6. torsion of closed section beams - hw 5_c

7/27/2019 6. Torsion of Closed Section Beams - Hw 5_c

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Department of Aerospace Engineering

AERSP 301Torsion of closed and open section

beams

Jose L. Palacios

July 2008


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Torsion of closed section beams

To simultaneously satisfy these, q = constant

Thus, pure torque const. shear flow in beam wall

A closed section beam

subjected to a pure

torque T does not in the

absence of axial

constraint, develop anydirect stress, z

Now look at pure torsion of closed c/s


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Torque produced by

shear flow acting on

element s is pqs

[Bredt-Batho formula]

Since q = const. &

Hw # 3, problem 3


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Already derived warping distribution for a shear loaded closed c/s

(combined shear and torsion)

Now determine warping distribution from pure torsion load

Displacements associated with Bredt-Batho shear flow (w & vt):

0 = Normal Strain


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In absence of direct stress,

Recall

No axial

restraint


7/43Department of Aerospace Engineering


To hold for all points around the c/s

(all values of)

c/s displacements have a

linear relationship with

distance along the beam, z




Earlier,

For const. q

Twist and Warping of closed

section beams Lecture

Also Needed for HW #5 problem 3




Starting with warping expression:

For const. q

Using



Twisting / Warping sample problem

Determine warping distribution in doubly symmetrical,closed section beam shown subjected to anticlockwise

torque, T.

From symmetry, center oftwist R coincides with

mid-point of the c/s.

When an axis ofsymmetry crosses a wall,

that wall will be a point ofzero warping.

Take that point as theorigin of S.



Sample Problem

Assume G is constant

abAt

a

t

bw

tds

tds

A

A

AG

Tww

ab

s

ssos

and,2,0

and

2

0

00s

00

From 0 to 1, 0 S1 b/2 and

4and, 10

1

0

10

1 asA

t

s

t

dss

b

s

s Find Warping Distribution



Sample Problem

Warping Distribution 0-1 is:

abo t

a

t

b

b

s

abG

T

w

1

1 4

ab t

a

t

b

abG

Tw

bs

8

2/@

1

1


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Sample Problem

The warping distribution can be deducedfrom symmetry and the fact that wmust be

zero where axes of symmetry intersect the

walls. Follows that: w2= -w1, w3 = w1, w4 = -w1

What would be warping fora square cross-section?

What about a circle?


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Sample Problem

Resolve the problem choosing the point 1as the origin fors.

In this case, we are choosing an arbitrary

point rather than a point where WE KNEWthat wowas zero.


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Sample Problem

In the wall 1-2

ab

a

ab

a

s

ss

t

a

t

b

abG

Tw

as

ts

abGTw

abAt

a

t

bw

t

s

t

ds

t

ds

AA

AGTww

42@

42'

and,2,0setting

and

2

2

1112

0

1

00s

00012

a

s

tt

a

t

bs

abGt

a

t

bT

w

a

ab

ab

42

2

2

' 1112


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Sample Problem

Similarly, it can be show that

2

2

23 4

11

2' sbbt

s

t

a

abG

T

wba

22

1

22

12

2

0

0s

as

baA

t

s

t

a

t

ds

os

ba

s

b

a

s2


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Thus warping displacement varies linearlyalong wall 2, with a value w2at point 2,

going to zero at point 3.

Distribution in walls 34 and 41 follows fromsymmetry, and the total distribution is

shown below:

Sample Problem

Now, we calculate w0which we

had arbitrary set to zero


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Sample Problem

We use the condition that for no axial restraint,

the resultant axial load is zero:

0 dstz

tds

dstww

dstww

dswt

s

o

os 0)(

0

zw


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Sample Problem

Substituting forw12andw23 and evaluating theintegral:

a b

sba

ba

o dstwdstwbtat

w0 0

23112 ''2

2

abo t

a

t

b

abG

Tw

8

Offset that need to be added to previously found warping distributions


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Torsion / Warping of thin-walled OPEN section

beams

Torsion of open sections creates a different type of shear distribution Creates shear lines that follow boundary of c/s

This is why we must consider it separately

Maximum shear located

along walls, zero in center

of member


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Torsion / Warping of thin-walled OPEN

section beams

Now determine warping distribution, Recall:

Referring tangential displacement, vt, to center or twist,

R:


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section beams

On the mid-line of the

section wall zs = 0,

Integrate to get warping displacement:

where

AR, the area swept by agenerator rotating about

the center of twist from

the point of zero

warping

Distance from wall to shear center


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section beams

S = 0 (W = 0)

ARR

R

The sign ofwsis dependent on the

direction of positive torque

(anticlockwise) for closed section

beams.

