

Therefore the equilibrium equations reduce to

=0

∂ +∂

∂

∂

y x

xz τyz

τ

This equation is identically satisfied if the stresses are derived from a stress function φ(x,y), so that

xz y

∂

= ∂φ

τ ,

yz x

∂

−∂

= φ

τ (3.1.2)

(c) From the displacement field and stress-strain relationship, we can obtain x y

w z u x w

xz θ

γ −

∂

= ∂

∂ +∂

∂

= ∂ (3.1.3)

y x w z v y w

yz θ

γ +

∂

= ∂

∂ +∂

∂

= ∂ (3.1.4)

So it forms the compatibility equation γ γ θ

=2

∂

−∂

∂

∂

y x

yz xz

, or in terms of Prandtl stress function φ φ θ

y G

x ₂ 2

2 2

2 =−

∂ +∂

∂

∂ (3.1.5)

(d) Boundary conditions,

=0 ds dφ

, or φ =const. But for a solid sections with a single contour boundary, this constant can be chosen to be zero. Then we have the boundary condition

=0

φ on the lateral surface of the bar.

(e) For a bar with circular cross-section, assume the Prandtl stress function as

) 1

( ₂

2 2 2

− +

= a

y a C x

φ which satisfies the boundary conditions stated above.

Substitute φ into (3.1.5), we obtain C a²Gθ 2

−1

=

Then ( )

2

2 2

2 y a

G x + −

−

= θ

φ

Using (3.1.2), we have y y

xz G φ θ

γ =−

∂

= 1 ∂

, and x

x

yz G φ θ

γ =

∂

− ∂

= 1 Comparing with (3.1.3) and (3.1.4), we have

y x y

w

xz θ θ

γ − =−

∂

= ∂ => =0

∂

∂ x

w . Thus, w= f( y)

x y x

w

yz θ θ

γ + =

∂

= ∂ => =0

∂

∂ y

w , Thus, w=g(x)

Hence we conclude . This means that the cross-section remains plane after torsion. In other words, there is no warping.

const w=

Therefore can be verified, and it successfully expresses the statement.

0 ) , (x y = w

--- ANS

3.1.2

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3.2 Show that the Prandtl stress function for bars of circular solid sections is also valid for bars of hollow circular sections as shown in Fig. 3.34. Find the torsion constant in terms of the inner radius and outer radius , and compare with the torsion constant obtained using (3.59) for thin-walled sections. What is the condition on the wall thickness for the approximate to be within 1 percent of the exact ?

J a_i a₀

J J

ai

a0

Figure 3.34 Bar of a hollow circular section

Solution:

Recall:

(a) Saint-Venant assumed that as the shaft twists the plane cross-sections are warped but the projections on the x-y plane rotate as a rigid body, then,

zy u=−θ

zx

v=θ (3.2.1)

) , (x y w=θψ

where ψ(x,y) is a function of x and y, called warping function, and θ is the angle of twist per unit length of the shaft and is assumed to be very small.

(b) From the displacement field above, it is easy to obtain that

=0

=

=

= _{yy zz xy}

xx ε ε γ

ε

So from the stress-strain relationship, we have

=0

=

=

= _{yy zz xy}

xx σ σ τ

σ

Therefore the equilibrium equations reduce to

=0

∂ +∂

∂

∂

y x

xz τyz

τ

This equation is identically satisfied if the stresses are derived from a stress function φ(x,y), so that

xz y

∂

= ∂φ

τ ,

yz x

∂

−∂

= φ

τ (3.2.2)

(c) From the displacement field and stress-strain relationship, we can obtain x y

w z u x w

xz θ

γ −

∂

= ∂

∂ +∂

∂

= ∂ (3.2.3)

y x w z v y w

yz θ

γ +

∂

= ∂

∂ +∂

∂

= ∂ (3.2.4)

So it forms the compatibility equation γ γ θ

=2

∂

−∂

∂

∂

y x

yz xz

, or in terms of Prandtl stress function φ φ θ

y G

x ₂ 2

2 2

2 =−

∂ +∂

∂

∂ (3.2.5)

(d) Boundary conditions,

=0 ds dφ

, or φ =const.

