solution ch3 (1)

8/10/2019 Solution CH3 (1)

1/46

Mechanics of Aircraft structuresC.T. Sun

3.1 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 thex-yplane rotate as a rigid body, then,

zyu =

zxv = (3.1.1)

),( yxw =

where ),( yx is some function of x and y, called warping function, and 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==== xyzzyyxx

So from the stress-strain relationship, we have

0==== xyzzyyxx

Therefore the equilibrium equations reduce to

0=

+

yx

yzxz

This equation is identically satisfied if the stresses are derived from a stress

function ),( yx , so that

yxz

=

,x

yz

=

(3.1.2)

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

yx

w

z

u

x

wxz

=

+

= (3.1.3)

xy

w

z

v

y

wyz

+

=

+

= (3.1.4)

So it forms the compatibility equation

2=

yx

xzyz

,

or in terms of Prandtl stress function

Gyx

22

2

2

2

=

+

(3.1.5)

(d)Boundary conditions,

0=ds

d, or .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= on the lateral surface of the bar.

(e)

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

3.1.1

8/10/2019 Solution CH3 (1)

2/46


)1(2

2

2

2

+=a

y

a

xC which satisfies the boundary conditions stated above.

Substitute into (3.1.5), we obtain GaC 2

2

1=

Then )(2

222ayx

G+=

Using (3.1.2), we have

yyG

xz

=

=

1, and x

xGyz

=

=

1

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

yyx

wxz =

= => 0=

x

w. Thus, )(yfw =

xxyw

yz =+

= => 0=

yw , Thus, )(xgw =

Hence we conclude . This means that the cross-section remains plane

after torsion. In other words, there is no warping.

constw =

Therefore can be verified, and it successfully expresses the

statement.

0),( =yxw

--- ANS

3.1.2

8/10/2019 Solution CH3 (1)

3/46


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 ia 0a

J

J

ia

0a

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,zyu =

zxv = (3.2.1)

),( yxw =

where ),( yx is a function of x and y, called warping function, and 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==== xyzzyyxx

So from the stress-strain relationship, we have

0==== xyzzyyxx


0=

+

yx

yzxz

This equation is identically satisfied if the stresses are derived from a stress

function ),( yx , so that

3.2.1

8/10/2019 Solution CH3 (1)

4/46


yxz

=

,x

yz

=

(3.2.2)

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

yx

w

z

u

x

wxz

=

+

= (3.2.3)

xy

w

z

v

y

wyz

+

=

+

= (3.2.4)

So it forms the compatibility equation

2=

yx

xzyz ,


Gyx

22

2

2

2

=

+

(3.2.5)

(d)Boundary conditions,

0=dsd , or .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

xC (here we use . Assuming0a )1( 2

2

2

2

+=ii a

y

a

xC

would be fine too).

Then substitute into (3.2.5), we have GaC2

02

1= . Thus we have

)1(2 2

0

2

2

0

22

0 +=a

y

a

xaG . (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

8/10/2019 Solution CH3 (1)

5/46

8/10/2019 Solution CH3 (1)

6/46


Substituting and into the above error equation, we haveappJ J

01.0)(2

)(

)(2

)(2

)()((4

220

20

44

0

44

003

0

+

=

+=

i

i

i

iiiapp

aa

aa

aa

aaaaaa

J

JJ

Because and are positive real number, we haveia 0a

01.0)(2

)(22

0

20

+

i

i

aa

aa => 01)(040816.2)(

0

2

0

+a

a

a

a ii

We can obtain the solution of the above equation as

2235.18174.00

a

ai

Since we have the solutioniaa >0 8174.00

a

ai

Therefore the condition on the wall thickness t is

0000 1826.08174.0 aaaaat i ==

(OR iiii aaaaat 2235.08174.0

10 == )

--- ANS

3.2.4

8/10/2019 Solution CH3 (1)

7/46


3.3 Consider the straight bar of a uniform elliptical cross-section. The semimajor

and semiminor axes are aand b, respectively. Show that the stress function of

the form

)1( 2

2

2

2

+= b

y

a

x

C

provides the solution for torsion of the bar.

Find the expression of Cand show that

22

33

ba

baJ

+=

3

2

ab

Tyzx

= ,

ba

Txzy 3

2

=

and the warping displacement

xyGba

abTw33

22 )(

=

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,

zyu =

zxv = (3.3.1)),( yxw =

where ),( yx is warping function, and 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==== xyzzyyxx

From the stress-strain relationship, we have

0==== xyzzyyxx


0=

+

yx

yzxz

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

),( yx , so that

yxz

=

,x

yz

=

(3.3.2)

3.

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

3.3.1

8/10/2019 Solution CH3 (1)

8/46


yx

w

z

u

x

wxz

=

+

= (3.3.3)

xy

w

z

v

y

wyz

+

=

+

= (3.3.4)

The compatibility equation becomes

2=

yx

xzyz ,


Gyx

22

2

2

2

=

+

(3.3.5)

4. The boundary condition along the bounding surface is

0=ds

d, or .const=

---

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

2

2

2

+=b

y

a

xC . 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( 2byC

yxz ==

, )2( 2axC

xyz ==

Substituting the above stress expressions into the equilibrium equations, we

have

000 =+=

+

yx

yzxz

(2)Compatibility equations

Substituting )1(2

2

2

2

+=b

y

a

xC into (3.3.5) we get

22

22

ba

baGC

+= . (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

d on the

traction free lateral surface.