For open section beams, pris positive if

the movement of the foot ofpralong the

tangent of the direction of the assumed

positive sprovides a anticlockwise areasweeping


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Torsion / Warping Sample Problem

Determine the warping

distribution when the thin-

walled c-channel section

is subjected to an anti-

clockwise torque of 10 Nm

SideNote:

G = 25 000 N/mm2


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BEGINNING SIDENOTE


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SideNote: Calculation of torsional

constant J

(Chapter N, pp 367 Donaldson,

Chapter 4 Megson)

Torsional Constants Examples and

Solutions


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Stresses for Uniform Torsion

z

x

y

MtMt

Assumptions:

1) Constant Torque Applied

2) Isotropic, Linearly Elastic

3) No Warping Restraint

All Sections Have Identical Twist per Unit Length:

No Elongation

No Shape Change

z


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St. Venants Constant For Uniform Torsion

(or Torsion Constant)

A

t dA

dz

dG

MJ

2

4

FuEA

MGJ t

F

Mt

z

y


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Torsion Constant

Jis varies for different cross-sections

#1 #2

#3


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EXAMPLE #1 (ELLIPSE)

Find S. Torsion Constant For Ellipse:

Find Stress Distribution (xyxz)

01

22

b

z

a

y

2b

2a

1) Eq. Boundary:

2) = 0on Boundary:

22

1),(b

z

a

yCzy o3) Substitute into GDE:

22

222 2),(babaGC

dzdGzy o

22

22

22

1),(

b

z

a

y

ba

baGzy

y

z


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EXAMPLE # 1

ab

pIba

baJ

22

33

y

zy

z

zy

xz

xy

),(

),(

2b

2a

4) J:

22

33),(2

ba

ba

G

dzdyzy

G

MJ t

5) Substitute into (y,z)J

M

Gt

22

1),(b

z

a

y

ab

Mzy t

y

z

Area Ellipse:

6) Differentiate 5)

Polar Moment of Inertia: 22

4

1baabIp


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EXAMPLE #2 (RECTANGLE)

m n

mnb

znCos

a

ymCosCzy

),(

b

a

1) Eq. Boundary: Simple Formulas

Do Not Satisfy GDE and BCs

NEED TO USE SERIES

For Orthogonality use Odd COS Series

(n & m odd)

2) Following the procedure in pp 391 and 3923abJ

y

z

2max

1

ab

Mtxs

))/((

1256222226 nabmnmb

af

Find S. Torsion Constant For Ellipse:

Find Stress Distribution (xyxz)


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31.0

10/

ba

3

1

3

1

/

ba

Stress and Stiffness Parameters

for Rectangular Cross-Sections (pp 393)

,


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a>>b Rectangle

0)2/( b

GzGz 22)( 2

22

2

2

2)(

bzGz

by

z

No variation in in y

BCs:

G

MJ

dAM

t

A

t 2

J

bMtxs max

3

3

1

abJ

Integrating

Differentiating


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Similarly: Open Thin Cross-Sections

t

S

J

1

3St

3

S is the Contour Perimeter


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Extension to Thin Sections with Varying

Thickness (pp 409)

2

2

2

)()(

bGz

ddbGdydzzyM

a b

bA

t0

2/)(

2/)(

22 )(4

12),(2

GJdbGM

a

t0

3 )(3

1

Thickness b()

z

y

By analogy to thin section

J

bM

dbJ

txs

a

max

max

0

3 )(31


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Torsional Constantsfor an Open and Closed CS


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END SIDENOTE


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Determine the warping

distribution when the thin-

walled c-channel section

is subjected to an anti-

clockwise torque of 10 Nm

Side Note:

G = 25 000 N/mm2


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433 mm7.316)5.2505.1252(31 J

Origin for s (and AR) taken at intersection of web and axis of symmetry,

where warping is zero

Center of twist = Shear Center, which is located at:(See torsion of beam open cross-section lecture)

42

mm04.8

1

3

h

bh

bs

In wall 0-2:104.8

2

1sAR

Since pRis positive

Positive

pR


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mm01.07.31625000

1010

04.82

1

2 1

3

102 ssw

Warping distribution is linear in 0-2 and:

mm25.02501.02 w


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In wall 2-1:

21

252

104.8

2

1ssA

R

dspdspA RRR 2102 21

2

1

??

mm04.8

21

02

R

R

p

p pR21

-25 mm

NegativepR

The are Swept by the generator

in wall 2-1 provides negative

contribution to AR


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mm04.803.07.31625000

101025

2

12504.8

2

12

2

3

221

s

sw

Again, warping distribution is linear in wall 2-1,

going from -0.25 mm at pt.2 to 0.54 mm at pt.1

The warping in the lower half of the web and

lower flange are obtained from symmetry

6. torsion of closed section beams - hw 5_c

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