---

1. To show that the Prandtl stress function for bars of circular solid sections is also valid for bars of hollow circular sections, we have to show that the Prandtl stress function for hollow circular sections satifies equilibrium equations, compatibility equations as well as traction boundary conditions.

(1) Equilibrium equations

Prandtl stress functions by their definition must satify equilibrium equations..

(2) Compatibility equations

Use the Prandtl stress function as it stated for bars of circular solid sections )

1

( ₂

0 2 2 0

2 + −

=

a y a C x

φ (here we use a₀. Assuming ( ₂ 1)

2 2

2 + −

=

i

i a

y a C x φ

would be fine too).

Then substitute φ into (3.2.5), we have C a₀²Gθ 2

−1

= . Thus we have

) 1

2 ( ₀²

2 2 0 2 2

0 + −

−

= a

y a

x a Gθ

φ . (3.2.6)

Therefore we have a stress function for bars of hollow circular sections satisfying the compatibility equation

(3) Traction boundary conditions

3.2.2

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To satisfy the traction boundary conditions we must show =0 ds dφ

on the traction free surfaces.

.

0 const

a

It shows that the B.C.’s have been satisfied.

(4) Since equilibrium equations, compatibility equations and traction boundary conditions are all satisfied, the Prandtl stress function for bars of circular solid sections is also valid for bars of hollow circular sections.

--- ANS 2. Compare torsion constant

(1) The torque produced by the stresses is

∫∫

^{− ∂}_∂ ^{− ∂}_∂ Substituting (3.2.6) into (3.2.7) and use polar coordinates to perform

integration, we have,

)] (2) Using (3.59) in the textbook for thin-walled sections, we have the

approximate torsion constant

=

∫

where A is the area enclosed by the centerline of the wall section.

2

Substituting J_app and J into the above error equation, we have

01 . 0 ) (

2

) (

) 2(

) 2(

) (

) 4((

2 2 0

2 0 4

4 0

4 4 0 0

3

0 ≤

+

−

= −

−

−

−

−

= +

−

i i

i

i i

app i

a a

a a a

a

a a a

a a a J

J J

π

π π

Because a_i and a₀ are positive real number, we have 01

. 0 ) (

2

) (

2 2 0

2

0 ≤

+

−

i i

a a

a

a => ( ) 2.040816( ) 1 0

0 2

0

≤ +

− a

a a

a_{i i}

We can obtain the solution of the above equation as 2235

. 1 8174

. 0

0

≤

≤ a a_i

Since a₀ >a_i we have the solution 0.8174

0

a ≥ a_i

Therefore the condition on the wall thickness t is

0 0

0

0 a a 0.8174a 0.1826a

a

t= − _i ≤ − =

(OR t a a_i a_i a_i 0.2235a_i 8174

. 0

1

0 − ≤ − =

= )

--- ANS

3.2.4

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3.3 Consider the straight bar of a uniform elliptical cross-section. The semimajor and semiminor axes are a and b, respectively. Show that the stress function of the form

) 1

( ₂

2 2 2

− +

= b

y a C x φ

provides the solution for torsion of the bar.

Find the expression of C and show that

2 2

3 3

b a

b J a

= π +

3

2 ab

Ty

zx π

τ = ⁻ ,

b a

Tx

zy 3

2 τ =π and the warping displacement

G xy b a

a b w T _{3 3}

2

2 )

( π

= −

Solution:

Recall:

1. Saint-Venant assumed that as the shaft twists the plane cross-sections are warped but the projections on the x-y plane rotate as a rigid body, then,

zy u=−θ

zx

v=θ (3.3.1)

) , (x y w=θψ

where ψ(x,y) is warping function, and θ is the angle of twist per unit length of the shaft and is assumed to be very small.