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

3.3.2

8/10/2019 Solution CH3 (1)

9/46


012

2

2

2

=+b

y

a

x,

it is easy to see that 0)1(2

2

2

2

=+=b

y

a

xC on the free surface and therefore

it satisfies the required condition 0=ds

d

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

=

A

dAy

yx

xT )( (3.3.7)

Substituting )1(2

2

2

2

+=b

y

a

xC into (3.3.7), then we have,

+==AA

dAb

y

a

xCdA

b

yCy

a

xCxT )())

2()

2((

2

2

2

2

22

Note that the integral part of the above equation is the area of the elliptical

cross-section. It can be easily obtained that abdA

b

y

a

x

A

=+ )( 22

2

2

So we have the torsion abCT =

By substituting C and utilizing GJT= , we have

22

3322

22

)(

ba

ba

G

abba

baG

G

abCJ

+=+

=

=

(3.3.8)

--- ANS

(2)322

2)

2()

2(

ab

Ty

b

y

ab

T

b

yC

yxz

=

==

= , (3.3.9)

andba

Tx

a

xC

xyz 32

2)

2(

==

= (3.3.10)

--- ANS

(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 have3

2

abG

Tyxz

= and

baG

Txyz 3

2

= .

Also we need to know33

22 )(

baG

baT

GJ

T

+==

So from (3.3.3) and (3.3.4), we can rewrite in

3.3.3

8/10/2019 Solution CH3 (1)

10/46


33

22

33

22

3

)()(2

baG

yabT

baG

ybaT

abG

Tyy

x

wxz

=

++=+=

(3.3.11)

33

22

33

22

3

)()(2

baG

xabT

baG

xbaT

baG

Txx

y

wyz

=

+==

(3.3.12)

From (3.3.11), we can obtain

)()(

),(33

22

yfxybaG

abTyxw +

=

(3.3.13)

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

)()(),(

33

22

yfxbaG

abT

y

yxw+

=

.

Comparing this equation with (3.3.12) we have 0)( = yf , that is .)( constyf =

For a symmetric cross-section 0)0,0( =w , that is, .0)( =yf Thus, the warping displacement is

.)(

),(33

22

xybaG

abTyxw

=

--- ANS

And it is easy to also find that the warping function

xyba

ba

baG

baT

xybaGabT

yxwyx

22

22

33

22

33

22

)(

.)(),(

),(+

=

+

==

3.3.4

8/10/2019 Solution CH3 (1)

11/46


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 isJ

=

tds

AJ

/

42

(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

tds / 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-sectionFor the elliptical cross-section, the cross-sectional area is

abAellp = , (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

22

sin14

dkaL ,

where tyeccentricia

bak =

=

22

For the purpose to just comparing the area enclosed by the centerline of the

wall section, We approximate the perimeter with

22

22ba

L +

= (3.4.3)

By changing the form of (3.4.3) se have

3.4.1

8/10/2019 Solution CH3 (1)

12/46


2222)2

(2 aCaL

b ==

, where 2)2

(2

LC= (3.4.4)

Substituting (3.4.4) into (3.4.2) we have,

22

aCaAellp =

We can find the optimum solution by 0=

a

A, by some operations leads to

02

22

22

=

=

aC

aC

a

A , therefore we have

2

2C

a= for 0, >ba

Substitute it back to (3.4.4), we have aC

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 2

)2

(2

2

2

22

LLCba ====

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

222cir L0796.0)

2

L(aA ===

--- ANS

(2)Rectangular section

For rectangular section, the perimeter is)(2 qpL += , (3.4.6)

where pand qare length and width, respectively.

The cross-sectional area of rectangular sections is simply,

pqArec = , (3.4.7)

Substituting (3.4.6) into (3.4.7), we have

)2

( pL

ppqArec ==

We use 0=

p

A to find the optimal solution,

022

==

p

L

p

A, we have

4

Lp= , and from (3.4.6), it is clear that

4

Lqp == , 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

22

squ L0625.0)4

L

(pqA ===

3.4.2

8/10/2019 Solution CH3 (1)

13/46


--- ANS

(3)Equilateral triangular section.

For a equilateral triangle, the length of the lateral side is3

Ll= .

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

222tri L048.0)

3

L(

4

3l

4

3A ===

--- ANS

(c)Comparison

From the results above we can easily tell

trisqucir AAA >>

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 variablesa

b and

p

q from 0~1.

(We here assume a>band p>q)

For ellipse, 222222

22

)1

(2

1))

2(2)(( L

b

a

a

b

L

ba

ab

ba

baababAellp

+

=+

+

+==

For rectangle, 2

2

2

22

2

)1(4

1)

2(

)()(

)(L

p

q

p

q

L

qp

pq

qp

qppqpqArec

+

=+

=+

+==

For equilateral triangle, 236

3LAtri =

We can illustrate2

L

A in terms of

a

b and

p

q, and have the plot of torsional rigidity

of different shapes vs. variable aspect ratios.

