department of aerospace engineering aersp 301 torsion of closed and open section beams jose l....

Department of Aerospace Engineering

AERSP 301Torsion of closed and open section

beams

Jose L. Palacios

July 2008


REMINDERS

• IF YOU HAVE NOT TURN IN HW# 4 PLEASE DO SO ASAP TO AVOID FURTHER POINT PENALTIES.

• HW #5 DUE FRIDAY, OCTOBER 3

• HW #6 (FINAL HW from me) DUE FRIDAY OCTOBER 10

• EXAM: OCTOBER 20 – 26 HOSLER – 8:15 – 10:15 PM

• REVIEW SESSION: OCTOBER 19 – 220 HAMMOND – 6 – 9 PM


Torsion of closed section beamsTorsion 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 any direct stress, z

• Now look at pure torsion of closed c/s


Torsion of closed section beams Torsion of closed section beams

• Torque produced by shear flow acting on element s is pqs

[Bredt-Batho formula]

• Since q = const. &

Hw # 3, problem 3


Torsion of closed section beamsTorsion of closed section beams• 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



• In absence of direct stress,

• Recall

No axial restraint



• 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


Torsion of closed section beamsTorsion of closed section beams

• Starting with warping expression:

• For const. q

• Using


Twisting / Warping sample problemTwisting / Warping sample problem

• Determine warping distribution in doubly symmetrical, closed section beam shown subjected to anticlockwise torque, T.

• From symmetry, center of twist R coincides with mid-point of the c/s.

• When an axis of symmetry crosses a wall, that wall will be a point of zero warping.

• Take that point as the origin of S.


Sample ProblemSample Problem• Assume G is constant

abAt

a

t

bw

t

ds

t

ds

A

A

AG

Tww

ab

s

ssos

and ,2 ,0

and

2

0

00s

00

From 0 to 1, 0 ≤ S1 ≤ b/2 and

4 and , 1

01

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

Tw 1

1 4

ab t

a

t

b

abG

Tw

bs

8

2/@

1

1


Sample ProblemSample Problem

• The warping distribution can be deduced from symmetry and the fact that w must be zero where axes of symmetry intersect the walls.

• Follows that: w2 = -w1, w3 = w1, w4 = -w1

What would be warping for a square cross-section?

What about a circle?



• Resolve the problem choosing the point 1 as the origin for s.

• In this case, we are choosing an arbitrary point rather than a point where WE KNEW that wo was zero.


Sample ProblemSample Problem• In the wall 1-2

ab

a

ab

a

s

ss

t

a

t

b

abG

Tw

a

s

t

s

abG

Tw

abAt

a

t

bw

t

s

t

ds

t

ds

A

A

AG

Tww

4 2@

42'

and ,2 ,0 setting

and

2

2

1112

0

1

00s

00012

a

s

tt

a

t

b

s

abG

t

a

t

bT

w

aab

ab

42

2

2

' 1112



• Similarly, it can be show that

2

223 4

11

2' sb

bt

s

t

a

abG

Tw

ba

22

1

22

12

2

00s

as

baA

t

s

t

a

t

ds

os

ba

s

b

a

s2


• Thus warping displacement varies linearly along wall 2, with a value w’

2 at point 2, going to zero at point 3.

• Distribution in walls 34 and 41 follows from symmetry, and the total distribution is shown below:


Now, we calculate w0 which we had arbitrary set to zero


Sample ProblemSample ProblemWe 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



Substituting for w’12 and w’23 and evaluating the integral:

a b

sbaba

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


Torsion / Warping of thin-walled OPEN section Torsion / Warping of thin-walled OPEN section beamsbeams

• 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


Torsion / Warping of thin-walled OPEN Torsion / Warping of thin-walled OPEN section beamssection beams

• Now determine warping distribution, Recall:

• Referring tangential displacement, vt, to center or twist, R:



• On the mid-line of the

section wall zs = 0,

• Integrate to get warping displacement:

where

AR, the area swept by a generator rotating about the center of twist from the point of zero warping

Distance from wall to shear center



S = 0 (W = 0)ARR

ρR

The sign of ws is dependent on the direction of positive torque

(anticlockwise) for closed section beams.