2. From the displacement field above, it is easy to obtain that

=0

=

=

= _{yy zz xy}

xx ε ε γ

ε

From the stress-strain relationship, we have

=0

=

=

= _{yy zz xy}

xx σ σ τ

σ

Therefore the equilibrium equations reduce to

=0

∂ +∂

∂

∂

y x

xz τyz

τ

which is identically satisfied if the stresses are derived from a stress function )

, (x y

φ , so that

xz y

∂

= ∂φ

τ ,

yz x

∂

−∂

= φ

τ (3.3.2)

3. From the displacement field and stress-strain relationship, we can obtain

x y w z u x w

xz θ

γ −

∂

= ∂

∂ +∂

∂

= ∂ (3.3.3)

y x w z v y w

yz θ

γ +

∂

= ∂

∂ +∂

∂

= ∂ (3.3.4)

The compatibility equation becomes γ γ θ

=2

∂

−∂

∂

∂

y x

yz xz

, or in terms of Prandtl stress function φ φ θ

y G

x ₂ 2

2 2

2 =−

∂ +∂

∂

∂ (3.3.5)

4. The boundary condition along the bounding surface is

=0 ds dφ

, or φ =const.

---

(a) Let the stress function be of the form ( ₂ 1)

2 2 2

− +

= b

y a C x

φ . In order to show this stress function provides the solution for torsion of the bar, we have to show that this stress function satisfies the equilibrium equations, compatibility equations and traction boundary conditions.

(1) Equilibrium equations 2 ) ( ₂

b C y

xz y =

∂

=∂φ

τ , 2 )

( ₂ a C x

yz x =

∂

−∂

= φ τ

Substituting the above stress expressions into the equilibrium equations, we have

0 0 0+ =

∂ = +∂

∂

∂

y x

xz τyz

τ

(2) Compatibility equations

Substituting ( ₂ 1)

2 2 2

− +

= b

y a C x

φ into (3.3.5) we get

2 2

2 2

b a

b G a

C=− θ + . (3.3.6)

Therefore we have a stress function satisfying compatibility equation (3) Traction boundary conditions

To satisfy the traction boundary condition we must show =0 ds dφ

on the traction free lateral surface.

Since the boundary of the cross section is given by the equation

3.3.2

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0

it satisfies the required condition =0 ds dφ

Since equilibrium equations, compatibility equations and traction boundary conditions are all satisfied, the stated stress function provides the solution for torsion of the bar.

--- ANS (b) Torsion constant J

(1) We have the torque produced by the stresses is

∫∫

^{− ∂}_∂ ^{− ∂}_∂

Note that the integral part of the above equation is the area of the elliptical cross-section. It can be easily obtained that dA ab

b (c) The warping displacement can be derived from (3.3.3), (3.3.4), (3.3.9), (3.3.10)

From (3.3.9) and (3.3.10), we have ² ₃

3

From (3.3.11), we can obtain )

Then differentiating (3.3.13) with respect to y, we have )

Thus, the warping displacement is ) .

And it is easy to also find that the warping function

b xy

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3.4 A thin aluminum sheet is to be used to form a closed thin-walled section. If the total length of the wall contour is 100 cm, what is the shape that would achieve the highest torsional rigidity? Consider elliptical (including circular), rectangular, and equilateral triangular shapes.

Solution:

(a) We denote as torsional rigidity, for the same material in comparison, only the torsion constant needs to be taken into consideration.

GJ

J

For the closed thin-walled section, the torsion constant J is

=

∫

t ds J A

/ 4 ²

(3.4.1)

where A is the area enclosed by the centerline of the wall section.

We now have a thin aluminum sheet with its thickness , all shapes of products made from this aluminum sheet will have the same thickness, . Also, the total length of the wall contour is 100cm. Then

t

t

∫

^{ds /t} is the same for all shapes of the cross-section. Consequently, only A needs to be taken into consideration in the evaluation of the torsional rigidity.