3.4.3

8/10/2019 Solution CH3 (1)

14/46


3.4.4

8/10/2019 Solution CH3 (1)

15/46


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.

cmLL 1021 ==

cmL 51= cmL 152 =

cmt 3.0=

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 isJ

=

tds

AJ

/

42

(3.5.1)

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

We have2

321 2001020)( cmLLLA ==+= . The torsion constant can be

simply derived as

J

3.5.1

8/10/2019 Solution CH3 (1)

16/46


42

321

2

321

2

1 8003.0)1020(2

)200(4

/)(2

])[(4

/

4cm

tLLL

LLL

tds

AJ

cell =

+=

++

+==

(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

1q

2q 2112 qqq =

Also, we have the torque for two-cell section

21 21 22 qAqAT += (3.5.2)

where 311 LLA = , 322 LLA =

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

have

))()2((2

1

2

1321311

311

11 LqqLLq

tLGLt

qds

AG cell++== (3.5.3)

and for the right cell

))()2((2

1

2

1321322

322

22 LqqLLq

tLGLt

qds

AG cell+== (3.5.4)

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

require the compatibility condition == 21 (3.5.5)

From (3.5.3) to (3.5.5), we derive the relation between and ,1q 2q

1

1

3

2

3

2

3

1

3

2

)22(

)22(

q

L

L

L

L

L

L

L

L

q

++

++

= (3.5.6)

Substituting (3.5.6) into (3.5.2) and usingG

TJ= and (3.5.3), we have

)22(

)(4

))()2((2

1

)22(

323111

23213131

321311

31

2211

LqLqLq

tqLLqLLLL

LqqLLqtLGL

G

qAqAJ

+

+=

++

+=

(3.5.7)

(2)Case 1: andcmLL 1021 == cmL 103 =

From (3.5.6), , then substituting into (3.5.7) we have12 qq =

4

323111

23213131

12 800)22(

)(4cm

LqLqLq

tqLLqLLLLJ

cell =

+

+=

3.5.2

8/10/2019 Solution CH3 (1)

17/46


--- ANS

(3)Case 2: ,cmL 51 = cmL 152 = and cmL 103 =

From (3.5.6), 112 25.1

)5

10

15

1022(

)15

10

5

10

22(qqq =

++

++=

Then substituting into (3.5.7) we have

4

111

1122 2857.814

)1025.110252(

3.0)25.11015105(1054cm

qqq

qqJ cell =

+

+==

--- ANS

(c)Comparison

From the results above we have

4

11222

4 8002857.814 cmJJJcm cellcellcell ==>=

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

--- ANS

3.5.3

8/10/2019 Solution CH3 (1)

18/46


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?

cmLL 1021 ==

cmt 3.0=

Figure 3.35 Two-cell thin-walled section

Solution:

We denote torsional rigidity by as.GJ

(a)Closed sidewall

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

case with

12 cellJ

cmLL 1021 == 4

323111

2321313112 800

)22(

)(4cm

LqLqLq

tqLLqLLLLJ cell =

+

+=

So we have the original torsional rigidity GGJ cell 80012 = (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

cutcellcutnotcellopencut GJGJGJ += (3.6.2)

(1) + (2)

Where

3.6.1

8/10/2019 Solution CH3 (1)

19/46


=tds

AJ

cutnot

cutnotcell

/

)(4 2, and 32LLA cutnot = (3.6.3)

=>

42

300)1010(2

3.0)1010(4

cmJ cutnotcell =+

=

and

=i

iicutcell tbJ3

3

1 (3.6.4)

=> 43 27.03.0)101010(3

1cmJ cutcell =++=

So, from (3.6.2) we get

GGJGJGJ cutcellcutnotcellopencut 27.300=+=

---ANS

(c)The reduction of torsional rigidity is obtained as

%5.62625.0800

27.300800

GJ

GJGJR

1cell2

opencut1cell2 ==

=

=

--- ANS

3.6.2

8/10/2019 Solution CH3 (1)

20/46


3.7 Find the torque capability of the thin-walled bar with the section shown in Fig.

3.36. Assume that the shear modulus GPaG 27= and the allowable shear

stress of MPaallow 187= .

cmt 3.0= Figure 3.36 Single thin-walled section

Solution:

Since the thickness of all walls are equal to cmt 3.0= , we can obtain the allowable

shear flow from allowable shear stress, that is

m/N1061.5003.010187tq 56allowallow ===

Then we have the torque capability as

mN224401061.5)2.01.0(2qA2T 5allowallow ===

--- ANS

3.7.1

8/10/2019 Solution CH3 (1)

21/46


3.8 A two-cell thin-walled member with the cross-section shown in Fig. 3.37 is

subjected to a torque T. The resulting twist angle is . Find the shear

flows of the applied torque, and the torsion constant. The material is aluminum

alloy 2024-T3.

m/3o

Figure 3.37 Two-cell section

Solution:

(a)Assume the material is linearly elastic under the twist angle . For aluminum

alloy 2024-T3, we have the shear modulus

GPa27)33.01(2

72

)1(2

EG =

+=

+=

(b)

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

1q2q 2112 qqq = , are the positive directions

as shown in the figure above.