For open section beams, pr is positive if the movement of the foot of pr along the tangent of the direction of the assumed positive s provides a anticlockwise area

sweeping


Torsion / Warping Sample ProblemTorsion / 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


BEGINNING SIDENOTE


SideNote: Calculation of torsional SideNote: Calculation of torsional constant Jconstant J

(Chapter N, pp 367 Donaldson, Chapter 4 Megson)

• Torsional Constants Examples and Solutions


Stresses for Uniform TorsionStresses 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 ElongationNo Shape Change

z


St. Venant’s Constant For Uniform Torsion St. Venant’s Constant For Uniform Torsion (or Torsion Constant)(or Torsion Constant)

A

t dA

dzd

G

MJ

2

4

FuEA

MGJ t

F

Mt

z

y

Φ


Torsion Constant

• J is varies for different cross-sections

#1 #2

#3


EXAMPLE #1 (ELLIPSE)EXAMPLE #1 (ELLIPSE)

• Find S. Torsion Constant For Ellipse:

• Find Stress Distribution (σxy σxz)

0122

b

z

a

y2b

2a

1) Eq. Boundary:

2) Ψ = 0 on Boundary:

22

1),(b

z

a

yCzy o3) Substitute Ψ into GDE:

22

222 2),(

ba

baGC

dz

dGzy o

22

22

22

1),(b

z

a

y

ba

baGzy

y

z


EXAMPLE # 1EXAMPLE # 1

ab

pIba

baJ

22

33

y

zyz

zy

xz

xy

),(

),(

2b

2a

4) J:

22

33),(2

ba

ba

G

dzdyzy

G

MJ t

5) Substitute into Ψ(y,z)J

MG t

22

1),(b

z

a

y

ab

Mzy t

y

z

Area Ellipse:

6) Differentiate 5)

Polar Moment of Inertia:

22

4

1baabI p


EXAMPLE #2 (RECTANGLE)EXAMPLE #2 (RECTANGLE)

m nmn b

znCos

a

ymCosCzy

),(

b

a

1) Eq. Boundary: Simple Formulas Do Not Satisfy GDE and BC’s

NEED TO USE SERIESFor Orthogonality use Odd COS Series(n & m odd)

2) Following the procedure in pp 391 and 392

3abJ

y

z

2max

1

ab

M txs

))/((

1256222226 nabmnmb

af

• Find S. Torsion Constant For Ellipse:

• Find Stress Distribution (σxy σxz)


31.0

10/

ba

3

13

1

/

ba

Stress and Stiffness ParametersStress and Stiffness Parametersfor Rectangular Cross-Sections (pp 393)for Rectangular Cross-Sections (pp 393)

,


a>>b Rectanglea>>b Rectangle

0)2/( b

Gz

Gz 22)(2

22

22

2)(

bzGz

by

z

No variation in Ψ in y

BC’s:

G

MJ

dAM

t

A

t 2

J

bM txs max

3

3

1abJ Integrating

Differentiating Ψ


Similarly: Open Thin Cross-SectionsSimilarly: Open Thin Cross-Sections

t

S

J 1

3St3

S is the Contour Perimeter


Extension to Thin Sections with Varying Extension to Thin Sections with Varying Thickness (pp 409)Thickness (pp 409)

22

2

)()(

bGz

ddbGdydzzyM

a b

bA

t 0

2/)(

2/)(

22 )(4

12),(2

GJdbGM

a

t 0

3 )(3

1

Thickness b(ξ)η

ξz

yBy analogy to thin section

J

bM

dbJ

txs

a

maxmax

0

3 )(3

1


Torsional Constants for an Open and Closed CS


END SIDENOTE



• 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



433 mm 7.316)5.2505.1252(3

1J

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

mm 04.81

3

hb

h

bs

In wall 0-2: 104.82

1sAR

Since pR is positive

PositivepR



mm 01.07.31625000

101004.8

2

12 1

3

102 ssw

Warping distribution is linear in 0-2 and:

mm 25.0 2501.02 w



In wall 2-1:

21 252

104.8

2

1ssAR

dspdspA RRR 2102 2

1

2

1

??

mm 04.8

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



mm 04.803.0

7.31625000

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

department of aerospace engineering aersp 301 torsion of closed and open section beams jose l....

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