(b) Comparison ofA

(1) Elliptical cross-section

For the elliptical cross-section, the cross-sectional area is ab

A^ellp =π , (3.4.2)

where a and b are the semimajor and semiminor axes, respectively.

Unfortunately, the length of the perimeter of an elliptical cross-section is much more complicated to evaluate. The formula for the length of the perimeter can be found from many math handbook It is

∫

⁻

= ^/2

0

2 2sin 1

4a ^π k φdφ

L ,

where eccentricity

a b k = a² − ² =

For the purpose to just comparing the area enclosed by the centerline of the wall section, We approximate the perimeter with

2 2

2

2 b

L= π a + (3.4.3)

By changing the form of (3.4.3) se have

2 2 2

)2

(2

2 L a C a

b= − = −

π ^{, where}

)2

(2 2 π

C = L (3.4.4) Substituting (3.4.4) into (3.4.2) we have,

2

2 a

C a Aellp =π −

We can find the optimum solution by =0

∂

∂ a

A , by some operations leads to

2 0

2 2

2

2 =

−

= −

∂

∂

a C

a C a

A π , therefore we have

2 C2

a= for a,b>0

Substitute it back to (3.4.4), we have C a

b= =

2

2

(3.4.5) That means the optimum cross-section for elliptical shapes is a circle.

Then from (3.4.5) we have

π π) 2 (2

2 2 2

2

2 L L

b C

a= = = =

Finally, for a circle, the area enclosed by the centerline is

2 2

cir 2 ) 0.0796L

2 ( L a

A = = =

π π π

--- ANS (2) Rectangular section

For rectangular section, the perimeter is )

( 2 p q

L= + , (3.4.6)

where p and q are length and width, respectively.

The cross-sectional area of rectangular sections is simply, pq

A^rec = , (3.4.7)

Substituting (3.4.6) into (3.4.7), we have 2 )

(L p p pq

Arec = = −

We use =0

∂

∂ p

A to find the optimal solution,

0 2−2 =

∂ =

∂ L p

p

A , we have

4

p= L, and from (3.4.6), it is clear that

4 q L

p= = , i.e., the optimal cross-section for rectangular shapes is a square.

Finally, for a square thin-walled section, the area enclosed by the centerline is squ )² 0.0625L²

4 (L pq

A = = =

3.4.2

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--- ANS (3) Equilateral triangular section.

For a equilateral triangle, the length of the lateral side is 3 l = L .

The area enclosed by the centerline of this triangular thin-walled section is

2 2

tri 2 ) 0.048L

3 (L 4 l 3 4

A = 3 = =

--- ANS (c) Comparison

From the results above we can easily tell

tri squ

cir A A

A > >

Consequently we can conclude that the shape achieving the highest torsional rigidity is a CIRCLE.

--- ANS

NOTE: It is interesting to compare in details with variables a b and

p

q from 0~1.

(We here assume a>b and p>q)

For ellipse, _{2 2 2 2} ^{2 2}

2 2

1 ) 2 ( ) 1 2 ) ( 2 )(

( L

b a a b L

b a

ab b

a b aba ab Aellp

+ + =

+ ≈

= +

=π π π π π

For rectangle, ²

2 2

2 2

2

) 1 4( ) 1 (2 ) ( ) (

)

( L

p q p q L

q p

pq q

p q pq p pq Arec

+ + =

+ =

= +

=

For equilateral triangle, ² 36

3 L Atri = We can illustrate ₂

L

A in terms of a b and

p

q , and have the plot of torsional rigidity of different shapes vs. variable aspect ratios.

3.4.4

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3.5 The two-cell section in Fig.3.35 is obtained from the single-cell section of Fig.3.36 by adding a vertical web of the same thickness as the skin. Compare the torsional rigidity of the structures of Figs. 3.35 and 3.36 with

and , , respectively.

cm L

L₁ = ₂ =10 cm

L₁ =5 L₂ =15cm

cm t =0.3

Figure 3.35 Two-cell thin-walled section

Figure 3.36 Single-cell section

Solution:

We denote as torsional rigidity. For the same material in comparison, only the torsion constant needs to be considered.