Also, we have the torque for two-cell sections

21 21 22 qAqAT += (3.8.1)

where 222

21 m098.08

)5.0(

8

dAA ====

,

The twist angle of the left cell is

))((2

1

2

121

12

121

1

1

11

11 qq

t

sq

t

s

AGt

qds

AG cell+== (3.8.2)

where m785.02

ds1 ==

is the length of the left side wall, and is

the length of the vertical web.

ms 5.012 =

The twist angle of the right cell is

))((2

1

2

121

12

122

2

2

22

22 qq

t

sq

t

s

AGt

qds

AG cell== (3.8.3)

3.8.1

8/10/2019 Solution CH3 (1)

22/46


Again, we have m785.02

ds2 ==

, the length of the right side wall.


require the compatibility condition

m/rad0524.0m/321 ==== o

(3.8.4)

From (3.8.2) to (3.8.4) and noting that 21 AA = , we derive the relation between

and by substituting all the known quantities,1q 2q

122121002.0

5.0

003.0

785398.0

002.0

5.0

001.0

785398.0qqqq =+

Substituting , in the equation above, we obtain2112 qqq =

12 q687.1q = (3.8.5)

Back substituting into (3.8.2) and (3.8.4), we have

191 q

)098175.0)(100677.27(2

))169732.1(250398.785(m/rad0524.0

==

From which we obtain

m/N500,453q1=

Subsequently from (3.8.5) we obtain

m/N765000q687.1q 12 ==

--- ANS

(c)

The applied torqueFrom (3.8.1), we compute the applied torque

mN10393.2mN239300

)765000453500)(098.0(2qA2qA2T

5

2211

==

+=+=

--- ANS

(d)The torsion constant J

From the fundamental relationship of torque and twist angle, we have GJT=

So the torsion constant can be derived as

44

9 m1069.1

)0524.0)(1027(

239300

G

TJ

=

==

--- ANS

3.8.2

8/10/2019 Solution CH3 (1)

23/46


3.9 For the bar of Fig. 3.37, find the maximum torque if the allowable shear stress is

MPaallow 187= . What is the corresponding maximum twist angle ?

Figure 3.37 Two-cell section

Solution:

(a)Assume the material is linearly elastic under the twist angle . For aluminum


GPa27)33.01(2

72

)1(2

EG =

+=

+=

(b)

We denote the shear flow on the left cell as and that on the right cell as .

The shear flow in the vertical web is

1q 2q

2112 qqq = . The positive directions for the

shear flows are shown in the figure above.

The torque for two-cell section is

21 21 22 qAqAT += (3.9.1)

where 222

21 m098.08

)5.0(

8

dAA ====

,

The twist angle of the left cell is

))((2

1

2

121

12

121

1

1

11

11 qq

t

sq

t

s

AGt

qds

AG cell+== (3.9.2)

where m785.02

ds1 ==

is the length of the left side wall, and is

the length of the vertical web.

ms 5.012 =

Also we have the twist angle of the right cell as

))((2

1

2

121

122

2

2

22

22 qq

t

sq

t

s

AGt

qds

AG cell== (3.9.3)

where m785.02

ds2 ==

is the length of the right side wall.

(c)Since the entire thin-wall section must rotate as a rigid body in the plane, we


3.9.1

8/10/2019 Solution CH3 (1)

24/46


== 21 (3.9.4)

From (3.9.2) to (3.9.4) and note that 21 AA = , we derive the relation between

and by substituting all the known quantities,

1q

2q

122121002.0

5.0

003.0

785398.0

002.0

5.0

001.0

785398.0qqqq =+

In view of the relation 2112 qqq = we obtain

12 q687.1q = (3.9.5)

Back substituting (3.9.5) into (3.9.2) and (3.9.4), we have

19 )098175.0)(100677.27(2

))169732.1(250398.785(q

= , and then

8662000q1=

Subsequently, (3.9.6)000,600,14q687.1q 12 == (3.9.7)

000,953,5qqq 2112 == (3.9.8)

Note the units are m/radin , and are in N/m.1221 q,q,q

(d)Stress in the wall

From the above quantities of shear flow, we can then compute the shear stress in

each wall byt

q= . We have

9

1

11 1066.8

001.08662028

tq === (3.9.9)

9

2

22 1087.4

003.0

14615612

t

q=== (3.9.10)

9

12

1212 1098.2

002.0

5953584

t

q=

== (3.9.11)

(e)From the above stresses (3.9.9) to (3.9.11), because the negative value just denote

the negative direction, the maximum absolute magnitude of shear stress is6

allow9

1 101871066.8 ==

Therefore the maximum twist angle ism/24.1m/rad0216.0max

o==

--- ANS

(f)

The maximum torque can be solved by using (3.9.1), (3.9.6), (3.9.7) and the

maximum twist angle, that is

mN98700

)0216.0)(146000008662000)(098.0(2qA2qA2T 2211

=

+=+=

--- ANS

3.9.2

8/10/2019 Solution CH3 (1)

25/46


3.10 Find the shear flow and twist angle in the two-cell three-stringer thin-walled

bar with the cross-section shown in Fig. 3.38. The material is Al2024-T3. The

applied torque is .mN 5102

Figure 3.38 Two-cell three-stringer thin-walled section

Solution:

(a)Assume the material is linearly elastic under the applied torque. For aluminum


GPaE

G 27)33.01(2

72

)1(2=

+=

+=

(b)

Denote the shear flow on the left cell as , and the shear flow on the right cell as; both are considered positive if counterclockwise. The shear flow in the

vertical web is , which is positive if it is in the same direction as .