GJ J

(a) Single-cell thin-walled section The torsion constant J is

=

∫

t ds J A

/ 4 ²

(3.5.1)

where A is the area enclosed by the centerline of the wall section.

We have A=(L₁+L₂)L₃ =20×10=200cm². The torsion constant can be simply derived as

J

4

(b) Two-cell thin-walled section (1) General Form

We denote the shear flow on the left cell by , and the shear flow on the right cell by . The shear flow in the vertical web is

q1

q2 q₁₂ =q₁−q₂

Also, we have the torque for two-cell section

2

The twist angle of the section is obtained from eirher cell. For left cell we have

and for the right cell

)

Since the entire thin-wall section must rotate as a rigid body in the plane, we require the compatibility condition

θ

Substituting (3.5.6) into (3.5.2) and using

θ

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--- ANS (3) Case 2: L₁ =5cm, L₂ =15cm and L₃ =10cm

From (3.5.6), _{2 1} 1.25 ₁

5) 10 15 210 2 (

15) 10 5 210 2 (

q q

q =

+ +

+

= +

Then substituting into (3.5.7) we have

4 1

1 1

1 1

2

2 814.2857

) 10 25 . 1 10 2 5 2 (

3 . 0 ) 25 . 1 10 15 10 5 ( 10 5

4 cm

q q

q

q

J _cell q =

⋅

−

⋅ +

⋅

⋅

⋅

⋅ +

⋅

⋅

== ⋅

−

--- ANS (c) Comparison

From the results above we have

4 1

1 2 2 2

4 800

2857 .

814 cm = J _cell− >J _cell− = J _cell = cm

Adding a vertical web does not significantly improve the torsional rigidity.

--- ANS

3.6 Find the torsional rigidity if the side wall of one of the two cells in Fig. 3.35 (with ) is cut open. What is the reduction of torsional rigidity compared with the original intact structure?

cm L

L₁ = ₂ =10

cm t =0.3

Figure 3.35 Two-cell thin-walled section

Solution:

We denote torsional rigidity by GJ as.

(a) Closed sidewall

From the solution of Problem 3.5, we have the torsion constant of the case with

1 2cell−

J cm

L L₁ = ₂ =10

4 3

2 3 1 1 1

2 3 2 1 3 1 3 1 1

2 800

) 2

2 (

) (

4 cm

L q L q L q

t q L L q L L L

J _cell L =

− +

= +

−

So we have the original torsional rigidity GJ_2cell−1 =800G (3.6.1) (b) With one side wall cut open

Assuming that the cell is cut open as shown in the figure, the torsional rigidity can be derived from

cut cell cut

not cell open

cut GJ GJ

GJ ₋ = _{− −} + ₋ (3.6.2)

(1) + (2)

Where

3.6.1

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∫

−

−

− =

t ds J_{cell not cut} A^{not cut}

/ ) (

4 ²

, and Anot−cut =L₂L₃ (3.6.3)

=> ⁴

2

) 300 10 10 ( 2

3 . 0 ) 10 10 (

4 cm

J_{cell not cut} = +

⋅

= ×

−

−

and

∑

− =

i i i cut

cell bt

J ³

3

1 (3.6.4)

=> (10 10 10) 0.3³ 0.27 ⁴ 3

1 cm

J_cell−cut = + + ⋅ = So, from (3.6.2) we get

G GJ

GJ

GJ_cut−open = _{cell−not−cut} + _cell−cut =300.27

---ANS (c) The reduction of torsional rigidity is obtained as

% 5 . 62 625 . 800 0

27 . 300 800 GJ

GJ R GJ

1 cell 2

open cut 1

cell

2 − = − = =

=

−

−

−

--- ANS

In document C T Sun Mechanics of Aircraft Structures Solution (Page 54-73)