1q2q

2112 qqq = 1q

We have the torque for the two-cell section as

21 21 22 qAqAT += (3.10.1)

where 222

1 565.08

)2.1(

8m

dA ===

,

and2

2 2.12

)2.1(2

2 mbh

A ===

The twist angle of the left. cell is

))((2

1

2

121

1

12

1

1

1

11

1

1 qqt

sq

t

s

AGt

qds

AG cell+== (3.10.2)

where md

s 88.12

1 ==

is the length of the left half circular wall, and

is the length of the vertical web.ms 2.112 =

The twist angle of the right. cell is

3.10.1

8/10/2019 Solution CH3 (1)

26/46


))((2

1

2

121

1

122

3

32

2

2

22

22 qq

t

sq

t

sq

t

s

AGt

qds

AG cell+== (3.10.3)

Again, we have , the length of the lower wall,ms 22=

and ms 33.22.1222

3 =+= , the length of the inclined wall of thickness t3 in

the right cell.



== 21 (3.10.4)

Solving the two equations, (3.10.2) and (3.10.4), we obtain

)005.0

2.1

007.0

332.2

007.0

2(

2.1

1)

005.0

2.1

005.0

885.1(

565.0

11222121 qqqqq +=+

Eliminating from the equation above using12q 2112 qqq = we obtain

12 132.1 qq = (3.10.5)

(c)To find the shear flow , we back substitute (3.10.5) into (3.10.1) and have

121 2121 )265.22(22 qAAqAqAT +=+=

=> mNAA

Tq /51966

)2.1)(265.2()565.0)(2(

102

265.22

5

211 =

+

=

+=

--- ANS

From (3.10.5),mNqq /58844132.1 12 ==

--- ANS

(d)For the twist angle, we can utilize the shear flows and equations (3.10.2) and

(3.10.4) to get,

mmrad

qqt

sq

t

s

AG

/0336.0/1086.5

)565.0)(1027(2

51966))132.11(005.0

2.1

005.0

885.1(

))((2

1

4

921

1

121

1

1

11

o==

+=+==

--- ANS

3.10.2

8/10/2019 Solution CH3 (1)

27/46

8/10/2019 Solution CH3 (1)

28/46


where is the length of the lower straight wall of thickness tm2s2 = 2, and

m33.22.12s 22

3 =+= is the length of the inclined wall of thickness t3.



== 21 (3.11.4)

From (3.10.2) to (3.10.4), we can derive the relation between and by

substituting all the known quantities,

1q 2q

)q005.0

2.1q

007.0

33.2q

007.0

2(

2.1

1)q

005.0

2.1q

005.0

88.1(

566.0

11222121 +=+

Substituting into the above equation, we obtain2112 qqq =

12 q13.1q = (3.11.5)

(c)

Since we have the condition , and, thus,m/rad035.0m/2allowable == o

m/rad035.0allowable21 =

8/10/2019 Solution CH3 (1)

29/46


3.12 The two shafts of thin-walled cross-sections shown in Fig. 3.39a and b,

respectively. Contain the same amount of aluminum alloy. Compare the

torsional rigidities of the two shafts without end constraints.

(a) (b)

Figure 3.39 Cross-sections of two shafts

Solution:

(a)

Fig. 3.39a is a cross-section of an open thin-wall, its torsional rigidity isa

GJ

43

i

3

iia mmG5400G)3)(200)(3

1(3tb

3

1GGJ ===

--- ANS(b)Fig. 3.39b is a cross-section of a closed thin-wall, its torsional rigidity is bGJ

=

t

ds

AGGJ

b

2

4, where

4

3 2bA= ,

46342

1064)3(4

34GmmG

tb

b

tbG

t

ds

AGGJ

b ====

--- ANS

(c)The ratio of the torsional rigidities is

1111G5400

G106

GJ

GJ 6

a

b =

=

--- ANS

3.12.1

8/10/2019 Solution CH3 (1)

30/46


3.13 Find the distributions of the primary warping displacement on the cross-sections

shown in Fig. 3.39b. Due to symmetry, the center of twist coincides with the

centroid of the section, and warp at the midpoint of each flat sheet section is zero.

Sketch the warping displacement along the wall.

(b)

Figure 3.39 Cross-sections of two shafts

Solution:

(a)Observation.

Because of the symmetry, the center of twist coincides with the centroid of the

section, and warp at the midpoint of each flat sheet section is zero.

So from the figure above we set 0=w at the midpoint of each flat sheet. First

we assume the warp at point A is positive of z-direction. While going from A to B,

we pass the midpoint and then the warp goes from positive into negative part, then

end at point B with the maximum negative warping. Using the same concept on

sheet BC will result in a maximum positive warping at point C. Now we consider

the sheet CA by using the same conclusion, we will surprisingly find the warping

at A is negative of z-direction. Hence it contradicts our assumption of A being

3.13.1

8/10/2019 Solution CH3 (1)

31/46


positive. By applying the assumption of A is negative direction we will conclude

in another contradiction. Therefore, we can confidently assure that there is no

warping in this equilateral triangular thin-walled cross-section. In the following

we will approve it by further derivatives.

(b)

For the closed thin-walled section, we have

+

=

+

=

s

w

z

u

s

w ssz , (3.13.1)

where is the distance from the center of twist to the tangent line of point P of

interest, wis the warping that we are seeking. Also, we have

Gt

q

G

ssz

sz ==

(3.13.2)

where is the shear flow along s-direction, tis the thickness of the wall and G

is the shear modulus.

sq

Again, recall from the relationship between applied torque and shear flow, we

haveA

Tqs

2= (3.13.3)

Combining (3.13.1) to (3.13.3) results in

GtA

T

s

w

2=+

, or =

GtA

T

s

w

2

=> ssss

Ads

GtA

Tdsds

GtA

Twsw 2

22

)0()(000

== (3.13.4)

Also the twist angle can be derived from

= tds

AG2

1 (3.13.5)

(c)Assume the applied torque is uniformly applied to the cross-section. Also, the

material is isotropic so that the shear modulus is constant.

For the equilateral triangular section, we have

4

3 2bA= (3.13.6)

And since the section is symmetric, we can just take the sheet CA into

consideration and applied to all other sheets. Assume the origin of sis on the

midpoint of sheet CA, so 0)0( =w , then we have

12

3bsAs = (3.13.7)

From (3.13.4) to (3.13.7), we obtain

))16/3(

)8/3(

3

2(

)(4

)3)(12

3(2

22

2)(

4

2

220 b

b

bGt

Ts

tAG

Tbbs

tAG

TsAds

GtA

Tsw

s

s

===

3.13.2

8/10/2019 Solution CH3 (1)

32/46


=> 0)( =sw

This approves our observation in part (a).

--- ANS

3.13.3

8/10/2019 Solution CH3 (1)

33/46

8/10/2019 Solution CH3 (1)

34/46


433 1729080)2)(2

(12

1)2)(

2(

12

1mmth

tbth

tbIx =++=

( Since the thickness of walls is relatively small, there are some approximated

solutions such as

4233

1728270]12[2)2(12 mmbtht

b

h

t

Ix =++= , or

423 1728000)(2)2(12

mmbthht

Ix =+= are all the acceptable approximations)

The eccentric distance mm49.221729080

)60(3

I

htbe

4

x

22

=== (3.14.5)

The torsional constant is

433

32160

3

3120)

3

360(2

3

1mmtbJ

i

ii =

+

==

and the twist angle per unit length can be obtained from

G

T

GJ

T

2160== (3.14.6)

(c)Break up the contour sinto two straight parts and , as shown below1s 2s

For contour , we have1s

11

2

1esAs = (3.14.7)

and for contour (the point2s 0s2 = is at upper left corner of the section) we have

22

2

1hsAs = (3.14.8)

(d)On the contour , the warping displacement wis calculated from equation

(3.14.2):

1s

G

Ts0104.0

G2160

TesA2ds)s(w 111s

s

01 ====

In which the condition 0)0(w = has been used. This is obvious since the warping

at the middle point of the vertical web is zero because of anti-symmetry. Also note

that

3.14.2

8/10/2019 Solution CH3 (1)

35/46


G

T62.0)mm60h(w)hs(w 1 ====

--- ANS

(e)On the contour , the warping can be obtained from equation (3.14.2) by

integrating from s

2s

1=0 to any point s2. Thus,

2s12 A2)hs(w)s(w ==

=>G

Ts

G

Tsw 22 028.062.0)( += --- ANS

So the warping displacement at the left upper corner is

G

Thsw 62.0)( 1 ==

and at the right upper edge is

G

Tbsw 04.1)( 2 ==

--- ANS

(f) Similar calculations show that the warping displacement is anti-symmetric with

respect to the x-axis. From the above calculations, the maximum warp (absolute

value) is

G

Tbsww 04.1)( 22max ===

and are located at both free edges.

--- ANS

G

Tssw 111 0104.0)( =

G

T

G

Tssw 62.00278.0)( 222 =

3.14.3

8/10/2019 Solution CH3 (1)

36/46


3.15 Consider the shaft of the channel section shown in Fig. 3.40. If one end of the

shaft is built in and the other end is free, find the effective torsional rigidity as a

function of the distance from the built-in end. Assume that the length Lof the

shaft is sufficiently large so that near the free end the Saint-Venant torsion

assumptions are valid. Compare the total twist angle with that for a free-free

shaft for mL 2= .

Figure 3.40 Dimensions of a channel section

Solution:

(a)

The government equation for the twist angle under an applied torque Tendconstraints is

GJ

T

dz

dk =

2

22

, (3.15.1)

whereGJ

Ek

=2 , =

1

0

24

s

ss tdsA and =

i

iitbJ3

3

1 (3.15.2)

The general solution is

)1( /2/

1kzkz

ph eCeCGJ

T ++=+= (3.15.3)

(b)Applying boundary conditions

We assume the shaft is built in at 0=z and free at Lz= where the torque Tis

applied.

(1) First, assume the length Lof the shaft is sufficiently large so that near the

free end the Saint-Venant torsion assumptions are valid, so that GJT=

(Saint-Venant torsion) when . To satisfy this condition, we require

that

Lz

01 =C , then (3.15.3) will converge to GJT= .

(2)

Second, at the built-in end ( 0=z ), warping suppressed and . From

the equation

0=w

)()(),( zswzsw s = , we conclude that 0= . Thus, we have

3.15.1

8/10/2019 Solution CH3 (1)

37/46


12 =C .

From (1) and (2), we have the solution becoming

eff

kz

GJ

Te

GJ

T== )1( /

Then the effective torsion constant is

kzeffe

JJ

/1 = (3.15.4)

(c)

In Fig. 3.40, the channel cross-section has the properties from (3.15.2).

49433

31016.22160

3

3120)

3

360(2

3

1mmmtbJ

i

ii

==

+

==

=1

0

24

s

ss tdsA

Because of the symmetric w.r.t x-axis, it is more convenient to measure distancefrom the middle point of the vertical web.

From the solution of Problem 3.14, we have

mmmI

htbe

x

02249.049.221729080

)60(3 422

cot ====

Therefore,

})](2

1

2

1[4)

2

1(4{24 1

21cot

01

21cot

1

0

2

+

+= bh

h

hs

ss tdshshhetdssetdsA

~~~~~~A~~~~~~ ~~~~~~~~~B~~~~~~~~~

Part A:

61032

cot0

3

1

2

cot0

12

1cot 1009.13

1|)

3

1()

2

1(4 mthestetdsse h

h===

Part B:

3.15.2

8/10/2019 Solution CH3 (1)

38/46


6103222cot

22

cot0

22

2cot

12

1cot

1031.2)3

1()(

)](2

1

2

1[4

2

mbhbhebhettdshshe

tdshshhe

bb

bh

h

+

=+==

=> 6101

0

2108.6)(24 mPartBPartAtdsA

s

ss

=+==

Taking andGPaE 70= GPaG 27= for aluminum, then

904.0=

=GJ

Ek

Thus, the effective torsional rigidity is obtained as

904.0/904.0/

99

/ 1

32.58

1

1016.21027

1 zzkzeff

eee

GJGJ

=

=

=

--- ANS(d)The total twist angle.

For the case with end constraints,

TT

ezT

dzeT

dz zz

L

fixed

0205.0)805.02(32.58

|)904.0(32.58

)1(32.58

20

904.0/2

0

904.0/

0

==

+===

For the case with free-free end,

TTGJTLLfree 0343.0

32.582 ====

The ratio of the two twist angles is

6.0=free

fixed

--- ANS

It is clear that the end constraints reduce the twist angle. In other words, end

constraints increase the torsional stiffness.

3.15.3

8/10/2019 Solution CH3 (1)

39/46

8/10/2019 Solution CH3 (1)

40/46


TseseTsz zz

zzz 10904.0/

1cot0 0299.0|)2

1(656.2|),( == =

=

--- ANS

At ,01 =s 0)0,0( =zz

At mhs 06.01 == , )/(1079.106.00299.0 23 mNTTzz ==

On the horizontal sheet, we can derive the normal stress with (3.16.3) and

(3.16.5)

Ts

ehsheTsz zz

zzz

)0797.01079.1(

|)2

1

2

1(656.2|),(

23

0904.0/

2cot0

+=

=

=

=

--- ANS

At ,02 =s Tzz31079.1 =

At mbs 06.02 == ,)/(1099.2)06.00797.01079.1( 233 mNTTzz

=+=

The distribution of normal stress on the cross-section is anti-symmetric. The

distribution can be illustrated as the figure below.

--- ANS

(d)

The distribution of shear flow at the built-in end ( 0=z ).

From the equation (3.85) in the textbook, the shear flow at any location s at the

built-in end ( 0=z ) is

== = s

s szz tdsw

dz

dEsq

002

2

0 ||)(

(3.16.6)

where ss Asw 2)( =

3.16.2

8/10/2019 Solution CH3 (1)

41/46


Substituting (3.16.1) in (3.16.6), we have

=

=

=

=

s

ss

s

ssz

z

z

dsAT

dsAeT

sq

0

3

00

904.0/20

1082.8

)003.02(|))904.0

1()(

32.58)(70(|)(

(3.16.7)

Here it is important to emphasize that the s-direction is measured from the point

where shear flow vanishes. Hence sbegins from the free end of the horizontal

sheet as shown in the figure below. Also, due to the symmetry w.r.t. x axis, we

only need to consider the part above x-axis. This allows us to modify equation of

sA from (3.16.4) and (3.16.5).

For the horizontal sheet, we have

12

1hsAs = , (3.16.8)bs 0:1

Moving along the vertical web, we have

2cot2

1seAs = (3.16.9)hs 0:2

On the horizontal sheet, we can derive the shear flow from (3.16.8) and

(3.16.7)

Tsx

T

hxdxTsq

s

s

z

2

14

0

24

0

301

10323.1|)2

(10646.2

2

11082.8|)(

1

1

=

==

=

--- ANS

At ,01 =s 0)0( =q

At mbs 06.01 == ,

mNTTq /10762.4)06.0(10323.1)06.0( 7241 ==

On the vertical web, we can derive the shear flow with (3.16.9) and (3.16.7).

Since the shear flow is continuous, we have

Tsx

T

dxxeTbsqsq

s

s

zz

2

252

0

25

2

0cot

3011022

10957.4|2

10914.9

)2

1(10818341.8|)(|)(

==

==

==

3.16.3

8/10/2019 Solution CH3 (1)

42/46


=> Tssq z )10762.410957.4(|)(72

25

022

= =

--- ANS

At ,02 =s Tq7

2 10762.4)0( =

At mhs 06.02 == ,

T

Tsq

7

72522

10977.2

)10762.4)06.0(10957.4()06.0(

=

==

The distribution of the shear flow at the fixed end is sketched in the figure below.

--- ANS

3.16.4

8/10/2019 Solution CH3 (1)

43/46


3.17 Assume that the shaft of the channel section of Fig. 3.40 is built in at both ends.

Find the torque that is necessary to produce a relative twist angle

between two ends. Assume that

o5=

mL 1= , Youngs modulus , and

shear modulus . Compare this with the free-free case.

GPaE 70=

GPaG 27=

Figure 3.40 Dimensions of a channel section

Solution:

(a)

Since both end of the channel are built-in, it allows us to set 0=z at the middle

of the channel as shown above.

(b)The governing equation for the twist angle under an applied torque Tend

constraints is

GJ

T

dz

dk =

2

22

, (3.17.1)

whereGJ

Ek

=2 , (3.17.2)

=1

0

24

s

ss tdsA and =

i

iitbJ3

3

1 for open thin-walled section. (3.17.3)

The general solution of differential equation (3.17.1) is

3.17.1

8/10/2019 Solution CH3 (1)

44/46


)sinhcosh1( 21k

zC

k

zC

GJ

Tph ++=+= (3.17.4)

(c)Compute =i

iitbJ

3

3

1 and =

1

0

24

s

ss tdsA .

49433

31016.22160

3

3120)

3

360(2

3

1mmmtbJ

i

ii

==+==

=1

0

24

s

ss tdsA

Because of symmetry w.r.t x axis, it is more convenient to set up the s contour

with the origin at the middle point of the vertical web as shown in the figure

below.

From the solution of Problem 3.14, we have

mmmI

htbe

x

0225.0485946.221729000

)60(3 422

cot ====

Therefore,

})](2

1

2

1[4)

2

1(4{24 1

21cot

01

21cot

1

0

2

+

+= bh

h

hs

ss tdshshhetdssetdsA

~~~~~~A~~~~~~ ~~~~~~~~~B~~~~~~~~~

Part A:

61032

cot0

3

1

2

cot0

12

1cot 10092.13

1|)

3

1()

2

1(4 mthestetdsse h

h===

Part B:6103222

cot22

cot0 22

2cot

12

1cot

1031.2)3

1()(

)](2

1

2

1[4

2

mbhbhebhettdshshe

tdshshhe

bb

bh

h

+

=+==

=>

6101

0

2

10804.6)(24 mPartBPartAtdsA

s

s s

=+==

3.17.2

8/10/2019 Solution CH3 (1)

45/46


Taking andGPaE 70= GPaG 27= for aluminum, then

904.0=

=GJ

Ek

(d)Applying boundary conditions

(1)

First, because of symmetry of with respect to , the odd function

should be dropped. This is accomplished by setting .

z

)/sinh( kz 02 =C

(2) Second, at the built-in end ( 2/Lz= ), warping is suppressed and .

Since

0=w

)()(),( zswzsw s = , we conclude that 0= . Thus, we have

0)2

cosh1( 1 =+k

LC

GJ

T

Then

k

LC

2

cosh

11 =

Since and , we havemL 1= mk 904.0=

864.0)

)904.0(2

1cosh(

11 =

=C

Thus the solution for the rate of twist angle is

)904.0

cosh864.01(32.58

)cosh864.01( zT

k

z

GJ

T==

The twist angle

related to distance z from the middle of the channel then is

)904.0

sinh78.0(32.58

)904.0

cosh864.01(32.5800

zz

T

duuT

dzzz

=

== (3.17.5)

This is the twist angle measured from the middle of the channel bar to the built-in

end.

(e) If we produce a relative twist angle , then the twist angle from the middle

of the channel to the built-in end (

o5=

mLz 5.02/ == ) is

rad0436.05.22

5=== o

o

.

From equation (3.17.5) we can determine the required torque to produce such an

angle.

TT 410795.7)

904.0

5.0sinh78.05.0(

32.580436.0 ==

Then mNT = 97.55

--- ANS

(f)

For the free-free end case

3.17.3

8/10/2019 Solution CH3 (1)

46/46


mNz

GJGJT endfree ==== 089.55.0

0436.032.58

.

The ratio of built-in ends case and free-ends case is

110894.5

974.55==

endfree

inbuilt

T

T

It is likely that the rigidity of the built-in ends case is enhanced eleven times more

than the free-ends case.

--- ANS

solution ch3 (1)

Documents